Subgroup Lattices of Groups - Department of Mathematics · Subgroup Lattices of Groups by Roland...

DE GRUYTER Roland SchmidtEXPOSITIONS

IN

MATHEMATICS

14

Subgroup

latticesof

Groups

de Gruyter Expositions in Mathematics 14

Editors

0. H. Kegel, Albert-Ludwigs-Universitat, FreiburgV. P. Maslov, Academy of Sciences, Moscow

W. D. Neumann, The University of Melbourne, ParkvilleR. 0. Wells, Jr., Rice University, Houston

de Gruyter Expositions in Mathematics

I The Analytical and Topological Theory of Semigroups, K. H. Hofmann,J. D. Lawson, J. S. Pym (Eds.)

2 Combinatorial Homotopy and 4-Dimensional Complexes, H. J. Baues

3 The Stefan Problem, A. M. Meirmanov

4 Finite Soluble Groups, K. Doerk, T. O. Hawkes

5 The Riemann Zeta-Function, A. A. Karatsuba, S. M. Voronin

6 Contact Geometry and Linear Differential Equations, V. R. Nazaikinskii,V. E. Shatalov, B. Yu. Sternin

7 Infinite Dimensional Lie Superalgebras, Yu. A. Bahturin, A. A. Mikhalev,V. M. Petrogradsky, M. V. Zaicev

8 Nilpotent Groups and their Automorphisms, E. I. Khukhro

9 Invariant Distances and Metrics in Complex Analysis, M. Jarnicki, P. Pflug

10 The Link Invariants of the Chern-Simons Field Theory, E. Guadagnini

1 I Global Affine Differential Geometry of Hypersurfaces, A.-M. Li, U. Simon,G. Zhao

12 Moduli Spaces of Abelian Surfaces: Compactification, Degeneration, andTheta Functions, K. Hulek, C. Kahn, S. H. Weintraub

13 Elliptic Problems in Domains with Piecewise Smooth Boundaries, S. A. Na-zarov, B. A. Plamenevsky

Subgroup Lattices of Groups

by

Roland Schmidt

WDE

GWalter de Gruyter Berlin New York 1994

Author

Roland SchmidtMathematisches SeminarUniversitat KielLudewig-Meyn-StraBe 4D-24098 KielGermany

1991 Mathematics Subject Classification: 20-01, 20-02

Keywords: Group, lattice, subgroup lattice, projectivity, duality

® Printed on acid-free paper which falls within the guidelines of the ANSI to ensure permanence and durability.

Library of Congress Cataloging-in-Publication Data

Schmidt, Roland, 1944-Subgroup lattices of groups / by Roland Schmidt.

p. cm. - (De Gruyter expositions in mathematics ; 14)Includes bibliographical references and index.ISBN 3-11-011213-21. Group theory. 2. Lattice theory. I. Title. II. Series.

QA174.2.S36 1994512'.2 - dc20

Die Deutsche Bibliothek - Cataloging-in-Publication Data

94-20795

CIP

Schmidt, Roland:Subgroup lattices of groups / by Roland Schmidt. - Berlin ; NewYork : de Gruyter, 1994

(De Gruyter expositions in mathematics ; 14)ISBN 3-11-011213-2

NE: GT

© Copyright 1994 by Walter de Gruyter & Co., D-10785 Berlin.All rights reserved, including those of translation into foreign languages. No part of this bookmay be reproduced in any form or by any means, electronic or mechanical, including photocopy,recording, or any information storage and retrieval system, without permission in writing fromthe publisher.Printed in Germany.Typesetting: ASCO Trade Ltd. Hong Kong. Printing: Gerike GmbH, Berlin.Binding: Liideritz & Bauer GmbH, Berlin. Cover design: Thomas Bonnie, Hamburg.

To Traute and

Alexander Katharina Viktoria Johannes Maria

Preface

The central theme of this book is the relation between the structure of a group andthe structure of its lattice of subgroups. The origin of the subject may be traced backto Dedekind, who studied the lattice of ideals in a ring of algebraic integers; hediscovered and used the modular identity, which is also called the Dedekind law, inhis calculation of ideals. But the real history of the theory of subgroup lattices beganin 1928 with a paper of Ada Rottlander which was motivated by the Galois cor-respondence between a field extension and its Galois group. Subsequently manygroup theorists have worked in this field, most notably Baer, Sadovskii, Suzuki, andZacher. Since the only book on subgroup lattices is still Suzuki's Ergebnisberichtof 1956, there is a clear need for a book on this subject to describe more recentdevelopments.

We denote the subgroup lattice of a group G by L(G). If G and Gare groups, anisomorphism from L(G) onto L(G) is called a projectivity from G to G. Since isomor-phic groups have isomorphic subgroup lattices, there is a correspondence betweenthe class of all groups and a certain class of (algebraic) lattices in which a group issent to its subgroup lattice. This raises the following questions:

(A) Given a class X of groups, what can we say about the subgroup lattices ofgroups in X?

(B) Given a class IV of lattices, what can we say about the class of groups G suchthat L(G) a J?

Since there are nonisomorphic groups having isomorphic subgroup lattices, wemay further ask:

(C) If X is a class of groups, what is the lattice-theoretic closure 1 of X? Here X isthe class of all groups G for which there exists G e X such that L(G) is isomorphic toL(G).

(D) Which classes of groups X satisfy _ X, that is, are invariant under pro-jectivities?

If X consists of just one group, Problem (D) becomes the following:(E) Which groups G are determined by their subgroup lattice, that is are isomor-

phic to every group G with L(G) isomorphic to L(G)?Every isomorphism between two groups induces a projectivity in a natural way;

our final question is:(F) Which groups are not only determined by their subgroup lattice but even

have the property that all their projectivities are induced by group isomorphisms?These are the main problems in the field and they are studied in the first seven

chapters of the book. In Chapter 8 we consider dualities between subgroup latticesand in Chapter 9 we briefly handle other lattices associated with a group, namelythe lattices of normal subgroups, composition subgroups, centralizers, andcosets.

viii Preface

Although the theory of subgroup lattices is just a small part of group theory, forreasons of space I have been unable to include all the contributions to this theory. Ihave tried to compensate for omissions by at least including statements and refer-ences wherever possible; also many interesting results are cited in the exercises at theend of each section. However, certain parts of the theory have not been covered atall. In writing what is after all a book on group theory I decided to omit those topicswhich would have led too far away from this subject. Therefore I do not coverembeddings of lattices in subgroup lattices, lattices of subsemigroups, homotopyproperties of subgroup lattices, inductive groupoids, or anything about topologicalgroups, although there are also interesting results in these fields. Furthermore I donot study homomorphisms between subgroup lattices. The reason is that when Istarted to write this book there had been no activity in this area for more than 30years, so that Suzuki's book remained an adequate reference. This situation changedin 1985 and especially since 1990, but by then there was no room in the book for achapter on lattice homomorphisms.

Nevertheless I hope that this book will serve as a basic reference in the subjectarea, as a text for postgraduate studies, and also as a source of research ideas. It isintended for students who have some basic knowledge in group theory and wish todelve a little deeper into one of its many special areas. It is assumed that the readeris familiar with the basic concepts of group theory, but references are given for manywell-known group theoretical facts. For this purpose I mainly use the books byRobinson [1982] and Huppert [1967]. I do not assume that the reader is familiarwith lattice theory, but the basic facts are developed in § 1.1 and at the beginning ofsome other sections. For a thorough introduction into lattice theory the reader isreferred to the books of Birkhoff, Crawley and Dilworth, or Gratzer. Since the bookis intended as a textbook for students, I have included a large number of exercises ofvarying difficulty. While these exercises are not used in the main text they often givealternate proofs or contain supplementary results for which there was no room inthe main text.

It is a pleasure to thank the many students and colleagues who have read parts ofthe manuscript and made helpful suggestions. In particular, I mention O.H. Kegeland G. Zacher who have followed the whole project with much interest. Thanksare also due to Ms. J. Robinson for checking the English text and to FrauG. Christiansen for typing parts of the manuscript. Finally, I thank Dr. M. Karbeand the publisher for their courtesy and cooperation.

Contents

Preface ............................................................ vii

Notation .......................................................... xiii

Chapter 1Fundamental concepts

1.1 Basic concepts of lattice theory .................................... 1

1.2 Distributive lattices and cyclic groups .............................. 11

1.3 Projectivities ................................................... 16

1.4 The group of autoprojectivities .................................... 231.5 Power automorphisms ........................................... 291.6 Direct products ................................................. 34

Chapter 2Modular lattices and abelian groups .................................... 41

2.1 Modular lattices ................................................ 41

2.2 P-groups ....................................................... 492.3 Finite p-groups with modular subgroup lattices ..................... 542.4 Groups with modular subgroup lattices ............................ 692.5 Projectivities of M-groups ........................................ 882.6 Projectivities between abelian groups .............................. 101

Chapter 3Complements and special elements in the subgroup lattice of a group ........ 114

3.1 Groups with complemented subgroup lattices (K-groups) ............. 1153.2 Special complements ............................................ 121

3.3 Relative complements ............................................ 1263.4 Neutral elements and related concepts ............................. 135

3.5 Finite groups with a partition ..................................... 145

Chapter 4Projectivities and arithmetic structure of finite groups ..................... 157

4.1 Normal Hall subgroups .......................................... 158

4.2 Singular projectivities ............................................ 1664.3 OO(G), O°(G), Fitting subgroup and hypercentre ...................... 1784.4 Abelian p-subgroups and projectivities ............................. 185

x Contents

Chapter 5Projectivities and normal structure of finite groups ....................... 199

5.1 Modular subgroups of finite groups ................................ 2005.2 Permutable subgroups of finite groups ............................. 2165.3 Lattice-theoretic characterizations of classes of finite groups ........... 2285.4 Projective images of normal subgroups of finite groups ............... 2355.5 Normal subgroups of p-groups with cyclic factor group ............... 2435.6 Normalizers, centralizers and conjugacy classes ...................... 272

Chapter 6Projectivities and normal structure of infinite groups ...................... 281

6.1 Subgroups of finite index ......................................... 2826.2 Permodular subgroups ........................................... 2886.3 Permutable subgroups of infinite groups ............................ 3036.4 Lattice-theoretic characterizations of classes of infinite groups ......... 3116.5 Projective images of normal subgroups of infinite groups and index

preserving projectivities .......................................... 321

6.6 The structure of N°/NU and N°/Na and projective images ofsoluble groups .................................................. 330

Chapter 7Classes of groups and their projectivities ................................ 343

7.1 Free groups .................................................... 345

7.2 Torsion-free nilpotent groups ..................................... 3607.3 Mixed nilpotent groups .......................................... 374

7.4 Periodic nilpotent groups ........................................ 385

7.5 Soluble groups .................................................. 395

7.6 Direct products of groups ........................................ 406

7.7 Groups generated by involutions .................................. 421

7.8 Finite simple and lattice-simple groups ............................. 439

Chapter 8Dualities of subgroup lattices ......................................... 452

8.1 Abelian groups with duals ........................................ 4528.2 The main theorem ............................................... 4588.3 Soluble groups with duals ........................................ 4628.4 Finite groups with duals ......................................... 4688.5 Locally finite groups with duals ................................... 476

Chapter 9Further lattices ..................................................... 4859.1 Lattices of normal subgroups ..................................... 4869.2 Lattices of subnormal subgroups .................................. 498

Contents xi

9.3 Centralizer lattices .............................................. 511

9.4 Coset lattices ................................................... 527

Bibliography ....................................................... 542

Index of Names .................................................... 561

Index of Subjects ................................................... 564

Notation

Throughout this list, G denotes a group, I a class of groups, L a lattice, p a prime, ita set of primes, n a natural number, x and y elements of G, u and v elements of L, andcp a projectivity from

IV, fro, 7L

0, IB, C

PGF(p")UpXS-= Y,XX\YIXI

(a, b)

P nCH=K

G to a group G.

Y

DrX,AeA

set of natural numbers, nonnegative integers, integersfield of rational numbers, real numbers, complexnumbersset of all primesfield with p" elementsgroup of p-adic unitsX is a subset, a proper subset of the set Yset of all elements in X which are not contained in Ycardinality of the set Xa divides b, a does not divide b, (a, b e 7L)greatest common divisor of a and b, (a, b e 7L)P\(p), P\nH is isomorphic with Kdirect product

Lattice theory notation

S,n,Uns,us

,

[u/v]

u mod LC(L)

M"

partial order, intersection, union in Lgreatest lower bound, least upper bound of the subset Sleast, greatest element of Linterval of all w e L such that v< w 5 uu is a modular element of Lset of all cyclic elements of Llattice of length 2 with n atoms

Group theory notation

Y-'xYx-'x' where s is an element or an automorphism of Gcyclic subgroup generated by xorder of x(y°)-'(x°)-'(xy)°, amorphy of the map a

xiv

HS G, H < GHgGHgaGH mod GH pmo GH per GIG:HI<XAIA E A)HuK, U Ht

AEA

HxHa, Ha

H'0z'0 -'

HG,HUCG(H), N6(H)CG(H/K)

J(H)<XIR>H wr KCr Gx.SEA

Aut G, Inn GPot GEnd GHom (A, B)P(G)PA(G), PI(G)

G"

((G), ("(G)U"(G)Z(G)

N(G)

[X, S]

G'=[G,G]Com GZ"(G), K"(G), Gc"'

Z.(G)F(G)

cD(G)

Notation

H is a subgroup, a proper subgroup of GH is a normal subgroup of GH is a subnormal subgroup of GH is a modular subgroup of GH is a permodular subgroup of GH is a permutable subgroup of Gindex of the subgroup H in Gsubgroup generated by subsets Xx of Gsubgroup generated by subgroups H, K, Hx of G

<Hxlx a X) for subsets H and X of Gnormal closure, core of the subgroup H in G((H0)z0-' for H S G and z e G((H4' )Gyp ', ((HP)-)'P -'for H < Gcentralizer, normalizer of H in G{xeGI [H,x]<K}orbit of a E ) under the operation of G on the set )stabilizer of the point a under the operation of GI(x): <x) n HI, order of x modulo the subgroup Hset of all x e G such that o(x, H) is infinite or a primepowerset of all x E G such that o(x, H) is a prime powergroup presented by generators X and relations R(standard restricted) wreath product of H with Kcartesian product of groups Gx

automorphism group, inner automorphism group of Ggroup of power automorphisms of Gset of endomorphisms of Gset of homomorphisms from A to Bgroup of autoprojectivities of Ggroup of all autoprojectivities of G induced by auto-morphisms, by inner automorphisms(x"lx a G>(xeGlxP= 1>,(xeGlx°"= 1) if G is a p-groupGP" if G is a p-groupcentre of Gnorm of G([x, s] I x e X, s E S> for subsets X of G and S of G orAut Gcommutator subgroup of Gset of commutators of elements of Gterms of the upper central series, the lower centralseries, the derived series of Ghypercentre of GFitting subgroup of GFrattini subgroup of G

F"(G), 1V"(G)

R(G)S(G)G.'G,Op(G), O"(G)Op(G), O"(G)

Gp

T(G)Sylp(G)Exp G7r(G)

c(G)d(G)1p(G)

ro(G)LX

W, 91, aP(n, p)Sym X, S.Alt X, AnC", CW

Cp=

D.Q2GL(V), GL(n, F), GL(n, q)SL(V), SL (n, F), SL (n, q)PGL(n, q), PSL(n, q)

PSU(V), PQ(V), PSU(V)Sz(q)L(G)L"(G)92(G)R(G)(!(G)9R(G)

[X/Y]N, [X/Y]K, [X/Y]c

Notation xv

terms of the upper Fitting series, the lower Fitting seriesof GHirsch-Plotkin radical of Glocally soluble radical of GI-radical of GI-residual of Gmaximal normal p-subgroup, 7r-subgroup of Gminimal normal subgroup of G with factor group a p-group, a n-groupp-component of the nilpotent torsion group Gtorsion subgroup of Gset of Sylow p-subgroups of Gexponent of Gset of all primes dividing the order of an element of Gnilpotency class of Gderived length of Gp-length of Gtorsion-free rank of the abelian group Gclass of all locally 1-groupsclasses of abelian, nilpotent, soluble groupsP-groupssymmetric group on the set X, on (1,...,n}alternating group on the set X, on (1, ... , n}cyclic group of order n, of infinite orderPrefer group, quasicyclic groupdihedral group of order ngeneralized quaternion group of order 2"general linear groupspecial linear groupprojective general linear and projective special lineargroupsprojective symplectic, orthogonal, unitary groupsSuzuki groupssubgroup lattice of Gset of all subgroups of G generated by n elementslattice of normal subgroups of Glattice of composition subgroups of Gcentralizer lattice of Gcoset lattice of Ginterval in 91(G), R(G), (1(G)

Chapter 1

Fundamental concepts

In this first chapter we briefly introduce the basic concepts of lattice theory anddevelop the elementary properties of subgroup lattices and projectivities of groupsthat will be needed later.

In § 1.1, in addition to the purely lattice-theoretic concepts, we define the sub-group lattice L(G) of a group G, projectivities between groups, and introduce certainsublattices and meet-sublattices of L(G), namely the lattices of normal subgroups,composition subgroups, centralizers, and cosets of G.

The main result of § 1.2 is Ore's theorem of 1938 which characterizes the groupswith distributive subgroup lattices as the locally cyclic groups. It yields the basic factthat every projectivity maps cyclic subgroups to cyclic subgroups. So it seems rea-sonable to try to construct projectivities between two groups G and G using mapsbetween the sets of cyclic subgroups or even element maps between G and G. Thiswe study in § 1.3, and then prove Sadovskii's approximation theorems which enableus to extend isomorphisms inducing a given projectivity cp on certain sets of sub-groups or factor groups of G to an isomorphism inducing cp on the whole group.

In § 1.4 we study the group P(G) of all autoprojectivities of G and introduce thesubgroups PA(G) and PI(G) consisting of all autoprojectivities induced by automor-phisms and inner automorphisms, respectively. The kernel of the natural map fromAut G to PA(G) is the group Pot G of all power automorphisms of G. In § 1.5 weprove Cooper's theorem that every power automorphism is central and determinethe power automorphisms of abelian groups.

Finally, in § 1.6 we investigate direct products of groups and prove Suzuki's theo-rem that the subgroup lattice of a group G is a direct product if and only if G is adirect product of coprime torsion groups. As a nice application we obtain that thenumber of groups whose subgroup lattice is isomorphic to a given finite lattice L isfinite if and only if L has no chain as a direct factor.

1.1 Basic concepts of lattice theory

A partially ordered set or poset is a set P together with a binary relation < such thatthe following conditions are satisfied for all x, y, z e P:

(1) x < X. (Reflexivity)

(2) If x < y and y < x, then x = y. (Anti symmetry)

(3) If x < y and y < z, then x < z. (Transitivity)

2 Fundamental concepts

An element x of a poset P is said to be a lower bound for the subset S of P if x < sfor every s e S. The element x is a greatest lower bound of S if x is a lower bound ofS and y < x for any lower bound y of S. By (2), such a greatest lower bound of S isunique if it exists; we denote it by inf S or n S. Similar definitions and remarks applyto upper bounds and the least upper bound; the latter is denoted by sup S or U S.

Lattices

A lattice is a partially ordered set in which every pair of elements has a least upperbound and a greatest lower bound. A partially ordered set in which every subset hasa least upper bound and a greatest lower bound is called a complete lattice. We canalso view lattices as algebras with two binary operations.

1.1.1 Theorem. (a) Let (L, <) be a lattice and define the operations n and u on L(called intersection and union, respectively) by

(4) x n y = inf {x, y} and x u y = sup {x, y}.

Then the following properties hold for all x, y, z e L:

(5) xny = y n x and x u y= y u x. (Commutativity)

(6) (x n y) n z= x n (y n z) and (x u y) u z= x u (y u z). (Associativity)

(7) x n (x u y) = x and x u (xny) = x. (Absorption identities)

Furthermore, we have x < y if and only if x = x n y (or y = y u x).(b) Conversely, let L be a set with two binary operations n and u satisfying (5)-(7)

and define the relation S on L by

(8) x 5yif and only if x =xny.

Then (L, <) is a lattice with x n y = inf{x, y} and x u y = sup{x, y} for all x, y e L.

Proof. (a) Clearly, the operations defined in (4) are commutative. Also it is easy tosee that (x n y) n z and x n (y n z) both are the greatest lower bound of {x, y, z}.Hence (x n y) n z = x n (y n z) and similarly (x u y) u z = x u (y u z). Furthermorex < y if and only if x = inf{x, y} = xny (or y = sup{y, x} = y u x). Since x < x u yand x n y < x, we therefore have x n (x u y) = x and x u (x n y) = x.

(b) Let x, y, z e L. We note first that the conditions x n y = x and y u x = yare equivalent; for, if x n y = x holds, then y u x = y u (y n x) = y and converselyy u x = y implies x n y = x n (y u x) = x. So by (8),

(9) x < y if and only if y = y u x.

We now show that (L, 5) is a poset. The absorption identities yield x n x =xn(xu(xnx))=x;by (8),x<x.If x<yand y5x,then x=xny=ynx=y.If x <y and y <z, then

xnz=(xny)nz=xn(ynz)=xny=xand hence x < z. So (1)-(3) hold and (L, <) is a poset.

1.1 Basic concepts of lattice theory 3

Since (xny)nx=xn(xny)=(xnx)ny=xny,xny<x. Similarly x r) y!5; y.If z is any lower bound of {x, y}, then z n x = z and z n y = z. Hence z n (x n y) =(z n x) n y = z n y = z and z < x n y. This shows that x n y = inf {x, y}. Replacingn by u and using (9) instead of (8) we get x u y = sup {X, y}. In particular, (L, S) isa lattice.

Lattices occur in abundance. For any set X, the set of all subsets of X partiallyordered by set inclusion is a complete lattice; the greatest lower bound of any collec-tion of subsets is the set intersection, and the least upper bound is the set union. Thesubject of this book is the following lattice.

Subgroup lattice

If G is any group, the set L(G) of all subgroups of G is partially ordered with respectto set inclusion. Moreover any subset of L(G) has a greatest lower bound in L(G),the intersection of all its elements, and a least upper bound in L(G), the join of all itselements. Thus L(G) is a complete lattice, the subgroup lattice of G. We denote theoperations of this lattice by n and u. So we shall write X n Y for the intersection,and X u Y but sometimes also <X, Y> for the join of the subgroups X and Y of G.Also, for a set .9' of subgroups of G, U .9' and <SIS e .9') both denote the groupgenerated by the subgroups in Y.

Minimal and maximal elements, chains and antichains

For elements x, y of a poset P we write x < y if x 5 y and x 0 y. If x < y and thereis no element z e P such that x < z < y, then we say that x is covered by y or that ycovers x. Let S be a subset of P and x e S. Then x is a minimal element of S if thereis no s e S with s < x, and x is a least element of S if x 5 s for every s e S. Certainlythe least element of S is a minimal element of S and is unique if it exists; on the otherhand, it may happen that S contains many minimal elements but no least element.Similar definitions and remarks apply to maximal elements and the greatest elementof S. We write 0 and I for the least and greatest elements of P, respectively (if theyexist). An element of P that covers 0 is called an atom, an element that is covered byI is an antiatom of P. If L is a complete lattice, then inf L is the least and sup L thegreatest element of L. In the subgroup lattice of the group G the trivial subgroup 1is the least element, G is the greatest element, the minimal subgroups are the atomsand the maximal subgroups are the antiatoms of L(G).

Two elements x and y in a poset P are called comparable if x < y or y 5 x. Asubset S of P is a chain if any two elements in S are comparable, S is an antichain ifno two different elements of S are comparable. The length of a finite chain S isISI - 1. The poset P is said to be of length n, where n is a natural number, if there isa chain in P of length n and all chains in P are of length at most n. A poset P is offinite length if it is of length n for some n e N. Similarly P is said to be of width n ifthere is an antichain with n elements in P and all antichains in P have at most nelements.

A poset P with the property that each of its nonempty subsets contains a maximalelement is said to satisfy the maximal condition or the ascending chain condition. Thesecond name comes from the following alternative formulation: P satisfies the maxi-mal condition if and only if it contains no infinite sequence of elements x1, x2, ...such that x1 < x2 < x3 < - The minimal condition or descending chain condition isdefined correspondingly.

Isomorphisms and projectivities

Let L and L be lattices. A map a: L -. L is called a homomorphism if

(10) (xny)° = x°ny° and (xuy)° = x°uy°

for all x, y c L. A homomorphism is an isomorphism if it is bijective, and an isomor-phism of a lattice with itself is called an automorphism. We write L - L if L and L areisomorphic. Finally, we introduce a shorter name for isomorphisms between sub-group lattices. If G and G are groups, an isomorphism from L(G) to L(G) is calleda projectivity from G to G. We also say that G and G are lattice-isomorphic if thereexists a projectivity from G to G.

In order to show that a bijective map between two lattices is an isomorphism itsuffices to prove that it has one of the two properties in (10) or that it preserves theorder relations of the lattices. In addition, every isomorphism preserves union andintersection of arbitrary subsets.

1.1.2 Theorem. Let a be a bijective map of a lattice L to a lattice L. Then the follow-ing properties are equivalent.

(a) For all x,yEL,x <y if and only if x' <y'.(b) (xny)°=x°ny°for all x,yEL.(c) (xuy)°=x°uy°forallx, yEL.

Furthermore, if a satisfies (a) and S is a subset of L such that n S exists, then alson S' exists and (n s), = n Sc; similarly (U S)° = U S° if U S exists.

Proof. Let x, y c- L. By 1.1.1, x < y if and only if x n y = x (or x u y = y). It followsthat (b) (and also (c)) implies (a). Conversely, if a satisfies (a) and if S is a subset of Lsuch that z = n S exists, then z° is a lower bound of S° and every lower bound of S°is of the form w° where w is a lower bound of S. Since z is the greatest lower boundof S, w < z and hence w° < z°. This shows that z° is a greatest lower bound of S° andthat (n S)° = n S°. In particular (b) holds. The corresponding result for the leastupper bound follows similarly.

An easy, but useful, consequence of this result is that a projectivity of a group isdetermined by its action on the cyclic subgroups.

1.13 Corollary. If q and 0 are projectivities of a group G such that X1 = X'' forevery cyclic subgroup X of G, then cp = 0.

Proof. Let H < G. Then by 1.1.2, HW _ (xU (x>)4= zU (x)- = xU <x> = H.

Hasse diagram

Every finite poset P, in particular every finite lattice, may be represented by a dia-gram. Represent each element of P by a point in the plain in such a way that thepoint py associated with an element y lies above the point associated with x when-ever x < y. Then, whenever y covers x, connect the points px and p, by a linesegment. We call the resulting figure a Hasse diagram of P. It is easy to see that it ispossible to construct such a diagram. For the finite poset P has a maximal elementz, P\ {z} is a smaller poset, and therefore, by induction, has a diagram; now choosea point p= above the points of this diagram and connect it with all the points pxassociated to elements x covered by z. Figure 1 shows the Hasse diagrams of alllattices with at most five elements.

A, A, A3 A, B, A5 B5 C5 D5 E5

Figure 1

Sublattices

A subset of a lattice is called a sublattice if it is closed with respect to the operationsn and u defined in I.I.I. It is evident that a sublattice is a lattice relative to theinduced operations. Examples of sublattices of a lattice L are, for x, y E L, the setsS = {x, y, x n y, x u y} or, if x < y, the intervals

[y/x] = {z E Llx < z < y}.

We mention two sublattices of the subgroup lattice of a group G.

Lattice of normal subgroups

SJR(G) = {NON L:3 G}.

Lattice of composition subgroups

We say that the subgroup H of G is subnormal in G, and we write H g a G, if thereare a nonnegative integer n and a series

(11) H=H, =G

of subgroups of G. We say that H is a composition subgroup of G, if there exists aseries (11) for H in which all the factor groups H./H;+1 are simple; we then call n thetype of H and (11) a composition series from G to H of length n. Clearly a composi-tion subgroup is always subnormal, but the converse is false.

1.1.4 Lemma. If H is a composition subgroup of type n in G, H < K < G andK :9:9 G, then there exist composition series of lengths less than n from G to K andfrom K to H.

Proof. Let K = K. a Krn_1 9 g KO = G and let (11) be a composition seriesfrom G to H. Then

(12) =G

is a normal series from G to H. By the Schreier refinement theorem, the normal series(11) and (12) have isomorphic refinements. There is no proper refinement of thecomposition series (11). Hence the isomorphic refinement of (12) is a compositionseries from G to H of length n. Since H < K < G, the two parts from G to K andfrom K to H of this series are composition series of lengths less than n. El

1.15 Theorem (Wielandt [1939]). Let G be a group. The set $1(G) of all compositionsubgroups of G is a sublattice of the subgroup lattice L(G).

Proof. Let A, B e R(G). Then there are composition series

and =G

from G to A and B, respectively. We use induction on min(m, n) to prove thatA n B e R(G), and we may assume that m 5 n. If m = 0, then A n B = B e R(G); solet m >_ 1. Since

Al SA1uB=A1uB,9A1uBn_1 =G

and G/A1 is simple, we have either Al u B = Al or Al u B = G. In the first case,Al n B = B; in the second, Al n B a B and B/A1 n B G/A1 is simple. In bothcases, Al n B e R(G) and 1.1.4 yields that Al n B e R(A1). By induction A n B =A n (A1 n B) e R(A1), and so A n B e R(G) since G/A1 is simple.

We use induction on the ordered pair (max(m, n), min(m, n)) to prove that A U B eR(G). Since this is clear if m = 0 or n = 0, we may assume that m, n >_ 1 and that thejoin of two composition subgroups of types r and s in a group is again a compositionsubgroup whenever r 5 s < max(m, n) or s = max(m, n) and r < min(m, n). Suppose

that A is not normal in A v B. Then there exists b e B such that A6 # A. SinceA1 :Q G, Ab < A, and there are composition series of lengths m - 1 < max(m, n)from A, to A and to Ab. By induction A u Ab e $1(A,) and hence A u Ab 9 9 G.By 1.1.4 there is a composition series of length less than m from G to A u A6.By induction A u B = (A u Ab) u B E R(G), as desired. So we may assume thatA 9 A u B and, similarly, B 9 A u B. Again by induction A._, v B E R(G). IfAm_, u B # G, then by 1.1.4 there exist composition series of lengths less than m andn from Am_, u B to A and B, respectively; by induction A u B E R(Am_, U B) andtherefore A u B E R(G). So we may finally assume that Am_, u B = G and, similarly,A u G. Then A 9 G since A is normalized by Am_, and B. Similarly B:9 Gand so AuB9G.By 1. 1.4, A u B e R(G).

Lemma 1.1.4 shows that in a group G with a composition series (from G to 1)every subnormal subgroup is a composition subgroup. Hence in these groups, inparticular in finite groups, subnormal and composition subgroups coincide. There-fore, in this case, R(G) is also called the lattice of subnormal subgroups of G. Forsubnormal subgroups A and B in an arbitrary group G, it is clear that A n B 9 9 G,but A u B in general is not subnormal in G. The reader will find examples and athorough investigation into the join problem in the book by Lennox and Stonehewer[1987].

Meet-sublattices

Let L be a lattice and M a subset of L which is a lattice relative to the induced partialorder. Clearly, M need not be a sublattice of L; the subgroup lattice of a group G asa subset of the lattice of all subsets of G is an example. If n and n denote theintersection in L and M, respectively, then for x, y e M, x A y is a lower bound of{x, y} in L and so x A y S x n y. We call M a meet-sublattice of L if x A y= x n yfor all x, y E M, and M is a complete meet-sublattice of L if M and L are completelattices and for every subset S of M the greatest lower bounds of S in M and Lcoincide. Actually, one need not require that M is a lattice here.

1.1.6 Lemma. If M is a subset of a complete lattice L such that for every subset S ofM the greatest lower bound n S of sin L is contained in M, then M is a completemeet-sublattice of L.

Proof. For a subset S of M let S* be the set of upper bounds of S in M. By hypothe-sis, inf S* E M. Every s E S is a lower bound of S*, hence s S inf S*. Thus inf S* is anupper bound of S and then, certainly, the least upper bound of S in M. Clearly, inf Sis the greatest lower bound of S in M. Hence M is a complete lattice and the greatestlower bounds of S in M and L coincide.

If G is a group, L(G) is an example of a complete meet-sublattice of the lattice ofall subsets of G. We introduce two further lattices of this kind.

Centralizer lattice

Let (1(G) be the set of all centralizers of subgroups of G. Since for Hi 5 G,

in C6(Hi) = Ca Hi I e I(G),

(4;(G) is a complete meet-sublattice of L(G); in general it is not a sublattice (seeExercise 7). We call E(G) the centralizer lattice of G. The least element of (f(G) is thecentre Z(G), the greatest element is G.

Coset lattice

Let 91(G) be the set of all right cosets of subgroups of G together with the empty set0. If 9 is a nonempty subset of 91(G), then either n .9' = 0 e 91(G) or there existsan element x en .. In the latter case, 9 = {HixIi e I} for certain subgroups Hiof G, and hence n.5" = (n Hi) x e 91(G). If .5" is the empty subset of 91(G), then

n .5" = G E 91(G). By 1.1.6, 91(G) is a complete meet-sublattice of the lattice of allsubsets of G. Since Hx = x(x-'Hx) for all H 5 G and x e G, every right coset is a leftcoset, and conversely. So 91(G) is the set of all (right or left) cosets in G and we call91(G) the coset lattice of G. The subgroups of G are precisely the cosets containingthe coset 1. Hence L(G) is the interval [G/1] in 91(G).

Direct products

Let (LA)A E A be a family of lattices. The direct product, L = Dr Lx, is the lattice whoseAEA

underlying set is the cartesian product of the sets Lx, that is the set of all functions fdefined on A such that f (A) e Lx for all ). a A, and whose partial order is defined bythe rule that f 5 g if and only if f(A) < g(A) for all A E A. Clearly, f n g and f ug arethe functions mapping each A to f(2) n g(A) and f(l) u g(.1), respectively, so that L isa lattice in which intersection and union are performed componentwise. It followsthat L satisfies a given lattice identity if and only if each Lx satisfies this identity. IfA is finite, say A = { I_-, n} for some n e ICI, we also use the simpler n-tupel notationfor the elements of L. Thus, in this case,

L= L, x x L. =

and (x...... x.) 5 if and only if xi < yi for all i = 1, ..., n.We shall need the following simple result which clarifies the structure of a direct

product of lattices with greatest and least elements.

1.1.7 Lemma. Let (LA)AEA be a family of lattices, and put L = Dr L.J. Suppose that LleA

contains a least element 0 and a greatest element I. If A a A, then 0(2) is the least

and I(A) the greatest element of LA. Define fA e L by ft(µ) = 0(µ) for A # µ E A andf(A) = 1(A1). Then

(a) U f,,=land U f, nf,=0,veA vEA\{.Z}

(b) [ ft/0] z LA, and(c) [x u y/O] [x/O] x [y/0] if x e [fx/O], y e [ f,/O] and A µ E A.

Proof. For z e LA, the axiom of choice implies that there is a function f E L withf(A) = z. From 0S f 5 I it follows that 0(A) S z S 1(.1). This shows that 0(.1) is theleast and 1(2) the greatest element of LA. Since union and intersection in L areperformed componentwise, (a) is clear. The map a: Lx - [ fZ/0] defined by z°(v) =0(v) for A # v e A and z°(.1) = z clearly is an isomorphism; thus (b) holds. Finally, themap

T: [x/0] x [y/O] - [x v y/0]; (u, v) i-- w

where w(1) = u(.1), w(µ) = v(µ) and w(v) = 0(v) for v e A with A # v # z also is anisomorphism, and this implies (c).

Embedding of lattices in subgroup lattices

We mention without proof three results on subgroup lattices that have little connec-tion with the topics treated in this book. They show how complex subgroup latticescan be.

1.1.8 Theorem (Whitman [1946]). Every lattice is isomorphic to a sublattice of thesubgroup lattice of some group.

The reader can find a proof of this result in Whitman's paper or in the book byGratzer [1978]. Whitman first shows that every lattice L is isomorphic to a sub-lattice of a partition lattice of some set X; this lattice then is isomorphic to a sub-lattice of the subgroup lattice of the symmetric group Sym X on X (see Exercise 9).The set X constructed by Whitman in general is infinite even when L is finite.However, with a different construction, it is possible to keep the set X finite in thiscase.

1.1.9 Theorem (Pudlak and Tuma [1980]). Every finite lattice is isomorphic to asublattice of the subgroup lattice of some finite group.

The situation is more complex if one tries to replace "sublattice" by "interval" inthese statements. An element c in a complete lattice L is called compact if wheneverS c L and c <_ U S there exists a finite subset T S S with c :!g U T; and a latticeis called algebraic if it is complete and each of its elements is a join of compact

elements. Now if H is a subgroup of a group G and g e G, then <H, g) clearlyis a compact element in [G/H]. Thus every interval [G/H] is an algebraic lattice,and so the following converse improves Whitman's theorem in this direction as faras possible.

1.1.10 Theorem (Tuma [1989]). Every algebraic lattice is isomorphic to an interval inthe subgroup lattice of some group.

The finite analogue, however, is an open problem even for the simplest lattices, forexample, for the lattice M of length 2 with n atoms. It is easy to construct finite(soluble) groups with such an interval if n - 1 is a prime power (see Exercise 11) andthe lattices M7 and M11 occur in the alternating group A31 (see Feit [1983], Palfy[1988]). However, if n > 13 and n - 1 is not a prime power, it is not known whetherM,. is an interval in the subgroup lattice of some finite group.

Exercises

1. LetLbealatticeandletx,y,zeL.(a) If y<z,then xny<xnzand xuy5xuz.(b) xu(ynz)S(xuy)n(xuz)and (xny)u(xnz)<xn(yuz).(c) IfxSz,then xu(yr z)<(xuy)r z.

2. Which of the Hasse diagrams displayed in Figure 1 represent subgroup lattices?3. Draw Hasse diagrams for the 15 lattices with 6 elements.4. Let G be a group and let A, B g g G.

(a) Show that A n B 9 a G.(b) If AB=BA,prove that AuBggG.

5. (Curzio [1953]) Show that the abelian group G = (a) x with o(a) = 8,o(b) = 2 and the group <c,dlc8 = d2 = 1, dcd = cs) have isomorphic cosetlattices.

6. If G = H x K, then (1;(G) C(H) x E (K).7. If G = <a, bIa15 = b2 = 1, bab = a-1) is the dihedral group of order 30, show

that E (G) is not a sublattice ofL(G). _8. If G and G are groups with 91(G) a- 91(G), show that L(G) L(G) and 161 = I G I9. Let X be a set and let 11(X) be the set of all equivalence relations (or parti-

tions) on X. For p, rr e 11(X) define p S T if and only if xpy implies xry for all x,yeX.(a) Show that II(X) is a lattice, the partition lattice of X.(b) Prove that 11(X) is isomorphic to a sublattice of L(Sym X).

10. Show that the compact elements in the subgroup lattice of a group are preciselythe finitely generated subgroups.

11. Let F be the field with p' elements, V a vector space of dimension 2 over F andG the group of all permutations x -o, ax + v over V where 0 # a e F and v e V. IfH = {x ax 10 # a e F}, show that [G/H] M,+1.

1.2 Distributive lattices and cyclic groups 11

1.2 Distributive lattices and cyclic groups

In this section we are going to study subgroup lattices of cyclic groups. Since theseturn out to be distributive, we shall first determine all groups with distributivesubgroup lattices.

Distributive lattices

A lattice L is called distributive if for all x, y, z e L the distributive laws hold:

(1) xu(ynz)=(xuy)n(xuz),(2) xn(yuz)=(xny)u(xnz).

1.2.1 Remark. Since distributivity is a property defined by identities, sublatticesand direct products of distributive lattices are distributive. Furthermore, a lattice Lis already distributive if it satisfies one of the two distributive laws. For example, if(1) holds in L, then for x, y, z e L we have

(xny)u(xnz) _ ((xny)ux)n((xny)uz) = xn(zu(xny))

= xn((zvx)n(zuy)) = xn(yuz);

thus (2) holds. The other implication follows similarly.

It is easy to see that every chain is a distributive lattice. We shall also need thefollowing distributive lattices. For a, b c No, the set of nonnegative integers, writea 5'b if b divides a. Clearly, < is a partial order on No for which sup{a,b} is thegreatest common divisor and inf{a, b} is the least common multiple of a and b. Thus(NO, S') is a lattice with greatest element 1 and least element 0. We denote this latticeby T., and for n e N, we write T, for the interval [1/n] in T.. Then T is the lattice ofall divisors of n. The reader may check that T. is distributive; this will, however, alsofollow from 1.2.2 and 1.2.3. As sublattices of T., all the T,. are distributive as well.

72

Figure 2: T72

Cyclic groups

Let n be a natural number or the symbol oo. We write n e N u {oo} to express thisand we denote the cyclic group of order n by C. Let C = <g>. When n = oo, it iswell-known that for every r e N there is a subgroup of index r in <g>, namely <g'),and every nontrivial subgroup of <g) is of this form. If n is finite, then for everydivisor r of n there is exactly one subgroup of index r in <g>, again <g'>. Hence, ifwe define v: T,. - L(<g)) by r° = <g'> for all r e T,,, then v is bijective. Furthermore,for r, s e T., we have r <'s if and only if s divides r, and this holds if and only if<g'> < <g'). By 1.1.2, a is an isomorphism. Thus we have shown the following.

1.2.2 Lemma. If n e N u {oo}, then L(C,,) ^- T,,.

Groups with distributive subgroup lattices

A group G is called locally cyclic if every finite subset of G generates a cyclic sub-group. Equivalently we can say that <a, b) is cyclic for every pair a, b of elements ofG. In particular, every locally cyclic group is abelian. The additive group a ofrational numbers and the group Q/7L of rational numbers modulo 1 are locally cyclicsince finitely many rational numbers have a common denominator n and there-fore are contained in the cyclic group generated by -,,. It is a familiar result thata group is locally cyclic if and only if it is isomorphic to a subgroup of 0 or of Q/7L.We come to our first major result.

1.2.3 Theorem (Ore [1938]). The subgroup lattice of a group G is distributive if andonly if G is locally cyclic.

Proof. Suppose first that L(G) is distributive and let a, b e G. We have to show that<a,b) is cyclic. Since <a> n is centralized by a and b, <a> n S Z(<a,b)).Also <ab) u <a> = <a,b) = <ab) u and so by (1),

<ab) u (<a> n ) = (<ab) u <a>) n (<ab) u ) = <a, b).

Thus <a, b)/<a) n <b) <ab)/<ab) n (<a> n <b)) is cyclic and therefore <a, b) isabelian, as a cyclic extension of a central subgroup. By the structure of finitelygenerated abelian groups (see Robinson [1982], p. 100) there exist c, d e G suchthat <a, b) = <c) x <d ). By what we have shown, <c, d)/<c) n <d) is cyclic. Since<c) n <d) = 1, <a, b) = <c, d) is cyclic.

Now suppose that G is locally cyclic and let A, B, C e L(G). By 1.2.1, we need onlyshow that the first distributive law holds, or, since G is abelian, that A(B n C) =AB n AC. Clearly, A(B n Q< AB n AC. Let x e AB n AC, hence x = ab = a'c witha, a' a A, b e B and c e C. Since G is locally cyclic, there exists g e G such that<a,a',b,c) _ <g>. Then ab = a'c implies that <g) = (A n <g))(B n <g)) =(A n <g))(C n <g>). If one of the three subgroups A n <g>, B n <g>, C n <g) istrivial, then either x = b = c e B n C or x e A. In both cases, x e A(B n Q. So sup-

pose that all these subgroups are nontrivial and let n, r, s be the respective indicesof A n (g>, B n <g), C n <g) in <g)'. Then (n, r) = 1 = (n, s), hence (n, rs) = 1 andtherefore

<g) = <g") <grs) = (A n <g>)(B n C n <g)) < A(B n Q.

Again it follows that x e A(B n Q. Thus AB n AC :!5; A(B n C), as required.

1.2.4 Corollary. The subgroup lattice of a finite group G is distributive if and only ifG is cyclic.

Ore's theorem, and perhaps even more its corollary, is one of the most beautifulresults on two of the basic problems mentioned in the preface: Which class of groupsis the class of all groups with a given lattice property and, conversely, which latticeproperty characterizes a given class of groups. Here an interesting class of lattices,the finite distributive lattices, belongs to a simple class of groups, the finite cyclicgroups. This example is rather atypical. We shall see that nice classes of latticesoften lead to fairly complicated classes of groups and, similarly, it is usually verydifficult or even impossible to characterize a given interesting class of groups.

Lattice-theoretic characterizations of cyclic groups

Using Ore's theorem, however, it is not difficult to characterize the class of cyclicgroups.

1.2.5 Theorem. The group G is cyclic if and only if its subgroup lattice L(G) is distrib-utive and satisfies the maximal condition.

Proof. It is well-known and easy to prove that a group whose subgroup latticesatisfies the maximal condition is finitely generated. Thus if L(G) is distributive andsatisfies the maximal condition, then G is locally cyclic, by Ore's theorem, andfinitely generated; hence G is cyclic. Conversely, if G is cyclic, then L(G) is distribu-tive, by Ore's theorem. Furthermore, if 1 < H < G, then [G/H] is finite. So, clearly,L(G) satisfies the maximal condition.

1.2.6 Corollary (Baer [1939a]). Let G be a group. Then G C. if and only ifL(G) T..

Proof. If G C., then L(G) Tom, by Lemma 1.2.2. Conversely, let L(G) T.. By1.2.2, L(G) L(C.). By 1.2.5, L(G) is distributive and satisfies the maximal condi-tion and therefore G is cyclic. Clearly, I GI is infinite.

Thus the infinite cyclic group C. is determined by its subgroup lattice: it is theonly group G such that L(G) L(CD). For finite cyclic groups the situation is differ-ent since for m, m' a N, the lattices T. and T", may be isomorphic although m 96 m'.

If m = p;' ... p;r is the prime power decomposition of m, then every divisor of m hasthe form pi' ... p;" with 0 < ki < n; for i = 1, ..., r. The map o defined by

k, k,. a k. k,.

'P1 Pr) =(P1 ,. ,Pr )

is an isomorphism from Tm to the direct product of the Tp.,,, and these TP., are chainsof lengths ni. Furthermore, Ti,, contains exactly r antiatoms, namely p,, ..., Pr, andto each pi exactly one chain of maximal length through 1 and p,, namely El/p"'] oflength ni. So if m' e R such that Tm. T. and r, say, is an isomorphism from T. toT,,.,, then the p; are distinct primes and m' = (pi)', ... (p; )"r.

1.2.7 Theorem (Baer [1939a]). Let n1, ..., nr E N. The group G is cyclic of orderpi' ... p;r with distinct primes pi if and only if L(G) is a direct product of chains oflengths n l , ..., nr.

Proof. If G is cyclic of order m = p;' ... p,r, then L(G) T. is a direct product ofchains of lengths nl, ..., nr. Conversely, suppose that L(G) is such a direct product,let q1, ..., qr be distinct primes and put m' = qi' ... q;r. Then L(G) Tm. is dis-tributive and finite. By 1.2.5, G is cyclic. Thus Tm. L(G) = Tic,, and hence IGI =pi' ... p" with distinct primes A.

1.2.8 Corollary. Let n1, ... , nr E F and let pl, ..., pr be distinct primes. If G is a cyclicgroup of order p;' ... p;, and G is any group, then L(G) L(G) if and only if G iscyclic of order qi' ... q;r with distinct primes q1, ..., qr.

Thus the finite cyclic groups are far away from being determined by their sub-group lattices. Indeed, to every finite cyclic group C,, there exist infinitely manynonisomorphic groups C with L(C) However, all these groups are cyclic.And that is the important fact. If H is a subgroup of a group G, L(H) is the interval[H/1] in L(G). Hence it is possible to decide in the lattice L(G) whether a givensubgroup H of G is cyclic or not.

Lattice-theoretic characterizations and invariance under projectivities

Let us pause for a general remark on the study of classes of groups and their lattices.Ore's theorem, or rather Theorem 1.2.5, is our first example of a lattice-theoreticcharacterization of a class I of groups. By this we mean the specification of a lattice-theoretic property, that is a class 9) of lattices, such that for every group G,

(3) G E L if and only if L(G) E 9).

We call a class L of groups invariant under projectivities if for every projectivitybetween two groups G and G,

(4) G E L implies 6 E X.

Suppose that a class I of groups has a lattice-theoretic characterization J. Then ifG E I and cp is a projectivity from G to G, L(G) L(G) c J and hence G E X, by (3).So we have the following.

1.2.9 Lemma. Let X be a class of groups. If X has a lattice-theoretic characterization,then I is invariant under projectivities.

The converse of this statement is also true. If X is invariant under projectivities,then the class IV of all lattices isomorphic to some L(G) with G E £ satisfies (3).However, we are looking for a lattice-theoretic description of this class J as, forexample, in Theorem 1.2.5. This result together with 1.2.9 yields the following.

1.2.10 Theorem. Every projectivity maps cyclic groups onto cyclic groups.

If H is a subgroup of G and cp is a projectivity from G to G, then the restriction cPHof p to [H/1] = L(H) clearly is a projectivity from H to H`°; we say that cp inducesthe projectivity (PH in H. Theorem 1.2.10 shows that if H is cyclic, so is H`° since it isthe image of H under the projectivity (PH.

Finite and locally finite groups

We conclude this section with a first application of our results on subgroup latticesof cyclic groups. Let I be a class of groups. We say that G is locally an X-group, ifevery finitely generated subgroup of G lies in X. We write LX for the class of alllocally X-groups.

1.2.11 Lemma. Let X be a class of groups. If X is invariant under projectivities, thenso is LX.

Proof. Let q be a projectivity from G to G and suppose that G E LX. Let H =<x 1i ... , be a finitely generated subgroup of G. By 1.2.10 there exist y1, ... , y E Gsuch that

H4'=(<x,>u...u <x.>)w=<xl>4U...u<x,. >4= <yj > u ... u <y.>-

Thus H4' is finitely generated and hence H`0 E X, since G E LX. Now cp-' induces aprojectivity from H4' to H, and since I is invariant under projectivities, H E 1. ThusG E LX.

1.2.12 Theorem. A group is finite if and only if its subgroup lattice is finite. Henceevery projectivity maps finite groups onto finite groups and locally finite groups ontolocally finite groups.

Proof. If JGI is finite, then G has only finitely many subgroups. Conversely, supposethat L(G) is finite. If C is a cyclic subgroup of G, then L(C) is an interval in L(G),

hence finite; by 1.2.2, ICI is finite. Since G possesses only finitely many such sub-groups and every element of G is contained in one of them, I G I is finite. By 1.2.9, theclass of all finite groups is invariant under projectivities. Then 1.2.11 implies that theclass of locally finite groups has the same property.

Exercises

1. Let p be a prime and write C,, for the Priifer group of type p°°, that is CP. = A/7Lwhere A is the additive group of rational numbers whose denominator is a powerof p. Show that L(CP=) is a chain.

2. If L(G) is a chain, show that G - CP., where p e P and n e N u {oo}.3. (Ore [1938]) The elements x, y of a lattice L are said to be a n-distributive ("meet-

distributive") pair in L if for all z e L the distributive law

zn(xuy)=(znx)u(zny)

holds. Show that the subgroups A, B form a n-distributive pair in L(G) if andonly if for every c e A u B, in neither A nor B, there-exist coprime natural num-bers m and n such that cm e A and c" E B.

4. If A, B is a n-distributive pair in L(G) with A n B a A u B, show that A and Bare normal subgroups of A u B. (Hint: For a e A, B = (B n A) u (B n B°) = B°since A, B and A, Ba are n-distributive pairs.)

5. Use Exercises 3 and 4 to give an alternate proof of Theorem 1.2.3.6. A n-distributive pair x, y is said to be trivial if x < y or y < x. Show that L(G)

has only trivial n-distributive pairs if and only if every element of G has primepower order.

7. (Bruno [1972]) If A, B is a n-distributive pair in L(G), show that A, B is au-distributive pair in L(A u B), that is satisfies

Cu(AnB)=(CuA)n(CuB)

for all C e L(A u B). Also show that the converse is false.8. Let £ be a class of groups and suppose that'd is a class of lattices such that (3)

holds. Find a lattice-theoretic characterization of the class Ll of all locally X-groups.

1.3 Projectivities

In this section we shall present some general methods to construct projectivitiesbetween groups and to find isomorphisms that induce a given projectivity.

1.3 Projectivities 17

Construction of projectivities

It is often possible to work with element maps. For such a map a: G - G and asubset X of G we define X° = {x°jx c X}. If a is bijective and satisfies

(1) X < G if and only if X° < G

for all subsets X of G, then by 1.1.2, the map

(2) i7: L(G) - L(G); H - H°

is a projectivity from G to G. We call it the projectivity induced by Q, and, moregenerally, we say that a given projectivity cp is induced by an element map a betweenthe two groups if H10 = H° for all H < G.

A useful concept in the study of element maps between groups is the amorphy ofa map. It measures the deviation from homomorphy. To be precise, if a is an elementmap from G to G, then the map

(3) a: G x G _9 G; (x,Y)i--,(Y°)-1(x°)-'(xy)°

is called the amorphy of Q. Clearly, (xy)° = x°y°a(x, y) and so a is a homomorphismif and only if a(x, y) = 1 for all x, y c G. Furthermore, we have the associativityidentities

(4) a(x, y)='a(xy, z) = a(y, z)a(x, yz) for all x, y, z e G.

For, we can write

((xy)z)° = (xy)°z°a(xy, z) = x°y°a(x,Y)z°a(xY, z)

and

(x(Yz))° = x°(Yz)°a(x,Yz) = x°y°z°a(Y,z)a(x,Yz),

and the associative law in G implies (4).If a: G - G is bijective and induces a projectivity, then (xy)° E <x,y>° _

<x>° u <y>° = <x°, y°> and hence also a(x,y) E <x°, y°>. We show that, conversely,this property and the corresponding one for o--' suffice to guarantee that a inducesa projectivity. This will be our most useful tool in constructing projectivities.

1.3.1 Theorem. Let a- be a bijective map from G to G such that

(5) (xy)° c <x°, y°> for all x, y c G, and

(6) (uv)°-' E <u°-', v > for all u, v c- G.

Then the map Q: L(G) -- L(G) defined in (2) is a projectivity from G to G.

Proof. Let H < G and let x°, y° E H°. By (6), (x°y°)°-' e <x, y> < H and hencex°y° E H. If we can show that also (x°)-' E H°, then H° is a subgroup of G. By

symmetry then also K°-' < G for every K S G. Thus a satisfies (1) and 6 is a projec-tivity. So it remains to show that (x°)-' a H.

To do this we study an arbitrary pair of elements a, b e G with (a°)-' = b° = u,say. By an obvious induction we get from (5),

(7) (a")° a <u) and (b")° a 

for all n e NI, and from (6), (u")°-' e and (u-")° e <a>, that is .

(8) u" e ° and u-" e <a>°.

Again by (5), (ab)° is contained in <a°,b°> = and hence in ° or in <a>°, by(8). Thus ab lies in or in <a> and so we get the following.

(9) If a, b e G such that (a°)-' = b°, then a e or b e <a>.

We want to show that <a> = in (9). For this purpose we suppose that <a> 0 and, without-loss of generality, we may assume that <a> < . Then (7) and (8)show that s and that a, b and u all have infinite order. Write (b-')° = v.Then (9), applied to u, v e G and v-', yields that u e <v> or v e .

Suppose first that v e and let v = u' with r e Z. Since o is bijective, Irl > 2. By(7), (b-")° a <u'> for all n e N and so c . Thus = . Hence for everyi e RI there exist k, t e 7L such that u" = (bk)° and (b -')a = u', and we can choose i sothat I k I > 2 and I t I > 2. Now (9), applied to u'", u' e G and a -', shows that u'" a <u'or u' a <u">. In both cases, d = (r, t) # 1 and (bl'l)° a <ud>. Since also (b-")° e<u'> < <u'> for every n e NI, we get by (5)

u = b° = (bIkIb-(Ik1-1))° a <udx

a contradiction. The case that v 0 is similar. Here < <v>, let u = vS whereIsi > 2. Again by (7), c S <v>. Hence for i e N there exist k, t e 7L such that(bk)° = vs" and (b-k)° = v', and we can choose i so that Ikl > 2 and Iti >_ 2. Nowreplace u by v and r by s to get the same contradiction as before.

Thus we have shown that <a> = in (9) and hence that (a°)-' e <a>'. Return-ing to the element x e H we see that (x°)-' e <x>" s H°, as required.

For n e N let L1(G) be the set of all subgroups of the group G that can be gener-ated by n elements. Since any group is the join of its cyclic subgroups, every projec-tivity is determined by its action on the set L1(G). In general the most accessiblesubgroups of a group are the cyclic ones. Therefore it is important to know underwhich conditions one can extend a bijection between L1(G) and L1(G) to a projec-tivity from G to G.

1.3.2 Theorem. Let G and G be groups and suppose that r is a bijective map from theset L1(G) of all cyclic subgroups of G to L1(G) such that for all X, Y, Z e L1(G),

(10) X 5 <Y,Z> if and only if XT < <YT,ZT>.

For H < G let HO be the set union of all the XT where X e L1(H). Then 9 is aprojectivity from Gto G.

Proof. We first show that H9' is a subgroup of G. So let a, b e H9. Then there existY, Z e L 1(H) such that a e Y` and b e Z'. Since T is surjective, there is an X E L, (G)such that X` = <ab-' ). Then X'< <a, b) 5 <Y',Zi) and by (10), X 5 <Y, Z) 5 H.Therefore X E L1(H) and ab-' a X-- s He'. Thus Hg' 5 G and cp is a map from L (G)to L(G). Since T-' satisfies (10) with G and G interchanged, there is a map 0: L(G) -L(G) defined to T-' in the same way as p to r. For x e H5 G, <x>' c- L1(H'') andhence x e H`0*; thus H < H". Conversely, the definitions of f and p imply thatto every y e H1*0 there exist a L1(H9) and Z E L1(H) such that y e <u)`-' andu c Zi. Hence < Zr and therefore y e '-' < Z, by (10). Thus y e H and soH'P" = H. Similarly, Kd'' = K for all K S G. It follows that q' and i are bijectivemaps and they certainly preserve inclusion. By 1.1.2, cp is a projectivity.

It does not suffice to assume in Theorem 1.3.2 that T is an isomorphism betweenthe partially ordered sets (by inclusion) L1(G) and L 1(G). This can be seen by takingany two groups of the same order p" and exponent p, p a prime, that are not latticeisomorphic. One needs the subgroups generated by two elements in (10). But forn > 2, every isomorphism between L"(G) and L"(G) can be extended to a projec-tivity. This result, of course, is most interesting for n = 2; however, it seems possiblethat one knows L"(G) and L"(G) for some n > 2 without being able to decide whichsubgroups are cyclic or lie in L2(G).

1.3.3 Theorem (Poland [1985]). Let G and G be groups and let n e N with n > 2. Ifa is a bijective map from L"(G) to L"(G) such that for all X, Y e L"(G),

(11) X < Y if and only if X ° 5 Y°,

then there exists a unique extension p of a to a projectivity from G to

Proof. We first show that a maps cyclic subgroups of G to cyclic subgroups of G. Solet H S G be cyclic. Then L"(H°) L"(H) = L(H) and therefore L"(H°) ^- T. wherem = I H I e N u {cc}, by 1.2.2. In particular, L"(H°) satisfies the maximal conditionand hence there exists a maximal cyclic subgroup <x> of H°. Suppose that H° #<x>, take y e H'\ <x> and put Z = <x,y). Since L"(Z) is the interval [Z/1] inL"(H°), we have L"(Z) a- T for some r e N u {co}, and if r = oo, then <x> n <y> # 1.Thus in any case, the interval [Z/<x) n <y)] in L"(Z) is isomorphic to some T.where s c N. Now y operates by conjugation on this interval in which <x> is the onlycomplement to <y>; it follows that <x)'' = <x>. Thus <x> 9 Z and Z/<x) is cyclic.Therefore, if K is any subgroup of Z, K n <x> is a cyclic normal subgroup of K withcyclic factor group and hence K e L2(Z) c L"(Z). This shows that L(Z) = L"(Z) -- Tand by 1.2.5, Z is cyclic, contrary to the maximality of <x>. We have shown thatH° is cyclic.

Since a-' satisfies the assumptions of the theorem with G and G interchanged, a-'maps cyclic subgroups of G to cyclic subgroups of G. The restriction r of a to L, (G),therefore, is a bijective map from L1(G) to L1(G). For k < n and X1, ..., Xk EL1(G), by (11) we have X, = X° S <X...... Xk)° for all i and so <X;,...,Xk) S<X 1, ... , Xk)°. The corresponding result for a-' yields the reverse inclusion. Thus

(12) <XI,...,Xk) _ <X1,...,Xk)0.

Now for X, Y, Z E L,(G), (11) and (12) imply that X < <Y,Z> if and only if X' =X° S <Y,Z)° = <Y`,Z`>. By 1.3.2 there is a projectivity cp from G to G whoserestriction to L,(G) is T. By (12), cp is an extension of a and it is unique since everyprojectivity is determined by its action on L,(G).

We shall quite often use 1.3.1 or 1.3.2 to construct projectivities. Theorem 1.3.3,however, is more of theoretical interest. Nevertheless it seems worthwhile to investi-gate more closely these partially ordered sets L°(G), in particular, L2(G).

Induced projectivities

An isomorphism a from G to G satisfies (1) and therefore induces a projectivity. It isone of the main problems about subgroup lattices to decide under what conditionsa given projectivity cp is induced by a group isomorphism a. Clearly, it suffices toknow that cp and a operate in the same way on L, (G). We shall need the followingmore general result.

1.3.4 Lemma. If cp is a projectivity from G to G and a: G -+ G is a homomorphism suchthat X° = X" for every cyclic subgroup X of G, then a is an isomorphism inducing cp.

Proof. If X is a cyclic subgroup of the kernel of o, then X`0 = X° = 1 and so X = 1.Thus u is injective. Let H S G. Since a is a homomorphism, He < G. Hence by 1.1.2,

(P

H`0 =\

U <x>J = U <x> = U <x>° = H°.E H XEH XEHX

In particular, G° = G. Thus a is surjective and induces cp.

The fact that a projectivity maps cyclic groups onto cyclic groups makes it possi-ble to construct element maps. A general result in this direction is the following.

1.3.5 Theorem._A projectivity cp from a group G to a group G is induced by a bijectivemap from G to G if and only if I <g>'* I = I <g> I for every g c- G.

Proof. The condition is clearly necessary. Conversely, let cp be a projectivity sat-isfying I <g>" I = I <g> I for all g e G. For any cyclic group C let C* be the set ofgenerators of C. If C < G is cyclic, then so is C", by 1.2.10, and since IC°I = ICI, wehave I(CP)*I = I C*I; let ac be a bijection from C* to (CP)*. Forge G define g° = gwhere C = <g>. Then, clearly, a is injective. If H < G and g e H, then <g> < H10and hence g° a H`0; conversely, for y e H0 the preimage <y>'P-' = D is cyclic andtherefore y = x = x° for some x e D < H. Thus He = H`0. This shows that o isbijective and that cp is induced by o. 11

The disadvantage of this theorem is that the map a constructed in its proof doesnot have any nice properties. It is much more interesting to know, for example, that

a given projectivity is induced by an isomorphism. We present two general resultsdue to Sadovskii that reduce this problem to the corresponding problem for finitelygenerated groups.

Sadovskii's approximation theorems

A family $ of subgroups of a group G is called a local system of subgroups of G if

(13) for any X, Y E . there exists Z E S/ such that X u Y < Z, and

(14) every element of G is contained in some X E Y.

1.3.6 Theorem (Sadovskii [1941]). Let cp be a projectivity from a group G to a groupG and let . ° be a local system of subgroups of G. For every X E. let cpx be theprojectivity induced by cp in X, and Ax be the set of all isomorphisms from X to X" thatinduce qx. If Ax is nonempty and finite for all X E., then cp is induced by anisomorphism from G to G.

Proof. Recall that a directed set is a partially ordered set in which every pair ofelements has an upper bound. By (13), . is a directed set and (Ax)x E , is a family ofsets indexed by 9. For each pair (X, Y) of elements of .' such that X < Y, we definea mapping irxy: Al - Ax in the following way. Take a E Ay. Then o induces cpy, sothe restriction 6jx of a to X induces cpx, and a E Ax; now we can define 7rxy(a) _aI x. Clearly, these mappings satisfy the following conditions:

(15) For every X E ., nxx is the identity mapping of Ax.

(16) If X < Y < Z, then rtxz = trxytryz

Since every Ax is nonempty and finite, the inverse limit A = lim Ax of the family(Ax)x,.v with respect to the mappings ttxy is not empty (see Bourbaki [1968], Chap-ter III, § 7.4). An element of A is an element of the cartesian product of the sets Ax,hence a function f defined on . such that f(X) E Ax for all X E 9, with the propertythat

(17) f(X) = nxy(f(Y)) = f(Y)Ix

for X < Y. We take such an f and define a mapping p: G G in the following way.For g e G there exists X E 9 such that g c- X, and we let gP = gf(x). If in additiong c- Y E 9, then there exists Z E. such that X u Y < Z, and hence gf(X) = gf(Z) _gf(I), by (17). Thus p is well defined. For x, y e G there are X, Y, Z E .' such thatx c X, y c Y and X u Y < Z. Since p1z = f(Z) is an isomorphism inducing the pro-jectivity cpz, (xy)P = xPyP and <x>P = <x>". By 1.3.4, p is an isomorphism induc-ing cp. O

If H = <x....., is a finitely generated group and ' is a projectivity from H tosome group H, then every isomorphism a: H - H that induces 0 has to map x;to some generator of <x>', i = 1, ... , n. Since <x1> ' has only a finite number of

generators and since a is determined by the images of the xi, there are only finitelymany isomorphisms inducing 0. So we get the following corollary to Theorem 1.3.6which usually suffices for the applications.

1.3.7 Corollary. Let cp be a projectivity from G to G and let . be a local system offinitely generated subgroups of G. If cp is induced by an isomorphism on every X E .9',then cp is induced by an isomorphism on G.

We now prove a result dual to Theorem 1.3.6. Clearly, if cp is a projectivity fromG to G and N a G such that N" a G, then the map

(P": L(G/N) --+ L(G/N"); H/N i--. H10/N11

is a projectivity from GIN to G/N'; we call gyp" the projectivity induced by q in G/N.

1.3.8 Theorem (Sadovskii [1965a]). Let p be a projectivity from a group G to agroup G and let 9 be a family of normal subgroups of G such that

(18) for any X, Y e. there exists Z e So with Z < X n Y, and

(19) for every g e G there exists X e .So such that <g> n X = 1.

For every X e., suppose that X`0 g G and let Ax be the set of all isomorphisms fromG/X to G/X0 that induce the projectivity tpx. If A. is nonempty and finite for allX e .9', then cp is induced by an isomorphism from G to G.

Proof. Let X, Y e 5° and define X* = {g e GI<g> n X = 1). Every a e Ax inducesa map a* from X* to G. For, if g e X*, then (gX)° e <g> X`°/X9', and since<g)" n X° = 1, there is a unique g' E <g> such that (gX)° = g'X°. Now we setg = g' and define Bx = {a*Ia e Ax}. If Y < X and /3 e A1, then the isomorphism/3: G/Y 6/Y`° induces in the natural way an isomorphism y : G/X --+ 6/X`°. Since /3induces qp on G/Y, it follows that y induces q on G/X and hence y e Ax. Now Y < Ximplies X* < Y*, and for g e X*, gQ' is an element of <g> satisfying (gX)7 = gP*X''.Thus gfl = gY' and hence y* = fl*Ix..

Let us write X <' Y if and only if Y < X. Then (Y, <') is a directed set, by (18),and we have just shown that for X <' Y there exists a map 7[,y: By -+ Bx defined bynxr(a) = aI x. for all o e By.. Clearly, these maps have the properties (15) and (16)with respect to and since every B. is finite and nonempty, the inverse limitB = lim B. with respect to the mappings nxy is not empty. An element f e B has theproperty that f(X) e Bx and

(20) f(X) = nxr(f(Y)) = f(Y)lx.

if X <' Y, i.e. Y<_ X. We take such an f and define p: G - G in the obvious way:if g e G, then by (19) there exists X e. such that g e X*, and we define g° = gf(X).If also g e Y*, then there exists Z e . such that Z < X n Y, and hence gf(x) =gf(Z) = gf(I), by (20). Thus p is well defined. Let x, y e G and consider the amorphya(x,y) = (y°)-`(x°)-`(xy)° e G. Since cp is a projectivity, there exists g e G such that<g)`0 = <a(x,y)>, and our assumptions on . yield the existence of a subgroup

1.4 The group of autoprojectivities 23

Z E Y such that x, y, xy, g are all contained in Z*. Then f(Z) = a* for some a e AZand, using the definitions of p and a*, we get <x> = <x) = <x>" and

(xy)°Z`° = (xY)a*Z* = (xYZ)° = (xZ)°(yZ)° = x' yPZ'D.

Therefore a(x, y) c -Z". On the other hand a(x, y) E <g> and <g)`° n Z( = 1. Thusa(x,y) = I and so (xy)P = x°y'. By 1.3.4, p is an isomorphism inducing cp.

Also in this theorem the assumption that Ax is nonempty and finite may bereplaced by the condition that G/X is finitely generated and cp is induced by someisomorphism on G/X.

Exercises

1. Let a: G - G be a map, a the amorphy of a and write A = <a(x, y)Ix, y e G> andB = <g°Jg a G>. Show that A a B and that the map a* defined by x = xA forall x a G is an epimorphism from G to B/A.

2. (Gaschutz and Schmidt [1981]) Let G be a finite group, a a permutation of G anda the amorphy of a. Show that the following are equivalent:(a) (Hx)° = H°x° and (xH)° = x°H° for all H < G and x e G,(b) a(x,y) e <x°) n <yc> for all x, y e G.

3. Let a be a bijective map from G to G such that(a) (xy-' )° E <x°, y°> for all x, y c- G, and(b) (uv-')°-' E <u°-', v°-') for all u, v a G.Show that the map d: L(G) -+ L(G) defined in (2) is a projectivity.

4. (a) If L2(G) is a sublattice of the finite group G, show that L2(G) = L(G).(b) Find a group G such that L2(G) is a meet-sublattice but not a sublattice of

L(G).(c) Find a group G for which L2(G) is not a lattice.

5. If L2(G) is a distributive lattice, show that G is locally cyclic.6. If G and G are lattice isomorphic p-groups, p a prime, and (P is a projectivity from

G to G, show that there exists a bijective map a: G -+ G inducing cp such that(xP)Q = (x°)P for all x e G.

1.4 The group of autoprojectivities

Let G be a group. A projectivity from G to G is called an autoprojectivity of G; thegroup of all autoprojectivities of G, that is of all automorphisms of L(G), is denotedby P(G). Clearly, every automorphism of G induces an autoprojectivity, as doesevery element g a G via the inner automorphism induced by g. We collect thesesimple facts in the following theorem and introduce important subgroups of G,Aut G and P(G).

1.4.1 Theorem. Let G be a group.(a) The map p: Aut G -+ P(G) defined by H" = Ha for H < G, a e Aut G is a homo-

morphism. The kernel of p is the group

Pot G = {aeAutGJH2=H for all H <G}

of power automorphisms of G, the image of p is the group PA(G) of all autoprojec-tivities of G that are induced by automorphisms. We have PA(G) Aut G/Pot G.

(b) The map n: G -+ P(G) defined by H9' = g-'Hg for H < G, g c- G is a homomor-phism. The kernel of rl is the group

N(G) = n NG(H),H<G

called the norm of G, the image of rl we denote by PI(G). We have PI(G) G/N(G).(c) If 7t: G - Aut G is the homomorphism mapping g e G to the inner automorphism

induced by g, that is x9' = g-'xg for x e G, then rl = 7tp and N(G)" = G" n Pot G =Inn G n Pot G.

0 P (G)

Aut G

G t

N(G)

Z (G)

1 6

n

Pot G

-0 1

Figure 3

PA(G)

PI (G)

1

Proof. Let a, /3 E Aut G and H < G. Then t' is the projectivity induced by Of andH1afi'° = Hae = Ha°P°. Hence p is a homomorphism. For g c G we have H9'° = g -'Hg =H9", and therefore rl = 7rp is the product of two homomorphisms. Thus rl is also ahomomorphism. The remaining assertions are trivial or follow from the isomor-phism theorem for groups. El

If two isomorphisms a and z induce the same projectivity between two groups Gand G, then o r ' is a power automorphism of G, and conversely. Therefore,

1.4.2 Corollary. If a projectivity of a group G is induced by an isomorphism, it is

induced by exactly I Pot GI isomorphisms.

There is a large class of groups in which the homomorphisms p and ry describedin Theorem 1.4.1 are injective.

1.4.3 Theorem (Baer [1956], Suzuki [1951a]). If G is a group with Z(G) = 1, thenalso N(G) = I and Pot G = 1, hence PI(G) G and PA(G) Aut G.

Proof. This is an immediate consequence of a more general result on power auto-morphisms of groups, so we shall postpone the proof until 1.5.2 below. But there isa short proof for finite groups which we want to present here. So let G be finite andsuppose first that N(G) : 1. Then there exists a nontrivial Sylow p-subgroup P ofN(G). Since N(G) normalizes every subgroup, P 9 N(G) and hence P a G. So ifSc Sylo(G), then P a S and P n Z(S) 1; let 1 a E P n Z(S). Then S< CG(a) andfor primes q # p, P normalizes every Sylow q-subgroup Q of G and hence QP =Q x P. Therefore also Q < CG(a) and CG(a) = G, contrary to Z(G) = 1. ThusN(G) = 1 and PI(G) G, by 1.4.1.

The second assertion now follows from the fact that G(Inn G) = 1 if G is agroup with trivial centre. Indeed, Pot G n Inn G = N(G)" = 1, by 1.4.1, and sincePot G and Inn G are normal subgroups of Aut G, they centralize each other. There-fore, if aEPotGandx,geG,

g lxag = xaa' = X9,a = (g-lxg)a = (ga)-'xaga

and so gag-l centralizes xa. Since x was arbitrary, gag-' E Z(G) = 1 and hencega = g. Thus Pot G = 1 and PA(G) Aut G.

In the remainder of this section, we shall study L(G) and P(G) for some well-known groups G; in particular we want to determine the subgroups PA(G) andP1(G) of P(G). Since every automorphism of a cyclic group is a power automor-phism, 1 for all n c N u {cc}. The results of Section 1.2 yield the descrip-tion of given in Exercises 1 and 2. The next example will explain those parts ofour terminology and notation that originated in projective geometry.

Elementary ahelian groups

Let p be a prime, n a natural number and let G be an elementary abelian group oforder p". Then it is well-known that we can regard G as a vector space of dimensionn over GF(p), the field with p elements, and that L(G) is the projective geometry ofthis vector space. An automorphism of the group G is an invertible linear map of thevector space G onto G, and an autoprojectivity of G is simply a projectivity of theprojective geometry L(G) onto itself in the sense of projective geometry. It followsthat PA(G) PGL(n,p). If n = 2, then every nontrivial, proper subgroup of G hasorder p and hence L(G)\{G, 1 } is an antichain with p + I elements. Every permuta-tion of these elements yields an autoprojectivity of G, and therefore P(G) is isomor-phic to the symmetric group Sp+1 on p + 1 letters. If n >_ 3, the Fundamental Theo-rem of Projective Geometry (see, for example, Baer [1952], p. 44) tells us that every


autoprojectivity of G is induced by a semi-linear mapping of the vector space G. Butsince GF(p) has only the trivial automorphism, this mapping is linear, i.e. an auto-morphism. Thus we have the following result.

1.4.4 Theorem. Let G be an elementary abelian group of order p", p a prime, n e N.Then PA(G) PGL(n, p); and P(G) = PA(G) for n >- 3, but P(G) ^-- SP+i if n = 2.

In Section 2.6 we shall show, without using the Fundamental Theorem of Projec-tive Geometry, that for a much larger class of abelian groups all autoprojectivitiesare induced by automorphisms.

Groups of order pq

Let p and q be primes such that q I p - 1 and let H be a nonabelian group of order pq.Then by Sylow's theorem there are p subgroups of order q and one subgroup oforder p in H and so L(H) consists of H, 1 and an antichain with p + 1 elements.Therefore, again, P(H) Sp+1 whereas PA(H) Aut H and PI(H) H, by 1.4.3.Thus PA(H) and PI(H) are subgroups of orders p(p - 1) and pq, respectively, inSp+1

This shows that in general PA(G) and PI (G) are not distinguished subgroups ofP(G); in fact they are not even normal in P(G). However, this is not very surprisingif, as is the case here, the subgroup lattice is structurally small, in particular if it doesnot determine the group. The situation is different in our next example.

The alternating group A4

Let G be the alternating group of degree 4. Then G has 3 subgroups of order 2,generating the Klein 4-group V, and 4 maximal subgroups of order 3. It is easy tosee that G is determined by its subgroup lattice. Indeed let G be a group withisomorphic subgroup lattice; then G operates faithfully by conjugation on the set Fof those atoms of this lattice which are also antiatoms. Hence G is a subgroup of S4and it follows that G G. Let A be the set of those atoms of L(G) that are notantiatoms, and let S, T be the kernels of the actions of P(G) on T, A, respectively.

zxFigure 4: L(A4)

Clearly,

P(G)=SxT=S3xS4.

By 1.4.3, PA(G) Aut G. An automorphism a of G fixing all H E I operates on eachof these groups in the same way as on G/V. Hence a either centralizes or inverts allH E r; and since one can easily check that the latter is impossible, a = 1. ThusPA(G) n S = 1 and PA(G) is isomorphic to a subgroup of T ^- S4. On the otherhand, S4 operates faithfully by conjugation on its normal subgroup A4 and inducesan S3 on V. It follows that PA(G) S4 and PA(G)S = P(G) = PA(G)T. An elementof S normalizing PA(G) would centralize PA(G) and therefore also PA(G)T = P(G);but Z(S) = 1. Thus N,,(G)(PA(G)) = PA(G) and PA(G) is one of 6 conjugate sub-groups of P(G).

Dihedral 2-groups

The results obtained so far show an obvious general phenomenon. The more struc-ture the subgroup lattice of a group has, the more likely it is that it determinesgroup-theoretic invariants. Compare, for example, the situation for A4 and S4 (seeExercise 3) or the cases n = 2 and n > 3 in 1.4.4. The dihedral 2-groups are anexception to this rule.

Let n>3 and let G = D2" _ <x,yIx2"-` = y2 = 1, yxy = x-`> be the dihedralgroup of order 2". Since (x'y)2 = x'x -' = 1 for all i c N, every element in G \ <x> isan involution; in particular, <x> is the only cyclic maximal subgroup of G. It followsthat <x> is invariant under P(G) and that P(G) operates faithfully on the set A ofsubgroups of order 2 not contained in <x>. Every power automorphism has tocentralize all these involutions and therefore Pot G = 1 and PA(G) Aut G. If n =3, it is easily seen that

P(D8) = PA(D8) Aut D8 nt D8.

Figure 5: L(D1e)

For n > 4 there are exactly three maximal subgroups in G, namely <x> and twosubgroups, M1 and M2, say, isomorphic to D2" Since M1 n M2 = <x2>, P(G)contains a subgroup isomorphic to P(M1) X P(M2), and because there are evenautomorphisms of G interchanging M, and M2, P(DAI = 21P(D2"-,)12. By an obvi-ous induction we get I 22"-'-1 for all n > 4. Since this is the highest powerof 2 dividing (2"-')! and IAA = 2"-', P(D2.) is isomorphic to a Sylow 2-subgroup ofthe symmetric group S2"-1. Every automorphism of G is determined by the imagesof y and xy, and for these there are 2"-' and 2n-2 involutions respectively available.Thus IAutGI = 22n-3 so that PA(D2") is properly contained in P(D2.,) for all n > 4.

Quaternion groups

Let n > 3 and let G = Q2" = <x, yIx2" = 1, y2 = x2"-2, y-'xy = x-' > be the gener-alized quaternion group of order 2". Then Z(G) is the only minimal subgroup ofG and G/Z(G) ^ - D2" Thus P(Q2") ^- P(D2"-,), and since every automorphism ofG/Z(G) can be lifted to an automorphism of G.

PA(Q2") PA(D2"-,) ^, Aut D2.,-,.

Figure 6: L(Q16)

In particular, P(Q,) = PA(Q,) ^- S31 P(Q16) = PA(Q16) De and PA(Q2") <P(Q2") for n > 5. For every homomorphism a: G -+ Z(G), the map a: G -i G definedby g° = gg° for g e G is a power automorphism of G. It follows that I Pot G 1 > 4, andsince G is generated by two elements of order 4, 1 Pot G I = 4.

Finally, we mention that because the quaternion groups are the only noncyclicfinite groups with just one minimal subgroup, they are determined by their subgrouplattices. We leave it as an exercise to show that this also holds for the dihedral2-groups.

Exercises

1. Show that P(C.) ^ Sym P.2. Let m e N and for i e t\J define n; to be the number of primes p such that p' is the

maximal power of p dividing m. Show that P(C,") Dr S. (where of course, So isthe trivial group). ' E "

1.5 Power automorphisms 29

3. Draw a Hasse diagram of L(S4) and show that P(S4) = PA(SO) ^ S4.4. Give details for those parts in the discussion of the dihedral and quaternion

groups that have only been sketched in the text.5. Show that D2., is determined by its subgroup lattice for n > 2.6. Let p be an odd prime and let G be the nonabelian group of order p3 and

exponent p. Determine P(G), PA(G) and Pot G.

1.5 Power automorphisms

In this section we shall prove Cooper's theorem that every power automorphism ofa group is central. This will have a number of interesting consequences. First we givean elementary result.

1.5.1 Theorem. Let G be a group. Then Pot G is abelian.

Proof. For g c G let F(g) = {a e Pot GIga = g}. Then F(g) a Pot G and Pot G/F(g),being isomorphic to a subgroup of Aut(g)., is abelian. Since n F(g) = 1, Pot G isabelian. a E c

Commutators

For elements x, y e G and a E Aut G, [x,y] = x-'y-'xy = x-'xy is the commutatorof x and y and, similarly, [x, a] = x-'x° is the commutator of x and a. Then ifX s G and S is a subset of G or Aut G, [X, S] = ([x, s] Ix e X, s e S). Finally, forelements x e G and s...... s" of G or Aut G we define inductively

[x, s,,.-.,s"] = [[x,s"--.,s"-,],s"]for n > 2;

for subsets X of G and S...... S. of G or Aut G we define [X, S,_., S"] similarly.For the convenience of the reader we list the basic properties of commutators whichwe shall require. So let x, y e G, s and t elements of G or Aut G, X, Y, Z subgroupsof G, S a subset of G or Aut G and let n e 101, m e Z. Then

(1) [x y, s] = [x, s]y,[ y, s] and [x, st] = [x, t] [x, s]',

(2) [xm, s] = [x, s]m if [x, s] and x commute,

(3) [x, sm] = [x, s]m if [x, s]S = [x, s],

(4) (xy)" = x"y"[y, x]()) if [x, y] commutes with x and y,

(5) [x, y]° = [x°, y°] for every homomorphism a of G,

(6) [X, S]'r = [X, S] and [X, Y] a X u Y,

(7) [X, Y, Z] = I = [Y, Z, X] implies [Z, X, Y] = 1.

The proofs of these results for s, t e G and S 9 G can be found in Robinson [1982],Chapter 5; for s, t e Aut G and S c Aut G, they are similar.

Cooper's theorem

15.2 Theorem (Cooper [1968]). Let G be a group. Then [G, Pot G] 5 Z(G).

This result has a number of immediate consequences. First of all, we get the

Proof of Theorem 1.4.3. If G is a group with Z(G) = 1, then [G, Pot G] = 1 and thusPot G = 1. Hence also N(G) = 1, by 1.4.1, (c). Then 1.4.1, (a) and (b) show thatPA(G) Aut G and PI (G) G.

Furthermore, we can show that power automorphisms centralize not only G/Z(G)but also the commutator subgroup G' of G. In addition they commute with the innerautomorphisms. Finally, we get a result of Schenkman's [1960] that the norm of agroup G is contained in the second centre Z2(G).

15.3 Corollary. If G is a group, then

(a) [G', Pot G] = 1,

(b) [Inn G, Pot G] = 1,

(c) N(G) < Z2(G).

Proof. Let g, x e G and a e Pot G. By 1.5.2 there exist elements z, w e Z(G) such thatg2 = gz and x2 = xw. Hence [g, x]° = [ga, xa] = [gz, xw] = [g, x] and this implies(a). As in 1.4.1 let 7C be the natural homomorphism from G to Aut G. Then

x° = (g-1xg)Q=

(9a)-lxaga = z-1g-lxagz = xaa`

and (b) follows. Finally, if g e N(G), then g" e Pot G and hence [x, g] = [x, g'] e Z(G),by 1.5.2. Thus g e ZZ(G) and (c) holds.

Proof of Theorem 1.5.2. Let x, y e G and a e Pot G. We have to show that [x, a, y] = 1.Since xa e <x)', the elements xa and [x, a] commute with x. By (2),

(8) [x"', a] = [x, a]' for all m e Z.

If H< G, then H" = (Hx)° = (H2)' = H' and hence [x, a] = x-1xa = xxx-1 e NG(H).Thus

(9) [x, a] e N(G).

Since xaya = (xy)° commutes with xy, we have [x, a, y] [y-1, a, x-1] _x-axy-1x-1 xaYy2y-1xyy-ax-1 = xx-ay-1x-1(xy)

(x ya)y-ax-1 = I. By (9),[x, a, y] e <y) and [y-1, a, x -1 ] e <x> and hence

(10) [X, a, Y] e <x) n W.

In particular, [x, a, y] commutes with [x, a] and y. Therefore the commutatorformulas (8), (2) and (3) yield

(11) [xm, a, y"] = [[x, a]m, y"] = [x, a, y]""" for all m, n e Z.

This implies [y -1, a, x -1 ] = [y, a, x] and so we finally get

(12) [x, a, y] [y, a, x] = 1.

Letx°=x',ya=ysand(yx)a=(y")` where r, s,te7L.Then (y`)"=(yx)`=(yx)(ya)"° = (ys)" and hence yS = (y`)x'-' and y' =

(ys)xr-'. It follows that ys2 = (,S)2 =((Y`)x, .)2

= (Y`S)"" =((Y`)""-.,.Y

= ((YS)',,

(YS2)x,)2 and this yields 1 =

[x"2,YS2] = [[x,a]r-l,YS2]. Using (11), we get

(13) [x, a, Y](r-1)S2 = 1.

Now suppose that the theorem is false. Then there exist elements x, y e G anda e Pot G such that [x, a, y] # 1. If x or y had infinite order, then so would [x, a, y],by (10), and (13) would imply that r = 1. But then [x, a] = xr-1 = 1 and [x, a, y] = 1,a contradiction. So x and y have finite order and we can choose x, y e G with[x, a, y] # 1 such that o(x) + o(y) is minimal. By (12), also [y, a, x] 1 and sowithout loss of generality we may assume that o([x, a]) z off y, a]). If p is a primedivisor of o(x) or o(y), then by our choice of x and y, [x°, a, y] = 1 or [x, a, y°] = 1;in both cases [x, a, y]" = 1, by (11). Hence o(x) and o(y) are powers of the sameprime p and

(14) [x, a, y]° = 1.

Again let r, s e Z such that xa = x' and ya = ys. Since a is an automorphism, s * 0(mod p). So from (13) we get [x, a, y]'-1 = 1 and then (12) and (11) yield

1 = [Y,a,x]'-1 = [Y,a,x'-1] = [[Y,a],[x,a]].

Thus A = < [x, a], [y, a] > is an abelian p-group and [x, a] is an element of maximalorder in A. Such an element generates a direct factor (see, for example, the moregeneral Lemma 2.3.11), that is we have A = ([x, a] > x B where B:!5; A. Now [y, a] e Aand so there exist b e B and n e 7L such that [y, a] = [x, a] -"b. Thus

(15) <[x, a]> n <[x, a]"[Y, a] > = 1.

Since < [x, a, y] > is a minimal and < [x, a] > a nontrivial subgroup of the cyclic p-group <x>, we have [x, a, y] e < [x, a] >. Hence there exists j e 7L such that [x, a, y] _{x,a]J;letm=n(1 - j). Then by (1),

rx mY, a] = [x -"jx"y, a] = [x -"', a]x"y [x"Y, a]

=([x,a]-"1)x"1[x",a]y[Y,a] (by(8)and(1))

_ [x, a, y] `[x", a] [x", a, y] [y, a] (since [x, a, y] e Z(<x, y>))

_ [x, a]"[y, a] (by (8) and (11)).

Now if [xmy, a] # 1, then <[x, a] > and <[xmy, a]> are nontrivial p-subgroups of <x>and <xmy>, respectively, and they have trivial intersection by (15). Since <x> is acyclic p-group and therefore contains only one minimal subgroup, it follows that<x) n <x'"y) = 1. By (10), [x, a, xmy] = 1 in this case; but if [xmy, a] = 1, then clear-ly [x'"y, a, x] = 1 and so [x, a, xmy] = 1, by (12). Hence in any case [x, a, xmy] = 1,and (1) implies that 1 = [x, a, xmy] = [x, a, y] [x, a, xm]y = [x, a, y], a contradiction.

Power automorphisms of abelian groups

Cooper's theorem does not say anything about abelian groups. However, it is notdifficult to determine all power automorphisms of these groups. We call a powerautomorphism a of G universal if there exists an integer n such that x° = x" forall x e G. We first remark that every power automorphism of a finitely generatedabelian group is universal. More generally we show the following.

15.4 Lemma. Let A be a finitely generated abelian group with A = <a1 > x x <a,>.If an automorphism a of A fixes <a1>, ..., <a,> and <a1...a,>, then there exists aninteger n such that a° = a"for all a e A.

Proof. By hypothesis there exist integers n, n, such that a; = a;' and

an1' ... anom = (a1 ... ar -a = (a1 ... ar-n- aln anr - r

It follows that a;=a;' =a'fori= 1, ..., r and then a' = a" for allaeA.

An abelian torsion group A is the direct product of its primary components APand therefore Pot A is the cartesian product of the groups Pot AP where p runsthrough P. So let A be an abelian p-group and let a e Pot A.

Consider first the case Exp A = pk, and take an x e A with o(x) = pk. Then ainduces an automorphism in <x> and so there is an integer n such that 0 < n < pk,(n, p) = 1 and x° = x". For every a e A, a induces a power automorphism in <x, a>and by 1.5.4 there exists an integer m such that a° = a' and xa = x'. It follows thatm = n (mod pk) and a2 = a". Hence a is universal and Pot A ^_- Aut <x>.

Now suppose that Exp A is infinite and consider KIk(A) = {a e Al a" = 11 for k cN. Then at is universal on ( (A), so there exist integers nk such that 0 < nk < pk anda° = a"k for all a E i2k(A). Since i2k(A) 5 ilk+1(A), we have a"k = a° = a""' and there-fore nk = nk+1 (mod pk). Hence there exist unique integers r, such that 0 < ro < p,

k-10< r, < p for i E 101 and nk = rip ; thus a determines a p-adic unit r = rip'.

i=o i=oConversely, such a p-adic unit yields a power automorphism of A.

1.55 Lemma. Let A be an abelian p-group and let r be a p-adic unit, r = j rip', say,i=o

where rie7L,0<ro<pand 05ri<pforie1J.For aeAwith o(a)=pk(keN)we

k-1

define a° = an,, where nk = rip'. Then a is a power automorphism of A; we shalli=o

write a' for a°.

Proof. First of all, a° is well-defined since the ri are uniquely determined by r.Furthermore n; = nk (mod pJ) for j S k and therefore a° = a"k for all a e S2k(A). Thusor induces a power automorphism on every S2k(A). It follows that a is a powerautomorphism of A.

Now if Exp A is infinite, the correspondence between power automorphisms andp-adic units described above clearly is bijective and multiplicative. Thus we have thefollowing result.

15.6 Theorem (Robinson [1964]). Let A be an abelian torsion group and for p e Plet AP be its p-component.

(a) Pot A _- Cr Pot Ap.pe P

(b) If Exp AP is infinite, then Pot AP ^-- Up, the multiplicative group of p-adic units;in fact, the map sending r e Up to the power automorphism or defined in 1.5.5 is anisomorphism.

(c) If Exp AP = pk is finite, then every power automorphism of AP is universal andPot AP ^-- Aut Cpk, the multiplicative group of units of the ring 7L/pk7L (or of p-adic unitsmodulo pk).

For nonperiodic abelian groups the situation is much simpler.

15.7 Theorem. If A is an abelian group with elements of infinite order, thenIPotAI=2anda°=a-1for1 #aePotAandallaeA.

Proof. Suppose that a e Pot A and let x e A with o(x) = oc. Since a induces anautomorphism in <x>, we have x° = x" where n e { + 1, - 1}. If a e A, then a inducesa power automorphism in <x, a> which is universal, by 1.5.4. Thus a" = a" for alla e A. The result follows.

Power automorphisms of nonabelian groups

Here we only need the following result; it is an immediate consequence of Cooper'stheorem.

15.8 Theorem. if G is a nonabelian group generated by elements of infinite order, thenPot G = 1.

Proof. Suppose that 1 # a e Pot G. Since G is generated by elements of infiniteorder, there exists x e G such that o(x) = oo and x" # x, and hence x" = x-1. By 1.5.2,x-2 = [x, a] e Z(G). Therefore, if g c G, <x2, g> is abelian and by 1.5.4, a induces a

universal power automorphism in this group. Thus g° = g-' for all g e G. But thisimplies that G is abelian, a contradiction. El

There are also nonabelian groups possessing nontrivial power automorphisms.Universal power automorphisms exist in metacyclic groups (see Exercise 1) and,more generally, in certain groups with modular subgroup lattices (see 2.3.10). Exam-ples of nonuniversal power automorphisms can be found in the quaternion groupsand in the groups of Exercise 2.

Exercises

1. If G = <a,bjaP" = bP"-' = 1, b-lab = al+P) where p E P, p > 2 and n >_ 2, showthat a: G -+ G defined by x2 = x'+P"-' for x e G is an automorphism of G.

2. Let G be a finite p-group and let 1 < K < H < G such that K S <x> for allxEG\H.(a) If a: G -+ K is a homomorphism with H < Ker a, show that a: G - G defined

by x° = xx° for x e G is a power automorphism of G.(b) If all elements in G \ H have the same order and if I G : HI > p2, show that

there exist power automorphisms of G that are not universal. (Groups withthese properties can be found in Huppert [1967], p. 334.)

3. (Baer [1934]) If N(G) contains an element of infinite order, show that N(G) _Z(G).

4. (Cooper [1968]) Let G be a nonabelian group with elements of infinite order anddefine W(G) = <x e GIo(x) = oo>. Show that every power automorphism of Gcentralizes W(G) and G/W(G).

5. (Huppert [1961]) If G is a nonabelian finite p-group, show that Pot G is a p-group.

1.6 Direct products

In this section we shall study the subgroup lattice of a direct product of groups; wealso investigate the structure of a group whose subgroup lattice decomposes into adirect product.

The subgroup lattice of the direct product of two groups

If G = H x K, then in general L(G) L(H) x L(K); see G = CP x CP for p e P.However, it is possible to construct L(G) out of L(H), L(K) and all isomorphismsbetween sections of H and K. This was discovered by Goursat as early as 1890.

1.6 Direct products 35

1.6.1 Theorem. Let H, K < G and G = H x K.(a) For U < G and x c- UK n H define xa = {y c- KIxy c- U}. Then a is an epimor-

phism from UK n H onto UH n K/U n K with kernel U n H.(b) Conversely, if U1 < H, W:9 UZ < K and a is an epimorphism from U, onto

U,/W, then

U = D(U,,a) = {xyIx c- U,,y c- xa}

is a subgroup of G with U, = UK n H, UZ = UH n K, W= U n K and Ker a =UnH.

(c) For U,, V, < H and epimorphisms a, fi as in (b), we have D(U,,a) < D(V,,fJ) ifand only if U, 5 V1 and xa9x0forall xeU,.

Proof. (a) Let U, = {x c- HIxy E U for some y c- K} be the set of H-components ofelements of U. If x c- U, and y c- K such that xy c- U, then x = (xy)y-' E UK n H;conversely, if x c- UK n H, x = uk with u c- U and k c- K, say, then u = xk-' andhence x c- U,. Thus U, = UK n H and, similarly, UH n K = U2 is the set of K-components of elements of U. For x c- UK n H, therefore, xa is a nonempty subsetof UH n K. Furthermore, if y, z c xa, then xy and xz are elements of U and soY-

1z = (xy)-'xz c- U n K. Hence y and z lie in the same coset of U n K, and if w isanother element of this coset y(U n K), then w = yu with u c- U n K implies xw =xyu c- U, that is w c- xa. Thus we have shown that x° E U2/U n K. For x,, x2 E U,and y; E x; we have x1xzyly2 = x1y1xzy2 E U and hence yty2 E (xlx2) Therefore(xt x2)a = y1 y2(U n K) = xi xz and a is an epimorphism. Finally, x c- Ker a if andonly if xa = U n K, i.e. 1 E x. It follows that Ker a = U n H.

(b) If x1, x2 E U1 and y; E x; , then y, y2-' E xi(xz)-' _ (x,xZ'r and hence(xty1)(xzy2)-' = x,x2'y1y2' E U. Therefore U is a subgroup of G and U, is its setof H-components. Thus U, = UK n H and a is the epimorphism defined in (a) forthe subgroup U of G. The other assertions in (b) now follow from (a).

(c) If U=D(U,,a)<D(V,,fl)=V, then U1= UKnH<VKnH=V1 andxa 9 x0 for all x c- U1. The converse is obvious.

The isomorphism theorem yields U1/U n H U2/U n K in 1.6.1 (b). Thereforewe are interested in direct products of isomorphic groups. These are characterizedby the existence of diagonals. Here, a subgroup D of G = H x K such that DH =G = DK and D n H = 1 = D n K is called a diagonal in G (with respect to H and K).

1.6.2 Theorem. Let H, K < G and G = H x K. If 6: H - K is an isomorphism, then

D(S) = D(H, S) = {xxalx c- H}

is a diagonal in G (with respect to H and K). Conversely, if D is a diagonal in G (withrespect to H and K), then there exists a unique isomorphism S: H --- K such thatD = D(S). Thus there is a bijection between diagonals (with respect to H and K) andisomorphisms of H to K, and also, if H K, between diagonals and automorphismsof H.

Proof. If 6: H - K is an isomorphism and U = D(H, 6), then it follows from 1.6.1 (b)that UnH=1=UnK,H=U1<UKand K=U2<UH.Thus UK =G=UHand U is a diagonal in G. Conversely, let D be a diagonal in G and considerthe epimorphism a defined for D in 1.6.1 (a). Then a maps DK n H = H ontoDH n K/D n K = K and Ker a = D n H = 1. Thus at is an isomorphism from H toK and D = D(H, a) = D(a). Finally, 1.6.1 (c) implies that different isomorphisms 6, elead to different diagonals D(S) and D(e) so that the correspondence b -+ D(b) isbijective. F-1

1.63 Remark. Every subgroup U of a direct product G = H x K is a diagonalin a certain section of G. More precisely, if U = D(U1, a), U2 = UH n K andUo = (U n H) x (U n K), then Uo g Ul x U2, Ul Uo = Ul x (U n K) and U2 Uo =(U n H) x U2. Thus

Ul X U2/Up = Ul U0/U0 X U2 U0/U0.

G

Figure 7

Furthermore, UUl >_ U2, UU2 >_ Ul and so UUl = Ul x U2 = UU2. NowDedekind's law yields that Ul U0 n U = (U1 n U)UO = (U n H) U0 = U0 and, simi-larly, U2 U0 n U = U0. Hence U/Uo is a diagonal in Ul x U2/Uo with respect toUl Uo/U0 and U2 Uo/U0. It is easy to see that the corresponding isomorphismb: Ul U0/U0 -+ U2 U0/U0 is the one induced by a, that is (xUo)" = x2U0 for x c U1.

Groups with decomposable subgroup lattices

Our description of the subgroup lattice of G = H x K shows that L(G) L(H) x L(K)if every epimorphism a in 1.6.1 (a) is trivial. Clearly, this holds if (and only if) theelements in H and K have finite, coprime orders. It is easy to see that this resultis also true for more than two factors. Let us say that the groups Gx (A E A) arecoprime if every G,A is a torsion group and (o(x), o(y)) = 1 for all x E G,l, y E Gµ withA :0 µ; for finite groups this is equivalent to (IGAl, 1 for a. µ.

1.6.4 Lemma. If G = Dr Gx with coprime groups Gx (A e A), then L(G) = Dr L(GA);AEA AEA

in fact, the map r: L(G) -+ Dr L(GA) defined by HT(A) = H n Gx for A e A and H < GAEA

is an isomorphism.

Proof. Let L = Dr L(GA). Recall that L is the set of all functions f defined on AAEA

such thatf(A)

e L(Gx) for all A e A; hence H' is a well -/defined element of L. Let

p: L - L(G) be defined by f ° = Dr f(.1). Then f °T(A) = I Dr f(µ) I n Gx = forAEA \UEA /

all A e A and hence pr is the identity on L. On the other hand, if H < G and x e H,then x is the product of certain xx e Gx and since the Gx are coprime, x = xxyxwhere xxyx = yxxx, o(xA) = n, o(yx) = m and (n,m) = 1. There exist integers s, tsuch that ns + mt=1; thus x""=xx`yx`=xx-'-=xx. Hence xxeH for all A andH= Dr (H n Gx)= H`°. Thus rp is the identity on L(G). It follows that T is bijective

AEA

and by 1.1.2, r is an isomorphism.

We come now to the main result of this section. We show that any group withdecomposable subgroup lattice is a direct product of coprime groups.

1.65 Theorem (Suzuki [1951a]). Let G be a group such that L(G) Dr Lx whereAEA

(Lx)x E A is a family of lattices, I A I > 2 and I L.I > 2 for all A e A; write L = Dr L. andIEA

suppose that or: L(G) --+ L is an isomorphism. For A e A let Ox be the least and Ix thegreatest element of Lx, define ft e L by fx(µ) = Oµ for A # µ e A and fx(A) = Ix and,finally, let Gx be the subgroup of G with G7 = fx. Then G = Dr Gx, the groups Gx(A a A) are coprime and L(GA) ^- Lx for all A e A. AEA

Proof. Clearly, G° is the greatest and 1° the least element 0 of L, so that by 1.1.7,every Lx has a least and a greatest element and L(GA) [fx/0] L. For A # µ e A,1 x e Gx and l y e Gµ, furthermore

L(<x> u ) = [<x>0 u <y>0/0] [<x>a/0] x [<y>0/0] L(<x>) x L(<y>).

By 1.2.3, L(<x>) and L(<y>) are distributive, as is their direct product. Therefore<x) u <y> is locally cyclic, and hence cyclic. Since x and y generate this cyclic groupand x # 1 y, <x> and <y> have finite, coprime indices in it, and as <x> n <y> = 1,o(x) and o(y) are finite and coprime. This shows that the groups Gx (A e A) arecoprime and that they centralize each other. By 1.1.7, G = M a A> and

A µ e A> n Gx = 1. It follows that Gx g G and G = Dr Gx.AEA

Suzuki's theorem and Lemma 1.6.4 have a number of important consequences. Inconjunction with Lemma 1.2.9 they show that the class of nontrivial direct productsof coprime groups is invariant under projectivities. Moreover, the direct factors aremapped onto direct factors.

1.6.6 Theorem. Let G = Dr G, with coprime subgroups Gk (A e A) and let cp be aAEA

projectivity from G to a group G. Then G = Dr Gx and the subgroups Gf (A a A) arecoprime. A E A

Proof. Let T: L(G) - Dr L(GA) be the isomorphism defined in 1.6.4. Then a = cp-1T

'I

CAis an isomorphism from L(G) to Dr L(GA). In the notation of 1.6.5, fA(A) = GA _

AEA

G1(A) and f1(µ) = 1 = for A µ e A. Hence fA = Gx and so (G9)' = ft. Now allthe assertions follow from 1.6.5.

We mention the following special case of 1.6.6 since we shall often use the resultin this form.

1.6.7 Corollary. Let H and K be coprime subgroups of a group G such that [H, K] = 1.If cp is a projectivity from G to some group d, then (HK)' = H' x K' and H`0, K"are coprime.

Proof. Apply 1.6.6 to the projectivity induced by cp in HK = H x K.

Note that neither of the properties [H, K] = 1 and H, K coprime in 1.6.7, ispreserved under projectivities. This, for example, is shown by the projectivities be-tween the elementary abelian group of order 9 and the nonabelian group of order 6.As a consequence of 1.6.7 we get a more general result in this direction.

1.6.8 Corollary. Let it be a set of primes and suppose that H and K are subgroups ofa group G such that H is generated by 7t-elements, K by n'-elements and [H, K] = 1.If 9 is a projectivity from G to some group G, then [H', K'] = 1.

Proof. For every 7t-subgroup X of H and every n'-subgroup Y of K, [X`0, Y"] = 1,by 1.6.7. Since H' is generated by these X', it follows that [H', Y'] = 1, and sinceK' is generated by these Y', [H', K'] = 1.

We have shown that L(G) is decomposable if and only if G is a'nontrivial di-rect product of coprime groups. For finite groups this property is inherited by theFrattini factor group. We remind the reader that the Frattini subgroup fi(G) of anarbitrary group G is defined to be the intersection of all the maximal subgroups ofG, with the stipulation that it shall equal G if G should have no maximal subgroups.If G is finite, then D(G) is nilpotent and cD(G)H < G for every proper subgroup Hof G.

1.6.9 Theorem. Let G be a finite group. Then the following properties are equivalent.(a) L(G) is directly decomposable.(b) There exist nontrivial subgroups H and K of G such that G = H x K and

(IHI, IKI) = 1.(c) L(G/b(G)) is directly decomposable.

Proof. That (a) implies (b) follows from 1.6.5. If (b) holds, then G/I(G) is the directproduct of HD(G)/1(G) and K b(G)/(D(G) and these groups are coprime. Since everyprime divisor of IGI also divides IG/O(G)I (see Robinson [1982], p. 263), both factorsare nontrivial. Now (c) follows from 1.6.4.

Finally, let L(G/b(G)) be decomposable, G/I(G) = H,/CD(G) x H2/I(G) wherelb(G) < Hi (i = 1, 2) and (IH,/cD(G)I, H2/D(G)I) = 1. If Di is the product of thoseSylow p-subgroups of (D (G) for which p divides IHi/D(G)I, then D; g G (i = 1, 2), ascharacteristic subgroups of b(G). Furthermore, D(G) = D, x D2 since every primedivisor of ID(G)I also divides IG/D(G)I. Thus D, is a normal Hall subgroup of H2. Bythe Schur-Zassenhaus theorem there exists a complement G2 to D, in H2 and allthese complements are conjugate in H2. Since D, and H2 are normal subgroups, Goperates by conjugation on the set S2 of these complements and H2 is transitiveon Q. By the Frattini argument, G = NG(G2)H2. As H2 = G2D G = NG(G2)D, <NG(G2)CD(G) and hence G = NG(G2) since 'D(G) is the set of nongenerators of G. ThusG2 a G with IG21 = IH2: D11 = IH2/c(G)IID21 and similarly, we get a normal sub-group G, of G such that IG11 = IHi/c(G)IID,1. Then (IG,1,1G21) = 1, 1G111G21 = IGIand so G = G, x G2. Now (a) follows from 1.6.4.

The number of groups with a given subgroup lattice

We finish this chapter with an application of our results on direct products. In 1.2.8we saw that for every finite cyclic group G there exist infinitely many nonisomorphicgroups that are lattice isomorphic to G. By 1.6.4 and 1.6.5, this still holds for finitegroups G whose subgroup lattice possesses a chain as direct factor; for then G =Cp x K where p E P, n e R, (p, IKI) = 1, and every group G= Cq., x K with q e Pand (q, I K 1) = 1 is lattice isomorphic to G. However, we can show that these are theonly finite groups with this property.

1.6.10 Theorem (Suzuki [1951a]). Let L be a finite lattice that has no chain as adirect factor. Then there are only finitely many nonisomorphic groups whose subgrouplattices are isomorphic to L. More precisely: If 1 is the length and w the width of L (seeSection 1.1), then IGI < w'" for every group G with L(G) ^- L.

Proof. It is clear that the first assertion follows from the second since there are onlyfinitely many groups of order at most w''". To prove the second assertion, we shallshow that p :!g w for every prime divisor p of IGI. Since a maximal subgroup of ap-group has index p, the subgroup lattice of a group of order p" contains a chain oflength n. Therefore it will follow that I P 1 < p' < w' for every Sylow p-subgroup P ofG, and since there are at most w primes p < w, we get IGI < (w')w = w''".

So let p be a prime divisor of I G I and let P be a Sylow p-subgroup of G. If P is notcyclic, then P contains at least p + 1 maximal subgroups and these form an an-tichain in L(G). If P is not normal in G, then, by Sylow's theorem, the conjugates ofP form an antichain containing at least p + 1 elements. In both cases p < w. So wemay suppose that P is a cyclic normal subgroup of G. By the Schur-Zassenhaustheorem, P has a complement K in G. If K were normal in G, then 1.6.4 would imply

that L(G) -- L(P) x L(K), which is impossible since L(P) is a chain. Thus K is notnormal in G and the conjugates of K form an antichain with p" (n c N) elements. Butthen p < w, as asserted.

Clearly, our bound on IGI is not best possible. It seems worthwhile to look forbetter bounds and also to ask how many groups G satisfy L(G) - L for a givenlattice L. Some remarks on the first problem and a characterization of infinitegroups with subgroup lattice of finite width can be found in Brand] [1988].

Exercises

1. Let G = H x K and let U be a subgroup of G. Show that the following propertiesare equivalent (the U, are defined as in 1.6.3).(a) U a G.

(b) [U H] < U n H and [U2, K] < U n K.(c) Uo a G and U/Uo 5 Z(G/U0).

2. (Rose [1965]) Let G = H x K, H ne K and let D be a diagonal in G with respectto H and K. Show that [G/D] is isomorphic to the lattice of normal subgroups ofH.

3. Show that for A, µ E A with A µ, the groups G,, and G defined in 1.6.5 form an-distributive pair in L(G) and use Exercises 1.2.3 and 1.2.4 to give an alternateproof of Theorem 1.6.5.

4. Let G, be a group having only finitely many nonisomorphic projective images-call them G1, ..., G,-and suppose that cp is a projectivity from G = G, x x G,to some group G. If Gr 9 G for all i = 1, ..., r, show that G f-- G.

5. (Brandl [1988]) Let p be a prime, n c N. Show that there are only finitely manynoncyclic finite p-groups whose subgroup lattices have width n.

Chapter 2

Modular lattices and abelian groups

In this chapter we are going to study subgroup lattices of abelian groups. Since theseare modular, we shall more generally investigate groups with modular subgrouplattices, called M-groups for short. Unlike the groups with distributive subgrouplattices, these do not form a nice class of groups. However, there are close connec-tions between M-groups and abelian groups.

In § 2.1 we give the basic properties of modular lattices and introduce one of themost important concepts of this book, the modular elements of a lattice. These willbe used later to study projective images of normal subgroups of groups.

In 1939 Baer determined the projective images of an elementary abelian p-group.These were called P-groups by Suzuki [1951a], who proved that every projectiveimage of an arbitrary noncyclic finite p-group either is a p-group or a P-group.These results are fundamental for the study of projectivities between finite groups;they are proved in § 2.2.

In 1941 and 1943 Iwasawa investigated the locally finite M-groups and those withelements of infinite order; he showed that they are metabelian and determined theirexact structure. His results are presented in §§ 2.3 and 2.4, together with the theoremof Schmidt [1986] which characterizes the periodic M-groups and thus completesthe characterization of groups with modular subgroup lattices.

The main results of § 2.5 are two theorems of Baer [1944] and Sato [1951]. Theseassert, respectively, that every nonhamiltonian locally finite p-group with modular sub-group lattice, and every M-group with elements of infinite order admit a projectivityonto a suitable abelian group. In the p-group case, there exists a crossed isomorphismwhich even induces an isomorphism between the coset lattices of the two groups.

In §2.6 we prove two important theorems of Baer [1939a] on projectivities be-tween abelian groups. He showed that every projectivity between two abelian p-groups is induced by an isomorphism if these groups contain with every element oforder p" at least three independent elements of this order. Also, if G is an abeliangroup with two independent elements of infinite order, then every projectivity of Gto an arbitrary group is induced by an isomorphism. Finally, we study projectivitiesbetween abelian groups of torsion-free rank 1.

2.1 Modular lattices

A lattice L is called modular if for all x, y, z e L the modular law holds:

(1) Ifx <z, then xu(ynz) =(xuy)nz.

42 Modular lattices and abelian groups

2.1.1 Remarks. Let L be a lattice and let x, y, z c L.(a) If x <z, then x <(xuy)nz and ynz <(xuy)nz, so that xu(ynz) <

(x u y) n z holds in any lattice. Thus only the reverse inclusion is required for themodular law.

(b) Every distributive lattice is modular. For, if L is distributive and x < z, thenxuz = z and hence xu(ynz) = (xuy)n(xuz) = (xuy)nz.

(c) If L is modular, then x u (y n (x u z)) = (x u y) n (x u z) since x < x u z. Con-versely, if this equation holds in L, then x < z implies x u z = z and hencex u (y n z) = (x u y) n z. This shows that modularity can be defined by an identity.Thus sublattices and direct products of modular lattices are modular.

It is easy to see that the only nonmodular lattice with 5 or less elements is the onecalled E5 in Section 1.1. This lattice can be used to characterize modularity.

2.1.2 Theorem. The lattice L is modular if and only if it does not contain a sublatticeisomorphic to E5.

Proof. If L is modular, then every sublattice of L is modular and therefore cannotbe isomorphic to E5. Conversely, we have to show that every nonmodular lattice Lcontains a sublattice isomorphic to E5. Since L is not modular, there exist x, y, z c Lwith x<zand xu(ynz)<(xuy)nz.Let a=ynz,b=xu(ynz),c=(xuy)nz,d = y, e = x u y and S = {a, b, c, d, a}. Then, clearly, a < b < c < e and a < d < e.Furthermore

cnd=(xuy)nzny=zny=a

and

bud=xu(ynz)uy=xuy=e;

a

Figure 8: E5

it follows that b n d = a and c u d = e. Thus S is a sublattice of L and as b < c andb u (d n c) = b c = (b u d) n c, S is not modular. Since all the other lattices withat most 5 elements are modular, S E5.

We shall also consider the modular law for single elements of a lattice. Thus wecome to one of the fundamental concepts of this book.

2.1 Modular lattices 43

Modular and permutable subgroups

We say that the element m of the lattice L is modular in L, and write m mod L, if

(2) xu(mnz)=(xum)r zforallx,zELwithx z, and

(3) mu(ynz)=(muy)nzfor ally,zeLwithm<z;a subgroup M of a group G is called modular in G if M is modular in L(G), we writeM mod G in this case.

Modular elements were introduced by Kurosh in 1940; see also Zassenhaus[1958], p. 74 where they are called "Dedekind elements". Clearly, a lattice L ismodular if and only if every element of L is modular in L. The following theoremshows why we use just (2) and (3), two of the three possible variants of the modularlaw, for the definition of a modular element.

2.1.3 Theorem (Ore [1937]). Let G be a group.(a) If NaG,then NH=HN for all H<G.(b) If M < G such that MH = HM for all H < G, then M mod G.

Proof. If N g G, then Nx = xN for all x E G and (a) holds. To prove (b), firstlet X< Z< G. Then, clearly, X u (M n Z) < (X u M) n Z. Furthermore, ifg c- (X u M) n Z, then, since X u M= XM, there exist x e X and m c- M such thatg = xm. Since X < Z, we have m = x-' g E Z and hence g = xm c X u (M n Z).Thus X u (M n Z) _ (X u M) n Z and (2) holds. Now let Y, Z < G with M < Z.Then Mu(YnZ)<(MAY)nZandfrom g=mye(MuY)nZand mEM<Zit follows as above that y c- Z. Therefore g E M u (Y n Z) and (3) holds. Thus M ismodular in G.

Following Stonehewer [1972] we call a subgroup M of G permutable in G, andwrite M per G, if MH = HM for all subgroups H of G. These permutable subgroupshad been introduced by Ore [1937] who had called them "quasinormal". Theorem2.1.3 shows that a normal subgroup is permutable and a permutable subgroup ismodular in G. Thus a normal subgroup is modular in G and the modular laws (2)and (3) are the main properties of a normal subgroup that are visible in the subgrouplattice. We shall use this fact later in our study of images of normal subgroups underprojectivities. In this chapter we investigate the connections between commutativityof the group and modularity of its subgroup lattice. Here Theorem 2.1.3 yields thefollowing.

2.1.4 Theorem. The lattice of normal subgroups of an arbitrary group and the sub-group lattice of an abelian group are modular.

Proof. By 2.1.3, every normal subgroup of a group G is modular in L(G) and hencealso in 91(G) since this is a sublattice of L(G). Thus 91(G) is modular. If G is abelian,then L(G) = 91(G).

Properties of modular elements

We give a useful characterization of modular elements in arbitrary lattices, whichleads to an important property of elements in modular lattices.

2.1.5 Theorem. The following properties of the element m of a lattice L areequivalent.

(a) m is modular in L.(b) For every a c- L, the map pa,m: [a/a n m] --+ [a u m/m]; x " x u m is an

isomorphism.(c) For every ac-L, the map Y'a.m: [a u m/m] -+ [a/a n m]; z H z n a is an

isomorphism.(d) For every a E L, coa.mll/a m = idiaianmj and Y'a.mcoa.m =

Proof. (a) (d). Let a E L. If x c- [a/a n m], then by (2),

xv°m'k°^" = (x u m) n a= x u (m n a) = x

and if z E [a u m/m], then by (3),

z0-9- _ (z n a) u m = (m u a) n z = z.

Thus cpa.mqia.m = idlaiatm) and Y'a.m(pa.m = idiajmim].(d) = (a). Let x, z c- L such that x < z. Then x u (m n z) E [z/z n m] and hence

xu(mnz) = (m n z)) u n z = (xum)nz,

that is, (2) holds. If y, z c- L such that m z, then (m u y) n z E [y u m/m] and hence

(muy)nz = ((muy)nz)'Oy.-wy.- _ (((muy)nz)ny)um = (ynz)um.

Hence (3) also holds and m is modular in L.To prove the equivalence of (b), (c) and (d), let a e L and write (p, = (P and

4/a.m= 0.(b) . (d). Let x c- [a/a n m]. Then x < (x u m) n a and hence x u m <

((x u m) n a) u m. The other inclusion is trivial since m and (x u m) n a are containedin x u m. So we get

x°=xuIn =((xum)na)um=((xum)na)".

Since cp is injective, it follows that x = (x u m) n a = x'". Now let z e [a u m/m].Then (z n a) u m < z and since cp is surjective, there exists b e [a/a n m] such thatz=b"=bum. Thenb<znaandz=bum <(zna)um.Thusz=(zna)um=zO" and (d) holds.

It is clear that (d) implies both (b) and (c). For, if (d) holds, then cp and I arebijective and satisfy (c) and (b) of 1.1.2, respectively. So it remains to show that

(c) (d). For x E [a/a n m], again, x < (x u m) n a. Since qi is surjective, thereexists c E [a u m/m] such that x= c o= c n a. Then x u m< c and (x u m) n a<c n a = x. Thus x = (x um)na = xl*O. Now let zn [a um/m]. Then (z a) urn <zand hence ((z n a) u m) n a < z n a. The other inclusion is trivial and so we getthat

z0 =zna=((zna)um)na=((zna)um)''.

Since 0 is injective, it follows that z = (z n a) u m = z 0". Thus (d) holds.

Theorem 2.1.5 shows that if m is a modular element of a lattice L, then

(4) [a u m/m] [a/a n m] for every a E L.

But this condition alone does not imply that m is modular in L. This is shown by thefollowing example. Take a circle K in the euclidean plane, choose two points I and0 on K dividing the circle into two arcs and for x, y E K, define x < y if and only ifx and y are on the same arc and x is nearer to 0 than y or x = y. Let a, m E K;then both [a u m/m] and [a/a n m] consist of a single element if a < m, and are

I=aum

0=anm

Figure 9

isomorphic to the real unit interval if a m. So m satisfies (4), but, clearly, m is notmodular in K if 0 V: m v: 1. If L is a finite lattice, it is easy to see that m is modularin L if it satisfies (4) (see Exercise 2). However, we shall not need this fact since weshall always use the mappings cpa., and t/ia,, to prove (4).

We collect the main inheritance properties of modular elements.

2.1.6 Theorem. Let L be a lattice, m, n, a c L and suppose that m is modular in L.(a) m n a is modular in [a/0] = {x E Lix < a}.(b) m u a is modular in [I/a] = {y E LIa < y}.(c) If m < n and n is modular in [I/m], then n is modular in L.(d) If n and m are modular in L, then so is m vn. _(e) If u is an isomorphism from L to a lattice L, then m° is modular in L.

Proof. (a) Let b e [a/0]. If X E [b/b n (m n a)] = [b/b n m], then by (2),

x`°b.- = x v (m n a) = (x u m) n a = xa1b.-O..M.

Hence cpb,m,.,° is the product of (pb,m and the restriction of 'n,m to [b u m/m]. Sincethese two maps are isomorphisms, (p6,mnn is also an isomorphism and by 2.1.5, m n ais modular in [a/0].

(b) Let c e [I/a]. If z c- [c u (m u a)/m u a] = [c u m/m u a], then z

z n c = Thus oc,m° is the restriction of 'tl m to [cum/m u a] and hence anisomorphism. By 2.1.5, m u a is modular in [I/a].

(c) Let d e L. Then for z e [d u n/n], we have z k&n= z n d = z n (d u m) n dz note that the right-hand side is well-defined because m S z n (d u m) <d u m. Since Iidum,n and the restriction of Od,m to [d u m/(d u m) n n] are isomor-phisms, Iid,n is also an isomorphism. By 2.1.5, n is modular in L.

(d) By (b), m u n is modular in [I/n] and therefore by (c), m u n is modular in L.(e) is trivial.

Note that m n n in general is not modular in L if m and n are modular elements inL. This, for example, is shown by the lattice represented by the Hasse diagram ofFigure 10. But there are also simple examples in subgroup lattices (see Exercise 3).

Figure 10

It is well-known (see Robinson [1982], p. 66) that the class of groups satisfying themaximal condition (or other chain conditions) is closed with respect to formingextensions. This is a more general property of intervals in lattices.

2.1.7 Lemma. Let a, b, m be elements of a lattice L such that b < m S a and supposethat m is modular in L. If [a/m] and [m/b] satisfy the maximal condition, then so does[alb].

Proof. Let (c;)j, N be an ascending chain in [a/b]. Then (c; u m);,-: N and (c; n m),E N

are ascending chains in [a/m] and [m/b], respectively. Since these two intervalssatisfy the maximal condition, there exists a natural number r such that c, u m =csumand c,nm=cSnmfor all s _! r. Then c, :!9 c, and by (2),

c,=c,u(mnc,)=c,u(mnc5)=(c,um)nc5=(c,um)nc,=c5

for all s > r. Thus the chain (c1)1 N is also finite.

Semimodular lattices

A lattice L is called upper semimodular if for all x, y e L,

(5) x u y covers x whenever y covers x n y;

and L is called lower semimodular if for all x, y a L,

(6) y covers x n y whenever x u y covers x.

By 2.1.5, a modular lattice satisfies (5) and (6) and is therefore upper and lowersemimodular. The following theorem shows that for subgroup lattices, property (6)is more important.

2.1.8 Theorem. The lattice of composition subgroups of an arbitrary group and thesubgroup lattice of a finite nilpotent group are lower semimodular.

Proof. Let G be a group and let X, Y be composition subgroups of G such thatX u Y covers X in the lattice R(G) of composition subgroups of G. By 1.1.4, thereexists a composition series from X u Y to X. Clearly, all members of this seriesbelong to R(G) and since X u Y covers X in R(G), the series must have length 1. SoX a X u Y and X u Y/X is simple. Hence also X n Y g Y and Y/X n Y is simple.Again by 1.1.4, if H e R(G) such that X n Y< H < Y, then there exists a composi-tion series from Y to H and it follows that H = X n Y. So Y covers X n Y in R(G)and R(G) is lower semimodular. In a finite nilpotent group G, every subgroup issubnormal and hence L(G) = R(G).

The Jordan-Dedekind chain condition

We say that a lattice L satisfies the Jordan-Dedekind chain condition if for everyx, y e L with x < y there exists a finite maximal chain x = xo < < xr = y from xto y and all maximal chains from x to y have the same length l(x, y).

2.1.9 Theorem. Every upper or lower semimodular lattice L of finite length satisfiesthe Jordan-Dedekind chain condition.

Proof. We give a proof in the case that L is lower semimodular; the proof for uppersemimodular lattices is similar. If x, y e L with x < y, then there exists a finitemaximal chain from x to y since L has finite length. We prove by induction on n thefollowing assertion which then, clearly, will imply the Jordan-Dedekind chain condi-tion for L.

(*) If x, y a L with x < y and if there exists a maximal chain from x to y of lengthn, then every maximal chain from x to y has length n.

This is clear if n = 1, that is if y covers x. So assume that (*) is true for n - 1,let x = xo < < x = y be a maximal chain of length n and suppose that x =yo < . . < y, = y is another maximal chain. If yrn_1, then by induction,n - 1 = m - 1. So assume that yrn_1 and let z = n yrn_1. Then y =

u and z is covered by x,,_1 and yrn_1 since L is lower semimodular. There-chain, then

x = xo < < xi_1 are maximal chains from x to xn_1. By the induction assump-tion, n - 1 = k + 1. Thus also x = zo < < zk < y._, is a maximal chain of length

n - 1 and the induction assumption applied to this chain and to x = yo < < yin_1yields n - 1 = m - 1. Thus n = m and (*) holds.

2.1.10 Theorem. The following properties of a lattice L of finite length are equiva-lent.

(a) L is modular.(b) L is upper and lower semimodular.(c) L satisfies the Jordan-Dedekind chain condition and 1(x, x u y) = !(x n y, y) for

all x, y e L.

Proof. That (a) implies (b) is clear. If (b) holds, then L satisfies the Jordan-Dedekindchain condition as we have just shown. Let x = xo < < x, = x u y be a maximalchain. Then

(7)

For every i e {0, ... , n - 11, since xi+1 covers xi, either xi n y = xi+1 n y orxi u (xi+1 n y) = xi+1 and then xi n y = xi n (xi+1 n y) is covered by xi+1 n y since Lis lower semimodular. Therefore (7) yields a maximal chain from x n y to y oflength at most n; that is, we have 1(x, x u y) > 1(x n y, y). The other inequality fol-lows similarly from the upper semimodularity of L. Thus (c) holds. Finally, supposethat L satisfies (c) but is not modular. Then by 2.1.2, there exists a sublattice{a, b, c, d, e} of L isomorphic to Es. If a < b < c < e and a < d < e, then

l(a, b) = l(d n b, b) = l(d, d u b) = l(d, e);

similarly, l(a, c) = l(d, e). Thus l(a, b) = l(a, c), but this is impossible since b < c. Thiscontradiction shows that (c) implies (a).

Exercises

1. Show that a modular subgroup of a group G is in general not permutable, and apermutable subgroup is in general not normal in G.

2. (Schenke [ 1986]) Let L be a lattice and let m e L such that [a u m/m][a/a n m] for every a e L. If [I/m] is finite, show that m is modular in L.

3. Let p be an odd prime, H = <a,bIa"2 = b° = l,ab = a1+p> the nonabelian groupof order p3 and exponent p2, K = <c> cyclic of order p2 and let G = H x K.Find two modular subgroups of G whose intersection is not modular in G.

4. Show that a lattice all of whose sublattices are lower semimodular is modular.Deduce that sublattices of lower semimodular lattices are in general not lowersemimodular.

5. Let L be the direct product of lattices Lx. Show that L is lower semimodular ifand only if every Lx is lower semimodular.

6. Show that for an arbitrary lattice L, no two of the conditions (a)-(c) in Theorem2.1.10 are equivalent.

2.2 P-groups 49

7. The element m of the lattice L is called semimodular in L if it satisfies (2). Showthat(a) m is semimodular in L if and only if (p",ml//a,m = id["i"nm, for all a e L;(b) if m is semimodular in L, then 0m,"(pm," = id[.,,.,.] for all a e L;(c) if m and n are semimodular in L, then [m u n/n] [n/m n n];(d) an antiatom of L is semimodular in L if and only if it is modular in L.

8. Let L be a lattice, m, n, a e L, and suppose that m is semimodular in L. Show that(a) m n a is semimodular in [a/0];(b) if m is modular in L, m < n and n is semimodular in [I/m], then n is semi-

modular in L;(c) if m 5 n and n is modular in [I/m], then n in general is not semimodular in L;(d) if [I/m] is a chain and m < n, then n is semimodular in L.

2.2 P-groups

By 1.6.6, every projective image of a periodic abelian group G is the direct productof the images of the primary components of G. However, in general these images areneither abelian nor primary: the nonabelian group of order pq, p and q primes withq dividing p - 1, and the elementary abelian group of order p2 have isomorphicsubgroup lattices. This is the first of a series of examples that we shall now investi-gate more closely.

The classes P(n,p)

Let p be a prime and n >- 2 be a cardinal number. We say that a group G belongs tothe class P(n, p) if G is either elementary abelian of order p", or a semidirect productof an elementary abelian normal subgroup A of order p"-1 by a group of prime orderq # p which induces a nontrivial power automorphism on A. Here, if n is infinite, byan elementary abelian group of order p" or p"-1 we shall mean a direct product of ncyclic groups of order p. We call G a P-group if G e P(n, p) for some prime p andsome cardinal number n >- 2.

2.2.1 Remarks. Let G be a nonabelian P-group, so G = A<t> with an elementaryabelian p-group A and an element t of order q that induces a nontrivial powerautomorphism on A. By 1.5.6, t is universal on A, that is, there exists an integer rsuch that

(1) t-tat=a'for all aeA.

Since t 0 CG(A) and t9 = 1,

(2) r # 1 (mod p) and r9 = 1 (mod p).

Hence q divides p - 1. In particular, the classes P(n, 2) only contain the elementaryabelian groups of order 2". For p > 2, however, to every prime divisor q of p - 1,

there exists an integer r satisfying (2); then the semidirect product G = A<t) of anontrivial elementary abelian p-group A by a cyclic group <t> of order q wheret-1 at = a' for all a e A is a nonabelian P-group. Any two such groups with the sameA and q (for different r) are isomorphic, as the reader can easily verify. Thus forp > 2, the classes P(n, p) contain the elementary abelian group of order p", and, forevery prime divisor q of p - 1, exactly one nonabelian P-group with elements oforder q; if n is finite, the order of this group is p"-' q.

We want to show that these P(n, p) are complete classes of lattice-isomorphicgroups. For this purpose and for later use, we collect some simple properties ofP-groups.

2.2.2 Lemma. Let G be a nonabelian P-group and suppose that p, q and A are as in2.2.1.

(a) G' = A = CG(a) for all 1 0 a e A.(b) Z(G) = 1.(c) The normal subgroups of G are G and the subgroups of A.(d) Every element x e G\A has order q. Hence the subgroups of order q generate G.

Proof. (a) Since G/A is cyclic, G' < A. If 1 a e A, in the notation of 2.2.1, then[a, t] = a'-' and <a'-') = <a> since r # 1 (mod p). Hence <a> < G' and a 0 Z(G).Thus G' = A and C6(a) = A since A is an abelian maximal subgroup of G.

(b) By (a), Z(G) < CG(A) = A and a 0 Z(G) for all 1 rA a E A. Thus Z(G) = 1.(c) Since A is abelian and t induces a power automorphism on A, every subgroup

of A is normal in G. If N:9 G and N A, then AN = G and GIN A/A n N isabelian. It follows that A = G' < N and hence N = G.

(d) If x e G\A, then G = A<x) and x e CG(A n <x>). By (a), A n <x> = 1. Hence<x) - G/A is of order q. Since a group cannot be the set-theoretic union of twoproper subgroups, G is generated by the elements in G\A.

2.2.3 Theorem (Baer [1939a]). For every prime p and every cardinal number n > 2,all groups in P(n, p) are lattice-isomorphic.

Proof. We show that every nonabelian group G E P(n, p) is lattice-isomorphic to theelementary abelian group in P(n,p). So let q, r and G = A<t) be as in 2.2.1. ThenG = A x <t), where o(t) = p, is the elementary abelian group in P(n, p). We want toconstruct a bijective map from the set L1(G) of cyclic subgroups of G to L1(G) thatinduces a projectivity from G to G. If x e G\A, then by 2.2.2, <x> has order q andtherefore contains exactly one element out of each coset of A, in particular out of At.Hence there exists exactly one element a e A such that <x> = <at>. We define themap r: L1(G) - L1(G) by <x>' = <x> if x e A, and <x> = <at> if <x> = <at> A.

Since every cyclic subgroup of the elementary abelian group G = A x <1) which isnot contained in A also contains exactly one element of the form at with a E A, wesee that r is bijective. By 1.3.2, we have to show that for all X, Y, Z E L1(G),

(3) X < <Y,Z) if and only if X' S <Y`,Z`).

2.2 P-groups 51

This is clear if Y and Z are contained in A since T is the identity on L,(A). If Y < Aand Z $ A, that is, Y = and Z = <ct> with b, c e A, then <Y, Z> n A = Y sinceevery subgroup of A is normal in G. Similarly, <Y`, Z=> n A = Y`. Thus, if X < A,then X < <Y,Z> if and only if X < Y, and this is the case if and only if X'<

Y`, Z`>. For X $ A, that is, X = <at> with a e A, we have X < <Y, Z> if and onlyif at e <b, ct> n At = (<b, ct> n A)ct = ct, i.e. a e c; moreover, this is the caseif and only if U` = <al> < x <ci-> = < YT, Z`>. Finally, if neither Y = <bt> norZ = <ct> is contained in A, then with W = <bc-' > < A we have <Y, Z> = (W, Z>and <Y`,Z`> = <bc-',cf> = <W`,Z`>. Thus (3) also holds in this case as we havejust shown.

We now want to show that the classes P(n, p) are closed under projectivities. Forthe case n = 2 of this assertion we prove a more general result that will also be usedin the next section.

2.2.4 Lemma. Let G be a finite group and suppose that M and N are two differentminimal subgroups of G which are modular in G. Then I M u NI = pq where p and q areprimes.

Proof. We may assume that G = M u N. If H is a maximal subgroup of G, then Hcannot contain both M and N. So if M H, say, then H n M = 1, H u M = G and,by 2.1.5,

[H/1] = [H/H n M] [H u M/M] = [N u M/M] [N/N n M] = [N/1].

Hence H is a minimal subgroup of G. If H a G, then I HI = p and I G/HI = q areprimes and I GI = pq. So assume that H is not normal in G. Then H = NG(H) and theconjugates of H in G together contain I G : HI (I HI - 1) = IGI - I G : HI nontrivialelements, that is, at least i IGI but less than IGI - 1. Since H is not normal, G is notcyclic and therefore every element of G is contained in a maximal subgroup of G.Hence there exists a maximal subgroup K of G that is not conjugate to H. But thenK a G and IGI = pq since otherwise the conjugates of K would contain at least i I GIfurther nontrivial elements of G.

2.2.5 Theorem (Baer [1939a]). For every prime p and every cardinal number n - 2,the class P(n, p) is invariant under projectivities.

Proof. Let cp be a projectivity from G to G and suppose that G E P(n, p). If n = 2,then the nontrivial proper subgroups of G form an antichain with p + 1 elements. By2.2.4, 1 GI = qr where q and r are primes with q > r, say. Since the antichain containsmore than 2 elements, G is not cyclic and hence elementary abelian of order q2 (ifq = r) or nonabelian of order qr (if q r). In both cases, G has exactly q + 1 minimalsubgroups. It follows that q = p and G E P(2, p).

Now let n > 3. By 2.2.3, we may assume that G is elementary abelian. So if x andy are nontrivial elements of G such that <x> <y>, then <x>' and <y>' are cyclic of

order p, and <x, y)`0 = <x)`0 u <y> is elementary abelian of order p2. By the casen = 2, already settled,

(4) <x, y> E P(2, p).

In particular, if P is the set of p-elements in G, then P 1 and we show that P is anelementary abelian normal subgroup of G. For 1 x c- P, <x> is cyclic of order pand hence also o(x) = p. If x, y E P with o(x) = p = o(y) and <x> : <y>, then by (4),<x, y> E P(2, p). And since a nonabelian group in P(2, p) has only one subgroup oforder p, I <x, y> = p2. Thus xy-' E P and xy = yx. This also holds if <x> = <y> andhence P is an abelian subgroup of G. Since it is the set of all elements of order p or1 in G, it is clearly normal and elementary abelian.

Now P and P° are lattice-isomorphic elementary abelian p-groups. If P = Dr Px.ZEA

where I P I = p for all A, then P' = U Px and Px n U P", = (Px n U Pµ ) = l forA E A

all A. Hence P`° = Dr Px and PO P. Therefore if P = G, we are finished. So assumeA E A

that P G. Since G/P does not contain elements of order p, 16/P01 = p by the casen = 2, which is already settled. Thus P`0 and P are elementary abelian of order p"-'and I G : PI = q for some prime q. Let t E G with o(t) = q. Then G = P<t>. For1 x E P, <x, t> E P(2, p) by (4) and therefore <x> g <x, t>. Thus t induces a non-trivial power automorphism on P and G E P(n, p).

Projective images of locally finite p-groups

By 2.2.3 and 2.2.5, the class P(n, p) is precisely the class of projective images of theelementary abelian group of order p". So for p > 2, this group has projective imagesthat are not even primary groups. We show that the elementary abelian groups arethe only finite p-groups with this property.

2.2.6 Theorem (Suzuki [1951a]). Let p be a prime, n a natural number, G a group oforder p", and suppose that cp is a projectivity from G to some group G. If 161 IGI,then either

(a) G is cyclic and G is cyclic of order q" where q is a prime different from p, or(b) G is elementary abelian, n >_ 2 and G is a nonabelian P-group of order p"-'q

where q is a prime dividing p - 1.

Proof. We use induction on n. Since IGI IGI, G is not a p-group. If G is cyclic, thenby 1.2.8, (a) holds. In particular our assertion is correct for n = 1 and we may assumethat n > 2 and that G is not cyclic. Then by the Burnside Basis Theorem, G/cD(G)is elementary abelian of order p' where m >_ 2. Since cb(G)`' = cb(G), cp induces aprojectivity from G/cD(G) to 61(l)(6) and by 2.2.5, G/cb(G) E P(m, p). Hence thereexists a subgroup P of G such that cD(G) < P and I P'/D(G)J = p'-' By the inductionassumption, P'0 is a p-group or lies in P(k, p) for some k; in the latter case by 2.2.2,D(G) is a p-group as a normal subgroup of P°. Hence in both cases P" is a p-group.Since G is not a p-group, OD(G) is nonabelian of order p"'-' q where q is a primedividing p - 1 and P'0 is the Sylow p-subgroup of G.

2.2 P-groups 53

1

Figure 11

If (D(G) = 1, then (b) holds and we are finished. So suppose for a contradiction thatD6(G) # 1. Let Q < G such that IQ"I = q, take a maximal subgroup R of G containingQ and put D = P n R. Since P`°Q° = G, R 0 P and so G/D is elementary abelian oforder p2. Hence there exists a maximal subgroup_S of G such that D < S and RS : P. Since P4' is the only Sylow p-subgroup of G, R0 and SP are not p-groups; onthe other hand, p divides the order of cD(G) < R'O n S". By induction, R and S areelementary abelian. This implies that D 5 Z(G) and that P is abelian as a cyclicextension of the central subgroup D. So for x, y E G, we have [y, x] e G' S D 5 Z(G)and by (4) of 1.5,

(xy)" = x"y"[y,x](z) = xryr

since D is elementary abelian and p > 2. Therefore the set of elements of orderdividing p is a subgroup of G containing R and S, hence is G. It follows that P iselementary abelian. Since P' is a p-group, it is also elementary abelian, and, byMaschke's Theorem, P" is completely reducible under the action of Q'P. Hence thereexists a Q'-invariant subgroup KP of G such that P' = K0 x cD(G). Then K0Q' is aproper subgroup of G, but every maximal subgroup of G containing K°Q0 alsocontains KocF(G)Qq' = P'Q' = G. This contradiction shows that D(G) = 1 and (b)holds.

The corresponding result for locally finite groups is an easy consequence.

2.2.7 Corollary. Let p be a prime and let G be a locally finite p-group. If there existsa projective image of G that is not a p-group, then G is locally cyclic (that is G - Cpfor some n e f\l u { oo }) or elementary abelian.

Proof. Let cp be a projectivity from G to G where G is not a p-group and supposethat G is not elementary abelian. Then there exist elements x, y, z c- G such that <x>wis not a p-group and <y,z> is not elementary abelian. Now if g1, ..., g E G, thenH = <g1,.. . , g,,, x, y, z> is a finite p-group, and since x, y, z e H, we see that I H' :IHI and H is not elementary abelian. By 2.2.6, H is cyclic and thus G is locally cyclic.

As another consequence of Suzuki's theorem it is possible to describe the projec-tive closure of the class of finite nilpotent groups, that is, the smallest class of groupswhich is invariant under projectivities and contains all finite nilpotent groups (seeExercise 7). Nilpotent torsion groups will be handled in Section 7.4.

Exercises

1. Show that there is at most one P-group of a given finite order (see Remark 2.2.1).2. Give an alternate proof of Theorem 2.2.3. With the notation used there show the

following.(a) Every subgroup H of G that is not contained in A has the form H = A, <at>

where A, = Hr'A <A and ac A.(b) For A, < A and a; E A, A, <a, t> < A2 <a2 t> if and only if A, < A2 and

a, a2 E A2'(c) The map cp: L(G) - L(G) defined by H0 = H for H < A and (A, <at>)" _

A, <at > for A, < A and a E A is a projectivity from G to G.3. Let p be a prime and 2 < n c N. If H and K are nonabelian groups in P(n, p) and

P is a p-group, show that P x H and P x K are lattice-isomorphic.4. Let G be a finite group such that any two different maximal subgroups of G

intersect trivially. Show that G is cyclic of prime power order or a semidirectproduct of an elementary abelian normal subgroup A by a group of prime orderoperating irreducibly on A.

5. Give an alternate proof of Theorem 2.2.6 (avoiding Maschke's Theorem). Showby induction that IG : D(G)l = p2 and JD(G)J = p if D(G) # 1. Then study theconjugacy classes of subgroups of order p of P`0.

6. Illustrate Theorem 2.2.6 by the following examples. That is, without using 2.2.6,show that for primes p and q with qlp - 1, the groups G and G are not lattice-isomorphic if _(a) G = Cpl x CP and G is the semidirect product of C,,2 by Cq to an automor-

phism of order q of CPz, _(b) G is nonabelian of order p3 and exponent p, and G = CP x H where H is

nonabelian of order pq, and(c) G is as in (b), G = <a, b, cI aP = bP = c9 = 1, ab = ba, c-' ac = a', c-' be = b'2

where r is an integer such that r 0- 1 (mod p) and r9 - 1 (mod p).7. Show that a finite group is lattice-isomorphic to a nilpotent group if and only if

it is a direct product of p-groups and P-groups with relatively prime orders.

2.3 Finite p-groups with modular subgroup lattices

In this section and the one following we wish to determine all groups with modularsubgroup lattices, called M-groups for short. We start with the most difficult part ofthis program and describe first the finite p-groups with this property. So, in this

2.3 Finite p-groups with modular subgroup lattices 55

section, p is a prime and every p-group considered is finite. Our aim is to prove thefollowing result.

2.3.1 Theorem (Iwasawa [1941]). A finite p-group G has modular subgroup lattice ifand only if

(a) G is a direct product of a quaternion group Q8 of order 8 with an elementaryabelian 2-group, or

(b) G contains an abelian normal subgroup A with cyclic factor group G/A; furtherthere exists an element b c G with G = A and a positive integer s such thatb-'ab=a1+F` for all aEA,with s>_2incase p=2.

The proof of Iwasawa's theorem will occupy nearly the whole section. In 2.3.3 weshall present a useful criterion for a p-group to have modular subgroup lattice. Fromthis it will easily follow that the groups in 2.3.1 are M-groups. The converse, how-ever, is much deeper. We first study p-groups with modular subgroup lattices inwhich the quaternion group Q8 is involved, and show in 2.3.8 that they are thegroups in (a) of Theorem 2.3.1. Recall that a group is called hamiltonian if it is

nonabelian and all of its subgroups are normal. By 2.1.4, these groups have modularsubgroup lattices, so the hamiltonian p-groups have to show up in Iwasawa's theo-rem. We shall see in 2.3.12 that they are also exactly the groups in (a) of Theorem2.3.1. Therefore it remains to consider p-groups with modular subgroup lattices inwhich Q8 is not involved, or which are not hamiltonian. We call these M*-groupsand establish a number of important properties that will also be used in latersections, the most noteworthy being 2.3.10, 2.3.11, 2.3.14 and 2.3.16. These proper-ties, of course, can also be proved as soon as one knows that the groups satisfy (b)of Theorem 2.3.1; however, they are not obvious consequences of Iwasawa's theoremand we need them in the proof of it. This proof will finally distinguish the two casesthat there exists a cyclic normal subgroup X in the M*-group G such that X J =Exp G, in which case we get the desired structure for G quite easily (2.3.15), and thatthere is no such X, where the computations become a little more involved (2.3.18).In both cases additional information will be obtained on how to choose the sub-group A, the element b and the integer s in Iwasawa's theorem. That these are notuniquely determined by G is shown by the abelian groups, for example. There anyinteger s > 2 with ps > Exp G and every subgroup A and element b with G = A will satisfy (b) of 2.3.1.

We start with an elementary remark. If H and K are subgroups of a p-group Gsuch that [H u K/K] [K/H n K], then I H u K: K I= p" = I K: H n K I, where nis the length of any of the two isomorphic intervals, and hence HK = H U K = KH.So 2.1.5 and 2.1.3 yield the following result.

2.3.2 Lemma. A finite p-group has modular subgroup lattice if and only if any two ofits subgroups permute.

The groups of order p3 are well-known. The dihedral group and the nonabeliangroup of exponent p for p > 2 are generated by two elements of order p and there-fore, by 2.2.4, cannot be M-groups. In the other groups of order p3 any two elements

of order p commute and every subgroup of order p2 is normal. By 2.3.2, these groupsare M-groups. For an arbitrary p-group, the sections of order p' decide whether ithas modular subgroup lattice or not.

2.3.3 Lemma. Let G be a finite p-group. Then G has modular subgroup lattice if andonly if each of its sections of order p' does. Therefore if G is not an M-group, thenthere exist subgroups H, K of G with K g H such that H/K is dihedral of order 8 ornonabelian of order p' and exponent p for p > 2.

Proof. The second assertion follows from the first as our remarks on groups of orderp3 show. And since sublattices of modular lattices are modular, one part of the firstassertion is trivial. So we have to show that G has modular subgroup lattice if everysection of order p3 does. We choose a minimal counterexample G to this statement.Then every proper section of G is an M-group. Since G is not an M-group, thereexist subgroups H and K of G with HK KH; choose H and K so that IHI IKIis minimal. If H, and K1 are maximal subgroups of H and K, respectively, thenH1K1 = K1H1, H1K = KH1 and HK1 = K1H. For X = HK1 and Y = H1K, itfollows that HK = HH1 K 1 K = HK 1 H1 K = X Y and KH = KH1 K 1 H = YX. ThusX Y 0 YX, and furthermore

IX:X r Yl <_ IHK,:H1K11 = IH:HnH1K1I <IH:H1I = p.

Similarly I Y: X n YI S p and hence X n Y:9 X v Y. Since all proper sections of Gare M-groups, X u Y = G and X n Y = 1. In particular, IXI = I YI = p. If N is aminimal normal subgroup of G, then, since INI = p and XY YX, we see thatXN/N and YN/N are different subgroups of order p generating the M-group GIN.By 2.2.4, IG/NI = p2 and hence IGI = p'. But then by assumption, G is an M-group,a contradiction.

As an easy application of 2.3.3 we get the following criterion.

2.3.4 Lemma. Let G be a finite p-group and let A be an abelian subgroup of G suchthat every subgroup of A is normal in G and G/A is cyclic. If p = 2 and Exp A > 4,assume further that there exist subgroups A2 < A, of A with A, /A2 ^- C4 and[A,,G] < A2. Then G is an M-group.

Proof. We use induction on IGI. Since G/A is cyclic, there exists b E G with G =A (b). By 1.5.4, b induces a universal power automorphism on A; that is, there existsan integer k such that b-1 ab = a' for all a e A. If p = 2 and Exp A > 4, our assump-tion implies that k =_ 1 (mod 4) and so every element of A with order at most 4 liesin the centre of G. Thus if H is a proper subgroup of G, every subgroup of A n H isnormal in H, and H/A n H AH/A is cyclic; also, if p = 2 and Exp A n H > 4, thereexist 1 = A2 < Al < A n H such that A 1 C4 and [A1, H] 5 [A G] = 1, By in-duction, L(H) is modular. So if G were not an M-group, then by 2.3.3 there wouldexist a normal subgroup N of G with GIN D8 or GIN nonabelian of order p' andexponent p for p > 2. Since G/AN is cyclic, we have I G : AN I < p in both cases. But

every subgroup of AN/N is normal in GIN, and this implies that p = 2 and AN/N isthe cyclic subgroup of order 4 in the dihedral group GIN. Hence A/A n N C4 andby assumption, [A, G] < A n N. It follows that [AN, G] 5 N, a contradiction. ThusG is an M-group.

It will follow immediately from Lemma 2.3.4 that the groups in Theorem 2.3.1 areM-groups. We turn to the proof of the more difficult part of this theorem. Recall thedefinition, for an arbitrary p-group G and i e N, of the characteristic subgroups

O,(G)=<xEGixP'= 1> and J(G)=<xP'IxEG>;

we put f2(G) = Q1(G). By 1.2.7, Q,(G)0 = fl,(G) and Ui(G)'P = Ji(G) for everyprojectivity cp from G to a p-group G.

23.5 Lemma. If G is a finite p-group with modular subgroup lattice, then (1(G) iselementary abelian and Q,(G) = {x E GIxP' = 1} for all i E N.

Proof. Let P = {x E G i xP = 11. If x and y are nontrivial elements of P with <x><y>, then by 2.3.2, <x, y> is elementary abelian of order p2 and hence xy_'

E P andxy = yx. This clearly also holds if <x> = <y>. Thus P is an abelian subgroup of G,and therefore ((G) = P is elementary abelian. This is the first assertion of the lemmaand also the case i = 1 of the second. So assume that this second statement is truefor i - 1. Then for x, y e G with xP' = 1 = yP', we have x"'-', y"'-' E ((G), and byinduction applied to G/11(G), we get (xy)P'-' E ((G). It follows that (xy)P' = 1 andthat fQi(G) only contains elements of order at most p'.

We first determine the 2-groups with modular subgroup lattices in which thequaternion group Q8 is involved. We start with a property of modular 2-groups thatwill be needed later.

23.6 Lemma. If G is a finite 2-group with modular subgroup lattice, then (2(G)normalizes every subgroup of G, that is f22(G) < N(G), the norm of G.

Proof. We use induction on IGI. Let H be a nontrivial subgroup of G and letZ = <z> be a normal subgroup of order 2 of H. By 2.3.2, if x E f22(G), thenI <x, z> : <x> I S 2 and hence <x>' = <x>. If z did not centralize x, then xZ = x-' and<x, z> would be a dihedral group of order 8; but this is not an M-group. Therefore<x,z> is abelian and it follows that Z a Hc22(G). The group H02(G)/Z has smallerorder than G, hence by induction, H/Z is normalized by f22(Hf22(G)/Z) and thisgroup contains (2(G)Z/Z. Thus (12(G) normalizes H.

23.7 Lemma. If G is a finite M-group of exponent 4, then G is abelian or G = Q x Awhere Q - Q8 and A is elementary abelian.

Proof. Suppose that G is not abelian, let x, y e G with xy yx and put Q = <x, y>.By 2.3.6, every subgroup of G is normal in G. Hence 1 # [x, y] E <x> n <y> and so

I <x, y> I < 8. Since the dihedral group D. is not an M-group, Q - Q8. As a subgroupof Pot Q, G/CG(Q) has order at most 4. On the other hand QCG(Q)/CG(Q) = Q/Z(Q)is of order 4. Thus G = QCG(Q). Force CG(Q), we have (xc) = x-'c : xc and sinceo(xc) < 4, it follows that x-'c = (xcv = (xc)-' = x-'c-'. Thus c2 = 1 and CG(Q)is elementary abelian. Then G = Q x A where A is a complement to Z(Q) inCG(Q)

2.3.8 Theorem. Let G be a finite 2-group with modular subgroup lattice. If thequaternion group Q8 is involved in G, that is, if there exist K a H < G withH/K Q8, then G = Q x A where Q ^-' Q8 and A is elementary abelian.

Proof. We use induction on I GI to show that Exp G = 4. Then the theorem willfollow from 2.3.7. So suppose that Exp G > 4 and take u e G with o(u) = 8. Assumefirst that there exists a proper subgroup of G in which Q8 is involved. Then byinduction, Q8 is also a subgroup of G and hence of i22(G). By 2.3.7, S22(G) = Q x Awhere Q Q8 and A is elementary abelian. Since u2 E S22(G), there exist x e Qand a c A with o(x) = 4 and u2 = xa. Let Q = <x, y>. Then u4 = x2 = y2, hence<u4> a <Q, u> and, by 2.3.6 applied to <Q, u>/<u4>, it follows that <y>u = <y>.Since o(y) = 4, y"2 = y; on the other hand, y" = y"° = y-', a contradiction.

We are left with the case that Q8 is not involved in any proper subgroup of G.Since Q8 is involved in G, there exists N a G with GIN Q8. As Exp G > 4, N :A 1and we take a minimal normal subgroup Z of G contained in N. By induction, Q8 isa subgroup of G/Z and, since it cannot be a proper subgroup, G/Z ^- Q8. ThusZ = N and IGI = 16. Since L(D8) is not modular and Q8 is not a subgroup of G,every subgroup of order 8 of G is abelian. Since GIN Q8, we see that N < and<u2> is the intersection of abelian maximal subgroups of G; thus <u2> :!5; Z(G). Onthe other hand there exists v e G with GIN = <uN, vN> and u° = u-'z for somez e N, and it follows that (u2)" = u-2. This contradiction shows that Exp G = 4.

M*-groups

We say that a finite p-group G is an M*-group if L(G) is modular and the quaterniongroup Q8 is not involved in G. Thus for p > 2, every p-group with modular subgrouplattice is an M*-group. Furthermore, every subgroup and every factor group of anM*-group is an M*-group. We want to show that every M*-group satisfies (b) ofTheorem 2.3.1; in view of 2.3.8 this will finish the proof of that theorem. We startwith a useful remark on cyclic permutable subgroups of maximal order in arbitraryp-groups.

2.3.9 Lemma. If X is a cyclic permutable subgroup of the p-group G such that I X I =Exp G, then f2(X) < Z(G).

Proof. Let y c- G and Y = <y>. If y E X, then y clearly centralizes f2(X). So assumethat y X and let N be a maximal subgroup of X Y containing X. Then N a X Y,

2.3 Finite p-groups with modular subgroup lattices

hence Xy < N and therefore

JX:XnX =IXX':Xyl <jXY:Xyl =JXY:XI=I Y:XnYj.

59

Since I XI = ExpG, we have I Y: X n YI < I X I and it follows that X n Xy 1. Butthen the minimal subgroup of X n X' is S2(X) and also Q(Xy), and we get that0(X) = Q(Xy) = S2(X)y. Since Q(X) is of order p, it is centralized by y. ThusS2(X) < Z(G).

2.3.10 Lemma. Let G be an M*-group of exponent p". Then the map u defined byx° = x°"-' for x e G is a homomorphism from G to i2(Z(G)).

Proof. That x° e S2(Z(G)) for all x e G follows from 2.3.9. So we have to show that(xy)""-' = x°"-'y°"-' for all x, y e G, and we do this by induction on n.

The case n = 1 being trivial, assume first that n = 2. Then if p = 2, G is abelian by2.3.7. For p > 2, we let H = <x, y> and show that H' < Q(H) n Z(H). This is clear ifIHI < p3. And if JHI > p3, then o(x) = o(y) = p2 and thus S2(<x>)Q(<y>) < Z(H) by2.3.9. Furthermore H/S2(<x>)i2(<y>) is generated by two elements of order p andhence is abelian. Therefore once again H' < f2(H) n Z(H) and by (4) of 1.5,

(xy)" = x"y"[y, x](2) = x"y".

Finally, let n >_ 3 and assume that the formula is correct for n - 1. By 2.3.5, G/f2(G)is an M*-group of exponent p"-' and it follows that

(xy)p°-2

= xp°-2ypi-'z wherez e Q(G). Since x""-', yp"-' and z are contained in the M*-group Q2(G) of exponentp2, the first part of our proof yields that

(xy)p" l

=(x"" 2y".

2Z)" =xp^-'yp-'Zp = xp- 'yp-'.

The last two results suggest studying elements of maximal order in M*-groups. Inabelian p-groups, a fundamental property of such elements is that they generatedirect factors. A similar result holds in M*-groups.

2.3.11 Lemma. Let X be a cyclic subgroup of the M*-group G such that I X j =Exp G. If H < G with H n X = 1, then there exists a complement C of X in Gcontaining H, that is, a subgroup C of G such that G = XC, X n C = 1 and H < C.

Proof. We use induction on I G1. If H contains a nontrivial normal subgroup N of G,then X N/N is cyclic of maximal order in GIN and X N n H = N(X n H) = N. Byinduction there exists a subgroup C of G containing N and H such that G = XNCand XNnC= N.HenceG = XCand XnC <XnXNnC=XnN= l,thatis,C has the desired properties.

Now let Ho = 1 and suppose first that there exists a normal subgroup Z of G suchthat IZI = p and Z n X = 1; let K = HZ. If K n X # 1, then Q(X) < K and henceQ(X)Z is an elementary abelian subgroup of K of order p2, which, by 2.3.9, iscontained in Z(G). Since I K: HI < p, it would follow that 1 j4 H n O(X)Z < Z(G),

contradicting HG = 1. Thus K n X = 1, and the first part of our proof shows thatthere exists a complement to X in G containing K and hence also H.

Finally assume that Q(Z(G)) < X, let p" = ExpG and put L = S2"_1(G). ThenIS2(Z(G)l = p and hence I G: LI = p since L is the kernel of the homomorphism

(g a G) defined in 2.3.10. It follows that G = XL. For h e H, we haveh" e H n Q(Z(G)) = 1 and hence H < L. Since X n L is cyclic of maximal order

pn-1 in L, by induction there exists a subgroup C of L such that L = (X n L)C,XnLnC= 1 and HSC.Then G=XL=XCand XnC= 1, thatis,Chasthedesired properties.

Hamiltonian p-groups

2.3.12 Theorem. All the subgroups of a finite p-group G are normal if and only if Gis abelian or G = Q x A where Q ^- Q8 and A is an elementary abelian 2-group.

Proof. If G is such a direct product, then the elements of order 2 in G are containedin Z(G) and subgroups of exponent 4 contain Z(Q) = G'. Hence every subgroup ofG is normal. Conversely, suppose that G is a hamiltonian p-group. We show byinduction that G has the desired structure. By 2.1.4, G is an M-group. If G were anM*-group, then for a cyclic subgroup X of maximal order in G, by 2.3.11 therewould exist a subgroup C of G such that G = X x C. Since G is not abelian, C wouldbe hamiltonian and, by induction, Q8 would be involved in C, a contradiction. ThusG is not an M*-group and by 2.3.8, G has the desired structure.

The structure of arbitrary hamiltonian groups is well-known (see Robinson[1982], p. 139), and is described in Exercise 1. Theorems 2.3.8 and 2.3.12 show thata p-group is an M*-group if and only if it is a nonhamiltonian M-group. So thesetwo notions are the same and we introduce a third, related one.

Iwasawa triples

The triple (A, b, s) is called an Iwasawa triple for the p-group G if A is an abeliannormal subgroup of G, b e G and s is a positive integer which is at least 2 in casep = 2 such that G = A and b-lab = a1+p° for all a c A. To finish the proof ofIwasawa's theorem, we have to show that every M*-group possesses an Iwasawatriple. We start with a sufficient condition for the existence of such a triple.

2.3.13 Lemma. Let A be a subgroup of the M*-group G such that every subgroup ofA is normal in G.

(a) Then A is abelian and G/CG(A) is cyclic.(b) If Exp A # 4 and B is a cyclic subgroup of G, then there exist a generator b of

B and an integer s such that (A, b, s) is an Iwasawa triple for AB.

Proof. By 2.3.12, A is abelian since QS is not involved in G. Furthermore, theelements of G induce power automorphisms on A which are universal by 1.5.4. So ifx e A with o(x) = Exp A = p", then CG(A) = CG(<x>) and hence G/CG(A) is isomor-phic to a subgroup of Aut(x).

If p > 2, Aut (x) is cyclic of order pn-1(p - 1) and for s = 1,... , n, the sub-group of order p"-' of Aut(x) is generated by the automorphism a with x° = xl+°'

(see Huppert [1967], p. 84). In particular, (a) holds in this case. Furthermore,IB/CB((x))l = p"-' for some s e {1,...,n}, and hence there exists b e B such thatB = and xb = xl+P. Since b operates as a universal power automorphism on Aand o(x) = Exp A, it follows that ab = a1+p' for all a e A. Thus (A, b, s) is an Iwasawatriple for AB.

Now assume that p = 2 and n >_ 3. Then Aut<x) _ (fi) x (y) where xP = x 1and x1 = x5. Suppose that some element g e G induces an automorphism in (x)that does not lie in (y). Then every section of order 4 of <x) is inverted by g.It follows that I(x) n (g)l < 2 and hence <x> n (g) S (x4). By 2.3.6 applied toG/<x4), <g, x4) is normalized by x and therefore [x,g] a (x) n <x4) (g) = (x4).Thus (x)/<x4) is centralized by g, a contradiction. This shows that G/CG(A) isisomorphic to a subgroup of (y), that is, (a) holds. And since the subgroups of (y)are generated by the automorphisms b such that x" = x1 "' for 2 S s < n, statement(b) also follows as in the case p > 2.

Finally, if p = 2 and n 5 2, then lAut<x>l < 2. In particular, G/CG(A) is cyclic. IfExp A # 4, it follows that A < Z(G) and then for every generator b of B, (A, b, 2) isan Iwasawa triple for AB.

If Exp A = 4 in 2.3.13, A need not be a member of an Iwasawa triple for AB (seeExercise 4). It is not difficult to determine the structure of AB in this case (seeExercise 5). However, we shall not need this in our proof of Iwasawa's theorem sincewe can use 2.3.6 if Exp A = 4.

2.3.14 Lemma. Let G be an M*-group, X a cyclic subgroup of G with I X I = Exp Gand suppose that H < G such that H n X = 1.

(a) If Hx = H, then every subgroup of H is normal in G.(b) H/HG is cyclic.

Proof. (a) Let X = <x> and p" = o(x) = Exp G. By 2.3.11 there exists a complementC to X in G with H S C. Every g e G therefore has the form g = x'y where i e N,y e C, and it follows that Ha = H'' S C. Thus the normal closure HG of H is con-tained in C. Therefore, if K 5 H and g e G with <g> n C = 1, then

K = K((g) n HG) = K<g) n HG 9 K<g),

that is, g c NG(K). In particular, x c NG(K). And for y c C, by 2.3.10, (xy)P"-' _XP"-' yP"-' 0 C. Thus (xy) n C = 1, hence xy a NG(K) and therefore y e NG(K). Itfollows that NG(K) >_ XC = G.

(b) Since our assumptions are inherited by G/HG, we may assume that HG = 1.Then by (a), CH(X) = 1 and hence NH(X) is cyclic by 2.3.13, (a). On the other hand,

every minimal subgroup K of H normalizes X since I X K: X I= I K I= p. Thus H hasonly one minimal subgroup and therefore is cyclic since QB is not involved in G..

2.3.15 Theorem. Let G be an M*-group and suppose that X is a cyclic normal sub-group of G with I X I = Exp G. Then every subgroup of C6(X) is normal in G andG/CG(X) is cyclic. If G = CG(X)B where B is cyclic, then there exists a generator b ofB and an integer s such that (CG(X), b, s) is an Iwasawa triple for G.

Proof. By 2.3.13, G/CG(X) is cyclic. If C is a complement to X in G, then CG(X) =XC n CG(X) = X x Co where Co = C n CG(X). Let g e G and X = <x>. Since X 9 G,there exists an integer r such that xa = x'. For c e Co take y E X with o(y) = o(c).Then <y, c> = <y> x <c> is abelian and < yc> n X = 1 = <c> n X. By 2.3.14, <yc>and <c> are normal subgroups of G. Since y c- X a G, also <y> 9 G and by 1.5.4, ginduces a universal power automorphism in <y, c>. Hence ca = c' and it follows thatas = a' for all a e CG(X) = X x Co. Thus every subgroup of CG(X) is normal in G.By 2.3.13, G has the desired structure except, possibly, when Exp CG(X) = 4. Butthen Exp G = o(x) = 4, G is abelian by 2.3.7 and the assertion holds trivially.

Of course, CG(X) is abelian in 2.3.15. In particular, as a useful consequence, wenote the following.

2.3.16 Corollary. If G is an M*-group containing an element z e Z(G) with o(z) _Exp G, then G is abelian.

Theorem 2.3.15 shows how to find an Iwasawa triple in the M*-group G if thereexists a cyclic subgroup X of maximal order that is normal in G: one can take CG(X)as the abelian normal subgroup A. If there is no such X, we cannot describe A soeasily. Furthermore, we have to do some unpleasant calculations.

2.3.17 Lemma. Let L = <x, a> be an M*-group, X = <x>, o(a) < o(x) = Exp L andsuppose that ax = a'+psxo where x0 e S2(X) and s is an integer which is at least 2 incase p = 2.

(a) If X is not normal in L, then R+1(X) 5 Z(L) and there exists x1 E Sts+1(X) suchthat d" = d'+e` and L = <x,d> where d = ax1.

(b) Suppose that there exists u e L such that L = X , X n = 1 and 9 L. Then Sts+,(X) 5 Z(L), u e <a>Q5+1(X) and ux = u'+e.

Proof. (a) First note that by 2.3.9, xo E Z(L) and therefore ax' = for alli e N. Indeed, this is clear by assumption if i = 1; and if it is true for some i e N, then

axi+i = (ax)cl+papx0 = all+p"1,+ix0i+1

Let o(x) = p". Since X is not normal in L, ae' o X and hence ps < o(a) < p", thatis, s < n - 1. It follows that (1 + p'')e"-'-' ° 1 (modp"-') and therefore ax°" '

' =

a('+e`)°,-lxo"

' ' = a. Thus Sts+1(X) = <xe"-'_'> 5 Z(L). Finally, x0 = x`e"-' for

some i e NJ and we put x, = x'P"-'-'. Then for d = ax,, we clearly have L = (x, a> _(x, d > and

d' = (ax'D^')x = al+P.xfP. x"'"-.-'

= (axl)l+t = dl+p

(b) By (a), I,+1(X) < Z(L) if X is not normal in L; and if X a L, then L =X x is even abelian. Since L/(u) X is cyclic, [a,x] = aP'xo e and henceaP'+` _ (aP'xo )P a <a> n . Therefore I <a, u> : I = I <a>: (a> n I << ps+l andso <a, u> < C,+, (X). Let u, a (u> and x, a 52,+, (X) such that a = ul x- j'. ThenL = X<ul> and therefore = <u1> < <a>i2,+,(X). Finally,

u; = (axl)x - al+P'xl - al+P'xl+P' = (axl)1+p' = ui+P' (modi)(X))

and hence ui+P'(ui)-1 a Q(X)r (u> = 1. It follows that ui = u;+P' and then alsoux = u1+P=

2.3.18 Theorem. Let G be an M*-group in which no cyclic subgroup of maximal orderis normal. Then there exist a subgroup A of G and an integer s with the property thatfor every cyclic subgroup B of maximal order in G,

(a) there is a generator b of B such that (A, b, s) is an Iwasawa triple for G, and(b) A contains every subgroup H of G that is normalized by B and intersects B

trivially.

Proof. We proceed by induction on IGL. Let X be a cyclic subgroup of maximalorder in G and let I X I = Exp G = p". By 2.3.9, Q(X) a G and hence by induction or2.3.15 (in the case where there exists a cyclic subgroup of maximal order in G10 (X)that is normal), there exists a subgroup S/i2(X) of G/i2(X) that is a member of anIwasawa triple for G/S2(X). In particular,

(1) S/i2(X) is abelian

and G/S is cyclic; let t e G with G = S(t>. If G : SX, then SX = S(SX n <t>) <S<t"> and hence at'P generates X for some a e S and i e N. By 2.3.10, (at'P)P"-` _aP"-`t'P" = aP"-' and hence CI(X) < (a> < S and o(a) = p". But then (a> would be acyclic subgroup of maximal order, normal in G, a contradiction. Thus

(2) G = SX.

Since X is not normal in G, 2.3.6 shows that Exp(S/S2(X)) 4. By 2.3.13 there exista generator x of X and an integer s which is at least 2 in case p = 2 such that for alla e S,

(3) ax = al+P'xa where xa e S2(X).

We want to show next that

(4) Exp S < p" = o(x).

In order to obtain a contradiction suppose that there exists an element a e S witho(a) = p". Since subgroups Y satisfying fl(X) < Y< S are normal in G, we see that

52(X) $ <a> and hence X n <a> = 1. Let H = <X, a>. Then

"xa)"j'(a°'x") = (a )Px =

a centralizes aP'xa, it follows that <aP'xa) 9 H. Now arxa = [a,x] impliesthat H/<aP°xa) is abelian and H' = <a"'xa) is cyclic. It follows that H' o X = 1 orH' r) <a> = 1. By 2.3.11 there exists a complement C to X or <a) in H containingH'. By 2.3.14, this complement is normal in G and, by 2.1.5, its subgroup lattice isisomorphic to [H/X] L(<a)) or to [H/<a)] L(X). Hence C is a cyclic normalsubgroup of order p" and this contradicts our assumption. Thus (4) holds.

We are now going to construct a subgroup A of G that together with the integers in (3) satisfies (a) and (b) for B = X; it will then be easy to show that A and s alsohave these properties for arbitrary B. Let M be a complement to X in G. If g e G,then, by (2), there exist a E S and i E N such that g = ax'. Then (3) and (4) show thatL = <x, g) = <x, a) satisfies the assumptions of 2.3.17. Since X is not normal in G,M is generated by the elements outside NM(X). By 2.3.14, M/Mc is cyclic and hencethere exists an element c E M\NM(X) with M = MG <c). By 2.3.17, c e Cc(523+1 (X))and there exist x, a 0,+, (X) and a, e S such that for d = a, x, , d" = d' +P' and<x, c) = <x, d ). Let

(5) A=<d,HI HSG,HnX=1,H"=H).Then AX Mc <d) X = Mc <c) X = MX and hence

(6) AX = G.

We show that any two of the generators of A given in (5) commute and that they areall mapped by x to their (1 + ps}th power. Then it will be clear that A is abelian and

(7) (A, x, s) is an Iwasawa triple for G.

So let u e H < G where H n X = 1 and H" = H. By 2.3.14, 9 G and by 2.3.17,u e Cc(525+1(X)), u = a2 x2 for some a2 e S and x2 e 52,+, (X) and ux = u'+". In par-ticular, f +, (X) is centralized by MG, c and X and hence lies in the centre of G. SinceS/52(X) is abelian, it follows that

[d, u] = [a,xi, a2x2] = [a,, a2] a Q(X) n <u) = 1.

If v e K 5 G where K n X = 1 and K" = K, then similarly v = a3x3 with a3 a S,x3 a 525+, (X) and [u, v] = [a2i a3] E 52(X) n <u) = 1. Thus A is abelian and (7)holds.

Finally, suppose that B is an arbitrary cyclic subgroup of maximal order in G. By(7), B = <az) where a E A and z E X, and, since there are no cyclic normal subgroupsof order p" in G, o(a) < p". By 2.3.10, (az)P"-' = aP"-'zP"-' = zP"-' and this impliesthat o(z) = p" and 52(B) = 52(X). Thus <x) = <z> < AB and G = AB. Similarly, ifa E A and b E B such that x = ab, then b is a generator of B and since it induces thesame automorphism on A as x does, (A, b, s) is an Iwasawa triple for G. Thus (a) holds.And if H is a subgroup of G with H n B = 1 and B 5 NG(H), then H a G by 2.3.14and H n X = 1 since 52(X) = 52(B). By (5), H < A and hence (b) too is satisfied.

Iwasawa's theorem can now be proved; in fact, we only have to quote the resultsof this section.

Proof of Theorem 2.3.1. If G satisfies (a) of 2.3.1, then G is hamiltonian by 2.3.12, andif (b) holds, then 2.3.4 shows that G is an M-group. Conversely, let G be a p-groupwith modular subgroup lattice. If the quaternion group Qs is involved in G, then2.3.8 yields that G satisfies (a) of 2.3.1. And if G is an M*-group, then by 2.3.15 and2.3.18 there exists an Iwasawa triple for G, that is, (b) holds.

A word of warning

There are a number of mistakes in the literature on Iwasawa's theorem. It starts withIwasawa [1941], p. 197 where in the proof of his theorem he chooses a certain basisof an abelian group which, in general, does not exist. This gap in the argument isreproduced in Suzuki [1956], p. 16; it was closed in a paper by Napolitani [1967a].Our method of proof is different. Another mistake that is found, for example, in thepapers by Jones [1945], p. 549 and Sato [1951], p. 220 is the claim that an M*-group always has an Iwasawa triple (A, b, s) with A n = 1. However, simpleexamples show that this is not so.

23.19 Example. Let p > 2 and G = <a,xIaP' = 1,X"' = a"2,ax = a"">. Since a"2is centralized by the automorphism u of <a> with a° = al+", G is an extension of <a>by a cyclic group of order p3. By Iwasawa's theorem, G is an M*-group of order p6and has exponent p4 by 2.3.5. Let <x> = X and suppose that Y is a cyclic subgroupof order p4 of G. Then I X: X n Yj _ I X Y : YI < p2 and hence X n <a> = i2(X) <X n Y. This implies that G' = <a"> Y and since IG: Yl = p2, Y cannot be normalin G. Now suppose that G = AB where B is cyclic and every subgroup of A is normalin G. Then Exp A < p4 and hence I BI = p4. It follows that Q (X) < B; but as G/A iscyclic also i2(X) < G' < A. Thus A n B :A 1.

Finally, we mention that in general there are many different Iwasawa triples ina given M*-group. Some examples and results in this direction can be found inExercises 7-9.

Characteristic subgroups in M*-groups

There are a number of interesting results on the existence of certain characteristicsubgroups in M*-groups. The following is due to Menegazzo.

23.20 Theorem (Busetto and Menegazzo [1985]). Let G be an M*-group of expo-nent p". If G/S2n_1(G) is not cyclic, then there exists an abelian characteristic subgroupA of G such that G/A is cyclic and [G, Aut G] < A.

Proof. We choose an Iwasawa triple (A, b, s) for G such that A has maximal order.Since G/A is cyclic and G/i2"_1(G) is not, A 4 i2"_,(G) and hence there exists anelement a e A of order p". For every such a, <a> 9 G and A < CG(a). By 2.3.15, CG(a)belongs to an Iwasawa triple for G and our choice of A implies that

(8) A = CG(a) for all a e A with o(a) = p".

We want to show that

(9) A = <X I X < G, X cyclic, X J = p" ).

Since A is generated by its elements of order p", it is certainly contained in the join ofall cyclic normal subgroups of order p" of G. Conversely, suppose that X is a cyclicnormal subgroup of order p" of G. If there exists an element a c- A of order p" with<a> n X = 1, then X < CG(a) = A and we are done. So assume that there is no sucha. Then t5"_,(A) < X is cyclic and hence t5"_,(A) = Q(X). By 2.3.10, t5"_,(G) =t5"_,(A)<bP"-') is isomorphic to G/S2n_1(G) and therefore is not cyclic. It followsthat o(b) = p" and n U,,-, (A) = 1. Let X = (x> and take c e A, i e 101 with x =b'c. Since

1 -A xP"-' = b'P"'cP"' EQ(X) = t5n_1(A),

b'P"-' = 1 and cP"-' 1. Thus o(c) = p" and S2(<c>) = t5"_,(A). Since X a G,

[x, b] = [b'c, b] = [c, b] = cP' E X

and since x and c have the same order, <cP'> = <x"> = <b'P'c'> for some k e N.Hence b'P' E n <c> = n U ,,-,(A) = 1 and so b' c- f2S(). Since ab = a' 'P'for all a e A and ExpA = p" = o(b), it follows that f2S() < CG(A) = A. Thus x =b'c e A and (9) holds.

In particular, A is a characteristic subgroup of G. Hence for a E Aut G and a E Awith o(a) = p", aTM e A and

(aTM)b' = (ab)a = (a'+P`)a = (aTM)'+p = (aa)b

Thus b-1bTM E CG(aTM) = A by (8). It follows that G/A is centralized by a and hencethat [G, Aut G] < A.

We mention an obvious consequence of Menegazzo's theorem that will be usedlater.

2.3.21 Corollary. Let N be an M*-group of exponent p" such that N/i2"_1(N) is notcyclic. If N is contained as a normal subgroup with cyclic factor group in a group G,then G is metabelian.

Proof. By 2.3.20 there exists an abelian characteristic subgroup A of N such thatN/A is cyclic and [N, Aut N] < A. Then A a G and N/A is a central subgroup ofG/A with cyclic factor group. It follows that G/A is abelian.

Another immediate application of Menegazzo's theorem is an older result of v.d.Waall's.

2.3.22 Theorem (v.d. Waall [1973]). Every nonabelian M*-group has a characteristicmaximal subgroup.

Proof. Let G be a nonabelian M*-group of exponent p". If G/Qn_1(G) is cyclic, thenfen_1(G) is a characteristic maximal subgroup of G. If G/fQn-1 (G) is not cyclic, then by2.3.20 there exists an abelian characteristic subgroup A of G with cyclic factor group.Since G is not abelian, A G and the maximal subgroup of G containing A ischaracteristic in G.

2.3.23 Theorem (Seitz and Wright [1969]). Let G be a nonabelian M*-group. Thenthere exist (characteristic) subgroups R and S of G such that c(G) < S < R and[R, Aut G] < S.

Proof. We proceed by induction on I G 1. Suppose first that there exists a nontrivialcharacteristic subgroup K of G such that G' K. Then G/K is a nonabelianM*-group and by induction there are subgroups R/K and S/K of G/K such that(D(G/K) < S/K < R/K and [R/K, Aut(G/K)] < S/K. Hence '(G) 5 S < R and sinceevery automorphism of G induces an automorphism in G/K, it follows that[R, Aut G] < S. So we are done in this case and therefore may assume that

(10) G' is the only minimal characteristic subgroup of G.

Let ExpG = p". If G/Q _1(G) is not cyclic, then by 2.3.20 there exists an abeliancharacteristic subgroup of G such that [G, Aut G] < A. Then R = G and a maximalsubgroup S of G containing A have all the desired properties. So let G/f2n_1(G) becyclic, that is, IG : fl _1(G)I = p. By 2.3.10, Jun-,(G)I = p and since G' is the onlyminimal characteristic subgroup of G, it follows that

(11) G' = Y,,_1(G) is of order p.

Let a e G with o(a) = p". Then G' = J _1(G) = <aP"-'> < <a> and hence <a> a G.By 2.3.15 there is an Iwasawa triple (A, b, s) for G in which A = CG(a). Since I G' j = p,[a, bP] = [a, b]P = 1 and hence bP e CG(a) = A. This implies that IG : A = p ands = n - 1. It follows that Qn_1(A) < Z(G). Since IA : Q.-1(A)I = jZ5 _1(A)I = P,

(12) IG:f2n-1(A)I = p2 and Q .-,(A) = Z(G)

So if we choose R = Qn_1(G) and S = Q.-, (A) = Z(G), then R and S are characteris-tic subgroups of G such that 1(G) < S < R. Since G = AR, there exist c e A andd e R such that b = cd. Then (A, d, s) is an Iwasawa triple for G. It follows thatd Z(G) = S and hence R = S<d>. For g e G, there exist a1 e A, i e IOI such thatg = d`a1 and since d e Q. -I (G),

[g,d] = [d'al,d] _ [a1,d] = af" = d`P"-'al"-I = g13"

Hence if a e Aut G, then [a, da] _ [aa-', d]a = ((a')"') a = aP"-' _ [a, d] and there-fore dad-1 e CG(a) = A. Since d e R = Ra, we have dad-1 e A n R = S, that is, Aut Gcentralizes R/S.

Power automorphisms of M*-groups

By 1.5.4, every power automorphism of a finite abelian group is universal. On theother hand, the quaternion group of order 8 has nonuniversal power automor-phisms of order 2. So it is interesting to note that M*-groups may also possessnonuniversal power automorphisms of order p (see Exercise 11), but only if p = 2.

2.3.24 Theorem (Napolitani [1971]). Let a be a power automorphism of order pof the M*-group G and assume that or fixes an element of order 4 in G if p = 2 andExp G >_ 4. Then a is universal.

Proof. We use induction on I GI. Let Exp G = p" and put H = L2.-, (G). If p = 2 andExp H > 4, then H contains every element of order 4 of G. Thus either aIH is theidentity or it satisfies the assumptions of the theorem. By induction, of is universal onH. Let x e G with o(x) = p". Then a induces an automorphism of order 1 or p in<x> and therefore x' = x1+rp"-' where 1 < r < p. We have to show that a inducesthe same power in every cyclic subgroup Y = (y> of G. If I Yl < p", then xp and yare contained in S2n_1(G) and, since a is universal on S2"_1(G), y" = y1+'p"-' = y

So let I Yl = p". If (x> n Y:11- 1, we may choose the generator y of Y such thatyp"-' = (xP"-')-1. By 2.3.10, (xy)p"-' = xp"-'yp"-' = 1 and therefore, as we have justshown, xy = (xy)' = x'y" = xl+rp"-'ya Thus y' = (xp"-')-'y = y1+rp"-' Finally, if<x> n Y = 1, then (xy)p"-' 1 and therefore (xy)2 = (xy)1+t1 ' and y° = y1+,p"-'

where 1 < s, t < p. Using 2.3.10 and (1) of 1.5, we get

xrp"-' ysp" ' = [x, a] [y, a] = [xy, a] = (xy)tp", =

xtp",

ytp",

and hence r = t = s. Thus y- = y' +'P'- 1.

Exercises

1. Show that a group is hamiltonian if and only if it is the direct product of aquaternion group of order 8, an elementary abelian 2-group and an abeliangroup with all its elements of odd order.

2. If G is an M*-group, show that U.(G) = {xp'"Ix e G} for every m e N.3. Let G be an M*-group with I G' J = p. Show that there exist elements a, b e G, an

integer m and an abelian subgroup H of G such that G = (a, b> x H, o(a) _p'"> ExpHanda'=a'+p'"-'.

4. Let G = (u,vlu8 = v2 = 1, u" = u5> and A = (uzv>. Show that A 9 G, G/A iscyclic, but A is not a member of an Iwasawa triple for G.

5. Let A be a subgroup of the M*-group G such that every subgroup of A isnormal in G and suppose that B is a cyclic subgroup of G. If the assertion of2.3.13 (b) does not hold in AB, show that there exist an element x of order 4, anelementary abelian 2-group E, and an integer t such that A = <x> x E and(EB, x, t) is an Iwasawa triple for AB.

2.4 Groups with modular subgroup lattices 69

6. Let G = <u,v,xluP'" = vP"' =1, UV = Vu, Xp2 = V-P, ux = ul+P'^-1, v" = V1+P--1 >

where m > 4. Determine a complement to <x> in G.7. Let G be an M*-group with nilpotency class c(G) > 3 and suppose that (A, b, s)

and (A 1, 61, s1) are Iwasawa triples for G.(a) Show that Exp A = Exp A 1 and hence s = s1.(b) Show that I A I = I A, I if A and A 1 are maximal among the subgroups that

belong to Iwasawa triples for G.8. Let n >- 5 and G = <u, vl up" = vP" = 1, u' = u1+p2>. Show that there are

Iwasawa triples (A, b, s) and (A 1, 61, s1) for G with A 1 < A; note that c(G) _n-2>3.

9. Let G = <u, vlup` = vp = 1, u" = u1+P'>. Show that (, v, 3) and (<upv>, u-1, 2)are Iwasawa triples for G. Deduce that in general (a) and (b) of Exercise 7 do nothold if c(G) = 2.

10. Construct a nonabelian M*-group G with [G, Aut G] = G. (Hence, in general,one cannot have R = G in Theorem 2.3.23.)

11. Let n > 3 and G = <a,bla2" = b2" = 1, ab = al+zn-1>. Show that there exists apower automorphism of order 2 of G that is not universal.

2.4 Groups with modular subgroup lattices

We continue our study of groups with modular subgroup lattices. We shall describethese groups completely modulo the so-called Tarski groups. First we introduce aclass of groups similar to the P-groups studied in Section 2.2.

P*-groups

We say that G is a P*-group if G is the semidirect product of an elementary abeliannormal subgroup A by a cyclic group <t> of prime power order such that t inducesa power automorphism of prime order on A. As for nonabelian P-groups, this powerautomorphism is universal so that there exist primes p, q and integers m, r such thatA is a p-group, o(t) = q' and t-1 at = a' for all a e A. Since t induces an automor-phism of order q in A, r it 1 (mod p) and r9 - 1 (mod p) so that q I p - 1.

2.4.1 Lemma. Let G = A<t> be a P*-group and let o(t) = q'". Then Z(G) = <t9> _D(G), G/Z(G) is a nonabelian P-group and L(G) is modular.

Proof. It is clear that t9 e Z(G) and G/< t9> is a nonabelian P-group. Thus D(G) <<ta> < Z(G). By 2.2.2, Z(G/<t9>) = 1 and hence <t9> = Z(G). Furthermore, forx e G\(A x <ta>), x<t9> is of order q in G/<t9> and hence <x, t9) G/A <t>. Itfollows that o(x) = q' and t9 E,<x>. Therefore, if H is a subgroup of G, then eitherH :!g A x <t"> or <t9> < H. In the first case, H a G. In the second, H is modular in[G/< t9>] since this lattice is isomorphic to the subgroup lattice of an abelian group.It follows from 2.1.6(c) that H is modular in G. Thus every subgroup of G is modular

in G, that is L(G) is modular. Finally, if M is a maximal subgroup of G, then<to><Mn<t>by 2.1.5and so <t">=D(G).

Conversely, for finite groups, we have the following result.

2.4.2 Lemma. If G is a finite M-group such that G/D(G) is a nonabelian P-group,then G is a P*-group.

Proof. Suppose that I G/'D(G)I = p"q where p and q are primes with p > q, let A/D(G)be the Sylow p-subgroup of G/D(G) and P a Sylow p-subgroup of G. Since A 9 Gand I G : Al = q, we have P< A and hence A = PD(G). By the Frattini argument,

G = NG(P)A = NG(P)PD(G) = NG(P)CD(G)

Since 'D(G) is the set of nongenerators of G, it follows that G = NG(P), that is, P 9 G.If Q is a Sylow q-subgroup of G, then G = AQ = cD(G)PQ and hence G = PQ as

before. Since G/(D(G), and therefore also G, has only one subgroup of index q, Q iscyclic; let Q = <t). As G/D(G) is not nilpotent, t operates nontrivially on P. ButI G/D(G)I = p"q implies that Q n D(G) = <t"> is the Sylow q-subgroup of D(G).Hence <t9) 9 G and P is centralized by t". Finally, for H < P,

H=Hu(QnP)=(HuQ)nPaHuQ

since P 9 G. It follows that t induces a power automorphism of order q on P.It remains to be shown that P is elementary abelian. Since <t9> a G, it fol-

lows that <t9) = Q n Q, for every Sylow q-subgroup Q, Q of G. By 2.2.4,I Q/<t 9) u Q, /<t9) I = pq and therefore IQ u Q, I = p I Q I. This implies thatQ u Q, < O.(P)Q. By 2.2.2, G is generated by its Sylow q-subgroups and henceG = f)(P)Q. It follows that P = 0(P) and by 2.3.5, P is elementary abelian.

Finite M-groups

2.4.3 Lemma. If G is a finite M-group with D(G) = 1, then G is a direct product ofP-groups with relatively prime orders.

Proof. We use induction on IGI. If H :!g G and M is a maximal subgroup of G, thenby 2.1.5, [H u M/M] [H/H n M] and hence H n M is H or a maximal subgroupof H. In both cases, b(H) < M. It follows that cb(H) < D(G) = 1. In particular,

(1) every Sylow subgroup of G is elementary abelian

and every subgroup of G satisfies the assumptions of the lemma.Let R and T be subgroups of prime order in G. Then by 2.2.4, IR U TI is the

product of two primes. Therefore if I R I > I TI, then I R u TI = I R I I TI and R a R u T.So we have that

(2) T < NG(R) if I R I > I TI are primes.

Denote by p the maximal prime dividing IGI and let S be a Sylow p-subgroup of G.By (1) and (2), every subgroup of S is normal in G. If S < Z(G), then the Schur-Zassenhaus theorem yields that G = S x C where C is a complement to S in G. Byinduction, C is a direct product of P-groups with relatively prime orders; henceG = S x C also has this structure. So we may assume that S is not central in G.Then by (1) there exists a subgroup Q of prime order q in G that does not centralizeS. Thus

(3) SQ is a nonabelian P-group

and we want to show that Q is a Sylow q-subgroup of G. Suppose that this is not thecase and let T be a subgroup of order q2 of G containing Q. Let P < S with I P I = pand put H = P u T. By (2), P:9 H and hence IHI = pq2 and H/P T. Since P is notcentral in H, at most one of the q + 1 subgroups of order pq containing P may becyclic. Hence H contains two different nonabelian subgroups U and V of orderpq. Let Q1, Q2 < U and Q3 < V be subgroups of order q such that Q, : Q2. IfI Q 1 U Q3I = pq, then U = PQ1 = Q, U Q3 = PQ3 = V, a contradiction. By 2.2.4,IQ1 U Q3I = q2 and similarly IQ2 U Q3I = q2. It follows that Q3 is centralized byQ1 U Q2 U Q3 = U U Q3 = H; but V = PQ3 is not abelian. This contradiction showsthat Q is a Sylow q-subgroup of G and that

(4) SQ is a Hall subgroup of G.

It remains to be shown that if r is a prime such that p # r q, then

(5) SQ is centralized by every subgroup R of G with I R I = r.

Indeed (1) implies that SQ g G, and the Schur-Zassenhaus theorem yields the exis-tence of a complement K to SQ in G and again by (5) and (1), G = SQ x K. Byinduction, K has the desired structure and therefore so does G.

If r < q in (5), then by (2), R normalizes every subgroup of SQ. Since Z(SQ) = 1,Cooper's theorem (or 1.4.3) shows that R centralizes SQ. If p > r > q and againP < S is of order p, then R is normalized by every subgroup of order q of PQ. HencePQ < NG(R) and so P and R are normal subgroups of L = PQR. Thus ILI = pqr andPR = P x R. Suppose that Q does not centralize R. Then NL(Q) = Q and L containsIL : QI = pr Sylow q-subgroups. For every Sylow q-subgroup Q1 : Q of L, by 2.2.4,Q u Q1 is a subgroup of order pq or rq of L and hence is PQ or RQ. But these twogroups contain only p + r - 1 Sylow q-subgroups. This contradiction shows thatL = PQ x R and that (5) also holds in this case.

2.4.4 Theorem (Iwasawa [1941]). A finite group has modular subgroup lattice if andonly if it is a direct product of P*-groups and modular p-groups with relatively primeorders.

Proof. Since a direct product of modular lattices is modular, 1.6.4 and 2.4.1 yieldthat L(G) is modular if G is a direct product of P*-groups and modular p-groupswith relatively prime orders. Conversely, we show by induction on I G I that everyfinite M-group G is such a direct product. If L(G) is directly decomposable, then by1.6.5 there exist nontrivial subgroups H and K of G such that G = H x K and

(I H I, I K I) = 1. By induction, H and K have the desired structure and then so does G.So suppose that L(G) is directly indecomposable. Then by 1.6.9, L(G/(D(G)) is alsodirectly indecomposable and 1.6.4 and 2.4.3 yield that G/(D(G) is a P-group. If thisP-group is nonabelian, then G is a P*-group by 2.4.2. And if G/cb(G) is an elementaryabelian p-group, then G is a p-group since every prime divisor of IGI also dividesIG/c1(G)I. In any case, G has the desired structure.

Theorems 2.4.4 and 2.3.1 give the precise structure of finite M-groups. As animmediate consequence we note the following.

2.4.5 Theorem. Every finite group with modular subgroup lattice is metabelian.

We shall need some simple properties of elements of prime power order in finiteM-groups. They can easily be proved directly but also follow immediately fromIwasawa's theorem.

2.4.6 Lemma. Let G be a finite M-group and x, y e G. If x is a p-element and y aq-element where p and q are primes with p > q, then <x> = (x).

Proof. By 2.4.4, G is a direct product of groups G1, ..., G, with relatively primeorders such that every G. is either a P*-group or an M-group of prime power order.If x and y are contained in different components of this decomposition, then xy = yxand hence <x> = <x>. So suppose that x, y e G; for some i. Then G. is a P*-groupand hence <x> G; since p is the larger prime dividing IG,I.

2.4.7 Lemma. Let p be a prime, G a finite M-group and suppose that x, y e G arep-elements.

(a) If o(x) # o(y), then <x, y> is a p-group.(b) If o(x) > o(y) and o(x) > p2, then i2(<x>) g (x,y>.

Proof. Let G = Gl x x G, as in the proof of 2.4.6. Since the orders of the G. arerelatively prime, x and y have to lie in the same component G.

(a) If G, is a p-group or a P*-group and p the larger prime dividing I G; I, then<x, y> is clearly a p-group. So suppose that G, is a P*-group and p the smaller primedividing IG,I. Then by 2.4.1, the maximal subgroups of the Sylow p-subgroups arecentral in G,. Since o(x) o(y), therefore x or y is contained in Z(G,). It follows that<x, y> is a p-group.

(b) If G, is a P*-group, then p is the smaller prime dividing 1G,1 since o(x) > p2.Again <xv> 5 Z(G,) and hence fZ(<x>) 9 <x,y>. And if G, is a p-group, then 2.3.5implies that Exp<x,y> = o(x) and 2.3.9 yields that f1(<x>) a <x, y>.

M-groups with elements of infinite order

The structure of M-groups with elements of infinite order is well-known. We firstshow that the torsion subgroup of such a group and its factor group are abelian.

2.4.8 Lemma. Let G be an M-group. Then the set T(G) of all elements of finite orderin G is a characteristic subgroup of G. If G contains an element of infinite order, thenT(G) is abelian and every subgroup of T(G) is normal in G.

Proof. For a, b e T(G), 2.1.5 yields that

[<ab>/<ab> n <a>] ^' [<ab> u <a>/<a>] = [<a,b>/<a>] [/<a> n ]

is a finite lattice. Since also <ab> n <a> is finite, ab has finite order. Thus T(G) is asubgroup and then of course a characteristic subgroup of G.

Let x e G be an element of infinite order. Then <x> n T(G) = 1 and if H :!S;T(G), then we see that H = H u (<x> n T(G)) = (H u <x>) n T(G) H u <x> sinceT(G) g G. Thus x e NG(H) and since G is generated by the elements outside T(G), itfollows that H a G. In particular, T(G) is abelian or hamiltonian. In the latter caseT(G) would contain a finite nonabelian p-subgroup since a hamiltonian torsiongroup is clearly the direct product of its primary components. By 2.3.12, G wouldcontain a subgroup Q isomorphic to the quaternion group Q8 (see also Exercise2.3.1). Since Q a G, the element x e G of infinite order would operate on Q andhence a nontrivial power y of x would centralize Q. Then <y> x Q/<y4> ^_ C4 X Q8would be an M-group, contradicting 2.3.8. Thus T(G) is abelian.

2.4.9 Lemma. If G is an M-group, then G/T(G) is abelian.

Proof. Since G/T(G) does not contain elements of finite order, we may assume thatT(G) = 1, that is, G is torsion-free. Let a, b c- G. We shall show in several steps thatab = ba.

(6) If b-t amb = an with integers n and m, then n = m.

Indeed, if <a> n = <ak> 4 1, then the assumption in (6) implies that akm =b-1 akmb = ak"; therefore n = m since G is torsion-free. In addition, if <a> n = 1,the modular law and the assumption in (6) yield that

<an> = (<an> u ) n <a> = (<am> u ) n <a> = <am>,

that is, n = m or n = -m. If n = -m, then <a"> is inverted by b and hence<a", b>/<a4n, b2 > is a dihedral group of order 8. But this group is not an M-group.Thus n = m and (6) holds.

(7) If <a> n <ab> 1, then ab = ba.

Suppose that this is false. Thus a # ab, let I <a>: <a> n <ab> I = n. Then an = (ab)m forsome integer m and by (6), n = m. So if we put c = ab and d = an, we have foundelements a 0 c and d 1 in G satisfying <a> n <c> = <d> and d = a" = c". Since<a, c>/<d> is generated by elements of finite order, it is a torsion group by 2.4.8.Hence there exists an integer r such that (ac-1r e <d> and we can choose k in{ + 1, -1 } such that (ac-')" = d' where s Z 0. Since a c and G is torsion-free,s > 0. Consider the subgroups U = <ad'> = <aru+1>, V = <cds> = <cns+l> and

W = <a>. Clearly, U S W and U u V contains ad'(cd')-l = ac-l; hence U U V alsocontains (ac-1)" = ds and so a and c. Thus U u V = <a,c> >_ W. Furthermore

V n W = <cn9+1> n <a> = <cns+l> n <c> n <a> = <cn,+1> n <c">

= <cn(ns+l)> = <an(ns+l)> S U.

The modular law yields <a> = W = (U u V) n W = U u (V n W) = U = <a"s+l >, acontradiction since ns > 0. Thus (7) holds.

Since <a> n = (<a> n )b < <a> n <ab>, an immediate consequence of (7)is the following.

(8) If <a> n 1, then ab = ba.

To deal with the case <a> n = 1, we strengthen property (7). Assume thatb-"amb" = a' for integers n and m with n # 0 m. Then a' e e <a> n <ab"> and by(7), ab" = b"a. This implies that b" e n <b"> and (7) yields that ba = ab. So wehave shown:

(9) If amb" = b"a' for integers n and m with n 0 0 m, then ab = ba.

The second main step in the proof of the lemma is the following.

(10) If ab # ba, then <a, b>/<a, b )' is a nontrivial finite group and <a, b)' = <c, d>with cd 0 dc.

To show this let H = <a, b>, K = <a, ab> and N = <a2, (a2)b>. Since H = <a> u _<ab> u <b),

K=<a>u(Kn)=<ab>u(Kn)=Kb

and hence K a H. Similarly, N g <a2, b> and N = <a2, (a2yb> g <a2, ab>. ThusN g <b, ab> = H. Now K/N is generated by the involutions aN and abN. Thereforeit is a dihedral group and hence finite since the infinite dihedral group is notan M-group. Furthermore, K n 1 since otherwise ab a <a> and (7) wouldimply that ab = ba. So K n = <b'> for some positive integer r and H/K/K n is cyclic of order r. It follows that H/N is finite. From (8) we have n <a> = 1 and hence <a2, b> n <a> = <a2 ). Thus N < <a2, b> < H and so wehave shown that H/N is a nontrivial finite M-group. By 2.4.5, H' H. Since H/K iscyclic, H' < K; on the other hand, if we put c = [a, b] = a-lab, then K = <a, ab> =<a, c). It follows that K = <a> u H' and H' = <c> u (<a> n H') = <c, a'> with<a') = <a> n H'. Suppose that <a> n H' = 1. Then H' = <c> is centralized by a2since I Aut <c> I = 2. Hence a2 e Z(K), a2 b' = b'a2 and (9) implies that ab = ba,a contradiction. Thus <a> n H' 1. Similarly, n H' = <b") 1. Then<a'",b"> 5 H'. It follows that H/H' = <aH',bH'> is finite and (9) implies that H' isnonabelian. So if we put d = a', then H' = <c,d> and cd 0 dc. Thus (10) holds.

To finish the proof of the lemma we suppose that ab # ba and put H = <a, b>.Then H' = <c, d > with cd # dc. Applying (10) two further times we get that H" _<c, d >' = <e, f > with of fe and H"' = <e, f >' = <g, h> where gh 0 hg. Further-

more all the commutator factor groups H/H', H'/H" and H"/Hare nontrivial finitegroups. But then H/H" is a finite M-group that is not metabelian, contradicting2.4.5. It follows that ab = ba.

We show next that for a nonabelian M-group G with elements of infinite order,the torsion-free abelian group G/T(G) has rank one.

2.4.10 Lemma. Let G be an M-group. If G contains two elements a and b of infiniteorder such that <a> n = 1, then G is abelian.

Proof. Let S be the set of elements of infinite order in G. For u, v e S, define u ' v

if and only if n <v> # 1. If u - v and v - w, then n <v> = <vt> 1 and<v> n <w> = <v") # 1; thus 1 vn' e n <w> and hence u - w. This shows thatis an equivalence relation on S. Let x, y e S. If x f y, that is if <x> n <y> = 1,then every element g of finite order in <x, y> satisfies

L(<g>) [<g> u <x>/<x>] = [<y"> u <x>/<x>] ^' L(<y"))

for some integer n and hence g = 1. Thus <x, y> is torsion-free and by 2.4.9, xy = yx.And if x - y, that is, if <x> n <y> # 1, then by assumption there exists an elementz e S that is not equivalent to x and y. Hence, as we have seen, xz = zx and yz = zy.It follows that <y,z> = <y> x <z> and <yz> n <y> = 1. So yz e S and yz f y,whence also yz f x. Thus x(yz) = (yz)x = yxz and hence in this case also xy = yx.We have shown that all elements of infinite order in G commute. By 2.4.8, G isgenerated by these elements and hence is abelian.

So far we have obtained a number of restrictions on the structure of a nonabelianM-group with elements of infinite order. We now give the precise structure of thesegroups.

2.4.11 Theorem (Iwasawa [1943]). Let G be a nonabelian M-group with elements ofinfinite order. Then T(G) is abelian and G/T(G) is a torsion free abelian group of rankone.

(a) If G/T(G) is cyclic, then G is the semidirect product of T(G) by an infinite cyclicgroup <z>, and for every prime p there exists a p-adic unit r(p) with r(p) __ 1 (mod p)and r(2) = 1 (mod 4) such that a' = a'(P) for all a e T(G)P, the p-component of T(G).(See 1.5.5 for the definition of a'(P).)

Conversely, for every prime p, let r(p) be a p-adic unit with r(p) = 1 (mod p) andr(2) = 1 (mod 4). If G is the semidirect product of an abelian torsion group T by aninfinite cyclic group <z> to the automorphism given by a' = a'(P) for all a e TP, thenHK = KH for all H, K S G and G is an M-group with G/T(G) infinite cyclic.

(b) Suppose that G/T(G) is not cyclic and let Exp T(G)P = p" where n e fN v {O, ao}.Then there exist elements z, e G, a, e T(G), primes pi and p-adic units r;(p) such thatG = <T(G), z 1, z2, ...) and for all primes p and all i n IN,

(11) r.(p) = 1 (mod p) and r,(2) - 1 (mod 4),

(12) r.+1(p)P' = r,(p) (mod p") (that is r;+1(p)P' = r1(p) if n = oo),

(13) o(zi) is infinite,

(14) a=' = a''(P' for all a e T(G)p,

(15) z,+, = ziai, and

(16) (z1a,)z1 = ziai

Conversely, let T be an abelian torsion group with Exp Tp = p", let ai E T and ri(p) bep-adic units satisfying (11) and (12) for all primes p and all i e N. If we extend Tsuccessively by cyclic groups <z,> according to the relations (13)-(16), we get anM-group G = <T, z,,z2, ... > in which any two subgroups permute.

Proof. By 2.4.8 and 2.4.9, T(G) and G/T(G) are abelian and G/T(G) is torsion-free.The rank of this group is one since otherwise there would exist two infinite cyclicsubgroups with trivial intersection in G/T(G), hence also in G; so G would be abelianby 2.4.10.

(a) Suppose that G/T(G) is cyclic and let z e G with G/T(G) = <zT(G)>. Then zhas infinite order and hence T(G) n <z> = 1. Thus G is the semidirect product ofT(G) by <z> and 2.4.8 shows that z induces a power automorphism in T(G). So forevery prime p, by 1.5.6 there exists a p-adic unit r(p) such that a2 = a'(P) for alla c- T(G)p. If T(G)p = 1 or Exp T(G)2 < 2, we may choose r(p) = 1 or r(2) = 1, re-spectively. So suppose that T(G)p 1 and Exp T(G)2 > 4 and take a e T(G) witho(a) = p. If r(p) # I (mod p), then a suitable power b of z would induce an automor-phism of prime order q pin <a>; but then <a, b>/<bP9> would be a group of orderp2q which, by 2.4.4, would not be an M-group. Similarly, if r(2) # 1 (mod 4), everyelement a of order 4 in T(G) would be inverted by z and hence <a, z>/<z2> would bea dihedral group of order 8 and not an M-group. Thus r(p) = 1 (mod p) and r(2) = 1(mod 4).

Conversely, suppose that G = T<z> as described in the theorem, let H, Kand take x e H, y e K. If o(x) is finite, then x c- T since G/T is infinite cyclic. Itfollows that <x> a G and xy = yx' a KH. So suppose that o(x) is infinite and writex = az' and y = bzj with a, b c- T and integers i, j. Then x and y are contained inL = <a, b, z> = A <z> with A = <a, b> < T Since A is finite, L/C<2>(A) too is finite,and therefore x' - C<2>(A) for some positive integer k. Thus if N = <x> n C(2)(A),then 1 N g L and L/N is finite. For every prime p, r(p) = 1 (mod p) implies that zinduces a power automorphism of order some power of p in the Sylow p-subgroupof A. It follows that L/N is a direct product of finite p-groups and since r(2) __ 1(mod 4), 2.3.1 shows that these p-groups have modular subgroup lattices. By 2.3.2,any two subgroups of these p-groups, and hence also of L/N, permute. In particular,there exist integers n, m and an element u_E N < <x> such that xy = y"xmu e KH.This shows that HK = KH and by 2.1.3, G is an M-group.

(b) Now suppose that G/T(G) is not cyclic. As a torsion-free abelian group of rankone, G/T(G) is isomorphic to a subgroup of the rationals (see Robinson [1982],p. 112); hence it is countable and locally cyclic. It follows easily that there existsubgroups Gi of G and primes pi such that T(G) < G, < G2 < ... < G, U G. = G,

LE N

Gi/T(G) is infinite cyclic and I Gi,, : Gi I = p, for all i e N. By (a) there exist z, e G with

G1 = T(G)<z1) and p-adic units r1(p) satisfying (11) and (14) for i = 1 and all primesp. We show that if for some integer i, there exist an element z; e G with G, =T(G) <zi> and p-adic units r;(p) satisfying (11) and (14) for i, then there are elementszi+1 e G, ai E T(G) and p-adic units r;+1(p) such that Gi+1 = T(G)<zi+1>, (11) and (14)hold for i + 1, and (12), (15), (16) are satisfied for i. This will prove the first assertionof (b). So suppose that G; = T(G) <zi> as described and let w c Git1 such that G;+1 =T(G) <w>. Since G; = T(G)(G; n <w>) has index pi in Gi+1, G; n <w> = <wP'> and w°iis congruent to z; or z,-' modulo T(G). Let z,+1 be the generator of <w> satisfyingz°+1 =_ zi (mod T(G)), that is z,°+1 = z;a; for some a; E T(G). By (a) there exist p-adicunits r;+1(p) satisfying (11) and (14) for i + 1 and all primes p. Since z° 1 and zi inducethe same automorphism in T(G), (12) holds; finally (16) follows from z;a; E <zi+1>.

Conversely, let G1 be the semidirect product of T by an infinite cyclic group <zl >to the automorphism given by a=' = a''(P) for all a e TP. If G; = <T,z1,...,z,> =<T,zi> is already constructed, then (14) and (16) define an automorphism of G; =<T,z;a;) that fixes zia, and whose p; th power, as (12) shows, is the inner automor-phism induced by zia,. Thus there exists the extension Gi+1 of Gi by a cyclic group<zi+1> with respect to this automorphism and the relation z?+1 ziai. Hence Gexists and is the set-theoretic union of the groups G;. So if x, y e G, there exists aninteger i such that x, y e G; = T<z,>. By (a), <x> <y> = <y> <x>. It follows that anytwo subgroups of G permute and by 2.1.3, G is an M-group.

It is possible to choose the p-adic units ri(p) in Theorem 2.4.11(b) in such a waythat

(12') r;+1(p)" = ri(p) for all i e N.

This has as a consequence the following result.

2.4.12 Theorem (Mainardis [1991]). Every M-group with elements of infinite ordercan be embedded in an M-group with divisible (abelian) torsion subgroup.

Proof. First let us prove that we may choose the r;(p) such that (12') holds. Sosuppose that G is an M-group, G/T(G) is not cyclic and let z;, pi, ri(p) be as inTheorem 2.4.11(b). If Exp T(G), = oo, then clearly (12') holds. And if Exp T(G), isfinite, then also Pot T(G), is finite. Let a; be the power automorphism induced by z;in T(G)P; so t = t''(p) for all t e T(G)p. If all the ai are trivial, we may chooser,(p) = 1 for all i and (12') holds. So suppose that aj # 1 for some j. Since r3(p) = 1

k-1

(mod p), o(a;) = p" for some u e N. For every k > j, zj = zk where d = p, and if vi=;

of these pi are equal to p, then o(ak) = p"+° Since Pot T(G)p is finite, it follows thatthere exists k E N such that p, p for all i >_ k.

For every prime q p and every p-adic integer y = 1 (mod p), the equation 13Q = y

has a unique solution 13 E RP with 1 = 1 (mod p). For, let y = Y_ cipi where co = 1M-1 i=o

and 0 < ci < p for all i and put y, = E c,p'. Let bo = 1 and assume that bo,..., brn-1i=o / m-1

are integers such that 0 S bi < p for all i and P. = bipi is the unique integeri=o

satisfying 1 < I'm < pm, I'm = 1 (mod p) and /3m = ym (mod pm). Since y = 1 (mod p),ym+1 lies in the Sylow p-subgroup of the multiplicative group of integers modulopm+1 In this abelian p-group, the map x - x9 is an automorphism and hence thereexists a unique integer I'm+1 such that 1 < Nm+1 < pm+1 N.+1 = 1 (mod p) and

Qm+1 = ym+1 (mod pm+1) If /3m+1 = > dip' where 0 < d; < p, then(M-

Y1 d; p')Q ym

(mod pm) and the uniqueness of /3m implies that d; = b; for i = 0, ..., m - 1. So if wem

define bm = dm, then/I'm+1 = Y b;p'. Thus, inductively, we obtain that there exists a

i=ooo

unique p-adic integer /3 = Y b;p' such that / __ 1 (mod p) and Q9 = y (mod pm) fori=o

allmaFk,that is/3"=y.Now we define p-adic integers s;(p) satisfying (12') in place of the r;(p). We put

sk(p) = rk(p) and, inductively, s;(p) = s;+,(p)PI for i < k; and if i >_ k, then p,: p andwe let s;+1(p) be the unique solution of the equation xP, = s;(p) in RP satisfyings;+I(p) = 1 (modp). Then, clearly, (12') holds for the s;(p) and we have to show that

(14') a'- = as") for all a e T(G)P;

then we can replace the r;(p) by the s;(p) in Theorem 2.4.11(b). So let a; be the powerautomorphism of T(G)P defined by a = as" for a e T(G)P. Clearly, ak = ak. If i < k

k-1 i-Iandu= Hp1,then a ak=ak=

J=i J=ka;°; since (v, p) = 1 and a;, a, are p-elements of Pot T(G)P, it follows that a; = a;. Thus(14') holds.

Finally, we prove the assertion of the theorem. Let G be an M-group withelements of infinite order. If G is abelian, the theorem is well-known (see Robinson[1982], p. 95). So we may assume that G is nonabelian and therefore has the struc-ture given in 2.4.11. In particular, T(G) is abelian and by the result just cited thereexists a divisible abelian torsion group T containing T(G) as a subgroup. If (a) of2.4.11 holds, the semidirect product of T and <z> with respect to the automorphismgiven by a' = a'1P1, a e T(G)P, p e P, is an M-group containing G = T(G) (z) as asubgroup. Similarly, if (b) of 2.4.11 holds and the r;(p) satisfy (12'), then by 2.4.11,the relations (13)-(16) define an M-group T<z1,z2,...> containing G as asubgroup.

We finally mention that the structure of G in Theorem 2.4.11(b) becomes simplerif T(G) is a p-group and also if T(G) is divisible. For, in the first case, we may choosepi j4 p for all i e N. And then, in either case, we may take a; = I for all i e J so thatG is the semidirect product of T(G) and Z where Z = <z1,z2,...> is a torsion-free,locally cyclic group and the z;, r;(p) satisfy (11)-(14) and zp I = z;. We leave the proofof the first assertion as an exercise and shall prove the second one in 2.5.13.

Locally finite M-groups

An abelian torsion group clearly is locally finite. Hence a projective image of such agroup is a locally finite M-group. Therefore the structure of these groups is of

particular interest and it will be described in the next two theorems; these may beregarded as extensions of Theorems 2.4.4 and 2.3.1.

2.4.13 Theorem (Iwasawa [1943]). The group G is a locally finite M-group if andonly if it is a direct product of P*-groups and locally finite p-groups with modularsubgroup lattices such that elements of different direct factors have relatively primeorders.

Proof. That such a direct product is an M-group follows from 1.6.4 and 2.4.1 as inthe finite case. Furthermore, it is obvious that every P*-group, and therefore a directproduct of such groups, is locally finite. Conversely, suppose that G is a locally finiteM-group. For every prime q, let Gq be the subgroup generated by all the q-elementsof G. Clearly, Gq is a characteristic subgroup of G. If any two q-elements generate aq-group, then Gq is the set of all q-elements in G and hence a locally finite q-groupwith modular subgroup lattice. Now suppose that there are q-elements u, v such that<u, v> is not a q-group. Then by 2.4.4, the finite M-group <u, v> must be a P*-groupof order p`q' where p is a prime with p > q. Since u 0 Z(<u,v>), there exists aninteger r with r * 1 (mod p) and rq = 1 (mod p) such that a" = a' for every p-elementa of <u, v>. Let x, y be arbitrary p-elements and z an arbitrary q-element of G. ThenH = <u, v, x, y, z> is a finite M-group and since <u, v> < H, it follows from 2.4.4 thatH is a P*-group of order p"q"' with n, m e N. The p-elements x and y are containedin the Sylow p-subgroup of H, hence they have order p and commute; furthermorex" = x' since u induces a universal power automorphism in this Sylow p-subgroup.Since x and y were arbitrary, it follows that GP is an elementary abelian p-group andthat Gp is a P*-group. By 2.4.1 and 2.2.2, this P*-group is generated by q-elements; on the other hand, z e H < Gp. Since z was an arbitrary q-element,Gq = Gp is a P*-group. If g e G with (o(g), pq) = 1, then <H, g> is a finite M-group and by 2.4.4, g centralizes x and z. This shows that GP cannot be contained ina second P*-subgroup GS of G. Clearly G is the product of the P*-groups Gq obtainedin this way and the t-groups G, that are not contained in these P*-groups. Since theorders of the elements in different factors of this type are relatively prime, the prod-uct is direct.

2.4.14 Theorem (Iwasawa [1943]). Let p be a prime. The group G is a nonabelianlocally finite p-group with modular subgroup lattice if and only if

(a) G is a direct product of a quaternion group Q8 of order 8 with an elementaryabelian 2-group, or

(b) G contains an abelian normal subgroup A of exponent p' with cyclic factorgroup G/A of order p' (k, m e N) and there exist an element b e G with G = A andan integer s which is at least 2 in case p = 2 such that s < k < s + m and b-' ab =a'+P`forall aeA.

Proof. If G satisfies (a), then the elements of order 2 of G are contained in Z(G) andsubgroups of exponent 4 contain G'. Thus G is hamiltonian and hence L(G) ismodular. Clearly, G is locally finite. Now suppose that (b) holds. For x 1, ..., x" e Gthere exist a, e A and b, e such that x, = a,b;. Then every x; is contained in

<a1...., a.> (b> and this group is a finite M-group by 2.3.1. Thus G is locally finiteand by 2.3.2, (x1> (xi> = (x,) (x1> for all i,j. It follows that any two subgroups ofG permute. Hence L(G) is modular and since s < k, G is not abelian.

Conversely, let G be a nonabelian locally finite p-group with modular subgrouplattice. If G contains a subgroup Q Q8, then for x e G and y e CG(Q), H =(Q, x, y> is a finite 2-group. By 2.3.8, H = QCH(Q) and CH(Q) is elementary abelian.It follows that x e QCG(Q) and o(y) = 2, that is, G = QCG(Q) and CG(Q) is elemen-tary abelian. Thus G = Q x A where A is a complement to Z(Q) in CG(Q) and (a)holds.

So suppose that G does not contain a subgroup isomorphic to Q8. Again by 2.3.8,every finite subgroup of G is an M*-group. Since G is not abelian, there exist x, y e Gwith xy yx. Then (x, y> is finite of exponent p", say. We claim that

(17) Exp G < pzn.

For, suppose that there exists an element z c- G of order pen. Then H = (x, y, z>is still finite and of exponent pen by 2.3.5. For w c H with o(w) = p2" there area e S2"(H) > (x,y> and zl e (z> such that w = azl. It follows that

I <w>: (z> n (w> I = I (w, z> : (z> I = I (a, z> : <z> I = I <a>: <a> n (z> I s p"

and hence I<z> n (w>I > p". Thus S2"((z>) < (w>. Since H is generated by itselements of maximal order, S2"((z)) < Z(H). In particular, f2"((z)) < Z(92"(H)) andby 2.3.16, applied to S2"(H), this group is abelian. But this is impossible sincex, y c S2"(H). Thus (17) holds.

Let So be the set of all finite subgroups of G containing x and y. Any H e .9' is anonabelian M*-group and hence every Iwasawa triple (A, b, s) for H satisfies p' <Exp G; we denote the maximal integer s occurring in an Iwasawa triple for H bys(H). We choose a subgroup M E .So with minimal s(M) and among these withmaximal Exp M', that is, satisfying

(18) s(M) < s(H) for all H e .9', let s(M) = s, and

(19) Exp M' > Exp H' for all H e . with s(H) = s(M).

We shall construct a special Iwasawa triple for M and from this an Iwasawa triplefor G. First we note the following.

(20) If M < H e 9' and (C, d, t) is an Iwasawa triple for H with t = s(H), then t = sand there is an element u e M such that (C, u, s) is an Iwasawa triple for H.

To show this let CM = K and IH: KI = p'. Since H = C(d>, K = C(K n (d>) _C(d°'>. If Exp C = p", then d induces an automorphism of order pk-` in C. Since Kis not abelian, d°' induces a nontrivial automorphism of order pk-`-' on C and thereexists a generator dl of (d°') such that (C, dl, t + j) is an Iwasawa triple for K. AsCM = K, there exist c e C and u e M such that dl = cu; hence (C, u, t + j) is anIwasawa triple for K. Now M = (C n M) (u) and so (C n M, u, t + j) is an Iwasawatriple for M. Hence t +j < s(M) = s, on the other hand s < s(H) = t by (18). Itfollows that t = s and j = 0. Thus (C, u, s) is an Iwasawa triple for K = H.

(21) There exists an Iwasawa triple (A*, b, s) for M with the property that for everyM < H E 9 there is an Iwasawa triple (C, b, s) for H with A* < C.

Suppose that this is false. If (A,, b,, s), ..., (A b s) are all the Iwasawa triples forM with third component s, then for every i e { 1, ... , r} there exists a subgroup Hi inS such that M < Hi and Hi does not contain an Iwasawa triple of the form (Ci, bi, s)with A. < Ci. Since G is locally finite, H = H, u u H, is finite; choose an Iwasawatriple (C, d, t) for H with t = s(H). By (20), t = s and there is an element u E M suchthat (C, u, s) is an Iwasawa triple for H. It follows that (C n M, u, s) is an Iwasawatriple for M, that is C n M = A, and u = bi for some i. But then (C n Hi, bi, s) is anIwasawa triple for Hi with A. < C n Hi. This contradiction yields (21).

(22) Let Hi E .9" and suppose that (Ai, b, s) are Iwasawa triples for Hi such thatA* < Ai for i = 1, 2. Then A, u A2 is abelian.

For, the definition of an Iwasawa triple implies that U,(A*) = M' and t5,(Ai)H. Furthermore by (20), s(Hi) = s and thus (19) shows that ExpA* > ExpAi fori = 1, 2. On the other hand, A* 5 A, n A2 and hence Exp Ai = Exp A* for i = 1, 2.By 2.3.5, A, u A2 has the same exponent as A* and therefore contains an elementa e A* of maximal order in its centre. By 2.3.16, A, u A2 is abelian.

Now let A be the join of all finite abelian subgroups C of G such that A* < Cand cb = c'+P° for all c c- C. For any such C, (C, b, s) is an Iwasawa triple for H =C<b) > A*<b) = M. So (22) shows that any two such C's centralize each other, thatis, A is abelian. It follows that ab = a'+P for all a e A. If g c- G, then H = <M, g) e Sand by (21), g e C<b) < A<b) for some Iwasawa triple (C,b,s) of H with A* < C.Thus G= A<b) and A a G. If Exp A = pk and I G : A l =p', then s < k since G is

not abelian. Furthermore b induces a power automorphism of order pk_, in A andsince bP'" E A operates trivially on A, k - s < m. Finally, s > 2 in case p = 2 since(A*, b, s) was an Iwasawa triple for M. Thus G satisfies (b).

Clearly, the groups in (a) of Theorem 2.4.14 are the hamiltonian locally finitep-groups. We shall need an additional property of the other locally finite M-groups.

2.4.15 Corollary. Let G be a nonabelian locally finite p-group with modular subgrouplattice. Then Exp G = p" is finite. If G is not hamiltonian, then there exists a triple(A, b, s) with the properties given in 2.4.14(b) and, in addition, o(b) = p" and s + m = n.

Proof. That G has finite exponent follows immediately from 2.4.14. Suppose that Gis not hamiltonian and let (A, b, s) be as in 2.4.14(b). Since G/A is finite, we maychoose A in such a way that no subgroup of G containing A properly occurs in atriple with these properties. Two elements of order at most pn-1 generate a finitesubgroup of G and by 2.3.5, their product has order at most pn-1. Thus fl,-,(G) isthe set of elements of order at most pn-1, also in the locally finite M-group G.Therefore G is generated by its elements of order p" and since G/A is cyclic, thereexists a cyclic subgroup C of order p" such that G = AC. Let ao e A and c c- C suchthat b = aoc. Then G = A<c) and C = <c>; hence o(c) = p" and ac = a'+P° for alla e A. Since ca'" e A, cr'" = (cP"')` = cP'"i'+' 1, that is p"+s = 0 (mod p") and hence

m + s >_ n. Suppose that m + s > n. Then c°'"-' e CG(A) since the automorphisminduced by c in A has order at most p"-s < p'. Thus A* = <A,c""-') would beabelian and (c°"'-' )` = cP--' = (c""')1 +p', that is, (A*, c, s) would have all the proper-ties given in 2.4.14(b). This would contradict the choice of A. Hence m + s = n andthe triple (A, c, s) has the desired properties.

Torsion groups with modular subgroup lattices

We say that a group G is a Tarski group if it is infinite but every proper, nontrivialsubgroup of G has prime order. For a long time it was not known whether Tarskigroups exist until Olshanskii [1979], using an extremely complicated inductive con-struction, produced the first examples of such groups. Clearly, the set of proper,nontrivial subgroups of a Tarski group G is an infinite antichain and hence L(G) ismodular. If one adds a chain to the zero element of this lattice, one clearly also getsa modular lattice. Therefore we call a group G an extended Tarski group if it containsa normal subgroup N such that

(23) GIN is a Tarski group,

(24) N is cyclic of prime power order p' 1, and

(25) for every subgroup H of G, H 5 N or N 5 H.

IExtended Tarski Group

Figure 12

It follows from (25) that for every minimal subgroup H/N of GIN, L(H) is a chainand hence H is cyclic of order p`+1 Thus N = Z(G) and G is a p-group with modularsubgroup lattice. Groups of this type do also exist (see Olshanskii [1991], p. 344).However, a complete description of all Tarski and extended Tarski groups is notknown and seems difficult to establish. So we have to take the following result as ourfinal description of periodic M-groups.

2.4.16 Theorem (Schmidt [1986]). Let G be a torsion group. Then G has modularsubgroup lattice if and only if G is a direct product of Tarski groups, extended Tarskigroups and a locally, finite M-group such that elements of different direct factors haverelatively prime orders.

In order to prove this theorem we need four lemmas. The first shows how theTarski groups get involved.

2.4.17 Lemma. Let x and y be elements of prime power order of an M-group G andsuppose that H = <x,y> is infinite.

(a) (Rudolph [1982]) If <x> n <y> = 1, then H is a Tarski group.(b) If <x> n <y> 0 1, then H is an extended Tarski group.

Proof. (a) Assume that the assertion is wrong and choose a counterexample inwhich o(x) + o(y) is minimal. Let X = <x> and Y = <y>. Suppose first that o(x) ando(y) are primes. A proper, nontrivial subgroup K of H cannot contain both X andY. And if X :C K, say, then X n K = 1 and by 2.1.5, [K/1] [X u K/X] is an inter-val in [H/X] [Y/1], that is, IKI is a prime. Thus H is a Tarski group, contradictingthe choice of G.

It follows that o(x), say, is not a prime; let A = f2(X) and B = f2(Y). Again by2.1.5, [X u B/B] is a chain and every minimal subgroup of X u B is contained inthe atom A u B of this chain. Thus A u B= .Q(X u B) 9 X u B. Similarly, A u B =(I(A u Y):9 A u Y and hence A u B:9 X u Y = H. Also by 2.1.5, X is a maximalsubgroup of X u B and therefore A <'D(X) < X n X6 for every b e B. It followsthat A a X u B; in particular, A u B is a finite group. Let Y, = Y n NH(A). Themodular law yields that NH(A) = (X u Y) n NH(A) = X u Y, and (A u Y,) n X =A u (Y, n X) = A. Thus NH(A)/A is generated by the cyclic subgroups X/A andAY,/A of prime power order intersecting trivially. Since NH(A)/A contains the nor-mal subgroup A u B/A of prime order, it is not a Tarski group. The minimality ofo(x) + o(y) implies that NH(A)/A is finite. Every conjugate of A in H is contained inthe finite normal subgroup A u B of H and hence also the number I H : NH(A) I ofthese conjugates is finite. So, finally, H is finite, a contradiction. Thus (a) holds.

(b) Let N = <x> n <y>. Then N is cyclic of prime power order and lies in thecentre of H. Therefore H/N is infinite and hence is a Tarski group by (a). Thus (23)and (24) hold for H. To show (25), suppose that there exists a subgroup K of H suchthat K -t N and N K; let X = <x>. Since L(X) is a chain and N:!5: X, it followsthat K X and D = K n X < N. As H/N is a Tarski group, X is a maximal sub-group of H. So X u K = H and [KID] [H/X], that is, KID is cyclic of primeorder. Therefore HID is generated by the cyclic subgroups X/D and KID of primepower order intersecting trivially, but is neither finite nor a Tarski group. Thiscontradicts (a). Thus (25) holds and H is an extended Tarski group.

The following lemma is the main step in the proof of Theorem 2.4.16. It shows inparticular that there are no "Tarski groups of higher dimension", that is torsiongroups G with L(G) modular of length at least 3 in which any two different minimalsubgroups generate a Tarski group. For any group T, we denote by n(T) the set ofall primes dividing the order of an element of T

2.4.18 Lemma. Let G be a torsion group with modular subgroup lattice and supposethat the subgroup T of G is a Tarski group.

(a) If x e G of prime power order q" (q a prime, n E N), then x e T for q e 7r(T) andx e CG(T) in case q 0 ir(T).

(b) T is the set of all rr(T)-elements in G and G = T x CG(T).

Proof. (b) is an immediate consequence of (a). For, every g E G is the product of itsprimary components. Therefore by (a), every n(T)-element of G lies in T and everyelement of G is contained in <T, CG(T)). Since T n CG(T) = Z(T) = 1, it follows thatG = T x CG(T).

To prove (a) we use induction on n and first consider the case n = 1. So let x E Gwith o(x) = q and suppose that x T; we have to show that x e CG(T) and q 0 ir(T).To do thisletX = <x>,H = TuXand 1 #aeTTakebeTwith T= <a,b>andput S = <b,x). Since and <x> are different minimal subgroups of H and L(H) ismodular of length 3, S is a maximal subgroup of H. If S = S°, then S n T = <b)would be invariant under a which is clearly not the case. Thus S # S° and S n S° isa minimal subgroup of H different from <a>. So, finally, K = <a> u (S n S°) is amaximal subgroup of H. Since a 0 S and a 0 S°, K n S = S n S° = K n S° and hence

(SnS)°=(KnS)°=KnS°=SnS°,

that is, S n S° is normalized by a. Thus S n S° 9 K and K is finite. As x E S,XnK = XnSnK = Xn(SnS°). So if X -A SnS°, then XnK = I and henceX u K = H. By 2.1.5, L(K) ^- [H/X] ^ L(T), a contradiction, since K is a finite andT an infinite group. Therefore X = S n S°. So X is normalized by a and as a was anarbitrary nontrivial element of T, X is normalized by T Since the simple group Tcan only operate trivially on the cyclic group X, it follows that x e CG(T). Finally, ifq e ir(T), then there would exist an a e T with o(a) = q. Then <a, x) would be ele-mentary abelian of order q2 and ax would be an element of order q in G notcontained in T But then ax e CG(T), as we have just shown, and hence alsoa e CG(T), a contradiction. Consequently, q ir(T) and this finishes the case n = 1.

Now let n > 2 and suppose that the assertion of the lemma is correct for n - 1;again let X = <x) and H = T u X, and put Z = f)(X). If T n X # 1, then T n X =Z and H = T u X= Y u X for some cyclic subgroup Y of T with T= Y u Z. As His neither finite nor a Tarski group, this would contradict 2.4.17. Hence T n X = 1and the case n = 1 already settled shows that Z 5 CG(T) and q 0 ir(T). Thus Z <Z(H) and, by induction, TZ/Z is centralized by X/Z. In particular, X a H and, asbefore, it follows that x e CG(T) since T cannot operate on X.

We prove the corresponding result for the extended Tarski groups.

2.4.19 Lemma. Let G be a torsion group with modular subgroup lattice and supposethat the subgroup T of G is an extended Tarski group; let rr(T) = {p}.

(a) If x e G of prime power order q" (q a prime, n e N), then x e T for q = p andx E CG(T) in case q A p.

(b) T is the set of all p-elements in G.

Proof. Since T is a p-group, (b) clearly follows from (a). To prove (a), we first showthat Z(T) a G. So suppose that Z(T) is not normal in G. Then there exists g E G

such that Z(T)' # Z(T). Take c e Z(T)a with <c"> = Z(T) n Z(T)a, let H = T v <c>and consider a maximal subgroup <a> of T Since a and c are p-elements of differentorders, <a, c) is neither a Tarski nor an extended Tarski group. By 2.4.17, <a, c) is afinite subgroup of G and then a p-group as 2.4.7(a) shows. Since cP E <a>, <a> is amaximal subgroup of <a, c> and hence normal in this finite p-group. It follows thatZ(T) g H. But then cZ(T) is an element of order p in HIT that does not lie in theTarski p-subgroup T/Z(T); this contradicts 2.4.18.

Thus Z(T) a G. By 2.4.18, xZ(T) E T/Z(T) for q = p, that is, x E Tin this case. Solet q # p. Then T/Z(T) is centralized by xZ(T). Therefore if <a> is a maximal sub-group of T, it follows that [x, a] e Z(T) = <aP). So x induces an automorphism in<a> centralizing <a)/I(<a)) and as o(x) is prime to p, x centralizes <a). Since <a>was an arbitrary maximal subgroup of T, we conclude that x c CG(T).

The last of our four lemmas will be used to show that the product of all Tarski andextended Tarski groups in G has a locally finite complement.

2.4.20 Lemma. Let G be an M-group and let x,, ..., x, c- G. If <x,, xj ) is finite forall i and j, then also <x,,..., x,) is finite.

Proof. We may assume that the x, are elements of prime power order; for, theprimary components of the xi satisfy the same assumptions as the x, and also gener-ate <x,,...,x,>. So we prove the lemma under this additional assumption and useinduction on o(x,) + + o(x,). Let H = <x1,...,x,> and let p be the largest primedividing the order of one of the x,.

Suppose first that there exists another prime dividing one of these orders. Withoutloss of generality, let x,, ..., x, be the p-elements among the x,. Then for i < s andj > s, the subgroup <xi, x3) is a finite M-group and by 2.4.6, <x, >'j = <x,). Thus<x,,...,x,) 9 H and H = <x,,...,x,)<x,+,,...,x,) is finite since, by induction,<x,,...,x,) and <x,+,,...,x,) are finite.

Now suppose that all the xi are p-elements and let x, be an element of maximalorder among the xi. If o(x,) > p2, then 2.4.7(b) shows that fl(<xl )) is normalized byevery xi and hence is normal in H. By induction H/S2(<x,) ), and hence H, is finite.

We are left with the case that o(xi) = p for all i. If all the <xi,xx) are p-groups, anytwo of the xi commute; it follows that H is abelian and hence finite. Finally, supposethat <x1, x2 ), say, is not a p-group. Then by 2.2.4, 1 <x,, x2 ) I = pq where q is a primewith q > p. Let Q be the subgroup of order q in <x,, x2 ). We claim that <Q, x,) isfinite for all i >_ 3. For otherwise <Q, xi) would be a Tarski subgroup of the torsiongroup <xi, X2, xi), by 2.4.17. By 2.4.18, <Q, xi) would have to contain the p-elementsx, and x2i but a Tarski group cannot have a subgroup <x,, x2) of order pq. So<Q, xi) is finite and by 2.4.6, Q a_ <Q, x, ). Thus Q a H and by induction, H/Q =<x2Q,...,x,Q) is finite. Therefore H too is finite.

Proof of Theorem 2.4.16. If G has the structure given in this theorem, then by 1.6.4,L(G) is isomorphic to a direct product of modular lattices and hence is modular.Conversely, suppose that G is a torsion group with modular subgroup lattice, let .lbe the set of all Tarski subgroups and extended Tarski subgroups of G and let it be

the union of all the n(T) with T e . By 2.4.18 and 2.4.19, every T e is the set ofall n(T)-elements of G and hence is a normal subgroup of G. Since the Tarski groupsare simple and every extended Tarski group has only one minimal subgroup, anytwo such groups intersect trivially and the tr(T) with T E J form a partition of it.Let K be the set of all n -elements in G. We shall show that K is a locally finitesubgroup of G. Then, clearly, K is a normal subgroup of G and since K and theT E are coprime subgroups, G is the direct product of K and all the T e . Thiswill prove the theorem. First we show that

(26) <x, y) is finite for all x, y e K.

For, if x = x, ... x, and y = yi ... y, are written as products of their primary compo-nents, then all the xi, yk are n'-elements and, clearly, <x;,x3) < <x> and <y;,yy) <<y) are finite for all i and j. If <xi, yy) were infinite, then <x;, y3) E by 2.4.17 andthen x; would be a n-element, a contradiction. Thus all the <x;, yy) are finite. By2.4.20, <x,,._ x y, , ... , y5) = <x, y) is finite and (26) holds.

We show now that K is a subgroup of G. For this let x, y e K and suppose thatxy-' 0 K. Then there exists a prime p e n dividing o(xy-' ); let T e .°l with p e n(T).By (26), <x, y) is finite and by 2.4.18 or 2.4.19, T n <x, y) is the set of n(T)-elementsin <x, y). Hence T n <x, y) is a normal Hall subgroup of <x, y) and the Schur-Zassenhaus theorem yields that there exists a complement S to T n <x, y) in <x, y).This centralizes T n <x, y), again by 2.4.18 or 2.4.19. Thus S is a normal subgroupof <x, y) and therefore contains the n(T)'-elements x and y. Hence also xy-' E S, butp divides o(xy-' ). This contradiction shows that xy-' E K and that K is a subgroupof G. By (26) and 2.4.20, K is locally finite.

Further topics

Theorems 2.4.11, 2.4.13 and 2.4.14 (or 2.4.4 and 2.3.1 in the finite case), and finally2.4.16 give the structure of nonabelian M-groups. These results are basic and willoften be used. In the remainder of this section we point out some immediate conse-quences and report on related results.

In 2.4.5 we noted that every finite M-group is metabelian. The results mentionedabove show that this is also true for locally finite M-groups and M-groups withelements of infinite order. Considering 2.4.16, we get the following result.

2.4.21 Theorem. Let G be an M-group. If G contains elements of infinite order or ifno Tarski group is involved in G, then G is metabelian. In particular, every projectiveimage of an abelian group is metabelian.

If any two subgroups of the group G permute, then G is an M-group by 2.1.3. And,clearly, G is locally finite if it is a torsion group. So 2.4.11 and the results on locallyfinite M-groups give the structure of these groups (see Exercise 3). In particular, wenote the following.

2.4.22 Theorem (Iwasawa [1943]). If any two subgroups of the group G permute,then G is metabelian.


In the theory of finite groups, for inductive proofs, it is often useful to know for acertain class X of groups the class of all groups in which every proper subgroup (orproper factor group) is an .1-group. For the class I of M-groups, these classes havebeen studied extensively. First Napolitani [1971] determined the finite groups all ofwhose proper subgroups are M-groups. Since every finite M-group is supersoluble,these groups are soluble and Napolitani gave a complete list of them. Then Fort[1975] investigated finite groups G in which

(27) [G/H] is modular

for every nontrivial subgroup H of G, that is, with weakly modular subgroup lattices.Also these groups are soluble, since the minimal simple groups do not have thisproperty and Fort was able to determine them completely. Finally, Longobardi[1982] studied finite groups G satisfying (27) for every nontrivial normal subgroupH of G, that is, in which all proper factor groups are M-groups. Since every finitesimple group has this property, Longobardi restricted herself to supersoluble groupsof this type and produced a complete list of them. Slightly smaller than the classstudied by Longobardi is the class of all finite groups in which (27) holds for everynontrivial modular subgroup H of G. This class is interesting since it is clearlyinvariant under projectivities. Longobardi and Maj [1985] showed that the nonnil-potent groups in both classes coincide and, using Longobardi's list, determined thenilpotent groups in the smaller class. All these results are much too technical to bepresented here.

Finally we mention a theorem of a quite different character. Fort [1983] provedthat a finite group G has modular subgroup lattice if and only if every maximalsubgroup of G is dually modular in L(G); here an element d of a lattice L is calleddually modular in L if it satisfies

(28) xn(duz)=(xnd)uzfor all x,zeLwith x>z,and(29) dn(yuz)=(dny)uzfor ally, zELwithd>z.

Note that these conditions are dual to (2) and (3) of 2.1 in the definition of modularelements of a lattice.

Exercises

1. (Jones [1945]) Let G be a P*-group of order p"q' with n, m e N and primes p > qand suppose that m >_ 2 (that is, G is not a P-group). Show that the group G islattice-isomorphic to G if and only if G is a P*-group of order p"r' where r is aprime dividing p - 1.

2. Let G be a nonabelian finite P-group. Determine Aut G and eliminate Cooper'stheorem (or 1.4.3) from the proof of Lemma 2.4.3.

3. Show that the following properties of a torsion group G are equivalent.(a) Any two subgroups of G permute.(b) G is the direct product of its p-components and these are locally finite p-

groups with modular subgroup lattices.(c) G is a locally nilpotent M-group.

4. Let G be an M-group with elements of infinite order. Show that G is locallynilpotent but in general not nilpotent.

5. Let G be a nonabelian M-group such that T(G) is a p-group and G/T(G) is notcyclic.(a) Show that in Theorem 2.4.11(b) we may choose p; # p for all i e N. (Hint: If

a e T(G) and k e N such that azk # a, show that there are only finitely manyj>kwith p;=p.)

(b) If p; # p for all i, show that there exist b, e T(G) such that the elements x, _z;6; satisfy xP+l = xt for all i e N. Conclude that G is the semidirect product ofT(G) and the torsion-free, locally cyclic group Z = <x1, x2, ... ).

6. (Napolitani [1971]) Let G be a finite group in which every proper subgroup is anM-group.(a) Show that G is soluble.(b) If I G I is divisible by 3 different primes, show that G is supersoluble.(c) If I G I is divisible by 4 different primes, show that G is an M-group.

7. Let G = H x K be an M-group. Show that G is abelian if to every x e H thereexists y e K and to every y e K there exists x e H such that o(x) = o(y). (In partic-ular, G is abelian if H K; for a short direct proof of this result see Lukacs andPalfy [1986].)

2.5 Projectivities of M-groups

It is a remarkable fact that large classes of nonabelian M-groups admit projectivitiesonto abelian groups. In particular, all M-groups with elements of infinite order andall nonhamiltonian locally finite p-groups with modular subgroup lattices have thisproperty. We shall construct these projectivities using special element maps, so-called crossed isomorphisms, in the locally finite case. More generally, we wantto determine all projectivities of an arbitrary M-group. We start with the torsiongroups and, in the light of 1.6.6, we may restrict our attention to the direct factorsthat appear in Theorems 2.4.16 and 2.4.13.

Tarski groups and extended Tarski groups

Clearly, every projective image of a Tarski group or an extended Tarski groupis a Tarski group or extended Tarski group, respectively. Since a Tarski groupis generated by two elements, it is countable. Therefore any two Tarski groupshave the same number of minimal subgroups and are lattice-isomorphic. And twoextended Tarski groups are lattice-isomorphic if and only if their subgroup latticeshave the same lengths. Finally, the group of autoprojectivities of a Tarski orextended Tarski group is isomorphic to Sym f\l, the symmetric group on a countableset.

2.5 Projectivities of M-groups 89

P*-groups

We saw in Section 2.2 that every P-group is lattice-isomorphic to an elementaryabelian group. It is not difficult to describe the projectivities of those P*-groups thatare not P-groups (see Exercise 2.4.1). Even without such a description, the results of2.2 show that a P*-group is lattice-isomorphic to an abelian group if and only if itis a P-group. For, if G is a P*-group with nonnormal Sylow subgroup of order q"and qp is a projectivity from G to an abelian group G, then a subgroup H of orderpq' of G has indecomposable subgroup lattice. Its abelian projective image H0therefore must be a primary group and by 2.2.6, H' is elementary abelian. Thusm = 1 and G is a P-group.

Autoprojectivities of elementary abelian groups, and hence also P-groups, weredescribed in 1.4.4; see also 2.6.7. If G is a P*-group and not a P-group, everyautoprojectivity of G fixes D(G) and the p-component A of G where p is the largerprime involved in G. It follows that P(G) is isomorphic to the subgroup of elementsof P(G/(D(G)) fixing AD(G), that is, to the group of autoprojectivities of an elemen-tary abelian group fixing a maximal subgroup.

Hamiltonian 2-groups

These groups clearly cannot be lattice-isomorphic to an abelian group since thequaternion group Q8 is determined by its subgroup lattice. In fact, this holds for anarbitrary hamiltonian 2-group and we prove an even stronger result.

25.1 Theorem (Baer [1939a]). Every projectivity of a hamiltonian 2-group is inducedby exactly four isomorphisms.

Proof. Let G be a hamiltonian 2-group, that is, G = Q x A where Q - Q8 and A iselementary abelian, and let q be a projectivity from G to some group G. Since Q0contains only one minimal subgroup, Q' Q$ ^- Q. If x e Q has order 4 and 1 0a e A, then <x2, a)9' is a four-group. Hence I <a>"I = 2 and A0 is an elementaryabelian 2-group (see also 2.2.5). Furthermore, <x, a>0 is a group of order 8 with 3minimal subgroups and therefore is abelian since Q8 has 1 and D. has 5 minimalsubgroups. It follows that Q' < CU(A'') and hence G = Q4' x A".

Let H = Z(Q) x A and define r: H - H0 by <at> = <a>0' for a e A. Then, clearly,r is the only element map inducing (p on H. If a, b E H, and <a, b> is a four-group,then (ab>0 = <a`bt> is the minimal subgroup different from <a>0 and 0 in<a, b)0. Thus (ab)` = a`b` and this, of course, also holds if I <a, b> I S 2. Now let Q =<x, y>, <x>0 = , <y>0 = <v> and let a: G -- G be the isomorphism satisfyingx° = u, y° = v and a° = at for all a E A. This isomorphism clearly induces q on Qand on H. And if g e G with g 0 Q and g 0 H, then there exist z e Q and a e A witho(z) = 4 and g = za. As <z>0 = <z°> and <a>0 = <a°>, also <z, a>" = <z, a)° andsince <g> is the only cyclic maximal subgroup different from <z> in <z, a>, it followsthat <g>0 = <g°). Thus a induces 'p. We had a choice of two elements each for uand v. So we have shown that there are 4 different isomorphisms inducing cp. On the


other hand, every isomorphism inducing rp has to map x and y onto a generator of(x> and (y> , respectively. Hence there are exactly 4 such isomorphisms.

Crossed isomorphisms

Let G be a group and let f be a map from G into the set End G of all endomorphismsof G. A bijective map from G to a group G is called an f-isomorphism if it satisfies

(1) x°y° = (xf(Y)y)° for all x, y e G.

A crossed isomorphism is an f-isomorphism for some map f : G - End G. Examplesof crossed isomorphisms are isomorphisms-here f(x) is the identity for all x E G-and antiisomorphisms. For, if f(x) is the inner automorphisminduced by x-', thatis, gf(x) = xgx-' for all g, x e G, then a bijective map a: G -+ G is an f-isomorphismif and only if x°y° = (yxy-' y)° = (yx)° for all x, y E G.

We want to use crossed isomorphisms to construct projectivities between locallyfinite, nonhamiltonian p-groups and abelian groups. So we have to find out forwhich maps f there exist f-isomorphisms and when they induce projectivities. Thefirst question is easy to answer.

2.5.2 Theorem. Let G be a group and f : G -+ End G. If there exists an f-isomorphismfrom G to some group G, then

(2) f (l) is the identity,

(3) f(x) E Aut G for all x E G, and

(4) f(x)f(y) = f(xf(Y)y) for all x, y E G.

Conversely, suppose that (2)-(4) hold and define a new operation o on G by

(5) x o y = xf(Y) y

for x, y c- G. Then G* = (G, o) is a group and the identity on G is an f-isomorphismfrom G to G*.

Proof. Let x, y, z e G. If a is an f-isomorphism from G to G, then 1°y° = (IJ(Y)y)° _y° and hence

(6) 1° = 1.

Therefore x° = x°1° = (xf('))° and since a is injective, x = xf('). Thus (2) holds. Toprove (3), suppose that xf(Y) = zf(Y). Then x°y° = (xf(Y)y)° = (zf(Y)y)° = z°y°. As a isinjective, x = z and hence f(y) is injective. Since a is surjective, there exists g E Gwith g° = (xy)°(y°)-' and hence (xy)° = g°y° = (gf(Y)y)° Again the injectivity of aimplies that x = gf(Y). Thus f(y) is also surjective. It follows that f(y) e Aut G and (3)holds. Finally,

(zf(xfly)y)xf(Y)y)° = Z°(xf(Y)y)° = Z°x°y° = (zl(x)x)°y° = (zf(x)f(Y)Xf(Y)y)°

and since a is injective, it follows that z1(x)f(Y) = zf(xf")Y). Thus also (4) holds.

2.5 Projectivities of M-groups

Conversely, suppose that (2)-(4) are satisfied. Then

(x o y) o z = (xf(y)y)f(z)z = xf(y)f(z)yflz)z = x-(yflllz)(Yf(=)Z) = Xo (y o Z),

91

that is, the operation o is associative. Furthermore, 1 o y = if(y)y = y. Since f(y) issurjective, there exists g e G such that gf(y) = y-' and hence g o y = gf(Y)y = 1. ThusG* = (G, o) is a group and (5) implies that the identity is an f-isomorphism from Gto G*.

2.5.3 Corollary. If a: G -p G is an f-isomorphism, then the map v: G - Aut G given byu" = f(u° ') for u e G is a homomorphism.

Proof. Let u, v c- G and take x, y c- G with x° = u, y° = v. By (3), u" E Aut G and (1)and (4) imply that

(uv)" = WY), = ((xf(y)y)°)V = f(xf(y)Y) = f(x)f(y) = u"v".

Thus v is a homomorphism from G to Aut G.

2.5.4 Lemma. Let G be a group, f : G -i End G and suppose that u: G -i G is anf-isomorphism. For H < G and x c- G,

(7) (Hx)° = H°x° if and only if Hf(x) = H, and

(8) H° is a subgroup of G if and only if Hf(=) = H for all z e H.

Proof. By (3), f(x) -' exists and (1) implies that for all z e H, z°x° = (zf (')x)' and(zx)° = (z1(x)-')°x°. Since Q is injective, it follows that (Hx)° = H°x° if and only if forall z c- H, zf(x) and Zf(x) are contained in H, that is, Hf(x) = H. Thus (7) holds. Toprove (8), suppose first that H° < G. Then for all z e H, (Hz)' = H° = H°z° andhence Hf(=) = H by (7). Conversely, let Hf(') = H for all z c- H. If y, z c- H, then (7)shows that y°z° E (Hz)" = H. Furthermore there exists w c- H such that wf(z) = z-1.By (6) and (1), 1 = 1° _ (wf(z)z)° = w°z° so that (z°)-' = w° e H. Thus H° is asubgroup of G.

The lemma shows that if an f-isomorphism a is to induce a projectivity, everyendomorphism f(x) has to fix all subgroups containing x. However, this conditionis not sufficient. Even if all the f(x) are power automorphisms, a need not induce aprojectivity.

2.55 Example. Let G be an abelian group and suppose that N is a nontrivial sub-group of index 2 in G. For a E N, let f(a) = id, the identity map on G; if x E G\N, letf(x) = - id, that is, gf(x) = g-' for g e G. Then (2) and (3) are satisfied and sinceI G: NI = 2, it is easy to see that also (4) holds. So if we define the operation o on Gby (5), then H = (G, o) is a group and u = id is an f-isomorphism from G to H.If xeH\N and aeN, then xox=xf(x)x=X-'x= 1 and xoaox=(xa)ox=a-' x-i x = a-'. Since the multiplication in N has not changed, a-' is also the inverse

to a in H and it follows that H is a generalized dihedral group. In particular, if G iscyclic, then H is a dihedral group and clearly not lattice-isomorphic to G. Note thatby (8), every subgroup of G is also a subgroup of H, but H may contain additionalsubgroups.

In our computations with crossed isomorphisms we shall need some simplenumber-theoretic facts.

2.5.6 Lemma. Let p be a prime and suppose that 1, k, n are natural numbers such that

2 < k < n. If ' divides n, then i+1 divides m =(n

k

) k-1,p p p except in the case p = 2,

k = 2 in which m is divisible just by p'.

Proof. The assertion certainly is correct if k >- i + 2; assume therefore that k <is k-1 is k-1 'a -

i + 1. Now n = p'a where a e N and m = Cpk/

pk-1 = pkp

[ p . Sincef i=1 j

j < i, the powers of p in the factor j of the denominator cancel with those in p'a - j.Therefore, if k = p's with (s, p) = 1 and r > 0, then m is divisible by p` wheret = p's - 1 +i-r. Clearly t> i+ 1, except in the case r = s = 1, p = 2 in whicht=i.

2.5.7 The Functions µr. For every integer r, define u,: N - Z byn-1

(9) u,(n) = I r'.i=o

Then for all n, m e N,n-1 m-1

(10) µr(n)r' + µr(m) = I ri+m + Y r' = Ur(n + m).i=o j=0

We shall use the functions µ, to compute powers of elements. In fact, if v is anf-isomorphism from G to G and xflx) = Xr where r e Z, then

(11) (x ")n = (xp,(n))a

For, if n = 1, then u,(n) = 1. And if (11) holds for n - 1, then (1) and (10) imply that

(x")n = (xp,(n-1))ax" = (xp,(n-1)r+1)" = (xµ,(n))ff

Let r = 1 + psa where p is a prime and a, s e N with (p, a) = 1 and s > 2 in case

p = 2. Then r" _ (n) pksak and hencek=o k

(k-1)sak-1(12) µ.(n) =r" - 1 = n + n (n)

pr - 1 k=2 k

Let i e N. If p']n, then by 2.5.6, every summand on the right hand side of (12) isdivisible by p' and hence p'Iµr(n). Conversely, suppose that p1 ,(n) and let n = p'b

with (p, b) = 1. By (12), µ,(n) - n ((mod p) and hence j > 1. Since s >_ 2 in case p = 2,

2.5.6 shows that every summand ()p_1ak_1 in (12) is divisible by p'+'. Then (12)

yields that i < j + 1. Thus p`I n. We have shown that for all i, n e N,

(13) p`ln if and only if p`lu,(n).

It follows that for all x, y, m e N,

(14) x = y (mod p') if and only if µr(X) - µr(y) (mod pm).

This is clear if x = y. If, say, x = n + y with n e N, then u,(x) = µr(n)r'' + µ,(y) by(10). Since r = 1 (mod p), therefore u,(x) = µ,(y) (mod p') if and only if u,(n) __ 0(mod pm) and by (13), this is the case if and only if n = 0 (mod pm). This proves(14). Now (14) shows that the map ar,m: 7L/p'"7L - 7L/pm7L given by µr,m(X + pm7L) =µ,(x) + pm7L is well-defined and injective. Since 7L/pm7L is finite, this map is alsosurjective and it follows that

(15) to every i e N there exists j e N such that µr(j) = i (modpm).

We leave open the question exactly when a crossed isomorphism induces aprojectivity, and content ourselves with the following result. Note that property (d)of the theorem only contains conditions on f that are easy to control in G.

2.5.8 Theorem (Baer [1944]). Let G be a group, f: G -+ End G and suppose thata: G - G is an f-isomorphism. Then the following properties (a)-(d) are equivalent.

(a) a induces a projectivity from G to G and satisfies (Hx)° = H°x° for all H < Gand x e G.

(b) For every x e G, f(x) is a power automorphism and <x)° _ <x°>.(c) For every x e G, f(x) is a power automorphism and o(x°) = o(x).(d) For all x, y e G, the following conditions are satisfied.

(i) f(x) is a power automorphism.(ii) If o(x) is infinite, then xf(x) = X.

(iii) If o(x) = 2m with m > 1, then xf(x) = x1+4' with i e N.(iv) If xy = yx and o(x) and o(y) are finite and relatively prime, then xf(v) = X.

Proof. (a) (b). By (7), every f(x) is a power automorphism. Since a induces aprojectivity, <x>° = <x°).

(b) = (c). Since a is bijective, o(x°) = I <x°> I = I <x)°I = o(x).(c) (d). Clearly (i) holds. If o(x) is infinite and xf(x) 0 x, then xf(x) = x-1 and

X°X° _ (Xf(x)X)° = 1° = 1 by (6). This would contradict the assumption o(x°) =o(x). Hence (ii) holds and (iii) follows similarly. For, if o(x) = 2m with m > 1 andXPX) = X-1+4; with j e N, then (x°)Z = (x4')° by assumption would have order 2m-1,on the other hand its order would divide 2m-2. Finally, if x, y e G with xy = yx and(o(x), o(y)) = 1, then <xy> = <x> x <y>. By (8), <xy>° is a subgroup of order I <xy>containing the element (xy)° of this order. Thus <xy)° is cyclic and since x°, y° e<xy)°, we have (xf('')y)° = x°y° = y°x° _ (yf(x)x)°. Hence xf(Y)y = yf(x)x and sincexf(r) E <x> and yf(x) a <y>, it follows that xf(Y) = X.

(d) - (b). By (8), <x>° is a subgroup of G; we have to show that it is generatedby x°. First let o(x) be infinite and write H = <x>. For 1 : y e H, xf(r) = x orxfly) = x-'. In the latter case, it would follow that yf(r) = y-', contradicting (ii).Thus f(y) is the identity on H. By (2), this also holds for y = 1 and therefore y°z° =(yf(z)z)° = (yz)° for all y, z e H. Thus a induces an isomorphism on H and it followsthat <x)° = <x°). Now let o(x) be finite and suppose first that o(x) = p" for someprime p and m e N. If m = 1, then by (6), x° is a nontrivial element in the group <x)°of order p and hence <x>° = <x°). So suppose that m > 2. Then I<x>'I = p" and by2.5.3, f(x) is the image of the p-element x° under a homomorphism. Thus f(x) isa p-element in Pot G and therefore xf(x) = x' with r - 1 (mod p); let r = 1 + p'awhere (p, a) = I and s >_ 1. The assumption (iii) implies that s >_ 2 in case p = 2.By (13), p' does not divide u,(p'') and then (11) and (6) yield that (x°)P'' =

0 1. It follows that x° generates <x)°. Finally suppose that x is of com-posite order, that is, x = yz with y : 1 : z, yz = zy and (o(y), o(z)) = 1, and assumethat the assertion is true for elements of smaller order. Then o(y°) = o(y), o(z°) _o(z) and by (iv),

y°z° = (yf( )Z)° = (yz)° = x° = (zy)° = (zf(v)y)° = z°y°.

It follows that o(x°) = o(y°)o(z°) = o(x) and this implies that <x>° = <x°>.(b) (a). By 2.5.4, H° < G and (Hx)° = H° x° for all H < G and x e G. We have

to show that preimages of subgroups of G are subgroups in G. Let S c G satisfyS° < G and take x, y e S. Then x°(y°)-' a S°. Hence there exists z e S such thatx°(y°)-' = z°, that is, x° = z°y° = (zf(Y)y)°. It follows that x = zf('')y and hencexy ' = zf(r) e <z> since f(y) is a power automorphism. Thus (xy-')° c- <z)° =<z°) < S°, that is, xy-' e S. Therefore S is a subgroup of G and a induces aprojectivity.

We remark that element maps satisfying condition (a) of Theorem 2.5.8 can beregarded as special types of isomorphism between the coset lattices of G and G, theso-called right affinities.

Nonhamiltonian locally finite M-groups

We use Theorem 2.5.8 to construct projectivities between abelian and nonabelianp-groups.

2.5.9 Theorem (Baer [1944]). Let G be a locally finite p-group with modular sub-group lattice. If G is not hamiltonian, then there exist an abelian p-group G anda crossed isomorphism a: G - G that induces a projectivity between G and G andsatisfies (Hx)° = H°x° for all H < G and x e G.

Proof. If G is abelian, then G = G and the identity a on G has the desired properties.So suppose that G is not abelian and take an Iwasawa triple (A, b, s) for G as de-scribed in 2.4.15. Then A is an abelian normal subgroup of 6 with IG : Al = p'", b is

an element of G with o(b) = p" = Exp G and s is an integer with s > 2 in case p = 2such that G = A , b-' ab = a' +p' for all a e A and

(16) m + s = n.

Let r = 1 + p5 and µ = µ, the function defined in 2.5.7. By (13), p-',U(p') is aninteger prime to p; let t e 101 such that 0 < t < p" and

(17) t = p-'"µ(p") (mod p").

Then t # 0 (mod p) and therefore the map a - at is an automorphism of A. Nowbp"' E A since I G : A I = p'", and hence there exists a unique element ao e A such that

(18) bp' = a'..

Let G be the abelian group obtained by adjoining to A an element c, subject to therelation

(19) cp"' = ao.

Then o(c) = o(b) = p" = Exp G. We want to construct a map f : G - End G sat-isfying (2)-(4) and property (d) of Theorem 2.5.8.

For this let a be the power automorphism of G defined by g° = g' = g'+p for all

g e G. By (16), o(a) = p'". Since IG: Al = p', every element x e G can be writtenuniquely in the form x = ac'(x) with a c A and I < i(x) < p'", By (15) there existintegers j such that µ(j) i(x) (modpt) and (14) shows that any two such integersare congruent modulo p'. Therefore, since o(a) = p',

(20) f(x) = a'

is independent from the choice of the integer j satisfying µ(j) = i(x) (mod p') andhence is a well-defined power automorphism of G. Now f (x) is the identity if andonly if p'"l j and by (13), this holds if and only if p' =_ µ(j) = i(x) (mod p'), that is, ifand only if x e A. Thus

(21) A={xeGlf(x)=id}.In particular, (2) holds and (3) is trivial. To prove (4), consider elements x = a, OX)and y = a2C'(i) of G, and let j, k e 101 satisfy µ(j) = i(x) (modpm) and µ(k) = i(y)(modpm). Then ce'" a A, and the definition of f(y) and (10) yield that modulo A,

xf(Y)y = (c i(x))Py)C i(Y) = CM(!)rk+µ(k) = Cµ(j+k).

Thus i(xf('')y) = µ(j + k) (mod p"') and hence f(xf(Y)y) = a'+k = a'ak = f(x)f(y).So if we write x o y = xf('')y for x, y e G, then by 2.5.2, G* _ (G, o) is a group and

the identity a is an f-isomorphism from G to G*. Since s >_ 2 in case p = 2, f has allthe properties in (d) of 2.5.8 and therefore a induces a projectivity from G to G* andsatisfies (Hx)° = H°x° for all H < G and x e G. If we can show that G* and G areisomorphic, then the theorem will be proved.

By (21), f(x) is the identity for all x e A and hence elements of A have the sameproduct in G and G*. Furthermore i(c) = 1 and µ(l) = 1 so that

(22) f(c) = a.

Since f (x o y) = f (xf ('') y) = f(x)f(y) for x, y E G*, the map f can be regarded as anepimorphism from G* to <a> < Aut G with kernel A. It follows from (22) that G* isgenerated by A and c. If d is the inverse to c in G*, then df(')c = d o c = 1. So (21)and (22) imply that for a e A,

doaoc=(df(a)a)oc=(da)f(c)c=df«)caf(c)=a'.

And if e is the pm-th power of c in G*, then (11), (17), (19) and (18) yield that

e=cv(p"')=cp'"'=ao' = bp"'.

Thus G and G* are extensions of A by elements b and c, respectively, inducing thesame automorphism on A and subject to the same relation zr'" = ao. Hence thesegroups are isomorphic and the proof of the theorem is complete.

By 1.6.4 and 1.6.5, the results of Section 2.2 and Theorem 2.5.9 give the structureof projective images of abelian torsion groups.

2.5.10 Theorem. The group G is lattice-isomorphic to an abelian torsion group if andonly if G is a direct product of coprime P-groups and nonhamiltonian locally finitep-groups with modular subgroup lattices.

M-groups with elements of infinite order

Every nonabelian M-group with elements of infinite order admits a projectivity ontoan abelian group. We shall prove this beautiful theorem of Sato [1951] in theremainder of this section. Our proof follows Mainardis [1991], [1992]; its main ideais to embed the given M-group into an M-group with divisible torsion subgroup.For such a group G, the infinite cyclic subgroups can be described quite easily andSato's construction of the projectivity between G and an abelian group becomesmuch simpler. To describe these cyclic subgroups, we have to compute powers ofelements in G and for this we need the following functions.

2.5.11 The Functions ,a, and v,. Let p be a prime and let U,, be the set of (p-adic)units in the ring RP of p-adic integers. For r e Up, the formula

"-1

(9) µ,(n) _ Y- r'1=0

again defines a function u,; this time, p,: FJ -+ R.Let s e R with s > 2 in case p = 2 and assume that r = 1 + p'a where 0 a E R.

Write n = p'b where t > 0 and (b, p) = 1; then b c- Up and hence there exists d e Upsuch that bd = 1. As in (12),

µ.(n) =r" - 1 = n + n (fl)p(k_1)sak_1r - 1 k=z \k

Again by 2.5.6, pt+1 divides (n)_1 )S for 2 < k 5 n and hence there exists f e Rpsuch that

p,(n) = n + p'+1 f = n + bdp'+1 f = n(1 + pdf).

Thus there exists one and, since Rp is an integral domain, only one element g e Rpsuch that µ,(n) = ng. That is, we have a second function v,: N - RP given by

(23) µ,(n) = nv,(n)

and satisfying

(24) v,(n) - 1 (mod p) for all n e N.

As mentioned above, we can use the functions µ, and v, to compute powers ofelements.

2.5.12 Remark. Let G be a nonabelian M-group and let z e G be an element ofinfinite order. Then by 2.4.11 there exist p-adic units r(p) with r(p) - 1 (mod p) andr(2) - 1 (mod 4) such that c' = c(P) for all c c T(G)P, the p-component of T(G). Andan obvious induction then shows that for every finite subset it of P, for all n e N andaP a T(G)P,

(25) z Li ap = z" fl ap'-,(n) = zn r7 apv-)(n)PER PER PER

For, since µ,(P)(1) = 1, the first equation is clear if n = 1. And if it holds for n, then

n+l

z jj ap) = ( 11 ap)(zn zn+1

PER PER PER

= Zn+1 T7P

PEN

TT a;(P)"+I"jP)(")jj PPER

The second equation now follows from (23).To simplify the notation, we define maps v(z, n): T(G) -, T(G) by

(26) (nPER

Li av-(") for aP e T(G)P, zC c P, it finite.PER

By (24) and 1.5.5, the map c - 0-0) is a power automorphism of T(G)p and hence

(27) v(z, n) e Pot T(G) for all z e G\T(G), n e N.

And if we use these maps, we can rewrite (25) in the form

(28) (za)" = z"a""(-,n) for all z e G\T(G), a e T(G), n e N.

We need a result on the structure of M-groups with divisible torsion subgroup,due to Mainardis [1991], which is of independent interest.

2.5.13 Lemma. Let G be a nonabelian M-group with divisible torsion subgroupT(G) = T

(a) For every z E G\T, a E T and n c- N, there exists b e T such that z"a = (zb)".(b) There exists a torsion free, locally cyclic subgroup Z of G such that G = TZ.

Proof. (a) Since T is divisible and v(z, n) E Pot T, there exists b e T such thatb""(=,n) = a. By (28), (zb)" = z"b"(Z") = z"a.

(b) The assertion is clear if G/T is cyclic. So suppose that G/T is not cyclic. ThenG has the structure given in 2.4.11(b). In particular, G = T<z1,z2,...> and z'41 =ziai with a, E T, pi e P. We show that there exist bi e T such that the elements xi =zibi satisfy x?+1 = xi for all i e N. For this we may take b1 = 1. And if bi exists, thenai ' bi e T and hence by (a) there exists bi+1 E T such that

x +1 = (zi+1bi+1)1" = z° la;'bi = zibi = xi.

Thus the xi exist and <x1, ... , xi) = (x1> is infinite cyclic for all i. So Z =<x1, x2, ... > is a torsion-free, locally cyclic subgroup of G satisfying TZ >_T<z1,z2,...> = G.

The main step in the proof of Sato's theorem is the following lemma.

2.5.14 Lemma (Mainardis [1992]). Let G = T<z> be a nonabelian M-group withdivisible torsion subgroup T(G) = T and o(z) infinite; let G =_T x <w> where o(w) isinfinite. For every fl E Pot T, we define a map T: L1(G) - L1(G) in the following way.If X E L1(T), let XT = X; and if X E L1(G) such that X T, that is, X = <(wa)">where n e N and a E T, define XT = <(zag)">. Then T = T(W,Z; fl) is a bijective mapfrom L1(G) onto L1(G) satisfying for all X, Y, Z E L1(G),

(29) X < <Y,Z> if and only if XT < <YT,ZT).

Thus, by 1.3.2, T induces a projectivity from G onto G.

Proof. First of all, note that T is well-defined. For, if X is an infinite cyclic subgroupof G, then just one of its two generators is of the form w"b with n > 0 and b e TSince T is divisible, there exists a c- T such that b = a"; thus X = <(wa)">. And ifX = <(wa)"> = <(wc)m> where n, m c- N and a, c E T, then n = m and a" = cm. So ifv(n) = v(z, n) is as in 2.5.12, then by (28),

(zag)" = z"aPnv(n) = z"anp"(n) = z"cnp"(n) = (zcP)".

This shows that T is a well-defined map from L1(G) to L1(G). If X is a cyclic sub-group of infinite order in G, again one of the two generators of X has the form z"bwhere n e N and b e T. By (a) of 2.5.13 and since j E Pot T, there exists a e T suchthat z"b = (zag)" and hence X = <(wa)" )T. Thus T is surjective. And if <(wa)" )T =<(wc)m)T, then <z"ap""(")) = <zm'cflm"(m)> and since n, m c- N, this implies that n = mand (a")fl"(") = (c")fl"("). Since the map t -- to"(") is an automorphism of T, it followsthat a" = c". Thus T is bijective.

It remains to be shown that r satisfies (29). So let X, Y, Z e L1(G). Note that by2.4.11, any two subgroups of G permute so that < Y`, ZT> = YZZ` and also, of course,<Y, Z> = YZ. If Y and Z are finite, then Y, Z < T and since r is the identity onL1(T), (29) clearly holds for X, Y, Z. Suppose next that Y is infinite and Z is finite.Then Y = <(wa)"> and Z = where n e t\! and a, b e T. Let X < YZ. If X is finite,X < Z and (29) holds. So let X be infinite, that is X = <(wa)"`b'> where i, j e N.Since T is divisible, there exists c e T such that c"` = b'. Thus Y` = <(zap)"> andZr = >_ <c`>; since fi and v(za', ni) are power automorphisms of T, using (28),we obtain

Xi = <(wac)"`>t = <(z(ac)')"`> = <(za,')n(cneev(--',")> < YTZ`.

Conversely, assume that XT < YTZT and X` ZT. Then XT = <(zafl)"kbm> wherek, m c N. Since T is divisible and # and v(nk) are power automorphisms of T, thereexists d e T such that bm = dP"k"("k). By (28),

XT = <(ZaP)nkbm> = <ZnkaPnkv(nk)dOnkv(nk)\ = <(z(ad)P)nk\

and since d"k = bm"(nk) '# ` e Z, it follows that X = <(wad)`> = <(wa)nkdnk> < YZThus (29) holds for X, Y, Z.

Finally suppose that Y and Z are infinite. We shall reduce this case to the onejust dealt with. Now YZ, being a subgroup of G = T x <w>, has the form H x Uwhere H = YZ r T is the torsion subgroup of YZ and U is infinite cyclic. Clearly,H HY/Y is cyclic. By the case already settled, Y` < HT UT and Z` < HT UT, so thatYTZ` = KT V` where K'< HT and VT < HT UT is infinite cyclic. Applying this caseonce again, we get that Y < K V and Z < K V. Thus H U = YZ < K V and thereforeagain H` < KT VT and U` < KT VT. So we finally get that YTZT = KT VT = HT UT. Since(29) holds in this case, X < YZ = HU if and only if XT < HT UT = YTZT. This finishesthe proof of (29) and of the lemma.

2.5.15 Theorem (Sato [1951]). Let G be an M-group with elements of infinite order.Then there exists an abelian group G lattice-isomorphic to G.

Proof. The assertion is clear if G is abelian; so let G be nonabelian. By 2.4.12, wemay assume that T(G) = T is divisible. By 2.5.13, G = TZ where Z is a torsion-free,locally cyclic subgroup of G. Finally, by 2.5.14, we may assume that Z is not cyclic.Thus there exist z, e Z and primes pi such that Z = <z1, z2, ... > and z°+1 = z, for alli e N. Take a group W = <w1, w2, ... > isomorphic to Z via the natural isomorphismmapping z; to w1, let G=Tx W,Gi=T<z1>SGand G1=Tx <we>foralliet N.Then the (G,)iE N form an ascending chain of subgroups of G whose set-theoreticunion is G; similarly for G. Inductively we define power automorphisms F'1 = 1 andA+1 = Av(zi+1, p,)-1 of T where v is as in (27). Then by 2.5.14, for every i e t\!, themap r, = r(wi, zi; $i): L1(Gi) - L1(G,) is bijective and induces a projectivity from G,onto Gi. All the Ti clearly coincide on L1(T). If X is an infinite cyclic subgroup of Gi,then X = <(wia)"> for some n e t\!, a e T and since T is divisible, there exists b e T

such that bP, = a; using (28), we obtain that

X"', = <(w1a)">' = <(w,+1b)P1")T"' = <(z1+1bv'.,)P;")

= <(zP+1 bP"Q'+' v(z-.P'))n> = <(z1aP')n) = X.

Thus Ti is the restriction of Ti+1 to L1(G1). If X is a cyclic subgroup of G, there existsi e N such that X 5 G; and we define X` = X". Then T is a well-defined bijectivemap from L1(G) onto L 1(G). And if X, Y, Z e L, (G), there exists Gk such thatX, Y, Z S Gk and it follows that (29) holds for X, Y, Z and t. By 1.3.2, T induces aprojectivity from G onto G.

Finally we mention that Sato's theorem also implies that every locally finitenonhamiltonian p-group with modular subgroup lattice is lattice-isomorphic to anabelian group (see Exercise 7). However, we chose to present Baer's proof since it isindependent of Iwasawa's theorem on M-groups with elements of infinite order;what is more, it even yields an isomorphism between the coset lattices of the twogroups. Unfortunately, as Exercise 6 shows, Baer's method cannot be used to proveSato's theorem.

Exercises

1. Let G be a group, f : G - End G, a an f-isomorphism from G to a group G andE= {xeGIf(x)=id}. Show that for all x, yeG and ac E,(a) E and E° are subgroups of G and G°, respectively,(b) f(x) = f(y) if and only if x-1 y e E,(c)

(x°)-1a°x° = (x-1at(:)x)°.

2. Let A be an abelian group, f : A - Pot A, a an f-isomorphism from A to a groupB and E = {a e A If(a) = id}. Show that every subgroup of E° is normal in B andthat E° and B/E' are abelian.

3. Prove Theorem 2.5.10.4. Show that an abelian p-group possesses a projectivity onto a nonabelian group if

and only if it is not cyclic and has finite exponent which is at least 8 in case p = 2.5. Let G = T x Z where T is an abelian torsion group and Z is infinite cyclic. Show

that G is lattice-isomorphic to a nonabelian group if and only if T contains anelement of order 8 or there is a prime p > 2 such that T contains an element oforder p2.

6. Let G = T x <w> where T = <a) x with o(a) = o(b) = p" for some prime p,n e t and o(w) is infinite. Show that if a is a crossed isomorphism from G to somegroup G inducing a projectivity from G to G, then G is abelian.

7. (Mainardis [1992]) Let G be a locally finite nonhamiltonian p-group with modu-lar subgroup lattice. Show that there exists an M-group G* with elements ofinfinite order such that G is isomorphic to a factor group of G*. (Hint: Show that,for arbitrary groups, if G = TZ, T:9 G and µ: X - Z is an epimorphism, then Gis isomorphic to a factor group of a semidirect product of T by X.)

2.6 Projectivities between abelian groups 101

2.6 Projectivities between abelian groups

We have seen that many abelian groups admit projectivities onto nonabelian groups(Exercises 2.5.4 and 2.5.5). In this section we shall show that, on the other hand,lattice-isomorphic abelian groups are nearly always isomorphic, and that frequentlyevery projectivity between them is induced by a group-isomorphism. Exceptions tothese classical results of Baer [1939a] are the groups of torsion-free rank 1. Theirprojectivities will be studied at the end of this section.

Element maps derived from projectivities

We want to construct group-isomorphisms from projectivities between abeliangroups. In the definition of the element maps, we do not require the image group tobe abelian. In general it will be an M-group and we shall use the following.

2.6.1 Lemma. Let H be an M-group, H = X u Y with X, Y cyclic and X n Y = Iand suppose that Z is a cyclic subgroup of H such that X u Z = H = Y u Z.

(a) If Z = <xy) where x e X and y e Y, then <x) = X and <y) = Y.(b) If either X is infinite or H is a finite p-group and IXI > I YI, then to every

generator x of X there exists exactly one generator y of Y such that Z = <xy).

Proof. (a) Since <x) u Y >_ Z u Y = H and H is an M-group, X = (<x) u Y) n X =<x) v (Y n X) = <x>. Similarly, Y = < y).

(b) Suppose first that X is infinite. If Y is finite, then Y H and Z is infinite by2.4.8. If Y is infinite, then H is abelian by 2.4.10 and Z clearly is infinite. In bothcases, H = X Y and hence Z = <xy) where x e X, y e Y. Since Z is infinite, xy and(xy)-' = y-'x-' = x-'yk, for some integer k, are the generators of Z. By (a), xor z = x-' and the assertion is clear.

Now suppose that H is a finite p-group and IXI >_ I YI. By 2.3.2, H = XY; letZ = <xy) where x e X, y e Y. By 2.3.5, Exp H = I X I = p", say, and 2.3.8 shows thatQ8 is not involved in H, that is, H is an M*-group. Since x and y are generators ofX and Y, respectively, o(y)5 o(x) = p". By 2.3.10, (xy)""-' = x" 'y""-' it Y and, simi-larly, (yx)""-' 0 Y. It follows that IZI = p" = IXI and <yx) n Y = 1. Suppose thatthere is another element yo e Y such that <xyo) = Z = <xy). Then there existsk e t\l such that xyo = (xy)'` = x(yx)k-' y and hence y0y' _ (yx)k-' e <yx) n Y = 1,that is, yo = y. This shows that to every x e X there is at most one element y e Ywith Z = <xy). Since Z and X have the same number of generators, it follows from(a) that to every generator x of X there is exactly one y E Y such that Z = <xy).

2.6.2 Lemma. Let q be a projectivity from the group G to the group G and let a, b e Gsuch that <a, b) = <a> x . Assume that either o(a) is infinite or G and G are finitep-groups and o(a) >_ o(b). Then if <a>" = <a>, there exists a unique element b e Gsuch that

(1) " = and <ab>" = <ab>.

We write b = f(b;a,a,cp) or, briefly, b = f(b;a,a) since we are never going to changethe projectivity.

Proof. Since <a, b> is abelian, H = <a, b)`° is an M-group and X = <a>9, Y= ",and Z = <ab> ° satisfy the assumptions of 2.6.1(b). It follows that to the givengenerator a of X there corresponds exactly one generator 6 of Y = " such that<ab) = Z = <ab>0.

2.6.3 Lemma. Let G = <a> x B and let cp be a projectivity from G to some group G.Assume that either o(a) is infinite or G and G are finite p-groups and o(a) > Exp B. Let<a>0 = <a> and define u: B -- B9' by b° = f(b; a, a) for b e B. Then a is bijective andinduces cp on B.

Proof. For every b E B, <a, b> satisfies the assumptions of 2.6.2. Hence f (b; a, a) is auniquely determined element of " < B' and thus a is a well-defined map fromB to B`0. If b,, b2 E B such that bi = bz, then <ab, )`° = <ab; > = <abZ> = <ab2>0.Since cp is bijective, <ab,> = <ab2> and there exists an integer k such that ab, _(abZ)" = a"bk. It follows that a = a'` and hence k = 1 if o(a) is infinite and k =_ 1(mod o(a)) if o(a) is finite. In both cases, b, = bZ = b2. Thus a is injective. To showthat u is also surjective take c c B`° and let Y, Z < G such that Yq' = <c> andZ0 _ <ac>. If X = <a>, then X`° u Y`° = <a, c> = X`° u Z`° = Y`° u Z`° and henceH = X u Y = X x Y satisfies the assumptions of 2.6.1(b). It follows that to thegenerator a of X there is a generator b of Y such that (ab> = Z. So (b)`° = Yq, =<c> and <ab> = Z`° _ <ac>, that is, c = f(b; a, a) = V. Thus a is surjective. Finally,(1) shows that <b°> _ <f(b;a,a)> = 0 for all b e B and it follows that H`0 = H°for all H < B. Thus a induces q on B.

2.6.4 Lemma. Under the hypotheses of 2.6.3, assume moreover that G and G areabelian and let b, c E B.

(a) If n <c> = 1, then (be)' = b°c°.(b) If n <c> = <bc> n <c> = 1, then (c")° = (c°)" for all n c- Z.

Proof. (a) Since G and G are abelian, <a, b, c> = <a> x x <c> and <a, b, c)4' =<a>`° x " x <c> °. Furthermore b° = f(b; a, a), c° = f(c; a, a) and (bc)° =f(bc; a, a) are by definition generators of 4', <c> and <bc)`°, respectively, suchthat <ab>° = (ab°>, <ac>0 = <ac°) and <abc>0 = <a(bc)°). If o(a) is infinite, thenalso o(ac) is infinite and if o(a) is finite, then o(ac) = o(a) > o(b). In both cases,(ac, b> = <ac> x so that by 2.6.2 there exists b = f (b; ac, ac°) E q, with<acb>0 = <ac°b). Similarly, we have c = f(c;ab,ab°) E <c>0 with <abc>0 = <ab°c).Hence

(2) <abc> ° = <a(bc)a> = <a6c°> = <ab°c>

and there exist integers r, s such that a(bc)° = (abccy = (ab°c)s. Since a e <a>°,b, b° E 0, c, c° E <c> P and the product of these three groups is direct, it followsthat a = a' = as, 6' = (b°)s and (bc)° = (6c°)'. If o(a) is finite, this implies thatr - s = 1 (mod o(a)) and, since o(a) = Exp G, that x' = x = xs for all x c- G. If o(a) isinfinite, r = s = 1. In both cases, it follows that b = b° and (bc)° = 6c° = b°c°.

(b) Clearly, (c")° = (c°)" holds for n = 0 and n = 1. Suppose it is true forn - 1 E N. Then by (a),

b°(c")" = (bc")" = (bcc"-')o = (bc)o(c"-1)" = baca(co)"-i = bo(co)n

and hence (c")° = (c°)". This proves the assertion for n > 0. Again by (a),b° = (bc)"(c-')° = b°c"(c-')" and hence (c-')° = (c°)-' Since b-' and c-' satisfythe same assumptions as b and c, it follows that for n > 0, (c-")° =

((C-1)")" = (Co)-n

2.6.5 Lemma. Let B and b be abelian groups and suppose that a: B -+ B is a bijectivemap satisfying (c")° = (c°)" for all c e B, n e 7L and (bc)° = b°c" for all b, c e B with n <c> = 1. Then a is an isomorphism.

Proof. If x, y e B, then by the structure of finitely generated abelian groups there areb, c c- B such that <x, y) = <b) x <c>. Hence x = b'c', y = bres where i, j, r, s areintegers and our assumptions yield that (xy)° = (b'+rC;+s)o = (bo)i+r(coy+s = xoyo

We summarize the results obtained so far.

2.6.6 Lemma. Let the hypotheses of 2.6.3 hold. Assume in addition that G and Gare abelian and that to every c e B there exists b e B such that n <c) _<bc) n <c) = 1. Then a is an isomorphism inducing cp on B.

Proof. By 2.6.3, a is bijective and induces cp on B. Then 2.6.4 yields that a satisfiesthe assumptions of 2.6.5 and therefore is an isomorphism.

Torsion groups

We can now prove our first main result on projectivities between abelian groups.

2.6.7 Theorem (Baer [1939a]). Let G be an abelian p-group with the followingproperty:

(3) If G contains an element of order p", then G contains at least three independentelements of this order.

Then any projectivity cp of G onto an abelian group G is induced by an isomorphism.

Proof. Let 9 be the family of all finitely generated subgroups of G that satisfy (3).Clearly, 9 is a local system of subgroups of G and by 1.3.7, it suffices to prove thatq is induced by an isomorphism on every X E 9. Thus we may assume that G itselfis finitely generated and hence a finite group.

Our assumptions then imply that G = <a> x B where o(a) = p" = Exp G and Bcontains two independent elements of order p". Hence if c E B, there exists b E B such

that o(b)_= p" and n <c> = 1, and therefore also <bc> n <c> = 1. Since G isabelian, G = <a>0 x B° and by 2.2.6, G is a p-group. So all the assumptions of 2.6.6are satisfied and if we choose a e G with <a> = <a>", then a: B - B0 given byb° = f(b; a, a) for b e B is an isomorphism inducing qp on B. We define r: G -, G by(a'.b)' = a'b° for b E B and i e N. Then r is clearly an isomorphism as well and wewant to show that it induces q,. Since every group is the join of its cyclic subgroups,it suffices to prove that

(4) <a'b>0=<(a`b)T>forallbEB,ieNl.

So let b e B and take c e B with o(c) = p" and <c)' n = 1. By 2.3.11 there existsa complement D of <c> in G containing <a, b>. Hence G = <c> x D and by 2.6.6,the map v: D -' D0 defined by d" = f(d; c, c") is an isomorphism inducing cp on D.Since a induces (p on B, <cb>0 = <(cb)°> = <c°b°> and the definition of f in 2.6.2shows that b° = f(b;c,c°) = by. Similarly,

<ca>0 = <ac>0 = <af(c; a, a)> = <ac°) = <c°a)

implies that a = f(a; c, c") = a". Since v induces cp on D, we finally get that

<a'b>0 = <(a`b)"> = <(a")'b"> = <a'b"> = <(a'b)`>.

Thus (4) holds and r induces q,.

Baer's theorem shows in particular that every autoprojectivity of an abelian p-group having property (3) is induced by an automorphism, that is, P(G) = PA(G) forsuch a group G. An elementary abelian group of order p" clearly satisfies (3) if n > 3.Thus Baer's theorem yields a proof of Theorem 1.4.4 that is independent of theFundamental Theorem of Projective Geometry; in fact, it is a wide generalization ofthis result.

For finite abelian p-groups, condition (3) means that the type of G is of the form(p",_., with r > 3, n1 = n2 = n3 = n and p" = ExpG. It is easy to see thatthis condition is almost necessary if G is to have every projectivity induced by anisomorphism. We illustrate this in Exercises 3-5 and mention that Holmes [1988]determines P(G) for a group G of type (p", p') completely. It is also clear that twolattice-isomorphic finite abelian p-groups are isomorphic, as one can tell from thesubgroup lattice the type of the group. In fact, it is a consequence of Theorem 2.6.7that any two lattice-isomorphic abelian p-groups for a fixed prime p are isomorphic.

2.6.8 Theorem (Baer [1939a]). Let G and G beabelian p-groups for some prime pand suppose that qp is a projectivity from G to G. Then there exists an isomorphisma: G G inducing cp on H = <S2m(G)I SZm(G)/0rn-1(G)I > p3).

Proof. If x e H is of order p", then 1,,(H) = S2"(G) is a direct product of cyclic groups(see Robinson [1982], p. 105) and since IS2"(G)/S2n_1(G)I > p3, at least three of thedirect factors are of order p". By 2.6.7 there exists an isomorphism r: H -+ H10inducing cp on H. If H = G, we are done. So suppose that H < G and take a basic

subgroup B of G. Then B is a direct product of cyclic groups and G/B is a directproduct of quasicyclic groups. Since H < G, B has finite exponent and therefore is adirect factor of G (see Robinson [1982], p. 106). Thus G is a direct product of cyclicand quasicyclic groups and it follows that G = X1 xX2 x Y where the Xi are cyclic(possibly trivial) or quasicyclic and Y< H. Then G = X' x Xz x Y' and clearlythere are isomorphisms a;: X; - Xr extending the restrictions of T to Xi n H. Thusthere is an isomorphism a: G - G extending T.

Using 1.6.6, it is not difficult to extend our results from p-groups to arbitrarytorsion groups (see Exercise 1).

Groups with elements of infinite order

We want to show that if G is an abelian group of torsion-free rank ro(G) >_ 2, thenevery projectivity of G is induced by an isomorphism. To prove this we must firsthandle the following special case.

2.6.9 Lemma. If G = <a> x where o(a) and o(b) are infinite, then every projec-tivity (p from G to some group G is induced by an isomorphism.

Proof. By 2.4.10, G is abelian and hence G = <a>" x ' -_ G. Let <a> = <d> andput b = fl b; a, a) where f is the function defined in 2.6.2. Then v = , G =<a> x and we want to show that the isomorphism T: G -+ G mapping a to a andb to & induces cp.

For k E N, q induces a projectivity from G/<b'"> = <a, b')/<bk> x /<b"> toG/<bk )`° and since <a, bk)/<bk) is infinite cyclic, I /<bk> I = I '/<bk> 0I by 2.6.3.It follows that <bk>w = <bk> and, similarly, <ak>,, = <ak>

Let n, m e 7L\{0}. Then by (1), x = f(bm; a", a") is a generator of <bm> satisfying<a"bm)'0 = <a"x>. Since bm and b-m are the only generators of <bm)`0, it follows thatthere exists a sign E(n, m) e { + 1, -1 } such that

(5) <anbm> = <anbe(n,m)m\

We have to show that E(n, m)) = 1 for all n, m since this will imply that X' = XT forall cyclic subgroups X of G and hence that cp is induced by T. First of all,

(6) E(n, m) = E(- n, - m) = E(n, - m).

For, <a-"b-m\N = <anbm> = <anbs(n.m)m> = <a-nbe(n.m)(-M)> yields E(n,m) =

E(-n, -m), and 9(n, m) = - E(n, -m) would imply that <a"b-m >11 = <a"be(" m)m \ _<a"bm)m, a contradiction. We show next that

(7) E(n, m) = E(n - 1, m) for n > 2.

For, since <a> n <a"-lbm> = 1, there exists y = f(a"-lbm;a,a) satisfying <y> =<an-lbm// \M = <an-1be(n-l,m)m> and <ay> = <a(an-lbm)>dl = <anbe(n.m)m\. Hencethere exist u, v e { + 1, -1} /such that al+"(n-1)b 1e(n-1,m)m = avnbve(n,m)m /It follows

that 1 - it = (v - u)n and ys(n - 1, m) = ve(n, m). Since n > 2, the first equation

implies that v = e and the second then yields (7). Finally we claim that

(8) e(n, m) = e(n, m - 1) form >- 2.

Indeed z = f(a"b'"-l;b,b) satisfies <z> = <a"b'n-1)`0= <a"be(R,m-')('"-')) and

<bz) = <ba"b'-' >0 = <a"be(" so that there exist u, v e (+ 1, - 1} such thataµnbµe(n,m-1)(m-1)+1 = avnbve(n.m)m This implies that µ = v and µe(n,m - 1) - I =µm(e(n, m - 1) - E(n, m)); since m > 2 it follows that e(n, m - 1) = e(n, m).

By induction, (7) and (8) yield that e(n,m) = E(1, 1) for all n, m e N. Sinceb = f(b; a, a), <ab>0 = <ab> and hence E(l,1) = 1. By (6), e(n, m) = 1 for alln,me7L\{0}.

2.6.10 Theorem (Baer [1939a]). Let G be an abelian group which contains twoelements a and b of infinite order such that <a> n = 1. Then any projectivity of Gis induced by exactly two isomorphisms.

Proof. Let p be a projectivity from G to some group G. We have to show that (P isinduced by an isomorphism; by 1.5.7, 1 Pot G I = 2, so that p will then be induced byexactly two isomorphisms. By 2.4.10, G is abelian.

We note first that 2.6.9 has a number of immediate consequences for thefunctions f( ; a, a) defined in 2.6.2. Let a, b e G be elements of infinite order such that<a> n = 1; let <a>0 = <a> and put b = fl b; a, a). If r is an isomorphism from<a, b> to <a, b>0 inducing q there and if x, y e <a, b> such that <x> n <y> = 1, then<y>'P = <y`> and <xy>0 = <(xy)`> = <x`y`> so that

(9) fly; x,xI) = y`.

Exactly one of the two isomorphisms inducing (p on <a, b> satisfies at = a. It followsfrom (9) that for this T, b` = fl b; a, a) = b and hence that

(10) a = fl a; b, b) and ab = flab; a,a).

Again by (9), since (x")` = (x`)" for every integer n,

(11) f (x'; a, a) = f (x; a, a)" for all x e <a, b> with <x> n <a> = 1.

Finally we claim that

(12) f(x; a, a) = f(x; b, b) for all x c- G with <x> n <a> = I = <x> n .

For, if 1 x" e <a, b> for some n E t%J, it follows from (9) that f (x'; a, a) = (x")` _f (x"; b, b) and then (11) applied to <a, x> and <b, x> yields that

f (x; a, a)" = fl x"; a, a) = fl x"; b, b) = f (x; b, b)".

This implies f (x; a,a) = f (x; b, b) since both elements generate the same infinite cyclicgroup <x>0. And if <x> n <a, b> = 1, then 2.6.4(a) and (10) yield that

f(xab;a,a) = f(x;a,a)f(ab;a,a) = f(x;a,a)ab.

By (1), this implies that <abx>0 = <abf(x; a, a)> and hence f(x; a, a) = f(x; ab, ab).Similarly, f (x; b,b) = f (x; ab, ab) and this yields (f2).

Now we define a map a: G -+ G that will turn out to be an isomorphism inducingcp. We choose a fixed pair u, v of elements of infinite order in G such that n <v> = 1, take u e G with ' = and put u = f(v; u, u). Let x e G. If<x> n = 1, we define x° = f(x; u, u); and if <x> n : 1, then <x> n <v> = 1and we define x° = f(x; v, v). By 2.6.3, a1, is a bijective map from X to X' for everycyclic subgroup X of G and since q is a projectivity, this implies that a is bijectiveand induces cp. We claim that for every w e G of infinite order with n <w> = 1,

(13) x° = f(x; w, w°) for all x e G with <x> n <w> = 1.

For, w° = f(w; u, u). So if <x> n = 1, then x° = f(x; u, u) = f(x; w, w°) by (12).And if <x> n # 1, there exist n, m e 7L such that x" = um and (11) and (10) yieldthat

R X; w, w°)" = fix"; w, w°) = ./ (um; W, W°) = f (u; W, W°)m = u'.

Thus f(x; w, w°) is the unique generator of <x>" whose n-th power is um, and sincethis description is also true of f(x; v, v°) = x°, we obtain x° = f(x; w, w').

Now we are able to prove that a is an isomorphism. Let x, y e G and suppose firstthat <x, y> Cm x C. Since in such a group there exist infinitely many infinitecyclic groups, every pair intersecting trivially, there is 1 # w e <x, y> such that<w> n <x> = <w> n <y> = <w> n <xy> = <w> n = 1. If r: <x, y> - <x, y>° isthe isomorphism inducing cp on <x, y> and satisfying wt = w°, then by (13) and(9), z° = f (z; w, w°) = f (z; w, w`) = zt for every z e {x, y, xy}. Thus (xy)° = (xy)T =x`y` = x°y°. Now suppose that <x, y> * C,,,, x C. Then by the structure of finitelygenerated abelian groups, there exists a subgroup B of G of torsion-free rank 1 suchthat <x, y> < B. If n B = 1, take w = u; if n B ; 1, take any w e G of infi-nite order satisfying <w> n B = 1. In both cases, c° = f(c; w, w°) for all c e B. Itfollows from 2.6.4 that (bc)° = b'c' for all b, c e B with n <c> = 1 and that(c")° = (c°)" for all c e B of finite order; if o(c) is infinite, then (c")° = f(c"; w, w°) _f(c; w, w°)" = (c°)" by (11). Thus 2.6.5 shows that ajB is an isomorphism and hencethat (xy)° = x°y°.

In view of Baer's theorem it remains to study projectivities between abeliangroups of torsion-free rank 1; we start with the torsion-free groups of this type.

Torsion-free groups of rank one

The structure of these groups is well-known (see Robinson [1982], p. 112); we reviewthe basic facts. In general, if G is an abelian group, written additively, an elementg e G is said to be divisible in G by a positive integer m if g = mg, for some gl a G. Ifp' is the largest power of the prime p dividing g, then h is called the p-height of g inG; should g be divisible by every power of p, we say that g has infinite p-height in G.Let pl, P2, ... be the sequence of primes written in their natural order. Then theheight of g in G is the vector l)(g) = (hl,h2,...) where h; is the p; height of g inG. Two such vectors b and f are called equivalent if hi = k, for almost all i and h, = ki

whenever hi or k, is infinite. This is an equivalence relation on the set -5 of all suchvectors, and the equivalence classes are termed types. The type of a group element gis defined to be the type of its height vector. If G is a torsion-free abelian group ofrank 1, then all nonzero elements of G have the same type, which is referred to as thetype of G, in symbols t(G). The structure theorem now states that two torsion-freeabelian groups of rank 1 are isomorphic if and only if they have the same type.Furthermore, every such group is isomorphic to a subgroup of the additive group 0of rationals containing the integers. So we only have to study projectivities of thesesubgroups of 0 and the basic tool here is the following.

2.6.11 Lemma. Let 71 < G < 0 and suppose that cp is a projectivity from G to somesubgroup G of 0. If ° = <p) for all p e P, then G = G and co is the trivial auto-projectivity of G.

Proof. We want to show that cp fixes every cyclic subgroup of G. First of all, Z10 = 71since 71 = <2> u <3>; in particular, 71 5 G. If p is a prime, r e IN and s a nonnegativeinteger such that p_5 a G, then <p') < 71 5 <p-') S G and <p-')/<p') is cyclic oforder p'+'. By 1.2.8, <p-')'/<p')' is cyclic of order q'+, for some prime q and asI DOI <p)" I = I71/<p) I = p divides the order of this group, q = p. Since 71' = 71, itfollows that <p' ),, = <pr > and <p-' )°' = <p-' ). Next, suppose that m = pr, ... pkk is

1

the primary decomposition of the positive integer m. Then <m> = <pil) n ... n <pkk )and hence <m)" = <m>. If m-1 e G, then also pi e G for all i, and <m-1) _<p'') u u <pk'k) implies that <m-1 )°' = <m-1 ). Finally, if m,

(!)n e71 are coprime

and e G, then <m> and are subgroups of G and the group is the unique

subgroup H of (!) satisfying (1) = H u 71 and H n 71 = <m>. Since (_), 71 andn n n

are invariant under(/m m m

<m> cp, it follows that\ =

(). Thus every cyclic sub-n n

group of G is fixed by cp and hence G 5 G. If we apply this result to c0-1, we get thatG = G and then cp clearly is the trivial autoprojectivity of G.

2.6.12 Corollary. If 71 < G < Q and cp and ,1i are projectivities from G to some groupG such that <p)IP = " for all p e P, then cp = 0.

Proof. Apply 2.6.11 to the autoprojectivity c*_' of G.

We are now able to determine the autoprojectivities of 0; the structure of P(G) foran arbitrary torsion-free abelian group G of rank 1 is given in Exercise 7. All thesegroups possess autoprojectivities that are not induced by group automorphisms.

2.6.13 Theorem (Gasparini and Metelli [1984]). The group P(Q) of autoprojectivi-ties of 0 is the semidirect product of PA(G) by S = (co e P(Q)I 1W = 71). Furthermore,PA(0) is isomorphic to the multiplicative group of positive rational numbers andS - Sym N.

Proof. It is well-known and easy to prove that Aut 0 is isomorphic to the multi-plicative group of rational numbers: for q e 0\{0}, the multiplication q: x - qx is anautomorphism of 0 and every automorphism is of this form. If we write pq for theautoprojectivity induced by q, then

(14) <x>Pa = <qx> for all x e 0

and by 1.4.1, the map p: q pq is an epimorphism from Aut 0 to PA(0). Sincekerp = (1, -1 }, it follows that PA(Q) is isomorphic to the multiplicative group ofpositive rational numbers.

Let cp e P(Q). Since 7L is cyclic, 7L'' = <q> = 7LP9 for some q e 0. Hence p, ,-'(p e Sand cp e PA(QQ)S. If cp e PA(Q) n S, then cp = p, for some r e QQ and 7L = ZPr = <r>.It follows that r e 1+ 1, -11 and cp is the trivial autoprojectivity. Thus P(Q) _PA(QQ)S and PA(Q) n S = 1.

We show next that S Sym N. For this recall that pl, pZ, ... is the sequence ofprimes written in their natural order. If a E Sym 101, we define a map 6: 0 - 0 by

(15) 0°=0,(ii = [1 P i)fork,aZ,naNand(-x)'= -x' for x > 0.°t=1 / i=1

Clearly, Q is bijective and satisfies (xy)° = x°y" for all x, y c- Q. Furthermore 7L° = 7Land therefore <x>° = (x7L)6 = x°7L = <x°> for all x e Q. Since every finitely gener-ated subgroup of 0 is cyclic, Theorem 1.3.3 shows that Q induces an autoprojectivity0° of 0 and as 7L° = 7L, 0° E S. We claim that the map 0: Sym N - S; a ' is anisomorphism. Clearly,

(16) <x>O° = <x°> for all x e 0;

in particular, <pi>"e = <p6> = <Pa(j)> for all i and this shows that 0 is injective.Every cp e S permutes the maximal subgroups of Z. So if we define T E Sym N by<Pt>' = <PT(i)>, then <pi>" = for all i and by 2.6.12, (p Thus 0 issurjective. And if a, r E Sym N, then

<A>11111 = <P.(i)> "= <P,(a(i)) ) = <Pt>0-

for all i and again by 2.6.12, '°7L Thus 0 is an isomorphism. Finally, forge0,aaSym101andpEP,

'-"- = <qP> "= <(qp)°> = <q°P°> =<P°>P = 1'.Pv,

and hence, again by 2.6.12, t/rs 1 pqO° = pq, e PA(QQ). Since P(O) = PA(U)S, itfollows that PA(Q) a P(0) and P(0) is the semidirect product of PA(O) by S.

We finish our study of torsion-free abelian groups of rank 1 by showing that twosuch groups are lattice-isomorphic if and only if their types can be obtained fromeach other by a suitable permutation of the primes. Note that Sym N operates onthe set , of height vectors by h° = (h°-,(1),h°-l(2),...) for 1) = (hl,h2i...) E.5 anda e Sym N. If h and t are equivalent, then so are If and t° and it follows that for anytype t, t° = {E)°14 E t} is again a type. Thus Sym N also operates on the set of types.

For example, if 7 < G 5 Q and t is an integer, then p; E G if and only if p"'m = (p; )° C

G° = GO-. This shows that h; = k°(i) if 1) = (hl,h2.... ) and t = (k, k,,...) are theheight vectors of 1 in G and GO respectively. Hence k. = h°-,(i) for all i, that is t = 1)°and it follows that

(17) t(GO°) = t(G)° for all 7L < G < 0 and a e Sym 101.

2.6.14 Theorem (Fuchs [1960]). Two torsion free abelian groups G and G of rank 1are lattice-isomorphic if and only if there exists a permutation a e Sym t\J such thatt(G)° = t(G).

Proof. Let G and G be torsion-free abelian groups of rank 1; we may assume that Gand G are subgroups of 0 containing Z. First suppose that G and G are lattice-isomorphic and consider a projectivity (p from G to G. Then r is a cyclic subgroupof G and hence l`° = (q> for some q e Q. Since (p maps every maximal subgroup of7L to a maximal subgroup of (q>, there exists a permutation a e Sym 101 such that(pi )'° = (p°(,)q> for all i e N. Let be the projectivity induced by (Ii° pq)-' in G.Then cpc is a projectivity from G to some subgroup G'14 of 0 satisfying

(pi>"' = (p°(1)q)°°'11e` _ (p°(i))'4°' = (pi)

for all i. By 2.6.11, G`° = G and hence G = G° = GO-°Q. Since pq is induced by anisomorphism, G GO- and therefore by (17), t(G) = t(G'") = t(G)°. For later use, werecord what has been proved so far:

(18) If Gand G are torsion-free abelian groups of rank 1 and cp is a projectivityfrom G to G, then t(G) = t(G)° where a e Sym N satisfies <x>11 = (y> and (pix>c _(p°)i)y> for a suitable nonzero element x e G and all i e N.

To complete the proof of Theorem 2.6.14, we suppose conversely that there existsa e Sym 101 such that t(G)° = t(G). Then t(GO°) = t(G)° = t(G). It follows that GO- isisomorphic to G and hence that G and G are lattice-isomorphic.

Theorem 2.6.14 shows that 0 is determined by its subgroup lattice, whereas, forexample, all the groups UP of rational numbers with denominator a power of theprime p are lattice-isomorphic but pairwise nonisomorphic.

Mixed groups of torsion-free rank one

It is an open problem to determine the projective closure of an arbitrary abeliangroup G of torsion-free rank 1. Using the methods developed in this section, we givesome necessary conditions for an abelian group to be lattice-isomorphic to G. Weresume the multiplicative notation and write Go for the p-component of the torsionsubgroup T(G) of G. Define

MP(G) = (S2i(Go)IS2i(Go)1S2i-1(Go) is not cyclic>

and put M(G) = (MP(G)l p E P>.

2.6.15 Theorem. Let G be an abelian group of torsion free rank 1 and let cp be aprojectivity from G to an abelian group G. Then

(a) T(G) T(G),(b) cp is induced by an isomorphism on M(G),(c) G/T(G) is of rank 1 and there exists u c Sym N fixing every i c N for which

Gp, 1 such that t(G/T(G))° = t(G/T(G)).

Proof. Since cyclic groups are preserved under projectivities, T(G)`° = T(G) andro(G) = ro(G) = 1. Let a E G be of infinite order, take a c G such that <a>" = <a>and consider the map v: T(G) - T(G) given by b" = f(b; a, a) where f is the functiondefined in 2.6.2. By 2.6.3, v is bijective and induces cp on T(G). In particular, I (b) I =I(b)'I for all b c T(G) and hence Gp is the p-component of T(G). By 2.6.8, Gp Gp

and therefore T(G) = Dr G`° T(G).pe P

If c c MP(G) and o() = p", say, then S2"(Gp) < MP(G) and S2"(GP)/S2"_,(GP) is notcyclic. Now S2"(GP) is a direct product of cyclic groups and it follows that thereare at least two direct factors of order p". Hence there exists b c S2"(GP) such that(b) n <c) = 1 = <bc) n <c). By 2.6.6, the restriction of v to MP(G) is an isomor-phism inducing cp on MP(G). It follows that VIM(G) is an isomorphism inducing cp onM(G) = Dr MP(G).

pE P

_ Finally, cp induces a projectivity i-p between the torsion-free groups G/T(G) andG/T(G) of rank 1. By (18), t(G/T(G)) = t(G/T(G))° where u c Sym N satisfies <x)`° _(y) and <x"> _ (yP-(-)> for a suitable nontrivial element x E G/T(G). Take u c Gwith (x) = <uT(G)>. Then o(u) is infinite and <y) = <uT(G)) where (u) = (u)`°. IfG contains an element b of order p, then <u,b)/<uP) is elementary abelian of orderp2 and by 2.2.5, <u,b)m/<uP)`°j = p2 as well. Hence <uP)`° = <u"> and it followsthat (xP)' = (y">. This shows that po(i) = p; for those i for which GP, 1.

If G does not split over T(G), then G and G = T(G) x (G/T(G)) satisfy (a) and (c)of Theorem 2.6.15; but clearly these are not lattice-isomorphic. For groups that splitover their torsion subgroups, however, our conditions are necessary and sufficient.

2.6.16 Theorem (Ostendorf [1991]). Let G be an abelian group of torsion free rank 1that splits over its torsion subgroup. Then the abelian group G is lattice-isomorphic toG if and only if G splits over its torsion subgroup and satisfies

(a) T(G) T(G),(b) G/T(G) is of rank 1 and there exists a E Sym 101 fixing every i c 101 for which

Gp, 1 such that t(G/T(G))" = t(G/T(G)).

Proof. If cp is a projectivity from G to Gand G = T(G) x_H, then G = T(G)`° X H`° _T(G) x H° since G is abelian. Thus G splits over T(G) and the other assertionsfollow from 2.6.15. Conversely, suppose that G and G satisfy the conditions of thetheorem. Then G = T x R and G= T x R where T, T are isomorphic torsiongroups, R, R are torsion-free of rank 1 and there exists u E Sym 10J fixing every i E 10Jfor which TP, 1 such that t(R)° = t(R). Then R is isomorphic to a subgroup H of0 containing 7L and by (17), t(HO") = t(H)° = t(R)° = t(R) where Wn is the auto-

projectivity of 0 defined in (15) and (16). Hence H10° R and, since we only want toshow that G and G are lattice-isomorphic, we may assume, using additive notationagain, that G = T e H and G= T e K where T is a torsion group, 7L < H< 0 andK = HO'- = W. We want to construct a bijective map r from the set L1(G) of cyclicsubgroups of G to L1 (G) such that for all X, Y, Z E L1(G),

(19) X < Y + Z if and only if X` < Y` + Z`.

Then by Theorem 1.3.2,,r will induce a projectivity from G to G. For this let X be acyclic subgroup of G. If X < T, we define X` = X, that is, r is the identity on L1(T).Let X 4 T. Then X can be written uniquely in the form X = <a + h> where a E Tand 0 < h E H. Since a fixes the primes involved in T, the definition of it in (15) shows

hathat = g;' ... q" where n; E 7L and the g; are primes not dividing o(a). It follows

that hQa = gi' ... g; "a is a well-defined element of <a>; here g"a, for n < 0, is theh

unique element b E <a> satisfying g-"b = a. We put X` _ ( ha + h° . This defines

a map from L1(G) to L1(G) which is bijective since there is an obvious inverse map

defined in the same way to Q-' = Q-', that is, sending (a + k> to (-j-a + k°-`) for

a E T and 0 < k E K. We want to show that r satisfies (19). So let X, Y, Z EL1(/G).

IfY and Z are finite, then Y, Z < T and since r is the identity on L1(T), (19) clearlyholds for X, Y, Z. Suppose next that Y = (b> is finite and Z = <a + h> is infinite.Then X = <ib + j(a + h) > where i, j c- Z. If j = 0, then X` = X < Y = Y`; and ifj 0, since Q(2-_(ibis multiplicative,

(--b / \ \X` = + ja) + (jh)°, _ + jI Qa + h° I) < Y` + Z`.

Similarly, X` < Y` + Z` implies that X < Y + Z and hence (19) /holds. Finally sup-pose that Y and Z are infinite. Since ro(G) = 1, Y + Z = U e R where U and R arecyclic and U is finite. As in the proof of Lemma 2.5.12, the result just proved showsthat Y` + Zr S U` + R`. Hence Y` + Z` = V` e S` where V` < U` and S` 5 U` + R`is infinite cyclic. It follows that U + R = Y + Z < V + S and hence U` + R` SV` + S`. Thus Y` + Zr = U` + R` so that this case is reduced to the one just settled.This finishes the proof of (19) and of the theorem.

The above theorem has been proved independently and generalized to largerclasses of mixed abelian groups of torsion-free rank 1 by Mahdavi and Poland[1992], [ 1993].

Exercises

1. Let G be an abelian torsion group with the property that if G contains an elementof order n, then G contains at least three independent elements of this order.

Show that every projectivity from G to an abelian group G is induced by anisomorphism.

2. Let p > 2, n > 2 and let G be the abelian group of type (p", p). Determine P(G)and PA(G).

3. Let G be an abelian group of type (2", 2, ... , 2). Show that every projectivity fromG to an abelian group G is induced by an isomorphism.

4. Suppose that G = <a> x B is an abelian group, o(a) = p" > p' = Exp B and thatm > 2 if p = 2. Let N = <a""'> and let a be the automorphism of G/N satisfying(bN)° = bN for all b c- B and (aN)° = (aN)1+p'"-' if m > 2 and (aN)° = a2N ifm = 1.(a) Show that there is a unique autoprojectivity cp of G that fixes every subgroup

of L2,-, (G) and is induced by a on G/N.(b) This projectivity is not induced by an automorphism of G.

5. Let p > 2 and G = A x B be an abelian group with A of type (p", p") andExp B = p' < p". For every subgroup N of order p of A let aN be an automor-phism of GIN fixing every subgroup U/N of G/N for which fl(A)5 U or U<n"1(G).(a) Show that there is a unique autoprojectivity cp of G that fixes every subgroup

of L2,-, (G) and is induced by QN on G/N for every N.(b) Find automorphisms aN such that cp is not induced by an automorphism of G.

6. Let Z < G S Q. Show that every projectivity cp from G to some subgroup of 0 isinduced by a unique autoprojectivity of 0.

7. (Gasparini and Metelli [1984]) Let Z < G 5 G. Show that P(G) is the semi-direct product of PA(G) and a group isomorphic to the subgroup{Q E Sym 101It(G)' = t(G)} of Sym N.

Chapter 3

Complements and special elementsin the subgroup lattice of a group

The main subject of this chapter is groups with complemented subgroup lattices.Since not much is known about these groups in general, various stronger comple-mentary conditions and related lattice-properties are also considered. All these arefar less important than distributivity or modularity since there are no natural classesof groups connected with them.

The group G is called a K-group if its subgroup lattice is complemented, that is, ifto every H < G there exists K < G such that G = (H, K> and H n K = 1. We showin § 3.1 that G is a K-group if and only if its Fitting subgroup F(G) is a direct productof abelian minimal normal subgroups of G and has a complement that is a K-group;from this we derive a characterization of soluble K-groups. In particular, we get thata finite soluble group G is a K-group if and only if 1(G/Fi(G)) = I for all i >_ 0; thiswas proved by Zacher in 1953. However, the structure of finite K-groups in generalis not known. Therefore, in § 3.2, we study groups G in which every subgroup H hasa complement K satisfying G = HK; groups with this stronger property are calledC-groups. As early as 1937 P. Hall proved that the finite C-groups are precisely thefinite supersoluble K-groups and determined the structure of these groups. We gen-eralize his result to infinite groups and also consider groups with the weaker prop-erty that every subgroup H has a sectional complement, that is a subgroup K suchthat X n K is a complement to H in X for every subgroup X of G containing H.

In § 3.3 we mainly investigate groups with relatively complemented subgrouplattices and IM-groups. A group G is called an RK-group (or RC-group) if for allL, H, M < G such that L < H < M there exists K < G such that H n K = L and(H, K> = M (or HK = M, respectively). In 1952 Zacher proved that a finite groupis an RK-group if and only if it has elementary abelian Sylow subgroups and nor-mality is a transitive relation in every subgroup; in particular, such a group issupersoluble. We further show that the RC-groups are precisely the soluble (orlocally finite) RK-groups and have a similar structure to finite RK-groups. An IM-group is a group in which every proper subgroup is the intersection of maximalsubgroups. Clearly, every RK-group is an IM-group. The classification of finitesimple groups yields that every finite IM-group is soluble. The structure of arbitrarysoluble IM-groups was determined by Menegazzo in 1970. It follows from his resultthat a finite group is a (soluble) IM-group if and only if it is an RK-group.

The last two sections of this chapter are concerned with some isolated topics. Anelement a of a lattice L is called neutral if every triple a, x, y of elements in Lgenerates a distributive sublattice; equivalently a is join-distributive (that is, satisfies

3.1 Groups with complemented subgroup lattices (K-groups) 115

a u (x n y) = (a u x) n (a u y) for all x, y c L), meet-distributive (defined dually) anduniquely complemented (that is, a u x = a u y and a n x = a n y implies x = y for allx, y c L). We investigate these and similar properties in subgroup lattices of groupsand, as our final result in § 3.4, present the characterization of neutral elements in thesubgroup lattice of a finite group that was given by Suzuki and Zappa, indepen-dently, in 1951.

A partition of a group G is a set E of nontrivial subgroups of G such that everynonidentity element of G is contained in a unique subgroup X E E; it is called non-trivial if X G for all X E L. Possession of a nontrivial partition is a lattice-propertyof a group. In 1961 Baer and Suzuki determined all finite groups having a nontrivialpartition; their results are presented in § 3.5. It is shown, among other things, that theprojective special linear groups PSL(2, p") and the Suzuki groups Sz(2") are the onlyfinite simple groups with this property.

3.1 Groups with complemented subgroup lattices (K-groups)

Let L be a lattice with least element 0 and greatest element I. For a E L, an elementc e L is called a complement to a in L if a n c = 0 and a u c= I; for a group G, wesay that K is a complement to H in G if it is a complement to H in L(G). The latticeL is said to be complemented if every element of L has a complement in L. In thissection we want to study groups with complemented subgroup lattices, K-groups forshort. First of all we note an obvious necessary condition for a group to have thisproperty.

3.1.1 Lemma. The Frattini subgroup of a K-group is trivial.

Proof. Let 1 x c- G and take a complement K to <x> in G. Consider the set 9 ofall subgroups of G containing K but not x. Then 9' is not empty since K is amember. Clearly .' is partially ordered by inclusion; moreover the union of anychain in . is likewise in Y. By Zorn's Lemma . has a maximal element M. IfM < H < G, then K < H and as H Y, x e H; it follows that H > <x> u K = G.

Hence M is a maximal subgroup of G and therefore contains the Frattini subgroupO(G). Since M e S x M and so, finally, x 0 (D(G). Thus D(G) = 1.

It follows from 3.1.1, for example, that a finite p-group has complemented sub-group lattice if and only if it is elementary abelian. There are also large classes ofnonsoluble K-groups.

3.1.2 Theorem. If G is a primitive permutation group and the stabilizer of a point is aK-group, then G is a K-group. In particular, every finite symmetric or alternatinggroup has complemented subgroup lattice.

Proof. Let G be primitive on the set f) and suppose that 1 H < G. Since G isfaithful on Q, there exists an a c- S2 such that H G.. The stabilizer of a point in a

116 Complements and special elements in the subgroup lattice of a group

primitive permutation group is a maximal subgroup and therefore H U G, = G.Since G. is a K-group, there exists a complement K to H n G. in G,. It follows thatHnK=HnG2nK= 1 and HuK=Hu(HnG,)uK=HuG,=G. Thus Kis a complement to H in G and G is a K-group. For n > 3, the symmetric group Sand the alternating group A. are primitive permutation groups and the stabilizer ofa point is Si_1 and Ai_1, respectively. Since S3 and A3 are K-groups, it follows byinduction that all finite symmetric and alternating groups are K-groups.

In general the structure of finite K-groups is not known. In particular, it is un-known whether every finite simple group is a K-group. Previato [1982] proved thisconjecture for the alternating groups, the projective special linear groups PSL(n, p")and the Suzuki groups Sz(q).

Theorem 3.1.2 shows that subgroups of K-groups need not be K-groups. Forfactor groups and direct products of K-groups, the situation is different.

3.1.3 Lemma. If N < G, N < H < G and K is a complement to H in G, then NK/Nis a complement to H/N in G/N. In particular, every epimorphic image of a K-group isa K-group.

Proof. Clearly, H u NK > H u K = G and by Dedekind's law, H n NK =N(H n K) = N. Thus NK/N is a complement to H/N in GIN.

3.1.4 Lemma. Let H < G = AB and assume that C and D are complements to H n Ain A and (H u A) n B in B, respectively. If CD = DC, then CD is a complement to Hin G.

Proof. Suppose that x = cd e H where c e C and d c- D. Then

d=c-'xE(HuC)nD<(HuA)nBnD= 1

and hence x=cc-HnC=HnAnC=1. Thus HnCD=1. Now HuCD>_(HnA)uC=A and therefore HuCD>HuAuD>B. Hence G=AB<H u CD and CD is a complement to H in G.

3.1.5 Corollary. The direct product of two K-groups is a K-group.

Proof. If H < G = A x B with K-groups A and B, then the complements C and Din 3.1.4 exist and they certainly satisfy CD = DC. Thus H has a complement in G.

The Fitting subgroup of a K-group

The Fitting subgroup F(G) of a group G is the subgroup generated by all the normalnilpotent subgroups of G. Since the product of two normal nilpotent subgroups isnilpotent, F(G) is the unique largest normal nilpotent subgroup of G if G is finite. Ingeneral, F(G) need not be nilpotent.

3.1.6 Lemma. The Fitting subgroup of a K-group is abelian.

Proof. Let G be a K-group. We first show by induction on the nilpotency class c(N)that every normal nilpotent subgroup N of G is abelian. This is clear if c(N) = 1.Thus we can suppose that c(N) > 2 and consider a complement K to Z(N) in G. Asa characteristic subgroup of N, Z(N) is normal in G. Hence G = Z(N)K andN = Z(N)(N n K) = Z(N) x (N n K). Furthermore N n K is normal in K and cen-tralized by Z(N). Thus N n K is a normal subgroup of Z(N)K = G and its nil-potency class is c(N) - 1 since N n K ^- N/Z(N). By induction, N n K is abelianand then N = Z(N) x (N n K) is abelian as well. Now F(G) is generated by abeliannormal subgroups of G. By Fitting's theorem, any two of these generate a nilpotent,hence abelian normal subgroup of G; thus they centralize each other. It follows thatF(G) is abelian.

We want to show next that the Fitting subgroup of a K-group is a direct productof minimal normal subgroups. We prove a more general result.

3.1.7 Lemma. Let A be an abelian normal subgroup of the group G. If every normalsubgroup of G that is contained in A has a complement in G, then A is a direct productof minimal normal subgroups of G.

Proof. We may assume that A 54 1. Clearly, every normal subgroup B of G con-tained in A satisfies the assumptions of the lemma. Furthermore, if K is a comple-ment to B in G, then A = A n BK = B(A n K) so that A n K is a complement to Bin A. Since Ais abelian,AnK aAandasAaG,AnK aK.Thus AnK aAKG and it follows that A = B x (A n K) is a completely reducible G-module.

We want to show that A contains a minimal normal subgroup of G. Take 1a e A and consider .9" = {N a G I a 0 N < A}. Clearly .91 is not empty and it containsthe join of any chain in Y. By Zorn's Lemma there exists a maximal element M inY. Since A is completely reducible, A = M x C for some normal subgroup C of G.And if C were not a minimal normal subgroup of G, then C = Cl x C2 with 1 54

C, a G. The maximality of M would imply that M x Ci 0 St for i = 1, 2; hencea E (M x C,) n (M x C Z) = M, a contradiction. Thus C is a minimal normal sub-group of G contained in A.

Now if A* is the subgroup generated by all the minimal normal subgroups ofG contained in A, then again A = A* x D where D a G. If D 54 1, then D wouldcontain a minimal normal subgroup N of G, as we have just shown; but then N <A* n D = 1, a contradiction. Thus D = 1 and A is generated by minimal normalsubgroups of G. By another straightforward application of Zorn's Lemma A is adirect product of minimal normal subgroups of G (see Robinson [1982], p. 83).

Conversely, we have the following results.

3.1.8 Lemma. Let A be an abelian subgroup of G generated by minimal normal sub-groups Ai (i e 1) of G and let B < A. If either B a G or all the Ai are cyclic, then thereexists a complement C to B in A that is normal in G.

Proof. Consider Y = IN a GIN < A and N n B = 1}. Again Y is not empty andcontains the join of any chain in .9' so that by Zorn's Lemma there is a maximalelement C in Y. Suppose that A 0 BC. Then since A is generated by the A,, there isan i c- I such that A. 4 BC. If B < G, then BC a G and BC n A. = 1 since A; is aminimal normal subgroup of G; if B is not normal in G, then A. is cyclic of primeorder and again BC n A. = 1. In both cases, BCAi = B x C x A; and thereforeC x A. e.91, contradicting the maximality of C. Thus A = BC and C is a comple-ment to B in A that is normal in G.

3.1.9 Lemma. If G contains an abelian subgroup A generated by minimal normalsubgroups of G and a complement K to A that is a K-group, then G is a K-group.

Proof. Let H < G. Since K is a K-group, there is a complement C to H n K in K.Since A is abelian, B = (H u K) n A < A and as A a G, B < H u K. HenceB g AK = G and by 3.1.8 there is a complement D to B = (H u K) n A in A that isnormal in G. By 3.1.4, CD = DC is a complement to H in G.

We summarise the results proved so far in the following.

3.1.10 Theorem (Zacher [1953], Emaldi [1969]). The group G is a K-group if andonly if its Fitting subgroup F(G) is a direct product of abelian minimal normal sub-groups of G and has a complement K that is a K-group.

Proof. If G is a K-group, then F(G) by 3.1.6 is abelian and by 3.1.7 is a direct productof minimal normal subgroups of G. Clearly, F(G) has a complement K in G and by3.1.3, K G/F(G) is a K-group. Conversely, if G satisfies the conditions of thetheorem, it is a K-group by 3.1.9.

Soluble K-groups

It is clear that Theorem 3.1.10 tells us nothing about the structure of semi-simpleK-groups. In soluble groups, however, F(G) : I and the Fitting series, defined in-ductively by F0(G) = I and Fi+, (G)/Fi(G) = F(G/Fi(G)) for i > 0, reaches G. There-fore we get the following characterization of soluble K-groups.

3.1.11 Theorem (Zacher [1953], Emaldi [1969]). The soluble group G is a K-groupif and only if for all i > 0, F;+, (G)/Fi(G) is a direct product of minimal normal sub-groups of G/Fi(G) and has a complement in G/Fi(G).

Proof. If G is a K-group, then 3.1.3 shows that also G/Fi(G) is a K-group. Thereforeby 3.1.10, its Fitting subgroup Fi+,(G)/Fi(G) has the stated properties. Conversely, ifG satisfies the conditions of the theorem, we prove by induction on the Fitting lengthn that G = is a K-group. For n = 0, there is nothing to prove. And if theassertion is true for n - 1, then G/Fl (G) is a K-group. Since the complement to F, (G)in G is isomorphic to this group, G is a K-group by 3.1.10.


It is a well-known theorem of Gaschiitz (see Robinson [1982], p. 131) that for afinite group G, the factor group F(G)/cF(G) of the Fitting subgroup modulo theFrattini subgroup is the product of all the abelian minimal normal subgroups ofG/D(G). We use this to give the following more elegant characterization of finitesoluble K-groups.

3.1.12 Theorem (Zacher [1953]). The finite soluble group G is a K-group if and onlyif (D(G/ F1(G)) = 1 for all i > 0.

Proof. If G is a K-group, then 3.1.3 shows that G/Fi(G) is a K-group and by 3.1.1,(D(G/Fi(G)) = 1. Conversely, suppose that (D(G/F1(G)) = I for all i >_ 0. Then again byinduction, G/F(G) is a K-group and by Gaschutz's theorem, F(G)/N(G) = F(G) is adirect product of abelian minimal normal subgroups of G. It remains to be shownthat F(G) has a complement in G: this will be isomorphic to G/F(G) and Theorem3.1.10 will yield that G is a K-group. For this purpose we take a subgroup L of Gwhich is maximal with the properties that it is contained in F(G) and possesses acomplement H in G. We want to show that L = F(G), so suppose that L = F(G).Then F(G) n H # 1 and F(G) n H a F(G)H = G since F(G) is an abelian normalsubgroup of G. Let N be a minimal normal subgroup of G contained in F(G) n H.Since V(G) = 1, there is a maximal subgroup M of G with N $ M and as N isabelian, it follows that N n M = I and NM = G. Now N is a modular subgroup ofG and hence

(NuL)n(MnH) =((NuL)nH)nM =(Nu(LnH))nM = N n M = 1,

(N u L) u (M n H) = L u (N u (M n H)) = L u ((N u M) n H) = L u H = G.

Thus M n H is a complement to NL in G. This contradicts the maximality of L.Hence L = F(G) and F(G) has a complement in G.

Zacher's theorem has some rather surprising consequences.

3.1.13 Corollary. Every normal subgroup of a finite soluble K-group is a K-group.

Proof. Let G be a finite soluble K-group and suppose that N a G. Then it is well-known that F;(N) = Fi(G) n N and hence N/F;(N) NF1(G)/F1(G) a G/F1(G). Since(D(G/Fi(G)) = I and the Frattini subgroup of a normal subgroup is contained inthe Frattini subgroup of the whole group, '(NFL(G)/FL(G)) = 1. It follows thatD(N/F1(N)) = 1 and by 3.1.12, N is a K-group.

It is not known whether the assertion of the corollary is also true for infinitesoluble groups or arbitrary finite groups; nor is it known if it is true for an arbitraryK-group. Another easy consequence of Zacher's theorem is that in order to find outif a finite soluble group G is a K-group, it suffices to know that every normalsubgroup of G has a complement, or even that every characteristic subgroup of Ghas a complement in G. Here is a more general statement.

3.1.14 Theorem (Napolitani [1967b], Emaldi [1970]). The following properties of asoluble group G are equivalent.

(a) Every subgroup of G has a complement in G, that is, G is a K-group.(b) Every normal subgroup of G has a complement in G.(c) Every characteristic subgroup of G has a complement in G and for every pair of

normal subgroups N, M of G with N < M there exists a minimal normal subgroup ofG/N contained in M/N.

Proof. That (a) implies (b) and (b) implies the first part of (c) is trivial. To show that(c) follows from (b), we therefore have to consider a pair of normal subgroups N, Mof G with N < M. Since G is soluble, there exists a nontrivial abelian normal sub-group A/N of GIN contained in M/N. By 3.1.3, every normal subgroup of G/N hasa complement in GIN and hence Lemma 3.1.7 yields that A/N contains a minimalnormal subgroup of GIN. Thus (c) holds.

It remains to show that (c) implies (a), which will be proved by induction on thederived length k of the soluble group G. If k = 0, then G = I and (a) holds. So wemay assume k > I and take A = G"", the penultimate term of the derived series ofG. Then A is a nontrivial abelian characteristic subgroup of G and, by assumption,A has a complement K in G. Since both conditions in (c) are preserved by factorgroups with respect to characteristic subgroups, by induction, K -_ G/A is a K-group. Finally, if B is the subgroup generated by all the minimal normal subgroupsof G contained in A, then B is characteristic in G and therefore has a complementC in G. Since A is an abelian normal subgroup of G, A n C < AC = G. Thus ifA n C 0 1. there would exist a minimal normal subgroup N of G contained in A n Cand it would follow that N < B n C = 1. This is impossible. Thus A n C = I andA = B(A n C) = B, that is, A is generated by minimal normal subgroups of G. By3.1.9, G is a K-group and (a) holds.

The second assumption in (c) clearly holds if G satisfies the minimal condition onnormal subgroups. That the condition cannot be omitted from the theorem is shownby the additive group CD of rational numbers. This has only trivial characteristicsubgroups since multiplication by an arbitrary nonzero rational number is an auto-morphism; hence every characteristic subgroup has a complement. On the otherhand, 0 is not a K-group since any two nontrivial subgroups of 0 have nontrivialintersection.

Exercises

1. (Previato [1982]) (a) Let G be a finite group and suppose that H < G is a K-group. If H is contained in exactly one maximal subgroup M of G and if MG =n M' = 1, show that G is a K-group.

xEG(b) Taking for H a Sylow p-subgroup, deduce that PSL(2, p") is a K-group for all

pEP.nEN.

3.2 Special complements 121

2. (Emaldi [1969]) Show that every soluble K-group is periodic. (Hint: use the factthat an abelian group A with a locally finite subgroup of Aut A operating irre-ducibly on A is periodic (see Baer [1964]).)

3. (Curzio [1960]) If G is a soluble group such that L(G) is complemented andsatisfies the maximal condition, show that G is finite.

4. (Christensen [1964]) Show that G is a finite metabelian K-group if and only if Gis the semidirect product of two abelian groups and has elementary abelian Sylowsubgroups.

5. (Dinerstein [1968]) Let G he a finite group (it suffices to assume that L(G) satisfiesthe minimal condition) in which every characteristic subgroup has a complement.Show that every normal subgroup of G has a complement in G.

3.2 Special complements

Much more can be said about K-groups in which the complements have additionalproperties. The most natural of these is that they permute with the given subgroups.We call G a C-group if to every subgroup H of G there is a subgroup K of G such that

(1) G=HK and HnK= 1;

we also say that H is permutably complemented by K in G if (1) holds. The structureof C-groups is well-known, and will be presented in this section. First of all we havethe following simple inheritance properties.

3.2.1 Lemma. (a) Subgroups and epimorphic images of 'C-groups are C-groups.(b) The direct product of two C-groups is a C-group.

Proof (a) Let H < X < G. If K satisfies (1), then X = X n HK = H(X n K) andH n (X n K) = 1. Thus X n K is a complement to H in X that permutes with H. IfN < G, N < H < G and K again satisfies (1), then by 3.1.3, NK/N is a complementto H/N in GIN and clearly (H/N)(NK/N) = HK/N = G/N.

(b) Let H < G = A x B with C-groups A and B. Then there exists a subgroup Cof A such that A = (H n A)C and (H n A) n C = 1 and a subgroup D of B with theproperty that B = (HA n B)D and (HA n B) n D = 1. By 3.1.4, CD is a complementto H in G and

HCD = H(H n A)CD = HAD = HA(HA n B)D = HAB = G.

Thus G is a C-group.

Another natural property a complement K to a subgroup H of G might have isthe following.

(2) If H < X < G, then X n K is a complement to H in X.

We call a subgroup K of G with this property a sectional complement, or S-comple-ment for short, to H in G. And a group in which every subgroup has an S-comple-ment is called an SC-group. The proof of Lemma 3.2.1 (a) shows that if G = HK andH n K = 1, then K is an S-complement to H in G. Hence

(3) every C-group is an SC-group.

The structure of SC-groups is not known. We shall, however, show that the classesof C-groups and of locally finite SC-groups coincide.

3.2.2 Lemma. Subgroups and epimorphic images of SC-groups are SC-groups.

Proof. Let G be an SC-group, H < X < G and suppose that K is an S-complementto H in G. Then clearly X n K is an S-complement to H in X. And if N < G suchthat N < H, then by 3.1.3, N(X n K)/N is a complement to H/N in X/N. SinceN(X n K) = X n NK, it follows that NK/N is an S-complement to H/N in GIN.

The main property of SC-groups which we shall require is the following.

3.2.3 Lemma. If N is an abelian minimal normal subgroup of an SC-group G, then Nis cyclic of prime order.

Proof. Suppose that M is a proper subgroup of N and take an S-complement K toM in G. Then N n K is a complement to M in N and hence 1 < N n K < N. SinceN is an abelian normal subgroup of G, N n K a NK = G and the minimality of Nimplies that N n K = N. Thus M = I and N is cyclic of prime order.

As early as 1937 P. Hall proved that the finite C-groups and the finite super-soluble K-groups coincide and determined their structure. He did not consider S-complements but his proof also works for SC-groups.

3.2.4 Theorem (P. Hall [1937]). The following properties of the finite group G areequivalent:

(a) G is a C-group,(b) G is an SC-group,(c) G is a supersoluble K-group,(d) G is isomorphic with a subgroup of a direct product of groups of squarefree

order.

Proof. That (a) implies (b) follows from (3). If G is an SC-group, then it clearly is aK-group and we use induction on I GI to prove that G is supersoluble. By 3.2.2, everyproper subgroup of G is supersoluble and a well-known result of Huppert's (seeRobinson [1982], p. 287) shows that G is soluble. So if N is a minimal normalsubgroup of G, then by 3.2.3, N is cyclic and 3.2.2 yields that G/N is an SC-group.By induction GIN and hence also G is supersoluble. Thus (b) implies (c).

Now let G be a supersoluble K-group and for every maximal subgroup M of G,let MG = n MX be the core of M in G. By 3.1.1, the intersection of all these MG is

xcGtrivial so that G is isomorphic with a subgroup of the direct product of the groupsG/MG. To prove that (c) implies (d), we therefore show that every such group G,/MGhas squarefree order. By 3.1.3, G/MG is a supersoluble K-group and hence we mayassume that MG = 1 and M 0 1. Then a minimal normal subgroup N of G is cyclicof prime order p, furthermore G = NM and C,(N) = 1 since M, = 1. So M isisomorphic to a subgroup of Aut N and hence cyclic of order dividing p - 1. There-fore every Sylow subgroup of M ^- GIN is an epimorphic image of G; hence it is aK-group and so has prime order. It follows that I G I = pIMI is squarefree.

Finally suppose that (d) holds. We want to show that G is a C-group, and sincedirect products and subgroups of C-groups are C-groups, we may assume that G hassquarefree order. It is well-known that groups of squarefree order are soluble (seeRobinson [1982], p. 281). Every subgroup H of G is a Hall n-subgroup for some setit of primes and if n' = P\n, any Hall n'-subgroup K of G satisfies H n K = 1 andHK = G. Thus G is a C-group.

It is now not difficult to give the exact structure of a finite C-group G. By (c) or (d)of Hall's theorem, G is metabelian and 3.1.7 then shows that G' = A 1 x . . . x A, withminimal normal subgroups A; of G. Since G is supersoluble, these A. have primeorder. Clearly G' has a complement B in G. And since B GIG' is an abelian K-group, again 3.1.7 yields that B = B, x x B., where the B. also have prime order.It was proved by Cernikova in 1953 that infinite C-groups have exactly the samestructure; the other two properties in Theorem 3.2.4 were later generalized to infinitegroups by Emaldi. We remind the reader that a group G is called hypercyclic if everynontrivial epimorphic image possesses a nontrivial cyclic normal subgroup. Equiva-lent is that G has an ascending normal series with cyclic factors (see Robinson[1972a], p. 14); thus hypercyclic groups are a natural generalization of supersolublegroups. If N and M are normal subgroups of G such that N < M, it follows fromZorn's Lemma that there exists a normal subgroup K of G maximal with the prop-erty that M n K = N. Then K 0 G and if G is hypercyclic, there exists a non-trivial cyclic normal subgroup Z/K of G/K. The maximality of K implies that H =M n Z > N and clearly H/N is cyclic. Thus:

(4) If G is hypercyclic and N < M are normal subgroups of G, then there existsH a G such that N < H < M and H/N is cyclic. In particular, chief factors of hyper-cyclic groups are cyclic.

3.2.5 Theorem (Cernikova [1956], Emaldi [1969], [1978]). The following propertiesof a group G are equivalent:

(a) G is a C-group,(b) G is a locally finite SC-group,(c) G is a hypercyclic K-group,(d) G is the semidirect product of A = Dr A; by B = Dr BB where all the A. and BB

have prime order and A . a G for all i e I.` E 1 J E J

Proof. Suppose first that G is a C-group. Then G is an SC-group by (3) and we wantto show that G is locally finite. Every subgroup of G is a C-group and, since thesubgroup lattice of an infinite cyclic group is not complemented, G is a torsiongroup. Let 1 0 x e G and take C < G such that <x> n C = 1 and G = <x>C. ThenI G : Cl = I<x>l is finite. Therefore by 3.2.1, G/CG is a finite C-group and Hall's theo-rem shows that G/Cc is metabelian. It follows that G" < C and hence x G". Sincex was arbitrary, G" = 1. So G is a soluble torsion group and therefore locally finite(see Robinson [1982], p. 147). Thus (b) holds.

Now suppose that G is a locally finite SC-group. Then G is a K-group and wewant to show that G is hypercyclic. If a, b, c, d c- G, then by 3.2.2, H = <a, b, c, d > isa finite SC-group and Hall's theorem shows that H is metabelian. It follows that[ [a, b], [c, d]] = 1 and hence G" = 1. So if N g G with GIN o 1, then there exists anontrivial abelian normal subgroup A/N of G/N. By 3.2.2, GIN is an SC-group.Therefore 3.1.7 implies that A/N contains a minimal normal subgroup of G/N whichis cyclic by 3.2.3. Thus G is hypercyclic and (c) holds.

Next suppose that G is a hypercyclic K-group. We want to show that (d) holds,and again we first claim that G is metabelian. To prove this, we show that G" iscontained in every maximal subgroup of G; then 3.1.1 will imply that G" = 1. So letM be a maximal subgroup of G and suppose that G' M. Then G = G'M andG' r) M g M. Consider N = (G' r) M)u. By (4) there exists a nontrivial cyclic normalsubgroup H/N of G/N contained in G'/N. If WIN = CG/N,(H/N), then G/W is isomor-phic to a subgroup of AutH/N and therefore abelian. Hence G' < W and H/Ncentralizes G' n M/N. Thus H normalizes G' r) M and since H M and M is maxi-mal in G, it follows that G' r) M a G and G'/G' n M is a chief factor of G. By (4),G'/G' n M is cyclic and hence G" < G' r) M < M. Thus G" < D(G) = 1. Now A = G'is an abelian normal subgroup of G and by 3.1.7, A = Dr A; with minimal normal

IEIsubgroups A, of G. By (4), every A, is cyclic of prime order. Clearly A has a comple-ment B in G. And since B G/G' is an abelian K-group, again 3.1.7 yields thatB = Dr Bj where the Bj also have prime order. Thus (d) holds.

jEJ

Finally suppose that (d) is satisfied and let H < G = AB. By 3.1.8 there exist acomplement C to H n A in A that is normal in G, and a complement D to HA n Bin B. By 3.1.4, CD is a complement to H in G and, as in the proof to (b) of 3.2.1,HCD = G. Thus (a) holds.

Note that by Exercise 1.2.8, property (b) of Theorem 3.2.5 is a lattice-property sothat the class of C-groups is invariant under projectivities, a result not obvious fromthe definition.

Property (d) of Hall's theorem can also be generalized to infinite groups: everyC-group is isomorphic with a subgroup of a cartesian product of finite groups ofsquarefree order. However, it is not possible to replace "cartesian" by "direct" in thisstatement and it is also not true that every periodic subgroup of such a cartesianproduct is a C-group. For details see Exercises 3 and 4 or Cernikova [1956].

As for K-groups, to prove that a soluble group G is a C-group it is not necessaryto show that every subgroup of G is permutably complemented; it suffices to look atthe subnormal subgroups. In fact one need only consider the subnormal subgroups

of defect at most 2, that is subgroups H of G such that H o N a G for some N. Thiswas shown for finite groups, using Hall's theorem, by Bechtell [1969] and for arbi-trary groups, using Cernikova's theorem, by Kochendorfer [1969]. However, thereis a simple direct proof by induction on the derived length of G.

3.2.6 Theorem (Kochendorfer [1969]). The group G is a C-group if and only if G issoluble and every subnormal subgroup ofdefect at most 2 is permutably complementedin G.

Proof: If G is a C-group, then by 3.2.5, G is soluble and, clearly, every subnormalsubgroup of defect at most 2 is permutably complemented in G. Assume converselythat G has these properties. If G is abelian, then every subgroup is complementedand G is a C-group. So suppose that the derived length of G is n > 2 and that theassertion is true for soluble groups of smaller derived length. Then A = 0"-1) is anabelian normal subgroup of G. And if S/A is subnormal of defect at most 2 in G,/A,then there exists a subgroup T of G such that S n T = 1 and ST = G and S/ A ispermutably complemented by AT/A in G/A; the induction assumption yields thatG/A is a C-group. Let H < G. Then there exists a subgroup C/A of G/A such thatHA n C = A and HAC = G. Hence H n C< A and since A is abelian, H n C aA o G. By assumption there exists a subgroup D of G such that (H n C) n D = 1 and(H n C)D = G; it follows that C = (H n C)(C n D). So if we put K = C n D, then

HnK=HnCnD=1

and

HK = H(C n D) = H(H n QC n D) =HC = HAC = G.

Thus G is a C-group.

Further topics

Since C-groups have such a restricted structure, it seems reasonable to weaken thecondition of complementability. This has mainly been done in two directions andthere is a vast literature on this subject. But all these results are far too special to bepresented here; we shall only give some hints for further reading. First of all, onestudies groups G such that for a certain class X of groups, every X-subgroup of G ispermutably complemented in G and tries to prove that G then (nearly) is a C-group.This was initiated by Cernikov [1954] who took for X the class of abelian groups.In the meantime many other classes have been considered, among them the classesof primary groups (Gorcakov [1960]), elementary abelian groups (Sysak [1977]),infinite abelian groups (Cernikov [1980]), nonabelian groups (Barysovec [1977],[1981 a], [1981 b]), infinite groups (Cernikov [1967]), and finite groups (de Giovanniand Franciosi [1987]). The second possibility is to assume that, again for a certainclass X of groups, to every subgroup H of G there exists a subgroup K such that


G = HK and H n K e X. This has been done in particular where X is the classof finite groups (Sergeev [1964]; Tsybanev [1973], [1980]) and of cyclic groups(Mozarovskaja [1978]).

Exercises

1. (Hall [1937]) Show that a finite group is a C-group if and only if it is supersolubleand its Sylow subgroups are all elementary abelian.

2. (Cernikova [1956]) Show that every group having an ascending normal serieswith cyclic factors of nonrepeating prime orders is a C-group.

3. Let G be an infinite nonabelian P-group. Show that G is a C-group which is notisomorphic with a subgroup of a direct product of finite groups of squarefreeorder.

4. Let I be an infinite set, Gi a nonabelian group of order 6 for every i e I and letG = Cr Gi. Show that G is periodic but not a C-group. (Hint: determine the

i E I

normal subgroups of order 3 in G and use 3.1.7.)5. (Emaldi [1970]) Show that G is a C-group if and only if G is a hypercyclic torsion

group in which every characteristic subgroup has a complement. (Hint: first showthat G' is abelian.)

6. (Emaldi [1985a]) Show that a finite group is a C-group if and only if everysubgroup of prime order has an S-complement.

7. (Emaldi [1985a]) Show that G is a C-group if and only if every subgroup H of Ghas an S-complement K such that K has finite index in G if H is finite.

3.3 Relative complements

In this section we shall study groups that satisfy various stronger complementaryconditions.

SK-groups

A group G is called an SK-group if every subgroup of G is a K-group, that is if L(G)is sectionally complemented. By 3.2.2, every SC-group is an SK-group but the con-verse does not hold, the alternating group A4, or even A5, is an example. Clearly,every SK-group is periodic and has all its elements of squarefree order; however thestructure of SK-groups is not known. For finite groups, at least, we have the follow-ing characterization.

3.3.1 Theorem (Bechtell [1965]). The following properties of the finite group G areequivalent.

3.3 Relative complements 127

(a) G is an SK-group.(b) 1(H) = 1 for every subgroup H of G.(c) Every Sylow subgroup of G is elementary abelian.

Proof. That (a) implies (b) follows from 3.1.1. And since a finite p-group with trivialFrattini subgroup is elementary abelian, (b) implies (c). So we finally have to showthat a finite group G with elementary abelian Sylow subgroups is an SK-group. Byinduction, every proper subgroup of G is a K-group and it remains to be shown thatG is a K-group. For this we first note that b(G) = 1. For, if 1(G) # 1, there wouldexist a minimal normal subgroup N of G contained in F(G). As b(G) is nilpotent, Nwould be a p-group and since the Sylow p-subgroups of G are elementary abelian, awell-known theorem of Gaschiitz (see Huppert [1967], p. 121) would yield a comple-ment C to N in G. But then (D(G)C = G would imply C = G and N = 1, a contradic-tion. Thus 1(G) = 1. So if 1 H < G, there exists a maximal subgroup M of G suchthat H M. Since M is a K-group, there is a complement K to H n M in M.Clearly, HnK=HnMnK=1 and HuK=Hu(HnM)uK=HuM=G.Thus K is a complement to H in G and G is an SK-group.

RK-groups

The group G is called an RK-group if its subgroup lattice is relatively complemented,that is if for all L, H, M c L(G),

(1) L < H < M implies the existence of K < G such that H n K = L andHuK=M.

It is easy to see that this property implies that normality is transitive in G. Recallthat G is called a T-group if L a H g G implies L a G, and that G is a T*-group if

(2) L< H g M< G implies L< M,

that is every subgroup of G is a T-group.

3.3.2 Lemma. Every RK-group is a T*-group.

Proof. If L a H a M< G and K is as in (1), then L = H n K a K since H< M andK < M. It follows that L g H u K = M.

Every simple group is a T-group and every Tarski group is a T*-group, so thestructure of these groups is not known. But clearly, every subgroup of a T*-group isa T*-group. So to prove that

(3) every finite T*-group G is supersoluble,

we may use induction on IGI and get that every proper subgroup of G is super-soluble. By Huppert's theorem (see Robinson [1982], p. 287), G is soluble. If H/N isa chief factor of G and N < L < H, then L < H < G and hence L < G. Thus H/N isof prime order and G is supersoluble. Now we can prove the following characteriza-tion of finite RK-groups.

3.3.3 Theorem (Zacher [1952]). A finite group is an RK-group if and only if it is aT*-group with elementary abelian Sylow subgroups.

Proof. That these conditions are necessary follows from 3.3.2 and 3.3.1. Converselysuppose that G is a T*-group with elementary abelian Sylow subgroups. We useinduction on I GI to show that G is an RK-group. First of all, since the assumptionsare inherited by subgroups and epimorphic images, every proper subgroup andevery proper factor group of G is an RK-group. Therefore (1) holds for subgroups L,H, M of G such that M -A G or LG 0 1, and it remains to show (1) in the case that

(4) M=GandLG= 1.

Let p be the largest prime dividing IGI. Since G is supersoluble, there exists a normalsubgroup N of order p in G (see Robinson [1982], p. 145) and N $ L as LG = 1.Since G/N is an RK-group, there exists K < G such that

(5) HNnK=LNandHNuK=G.IfN H,then HuK=HuNuK=GandL<HnK<HNnK=LN. SinceN H and ILN : LI = p, it follows that H n K = L and K is the desired relativecomplement. So suppose that N < H. Then (5) yields that H nK = LN andH u K = G. If K < G, then K is an RK-group and there exists R < K such thatLNnR=LandLNuR=K.Itfollows thatHnR=HnKnR=LNnR=Land H u R = H u N u L u R= H u K = G, thus R has the desired properties. Itremains to consider the case where N < H and K = G. Here (5) yields that H = LN.

Case 1:N H

1

H

Case 2:N<H,K9 G Case 3: N <H, K = GFigure 13

Every Sylow subgroup of G is elementary abelian and hence by 3.3.1 (or Gaschiitz'stheorem) there exists a complement S to N in G. It follows that H = (S n H)N, thatis L and S n H are complements to N in H. Now G is supersoluble and p is thelargest prime dividing IGI. Therefore the Sylow p-subgroup P of G is normal in G.And if Q is any p-subgroup of G, then Q a P a G and hence Q a G as G is aT-group. Since LG = 1, it follows that ILI is prime to p and N is a normal Hallsubgroup of H. By the Schur-Zassenhaus Theorem, L and S n H are conjugate in H,that is there exists an x e H such that L = (S n H)X = SX n H. Clearly SX u H =SX u N u H = G and hence SX is a complement to H in [G/L].

The structure of infinite RK-groups is not known. However, it is possible toextend Zacher's theorem to infinite soluble groups. Using Zacher's theorem, it is not

difficult to show that finite RK-groups have the following property:

(6) if L, H, M E L(G) and L < H < M, then there exists K < G such thatHnK=LandHK=M.

This property is slightly stronger than (1). We call groups of this type RC-groups andwe shall show that the RC-groups are precisely the soluble or locally finite RK-groups, and these have a similar structure to the finite RK-groups. It is difficultto attribute this result correctly. There are two slightly erroneous versions of itin Emaldi and Zacher [1965] and Abramovskii [1967]; they were corrected byMenegazzo [1970a]. Emaldi [1971] added the characterization of RC-groups. Forthe proof we shall need the following technical results.

3.3.4 Lemma. If G is a metabelian periodic T-group with all Sylow subgroups elemen-tary abelian, then G' is a Hall subgroup of G (that is n(G') n n(G/G') = 0).

Proof. Suppose that p e n(G') n n(G/G'). If N is the p-complement in G', then GINsatisfies the assumptions of the lemma. Therefore we may assume that N = 1. Let Pbe a Sylow p-subgroup of G. Then G' < P and hence P a G. Since G is a T-groupand P is elementary abelian, every subgroup of P is normal in G and P = G' x Kwhere 1 < K < P. By 1.5.6, every g e G induces a universal power automorphism inP and if x9 = x' for x e P, then xr"1 = [x, g] e G' r) K = I for every x E K. It followsthat r = I (mod p) and hence P < Z(G). There exist u, v e G of prime power ordersuch that [u, v] 0 1. Then l = [u, v]P = [uP, v] by (2) of 1.5. Since P < Z(G), u 0 Pand hence u e <uP> < CG(v), a contradiction. Thus G' is a Hall subgroup of G.

3.3.5 Lemma. Let G be a locally finite T*-group with all Sylow subgroups elemen-tary abelian. Then G' is an abelian K-group and every subgroup of G' is normal in G.Let it = n(G/G') and suppose that every maximal n-subgroup of G is a complement toG' in G.

(a) If N < G', then every maximal n-subgroup of GIN is a complement to G'/N inGIN.

(b) If H < G, then every maximal n-subgroup of H is a complement to G' n H in H.

Proof. Since finite supersoluble groups with elementary abelian Sylow subgroupsare metabelian, also G is metabelian. Hence G' is a direct product of its elementaryabelian p-components and therefore an abelian K-group. Since G is a T-group, everysubgroup of G' is normal in G.

(a) Let S/N be a maximal n-subgroup of GIN, choose a maximal n-subgroup S,of S and a maximal n-subgroup T of G containing S,. By assumption, T is a comple-ment to G' in G and since G' is an abelian K-group, G' = M x N for some M < G'.Then MT nSnG'=MnS= 1and MTnSis an-group. From S, <MTnS<Sand the maximality of S, it follows that MT n S = S, and NS1 = N(MT n S) =NMT n S = S. Thus S/N is contained in the 7t-group NT/N. The maximality of S/Nyields that S/N = NT/N and so by 3.1.3, S/N is a complement to G'/N in G/N.

(b) Let S be a maximal n-subgroup of H and choose a maximal n-subgroupT of G containing S. Then once again T is a complement to G' in G, and hence

(G' n H) n S = 1. From S < T n H and the maximality of S it follows that S = T n H.Hence if G' < H, then H = G'T n H = G'(T n H) = G'S and S is a complement to G'in H. Now suppose that G' H and let N < G' such that G' = N x (G' n H). Thenby (a), GIN satisfies the assumptions of the lemma and NH/N = G'H/N is a sub-group of GIN containing G'/N = (G/N)'. Since N n H = 1, NS/N is a maximal n-subgroup of NH/N; therefore, as we have just shown, it is a complement to G'/N inG'H/N. In particular, G'H = G'NS = G'S and then H = (G' r) H)S. Thus S is a com-plement to G' r) H in H.

3.3.6 Theorem (Menegazzo [1970a], Emaldi [1971]). The following properties of agroup G are equivalent:

(a) G is an RC-group,(b) G is a soluble RK-group,(c) G is a locally finite RK-group,(d) G is a locally finite T*-group with all Sylow subgroups elementary abelian such

that.for it = n(G/G'), the set of primes dividing the order of an element of G/G', everymaximal n-subgroup of G is a complement to G' in G.

Proof. Every RC-group is a C-group and hence soluble by 3.2.5. Thus (a) implies (b).Since every RK-group is periodic and a soluble periodic group is locally finite, (c)follows from (b). Now suppose that (c) holds. Then by 3.3.2, G is a T*-group. By3.3.3, every finite subgroup of G is supersoluble with elementary abelian Sylowsubgroups and hence metabelian. Since G is locally finite, G is metabelian and haselementary abelian Sylow subgroups. By 3.3.4, G' is a Hall subgroup of G. So if Sis a maximal n-subgroup of G, then S n G' = 1. Since G is an RK-group, there existsa subgroup T of G such that G'S n T = S and G = G'S u T = G'T. Hence T n G' =T n G'S n G' = S n G' = I and therefore T TG'/G' is a n-group. The maximalityof S implies that S = T is a complement to G' in G.

Finally suppose that (d) holds. We want to show (6) and take L, H, M E L(G) suchthat L < H < M. By 3.3.5, L n G' a G and G,IL n G' satisfies (d). Therefore we mayassume that L n G' = I so that L is a n-group. Let S be a maximal n-subgroup of Hcontaining L and let T be a maximal n-subgroup of M containing S. Then Lemma3.3.5 (b) shows that H = (G' r) H)S and M = (G' n M)T. Furthermore, since TG'T/G' and G' n M are abelian K-groups, there exist subgroups N, Q, R of G suchthat G' n M= (G' n H) x N, S = L x Q and T= S x R= L x Q x R. We claimthat K = N(L x R) is the relative complement we are looking for. Since N a G,

HK =(G'nH)SN(L x R)=(G'nH)NS(L x R)=(G'nM)T=M

and if x e H n K = (G' r) H)S n N(L x R), that is x = as = bt where a e G' r-) H,snS, bnN, tnL x R, then b-'a =ts-'E(G'nM)nT=1. Hence a = b e(G'nH)nN= 1 and x=s=tESn(L x R) = L. Thus HnK=Las desired.

IM-groups

A group G is called an IM-group if every proper subgroup of G is the intersection ofmaximal subgroups of G. Clearly, every RK-group is an IM-group (see also 3.3.12),

but one can show (see Exercise 6) that the converse is wrong. A basic property ofIM-groups is the following.

3.3.7 Lemma. If N is a normal subgroup of the IM-group G, then N and GIN areIM-groups.

Proof. That GIN is an IM-group is trivial. Suppose that H is a proper subgroup ofN and take x e N\H. Then by Zorn's Lemma, the set of all subgroups R of G suchthat H < R < N and x 0 R has a maximal element S. Since x 0 S, we haveT = <S, x> > S, and for every subgroup X of N containing S properly, themaximality of S implies that x e X and hence T< X. Thus

(7) T=n{XiS<X<N}and S is a maximal subgroup of T. Since G is an IM-group, there exists a maximalsubgroup M of G such that S < M but T M. Then (7) implies that M n N = S.Thus N and S, and therefore also T, are normalized by M. It follows that G = TMand hence N = TM n N = T(M n N) = T, that is, S is maximal in N. We haveshown that to every x e N\H there exists a maximal subgroup S of N containing Hsuch that x 0 S. In particular, x is not contained in the intersection K of all themaximal subgroups of N containing H. It follows that K = H and N is anIM-group.

By the lemma the existence of a finite nonsoluble IM-group implies that of a finitenonabelian simple IM-group. But Bianchi and Tamburini Bellani [1977] and DiMartino and Tamburini Bellani [1980], [1981] checked the list of finite nonabeliansimple groups and proved that none of them is an IM-group. Thus, modulo theclassification of finite simple groups (see Gorenstein [1982]), we have the followingresult.

3.3.8 Theorem. Every finite IM-group is soluble.

The structure of soluble IM-groups was determined by Menegazzo [1970b]. Firstof all we have the following general result.

3.3.9 Lemma. Every IM-group is a T-group.

Proof: Suppose that G is an IM-group and let L g H < G. If .9' is the set of all

maximal subgroups of G containing L but not H, then L = H n ( n M I =ME Y

n (H n M). Every M e Y satisfies G = MH and H n M < M. So for every x e G,ME .!if we write x = mh with m e M and h e H, it follows that (H n M)" = (H n M)" >-L" = L and hence Lx = n (H n M)' > L. Thus L < G.

MES

Let us call a group elementary abelian if it is an abelian torsion group with allp-components elementary abelian. Then we can state Menegazzo's theorem in thefollowing form.

3.3.10 Theorem (Menegazzo [1970b]). The following properties of a group G areequivalent:

(a) G is a soluble IM-group,(b) G is a metabelian periodic T-group with all Sylow subgroups elementary abelian,(c) G contains a normal elementary abelian Hall subgroup N with elementary

abelian factor group such that every subgroup of N is normal in G.

Proof. Suppose first that G is a soluble IM-group. Then by 3.3.9, G is a T-groupand a well-known theorem of Robinson's (see Robinson [1982], p. 389) yields thatG is metabelian. So by 3.3.7, G' and GIG' are abelian IM-groups. Since every sub-group of an abelian IM-group is an IM-group and the infinite cyclic group clearlyis not an IM-group, G' and G/G' and hence also G are periodic. If P is a Sylowp-subgroup of G, then P PKIK where K is the p-complement in G'. ThenPK/K a G/K, and again by 3.3.7, PK/K is an IM-group. In particular, 4)(P) = 1.But a metabelian p-group is locally nilpotent and therefore has all its maximalsubgroups normal (see Robinson [1982], p. 345). Hence P is elementary abelian and(b) holds.

If G satisfies (b), then 3.3.4 shows that N = G' is a Hall subgroup of G; clearly italso has the other properties in (c). Finally, suppose that (c) holds. Then G certainlyis soluble. Let H be a proper subgroup of G. We have to show that H is the intersec-tion of maximal subgroups of G. Since H n N _ G and G/H n N also satisfies (c), wemay assume that H n N = 1. We shall prove that for every x E G\H of prime orderp there exists a maximal subgroup M of G that contains H but not x; since all theSylow subgroups of G are elementary abelian, the assertion will then follow. Sup-pose first that x e N and let P be a Sylow p-subgroup of G. Since G is locally finite,every pair x, y of p'-elements of CG(P) generates a finite subgroup of CG(P) in whichP n <x, y> is a direct factor; thus xy-' is a p'-element. Therefore the set L of p'-elements in CG(P) is a subgroup and CG(P) = P x L. Since G induces power auto-morphisms in P, IG/CG(P)I is finite and clearly prime to p. So Gaschiitz's theorem(see Huppert [1967], p. 121) shows that P has a complement K in G and all thecomplements to P in G are conjugate. Then H and PH n K are complements to P inPH and hence there exists g e G such that H < K9. So if P = (x> x Q, then M =QK9 is a maximal subgroup of G that contains H but not x. Now suppose that x 0 Nbut xN E HN; let x = by where h E H and 1 0 y E N. As we have just shown, thereis a maximal subgroup M of G containing H but not y; then x = by 0 M. Finally, ifxN 0 HN, then since GIN is elementary abelian, GIN = <xN> x M/N for somesubgroup M of G containing HN. Again M is a maximal subgroup of G containingH but not x. Thus G is an IM-group.

3.3.11 Corollary. A finite group is a (soluble) IM-group if and only if it is anRK-group.

Proof. By a theorem of Gaschiitz (see Robinson [1982], p. 392), every finite solubleT-group is a T*-group. Therefore the conditions in Theorem 3.3.3 and (b) in Theo-rem 3.3.10 are equivalent for finite groups.

Further topics

A lattice L is called weakly join-complemented if it contains a greatest element Iand for every pair x, y of elements of L such that x < y there exists an element z a Lsuch that x u z 0 1 and y u z = I. A complement z to y in the interval [I/x]would clearly satisfy this, so that every relatively complemented lattice is weaklyjoin-complemented.

3.3.12 Theorem (de Giovanni and Franciosi [1981]). The subgroup lattice of a groupG is weakly join-complemented if and only if G is an IM-group.

Proof. If G is an IM-group and H < K < G, then there exists a maximal subgroupM of G such that H< M but K M. It follows that K u M = G and H u M =M < G. Thus L(G) is weakly join-complemented. Conversely, assume that L(G) isweakly join-complemented and let H be a proper subgroup of G. Let K be theintersection of all the maximal subgroups of G containing H and suppose for acontradiction that H < K. Then there exists a subgroup L of G such that K U L = Gand H u L G; take x a G\(H u L). By Zorn's Lemma there exists a subgroup Mmaximal with the properties H u L < M and x 0 M. Every subgroup of G contain-ing M properly contains x and so M is a maximal subgroup of <M, x>. Againsince L(G) is weakly join-complemented, there exists a subgroup N of G such thatM u N 0 G and (M, x> u N = G. If N M, then M < M u N and the maximalityof M would imply that x a M u N and M u N = <M, x> u N = G, a contradiction.Hence N < M and <M, x> = G. Thus M is a maximal subgroup of G containing H.The definition of K implies that K < M; since also L< M, it follows that M con-tains K u L = G. This contradiction shows that K = H and G is an IM-group.

A subgroup H of a group G is called maximal sensitive in G if every maximalsubgroup of G that does not contain H intersects H in a maximal subgroup of Hand, conversely, every maximal subgroup of H is the intersection of H with a suit-able maximal subgroup of G.

3.3.13 Theorem (Venzke [1972]). Let G be a finite soluble group. Then every normalsubgroup of G is maximal sensitive in G if and only if G is an IM-group.

Proof. First assume that every normal subgroup of G is maximal sensitive. We wantto show that G is a T-group. So let L a H a G and suppose for a contradiction thatL is not normal in G. Then L:9 La, the normal closure of L in G, and LG/L is anontrivial finite soluble group. Hence there exists a normal subgroup N/Lof primeindex in LG/L. Since LG is maximal sensitive in G, there is a maximal subgroup M ofG such that N = LG n M. It follows that N a LG u M = G, a contradiction since LGis the smallest normal subgroup of G containing L. Thus G is a soluble T-group andhence metabelian (see Robinson [1982], p. 389). Let P be a Sylow p-subgroup of Gand let N be the p-complement of G'. Then P PN/N and PN a G; let Q/N =fi(PN/N). If Q > N, then since Q is maximal sensitive in G, there exists a maximalsubgroup M of G such that Q n M is a maximal subgroup of Q containing N. Since

PN is maximal sensitive in G, PN n M is maximal in PN and it follows that Q <PN n M < M, a contradiction. Thus Q = N and P PN/ N is elementary abelian.By 3.3.10, G is an IM-group.

Conversely assume that G is an IM-group and take a normal subgroup H of G.If M is a maximal subgroup of G that does not contain H, then G = HM andI H: H n MI = I G: MI is a prime since G is supersoluble. Thus H n M is a maximalsubgroup of H. On the other hand, every maximal subgroup K of H is the intersec-tion of maximal subgroups of G and hence there exists a maximal subgroup M of Gsuch that K < M but H M. It follows that K = H n M and H is maximal sensitivein G.

Emaldi [1985b] generalized Venzke's theorem to infinite groups. A number ofinteresting results on chief factors and the socle of an IM-group are contained inCurzio and Rao [1981].

Exercises

1. (Bechtell [1965]) Show that a finite group G satisfies b(H) < D(G) for everysubgroup H of G if and only if GIN(G) is an SK-group.

2. Find a supersoluble K-group that is not an RK-group.3. (Di Martino and Tamburini Bellani [1980]) Let n > 7 and G = A. be the alter-

nating group of degree n. Show that if H = <(123)>, then [G/CG(H)] is a chain oflength 2; thus G is not an IM-group. (Hint: show that a transitive subgroup of Gcontaining CG(H) is primitive and use the fact that a primitive permutation groupof degree n containing a 3-cycle contains A..)

4. (Di Martino and Tamburini Bellani [1980]) Suppose that G contains propersubgroups B and N such that G = BNB, H = B n N a N and B = UH whereH < NG(U). If G is simple, show that every maximal subgroup of G containing Ualso contains H; thus if H 1, G is not an IM-group. (Note that all simpleChevalley groups and all twisted groups over a field K have subgroups B, N withthe given property and that H = I only if K is the field with 2 elements orG = PSU(3.4).)

5. (Menegazzo [1970a]) Let G be a periodic group and w a set of primes. A Syloww-basis of G is a collection . " of subgroups of G consisting of a maximal w-subgroup and some Sylow p-subgroups of G for primes p 0 w such that ST = TSfor all S, T e Y and U S = G. Show that G is a soluble RK-group if and only if

sEG is a locally finite T*-group with all Sylow subgroups elementary abelian suchthat for every set w of primes, every maximal w-subgroup of G is a member of asuitable w-basis of G.

6. (Menegazzo [I970a]) For every i e N let pi, qi be odd primes such that g1Jpi -and i pi, qi } n { pj, qj } = 0 if i 0 j; let Hi = <ai, bi) be a nonabelian group oforder pigi where o(ai) = pi and o(bi) = q,. Let H = Dr H; and G = H(z> where

i6 Zo(z) = 2 and zaiz = a;', zbiz = b; for all i c N. Show that

3.4 Neutral elements and related concepts 135

(a) G is a locally finite T*-group such that every Sylow p-subgroup of G iselementary abelian and is a member of a suitable Sylow {p}-basis of G, inparticular,

(b) G is an IM-group, but(c) G is not an RK-group.

7. (de Giovanni and Franciosi [1981]) Show that the subgroup lattice of a group Gis weakly meet-complemented (that is, for every pair H, K of subgroups of G suchthat H < K there exists a subgroup L of G such that H n L= 1 and K n L 0 1)if and only if G is periodic and every element of G has squarefree order.

8. Show that every subgroup of a group G is maximal sensitive if and only if G is anIM-group and every maximal subgroup of G is modular in G.

3.4 Neutral elements and related concepts

Distributive, standard and neutral elements were introduced into lattice theory byOre, Gratzer and Birkhoff. We shall study these and some further related conceptsin subgroup lattices of groups.

An element a of a lattice L is called neutral if every triple a, x, y of elements in Lgenerates a distributive sublattice. The reader can easily verify that greatest and leastelements of L are neutral and that if L is a direct product of lattices L, and L2, thenan element a = (a, a,) of L is neutral if and only if both a; are neutral in L.

3.4.1 Theorem. The element a of the lattice L is neutral if and only if for all x, y a L,

(1) au(xny) = (aux)n(auy),that is the map cpo: L -+ [I/a]; x - a u x is an epimorphism,

(2) an(xuy) _ (anx)u(any),that is the map via: L -, [a/0]; x -- a n x is an epimorphism, and

(3) aux=auyandar x = anyimplies that x= y.

Proof. Since (1) and (2) are just distributive laws involving only a, x and y, theyclearly are satisfied if a is neutral; in addition, a u x = a u y and a n x = a n y to-gether with the distributive laws imply that

x = xn(aux) = xn(auy) _ (xna)u(xny)

_ (any)u(xny) _ (aux)ny = (auy)ny = y.

Conversely, suppose that (1)-(3) hold. Then the map a: L [I/a] x [a/0] definedby x° = (a u x, a n x) for x a L is a monomorphism. For, (1) implies that

(xny)° _ (au(xny),an(xny)) _ ((aux)n(auy),(anx)n(any))

_ (a u x, a n x) n (a u y, a n y) = x°ny°,

similarly (2) yields (x u y)° = x° u y°, and r is injective by (3). Now a is neutral in[I/a] and in [a/0]. Therefore a° = (a, a) is neutral in [I/a] x [a/0] and hence also inthe sublattice L° of this direct product. It follows that a is neutral in L.

All the properties (1)-(3) in Theorem 3.4.1 have been studied separately. An ele-ment a of the lattice L is called join-distributive if it satisfies (1), meet-distributive if (2)holds, and uniquely complemented if (3) is satisfied for all x, y e L. Furthermore, a iscalled join-quasi-distributive if (1) holds for all x, y e L such that a u x = a u y;dually, a is meet-quasi-distributive if (2) is satisfied whenever a n x = a n y. Finally, ais a standard element if it is join-distributive and uniquely complemented, that is, itsatisfies (1) and (3).

General properties of distributive elements

In arbitrary lattices there are only the obvious interrelations between the conceptsintroduced above. For subgroup lattices, however, we have some unexpected results.

3.4.2 Theorem (Ivanov [1969]). Every join- or meet-distributive element in the sub-group lattice of a group G is normal in G and uniquely complemented in L(G).

Proof. Let N< G, z e G and Z = <z>. Then N' u Z = N u Z and NZ n Zz-' (N n Z)z = N n Z. Therefore if N is join-distributive, then

N = Nu(NZnZ)=(NuNZ)n(NuZ)= NuNZ

and hence N= < N. Similarly, if N is meet-distributive, then

N=Nn(NZuZ)=(NnNZ)u(NnZ)=NnNZ

and hence N < NZ. Since z is arbitrary, it follows in both cases that N a G.Now let X, Y< G such that N u X= N u Y and N n X= N n Y. If N is join-

distributive, then since N a G, NX = NY= N(X n Y) and therefore

X = NX n X = N(X n Y) n X =(NnX)(XnY)=(NnY)(X r) Y)

=N(XnY)nY=NY nY=Y.

If N is meet-distributive, then N n (X u Y) = N n X = N n Y< X n Y and there-fore since N a G,

X =(Nn(XuY))X =NXn(XuY)=NY n(XuY)=(Nn(XuY))Y= Y.

In both cases, N is uniquely complemented in L(G).

3.4.3 Theorem (Curzio [1964a], Napolitani [1965]). Let N be a subgroup of thegroup G. Then N is join-quasi-distributive in L(G) if and only if N is join-distributivein L(G).

Proof. Clearly every join-distributive element is join-quasi-distributive. So supposethat N is join-quasi-distributive in L(G). We want to show first that N a G. If x e Nand g e G\N, then N u <g> = N u <xg) = N u (<g) n <xg)) as N is join-quasi-distributive. Hence there exists an integer r such that

(4) <g) n <xg) = <g') N and N u <g> = N u <g'); furthermore g' E CG(x)

since <g) n <xg) is centralized by g and xg and hence also by x. We use the firstassertion in (4) to show that if N contains an element of finite order n, then everyelement of G of this order is contained in N. For, if this were false this statementwould fail for a smallest integer n; choose x e N and g c G\N such that o(x) = n =o(g). Then every proper subgroup of <g) would be contained in N and therefore by(4), <g) n <xg) = <g>. It would follow that g c <xg), hence also x e <xg) and sincethe cyclic group <xg) contains only one subgroup of order n, <g) = <x> < N, acontradiction.

The result just proved implies that N is normal in G if it is periodic. So supposenow that N contains an element x of infinite order and take g e G\N. Then by (4),N u <g) = N u <g') and g'x = xg' for some integer r. Again by (4), <g') n <xg') Nand hence there exist nonzero integers s, t such that g'S = (xg')' = x'g't. Since xhas infinite order, I : x' = g'(S-') and therefore I<g>: N n <g)I is finite. Now chooseg, e <g) satisfying N u <g,) = N u <g> for which I<g, ): N n <g, )I is minimal. Ify e N, then again by (4) there exists 92 e <g,) such that N u <g2) = N u <g,) _N u <g) and y92 = g2y. The choice of g, implies that

<9i): N n <91>I < I<92): N n <92)I = I<92)(N n <gI>): N n <g,>I

S I<gl>: N n <9,>1

and it follows that <g,) = <92)(N n <g,)). Thus g, = g3z with g3 E <9z>, z e Nand hence y9i = y93= = y' E N. Since y was arbitrary, N9' < N. Similarly N91' < Nso that g, e NO(N). Therefore g e N u <g,) < NG(N) and since g was arbitrary,NaG.

Thus N < G in all cases. For arbitrary subgroups X and Y of G therefore, byDedekind's law, NX n NY= N(NX n Y) < NX and hence

N(NX n Y) = N(X n N(NX n Y)) = N(NX n Y n X n N(NX n Y)) = N(X n Y)

since N is join-quasi-distributive. Thus NX n NY = N(X n Y) and N is join-distributive.

3.4.4 Corollary. In the subgroup lattice of a group, the join-quasi-distributive ele-ments, the join-distributive elements and the standard elements coincide.

Note that the nonmodular lattice with 5 elements contains an element that isjoin- and meet-distributive but not uniquely complemented, another one that isjoin-quasi-distributive but not join-distributive, and a third one that is meet-quasi-distributive but not meet-distributive. This last situation also occurs in subgrouplattices so that the dual to Theorem 3.4.3 does not hold.

3.4.5 Example. Let G = <a,b,cla' = h' = c4 = 1, ab = ba, c-'ac = b, c-'bc = a-'>.Then c operates fixed-point-freely on P = <a, b> and hence G is a Frobeniusgroup of order 72 - 4. We claim that N = P<c2> is meet-quasi-distributive in L(G).To show this let X, Y< G such that N n X = N n Y. If X u Y < N, thenN n (X u Y) = X u Y = (N n X) u (N n Y). And if X, say, is not contained in N,then X contains a Sylow 2-subgroup <c>9 of G. Since c operates irreducibly on P,<c >9 is a maximal subgroup of G and hence X = G or X = <c>9. In the first case,clearly X u Y = X; in the second, N n Y = N n X = <C2> implies that <C2> <Y < <c>9 = X and therefore also X u Y = X. In both cases

/N

n (X u Y) = N n/X

=N n Y and N is meet-quasi-distributive. Since <a> u <c> = G and a`2 = a-'

N n (<a> u <c>) = N 0- <a> u <C2> = (N n <a>) u (N n <c>)

and hence N is not meet-distributive in L(G).

Uniquely complemented elements

We want to determine these elements in the subgroup lattice of a finite group. So inthe sequel, let G be a finite group and N < G a uniquely complemented element inL(G). In the first place the following is evident:

(5) if M < N and M a G, then N/M is uniquely complemented in L(G/M).

Furthermore, if H < G and X, Y e L(H) such that (N n H) u X = (N n H) u Y and(NnH)nX = (NnH)nY, then NuX =Nu(NnH)uX = Nu(NnH)uY=NuYandNnX=NnHnX=NnHnY=NnY;since Nis uniquely com-plemented in L(G), it follows that X = Y. We have now shown the following which,of course, also holds more generally in arbitrary lattices.

(6) If H < G, then N n H is uniquely complemented in L(H).

If X <G and zEN,N(NnX), then NnX =(NnX)Z=NnX2 and NuX =(N u X)= = N u XZ. Since N is uniquely complemented, X = XZ and we have shownthat

(7) N,,(N n X) < NG(X) for all X < G.

In particular, if p is a prime dividing IN) and X is a subgroup of order pin G that isnot contained in N, then N n X = 1; hence P < N0(X) for every subgroup P of orderp of N. It follows that PX is elementary abelian of order p2 and PX n N = P.Therefore every minimal subgroup Y of PX different from P and X satisfies N n Y =

1 = N n X and N u Y = N u X. This contradiction shows that there is no such X,another important property of N.

(8) If X < G such that IX I = p is a prime dividing I NI, then X < N.

It follows easily that

(9) N is a characteristic subgroup of G.

For, if we choose a prime p dividing INI and let M = (P < GIIPI = p), then M is acharacteristic subgroup of G contained in N and by (5) and induction, N/M is char-acteristic in G/M. Thus N is characteristic in G. Finally (7) and (9) yield that

(10) N<CG (X)for all X<Gsuch that NnX= 1;

for, N normalizes X and X normalizes N and therefore [N, X] < N n X = 1.Note that if N is a Hall subgroup of G, then (9), (10) and the Schur-Zassenhaus

Theorem imply that G = N x K where K is a complement to N in G; therefore N isneutral in L(G) since it is the greatest element of a direct factor of L(G). On the otherhand, the following lemma will show that if G is a p-group and I < N < G, then Gis cyclic or generalized quaternion and again N is neutral in L(G).

3.4.6 Lemma. Let N be a uniquely complemented element in the subgroup lattice of afinite group G and suppose that p is a prime dividing INI and IG/NI. Then every Syloivp-subgroup S of G is cyclic or a generalized quaternion group, and in the latter caseI NI - 2 (mod 4). Furthermore, NN(N n S) < CG(S) and N has a normal p-complement.

Proof. By (6), R = N n S is uniquely complemented in L(S). Since N < G, R is aSylow p-subgroup of N and therefore 1 < R < S. Hence there exists H < S such thatR is a maximal subgroup of H, and as R is uniquely complemented in L(S) it is theonly maximal subgroup of H. Thus H is cyclic, but by (8), its subgroup R containsevery minimal subgroup of S. It follows that S has only one minimal subgroup andtherefore is cyclic or a generalized quaternion group (see Robinson [1982], p. 138).In the latter case, (5) shows that R/Z(S) is uniquely complemented in L(S/Z(S)) andtherefore is trivial since S/Z(S) is a dihedral group; thus R = Z(S) and INI = 2(mod 4).

To prove that NN(R) < CG(S) take z E NN(R). If z is a p-element, then z is containedin R, the only Sylow p-subgroup of NN(R). Since R = Z(S) if S is generalized qua-ternion, it follows that z E CG(S). Now suppose that z is a p'-element. By (7), znormalizes S and as N < G, [z, S] < N n S = R. Since R < (D(S), it follows thatz centralizes S/ED(S) and hence also S. Thus z E CG(S) in this case also and so, final-ly, NN(R) < CG(S). In particular, NN(R) = CN(R) and by Burnside's theorem (seeRobinson [ 1982], p. 280) there exists a normal p-complement in N.

Now we can give the desired characterization of uniquely complemented elementsin the subgroup lattice of a finite group.

3.4.7 Theorem (Napolitani [1968]). Let N be a subgroup of a finite group G. ThenN is a uniquely complemented element in L(G) if and only if there exist subgroups Land K of G with the following properties.

(a) G = LK, L a G, (I LI, I K I) = 1, that is, L is a normal Hall subgroup of G withcomplement K.

(b) N = L(N n K) and N n K < Z(K).(c) For every prime p dividing IN n KI, the Sylow p-subgroups of G are cyclic or

generalized quaternion groups.(d) For every Sylow subgroup S of K, CL(N n S) = CL(S).

Proof. Suppose first that N is uniquely complemented in L(G). Then by (9), N a Gand Lemma 3.4.6 shows that N has a normal p-complement for every prime pdividing I N I and I G/N I. Let L be the intersection of these normal p-complements ofN or L = N if there is no such prime p. Then L is a characteristic subgroup of N andhence normal in G. Furthermore the construction of L shows that a prime q dividingILI cannot divide I G/N I. It follows that Lisa normal Hall subgroup of G and by theSchur-Zassenhaus Theorem there exists a complement K to L in G. We claim thatL and K satisfy (a)-(d). Clearly (a) holds and N = L(N n K). Furthermore N/L isnilpotent and has cyclic Sylow subgroups by 3.4.6. Thus N/L N n K is cyclic. LetS be a Sylow subgroup of K. If N n S -A 1, then N n K < NN(N n S) < CG(S) by 3.4.6,since every prime dividing I N n K I = I N/LI also divides I G/N I. And if N n S = 1,(10) shows that N n K < CG(S). Thus N n K centralizes every Sylow subgroup of Kand hence K. Therefore (b) holds and (c) follows again from 3.4.6. Finally, for everysubgroup X of K, CL(N n X) < NG(X) by (7). Hence [CL(N n X), X] < L n X = Iand CL(N n X) = CL(X). In particular, (d) holds.

Suppose conversely that G contains subgroups L and K satisfying (a)-(d). By (b),N a G and we claim that

(11) K is the only complement of L in G intersecting N in N n K.

To show this, suppose that H is such a complement. Let p be a prime dividingK I = I H I and take Sylow p-subgroups So and T of K and H, respectively. By (a),

So and T are Sylow p-subgroups of G and there exist x e L, y e K such that T =Sox = SX for S = So, a Sylow p-subgroup of K. Now N n S and N n T are Sylowp-subgroups of the abelian group N n K = N n H. It follows that N n S =NnT=NnSX=(NnS)Xand therefore [x,NnS] <LnS= 1. By(d),S=SX= Tand so T < K. Since p was arbitrary, H < K and hence H = K. This proves (11).

Now consider subgroups X and Y of G such that N n X = N n Y and NX = N Y.We have to show that X = Y and may assume that

(12) NX =G=NY

since L and Ko = K n NX satisfy (a)-(d) in Go = NX. If p is a prime dividingI N n K I but not I G/N I and P is a Sylow p-subgroup of N n K, then by (b), P is acentral Sylow subgroup of K and hence K = P x K* where K* is a p'-group. Itfollows that L* = LP is a normal Hall subgroup of G with complement K* sat-isfying (a)-(d). We may therefore also assume that

(13) every prime dividing IN n KI also divides I GIN 1.

Since L n X is a normal Hall subgroup of X, we can write X = (L n X)X1 where X1is a complement to L n X in X; similarly, Y = (L n Y)Y1 with a complement Yl to

LnYin Y.ThenLnX =LnNnX =LnNnY=LnYandhence(14) X=MX1andY=MY1

where M=LnX=LnY.Since NnX=NnMX1=M(NnX1)and Nn Y=M(N n Y1), N n X 1 and N n Y1 are complements to the normal Hall subgroup Min N n X = N n Y. By (b), N/L and hence also N n X/M is abelian. Therefore bythe Schur-Zassenhaus Theorem there exists z e M such that N n X 1 = (N n Y1 )Z =N n Y,. Since also NX 1 = G = N Y, , and, in view of (14), it is sufficient to show thatX1 = Y1=, we may finally assume that

(15) LnX=1=LnY.We claim that X and Y then are complements to L in G. For, since G/N X/N n X,it follows from (13) that there exists a p-element x e X \N for every prime p dividingN n K 1. By (c), a Sylow p-subgroup S of G containing x is cyclic or generalized

quaternion, and in the latter case, IN n SI = 2 since N n S < Z(S). Hence N n S <<x> < X and therefore the normal subgroup LX of N contains every Sylow p-subgroup of N. It follows that N n K < LX and G = NX = L(N n K)X < LX. To-gether with (15) this shows that X is a complement to L in G; similarly for Y. Itfollows that N n K and N n X = N n Y are complements to L in N and the Schur-Zassenhaus Theorem implies that there exists z e L such that N n K = (N n X)2 =N n X` = NnY=. By (11), X2 = K = Y= and hence X = Y as desired.

No characterization of uniquely complemented elements in the subgroup latticeof an arbitrary group is known.

Join-distributive elements

3.4.8 Theorem (Zappa [1949]). Let N be a subgroup of a finite group G. Then N isa join-distributive (or standard) element in L(G) if and only if there exist subgroups Land K of G with the following properties.

(a) G = L x K and (ILI, IKI) = 1.(b) N = L x (N n K) and N n K< Z(G).(c) For every prime p dividing IN n K1, the Sylow p-subgroups of G are cyclic or

generalized quaternion groups.

Proof. First suppose that N is join-distributive in L(G). Then by 3.4.2, N is uniquelycomplemented in L(G) and hence there exist subgroups L and K with properties(a)-(d) of Theorem 3.4.7. As in the proof of this theorem, if p is a prime dividingIN n KI but not IG/NI and P is a Sylow p-subgroup of N n K, then K = P x K* forsome p'-subgroup K* and L* = LP is a normal Hall subgroup of G with comple-ment K* satisfying (a)-(d) of 3.4.7. So we may assume that every prime dividingI N n KI also divides IG/NI; in addition we want to show that under this assumptionK a G. Then it will follow that G = L x K, N = L x (N n K) and L < CG(N n K)so that N n K < Z(G); thus (a)-(c) will be satisfied.

Let p be a prime and S a Sylow p-subgroup of K. If N n S = 1, then (d) of 3.4.7shows that S is centralized by L. So suppose that N n S : 1. Then p divides I G/ Nand therefore S $ N. For x e L, NS = (NS)' = NSX and since N is join-distributive,N(S n SX) = NS n NSX = NS. In particular, S n S' $ N. By (c) of 3.4.7, S is cyclic orgeneralized quaternion and in the latter case, IN n S1 = 2 since N n S <Z(S). Itfollows that N n S < S n SX < S. But then N(S n SX) = NS implies that S n SX = Sand hence S = SX. This shows that L normalizes every Sylow subgroup of K. ThusK<LK=G.

Suppose conversely that G contains subgroups L and K satisfying (a)-(c) and letX < G.

(16) If g e NX is an element of prime power order, then g c N or g e X.

To prove this, suppose that o(g) = p" and let g 0 N. Let a E N, x e X such thatg = ax and write a = be with commuting p-element b and p'-element c. Since g 0 N,p divides S K I and hence g e K and h e N n K< Z(G). Hence <b, g> is a p-group and(c) shows that <h> < <g>. It follows that <g> = <b-'g>. Now x = a-'g = c-'b-'gwhere c-' E N is a p'-element and b-' g e K is a p-element. Since [N, K] = 1, b-' ytherefore is a power of x. Thus <g> = <h-'g> < X and (16) holds.

Now we prove that N is a join-distributive element in L(G). Since N < G, we haveto show that N(X n Y) = NX n NY for all X, Y!59 G. Let g e NX n NY be anelement of prime power order. If g e N, then g e N(X n Y); on the other hand,if g 0 N, then by (16), g e X and y e Y, so that again g e N(X n Y). It followsthat NX r) NY < N(X n Y). The other inequality is obvious. Hence N(X n Y) =NX n NY, as desired.

The join-distributive elements in the subgroup lattice of an infinite group werecharacterized by D.G. Higman [1951] in the case of a torsion group and by Sato[ 1952] for groups with elements of infinite order. We also refer the reader to Suzuki[1956], Chapter IV for a systematic treatment of lattice-homomorphisms of groups.Since any join-distributive element N in L(G) is normal in G, the map sending everysubgroup X of G to NX/N is a homomorphism from L(G) to L(G/N) induced by thenatural group-homomorphism from G to GIN. So Suzuki is able to use his generaltheory of lattice-homomorphisms to give shorter proofs of Zappa's, Higman's andSato's results.

Neutral elements

3.4.9 Theorem (Suzuki [1951b], Zappa [1951c]). Let N he a subgroup of a finitegroup G. Then N is a neutral element in L(G) if and only if there exist subgroups L, Hand M of G with the .following properties.

(a) G = L x HM,M<Gand (ILI,IHI)=(I LI,IMI)=(IH1,IM1)= 1 (so that L, Hand M are Hall subgroups of G).

(b) N = L x (N n H) and N n H< Z(G).(c) H is nilpotent with Sylow subgroups cyclic or generalized quaternion groups.

Proof. First suppose that N is neutral in L(G). Then by 3.4.1, N is join-distributivein L(G) and hence there exist subgroups L and K with properties (a)-(c) of Theorem3.4.8. Let M = {x E GI(o(x), INI) = 11. If x, y c- M, then N n <x> = I = N n <y>and, since N is meet-distributive, also N n(<x> u <y)) = 1. It follows from (8) that(INI,I<x,y>I) = 1. In particular, (o(xy-'),INI) = 1 and hence xy-' E M. This showsthat M is a subgroup, and it is clearly a normal Hall subgroup of G. Now M < Kand by the Schur-Zassenhaus Theorem there is a complement H to M in K. SinceN a G and LH is a Hall subgroup of G, N < LH and N n H = N n K. Thus (a) and(b) are satisfied and since every prime dividing IHI also divides I N n HI, it remainsto be shown that H is nilpotent. So let p be a prime and suppose for a contradic-tion that S and T are different Sylow p-subgroups of H. Since N a G, N n S andN n T are Sylow p-subgroups of N n H. An abelian group has only one Sylowp-subgroup, hence N n S = N n T. Since N is meet-distributive, N n (S u T) =(N n S) u (N n T) = N n S is a p-group. But as S T, there exists a subgroup Q ofprime order q 0 p in S u T< H. The definition of M implies that q divides I N I andtherefore Q < N by (8). Thus Q < N n (S u T), a contradiction. It follows that Hcontains only one Sylow p-subgroup for every prime p and hence is nilpotent.

Suppose conversely that G contains subgroups L, H, M satisfying (a)--(c) and letX, Y < G. We have to show that

(17) Nn(XuY)=(NnX)u(NnY).Then N will be meet-distributive in L(G); by 3.4.8, N is join-distributive, and 3.4.2and 3.4.1 will imply that N is a neutral element of L(G). If L(G) is a direct productand (17) is true for the components of N, X and Y, then it certainly also holds for thesubgroups themselves. By 1.6.4, L(G) ^-, L(L) x L(HM) and since L < N, we maytherefore assume that L = 1. Then N < Z(H) and (17) holds for X, Y < H sinceH is nilpotent and (17) is satisfied for the Sylow subgroups of N, X and Y. NowG%M H and it follows that MN is meet-distributive in [G/M]. Furthermore(I M I,1 N I) = I and therefore for every subgroup Z of G,

M(N n Z) = M((M n Z) x (N n Z)) = M(MN n Z) = MN n MZ.

Hence for arbitrary X, Y < G,

M(N n (X u Y)) = MN n M(X u Y) = MN n (MX u MY)

= (MN n MX) u (MN n M Y) = M(N n X) u M(N n Y)

= M((N n X) u (N n Y)).

It follows that N n (X u Y) = (N n X) u (N n Y) as required.

No characterization of neutral elements in the subgroup lattice of an arbitrarygroup is known, for torsion groups they were determined by Martino [1973]. Themeet-distributive and meet-quasi-distributive elements in subgroup lattices of finitegroups were characterized by Zappa [1951a] and Napolitani [1968], respectively

(see Exercises 2 and 4 or Suzuki [1956], Chapter IV). Ivanov [1969] showed thatmeet-distributivity is a local property and thus extended Zappa's result to locallyfinite groups. Meet-distributive elements in subgroup lattices of non-periodic locallysoluble groups have been considered by Ivanov [1985], Stonehewer and Zacher[1990a], [1990b], [1991b]. A characterization of meet-distributive or meet-quasi-distributive elements in subgroup lattices of arbitrary groups is not known.

Modular elements

It is natural to look for elements m of a lattice L that stand in the same relation tomodularity as neutral elements do to distributivity; thus we introduce:

(18) every triple m, x, y of elements in L generates a modular sublattice.

This was done by Zacher [1955a], [1955b], [1956] who used the slightly strongercondition that m generates a modular sublattice with every modular sublattice of L.Clearly, if m satisfies (18), then

xu(mny)=(xum)ny forallx,yELwith x<y,and

mu(xny)=(mux)ny forallx,yELwithm<y,

that is, m is a modular element of L in the sense of Section 2.1. The converse is nottrue; in fact, if G is the nonabelian group of order p3 and exponent p and X and Yare subgroups of order p of G with G = <X, Y>, then the sublattice of L(G) generatedby Z(G), X and Y is not modular. We shall study modular elements in the subgrouplattice of a finite group Gin Section 5.1 and will get a fairly good description of thesemodular subgroups. While characterizations of subgroups of G satisfying either (18)or Zacher's stronger condition in L(G) are not known, Zacher [1956] determined theHall subgroups of finite groups with the second property.

Exercises

1. Let L be a lattice.(a) Show that greatest and least elements of L are neutral.(b) Show that if L is a direct product of lattices L, and L2, then an element

a = (aI,a2) of L is neutral if and only if both a; are neutral in L.2. (Zappa [1951 a]) Show that a subgroup N of a finite group G is meet-distributive

in L(G) if and only if there exist subgroups L, H and M of G with the followingproperties.(a) G = LHM, L a G, M a G and (ILI, IHI) = (ILI, IMI) = (IHI, IMI) = 1.(b) N = L(N n H) and N n H< Z(HM).(c) H is nilpotent with Sylow subgroups cyclic or generalized quaternion.

3.5 Finite groups with a partition 145

(d) For every Sylow subgroup S of H, CL(N n S) = CL(S).(e) For every Sylow subgroup S of H, every subgroup of L that is normalized by

N n S is normalized by S.Using the classification of finite simple groups, Stonehewer and Zacher [1991a]showed that the group [L, H] in Exercise 2 is nilpotent.3. (Stonehewer and Zacher [1991a]) If M = 1 in Exercise 2 and the Sylow 2-

subgroup Q of H is a generalized quaternion group, show that Q is a directfactor of G.

4. (Napolitani [1968]) Show that a subgroup N of a finite group G is meet-quasi-distributive in L(G) if and only if it satisfies properties (a)-(d) of Exercise 2.

5. Show that a subgroup N of a finite group G is neutral in L(G) if and only if it isboth join- and meet-quasi-distributive.

3.5 Finite groups with a partition

A partition of a group G is a set E of nontrivial subgroups of G such that everynonidentity element of G is contained in a unique subgroup X E E. Equivalent tothis condition clearly is that

(1) X n Y = 1 for all X, Y E E such that X -A Y, and

(2) for every cyclic subgroup C of G there exists Z E E such that C < Z.

The elements of E are termed components of the partition; E is called nontrivial ifX G for all X E E. By 1.2.5, a subgroup C of G is cyclic if and only if [C/1] isdistributive and satisfies the maximal condition. Therefore (1) and (2) show that fora group G to possess a nontrivial partition is a lattice-theoretic property. We giveexamples for partitions of finite groups.

3.5.1 Examples. Let G be a finite group, p a prime, n e N.(a) If 1 < N < G such that every element in G\N has order p, then any partition

of N together with the set of cyclic subgroups of G not contained in N is a nontrivialpartition of G.

(b) In particular, if N is a nontrivial normal subgroup of index p in G such thatevery element in G\N has order p, then N together with the set of cyclic subgroupsof G not contained in N is a partition of G.

(c) A Frobenius group is a group G containing a proper nontrivial subgroup Hsuch that H n Hz = I for all x e G\H. It is well-known (see Robinson [1982], pp.243 and 299) that the set N of all elements of G not contained in the conjugates of Htogether with I is a nilpotent normal subgroup of G. This is called the Frobeniuskernel and H a Frobenius complement. Clearly, N together with the set of conjugatesof H is a nontrivial partition of G.

(d) If G is the symmetric group on 4 letters, then the set E of maximal cyclicsubgroups of G is a partition; for in any finite group this set satisfies (2). And ifX, Y E E such that X j4 Y and X n Y 1, then, since every element in S4 has order

at most 4, it would follow that JXI = I Yl = 4 and IX n YJ = 2. But S4 has only 3cyclic subgroups of order 4 and any two of them intersect trivially. Thus (1) alsoholds and E is a partition of G.

(e) Also for G = PGL(2, p"), the projective general linear group of dimension 2over the field with p" elements, p" > 3, the set of maximal cyclic subgroups is apartition. It consists of groups of order p, p" + 1 and p" - 1. The Sylow p-subgroups(of order p"), together with these maximal cyclic subgroups of order p" + 1 andp" - 1, form another partition E of G, and E' _ {X n HEX e E} is a partition for thesubgroup H = PSL(2, p") of G (see Huppert [1967], p. 193).

(f) There is another class of finite simple groups with a nontrivial partition, theSuzuki groups Sz(q), q = 22n+1. This consists of the Sylow 2-subgroups and cyclicsubgroups of order q - 1, q + 2n+1 + 1 and q - 2n+1 + 1 (see Huppert andBlackburn [1982b], p. 190).

Our aim in this section is to show that every finite group with a nontrivial parti-tion is one of the examples given in 3.5.1. This was proved in 1961 for soluble groupsby Baer and in the general case by Suzuki. The main problem in Suzuki's proof wasto show that a finite nonsoluble group with a nontrivial partition has even order. Weshall use the Feit-Thompson Theorem to achieve this, and then the classification ofZassenhaus groups to identify the group. In this section, all groups considered arefinite.

Elementary results

A partition E of G is called normal if X9 E E for every X E E and g E G. If E and Aare partitions of G, 1 H < G and g e G, then clearly

(3) H n E :_ J H n X (X E E, H n X 1} is a partition of H,

and E n A:= {X n YIX E E, Y E A, X n Y 0 1} and I-:_ {X91X E E} are parti-tions of G. Thus if E is a nontrivial partition of G, then

(4) EG := n Y-' is a nontrivial normal partition of G.xEG

Therefore we may usually assume that the partitions we consider are normal. Thefollowing simple results will be used over and over again.

3.5.2 Lemma. Let Y. be a partition of G and let a, b e G such that ab = ba and either(i) o(a) o(b) or(ii) o(a) = o(b) is not a prime.

Then there exists a component X of E containing a and b.

Proof. We may assume that a 0 1: b and take X, Y e Y. such that a e X and b E Y;we have to show that X = Y. Suppose first that (i) holds and take Z E I such thatab e Z. Then o(a) > o(b), say, and since ab = ba, 1 a°(bl = (ab)°(b) E X n Z. By (1),X = Z. So Z contains a and ab and therefore also b; thus X = Z = Y as desired.

Now suppose that (ii) holds and take a prime p dividing o(a). Then p 0 o(a) andhence aP 0 1, aPb = baP and o(aP) 0 o(b). Thus (i) holds for aP and b and, as we havejust shown, this implies that aP and b lie in the same component of E. Since aP e X,it follows from (1) that X = Y.

The lemma shows that a finite abelian group has a nontrivial partition only if it isan elementary abelian p-group. More generally, if E is a partition of G and 1 : a eX E E such that o(a) is not a prime, then every element b c- CG(a) satisfies (i) or (ii) of3.5.2 and hence CG(a) < X. In particular, we have the following.

3.5.3 Corollary. If E is a partition of G and H is a nilpotent subgroup of G that is notcontained in a component of E, then H is a p-group for some prime p.

3.5.4 Lemma. Let E be a normal partition of G and let X E E. If 1 H < X, thenNG(H) < NG(X). In particular, CG(x) < NG(X) for every I x e X.

Proof. Let g c- NG(H). Then 1 0 H = H9 < X n X9 and X9 E E since E is normal.By (1), X = X9 and hence g e NG(X). In particular, CG(x) < NG((x>) < NG(X) for1:xeX.

Admissible subgroups

Let E be a partition of G. A subgroup H of G is called E-admissible if for every X E E,either X < H or X n H = 1. Clearly, every component of E is E-admissible. Wewant to show that every E-admissible subgroup of a finite group G is either self-normalizing or nilpotent. For this we consider the Hughes subgroup HP(G) of Gdefined by

(5) HP(G) = (x e GIo(x) # p>.

Suppose that HP(G) < G. Then every element in G\HP(G) has order p. So ifa c- G\HP(G) and x e HP(G), then

1 = (xa-1)P = xx°... x°"a-P

By a well-known theorem of Hughes, Kegel and Thompson (see Huppert [1967], p.502), a finite group with such an automorphism is nilpotent. Thus

(6) HP(G) is nilpotent if H,(G) < G.

We shall also need the following result due to Hughes and Thompson [1959].

3.5.5 Lemma. If HP(G) < G and G is not a p-group, then I G: HP(G)I = p.

Proof. We use induction on I G I. Let H = HP(G) and P be a Sylow p-subgroup of G.Then G = HP and by (6), H = Q x (H n P) where Q 1 is a p-complement in G. Ifx e P and 1 0 y c- CQ(x), then o(xy) : p and hence xy e H; since also y e H, it fol-

lows that x e H n P. Thus if H n P = 1, then P operates fixed-point-freely on H andby a well-known theorem of Burnside (see Robinson [1982], p. 298), P is cyclic orgeneralized quaternion. Since every element in G\H has order p, it follows thatIPI = p and I G : HI = p. If H n P 0 1, there exists a subgroup N of order p inH n Z(P). Clearly N G and every element in GIN that is not contained in H/N hasorder p. Thus HP(G/N) < H/N and by induction, I GIN : HP(G/N)I = p. It followsthat HP(G/N) = H/N and IG: HI = p.

It was conjectured, and then proved for p = 2 and 3, that IG : HP(G)I < p if G is ap-group; however Wall [1965] gave a counterexample for p = 5. The reader can findfurther results on the Hughes subgroup and on partitions of p-groups in the bookby Khukhro [1993]. In any case, Lemma 3.5.5 suffices to prove the result on admis-sible subgroups announced earlier.

3.5.6 Lemma (Kegel [1961]). If E is a partition and H a E-admissible subgroup of G,then H = NG(H) or H is nilpotent.

Proof. Suppose that H < NG(H). Then there exists a subgroup M of NG(H) such thatIM: HI = p for some prime p. Let x e M\H and take X e E such that x e X. SinceH is E-admissible, X n H = 1. As M = H<x>, it follows that <x> M/H and henceo(x) = p. Thus HP(M) < H. If H is a p-group, it is nilpotent. And if H is not ap-group, then by 3.5.5, IM : HP(M) I < p. Thus HP(M) = H and by (6), H is nilpotent.

In our first main result on a finite group with a partition we deal with the situationthat G has a nontrivial admissible normal subgroup. This clearly holds in Examples3.5.1, (b) and (c). The following theorem shows the converse.

3.5.7 Theorem (Baer [1961a]). Let G be a finite group with a normal partition E andsuppose that N is a nontrivial proper E-admissible normal subgroup of G. Then one ofthe following holds.

(a) G is a Frobenius group, the Frobenius complements are components of E and Nis contained in the Frobenius kernel of G.

(b) There exists a prime p such that all elements in G,,N have order p, and if G isnot a p-group, then N = HP(G), I G : NI = p and N is a component of E.

Proof. Suppose first that there exists X E E such that X N and CN(X) = 1. Thenas N is E-admissible, N n X = 1 so that N n NG(X) and X are normal subgroups ofNG(X) intersecting trivially. Therefore N n NG(X) < CN(X) = 1 and by Dedekind,

(7) NNX(X) = NX n NG(X) = (N n NG(X))X = X.

Thus NX is a Frobenius group with complement X and kernel N. It follows thatNX is the set-theoretic union of N and all the X9 where g e NX, and hence isE-admissible since all these X9 are components of E. Furthermore NX is not nil-potent and by 3.5.6, NG(NX) = NX. But as N :!g G, NG(X) < NG(NX) and then (7)shows that NG(X) = X. Thus G is a Frobenius group with complement X, N is

contained in the Frobenius kernel and (a) holds. So we may assume that

(8) CN(X) -A 1 for every X e E such that X .4 N

and want to show that then (b) is satisfied. If 1 a E C,N(X) and 1 b e X, then aand b lie in different components of E. By 3.5.2, o(a) = o(b) is a prime p and everynontrivial element in CN(X) and in X has order p. In particular,

(9) p divides I NI and ExpX = p.

It follows that every element in G\N has prime order. Therefore (b) holds if G is ap-group and we may assume that G is not of prime power order. Then (9) shows thatN too is not of prime power order. If p is a prime dividing I G : N I and P is aSylow p-subgroup of G, then INP: NI is a power of p and hence every element inNP\N has order p. Thus HP(NP) < N < NP. By 3.5.5, I NP : HP(NP) I = p and henceHP(NP) = N and INP: NI = p. It follows that all Sylow subgroups of GIN are cyclicof prime order and by a well-known theorem of Zassenhaus (see Robinson [1982],p. 281), GIN is soluble. Hence there exists a normal subgroup M/N of GIN ofprime index p. Again every element of G\M has order p and hence HP(G) = M by3.5.5. By (6), M is nilpotent and not of prime power order since N < M. But then3.5.3 shows that M is a component of E and the E-admissibility of N implies thatM = N. Thus (b) holds.

3.5.8 Remark. If p = 2 in (b) of Theorem 3.5.7, then G is either an elementaryabelian 2-group or a generalized dihedral group, that is, N is abelian, G = N<t>where o(t) = 2 and tat = a-' for all a e N. For, if we take t E G\N and a e N, thenat N and hence t2 = 1 and atat = 1; it follows that tat = a-' for all a E N and thisimplies that N is abelian. If Exp N = 2, then every element in G has order at most 2and G is elementary abelian. And if Exp N 2, then t 0 CG(N). Since t was arbitrary,it follows that CG(N) = N and that the automorphism group induced by G in N hasorder 2. Thus I G : NI = 2 and G = N<t> as desired.

A characterization of S4

In view of Theorem 3.5.7 we may assume in the remainder of this section that ourgroups do not possess admissible normal subgroups. We first establish the followingnice characterization of the symmetric group on four letters.

3.5.9 Theorem (Baer [1961b]). Let E be a nontrivial normal partition of the finitegroup G and suppose that there is no nontrivial proper E-admissible normal subgroupof G. If the Fitting subgroup F(G) : 1, then G S4 and E consists of the maximalcyclic subgroups of G.

Proof. Since F(G) 1, there exists a prime p such that G contains a nontrivialnormal elementary abelian p-subgroup P. We shall show in a number of steps thatp = 2 and IPI = 4 for any such P; it will follow easily that G S4. First of all,

(10) Z(G) = 1.

For, if 1 0 z c Z(G) and Z e E such that z e Z, then Z:9 G by 3.5.4. Since E isnontrivial, Z would be a proper nontrivial E-admissible normal subgroup of G, acontradiction. Similarly, if P < X e E, then X g G by 3.5.4, the same contradiction.Thus

(11) P is not contained in a component of E.

But if 1 c e CG(P) and o(c) 0 p, then by 3.5.2, P< X where X is the component ofE containing c. Thus (11) implies that

(12) Exp CG(P) = p.

Let X be a component of E. Suppose first that P n X 1. Then by 3.5.4, P <_No(P n X) < NN(X) so that X a H = PX. By (11), X < H and hence 3.5.6 impliesthat X is nilpotent. Thus H = PX is the product of nilpotent normal subgroups andtherefore, by Fitting's theorem, is nilpotent. Hence every p'-element of H centralizesP and (12) implies that H is a p-group. Furthermore X is a component of the normalpartition H n E of H and Theorem 3.5.7 and Remark 3.5.8 yield the remainder of thefollowing assertion.

(13) If X E E such that P n X 1, then PX is a p-group, X < PX and every ele-ment in PX \X has order p; if p = 2, then PX is elementary abelian or I PX : X I = 2.

Now suppose that P n X = I and that p divides I X 1. Let S be a Sylow p-subgroupof X. Then S 0 1 and since P is a nontrivial normal subgroup of the p-group PS,there exists t c- P n Z(PS) such that o(t) = p. By 3.5.4, t c CG(S) < NG(X) and hence[t, X] < P n X = 1. Since t 0 X, 3.5.2 shows that every nontrivial element in X hasorder p. We have thus proved:

(14) If X e E such that P n X = 1 and p divides JXI, then Exp X = p.

In particular, X is a p-group, and (13) and (14) show that

(15) every component of E is either a p-group or a p'-group.

We want to show next that G contains elements of order p2. So suppose that this isfalse. Then every nontrivial p-element of G has order p and by (15), the set A ofsubgroups of order p of G and of components of order prime top of E is a nontrivialnormal partition of G. Clearly, P is a A-admissible normal subgroup of G and by(10), G is not a p-group. It follows from 3.5.7, that either G is a Frobenius group withcomplement X E A and kernel K containing P, or P is a component of A. In the firstcase, since K as a Frobenius kernel is a normal Hall subgroup of G, X is a p'-groupand hence X c E; this implies that K is E-admissible, contradicting our assumption.In the second case, IPI = p and hence P is contained in a component of E, contra-dicting (11). This contradiction shows that

(16) G contains elements of order p2.

In the remainder of the proof let x e G be an element of order p2, X E E such thatx E X and H = PX. We want to study the automorphism 6 induced by x in P whichis given by a° = x-lax for a e P. By (14), P n X 0 1 and (13) then yields that X a H.It follows that [P, X] < P n X. If a, b e P, then since P is abelian, by (1) of 1.5,

[ab, x] = [a, x]b[b, x] = [a, x] [b, x]

so that the map a: P -+ P n X defined by as = [a, x] = a°-' for a e P is a homomor-phism. If a e Kera, then ax = xa and by 3.5.2, a e X. Thus Kera < P n X and thehomomorphism theorem yields that

(17) IP1= IP°IjKeral < IPnXI2.

If P n X a G, then by 3.5.4, X a G, a contradiction. So there exists g e G suchthat P n X (P n X)9 = P n Xg and since X and X9 are components of E,(Pr'X)n(Pr X)9= 1. But as Pa G, (Pr X)(PnX)9<Pand hence lP1> IPnXI2.Thus

(18) [pi = IPnXI2

and since P" and Kera are contained in P n X, it also follows from (17) thatP° = P n X = Kera. Hence a82 = 1 for all a e P, that is (a - 1)2 = 0 in the endo-morphism ring of P. On the other hand, by (11) there exists a e P such that a 0 X.Then ax-1 E H\X so that o(ax-1) = p by (13). It follows that

I = (ax-1)p =aax...axp-'x-p = a1+o+..+"x-p

p-1

Since x-p 1, we have al""" # I and hence y a' 0. In the polynomial=o

P-'ring GF(p)[z], the identity (z - 1)p = zp - I =(z - 1) E z' implies that (z - 1)p-1 ==o

P-1Y- Z'. Now there is an obvious ring epimorphism from GF(p)[z] to the commuta-

i

tive subring of End P generated by I and a in which z maps to a; hence (a - 1)p-1 =P-1Y, a'. Thus (a - 1)p-' 0 and, because (a - 1)2 = 0, it follows that p = 2. Since=o

o(x) = 4, X is not elementary abelian and by (13), 1 P: P n X I = I PX : X 1 = 2. On theother hand, (18) shows that I P: P n X I= P n X 1. It follows that I P n X I= 2 and

(19) 1P1 = 4.

By (12), CG(P) has exponent 2 and therefore is an elementary abelian normal 2-subgroup of G. So we may apply our result (19) to CG(P) instead of P, obtainingICG(P)l = 4. Hence

(20) P = CG(P)

and G/P is isomorphic to a subgroup of Aut P S3. Since G has elements of order4, 1 G/PI is even and since Z(G) = 1, G is not a 2-group. Thus G/P S3 and I GI =24. Let Y be a component of E containing an element of order 3; by (15), 1 Y1 = 3.

Hence G does not contain elements of order 6 and it follows that Y = CG(Y). IfY = NG(Y), then G is a Frobenius group and its Frobenius kernel is a E-admissiblenormal subgroup of G, a contradiction. Thus I NG(Y)l = 6 and since G operatesfaithfully on the set of its Sylow 3-subgroups, G 2 S4. By (15) and (11), all cyclicsubgroups of order 3 and 4 of G belong to E and a noncyclic component W of E has


to be elementary abelian of order 4. But since the Sylow 2-subgroups of G aredihedral, there would exist a cyclic subgroup of order 4 intersecting W nontrivially,a contradiction. Thus E consists of the maximal cyclic subgroups of G.

The main theorem

We collect the results proved so far to give the structure of a finite soluble groupwith a partition.

3.5.10 Theorem (Baer). Let E be a nontrivial partition of the finite soluble group G.Then one of the following holds.

(a) G is a p-group for some prime p and E contains a component X such that everyelement in G\X has order p; furthermore IEI = 1 (modp).

(b) G possesses a nilpotent normal subgroup N such that N e E, IG : NI is a prime pand every element in G\N has order p.

(c) G is a Frobenius group.(d) G^- S4.

Proof. First suppose that G is a p-group. If 1 A x e Z(G) and X E E such that x e X,then by 3.5.2, X contains every element of order different from p in G; thus o(g) = pfor all g e G\X. Every subgroup of order p of G is contained in a unique componentof E. Since k(G) = I (mod p) if k(G) is the number of subgroups of order p in G (seeHuppert [1967], p. 33), it follows that 1 = k(G) _ k(X) - 111 (modp).

Now suppose that G is not a p-group. By (4), EG is a nontrivial normal partitionx6yof G. Therefore 3.5.7 and 3.5.9 show that (c) or (d) is satisfied or there is a prime psuch that N = HP(G) E EG and I G : NI = p. Since every component of E, is con-tained in a component of E, N E E in this case. By (6), N is nilpotent and (b) holds.

Note that there are nonnormal partitions of S4 and of many Frobenius groups.Finally we determine the nonsoluble groups with a partition.

3.5.11 Theorem (Suzuki [1961]). Let G be a nonsoluble finite group with a nontrivialpartition. Then one of the following holds.

(a) G is a Frobenius group.(b) G PGL(2, q), q a prime power, q > 4.(c) G PSL(2, q), q a prime power, q > 4.(d) G Sz(q), q = 2"', n e N.

Proof. We cannot give a complete proof of this beautiful theorem here since someclassification theorems are required to identify the groups. As a matter of fact,Suzuki's theorem is a consequence of the classification of finite simple groups; how-ever, a proof using this result might not be appropriate. We decided to use theFeit-Thompson Theorem on the solubility of groups of odd order, the classification

of Zassenhaus groups (a complete proof of which can be found in Huppert andBlackburn [1982b], pp. 160-286) and elementary parts of Bender's classification ofgroups with a strongly embedded subgroup and of the Brauer-Suzuki-Wall Theo-rem (both to be found in Suzuki [1986], pp. 391-404 and 494-504, respectively).Thus by consulting these books, the reader can find a complete proof of Theorem3.5.11 modulo the Feit-Thompson Theorem; this is avoided in the original paper ofSuzuki by the use of the theory of exceptional characters to prove that the order ofthe group is even.

So let G be a finite nonsoluble group with a nontrivial partition E. By (4), we mayassume that E is normal and we may clearly also assume that G is not a Frobeniusgroup. It follows that for every X e E,

(21) X < NG(X) and X is nilpotent

by 3.5.6. Furthermore, 3.5.7 and the nilpotency of HP(G) imply that there is nonontrivial proper 1-admissible normal subgroup of G. By 3.5.9,

(22) F(G) = 1.

Let S be a Sylow 2-subgroup of G; the Feit-Thompson Theorem implies that S 0 1.We divide the remainder of the proof into four parts.

(1) If there exists a component X e E such that S < X, then G is isomorphic toPSL(2, 2") or Sz(2") for some n e N.

Proof. Recall that a subgroup H of G is called strongly embedded in G if H < G, I H Iis even and 1H n H91 is odd for every g e G\H. Now if H = NG(S), then H < Gby (22), and clearly IHI is even. By 3.5.4, NG(S) < NG(X) and since X is nilpotent,NG(X) < NG(S). So if g e G\H, then X9 # X so that S n S9 < X n X9 = 1. But theSylow 2-subgroup of H n H9 is contained in S n S9 and hence is trivial. ThusIH n H91 is odd and

(23) H = NG(S) = NG(X) is strongly embedded in G.

If S were cyclic or generalized quaternion, then H would centralize the involution inS and since S < X < H, 3.5.2 would imply that every element in H\X would becontained in X; this is impossible. Thus S is not cyclic or generalized quaternion andit follows (see Suzuki [1986], p. 396) that the action of G on the cosets of H is doublytransitive and that a maximal p-subgroup P of H, p an odd prime, stabilizing threepoints is contained in a conjugate L of H for which INL(P)I is even. We want to showthat P = 1; then the stabilizer of three points will be trivial and since, by (22), G hasno regular normal subgroup, G will be a Zassenhaus group.

So suppose that P : I and that T is a nontrivial 2-subgroup of NL(P); let L = H"where u c G. Since P is contained in three different conjugates of H, we may assumethat H" 0 H; let D = H n H". Then T < S" < H" and hence [T, P] < S" n P = 1.

Since T< S" < X", it follows from 3.5.2 that P < X". By (23), X : X" and thereforeP n X = 1. Thus p divides I X I and I H: X 1. Clearly, X is an (H n E)-admissible prop-er normal subgroup of H. If H were a Frobenius group containing X in its kernel K,then 2p would divide IKI and since K is nilpotent, K < X by 3.5.3. Thus K = X and

this would contradict the fact that a Frobenius kernel is a normal Hall subgroup.Therefore 3.5.7 shows that I H : X I = p. Since P n X = 1, 1 PI = p and H = PX. Itfollows that D = H n H" = P(X n H") and I X n H"I divides I H": X"J = p. We mayassume that u is an involution (see Suzuki [1986], p. 395). Then P" < (H n H")" _H n H" and P 0 P" since otherwise P :!g X n X" = 1. Thus

(24) D = H n H" = PP" is elementary abelian of order p2

and the involution u interchanges the subgroups P and P". It follows that D =A x B where IAI = IBI = p, A is centralized and B is inverted by u. Let Y bethe component of E containing B. Then D<u) < NG(B) < NG(Y) by 3.5.4. By (21),NG(Y) > Y and so 3.5.7 shows that either NG(Y) is a Frobenius group containing Yin its kernel or there exists a prime q such that JNG(Y) : Yl = q. In the first case, theFrobenius kernel is a nilpotent normal Hall subgroup of NG(Y) and therefore con-tains D and does not contain u; this contradicts the fact that A is centralized by u. Inthe second case, again u 0 Y since Y is nilpotent and contains B. Hence q = 2 andD = PP" < Y. Since P < X" and P" < X, it follows that X n Y 1 0 X" n Y andhence X = Y = X". This contradiction shows that P = 1 and that G is a Zassenhausgroup of degree IG: HI = 1 (mod 2). It follows (see Huppert and Blackburn [1982b],p. 286) that G PSL(2, 2") or G ^- Sz(2") for some n e N.

(II) If CG(x) is nilpotent for every involution x e Z(S), then either(a) G is isomorphic to PSL(2, 2") or Sz(2") for some n e N, or(b) S is dihedral of order at least 8 and if Z(S) = <x) and X E E is the com-ponent containing x, then S = CG(x) = NG(X) = X<t) where o(t) = 2 andtat = a-' for all a e X.

Proof. Let x e Z(S) be an involution and let X E E such that x e X. If CG(x) is not a2-group, then by 3.5.3, CG(x) < X; it follows that S < X and by (I), (a) holds. So wemay assume that

(25) CG(x) = S for every involution x e Z(S).

Since X is nilpotent, its 2'-component centralizes x and hence is trivial; thus X is a2-group. By 3.5.4, S = CG(x) < NG(X) and hence XS is a 2-group. It follows thatXS < S and therefore

(26) X a S.

If X = S, we are done by (I); so we may assume that X < S. Then X is a proper(S n E)-admissible normal subgroup of S and by 3.5.7 and 3.5.8, either S is elemen-tary abelian or

(27) S = X <t> where X is abelian, Exp X > 4, o(t) = 2 and tat = a-' for everyaEX.

If S is elementary abelian, (25) and (26) show that X < S = CG(x) for every 1 0 x e Sand this implies that S n S9 = 1 for all g e G\NG(S). Therefore the conjugates of Stogether with the 2'-subgroups in E form a nontrivial normal partition of G whichsatisfies the assumptions of (I), and again (a) is satisfied. So we may assume that (27)holds.

Then Z(S) = CI(X) and every element of S\X is an involution. If none of these isconjugate to an element of X, then they all operate as odd permutations on the setof cosets of X in G (see Suzuki [1986], p. 128). It follows that G has a normalsubgroup N of index 2 such that N n S = X. Then N n E is a normal partition of Nhaving a Sylow 2-subgroup X of N as a component. By (I), N PSL(2, 2") orN Sz(2") for some n E N. Since X is abelian, N 2w PSL(2, 2") and then X is elemen-tary abelian, contradicting (27).

Thus there exist s E S\X and g e G such that s9 E X. Then s e S2(X9) = Z(S9)and therefore S2(X) = Z(S) < CG(s) = S9 by (25). Since IS9 : X9I = 2 andS2(X) n X9 = 1, IS2(X)I = 2. Hence X is cyclic, S is dihedral and NG(X) < CG(x) = S;thus (b) holds.

(III) If there exists an involution x e Z(S) such that H = CG(x) is not nilpotent, thenS is a dihedral group and if X E E is the component containing x, then H = CG(x) _NG(X) = X<t) where X is abelian, o(t) = 2 and tat = a-' for all a E X.

Proof. First let x be any involution in S such that CG(x) is not nilpotent, let X E Ebe the component containing x and put H = NG(X); by 3.5.4, CG(x) < H. Again (21)shows that X is a proper (H n E)-admissible normal subgroup of H and by 3.5.7,either H is a Frobenius group containing X in its kernel K or there exists a prime psuch that all elements in H\X have order p. In the first case, x e K and therefore alsoCG(x) < K, a contradiction since K is nilpotent. Thus the second alternative holds.Since X is nilpotent, CG(x) 4 X and for z E CG(x)\X, also xz 0 X so that 1 = (xz)P =x°zP = XP. Thus p = 2 and 3.5.8 shows that

(28) H = X<t) where X is abelian, o(t) = 2 and tat = a-' for all a e X;

in particular, H = CG(x). We now choose an involution x E Z(S) such that CG(x) isnot nilpotent. Then S < H = X<t) and we have to show that X n S is cyclic; thenS = (X n S)<t), say, will be a dihedral group.

Since H is not nilpotent, X is not a 2-group and Z(H) = O(X n S). Let Z = Z(H)and suppose for a contradiction that IZI >_ 4; let y E Z such that 1 # y # x and takeg E G\H. If x" E Z for some u E G, then x" E X n X" and hence X = X"; it followsthat u E NG(X) = CG(x) and x" = x. This shows that x is not conjugate to y andto y9. Therefore <x, y9) is a dihedral group of order divisible by 4 and thereexists an involution z E <x, y9) that centralizes x and y9. Thus Z E CG(x) = H andz e CG(y9) = H9 and hence <Z, Z9> < CG(z). If CG(z) is nilpotent, then Z and Z9 arecontained in the Sylow 2-subgroup of CG(z) and hence in a Sylow 2-subgroup S° ofG. Since (X n S)" is a normal subgroup of index 2 in S" and IZI > 4, Z n X° r 1

Z9 n X°. It follows that X = X° = X9, a contradiction since g H. If CG(z) is notnilpotent and W E E such that z E W, then by (28), W :!g CG(z) of index 2. It followsthat W n Z 1 W n Z9 and X = W = X9, the same contradiction as before.Thus IZI = 2, X n S is cyclic and S is a dihedral group.

(IV) G is isomorphic to PGL(2, q), PSL(2, q) or Sz(q) for some prime power q.

Proof. Suppose first that G has no normal subgroup of index 2. Then by (II) and(III), either G is isomorphic to PSL(2, 2") or Sz(2") for some n e N, or an involu-


tion x c- Z(S) and its centralizer satisfy the assumptions of the essential part of theBrauer-Suzuki-Wall Theorem (see Suzuki [1986], p. 497). It follows that G is aZassenhaus group of degree q + 1 and order (q + 1)q(q - 1)/2 for some primepower q. This implies (see Huppert and Blackburn [1982b], p. 286) that G PSL(2, q).

In general, let N denote the smallest normal subgroup of G such that IG: NI is apower of 2 and suppose that N G. Then N is a nonsoluble group with a nontrivialnormal partition N n E. Clearly, N has no normal subgroup of index 2 and F(N) =I so that N is not a Frobenius group. Thus N PSL(2,q) or N Sz(q) for someprime power q, as we have just shown. By (II) and (III), the Sylow 2-subgroup S ofG is dihedral and N n S is a proper noncyclic normal subgroup of S. ThereforeN n S is a dihedral group and IS: N n S I = 2. Thus I G : NJ = 2, N PSL(2, q) and Gis isomorphic to a subgroup of Aut N PFL(2, q). Since G has dihedral Sylow2-subgroups, it follows that G PGL(2,q) (see Suzuki [1986], pp. 507-510). Thiscompletes the proof of Theorem 3.5.11.

3.5.12 Corollary. The only finite simple groups with a nontrivial partition are thegroups PSL(2, p"), p" > 3, and Sz(2z"+i) n e N.

Exercises

1. (Baer [1961a]) Show that a partition is trivial if it consists of the conjugates ofone subgroup.

2. Show that a p-group has a nontrivial partition if and only if it has order at leastp2 and is not generated by its elements of order different from p.

3. Show that a 2-group has a nontrivial partition if and only if it is a generalizeddihedral group of order at least 4.

4. Determine all partitions of S4.5. Let G be a P-group of order pzq, with p and q primes, p > q. Find a nonnormal

partition of G. (Note that G is a Frobenius group.)6. Show that any of the groups PGL(2, p"), PSL(2, p") and Sz(2"), where p" > 3 for

the linear groups, is determined by its subgroup lattice.

Chapter 4

Projectivities and arithmetic structure of finite groups

In this chapter we investigate the influence of the subgroup lattice on the arithmeticstructure of a finite group G. By this we understand first of all the order of G, thenumber, structure and embedding of Sylow subgroups, and the number of Hallsubgroups or, more generally, of subgroups of any given order in G.

Finite groups with proper normal Hall subgroups occur in abundance. For exam-ple, every Sylow subgroup that is not self-normalizing is a proper normal Hallsubgroup in its normalizer. Therefore in §4.1 we describe the subgroup lattice of afinite group G with a normal Hall subgroup N in terms of L(N), L(G/N) and conju-gation by elements of N. This description is used to construct projectivities betweenfinite groups with isomorphic normal Hall subgroups that will be used quite often.

A projectivity cp from G to G is called index preserving if it satisfies I H`° : K1* 1 =

I H : K I for all subgroups K < H of G. It is easy to see that if cp is not indexpreserving, then there exist a prime p and a Sylow p-subgroup P of G such thatP`°I J P 1. By 2.2.6, P then is cyclic or elementary abelian and we show in §4.2

that G either is a direct product of a P-group with a group of coprime order(P-decomposable) such that P is contained in the P-group, or has a normal p-complement N such that N° is a normal and, most often, a Hall subgroup of G. Thisresult was proved by Suzuki in 1951 and is one of the fundamental theorems onsubgroup lattices of groups. First of all, it shows that many groups only possessindex preserving projectivities. And secondly, it reduces most problems on projec-tivities of finite groups to index preserving ones since, using the results of § 4.1, it isusually not difficult to handle the exceptional situations described above.

Many characteristic subgroups of finite groups can be defined by arithmeticconditions. We show in §4.3 that a projectivity between P-indecomposable groupsG and G maps the maximal normal p-subgroup OP(G) to Oq(G) for some prime q andhence the Fitting subgroup F(G)to F(G). This, of course, is not true for P-groups;however, for k>_ 2, F,(G)° = Fk(G) for every projectivity cp between arbitrary finitegroups G and G. The corresponding result for the iterated nilpotent residuals is alsotrue, but we will not be able to prove this until § 5.4. The theorem on the Fittinggroups is due to Schmidt [ 1972b]; as a consequence we get the fact, proved indepen-dently in 1951 by Suzuki and Zappa, that projective images of finite soluble groupsare soluble. We shall give other proofs of this theorem in Chapter 5. Finally we showthat index preserving projectivities map the hypercentre of G to that of G.

In § 4.4 we study images of abelian Sylow p-subgroups under projectivities andprove two theorems due to Menegazzo [1974] on the image of the centre of a Sylowp-subgroup P under an index preserving projectivity cp of a finite group G generated

158 Projectivities and arithmetic structure of finite groups

by p'-elements where p > 2: If G is p-normal, then Z(P)" < Z(P'0); if G is p-soluble,then Z(P)'0 = Z(P°). These are the only general results that are known about theprojectivities induced in the Sylow subgroups of a finite group.

4.1 Normal Hall subgroups

In this section we study the subgroup lattice of a finite group G with a normal Hallsubgroup N. Basic for this is the Schur-Zassenhaus Theorem which asserts that Nhas a complement H in G and that all the complements of N in G are conjugateunder N. To prove this last assertion the Feit-Thompson Theorem on the solubilityof groups of odd order is necessary. This can be avoided if one assumes that N orG/N is soluble. This will always be the case in our applications of the results ofthis section. Therefore, for these applications, the Feit-Thompson Theorem is notneeded. However, we state our results without the assumption that N or G/N issoluble, that is we freely use the Feit-Thompson Theorem in this section.

Normalizers, centralizers and commutator subgroups

We want to describe group-theoretic concepts in the subgroup lattice of G and westart with the core HG = n HX and the normal closure H' = U HX of a comple-x" XEG

ment H to N in G. For this we do not need that (I NJ, CHI) = 1.

4.1.1 Lemma. Let N g G and H < G such that G = NH and N n H = 1.(a) If H,, H2 < H and x e N such that Hi < H2, then H, < H2 and x e CN(H, ).(b) For x e N, H n HX = CH(x) and H u HX = [x, H] H.(c) HG = CH(N), HG = [N, H] H, HG n N = [N, H].

Proof. (a) Let a E H,. Then aX E H2 and hence [a,x] = a-'a' E H n N = 1 sincex e N a G. It follows that x E CN(H,) and H, = Hi < H2.

(b) Clearly, CH(x) = CH(x)x < H n Hx. Conversely, (H nHx)x H and hence (a)

yields that x e CN(H n Hx), that is, H n HX < CH(x). This proves the first assertion.By (6) of § 1.5, H normalizes [x, H] so that [x, H] H is a subgroup of G containing H.The trivial identities [x, a-'] a = ax and [x, a] = (a-' )Xa show that [x, H] H alsocontains HX and that [x, H] < H u Hx. Thus [x, H] H = H u HX and (b) holds.

(c) Now (b) and G = HN yield that HG = n Hx = n Hx = n CH(x) _XEG xEN xEN

CH(N) and, similarly, HG = U [x, H]H = [N, H]H. By Dedekind, HG n N =xEN

[N, H] (H n N) = [N, H].

4.1.2 Theorem. Let G be a finite group, N a normal Hall subgroup and H a comple-ment to N in G. Suppose that cp is a projectivity from G to a group G such that N' isa normal Hall subgroup of G. Then for all N, < N and H1 < H,

4.1 Normal Hall subgroups 159

(a) H1 < NG(N,) if and only if HI < Ni(N;),(b) CN(H1)' = CN,(HI) and CH(N1)" = CH..(Nr),(c) IN:CN(H1)I =IMP: CN-(Hf)I,(d) [N1, H1 ]10 = [Ni, Hi] if H1 <_ Nc(N1)

Proof. (a) If H1 < NG(N1), then N1 H1 is a subgroup of G and N1 H1 n N = N1;conversely, if N1 = (N1 u H1) n N then N1 a N1 u H1 since N a G. ThusH1 < NG(N1) if and only if N1 = (N1 u H1) n N. This is equivalent to N; _(Nw u H;) n N° and hence to HI <

(b) By 1.6.7, H; is centralized by CN(H1)`0, that is, CN(H1)`° < The corre-sponding assertion for (p-1 yields the other inclusion. Similarly, CH(N1)`° = CHm(N;).

(c) By the Schur-Zassenhaus Theorem, IN: CN(H1)I = IN: NN(H1)I is the num-ber of complements to N in NH1. This implies (c).

(d) Since H1 < NG(N1), N1 is a normal Hall subgroup of N1 H1 and by (a), N; is anormal Hall subgroup of NPHI*. By (c) of 4.1.1 and the Schur-Zassenhaus Theorem,[N1,H1] is the intersection of N1 with the join of the complements to N1 in N1 H1,and is therefore mapped by (p to the intersection of Nl with the join of the comple-ments to N" in N;HI, that is, to [N;, HI]. El

The subgroup lattice of a finite group with a normal Hall subgroup

To construct the subgroup lattice, we need some further properties of a group witha normal Hall subgroup.

4.1.3 Lemma. Let N be a normal Hall subgroup and H a complement to N in G.(a) For every U < G satisfying (I U 1, I N 1) = 1, there exists x e N such that Ux < H.(b) N = [N, H] CN(H) and [N, H, H] = [N, H].(c) If N is abelian, then N = [N, H] x CN(H).(d) For every prime p, there exists an H-invariant Sylow p-subgroup of N.

Proof. (a) Since NU = N(NU n H), U and NU n H are complements to N in NU.By the Schur-Zassenhaus Theorem there exists x e N such that Ux = NU n H < H.

(b) By 4.1.1, [N, H] is a normal Hall subgroup in H' = [N, H] H. For x E N, Hand Hx are complements to [N, H] in HG and hence there exists y c [N, H] suchthat Hx = Hy. Then x = (xy-1)y e NN(H) [N, H] and, since x was arbitrary, N =[N, H] NN(H) = [N, H] CN(H). Furthermore, again by 4.1.1,

[N, H, H] H = U Hx = U Hx = [N, H] Hxe[N,H] xeN

and hence [N, H, H] = [N, H].(c) By (b), we only have to show that [N, H] n CN(H) = 1. For this we study the

endomorphism r of N defined by x` = fl xh for x e N. If a E H and x e N, thenheH

(Xa)t = l Xah = fl xh = xT and hence [x, a]' = (X-1)t(Xa)t = (xr)-1xt = 1. SinceheH bEH l

[N, H] is generated by these [x, a] and r is a homomorphism, yT = 1 for all

y c [N, H]. So if x e [N, H] n CN(H), then 1 = x` = xini and x = I since (o(x), I H 1) = 1.Thus [N, H] n CN(H) = I and (c) holds.

(d) Let P E Sylp(N). By the Frattini argument, G = NG(P)N and hence N n NG(P)is a normal Hall subgroup of NG(P) having a complement K isomorphic to H. By (a)there exists x e N such that KX < H and it follows that H = KX < NG(P)X = NG(Px).Thus PX is an H-invariant Sylow p-subgroup of N.

4.1.4 Theorem (Paulsen [1975]). Let G be a finite group, N a normal Hall subgroupand H a complement to N in G. Then L(G) = {Nx H; Jx E N, N, < N, H, < NH(N, )}and N; Hi < NzHz (for x, y e N, N,, N2 < N, H. < NH(N;)) ifand only if H, <_ H2,N; `' N2 and xy ' E CN(H, )N2. (Note that, in general, CN(H, )N2 is not a subgroupoJ' N.)

Proof'. If H, < N,,(N, ), then N; Hi = (N, H, )X is a subgroup of G. Conversely,suppose that U < G. Then U n N is a normal Hall subgroup of U and thereforepossesses a complement K in U. Since JKj = I U: U n NI = I UN: NI is prime to INI,it follows from 4.1.3(a) that there exists v c N such that K` < H. So if we put N, _(U n N)`, H, = K" and x = v-', then U = (U n N)K = N; H; as desired.

To prove the second assertion of the theorem, assume first that N; H; < N2Hz,that is, N;' 'H;'' N2H2. Then clearly N;'' N2H2 n N = N2. Furthermore,applying 4.1.3(a) to the subgroup H;y ' of N2H2, we see that there is z e N2 suchthat H;'' < H2. Then 4.1.1(a) shows that H, < H2 and xy-' z E CN(H, ), thatis, xy-' E C,, (H,)N2. Conversely, suppose that H, < H2, Nil' ' < N2 and xy-' EC^,(H, )N2. Then there are s e CN(H1) and t e N2 such that xy-' = st. It follows that

Nx,. Hi,., =Nii. 1H'1 SN2Hi=N2H2

and hence N; H; <NZ Hz.

4.1.5 Remark. In certain situations our description of L(G) in 4.1.4 becomessimpler. For example, if N is abelian or hamiltonian, then N; = N, for all x E N.Therefore in this case, L(G) = IN, H; l x E N, N, < N, H, < NH(N,) } and N, H; <_N2Hz if and only if H, < H2, N, < N2 and xy ' e CN(H, )N2i here CN(H, )N2 is asubgroup of N.

Projectivities of finite groups with normal Hall subgroups

We use Theorem 4.1.4 to construct projectivities between groups with isomorphicnormal Hall subgroups.

4.1.6 Theorem. Let G = NH and G = MK he finite groups with normal Hall sub-groups N and M and complements H and K, respectively. Suppose that 6: N -a M isan isomorphism and r: L(H) -+ L(K) is a projectivity such that for every N, < N andfor every cyclic subgroup H, of prime power order of H,

(1) C,v(H, )° = CM(H,) and

(2) H, <_ NH(N1) if and only if Hi < NK(Ni ).

Then the map cp: L(G) -. L(G) defined by (N, Hi )v = (N;)X°(H'j)X°for x e N, N, < N,H, < NH(N1) is a projectivity from G to G.

Proof: Since rr is a projectivity, it is clear that if (1) and (2) hold for two subgroupsH, and H2 of H, then they also hold for H, u H2. Therefore our assumptions implythat (1) and (2) are satisfied for every subgroup H, of H. Now suppose that x, y c N,N1, N2 < N and Hi < NH(N;). Then by (2), H, < NK(N') so that (Nl )X°(H,)X° =(Ni H; )X° is a subgroup of G. And by 4.1.4, NxHi < NZHZ if and only if H, < H2,N;'' ' < N2 and xy-' E C,,,(H1)N2. Since a is an isomorphism and r a projectivitysatisfying (1), this is the case if and only if H; < Hz, (N' )X°('°) I = (Ni' )° < NZ andxY°) ' = (xY E CN(H, )° Nz = CM(H; )Nz, that is if and only if (Nl )X°(H' )X° <(N2)1°(HZ)'°. This shows that cp is well-defined and preserves inclusion. The sameholds for the map : L(G) - L(G) defined with respect to a-' and r-' in the sameway as cp with respect to o and T. Of course, U"y = U for all U < G and V101* = Vfor all V < G. Thus cp is bijective and, by 1.1.2, it is a projectivity from G to G.

To illustrate Theorems 4.1.4 and 4.1.6, we study the simple but interesting situa-tion where N is elementary abelian and H is cyclic of prime order.

4.1.7 Example. Let p and q be different primes, F = GF(p) and suppose that I H I = q.(a) Let G = NH where N is an elementary abelian normal p-subgroup of G. Then

it is well-known that we can regard N as an FH-module and by Maschke's Theorem(see Robinson [1982], p. 209), this FH-module is completely reducible. Thus

(3) N = CN(H) x N, x x N,

where the Ni are nontrivial irreducible FH-modules.(b) It is also well-known (see Huppert [1967], p. 166) that all nontrivial irreduc-

ible FH-modules N have the same order p'", where m is the smallest natural numbersuch that glpm - 1, and all these modules lead to isomorphic semidirect productsG = NH. By 4.1.5, the subgroups of G are of the form N, H; where x e N, N, < Nand H, < NH(N,). Since I H I = q, only H, = 1 or H, = H is possible. If H, = 1, then

L(G) for p = 5, q = 3, m = 2Figure 14

N, Hi = N, < N; for H1 = H, since N is FH-irreducible, either N, = N or N1 = 1,and then N1 H; = NH" = G or N, Hi = Hx. Therefore L(G) consists of G, the sub-groups of N and the INI conjugates of H. In particular, this lattice does not dependon the prime q. Therefore, if we take another prime r such that rl p' - 1 and r,( p' - 1for j < in, a group Kof order r and a nontrivial irreducible FK-module M, then thesemidirect product G = MK is lattice-isomorphic to G. For m = 1, G and G arenonabelian groups in P(2, p). But for m > 2 we get, in addition to the cyclic groupsand the P-groups, another class of examples of lattice-isomorphic groups with differ-ent orders. The smallest primes are p = 29, q = 3, r = 5 for m = 2 and p = 11, q = 7,

r = 19 form=3.(c) In the case m = 1, that is, q1 p - 1, a fixed generator a of H induces power

automorphisms x -+ x` in the irreducible FH-modules Ni. Since Exp N = p, we mayregard these:: as (nonzero) elements of the field F = l/(p) and by (3), N is the directproduct of the eigenspaces

E(i.)=E(...,a,N)= {xENIx°=x'-}

of a. In general, for an automorphism x of N of order dividing p - 1, we define thetype of x to be (do;d...... dk) if pdo = IE(1,x,N)I = IC,v(x)I and d1 > - - - > dk are thedimensions of the eigenspaces of x to eigenvalues ., 1. We shall show in 4.1.8 thattwo groups N<a) and N<b) of the form considered are lattice-isomorphic if theautomorphisms induced by a and b in N have the same type. It is not so easy todecide whether they are isomorphic; this is the case if and only if <a) and <b) induceautomorphism groups in N that are conjugate in Aut N (see Exercise 3).

(d) If there are two different primes q and r dividing p - 1, then we can constructgroups N <a> and N <b) with o(a) = q and o(b) = r such that a and h induceautomorphisms of the same type in N. So we get many further examples of lattice-isomorphic groups having different orders of the form p"q and p"r. The followingtable shows all types for n < 2 and the resulting groups of order p"q; in these smallcases it is also often clear from earlier results, for example 1.2.8, 1.6.4 or 2.2.3, thatall groups of the given type are lattice-isomorphic.

type group

CP X Cq

nonabelian group of order pqCPxC,,xCqCP x F where F is nonabelian of order pqnonabelian P-group of order p2q<x,y,alxP=yP=a"=[X,y]= 1,xa=XS,y°=}`)where 1 <s<tp - 1 and s`' = I = tQ (mod p)

The last groups in the table are the Rottlander groups which we shall study moreclosely in 5.6.8. There are (q - 1)/2 nonisomorphic groups of this type and orderp2q

4.1.8 Theorem. Let q and r be (not necessarily distinct) primes dividing p - 1 andsuppose that G = N<a> and G = M with normal elementary abelian p-subgroupsN, M and o(a) = q. o(h) = r. Then there exists a projectivity cp from G to G satisfyingN`° = M and <a>" = if and only if the automorphisms induced by a on N and byh on M have the same type.

Proof. Let of and /3 be the automorphisms induced by a and b in N and M, respec-tively. Suppose that the type of x is (do; d1 . ... .dk), that of /3 is (eo; e1.... . e,), and letN = Do x x Dk and M = Eo x x E, be the decompositions of N and M intoeigenspaces of x and /3, respectively, such that Do = CN(a), Eo = CM(b), I D; I = pe, and

I E; I = pe- for all i. _Suppose first that cp is a projectivity from G to G satisfying N" = M and <a>" _

. Then C,N,(a)' = CM(b) by 4.1.2(b). If D is an eigenspace of x, then a normalizesevery subgroup of D. Conversely, if F is a subgroup of N such that a normalizesevery subgroup of F, then by 1.5.4, a induces a universal power automorphism on F,that is, F is contained in an eigenspace of x. Thus the eigenspaces of a are preciselythe subgroups of N maximal with the property that all their subgroups are normal-ized by a. By 4.1.2, these subgroups are mapped by (p to the corresponding sub-groups of M with respect to b, that is, to the eigenspaces of /3. It follows that d; = e;for all i and of and /3 have the same type.

Conversely, if x and /3 have the same type, then k = l and J DJ = I E; I for all i. Hencethere exists an isomorphism a: N - M such that D; = E. for all i. This isomorphisma and the trivial autoprojectivity t from <a> to satisfy (1) and (2). For, CN(a)° _Do = Eo = CM(b) and, as we have seen in 4.1.7, a subgroup F of N is normalized bya if and only if F is a direct product of subgroups of the eigenspaces of a, that is, ofthe Di. These subgroups are mapped by a onto the direct products of subgroups ofthe E,, that is, onto the b-invariant subgroups of M. By 4.1.6 there is a projectivitycp from G to G satisfying N' = M and <a>'* = <h>.

As another application of Theorem 4.1.4, we show how a projectivity ofG/ CH(N) = G/ HG can be lifted to a projectivity of G.

4.1.9 Theorem. Let G = NH and G = MK be finite groups with normal Hall sub-groups N and M and complements H and K, respectively. Let u he a projectivity fromG/CH(N) to G/CK(M) satisfying (NCH(N),/CH(N))µ = MCK(M)/CK(M), let r be aprojectivity from H to K such that CH(N)` = CK(M) and suppose that p and t inducethe same projectivity in H/CH(N). Then there exists a projectivity cp from G to Ginducing p in G/CH(N) and t in H.

Proof. For X < G let X = XCH(N) and let j be the map induced by p in the interval[G/CH(N)], that is, (X/CH(N))" = X"/CK(M) for CH(N) < X < G.

Let U < G. Then by 4.1.4 there exist x c N, N1 < N, H1 < NH(N,) and x c M,M1 < M, K 1 < NK(M,) such that

(4) U = N' H', U = Ni (Hl CH(N))X = Nl Hi and U" = M' K'*

We want to define U1* as (M, H',)'. In the first place this is a subgroup of G; indeedCK(M) < K, since CK(M) < U"; therefore

Hi = Hi = (NH1 n H)" = (NU n H)" = Nµ Uµ n H" = MCK(M)Mf K' n K

=MKinK=K1,

that is,

(5) Hi < Hi = K1 < NK(M1)

Furthermore, N; Hi = U < V = NzHz (y E N, N2 < N, H2 < NH(N2)) implies that(M, K1)x = U" < V" _ (M2K2)y and, by 4.1.4, H, < H2, K, < K2, MV-' < M2 andxy 1 E CM(K1)M2. It follows that H; < HZ and xy ' e CM(H;)M2 since H, < K1.Again by 4.1.4, (M1 HT)' < (M2 H2T)Y. So if we put U" = (M1 H')', then cp is well-defined and preserves inclusion. The same holds for the map 0: L(G) L(G) definedwith respect to p-1 and T-1 in the same way as cp with respect to p and T. NowCH(M)T = CK(M) and (5) imply that

(6) U`°CK(M) = Mi (Hi CK(M))x = Mi (Hi )x = M1 K1

= U",

hence (U°CK(M))" ' = U = N1 Hi and therefore, finally, (Ul')O = (N1(H;)T-')x = U.Since the situation is symmetric in cp and 0, it follows that also (W'')° = W for allW< G. Thus cp is bijective and, by 1.1.2, it is a projectivity from G to G. If CH(N) < U,then U = U and, by (6), UP = U". Hence cp induces p in G/CH(N) and it triviallyinduces r in H. F1

Autoprojectivities

The group P(G) of autoprojectivities of a group G with normal Hall subgroup N andcomplement H was studied by Paulsen [1975]. His results are rather technical. SinceN is the only subgroup of its order, every autoprojectivity t/i of G that does not fixN satisfies IN''I 0 INS; in the next section we shall study this rare situation to obtaincomplete information about autoprojectivities of this type. If NO = N, then 0induces autoprojectivities in N and in G/N and operates on the set _t- of com-plements to N in G. And i/i is determined by its action on these three subsets of L(G),that is on L(N), [G/N] and '; for, if x e N and H1 < H, then H; = NH1 n Hx.There is not much more that can be said in general. If CN(H) = 1, the operation of clion . '' yields via the Schur-Zassenhaus Theorem a permutation of N which we studyfor two special cases in the last theorem of this section and in Exercise 5.

4.1.10 Theorem. Let G be a finite group, N a normal Hall subgroup and H a comple-ment to N in G such that CN(H) = 1. Suppose that 0 is an autoprojectivity of G whichsatisfies NO = N and induces the trivial autoprojectivity in H; define p: N -+ N by(H')0 = Hx" for x e N.

(a) Then p is a permutation of N and 1° = 1.(b) For every H1 H and x e N, (Hi)p' = Hi°; thus li is determined by p and its

action on L(N).

(c) If H1 < H and L is an H1-invariant normal subgroup of N such that CN(H,) < Land L' = L, then (xL)P = xPL for all x e N; in particular, L° = L.

Proof. (a) For x, y e N, H" = H'' if and only if xy-' E NN(H) = CN(H) = 1, that is,x = y. Since 0 maps complements of N onto complements of N, the Schur-Zassenhaus Theorem shows that p is well-defined and bijective. As H" = H, 1P = 1.

(b) Since N" = N and Hi = H1 we have (Hi)V' = (NH1 n Hx)4' = NH1 n Hx°H;°; by 4.1.4, 0 is determined by p and its action on L(N).

(c) For x e N, let S(x) y e NILHi n H'' = H; }. Then y e S(x) if and only if(LHi n H'')'' = (Hi )'' and (b) shows that this is the case if and only if LHi° n Hy' _Hi', that is, yP E S(xP). Hence S(x)P = S(xP), and it suffices to show that S(x) = xL.If y e S(x), then H; < LHi and hence LHi = LHi; by the Schur-Zassenhaus Theo-rem there exists z e L such that Hi= = H; . It follows that xzy-' E CN(H,) < L andtherefore y e xL since L a N. Conversely, if y = xz where z e L, then LHi n H'° =(LHi )z n H'' = LHi n H'' = Hi and hence y e S(x). Thus S(x) = xL and (c) holds.

1-1

Exercises

1. Let G be a finite group, N a normal Hall subgroup and H a complement toN in G. Show that L(G) = {N, Hi I x e N, N, < N, H, < H, Hi < NG(N,) } andN, Hi < N2 H2 (for x, y e N, N1, N2 < N, Hl, H2 < H, Hi < NG (N,) and HZ <NG(N2)) if and only if H1 < H2, N1 < N2 and x e CN(H,)yN2.

2. For p > 2, let N= (x, y I xp2 = yP = [x, y] = 1 > be the abelian group of type(p2, p) and M = (u, vluP2 = vP = 1, u" = u' {'P> the nonabelian group of order p3lattice-isomorphic to N. Let G = N<a> and

G/=M be semidirect products of

N and M by cyclic groups of order 2 with respect to the automorphisms given by' b = '1' =x'=x- 'y° = y and uu 'V v. Show that there are projectivites a from N

to M and r from H = <a> to K = satisfying (1) and (2), but that G and G arenot lattice-isomorphic. (This shows that we cannot replace the isomorphism a inTheorem 4.1.6 by a projectivity.)

3. Let N be a group, H and K subgroups of Aut N and G = NH and G = NKthe corresponding semidirect products. Show that there exists an isomorphisma: G -+ G satisfying N° = N and H° = K if and only if H and K are conjugate inAut N.

4. Let G = NH and G = MK be finite groups with normal Hall subgroups N andM and complements H and K, respectively, and suppose that CN(H) = I =CM(K). Show that there exists a projectivity cp from G to G satisfying N`0 = M ifand only if there are projectivities a from N to M and r from H to K and abijective map 7t: N -. M such that for all x, y e N, N, < N and H, <_ H,(a) H, < NH(N,) if and only if Hi < NK(Nl ), and(b) if H, < NH(N,), then xy-' e CN(H,)N, if and only if x"(y")-' e CM(Hi)N'

5. Let G be a finite group, N an abelian normal Hall subgroup, H a complement toN in G and suppose that CN(H) = 1.

(a) Let it be a permutation of N such that if N, < N and H, < NH(N,) thenxy-' c CN(H, )N, if and only if x" (y")-' e CN(H, )N,. Show that q : L(G) -L(G) defined by (N, Hi )`0° = N, Hf is an autoprojectivity of G. Denote thegroup of all these autoprojectivities by F.

(b) Let r be an autoprojectivity of H such that for all H, < H, CN(Hi) = CN(H, )and Hi and H, normalize the same subgroups of N. Show that ,: L(G) --9L(G) defined by (N, H' )O, = N, (H')' is an autoprojectivity of G. Denote thegroup of all these autoprojectivities by A.

(c) Let A = {0 c- P(G) I U0 = U for all U < N}. Show that A = I x A.

4.2 Singular projectivities

We already know many projectivities that map a group G onto a group of differentorder; this may happen, for example, if G is cyclic (see 1.2.8), a P-group (see 2.2.3),or for the groups in 4.1.7. On the other hand, Theorem 2.2.6 states that for aprojectivity 9 of a nonabelian p-group G, I G" I= I G I and then also I H4'I= I H I forevery subgroup H of G. In this section we want to show that this last property holdsfor projectivities of many more groups and, in fact, reflects the normal behaviour ofa projectivity. We first of all give it a name.

Index preserving and singular projectivities

A projectivity cp of a group G is called index preserving if it satisfies IHI: K`°I =I H : K I for every pair H, K of subgroups of G such that K < H; cp is called singularif it is not index preserving. An index preserving projectivity in particular satisfiesH"=IHI for every subgroup H of G, and from Lagrange's Theorem it is clear that

for a finite group G the converse holds: cp is index preserving if IHII = IHI for allH < G. But we can say a little more.

4.2.1 Lemma. If cp is a projectivity of a finite group G satisfying IS`°I = I S I for everySylow subgroup S of G, then cp is index preserving.

Proof. Assume that K < H < G, let p be a prime and let P e Sylp(H); takeS E Syl, (G) such that P < S. Since SP is a p-group, I P`° I = I P I. If P° would beproperly contained in a p-subgroup T10 of Hw, then P < T < H and T would containan element x of prime order q p. For a Sylow q-subgroup Q of G containing x,p = I <x> I would divide IQ PI = I Q I, a contradiction. Thus P° is a Sylow p-subgroupof HP and hence the p-part of I HP I is the same as that of IHI. Since this holds forall primes p, it follows that IH`°I = IHI. Similarly, IK'I = IKI and by Lagrange'sTheorem, IHI*: K10I = IH`°I/IK`'I = IHI/IKI = IH: KI.

The above lemma suggests the introduction of the following concepts.

4.2 Singular projectivities 167

p-singular projectivities

Let p be a prime and G a finite group. A projectivity cp from G to a group G is calledp-singular or singular at p if there exists a p-subgroup (or, equivalently, a Sylowp-subgroup) P of G such that IP'I IPI; cp is called p-regular or regular at p if it isnot p-singular. Lemma 4.2.1 shows that if cp is not index preserving, then there existsa prime p such that cp is singular at p. By 2.2.6 the Sylow p-subgroups of G are thencyclic or elementary abelian and we are looking for further restrictions on the struc-ture of G and G if G has composite order. However, if T is any group of order primeto I G I and 161, then by 1.6.4 there exists a projectivity from G x T to G x T whichinduces cp in G and therefore is singular at p. Hence there is no global restriction onthe structure of a group with a p-singular projectivity. We can only hope for restric-tions on its p-structure, for example on the embedding of its Sylow subgroups. Wedistinguish two kinds of singularities at p.

4.2.2 Lemma. Let q be a projectivity from G to G and suppose that P is a Sylowp-subgroup of G such that I P°I 0 P1. Assume further that there is no P-group S < Gcontaining P as a proper normal subgroup (we call cp singular of the first kind at p inthis case). Then P < Z(NG(P)) and hence G has a normal p-complement.

Proof. We have to show that P< Z(NG(P)); then Burnside's theorem (see Robinson[1982], p. 280) yields that G has a normal p-complement. So we suppose that thisassertion is wrong and take a counterexample G of minimal order. If P < H < G,then H and the projectivity induced by cp in H satisfy the assumptions of the lemma;the minimality of G implies that P < Z(Nf(P)). In particular, if NG(P) < G, thenP< Z(NNG(p)(P)) = Z(NG(P)), contradicting the choice of G. Thus

(1) P a G,

(2) P <_ Z(H) for all P < H < G, and

(3) P Z(G)

since G is a counterexample. It follows that there exists a cyclic subgroup Q of primepower order in G such that Q CG(P). By (1) and (2) G = PQ, and, by 2.2.6, P iscyclic or elementary abelian. Thus P G and hence Q is a q-group for some primeq 96 p. We want to show that I Q I = q and assume, for a contradiction, that IQI > q2.If D is the maximal subgroup of Q, PD < G and, again by (2), D is centralized by P.By 1.6.7, [D", P"] = 1 and since Q and QO are cyclic, it follows that D < Z(G) andD" < Z(G). Thus q induces a projectivity 0 from G/D to 6/D" that maps the Sylowp-subgroup PD/D of G/D onto the subgroup P`0D°/D`0 of order I P°I 0 JPD/DJ ofG/D`0. If G/D were a P-group and G/D e P(m, p), then also G/D° e P(m, p) by 2.2.5.Since iP is singular at p, G/D`° would be nonabelian and since the Sylow p-subgroupof G/D is not mapped onto the Sylow p-subgroup of G/D', there would exist asubgroup Q 1 /D of order q in G/D such that I QI*/D" I = p. Since G/D is generated byits subgroups of order q, there would also exist Q2/D < G/D of order q such thatQz/D101 96 p. Then Q1 and Q2 would be Sylow q-subgroups of G and hence Q`1° and

Qz would be cyclic of prime power order. Since 1 D" < Q1' n QZ, the primes have

to be the same; but IQ;/D`°I = p 0 IQZ/D`°1. This contradiction shows that G/D is nota P-group and the minimality of G implies that PD/D < Z(G/D). It follows thatQ < G and hence G = P x Q is abelian, contradicting (3). We have shown that

(4) G = PQ andJQI=gwhere p0gEP.

We want to show next that P is not cyclic. So suppose, for a contradiction, that P iscyclic. If JP1 = p, then P would be a proper normal subgroup of the P-group G,contradicting our assumption. Thus IPI = p" where n > 2. If x e G then ((P")')"-' isa cyclic subgroup of G of order t" for some prime t. Since I G I = p"q and n > 2, t = pand hence ((P°)X)`0-' = P, the only Sylow p-subgroup of G. Thus P' < G and cpinduces a projectivity from G/D(P) to G/D(P4'). Since Q operates nontriviallyon P, the group G/D(P) is not cyclic; hence G/(D(P) E P(2, p) and therefore alsoG/ b(P`0) E P(2, p). Now P" < G implies that I P"/'(P`°)I = p and since P° is alsocyclic of prime power order, it follows that IP`PI = p" = JPJ, a contradiction. Thus Pis not cyclic and by 2.2.6,

(5) P is elementary abelian of order p" and P° is a nonabelian P-group of orderp"-'r where n>2andp>rcP.

Let Po be the Sylow p-subgroup of P". If P" < G, then Po g G since it is characteris-tic in Pq'. It follows that Po = Po Q" n P°; hence PO = (P0 U Q) nP a PO U Q andtherefore Po < G. If P° is not normal in G, then there exists x e G such that P' 0P°'. By 2.2.5, Pl"'*-' lies in P(n,p) and since it is different from P, !P`°,`' ' I = p"-' qThen I P n P' I = p"-' and P n P`""-' < P u P°X`°-' = G. Therefore, in bothcases, there is a maximal subgroup M of P such that M < G. By Maschke's theorem(see Robinson [1982], p. 209) there exists a Q-invariant subgroup T of P such thatP = M x T; clearly, also T:9 G. Suppose first that I T'1 = p. Then T< P0; henceM Po and therefore M10 is not a p-group. Thus cp induces a p-singular projectivityfrom MQ < G to (MQ)" and the minimality of G implies that MQ is a P-group orM < Z(MQ). In both cases, every subgroup of M is normalized by Q and hence is anormal subgroup of G. Since r divides IM101,

(6) there exists a normal subgroup R of G contained in P such that JR101 = r.

This is trivial if I P : p, since we can take R = T in that case; thus (6) holds ingeneral. Now if X is any subgroup of P such that IXP I = r, then by Sylow's Theoremthere exists z e PP such that R"' = X". Since R a G, either RQ E P(2, p) or RQ iscyclic; hence also (RQ)`°ZP = X u QQZ'P ' E P(2, p) or this is cyclic. Since I X I = p, itfollows that X < X u QWZ`°-'. As z e P°, Q`"`°-' P and hence X < G. If Y < P suchthat I Y`°I = p, then (R Y)`0 E P(2, p) has order pr and every minimal subgroup of R Ydifferent from Y is mapped by q to a subgroup of order r in G. These groups arenormalized by Q, as we have just shown, and hence also R Y and Y are normalizedby Q. Thus Q induces a power automorphism in P and, since P Z(G), it followsthat G is a P-group. This contradicts our assumption on G.

4.2.3 Lemma. Let cp be a projectivity from G to G and suppose that P is a Sylowp-subgroup of G such that IP`°I:IPI. Assume further that there exists a P-groupS < G containing P as a proper normal subgroup (we call cp singular of the second kind

at p in this case). Then there exists a subgroup K of G such that G = S x K and(ISI>IKI)= 1.

Proof. Again we suppose that the assertion is wrong and consider a counterexampleG of minimal order. By 2.2.1, ISI = p"-'q where n > 2 and p > q E P. ThenS`° E P(n, p) and, since IP"I IPI, we see that S° is not abelian and (p does not mapthe Sylow p-subgroup of S to that of S°. It follows that there exists a subgroup Q oforder q in S such that IQ"I = p. Thus cp is singular at q and (p -' is singular at p, hencethe Sylow q-subgroups of G and the Sylow p-subgroups of G are cyclic or elementaryabelian. If Q were contained in a P-group H as a normal subgroup, then H E P(m, q)for some m E N. Hence also H° E P(m, q) and q would be the largest prime dividingIHI1; but IQcI = p > q divides IHOI, a contradiction. Thus:

(7) There is no P-group H < G containing Q as a normal subgroup.

We want to show next that Q is a Sylow q-subgroup of G. Suppose that this is wrongand take T E Sylq (G) such that Q < T. By (7), T is not elementary abelian and hencecyclic. Since IQc1 = p, T° is a cyclic p-group of order at least pz and therefore theSylow p-subgroups of G are not elementary abelian. It follows that cp-' is singularof the first kind at p; by 4.2.2 the group G, and hence also every subgroup of G, hasa normal p-complement. But Sc has none, a contradiction. Thus

(8) Q E Sylq (G).

By (7), qp is singular of the first kind at q and by 4.2.2 there exists a normal q-complement N in G, that is

(9) NaGsuch that G=NQ and NnQ= 1.

Certainly P < N and hence cp induces a p-singular projectivity from N to N`°.We want to show that this singularity is of the first kind. So suppose, for a contra-diction, that T< N is a P-group containing P as a proper normal subgroup.Then T E P(n, p) and T and S are contained in NG(P). Therefore, if x E S`° thenT"""' ' < NG(P) and T""I" e P(n,p). Since P is the only Sylow p-subgroup of NG(P),it follows that P < T"`° '. Furthermore S`"`' ' = S and since S N, S n T = P.Therefore Pwxc-' = S`0x`0-' n Tcxc-' > P, that is, (P`°)x = P°. Thus P" :9 S4' and by2.2.2, 1PcI = p"-' = IP1, contradicting the assumption of the lemma. Hence the pro-jectivity induced by cp in N is singular of the first kind at p and by 4.2.2 there existsa normal p-complement K in N. Thus N = KP and K n P = 1, and since K ischaracteristic in N g G, it follows that

(10) K aGsuch that G=KS and KnS= 1.

If S a G, then G = S x K is not a counterexample since (ISI, IKI) = 1. Thus

(11) S is not normal in G.

Suppose that K is not a primary group, and let r be a prime dividing I K I. By 4.1.3(d)there exists an S-invariant Sylow r-subgroup R of K and cp induces a projectivityfrom RS to (RS)° that is singular of the second kind at p. Since K is not an r-group,RS < G and the minimality of G implies that S :!g RS. Now K is generated by its

Sylow subgroups and it follows that K < NG(S). Hence S a G, contradicting (11).Thus

(12) K is an r-group for some prime r.

Let M be a maximal subgroup of G containing S. By (10), M = LS where L =M n K and the minimality of G implies that M = L x S. Since K is an r-group andL < K, there exists g e NK(L)\L. Now M9 = M would imply that M g <M, g> = Gand hence that S < G as a characteristic subgroup of M; this would contradict (11).Thus M90- M and M9 = Lq x S9 = L x S9 implies that L a M u M9 = G. By 1.6.7,M" = L" x S" and (M9)" = L" x (S9)"; hence also L" g M" u (M9)" = G. It followsthat cp induces a projectivity from GIL to G/L" that maps PL/L to P"L"/L" P"and SL/L is a P-group containing PL/L as a proper normal subgroup. Therefore ifL 1, the induction assumption would imply that SL/L o GIL and hence thatM = SL a G, contradicting M9 0 M. Thus L = 1, that is,

(13) S is a maximal subgroup of G.

Let Po be the Sylow p-subgroup of S". Then P n Po is a maximal subgroup ofP. We want to show that I P I = p and suppose, for a contradiction, that I P I > p2.Then there exists a maximal subgroup D of P different from P n P0. And sinceSID e P(2, p), there are two different maximal subgroups Sl and S2 of S such thatS1 # P : S2 and S1 n S2 = D. Then D = Si n P is a Sylow p-subgroup in G. = KS;for i = 1, 2. Since D PO, we see that D" is not a p-group and S. < G; is a P-groupcontaining D as a proper normal subgroup. Thus cp induces projectivities in the G.which are singular of the second kind at p and the minimality of G implies thatSi a G. for i = 1, 2. Hence St, S2 and therefore also S = S1 U S2 are normalized by K.It follows that S a G, a contradiction. Thus I P I= p and

(14) ISI = pq.

Let x e S" such that (P")" # P". Then z: L(G) - L(G) defined by X` = X"""' ' forX < G is an autoprojectivity of G satisfying P 0 PT < S` = S. By (14), P` is a Sylowq-subgroup of S and S = P u PT. Since S cannot operate fixed-point-freely on K (seeRobinson [1982], p. 298), there exist nontrivial subgroups H of K and U of S suchthat [H, U]= 1, that is,HU=H x U.IfP<U,thenPgHuS=Gby(13).Andif P U, then U is a Sylow q-subgroup of S and there exists y e S such that Uy = PT.Then (HU)y = Hy x PT and, by 1.6.7, (HU)yT-' = Hyt ' x P. Since CS(P) = P, wehave Hy' ' S and hence P a HyT-' u S = G. Thus in both cases P 9 G, hencePK = P x K and (PK)T = PT x KT. Consequently KT normalizes P and PT and there-fore also P u PT = S. Since K' u S = (K u S)T = G, it follows that S g G, a finalcontradiction.

The following terminology will simplify our discussion.

P-decompositions

A P-decomposition of a finite group G is a pair (S, T) of subgroups of G such thatG = S x T, (BSI, 1 TS) = 1 and S is a P-group. We call G P-decomposable if there

exists a P-decomposition of G; otherwise G is called P-indecomposable. Note thatT is allowed to be trivial so that a P-group is P-decomposable. Note furtherthat P-decompositions are preserved by projectivities; they are even fixed byautoprojectivities.

4.2.4 Lemma. Let (S, T) be a P-decomposition of G. _(a) If cp is a projectivity from G to G, then (S`°, T`°) is a P-decomposition of G.(b) If a is an autoprojectivity of G, then S° = S and T° = T.

Proof. (a) By 1.6.6, G = S° x T`° where (IS" I, I T" I) = 1 and, by 2.2.5, S" is a P-group.Thus (S`°, T") is a P-decomposition of G.

(b) Suppose that S e P(n, p) and let P be the Sylow p-subgroup of S. By (a), G =S° x T" where (I S% I T°I) = I and S° e P(n, p). Therefore p divides I S' I and hence T"is a p'-group and P < S". It follows that T° < C0(P) = P x T and, as a p'-group,T° < T. Since a finite group has no projectivity onto a proper subgroup, T° = T. By1.6.4, T has only one complement in G and this implies that S° = S.

We show next that autoprojectivities of P-indecomposable groups map Sylowsubgroups onto Sylow subgroups.

4.2.5 Theorem. Let p be a prime, G a finite group, P a Sylow p-subgroup of G,I P I = p" where n e f\i and suppose that there is no P-decomposition (S, T) of G suchthat P < S.

(a) Then P° is a Sylow subgroup of G for every a a P(G).(b) Let it be the set of primes dividing I P"I for some a e P(G). If it {p}, then P is

cyclic and there exists a normal n-complement D in G with cyclic factor group G/ D

of order n q"). Furthermore D = D` for all r e P(G) and P(G) operates transitivelyqE"

on the set of normal q-complements (q e it) of G.

Proof. We first prove the following assertion.

(15) Let q be a prime, Q a Sylq(G) and r a P(G) such that P = Q or P < Q`. IfI Q`I - IQ I, then Q is cyclic and has a normal complement in G.

For, if r were singular of the second kind at q, then by 4.2.3 there would exist aP-decomposition (S, T) of G such that Q < S. Then also Q' < S` = S by 4.2.4. SinceP = Q or P < Qt, it would follow that P < S, contradicting our assumption. Thus ris singular of the first kind at q and by 4.2.2 there exists a normal q-complement inG. It follows that Q` also has a normal q-complement and therefore cannot be aP-group of order qm'r where q > r e P; by 2.2.6, Q is cyclic. This proves (15).

(a) Suppose, for a contradiction, that P° is not a Sylow subgroup of G. Then, inparticular, IP"I # IPI and, by (15), P is cyclic. Hence P° is a q-group for some primeq and there exists a Sylow q-subgroup Q of G such that P° < Q. Since P a Syl°(G),the subgroup Q" ' is not of prime power order. In particular, IQ' '

IIQI and there-

fore r = a-' and Q satisfy the assumptions of (15); it follows that Q is cyclic. But thenQ" ' has prime order, a contradiction.

(b) Let p : q e it and let a- e P(G) such that q divides I P°I. By (15), P is cyclic andhas a normal complement NP in G. By (a), P° = Q is a Sylow q-subgroup of G andagain (15), applied to Q and r = a ', shows that Q is cyclic and has a normalcomplement Nq in G. The intersection D of all these normal q-complements N. whereq E it is a normal rr-complement of G with cyclic factor group G/D of order fl q".

gER

Suppose, for a contradiction, that D` D for some r e P(G). Then D`D/D 0 1 andhence there exists r E it dividing ID'I; let R < D` such that I R I = r. By (a) and thedefinition of it there exists v e P(G) such that P" E Syl,(G) and hence R < (P')' forsome x E G. By (a), ((P")x)` is a Sylow s-subgroup of G for some s e it and since1 0 R` ' < D n ((P")x)' it follows that s divides IDI. This is a contradiction sinceD < N,, a normal s-complement in G. Thus D` = D for all r e P(G).

Therefore every r e P(G) induces an autoprojectivity i in the cyclic group G/D. Ifq e it and u e P(G) such that P° E Sylq (G), then 6: maps the Sylow p-subgroup PD/Dof G/D onto the Sylow q-subgroup P°D/D and hence maps the unique complementNP/D of PD/D in G/D to the unique complement Nq/D of P°D/D. It follows thatNP' = N. and P(G) operates transitively on the set of normal q-complements (q e ir)of G.

Simple examples for the situation described above can be found in cyclic groups.It is easy to see that if in, n e f\J, G = C. and {p1,...,p,} is the set of primes p forwhich p" is the largest power of p dividing in, then P(G) induces the full symmetricgroup S, on the set of p;-complements of G (cf. Exercise 1.4.2). Further examples aregiven in Exercises 2 and 3.

Suzuki's theorem

We collect the results proved so far in what is our main theorem on finite groupswith singular projectivities.

4.2.6 Theorem (Suzuki [1951a]). Let cp be a projectivity from the finite group G tothe group G and suppose that P is a Sylow p-subgroup of G such that IP1*I IPI; letI P I = p". Then (a) or (b) holds.

(a) There exists a P-decomposition (S, T) of G such that P < S; that is, G = S x Twhere S is a P-group properly containing P and (ISI,ITI) = 1.

(b) There exists a subgroup N of G with the following properties.(b I) N 9 G, G = NP and N n P = 1; that is, N is a normal p-complement in G.(b2) Let K= n N". If P is cyclic and it is the set of primes dividing IP"I for

ac P(G)some a e P(G), then K < G and G/K is cyclic of order fl q". If P is not cyclic, then

gErz

P is elementary ahelian, P' is a nonabelian P-group of order p"- lq where p > q e Pand K = N.

(b3) N`P < G.(b4) If (I G/NPI, I Nwl) : 1, then there exist a P-group S > P and a subgroup M < N

satisfying MS = G and M n S = 1 such that M and M`0 are normal Hall subgroups ofG and G, respectively, and ISI = rm-lp, IS`°I = rm where p < r e P and 2 < in E

(b5) If I PI > p or I P I = p > I P" 1, then NP is a normal Hall subgroup of G.

Proof. Suppose that (a) does not hold. Then by 4.2.3 and 4.2.2, (p is singular of thefirst kind at p and there exists a normal p-complement N in G. Thus N satisfies (bl)and we want to show that it also has the other properties in (b).

By 2.2.6, P is cyclic or elementary abelian and, in the latter case, P`° is a non-abelian P-group; hence in (b2) we only have to prove the statements on K. These areclearly satisfied if N is invariant under P(G); for, in that case, IP°I = IG : NI = I P I forall v E P(G) and hence it = { p}. So suppose that there exists r c- P(G) such thatN' # N. Then p divides I NT N/NI = IN': N n N T I and there exists a Sylow p-subgroup P" of G such that P" n NT 1. Therefore (P")T-' n N # 1, hence (P')' ' isnot a p-group and it 0- { p}. Since (a) does not hold, the assumptions of 4.2.5 aresatisfied. It follows that P is cyclic and G has a normal n-complement D with cyclicfactor group G/D satisfying D° = D for all a c- P(G). Since D < N, this implies thatD < K. But P(G) operates transitively on the set of normal q-complements (q E n),hence every such complement is of the form N° where a- E P(G) and it follows thatD = K. Thus (b2) holds.

If r c- P(G), then tprg-' E P(G). Since K is invariant under P(G), it follows thatK°' = K P; in particular, K`' a G. Thus N" -< 6 if K = N; and if K < N, we have justshown that G/K is cyclic. Then 6/K'° is cyclic and hence N10 a G in this case also.Thus (b3) holds.

Now let r be a prime dividing (IG/N'°I, I NI°I), let R`° E Sylr(G) and R`°I = r'.Then R ° n N`° is a Sylow r-subgroup of NP and hence 1 < R ° n N`0 < RP; thusI < R n N < R. Since R n N is a normal p'-subgroup of R and R/R n N RN/Nis a p-group, it follows from 2.2.6 that R ° is elementary abelian and R E P(m, r). By2.2.2, R n N is an r-group and I R I = r°'-' p where r > p. If P, E Sylp(G) such thatR n P, # 1, then r divides IPfl. Since r > p, we have P; 0 P(n,p) and hence P; is acyclic r-group. But the Sylow r-subgroups of G are elementary abelian; hence IP;I = rand I P, I = p. We have shown:

(16) If N`° is not a Hall subgroup of G, then IPI = p.

We now apply the results proved so far to R" and cp-'. Clearly, q is singular at r.If (S°, T`°) were a P-decomposition of G such that R`' < S`°, then by 4.2.4, (S, T)would be a P-decomposition of G satisfying R < S. Therefore p would divide ISI andit would follow that P < S, a contradiction since we assume that (a) is not true forP. Thus (a) does not hold for RP and hence R4' satisfies the parts of (b) we havealready proved. In particular, G has a normal r-complement M`° such that M o Gand since r2 divides IR`°I, (16) shows that M is a Hall subgroup of G. Now p dividesI R I and R is a complement to M in G. Therefore (p, I M I) = 1 and hence M < N andP n M = 1. Thus P° n M° = I and 1P10I = r; let Sg0 E Sylr(G) such that P`° < S`°.Then R" and S`° are complements to M`° in G, hence R and S are complements to Min G and it follows that S -- G/M R is a P-group of order r'-'p and IS`°I = r'".Thus (b4) holds and (b5) follows from the fact that I P"I = r > p =IPI in (b4).

We emphasize the following consequences of Suzuki's theorem since they areoften sufficient to obtain the applications.

4.2.7 Theorem. If cp is a projectivity from the finite group G to the group G, thenthere exists a normal Hall subgroup N of G such that

(i) N" a G,(ii) GIN is a direct product of coprime groups which are P-groups or cyclic p-

groups, and(iii) for every prime p dividing IN 1, (p is regular at p; in particular, cp induces an

index preserving projectivity in N.

Proof. For every prime p for which (p is p-singular, we define a subgroup NP of G: If(a) of 4.2.6 holds and (S, T) is the P-decomposition appearing there, we put NP = T;if (a) does not hold, then (b) is satisfied and we let NP be the normal p-complementin G. By 4.2.4 and 4.2.6, every NP is a normal Hall subgroup of G such that Np 9 Gand G/NP is a P-group of order divisible by p or a cyclic p-group. Therefore theintersection N of all these NP is a normal Hall subgroup of G satisfying N'' a G, thatis, (i) holds. If G/NP is a nonabelian P-group and q the second prime dividing IG/NPI,then G = S x T where S is a P-group of order p"q and (IS1,1 TI) = 1. Thus cp is alsosingular at q but Nq = T = NP. Therefore the factor groups of G with respect todifferent NP are coprime and it follows that GIN is isomorphic to the direct productof the different G/NP, that is, (ii) holds. Finally, by definition, cp is regular at pfor every prime p dividing INI and hence satisfies IS"I = BSI for all Sylow sub-groups S of N. By 4.2.1, (p induces an index preserving projectivity in N and (iii)holds.

If cp is not index preserving, then the normal Hall subgroup N constructed in 4.2.7is different from G and (ii) shows that there exists a normal Sylow complement in Gwith cyclic or elementary abelian factor group. Also by (ii), G" < N. Thus we get thefollowing two corollaries.

4.2.8 Corollary. If G has no normal Sylow complement with cyclic or elementaryabelian factor group, then every projectivity of G is index preserving.

4.2.9 Corollary. Every projectivity of a finite group G induces an index preservingprojectivity in the second commutator subgroup G".

We remark that the results of this section can be generalized to locally finitegroups. This is fairly straightforward for the assertions on p-singular projectivities,but the argument requires the Zacher-Rips Theorem 6.1.7 for the other results. Weshall carry this out for Corollary 4.2.9 in 6.5.9.

The structure of a finite group with a p-singular projectivity

Suzuki's theorem and its consequences are the main tools for handling singularprojectivities. In the remainder of this section we are looking for restrictions on thestructure of the normal p-complement N in (b) of 4.2.6 and the operation of theSylow p-subgroup P on N. The basic result here is that the operating group has tobe cyclic.

4.2.10 Theorem (Schmidt [1972b]). Let (p be a projectivity from the finite group G tothe group G, let P be a Sylow p-subgroup and N a normal p-complement in G. Supposethat P° is a nonabelian P-group, let Po be the Sylow p-subgroup of P° and Q be asubgroup of P such that IQI = p :A IQ4I, that is, P = PO x Q. Then PO < Z(G) andevery subgroup of RP is normal in G; furthermore, G = NQ x P. and G =(Nw x Po )Q` .

Proof. Since P° is a nonabelian P-group, P is elementary abelian of order at leastp2. It follows that there is no P-group S such that P < S < G; for, since G has anormal p-complement, p must be the smallest prime dividing ISI and hence IPI = p,a contradiction. We conclude that (b) of 4.2.6 holds for P; in particular, N4' is anormal Hall subgroup of G. By 4.1.2, CN(A)4' = CN,(A4') for all A < P.

Let A be a maximal subgroup of P different from PO and take B< P such thatP = A x B. If b c B, then CN(A)° = CN(Ab) = CN(A), that is B normalizes CN(A). By4.1.2, CN(A)4' = CN,(A4') is normalized by B`°. Since A" is not a p-subgroup of theP-group P° = A" u B4', it is not normalized by P4' and there exists x E B`0 such that(A`')' # A". Hence P4' = A4' u (A")' and

CN®(P4') = CNm(A4) n CNm((A4')x) = CN,(A4') n CN,(A`°)x = CNm(A4'),

it follows that CN(P) = CN(A). It is well-known (see Suzuki [1986], p. 216) that

N=<CN(A)IA<P,IP:AI =p>

and we have just shown that CN(A) = CN(P) is contained in CN(PO) if A P0. HenceN = CN(Po) and since P is abelian, it follows that PO is centralized by NP = G. By1.6.7, (NPO)4 = NIP x P04' and, as every subgroup of Po is normal in the P-group P°,every such subgroup is normalized by N4P4' = G. Finally, since P = Q x PO, weconclude that G = NP = NQ x Po and G = N4P° = (N4' x Po )Q`°.

The above theorem reduces the problem of determining the structure of a groupwith a p-singular projectivity to the study of groups with cyclic Sylow p-subgroups.

4.2.11 Theorem. Let G be a finite group with elementary abelian Sylow p-subgroupsof order p", n > 2. Then G has a p-singular projectivity if and only if either (a) or (b)holds.

(a) G= S x T where S is a P-group of order p"q, p> q e P and (ISI1I TI) = 1.(b) There exist subgroups Go and PO of G satisfying G = Go x P. and IPO I = p"-1

and then a projectivity tli from GO to a group G, and a Sylow p-subgroup Q of GO suchthat (p, IG, I) = 1 and I Q" I divides p - 1.

Proof. Let qp be a p-singular projectivity from G to a group G and let P E Syl, (G)such that IP4I 0 IPI. If there exists a P-decomposition (S, T) of G such that P < S,then, since I P I > p2, p is the larger prime dividing I S I and (a) holds. If there is no suchP-decomposition, then by 4.2.6 there exists a normal p-complement N in G such thatN4' is a normal Hall subgroup of G. Let Po be the Sylow p-subgroup of P4', and let

Q < P satisfy IQI = p 0 IQ9I and Go = NQ; let G1 = Gg and c be the projectivityinduced by 'p in Go. By 4.2.10, G = Go x P0 and, since PP is a nonabelian P-group,I Po I = p"-' and Q111 divides p - 1. Finally, since N" is a normal Hall subgroup of Gand p divides IG/N4I, p does not divide IN4I IQ4I = IG11. Thus (b) is satisfied.

Conversely, if (a) holds, then there exist p-singular autoprojectivities of G; for,L(G) L(S) x L(T) and hence every autoprojectivity of S can be extended to anautoprojectivity of G. So, finally, suppose that G = Go x Po and iti is a projectivityfrom Go to a group G1 with the properties given in (b). Then cli is singular at p andwe show first that we may assume,

(17) there is a normal p-complement N in Go such that NO is a normal Hallsubgroup in G1.

This is clear if (b) of 4.2.6 holds for ; for, since I Q O I divides p - 1 , Q j = p > I Q O

and the assertion follows from (b5). If (a) if 4.2.6 holds, then Go = S x T and G1 =SO x TO where S is a P-group containing Q and (ISI, I TI) = I = (ISOI, I TOI). Since pdoes not divide G1 , p is the smallest prime dividing ISI and hence ISI = r'p wherep < r c P; also SO = r'"q since I QOI = q < p < r. Because L(G) L(S) x L(T),there exists a projectivity from Go to G1 mapping Q to Q'', the Sylow r-subgroup Rof S to the Sylow r-subgroup of S'' and inducing t' in T. We may assume that c isthis projectivity and then N = R x T is a normal p-complement in Go and NO _RO x TO is a normal Hall subgroup of G1. Thus (17) holds.

Now, clearly, N is also a normal p-complement in G and H = Q x Po is a Sylowp-subgroup and hence a complement to N in G. Let G be the semidirect product ofan elementary abelian p-group P1 of order p"' by G1 such that P1 is centralized byNa', and Q'" induces a nontrivial power automorphism in P1. Then K = P1Q'" is aP-group of order p"-' IQ0I and is a complement to the normal Hall subgroup N0 of6. If N is centralized by Q, then by 1.6.7, [NO, Q 0] = 1 and there exists a projectivityfrom G = N x H to G = NO x K which satisfies H`° = K and therefore is singularat p. So suppose that [N, Q] : I and hence also [NO, Q''] : 1. Then CH(N) = P0,CK(NO) = P1 and G/Po Go, G/P1 G1. Thus c induces a projectivity µ fromG/CH(N) to G/CK(N'') mapping NCH(N)/CH(N) to N''CK(N'")/CK(N'G) and H/CH(N)to K/CK(N'"). Of course there is a projectivity t from H to K mapping P0 to P1 andQ to Q''. By 4.1.9 there exists a projectivity tp from G to G which satisfies Hg, = K,and therefore is singular at p.

If p and q are different primes and (IN I, pq) = 1, then by 1.6.4 there exists a p-singular projectivity from G = N x CP to G = N x CQ. So it only makes sense tolook for restrictions on the structure of the normal p-complement N in 4.2.6(b) if Gis not the direct product of N and P. By 4.1.3, N = [N, P]CN(P) and the examples in4.1.7(d) show that [N, P] and CN(P) can be elementary abelian groups of arbitraryorder. We give examples of directly indecomposable groups in which CN(P) is notsoluble.

4.2.12 Example (Paulsen [1975]). (a) Let A, F be finite groups and N = AF be asemidirect product of A by F such that F induces power automorphisms on A. Letp and q be primes satisfying (pq, I NI) = 1, let P CP and Q ^- Cq, and suppose that

G = NP and G = NQ are semidirect products of N by P and Q, respectively, suchthat CN(P) = F = CN(Q) and P and Q also induce power automorphisms on A. Thenthere exists a projectivity p from G to G satisfying N° = N and P`0 = Q.

(b) It is possible to fulfil the assumptions of (a) for the following groups.(i) F = S,,, n < p, n < q, A is a nontrivial abelian r-group for some prime r such

that pgIr - 1 (for example n = 5, p = 7, q = 11, r = 463), and CF(A) = A,,.(ii) F = PGL(2,t) where (pq,(t - 1)t(t + 1)) = 1, A is a nontrivial abelian r-group

for some prime r satisfying pgIr - 1 (for example t = 7, p = 5, q = 13, r = 131), andCF(A) = PSL(2, t).

Proof. If (i) or (ii) is satisfied, then p 0 2 0 q and hence A has fixed-point-free powerautomorphisms of order 2p and 2q. Therefore S x P and S. x Q operate on Awith kernel A. inducing theseautomorphisms, and the corresponding semidirectproducts G = A(S. x P) and G = A(S. x Q) have the structure described in (a)where N = AS.; similarly for PGL(2, t). It remains to prove (a). For this we apply4.1.6 to the normal Hall subgroups N = M of G and G, the identity a: N -+ M andthe trivial autoprojectivity r from P to Q. By assumption, CN(P) = F = CM(Q) andhence (1) of 4.1 holds. Moreover, CA(P) 5 CN(P) = F and CA(P) = 1 since F n A = 1.By 4.1.3, A = [A, P] CA(P) = [A, P] and hence [N, P] = [AF, P] = [A, P] = A.(Note that since P induces power automorphisms on A = [A, P], Cooper's theoremimplies that A is abelian.) Therefore if U < N is normalized by P, then U =[U, P] CU(P) where [U, P] < [N, P] = A and Cu(P) < CN(P) = F. Conversely, ifVl < A and V2 < F, then V = Vl V2 is a subgroup normalized by P since F and Pnormalize every subgroup of A and P centralizes F. Thus the groups Vl V2 whereVl < A and V2 < F are precisely the P-invariant subgroups of N. The same holds forQ and hence also (2) of 4.1 is satisfied. It follows that there is a projectivity from Gto G mapping N to N and P to Q.

Note that in the above examples, [N, P] is abelian. We can show (see Exercise 6)that in general [N, P] need not be abelian but leave the question open whether thereare any restrictions on the structure of [N, P].

Exercises

1. Let G be a finite group and qp be a projectivity from G to a group G. Show that (pis index preserving if(a) CHI l = JHJ for every minimal subgroup H of G, or(b) IG : M`°I = IG: MI for every maximal subgroup M of G.

2. Let p .... p, be different primes, n E N, m = (p, ... P,)", t a prime such that mdivides t - 1 and suppose that G is the semidirect product of a cyclic group oforder t by the subgroup of order m of its automorphism group. Show that P(G)induces the full symmetric group S, on the set of p,-complements of G.

3. Let t be a prime and p1, ..., p4 be different primes dividing t - 1. Let N =N, x . x N4 where I Ni I = t for all i, suppose that H = <a 1 > x x <a4 > whereo(a;) = pi and let G be the semidirect product of N by H in such a way that for alli e { 1, ... , 4}, CN(a;) = 1 and the eigenspaces of the automorphism induced by a;in N are N; x Ni,,, N1 2, N+3 (the indices taken modulo 4). Show that P(G)operates transitively on the set of pi-complements of G but does not induce thefull symmetric group S4. _

4. Let cp be a projectivity from the finite group G to the group G and writeG = S, x x S, x T where the S. are P-groups (i = 1, ... , r; r >- 0), T is P-indecomposable and (1S11, I S.11) = I = (1S11, I TI) for all i, j E { 1, ... , r} such that i j.Show that there exists a normal subgroup N of T such that T/N is cyclic,N" :!9 6 and cp induces an index preserving projectivity in N. _

5. Let rp be a projectivity from the finite group G to the group G and supposethat P1, PZ E SyIP (G) such that P1" I 0 PZ I. Show that there exist a P-group S(containing P1 and P2) of order r'p where r > p and a subgroup T of G such thatG=S x T and (ISI, I TI) = 1.

6. Let r > 3 be a prime, N the nonabelian group of order r3 and exponent r and letp, q be different odd primes dividing r + 1.(a) Show that there exist automorphisms u and v of N such that o(u) = p, o(v) = q

and [N, u] = N = [N, v].(b) Show that the semidirect products N and N<v> are lattice-isomorphic.

7. Let G and G be groups such that L(G) ^t L(G) and recall that for any group X, wedenote by 7r(X) the set of all primes dividing the order of an element of X.(a) If G is finite, show that I7r(G)l < 21 ir(G)1.(b) If G is locally finite and of finite exponent, show that Exp G is finite. _(Note that there are Tarski groups G and G such that Exp G is finite and Exp Gis infinite.)

8. Generalize 4.2.2, 4.2.3, and 4.2.6 to locally finite groups.

4.3 Op(G), OJI(G), Fitting subgroup and hypercentre

In this section we study projective images of characteristic subgroups of a finitegroup G that are defined, or at least definable, by arithmetic conditions.

O,,(G) and the Fitting subgroup

We begin with OP(G), the largest normal p-subgroup, and F(G) = Dr OP(G), thePC P

Fitting subgroup of G. Clearly, OP(G) is the intersection of all the Sylow p-subgroupsof G since this intersection is a normal p-subgroup of G and any normal p-subgroupof G is contained in every Sylow p-subgroup.

Now let rp be a projectivity from G to G. If cp and cp-1 are regular at p, then theSylow p-subgroups of G and G are mapped onto each other by cp and cp-1 and hencealso every characteristic subgroup of G that is defined lattice-theoretically by the

4.3 OP(G), O°(G), Fitting subgroup and hypercentre 179

Sylow p-subgroups is mapped to the corresponding subgroup of G. In particular,OP(G)`0 = OP(G). If (p is singular at p, we can use the results of § 4.2 to see whathappens to OP(G). It remains to consider the case that (p is regular and (p-' singularat p.

4.3.1 Lemma. Let p be a prime dividing I GI and (p a projectivity from G to a group Gsuch that (p is regular at cp-' singular at p. Then there exist a normal Hall subgroup Nand a complement S to N in G such that S is a nonabelian P-group of order p"q forsome prime q < p, N° is a normal p-complement and S`0 an elementary abelian Sylowp-subgroup of G. Moreover, the Sylow p-subgroup P of G is an elementary abeliannormal subgroup of G, P" E Z(G) and, if Q < S satisfies IQI = q, then G = (N x P)Qand G = N`°Q" x P`°.

Proof. Since cp-' is singular at p, there exists a Sylow p-subgroup T`° of G such thatI TI # I T`0I. Suppose, for a contradiction, that there is a P-decomposition (U`°, V°) ofG such that T`0 < U`0. By 4.2.4, (U, V) is a P-decomposition of G. Since cp is regularat p and (p, I V"I) = 1, p does not divide I VI and hence p divides I UI. Suppose firstthat U and U`0 lie in P(n, p) for some n E N. Then I U`° I = p"-' q for some prime q < psince T° is a proper subgroup of U". Now cp is regular at p and this implies thatI UI = p"-'r for some primer < p and that the Sylow p-subgroup of U is mapped tothe Sylow p-subgroup T° of U°. Hence I TI = I T`°I, a contradiction. It follows that pis the smaller prime dividing I UI and I U" 1; hence I U I = s'p = I U`°I for some primes > p. Since (p is regular at p, again the Sylow p-subgroups of U have to be mappedonto those of U`° and we get the same contradiction I TI = I T`°I as before. Thereforethere is no P-decomposition (U", V`°) of G satisfying T`0 < U`0 and by 4.2.6 thereexists a normal p-complement N" in G such that N a G.

Let P E Sylp(G). Then IP`°I = IPI since cp is regular at p; let S" be a Sylow p-subgroup of G containing P. Then S`° and T' are complements to N", hence S andT are complements to N and therefore I S I = I G : N I = I TI # I TP I = IS" I. It followsthat P < S so that S is not a primary group. By 4.2.6, S`0 is elementary abelian, S isa nonabelian P-group of order p"q for some prime q p, N is aHall subgroup of G. The remaining assertions of the lemma follow from 4.2.10applied to (p-' and the Sylow p-subgroup S`° of G.

It is clear that if G = S x T and G = S x T where S and S lie in P(n, p), S isnonabelian and (ISI, I TI) = (ISI, I TI) = 1, then there exists a projectivity (p from G toG mapping OP(G) to a subgroup of G which is not nilpotent. In particular, F(G)° j4F(G). If G is P-indecomposable, the situation is better.

4.3.2 Lemma. Let G be a P-indecomposable finite group and let cp be a projectivityfrom G to a group G.

(a) For every prime p there exists a prime q such that OP(G)° = O9(G).(b) F(G)`° = F(G).

9Proof. Clearly, (a) implies that F(G)" = (PC U OP(G) U O9(G) = F(G). By

P / qc P

4.2.4, 6 too is P-indecomposable and hence we similarly get that F(G)A-' < F(G).

Thus F(G)° = F(G) and (b) holds. For the proof of (a), we may assume thatOP(G) 1; otherwise every prime q not dividing 161 satisfies Oq(G) = 1 = OP(G)°.Furthermore, if (p and cp-' are regular at p, then the Sylow p-subgroups of G aremapped by cp onto those of G and hence their intersection OP(G) is mapped to theintersection OP(G) of the Sylow p-subgroups of G.

Now let cp be singular at p and P E Sy1P (G) such that I P`01 I P1. Since G is notp-decomposable, (b) of 4.2.6 is satisfied. In particular, G has a normal p-complementN and by 4.1.1, Cp(N) = PG = OP(G). If IPI = p, then, since 1 < 0,(G)< P,P = OP(G) and hence G _= N x P. By 1.6.7, G = N° x P`0 where (IN''I, IP`0I) = 1 andthus Op(G)° = PO = Oq(G) for some prime q. So let IPI > p2. Then N0 is a normalHall subgroup of G and therefore by 4.1.2 and 4.1.1, OP(G)`0 = Cp(N)0 = Cp.(N'') _(P°)G. If P is cyclic, then P° is a cyclic q-group for some prime q and hence(P'P)G = Oq(G). If P is elementary abelian, then P" is a nonabelian P-group and sinceG is P-indecomposable, P`0 is not normal in G. Therefore (P0)d is contained in theSylow p-subgroup of P" and hence (P0)G- = OP(G). (In fact, by 4.2.10, (P0)G- is the fullSylow p-subgroup of G.) In both cases, (a) holds.

Finally, let cp be regular and cp-' singular at p. Then, in the notation of 4.3.1, theSylow p-subgroup P of G is normal in G, that is, P = 0,(G), and the Sylowp-subgroup Sc of G is not normal in G since G is P-indecomposable. But P° < Z(G)and hence OP(G) = P" = OP(G)`0.

The Fitting series and soluble groups

Recall that the Fitting series of a finite group G is defined inductively by F0(G) = 1and Fk(G)/Fk_I(G) = F(G/Fk_,(G)) for k e N.

4.3.3 Theorem (Schmidt [1972b]). If (p is a projectivity from the finite group G to thegroup G, then Fk(G)`0 = Fk(G) for all k > 2.

Proof. Suppose that the theorem is false, let G be a minimal counterexample andtake k > 2 such that Fk(G)`0 FF(G). If G = S x T where S is a P-group and(IS1,1TI) = 1, then G = S1* x T`0 and the minimality of G would imply that

Fk(G)0 = (S x Fk(T))`° = S`° x Fk(T0) = Fk(G),

a contradiction. Thus Gis P-indecomposable. _By 4.3.2, F(G)0 = F(G). Hence if F(G) = 1, then F(G) = 1 and Fk(G) = I = Fk(G)0,

a contradiction. Thus F(G) 1; put F(G) = N and write i-p for the projectivity in-duced by cp in G/N. Then

Fk-,(G/N)l = (Fk(G)/N)'° = Fk(G)0/N0 # Fk(G)/N`° = Fk-,(G/N0)

and since the assertion of the theorem is true in G/N, it follows that k = 2 andF(G/N)" F(6/N`°). By 4.3.2 there exists a P-decomposition (S/N, T/N) of G/N.Since F(S) and F(T) are nilpotent normal subgroups of G, F(S) = N = F(T) and

4.3 OP(G), OP(G), Fitting subgroup and hypercentre 181

hence

F2(G)/N = F(G/N) = F(S/N) x F(T/N) = F2(S)/N x F2(T)/N,

that is, F2(G) = F2(S)F2(T). By 4.2.4, (SW/N",TW/NW) is a P-decomposition of GIN"and hence also F2(G) = F2(SW)F2(TW). Therefore if S < G, the minimality of G wouldimply that F2(S)W = F2(SW) and F2(T)W = F2(TW), and hence F2(G)W = F2(G), a con-tradiction. Thus S = G, that is, G/N is a P-group in P(n, p), say.

Then the Fitting subgroups of G/N and of G/NW are their Sylow p-subgroups and,since F(G/N)4' 0 F(G/NW), 9 or qp-' is singular at p. If cp were regular and qp-'singular at p, then by 4.3.1, the Sylow p-subgroup of G would be normal in G andhence would lie in F(G); this is impossible since p divides IG/NI. Thus (p is singularat p and since G is not P-decomposable, (b) of 4.2.6 holds. In particular, G has anormal p-complement. Then G/N also has a normal p-complement and it followsthat G/N is elementary abelian of order p" where n > 2. On the other hand, thisimplies that the Sylow p-subgroups of G are not cyclic and then 4.2.10 shows that p2does not divide IG/F(G)I. This is a contradiction.

As a consequence of the above theorem we get a classical result that was provedindependently by Suzuki and Zappa.

4.3.4 Theorem (Suzuki [1951 a], Zappa [1951b]). If G is a finite soluble group andq a projectivity from G to a group G, then G is soluble.

Proof. Since G is soluble, there exists k >_ 2 such that Fk(G) = G. By 4.3.3, Fk(G) _Fk(G)'° = G and hence G is soluble.

We shall give a simpler proof of this theorem in § 5.3, together with a lattice-theoretic characterization of the class of finite soluble groups. Suzuki's proof wasalso shorter. He only needed 4.2.6 and the following lemma that is often useful; weleave it to the reader to construct a proof of Theorem 4.3.4 from 4.2.6 and 4.3.5.

4.3.5 Lemma. If M is a normal subgroup of index p in G and q a p-regular projecti-vity from G to G, then M4 a G.

Proof. Let XW be a p'-subgroup of G. Then X" < MW since otherwise p =I X M: M I = I X: X n M I would divide I X I and cp would be singular at p. Thereforethe subgroup T generated by all the p'-subgroups of G is contained in M`0 and henceM° a G since M4/T is a maximal subgroup of the finite p-group G/T

OP(G) and the nilpotent residual

Recall that OP(G) is the intersection of all normal subgroups N such that G/N is ap-group and G.t = n OP(G) is the nilpotent residual, the smallest normal subgroup

PCP


with nilpotent factor group of G. Clearly OP(G) is the join of all the p'-subgroups ofG; indeed this join is a normal subgroup whose factor group is a p-group and it iscontained in any such normal subgroup of G.

4.3.6 Lemma. Let q be a projectivity from G to G. If G/OP(G) is not elementaryabelian, then there exists a prime q such that OP(G)° = OQ(G).

Proof. If cp is regular at p, then (p -' also is regular at p since otherwise, by 4.3.1, theSylow p-subgroup of G and hence also G/OP(G) would be elementary abelian. SinceOP(G) is the join of all the p'-subgroups of G, it follows that OP(G)" = OP(G) in thiscase. So suppose that q is singular at p and let P E Sylp(G) such that IP"I 0 IPI.Again P cannot be elementary abelian and hence P is cyclic of order p" where n > 2.Therefore (b) of 4.2.6 holds; in particular, there exists a normal p-complement N inG such that N° is a normal Hall subgroup of G. Then IP°I = q" for some prime q andO' (G) = N' = OP(G)''.

4.3.7 Remark. The assumptions in 4.3.6 are stronger than those in 4.3.2, thecorresponding result for OP(G), but they are necessary. It is clear that in certainP-decomposable groups, for example G = S x T where S is a nonabelian P-groupand p the smaller prime dividing IS 1, there are even autoprojectivities mapping OP(G)to a nonnormal subgroup of G. But with the hypothesis of Theorem 4.2.10, OP(G) =N and GIN" is a nonabelian P-group; hence OP(G)° O'(G) for every prime r. AndO9 (G) = (NP0)° if q is the smaller prime dividing IP`°I; however NPo is not an O'(G).That this situation actually occurs is easy to see. If we take primes p, q, r such thatqlp - I and pglr - 1, then, by 4.2.11 or 4.1.6, the semidirect product G = NP of acyclic group N of order r by an elementary abelian group P of order p2 inducing anautomorphism of order p in N has a projectivity (p onto the semidirect product G ofN by a nonabelian group P of order pq inducing an automorphism of order q in N.Here G is P-indecomposable, but OP(G) = N = Gn is mapped by (p to the normalsubgroup N of G with nonnilpotent factor group G/N P, that is, neither to G,i norto any 0`(G) for t e P.

This example shows that we cannot prove the analogue of 4.3.2 for OP(G) or thenilpotent residual G %. However, we shall show in § 5.4 that nevertheless the iteratednilpotent residuals behave under projectivities like the iterated Fitting subgroups,that is, they are mapped to the corresponding subgroups in the image group.

Op, (G), OP '(G) and p-soluble groups

If we replace the prime p in the definition of OP(G) and OP(G) by a set it of primes,we obtain the largest normal n-subgroup and the smallest normal subgroup0"(G) with factor group a n-group. Also these are, respectively, the intersection of allthe maximal n-subgroups and the join of all the rt'-subgroups of G; thus they aremapped by index preserving projectivities to the corresponding subgroups in theimage group. We are mainly interested in the largest normal p'-subgroup and

4.3 OP(G), OP(G), Fitting subgroup and hypercentre 183

the smallest normal subgroup OP'(G) with factor group a p'-group. They are mappedto and OP'(G), respectively, by projectivities cp from G to G such that cp andcp-1 are regular at p. Recall that G is p-soluble if its upper p-series

1 =PO <No aP1 :g a N, PmaNm...

defined by N,/Pi = and Pi+1/N; = OP(G/N;) reaches G; the length of thisseries, the number of p-quotients in it, is called the p-length of G and denoted by1,(G)-

4.3.8 Theorem. If G is a finite p-soluble group and cp a projectivity from G to a groupG, then G is p-soluble and IP(G) = lP(G) if l,(G) > 2.

Proof. Suppose first that 1,(G) 2. Then, by 4.2.6, (p is regular at p and, by 4.3.1, cp 1

is also regular at p. Therefore every term of the upper p-series of G is mapped ontothe corresponding term in G. In particular, G is p-soluble and 1,(G) = 1,(G).

Now suppose that lP(G) 5 1. Again if cp and cp-1 are regular at p, then G is p-soluble and lP(G) = 1,(G). If (p-1 is singular at p, then, by 4.2.6, G is p-soluble and!P(G) = 1. Finally, if q-1 is regular and (p singular at p, then either G is a p'-group orwe may apply 4.3.1 to cp-1 and conclude that G has a normal Sylow p-subgroup. Inboth cases, G is p-soluble of p-length at most 1.

We shall need the following result that is similar to 4.1.2(a).

4.3.9 Lemma. Let A and B be subgroups of G and suppose that A is generated byp-elements and B by p'-elements. If (p is a projectivity from G to a group G such thatpp and cp-1 are regular at p, then B < NG(A) (or A < NG(B)) if and only if B° < N-(A`°)(resp. A" < N-(B°)).

Proof. Suppose that B < NG(A), take a p'-element b e B and let H = <A, b>. Then Ais a normal subgroup of H generated by p-elements and H/A is a p'-group; thusA = OP'(H) and hence A" = OP'(H°) since cp and cp-1 are regular at p. Therefore A`°is normalized by < H`° and, since Bc is generated by these subgroups (b> 'P,it follows that B° < Ne;(A`°). Now Ac" is generated by p-elements and B° by p'-elements. Thus if we apply the result just proved to cp-1, we obtain the oppositeimplication. The equivalence of A < NG(B) and A" < Nd(B'') is proved in an entirelyanalogous way.

The hypercentre

Finally we discuss a characteristic subgroup that has not so much to do with thearithmetic structure of a group. It is the hypercentre Z.(G), the final term of theascending central series of G. Unlike the iterated Fitting subgroups F,k(G), the n-thcentre Z,,(G) in general is not mapped to Z (G) by a projectivity cp from G to G. Forexample, if G is an abelian P-group, then G for all n, but 1 in a

lattice-isomorphic nonabelian P-group G. Even if G is P-indecomposable and cpindex preserving, the situation is not much better, as is shown by the abelian groupG of type (p",p"-1) and the lattice-isomorphic M-group G = <a,bla°" = bP" ' = 1,an = a"P>; here again Zk(G) = G for all k, but Zk(G) A G for k = 1, ..., n - 1. Soonly under rather restrictive assumptions can we expect to be able to prove thatZk(G)`° = Zk(G); an example of such a result is contained in Theorem 5.3.4. But thehypercentre behaves better.

4.3.10 Theorem (Schmidt [1975a]). If (p is an index preserving projectivity from thefinite group G to the group G, then Zx(G)4' = Z, (G).

Proof. Let p be a prime and T= OP(OP'(CG(O°(G)))). We want to show that T<ZX (G) and T'° = OP(OP'(CG(OP(G)))) < Z, (G). First of all, T is a p-group and there-fore contained in a Sylow p-subgroup P of G. Hence there exists n e f\J such thatT < Z"(P) and, since T< CG(OP(G)) is centralized by every p'-element of G, itfollows that T< Z"(G) < Zj(G). Now H = O°'(CG(O°(G))) and K = OP(G) satisfythe assumptions of 1.6.8 for it = {p}, and hence K° = OP(G) is centralized byH". Since H'° is generated by p-elements, OP'(CG(OP(G)))° = HIP < O°'(CG(O°(G))).This statement for cp-' yields the other-inclusion and it follows that T° _OP(OP'(CG(OP(G)))); in particular, T`° < 4(G).

To prove the theorem, we use induction on G1 + G1 and we may assume thatZ(G) A 1, since otherwise we could consider q 1. Let p be a prime dividing JZ(G)J.Then the Sylow p-subgroup of Z(G) is contained in T = OP(OP'(CG(OP(G)))) andhence T A 1. Since T" = OP(OP'(CG(OP(G)))) a G, cp induces a projectivity fromG/T to G,IT`° and, by induction, this projectivity maps Zx(G/T) = Z,(G)/T ontoZ,(G/T`°) = Z,(G)/T`°. It follows that Z,,(G)`° = Z,,(G), as required.

The example constructed in 4.3.7 shows that even projectivities between P-indecomposable groups in general do not map the hypercentres onto each other; inthis example, IZ,(G)I = p whereas Z,, (G) = 1. Therefore the assumption that cp isindex preserving cannot be replaced, as in 4.3.2 for the Fitting subgroup, by theP-indecomposability of the groups involved.

Exercises

1. If cp is an index preserving projectivity from G to G, show that F(G)° = F(G) andG;;, = G.

2. Let cp be a projectivity from G to G.(a) If IG : OP.(G)l : p, show that there exists a prime q such that Oq,(G).

(b) Show that the assertion in (a) need not hold if IG: 0 p is replaced bythe assumption that G is P-indecomposable.

3. If G is P-indecomposable and cp a projectivity from G to G, show that O°(G)IO a Gfor all p e P.

4. Prove Theorem 4.3.4 using only 4.2.6 and 4.3.5.

4.4 Abelian p-subgroups and projectivities 185

5. (Suzuki [1951a]) Let (p be a projectivity from G to G.(a) If cp is index preserving and N a maximal normal subgroup of G, prove that

NcaG. _(b) Show that if G is perfect (that is G' = G), then G is also perfect.

6. Let it be a set of primes and let A and B be subgroups of G such that A isgenerated by n-elements and B by n'-elements. If (p is an index preserving projec-tivity from G to G, show that B < NG(A) if and only if Bc < NU(A`°).

7. Show that the assertions of 4.3.9 do not hold without the assumption that cp andcp-' are p-regular.

4.4 Abelian p-subgroups and projectivities

In the last section of this chapter we investigate the projectivity induced in a Sylowp-subgroup P by an index preserving projectivity q of G. If G = P x Q, then L(G)L(P) x L(Q) and every projectivity from P to a p-group P can be extended to aprojectivity of G (to G = P x Q). This will no longer be true if P is not a direct factorof G or even if G is generated by its p'-elements; but very little is known about thisproblem. We shall study it first for abelian Sylow p-subgroups and then go on toshow that for arbitrary P, under suitable assumptions on p and the p-structure of G,Z(P)`° = Z(P`°).

Abelian Sylow p-subgroups

4.4.1 Lemma. Let A be an abelian p-subgroup of G, B a p'-subgroup of N6(A) andsuppose that (p is a projectivity from G to G such that A`° is a normal Sylow p-subgroupof (AB)". Then

(a) A`' = [A'*, Bc] x C,,.(B'*),(b) C,(B)c,(c) [A`0, Bc] _ [A, B]`0 is abelian and hence isomorphic to [A, B].

Proof. By 4.1.3, A = [A, B] x CA(B) and 4.1.2 applied to the projectivity 0 inducedby cp in AB shows that [A, B]`° = [A`°, B4'] and CA(B)" = CA.(B`°). Thus A" =[A`°, B4'] u CA.,(B`°) and [A`°, B`°] n 1. By (6) of § 1.5, [A`°, B`°] a A`D u B`°,and we want to show that CAm(B`°) a A`°; then (a) and (b) will hold. Since CA(B) <Z(AB) we have [CA(B), OP(AB)] = 1 and, by 1.6.8, [CA(B)c, OP(AB)c] = 1. Our as-sumption on Ac' implies that >li and 0' are regular at p and hence OP(AB)c =OP((AB)`°). It follows that CA.(B`°) = C4(B)c < CA.(OP((AB)`°)) < CA®(Bc) and there-fore CA.(B`°) = C4.(OP((AB)c)) g A'*. Thus A'* = [A'*, B`°] x CA.(B`°).

It remains to be shown that [Ac', Bc] is abelian. As a projective image of anabelian group, [Ac', B`°] is a p-group with modular subgroup lattice in which thequaternion group is not involved, that is, an M*-group according to our definitionin § 2.3. If [A`°, B`°] were not abelian, then, by 2.3.23, there would exist characteristicsubgroups R > S of [AP, BP] such that [R, Aut[A`°, Bc]] < S. It would follow

that R a AcBQ and hence [R, B`0] < S; on the other hand, by 4.1.3, R =[R, B`']CR(B°) = [R, Be'] since CR(B`0) = R n CAO(B°) = 1. This contradiction showsthat [A`0, Bc] = [A, B]" is abelian and therefore isomorphic to [A, B] since the typeof a finite abelian p-group is determined by its subgroup lattice.

4.4.2 Theorem (Paulsen [1975]). Let G be a finite group with an abelian Sylow p-subgroup P and let cp be a p-regular projectivity from G to a group G. Then P =P, x P2 where P, = P n OP(G) and P2 = P n Z(NN(P)), P° = Pl x PZ and P? P, isabelian. If P`° is not abelian, then Exp P, < Exp P2.

Proof. Since P is a normal Hall subgroup of L = NG(P), there exists a complementB to P in L. By 4.1.3, P = [P, B] x Cp(B); let P, = [P, B] and PZ = Cp(B). Since P isabelian, Pz = P n Z(NG(P)). Furthermore, clearly, P, < P n L'; on the other hand,L,/'B` P/P n BL is abelian, hence L' < B' and therefore P n L' < P n Bt' = [P, B]by 4.1.1. Thus P, = P n L'. Since P is abelian, G' < OP(G) and OP(G)/G' is a p'-group;therefore P n OP(G) = P n G'. By Grun's First Theorem (see Robinson [1982],p. 283), P n G' = P n L'. It follows that P, = P n G' = P n OP(G), as desired. Now,by assumption, cp is regular at p and hence P° is a p-group. If cp-' is singular at p,then the Sylow p-subgroups of G are abelian and the remaining assertions of thetheorem hold trivially. So suppose that cp-' is regular at p. Then P° is a normalSylow p-subgroup of (RB)A and 4.4.1 implies that P° = Pf x P210, P, is abelian andP; P,. Finally, as a projective image of an abelian group, P10 again is an M*-group. Therefore if Exp P, > Exp P2, then Exp Pr = Exp P4' and, by 2.3.16, P° isabelian.

We note some immediate consequences of Paulsen's theorem.

4.4.3 Corollary. If G is a perfect finite group with an abelian Sylow p-subgroup P,then P° P for every projectivity cp of G.

Proof. Recall that G is perfect if and only if G = G'. Thus G = OP(G), hence P, = Pand, by 4.2.9, cp is index preserving. Now 4.4.2 yields that P' = Pt is isomorphicto P.

4.4.4 Corollary. Let G he a P-indecomposable finite group with an abelian Sylowp-subgroup P and let cp be a projectivity from G to G such that P° is not isomorphic toP. If P is homocyclic or generated by two elements, then G has a normal p-complement.

Proof: If cp is singular at p, then, by 4.2.6, G has a normal p-complement; so supposethat cp is regular at p. By 4.4.2, P = P, x P2 where P, = P n OP(G), P° = P; x P21',Pr is abelian and Exp P, < Exp P2. We have to show that P, = 1; then it will followthat G/OP(G) P/P n OP(G) = P and OP(G) is a normal p-complement in G. If P ishomocyclic of type (p", ... , p"), say, then every nontrivial direct factor of P containsan element of order p"; since Exp P, < Exp P, it follows that P, = 1. If P is generatedby two elements and P, 0 1, then P2 and hence also Pz would be cyclic; thereforeP10 = P; x Pz would be abelian and so P`0 P, a contradiction. Thus in this casealso, P, = 1.

In general, a finite group G with abelian Sylow p-subgroup P need not possess anormal p-complement if P is mapped to a nonabelian group by an index preservingprojectivity of G. This is shown by the following example, that, as Lemma 4.4.1, canbe found both in Menegazzo [1974] and in Paulsen [1975].

4.4.5 Example. Let p and q be primes such that qtp - 1, R = <a> x whereo(a) = p2, o(b) = p the abelian group of type (p2, p), S = <c, d1 c" = d° = 1, c' = c'+°)the nonabelian group lattice isomorphic to R and T a nonabelian group of orderpq. Then there exists a projectivity from G = R x T to G = S x T mapping theabelian Sylow p-subgroup of G to the nonabelian Sylow p-subgroup of G; clearly, Ghas no normal p-complement.

Proof. The map a: G G defined by (a`b't)° = c'd't for i,j e 7L and t e T is certainlybijective. If x = a'b't and y = a'b't' where i, j, k, I E 7L and t, t' E T, then

r°y" = (c`d't)(ckdtt') = c'+kdj+ttt'[dI ck] = (xy)°[d' ck]

Therefore to see that (xy)° e <x°, y°), we only have to show that [d', ck] E <x°, y" ).If k - 0 (mod p), then [d', ck] = 1 and, if k * 0 (mod p), then [d', ck] E <c°) _<(ckdtt')P9) < <x",y°). Similarly, (uv)' ' E for u, r e G and, by 1.3.1, Qinduces a projectivity from G to G with the desired properties. Since T has nonormal p-complement, neither does G.

Further results on projective images of groups with abelian Sylow p-subgroupsare contained in Huppert [1960] and Seitz and Wright [1969]. Both papers areconcerned with groups G having a Sylow p-subgroup P such that L(P) is modular.Corollary 4.4.3, for example, also follows from a theorem of Seitz and Wright whichasserts that O°(G) i4 G if P is a nonabelian M*-group (see Exercise 1).

Normal abelian p-subgroups

The crucial step in the proof of the main results of this section is to show that ifp > 2, G = O°(G) and Z is a normal subgroup of G contained in the centre of aSylow p-subgroup P of G, then Z'° < Z(P°) for every index preserving projectivity ofG. This will then be applied to NG(Z(P)) in place of G. The proof of this result is quitelong and will be given in the next three lemmas. The first two deal with the structureof P° and its normalizer in a minimal situation which will emerge in the course ofthis proof.

4.4.6 Lemma. Let H be a p-group with modular subgroup lattice, Z a homocyclicnormal subgroup of H of exponent p", and Y a cyclic subgroup of H that centralizesZ/S2(Z) and S2(Z) but does not normalize every subgroup of Z. For y C- Y, writeDZ(y) = {z e Ztz'' = z'+°"-'}; note that Dz(y) is a subgroup of Z.

(a) If Y n Z < W < Z, then Y < NH(W).(b) Either

(1) Z:C7(Y)I =P

or there exist a generator y of Y and an element v e Z of order p" such that

(2) Z : DZ(y)I = p and v'' = vl+z°" , where ). * I (mod p), or

(3) Z : Dz(y)I = p and v'' = v'+0" `w1 ' where w e Dz(y) and o(w) = p".

Proof. We may assume that H = ZY. If Y n Z< W< Z, then W = W u (Y n Z) =(W u Y) n Z < W u Y, that is, Y < NH(W), since L(H) is modular and Z < H. Thus(a) holds.

To prove (b), first note that since Y does not normalize every subgroup of Z butcentralizes Q(Z), Z is neither cyclic nor elementary abelian. In a direct product of Q8by an elementary abelian 2-group, however, every homocyclic subgroup is cyclic orelementary abelian. Therefore, by 2.3.8, Qg is not involved in H, that is,

(4) H is an M*-group.

Since Z/ f2(Z) is centralized by Y, H/S2(Z) is abelian and since Q(Z) is centralized byZ and Y,

(5) H' < Q(Z) < Z(H).

Therefore (1) of § 1.5 implies that

(6) [z, z2, y] = [z1, y] [z2, y] for all z; E Z, y c- Y

and hence [z', y] = [z, y]° = 1, that is, (z°)'' = z° = (z°)1+° ' for all z e Z. Thus

(7) (D(Z) <Cz(y)nDz(y)for all ye Y.

Suppose, for a contradiction, that I YI < p" = Exp Z. Then Exp H = p" and, by anobvious extension of 2.3.11, there exists a complement C to Z in H. For, given z E Zwith o(z) = p", by 2.3.11 there exists a complement K to <z> in H; hence Z =<z) (K n Z), K n Z is a homocyclic normal subgroup of K and, by induction, it hasa complement C in K such that H = <z>K = <z>(K n Z)C = ZC and Z n C =Z n K n C = 1. Now C in place of Y satisfies the assumptions of the lemma and, by(a), every subgroup of Z is normal in H, a contradiction. It follows that

(8) Exp Z = p" < p' = I YI = Exp H.

Now suppose first that there exists h e H such that o(h) = p'° and <h> < H. Thenh = vy where v c Z, y c Y and o(y) = p', that is, Y = <y). Thus Z<h) = H andHI<h) - Z/ Z n <h> is abelian. Therefore H' < <h> n Q(Z) and hence I H'I = p. By(6), the map z -+ [z, y] is a homomorphism from Z to [Z, y] < H' with kernel CZ(Y).It follows that IZ : Cz(Y)I = p and (1) holds.

It remains to consider the case that no cyclic subgroup of maximal order is nor-mal in H. Then by 2.3.18 there exists an abelian normal subgroup A of H, a genera-tor v of Y and an integer s which is at least 2 in case p = 2 such that H = A<y) anda'' = a' +°' for all a e A. Then US(A) = a $a E H} = H' < Q(Z) and hence Exp A =ps 1. Therefore, if s < n - 1, then A < S2n_1(H) and Z/S2"_,(Z) ZS2"_,(H)/S2"_,(H)would be cyclic; but Z has two independent elements of order p", a contradiction.

Thus s > n - I. If s > n - 1, then ExpZ = p" implies that a'' = a'+P, = a forall a E Z n A. Hence Z n A < Cz(Y) and Z/Z n A ZA/A is cyclic. Sincei(Z) < CZ(Y), it follows that I Z : CZ(Y)l = p and again (1) holds. So, finally, lets = n - 1, that is,

(9) a'' = a'+P"-' for all a e A.

Now Z n A < DZ(y), Z/Z n A is cyclic and (D (Z) < DZ(y). It follows thatJZ : D7(y)I = p, and we have to find an element v e Z of order p" for which theremaining assertions of (2) or (3) hold. If Z5"-,(D,(y)) n Y = 1, we choose v c- Z suchthat o(v) = p" and Y n Z < <v>. By (a) and (5), [v, y] E <v> n Q(Z) = <vP" ' > andhence v'' = v'+"P"-' where i. E Z. Since y does not normalize every subgroup of Z,Y n Z j4 1, and it follows that v 0 DZ(y). Thus ; -* I (mod p) and (2) holds. IfZ5"-I(D7(y)) n Y 1, we take v E Z\Dz(y). Then o(v) = p", v = ab where a c- A, b c- Yand o(b) < p" since Exp A = p". By (9) and 2.3.10 applied to S2"(H),

[v, y] = [a, y] =aP" , = vP"

' b-P"

and since v D7(y), h-P" 1. Now b-P"-' E Q(Y) < Z5"_,(DZ(y)) and hence thereexists w e DZ(y) such that w"'-' = b-P" '. It follows that o(w) = p" and vy _v`+P" ' w"" ', that is, (3) holds.

4.4.7 Lemma. Let p > 2 and Z be a homocyclic group of exponent p" and order p'"where n > 2, r > 2. Let A = BC < Aut Z, B a nontrivial elementary abelian normalp-subgroup of A and C = <> a cyclic q-group where q : p is a prime and suppose that

(10) y operates irreducibly on B, [B,,,,] = B, [B, y4] = 1,

(11) y operates irreducibly on S2(Z),

(12) [Z, B] < S2(Z), and

(13) every cyclic subgroup Y -A 1 of B satisfies (b) of 4.4.6.

Then r = 2, IBS = p, q = 2 and there exist a generator a of B and a basis {u, v} of Zsuch that u' = uP" ', v' = v' - P" ' and a'' = a-1.

Proof. To simplify notation, we write Cz(a) = C. and D7(x) = D. for x c- B. IfIZ : C,I < p, then clearly iV(Z) < Ca. And if IZ : C,I > p, then, by (13) there existsi E < > such that IZ: Dpi = p; hence zfl = z1+P"-' = z for all z E N(Z) and againN(Z) <_ C, = C. It follows that

(14) .D(Z) < CZ(B).

Since n > 2, (12) implies that [Z, B] is contained in 4>(Z) and hence is centralized byB and, of course, also by Z. Therefore (1) of § 1.5 yields that

(15) [uv, x] = [u, a] [v, x] and [z, a/3] = [z, a] [z, /3] for all u, v, z E Z and x, J3 E B.

In particular, the map z --+ [z, x] is an epimorphism from Z to [Z, x] with kernel C,and hence

(16) I[Z,x]I = IZ:C,I for xeB;

similarly,

(17) 1 [z, B] I = I B : CB(z)I for z e Z.

If a e B such that IZ : C,I = p, then 1 U, , < S2(Z); furthermore I [Z, a] I = pand hence also I < [Z, x] < S2(Z). Since y operates irreducibly on S2(Z), we have[Z, x] [Z, a]Y = [Z, 2Y] and C. 0 (C,)' = C,,. For z e Z and i = 1, ..., p - 1,

[z,a'aY] = [z,a]'[z,aY] e [Z, x] x [Z,a'']. If z e C,\C,,, then [z,a'a"] = [z,aY] is anontrivial element in [Z, al] and for z e C2,\C [z, a'ff] = [z, a]' is a nontrivialelement in [Z, x]. It follows that [Z, a'a'] = [Z, a] x [Z, a Y]. In view of (16), wehave shown:

(18) IfaEBsuch that IZ:C,,1=p, then IZ:CZ(a'aY)I=p2for i=l,...,p-1.

If a E B such that IZ : D,I = p, then every z e D. n (D,)' satisfies z'' = zY-'a' _z'+P" ' = z', and it follows that D, n (D,)" < CZ(aYa-'). Thus we have:

(19) If a e B such that IZ : D,I = p, then IZ : CZ(a1a-1)I <- p2.

From (18) and (19) we conclude that there exists /3 e B such that

(20) IZ:DOI = p and IZ:Cp1 =p2.

For, if there exists a e B with IZ : Cj = p, then IZ: CZ(aaY)I = p2, and if there is nosuch x, then by (13) there exists S E B with IZ: Dbl = p and henceIZ : C,(SYS4') = p2. Again (13) implies that a suitable generator i of (aa') or<6`6`> satisfies IZ : DPI = p and, of course, I Z : Cal = p2.

By (14) and (20), Qt(Z) < Cp n D. On the other hand, z e Cp n Dp implies thatz = zp = z'+P"-' and hence z e V(Z). Thus I(Z) = Cp n Dp and, by (20),pr = IZ : (D(Z)l < p3, that is, r < 3. Suppose, for a contradiction, that r = 3. ThenZ/.D(Z) = C,,/V(Z) x D,/I(Z) and hence there exists a basis {u, v, w} of Z such thatup = u, VP = v1+P"-', and wp = W"P"-'. Now Z is a free 7L/(p")-module and we mayconsider A as a subgroup of GL(Z) GL(3,7L/(p")). The determinant defined thereinduces a homomorphism from A to the group of units of 7L/(p"). By (10), A has nonormal subgroup of index p and this implies that the determinant is trivial on thep-elements of A. But det/3 = I + 2p"-' + (p") 1 + (p") since p > 2. This contra-diction shows that

(21) r = 2.

By assumption, y is irreducible on Q(Z). Therefore the order of the operating groupdivides p2 - 1; in particular, qI p2 - 1. This implies that every irreducible GF(p)-module of the cyclic group of order q has order at most p2 (see Huppert [1967],p. 166). Since (y)/(y9) operates irreducibly on B, it follows that

(22) IBI < p2.

Suppose, for a contradiction, that there exists a e B such that IZ : C,I = p. Then forevery 6 E (y), (C,)° = C,8 and hence IZ: CZ(a6)1 = p. By (18), IZ: CZ((X'a)I = p2 fori = 1, ..., p - 1. Therefore B = (a) x (al) has order p2 and among the p + 1

minimal subgroups of B, only <a> and (a') have their centralizers of index p in Z.Hence <y> operates on { <a), <a Y> } and it follows that q = 2; but this contradicts(10) because a group of order 2 cannot operate irreducibly on an elementary abeliangroup B of order p2. Thus there does not exist a e B with I Z : C2 = p; since (D(Z) iscentralized by B and has index p2 in Z, it follows that

(23) C2 = (D(Z) for all 1 a e B.

Let Bo be a subgroup of order p of B. By (13) there exist a e Bo and v e Zsuch that o(v) = p", IZ : Dj = p and, either v° = v'+"P"-' with 2 * 1 (mod p) orv° =

v'+P"-,

WP"-' with w e D2, o(w) = p". In the latter case, {v, w} would be a basisof Z and since w" = w`+P"-`, deta = 1 + 2p`1 + (p") 1 + (p"), a contradiction.Thus v2 = v'+"P"-' where 2 0 1 (modp). If u e D. with o(u) = p", then ua = u'+P"-'and {u, v} is a basis of Z. Then 1 + (p") = det a = 1 + (2 + 1)p"-' + (p") implies

- I (mod p) and hence our basis {u, v} of Z satisfies

(24) u° = u'+P" v" = v'

Now suppose, for a contradiction, that I BI = p2. Then by (24), any of the p + 1minimal subgroups of B fixes at least 2 cyclic subgroups of order p' of Z. Since Z hasonly p + 1 maximal subgroups, there must exist subgroups <a> <x > of order p ofB, and elements x, y e Z of order p" satisfying [x, a] e <x> and [y, x'] e (y> suchthat x and y lie in the same maximal subgroup <x).D(Z) = <y)D(Z) of Z. By (12),[x, a] and [y, a'] are contained in n(<x>) = n(<y>). Furthermore x = y`z for somei e Z, z e (D(Z) and hence (15) and (14) show that [x,a] = [y,a']' e Q((x>) also.It follows that [x, B] < SZ((x>) and hence p >- I [x, B] I = JB : CB(x)l by (17). Thisimplies that I CB(x)l > p and x e C. if 1 S e CB(x), contradicting (23). Thus

(25) IBI = p.

Since y does not centralize B, qlp - 1. And since y operates irreducibly on Q (Z), o(y)divides p2 - 1 but not p - 1. Therefore q also divides p + 1 and it follows that q = 2.By (24), there exist a generator or of B and a basis {u, v} of Z such that u' = u'+P"-'and v" = v'-P"-'. Finally, y induces an automorphism of order 2 in B and hencea''=a-'.

4.4.8 Lemma. Suppose that p > 2, G is a finite group with OP(G) = G, P e Sylp(G),Z is a normal subgroup of G contained in Z(P) and cp is an index preserving projectivityfrom G to a group G. Then Z" < Z(P').

Proof. Suppose that the lemma is false and consider a counterexample G of minimalorder. By 4.3.9,

(26) Zq' a G.

Furthermore, if A and B are nontrivial normal subgroups of G contained in Z suchthat A n B = 1, then A'* a G and B" a G, co induces projectivities in G/A and G/B,and the minimality of G implies that [P'*, Z1*] < A'* n B1* = 1. Hence Z`° 5 Z(P'*),contradicting the choice of G. We have shown:

(27) If A,B<Zsuch that AaG,BaGand AnB=1,then A=lor B=1.

Since P < CG(Z), the group of automorphisms induced by G in Z is a p'-group.Therefore Z = [Z, G] x CZ(G) (apply 4.1.3 to the semidirect product of Z by thisgroup of automorphisms) and, by (27), [Z, G] = I or CZ(G) = 1. In the first case,1.6.8 would imply that [Z`°, G] = 1 and hence Z`0 < Z(P°), contradicting the choiceof G. Therefore

(28) C,(G) = Z n Z(G) = 1 and [Z, G] = Z.

Furthermore, Maschke's Theorem (see Robinson [1982], p. 209) shows that to everyG-invariant subgroup A of Q(Z), there exists B a G such that )(Z) = A x B; then(27) implies that A = 1 or A = S2(Z). Thus

(29) G operates irreducibly on S2(Z).

In particular, Z5"_, (Z) = Q(Z) if Exp Z = p" and hence

(30) Z is homocyclic.

Take any p'-subgroup B of G. By 4.4.1, Z° = [Z`P, B"] x CZ(B)`° and [Z`0, B"] isabelian. Therefore (Z`°)' < C7(B)" and hence ((Z`°)')"-` < CZ(B). Since G = OP(G) isgenerated by these p'-subgroups B, it follows that ((Z")')`° C7(G) = 1. Thus(Z'*)' = I and

(31) Z`0 is abelian.

Put L = OP'(G) and M = OP(L). Then P< L and MP = L. Since cp is indexpreserving,

(32) L'° = OP'(G) and M* = OP(L`°).

Since GjCG(Z) is a p'-group, L < CG(Z). In particular, [Z, M] = 1 and, by 1.6.8,

(33) [Z`0, M°] = 1.

IZ

P

PI M

Figure 15

Hence L10/C,m(Z10) is a p-group operating on the p-group Z`0 and therefore it central-izes a nontrivial element in this group. Thus fl(Z`0) n Z(LP) 0 1 and, of course, isnormal in G. By 4.3.9, (fl(Z`0) n Z(L`0))'0-' < G and since G operates irreducibly onS2(Z), it follows that

(34) S2(Z`0) < Z(L"); in particular, fl(Z") < Z(P10).

Since G is a counterexample to the lemma, S)(Z) < Z and the minimality of G yieldsthat Z`0/S2(Z°) is contained in the centre of P"/Q(Z`0). Thus

(35) Exp Z = p" > p and [P`0, Z10] < fl(Z`0).

Now (35), (34) and (31) show that for x c P° and z c Z`0, [x, z] has order p andcommutes with x and z. Therefore 1 = [x, z]° = [x°, z] = [x, z°], that is, P° is cen-tralized by Z5'(Z") = 4t(Z`0) and Z`0 is centralized by ZS(P°). Since [P°, Z`°, P10] = 1 =[Z`0, P°, P°], we have [(P")', Z] = [P`0, P10, Z10] = 1 also (see (7) in § 1.5); hence Z° iscentralized by Z3'(P10)(P(0)' = cb(P`0). Thus

(36) [P`0, 1(Z")] = 1 = [(D(PP), ZP].

Put G* = G/M10 and G = G/Cj;(Z0). For x e G and H < G, write x* = xM, . =xCi;(Z`0), H* = H`0M`0/M`° and H = H10Ci;(Z°)/CG(Z10) to denote the images of xand H`0 under the natural epimorphisms from G to G* and G. By (33), M`° < CZ;(Z`0)so that G is an epimorphic image of G*. Since L`° = O°'(6) = PWMQ, we haveP* = L10/M10 < G*. Hence the Sylow p-subgroup P of the epimorphic image G of G*is normal in G and P 1 because G is a counterexample. Since Z`° is centralized by(D(P`0), finally,

(37) P is a nontrivial elementary abelian normal subgroup of G.

The Frattini argument shows that G = L°NN(P°) = M0NU(P°), and by the Schur-Zassenhaus Theorem there exists a complement QO to P° in Nd(P`0). Since P10 andQ`0 have relatively prime order, Q* and Q are also complements to P* and P,respectively, in the epimorphic images G* and G of Nj (P`0). Then G = O°(G) impliesthat G* = OP(G*) = (Q*)G' and, by 4.1.1, p* = P* n (Q*)G; = [P*, Q*]. The sameapplies for G; since P is abelian, 4.1.3 implies that C?(Q) = 1. Thus

(38) [P*, Q*] = P*, [P, Q] = P and CP(Q) = 1.

Our aim is to show that the assumptions of Lemma 4.4.7 are satisfied for Z`° in placeof Z, G = A, P = B and Q = C. By (30), (31) and (35), Z" is homocyclic of exponentp" where n > 2, and we may regard G = G/C6(Z") as a subgroup of Aut Z° in whichP, by (37), is a nontrivial elementary abelian normal p-subgroup. If Y # I is a cyclicsubgroup of P, then Y = <y> for some y e P°; let H = <y>Z`0. Since Z < Z(P), thesubgroup H`°-' = <y>w 'Z is abelian and hence L(H) is modular. By (35) and (36),Z`0/Q(Z") and S)(Z°) are centralized by <y>. If y normalized every subgroup of Z`0,then, by 1.5.4, y would induce a universal power automorphism in Z`°, which, ofcourse, would centralize every automorphism of Z`0; it would follow that y c- CP(Q),contradicting (38). Thus Y does not normalize every subgroup of Z'°; in particular,Z1* is not cyclic, that is, jZPj = p'" where r > 2, and since we shall need it later, wenote that we have shown:

(39) If y c P° such that y 0 1, then H = (y>Z`0, Z`° and (y> satisfy the assump-tions of 4.4.6.

Therefore, by 4.4.6, Y satifies (b) of this lemma, that is, (13) holds for P = B and (12)follows from (35). It remains to be shown that Q is a cyclic q-group for some primeq satisfying (10) and (11). For this we need:

(40) If S < G such that S`° < Qc and S* < Q*, then ELI, Sc'] < CG-(Z`°).

To prove (40), first note that (IS`°j, p) = 1 and hence Z`' = [Z"', Sc] Cz.(S"'); we showthat [L`°, S'] centralizes the two factors in this product. Now LS = G wouldimply that G* = L*S* = P*S*, contradicting S* < Q*. Thus LS < G, hence Go =OP(LS) < G and OP(G0) = Go. Since P c Sylp(LS), P n Go is a Sylow p-subgroup ofGo containing the normal subgroup Z n Go of Go in its centre. The minimality of Gimplies that (Z n G0)`° is centralized by (P n G0)`' and, by (33), it is also centralizedby MI. Since q is index preserving, Go = OP(LISI) and so,

(41) DO n OP(L'S`°) is centralized by M°(P`0 n OP(L`°S`°)).

Now Z`0 a LcSc and the p'-group S"' <OP(L`0S"); hence [Z`0, S`°] < Z`° n OP(LcS`°).Furthermore S" < OP(LcS') implies that [L`°, S`°] < OP(LISI) and so fromELI, S`°] < L`° = McP"' it follows that Mc'[L"', S`0] = Mc(P`° n M`'[L`°, S°]) <M`0(Pc n OP(L`0S`0)) since M`0 = OP(L`°) < OP(LISI) also. Thus (41) shows that[Z`°, S`°] is centralized by [L"', S`°].

The other factor of Z`° is easier to handle. Since L < CG(Z), [CZ(S), LS] = I andhence, in particular, [CZ(S), OP(LS)] = 1. By 1.6.8, [CZ(S)`°, OP(LS)m] = 1 and, by4.1.2, CZ(S)`° = CZ,(S`°). Since [L`°, S`°] < OP(LISI) = OP(LS)10, it follows that[L"', Sc'] centralizes CZm(S`0) and hence also [Z`°, S`0]CZv(S`°) = Z°. This proves (40).

We show next that Q* is cyclic of prime power order. If this were not the case,there would exist elements s; e Qc such that <s'> < Q* for all i and Q* is generatedby the' s*. Applying (40) to SIP = <si >, we obtain ELI, s1] < CU(Z`°) and hence[P*, s*] = [L °/M°, s*] < Cd(Z`°)/M"'. By (38), P* _ [P*, Q*] < CU(Z"')/M`° and thiswould contradict (37). Thus Q* is cyclic of prime power order and therefore also itsepimorphic image

(42) Q is a cyclic q-group for some prime q.

Lets e Q10 such that Q = (. >. We want to show that 9 = satisfies (10) and (11). By(34), S2(Zc) is centralized by P; but, by (29) and 4.3.9, S2(Z`') is irreducible underG = P<>. It follows that

(43) s operates irreducibly on S2(Z`°),

that is, (11) holds. By (38), [P,9] = P and, applying (40) to S10 = <s'>, we obtain[P`0, s9] < ELI, S1] < CU(Zc). It follows that [P, s9] = 1 and it remains to be shownthat s operates irreducibly on P. So suppose that Pl is a 0-invariant subgroup of Psuch that Pl < P. Let T < G such that T'°/Cd(Zc) = Pl Q. Then T < G, hence R =OP(T) < G and OP(R) = R. By 4.1.3, Z`° = [Z", Q`°] x CZm(Q`°); since IS2(Zc)l > p2and Q`0 operates irreducibly on S2(Z'*), it follows that Z`° = [Zc', Q`0]. Since Q`° <OP(TI) = R`° and Z'° < Tc, we have Z`0 = [Zc,Q`0] < R`'. Now I G: RI is a power of

p, hence G = PR and therefore Z is a normal subgroup of R contained in the Sylowp-subgroup P n R of R. The minimality of G implies that Z" is contained in thecentre of the Sylow p-subgroup (P n R)`° of RO. Hence R"/R`° n Q(Z') is a p'-group.It follows that R = R°CG(Z4')1Cd(Z4) is a p'-group and sinceT`°/0"(Tc)CU(Z`') is a p-group, R is a normal p-complement in T = P1Q. ThereforeT = P1 x Q and, by (38), P1 = 1. Thus s operates irreducibly on P and all the as-sumptions of 4.4.7 have been verified. By 4.4.7,

(44) r = 2, IPI = p, q = 2 and there exist y e P° satisfying <y) = P and a basis{v1,v2} of Z`° such that <v1> = <v1>, <V2> = <v2> and

Let W = n(<vi>) for i = 1, 2. If x e P°\Cpm(Z°), then <z> = P zA 1 and therefore,by (39), H = <x>Z' satisfies the assumptions of 4.4.6. If W1 S2(<x>) W2, thenthe cyclic subgroup <x> n Z" of the homocyclic group Z° would be contained ina maximal cyclic subgroup <v> of Z" such that <v1 > n <v> = 1 = <v2 > n <v>. By (a)of 4.4.6, <v>' = <v> and since also <vi>' = <vi> for i = 1, 2, it would follow from1.5.4 that x induces a power automorphism in Z". This is not the case; thus

(45) i2(<x>) e { W1, W2} for every x e P`°\Cpm(Z°).

By (44), (yCG-(Z°))S = y-'Cj;(Z1') and, since s normalizes P°, it follows thaty-'Cpm(Z°). By (45), s operates on {W1, W2} and, since s is irreducible

on Q(Z''), Wl = W2 and Wz = W1. For i = 1, 2 define

S i = {x e yCp.(Z1')IS2(<x)) = W} and T = {x e y-' Cpm(Z`°)Ii2(<x>) =

Then S._' = {x-'Ixe S1} = Ti and hence SiI = ITI for i = 1,2. If xeS1, thenW2 = Wl = n(<x>)S = n(<xs)) and so xs e T2; similarly, xs e T1 for x e S2. By (45),yCp-(Z°) = S1 U S2 (disjoint set-theoretic union) and y-' Cpm(Z") = T1 Q) T2; it fol-lows that S; = T2. Therefore IS1I = I T21 = IS2I and hence I Cpo(Z')I = IyCp®(Z")I =2IS1I. But this contradicts the assumption that p > 2.

The centre of a Sylow p-subgroup

We remind the reader that a finite group G is called p-normal if Z(P)9 < P impliesthat Z(P)9 = Z(P) whenever P e Syl"(G) and g e G. In particular, every finite groupwith abelian Sylow p-subgroups is p-normal. Therefore the following result is ageneralization of Theorem 4.4.2 in the case p > 2 and OP(G) = G.

4.4.9 Theorem (Menegazzo [1974]). Let p > 2 and suppose that the finite group G isp-normal and satisfies 0"(G) = G. If cp is an index preserving projectivity from G to agroup G, then Z(P)`° < Z(P°) for every Sylow p-subgroup P of G.

Proof. By Griin's Second Theorem (see Robinson [1982], p. 285), the largest abelianp-quotient of G is isomorphic to that of N = NG(Z(P)); hence O"(G) = G implies thatOP(N) = N. Now 4.4.8 applied to Z = Z(P) a N and the projectivity induced by qPin N yields that Z(P)" < Z(P4).

4.4.10 Theorem (Menegazzo [1974]). Let p > 2 and suppose that the finite group Gis p-soluble and satisfies OP(G) = G. If cp is an index preserving projectivity from G toa group G, then Z(P)4' = Z(P`0) for every Sylow p-subgroup P of G.

Proof. Since cp is index preserving, G is p-soluble (see also 4.3.8) and OP(G) = G.Therefore we only have to show that Z(P)" < Z(P`0); this result for gyp-' will yield theother inclusion. So suppose that G is a minimal counterexample to the assertionZ(P)'P < Z(P`0) and put Z = Z(P). For every nontrivial normal subgroup N of Gsuch that N° < G, qp induces a projectivity from GIN to G/N`0, and ZN/N is con-tained in the centre of the Sylow p-subgroup PN/N of G/N. The minimality of Gtherefore implies that

(46) [PQ, Z`'] < NP.

Hence if Op (G) : 1, it would follow that [P°, Z°] < OP.(G) n P`° = 1, contradictingthe choice of G. Thus

(47) 1.

Let Q be a complement to P in N°(P). By 4.1.1, T = [P, Q]Q is the normal closureof Q in NG(P); in particular, OP(T) = T. Furthermore, Q normalizes [Z, Q] and[Z, Q] < Z since Z a N°(P). Therefore [Z, Q] is a normal subgroup of T containedin the centre of the Sylow p-subgroup [P, Q] of T. By 4.4.8, [Z, Q]" < Z([P, Q]").Clearly, CZ(Q) is centralized by [P, Q] Q = T and, by 1.6.8, [CZ(Q)`P, T`0] = 1; inparticular, [Cz(Q)`0, [P, Q]`°] = 1. Since Z = [Z, Q] CZ(Q), it follows that

(48) Z10 is centralized by [P, Q]'*.

Let M be a maximal normal subgroup of G. Since OP(G) = G, G/M is a p'-group; letH = OP(M). Then M = PH and the Frattini argument yields that G = NG(P)M =NG(P)H. Now T = [P, Q]Q a NG(P) and hence [P,Q]QH is normalized byNG(P)H = G. Since QH/H contains a p-complement of G/H, G/[P,Q]QH is ap-group. But OP(G)=G then implies that G = [P,Q]QH. By Dedekind's law,P = P n [P, Q] QH = [P, Q] (P n QH) and, since (IQH : HI, p) = 1, P n H is a Sylowp-subgroup of H and of QH. Therefore P n QH = P n H and

(49) P = [P, Q] (P n H).

If Z < H, then Z would be contained in the centre of the Sylow p-subgroup P n Hof H and, since OP(H) = H, the minimality of G would imply that[Z`*, (P n H)`°] = 1. But then (48) and (49) would show that [Z", P°] = 1,contradicting the choice of G. Thus

(50) Z H.

Since G is p-soluble and OP.(G) = 1, we may conclude that CG(OP(G)) < OP(G) (seeRobinson [1982], p. 261). Of course, Z centralizes OP(G) < P and henceZ < CG(OP(G)) = Z(OP(G)). It follows that

(51) Z° is an abelian normal p-subgroup of G.

Let 1 = Ho < H1 < - - - < H. = H be part of a chief series of G. Then every H;/H;_,is a p-group or a p'-group, and we want to show by induction that

(52) [Z',H;] = l fori =0,...,n.This is certainly true for i = 0. So let [ZG, Hi_,] = 1 and suppose first that H;/Hi_1is a p-group. Since H; a G, P n H; is a Sylow p-subgroup of H; and thereforeH. = H;_, (P n H;). Hence for every x e G, there exists y e Hi_, such that(P n Hi)'-' = (P n H;)Y and ZY_' = Z since [ZG, Hi_1] = 1. Therefore

[Z', P n H;] = [ZX, (P n H;)YX] = [ZY-', P n H_]YX = [Z, P n H1]YX = 1

since Z = Z(P). It follows that [Zc, P n H;] = I and then also [ZG, H;] =[ZG, H;_, (P n H;)] = 1. Now suppose that H;/Hi-1 is a p'-group. Then, since[Zc, Hi-,] = 1, this p'-group operates on the abelian p-group Zc, and 4.1.3applied to the semidirect product of ZG by the operating group, shows that ZG =[Zc, H;] x Cz.(H;). Both factors are normal p-subgroups of G = O"(G). By 4.3.9,their images are normal in G and one of them has to be trivial since otherwise (46)would yield that [P'°, Z°] < [Zc, H;]`° n Cz.(H;)'° = 1, contradicting the choice ofG. But Cz6(H;) = 1 would imply that ZG = [ZG , H;] < H; < H, contradicting (50).Hence [ZG , H;] = 1, as desired. This proves (52).

Now (52) shows that [Zc, H] = [Zc, H = O°(H), 1.6.8 yieldsthat [(ZG)°, H °] = 1; in particular, [Z`°, (P n H)°] = 1. But then (48) and (49) implythat [Z", P`°] = 1, contradicting the choice of G.

We finally remark that Menegazzo states Theorems 4.4.9 and 4.4.10 without theadditional assumption that p > 2. Unfortunately his proof is incomplete. In theproof of his main lemma he verifies the assumptions of 4.4.6 and then claims that asuitable generator y of Y has to operate on Z in one of four explicitely given waysthat correspond to the assertions (1) and (2) in (b) of 4.4.6. The possibility (3), how-ever, is overlooked. Exercise 4 gives an example for an operation of this type forwhich (1) and (2) do not hold. Exercise 5 describes the situation that occurs in 4.4.7for p = 2. If one tries to prove Lemma 4.4.8 (and hence Menegazzo's theorems) forp = 2, this situation will occur in (44) in addition to the one considered there. Onehas to decide if it leads to a contradiction or to a counterexample to 4.4.8, andpossibly also to Theorems 4.4.9 and 4.4.10 for p = 2.

Exercises

1. (Seitz and Wright [1969]) If a Sylow p-subgroup of G is a nonabelian M*-group,show that O (G) : G. (Hint: Use 2.3.23.)

2. (Paulsen [1975]) Let G = NP where P is an abelian Sylow p-subgroup and N anormal p-complement of G. If cp is a projectivity from G to G, show that((P'*)')'° < Z(G) and (P`°)' centralizes N`°.

3. (Paulsen [1975]) Let G = NP where P e Syl°(G) is an M*-group and N is anormal p-complement in G. Show that there exists a projectivity from G to agroup with abelian Sylow p-subgroups if and only if P' < CP(N).

4. Let N = (a,) x x (a,> x (b> x (c> where o(a;) = o(b) = p", o(c) = p, n > 2,and n > 3 in case p = 2. Let x e Aut N be such that a; = a1 +P"-' (i = 1,... , r),b° = b'+°" 'a°and ca = ca1-'"-'.

1

(a) Show that a is an automorphism of order p of N and there exists an extensionH of N by (y> such that yP = a, c and xy = x° for all x e N.

(b) H is an M*-group and Z = (a...... a b> is a homocyclic normal subgroupof H.

(c) Show that I Z : CZ(y)l = p", JZ : DZ(y)l = p and there is no v e Z of order p"and 1 (mod p) such that vy = v'+ °"

5. Show that if p = 2 instead of p > 2 in the assumptions of 4.4.7, then r = 2, q = 3,A A4, and there exist a basis {u, v} and generators x, fi of B and y of C whosematrices with respect to the basis {u, v} are

1 + 2n-1 2n-' 1 + 2n-1 0 _ 0 1

0 1+2n 1' 2"-' 1+2n' '' -1 -1

Chapter 5

Projectivities and normal structure of finite groups

Many classes of groups are defined via properties of normal subgroups. To decidewhether such a class is invariant under projectivities, we have to know somethingabout projective images of normal subgroups; and in order to get lattice-theoreticcharacterizations of these classes, we have to be able to describe normal subgroupsin the subgroup lattice of a group. The basic tools to accomplish this, at least in finitegroups, are the modular subgroups introduced in § 2.1. Every normal subgroup N ofG is modular in L(G) and for every projectivity cp of G, NP is a modular subgroup ofG. Therefore we want to find out how far these modular subgroups can deviatefrom normality. How to measure this? We know that to every subgroup H of G thereexist two normal subgroups of G enclosing H,

the core HG = n Hx and the normal closure HG = U HxxEG xEG

of H in G. Obviously, HG is the largest normal subgroup of G contained in H, HG isthe smallest normal subgroup of G containing H and the structure of HG/HG and theembedding of this factor in G describe the deviation of H from normality.

For this reason we investigate the structure of MG/MG and G/MG for a modularsubgroup M of G, and in § 5.1 reduce this problem for finite groups to the case whereM is permutable in G. In § 5.2 we show that then MG/MG is hypercentrally embeddedin G and hence is nilpotent. Together with the results of § 5.1 this yields a structuretheorem for G/MG for modular subgroups of finite groups which, in particular,implies that MG/MG is hypercyclically embedded in G.

Our first nontrivial result on modular subgroups, Lemma 5.1.2, already leads tolattice-theoretic characterizations of the classes of simple, perfect and soluble finitegroups. These we present in § 5.3 together with two characterizations of the class ofsupersoluble groups. The first of these depends on the main result of § 5.2, the secondasserts that a finite group is supersoluble if and only if its subgroup lattice satisfiesthe Jordan-Dedekind chain condition. This characterization, given by Iwasawa in1941, and Ore's theorem on cyclic groups are the only results known in which aninteresting class of groups is characterized by a property that is important in latticetheory. We finish this section using our results on modular subgroups to determinethe finite groups with lower semimodular subgroup lattice.

In § 5.4 we come to the main topic of this chapter. Let N be a normal subgroup ofthe finite group G, cp a projectivity from G to a group G, H`° the normal closure andK4' the core of N" in G. We first show that H and K are normal subgroups of G, andthen use the results of §§5.1 and 5.2 to prove a structure theorem due to Schmidt

200 Projectivities and normal structure of finite groups

[1975a] for G/K and G/KcO. This theorem, in particular, implies that H/K andH°/K' are hypercyclically embedded in G and G, respectively, and this yields usefulcriteria for projective images of normal subgroups to be normal in the image group.We mention theorems of Suzuki [1951a] and Schmidt [1972b] which assert thatnormal subgroups with perfect factor groups or without adjacent cyclic chief factorshave this property. Using results in Chapter 4 on singular projectivities, we showthat subnormal subgroups of finite groups are mapped by projectivities to subnor-mal subgroups, modulo the obvious exception of P-groups.

The results of §§ 5.2 and 5.4 reduce most problems on projective images of normalsubgroups to the study of projectivities between p-groups. This, in particular, is thecase for the investigation of the finer structure of H/K, not only for finite but also forinfinite groups, as we shall see in Chapter 6. Therefore in § 5.5 we study this situa-tion; more precisely, we handle the crucial case that G is a finite p-group, GIN iscyclic and K = 1. In this case we show that IN'J < 2, G is metabelian and G is solubleof derived length at most 3. In §6.6 we shall use this together with some ratherspecial properties of permutable subgroups, proved in § 6.3, to show that forarbitrary groups, H/K and H`°/K`° are soluble of derived length 4 and 5, respectively.This whole theory was developed in the 1980's by Busetto, Menegazzo, Napolitaniand Zacher.

In § 5.6 we show that normalizer preserving projectivities are also centralizerpreserving, and present the classic examples, due to Rottlander, Honda, Yff, andHolmes, of projectivities between nonisomorphic groups that preserve indices, con-jugacy classes and the isomorphism types of proper subgroups. Finally, we begin tostudy an important problem that can be regarded as a generalization of the maintheme of this chapter: When can we say that a projectivity maps a conjugacy classA of subgroups of G to a conjugacy class of G? This is interesting because in this casenot only G but also G operates on A. We show that, modulo the obvious exceptionof P-groups, conjugacy classes of maximal subgroups of finite soluble groups aremapped to conjugacy classes by any projectivity.

All groups considered in this chapter are finite except for a few general remarkson modular and permutable subgroups in §§ 5.1 and 5.2, and on normalizer pre-serving projectivities in § 5.6.

5.1 Modular subgroups of finite groups

As already indicated, we shall investigate the structure of MG/MG and G/MG for amodular subgroup M of a finite group G. In this section we reduce this problem tothe case that M is permutable in G. More precisely, we show that if M is notpermutable in G, then G/MG is P-decomposable.

Modularity and permutability

First of all we give a useful criterion for a modular subgroup of a finite group to bepermutable; it will be generalized to arbitrary groups in 6.2.10.

5.1 Modular subgroups of finite groups 201

5.1.1 Theorem. The subgroup M of the finite group G is permutable in G if and onlyif M is modular and subnormal in G.

Proof. First suppose that M is permutable in G. Then by 2.1.3, M is modular in G.We want to show that M is subnormal in G and use induction on JGJ. Let N be amaximal permutable subgroup of G containing M. We show that N < G; by induc-tion, M !9!9 N, and it will follow that M < < G. So suppose, for a contradiction,that N is not normal in G. Then there exists x e G such that N" N. For H < G,

HNN" = NHN" = Nx-'H"-'Nx = Nx-'NH"x = NN"H;

hence NN" is permutable in G. Since N is a maximal permutable subgroup of G, itfollows that NN" = G. But then there exist a e N, b E N" such that x = ab and weget the contradiction N = N" = Nsb-' = N". Thus N < G and M !!9:!9 G.

Conversely suppose that M is modular and subnormal in G. We show that M ispermutable in G, again using induction on I G I. Clearly M is modular and subnormalin every subgroup which contains M. Let H < G. If H < M, then HM = M = MH.If H M, then M < H u M and there exists a proper normal subgroup N of H U Msuch that M < N. It follows that N < G and, by induction, M(H n N) = (H n N) M.The modularity of M implies that N = (H u M) n N = M u (H n N) and, sinceN < H u M, we finally get that

HM = H(H n N)M = HN = NH = M(H n N)H = MH.

Thus M is permutable in G.

Since we shall use them constantly, we collect for reference the basic properties ofmodular subgroups; these were proved more generally for modular elements oflattices in 2.1.5 and 2.1.6. So let M be modular in the (not necessarily finite) group G.Then

(1) [H u M/M] ^ [H/H n M] for all H < G,

(2) H n M mod H for all H< G.

(3) If N 9 G, then MN/N is modular in GIN.

(4) If M < M1 < G and M, mod [G/M], then M1 mod G. In particular, if N < Gand M1/N mod GIN, then M1 mod G.

(5) If M1 and M2 are modular in G, then so is M1 U M2.

(6) If cp is a projectivity from G to a group G, then M4' mod G.

Permutable subgroups have similar properties. We prove these and also somegeneral results on permutability of subgroups, again for arbitrary groups. So letH.(i c- I), H and K be subgroups of G.

(7) If G is finite, then HK = KH if and only if IH u K: KI _ IH: H n KI.

Indeed this equation is equivalent to I HK I = I H u K I = I KH I.

(8) If HiK=KH;for all H=<HiliEl>,then HK =KH.

For, every x E H has the form x = x,... x where xi E HH for some j E I; by inductionon n, we see that to every y c- K and x = x, ... x E H there exist y' E K and x' E Hsuch that xy = y'x'. This is clear by assumption if n = 1. And if it is true for n - 1,then there exist y, y' E K and x;,, z E H such that

where x' = zx,, e H. Thus HK c KH and the other inclusion is proved in an entirelyanalogous way. Since every subgroup of G is generated by its cyclic subgroups, animmediate consequence of (8) is the following.

(9) If XK = KX for all cyclic subgroups X of prime power or infinite order of G,then K is permutable in G.

We come to the basic inheritance properties of permutable subgroups.

(10) If Hi per G for all i El, then <H1i El> per G.

(11) If H per G and K < G, then H n K per K.

(12) If N a G and H per G, then HN/N per GIN.

(13) If N a G and N < H < G such that H/N per G/N, then H per G.

Clearly (10) immediately follows from (8). To prove (11), note that for every X < K,by Dedekind's law, (H n K)X = HX n K = XH n K = X(H n K). Finally, (12) and(13) are trivial. We remark that for finite groups (10)-(13) also follow from 5.1.1 andthe corresponding properties of modular or subnormal subgroups. Note further that5.1.1 and Exercise 2.1.3 show that the intersection of two permutable subgroups ingeneral is not permutable.

Maximal modular subgroups

By (5), the join of two modular subgroups of a group is again modular. Therefore, infinite groups, it seems reasonable to study maximal modular subgroups, that is,modular subgroups which are not contained in any larger proper modular subgroupof the group. We are able to show that such a subgroup is nearly normal. This resultis basic for everything that follows; for example, it immediately yields the lattice-theoretic characterizations of the classes of simple, perfect and soluble finite groupswhich will be presented in § 5.3.

5.1.2 Lemma (Schmidt [1969]). If M is a maximal modular subgroup of the finitegroup G, then either M g G or G/MG is nonabelian of order pq for primes p and q.

Proof. Let G be a minimal counterexample to the lemma, and take a maximalmodular subgroup M of G for which the assertion of the lemma does not hold. By

(3) and (4), M/Mc is a maximal modular subgroup of G/Mc. Hence G/MG is also acounterexample to the lemma, and the minimality of G implies that

(14) M6 = 1.

Let S be a maximal subgroup of G containing M. For x e S, M < M u Mx < S and,by (5) and (6), M u Mx mod G. The maximality of M implies that M = M" and henceM a S. Since M is not normal in G, S is the only maximal subgroup of G containingM. But then 2.1.5 shows that S mod[G/M] and, by (4), S mod G. The maximality ofM yields S = M, that is,

(15) M is a maximal subgroup of G.

Let N : M be a conjugate to M and put D = N n M. By (1), [G/M] L-t [N/D] andhence D is a maximal subgroup of N, and, similarly, of M. If H M is a maximalsubgroup of G containing D, then H u M = G and H n M = D; by (1), D is a maxi-mal subgroup of H. Thus

(16) [G/D]\{G, D} is an antichain.

We want to show next that D is normal in neither M nor N. So suppose, for acontradiction, that D 9 M, say. Then IM: DI = p is a prime and every p'-element ofM is contained in N. Since M and N are isomorphic, N cannot contain morep'-elements than M. It follows that

O"(M)_ <xEMJ(o(x),p)= 1>=O"(N)aMuN=G.

By (14), O (N) = 1. So D is a maximal subgroup of the p-group N and hence D :!g N.Therefore D a G and, by 2.2.4, IG/DI = pq where q is a prime. This contradicts thechoice of G. Thus

(17) D is neither normal in M nor in N.

By (2) and (6), D is modular in M and in N; the minimality of G implies that M/D,Mis nonabelian of order pq where p, q are primes and p > q, say. Since D is not normalin M, IM : DI = p, hence I N : DI = p and therefore N/DN is nonabelian of order pr

G

Figure 16

where r < p is a prime. Let it = {q, r}. Then

O"(D)=O"(DM)=O"(DN,)<MuN=G

and, by (14), 0"(D) = 1. This shows that I M: M n NI = p is the largest prime divid-ing I M I and M n N is a Hall p'-subgroup of M. This holds for every conjugateN # M of M. The normal p'-subgroup DM of M is contained in every Hall p'-subgroup of M, hence in every conjugate of M, and therefore in MG = 1. ThusDM = 1, that is,

(18) IMO=pgandIDI=q<p.

Let x c G such that N = Mx. Then D and D" are Sylow q-subgroups of MX andhence there exists y E Mx such that D = DXY. Since M 0 Mx, we have x 0 MX andtherefore xy 0 D. Thus T = NG(D) > D. By (14), T < G and then (16) implies that Dis maximal in T, that is,

(19) IT: DI = t is a prime.

Since MG = 1, there exists a conjugate L # M of M such that D T n L. By (1),T n L is maximal in T and hence of prime order. Since D < T of order q < p,I T n LI = p would imply that T were cyclic of order pq and hence T n L < T U L = G,

contradicting (14). Thus I T n LI # p and, since I LI = pq, it follows that I T n LI = qand I TI = q2. By (1), M intersects every conjugate of Tin a subgroup of order q. IfT1, T2 are different conjugates of T, then T, n M # TZ n M; for, otherwise T, n M =T2 n M < T, u T2 = G, contradicting (14). Therefore, since M has only p subgroupsof order q, T has at most p conjugates; on the other hand, T is a maximal subgroupof G and p divides I G : TI. It follows that I G : TI = p and hence G I = pq2. But thenthe subgroup of order p of M is a Sylow p-subgroup of G, its normalizer is M, andSylow's theorem implies that q = I G : M I - I (mod p). This contradicts (18).

The groups <M, x>

In addition to the structure theorem for G/MG we want to prove a characterizationof modular subgroups in finite groups, due to Previato [1975], that is similar to theuseful criterion (9) for permutability. It asserts that M is modular in the finite groupG if and only if M is modular in every subgroup <M,x> where x runs through theelements of prime power order in G. Of course, every modular subgroup has thisproperty, and indeed it is equivalent to modularity. Both results will follow from thefact that a subgroup M with this property has the desired structure. This we shallprove in a number of steps in the remainder of this section. It should be clear to thereader that all the results we derive in the course of this proof for subgroups havingPreviato's property, for example (b) of 5.1.5, 5.1.6 and 5.1.12, in particular hold formodular subgroups in finite groups and are useful properties of these.

First of all we have to handle the groups <M,x>. By (1) and 1.2.7, the interval[M u <x> M] [<x>/<x> n M] is a chain if M mod G and o(x) is a prime power.Therefore we shall study this situation.

5.1.3 Lemma. IJ' M is modular in the finite group G and [G/M] is a chain, then eitherG/MG is a p-group or M is a maximal subgroup of G and G/MG is nonabelian of orderpq for primes p and q.

Proof. Again let G be a minimal counterexample to the lemma and take a modularsubgroup M of G for which the assertion of the lemma does not hold. Then G/MG isalso a counterexample and the minimality of G implies that

(20) MG = 1.

By 5.1.2, [G/M] is a chain of length n >_ 2; let H be the maximal subgroup of Gcontaining M. Then H mod[G/M], therefore H mod G, by (4), and again 5.1.2 yieldsthat IG : HI = p is a prime. The minimality of G, furthermore, implies that eitherH/MH is a q-group or M is a maximal subgroup of H and IH: MI = q for someprime q; in both cases, JH : MI = q"-'. Let P, E Sylp(M) and take P E Sylp(G) suchthat P, < P. Since I G : H I = p, we have P $ H, and, since H is the only maximalsubgroup of G containing M, it follows that P u M = G = P u H. If q 0 p, then P,would be a Sylow p-subgroup of H contained in the p-subgroup P n H. It wouldfollow that P n H = P, = P n M and then by (1),

[G/ M] = [P u M/M] [P/P n M] = [P/P n H] [P u H/H] = [G/H],

contradicting n > 2. Thus p = q and

(21) IG:MI =p"wheren> 2.

Suppose, for a contradiction, that H is not normal in G. Then, by 5.1.2, G/HG isnonabelian of order pr for some prime r < p. If N/HG is the normal subgroup ofindex r in G/HG, then MN = G and hence [N/N n M] [G/M] is a chain of lengthn. Since IN: N n MI = I G: MI = p", the minimality of G implies that N/(N n M)N, isa p-group. It follows that OP(N) < M and, by (20), O°(N) = 1, that is, N is a p-group.If xEG'\H,then MnMX<HnHX=HG <Nis ap-group andMuMX=GsinceM HX and H' is the only maximal subgroup of G containing MX. As MX is modu-lar in G, M n MX mod M and [M/M n MX] [G/MX] is a chain of length n. Againthe minimality of G implies that M/(M n MX)M is a p-group since r2 does not dividethe order of G. But then I G : M 1, 1 M: M n MX I and I M n MX I are powers of p andhence G is a p-group. This contradicts the choice of G. Thus we have shown that

(22) H a G.

Since G is not a p-group, there exists a prime s p dividing I G 1; let S E Syls(G) andT= NG(S). If S a G, then (21) would imply that S < M, contradicting (20). Thus S isnot normal in G, that is, T 0 G. Since H is a normal subgroup of index pin G, S < Hand the Frattini argument yields that G = HNG(S) = HT. It follows that T H andhence T u M' = G for all x e G since H is the only maximal subgroup of G contain-ing M. Again M' mod G implies that T n M' mod T and [T/T n M'] [G/MX]is a chain of length n; furthermore I T : Tn H I = p divides I T : T n M' J. Thereforethe minimality of G implies that T/(T n M')T is a p-group and it follows thatS < (T r) MX)T < M' for all x c- G. But then S < MG = 1, a final contradiction.

We come back to the groups <M, x> for x of prime power order or, more gener-ally, such that

o(x, M) = I<x>: <x> n MI

is a power of a prime. In particular, we are interested to know when M and <x>permute.

5.1.4 Corollary. Let M be modular in the finite group T and suppose that T =<M, x> where o(x, M) = p", with p a prime, and n e N. Then one of the following holds.

(23) TIMT is a p-group and M is permutable in T; in particular, M<x> = <x>M.

(24) T/MT is nonabelian of order pq for some prime q and I M : MTI = q < p; hereM<x> = <x>M.

(25) T/MT is nonabelian of order pq for some prime q and IM: MTI = p < q; hereM<x> : <x>M.

Proof. If x e M, (23) holds. So suppose that x 0 M. By (1) and 1.2.7, [T/M] -[<x>/'<x> n M] is a chain and, since x 0 MT, p divides I T/MTI. Therefore 5.1.3 showsthat T/MT is a p-group or nonabelian of order pq where q is a prime. In the first case,M a a T; therefore, by 5.1.1, M is permutable in T and (23) holds. In the secondcase, M is not normal in T and hence I M : MTI is the smaller of the primes p and q.So if p > q, IM: MTI = q and hence I<x> u M: MI = p = I<x>: <x> n MI; by (7),M<x> = <x>M and (24) is satisfied. Finally, if p < q, then IM: MTI = p andJ <x) u M : MI = q : p = I <x> : <x> n MI; again by (7), M <x> : <x> M and (25)holds.

An immediate consequence of 5.1.4 is the second part of the following useful result.

5.1.5 Lemma. Let G be a finite group and suppose that M, H < G such that(IMI,IHI) = 1.

(a) If M is permutable in G, then H < NG(M).(b) If M is modular in <M, x> for all x e H of' prime power order, then MH = HM.

Proof (a) Clearly, I MH : M I = I H: H n M I is prime to I M I; on the other hand, forx c H, I MMX : M I = I MX: M n MX I divides I M O. Since M < MMX < MH, it followsthat M = MMX and hence M = MX.

(b) If x e H has prime power order, then M mod<M,x> and, since (IMI, IHI) = 1,(25) cannot hold for T = <M, x>. Therefore 5.1.4 implies that M<x> = <x>M and,by (8), MH = HM.

The structure of M/MG

To determine the structure of M/MG we need a description of MG in terms of thegroups M,,M.X, for x c G of prime power order. To obtain this we define for any

subgroup H and subset X of a (not necessarily finite) group G,

(26) Hx = <HXIx e X).

If Y G, then (Hx)Y = <HXlx e X)Y = <HX''Ix e X, y e Y) and hence

(27) (Hx)r = Hxr for H < G and arbitrary subsets X, Y of G.

If X is a subgroup of G, then clearly X < NG(Hx); furthermore, (27) implies that

(28) (Hx)Y = (HY)x for X, Y < G such that XY = YX.

5.1.6 Lemma. Let Go he the set of elements of prime power order in the finite groupG. /f M is modular in <M,x) for all x E Go, then MG =n e Go}.

Proof. Since L = n {M<M.X>Ix e Go} is an intersection of conjugates of M, it clearlycontains the intersection MG of all these conjugates. We have to prove the otherinclusion L < MG, that is, L < M9 for all g e G. We use induction on the number r ofprimes dividing o(g) to show that if 1 0 xi E Go such that <y) = (xl) x x <xr),then

(29) MCM.9) =n M<M,x,>-

Since the left hand side of this equation is contained in M9 and the right hand sidecontaines L, it will follow that L < M9.

For r = 1, (29) is trivial. So let r > 2 and assume that the assertion is true for r - 1.For every i = 1, ..., r, M is modular in <M, xi) and hence, by 5.1.4, any of thesegroups must have one of the properties (23)-(25). Without loss of generality we maychoose the notation in such a way that <M, xr) = T satisfies (24) or, if there isno such xi, it satisfies (25) if there is an xi with this property. Furthermore write<M,9) = H, <M,x1,...,xr-1) = S, xi ...xr-i = X, Xr = Y, X = <x> and Y = <y>.Then <x> = <x l) x . . X <xr_ 1 ), hence by our induction assumption,

r-1

(30) Ms =n M<M,X,)i=1

and we have to show that

(31) MH = Ms n MT.

This is clear if T < S; for, then S = H and hence MH = Ms = Ms n MT. Thereforesuppose that T S and let D = Ms n MT. Since MH is normal in S and T, MH < Msand M. < MT and hence

(32) MH < D.

If M permutes with all the <xi>, then, by (8), M also permutes with <x 1) ... <xr_ 1) _X and Y. It follows that I S: M I= I MX : M I= I X: X n M I is coprime toI Y: Y n M I _ I T : M I and hence S n T = M. If M does not permute with <xj ), say,then, by 5.1.4, <M, xj> satisfies (25) and our choice of Xr implies that M is a maximalsubgroup of T Since T S, again it follows that S n T = M. Thus in any event

<M,x>=S

Ms

Figure 17

(33) S n T = M.

Since D < Ms < S, we see that (Dx)Y < (Ms)' < T and, similarly, (DY)x < S. As Xand Y are contained in (g>, they permute and, by (27) and (28), DXY = (Dx)Y =(DY)x < S n T, that is,

(34) DxY < M.

Now suppose first that MY = YM. Then (DXY)M = (DX)YM is invariant underYM = T and contained in M. Thus (Dx)YM < MT; in particular, Dx < MT. SinceD < Ms < S, it follows that Dx < Ms and hence Dx < Ms n MT = D. ThusDx = D and (DY)x = (Dx)Y = D. Since also (DY)M = (DM)Y = DY, it follows thatDY (M, X, Y> = H. This implies that D < DY < M. and, by (32), D = MH, asdesired.

So, finally, suppose that MY 0 YM. Then (25) holds for T and our choice of x,implies that none of the <M, x1> satisfies (24). Therefore, by 5.1.4, for everyi E { I__ r}, the group M/M<M,X,

>is a p-group if x; is a p-element. Since the orders of

the x; are pairwise coprime, it follows from (30) that (M/M1 is prime to IM/MTIand hence Ms u MT = M. If Dx A D, then D < Dx < Ms and hence Dx = Ms sinceMs : D I = IM: MTI is a prime. By (34), Ms < DXY < M and so M = DxY U MT

would be invariant under Y; but M does not even permute with Y. Thus Dx = D.Again (DY)x = (Dx)Y = DY and, clearly, (DY)MT = (DMT)Y = D1. Since D < DY < MTand M/D = MS/D X MT/D, we obtain DY < M and hence DY < M u X = S. AsDY < M, it follows that DY < M. and then DY < Ms n MT = D. Thus DY = D andso, finally, D < <M, X, Y> = H. This implies that D < MH and (32) shows that D =MH, as desired. El

It is clear that MG < n {M<M.X>Ix c- Go} < n {MXIx c Go} holds for arbitraryM < G. We remark that for a modular subgroup M of a finite (or periodic) group G,

the proof of even the assertion MG = n {MXIx e Go} is much easier than that of5.1.6; one can assume that G = <M, x> for some x e G and then use induction ono(x, M) (see Stonehewer [1976b]). We shall see in 6.2.3 that this result generalizes tomodular subgroups of arbitrary groups; also this proof is easier than that of 5.1.6.

If M mod<M, x> and x E G is of prime power order, then, by 5.1.4, M/M<M,X> isnilpotent. It follows that M/n {M<M,X> Ix e Go} and therefore, by 5.1.6, that

(35) M/MG is nilpotent if M mod<M, x> for every x e Go.

In particular, we have our first general result on the structure of MG/MG for amodular subgroup M of G.

5.1.7 Theorem (Schmidt [1969]). If M is modular in the finite group G, then M/MGis nilpotent.

By 5.1.7, the proof of our main theorem divides into two parts. We first prove theresult for modular subgroups of prime power order, and then show that for anarbitrary modular subgroup M of G, the Sylow subgroups of M/MG are modular inG/MG. The first part, the proof of Lemma 5.1.9, is a combination of ideas of Previato[1975] and Schmidt [1970a]; in the second part we shall mainly follow Previato'spaper.

Modular subgroups of prime power order

We shall need the following simple lemma to establish the modularity in G of asubgroup M for which G/MG has the correct structure.

5.1.8 Lemma. Let G = G1 x G2 and (I G11,1 G21) = 1. If Mi mod Gi for i = 1, 2, thenM1 X M2 mod G.

Proof. By 1.6.4, L(G) L(G1) x L(G2). Since the modular laws (2) and (3) of § 2.1hold for Mi in L(G;), they also hold for (M1,M2) in L(G1) x L(G2) and hence forM1 X M2 in L(G).

We come to the first part of the program described above.

5.1.9 Lemma. If M is a subgroup of prime power order of the finite group G, then thefollowing properties are equivalent.

(a) M is modular in G.(b) M is modular in <M, x> for all x e G of prime power order.(c) M is permutable in G or G/MG = MG/MG x K/MG where MG/MG is a non-

abelian P-group of order prime to IK/MG I.

Proof. If the first alternative in (c) holds, then, by 2.1.3, M is modular in G; if thesecond one is satisfied, then 5.1.8 shows that M/MG is modular in G/MG and, by (4),M is modular in G. Thus (a) follows from (c) and, clearly, (b) follows from (a). Itremains to be shown that (b) implies (c).

Let G be a minimal counterexample to this assertion and take a q-subgroup M ofG, q a prime, satisfying (b) but not (c). If N is a normal subgroup of G contained in Mand gN e GIN of prime power order, then there exists g' e G of prime power ordersuch that gN = g'N. Because M mod<M, g'>, we obtain M/N mod<M, g' )/N =<M/N, gN>; hence M/N satisfies (b) in GIN. The minimality of G implies that

(36) MG = 1.

There exists x E G of prime power order such that M is not permutable in <M, x>;otherwise, by (9), M would be permutable in G. Put S = <M, x>. Since M is aq-group, 5.1.4 shows that S/Ms is nonabelian of order pq where q = <x>M so that

(37) S = MX where I X I = p > q and S/Ms is nonabelian of order pq.

It follows that S < MG and hence M is not permutable in MG. Therefore if M' <H < G, then (37) implies that p and q divide I M"/MHI and the minimality of G yieldsthat H/MH = M"/MH x L where M"/MH is a P-group of order p"q (n E N) and(I LI, pq) = 1. Then M is a Sylow q-subgroup of H, hence of M', and it follows thatthe core and the normal closure of M in MG are characteristic subgroups of M' andhence normal subgroups of G. Now MG < H < G trivially implies that

MG <MH <MMG <M <MMG <M" <MG

and the normality of MM,; and Mm(' yields that MMG = MH = MG = 1 andMMG =

M" = M'. Thus we have shown:

(38) If MG < H < G, then MG is a P-group of order p"q (n a N) and H = MG x Lwhere (IMGI,ILI)= 1.

Now we distinguish two cases.

Case]: I MI > q2.Then Ms 0 1 and hence NG(Ms) G. If O9 (G) < NG(Ms), there would exist a propernormal subgroup N of G such that M < NG(MS) < N. It would follow that MG < Gand (38) would show that I MI = q, a contradiction. Therefore O9 (G) is not containedin NG(Ms), that is, there exist q'-elements in G\NG(Ms) and then also such elementsof prime power order.

So let y E G\NG(Ms) such that o(y) = r'° where r q is a prime. Since x e NG(Ms),also yX e G\NG(Ms) of order r'". And if y E NG(M), then yX 0 NG(M) since otherwiseyX would normalize M and MX and hence M n MX = Ms. Therefore we may assumewithout loss of generality that y 0 NG(M). Then H = <M, x, y> is a counterexampleto our assertion; for, M is not permutable in S < H and, since MH < Ms, we con-clude that IM: MHI > q2 so that H/MH cannot have the structure given in (c). Theminimality of G yields that G = H = <M, x, y). Now 5.1.5 first of all shows that (y)permutes with M and MX and hence also with M u MX = S; thus G = S<y>. Sec-ondly it yields that M is not permutable in T = (M, y> since otherwise y wouldnormalize M. By 5.1.4, T/MT is nonabelian of order rq, hence o(y) = r and so, finally,

(39) IGI = ISI I<y>I = I MI pr.

Since MS is not normalized by y, MS 0 MT and hence M = MS U M. It follows that

T = MUM''= MU(MT)''U(MS)''= MU(MS)' = MU(MS)XY <MUMXY

and MX T since Sr T = M; therefore also MXy T. Thus T < M U MXY andsince I G : T J = p, it follows that

(40) MUM'Y = G.

If r = p, we could choose y from a Sylow p-subgroup of G containing x. Theno(.xy) = p and our assumption (b) together with (40) would imply that M is modularin (M, xy> = G. It would follow that [G/M] z [(xy>/ 1] and hence M would be amaximal subgroup of G, a contradiction. Thus r -,P-4 p. Now 5.1.5 shows that everysubgroup of order p permutes with M and MX, and hence with M U MX = S; there-fore it is contained in S since I G : SI = r. Similarly, every element of order r of G iscontained in T and it follows that S and T are the only maximal subgroups of Gcontaining M. By (40), MXY n S and MXY n T are proper subgroups of MXy and hencethere exists z e MXy such that z 0 S and z 0 T. Since o(z) is a power of q, again (b)implies that M mod(M, z> = G, but L((z>) z [G/M] is not a chain. This contradic-tion shows that Case I cannot occur.

Case 2: IMI = q.By (37), I MX I = pq and we first want to show that M° is a P-group of order p"q forsome n c N. For this let M = M0, M....., Mk be the conjugates of M in G. If R < Gof prime order r > q, then by 5.1.5, R permutes with every Mi; hence I Mi U R I = rqand R a Mi u R. Thus

(41) MG < NG(R) for every R < G of prime order r > q.

For i = 1, ... , k, furthermore, M is modular in M u M, and hence, by 5.1.4,1 M u Miff = qri for r; E P. Since MUM. contains two different subgroups of order q,ri >- q. Suppose, for a contradiction, that ri = q for some i. Then, by (41), H =X (M U Mi) has order pq2 and X M is the only subgroup of order pq of H containingM. Since M is not normal in XM, there exists a Sylow q-subgroup Q 0 M U Mi ofH and in Q an element z not contained in any of the two proper subgroups Q n X Mand Q n (M u Mi) of Q. Then M is modular in (M, z> XM, hence I(M, z>I = q2and M a <M,z> u Mi = H, a contradiction. Thus ri > q for all i. Now suppose, fora contradiction, that ri 0 p for some i; let ri = r. If R is the subgroup of order r inM u Mi and again H = X(M u Mi), then H = XRM has order prq and, by (41), Xand R are normal in H. Therefore XM and RM are the only maximal subgroups ofH containing M and it follows that M = NN(M). Thus H contains IH : MI = prconjugates of M and all of these have to lie in one of the two maximal subgroups ofH containing M. But these two groups only contain p + r - 1 subgroups of order q,a contradiction. So we have shown that I M u Mi l = pq for all i = 1, ... , k. If Pi is thesubgroup of order p in M u Mi, then, by (41), Pi a M° and hence P = Pt u u Pk,as a product of normal subgroups of order p of M°, is an elementary abelian normalp-subgroup of M°. Furthermore

M°=MUMIU.UMk=M(P1u...uPk)=MP

and, again by (41), every subgroup of P is normal in MG. It follows that

(42) MG is a P-group of order p"q for some n c N.

We want to show next that MG is a Hall subgroup of G. For this let t be a prime andg c G\MG be a t-element. Every conjugate M; of M is modular in <M;,g> = W. IfW/(M;)µ, were a P-group, we would have g c- W = Mi U Mg < MG, a contradiction.Therefore, by 5.1.4, M. is permutable in W. Now suppose that t 0 q. Then 5.1.5implies that M; a W so that in this case, g normalizes every subgroup of order q ofMG. If Y is a p-subgroup of MG, then YM is a P-group; hence it is normalized by gand it follows that Y9 = Y. Thus g induces a power automorphism in MG which, by1.4.3, has to be trivial since Z(MG) = 1. We have shown:

(43) If g e G\MG is of prime power order t'", then M<g> = (g>M and, fort q,

g E CG(MG).

Now suppose, for a contradiction, that MG is not a Hall subgroup of G. Then thereexists a subgroup H of G such that MG < H and IH : MGI = t where t = p or t = q.By (38), H = G and hence I G : MGI = t. Let T be a Sylow t-subgroup of G andg E T\MG. Then (43) shows that g E CG(MG) for t = p, and that <M,g> = M<g>is abelian for t = q since it is a q-group and IGI = p"qz in that case. Thus in bothcases, g e CG(M) and, since T is generated by the elements in T \M', it follows thatT < CG(M). Since T was an arbitrary Sylow t-subgroup, this implies that M < Z(MG)if t = p, and that M < 0Q(G) if t = q; both assertions contradict (42).

Thus MG is a normal Hall subgroup of G. By the Schur-Zassenhaus Theoremthere exists a complement K to MG in G and, by (43), K < CG(MG). But thenG = MG x K satisfies (c), a final contradiction.

The Sylow subgroups of M/MG

We want to show next that for a modular subgroup M of a finite group G, the Sylowsubgroups of M/MG are modular in GIMG. This is quite simple for permutablesubgroups; in fact, we prove a more general result for these, which we shall extendin 6.2.16.

5.1.10 Lemma. If M is a nilpotent permutable subgroup of the finite group G, thenevery Sylow subgroup of M is permutable in G.

Proof. Let q be a prime and Q E Sylq(M). By (9), we have to show that Q permuteswith every (cyclic) subgroup X of G of prime power order p". Let H = MX. If p = q,we consider a Sylow p-subgroup P of H containing X. Since I M: M n PI = IMP: PIis prime to p, M n P is a Sylow p-subgroup of M, that is, M n P = Q. By (11), Qis permutable in P and therefore QX = XQ. Now let p:1-1 q. Then IH: MI =J X: X n MI = p'° for some m e N and hence IM: M n WI _ I MME : MI is a powerof p for every y e H. It follows that Q < MH; since it is a Sylow subgroup of thenilpotent normal subgroup MH of H, we conclude Q a H. Thus QX = XQ in thiscase too.

The corresponding statement for modular subgroups is not true in general (seeExercise 3); however it holds if M is a Hall subgroup of G.

5.1.11 Lemma. If M is a nilpotent modular subgroup of the finite group G and(I G : MI,1 MI) = 1, then every Sylow subgroup of M is modular in G.

Proof. Let G be a minimal counterexample to the lemma, let M be a nilpotentmodular subgroup of G such that (I G : MI, I MI) = 1, and suppose that Q is a Sylowq-subgroup of M which is not modular in G. By 5.1.9 there exists x E G of primepower order such that Q is not modular in <Q, x> and the minimality of G impliesthat (M,x> = G. By 5.1.10, M is not permutable in G and then 5.1.4 shows that Mis a maximal subgroup of G and G/MG is nonabelian of order pt where p and t areprimes such that p > t, say. It follows that I G : MI = p is prime to I MI and thereexists Y < G such that I YI = p and G = MY.

Since Q is not modular in G but L(G/MG) is modular, MG 1; let N be a minimalnormal subgroup of G contained in MG. Then M/N is a nilpotent modular Hallsubgroup of G/N and the minimality of G implies that QN/N mod GIN. By (4),QN mod G. As a Sylow subgroup of the nilpotent normal subgroup MG of G,Q n MG 9 G. So if Q n MG 1, we could choose N < Q n MG and would get thecontradiction Q = QN mod G. Therefore Q n MG = 1 and hence t = q and I QI = q.

We want to show next that there exists a subgroup P of order p in G such that Qis not modular in Q u P. Since x 0 MG and I G : MGI = pq, x is either a p-element or aq-element. As p2 X IGI, we may take P = <x> if x is a p-element. And if x is a q-element, there exists a Sylow q-subgroup of G containing x. Since Q E Sylq(G)and G = MY, we can find a e M, y e Y such that x e Q°y = Qy, and then<Q, x> < <Q, Qy> < <Q, y). Thus Q is not modular in <Q, y> and we may takeP = <y> in this case.

Now QN mod G and (I QNI, J PI) = 1. By 5.1.5, QN permutes with P and henceIQNP: QNI = p is prime to IQNI. So if QNP < G, the minimality of G wouldimply that Q mod QNP, a contradiction. Thus QNP = G and therefore N = MG andM = Q x N < CG(N) a G. It follows that CG(N) = G and, since N is a minimalnormal subgroup of G, I N I = r is a prime and I G I = pqr. The subgroup NP of indexq in G is normal in G and cyclic since N < Z(G). It follows that P a G and henceIQ u P I = pq. But then, of course, Q mod Q u P, a final contradiction.

5.1.12 Lemma. Let G be a finite group and M < G such that M is modular in <M, x>for every x c- G of prime power order. If Q/MG is a Sylow subgroup of M/MG, then Qis modular in G.

Proof. Let G be a minimal counterexample to the lemma, let M be modular in<M, x> for every x E G of prime power order, and suppose that Q/MG is a Sylowq-subgroup of M/MG such that Q is not modular in G. Then G/MG too is a counter-example since the assumptions on M are also satisfied by M/MG in G/MG, as wehave shown at the beginning of the proof of 5.1.9, and since, by (4), Q/MG is notmodular in G/MG. The minimality of G implies that

(44) MG = 1.

Then by (35),

(45) M is nilpotent

and, by 5.1.9, there exists x e G of prime power order such that Q is not modular in<Q, x). Again 5.1.10 shows that M is not permutable in <M, x> = S and, by 5.1.4,S/MS is nonabelian of order pI M : MSI where p > IM: M51 are primes. Since everySylow subgroup of the nilpotent normal subgroup MS is normal in S, Q $ MS andhence IM : MSS = q. Furthermore we claim that we may assume that o(x) is a powerof p and M permutes with <x>, that is, we have the following situation:

(46) There exists x e G such that o(x) = p", S = <M,x> = M<x>, S/MS is non-abelian of order pq, p > q and Q is not modular in <Q, x>.

Indeed, since IS: MI = p and IM : MSS = q p, there exist cyclic p-subgroups Z of Ssuch that S = M u Z, and 5.1.4 shows that any such Z permutes with M. So if x is ap-element, then S = M<x> and (46) holds. If x is not a p-element, then x is a q-element, since IS/MS1 = pq, and therefore x is contained in a Sylow q-subgroupof S. As S = MZ, there are a e M, z e Z such that x e Q°Z = QZ and hence<Q, x> < <Q, QZ> < <Q, z). Then Q is not modular in <Q, z> and (46) holds with xreplaced by z.

Now if (p, I M 1) = 1, then the assumption of 5.1.11 would be satisfied in S andhence Q mod S, contradicting (46). Thus p divides IMl; let P be the Sylow p-subgroupof M. Now NG(P) # G since MG = 1; let y e G\NG(P) of prime power order and putT = <M, y). If P < MT, then as a Sylow subgroup of the nilpotent normal subgroupMT of T, P a T, contradicting the choice of y. Thus p divides I M : MTI and, by 5.1.4,

(47) M/MT is a p-group.

Suppose, for a contradiction, that H = <M, x, y> < G and let N = MH. Then theminimality of G yields that QN/N is modular in H/N and hence QN mod H, by (4).If QN per H, then Q per H, by 5.1.10; but Q is not even modular in H. There-fore QN/N is not permutable in H/N and, by 5.1.9, H/N = A/N x B/N where(IA/NI, IB/NI) = 1 and A/N = (QN/N)HI" is a nonabelian P-group. By (47), p dividesIM/NI. Since M is nilpotent, QN/N is centralized by PN/N and so 2.2.2 implies thatPN/N < B/N. It follows that IH : BI is prime to p, hence x e B and then (QN)" =QN. Since QN is nilpotent, this implies that Qx = Q, contradicting (46). Thus H = G,that is,

(48) G = <M, x, y> = S u T

Let L be the {p, q}-complement of M, that is, M = P x Q x L. Then L:' MS n MTand, as a characteristic subgroup of Ms :!g S, L a S; similarly, L o T and soL o S u T = G. Since MG = 1, it follows that L = 1, that is,

(49) M = P x Q.

Now let R be any Sylow subgroup of G and z e R. Then by assumption, M ismodular in <M, z> = U and, by 5.1.4, M/MU is a group of prime power order.Therefore either P< MU or Q < MU and, since MU is nilpotent, this again impliesthat P o U or Q a U, that is, z e NR(P) or z e NR(Q). Since R cannot be the set-

theoretic union of two proper subgroups, it follows that R = NR(Q) or R = NR(P).Thus:

(50) If R is a Sylow subgroup of G, then R < NG(P) or R < NG(Q).

By (46), Q and Q' are different Sylow q-subgroups of S. Therefore a Sylow q-sub-group of G containing Qx cannot normalize Q, nor can a Sylow p-subgroup of Gcontaining x. Thus (50) shows that NG(P) must contain these two Sylow subgroupsof G and, since MG = 1, NG(P) G. Hence there must exist a prime r dividing IGIsuch that p r q and a Sylow r-subgroup R of G that is not contained in NG(P).Then we may choose y e R\NG(P). For this y, M is not permutable in T = <M, y)since 5.1.5 (a) would imply that My = M and hence Py = P if M per T. By 5.1.4,T/MT is a P-group and so o(y) = I T : M I = r. Again by 5.1.5, < y> permutes with Mand Ms, hence with S, and (48) yields that IGI = I S I I<y>I = I MI pr. It follows that Sand T are maximal subgroups of G and hence S = NG(P) and T = NG(Q). SinceI G : T j = p, T cannot contain any Sylow p-subgroup of G; by (50), all these Sylowp-subgroups have to lie in S. Since P g S, it follows that P is contained in the inter-section OP(G) of all the Sylow p-subgroups of G, and, since P is a Sylow p-subgroupof T, we have P = OP(G) n T a T. This contradicts the choice of y e G\NG(P).

The main theorems

5.1.13 Theorem (Previato [1975]). The subgroup M of the finite group G is modularin G if and only if M is modular in <M, x> for every x e G of prime power order.

Proof. If M is modular in G, then it is also modular in every group <M, x>. Con-versely, if M is modular in <M, x> for every x E G of prime power order and Q/MGis a Sylow subgroup of M/MG, then, by 5.1.12, Q is modular in G. Since M isgenerated by these subgroups Q, it follows from (5) that M is modular in G.

5.1.14 Theorem (Schmidt [1970a]). The subgroup M of the finite group G is modularin G if and only if

G/MG = Sl/MG x ... X S,/MG x T/MG

where 0 < r e 7L and for all i, j e { 1, ... , r},(a) S;/MG is a nonabelian P-group,(b) (ISI/MGI,IS,/MGI)= 1 =(ISI/MGI,IT/MGI)fori 0j,(c) M/MG = QI/MG x x Q,/MG x (M n T)/MG and Q./MG is a nonnormal

Sylow subgroup of S;/MG, and(d) M n T is permutable in G.

Proof. If G/MG has the given structure, then L(S;/MG) is modular and henceQ./MG mod S;/MG for all i. Furthermore (M n T)/MG mod T/MG so that, by 5.1.8,M/MG mod G/MG; by (4), M is modular in G.

Conversely, we use induction on IGI to show that for a modular subgroup M ofG, G/MG has the structure given in the theorem. Of course, we may assume thatMG = I since otherwise G/MG by induction has the right structure. Then, by 5.1.7,M is nilpotent. If all the Sylow subgroups of M are permutable in G, then by (10), Mis permutable in G and the assertion holds with r = 0 and T = G. If Q is a Sylowsubgroup of M that is not permutable in G, then, by 5.1.12, Q is modular in G and5.1.9 yields that G = QG x L where QG is a nonabelian P-group of order prime toILI. By (2), M n L is modular in L and, clearly, (M n L)L = 1. Hence, by induction,L has the right structure. Since M n QG is nilpotent, 2.2.2 shows that Q = M n QG isa nonnormal Sylow subgroup of QG. Thus G has the right structure.

Exercises

1. Show that the subgroup M of the finite group G is a maximal modular subgroupof G if and only if either M is a maximal normal subgroup of G or M is a maximalsubgroup of G and G/MG is nonabelian of order pq where p and q are primes.

2. (Schmidt [1969]) Let M be modular in the finite group G and suppose thatI G: MI = p" where p e P and n e N. Show that G/MG is either a p-group or aP-group of order p"q where p > q e P.

3. Let G = PQ x R where I PI = I R I = p e P and PQ is nonabelian of order pq,p > q e P. Show that M = QR is a nilpotent modular subgroup of G whoseSylow q-subgroup is not modular in G.

4. (Schmidt [1970a]) Show that if M is a nilpotent modular subgroup of the finitegroup G, q the largest prime dividing I M I and Q the Sylow q-subgroup of M, thenQ is modular in G.

5. (Whitson [1977]) Let G be a finite group and let M < G be such that (X U M) n Z =X u (M n Z) for all subgroups X and Z of G satisfying X < Z, that is, M is semi-modular in L(G) as defined in Exercise 2.1.7. Show that if [G/M] is a chain oflength 2 and I G : MI = p2 for some prime p, then G/MG is a p-group.

5.2 Permutable subgroups of finite groups

Since we know the structure of P-groups very well, Theorem 5.1.14 reduces ourproblem of determining the structure of G/MG for a modular subgroup M of a finitegroup G to the study of the factor T/MG in this theorem, that is, to the case that Mis permutable in G. Our main result in this section is that MG/MG is then hyper-centrally embedded in G, in particular is nilpotent. In the remainder of this sectionwe shall study some special problems on permutable subgroups in order to prepareour investigations on projective images of normal subgroups. We shall not alwaysinsist on the finiteness of the groups considered since we shall be able to applythe more general results also in the study of images of normal subgroups underprojectivities of infinite groups in Chapter 6. For the same reason we introduce the

5.2 Permutable subgroups of finite groups 217

concepts of hypercentral and hypercyclic embedding in the most general form,although, at the moment, we are only interested in projectivities between finitegroups.

Hypercentral and hypercyclic embedding

Let K < H be normal subgroups of the group G. We say that H/K is hypercentrally(respectively hypercyclically) embedded in G if to every normal subgroup N of G suchthat K < N < H there exists a subgroup M of G satisfying N < M < H such that[M, G] < N (respectively such that M 9 G and M/N is cyclic). For groups satisfyingthe maximal condition for normal subgroups, in particular for finite groups, this isequivalent to the existence of a chain of subgroups N of G such that

K=Na<N,

and [Ni, G] < Ni-, (respectively Ni a G and N/N_1 is cyclic) for i = 1, ..., r. Inparticular, every hypercentrally (respectively hypercyclically) embedded factor of afinite group is nilpotent (respectively supersoluble). Since for N < M < G the inclu-sion [M, G] < N is equivalent to N a_ G and M/N < Z(G/N), it is clear that everyhypercentrally embedded factor of an arbitrary group is hypercyclically embedded.We shall need the following simple inheritance properties of the concepts introducedabove.

5.2.1 Lemma. Let F, H, K, L, M, N be normal subgroups of G such that K < M < H,L<HandK<F.

(a) If H/K is hypercentrally (hypercyclically) embedded in G, then so are HN/KNand H n N/K n N.

(b) H/K is hypercentrally (hypercyclically) embedded in G if and only if H/M andM/K are hypercentrally (hypercyclically) embedded in G.

(c) If F/K and H/K are hypercentrally (hypercyclically) embedded in G, then so isFH/K.

(d) If H/K and H/L are hypercentrally (hypercyclically) embedded in G, then so isH/K n L.

Proof. We sketch the proofs of these properties for hypercentral embedding; theproofs for hypercyclic embedding are similar. First of all, (a) follows from the factthat for normal subgroups S, T of G such that [T, G] < S, (1) of § 1.5 yields[TN, G] < [T, G]G[N, G] < SN; furthermore, of course, [T n N, G] < S n N. Toprove (b), observe first that if H/K is hypercentrally embedded in G, then from (a) itfollows that HM/KM = H/M and H n M/K n M = M/K are hypercentrally em-bedded. Conversely, if these two factors are hypercentrally embedded and S a Gsuch that K < S < H, then H/MS and M/M n S are hypercentrally embedded andtherefore, by (a), MS/S is also. Hence there exists T < G satisfying [T, G] < S suchthat S < T < MS < H if S < MS and such that S = MS < T < H in the othercase.

Thus (b) holds. In (c), F/K and FH/FK = FH/F are hypercentrally embedded; in (d),H/K and H n K/ L n K = K/L n K are hypercentrally embedded. In both cases, (b)yields the desired result.

We need a criterion for a normal subgroup of prime power order to be hyper-centrally embedded in a finite group.

5.2.2 Lemma. If p is a prime and N a normal p-subgroup of a finite group G, then Nis hypercentrally embedded in G if and only if G/C°(N) is a p-group.

Proof'. First assume that N is hypercentrally embedded in G and let Q be a q-subgroup of G for some prime q p. Then N and the q-group QN/N are hyper-centrally embedded in QN. Thus QN is hypercentrally embedded in QN, that is QNis nilpotent. It follows that Q < C°(N) and, since q was arbitrary, GIG(N) is ap-group.

Conversely, suppose that G/C°(N) is a p-group and take S g G such that S < N.Then G operates as a p-group on the p-group N/S and therefore has a nontrivialcentralizer T/S in this group. It follows that S < T and [T, G] < S. Thus N is hyper-centrally embedded in G.

The structure and embedding of M°/MG

5.2.3 Theorem (Maier and Schmid [1973]). If M is permutahle in the finite group G,then M'/M° is hypercentrally embedded in G.

Proo/. Let G be a minimal counterexample to the theorem and let M be minimalamong the permutable subgroups of G for which M°/M° is not hypercentrallyembedded in G. Then G/M° is a counterexample and the minimality of G impliesthat

(1) M° = 1.

By 5.1.7 and 5.1.10, M is nilpotent and its Sylow subgroups are permutable in G.Hence if M had composite order, the minimality of M would imply that pG wouldbe hypercentrally embedded in G for every Sylow subgroup P of M. By (c) of 5.2.1,the product of these normal closures, that is M°, would also be hypercentrallyembedded in G, a contradiction. Thus M is a p-group for some prime p and thereforealso

(2) M° is a p-group

since the product of p-groups is a p-group. Now M° is not hypercentrally em-bedded in G and therefore, by 5.2.2, there exists a prime q # p dividing I G/C°(M°)I.Let Q E Sylq(G) and put H = N°(Q)M. The Frattini argument shows that the Sylownormalizer N°(Q) cannot be contained in a proper normal subgroup of G. ThusO"(G)H = G and every x e G can be written in the form x = yh where y E O°(G).h E H. Since O°(G) is generated by p'-elements, 5.1.5 shows that M is normalized by

0°(G). Therefore MX = Mh and hence M" = MG and MH = MG = 1. So if H were aproper subgroup of G, the minimality of G would imply that .MG = M"/MH wouldbe hypercentrally embedded in H. But then 5.2.2 would imply that Q < CH(MG) andthis would contradict the fact that q divides 1G/CG(MG)I. Thus H = G, that is,

(3) G = NG(Q)M.

This yields QG = Q'N < QM and hence QG n MG < QM n MG = (Q n MG)M = Msince Q is a q-group and MG a p-group. Since MG = 1, it follows that QG n MG = 1and we get that Q < CG(MG), the same contradiction as before.

Recall that if K < H are normal subgroups of a group G, then G operates byconjugation on H/K and the kernel of this operation is

CG(H/K) = {x e GI[H,x] < K}.

Thus G/CG(H,'K) is isomorphic to the automorphism group induced by G in H; K;furthermore, CG(H/K)/K = CG/K(H/K).

5.2.4 Corollary. If M is permutable in the finite group G, then G/CG(MG/MG) isnilpotent and the princes dividing the order of this group are precisely those dividingIM/MGI.

Proof. We may assume that MG = 1 so that MG is hypercentrally embedded in G.Then MG is nilpotent and a prime divides IMGI if and only if it divides IMI. If p issuch a prime and P the Sylow p-subgroup of MG, then P < G and (b) of 5.2.1 showsthat P is hypercentrally embedded in G. By 5.2.2, G/CG(P) is a p-group which is nottrivial since otherwise P n M would be a nontrivial normal subgroup of G containedin M. Since CG(MG) is the intersection of all the CG(P), the assertion follows.

Our main results, Theorems 5.1.14 and 5.2.3, yield a rather good but somewhatcomplicated description of the structure and embedding of MG/MG for a modularsubgroup M of a finite group G. Therefore we prove a handier result that followsimmediately from these theorems and that will often suffice for the applications.

5.2.5 Theorem (Schmidt [1975a]). If M is modular in the finite group G, thenMG/MG is hypercyclically embedded in G and G/CG(MG/M(;) is supersoluble.

Proof. Again we may assume that MG = 1. Then, by 5.1.14, G = Sc x x S, x Twhere the S; are nonabelian P-groups, M n T is permutable in G, and MG =Sc x . x S, x (M n T)G. Clearly, all the Si are hypercyclically embedded in G andso is (M n T)G, by 5.2.3; thus (c) of 5.2.1 shows that M' is hypercyclically embeddedin G. Furthermore G/CG((M n T)G) is nilpotent and G/CG(S;) S. is supersolublefor every i = 1, ..., r. Since CG(MG) is the intersection of all these centralizers, itfollows that G/CG(MG) is supersoluble.

5.2.6 Corollary. if M is modular in the finite group G and G or M is perfect, thenM a G.

Proof. Recall that a group is perfect if it has no proper normal subgroup withsoluble factor group. Therefore, if G is perfect, Theorem 5.2.5 implies that G =CG(Mo/MG); and if M is perfect, it follows that M = MG. In both cases, M:9 G.

The results established so far will suffice to prove a number of important criteriafor projectivities of finite groups to map normal subgroups onto normal subgroups;Corollary 5.2.6 is a first example. But they say nothing if the groups considered arenilpotent groups or p-groups; rather they reduce the problems to the investigationof p-groups. As mentioned before, we shall be able to prove similar reduction theo-rems also for infinite groups. Therefore, in the remainder of this section, we shallstudy permutable subgroups of p-groups and cyclic permutable subgroups of arbi-trary groups as far as necessary for the applications to projective images of normalsubgroups.

Elements of infinite order

First of all we show that elements of infinite order cause little trouble in the prob-lems we consider. The reason for this is the following basic result which we shallgeneralize in 6.2.2.

5.2.7 Lemma. If M per G and g e G such that o(g, M) is infinite, then g e NG(M).

Proof. We may assume that M<g> = G. Since o(g, M) = I<g>: <g> n MI, it followsthat o(g) is infinite and M n <g> = 1. If p is a prime, S = M<g"> is a subgroup ofindex p in G; let N = SG. Then IG/NI divides p! and hence MN/N:5 S/N is a p'-subgroup of GIN. Since MN/N is permutable in GIN, we see that (MN/N)GI" =MGN/N is a p'-group and hence MGN/N < S/N. Thus MG < S. This shows that MGis contained in M<g"> for every prime p and, since, by 2.1.5, the intersection of thesesubgroups is M, it follows that M = MG a G.

Permutable subgroups of finite p-groups

The structure of M/MG may be rather complicated even if M is a permutable sub-group of a finite p-group. That M/MG need not be abelian is shown by Exercise 1.Papers by Bradway, Gross and Scott [1971], Gross [1971], [1975b], Stonehewer[1973], [1974], and Berger and Gross [1982] contain examples in which the nil-potency class, derived length and Frattini length of M/MG are arbitrarily large.Although all these structural complications, by 5.1.6, have to occur in groups of theform G = M<x>, these groups at least have a decent power structure.

5.2.8 Theorem (Gross [1971]). Let M be a core free permutable subgroup of thefinite p-group G = MX and suppose that X is cyclic of order p". Then the followingproperties hold.

(a) Every X-invariant subgroup of M is trivial.(b) M n X = 1, S1(X) < Z(G) < X.(c) S2(G) is elementary abelian.(d) For r = 1, ..., n, S2,(G) = 0,(M)S2r(X) has exponent p' and is soluble of derived

length at most r; furthermore, the core of MS2,(G)/0,(G) in G/S2r(G) is trivial.(e) ExpG=p">ExpM.(f) For p = 2, SI2(G) is abelian and hence S1r(G)/S1r_2(G) is abelian for r >_ 2.(g) If p = 2 and M 1, then M < S2n_2(G).

Proof. If n = 1, M a G and hence M = 1. Thus G = X is of order p and all theassertions hold trivially. So assume that n > 2. If H is an X-invariant subgroup ofM, then H G = HXM = HM < M and, since MG = 1, it follows that H = 1. In particu-lar, M n X = 1 and CG(X) = X(CG(X) n M) = X implies that Z(G) < X. ThereforeCI(X) < Z(G) since Z(G) 1. Thus (a) and (b) hold.

For g e G with o(g) = p, I M< g> : MI is contained in theatom MSI(X) of the chain [G/M] L(X). It follows that KI(G):!5; M11(X) and soS2(G) = (M n S)(G))S1(X). Therefore M n 1(G) is of index p in S1(G) and contains(D(S)(G)), a normal subgroup of G. Since MG = 1, it follows that D(Q(G)) = 1. ThusS1(G) is elementary abelian and (c) holds. This implies that S1(M) = M n O(G) andhence 1(G) = S1(M)SI(X). If K/1(G) is the core of M1(G)/S1(G) in G/S1(G), thenIK: M n KI = p since M is a maximal subgroup of MS2(G) = MS1(X). So this timec(K) is a normal subgroup of G contained in M and is therefore trivial. Thus Kis elementary abelian and hence K = 1(G). This shows that (d) holds for r = 1

Figure 18

and the general assertion follows by induction on r. For, if N = S2(G), then MN/Nis a core-free permutable subgroup of GIN, as we have just shown, and G/N =(MN/N)(XN/N); since S2(G) is elementary abelian, 1,_,(G/N)=52,(G)1N,S2,-,(MN/N) = S2,(M)N/N and S2,_,(XN/N) = Sl,(X)N/N. Therefore by induction,Q, -I (G/N) = S2,(G)/N is of exponent p'-' and soluble of derived length at most r -1,and hence S2r(G) is of exponent p' and soluble of derived length at most r. Further-more Q,(G) = 1r(M)N1,(X) = f1,(M)S2r(X) and the core of (MN/N)f2,_,(G/N)/S2,_, (G/N) in (G/N)/S2,_1 (G/N) is trivial. Thus the core of M1,(G)/f2r(G) in G/S2r(G)is also trivial and (d) holds.

Since MS2"_, (X) < M1"_, (G) is a maximal subgroup of G, we have MS2"_, (G) 9 G;on the other hand, by (d), is core-free. It follows thatM12,_1(G)/S2"_1(G) = I and hence M < S2"_,(G). Again by (d), ExpM < pn-' andG = MX = c2 (M)1"(X) = Q,(G) has exponent p". Thus (e) holds.

To prove (f) and (g), suppose that p = 2 and first note that I12(G) :12(M)I _A(X)I = 4; let 12(M) < T < 112(G). Then 112(M) g T !!g 12(G) and S22(M) has atmost two conjugates in 12(G), both contained in T. So if 12(M) were not normal in12(G) and L is the core of S22(M) in S22(G), it would follow that 112(G)/LI = 8 and,by (11) of § 5.1, S22(M)/L = (M n S22(G))/L would be a nonnormal permutable sub-group of order 2 in f'2(G)/L. But such a group of order 8 does not exist. This showsthat S22(M) a S22(G) and f12(G)/S22(M) is abelian. Therefore S22(G)' is a normal sub-group of G contained in M and hence is trivial. Thus 112(G) is abelian. By (d),G/1,_2(G) satisfies the assumptions of the theorem and hence 12(G/f1,-2(G)) =1.(G)/S2,_2(G) is also abelian. In particular, G/S2n_2(G) is abelian and, by (d),M1,_2(G)/S2"_2(G) is core-free in this group. It follows that M < S2"_2(G). Thus (f)and (g) are satisfied.

There is a vast literature on permutable subgroups of finite groups, especiallyfinite p-groups. We refer the reader to the papers mentioned above and in additionto Ore [1937], Ito and Szep [1962], Deskins [1963], and Nakamura [1966], [1970],[1979]. However, most of the results in these papers are not relevant to our pur-poses. For example, Gross [1971], [1975b] exhibits bounds for the nilpotency classand derived length of M/MG for a permutable subgroup M of a finite p-group G. Butall these bounds depend on the exponent of G, tend to infinity with Exp G, and arebest possible as is shown by the examples mentioned above. In contrast to this weshall prove in 6.6.1 that M/MG is metabelian if M is a projective image of a normalsubgroup. For this, the bounds of Gross are clearly not usable. In addition, thisdifferent behaviour of permutable subgroups and of projective images of normalsubgroups shows that the former do not sufficiently approximate the latter. It isclear that images of normal subgroups under projectivities between p-groups (called"images" for short) have a number of additional properties, for example:

(4) If M is an image, there exist images M; such that 1 = Mo < . . . < Mk =and IMi: Mi_,J = pfor i = 1, ..., r.

(5) If M is an image and T is a maximal subgroup of G, then M n T is an image.

(6) If M is an image and A < M is invariant under P(M), for example A = (D(M)or 1(M), then A is an image.

It would be desirable to use these or similar simple properties to define a class ofsubgroups in finite p-groups that lies between the projective images of normal sub-groups and the permutable subgroups, and to prove that M/MG is metabelian forany member M of this class. This would suffice to prove the same result for imagesof normal subgroups under projectivities between arbitrary groups as we shall see inChapter 6. However, such a theory does not exist.

Cyclic permutable subgroups

We want to show that MG/MG is hypercentrally embedded in G if M is a finite cyclicpermutable subgroup of an arbitrary group G. If I M I is a prime, we can say a littlemore.

5.2.9 Theorem. Let M be a permutable subgroup of prime order p in the group G andsuppose that M is not normal in G.

(a) Then MG is an elementary abelian p-group.(b) If x c G has infinite order or finite order prime to p, then x c CG(MG).(c) MG < N(G) < Z2(G).(d) MG = M X (MG n Z(G)).(e) MG n Z(G) n <x> : 1 for every x e G\CG(M).

Proof. (a) Since every conjugate of M is permutable in G, the subgroup <Mx, M''>has order p or p' and is therefore abelian for all x, y c G. It follows that MG =<Mxix E G> is abelian and hence elementary abelian as IMI = p.

(b) By 5.1.5 or 5.2.7, x normalizes every conjugate of M. Since M is not normal inG, there exists y e G\NG(M) such that o(y) = p' for some m e N. Put H = <M, x, y>and T = M(y). Since MT is normalized by x, we have MT = MH. FurthermoreM < Q (T) a T and IS2(T)J = p2. Since M is not normal in T, it follows that MH =MT = Q (T) is elementary abelian of order p2 and I MH n <y>l= p. Let N=MH n <y>. Then y e CH(N) and, since M" has only p + I subgroups of order p, N isthe only subgroup of order p of MH that is normalized by y. The other subgroups oforder p of MH are conjugates of M, and are therefore normalized by x; it follows thatNF = N. Since x e NG(M) and y NG(M), clearly xy 0 NG(M) and hence o(xy) is finite.Let <z> be the Sylow p-subgroup of (xy>. Then MZ M and hence MH = Mlm,z>and I MH n <z>I = p as for y. Therefore if L = MH n <z> = MH n <xy>, then<xy> < CH(L) and so y e NH(L). Since N is the only subgroup of order p of M"normalized by y, L = N. Thus N is centralized by y and xy and hence also by x. Butx normalizes every subgroup of M" and, by 1.5.4, induces a universal power auto-morphism in this group. It follows that x e CG(MH) < CG(M). Now x was arbitraryand hence also x9-' E CG(M), that is, x e CG(Mg) for all g c G. Thus x e CG(MG), asdesired.

(c) If X is a cyclic p-subgroup of G and g c G such that M9 $ X, then I M9X I _p 1 X I and hence Mg < NG(X). Thus M and its conjugates normalize every cyclicp-subgroup and centralize every element of infinite order and of finite order prime top. It follows that MG is contained in the norm N(G) of G and, by 1.5.3, N(G) < Z2(G).

(d) By (c), Z(G) < MZ(G) < Z2(G) and hence MZ(G) a G; it follows thatMG < MZ(G). Then Dedekind's law shows that MG = M(MG n Z(G)) and, sinceM Z(G), the product is direct.

(e) Let X = <x>. Since MG normalizes every subgroup of G, 10 [M, X] < MG n X.Therefore if MG n Z(G) n X = 1, it would follow that

M< MG = (MG n Z(G))(MG n X) < CG(X),

a contradiction. Thus MG n Z(G) n X 0 1, as desired.

To prove the result announced above we have to show first that every subgroupof a cyclic permutable subgroup is permutable in G. For this we need the following.

5.2.10 Lemma. Let K < H be permutable subgroups of the p-group G. If [H/K] is afinite chain and K < M < H, then M is permutable in G.

Proof. We use induction on the length of the chain [H/K] and have to show thatX M = MX for every cyclic subgroup X of G. This is clear if M = H, so assume thatM < H. If M< KX, then XM = XKM = KXM = KX = MKX = MX. So sup-pose that M KX and hence KX < HX. Since [H/K] is a finite chain, there existsg E G such that H = K(g>, and it follows that IH : KI = I(g>: (g) n KI andI HX : HI = J X: X n HI are finite. Then I HX : KI is also finite, thus HX/KHX is afinite p-group and hence there exists a normal subgroup T of index p in HX contain-ing KX. Then KS H n T< H and J H: H rn TI = I HT : TI = I HX : TI = p. Since[H/K] is a chain, it follows that M < H n T and (11) of § 5.1 yields that H n T andK are permutable subgroups of T By induction, M is permutable in T and henceXM=MX.

It is not difficult to show that the above lemma holds for arbitrary groups G; fora still more general result see Exercise 6.

5.2.11 Lemma. If M is a cyclic permutable subgroup of the group G, then everysubgroup of M is permutable in G.

Proof. Let 1 H S M and assume first that G is finite. Let p be a prime dividingI HI, P E Sylp(H) and Sc Sylp(M). Then P < S and S is permutable in G by 5.1.10.Therefore, if X is a subgroup of prime power order q" of G and p q, then, by 5.1.5,S a SX and hence P a SX; also if p = q, then SX is a p-group and P per SX by5.2.10. In both cases, PX = XP. Thus P is permutable in G and, since this holds forevery prime dividing IHI, it follows that H is permutable in G.

Now let G be arbitrary and take x e G. We have to show that (x>H = H(x>. Ifx e NG(M), then also x c NG(H) and we are done; so let x 0 NG(M) and considerL = M(x). By 5.2.7, I L : MI = o(x, M) is finite, hence IL: HI is also finite and so,finally, L/HL is a finite group. Then H/HL permutes with (x>HL/HL, as we have justshown, and it follows that (x>H = <x>HLH = H(x>HL = HHL(x> = H(x>.

5.2.12 Theorem (Busetto [1980b]). Let M be a finite cyclic permutable subgroup ofthe group G and suppose that MG = 1; let I M I = pi' ... p; be the prime factor decom-position of I M I and put n = max{ri;Ii = 1,...,r}. Then MG < Zzn(G).

Proof. By 5.2.11, the Sylow subgroups of M are permutable in G and, clearly, MG isthe product of their normal closures. Therefore we may assume that M is a p-group.Then I M I = p" and we show by induction on n that MG is a p-group of exponent p"and is contained in Zzn(G). If n = 1, this is true by 5.2.9. So assume that n > 2 andthe assertion holds for n - l; let N = S2n_1(M)G . Again by 5.2.11, S2"_1(M) is per-mutable in G and the induction assumption yields that N < Z2n_2(G) and Exp N =p"-'. Since M, = 1 and [M/1] is a chain, there exists x e G such that M n M" = 1.By 5.2.7 and 5.1.5, M is normalized by elements of infinite order or finite order primeto p; therefore we may choose x as a p-element. Then H = M<x> is a finite p-groupand M. = 1. By 5.2.8, On_1(H) = Qn_1(M)S2n_1(<x>) has order at most p2n-z; on theother hand, S2n_1(M)S2n_1(Mx) is contained in f2,-,(H) and has this order sinceM n Mx = 1. Thus n,,-,(H) = Qn_1(M)(2n_1(Mx) < N n H and, since Exp N = pn 'we obtain S2n_1(H) = N n H. Now (d) of 5.2.8 shows that MN n H = M(N n H) =MS2n_1(H) is not normal in H and hence MN/N is a nonnormal permutable sub-group of order p in GIN. By 5.2.9, (MN/N)GI" = MG IN is elementary abelianand MG IN < Z2(G/N). It follows that Exp G = p" and, as N < Z2 _2(G), thatMG < Zzn(G)

Metacyclic p-groups

A group G is called metacyclic if it has a normal subgroup N such that N and G/Nare cyclic. For p-groups a seemingly weaker condition suffices.

5.2.13 Theorem. The p-group G is metacyclic if and only if it is the product of twocyclic subgroups one of which is permutable in G.

Proof. If G is metacyclic, it is the product of a cyclic normal subgroup and a cyclicsupplement. So assume that, conversely, G satisfies the condition of the theorem.Then, as a product of two cyclic p-groups, G is finite and we use induction on I G I toshow that G is metacyclic. First of all we note that the assumption on G is equivalentto the following condition.

(7) There exists a modular subgroup M of G such that [G/M] and [M/1] arechains.

For, since G is a p-group, a cyclic permutable subgroup M with cyclic supplementclearly satisfies (7); and if (7) holds and g e G is not contained in the maximal sub-group of G containing M, then G = M<g>, M is permutable in G and M is cyclic.Now (1), (2) and (3) of §5.1 immediately show that property (7) is inherited bysubgroups and factor groups; for, if (7) holds and H < G or N 9 G, then M n H andMN/N satisfy (7) in H and GIN, respectively. Therefore, by induction,

(8) proper subgroups and factor groups of G are metacyclic.

We show next that we may assume that IG'I = p. If G' = 1, then G is clearly meta-cyclic. So assume that G' 1 and let N be a normal subgroup of order p containedin G'. By (8), GIN is metacyclic; let K/N a GIN such that KIN and G/K are cyclicand let t e K such that K = N<t). If K is cyclic, we are done; so assume that K isnot cyclic. Then N (t) and hence K = N x <t> since N < Z(G). If L = N(tP) =NO(K) is the maximal subgroup of K containing N, then K/L < Z(G/L) and G/K iscyclic. It follows that GIL is abelian and hence N < G' < L = N t (K). Thus

(9) G' = N x (G' n (D(K)).

Suppose, for a contradiction, that G' n s(K) : 1. Then the minimal subgroup N, ofthis cyclic normal subgroup is normal in G and G'/N, is cyclic as in the case of N. Itfollows that IG' n b(K)I = p and hence G' < Z(G). It is well-known (see Huppert[1967], p. 258) that for a group G generated by two elements, G'/[G', G] is cyclic;in our case, [G', G] = I and G' = N x Ni is not cyclic, a contradiction. ThusG' n D(K) = I and by (9),

(10) IG'I = p.

Since G is generated by two elements, GIG' = (aG') x <bG') for suitable a, b e G.Let o(aG') = p" and o(bG') = p'°. If aP" : 1, it generates G' and <a> > G' is a cyclicnormal subgroup of G with cyclic factor group. Thus G is metacyclic and we aredone. So assume that aP" = 1 and, similarly, b"" = 1. We want to show that thenn = 1 = m. Suppose that in fact n > 1. Of course, [a, b] = c is a generator of G'and, since G' < Z(G), we have [aP, b] = [a, b]P = 1 = [a, bP]. Thus H = (aP, b, c) isabelian and metacyclic by (8). Hence If)(H)I < p2 and since <aP"-'> and (c) aredifferent minimal subgroups of H, it follows that b""' ' e SI(H) = <aP"-', c). In partic-ular, b""' ' e <a>G'; but b has order p'° modulo <a)G'. This contradiction shows thatn = 1. In exactly the same way we get that m = 1 and hence I G I = p'. Since G is theproduct of two cyclic groups, one of these must have order at least p2 and is then acyclic normal subgroup with cyclic factor group. Thus G is metacyclic.

5.2.14 Corollary (Napolitani and Zacher [1983]). The p-group G is a metacyclicM-group if and only if it is the product of two cyclic subgroups which are bothpermutable subgroups of G.

Proof. If G is a metacyclic M-group, every subgroup is permutable in G and henceG is a product of two cyclic permutable subgroups. Conversely, let G = AB where Aand B are cyclic permutable subgroups of G. By 5.2.13, G is metacyclic; let N g Gsuch that N = (x) and GIN are cyclic. Since G = AB and GIN is cyclic, G = NA orG = NB; say that G = NB and let B = . Then, by 2.3.4, G is an M-group exceptwhen p = 2, INS > 4 and N/<x4) is not centralized by b. In this case, T = <x4, b2) g G,G/T would be dihedral of order 8 and BT/T would be a nonnormal permutablesubgroup of order 2 in G/T. But such a subgroup does not exist in a dihedral group.Thus G is an M-group.

We can regard Theorem 5.2.13 as a lattice-theoretic characterization of the classof metacyclic p-groups within the class of all finite p-groups; this restriction, of

course, is needed since the metacyclic, noncyclic group of order p2 is lattice-isomorphic to the nonabelian group of order pq for any prime q dividing p - 1. Thelattice-theoretic property (7) appearing in the proof of Theorem 5.2.13, that thereexists a modular subgroup M of G such that [G/M] and [M/1] are chains, charac-terizes in the class of finite groups those groups G which possess a normal subgroupN such that N and GIN are cyclic of prime power order (see Exercise 7). Finally, weremark that the class of finite metacyclic groups is not invariant under projectivities(see Exercise 8).

Exercises

1. (Thompson [1967], Nakamura [1968]) Let p > 2 and A = <a0,...,aP> be anelementary abelian group of order pP+l(a) Show that there exists an extension H = A<x> of A by a cyclic group (x>

satisfying x"2 = ao, x-'aox = ao and x-la;x = a;a;_, for i = 1, ..., p.(b) Show that there exists an extension G = H(y> of H by a cyclic group <y>

satisfying yP = al', y-'apy = a,ai 1, y-'a;y = ai for i = 0, ..., p - 1 andy-1xy = x1+p.

(c) Show that M = <y, a2, ..., ap> is a core-free nonabelian permutable subgroupof G.

2. (Nakamura [1979]) If G = MX is a finite p-group, M permutable in G and Xcyclic, show that there exists a permutable subgroup H of order p of G such thatH < Z(M).

3. (Gross [1971]) If G = MX is a finite p-group, M a core-free permutable subgroupof G and X cyclic, show that the nilpotency class of 02(G) is at most p - 1(compare with 5.2.8 (f)).

4. (Gross [1971]) Suppose that G = MX is a finite p-group, M a core-free per-mutable subgroup of G and X cyclic of order p", n > 2.(a) Show that IS2(G)l < p'° where m = pn-2(p - 1).(b) Show that the nilpotency class of G is at most pn-1 - 1.

5. Prove properties (4)-(6).6. (Busetto [1980b]) Let H and K be permutable subgroups of the group G and

suppose that there exists g c G such that H = K(g). Show that every subgroupM of G such that K < M < H is permutable in G.

7. Show that a finite group G contains a normal subgroup N such that N and GINare cyclic of prime power order if and only if L(G) contains a modular element Msuch that [G/M] and [M/1] are chains.

8. Let p and q be primes such that 2 < q I p - 1, and let G = NH and G = NK whereINS = p, H is the abelian and K the nonabelian group of order q3 and exponentq2, CH(N) = S2(H) and CK(N) = S2(K). Show that(a) G and G are lattice-isomorphic,(b) G is metacyclic, but 6 is not.


5.3 Lattice-theoretic characterizations of classes of finite groups

A rather natural way of obtaining a lattice-theoretic characterization of a class ofgroups is to replace concepts appearing in the definition by lattice-theoretic con-cepts that are equivalent to them or nearly so. For example, we should try to replace"normal subgroup" by "modular subgroup", "cyclic factor group" by "distributiveinterval in the subgroup lattice", "abelian factor group" by "modular interval in thesubgroup lattice", and so on.

We shall show that this method indeed yields lattice-theoretic characterizationsfor the classes of simple, perfect, soluble and supersoluble finite groups; and we shalluse this idea for many other lattice-theoretic characterizations. For the first threeclasses we only need Lemma 5.1.2 on maximal modular subgroups; in the case ofsupersoluble groups we shall use 5.2.5, but we remark that it is possible to give amore elementary proof using 5.1.4 and 5.2.3 instead (see Exercise 1). We remarkfurther that in § 6.4 we shall extend nearly all these results to infinite groups. Finally,we shall present Iwasawa's characterization of finite supersoluble groups and discusscertain classes of groups that are related to nilpotent groups, the most interestingbeing the class of groups with lower semimodular subgroup lattice.

Simple groups

5.3.1 Theorem. The finite group G is simple if and only if I and G are the onlymodular elements in L(G).

Proof. If 1 and G are the only modular elements in L(G), then G in particular has noproper nontrivial normal subgroup, that is, G is simple. Conversely, let G be simpleand suppose, for a contradiction, that there exists a modular element in L(G) differ-ent from I and G. Then there also exists a nontrivial maximal modular subgroup Min G and, by 5.1.2, M a G or G/Mc is nonabelian of order pq with primes p and q. Inboth cases, G is not simple, a contradiction.

Since projectivities map modular subgroups onto modular subgroups, Theorem5.3.1 has the following immediate consequence.

5.3.2 Corollary (Suzuki [1951a]). If G is a finite simple group and cp is a projectivityfrom G to a group G, then G is simple.

Perfect groups

Recall that a group is perfect if it coincides with its commutator subgroup. Forfinite groups, an equivalent condition is that no maximal subgroup is normal.To this definition we can apply the method described at the beginning of thissection.

5.3 Lattice-theoretic characterizations of classes of finite groups 229

5.3.3 Theorem. The finite group G is perfect if and only if no maximal subgroup of Gis modular in G.

Proof. If no maximal subgroup of G is modular in G, then, clearly, no maximalsubgroup of G is normal, that is, G is perfect. Conversely, let G be perfect andsuppose, for a contradiction, that there exists a maximal subgroup M of G that ismodular in G. Then again by 5.1.2, M a G or G/Mc is nonabelian of order pq withprimes p and q. In both cases there exists a normal subgroup of prime index in G,and this contradicts the assumption that G is perfect.

Using 5.2.6, we again obtain as a corollary an older result of Suzuki's.

5.3.4 Theorem (Suzuki [1951a]). If G is a finite perfect group and pp a projectivityfrom G to a group G, then G is perfect, cp induces an isomorphism from the lattice ofnormal subgroups of G to that of G, and Z (G)" = Z.(6) for all n e N.

Proof. By 5.3.3, G is perfect. If N a G, then N'' is a modular subgroup of G andhence N" a G by 5.2.6. Similarly, preimages of normal subgroups of G are normalin G so that qp induces an isomorphism from 9t(G) to 9t(G). Finally, let P e Syl,,(Z(G))for some prime p. Then [P, G] = 1 and, since O"(G) = G, 1.6.8 yields [Pp, G] = 1,that is, P" < Z(G). It follows that Z(G)4' < Z(G) and this statement for cp-1 yields theother inclusion. Thus Z(G)" = Z(G) and, since G/Z(G) is perfect, a trivial inductionyields that ZZ(G)4' = Z,(6) for all n e N.

Soluble groups

There are many ways of defining solubility and some of them are suitable for appli-cation of the method described at the beginning of this section. We present threelattice-theoretic characterizations of the class of finite soluble groups obtained thisway.

5.3.5 Theorem (Schmidt [1968]). The following properties of the finite group G areequivalent.

(a) G is soluble.(b) There are subgroups G; of G such that I = Go < < Gr = G, G. is modular in

G and [G;+1 /G] is modular (i = 0, ... , r - 1).(c) There are subgroups G. of G such that 1 = Go < < GS = G, G is modular in

Gi+1 and [Gi+1 /G] is modular (i = 0, ... , s - 1).(d) There are subgroups G of G such that 1 = Go < ... < G, = G, G. is maximal

and modular in G1+1 (i = 0,..., t - 1).

Proof. That (a) implies (b) is clear since, by 2.1.3 and 2.1.4, the members G; of a chiefseries of the soluble group G have the desired properties. Trivially, (b) implies (c) and(4) of § 5.1 shows that if (c) holds, then every subgroup between G and Gi+1 ismodular in Gi+1, so that every maximal chain in L(G) containing all the G. has the

properties required in (d). Finally, we prove by induction on I G I that (d) implies (a).The induction assumption yields that M = Gt_1 is soluble and, by 5.1.2, either M a Gand G/M is cyclic of prime order or IG/MGI = pq with primes p and q. In bothcases, MG and G/MG are soluble and hence G is soluble.

The above theorem yields a second proof of the fact that the class of finite solublegroups is invariant under projectivities; in 4.3.4 we used Suzuki's results on singularprojectivities to prove this. Another lattice-theoretic characterization of the class offinite soluble groups was given by Suzuki [1956]. He proved that the finite group Gis soluble if and only if there exists a maximal chain 1 = Go 5 < G, = G ofsubgroups Gi invariant under P(G) such that [Gi+1/Gi] is modular for all i =0, ,r- 1.

Recall that in every finite group G there exist the soluble residual Ga and thesoluble radical G a. The residual is defined as the intersection of all the normalsubgroups with soluble factor group and is therefore the smallest normal subgroupof G with soluble factor group; the radical is the product of all the soluble normalsubgroups and hence is the largest soluble normal subgroup of G. Let R = Ga. ThenR' is a normal subgroup of G with soluble factor group and hence R = R', that is, Ris perfect. On the other hand, if H is any subgroup of G, then R n H a H andH/R n H HR/R is soluble. It follows that every perfect subgroup of G is containedin R. Thus

(1) Ga is the largest perfect subgroup of G.

Now let T = G `. Then if M is a maximal soluble subgroup of G, TM/T ^ M/M n Tis soluble, hence TM is soluble and the maximality of M implies that TM = M.Thus T is contained in every maximal soluble subgroup of G and, since the intersec-tion of all these clearly is a soluble normal subgroup of G,

(2) G a is the intersection of all the maximal soluble subgroups of G.

By 5.3.3 and 5.3.5, we can detect perfect and soluble subgroups in the subgrouplattice of G. Thus (1) and (2) yield lattice-theoretic characterizations of the solubleresidual and the soluble radical in finite groups. We only note the usual corollary.

5.3.6 Theorem (Suzuki [1951a]). If q is a projectivity from the finite group G to agroup G, then (Ga)4' = G: and (G°)`° = G`.

Supersoluble groups

The finite supersoluble groups can be characterized like the soluble groups.

5.3.7 Theorem (Schmidt [1968]). The following properties of the finite group G areequivalent.

(a) G is supersoluble.(b) There are subgroups Gi of G such that 1 = Go < < G, = G, G. is modular in

G and [Gi+1 /G] is distributive (i = 0, ... , r - 1).

(c) There are subgroups G. of G such that 1 = Go < < G., = G, G; is modular inG and maximal in Gi+1 (i = 0, ... , s - 1).

Proof. That (a) implies (c) is clear since the members Gi of a chief series of thesupersoluble group G have the desired properties. And (c) implies (b) since everyinterval [Gi+1/G;] in (c) is a chain and hence is distributive. We prove by inductionon I G I that (b) implies (a). The property (b) is inherited by factor groups GIN; forNG;/N is modular in GIN and (1) of § 5.1 shows that

[(NGi+i/N)/(NG1/N)] ^' [NG,G1+i/NGi] ^-' [Gi+i/G;+i n NGi]

is an interval in [G;+,/Gi] and is therefore distributive. We may assume that1 = Go < G, = M. Then M is modular in G and L(M) ^- [Gl/Go] is distributive. By1.2.4, M is cyclic; in particular, MG is cyclic. By 5.2.5, MG/MG is hypercyclicallyembedded in G and, since MG 1, the induction assumption yields that G/MG issupersoluble. Thus MG, MG/MG and G/MG all are hypercyclically embedded in G,and (b) of 5.2.1 shows that G is supersoluble.

5.3.8 Corollary. If G is a finite supersoluble group and cp is a projectivity from G toa group G, then G is supersoluble.

The first lattice-theoretic characterization of the class of finite supersoluble groupswas given by Iwasawa. If G is supersoluble, it is easy to prove by induction on I GIthat every maximal subgroup M of G has prime index in G. This is clear if MG 0 1;for then M/MG is a maximal subgroup of the supersoluble group G/MG of smallerorder than IGI. And if MG = I and N is a minimal normal subgroup of G, thenMN = G and IG: MI = INS is a prime. Since every subgroup of G is supersoluble itfollows that for every pair of subgroups K < H of G, the length of a maximal chainin L(G) from K to H is the number of prime divisors of I H : K 1, including multiplici-ties. Thus L(G) satisfies the Jordan-Dedekind chain condition and this is Iwasawa'scharacterization.

5.3.9 Theorem (Iwasawa [1941]). A finite group is supersoluble if and only if itssubgroup lattice satisfies the Jordan-Dedekind chain condition.

Proof. It remains to be shown that a finite group G satisfying the Jordan-Dedekindchain condition is supersoluble. We use induction on IGI. Then every proper sub-group of G is supersoluble and, by a theorem of Huppert's (see Robinson [1982], p.287), G is soluble. Therefore a composition series of G is a maximal chain in L(G) andits length is the number d of prime divisors of I G1. If M is a maximal subgroup of G,the Jordan-Dedekind chain condition implies that L(M) has length d - 1 and, sinceM is supersoluble, M I has d - I prime divisors. It follows that I G: MI is a primeand, by another well-known theorem of Huppert's (see Robinson [1982], p. 268), Gis supersoluble.

We remark that Iwasawa's theorem (and the above proof) by now can be foundin many textbooks on group theory; see, for example, Robinson [1982], p. 288 or

Huppert [1967], p. 719. Huppert's theorem, used to show that G is soluble, dependson Gran's Second Theorem or a similar transfer argument. The reader can find adirect proof of Iwasawa's theorem, using only Gran's Second Theorem, in Suzuki[ 1956], p. 9.

Finally we mention that Berkovic [1966] and Fort [1978] study finite groups inwhich for every 1 : H < G the interval [G/H] satisfies the Jordan-Dedekind chaincondition. Both authors determine the structure of soluble groups with this property(see Exercise 4); using the classification of minimal simple groups, Fort, in addition,shows that any group with this property is soluble.

Groups with semimodular subgroup lattice

For the class of finite nilpotent groups, our method of obtaining lattice-theoreticcharacterizations does not work: the P-groups show that this class is not invariantunder projectivities. Nevertheless it is interesting to examine the properties thatemerge if we apply the usual process to a given characterization of nilpotency. Themost suitable one for our purpose is Wielandt's well-known theorem which assertsthat a finite group is nilpotent if and only if every maximal subgroup is normal.Using this, we get the following class of groups.

5.3.10 Theorem (Schmidt [1970b]). Every maximal subgroup of the finite group G ismodular in G if and only if G is supersoluble and induces an automorphism group of atmost prime order in every complemented chief factor of G.

Proof. Suppose first that every maximal subgroup M of G is modular in G. Then itfollows from 5.1.2 that IG : MI is a prime and, by Huppert's theorem (see Robinson[1982], p. 268), G is supersoluble. Let H/K be a complemented chief factor of G; wehave to show that IG: C,(H/K)I is a prime or 1. Since H/K is complemented, thereexists M<Gsuch that G=HM and HnM=K.Then JG:MI =JH:KI=pisaprime and hence M is a maximal subgroup of G. By 5.1.2, M:9 G or I G : MGI = pqfor some prime q p. In both cases, HMG < CG(H/K) and thus IG : Cc(H/K)I is qor 1.

Conversely, suppose that G satisfies the conditions of the theorem and take amaximal subgroup M of G; we have to show that M mod G. If M a G, we are done;so assume that M is not normal in G. Then G/MG is a primitive soluble group andtherefore (see Robinson [1982], p. 192) contains a minimal normal subgroup N/MGsuch that N = CG(N/MG). Since G is supersoluble, I N/MGI = p is a prime and, sinceN $ M, M/Mo is a complement to N/MG in G/MG. Thus N/Mo is a complementedchief factor of G and our assumption yields that I G: NJ = I G : CG(N/MG)I = q is aprime. Thus IG/MGI = pq and hence G/MG has modular subgroup lattice. By (4) of§ 5.1, M is modular in G.

A condition that appears to be slightly stronger than Wielandt's is that everymaximal subgroup of every subgroup H of G is normal in H. This clearly is also

equivalent to nilpotency and, if we translate this, we get the condition that everymaximal subgroup of every subgroup H of G is modular in H. By 2.1.5, a maximalsubgroup M of H is modular in H if and only if for every K:< H such that M u K = H,M n K is maximal in K. This shows that our condition is equivalent to the follow-ing: if K, M < G and M is maximal in M u K, then M n K is maximal in K. This isequivalent to L(G) being lower semimodular as defined in § 2.1. These groups werecharacterized by Ito; they are still closer to nilpotent groups than those in Theorem5.3.10.

5.3.11 Theorem (Ito [1951]). The subgroup lattice of the finite group G is lowersemimodular if and only if G is supersoluble and induces an automorphism group of atmost prime order in every chief factor of G.

Proof. Suppose first that L(G) is lower semimodular. Then by 5.3.10 or Iwasawa'stheorem, G is supersoluble. Let H/K be a chief factor of G; we have to show thatIG : CG(H/K)l is a prime or 1 and may assume that K = 1. Then JHI = p is a prime.Since G is soluble, there exists a Hall p'-subgroup Q in G. Then every maximalsubgroup of HQ is modular in HQ and H is a complemented chief factor in HQ. By5.3.10, GHQ: CQ(H)I q or divides

G is a of P.It follows that IG : CG(H)I divides q, as desired.

Conversely, suppose that G is supersoluble and IG : CG(H/K)I is a prime or I forevery chief factor H/K of G. Let S < G and take a chief series 1 = Go < < G, = Gof G. Then for every i, S n Gi a S and IS n G; : S n G;_, I divides I G; : G;_, I e P. Bythe Jordan-Holder Theorem, every chief factor of S is isomorphic to one of theS n G;/S n G;_, . Now CG(GI/Gi_1) n S centralizes S n G;/S n Gi_1 and has index in Sequal to 1 or a prime. Thus S satisfies the conditions of the theorem, in particularthose of Theorem 5.3.10. It follows that every maximal subgroup of S is modular inS and L(G) is lower semimodular.

It is easy to see that the two classes of groups in Theorems 5.3.10 and 5.3.11 aredifferent (see Exercise 5). An immediate consequence of 5.3.11 is the following resultwhich is a little surprising.

5.3.12 Corollary. If H and K are finite groups with lower semimodular subgrouplattice, then so is their direct product H x K.

Finite groups with upper semimodular subgroup lattice are less important sincethere is no related class of groups which is significant from the group-theoretic pointof view. The structure of these groups was determined by Sato, who also studiedinfinite groups with this property; see Suzuki [1956], pp. 24-26.

We shall give lattice-theoretic characterizations of certain classes of infinite nil-potent groups in §§7.2-7.4. However, not much is known about infinite groupswhich satisfy the Jordan-Dedekind chain condition or have lower semimodularsubgroup lattice.

Exercises

1. Use 5.1.4 and 5.2.3 instead of 5.2.5 to prove Theorem 5.3.7. (Hint: In a minimalcounterexample to the implication (b) (a) there is exactly one minimal normalsubgroup N, and GIN is supersoluble; furthermore, (G1)G = 1 for G1 > Go. If Mis maximal with respect to the properties G1 < M mod G and MG = 1, show thefollowing:(a) M < NM < D(M) = n {M u M"jx E G\NG(M)} and [D(M)/M] is a chain;

let F be the atom of this chain.(b) If M a F, then M per G; by 5.2.3, INI is a prime.(c) If M is not normal in F, then NM = D(M) = MG is a nonabelian group of

order pq (p, q c P) and hence again IN I is a prime.)2. Let us call a lattice L polymodular if it contains a least element 0, a greatest

element I and elements a, such that 0 = ao < < a, = I, ai is modular in L and[ai/ai_1 ] is modular (i = 1, ... , r). If G is a finite group, show that(a) G` is the largest modular subgroup of G with polymodular subgroup lattice

and(b) Gs is the smallest modular subgroup M of G with polymodular factor inter-

val [G/M].3. Let us say that the element a of the lattice L with least element 0 is polymodular

in L if there exist a1 E L such that 0 = ao < < a, = a, ai is modular in L andai_ 1 is maximal in ai (i = 1, ... , r). For a finite group G, let T(G) be the largesthypercyclically embedded normal subgroup of G, and G= the supersoluble resid-ual, the smallest normal subgroup of G with supersoluble factor group; by 5.2.1,these subgroups exist. Show that(a) T(G) is the largest polymodular element of L(G) and(b) GI is the smallest modular subgroup M of G for which G is polymodular in

[G/M].It follows that T(G)" = T(G) and (GI)`° = GI for any projectivity q from G to agroup G.

4. (Berkovic [1966], Fort [1978]) Let G be a finite soluble group. Show that [G/H]satisfies the Jordan-Dedekind chain condition for every nontrivial subgroup H ofG if and only if G is either supersoluble or is a semidirect product of an elemen-tary abelian group N by a cyclic group K of odd order such that every nontrivialsubgroup of K operates irreducibly on N.

5. Let p, q, r be primes such that q r and qr1 p - I and let 0 < s, t be the nonabelian group of order p3 and exponent p and define a E Aut P by(cib.akk)° = cisbitakst (0 < i, j, k < p). Show that every maximal subgroup of thesemidirect product G of P and <a> is modular in G, but that L(G) is not lowersemimodular.

6. (Venzke [1972]) Show that every subgroup of a finite group G is maximal sensi-tive in G (for the definition see § 3.3, Further Topics) if and only if G is isomorphicto a subgroup of a direct product S1 x x S, of P-groups Si E P(ni, pi) such thatpi0p;fori0j.

5.4 Projective images of normal subgroups of finite groups 235

7. (Venzke [1975]) If M r N is maximal in M (and in N) for any two differentmaximal subgroups M and N of the finite group G, show that every maximalsubgroup of G is modular in G. (Hint: In a minimal counterexample G choose amaximal subgroup M of minimal order which is not modular in G. If G is notsoluble, use the Feit-Thompson Theorem to get an involution t e G such thatG = <M, t>; show that (M r M`)<t> is a maximal subgroup of G and hence ismodular in G.)

5.4 Projective images of normal subgroups of finite groups

In the next two sections and in §§ 6.5 and 6.6 we shall study images of normalsubgroups N of a group G under projectivities cp from G to groups G. We want toshow that N° is near to being normal in G. The main result of this section, Theorem5.4.7, will reduce this problem for finite groups to the case that N° is permutable inG. Together with the results of § 5.2 this will yield a number of useful criteria for N°to be normal in G. In the course of our analysis we shall very often have to considerthe core or the normal closure of N`° and the preimages of these subgroups. There-fore it is worthwhile introducing the following general notation for arbitrary groups.

5.4.1 Notation. If p is a projectivity from G to G, H < G and x E G, we put H = H"and denote the preimages of the core and the normal closure of H`° in G by Hz; andHG, respectively, that is,

Hz; = ((HI)G-)'-' = (Hz;)'P-` and Ho = ((Hc)G)p 1 _ (HG)p

furthermore, as in Chapter 4, we shall write H'"`°-' = ((HP)")v

Minimal normal subgroups

It seems reasonable to start our investigations with minimal normal subgroups. Weshall show that these are mapped to minimal normal subgroups if they are not cyclicand we begin with the cyclic case.

5.4.2 Lemma. Let p be a prime, G a finite p-group and cp a projectivity from G to agroup G. _

(a) Then there exists a normal subgroup Z of order p in G such that Z`1 a G.(b) If N is a minimal normal subgroup of G, then NG < Q(Z(G)).

Proof. If G is not a p-group, then, by 2.2.6, G and G are cyclic or P-groups and bothassertions of the lemma hold trivially. So we may assume that G is a p-group.

(a) Suppose, for a contradiction, that there is no minimal normal subgroup Z ofG_such that 2:9 G. Let N be a minimal normal subgroup of G and M < G such thatM is a minimal normal subgroup of G. Then N and M are nonnormal permutable

subgroups of order p in Gand G, respectively. Let x c G\CG(M). By (e) of 5.2.9,Z(G) n <x) 0 1. If also Z(G) n <x)" 1, then Z = S2(<x)) < Z(G) and 2< Z(G);butt we assume that such a normal subgroup does not exist. ThereforeZ(G) n <x)4' = 1, and (e) of 5.2.9 now shows that <x)`° < CU(N). Since G is generatedby the elements x E G\CG(M), G is generated by these <x)4' and it follows thatN < Z(G). This contradiction proves (a).

(b) Let G be a minimal counterexample and N a minimal normal subgroup of Gsuch that H = NG is not contained in S (Z(G)). Then H N and hence again Nis anonnormal permutable subgroup of order p in G. By 5.2.9, H = N x (H n Z(G)) iselementary abelian. Since N < Z(G) and H Z(G), there exists a subgroup M oforder p in (H n Z(G))"-' that is not contained in Z(G). Therefore M is a nonnormalpermutable subgroup of order p in G. Finally, by (a) there exists a normal subgroupZ of order p in G such that Z:9 G. Then M Z N and the minimality of Gimplies that the preimage LIZ of the normal closure of (NZ)"/Z in G/Z with respectto the projectivity induced by qp in G/Z lies in the centre of G/Z. Since N < L 9 G,we have H <_ L and hence Z < MZ < L. Thus MZ a G and, since M is not normalin G, it follows that MZ = MG and MG n Z(G) = Z. For x e G\CG(M), again by (e)of 5.2.9, 1 0 MG n Z(G) n <x) = Z n <x> and hence Z < <x>. Then Z 0 N < Z(G)implies that N<x) = N x <x> and f (N<x)) = NZ. Since cp is a projectivity,

(NZ)° = (f)(N<x>))° = c2((N<x>)') 9 (N<x))',

that is, <x)`° normalizes (NZ)P. Now G is generated by these x e G\CG(M) and itfollows that (NZ)O a G. But then H = NZ < S2(Z(G)), a contradiction.

We need the following simple result on projective images of hypercentrally em-bedded normal p-subgroups.

5.4.3 Lemma. Let cp be a projectivity from the finite group G to a group G andsuppose that H < G such that H° is a normal p-subgroup of G which is hypercentrallyembedded in G but not contained in the centre of G.

(a) If P`° e Sylp(G), then P is a p-subgroup of G containing H and G = P U CG(H).(b) If H 9 G, then H is hypercentrally embedded in G.

Proof. Clearly, G = POP(G) and H< P since H is a normal p-subgroup of G. By5.2.2, [H, OP(G)] = 1. Since H $ Z(G), P is not abelian and hence P is a p-groupby 2.2.6. Furthermore, 1.6.8 yields that [H,OP(G)'P-'] = I and it follows that G =P U CG(H). Finally, if H 9 G, then H < P is a normal p-subgroup of G andG/CG(H) ^- P/P n CG(H) is a p-group. By 5.2.2, H is hypercentrally embeddedin G.

5.4.4 Theorem (Schmidt [1975a]). Let N be a minimal normal subgroup of the finitegroup G and suppose that cp is aprojectivity from G to a group G such that N' is nota minimal normal subgroup of G. Then N is cyclic of prime order p and for S = NG,one of the following two assertions holds:

(a) S and S`° are P-groups in P(n, p) for some n c- N and there exists T < G such thatG = S x T,G=S`° x T`°and(ISI,ITI)= I =(IS"I,IT`°I);

(b) S and S' are elementary abelian p-groups, furthermore S < Z(G) andS`° = N'° x (S`° n Z(G)) < Z2(G).

Proof. Suppose, for a contradiction, that there exists a subgroup M of G such thatI < M < N and M :!g G. Then M is modular in G and, since N is a minimal normalsubgroup, MG = N and MG = 1. By 5.2.5, N is hypercyclically embedded in G andhence INI is a prime, contradicting 1 < M < N. Thus there is no normal subgroupof G lying properly between l and N. In particular, if N:9 G, then N would be aminimal normal subgroup of G, contradicting our assumption. It follows that N isnot normal in G and NG = 1.

By 5.1.7, N is nilpotent and hence N is a soluble minimal normal subgroup of G.Thus N is an elementary abelian p-group and, by 2.2.6, its nilpotent projective imageN is a q-group for primes p and q. By 5.1.9 (or 5.1.14), either N is permutable in G orS is a P-group and G = S x T where (ISI, I TI) = 1. In the latter case, I NI = q, henceN is cyclic of prime order p and, by 1.6.6, G = S x T where (I S I, I TI) = 1. By 2.2.5, Sand S lie in the same class P(n, r) and, since N is a normal subgroup of order p in S,r = p. Thus (a) holds. _

Now suppose that N is permutable in G. As a product of q-groups, S is a normalq-subgroup of G and is hypercentrally embedded in G by 5.2.3. Then 5.2.2 shows thatG/CA(S) is a q-group and hence G= PCG(S) for some P E Sylq(G). FurthermoreS Z(G) since N is not normal in G. By 5.4.3, P is a q-subgroup of G containing S,hence p = q as N < S, and G = P u CG(S). Therefore 1 : Z(P) n N < Z(G) and,since N is a minimal normal subgroup of_G, it follows that N is cyclic of order p.Now G = CU(S)P implies that S = NG = N" and 5.4.2, applied to the projectivityinduced in P, yields that S < S2(Z(P)). Since G = P U CG(S), it follows that S is anelementary abelian p-subgroup of Z(G). Finally, N is a nonnormal permutable sub-group of order p in G_and, by 5.2.9, S is an elementary abelian p-group and S =N X (S n Z(6))< Z2(G). Thus (b) holds.

Our theorem in particular asserts that every projectivity between finite groupsmaps noncyclic minimal normal subgroups onto minimal normal subgroups. Weshow that the centralizers of these minimal normal subgroups are also mapped ontoeach other.

5.4.5 Theorem. Let N be a minimal normal subgroup of the finite group G and let ppbe a projectivity from G to a group G. If N is cyclic, of order p, say, assume furtherthat N4 a G and that p and qp` are regular at p. Then CG(N)4' = CG(N4').

Proof. By 5.4.4, N is a minimal normal subgroup of G if N is not cyclic. Therefore itsuffices to show that CG(N)' < ;(N); application of this result to cp-' and N willyield the other inclusion.

First assume that N is not abelian, let q be a prime and X a q-subgroup of CG(N).Since N is a nonabelian minimal normal subgroup, N = OQ(N) and 1.6.8 shows thatX < CU(N). Since CG(N) is generated by its subgroups of prime power order, itfollows that CG(N)4' < CG(N).

Now suppose that N is abelian, hence INS = p" for some prime p. Again by 1.6.8,[N, O°(CG(N))'] = 1 and we have to show that P < CiT(N) for some P E Sy1P(CG(N)).By the Frattini argument, G = CG(N)NG(P). This shows that N is a minimal normalsubgroup of H = NG(P) and, by 5.4.4 or assumption, N is a minimal normal sub-group of H. We claim that cp and q p` are regular at p. This is clear by assumption ifN is cyclic; and if N is not cyclic, it follows from the fact, see 4.2.6, that a group witha p-singular projectivity has only cyclic p-chief factors. Therefore P" < OP(H)° =OP(H`°) and hence k a OP(H). Then N n Z(OP(H)) is a nontrivial normal subgroupof H and, since N is a minimal normal subgroup of H, it follows that N < Z(OP(H)).In particular, P < OP(H) centralizes N, as desired.

The structure of G/Ne; and 6/N6

We come to our main theorem on images of normal subgroups under projectivitiesof finite groups. Our results on modular and permutable subgroups yield the struc-ture of G/&. To use this efficiently in the study of G, we need to know that thepreimage Nj of Rd is a normal subgroup of G so that cp induces a projectivity fromG/Nj; to G/NZ;. In fact we prove a little more than this.

5.4.6 Lemma. Let N be a normal subgroup of the finite group G, cp a projectivityfrom G to a group G, H = NG and K = that is, H`° the normal closure and K' thecore of N4' in G.

(a) Then H and K are normal subgroups of G.(b) If N`° is permutable in G, then H/K and H4'/K4' are hypercentrally embedded in

G and G, respectively.

Proof. (a) We use induction on I G I and show first that H a G. For this letM be a minimal normal subgroup of G contained in N and put S = MG. ThenS < H and, by 5.4.4, S 1 G. Therefore (p induces a projectivity i-p from G/S toG/S and, for NS/S a G/S, H/S is the normal closure of (NS/S)4' in G/S. By induction,H_aG.

Now we show that K a G. This is clear if K = 1. So let K 0 1 and take M < Gsuch that M is a minimal normal subgroup of contained in K. Then if S is thenormal closure of M in G, S < N since M < N a G. By 5.4.4, 9:9 G and, as k is thecore of N in G, it follows that S < K. Again cp induces a projectivity from G/S to G/Sand by induction, K a G.

(b) By (a), cp induces a projectivity from G/K to G/K and therefore we may as-sume that K = 1. Then by 5.2.3, H is hypercentrally embedded in G. In particular,H is nilpotent and hence H H' x x H4 with Sylow pi-subgroups HT of H.Every H4' is characteristic in H and hence normal in G; thus H? is a hypercentrallyembedded normal pi-subgroup of G that is not contained in Z(G) since Hi" n N : 1.By 1.6.6, H = H1 x x H, where (IH,I, IH,I) = 1 for i zA j; hence Hi is characteristicin H a G and thus Hi a G. By (b) of 5.4.3, every Hi is hypercentrally embedded in G,and then (c) of 5.2.1 shows that H = H1 ... H, is hypercentrally embedded in G.

5.4.7 Theorem (Schmidt [1975a]). Let N be a normal subgroup of the finite group G,(p a projectivity from G to a group G, H = Nc and K = Ni;, that is, H° the normalclosure and K° the core of N0 in G. Then H and K are normal subgroups of G andthere are primes p;, q; and ni c- N such that

G/K = Sl/K x x S,/K x T/K and G/K'' = ST/K" x x S°/K0 x TO/K0

where 0 < r e Z and for all i, j e { 1, ... , r},(a) S?/K° is a P-group of order p"iqi, pi > qi and (IS;r/K0I, IS7/K0I) = 1 =

(ISr/K0I, I Tw/KIOI) for i j,(b) Si/K E P(n, + 1, pi) and (ISi/KI, IS;/KI) = 1 = (IS/K I, I T/KI ).for i 0 j,(c) N/K = (N n S, )/K x - x (N n S,)/K x (N n T)/K, I (N n Si)/K I = pi,

I(N n Si)'Q/K0I = qi and (N n T)0 is permutable in G,(d) H/K = Sl/K x . . . x S,/K x (H n T)/K, (H n T)/K and (H n T)0/K° are hyper-

centrally embedded in G and G, respectively.

Proof. We have just shown that H and K are normal subgroups of G. Therefore toprove that G/K and G/K have the required structure, we may assume that K = 1. IfN a G, then all assertions hold trivially for H = N = K and r = 0. So assume thatN is not normal in G. Then by 5.1.14, G = S1* x . . x S' x T where (a) is satisfied,IN n S°I = qi and N n T is permutable in G. By 1.6.6 and 2.2.5, G = S1 x x S, x Tand (b) holds. Furthermore IN n SiI = p; since N n S; g Si; thus (c) is also satisfied.Since H = S1* x x S' x (H n T), H has the structure required in (d). Finally,N n T g G and (N n T)I* is permutable in G; therefore (b) of 5.4.6 yields that H n Tand (H n T)0 are hypercentrally embedded in G and G, respectively.

Since all the Si/K, Sr/K0, (H n T)/K and (H n T)0/K0 in Theorem 5.4.7 are hyper-cyclically embedded in G or G, (c) of 5.2.1 shows that the same is true for H/K andH",/K0. Thus we get the following result.

5.4.8 Corollary. Under the assumptions of Theorem 5.4.7, H/K and H0/K0 are hyper-cyclically embedded in G and G, respectively.

Furthermore, 5.4.7 and 5.4.8 yield some useful criteria for a projective image of anormal subgroup to be normal in the image group. We collect these in the followingtheorem. Parts (a) and (b) of it are due to Schmidt who improved similar criteriawith "cyclic" replaced by "abelian"; these, as part (c) of the theorem, had beenproved by Suzuki using his results on singular projectivities.

5.4.9 Theorem (Suzuki [1951a], Schmidt [1972b]). Let N be a normal subgroup ofthe finite group G and suppose that one of the following is satisfied:

(a) G/N has no nontrivial cyclic normal subgroup.(b) There is no normal subgroup M of G such that M < N and N/M is cyclic.(c) GIN is perfect.

Then N0 i 6 for every projectivity cp from G to a group G.

Proof. By 5.4.8, N' /NG- and hence also N°/N and N/NG- are hypercyclically em-bedded in G. Therefore (a) implies that N' = N and (b) yields NZ; = N; in both cases,N a G. Finally, if (c) is satisfied and R = G = is the soluble residual of G, thenG = NR since GIN is perfect. Furthermore, by 5.3.6, R is the soluble residual of G.By 5.2.5, G/C;(NG/Nj;) is supersoluble and henceis contained in this centralizer.In particular, N is normalized by R and, since G = N u k, it follows that N a G.

Iterated nilpotent residuals

As a first application of our results we prove the theorem on the iterated nilpotentresiduals announced in § 4.3. Let us write G for the smallest normal subgroup withnilpotent factor group in the finite group G and define inductively

No(G) = G and Nk(G) = Nk_t(G) n for k c N.

Then Nk(G) is the smallest normal subgroup of G with factor group soluble ofnilpotent length at most k.

5.4.10 Theorem. If (p is a projectivity from the finite group G to the group G, thenNk(G)4' = Nk(G) for all k > 2.

Proof. Let N = Nk(G) and K = N. By 5.4.7, G/K = S1/K x . . . x S,/K x T/Kwhere the S;/K are P-groups and N n T/K is hypercentrally embedded in G. NowT/N n T NT/N is soluble of nilpotent length at most k. If FIN n T is the Fittingsubgroup of T/N n T, then FIN n T is nilpotent and N n T/K is hypercentrallyembedded in F; by 5.2.1, F/K is nilpotent. It follows that T/K is soluble of nilpotentlength at most k. Since k > 2, the same holds for all the S;/K and hence also for G/K.The definition of N = Nk(G) implies that N < K and therefore N = K. This showsthat N a G and cp induces a projectivity from GIN to_G/N. By 4.3.3, GIN is solubleof nilpotent length at most k. It follows that Nk(G) < N = Nk(G)", and this result forcp - ' yields the other inclusion.

Projective images of subnormal subgroups

In the remainder of this section we show that subnormal subgroups of finite groupsare mapped by projectivities onto subnormal subgroups, with the obvious exceptionof P-groups. Since it is immediate that index preserving projectivities behave wellwith respect to subnormal subgroups, we use the results of Chapter 4 on singularprojectivities and not the methods developed in this section. In this way we alsoobtain another approach to our results on projective images of normal subgroups.

5.4.11 Lemma. Let cp be an index preserving projectivity from the finite group G tothe group G.

(a) 1J 'N a G, then N4' is permutable in G.(b) If N < < G, then N4 !9!9 G.

Proof. (a) For every H < G, I HuN : N I = H: H n N I. Since cp is index preserving,it follows that I H u N : NI = I H: H n NI and, by (7) of § 5. 1, HN = NH. Thus N ispermutable in G. (Of course, the assertion also follows from Theorem 5.4.7.)

(b) We use induction on IG : NI. For N = G there is nothing to prove, so supposethat N i4 G. Then there exists a normal subgroup M of G such that N < M < G. By(a), M is permutable in G and hence, by 5.1.1, M << G. (In fact, Exercise 4.3.5 showsthat M < G if M is a maximal normal subgroup of G.) Since N < < M, the inductionassumption yields that N < < M and it follows that N < < G.

5.4.12 Theorem (Schmidt [1975a]). Let N be a subnormal subgroup of the finitegroup G, cp a projectivity from G to a group G and K the largest normal subgroup ofG contained in N such that KIP < G. Then there exist primes pi, qi and ni e 1 such that

G/K = S, /K x x S,/K x T/K and G/KIO = S; /K4' x . x Sr/K4' x T4'/K4'

where 0 < r e Z and f o r all i, j e { 1, ... , r},(a) S;°/K`° is a P-group of order p"-qi, pi > qi and (ISr/K4I,IS`°/K4I) = 1 =

(ISi°/K4' I, I T4'/K4' I) for i j,(b) Si/K e P(ni + 1,p,) and (ISi/KI,IS1/KI) = I = (ISi/KI, IT/KI)for i 0 j,(c) N/K = (N n S1)/K x x (N n S,)/K x (N n T)/K, I(N n Si)/KI = pi,

I(N n Si)`°/K4I = qi, and(d) (NnT)4g_<G.

Proof. Let G be a minimal counterexample to the theorem. Then G/K is also acounterexample and hence

(1) K = 1.

If k < < G, then all assertions would hold trivially for r = 0 and T = G. Thus

(2) N is not subnormal in G.

If G=A x B where (IA I, IBI) = 1 and A0 1 B, then N = (N n A) x (N n B),N n A a< A and N n B< a B. The minimality of G would imply that A and Bwould have the right structure and so would G, a contradiction. Hence if (A, B) werea P-decomposition of G, then B = 1 and G would be a P-group in P(n, p), say. As asubnormal subgroup of this P-group, N would have order pk for some k e N and,since N is not subnormal in G, I NI = pk-'q for p > q e P. But K = 1 would implyk = 1, that is, I N I = p and I N I = q, and the assertions of the theorem would hold forSl = G and T = 1. This would contradict the choice of G. Thus

(3) G is not P-decomposable.

Since N is not subnormal in G, 5.4.11 shows that cp is not index preserving. Let rt bethe set of primes p for which cp is singular at p. Since G is not P-decomposable, (b)of Theorem 4.2.6 holds for all these primes. In particular, for every p e 7C,

(4) G has a normal p-complement Cp such that Cp a G.

Let D = n Cp. By 4.2.1, cp induces an index preserving projectivity from D to D.

Hence if N < D, then by 5.4.11, N !9:9 D; therefore N a a G since b a G. Thiswould contradict (2); thus N D and

(5) there exists a prime p e it dividing IN I.

Let C Cp. Then the Sylow p-subgroups of G are just the complements to C in G.Since C a G, the inner automorphisms of G permute the images of the Sylow p-subgroups of G. Therefore if V = <PIP E Sylp(G)> and W = <S2(P)IP E Sylp(G)>,then

(6) V-gG,W<G,V< and WgG.

If P E Sylp(G), then N n P E Sylp(N) since N a a G. By (5), N n P 96 1. Therefore ifthe Sylow p-subgroups of G were cyclic, it would follow that n(P) < N for everyP E Sylp(G) and hence W< N. But then (6) would contradict (1). Thus the Sylowp-subgroups of G are not cyclic and therefore are elementary abelian since cp issingular at p. Let P, E Sylp(G) such that IPri : IP1 I and let Po be the Sylow p-subgroup of the P-group Pr. By 4.2.10, Po Z(G) and every subgroup of P, isnormal in G; again (1) implies that N n PO = 1. But if P is an arbitrary Sylowp-subgroup of G, Po < P of index p and N n P : 1. It follows that P = Po(N n P)PoN and this shows that

(7) V < Po N.

Now PoN n C is the normal p-complement in PoN. Since Po < Z(G), N is a normalsubgroup of PoN and therefore contains this normal p-complement; in particular,V n C < N. By (4) and (6), V n C a G and (V n C)`0 = V10 n C° a G. Thus by (1),V n C = 1 and VC = G since P, < V and P1C = G. It follows that G = V x C andV = P, is a P-group, contradicting (3).

5.4.13 Remark. In 5.4.12, if N is not only subnormal but normal in G, thenN n T a G; hence (N n T)`0 is subnormal and modular in G, that is, by 5.1.1,(N n T)" per G. Thus Theorem 5.4.12, like 5.1.14 for modular subgroups, reduces thestudy of projective images of normal subgroups of finite groups to the case of per-mutable subgroups. In particular, via 5.4.12 we can obtain a second proof of themain theorem of this section which does not use Theorem 5.1.14 on modular sub-groups. For the first basic step in our discussion was Theorem 5.4.4 on minimalnormal subgroups. In the proof of this we need, apart from 5.4.12, just 5.1.1, 5.2.9and the Maier-Schmid Theorem 5.2.3. The proof of the latter theorem uses 5.1.5(a)and 5.1.10, which are independent of the results on modular subgroups, and 5.1.7.However, it is easy to show directly that M/MG is nilpotent whenever M is a per-mutable subgroup of G. The proof of 5.4.6 remains unchanged (it uses 5.4.4, 5.4.3 and5.2.3) and 5.4.6 (a) yields that for N a G, the subgroup K defined in 5.4.12 is just N.Then 5.4.7 follows from 5.4.12 and 5.4.6 (b).

If r > 0 in 5.4.12, that is, if there is a P-group S, /K, then neither G nor N is perfect;therefore we get the following corollary.

5.5 Normal subgroups of p-groups with cyclic factor group 243

5.4.14 Corollary. Let (p be a projectivity from the finite group G to the group G andlet N :!9:9 G. If G or N is perfect, then Nm :!9:!9 G.

Incidentally the first assertion in 5.4.14 also follows immediately from 5.4.11 andthe fact that projectivities of perfect groups are index preserving; the second asser-tion follows by induction from 5.4.9(b) (observe that if N is a perfect subgroup of G,then NG is perfect).

Exercises

In Exercises 1, 3-5, and 8, cp is a projectivity from the finite group G to a group G.1. (Schmidt [1975a]) If G is a p-group, show that S2(Z(G))m a G. (The assertion

holds more generally; see Zacher [1982a].)2. (Schmidt [1975a]) Let G=(x,y,zlxP'=yP'=zP'=1, [x,y] =zP, [x,z]=[y,z]=1)

and G = <x,y,2I 2 = yP' =2P'

= 1, [X,. ] = 2P, [x,2] = xP, [y,z] =y'> wherep > 5 is a prime. Show that there exists a projectivity cp from G to G such thatZ(G)4' = (z)P = (z) is not normal in G.

3. Show that Zx,(G)" g G. (Hint: Z,,(G) is the largest hypercentrally embedded nor-mal subgroup of G.)

4. If there exists no minimal normal subgroup N of G such that N" is a minimalnormal subgroup of G, show that G = S, x x Sr with nonabelian coprimeP-groups S; of order IS;I = p;q; where p;, q; e P. Conversely, if G is such a directproduct, show that there exists an autoprojectivity of G mapping every minimalnormal subgroup of G to a nonnormal subgroup.

5. For N a G, let SG(N) be the largest and TG(N) the smallest normal subgroup of Gsuch that SG(N)/N and N/TG(N) are hypercyclically embedded in G. Show thatSG(N)" and TG(N)`0 are normal subgroups of G and that SG(N)"/TG(N)`' is hyper-cyclically embedded in G.

6. (Ito and Szep [1962]) If M is a permutable subgroup of the finite group G, showdirectly (not using the results of § 5.1) that M/MG is nilpotent.

7. Use 5.4.12, 5.1.1, 5.2.3 and 5.2.9 to prove Theorem 5.4.4.8. Show that the following properties of (p are equivalent.

(a) NP a G for every normal subgroup N of prime index in G.(b) N4 a a G for every normal subgroup N of G.(c) N' :9:9 G for every subnormal subgroup N of G.

5.5 Normal subgroups of p-groups with cyclic factor group

We continue with the study of the structure of NG/NG and NG/NG- for a normal sub-group N of a finite group G. The main result of the last section, Theorem 5.4.7,reduces this to the case where N is permutable in G and Ni; = 1. Then 5.1.6 and 5.1.3,or the corresponding results for permutable subgroups (which are much easier to

prove), reduce the investigation of N/NG- and N/NG- further to the situation where Gis a p-group, GIN is cyclic and NG- = 1. Moreover the results of § 6.5 will do the samefor normal subgroups in arbitrary groups. Therefore we shall now study this situa-tion aiming to show that N is abelian and G and G are metabelian if p > 2; if p = 2,we shall show that N has modular subgroup lattice, I N' J < 2, G ismetabelian and Gis soluble of derived length at most 3. The results on N/NG- and N/NG- will be general-ized to arbitrary groups in 6.6.1.

Basic properties of G and G

We continue to use the notation introduced in 5.4.1, and throughout this section weassume the following.

5.5.1 Hypothesis. Let p be a prime, G a finite p-group, cp a projectivity from G to thep-group G, and N a normal subgroup of G such that GIN is cyclic and NG- = 1; writeExpN = p` and IS2(N)I = p'.

Let a E G be such that G = N<a>. Then <a> is cyclic, <a> = , say, N is acore-free permutable subgroup of G and G = N. Therefore the assumptions ofTheorem 5.2.8 are satisfied by N and we first of all note the consequences of thistheorem for our situation. By (a) and (b) of 5.2.8, we have:

(1) Every b-invariant subgroup of N is trivial; in particular N n (b> = 1 andhence also N n <a> = 1.

This immediately implies:

(2) If H is an a-invariant subgroup of N, then G* = H<a>, N* = H, and theprojectivity induced by cp in G* satisfy Hypothesis 5.5.1.

By (d) of 5.2.8,

(3) Nf k(G)/i2k(G) is core-free in G/Qk(G) for every k _> 0. Therefore cp inducesprojectivities (pk from G/S2k(G) to G42k(G) for which the images of NS2k(G)/f2k(G) arecore-free; thus G/S2k(G), NS2k(G)/S2k(G) and (Pk satisfy Hypothesis 5.5.1.

(4) S2(G) and S2(G) are elementary abelian. For k = 1_.., r, 92k(G) = S2k(N)S2k(<a))and S2k(G) = S1k(N)S2k() have exponent pk; furthermore, S2k(N) = S2k(G) n N.

Proof. The statements about G follow directly from 5.2.8(c) and (d). Thus fl(G) iselementary abelian and, since the exponent of a subgroup is invariant under cpand i2k(N) a G, we observe that S2k(G) = Qk(N)f2k(<a>) has exponent pk. Finally, byDedekind's law, f1k(G) n N = S2k(N)(S2k(<a>) n N) = 0,(N). El

(5) If p = 2, then Hk(G)/Qk-2(G), Qk(G)/Qk-z(G), Hk(N)/Qk-2(N) andQk(N)/S2k_2(N) are abelian for k > 2; furthermore, IG :1,(G)I > 4 if N 1.

Proof. By (f) of 5.2.8, f1k(G)/Qk_2(G) is abelian. As a projective image of this group,S2k(G)/S1k_2(G) is an M-group of exponent at most 4 in which the quaternion

group Q$ is not involved; by 2.3.7, S2k(G)/S2k_2(G) is abelian. By (4), 0k_2(N) =ilk-2(G) n N = S2k_2(G) n %(N) and hence S2k(N)/Qk_2(N) is isomorphic to thesubgroup Qk(N)Ok_2(G)/S2k_2(G) of the abelian group S2k(G)/S2,_2(G). ThusS2k(N)/S2k_2(N) is abelian, and in the same way one shows that Qk(N)/0k_2(N) isabelian. Finally, by (g) of 5.2.8, N< fl.-2(G) if 2" = Exp G and N 0 1. It follows thatr<n-2andhence IG:S2,(G)I>4.

In particular, f12(G) and S22(G) are abelian for p = 2 and hence the quaterniongroup QB is not a subgroup of G or G. Therefore if H is a subgroup of G or G withmodular subgroup lattice, then by 2.3.8, Qg is not involved in H and so H is anM*-group. Thus we have shown:

(6) Every subgroup of G or G having modular subgroup lattice is an M*-group.

As for M*-groups, the commutativity of S22(G) and 122(G) for p = 2 implies:

(7) If p = 2 and k e N, the maps x - x2k are endomorphisms of S2k+1(G) andQk+1 (G). Hence if x, y c G such that I : x2k = y2k c Q(G), then x = y (mod S2k(G)); ifin addition x, y c N, then x = y (mod b2k(N)). If X, Y < G such that X n Y = 1 andX Y = YX, then S2k(X Y) = S2k(X)S2k(Y).

Proof. We use induction on IGI to prove the first assertion. For k = 1, it is trivialsince C12(G) and S22(G) are abelian; so suppose that k > 2. Since G/52(G) satisfiesHypothesis 5.5.1, we may apply the induction assumption to b2k(G/Q(G)) =S2k+i(G)/S2(G) and deduce that for x, y E Ilk+,(G) there exists z c 92(G) such that

Zk-1 2k-1 2k-1 2k-1 1

(xy) = x y z. Now x , y21- and z are contained in the abelian groupS22(G) and hence

(xy)21 = (x2k 'y2k 'z)2 = x2ky2k,

as desired. The proof for S2k+, (G) is similar. Now if x, y c G such that 1 : x2k =y2k c 52(G), then x, y c S2k+, (G) and have the same image under the endomorphismv: g - g2k of S2k+i(G). Therefore x-'y c Ker v = S2k(G). If in addition x, y c N, thenx-'y E S2k(G) n N = Qk(N), by (4). Finally, for g e X Y there exist x c X and y c Ysuch that g = xy. If o(x) = 2', o(y) = 2', o(g) = 2k and h = max {i, j, k}, then

2h-. 2g = x y # 1 since <x> n (y) = 1. This shows that h = k and that b2k(X Y) _{z E XYIz2k = 1} g S2k(X)S2k(Y). The other inclusion is trivial.

We return to the general situation with arbitrary prime p and assume in additionthat N o 1; for, if N = 1, then G and G are cyclic p-groups and all the assertions wewant to prove hold trivially.

(8) Let T be an autoprojectivity of G and T = N n N'. Then T a N' with cyclicfactor group N'/T and T is permutable in N with [NIT] a chain.

Proof. Clearly, N:9 G and GIN cyclic imply that T = N n N':9 NT with cyclic fac-tor group NT/T. Since NT is a projective image of N, it is permutable in G and [GIN']is a chain. It follows from (11) and (1) of § 5.1 that T = N n N' is permutable in Nand [NIT] [NNT/NT] is a chain.

Since we shall study numerous autoprojectivities r of G, we shall frequently usethe following argument.

(9) Let H be a p-group of exponent p" and M a permutable subgroup of H suchthat [H/M] is a chain. Then H = M<x> for any element x e H that does not lie in themaximal subgroup of H containing M and hence JH : MI = I<x> : <x> n MI < p".If z e H such that o(z) = p" and M n <z> = 1, then I M<z> : MI = I<z>I = p" andhence H = M<z).

(10) ICO(N)(a)I = p; thus CN(a) is cyclic.

Proof. Let H = Cn(N)(a). Since S2(N) is elementary abelian, H<a> is abelian. By (6),(H<a>)O = H u is an M*-group and hence, by Iwasawa's theorem, contains anabelian normal subgroup A with cyclic factor group inducing power automorphismsin A. Then H n A is a b-invariant subgroup of N, hence is trivial by (1) and thereforeH HA/A is cyclic. It follows that H is cyclic and elementary abelian, hence I H I < p;on the other hand, N a G implies that S2(N) n Z(G) 0 1 and hence I H I = p. ThusCN(a) is a p-group with only one minimal subgroup and therefore is cyclic since, by(6), Q8 is not a subgroup of G.

(11) There exist bases of SI(G) and { fo, f1,.... fm} of S2(G) with thefollowing properties:

(1 la) <eo> = SI(<a>) <- Z(G), (fo> = S2() <- Z(G);

(l l b) SI(N) = <e,) x x ;

(11 c) <e1, ... , ei> is the unique a-invariant subgroup of order p' of n(N)(i = 1, ... , m); in particular,

(lid) e1 e H for every a-invariant subgroup H of G such that H n N 0 1;

(11 e) <e,> = Z(G) n SI(N) and e° = e;e;_, for i = 2, ... , m;

(llf) <f>=<ei>°fori=0,...,m;

(1lg) .Ti =.f1fo and <f1fo> = <e1eo>';

(llh) is the unique b-invariant subgroup of order p1+1 of n(6)(i = 0'...'M).

Proof. For every a-invariant subgroup L of S2(N), the map rL: x - [x, a] (x e L) isan endomorphism of L with kernel CL(a) and image [L, a]; for (1) of § 1.5 shows thatfor x, y e L, [xy, a] = [x, a]'' [ y, a] = [x, a] [y, a] since L is abelian. Now let 1 = Lo <L1 < < Lm = O(N) be part of a chief series of the group fl(N)<a> and let rL, = Tifor i = 1, ..., m. Then L;/L1_1 is centralized by a and hence L;' = [L,,a] < Li_1; onthe other hand, by (10), I Ker Ti I < p and the homomorphism theorem yields thatLi_1 = L;i = [L1,a]. In particular, Lm_1 = [S2(N),a] is uniquely determined by aand, similarly, Li = [S2(N), a,.. . , a] where a is written m - i times. Since the chiefseries was chosen arbitrarily, it follows that Li is the unique a-invariant subgroup oforder p' of S2(N). In particular, L1 = Cn(N)(a) = Z(G) n S2(N); let L1 = <e1>. For

i = 2, ... , m, we define inductively elements ej e SI(N) such that e° = e;e;_ 1 and<e1,...,e;) = L;: If e1, ..., e;_1 are defined, then since Li_1 = L;! there exists e, a Lisuch that ei_1 = e,i = [e;,a] and hence e° = e;ei_1; as [L;_1,a] = L1_2, e; L;_1 andhence Li = <e,,..., e.>. For i = m, we get SI(N) = <e 1, ... , em> = <e,> x x and hence (l Ib), (l lc) and (l Ie) are satisfied. Furthermore, if H < G is such thatH° = H and H n N : 1, then H n Q(N) is a nontrivial a-invariant subgroup of SI(N),and so, by (l lc), it contains e1; thus (1 Id) holds. Now let f be any generator of <e1>for i = 15 ..., m. Since e1 e Z(G), it is clear that <el,a> = <e1> x <a) and hence<e1,S2(<a))) = SI(<e1,a)). Thus <e1,S1((a)))1' = <f1,S1(<b))) is invariant under b.Furthermore, is invariant under f1 and <fl)b 0 <f1> by (1). It follows that1 : [f1,b] e <f1,SI((b))) n = S)((b>). So if we choose fo = [f1,b], then f; _f1 fo and < fo) = S1() < Z(G); and for a suitable generator eo of S2(<a)), we canwrite <f1 fo)`0 ' _ <eleo). Thus (1lf) and (I 1g) are satisfied. By (1), <f0> is the onlyminimal normal subgroup of 6 and hence 5.4.2 shows that <eo) = <f,>"-' is a normalsubgroup of G. Therefore (I la) also holds and, by (4), {eo,...,em} and {fo,...,fm}are bases of !Q(G) and SI(G), respectively. Finally, by (11c), <e0..... e;) =Q(<a, e1, ... , e; )) and hence < fo, ... , f) is invariant under b for i = 0, ..., m. SinceS2(G) is abelian, for every b-invariant subgroup L of S1(G) containing fo, the mapr: x --+ [x, b] (x a L) is an endomorphism with kernel CL(b) = < fo ), by (1). ThereforeIL: [L, b] I = IL: L' I = I Ker i I = p and hence [L, b] is the only b-invariant maximalsubgroup of L. Now starting with L = SI(G), a trivial induction yields (I Ih).

(12) If p > 2 and N' < <e1>, then (xy)pk = Xpkypk for all x, y e N and k e N.

Proof. By (1 Ie), <e1> < Z(G) and hence (4) of § 1.5 shows that (xy)pk = xpkypk[y, x]`k

where t = (') 0 (mod p). Since I N' l < p, it follows that [y, x]1 = 1.

(13) If p = 2 and m > 2, the bases {e0,. .. , em } and { fo, ... , fm } can be chosen withthe additional property that f2 = f2f1.

Proof. By (llh), <fo,f1) and [<fo,f1),b] = <fo) so that[f2,b] e <fo, f1)\(f0 ). Thus [ f2i b] =f&2 where a e {0,1 }. If a = 0, we are done;so suppose that a = 1. If we then put e* = e1 for i e {0, 1}, e* = e;ei_1 for i > 2 and<f *) = <e'> for i e {0,...,m}, it is clear that all the assertions of (11) will hold forthe a*, f * in place of e;, f except, possibly, (l le). But for i > 3,

(e*)° = (eie,-1)° = eiei-1ei-let-2 = e*e* 1,

furthermore (e2)° = e2e1e1 = e*e* and therefore (l le) holds. Since <e1, e2> is a four-group, <e*)" = <e2e1)4' = <f2f1>. Hence f2 = f2f, and so, finally,

(f2* )b = (f2 f1 )b = f2 fl f0f1 f0 = f2 fl fl = f2f1*,

as desired.

In the remainder of this section we fix bases {e ...... em } and If ..... , f. J with theproperties given in (11) and (13) and study the autoprojectivity a induced by b in G,that is, the map a: L(G) -- L(G) defined by

(14) H°=H°"°' for allH<G.

We first note that

(15) H°=Hif acH<G;for, then b e H4' and H° = HI""-' = H. In particular, <a>° = <a> and so G = N<a>implies

(16) G = N°(a>.

By (11), fo = f0 and f,' = f fo so that

(17) <eo>Q = <eo> and <el>° = <eieo>.

Since <e1> < N, therefore <eleo> < N°; on the other hand, <eo> N implies that<eo> = <eo>° $ N°. It follows that

(18) N° n <eo,el> = <e1eo>; in particular, e1 0 N°.

We study the core K° of N° in G and the intersection Q of N and N.

(19) Let K < G such that K° = (N°)c. Then N n K° = 1, K and K° are cyclicnormal subgroups of G, fl(K) = <e1> and Q(K°) = <el eo>.

Proof. Since e1 0 N°, e1 0 K° a G and hence N n K° = 1 by (1 ld). Therefore K° a-K°NIN is cyclic, hence also K is cyclic and, by 5.4.6 applied to the projectivity a,K a G. By (11) and (18), <eIeo> < Z(G) n N It follows that <eteo> = S2(K°) andthen (17) implies that f)(K) = <el>.

(20) K = CN(a) and K° <_ Z(G); if p = 2, then Q2(K) <_ Z(G).

Proof. By (15), K<a> = (K<a>)° = K°(a>° = KG<a> so that both K and K° arenormal subgroups of K<a> with cyclic factor groups. Thus (K<a>)' < K n K° = 1,by (19). This shows that K<a> = KG<a> is abelian so that K and K° are centralizedby a. On the other hand, by (10), CN(a) is cyclic and hence CN(a)<a> is abelianand invariant under Q. Therefore CN(a)' < CN(a)<a> is centralized by a. Since G =N°(a>, every g c- G can be written in the form g = a'z where z e N° and (CN(a)°)g =(CN(a)°)Z < N°. It follows that (CN(a)°)G < (N°)G = K° and hence CN(a) < K. ThusCN(a) = K, as desired.

If x e N, then G = N(xa>, and we may apply our results with xa in place of a.Since G = N<xa>4', there exists y E N and b' e <xa>" such that b = y-'b'; it followsthat G = N<b'> and hence b' = yb generates <xa>4'. So if we let d = (pb'cp-', thenN°' = Nwrbw-' = N° and (N°')o = K° is centralized by xa, as we have just shown.Thus K° is centralized by a and every x E N so that K° <_ Z(G). In particular,<xa, K°> is abelian and hence <xa, K°>°-' = <xa>°-'K is an M-group. If p = 2, thenby (5) and (7), o(x) < o(a) = 2" and (xa)2"-' = a2"-' = eo; thus <xa> n K° = 1. By

2.3.6, S22(K) normalizes <xa)°-' and hence [(xa) ',I 2(K)] < (xa)°-' n K = 1.Since G is generated by the <xa> (x e N), it follows that 122(K) < Z(G).

(21) Let Q = N n N°. Then QG = QQ°.

Proof. Since Q < N a G, we have Q < QG < N and hence [QG/Q] is a chain by (8).Since Q is permutable in N, Q u Q° = QQ°; let IQG : QQai = p`. Then QQ° is theunique subgroup of index p` in QG between Q and QG and also between Q° and QG.It follows that (QQ°)" = QQ°. By (16) and (8), G = N°(a) and Q a Na. Thus QG =Q<°> << (QQ°)<°> = QQ° and hence QG = QQ°.

(22) If p = 2 or N' < (e, ), then QG = S25(N) where p5 = Exp Q and s is thesmallest integer such that N/S25(N) is cyclic.

Proof. Let p5 = Exp Q. Then Q < Q5(N) g G so that QG < f25(N). By (7) or (12),ZSS_,(QG) = {xP9-'Ix e QG} is a nontrivial normal subgroup of G and therefore, by(l ld), contains e,. Thus there exists x e QG such that xP,-'

= e,. Let y e S25(N). SinceQ is permutable in N, we deduce that I Q<Y> : QI = I <y>: <y> n QI < p5; by (18),e, 0 Q and hence <x> n Q = I so that IQ<x) : QI = p5. Now [N/Q] is a chain and itfollows that Q<y) < Q<x) < QG. Thus QG = S25(N) and N/S25(N) is cyclic. Suppose,in order to obtain a contradiction, that

N//

SL5_, (N) is cyclic. Then If25(N) : Q 5_, (N) I =p and therefore S25(N) = QS25_,(N). Hence there exist z e Q, w e S25_1(N) such thatx = zw; by (7) or (12), e, = xP'-' = zPs-'wPs-' = zP'-' E Q, a contradiction. Thus s isthe smallest integer such that N/Q5(N) is cyclic.

(23) If p = 2 or N' < <e,), there exists u e N such that u" e,; let U = (u)and V = U°. For any such pair U and V, we have:

(23a) N=QUand N°n U= 1, hence also QnU= 1;

(23b) N° = QV and N n V = 1, hence also Q n V = 1;

(23c) NV = S2r(G) = N°U; furthermore

(23d) NS2r(<a)) = S2r(G) = N°S2r(<a)).

Proof. By (7) or (12), this time Z5r_,(N) = {xPr-'Ix E N} is a nontrivial normal sub-group of G and therefore, by (lid), contains e,; thus there exists u E N such thatuP'-' = e,. By (18), Q(U) = <e,).4 N° and hence N' r) U = 1; in particular,Q n U = 1 . By (17) and ( 1 1 ), 1(V) = f2(U°) = <e,eo) N and hence N n V = 1;in particular, Q n V = 1. By (9), N = QU and N° = QV so that (23a) and (23b)are satisfied. By (4), Qr(G) = NS2r(<a)) and hence S2r(G) = I r(G)° = N°S2r(<a))° =N°S2r(<a)) since b normalizes (S2r(<a)))11. Thus (23d) holds and again by (9), S2r(G) =NV = N°U.

(24) If S2r(<a)) < NG(Q), then N is abelian.

Proof. By (23d), S2,(G) = N°S2,(<a>). Since Q = N n N°< N° and 0,(<a>) normal-izes Q, it follows that Q < Q,.(G); in particular, Q < N. Then N/Q is cyclic and henceN' < Q < Na. Since N' < G, we get that N' < (N°)c = K°. But by (19), N n K° = 1and thus N' = 1.

(25) If N' < <e,>, then Q is abelian and every subgroup of Q is normal in N';furthermore, N and N° are M-groups.

Proof. Let U and V be as in (23). Since N' < (el> < U, we see that U < N andQ N/U is abelian. Furthermore, U° = V is permutable in Na. For every X < Q,therefore

X =X(VnQ)=XVnQ<XV

since Q < N°; thus X < QV = Na. Finally, as a projective image of the abeliangroup N/(e,>, the group N°/<e1eo> is an M-group, and, since <e1e0> < C1,(X),we see that XV/Cv(X) is an M-group. Thus if p = 2, ExpQ > 4 and X is cyclic oforder 4, then X < Z(N°) since otherwise x° = x-' for X = <x>, V = <v> and henceX V/CV(X) would be dihedral of order 8 and not an M-group. By 2.3.4, N° is anM-group and then so is N.

The case p > 2

We come to our first main result.

5.5.2 Theorem (Menegazzo [1978]). Suppose that Hypothesis 5.5.1 is satisfied andp > 2. Then N is abelian.

Proof. We proceed by induction on INI and keep the notation introduced above.If m = 1, N is cyclic; if r = 1, N is elementary abelian, by (4). Thus we may assumethat in, r > 2. We now apply the induction assumption to three different projectiv-ities. First of all, by (2), cp induces a projectivity in Qr_,(N)<a> satisfying Hypothesis5.5.1; by induction,

(26) Q_, (N) is abelian.

By (3), the projectivity induced by cp in G/S2(G) satisfies Hypothesis 5.5.1; thusNO(G)/S2(G) is abelian. Finally, a induces a projectivity from G/K to G/K° and,by (19), f2(K) _ <e1> so that K 1. Thus N/K is abelian and it follows thatN' < O(G) n K = S2(K) since K is cyclic and S2(G) is elementary abelian. So

(27) N' < (e,

and all the assertions in (22)-(25) hold for N; let ps = Exp Q, u e N such that uPr ' = eU = and V = U°. If s < r, then Q < 0,_,(N) and, since UQ = N, we haveUS2r_,(N) = N < G. If s = r, then by (12), the map v: N -+ N defined by x" = xP'-'(x c N) is an endomorphism of N with Kerv = Q_, (N) and, since uP'-' = e1, the

subgroup US2,_,(N) = {x e NIx°'-` e <e,>} is the preimage of <e,> with respect tothis homomorphism. Since N and <e,> are normal subgroups of G, it follows thatUS2,_,(N) g G. Thus in any case

(28) US2,_,(N) a G.

To simplify notation, put T = S2r_,(G). Then UT = Uf2,_,(N)T g G and hence UTis permutable in G. Since o(u) = p', UT/T is a permutable subgroup of order p inG_/T and hence normalizes the cyclic group T/T By (3) and (1) applied to G/T,NT/T only contains trivial b-invariant subgroups and hence (UT)b 96 UT; by (14),Ub = U`°b° 1`° = U°° = V so that (UT)b = VT Therefore if SIT = S2(UT(b)1T),then

(29) S/T has order p2 and contains UT/T, (UT)b/T = VT/T and TQ,(<a>)`°/T =S2(T/T) as three different minimal subgroups.

Figure 19

We want to show that VT normalizes every subgroup of Q5(N). By (4),VT: f2r-,(N)I = I VT : TI I T : S2,-,(N)I = PI(2r-L(<a>)I = pr and, by (23),VQ,_, (N) : Q,_, (N)I = I V: V n S2r-, (N)I = I VI = p'. Since S2r_, (N) < T, it follows

that

(30) VT = VQ,_,(N).

We show first that Q5(N) is centralized by f2,_1(N). This is clear ifs < r since thenS25(N) < S2,_, (N) and this group is abelian. So suppose that s = r. Then by (25),N = QU is an M-group and since Q n U = 1, IN: "r-l (N)I >_ p2. On the otherhand, N'< Q(U) implies that <u">CQ(U) < Z(N) and hence IN : Z(N)I <IN: <uo>CQ(U)I < p2. By 2.3.16, N is abelian or Z(N) < (2r-1 (N) and, in the latter

case, it follows from the above inequalities that Z(N) = f2,_,(N). Thus in any event

(31) f2,-,(N) < CG (c2S(N)).

Since VT = VS2r_1(N), it remains to be shown that V normalizes every subgroup off2s(N). By (25), every subgroup of Q is normal in the M*-group N° = Q V and henceby 2.3.13 there exists a generator v c V and k c N such that (Q, v, k) is an Iwasawatriple for N°, that is, x° = x'+pk for all x E Q. By (29) and (4), v e Qr(<a>)TU =Qr(<a>)Qr_, (N) U and hence there exist a, c f2r((a>) and x, E f)r_, (N) U such thatv = a , x, . Then [v, a] = [x,, a] E fQr_l (N) since, by (28), U0, _I (N)/0,_, (N) is a nor-mal subgroup of order pin G/c2r_, (N). By (31), v-'v° = [v, a] e C6(f2 (N)) and there-fore v and v° induce the same automorphism in Qs(N). By (21) and (22), fzs(N) =QQ". Hence for z E f2s(N) there exist x, y e Q such that z = xy° and, since()'°) . (y°)`l = (}t:)° (J l+pk)° = (J°)'+pk, it follows that z° = (xy°)V = x1+Ck(,) J°)l+pk

/Since p > 2,

l + pI 2 I ° 0 (mod p) so that I N'I < p and (4) of § 1.5 imply that

Z'+pk = xl+pk(y,°)l+pk /Thus z' = z'+pk, that is, v induces a power automorphism inR(N). Now (30) and (31) yield that

(32) VT normalizes every subgroup of fl,(N).

All assertions proved so far also hold with b replaced by b-'. Thus if r is theautoprojectivity of G induced by b-', that is, H` = H106- "° ' for all H < G, and W =U`, then by (29), WT/T is a minimal subgroup of S/T and, by (32), every subgroupof 0,(N) is normalized by WT where, by (22), s is still the smallest integer suchthat N/05(N) is cyclic. If VT = WT, then (UT)" = VT = WT = (UT)" ' and hence(UT)" = UT; since p > 2, it would follow that h c N (UT), contradicting (29). ThusVT/ T and WT/T are different minimal subgroups of S/T and hence generate SIT.Therefore U < VTW normalizes every subgroup of f2s(N). In particular, Qv = Qand so QaQU=N.Since N'<<e1><U,also UaNand, by(23),N=Q x U.By (25), Q is abelian and so, finally, N is abelian.

Since N is abelian and G/N is cyclic, G is clearly metabelian if p > 2. An examplein which N is not abelian is given in Exercise 2. Nevertheless, G is also metabelianas we are now going to show. In fact, we prove a more general result that will alsobe used in the case p = 2.

5.5.3 Lemma. Suppose that Hypothesis 5.5.1 is satisfied and N is abelian. Then f2r(G)is an .M*-group and G is metabelian; for every c E f2r(G) there exists k c- IN such thatk- 1(mod 4)if p=2and x`=x'"forall

Proof. We keep the notation introduced above and let f2r(<a>) = <c>. Since N' <<e,>, all the assertions in (22)-(25) hold and we use induction on INI to prove firstthat c induces a universal power automorphism v -+ v' in N. By (23) and (25), everysubgroup X of Q is normal in N° = QV. Since N is abelian, X a NV = f2,(G). By1.5.4, c induces a universal power automorphism in the abelian group Q, that is,there exists h e 7L such that x` = x' for all x c- Q. Since c e <a>, we obtain (x°)` =(x`)° = (x'')' = (x°)", so that c induces the automorphism x -+ x' in QQ°. By (21) and

(22), QQ° = f2,(N) for some s e N. If s = r, then f2,(N) = N and we are done; thus wemay assume that s < r. By (23), N = Q x (u) where uPr-' = e,. The induction as-sumption applied to G/S2(G) yields j c 77 such that ccl(G) induces the automorphisme - o' in NS2(G)/S2(G); thus u` = u'z where z e fl(G) n N = f2(N). It follows that(ups-')` = (u` )P' ujP'-5; on the other hand, (uP" ')` = u''p'-' since u°'-' e I(N). Thush - j (mod ps) and hence x` = x' for all x e Q. Since u° e N = (u) x Q, there existi e N and y e Q such that u° = u'y. It follows that (u°)` _ (u'y)` = (u`)'y` = (u'z)'y' _(u'y)'z' = (u')-1z'; on the other hand, (u°)` = (u`)° = (u'z)° = (u°)'z° since c c (a).Thus z° = z' E (z), hence (z)° = (z) and (l Id) implies that (z) < (e,). SinceuP' ' = e,, there exists t e N such that z = u'P' '. It follows that u` = u'+'P'-' andXC = x' =

x'+`P.-' for all x c Q so that c induces the power automorphism v -, v' inN where k = j + tpr-1

By (5), f2r(G)/f2r_2(G) is abelian if p = 2 and so [N, c] < f2,-2(G) n N = f2,_2(N),by (4); it follows that k = I (mod 4) in this case. By 2.3.4, f2r(G) = N(c) is an M-group and, by (6), f2r(G) and S2r(G) are M*-groups. Of course, every element of Q,(G)operates on N as a power of c and hence induces an automorphism v v' where1 - I (mod4) if p = 2. Finally, S2r_,(G) < NS2r_,(G) < S2r(G) and NS2r_,(G)/S2r_,(G)is core-free in It follows that f2r(G)/S2r_, (G) = S2r(G)/S2r_, (S2r(G)) is notcyclic. Since G/f2r(G) is cyclic, 2.3.21 yields that G is metabelian.

Now 5.5.2 and 5.5.3 immediately imply:

5.5.4 Theorem (Busetto and Menegazzo [1985]). Suppose that Hypothesis 5.5.1 issatisfied and p > 2. Then G is metabelian.

The case p = 2

In the proof of Theorem 5.5.2, the assumption p > 2 is used only in the last few lines.All the assertions up to (32) could be proved similarly for p = 2 with simple alter-ations; but for p = 2, of course, VT = WT so that we would not get U < VTW.Indeed, there is an example of a 2-group satisfying Hypothesis 5.5.1 in which N isnonabelian. We shall describe it briefly since the details are rather complicated andtedious.

5.5.5 Example (Busetto and Stonehewer [1985]). It is easy to see that the group

G = (a u w1a64 = U16 = w8 = 1, uw = U9, ua = u-Iw4 Wa = u2w-1\

is the semidirect product of the normal subgroup N = (u, wIu16 = w8 = 1, u'° = u9)of order 2' by the cyclic group (a) of order 26. Clearly, N is metacyclic and N' =(u8) has order 2.

The group G is best defined in three steps. First take H = (c) x (u) where o(c) =o(u) = 16 (note that H <a 4> x (u) < G). The group H has an automorphism f3 oforder 4 satisfying cp = c13u8 and uft = u5; hence we may form the semidirect prod-uct M of H by a cyclic group (w) of order 8 inducing the automorphism /3 in H; thus

IMI = 2''. Finally, there is an automorphism x of M satisfying c" = c, V =and w' = u14w-'; also it is not difficult to show that x4 is conjugation by c. There-fore there exists the extension G = M(a> where M a G, G/M is cyclic of order 4, aoperates as x on M, and a4 = c.

Every element of G can be written uniquely in the form akujw' where 0 < k < 63,0 < j < 15, 0 is generated by anelement of the form aujwi. Similarly for G, and we define v: G -+ G by

(akuiwi)v = ak(1+4i)uj(1+4i)Wi+e

where c = 2 if i is odd and e = 4jk if i is even. It is easy to check that v is bijec-tive; but it is less obvious-and we refer the reader to the paper of Busetto andStonehewer-that the map r defined by X` = X" for X < N(a2> and X' _((aujwi)"> with cyclic X = (aujwi> 4 N(az> can be extended to a projectivitycp from G to G. It is clear that N`° = N = (u, w> is core-free in G sinceS2(N) n Q(N)a n Q(N)a2 = 1. Therefore Hypothesis 5.5.1 is satisfied. Since [u, a]C ' and [w, a] = u6w6 do not commute, G" 1; on the other hand, M = H<iv>is metabelian so that G is soluble of derived length 3. Finally, N' = (u4> is cyclic oforder 4.

Busetto and Stonehewer also proved that the above example is the smallest possi-ble one for which Hypothesis 5.5.1 is satisfied and N is nonabelian. But N is still anM-group and JN'J < 2. It was shown by Busetto and Napolitani [1991] that thisis true in general. The basic breakthrough, however, had been accomplished byBusetto [1984] who had almost determined generators and relations for G to showthat N and N have derived length at most 3 and 4, respectively. Busetto andNapolitani improved this argument to obtain the results we shall present here. Westart with a preliminary result due to Menegazzo.

5.5.6 Lemma. Suppose that Hypothesis 5.5.1 is satisfied and p = 2. ThenS2(G) < Z(f),(G)).

Proof. We use induction on INI and may assume that r > 3 since f'2(G) is abelian.By (19), 6 induces a projectivity from G/K to G/K° and the induction assumptionyields that S2(G/K) < Z(S2,(G/K)) if 2` = Exp(N/K). Since S2(G)K/K < S2(G/K) andN1 K < S2,(G/K), it follows that [S2(G), N] < K and hence

(33) [S2(G), N] < S2(G) n K = i2(K) = (e1>.

Let U, V be as defined in (23). Then U < N implies that [S2(G), U] 5 <e1> < U andhence U is normalized by S2(G). It follows that I UX : UI < 2 for every minimalsubgroup X of G; hence I U° u X°: U°l < 2 and therefore V = U° is normalized byS2(G). The same holds for the projective image N° of the normal subgroup N. Thus[9)(G), Q] = [S2(G), N n N°] < <e1> n N° = 1, by (18), and [S2(G), V] < G' n V <N n V = 1, by (23b), so that

(34) [i2(G), N°] = [92(G), QV] = 1.

Since S2,(G) = NV = NN°, it follows from (33) and (34) that

(35) [n(G), n,(G)] < <el ).

To prove the lemma we therefore suppose, for a contradiction, that [S2(G), f2,(G)] _<e 1 >. By (18), eo N. So if W is a cyclic subgroup of order 2' in G such that eo e W,then W n N° = 1 and S2,(G) = N°W since Is2,(G) : N°I = 2'. If W were normalizedby f2(G), then (35) and (34) would imply that [S2(G), W] < W n <e1> = I and[f2(G), S2,(G)] = [S2(G), N°W] = 1; this would contradict our assumption. Thus

(36) S2(G) NG(W) for every cyclic subgroup W of order 2' of G containing eo.

In particular, S2,(<a>) is not normalized by Q(G) so that, by (35),

(37) [Q(G), Q,(<a>)] = <el ).

By induction on j e N we obtain that

(38)a2J = e; for 2 < i < 2' and e2' = e;ei_2 for 2' < i < in.

For, if j = 0, this holds by (l Ie); and if (38) holds for j, then

Q2'- Q2' Q2l 2'

e = (e ) = (eiei-2J)Q

= eiei-2jei-2jei-2i" = eiei-2r"

j+1 Q2J-' Q2l-' Q2,where 2j" ) = <a2k>. By (38), [ei,a2k] = I for i < 2k, and [ei,a2k] = e1_2k for2k < i. But by (37), [f2(G), <a2k>] = <e1> and it follows that

(39) m = 2k + 1.

Then <e0,..., em_, > is centralized by <a2k) = S2,(<a>) and by N°, as (34) states, andhence by S2,(<a>)N° = S2,(G). Thus

(40) <eo,...,em_i> = Z(Ir(G))nS2(G).

We want to show that, on the other hand, U is not centralized by <fo,..., fm_,). Forthis let U = <a) and W = <aab> '. By (7), (23), (1 If) and (1 Ig),

(iiiih)2 = U2' '(U2' ' )b =J 1 J lb = J 1J 1J0 = f

and hence W is a cyclic subgroup of order 2' of G containing eo. Since.fo E Z(G),

(41) U1+2' t(ab)1+2'' = i (ilUb)2rtub = (aab)1+2 t

By (33), U is normalized by S2(G). Thus if x E f2(G) and <y> = <x>11, then UY = Uand, since f22(G) is abelian, the automorphism induced by y in U has the form

(42) aY = u1 where I = I or I = 1 + 2-1

By (11 h), <f0. .. , fm_i > is b-invariant and of index p in Q (G); thus y' = yz wherez E < fo, . , fm _, >. Therefore if U were centralized by <J0,. - -, fm _, >, it would followthat u'' = aY2 = al; since (ab)Y" = (aY)b = (a)b = (ub)l we would get that (aab)Y6al(i b)l = (aab)1 where the last equation is trivial if 1 = I and is (41) for 1 = I + 2-1r .

Thus W = <uub> would be normalized by < yb> = <x> and so W by <x>" =<x)°. Since x e S2(G) was arbitrary, it would follow that W would be normalized byS2(G)° = S2(G), contradicting (36). Thus, as fo e Z(G),

(43) <f1,...,.f._1> CU(U)

Since <e1,. .. , em _ 1) is generated by the elements outside <e 1, ... , em _ 2 ), there existsx e <e1,...,em_1>\<e1,...,em_2> such that <y> = <x> Cd(U). Then 10 1 in (42)and hence

(44) [u, y] =u211 = .fl

By (40), <a 2k> = 0,(<a>) centralizes <e1,...,em_1>. Thus <x°'!0 is ana-invariant subgroup of <e1,...,em_1> and hence, by (I lc), is equal to <e...... em_1since x 0 <e1,...,em_2>. It follows that the 2' = m - I elements x°' (0 ,

then L° = <x°'I0 < j < 2k - 1, j odd> and L n L° = 1. In particular,

(45) CL(a) = 1 and L < CG(N),

by (40). Since L is generated by the conjugates of x under <a2>, we see that L isnormalized by this group. Thus if h e N and i e t\4 is odd, then L a <x, ha21> =L<ha2L>. By (5) and (7), o(a2`) > o(h) and S2((ha21>) = S2(<a2'>) = <e0>. ThusL n <ha2'> = I and hence

(46) S2(<x, ha21>) = L x <e0> for all h c N and i * 0 (mod 2).

Since ub2 e N<b2> = (N<a2>)`,, there exist h e N and i e N such that <ubZ) _<ha2i>". Again by (5) and (7), o(u) < o(b2) = o(ub2) and hence i is odd. By (46),Q(<y, ub2>) = S2(<x, hat'))"' = (L x <eo>)" and S2(<y, b2)) = S2(<x, a2))"' =(L x (eo>)". Since o(y) = 2, it follows that [5b 2, y] e Q(< y, ub2>) = (L x <e0>)"and[b2,y] e (L x <eo>)"'. By (1) of § 1.5, (44) and (11g),

[ub2, y] _ [u, y]b2 [b2, y] = fl" [b2, y] = f, [b2, y]

and hence fl = [ub2, y] [b2,y]-1 e (L x <eo>)"'. Thus e1 e (L x <e0>) n N = L; butby (45), CL(a) = 1. This is the desired contradiction.

Modularity in 2-groups

We interrupt our investigation of the case p = 2 to prove some lemmas on 2-groupswith modular subgroup lattices and permutable subgroups in 2-groups; these will beused in the sequel. We start with two rather trivial properties which hold for arbi-trary primes p.

(47) If H is a p-group of exponent p°, 1 < k< n, S2rt_1(H) and H/Z5k(H) are M-groups and TSk(H) is cyclic, then H is an M-group.

Proof. By 2.3.2, we have to show that X Y = YX for any two subgroups X, Y of H.This is clear if X, Y < S2n_1(H); and if X $ S2"_1(H), say, then X contains an elementx of order p" and K = 75k(H) = <X"3 < X. It follows that XY = XKY = KYX =YKX=YX.

(48) If H = M<z> is a p-group with modular subgroup lattice, M abelian, M a Hand M n <z> = 1, then there exists k e N, k = 1 (mod 4) in case p = 2, such thatx`'=x"for all xeM.

Proof. IfX <M,then X=Xu(<z>nM)=(Xu<z>)nM aXu<z>.By 1.5.4,zinduces a universal power automorphism x -+ xk in M. If Exp M < 2, k = 1; and ifx e M such that o(x) = 4, then by 2.3.6, xk-1 = [x, z] E <x> n <z> = 1 so that k - 1

(mod 4).

In the following four assertions let M be a subgroup and z an element of the groupH and let k E J. For short, let us say that z induces the power k in M if x` = xk forall xeM.

(49) If z induces the power kin M and a E H such that [a-1, z-1] e CH(M), then zalso induces the power k in M.

Proof. By assumption there exists c e CH(M) such that az = cza. Hence (x°)z =xCZ" = (xk)" = (x")k for every x e M.

(50) Let H be a 2-group and k = 1 (mod 4). If I M' j < 2, then (xy)k = xkyk for allx, y c M. So if z induces the power k in subgroups A, B of M and AB < M, then zalso induces the power k in AB.

Proof. By (4) of § 1.5, (xy)k = xkyk [y, x]` where t = 12 I is even, since k =- 1 (mod 4);

thus [y, x]` = 1. For a E A, b e B, therefore (ab)z = azbz = akbk = (ab)k.

(51) Let H = M<a>, M a H, z e <a> and L < M where L g M<z>. If z inducesthe power k in M/L, then it does likewise in M/LH.

Proof. By assumption, x-kxz E L for every x e M. For c e <a>, therefore L` contains(x-kx2)C = (x`)-k(x`)z and, since M a H, it follows that x` is an arbitrary element of

M. Thus y-ky2 e n Lc= n Lk = LH for every y e M.ce<a> hell

(52) Let H be a 2-group, l2(M) <_ Z(<M, z>) and suppose that z induces the powerk in M/Q(M). Then (x2)2 = (x2)k and xz: = xkx for all x e M; in particular, z2 inducesthe power k2 in M.

Proof. Let x2 = x k y where y e S2(M). Then (x2)2 = (xz)2 = (xky)2 = (xk)2y2 = (x2)kand, since yk = y = y2, we conclude that xzz = (xky)2 = (xky)kyz = xkzy2 = xk2

We have seen in § 2.3 that elements of order 4 play a special role in 2-groups withmodular subgroup lattices. One reason for this is the following more general result.

(53) If M is a permutable subgroup of the 2-group H and M < X < H such thatJX : MI < 4, then M -A X; in particular, M:9 MS22(H).

Proof. If M were not normal in X, then IX : MI = 4 and X/MX would be isomorphicto a 2-subgroup of order at least 8 of the symmetric group S4. But D8 E Sy12(S4) doesnot possess a nonnormal permutable subgroup of order 2. Thus M g X; in particu-lar, every element of order at most 4 of H normalizes M.

(54) Let H = <x, y) be a 2-group with modular subgroup lattice and suppose thato(x) = 4 and <x> is core-free in H. Then <[x, y]) = S22(<y)), <[x, y2]) = S2(<y))and <x)'' <x)y'-'.

Proof. By 2.3.6 or (53), x normalizes <y> and hence [x, y] E <y> n S22(H) =Q2(<)')). If [x, y] E S2(<y)), then [x2, y] = [x, y]2 = 1 since S2(<y)) < Z(H) andhence x2 e Z(H), contradicting <x)H = 1. Thus <[x, y]) = f22(W) and hence[x, y2] = [x, y]2 generates S2(<y)). Since <x> n <y) = 1, therefore <x)y2 # <x) andhence <x)y <x)y '.

(55) If H is a 2-group such that H' is cyclic of order at most 4, then[x2, H] < V(H') and x4 E Z(H) for every x E H.

Proof. Since H' is cyclic and I H' I < 4, we have [x, y]x = [x, y]-' or [x, y]x = [x, y]where y e H. Then (1) of § 1.5 shows that in the first case [x2, y] = [x, y]x [x, y] = 1and hence also [x4, y] = 1; in the second case, [x2, y] = [x, y]2 E D(H') and [x4, y] _[x,1']4 = I-

In the last two lemmas of this subsection, H again is a p-group with modularsubgroup lattice for an arbitrary prime p. And since both assertions trivially hold ina hamiltonian 2-group, that is, in a direct product of Q8 with an elementary abelian2-group, we may assume that H is an M*-group. Then by Iwasawa's Theorem 2.3.1there exist A a H, b e H and an integer s which is at least 2 in case p = 2 such thatH = A<b), A is abelian, and b-'ab = a'+"5 for all a c A. Clearly, H' = Z55(A).

(56) If the M*-group H contains a maximal subgroup M that is abelian, then H'is elementary abelian.

Proof. We keep the notation introduced above and assume first that A < M. Thenb" be M < CH(A) so that (I + p')" - I (mod Exp A). But since s > 2 in case p = 2,(1 + ps)" = 1 + p'+' (mod ps+2) and hence ps+' > Exp A. It follows that H' = Z55(A)is elementary abelian. If A M, then H' < A n M < Z(AM) = Z(H) andH = M<z) for some z c H. Since z" E M, we have [x, z]" = [x, z"] = 1 for everyx e M and therefore again H' is elementary abelian.

(57) If His an M*-group of exponent p' such that H' is cyclic and I H'11,_, (H)I > p3,then H is abelian.

Proof. Suppose, for a contradiction, that H' = UJA) o 1. Then A = <a> x B whereo(a) > Exp B. It follows that JA : f2r_, (A) I < p and hence also I AS2,_, (H) : S2,_, (H)l _I A: A n'2r_,(H)I < p. Since H/AS2r_,(H) is both cyclic and elementary abelian,I H : Af2r_, (H)I = p. Thus JH : Qr_, (H)I < p2, a contradiction.

Finally, we mention that we sometimes use Baer's Theorem 2.5.9 saying that anM*-group H of order p" is lattice-isomorphic to an abelian group H of the sameorder and therefore of type (pa',..., pa') for integers x, >_ xz ;; we also callthis the type of H.

The case p = 2, continued

We want to show that under Hypothesis 5.5.1, N is an M-group, IN'I < 2, G ismetabelian and G has derived length at most 3. Recall that IS2(N)I = 2'. So if m = 1,N is cyclic since G, by (6), does not contain subgroups isomorphic to Q8; then G ismetacyclic and 5.2.13 shows that G is metacyclic. Thus all the assertions we want toprove hold in this case and hence we shall assume in the sequel that

(58) m > 2.

For the proof of our result we have to study three autoprojectivities of G in somedetail: first of all a = cpbtp-', as before, but also a2 = (pb2cp-1 and Qaa-1. For theconvenience of the reader we recall from (11) and (13) the operation of a and bon thebases {e0,. .. , e,,, } and { f0, ... , fm } of fl(G) and S2(G), respectively. In addition wenote that, by Theorem 2.6.7, cp is induced on S2(G) by an isomorphism 1. Thus

(59) e' = f for i = 0, ..., m; eo, e, e Z(G); e° = e;e;_, for I = 2, ..., m; fo = fo,f 1b = .fl .fo and fi = fzfl

We shall need the following formulae (see also (17)).

(60) <eo)" _ <eo>, <er>" = <e1eo>, (e2>" = <eze1>, <eze,>" _ <ezeo>.

(61) <e1>"2 = <e1>, <e2>"2 = <ezeo>,<eze,>"2

= <eze,eo>, <ezeo>"2 = <e2>.

(62) <e1>""" ' = <e1>, <eze,>oa" = <ezeo>, <e2eie0>"a" = <ez>

Proof. By (59), a = cphcp-' is induced on S2(G) by the isomorphism 1bi-1, which weshall also call a. Then we can compute as follows: eo = fob, 1 = fo

1

= eo, ei = fob"1

=(.fifo)'

1 = e1eo, ei =fir 1

= (f2fI)' 1 = eze,, (eze,)" = eiei = ezeo. This proves (60);2 2 2 2 2using (60), we obtain: e; = (e,eo)" = e,, e2 = (eze,)" = ezeo, (eze,)° = e2 ei =

)"a =e e e (ez e)"2 = ea2e"2 = e and a"a = (e,eo)a" ' = (eieo)"-' = ei, (ezeiz i o, o z o z 1

(ezeo)°" = (e2 e1eo)"1

= ezeo, (ezeIeo)ff"-' = (ezeoeo)"" 1 = (e2 e1)"1

= e2.

(63) Again let Q = N n N" and define R = N n N"2 and S = N n N"a"-'. ThenQ n <eo,el,e2> = <eze1> and R n <eo,e1,e2> = <e1> = S n <eo,e,,ez>.

Proof. Let E _ <eo, e, , ez >. By (18), N" n <eo, e, > = <e, eo > and since e, eo 0 N,Q n <eo,e,) = 1; it follows that IQ n El < 2. On the other hand, <e2>" = <eze,

and hence <e2e1) < N n N° = Q. Thus Q n E = <e2e1). Similarly, sinceNn <e2, eo) = <e2 > but (e2) = (e2 eo)°2 4 N2 and <e2> = <e2 e 1

eo )°°°-1

N°°° 1, it follows that R n (eo, e2) = 1 = S r) <eo, e2 ). Thus R n Ej < 2 andjS n Ej < 2; on the other hand, <e1)°2 = <e1> = <e1)°°°-1 implies that <e1> <NnN°2nN°°°-1 =RnSandRnE= <el)=SnE.

(64) K< S and R n K = (e1).

Proof. Recall that K° = (N°)G. Thus S° = (N n N"'-')' = N° n (N°)° >- K° andhence S > K. Suppose, for a contradiction, that R n K > <e1>. Then R n K >(22(K) _ <g) since K is cyclic; let <g)'* = <h). By (20), <g,a) is abelian and hence<g, a)" _ <h, b) is a 2-group with modular subgroup lattice and o(h) = 4. Moreover,by (1), <h> is core-free in <h,b) and so (54) yields that <[h,b2]) = S2(<b)) = <fo).Now g E R= N n N°2 implies that h e (N n N2), = N° n N*b2 and hence fo =h_1hb2

E (N`°)b2. Since fo E Z(G), it follows that fQ E N41, a contradiction.

Recall that by (23) there exists u e N such thatu2r-1

= e1. For any such u,

(65) N = Q<u), Q n <u) = 1, u2r-1 = e1, u° = u (mod fl,-, (N)) and, more generally,(u2k)° = u2k (mod S2r_k_1 (N)) for k = 0, ... , r - 1.

Proof. The first three assertions have been proved in (23). For k = 0, ..., r - 1,

((u2k)°)2r-k-1 = (u2r-1)° = e = e1 = (u2k)2r-k-11

and hence by (7), (u2k)° = u2k (mod S2r_k_1(N)).

(66) Let 25 = ExpQ. Then there exists w e Q such that w2s-1 = e2e1. For anysuch w, Q = (Q n R)<w) = (Q n S)<w), (Q n R) n <w) = 1 = (Q n S) n <w) andN = (Q n R)(w)<u); furthermore w° = wu2r-` (mod (2s_1(N)).

Proof. To simplify notation, let r be one of the two autoprojectivities a2 or Qau-1of G and let T = N n NL; thus T = R or T = S and therefore T n (eo, e1, e2) _<e1), by (63). Now (65) and (7) imply that S25(N) = QK2,() and U,-, ((2,(N))Z5s_1(Q)(e1). Since e1 0 Q and ts_1(Q) 0 1, it follows from (l lc) that (e1,e2) <Vs_1(SL5(N)) and hence (e1,e2) n?55_1(Q) 1. Since Qn <e1,e2) = <e2e1) andZSS_1(Q) = {x21 Ix E Q}, there exists w e Q such that w2a-1 = e2e1. By (8), T ispermutable in N and [N/T] is a chain. Thus T n Q is also permutable in Q and[Q/Tn Q] is a chain. Since T r) <eo, e1, e2 > = <e1>, we deduce that (Q n T) n <w> = 1and, by (9), Q = (Q n T)<w). It follows that N = Q(u) = (Q n T)<w)(u). For thelast assertion in (66), note that

(w°)251

= (w251)° = (e2e1)° = e2e1e1 =

W21-1U21-1 = (wu2r ')2' 1

so that w° - wu2r-' (mod S2s_1(N)), by (7).

(67) If N/Qr_2(N) is not cyclic, we may choose the element u in (65) from R;similarly, there exists u' e S satisfying (65) and (66) in place of u. For these elements,

(67a) R = (Q n R) and S = (Q n S)<u'>,(67b) N = R<w> = S<w> and R n <w> = S n <w> = 1,(67c) N`2 =

R<w>a2 = R<wU2. s>al, R n <w>a2 = 1= R n <wU2-=>°2f

(67d) N°°° ' = S(w>°°°-'

Proof. Again let r be a2 or uacr' and T = N n N` so that T = R or T = S. For thepresent, choose u as in (65). Then N = Q = T<z> for some z e N, by (8) and (9),and hence by (7), t5r_,(N) = t5r_1(Q)t5r_1() = t3r_,(T)t5r_,(<z>). We are lookingfor t e T such that t2'-' = e,. By (11c), U,-,(N) = <e,,...,e1> for some i. If i > 2,then I?Sr_,(<z>)I < 2 implies that J,_,(T) n <e e2> 1. By (63), T r) <eo, e,, e2><e,>, therefore U,-, (T) n <e,, e2> = <e,>, and hence there exists t E T such that t21-' _e,. This, of course, is also the case if i = 1 and Jr_,(T) 1. So suppose, for acontradiction, that i = 1 and Jr_, (T) = 1. Then Exp T < 2r-' and t5r_, (N) =<e,> = Q(<z>) = Q((u>). Since Q n = 1, it follows that also ExpQ < 2'-'. IfQ < Qr_2(N), then N/Qr_2(N) would be cyclic, contradicting our assumption; thusExpQ = 2r-'. Now IN: TI = I<z>: <z> n TI < 2r-' since Q(<z>) _ <e,> < T; onthe other hand, I T<w>: TI = I(w>I = 2r-' since Q((w>) = <e2e,> T. Thus N =T <w> < Q,_, (N), which is a contradiction. Thus in any event there exists t e T suchthat t2"-' = e,. By (65), for any such t, N = Q(t> and Dedekind's law yields thatT = (Q n T)(t>. By (66), N = <t>Q = (t>(Q n T)<w> = T<w> and T n <w> _T n Q n <w> = 1; so (67a) and (67b) are satisfied.

For the proof of (67c), note that I N' 2: R I = IN: R I = I(w>I and that

Q(<w>) = <e2e,> and Q(<wu21-°>) = (e2>, by (7). Hence by (61) and (63),(w>Q2 n R = <e2e,eo> n R = 1 and <wu2"-=>Q2 n R = <e2eo> n R = 1. It follows

that N.2 = R<w>a2

= R<wu2r-,>a2. Similarly, I N"' : SI = IN: SI _ I(w>I and, by(62) and (63), <w)°°°-' n S = <e2e0> n S = 1; thus N"'-' = S<w)°°°-' and (67d)holds.

(68) If x e N such that [Q2(<x>), a] = 1, then [Qi+2(<x>), a] < n;(N) for all i > 0.

Proof. We use induction on o(x). If o(x) > 24, then the induction assumption ap-plied to x2 yields that [Q3(<x>), a] < O(N). Therefore, by (3) and (4), xQ(G) satis-fies the assumptions of (68) in G/Q(G) and the induction assumption implies that[QI+2(<x>), a] < QI(G) n N = Q1(N) for all i > 1. Thus we may assume that o(x) = 8and have to show that [x, a] E Q(N). Let H = <x, Q2(N)>. By (5), Q3(N)/Q(N) isabelian and hence H' < U(N). Let K be as defined in (19). Since Q(<x>) = <x4> iscentralized by a, we obtain x4 = e, e K, by (l ic). Thus HK/K < Q2(G/K) is abelian,by (5) applied to G/K. So H' < Q(N) n K = <e,>. Since (x°)4 = (x4)° = ei = e, = x4,(7) implies that xa = xz for some z E Q2(N). By assumption, x2 = (x2)° _ (Xa)2 =(xz)2 = x2z[z,x]z and hence z2 = [x,z] E H' < <e,> = <x4>. Thus z2 = x4k for somek, hence (zx-21)2 = I since Q2(N) is abelian and therefore zx-2' E Q(N) <_ Z(N),by Lemma 5.5.6. It follows that z centralizes x and hence z2 = [x, z] = 1. So[x, a] = z e U(N), as desired.

We now reduce the proof of our main theorem to a special case which is applica-ble to the different situations we have to consider.

5.5.7 Lemma. Suppose that Hypothesis 5.5.1 is satisfied and that in addition p = 2, Nis an M-group with cyclic commutator subgroup N' and one of the following holds:

(a) N/S2,_2(N) is not cyclic, or(b) Z5r_2(N) < Z(G)-

Then I N'I < 2.

Proof: We argue by induction on G I and first note that the assumptions of thelemma are inherited by G/S2(G). Indeed N1(G)/1(G) N/S2(N) is clearly an M-group of exponent 2'-' with cyclic commutator subgroup. If (a) holds for N, then italso holds for N/Q(N) except possibly when r < 2; but in this case, by (5), N isabelian and we are done. And if (b) holds, then Z5r_2(N) < CN,(a) is cyclic by (10). Thisis equivalent to the type of N being of the form (2', 2k, ...) where k < r - 2 and thisis preserved in N/S2(N); hence Z5,_3(N/S2(N)) is cyclic. Thus Z5,_2(N) = <U2'-'> forit e N with o(u) = 2' and Z5r_3(NS2(G)/S2(G)) = <u2r-'S2(G)>. Applying (68) with x, areplaced by u, ya (y c- N), we obtain [u2r-', ya] E Q(G) for all y E N and hence, finally,that <u2' 'f2(G)> < Z(G/S2(G)). Thus (b) holds in G/S2(G). The induction assumptionimplies that N'f2(G)/S2(G)I < 2 and, since N' is cyclic, it follows that IN' l < 4. Wewant to show that I N' l < 2; so we suppose, for a contradiction, that IN' I = 4. By(l Id), e, e N'; so

(69) N' is a cyclic subgroup of order 4 of N containing e,.

Define Q as in (21) and choose u and w as in (65) and (66); then N = Q(u> andu2' ' = e, . Since e, Q and N' is cyclic, Q n N' = I and hence Q QN'/N' isabelian. Then (55) implies that S2r_2(N) < Q<u'> centralizes Q. Thus

(70) Q is abelian and S2r_2(N) < CN,(Q).

We show next that we may choose u with the additional property that

(71) [u, a] E 52,-2(N).

This is clear if (b) holds; since then S22() < Z5r_2(N) < Z(G) and (68) implies (71).So suppose that (a) is satisfied. Then N/S2r_2(N) is not cyclic and Q S2,_2(N)since N = Q(u>. Thus 2'-' < o(w) = Exp Q. If o([u, w]) < 2, then [u, w] E Z(N)and hence [u2, w] _ [u, w]2 = I. It would follow that W E Z(<Q, u2>) and o(w) =Exp(Q, u2>. By 2.3.16, <Q, u2> would be abelian and (56) would imply that N' iselementary abelian, contradicting (69). Thus o([u, w]) = 4 and hence N' = <z>where z = [u, w]. Since N' a G, either z' = z or z° = z-' = ze,; in the latter case,(ze2)° = ze,e2e, = zee. Since e, e and <e2e,> E <w>, we obtain e2 E <u, w>, andhence H = <u, w> contains an element x, namely x = z or x = zee, such that x' = xand x2 = e,; the latter follows from (ze2)2 = z2ez = z2 = e, where we use 5.5.6 to gete2 E Z(N). Since n <w> = 1, the subgroup H is of type (2', 2') or (2', 2'-') and weclaim that there exists v e H such that

v2i-2

= x. This is clear if the type of H is (2', 2')since then ir_2(H) = 02(H). And if s = r - 1, then H/L is of type (2'-', 2'-') whereL = Vr_, (H) = <x2>; thus xL E S2(H/L) = Z5,-,(H/L) so that there exists g e H such

that g2 2 = x (mod L). It follows that g2 2 = x org2"-'

= x-', and in both casesthere exists v e H such that v2" ' = x. Then v21_1 = e1 and, by (68), [v, all E S2r_2(N);so if we replace u by v, (71) is satisfied and we may assume that (71) holds for u.

Now (71) implies_ that the group <u,a>S2r_2(G)/S2,_2(G) is abelian; hence<u,a>'°S2r_2(G)/S2r_2(G) is an M*-group containing "S2,_2(G)/S2r_2(G), which isa core-free cyclic subgroup of order 4, by (1) applied to G/f2r_2(G). By (54),"'12,-2(G): "b Hence if <u1> = ° = "b" ' and <u2> =° ' = "b-',p1' then <u1>K2,-2(G)/K2,-2(G) and <U2>K2,-2(G)/J2, -2(G) are twodifferent subgroups of order 4 in the abelian group <u,a>S2r_2(G)/f2,_2(G), andtherefore their join contains the socle of this group. Let <c> = S2,(<a>). Then u2 andc2 have order 2 modulo fl,-2(G) and it follows that

(72) <u2,c2> < <ui,u2>Qr-2(G)

By (23), N° = Q° = Q<u1>, Q n <ul> = 1 and Q= N r) N°a N. As a projectiveimage of N, the subgroup N° is an M-group. By (48) there exists k E N such thatk = I (mod 4) and x = xk for all x c- Q. Since u1 E <u, a>S2r_2(G) _ <u, a>S2r_2(N)and, by (71), this group is abelian modulo Qr_2(N), we obtain [a-', u, '] E Qr_2(N) <CG(Q), by (70). Thus (49) implies that conjugation by u1 also raises elements to thepower k in Q°. Since [N/Q] is a chain of length r, Qf2r_1(Q°) lies in the maximalsubgroup Q<u2> of N containing Q and, since [u2, x] = [u, x]°[u, x] = [u, x]2 or 1for x e Q, it follows that (Q<u2>)' has order at most 2. By (50), x = uk for allx c- QS2r_1(Q°). Similarly, if we apply our results to a-' in place of a, we get that u2induces a power automorphism x -+ x' in Q' = N n N° ' and Q'° and therefore also inQ°S2r_1(Q). We claim that

(73) QQ,-1(Q°) = Q°n,-JQ);

then it will follow that, in particular,

(74) <u1,u2> induces a group of power automorphisms in Q.

To prove (73), first note that by (21) and (22), Q' = QQ° = Q,(N) where p'=Exp Q and s is the smallest integer such that N/Q,(N) is cyclic. If s < r - 1, thenQS2r_1(Q°) = QQ° = S2,(N) and, similarly, Q°S2r_1(Q) = Q'°Q = S25(N). So supposethat s = r, that is ExpQ = 2'. Since N' is cyclic, IN/S2r_1(N)J < 4, by (57); on theother hand, u and w are independent elements of order 2' and hence IN/Qr_1(N)I = 4and Q = Q1 x <w> where Exp Q1 < 2'. So if v: N - N is the homomorphismdefined by x" = x2" ' for x e N, then by (l lc), N' = <e1,e2>. Since QQ° = Nand [N/Q] is a chain of length r, there exists x c- Q° such that Q<x> = N andQn <x> = 1. As (Qf2,-1(Q°))°=Q <w>"=<e2e1>, we have Q<x2><QS2,-1(Q°)<-v-'(<e2e1>)< N. But Q<x2> is a maximal subgroup of N and it follows thatQ"r_1(Q°) = v-'(<e2e1>). The same argument with Q, Q° replaced by Q°, Qshows that Q°S2r_1(Q) = v '((Q'°)"). Now Q n <e1,e2> = <e2> since, by (60),<e2e1>° = <e2> < Na n N = Q and <eleo>° = (e1> N° '. Therefore V<N'n Q = <e2> and, since Exp Q' = 2', we obtain 0'= <e2>. Thus (0°)" = <e2>° _(e2e1> and so Q°S2r_1(Q) = v-'(<e2e1>) = QK2,-i(Q°) This proves (73) and (74).

By (74) and (70), every subgroup of Q is normalized by <u1, u2 >f2,_ 2(N). Nowf)r-2(<a>) = <c4> and, by (72), <u1, u2>Q,-2(G) = <ui, u2>Q,-2(N)<c4> contains c2;

therefore c4 E'b(<c2)) lies in the Frattini subgroup of this group. Thus<u1,u2)S2r_2(G) = <u1 ,u2)Qr_2(N) and now (72) implies that u2 normalizes everysubgroup of Q. Thus u2 induces a universal power automorphism in the abeliangroup Q; since [u2, w] e N' n <w> = I and o(w) = Exp Q, u2 centralizes Q andQ<u2) is abelian. Again by (56), N' is elementary abelian, a final contradiction.

We come now to the main result of this section.

5.5.8 Theorem (Busetto and Napolitani [1991]). Suppose that Hypothesis 5.5.1 issatisfied and p = 2. Then N is an M-group and I N' < 2.

Proof. We use induction on N and keep the notation introduced above. By (5) and(6), N is abelian if Exp N = 2' < 4 and cyclic if m < 1. Thus we may assume thatr > 3 and m > 2. As in the proof of Menegazzo's theorem we apply the inductionassumption to various projectivities. First of all, by (3), cP induces a projectivity inG/S2(G) satisfying Hypothesis 5.5.1 and hence N'f2(G)/f2(G)J < 2; it follows thatN' < f22(G). Now we apply the induction assumption to the projectivity induced bya in G/K and obtain that I N'K/K I < 2. By (1 I c) applied to G/K, N/K contains onlyone a-invariant subgroup of order 2 and since, by (19), K is cyclic and contains e, ,this subgroup is <e2)K/K. Thus N' < <e2)K n 02(G) = <e2)S22(K). By 5.5.6 and(20), both factors are central in N and hence

(75) N' < (e2> x S22(K) < Z(N).

Again let S2,(<a>) = <c>. Since N' < Q2(N), it follows that NS22(G)/02(G) is abelian;therefore by (3) and 5.5.3 there exists k e N such that c induces the power k inNS22(G)/S22(G). By 5.5.6, S2(G) < Z(S2,(G)) and S22(G)/f2(G) < Z(Q,(G)/52(G)) andhence (52) implies that c4 induces the power k4 in NS22(G). In particular,

(76) c4 acts as a power automorphism on N.

We make another remark that will be used later. Suppose that z e N such thato(z) = 4 and z° = z-'. Then (az)2 = a2 so that <az) and <a> are cyclic subgroupsof <a, z> generating this group and intersecting in <a2>. Their images under tpare cyclic groups generating <a, z)`° and intersecting in <b2 > and it follows that<b2) < Z(<a, z)'°). In particular, b2 centralizes <z)11 and therefore every conjugate of<z)`° in N<b2) is contained in N. This shows that the normal closure of <z>° inN<b2) is contained in N and hence <z>4' < L`° if L" = Nn.<n:,; thus z E L. SinceL < N n N`°''"° R, (63) shows that e2e, L. Finally note that if N' is cyclic andN' K, then by (1 ld), e, e N' and hence N' = <z> has order 4; furthermore, by (20),C,(a) = K and therefore z" = z-'. Thus we have shown:

(77) Let M = N<a2> and L = NM. If z e N such that o(z) = 4 and z° = Z_', thenz e L; in particular, N' < L if N' is cyclic and not contained in K. Furthermoree2e, 0 L.

Now let Q, R, u, w be as in (65) and (66). Then Q' < N' r) Q < (<e2> x S22(K)) n Q =<e2e,), by (63). Since w2' 1 = e2e,, we see that Q' < <w> and hence <w) a Q andQ n R 2t Q/<w> is abelian. Thus

(78) Q' < <e2e1>, Q = <w>(Q n R), <w> n (Q n R) = 1, <w> a Q and Q n R isabelian.

Since Q = N n N° and R = N n N°2, it follows that Q n R = N n N' n N°2 <N° n N°2 = Q° and IQ°: Q n RI = IQ: Q n RI = I<w)I = 25. Now <w> Q, so <w>°is permutable in Q° and since, by (60), <e2e1)° = <e2eo), we have 52((w)°) N.Thus (Q n R) n (w>° = I and so Q' _ (Q n R)<w> It follows that N n Q° =(Q n R)(N n (w)°) = Q n R < Q° and hence for every X < Q n R, there resultsX<w)° n (Q n R) = X(<w>° n (Q n R)) = X < X<w>°. So we have shown:

(79) Let <w>° _ <w1>. Then Q° = (Q n R)(w1>, (Q n R) n <w1> = 1, Q n R a Q<w1> is permutable in Q° and induces a group of power automorphisms in Q n R.

By (7),(tiawu2...1,2' )21 I = w2a 1 u2r 1`2. I = e2e1e1eo =

e2eo = wia ' and

(80) w1 =wu2.,c2' (mod 525-1(G))

Now we distinguish the cases s = r, s = r - 1, s < r - 2 and split the first of theseinto two.

Case 1: s = r, Exp(Q n R) = 2'.

In this case we show that N is even abelian. Let x a Q n R of order 2'. Then by (65)and (66), Ur_1(N) > S2(<x>) x S2(<w>) x S2() and hence by (7), IN/52,_1(N)I > 8.By induction, however, N/52(N) N52(G)/52(G) is an M-group with cyclic commuta-tor subgroup; it follows from (57) that N/S2(N) is abelian. Thus N' < 52(N) and (75)yields that

(81) N' S <e1,e2>.

If IN/f2r_1(N)I > 16, then IN/KS2r_1(N)I > 8; again by induction, H/K is anM-group with cyclic commutator subgroup and hence is abelian. ThenN' < <e1,e2> n K = <e1> and, by (25), N is an M-group. But then again (57) yieldsthat N is abelian. Thus we are left with the case that IN;S2r_1(N) = 8 in which

(82) N = <x, w, u>Q,-1(N).

Now (65) and (66) show that a normalizes <u)52r_1(N) and <w,u)52r_,(N).Thus, applying (11c) to G/S2r_,(G), we see that <u)Or_1(G)/S2,_1(G) and

u)S2r_1(G),%S2r_1(G) are the only normal subgroups of order 2 and 4, respectively,of G/52,_1(G) contained in the elementary abelian group NS2r_1(G),/f2,_1(G) of order8; in particular, ((x)52r_1(G))° = NS2r_1(G). Since N = (Q n R)<w>, (7) showsthat <x>92,-, (N) < (Q n R)<w2><u2> so that ((Q n R)<w2><u2>)° = N. Now N' <Z(N) n 52(N) implies that u2, w2 a Z(N). Therefore (Q n R)<w2>(u2> is abelian andevery conjugate of it contains and hence centralizes the normal subgroup S2r_,(N).It follows that

(83) fl,-,(N) < Z(N).

By (65) and (66), u° - u and YCa - wu (mod flr_1(N)) and by (I le) applied toG;52(G) there exists z a N such that z° = zw (mod Q,-, (N)). Then z°2 - (zw)° = zu

and w°2 _ (wu)° - w (mod fl,-, (N)); since N's <e1, e2> < CN(a2) and N' < S2,_, (N) <Z(N), it follows that [z, w] = [z, w]°Z = [z°Z, w'2] = [zu, w] = [z, w] [u, w]. Thus[u, w] = I and <w> g N since <w> a Q and N = Q. Therefore [w, z] lies in<w> n <e,, e2) = <w2' '> and hence wZ = w'+e2' where a E {0, 11. Since w° _ wu(mod f2r_1(N)), we see that H = <w, u)S2r_1(N) <w, w°)S2,_, (N) is an abeliannormal subgroup of G and has index 2 in N; therefore it contains [a-', z-']. By (49),(w°)` = (w°)1+f2r '; by (83), h= = h = h1 '2r ' for all h E Sr_1(N) and then also for allh e H. But H<z> = N and Iwasawa's theorem implies that N is an M-group. By (6),N and its projective image N° are M*-groups. By (23), N° = Q<u)° andQ n (u)° = l; furthermore Q = N n N° a N°. So 2.3.14 shows that every subgroupof Q is normal in N°; in particular, Q n R 9 Q and by (78), Q = (Q n R) x <w> isabelian. By (21) and (22), QQ° = N and hence Q n Q° < Z(N). Since [N/Q] is a chainand IN/s2r_1(N)I = 23, there exists an element of order 2' in Q n Q° and, by 2.3.16, Nis abelian. This finishes Case 1.

Case 2: s = r, Exp(Q n R) < 2'.

By (65), N1 = Sr_1(N) is an a-invariant subgroup of G so that, by (2),G1 = N,<a>, N1 and the projectivity induced by cp in G1 satisfy Hypothesis 5.5.1. Byinduction, Ni is an M-group and IN1 'I < 2. Since N, is a-invariant, (I lc) yields thatN1' < <e1> < . By (23) applied to N1 and Q1 = N1 n N', we get that N1 = Q1,Q, n = I and N; = Q1<u)°, Q1 n ° = 1. Since IN1 : S2,_1(N1)I = 2, it followsthat S2r_,(N) = S2,_,(N,) = Q1<u2>. Furthermore Q1 n N,' = I implies thatQ1 _- Q1 N1'/N1 is abelian, and IN; I < 2 yields that u2 E Z(N,). Thus

(84) S2r_1(N) = Q1<u2> is abelian.

Let ° = (u1). Then by (48), u1 induces a power automorphism in Q1 a Ni.From <e1> = <e1eo>, there follows u;' (uc)2' ', and hence by (7), uc = u,y forsome y e Q ,_,(G). By 5.5.3 applied to G* = S2r_,(N)<a> = S2r_1(G)<a>, elements ofS2,_1(G) induce power automorphisms in S2,_1(N), and it follows that uc = my in-duces a power automorphism in Q1. By (65), [u,a] E S2r_,(N) and [c,a] = 1. Thus[a-', (uc)-' ] E Sr_, (N) < CG(Q,) and hence by (49), uc induces in Q the same poweras in Q1. Since s = r, we see that Ni/flr_2(Nl) is not cyclic; hence (21) and (22)applied to G1 yield that Q1Qi = nr-i(Ni) = or-1(N) and this group is abelian. So ucinduces a power automorphism in or-I(N). By (79), w1 induces a power automor-phism in Q n R and, by (80), w1 = wucz for some z E or-1(G). Then Q n R < Qr_1(N)since Exp(Q n R) < 2'. Thus w1, uc, z induce power automorphisms in Q n R andhence so does w. In particular, (Q n R)'° = Q n R and (78) shows that

(85) Q is abelian.

Since S2(<w' >) = <e2e1>° = <e2> and Q n (e2> = 1, it follows that N = Q _Q<w°>, by (23). Similarly, by (63), (Q n Q°) n <eo, e,, e2> = 1 and sine Q n Q° ispermutable in Q with [Q/Q n Q°] a chain, (9) shows that Q = (Q n Q°)<w>; thusN = (Q n Q°)<w> <w°). Now Q and Q° are abelian and QQ° > Q<w°> = N so thatQ n Q° < Z(N). By (75), N' < Z(N) and it follows that

(86) N' = ([w, w°]> is cyclic of order at most 4.

Since r = s, the group N/Qr_2(N) is not cyclic, and therefore it remains to be shownthat N is an M-group; then the assumptions of 5.5.7(a) will be satisfied and it willfollow that I N'I < 2.

For this, suppose first that N' < K. Since Q n K = 1, we see that N/K is abelianof exponent 2r, and by 5.5.3 there exists k c N such that k - I (mod 4) and x" = xk(mod K) for all x e N. Now Q:9 N° = Q(u1> implies that x-'x" E Q n K = I forx e Q. Thus x = xk for all x e Q and, by 2.3.4, N° is an M-group. Thus N is anM-group, as desired.

Now suppose that N' K. Let M = N(a2> and L = N. By 5.4.6, cp induces aprojectivity from M/L to M/L for which Hypothesis 5.5.1 is satisfied. By (77), N' < Land e2e1 0 L. Thus <w> n L = I and hence NIL is abelian of exponent 2r. By 5.5.3applied to M/L, the element c induces a power h - 1 (mod 4) in NIL and hence, by(51), also in N/LG. Since u1 E S2r(G) = N(c> and N/LG is abelian, u1 induces a powerk - 1 (mod4) in N/LG. Now e2e1 0 L together with (1 lc) implies that f2(LG) < (el)so that Q n LG = 1. Then again Q a N° = Q(u1> implies that x-kx E Q n LG = 1for all x e Q and that NQ is an M-group. Thus also N is an M-group and this finishesCase 2.

Case 3: s = r - 1.

We observe that N/S2r_2(N) is not cyclic and, by (67), we may assume that

(87) u e R so that R = (Q n R)(u> and N = R(w> = (Q n R)(u>(w>.

Since N = Q(u>, we have S2r_1(N) = Q(u2> and the induction assumption yieldsthat

(88) K2,-,(N) is an M-group and f2r_1(N)' < <e1>,

by (1lc). Since Q n (e1> = 1, it follows that

(89) Q is abelian.

We want to show next that R' < (e1>. If Exp(Q n R) = 2r-1, then K2,-, (N) has threeindependent elements of order 2r_' and cyclic commutator subgroup and therefore,by (57), it is abelian. Then N = f2,_1(N) and (75) imply that N' is elementaryabelian and N' < (e1,e2>. Thus R' < (e1,e2> n R = <e1>, by (63). Now supposethat Exp(Q n R) < 2r_' and consider S as defined in (63). Then (66) implies thatQ n R < S2r_2(Q) = (Q n S)(w2> and by (67) there exists u' E S such thatS = (Q n S)(u'> and Q(u'> = N. Now R = (Q n R)(u>, the commutativity of Q andN' < Z(N) imply that

R' _ [Q n R, (u>] < [(Q n S)(w2>, Q(u'>]

<- [Q n S, <u'>] [(w2>, (u'>] <- S'(ei>,

since [w2, u'] = [w, u']2 E 5(N') = <e, >, by (75). By (63) and (64), S' < (e2) K n S = Kand so R' < R n K = (e1>. Thus in any case,

(90) R' < (e1>.

We wish to apply Lemma 5.5.7(a) and therefore have to show that N' is cyclic andN is an M-group. Now N = (Q n R)<w>, u e R, Q abelian and N'< Z(N) implythat

N' = [Q n R, <w>] [Q n R, ] [<w>, ] < Q'R'<[w, u]) < <e1 ><[w, u]>.

Suppose, for a contradiction, that N' is not cyclic. Then e1 0 <[w, u]> and it followsfrom (75) that o([w, u]) = 2 and [w, u] = e2e1 or [w, u] = e2; in any case, N' < f2(N).Thus Nn(G)/fl(G) is abelian and, by 5.5.3, c acts as a power automorphism onNO(G)/fl(G) and hence also on N/S2(N). By (52) and 5.5.6, c2 acts as a power auto-morphism on N. Since N = R<w> and R n <w> = 1, it follows that 0r_2(N) < R(w2>.Clearly [w2, u] = [w, u]2 = 1 and R'< <e1) < implies !g R; thus is nor-malized by 0r-2(N)<c'> = flr-2(G)

Now assume first that [w,u] = e2e1; the other case will be similar. Then[w, u] a <w> and hence <w> g Q<u) = N since Q is abelian. By (67c), R =N n N' 2 g N°2 = R<w>°2 and R n <w>°2 = 1; let

<w>°2 = <w2>. Since <w> a N, weobtain that <w2> is permutable in N2, and hence (u><w2) n R = (<w2> n R) _ g (u)<w2>. Because of w2r-2 = e2eleo = (wc2)2r-2, we have w2 = wc2x for somex E S2r-2(G), by (7). Now w2, c2 and x normalize , as we have seen, and hence sodoes w. But this yields e2e1 = [w, u] E <u), a contradiction.

Now assume that [w, u] = e2. Then we use wu2 instead of w for our argument. As[wu2,u] = [w,u] = e2 = (wu2)2r-2, the subgroup <wu2> is normalized by u; further-more it is centralized by Q since Q is abelian and u2 a Z(N). Thus (wu2 > g N. By(67c),

N2

= R<wu2 >°2 and hence <v> = <wu2>Q2 normalizes . Finally, v2' 2 =

e2eo = (wu2c2)2r-2 implies that v = wu2c2y for some y E fl,-2(G). Now v, c2, y nor-malize and hence so does wu2. But [wu2, u] = e2 0 (u>, a contradiction. Thus

(91) N' is cyclic

and it remains to be shown that N is an M-group. To prove this, by (47), it willsuffice to show that

(92) N/<e1> is an M-group;

note that, by (88), 0,-, (N) is an M-group and U,-, (N) = <e1> is cyclic.To prove (92), suppose first that N' < K and again let <w>°2 = <w2>. Since

Q n K = 1 and s = r - 1, Exp(N/K) >- 2r-'; on the other hand, U,-, (N) _ <e1> < K.Thus N/K is an abelian group of exponent 2r-1 and, by 5.5.3, w2 e 0r_1(G) inducesa power automorphism x --+ xk, k = I (mod 4), in N/K. By (67c), R = N n Nat gNQ2 = R<w2> and hence x-'x'2 e R n K = (e1>, by (64), for every x e R. SinceR/R n K RK/K is abelian, it follows from 2.3.4 that R<w2>/<e,> = N62/(e,> is anM-group. By (61), (e1>°2 = <e1> so that N/<e,>, as a projective image of N°2/(e1>,is also an M-group.

Now suppose that N' K. Here, as in Case 2, we first study the projectivityinduced by T in M = N<a2> and then use aaa-' instead of a2; let (w3> = <w>°°°-`.By (77), N' < L and e2e, L for L = NM. Since U,-,(N) = (e1> < N' < L, we haveExp(N/L) < 2'-'; on the other hand, <w> n L = I and hence NIL is abelian ofexponent 2r-1. By 5.5.3 applied to M/L, the element c2 induces a power h = 1

(mod 4) in NIL and hence, by (51), also in N/LG. Since w3 E Q,-, (G) = N<c2) andN/LG is abelian, w3 also induces a power k = 1 (mod 4) in N/LG.

Consider S = N n N°°° '. By (63), e2 S and e2 L since e, e L and e2e, L.Therefore by (l lc), 1(SG) = <e,) = S2(LG) so that S. and LG are cyclic. Since N' < Land, by (64), K< S, we see that C 2(LG) = N' and S22(SG) = Q2(K). So N' K impliesthat N' S and hence S n LG = S n N' = <e,). By (67d), S = N n N°°°-' g N°°° 'S<w3) and hence x-'x W3 E S n L. = <e,) for every x c- S. Since S' < S n N' = <e, ),it follows that S/<e,) is abelian and, by 2.3.4, S<w3)/<e,) = N' aa- '/<e,) is an M-group. By (62), <e, >,a,-' = <e, ), so that N/<e, ), as a projective image of N°°°-'/<e, ),is also an M-group. This finishes the proof of (92) and Case 3.

Case 4: s<r-2.

Then N = Q<u) where Q:5; Qr_2(N) and hence U,-2(N) _ <u2"_2) is cyclic. Letur-2(N) = <z) and suppose first that za = z. Then z e CN(a) = K, by (20), and henceU,-2(N) = S22(K); thus Exp(N/K) < 2r-2. By (76), c4 induces a power automorphismin N and hence also in N/K. Therefore (24) applied to G/K yields that N/K is abelianand (75) implies that N' < Q2(K) = Jr_2(N). By induction, (I,-,(N) is an M-groupand so (47) shows that N is also an M-group. Since N' < K is cyclic and, by (20),ur-2(N) < Z(G), all the assumptions of Lemma 5.5.7(b) are satisfied and it followsthat J N' I < 2, as desired.

Now suppose that z" = z-1. Then by (77), <z) < L and e2e, 0 L if M = N<a2)and L = NM. Thus 'Z5'r_2(N) < L and hence Exp(N/L) < 2r-2. Again c4 induces apower automorphism in N and hence also in NIL; therefore (24) applied to theprojectivity induced by q in M/L yields that NIL is abelian. Now z2 = e, and hence(ze2)° = ze,e2e, = zee. As CN(a) = K, it follows that <ze2) = Q2(K) and so N' <(<e2) x <ze2>) n L = <z), since e2e, 0 L. Thus

(93) N' < <u2r-2) _ 't5r_2(N); furthermore e2 E K and n K = <e,)

since <z) = Q2(<u)) and <ze2) = (12(K). Again by induction, 0,_1(N) is an M-group and since N/t r_2(N) is abelian and U,_2(N) is cyclic, N is an M-group.It remains to be shown that IN'! < 2. Unfortunately, since U,-2(N) $ Z(G), wecannot apply Lemma 5.5.7 in this case. However, we can proceed as follows. SinceN' < Z(N) n

<u2'-,

) and Q n <u) = 1, the group Q is abelian, and [u2, x] _[u, x]2 E S2(<u)) for all x E N. Therefore u4 E Z(N) and

(94) S2r_2(N) = Q<u4) and Q<u2)/<e,) are abelian.

Since G/S2r_2(N) = N<a)/S2,_2(N) and N/Q,_2(N) is cyclic of order 4, (55) showsthat a4 centralizes N/S2r_2(N). By (5), S2r(<a)) < <a4) and it follows thatS2r(G)/S2r_2(N) is abelian. By (23),

-0r(G) = <u)<u)QQr-2(N) = <u)Qr(<a))Qr-2(N) = <u)QQr(<a))Qr-2(N)

and hence there exist v c <u)° and c e 0,(<a)) such that u = vc-1 (mod Q,-2(N)).Since S2r(G)/S2,((a))Qr_2(N) is cyclic, <v) = <u)° and, similarly, <c) = S2,(<a)).Finally, since Q,(G)/Qr_2(N) is abelian, we have

(95) v - uc and v2 = u2c2 (mod S2,-2(N)).

Now N/S22(N) is abelian and hence by 5.5.3 there exists 2 E N such that 2 = 1(mod 4) and c induces the power automorphism x -> x` in N/922(N). Since N central-izes N/Q,(N), this automorphism is also induced by v in N,/S22(N). By (52),

(96) v2 and c2 induce the power 22 in N/S2(N).

Since Q a N° = Q<v) and N° is an M-group, v induces a power u in Q, by (48). Thenv2 induces p2 in Q and, by (94) and (95), u2c2 induces the same power in Q. SinceQr-1(N)1fI.-2(N)I = 2, we have [u2, a] E fl,-,(N) and so [a-1 (u2c2)-1] E S2r-2(N) <CO(Q); by (49), u2c2 induces the power p2 in Q° and then also in the abelian groupQQ° = QJN), by (21) and (22). By (96), c2 induces the power 22 in N,12(N) and then,by (52), also in <u2 ). Therefore u2c2 induces in S2,(<u)) = <u2) n U(IV) the powers22 and p2 and it follows that 22 = p2 (mod 25). So u2c2 induces the power 22 in Qand, since Q<u2 )/ <e,) is abelian, u2c2 and c2 induce the same automorphism inQ<e 1)/<e, ); thus

(97) c2 induces the power 22 in Q<e,)/<e,).

We want to show, finally, that c2 induces a power automorphism in N/K. By (96),U' = for some e c- 92(N). Since the commutator subgroup of N/<e,, e> has orderat most 2, c2 induces the power 22 in Q<u)/<e,,e) = by (50). And by (51),c2 induces this power in N/<el,e)G. So if <e1,e)0 = <e1>, we are done. And if<el,e>G 0 <e1>, then <e1,e) = <e1,e2), by (l lc), and (93) shows that K ><e1, e) and o(uK) = 2r-1. Therefore e = up2r 2 (mod K) for some p e 71 and, sinceExpQ < 2r-2, c2 induces the power 22 + p2r-2 in QK/K and in <u)K/K. Nowo(z) = 4 implies that r > 4; hence 22 + p2r-2 = I (mod4) and, since I(N/K)'I < 2,(50) implies that c2 also induces the power 22 + p2r-2 in Q<u)/K = N/K. Thus inany case, c2 induces a power automorphism in N/K. Applying (24) to the projectivityo from G/K to G,/K°, we conclude that N/K is abelian; finally, (93) shows thatN' < n K = <e1>. Thus IN') < 2, as desired. This finishes Case 4 and the proofof Theorem 5.5.8.

By 5.5.2 and 5.5.8, the nilpotency class c(N) < 2 under Hypothesis 5.5.1. A boundfor c(N) is not known; in Example 5.5.5 and in the known examples for p odd,c(N) < 2. The projectivities between metacyclic groups show that neither c(G) norc(G) is bounded. Since N is an M-group, N and N are metabelian. For the derivedlengths of G and G, we obtain the following final result; note that d(G) = 3 in Exam-ple 5.5.5.

5.5.9 Theorem (Busetto and Napolitani [1991]). Suppose that Hypothesis 5.5.1 issatisfied and let M be the maximal subgroup of G containing N. Then G is metabelianand G has derived length at most 3; more precisely, M`° is metahelian and G" is elemen-tary ahelian.

Proof: We first show that G is metabelian. This is clear if N is abelian, so assumethat N' A 1. Then by 5.5.2, p = 2; 5.5.8 and (l lc) yield that N is an M-groupand N' = <e1). Thus I.N/S2r_1(N)t < 4, by (57). Again let N = Q(u) as in (23).Since N' < S2(Z(N)) n and Q n = 1, the group Q is abelian and u2 E Z(N);

thus Q<u2> is abelian. From G = N<a> follows G' = [N, <a>] N'; and [e2, a] = e1shows that N' < [N, <a>]. So if IN/f2r_,(N)I = 2, then G' = [N, <a>] < S2,_1(N) andS2,_1(N) = Q<u2> is abelian; thus G is metabelian. It remains to consider the casethat IN/f2r_,(N)j = 4. Then by (65), f2,_1(N) a G of index 2 in N so that G' =[N, <a>] < f2r_1(N). By (66), Q<u2> = Q,_,(N)<w> is not normal in G; notethat ExpQ = 2' since N/1 r_, (N)j = 4. Since Q<u2> is abelian, it follows thatQ.-1(N) = (Q<u2>)G < Z((Q<u2>)G) = Z(N). Therefore S2r_,(N) is abelian and Gis metabelian.

The assertions on G both follow from 5.5.3. For, by 5.5.2 and 5.5.8, N' S <e1>.Thus NQ(G)/Q(G) is abelian and 5.5.3 applied to G/Q(G) yields that G;f2(G) is meta-belian. So G" < f2(G), and this group is elementary abelian. By (11), f b2 = f1 andsince M = N<b2>, it follows that <f1>M < N and so f, e NM. This shows that<e1> < NM and hence N/NM is abelian. Therefore 5.5.3 applied to the projectivityinduced by p in M/NM yields that M/NM is metabelian. Thus M" < N and M" a Gsince k is a maximal subgroup of G. So M" < N- = I and M is metabelian.

Using 5.4.6 and 5.1.6, it is not difficult to extend the results on N/NG- and N/Neproved in this section to projectivities between arbitrary finite groups. However, weleave this and an application to projectivities between finite soluble groups as anexercise for the reader (Exercises 4-6) since we shall extend these results to arbitrarygroups in § 6.6.

Exercises

I . Let E = <eo, ... , em> and F = <f .... , f,,,> be elementary abelian p-groups oforder p'";1 and let n e N such that pn-2 > m and 2n-2 > m if p = 2. Define G =E<a> and G = F by aP"-' = eo, a-leoa = eo, a-'e1a = e1, a-1e;a = ejej_, fori = 2, ... , m and bP"-' = fo, b-'fob = fo, b-'f b = f f_1 for i = 1, ... , m. Constructa projectivity from G to G mapping the normal subgroup N = <e,,..., e,n) of Gto M = <f 1 . ... . J,> and show that ME; = 1.

2. (Menegazzo [1978]) Let p > 2 and E = <e0,...,ep> be an elementary abelianp-group of order pp+'(a) Construct extensions H* = E<a> satisfying a"2 = eo, a-'eoa = eo, a-'ela =

e1, a-'e,a = e;ei_1 for i = 2, ..., p and G* = H* where [H*,u] = I anduP = e-'1

(b) Show that there exists a projectivity from G* to the group G of Exercise 5.2.1mapping the abelian normal subgroup N = <u, e2, ... , eP> of G* onto thenonabelian, core-free subgroup M = (y, a2, ..., a p> constructed there.

3. (Busetto and Menegazzo [1985]) If Hypothesis 5.5.1 is satisfied and p = 2, showthat f2(G) < Z(S2r(G)).

4. Let N be a normal subgroup of an arbitrary finite group G and cp a projectivityfrom G to a group G. Show that(a) N/ Nc and N/Nd are metabelian, and(b) N/Nd is nilpotent of class at most 2 with commutator subgroup of exponent

at most 2; in particular, N/NG is abelian if 1 GI is odd.

5. Let N be a normal subgroup of the finite group G such that G/N is soluble ofderived length n c- N. If cp is a projectivity from G to a group G, show that6(3n) < N.

6. If G is a finite soluble group of derived length d(G) = n and cp is a projectivityfrom G to a group G, show that d(G) < 3n - 1.

5.6 Normalizers, centralizers and conjugacy classes

Apart from the index, studied in §4.2, there are other important interrelations be-tween subgroups that may be preserved by a projectivity. We briefly treat three ofthem and present the classical examples of particularly decent projectivities betweennonisomorphic groups.

A projectivity cp from a group G to a group G is called centralizer preserving ifCG(H)`° = C6(H`°) for all H < G, normalizer preserving if NG(H)`° = NZ(H`°) for allH < G, and conjugacy preserving if for all X, Y < W < G, X is conjugate to Y in Wif and only if X`° is conjugate to Y' in We'. These properties are not independent. Forexample, if cp is conjugacy preserving and X < G, then X is the only subgroup ofW = NG(X) conjugate to X in W; therefore the same holds for X10 in W10 and henceV' W. Thus NG(X)° < NZ;(X`0) and, since cp-' like cp is conjugacy preserving, weobtain the other inclusion. We have shown:

5.6.1 Theorem. Every conjugacy preserving projectivity is normalizer preserving.

Normalizer preserving projectivities

It is less obvious that every normalizer preserving projectivity is centralizer pre-serving, as we shall see for arbitrary (possibly infinite) groups. For this we givecharacterizations of both properties via images of commutator subgroups.

5.6.2 Theorem (Barnes and Wall [1964]). The following properties of the projec-tiuity (p from the group G to the group G are equivalent:

(a) cp is normalizer preserving,(b) (H')`° - (H°)' for every subgroup H of G,(c) [H, K]`0 = [H`0, K`0] for all H, K < G.

Proof. Suppose first that (a) holds and let H< G. If H' < M < H, then N0(M) > Hand therefore NZ;(M') = N0(M)`° > H. Thus (H')° a H° and H`°/(H')' is abelian orhamiltonian. Since every hamiltonian group contains a subgroup isomorphic to Q8and cp induces a projectivity from the abelian group H/H' to H'/(H')', it follows thatH`°,,'(H')° is abelian. Thus (H`°)' < (H')°, and this result applied to H° and the nor-malizer preserving projectivity cp-' yields the other inclusion; so (b) holds.

Now suppose that (b) is satisfied. If X = <x> and Y = (y) are cyclic subgroupsof a group, then [X, Y] a <X, Y> and the factor group <X, Y>/[X, Y] is generated

5.6 Normalizers, centralizers and conjugacy classes 273

by x[X, Y] and y[X, Y]. These elements commute modulo [X, Y] and hence<X, Y>/[X, Y] is abelian. Thus <X, Y>' < [X, Y] and, since the other inclusion istrivial, <X, Y>' = [X, Y]. Therefore if H and K are cyclic subgroups of G, thenH`, and K`° also are cyclic and

[H, K]' = ((H u K)')4' = ((H u K)")'= (H4' u K4')' = [HP, K`°].

For arbitrary subgroups H, K of G, [H, K] = U { [<x>, <y>] I <x> < H, <y> < K}and this is mapped by cp to U { [<x>`°, <y>4'] I <x>`' < H`°, < y>4' < K4'} = [H`°, K`°].Thus (c) holds.

Finally, if (c) is satisfied and H < G, then [H`°, NC(H)"] = [H, NG(H)]`° < H" andhence No(H)4' < Nd(H`°). This result applied to H`° and q' yields the other inclu-sion. Thus (a) holds.

5.6.3 Theorem. The following properties of the projectivity tp from the group G to thegroup G are equivalent:

(a) cp is centralizer preserving,(b) Z(H)`° = Z(H4') for every subgroup H of G,(c) for all H < G, H' = 1 if and only if (H4')' = 1.

Proof. First of all, (a) implies that Z(H)`° = (H n Cc(H))4' = H° n CG(HP) = Z(H4')for all H < G. Thus (b) follows from (a) and, since H' = 1 if and only if H = Z(H), (b)clearly implies (c). Finally, if (c) holds, then for x e H and Y E CG(H), the subgroup<x, y> is abelian, as is <x, y>" = <x>4' u <y>4'. It follows that <y>" < Cd(H') andCc(H)4' < CZ;(H4'). This result applied to H`° and cp-' yields the other inclusion; thus(a) holds.

Now 5.6.3(c) follows from 5.6.2(b) and therefore we obtain the result announcedabove.

5.6.4 Corollary. Every normalizer preserving projectivity is centralizer preserving.

Of course, our theorems can also be used to prove that a given projectivity isnormalizer or centralizer preserving. The commutator subgroup of a group is thejoin of the commutator subgroups of all its 2-generator subgroups, and a projec-tivity maps 2-generator groups to 2-generator groups. Therefore 5.6.2(b) yields thefollowing useful criterion.

5.6.5 Remark. A projectivity cp from G to G is normalizer preserving if (H')4' = (H4')'for every 2-generator subgroup H of G.

We use this to generalize a result of Spring [1963] to locally finite groups.

5.6.6 Theorem. If G is a nonabelian locally finite p-group of exponent p, then everyprojectiuity of G is normalizer preserving.

Proof. Let (p be a projectivity from G to a group G. Since G is nonabelian, 2.2.7yields that G is a p-group. Thus if H is a 2-generator subgroup of G, then H and H4'are finite p-groups of exponent p. It follows that (H')4' = (H)4' = ?(H4') = (H4')'. By5.6.5, q is normalizer preserving.

We give examples of normalizer preserving projectivities between nonisomorphicfinite p-groups.

5.6.7 Example (Barnes and Wall [1964]). Let p be a prime, 4 < n where o(s,) = p2 and o(si) = p for i > 2; thus N is an abelianp-group of type (p2, p, ... , p) and order pn-1. For y E A = { 1, ... , p - 1 }, let s = s(y) bethe automorphism of N satisfying

(1) s; = sisi+, (1 5 i 5 n - 3), sn_2 = n-2SIP

and let G = N(s> be the semidirect product of N and the subgroup of Aut Ngenerated by s = s(y). By (2) of § 1.5, [s,°, s] = [s, , s]° = sZ = I so that if we define

k

k (k)

=sip and s, = I for m > n, an obvious induction yields that s; = fl s;+; forj=0

all i, k > 1. Since n < p, it follows that sP = 1. Thus I<s> = p and I G I = p". Since(s°,, s2, ... , sn-2 > < G' has index p2 in G,

(2) G' = <s,°,s2,...,Sn .2> _ ID(G).

Then K.(G) = [Ki_, (G), G] _ (Si, ... , Sn_2, s;) for i = 2, ..., n - 2 and (G) _(sc>. so that G is a p-group of maximal class. As n < p, the group G is regular (seeHuppert [1967], p. 322) and, since G' is elementary abelian, (xy)P = xPyP for allx, v c G. In particular, (ssl )P = s; P for all i e 7L and since G = (s, s, >, the maximalsubgroups of G are N and the groups

(3) M; = <Ss;,s2,...,sn_2> where 0 < i < p.

We show that all the groups (y e A) are lattice-isomorphic but, in general, notisomorphic. More precisely, we claim that for G = G =

G G if and only if there exists E 7L such that d - Sn-2y (mod p), and(b) there exists a bijective map a: G -* G inducing a normalizer preserving projec-

tivity from G to G such that a1H is an isomorphism for every proper subgroup H of G.

Proof. Let G = N<t> satisfy (1) with s and y replaced by t and S, respectively.(a) Suppose first that r: G -b G is an isomorphism. Since N is the only abelian

maximal subgroup of G and G, we have N` = N and hence si = s;' ... s,", whererl, * 0 (mod p); let t` = s s; E K,(G), a trivial induction yieldsthat fort<i<n-2,

Si = [si-1 , t]t = [si-1 , tt] = ES; 1 1 , s5] = S" 11, (mod Ki+1(G))

Thus s_2 = sn 2 (mod Kn_1(G)) and since s] E K.-1(G) < Z(G), we get

s7" = (Sl )ap = (SbP)t = [Sn-2, t] = [Sn-2, tt] _ 2R' SS] = (sl')5" zRt

Since o(s,) = p2 and q, * 0 (mod p), it follows that b -n zy (mod p), as desired.Conversely, suppose that there exists S E 7L such that S - c" zy (mod p). Then if wedefine s* = s5, s* = .s, and s* _ [.s* ,,s*] for 2 - Gn(b) = G.

(b) Since y, d E A = { 1,... , p - 11, there exists k c 1L such that b - ky (mod p). For0<i.<p,letM;_=<tS',82,.... Sn_2>anddefineM,=N= MP. By (3), the M,' andM; (0 x ... X <Sn-2> x <x >)<x> and MkA =(<s2> x ... X <Sn-2> X <YP>)<Y>- Since [si,.x] = si+, _ [si,y] for 2 < i < n - 3,[sn_z, x] = sj" = (xP)"' and also [sn_2,y] = SIP = s; ''P = (yP)PY there exists an iso-morphism vA: MA - MkA fixing all the si (2 . Clearly, we may extend v to an isomorphism vP: N - Ndefining s`r" = s; and, since [sn_2,t] = SIP = SI''P = IS,-2,S] there is an extension ofv to an isomorphism vo: Mo - Mo satisfying s = t. Now all the vA agree on 4)(G)and therefore the map a: G - G defined by g° = g if g c MA (0 < A < p) is well-defined, induces an isomorphism in every maximal subgroup of G and hence satisfies(5) and (6) of § 1.3. By Theorem 1.3.1, o- induces a projectivity cp from G to G. Since(G')'P = G' and a induces an isomorphism in every maximal subgroup of G, 5.6.2implies that cp is normalizer preserving.

The simplest examples of nonisomorphic groups obtained in this way are G4(1)and G,(;) of order p° (p > 5), where y is a nonquadratic residue (mod p). Barnes andWall give similar examples with 3 < p < n, and also of groups of exponent p (seeExercise 3).

Finally, we remark that the converses of 5.6.1 and 5.6.4 do not hold. For example,the nonabelian group of order pq with p and q primes has only centralizer preservingautoprojectivities and its singular autoprojectivities are not normalizer preserving.Examples of index and centralizer preserving projectivities that are not normali-zer preserving are given in Exercise 1. The Rottlander groups, which we shall shortlystudy, possess normalizer preserving projectivities that are not conjugacy preserving.

Groups with identical subgroup structures

There is no general theory of conjugacy preserving projectivities giving results thatgo beyond those known for arbitrary projectivities. One reason for this is probablythe many examples of projectivities between nonisomorphic groups which preservealmost every structural property. We say that two groups G and G have identicalsubgroup structures if there exists an index and conjugacy preserving projectivity (pfrom G to G such that H`° H for every proper subgroup H of G. The first exampleof nonisomorphic groups with identical subgroup structures was published by AdaRottlander in the earliest study of subgroup lattices of groups.

5.6.8 Example (Rottlander [1928]). Let q > 5 and p be primes such that ql p - 1and let r E N such that r * 1 (mod p) and r9 - I (mod p). For every A E A ={2,..., q - 1 }, we define the Rottlander group

GA = <x,y,alxv = y" = aQ = [x,y] = 1, a-'xa = xr a-iya = yr">

and claim that for A, It e A,(a) GA and G have identical subgroup structures, and(b) GA Gµ if and only if A = It or AM = 1 (mod q).

Thus we obtain (q - 1)/2 pairwise nonisomorphic groups with identical subgroupstructures. The smallest primes occuring are p = 11, q = 5 with I GAl = 605.

Proof. (a) Let N = <x, y> so that GA = N<a> and let G = N where xb = xrand yb = yr". Since A # 0, 1 (mod q), the element a induces different nontrivial powerautomorphisms in <x> and <y>. Thus <x> and <y> are the eigenspaces of theautomorphism induced by a in N and hence the type of this automorphism is (0; 1, 1).By 4.1.8 there exists a projectivity (p from G. to G satisfying N ° = N and <a>' =. Clearly, cp is index preserving and, since the maximal subgroups of GA differentfrom N are all nonabelian of order pq, &' ^" H for every proper subgroup H of GA.The set f) of subgroups of order p of N different from <x> and <y> falls into (p - 1)/qconjugacy classes of length q. For every X c -Q, the subgroup N is the only maximalsubgroup of G. containing X. Hence there exists an autoprojectivity p of G fixingall the Y E and permuting the subgroups in S2 in such a way that ji = cppmaps conjugacy classes of subgroups of order p to conjugacy classes. Since any twoSylow q-subgroups are conjugate in every subgroup containing them, and since twosubgroups of order pq of GA are conjugate if and only if they intersect in a subgroupof order p, it follows that 0 is conjugacy preserving. Thus G. and G, have identicalsubgroup structures.

(b) Let a: GA -+ G," be an isomorphism and a° = x'y'bk where i, j, k e N. Then ahas to map the normal subgroups <x>, <y> of order p of GA to the normal subgroups<x>, <y> of order p of G.. If <z> is one of these normal subgroups, z° = zs and(z°)6 = (z°)` (s, t e N), then (z°)s = (z')" = (z°)° = (z°) = (z°)bk

= (zQ),k and hences - tk (mod p). Therefore if <x>" _ <x> and <y>" = <y>, then r = r' (mod p) andrA = rNk (mod p); thus k - 1 (mod q) and µk - µ (mod q). Since i.,,u e {2, ... , q -1 },

it follows that d = M. If <x>" = <y> and <y>" = <x>, then r - rµk (mod p) andrA = rk (mod p); hence µk - 1 (mod q) and ; - k (mod q) so that µi! - 1 (mod q).Conversely, if ::µ - 1 (mod q), then y" = yr"" = yr and xb" = xr" so that y, x, bAsatisfy the defining relations of GA. It follows that G. ^- GA and (b) holds.

For any A e A, there exists a unique µ e A such that Aµ - 1 (mod q). If A # q - 1,then A2 # 1 (mod q); thus µ A and hence these i in pairs yield (q - 3)/2 non-isomorphic groups. Together with Gq_, we obtain (q - 1)/2 pairwise nonisomorphicRottlander groups of order p2q.

Other examples of nonisomorphic groups with identical subgroup structures weregiven by Yff and Holmes; those of Yff can also be found in an earlier paper byHonda on subgroup lattices of metacyclic groups.

5.6.9 Example. Let p and q be primes such that qI p - 1 and let(a) (Honda [ 1952], Yff [ 1967]) t e P such that t 0 p and q I t - 1, or(b) (Holmes [1976]) t = q2.

Let N = P x T with I P1 = p and I TI = t be the cyclic group of order pt. ThenAut N = A, x A2 where A, = N(T) ^- AutP and A2 = N(P) Aut T, andtherefore Aut N contains an elementary abelian subgroup Q = Q, X Q2 of order q2with Q; < A. and I Q;1 = q. For every subgroup R of order q of Q different from Q,and Q2, let GR = NR be the semidirect product of N and R; thus

GR = <x alxp' = aQ = 1, a-'xa = x')

for some r e 1l satisfying rQ - 1 (mod pt), r # I (mod p) and r # I (mod t). Then allthe GR have identical subgroup structures. For, if S is another subgroup of order qof Q different from Q, and Q2, then by 4.1.6 there exists a projectivity cp from GR toGS. This is clear if (a) holds; in case (b), we take P to be a normal Hall subgroup, thenonabelian groups TR and TS of order q3 and exponent q2 as complements andr the projectivity induced by an isomorphism a: TR -* TS satisfying Ta = T andR' = S. It is rather obvious that cp preserves indices, conjugacy and satisfies HO -_ Hfor every proper subgroup H of G.

An isomorphism v: GR -+ Gs maps the cyclic normal subgroup N of order pt of GRto the cyclic normal subgroup N of order pt of Gs and satisfies

(Xv)al =(Xa)v = (Xr)v = (Xv)r

so that a' induces the same power as a on N; it follows that S = R. Thus we obtainq - 1 pairwise nonisomorphic groups with identical subgroup structures. Thesmallest primes occurring are q = 3, p = 7, t = 13 with I GR I = 273 in case (a), andq = 3, p = 7 with I GR I = 189 in case (b).

The examples constructed in 5.6.8 and 5.6.9 can be generalized to the situationwhere q does not divide p - 1, while those of Rottlander and Honda-Yff extendalso to the case where q is not a prime. For this the nonabelian groups of orderpq must be replaced by groups of order p'q in which the cyclic group of orderq operates irreducibly on an elementary abelian group of order p'"; for q E P,these groups have been described in 4.1.7(b). In general, however, the projectivitiesbetween these groups are not conjugacy preserving (Holmes [1984], Schmidt[1982]).

There are many other examples of nonisomorphic but lattice-isomorphic groups.We mention those due to Niegel [1973a], [1973b] of groups whose orders containat most three prime divisors or are square-free, Daues and Heineken [1975] ofgroups of order p6, Caranti [1979] of p-groups of maximal class, Holmes [1971] ofdirect products of Rottlander groups with certain other groups, Schmidt [1981] ofdirect products of two isomorphic Rottlander groups, and Schmidt [1982] of groupswith elementary abelian normal Hall subgroups.

Projective images of conjugacy classes

More important than the investigation of conjugacy preserving projectivities, whichare relatively rare, is the question when a single conjugacy class A of G is mapped toa conjugacy class of G by a projectivity (p from G to G. For then, not only G but alsoG operates on A via the map ro: G Sym A defined by

(4) H"iX) = HP" ' for H e A;

we shall use this fact quite systematically in § 7.7. Of course, we may assume thatJAI > 2 since the operation described above becomes trivial if JAI = 1. The firstobservation one makes is that in many groups there are conjugacy classes of rela-tively small subgroups which are not mapped to a conjugacy class by some projec-tivity. For example, the conjugacy classes of subgroups of order p in the Rottlandergroups, in the groups of Example 4.1.7(b), or in Exercise 5 have this property.Furthermore, as in the case of projective images of normal subgroups, P-groups willcause difficulties. If we exclude these and restrict our attention to maximal sub-groups, the situation is better. To show this, we first prove a useful general propertyof the autoprojectivities induced in G by the elements of G.

5.6.10 Lemma. Let cp he a projectivity from the finite group G to the group G, letx c G and r(x) he the autoprojectivity of G defined by HT(XJ = Hl"" ' for all H< G. IfG is P-indecomposable, then r(x) is index preserving.

Proof. Suppose, for a contradiction, that r(x) is not index preserving. By 4.2.5, r(x)maps Sylow subgroups to Sylow subgroups since G is P-indecomposable. Therefore4.2.1 implies that there exist primes p q, P E Sylp(G) and Q E SyIQ(G) such thatPT'x) = Q. Thus P"X = Q" and since IP"I = IP"I = IQ"I, it follows that cp is singularat p or at q. If p is singular at p, then by 4.2.6 there exists a normal p-complement Nof G such that N" a G. But then N"""-' = N, hence (PN)T'X' = Gand P"X' n N = (P n N)T'") = I so that IPT(x) I = IG : NI = IPI, a contradiction. If cp issingular at q, we argue in exactly the same way using Q and x"' instead of P and x.

If G is P-decomposable, r(x) in general is not index preserving: consider G = S3 =G, an autoprojectivity cp of G satisfying A3 A3 and x = (123). It is also clear thatfor P-decomposable groups, conjugacy classes A of maximal subgroups in generalare not mapped to conjugacy classes by projectivities cp. However, we show that forfinite soluble groups, this is the only situation in which All is not a conjugacy class.

5.6.11 Theorem (Schmidt [1977b]). Let G be a finite soluble group, A a conjugacyclass of maximal subgroups of G such that Al I> 2, and cp a projectivity from G to agroup G. If there is an element g e G such that (011)9 A", then there are subgroups Sand T o f G such that G = S x T, (ISI, ITI) = 1, S is a P-group and T< n H.

Proof. We use induction on IGI and first note that it suffices to show that G is P-decomposable. For then G = S x T where (ISI, ITI) = I and S is a P-group, so that

L(G) 2t L(S) x L(T) by 1.6.4. Every X E A is a maximal subgroup of G and henceit follows that either S < X or T< X. If T< X, then T < Xo = n H and we

HEDare done; so suppose that S < X. Then X = S x (X n T) and A* = {H n TIH e Alis a conjugacy class of maximal subgroups of T satisfying the assumptions of thetheorem. By induction, T = A x B where (IAI,IBI) = 1, A is a P-group andB < n (H n T). Then S* = A and T* = S x B satisfy the assertions of the theorem.

HEeThus suppose, for a contradiction, that G is P-indecomposable. Since JAI > 2, we

have G = <K`°IK e A>, and therefore the assumption (A4')9 A`0 implies that thereexist H, K e A and x e K`0 such that (H4')" 0 A`°. If p = r(x) is defined as in 5.6.10,then H° = H`0"I ' 0 A and K° = K e A. Thus H° and K° are maximal subgroups ofthe soluble group G which are not conjugate in G. By a theorem of Ore (see Huppert[1967], p. 165), H° and K° permute so that IG: H°I = IK°: H° n K°I. Since p is indexpreserving, it follows that I G : HI = I K: H n K I and hence G = HK. This is impossi-ble since H and K are conjugate. Thus G is P-decomposable.

5.6.12 Corollary. Let A be a conjugacy class of maximal subgroups of the finitesoluble group G. If cp is an index preserving projectivity from G to a group G, then Allis a conjugacy class in G.

Proof. For J A I = 1 , this follows from 4.3.5; and since an index preserving projec-tivity of a P-group is conjugacy preserving, 5.6.11 yields the result if I > 2.

It is an interesting open problem whether Theorem 5.6.11 and its corollary alsohold without the assumption that G is soluble.

Exercises

1. Let G = <a, h a°z = h 2 = 1, b-'ab = a'+°) Show that every autoprojectivity ofG is centralizer preserving and find an autoprojectivity which is not normalizerpreserving.

2. Find a group G and a projectivity (p of G which(a) is conjugacy preserving but does not satisfy H'° H for every H < G,(b) satisfies H" z- H for every H < G but is not centralizer preserving.

3. (Barnes and Wall [1964]) Let 8 < n x ... x <s"_,> whereo(s;) = p be an elementary abelian group of order pn-2. For c c- { + 1, - 11, lets, bethe automorphism of N satisfying s" = s;si+2 (i > 2), where we put S. = s"+, = 1,and let ME = N<s,> be the semidirect product of N and <s,>. Finally, letGE = ME<s> be the semidirect product of M, and <s> where s, = sis;+, (i >- 1).Show that G, and G_, are lattice-isomorphic but nonisomorphic groups of orderp", class n - I and exponent p.

4. Draw a Hasse diagram of the subgroup lattice of a Rottlander group of order605.

5. Let p > 2 and G = C,wrC, be the wreath product of two cyclic groups of order p-thus G = A where A = <a,> x ... x <a,>, o(a;) = o(b) = p, b-la;b = a;+t(1 ,..., <ap>} be the conjugacyclass of <a,> in G. Let a be the automorphism of A satisfying ai = az and a° = afor all a E G' and define cp: L(G) L(G) by H`° = H° for H < A and H`° = H forH 4 A. Show that qp is an autoprojectivity of G such that all the groups in All arepairwise nonconjugate.

Chapter 6

Projectivities and normal structure of infinite groups

We have already mentioned that it is possible to generalize most of the results ofChapter 5 to infinite groups. Basic for this is a theorem that was proved indepen-dently by Zacher and Rips in 1980. It states that a subgroup of finite index of agroup G is mapped to a subgroup of finite index in G by any projectivity from G to6. This will be proved in § 6.1; in fact, we shall even give a lattice-theoretic character-ization of the finiteness of the index of a subgroup in a group. This theorem reducesnearly every lattice-theoretic question on subgroups of finite index to the corre-sponding question for finite groups.

In the subgroup lattice of a finite group we used modular elements to approxi-mate normal subgroups. The Tarski groups show that this is not reasonable ininfinite groups, so we use here the so-called permodular subgroups, introduced byZacher [1982b]. These are modular subgroups with an additional lattice-theoreticproperty possessed by projective images of normal subgroups and also by permu-table subgroups. So in §§6.2 and 6.3 we study permodular and permutable sub-groups of groups as far as necessary for our applications. In contrast to the situationfor finite groups, M°/MG in general is not hypercentrally embedded in G if M ispermutable in G. There is no general structure theorem for permodular subgroups,like the one for modular subgroups of finite groups, which reduces all our problemsto the investigation of permutable subgroups. In our applications, however, we mayusually assume that G is finitely generated modulo M. In this situation, it is possibleto prove that 1 M° : MI is finite and that the structure theorem holds if M is per-modular in G; and if M is even permutable in G, then MG/MG is nilpotent of finiteexponent.

In § 6.4 we use our results on permodular subgroups to generalize the lattice-theoretic characterizations of § 5.3 to infinite groups. In particular, we get character-izations for the classes of simple, perfect, hyperabelian, polycyclic, finitely generatedsoluble, hypercyclic, and supersoluble groups. However, we do not obtain a charac-terization of soluble groups; in fact, it is one of the most interesting open problemsto find a lattice-theoretic characterization for this class of groups.

In § 6.5 we again turn to our main problem. First of all we generalize some of thecriteria of § 5.4 to infinite groups. In particular, we prove that, as in the finite case: ifN a G, cp is a projectivity from G to G, H° is the normal closure and K° is the coreof N4' in G, then H and K are normal subgroups of G. In § 6.6 we obtain a structuretheorem for G/K and G/K4', mostly due to Zacher [1982a], that is similar to thecorresponding result for finite groups. Finally, we prove a number of properties ofH/K and H`°/K`0, the most important being that H/K and H4'/K4 are soluble of

282 Projectivities and normal structure of infinite groups

derived length at most 4 and 5, respectively. These results emerged from joint effortsof Busetto, Menegazzo, Napolitani and Zacher in the 1980's. As an important appli-cation we prove that the projective image of a soluble group (of derived length n) issoluble (of derived length at most 3n - 1), another result of Busetto, Menegazzo andNapolitani that improves older bounds of Yakovlev [1970] who was the first toprove that projective images of soluble groups are soluble.

6.1 Subgroups of finite index

It is easy to prove (see Theorem 1.2.12) that a group is finite if and only if itssubgroup lattice is finite. We wish to show that the finiteness of the index of asubgroup in a group is also a projective invariant and can be recognized in thesubgroup lattice. For K < H < G, of course, IH : KI is finite if and only if there existsa finite chain of subgroups K = Ho < H1 < < H = H such that Hi is a maximalsubgroup of Hi,, and jHi+1 : Hit is finite for all i = 0, ..., n - 1. Therefore we onlyhave to consider maximal subgroups.

Groups covered by cosets

The main tool in the proofs of our results will be a quite elementary lemma of B. H.Neumann's on set-theoretic unions of subgroups of a group. To prove this, we haveto study groups that are covered by a set of cosets of subgroups. To simplify nota-tion, in this section, we shall write A 1 u u An for the set-theoretic union of thecosets A 1, ..., A. of G.

6.1.1 Lemma. Let G be a group n, m e N, and suppose that H; (i = 1,..., n) and D aresubgroups of G such that D 0 Hi for i = 1, ..., n. If there exist elements xi, yy e G suchthat

(1) G

(2) G Dyl u u Dym,

then G is the union of finitely many right cosets of the subgroups H1...., Hn.

Proof. By (2) there exists g e G such that g 0 Dy1 u u Dyn. It follows thatDg n (Dy, 1 u . u Dy,n) = 0 and then (1) shows that Dg S H1 x 1 u u Hn xn.Hence

Dyi c (H1x1 u ... u Hnxn)g-ly; = H1(xlg-%) u ... u Hn(xng-ly;)

for j = 1, ..., m. By (1), G is the union of finitely many right cosets of the Hi.

6.1.2 Lemma. If the group G is covered by finitely many right cosets of subgroups, atleast one of these subgroups has finite index in G.

6.1 Subgroups of finite index 283

Proof. Let Hi < G and xi e G (i = I__, k) such that G = H, x 1 u u Hkxk; wehave to show that IG : Hit is finite for some i. We prove this by induction on thenumber r of different subgroups among the Hi. If r = 1, then G is the union of k rightcosets of one and the same subgroup which then clearly has index at most k. Sosuppose that I {H1..... Hk } I = r > 1, and assume that the assertion is true for r - 1.Without loss of generality we may further assume that the cosets are arranged insuch a way that Hi 0Hkfori= 1,...,nandHi=Hkfori> n. Then H,,..., H. andD = Hk satisfy (1). Now if IG : DI is Finite, we are done. On the other hand, if IG : DIis infinite, then (2) holds and by 6.1.1, G is the union of finitely many cosets of ther - I subgroups H1, ..., H. By induction, one of these has finite index in G.

6.1.3 Lemma (Neumann [1954]). Let Hi be subgroups and xi elements of the groupG (i = I__ , n) such that G = H1 x, u u If H. has infinite index in G, thenG = H1 x1 u ... u Hn-, xn-,

Proof. By 6.1.2, at least one of the Hi has finite index in G. So suppose without lossof generality that IG:H;1 is Finite for i = 1, ..., k and infinite for i = k + 1, ..., n; letD = H1 n n Hk. Since every subgroup of finite index contains a normal subgroupof finite index, the intersection of a finite number of subgroups of finite index againhas finite index. Therefore I G : D I is finite and every Hi (i = I__ , k) is the union offinitely many cosets of D. It follows that H1 x1 u . . u Hkxk = Dy1 u . u Dym forcertain yJ e G. If G 0 Dy, u . . . u Dym, then by 6.1.1, G would be the union of finitelymany right cosets of the subgroups Hk+1, ... , however, by 6.1.2, this is impossiblesince all these Hi have infinite index in G. Therefore

G=Dy1u...uDym=H1x,u...uHkxk9H1x1u...uHn-,xn-,

We are interested in the special case of Neumann's lemma in which all the x; = 1,so that the cosets are subgroups of G. The lemma states that all the subgroups ofinfinite index can be omitted from the covering of G. We say that, conversely, G iscovered irreducibly by subgroups Hi (i e I) if G is the set-theoretic union of the H; andnone of these subgroups can be omitted from the covering.

Maximal subgroups

An immediate consequence of Neumann's lemma is the following lattice-theoreticcharacterization of the finiteness of the index of a nonnormal maximal subgroup.

6.1.4 Lemma. A nonnormal maximal subgroup M of the group G has finite index in Gif and only if there are finitely many subgroups H. (i = 1, ... , n) of G such that G is

n

covered irreducibly by the Hi and n H; < M.itProof. If the Hi exist, then by 6.1.3, IG : Hit is finite for all i. Hence the intersection ofall the Hi has finite index in G and it follows that I G : M I is finite. Conversely,

suppose that M has finite index in G. Then GIN is a finite group if N = M0 is thecore of M in G. Let H;/N (i = 1, ... , n) be the maximal cyclic subgroups of GIN. Ifx e G, then <x>N/N is cyclic and therefore is contained in one of the H;/N; thusx e H. and G is covered by the Hi. And if g c- G such that H;/N = <g>N/N, themaximality of Hi/N implies that g 0 H; for all j i. Thus Hi cannot be omitted fromthe above covering of G. It remains to be shown that the intersection D of all the Hiis contained in M. If this is false, then G = <M, D> since M is a maximal subgroupof G. Since D/N is centralized by all the Hi/N, it follows that DIN < Z(GIN) andhence that M a G, a contradiction. Thus D < M, as required.

6.1.5 Remark. (a) Note that the property in Lemma 6.1.4 can be recognized in thesubgroup lattice. For, since a subgroup H of G is cyclic if and only if L(H) isdistributive and satisfies the maximal condition, G is covered irreducibly by sub-groups Hi if and only if

(3) for every H E L(G) for which [H/1] is distributive and satisfies the maximalcondition, there exists Hi such that H < Hi, and

(4) for every H, there exists H e L(G) for which [H/1] is distributive and satisfiesthe maximal condition such that H < Hi and H HH for all j : i.

In particular, this property is preserved by projectivities.(b) Therefore if M is a nonnormal maximal subgroup of G of finite index and cp a

projectivity from G to a group G, then IG : M1*I is finite. For, M'D inherits the lattice-theoretic property of Lemma 6.1.4 from M. Thus if M`° is not normal in G, thenI G : M"I is finite by 6.1.4; and if M10 a G, then 16: M10I is a prime.

We now handle the case of a normal maximal subgroup and first consider projec-tive images.

6.1.6 Lemma. Let N be a normal subgroup of prime index in the group G. If cv is aprojectivity from G to a group G, then 16: N"I is finite (and hence a prime).

Proof. The assertion of the lemma clearly holds if N1* < G; so suppose that N4' is notnormal in G. Let a E G \N and = <a>4. Then <a> : N n (a> J = I G : NJ = p andhence I : M° n l = q for some prime q. Consider the autoprojectivity a of Gdefined by X° = ((X4')6)`0 -I for X < G. Since bQ e N`1, we have N°" = ((N4)b"p° = Nand hence

(5) T = N n N° n . . . n N°"-' is invariant under a.

To simplify notation, we put N°' = Ni (i = 0, ... , q) and define for i = 0, ..., q - 1,

(6) and Si=R,,

Finally, let m be the smallest integer such that Rm = T. By (5), m < q - I and

(7) Sm=Rm=T.

Since N°" = N, we have Nj°" = N, for all j, and it follows that

(8) R°"=RiandS°"=Sifori=0,...,m.

We show next that for i = 0,..., m - 1,

(9) R. Si, Ri+1 = N n Si is a normal subgroup of index p in Si and S,+1 =Si n N;+2 is a maximal subgroup of Si; thus

(10) Si = R;+1 Si+1 for i = 0, ... , m - 2

and the following section of the Hasse diagram of L(G) illustrates the situation.

G

Rm=T=Sm

Figure 20

To prove these assertions, suppose first, for a contradiction, that Ri = Si, that isRi = R', for some i < m - 1. Then Ri = R°' < NN for all j = 0, ..., q and henceRi < T, contradicting the choice of m. Thus R. 0 Si for i = 0, ... , m - 1 and, by (6),Ri+1 = N n Si a Si and Si+1 = Si n Ni+2. To complete the proof of (9), it remains tobe shown that Ri+1 Si and Si,, Si. Then ISi : Ri+1 I = IG : NI = p and Sin Ni+2is maximal in Si since Ni+2, as a projective image of N, is modular in G so that[Si/S n Ni+2] = [<Si, Ni+2>/Ni+2] = [G/N1+2]. So suppose, for a contradiction,that Ri+1 = Si or Si+1 = Si. The first assumption yields that Ri > Ri+1 = Si = R°,and the second that R1+1 = R, and hence R° = Si >- Ri+1 = Ri. By (8), R°Q = Ri andso in both cases it would follow that R° = Ri, a contradiction, as we have justshown. Thus (9) holds and, since for i = 0, ..., m - 2, Ri+1 and Si,, are differentmaximal subgroups of Si, we clearly get (10).

Our aim is to show that T is a normal subgroup of finite index in G. For this wefirst prove by induction that for i = 0, ..., m - 1,

(11) <S°'Ij=0,...,q- 1>=G.This is clear if i = 0 since So = N° 0 N is a maximal subgroup of G. So suppose that( 1 1 ) holds for some i < m - 2 and let H = <5,+1 I j = 0,..., q - 1 >. Then (8) and (10)show that Ri+1 = R°+1 = S +1' S H and hence Si = Ri+i Si,, < H. Again by (8), H is

invariant under a. Therefore all the S°' (j = 0,..., q - 1) are contained in H and, byinduction, their join is G. Thus H = G and (11) holds.

By definition, for any subgroup X of G, (X°')`0 = (X`0)6' < <X°,b). Thus (11) im-plies in particular that

(12) <Sm-,,b) = G.

Again by definition, N+, = N°' " = N j " and hence (N`0)6 = N`', for all j = 0,... , q - 1.Since bQ e N`0 = N,", it follows that bQ e Ni!" P for all j. Thus

(13) b9 e T°.

By (5), T' is normalized by b and hence I (b>T 0: T"I = I : (b> n T"I = q. Itfollows that T`0 = N`0 n T" and T = N n (a, T> a <a, T>. Now (9) implies thatT = N n g and hence NG(T)" contains <Sm_,, b> = G. Thus

(14) T < G.

Again by (9), IS.-, /TI = p. If S.-;/T is finite for some i e { I__, m - 11, then also itsprojective image R.-;/T is finite and by (10), Sm_;_,/T is finite. It follows that So/Tis finite, as must be Ro/T = NIT; finally,

(15) G/T is finite.

Let H;/T (i = I__, n) be the maximal cyclic subgroups of G/T. Then G is coveredirreducibly by the Hi and hence, by 6.1.5, G is covered irreducibly by the Hr". Sup-pose, for a contradiction, that the intersection D of the Hi is not contained in N.Then ID: D n NJ = I G: NJ = p and hence D/T would contain a subgroup E/T oforder p. From bQ e T`° and <a"> = (N n <a>)* = Nc" n (b> = <ba>, it follows thata° e T Therefore <a> TIT and Sm_,/T are two different subgroups of order p in G/T.One of them is different from E/T and lies in a maximal cyclic subgroup Hj/T of G/T.Since D < N, the cyclic group HJ/T would contain two different subgroups of orderp, a contradiction. Thus D < N and it follows that the intersection of the H' iscontained in N". By 6.1.4, 16: N"'I is finite. Then G/(Nl)d is a finite group and, by5.1.2, 1G:N4'Iisaprime.

The Zacher-Rips theorem

We come to the main result of this section.

6.1.7 Theorem (Zacher [1980], Rips). Let (p be a projectivity from the group G to thegroup G and suppose that H and K are subgroups of G such that K < H. Then K hasfinite index in H if and only if K`° has finite index in H°.

Proof. Suppose that I H: KI is finite and let K = Ho < < H = H such that Hiis a maximal subgroup of H;+, for i = 0, ..., n - 1. Then IH;+1 : H;I is finite and,by 6.1.5 and 6.1.6, also IH , : Hr! is finite for all i. It follows that IH4: K" is finite.

Recall that a group is residually finite if it contains subgroups of finite indexintersecting trivially. The above theorem clearly shows that this property is pre-served by projectivities.

6.1.8 Corollary. Let cp be a projectivity from the group G to the group G. If G isresidually finite, then so is G.

Lattice-theoretic characterizations

The Zacher-Rips Theorem also yields lattice-theoretic characterizations of the fi-niteness of the index of a subgroup in a group. A first such characterization wasgiven by Zacher [1980]; we present a simpler one and for this we need anotherconsequence of Neumann's lemma.

6.1.9 Lemma. Suppose that M is a maximal subgroup of the group G such that forevery g e G, M n Mg is modular in G and [G/M n M9] is a finite lattice. Then M hasfinite index in G.

Proof. This is clear if M a G; so suppose that M is not normal in G and take x e Gsuch that M : Mx. By assumption, D = M n Ms and M are modular subgroupsof G; furthermore [G/D] is finite. By 2.1.6, M" is modular in G and by 2.1.5,[ <M, Ms)/Ms] 2 [M/M n M']. Since M" is a maximal subgroup of G, it followsthat D is a maximal subgroup of M. If H j4 M is another maximal subgroup of Gcontaining D, then M n H is maximal in H and D< M n H < M. Thus D = M n Hand [G/D] is a lattice of length 2. Clearly, x M and x Ms. Therefore M n <x> isa maximal subgroup of <x> and is contained in M n Mx = D. Since D mod G, andhence [<g>/D n <g>] ^_ [<D,g>/D] for all g e G, we see that D is also maximal in<D, x> and therefore <D, x> is a maximal subgroup of G different from M and M".This shows that [G/D] has at least 3 atoms and so is not isomorphic to the subgrouplattice of a cyclic group. But [<D,g>/D] is isomorphic to the subgroup lattice of<g)/D n <g> and it follows that <D, g> < G for all g c G. Thus G is covered by theatoms of [G/D] none of which can be omitted from the covering. By 6.1.4, 1G : MI isfinite.

6.1.10 Theorem (Schmidt [1984b]). It H and K be subgroups of the group G suchthat K < H. Then K has finite index in H if and only if there exists a finite chainK = Ha < < H = H of subgroups H. such that for every i e 10,..., n - 11, Hi is amaximal subgroup of Hi,, and satisfies one of the following two conditions:

(a) Hi+1 is covered irreducibly by finitely many subgroups whose intersection iscontained in H.; or

(b) for every automorphism a of the lattice [Hi,, /I], the subgroup H. n H' is modu-lar in [H,..1/1] and [Hi+1/Hj n H°] is finite.

Proof. If there exists such a chain, then by 6.1.4 and 6.1.9, 1H;,1 : H;I is finite forevery i and hence J H : K I is finite. Conversely, suppose that K has finite index in H

and let K = Ho < < H. = K be a maximal chain of subgroups connecting K andH. If i E {0, ... , n - 11, then H. is clearly a maximal subgroup of Hi+1. If there existsan autoprojectivity a of Hi+1 such that H° is not normal in Hi,,, then by theZacher-Rips Theorem, H° has finite index in Hi,,. By 6.1.4, Hi+1 is covered irre-ducibly by finitely many subgroups whose intersection is contained in H. Then H;+1is also covered irreducibly by the images of these subgroups under a-1 and theintersection of these is contained in H,; thus (a) holds. If there is no such autoprojec-tivity, then H° is a normal subgroup of prime index in Hi,, for every automorphisma of [Hi+1/1]. It follows that H, n H° < H;+1, and H;+1 /Hl n H' is finite for everysuch a. Thus (b) is satisfied.

As observed in 6.1.5, property (a) can be described in L(G), so that the abovetheorem is indeed a lattice-theoretic characterization of the finiteness of JH : K1.Another such characterization is given in Exercise 2.

Exercises

1 . If a group G is covered by n cosets of subgroups H., show that I G : H,I < n forsome i.

2. (Schmidt [1984b]) Show that a maximal subgroup M of a group G has finiteindex in G if and only if it has one of the following properties:(i) M is not modular in G, but there exists a modular subgroup S of G such that

S < M and [G/S] is finite.(ii) There exists r c- P(G) such that M n M' mod G and [G/M n M'] is a lattice

of length 2 with at least 3 atoms.(iii) If o f P(G), then M n M° mod G and [G/M n M°] _ {G, M, M°, M n M°}.

6.2 Permodular subgroups

We call a subgroup M of the group G permodular in G and write M pmo G if

(1) M is modular in G, and

(2) for every g c- G and Y:!5; G such that M < Y< <M, g> and [ <M, g>/Y] is afinite lattice, the index I <M, g> : YI if finite.

6.2.1 Remark. Since the finiteness of the index of a subgroup can be recognized inthe subgroup lattice, and a subgroup X of G containing M has the form X = <M,g)for some g E G if and only if there exists H < G such that [H/1] is a distributivelattice satisfying the maximal condition and X = M u H, condition (2) is a lattice-theoretic property. Thus the permodular subgroups can be recognized in the sub-group lattice, and if cp is a projectivity from G to a group G and M is permodular inG, then M° is permodular in G.

6.2 Permodular subgroups 289

Permodular subgroups were introduced by Zacher [1982b] who called them "D-embedded" since he used the term "Dedekind subgroup" for a modular subgroup.Our name is intended to express the close connection to modular and permutablesubgroups.

6.2.2 Lemma. Every permutable subgroup is permodular.

Proof. Let M be permutable in G. Then by 2.1.3, M is modular in G. If g e G andY< G such that M< Y!5; <M,g> = X and [X/Y] is finite, then X = M<g> and itfollows that Y = M(<g> n Y). By 2.1.5, [<g>/<g> n Y] [X/Y] is a finite latticeand hence I <g>: <g> n YI is finite. But then IX: YI = I Y<g>: YI = I <g>: <g> n YIis finite and M is permodular in G.

Since permodularity is a lattice-theoretic property, the lemma implies that projec-tive images of permutable subgroups, in particular of normal subgroups, are per-modular. We want to use these permodular subgroups to approximate normalsubgroups in the subgroup lattice. Therefore we have to investigate the structure ofMG/MG and G/MG for such a subgroup M of G and, as in the finite case, we want toreduce this problem to the case that M is permutable in G. In fact, we shall be ableto do this in the two situations which are important for us, namely when M/MG isperiodic and when G is finitely generated modulo M.

Basic properties of permodular subgroups

First of all we generalize Lemma 5.2.7 to permodular subgroups.

6.2.3 Lemma. If M pmo G and g e G such that o(g, M) is infinite, then g e NG(M).

Proof. Let X = <M, g> and T be a maximal subgroup of X containing M. Since[X/M] ^- L(<g>), there exists Y < X such that M < Y<_ T and [X/Y] is a chain oflength 2. By (2), IX : YI is finite and hence X/YX is a finite group. Since M is modularin X and [X/M] is a modular lattice, it follows from (c) of 2.1.6 that Y is modular inX. By 5.1.3, X/YX is a p-group for some prime p and hence T:9 X. Now M is theintersection of all the maximal subgroups of X containing M. Thus M a X, asdesired.

Our next aim is to replace condition (2) in the definition of a permodular sub-group by properties that are easier to work with. One of them will be that g e NG(M)if o(g, M) is infinite. The others are concerned with elements of finite order moduloM. First of all note that for a modular subgroup M of G and g e G, by 2.1.5,[<M,g>/M] [<g>/<g> n M] and hence

(3) [ <M, g>/M] is finite if and only if o(g,M) is finite.

We have to compute MG in a similar way as we did for finite groups in 5.1.6.

6.2.4 Lemma. For M < G, let I (M) he the set of elements x e G such that o(x, M) isa prime power or infinite, and let J(M) = {x E I(M)Io(x,M) < oc}.

(a) if M mod G, then MG =n {MXIx e 1(M)} = n {M<M.X>Ix e 1(M)}.(b) If M pmo G, then M, =n {wx e J(M)} = n {M;M.X>Ix e J(M)}.

Proof. It is clear that (a) implies (b) since for a permodular subgroup M of G, by6.2.3, MX = M = M,M,X if o(x, M) is infinite. So we have to prove (a). To accomplishthis, first note that

(4) MG < n {M<M.X>I x e 1(M)} < n {MXIx e I(M)} =: D(M)

since the successive intersections involve fewer conjugates of M. We have to showthat D(M) < M9 for all g e G; then it will follow that D(M) < n Mg = M, and

gEG

equality will hold in (4). So let g e G. If g e 1(M), then clearly D(M) < M. Thussuppose that g 0 1(M) and let o(g, M) = n = p;' ... p<' be the prime factor decom-position of o(g, M). For i = 1, ..., t, put ni = n/p°, and take ri, si e 77 such that1 = p°-ri + nisi. Since (g",',)",' = g"', c M, we have gi,Si a I(M), and hence by 2.1.5,

D(M) < n M9"" < n (M u<ge>)9" = n

((M u<gi'',>)9°".' )y.,,,

i=1 i=1

= n (M u (g°,,>)9 = I n (M u <g°..,>)) = (M u ngr''>19

= (M u <g"))9 = M9, as desired.

6.2.5 Lemma. Let M he a modular subgroup of the group G normalized by every g e Gwith o(g, M) infinite. Then the following properties are equivalent:

(a) M is permodular in G;(b) <M, g> : MI is finite for every g e G such that o(g, M) is finite;(c) H : MI is finite for every H < G such that M < H and [HIM] is finite;(d) X : M I is finite for every X < G such that M < X and [X/M] is a finite chain.

Proof. If M pmo G and g c- G such that o(g,M) < then by (3), [<M,g>/M] isfinite and by (2), 1 <M, g> : M I < oc. Thus (a) implies (b). If (b) holds and M < H < Gsuch that [HIM] is finite, then again by (3), o(x, M) < oc for all x e H. So if X ={ <M, x> I x E H}, it follows from (b) that IX: MI < oc for every X E X. Clearly, H iscovered irreducibly by the set 3`* of maximal elements of X and, by 6.1.3, I H : X I < ocfor every X e 3r*. It follows that JH : MI is finite. Thus (b) implies (c) and we wantto show next that (c) implies (a). So suppose that (c) holds and let g e G andM < Y< <M, g> such that [<M, g>/Y] is finite. If o(g,M) is infinite, then byassumption, M o <M, g> and <M, g>/M is cyclic. Thus I <M, g> : YI < x. And ifo(g,M) < x, then [<M, g>/M] is finite and (c) implies that I <M, g> : MI < oc. Inparticular, I <M, g> : YJ is finite and M is permodular in G.

We have shown that (a), (b), and (c) are equivalent, and it is clear that (c) implies(d). To complete the proof of the lemma, we finally show that (d) implies (b). So

suppose that (d) holds and let g E G such that o(g, M) < or- Then by (3), [<M, g>/M]is finite and we show by induction on 1 [<M, g>/M] I that I< M,g>: MI < X. For thiswe may assume that <M, g> = G and that [G/M] is not a chain. Since [G/M] isisomorphic to the subgroup lattice of a finite cyclic group, there exist two differentmaximal subgroups R and S of G containing M such that [G/T] = { T, R, S, G}where T = R n S. By (c) of 2.1.6, every subgroup of G containing M is modular in G.In particular, T mod G and, by 6.2.4, TG = TR n Ts since for x e I(T), [<T,x>/T] iseither a chain or infinite and hence <T, x> E J R, S, T1. By 2.1.5, R = <M, h> forsome h e <g> and the induction assumption implies that IR : MI < X. In particular,I R : TI < X and hence R/TR is a finite group. Similarly, S/Ts is finite and thereforeI T/TR n T51 < x. Since TG = TR n Ts, we see that R/TG and S/To are finite groups.Since T/T0 and an element of prime power order of G/TG cannot generate G/TG, thiselement has to lie in R/TG or S/TG. Therefore G/TG contains only finitely many suchelements and, of course, no element of infinite order. But then every element of G,/Tcis the product of its primary components and it follows that G/TG can only containa finite number of elements. In particular, I G : R I < oe and hence I G : M I < -)c., asdesired.

One disadvantage of permodular as compared with modular subgroups is thatthey do not, at first sight, have the nice inheritance properties of the latter. Forexample, it is not so easy to see that the join of two permodular subgroups is againpermodular. This is proved in 6.2.21 after some detours; however we shall needearlier the special cases of this result given in the following lemma. We do not knowa simple, direct proof of 6.2.21.

6.2.6 Lemma. Let M he a permodular subgroup of the group G.(a) If H < G, then M n H is permodular in H.(b) If N g G, then MN is permodular in G and MN/N is permodular in GIN.(c) I/ L is permodular in G such that [M u L/M] and [M u L/L] are finite, then

M u L is permodular in G.(d) For every g c G, M u M9 is permodular in G.

Proof. (a) By 2.1.6, M n H mod H. If g e H, then o(g, M n H) = o(g, M). There-fore if o(g, M n H) is infinite, then M9 = M and hence also (M n H)9 = M n H.If o(g, M n H) < or, and <M, g> = T, then by (2), 1 T : MI < oc and hence T/MTis a finite group. Thus <M n H, g>/<M n H, g> n MT is finite and, since<MnH,g>nMT<MnH, it follows that I<MnH,g>:MnHl<oc. By 6.2.5,M n H is permodular in H.

(b) Again by 2.1.6, MN mod G. For g c- G such that o(g, MN) is infinite, o(g, M) isalso infinite and hence (MN)9 = MN. Now suppose that o(g, MN) < oc. If o(g, M) isinfinite, then again g e N6(MN) and hence I <MN, g> : MNI < oc. And if o(g, M) < x,then by (2), I <M, g> : MI < or- and therefore I <MN, g>: MN I = I N <M, g> : NM Iis finite. By 6.2.5, MN is permodular in G and hence MN/N is permodular in G/N.

(c) By 2.1.6, H = M u L mod G. If g e G and o(g, H) is infinite, then o(g, M) ando(g, L) are infinite and hence H9 = M9 u L9 = M u L = H. We want to verify (d) of6.2.5 for H and use induction on the length of the chain [X/H]. So suppose that

H < X < G, [X/H] is a finite chain and I Y: HI < oc for all chains [Y/H] of smallerlength. If Hx < X, then X/Hx is a finite group, JH' : HI < oo by induction andhence I X : HI < oc; so assume that Hx = X. Then since [X/H] is a chain, thereexists x e X such that X = H u HX. As H = M u L and [X/Hx] [X/H] is a chain,it follows that X = M u HX or X = L u HX; without loss of generality supposethat X = M u HX. Then [X/M] [HX/HX n M] [H/H n MX-`]. By 6.2.3 and(b) of 6.2.5, either MX = M or (M, x>/M(M,X> is a finite group. In both cases,I M: M n MX `I < oo. By (c) of 6.2.5, I H : MI < oo and hence I H: M n MX-'I < oo.Since H n MX ` > M n MX-`, it follows that IH: H n MX-'I < oo and therefore[H/H n MX `] is a finite lattice. Thus [X/M] is finite and, by 6.2.5, IX : MI < x. Inparticular, IX: HI < oc and H pmo G.

(d) Clearly, Mg pmo G and M < M u M9 < (M, g>. By 6.2.3 and 6.2.5, eitherMg = M or I (M,g> : MI < oo. In both cases, [M u Mg/M] and [M u M9/M9] arefinite and, by (c), M u Mg is permodular in G.

We often have to consider elements or finite subsets of MG. We show next that insuch a situation we may usually assume that G is finitely generated modulo M, andthen prove the basic result that in this case IM° : MI is finite for a permodularsubgroup M of G. Recall that a group G is finitely generated modulo its subgroup Hif there exists a finite subset F of G such that G = <H, F>.

6.2.7 Lemma. If M < G and F is a finite subset of MG, then there exists a subgroupL of G containing M such that L is finitely generated modulo M and F c ML.

Proof. For any x e F c MG, there exist z; E G and y; e M=1 such that x = y, ... y,,.Therefore if L is generated by M and all these z;, then L is finitely generated moduloM and all the y, e ML. Thus x e ML and it follows that F = ML.

6.2.8 Lemma. If M is permodular in G and G is finitely generated modulo M, thenCMG : MI is finite.

Proof. Let G = (M, g, , ... , gm >. We prove the assertion by induction on the sum sof all the o(g,, M) (i = I__ , m) which are finite. If Mgi = M for all i, then M a G andIMG : MI = 1 clearly is finite. So suppose that Mgt M, say, and consider H =M u M9'. By 6.2.3 and 6.2.5, o(g, , M) and I H : MI < (M, g,) : MI are finite. By (d)of 6.2.6, H is permodular in G = (H,g...... gm> and the sum of all the o(g1,H)(i = 1, ... , m) which are finite is less than s. For, since M < H < (M, g 1 >, it fol-lows that M n <g, > < H n <g, > and hence o(g, , H) < o(g, , M); furthermore, clearly,o(g, H) < o(g, M) for all g c- G and if o(g, M) is infinite, then so is o(g, H) since IH : MIis finite. The induction assumption implies that 111°: HI < oc. Since M° = H° andH : M I < oc, it follows that I M° : M I < oc.

Permodularity and permutability

In 5.1.1 we proved that a subgroup of a finite group is permutable if and only if it ismodular and subnormal. However, it follows from the results of Chapter 2 that a

permutable subgroup of an infinite group need not be subnormal. The followingexample will also show that some other conceivable statements are not true.

6.2.9 Example. Let p be a prime, T an abelian group of type p°°, and G = T(z> thesemidirect product of T and an infinite cyclic group <z> with respect to the auto-morphism t -+ t'+P (t E T) of T (or t - t5 for p = 2). By 2.4.11, every subgroup of Gis permutable in G. Since the automorphism induced by z in the cyclic subgroup oforder p" of T has order p"-', we have <z> = 1; also a normal subgroup H of Gcontaining <z> has abelian factor group and therefore contains G' = T, thus <z>° =G. In particular, <z> is not subnormal in G. For later use, note that if N is an infinitecyclic group, then by 2.5.14 there exists a projectivity 9 from the abelian groupG = T x N to G mapping the normal subgroup N of G to the core-free, infinitecyclic subgroup <z> of G.

The example shows that in order to generalize Theorem 5.1.1 and to give a crite-rion for a modular or permodular subgroup M of an arbitrary group to be permu-table, we have to weaken the assumption that M is subnormal. For the relevantgeneralizations of subnormality we refer the reader to Robinson [ 1982], pp. 344 and364. Note that every subnormal subgroup is ascendant and every ascendant sub-group is serial.

6.2.10 Theorem (Stonehewer [1972], [1976a]; Zacher [1982b]). The following pro-perties of a subgroup M of the group G are equivalent:

(a) M is permutable in G,(b) M is permodular and serial in G,(c) M is modular and ascendant in G.

Proof. We first show that (b) and (c) follow from (a). So let M be permutable in G.By 6.2.2, M is permodular and hence also modular in G. Since every ascendantsubgroup is serial, it remains to be shown that M is ascendant in G. For this weconsider the set X of ascending series (for ordinals p) of the form

(5) M = M0 M1 MP < MG

such that M,, per G and Ma,,/Ma is a finite cyclic group for every ordinal x < p. Ifwe order X in the usual way-two series are comparable if one of them is an initialpart of the other-Zorn's Lemma yields a maximal element in I; let (5) be thisseries and put H = M. We show that H = M°; then, clearly, M is ascendant in G.So suppose, for a contradiction, that H M°. Then there exists g e G such thatH < HH9. Since H is permutable in G, we can assert HH9 < H(g> and by 6.2.3,o(g, H) < oo. It follows that I HH9 : HI is finite and hence H :9:!g HH9 by 5.1.1.Therefore if we put Lo = HH9 and form the successive normal closures Li+1 = HL,(i = 0, 1, ...) of H, this series reaches H. For the last member L, different from H, ofthis series, H a L and L/H is cyclic since H < L< H(g>. Furthermore L is permu-table in G since it is a join of conjugates of H. Thus we can extend the series (5) byputting Mp+1 = L, and this contradicts the maximality of this series.

Now suppose conversely that (b) or (c) is satisfied. Thus M is modular and ascen-dant, or permodular and serial in G, and we have to show that M is permutablein G, that is, <M,x) = M<x) for all x e G. For this we may assume thatM:9:9 <M, x) = T. For, if M is permodular in G, then M:9 T for o(x, M) infinite;in addition, if o(x,M) < oo, then [TIM] is finite and M serial in T implies thatM :!9!9 T. If M is not permodular in G, then M is ascendant in T and hence M a g Tsince [TIM] [<x>/<x> n M] satisfies the maximal condition. So suppose that

(6) M=M"aMn_1

Since M mod T. we have M1 = M u (<x> n M1) and by induction on the length ofthe chain (6), M1= M(<x) n M1). It follows that T= M1 <x> = M(<x) n M1) (x) =M <x ).

We want to show next that a permodular subgroup M is permutable if there is anelement of infinite order modulo M. For this we need a simple lemma.

6.2.11 Lemma. If M pmo G and g e G such that o(g, M) = oc, then M" n <g) = 1for all x e G.

Proof. Suppose, for a contradiction, that there exists x c G such that M" n <g) =<g") for some n e k. Then <M, g") < M u M" < <M, x) and, by assumption,o(g", M) = oc. Thus by (3), [ (M, g")/M] is infinite, therefore [ <M, x)/M] ando(x, M) are infinite as well. By 6.2.3, Mx = M and then M n <g) = <g"). But thiscontradicts the assumption that o(g, M) is infinite.

6.2.12 Theorem (Zacher [1982b]). Let M be a permodular subgroup of the group Gand suppose that there exists g e G such that o(g, M) is infinite. Then

MaA(M):= <xEGIo(x,M)= oc> aG

and hence M is permutable in G.

Proof. Let g c G such that o(g, M) is infinite. For z e M,

[<zg)l<zg) n M] "' [<M,zg)lM] = [<M,g)/M] "' [<g)l<g) n M].

It follows that o(zg, M) is infinite and hence z E <g, zg) < A(M). Thus M < A(M)and M g A(M) by 6.2.3. Let x e G. Then by 6.2.11, Mx n <g) = 1 and henceo(g, Mx) is infinite. Thus g e A(Mx) and it follows that A(M) < A(MX). Since Mx ispermodular in G, we get the other inclusion and hence A(M) = A(Mx) = A(M)".Thus A(M) a G; in particular, M:9!9 G and, by 6.2.10, M is permutable in G.

An example of a nonnormal permutable subgroup M with an element of infiniteorder modulo M is given in Exercise 6.3.2.

Periodic permodular subgroups

We come to the main subject of this section, the structure of G/MG and MG/MG fora permodular subgroup M of G. We first of all collect the results proved so far onMG and M/MG.

6.2.13 Remark. Let M be permodular in G and let J(M) be the set of elements x E Gsuch that o(x, M) is a prime power. By 6.2.4,

(7) MG = n {M<M,x>I x e J(M)}.

Let x e J(M) and T = (M, x). Then [T/M] [(x>/(x) n M] and hence

(8) [T/M] is a finite chain.

By 6.2.5, 1 T : M I is finite and therefore T/MT is a finite group. By 5.1.3,

(9) T/MT is nonabelian of order pq or a finite p-group;

and in the latter case, M is permutable in T, by 6.2.10. In any case,

(10) M/MT is a finite primary group.

In particular, M/MG is a subdirect product of finite nilpotent groups.

We now study permodular subgroups M for which M/MG is periodic.

6.2.14 Theorem. Let M be a permodular subgroup of the group G such that M/MG isperiodic. Then M/MG is the direct product of its Sylow subgroups and MG/MG islocally finite. In particular, M/MG is locally nilpotent.

Proof. Of course, we may assume that MG = 1, and we show first that with thiscondition M is p-closed for every prime p. To do this we let x, y be p-elements of Mand suppose that z is a p'-element in (x, y). In every finite nilpotent factor groupM/N of M, the elements xN and yN generate a p-group and zN is a p'-element inthis p-group; thus z e N. By 6.2.13, M is a subdirect product of finite nilpotentgroups and it follows that z = 1. Thus (x, y> is a p-group and M is p-closed. Thisholds for every prime p and hence M is the direct product of its Sylow subgroups.Now let g...... g,, E M and H = <g...... gm>. We show that

(11) H is finite.

Since H < M is the direct product of its Sylow subgroups, it will follow that H is afinite nilpotent group and that M is locally finite and locally nilpotent. As everyelement is the product of its primary components and M is the direct product of itsSylow subgroups, we may assume in the proof of (11) that all the g, are p-elementsfor a fixed prime p. Let p" be the maximal order of such an element g;. We show that

(12) H(") = 1;

then H is a finitely generated soluble torsion group and hence finite (see Robinson[1982], p. 147). To prove (12), we may assume that r > I and, by 6.2.13, we have to

show that W1 < MT for all the groups T = <M, x> with x e J(M). If T/MT is non-abelian of order pq, then H(') < M' < MT; and if T/MT is a q-group for some primeq p, then all the g; are contained in MT and hence H < MT. Finally, suppose thatT/MT is a finite p-group. Then HMT/MT < S2,(T/MT) and, by (d) of 5.2.8, this groupis soluble of derived length at most r. Thus again H(') < MT and (12) holds.

It remains to show that MG is locally finite. For this let F be a finite subset of MG.By 6.2.7 there exists a subgroup L of G containing M such that L is finitely generatedmodulo M and F c ML. By 6.2.8, I ML: MI is finite. Therefore <F> n M is a sub-group of finite index in <F> and hence is finitely generated (see Robinson [1982], p.36). Since M is locally finite, <F> n M, and hence <F>, is finite. Thus MG is locallyfinite.

We want to show that if M is permutable in G, then the Sylow subgroups ofM/MG appearing in 6.2.14 are permutable in G/MG and their normal closures are theSylow subgroups of MG/MG. In fact, we prove more general assertions.

6.2.15 Lemma. Let it be a set of primes and M a n-subgroup of the group G. If(a) M !9:!g G or(b) M per G,

then MG is a n-group.

Proof. (a) Let M = Mo < . . . 9 M. = G. We prove the assertion by induction on n.Let H = Then the induction assumption yields that MH is a n-group, andsince H 9 G, we have M" < MG < H. It follows that M" < MG and MG = (MH)G isa product of normal n-subgroups. Then every element of MG lies in a product offinitely many normal n-subgroups and hence is a n-element. Thus MG is a n-group.

(b) Let x e MG. By 6.2.7 there exists a subgroup L of G containing M that isfinitely generated modulo M such that x e M' and, by 6.2.8, 1 ML: M I < cc. By6.2.10, M is ascendant in G, hence subnormal in M' and therefore also in L. By (a),ML is a n-group. Thus x is a n-element and MG is a n-group.

Note that since the product of normal n-subgroups is a normal n-subgroup, itfollows from 6.2.15 that, quite generally, the join of permutable (or subnormal)n-subgroups is a n-group.

6.2.16 Lemma. Let M = A x B be a periodic permutable subgroup of G and supposethat (o(a), o(b)) = 1 for all a e A, b e B. Then A and B are permutable in G.

Proof. We show that A per G. Let n = n(A) be the set of primes dividing the orderof an element of A and let x n G. We have to prove that A<x> = <x>A. If o(x) isinfinite, then by 6.2.3, x e NG(M) and hence x e NG(A). So we may assume thato(x) < oc and then, of course, that o(x) = p" for some prime p. Let H = M<x>. ThenIH:MI = <x>: <x> n MI and hence IM : Mn MI = lMM: MI is a power ofp forevery y e H. Therefore if p 0 n, then A <_ MI and, as a characteristic subgroup of thisgroup, A 9 H. Finally, suppose that p e n. By 6.2.10, M <9 H, hence A :9!!g H, and

(a) of 6.2.14 shows that A" is a it-group. Hence A"<x) is a n-group and it followsthat A"(.x) n M = A. Now (11) of 5.1 yields that A is permutable in A"(x) and thisimplies that A(x) = <x> A.

The above results sufficiently elucidate the structure of MG/MG for periodic per-mutable subgroups M of G (see Theorem 6.3.1). We turn to the more general case ofperiodic permodular subgroups; we wish to prove a structure theorem similar to5.1.14 for finite groups. For this purpose it is convenient to introduce a notation forthe situation that appears in this theorem. In the literature the term "Schmidt struc-ture" has been used (see Stonehewer [1976b], Zacher [1982a]), but we prefer tospeak of P-embedded subgroups instead.

P-embedding

The subgroup M of the group G is called P-embedded in G if G/MG is a periodicgroup of the form

(13) G/MG = (S1 /MG X S2/MG x ) x T/MG,

where the number of factors Si/MG may be finite (nonzero) or infinite, and for all i, j,

(14) Si/MG is a nonabelian (possibly infinite) P-group,

(15) (o(x), o(y)) = 1 for all x E Si/MG and y E SJ/MG with j 0 i or y E T/MG,

(16) M/MG = (Q1/MG X Q2/MG x ) x (M n T)/MG and Qi/MG is a nonnormalSylow subgroup of Si/MG, and

(17) M n T is permutable in G.

Every P-embedded subgroup is permodular in G, as the reader may easily prove.Our aim is to show the converse, at least for nonpermutable subgroups in periodicgroups. The situation in arbitrary groups is more complicated; see Exercises 3 and 4.

6.2.17 Theorem. Let M be a permodular subgroup of the group G such that M/MG isperiodic. Then M is permutable or P-embedded in G.

Proof. We assume that M is not permutable in G and also, of course, that

(18) MG=1.

By 6.2.12, o(g, M) is finite for every g c- G and, since M is periodic, it follows that

(19) G is periodic.

By 6.2.14, M is the direct product of its Sylow subgroups. Since M is not permutablein G, there exist Sylow subgroups of M that are not permutable in G. Let Q be oneof these, q the corresponding prime and K the q-complement of M, so that

(20) M = Q x K.

We shall show in a number of steps that IQI = q, QG is a nonabelian P-group,G = QG X CG(QG) and the groups QG and CG(QG) are coprime.

Since Q is not permutable in G, there exists x e G of prime power order such thatQ <x> <x) Q. Let x be such an element and X = <M, x). Since M is permodularin G, IX : MI is finite. If M were permutable in X, then by 6.2.16, Q would also bepermutable in X and hence Q<x) = <x>Q, a contradiction. Thus M is not per-mutable in X, but x c- J(M); by 6.2.13, X/Mx is nonabelian of order rs for primes rand s. If Q < Mx, then Q would be a characteristic subgroup of Mx a X and there-fore x c- NG(Q). This is not the case; hence QM, = M and it follows that IM: MxI =q is the smaller prime dividing IX/Mx I. Thus we have shown:

(21) If x e G is of prime power order such that Q<x) <x)Q, and if X = <M, x>,then IM : MxI = q and X/Mx is nonabelian of order rq for some primer > q.

Suppose that x and X are as before, let y c- G and consider H = <M, x, y> = <X, y).Since M is a maximal subgroup of X, we see that X = M u M" is permodular in Gby (d) of 6.2.6. It follows that I H : XI, and hence also I H : MI, is finite. Then H/MHis a finite group containing the core-free modular subgroup M/MH with Sylowq-subgroup QMH/MH. By 5.1.14, M is P-embedded in H. If QMH were permutablein H, it would also be permutable in X, and then M = (QMH)Mx would be permu-table in X, contradicting (21). Thus QMH/MH is not permutable in H/MH and henceit cannot lie in the permutable part of the decomposition (16) of M/MH. Therefore:

(22) If x is as in (21), y e G and H = <M, x, y), then H/MH is a finite group, M isP-embedded in H, (QMH)H/MH is a nonabelian P-group, and QMH/MH is aSylow q-subgroup of order q of H/MH.

In particular, IQ : Q n MHI = q = I Q : Q n M, 1. But X < H implies MH < Mxand it follows that Q n M. = Q n MH < MH < My. Since y c- G was arbitrary,Q n M,< MG = I and hence

(23) IQI = q.

Now suppose that y is a q-element. If <Q,y) is a q-group and again H = <M, x, y),then by (22), QMH/MH is a Sylow q-subgroup of H/MH; on the other hand,(Q, y)MH/MH is a q-group. It follows that (Q, y) < QMH < M and then y e Q sinceQ is the only Sylow q-subgroup of M. This shows that Q is a Sylow q-subgroup ofG. If (Q,y) is not a q-group, then Q(y> # <y)Q and we may choose y = x in (21).We obtain that y is contained in a conjugate of M in X = <M,x) and then, clearly,y is contained in a conjugate of Q. Thus

(24) Q E Sylq(G) and every q-element of G is contained in QG.

Now let g e G\NG(Q), so that Q9 0 Q. Since Q E Sylq(G), we have Q<x) # <x>Q forQ9 = <x>. Therefore if X = (M,x) = <M,Q9), then by (21), X/Mx is a nonabeliangroup of order rq for some prime r > q. By (23), Mx is the q-complement K inM = Q x K. Furthermore, Q9K/K is a subgroup of order q in X/K and hence isconjugate to M/K. Thus Q9K is conjugate to M and it follows that Q9K = Q9 x K.Thus K is centralized by Q9 and, since g was arbitrary,

(25) QG < CG(K).

The nonabelian group <M, Q9 )/K of order rq has a unique minimal normal sub-group S9/K, say; clearly, <M,Qg) = (Sg,M). Let S = (SoIg e G\NG(Q)>; thenQGK/K = (S,M)/K. For g,h e G\NG(Q), again write X = <M,Qg) and H =<M, Qg, Qh) = (Sg, Sh, M ). Then K = MH and, by (22), H/K is a finite group inwhich M/K is a modular, nonpermutable subgroup of order q that together with itsconjugates QgK/K and QhK/K generates H/K. By 5.1.9, H/K is a nonabelian P-group. Thus ISg/KI = ISh/KI, [Sg/K,Sh/K] = 1 and M/K induces a universal powerautomorphism in S9Sh/K. This shows that S/K is an elementary abelian p-group forsome prime p, and SM/K = QGK/K is a nonabelian P-group. By (25), QG n K <Z(QG) and since QG/QG n K QGK/K, as a nonabelian P-group, has trivial centre,it follows that QG n K = Z(QG) a G. Now MG = 1 implies that QG n K = I andhence that

(26) QG is a nonabelian P-group.

Let P be the normal Sylow p-subgroup of Q'. We want to show that P is the set ofp-elements of G. For this we choose a conjugate Qg = <x) 0 Q of Q and put X =(M, x). If M contained an element of order p, then, since n M<M.y> < n m,, = 1,

yeG veGthere would exist a subgroup H = <X, y) of G such that M/MH contains an elementof order p. By (22), this implies that M is P-embedded in H and (QMH)H/MH is anonabelian P-group. By (26), p would divide I(QMH)H/MHI and hence, by (15), couldnot divide IM: MHI, a contradiction. Thus M does not contain an element of orderp. Now let y be an arbitrary p-element of G and, this time, H = <X, y). Then againby (22) and (26), (QMH)H/MH contains the unique Sylow p-subgroup of H/MH. Itfollows that y E QHMH < QGM and, since M does not contain p-elements, y E QG.Thus y e P and hence

(27) P is the unique Sylow p-subgroup of G.

We finally show that

(28) G = QG X CG(QG).

Every element is the product of its primary components and, by (24) and (27), QGcontains every p-element and every q-element of G. Therefore to prove (28), we onlyhave to show that for every prime t such that p 0 t 0 q, every element of order t" ofG centralizes QG. So let x be such an element. If Q<x) (x> Q, we would considerX = <M,x) and obtain from (21) and (26) that IX/M,rI = pq. It would follow thatx e MX and hence Qx = Q, a contradiction. Thus Q<x) = <x>Q and henceIQ<x)I = qt". By (26), Q = Q<x) n QG a Q<x) and so x e NG(Q). If g e G, then xg-'also is a t-element; therefore xg-' e NG(Q), that is, x e NG(Qg). Thus x normalizesevery q-subgroup of QG and, since in this P-group every subgroup is the join ofq-subgroups or is the unique Sylow p-subgroup of such a join, every subgroup of QGis normalized by x. Hence x induces a power automorphism in Q' which, by 1.4.3,is trivial since Z(QG) = 1. Thus x e CG(QG) and (28) holds.

To complete the proof of Theorem 6.2.17, let Q1, Q2, ... be those Sylow subgroupsof M which are not permutable in G and let R be the join of the other Sylowsubgroups of M. Then R is permutable in G and M = (Q 1 x Q2 x . . ) x R. Let Q.

be a qi-group. By (26), for every qi, the group Q? is a nonabelian P-group in whicha prime pi > qi is involved. By (28), all these pi are distinct and also different from the

= QC X QZ x is the set of n-qi; let n = {pi, qiJi > 1}. Then (Q1 X Q2 X Pelements, and CG(QG X QZ x ) = n Co(Qq) = T is the set of n'-elements of G. Inparticular, i'-1

G x QZ x ... ) x T

and R = M n T is permutable in G. Thus M is P-embedded in G.

As an immediate consequence of Theorem 6.2.17 we get the corresponding resultin the case that G is finitely generated modulo M.

6.2.18 Theorem (Zacher [1982b]). Let M be a permodular subgroup of the group Gand suppose that G is finitely generated modulo M. Then IMG : MI is finite, Mc/Mc isa locally finite group of finite exponent and, if M is not permutable in G, then M isP-embedded in G.

Proof. We may assume that MG = 1; let MG = H. By 6.2.8, I H : MI < oc and henceHIM is a finite group of order n, say. For g e G, also (MH)9 a H and JH/(MH)91 = n.Therefore if x e H, then x" E (MN)9 < M9 and it follows that x" E n Mg = MG = 1.

ge GThus H is periodic of exponent dividing n. The other assertions now follow from6.2.14 and 6.2.17.

The join of two permodular subgroups

We finally want to show that the join of two permodular subgroups is permodular.For this we need a general result on permodular subgroups that is a consequence ofTheorem 6.2.18 and is of independent interest.

6.2.19 Theorem. If M is permodular in G, then M is permutable in MG".

Proof. Let Per(M) = {x E GI M per (M, x> }. We want to show that Per(M) is asubgroup of G containing MG", that is

(29) M < MG" < Per(M) < G.

This will prove the theorem since every x e Per(M) satisfies M(x) = <x>M and so(29) implies that M is permutable in MG". To prove (29), we take x, y E Per(M),a, b, c, d e G and show that

(30) xy-' E Per(M) and [[a,b],[c,d]] E Per(M).

This will imply that Per(M) < G and G" < Per(M); since obviously M c Per(M),(29) will follow. To prove (30), consider L = (M, a, b, c, d, x, y>. Then L is finitelygenerated modulo M and, by 6.2.18, either M is permutable in L or M is P-em-

bedded in L and ML/ML has finite exponent. In the first case, L s Per(M) and (30)holds; in the second case, let

L/ML = S, /ML x ... X S,/ML X T/ML

and

M/ML = Q 1 /ML x ... x Q,/ML x (M n T )/ML

be the decompositions described in (13) and (15). We claim that

(31) Per(M) n L = MT

Indeed M n T is permutable and the Q; are normal in MT, so M = Q1 ... Qr( M n T)is permutable in MT and hence MT c Per(M) n L. And if z e L\ MT, then z =zl ... z,t where z; E S;, t c T and zj Qj for some j. Since the orders modulo M. ofthese elements are coprime, there exists m e N such that zmML = zJML. It followsthat <zjML> < <zML) and hence (zz, QQ> < <z, M>. Since <zj, QJ>/ML is nonabelianof order pq for primes p, q and p > q = JQJ/MLA, the subgroup M n <zz, QQ> = Qjis not permutable in <zz, QQ>. In particular, M cannot be permutable in <M, z> andthus z 0 Per(M). This proves (31), and (30) is an immediate consequence. For, sincex, y e Per(M) n L = MT, we have xy-' E MT 9 Per(M) and since the factor groupLIT n- S1/ML x x Sr/ML is metabelian, [ [a, b], [c, d ] ] is contained in T andhence in Per(M).

6.2.20 Lemma. Let G be a soluble group. If M is a maximal subgroup of G that ismodular in G, then I G : M I is a prime.

Proof. We use induction on the derived length of G. If G' < M, then M a G andI G: MI is a prime; so suppose that G' M. Since M is a maximal subgroup of G, itfollows that G = G'M and, since M is modular in G, M n G' is maximal and modu-lar in G'. By induction, I G': M n G' j = I G : M I is a prime.

6.2.21 Theorem (Zacher [ 1982b] ). If M and L are permodular subgroups of the groupG, then M u L is permodular in G.

Proof. By 2.1.6, H = M u L is modular in G. If g e G and o(g, H) is infinite, theno(g, M) and o(g, L) are infinite and hence H9 = M9 u L9 = M u L = H. We want toverify (d) of 6.2.5 for H. So let H < X < G such that [X/H] is a finite chain. We haveto show that JX : HI is finite and may assume that X = G. Let R = G"H. Since[R/H] is a chain, there exists x e G" such that R = (H, x). By 6.2.19, M is per-mutable in G"M and hence M(x) = <x>M. Similarly, L<x> = <x>L and thus R =H<x>. It follows that JR : HI = I<x>: <x> n HI and [<x>/<x> n H] [R/H] isfinite since H is modular in G. Thus JR : HI is finite. By (c) of 2.1.6, every subgroup ofG containing H is modular in G; in particular, every subgroup of GIG" contain-ing R/G" is modular in G/G". By 6.2.20, any two successive members of the chain[(G/G")/(R/G")] have prime index in each other and it follows that IG: RI is finite.Thus I G : HI is finite and, by 6.2.5, H is permodular in G.

We mention without proof three further results of Zacher [1982b]. First of all,Theorem 6.2.21 is the main step needed to prove that the relation p on the subgrouplattice of a group G defined by

(32) HpK if and only if H < K and H pmo K

is a normality relation on L(G). This concept was introduced by Zassenhaus [1958],p. 76 who proved a Jordan-Holder Theorem for lattices with a normality relation.However, since every permutable subgroup is permodular, the remark following5.2.8 also applies here. Permodular subgroups are the lattice-theoretic approxima-tion to normal subgroups (and their projective images) which we shall use; however,in fact, they approximate permutable subgroups more precisely. It would be prefera-ble to have a better lattice-theoretic approximation of normal subgroups to workwith.

The second result is that if (Mi)iEl is a local system of subgroups of M such thatevery M; is permodular in G, then M is permodular in G. Using this, Theorem 6.2.21implies that the join of an arbitrary family of permodular subgroups is permodularin G.

Thirdly, Theorem 5.1.13 also holds for permodular subgroups of infinite groups:Zacher shows that M pmo G if and only if M pmo <M, g) for every g e G.

We finally remark that quite a few of the results in this section, in particular 6.2.3,6.2.8, 6.2.14 and Theorem 6.2.17, were inspired by and are more or less due toStonehewer [1976b], who proved them for modular subgroups in Tarski-free groups.

Exercises

All the exercises in this section are due to Zacher [1982b].1. Let M be permodular in G and suppose that there exists g E G such that o(g, M)

is infinite. Show that M/MG is a subdirect product of finite cyclic groups, hence isabelian, and MG/MG is nilpotent of class at most 2.

2. If M is P-embedded in G, show that M is permodular in G.3. We say that M is locally P-embedded in G if there exists a family (Hi)i., of

subgroups Hi of G containing M with the following properties.(i) M is P-embedded in Hi for every i e I.

(ii) For all i, j e I there exists k c I such that Hi u H. < Hk.(iii) G is the set-theoretic union of the Hi.Show that a subgroup M of G is permodular and not permutable in G if and onlyif M is locally P-embedded in G.

4. For every i e 1V, let pi and qi be primes such that qiI pi - 1 and { pi, qi} n { pj, qj} _ 0for i -A j; let Si = (a,,bi) be a nonabelian group of order pigi and bi-taibi =with ri c- Z. Put S = Dr <ai> and let r be the automorphism of S satisfying a, = a;

iE M

for all i. Finally, let M = (r) and G = SM be the semidirect product with respectto this automorphism.(a) Show that M is permodular and not permutable in G.(b) Show that M is infinite cyclic and MG = 1; hence M is not P-embedded in G.

6.3 Permutable subgroups of infinite groups 303

5. Let M be a subgroup of a group G such that M is permodular in (M, g> for everyg e G and suppose that G is finitely generated modulo M.(a) Show that (MG: MI is finite.(b) If M is not permutable in G, show that M is P-embedded in G.

6. Show that a subgroup M of the group G is permodular in G if and only if M ispermodular in <M, g> for every g e G. (Hint: Use Exercises 3 and 5.)

6.3 Permutable subgroups of infinite groups

Theorems 6.2.17 and 6.2.18 reduce the problem of determining the structure ofG/MG for a permodular subgroup M to the case where M is permutable in G; at leastthis is so in the two situations that are important for us, namely when M/MG isperiodic and when G is finitely generated modulo M. The results of § 6.2 also give thestructure of MG/MG in these situations; the only additional result proved in thissection is that MGIMG is nilpotent if G is finitely generated modulo M. Again nosystematic treatment of permutable subgroups is intended, but a number of specialresults will be proved in the remainder of this section to prepare the way for thestudy of the structure of NG/NG and NG/NG for a normal subgroup N of G and aprojectivity from G to G. Basic for this is a theorem of Busetto and Napolitani[1990] stating that if G is generated by a permutable subgroup M and a subset F,then MG is generated by the M(x) for x e F.

Periodic permutable subgroups

To give the structure of MG/MG for a permutable subgroup M of G with M/MGperiodic, we only have to collect the results proved in 6.2.14-6.2.16.

6.3.1 Theorem. Let M be a permutable subgroup of the group G such that M/MG isperiodic. Then MG/MG is locally finite and the direct product of its Sylow subgroups,the Sylow subgroups of M/MG are permutable in G/MG and their normal closures arethe Sylow subgroups of MG/MG.

Proof. We may assume that MG = 1. By 6.2.14, MG is locally finite and M is thedirect product of its Sylow subgroups MP (p c P). Then M' is the product of thenormal closures Mp P. By 6.2.16, MP is permutable in G, and 6.2.15 shows that Mp isa p-group. It follows that the product of the MG is direct and that MSG is the Sylowp-subgroup of MG. El

The above theorem in particular implies that for a permutable subgroup M of Gwith M/MG periodic, MG/MG is locally nilpotent. It need not be nilpotent; in fact,M/MG need not even be soluble (see Exercise 4). If G is finitely generated modulo M,then by 6.2.18, M/MG is periodic and we want to show that MG/MG is nilpotent inthis case. For this we need the following lemma.

6.3.2 Lemma. Let M be permutable in G, g1..... g,,, E G and H = M"... M9-. ThenH/(Mg,)H is finite and nilpotent for every i = 1, ..., m.

Proof. If o(g;, M) is infinite, M9i = M. Thus we may assume that gi = 1, o(g,, M) < 00for all j and G = (M, g...... g). Then by 6.2.8, 1 Mc : MI < ac; in particular, H/MHis a finite group. It remains to be shown that this group is nilpotent. For this letg be one of the g; with j : i and L = H(g). Since H per G and o(g, M) < 00, theindex IL: HI is finite, and hence L/ML is a finite group. By 5.2.3, ML/ML is nilpotentand therefore H n ML/ML is nilpotent as well. Since ML < MH, finally, H n ML/MHis a nilpotent normal subgroup of H/MH containing MMg/MH. Thus MMg/MH iscontained in the Fitting subgroup of H/MH and, since H is generated by these MM9,it follows that H/MH is nilpotent.

6.3.3 Theorem (Lennox [1981]). If M is a permutable subgroup of the group G suchthat G is finitely generated modulo M, then MG/MG is nilpotent.

Proof. Let H = Mo. By 6.2.8, I H : MI is finite and hence there exist g2, ..., g,,, E Gsuch that H = MM92 ... Mg-. By 6.3.2, H/MH is nilpotent. Hence some term K of thelower central series of H is contained in MH, and K < G since K is a characteristicsubgroup of H = M'. It follows that K < MG and hence Mo/Ms is nilpotent.

We remark that although Theorem 5.2.12 and Busetto [1979] contain partialresults in this direction, there is no analogue of the Maier-Schmid Theorem forinfinite groups. In fact, Exercise 6 shows that MG /Ms need not be hypercentrallyembedded in G even if M is a permutable subgroup of a finitely generated p-groupG. In this example, M/Ms is infinite and it remains an open problem whether a finitecore-free permutable subgroup of a group G is contained in the hypercentre of G.Furthermore, there are examples of (nonperiodic) core-free permutable subgroupswhich are not locally soluble (see Gross [1982]).

The normal closure of a permutable subgroup

In the remainder of this section we shall study rather special aspects of permutablesubgroups which are necessary for the applications to projective images of normalsubgroups. First we want to describe the normal closure of a permutable subgroupM in G and in Mo. For this we need a very simple technical lemma.

6.3.4 Lemma. If M per G and g c- G, then M<g> = M(M<g> n (g)) = MMg.

Proof. Since M is permutable in G, we have M<g> < M(g), and Dedekind's lawyields that M<g> = M(M<g> n (g)). Furthermore, MM9 < M(g) = M9(g) andagain Dedekind's law implies that MM9 = M(MM9 n (g)) = M9(MM9 n (g)). Itfollows that MM9 is normalized by g. Thus MM9 < M(g) and hence MM9 = M<9>.

6.3.5 Theorem (Busetto and Napolitani [1990]). Let M be a permutable subgroupand F a subset of the group G such that G = <M, F>. Then

MG = (M<X>Ix E F> = M(M<x> n (x>Ix e F>.

Proof. First of all note that we may assume IF1 = 2 and only have to show that

(1) MG = M<x>M<Y> if G = (M,x,y>.

For, if this holds, then for every x, y e F,

(M<X>)<Y> < M<M,X,y> = M<X>M<Y> < (M<X>)<y>

and hence (M<x))<y> = M<x>M<y>. Thus if H = (M<x>I x E F>, then

H<y> = (M<x>(x E F><Y> = ((M<X>)<Y>lx e F> = (M<x>M<y>(x E F> = H

and hence y e NG(H). This holds for arbitrary y E F and, since G = (M, F>, it followsthat H a G. Thus MG < H, and the other inclusion is trivial since H is generated byconjugates of M. Therefore MG = H = (M<x>lx E F> = M(M<x>n(x> xeF>, by6.3.4, and the theorem will be proved.

So assume that G = <M, x, y>, and consider T = M<x> = (M9ig e (x>>. As a joinof permutable subgroups, T is permutable in G and, if we apply 6.3.4 to T, we obtainthat T<x''> = TTxy = TTY = T<y). Therefore T<'> = (M<x))<3 is normalized by yand xy, hence is normal in (M, xy, y> = G and so contains M'; on the other hand,(M<x>)<y> is generated by conjugates of M. It follows that

(2) MG = (M<X>)<Y>

and by 6.3.4 applied to the permutable subgroup M<x>,

MG = (M<x>)<y> = M<X>(M<X> n (y)) <_ M<X>(MG n (y>).

The other inclusion is trivial and again by 6.3.4,

(3) MG = M(M<x> n (x>)(MG n (y>)

For the proof of (1) we may assume that MG = 1. By (2), we have to show that

(4) (M<x>)<y> = M<x>M <y>

Since G is finitely generated modulo M, by 6.2.18 and 6.3.1, MG has finite exponentand M is the direct product of its Sylow subgroups which are permutable in G.Now if M = AB with A and B permutable in G, then MX = AXBX for every sub-group X of G; so if (4) holds for A and B, it follows that

(M(x>)<y> = (A<x>)<Y>(B<x>)<Y> = A<x>A<y>B<x>B<Y> = M<X>M<y>

as desired. Therefore we may assume that M is a p-group for some prime p and thatG = (M, x, y> and MG = 1. By 6.2.15 and 6.2.8,

(5) MG is a p-group and 1 MG : MI is finite;

write H = MG. Then H/MH is a finite p-group; let v be the natural epimorphism fromH to HIM,,. If H n (x> = M<x> n (x>, then (3) and (4) with x and y interchangedyields

MG = M(M<y> n (y>)(MG n (x>) = M(M<y> n (y>)(M<x> n (x>) < M<Y>M<x>

and hence MG = M<x>M<y>, as desired. So suppose that H n (x) -A M<x> n <x>.Then M<x> n <x> is contained in the maximal subgroup of the cyclic p-groupH n <x>. Since MH < M<x> < H, this also holds for the images under v and, since H"is a finite p-group, it follows that (M<x> n (x>)" is contained in the Frattini subgroupof H". Then (3) and the main property of the Frattini subgroup yield that

H' = (M<x> n (x))"M"(H n (y))" = M"(H n (y>)"

and hence H = MMH (H n (y>) = M(H n (y>). Since H n (y) is a cyclic p-group,it follows that [H/M] is a chain and therefore either M<x> < M<'> or M<Y> < M<x>.

In the first case, as M<Y> < M(y), it follows from (2) that MG = (M<x>)<Y> < M<y>,and hence MG = M<Y> = M<x>M<Y>; in the second case, the same argument with xand y interchanged yields that MG = M<x>, but this contradicts our assumption thatH n (x> M<x> n <x>. Thus MG = M<x>M<Y>, as desired.

6.3.6 Corollary. MMG = e F> if M is permutable in G = <M, F>.

Proof. By 6.3.5, MG = M(M<x> n (x> lx e F>. Now apply Theorem 6.3.5 to MGand the set-theoretic union of the M<x> n <x> (x e F) in place of G and F.

The above result will be used to determine the structure ofNNG/N

for a normalsubgroup N of G and a projectivity from G to G. The structure of MG/MM" is rathersimple, even for arbitrary permutable subgroups.

6.3.7 Theorem. If M per G, then MG/MMG is nilpotent of class at most 2.

Proof. Let H = MG and S = MH. Then S is generated by conjugates of M and henceis permutable in G. If x e G, then Sx a Hx = H and SS' < S(x>, so that SSx/S is acyclic normal subgroup of HIS. Since the automorphism group of a cyclic groupis abelian, SSx/S is centralized by the commutator subgroup (H/S)' of HIS. ButH = MG = SG and therefore HIS is generated by these SS'/S. It follows that(H/S)' < Z(H/S). Thus HIS is nilpotent of class at most 2.

As an application of Corollary 6.3.6 we prove the following rather technical resultthat will be needed in § 6.6.

6.3.8 Lemma. Let p e P, H a core free permutable p-subgroup of G and M = fl(H).(a) Then MG is an elementary abelian p-group.(b) If M is permutable in G and H is not normal in HG, then MG < HHG.

Proof. (a) By 6.2.10, a permutable maximal subgroup of a group is normal. There-fore every element of order p of G normalizes H9 and hence also its characteristicsubgroup S2(H9) = Ma; it follows that M9 < MG for every g E G.

Let X E G\NG(H) and T = H(x). By 6.2.3, o(x,H) = IT : HI < oo and thereforeT/HT is a finite group in which H/HT is a permutable p-subgroup. By 5.1.5, thisp-subgroup is normalized by the p'-component of xHT, and so 5.2.8 shows that0(H/HT) is elementary abelian. This group contains MHT/HT and hence M/M n HTis elementary abelian. Since HG = 1, the intersection of all these M n HT is trivialand so M is elementary abelian.

It follows that M9 n H = M9 n M and MM9/M M9/M9 n M = M9/M9 n HM9H/H for every g E G. Now M9/M9 n H is elementary abelian and M9H/H iscyclic since its subgroup lattice is isomorphic to an interval in [H(g)/H][(g)/(g) n H]. Thus MM9/M is cyclic of order dividing p. Since MM9 a MG, thegroup MG/M is generated by cyclic normal subgroups of order p and hence is anelementary abelian p-group. Since MG = 1, the same holds for MG.

(b) Let S = H"` and x c G. We have to show that M<X> < S. If x c NG(M), thenclearly M<X> = M < S; so suppose that x 0 NG(M). Then since M is permutable inG, 5.2.7 and 5.1.5 imply that o(x) is finite and divisible by p. Let P be the subgroupof order p in (x). By (a), M<x> < M(x) is elementary abelian and hence M<X) =MP. Let X = H<X> n (x). If X NG(H), then HP < Hx < S since X < HG; in par-ticular, M<X> = MP < S, as desired. So we may assume that

(6) X = H<X> n (x) < NG(H).

By assumption and 6.3.5, H is not normal in HG = H<H<Y) n (y)Iy c G) and hencethere exists y c G such that

(7) Y = Ho'> n (y) NG(H).

Let L = (H, x, y) and Z = H<Xy> n (xy). Now M g H and therefore 6.3.5 and 6.3.4applied to the permutable subgroup M of L = <H, y, xy) yield that

(8) M<X> < ML = M<y>M(Xy> = (M<y> n (Y))M(M<XY> n (xy>).

Of course, HL < HG and by 6.3.6, S = H"G > HHL = HXHY = HXHZ = HY = Hzsince X normalizes H. By (7), H < HY < H(y) and as HY is a p-group, it containsthe subgroup of order p of (y). Since M<)'> is elementary abelian, M<Y> n (y) is con-tained in this subgroup of order p of (y) and it follows that M<Y> n (y) < HY < S.Now Z NG(H) since HY = Hz, and the same argument with xy in place of yshows that M<Xy> n (xy) < Hz < S. By (8), M<x> < S, as desired.

We do not know whether the hypothesis that Q(H) is permutable in G is redun-dant in Lemma 6.3.8(b).

Groups generated by cyclic permutable subgroups

If N < G and (p is a projectivity from G to G, then NG/N is generated by subgroupsof the form ()V v NX)c '/N for x e G. These groups are cyclic permodular subgroups

of G/ N since N u NX < <N, x>. Therefore we shall need two further technical resultson groups containing many cyclic permutable subgroups.

6.3.9 Lemma. Let G be a p-group generated by a family X of cyclic permutablesubgroups, and suppose that S is a nonabelian subgroup of G such that every subgroupof S is permutable in G and S does not contain subgroups isomorphic to the quaterniongroup Q8. If S is not generated by two elements, then IS n f2(Z(G))I >_ p2.

Proof. Since any two subgroups of S permute, S is a nonabelian locally finite p-group with modular subgroup lattice. By 2.4.15, Exp S is finite. Furthermore, S isnot hamiltonian since it does not contain subgroups isomorphic to Q8, and by 2.5.9there exists an abelian p-group lattice-isomorphic to S. An abelian group of finiteexponent is a direct product of cyclic groups (see Robinson [1982], p. 105) and, sinceS is not generated by two elements, it follows that

(9) Q(S) is elementary abelian of order at least p'.

Let 1 g e f(S). Since <g) per G, we have I <x> <g> : <x> l < p, and hence <x> _<x> for every x e G. Thus g induces a power automorphism in G that centralizesS22(G) if p = 2; for, an element of order 4 inverted by g would generate, together withg, a dihedral group of order 8 in which <g> is not permutable. Let C, D e X. By5.2.14, CD is a metacyclic M-group and 2.3.24 shows that the power automorphisminduced by g in CD is universal if CD is an M*-group. If CD is not an M*-group, itis a metacyclic hamiltonian M-group and, by 2.3.8, the only such group is Q8i thenCD < S2,(G) is centralized by g. In any case, there exists r E 1L such that x9 = x' forall x e CD and, since o(g) = p and g centralizes S22(G) if p = 2,

(10) r - I (mod p), r - I (mod 4) if p = 2, and r° - I (mod I D I).

We have to consider two cases. If {ICI IC e 3:} is not bounded, we claim thatg e Z(G). If this is false, there is a C E X such that C = <c> and c9 i4 c. Then thereexists D = <d> e X such that o(d) > o(c) = p" and for the universal power auto-morphism x -+ x' induced by g in CD, we have r° - 1 (mod p""). It follows thatr = 1 (mod p") and hence cQ = c, a contradiction. This shows that g centralizesevery C e X and thus g e Z(G) since G = <C I C e I ). Thus S2(S) < Z(G) andIS n f(Z(G))I > p3.

Now assume that { I C I I C e X} is bounded and take C E X such that ICI is maxi-mal. This time, at least Co(s)(C) < Z(G). For, if g e C0(S)(C) and D E X, then thepower automorphism x -+ x' induced by g in CD satisfies r - 1 (mod ICI) and, sinceICI > IDI, it follows that g centralizes D; again G = <DID E I) implies g e Z(G).Now for p > 2, Aut C is cyclic and for p = 2, (10) shows that the automorphismsinduced by f(S) in C lie in the cyclic subgroup of Aut C generated by x -+ x5. Thusf2(S)rC0(5)(C) is cyclic and, by (9), IS n f(Z(G))I > p2.

6.3.10 Lemma. Let M be a permutable subgroup and F a subset of the group G suchthat H = <M, F) is a p-group for some prime p and I H : MI is finite. Assume furtherthat for all x, y e F, <M, x) is permutable in G and the interval [<M, x, y)/M] isisomorphic to the subgroup lattice of some p-group. Then every subgroup T such thatM < T < MH is permutable in G.

Proof. Our first aim is to show that we may assume that G is a finite p-group. Wehave to show that

(11) T <c> _ (c) T for every c e G

and for this we clearly may assume that G = (H, c> and M° = 1. As a join of thepermutable subgroups (M,x)', (x e F), H is permutable in G. So G = H(c> and wefirst consider the case that o(c, H) is infinite. Then [G/H] [(c)/(c) n H] is iso-morphic to the subgroup lattice of an infinite cyclic group and therefore o(g, H) isinfinite for every g e G\H. In particular, o(g, M) is infinite and by 6.2.3, g E NG(M)for all g e G`,,H. Since these elements generate G, it follows that M < G. But thenM" = M and the assertion is trivial.

So suppose that o(c, H) = I G : HI < ac. Then our hypothesis yields that IG : MI < ccand G is a finite group since M° = 1. To prove (11), we now may assume that o(c)is a power of some prime q. By 5.2.4, G/C°(M°) is a p-group. So if q 0 p, it followsthat c e C°(M°) and hence T(c> = (c> T. Therefore it remains to consider the caseq = p and, since H is a p-group, we are finally reduced to the situation that

(12) G is a finite p-group.

Now we argue by induction on IM": MI. If x, y c- F, then M<"> < M(x) and hence[M "'/M] is a chain. Furthermore, by assumption, [(M, x, y>/M] L(W) for somep-group W. Since (M, x> and (M, y> are modular in [ (M, x, y>/M], the group W isthe product of two cyclic permutable subgroups and hence is an M-group by 5.2.14.Then 2.5.9 shows that W is lattice-isomorphic to an abelian p-group or is hamil-tonian, and in the latter case, W Qg since it is generated by two elements. Thus forall x,yEF,

(13) [<M, x, y>/M] is isomorphic to L(Q8) or to the subgroup lattice of an abelianp-group generated by two elements.

Let x c F such that IM''> : MI is maximal. If M<''> < M<z> for every y e F, thenby 6.3.5, M" = (M<`'>ly e F> < Mos>, and hence [M"/M] is a chain. M" is permu-table in G being a join of conjugates of M and, by 5.2.10, T is permutable in G. Sosuppose that there exists y c F such that M`''' M<`l and let L = (M, x, y). Weclaim that there exists N < G such that

(14) M < N < M", NIM is elementary abelian of order p2 and every R < G sat-isfying M < R < N is permutable in G.

To see this, first note that by 6.3.5, ML = M<x>M<'", so that [MLIM] is not a chain.Therefore, if [ML,,M] had only one atom, it could not be isomorphic to the sub-group lattice of an abelian group and, by (13), it would follow that [L;'M] L(Q8).Since L is a p-group, ML < L and hence [ML/M] would be a chain, a contradiction.Thus [ML/ 114] has more than one atom and, if N is the join of two of these, M < Nand N, M is elementary abelian of order p2. again by (13). Let M < R < N. Anyelement w in an abelian p-group W = <a> x (b> has the form w = a"b' withintegers n = ip', m = jp', i _* 0 j (mod p) and r < s, say. So if a = a'b'ps thenw = a°' e <a> and W = (a,b>. If we apply this via (13) to [L/M] L(W), we seethat there exist u. t' E L such that L = <M, u, v> and R < (M, u). Since [ML/M] is

not a chain, M' = M<">M(L) M<`> and hence M < M(">. Now [M(u)/M] is achain containing R and M("> and, since IR : M = p, it follows that R < Me"'. Againby 5.2.10, R is permutable in G and (14) holds.

It is easy to see now that every R such that M < R < N satisfies the hypotheses ofthe lemma. Clearly, (R, F) = H is a finite p-group, and (R, x) = (M, x) R is per-mutable in G where x e F; if R < (M, x, y) with y c- F, then by (13), [(R, x, y)/R] =[<M, x, y)/R] is isomorphic to the subgroup lattice of an abelian p-group and, ifR (M, x, y), then [ (R, x, y)/R] [ (M, x, y)/M]. Furthermore RH = MH so thatRH : R1 = <c>TR is a subgroup of G. If there exists such an R satisfying<c, T) n R = M, then Dedekind's law implies that (c, T) _ <c, T) n (c> T R =(c) T((c, T) n R) = (c) TM = <c> T and therefore <c) T = T <c>. So assume thatR < (c, T) and hence

(15) (c, T) = TR (c) for all R such that M < R < N.

Then N < TR(c) = T(c)R and, again by Dedekind's law, N = (N n T<c>)R. Itfollows that N n T(c) R; let z e (N n T(c>)\R. Then z = td with t e T, andd e (c), and since N/M is elementary abelian, S = <M, z> satisfies M < S < N.By (14), S is permutable in G and, since M < S n T has index p in S, we have1ST: TI < p. Thus z e NG(T) and hence T = Tz = T`d = Ta. It follows thatS= (M, td><T(d)s T(c) and (15) yields that <c, T) = TS(c> s TT(c)(c> =

Thus <c, T) = T(c) and so, finally, T(c) = <c> T, as desired.

Using similar methods, Napolitani and Zacher [1983] and Cardin [1984] deter-mined the structure of groups G generated by a family X of (cyclic) permutablesubgroups such that for all C, D E 1, every subgroup of CD is permutable in G. Sucha group G is locally nilpotent (see Robinson [1982], p. 344, and recall that everypermutable subgroup is ascendant), so the problem divides into the nonperiodic caseand the investigation of p-groups. We give their results without proof.

6.3.11 Theorem. Let G be a group generated by a family X of cyclic permutable sub-groups such that. for all C, D E X, every subgroup of CD is permutable in G.

(a) (Napolitani and Zacher [1983]) If G is periodic, then G is metabelian.(b) (Cardin [1984]) I/'G is not periodic, then L(G) is modular.

The exact structure of p-groups with this property is given in Exercise 7. Napoli-tani and Zacher used their theorem to show that N"7'/N is metabelian for N < G anda projectivity from G to G. Following Busetto and Napolitani [1990], we shallsimilarly use 6.3.9 and 6.3.10 to prove the stronger result that NG/N is metabelian.

Exercises

1. (Gross [1971]) Let M be a core-free permutable subgroup of the group G.(a) If n e N and x, y e M such that x" = I = y", show that (xy)" = 1.(b) Show that T(M) = {x e Mlo(x) < oc} is a locally nilpotent and locally finite

subgroup of G. (Hint: Use 5.2.8 and Exercise 5.2.4.)

6.4 Lattice-theoretic characterizations of classes of infinite groups 311

2. (Gross [1975a]) Let M be a permutable subgroup of the group G and n e N.(a) If M is subnormal in G of defect at most n, that is, there exist M; < G such that

M = MO a M = G, show that M/MG is soluble with derived length atmost n - 1.

(b) Deduce from 6.2.12 that M/Mc is abelian if there exists g c G such thato(g, M) is infinite; compare with Exercise 6.2.1.

(c) Let G = <a, b, c l az = b8 = [a, c] = 1 , [a, b] = 64, b -' cb = c-'> and H = <a>.Show that H is permutable in G, o(c, H) is infinite and IH : Hot = 2.

3. Let (G;); E, be a collection of coprime (periodic) groups and suppose that for everyi e I, M. per G.. Show that M = Dr M. is permutable in G = Dr G..

IEI IEI

4. Let P = [P1,P2, . } be the set of primes and for every n c N, let G be a finitecontaining a core-free permutable subgroup M. with derived length n

(by Stonehewer [1974], such G. exist). Show that G = Dr G. contains a non-soluble core-free permutable subgroup. "E N

5. (Busetto [1980b]) Let n > 3 and G = <x,y!y2° = 1, y-'xy = x-'>. Show thatM = <xy`> is permutable in G, infinite cyclic and I M/MGI = 2n z.

6. (Busetto and Napolitani [1992]) Let p be a prime such that there exists an ex-tended Tarski group T of exponent p3 (see Olshanskii [1991], p. 344 for theexistence of such a prime). Let A be the group algebra of the Tarski groupT/Z(T) over the field with p elements, N the augmentation ideal of A and let Gbe the semidirect product of N by Twhere the action of T on N is induced byT;Z(T) via right multiplication in A.(a) Show that G is a finitely generated p-group (note that since T/Z(T) is finitely

generated, N is a finitely generated right ideal of A) with S2(G) = NQ(T)elementary abelian.

(b) Show that [S2(G), G] = [N, G] = N (note that since T/Z(T) is perfect, N2 = Nas an ideal of A) and hence G is perfect.

(c) If y e G\NZ(T), show that <gP2> = fl(T) (note that g = xy with x e T'\Z(T),yENandhence g"=xPwwith wcN).

(d) If M is a maximal subgroup of S2(G) such that S2(T) M N, show thatM is permutable in G, M/Mc is infinite and

.MG/MGis not hypercentrally

embedded in G.7. (Cardin [1984]) Show that the p-group G is generated by a family X of cyclic

permutable subgroups such that for all C, D E it, every subgroup of CD is permu-table in G, if and only if L(G) is modular or p = 2 and G = H x A where A isan elementary abelian 2-group and H is a central product of the form Q' De,Q8 D'' C4, Q8+' DS for some n e N in which the central subgroups of order 2 of the2n or 2n + I factors are identified.

6.4 Lattice-theoretic characterizations of classes of infinite groups

In this section we want to extend the lattice-theoretic characterizations of § 5.3to arbitrary groups.

Simple groups

6.4.1 Theorem (Zacher [1982b]). The group G is simple if and only if I and G arethe only permodular subgroups of G.

Proof If 1 and G are the only permodular subgroups of G, then G in particular hasno proper nontrivial normal subgroups, that is, G is simple. Conversely, let G besimple and suppose, for a contradiction, that there exists a permodular subgroup Mof G different from I and G. Then MG" = G and, by 6.2.19, M is permutable in G;furthermore, of course, MG = I and MG = G. We now follow an argument due toStonehewer [1972] and show that every finitely generated subgroup of G is residu-ally nilpotent and therefore has a normal series with abelian factors; such groups arecalled SI-groups (see Robinson [1982], p. 365). A theorem of Malcev (see Kurosh[1956], p. 183) will then imply that G is an SI-group, contradicting the simplicityof G. So let X = G = M°. Then there exist finitely many elementsq ,. . . , , gm E G such that X < My' ... M9"'. Let g E G and H = M9M9' ... M9^-. By6.3.2, H/(My)H is nilpotent and hence so is X/X n (M9)H. Since X n (M9)H < My, theintersection of all these normal subgroups of X is contained in MG = I and it followsthat X is residually nilpotent. F1

Since projectivities map permodular subgroups onto permodular subgroups, weget the following corollary.

6.4.2 Corollary (Zacher [1982a]). If G is a simple group and cp is a projectivity fromG to a group G, then G is also simple.

Perfect groups

A group is perfect if it equals its commutator subgroup, that is, if it has only trivialabelian factor groups. To express this property in the language of lattice theory,we have to find a good translation for "abelian group". The Tarski groups showthat "modular lattice" is inadequate for infinite groups; on the other hand, ourresults on groups with modular subgroup lattice show that we only have to avoidthe Tarski groups. So the following definition should work:

We call a lattice L permodular if it is modular and for all a, h e L such thath < a, [a/h] is a finite lattice whenever it has finite length.

6.4.3 Theorem. The following properties of the group G are equivalent.(a) L(G) is permodular.(b) L(G) is modular and G is soluble (or metahelian).(c) Every subgroup of G is permodular in G.

Proof. If L(G) is permodular, then no Tarski group can be involved in G and, by2.4.21, G is metabelian.

Now let L(G) be modular and G soluble. Then every subgroup M of G clearlyis modular in G. Let g E G and M < Y< (M,g> such that [(M,g>/Y] is a finitelattice. Then there exist subgroups X; such that Y = X0 < < X = (M, g> and Xiis maximal in Xi+, for all i. By 6.2.20, lXi+, : Xj is a prime and it follows that(M,g> : YJ < x. Thus M is permodular in G.

Finally, suppose that every subgroup of G is permodular in G. Then L(G) is clearlymodular. If B < A < G are such that [A/B] is of finite length, then there exists achain of subgroups B = AO < < A. = A with Ai maximal in Ai+1 for all i. SinceAi is permodular in G, the index 1Ai+1 : Ail is finite. It follows that JA : BI < x andhence [A/B] is a finite lattice. Thus L(G) is permodular.

Theorem 6.4.3 in particular implies that the subgroup lattice of an abelian groupis permodular. We can now characterize the class of perfect groups; however, it issomewhat easier to state this in terms of nonperfect groups.

6.4.4 Theorem (Zacher [ 1982b]). The group G is not perfect if and only if there existsa proper permodular subgroup M of G having one of the following three properties.

(1) [G/M] is a finite chain.(2) [G/M] is permodular and contains an element X such that [X/M] is isomorphic

to the subgroup lattice of an infinite cyclic group.(3) [G/M] is isomorphic to the partially ordered set (N, <) of positive integers

and every subgroup of M that is invariant under P(M) is permodular in G.

Proof. First suppose that G is not perfect. Then GIG' is a nontrivial abelian group.If there exists an element of infinite order in GIG', then M = G' satisfies (2). Sosuppose that GIG' is a torsion group. It is well-known (see Robinson [1982], p. 107)that an abelian group which is not torsion-free has a nontrivial direct factor whichis either cyclic of prime power order or quasicyclic; let M/G' be a complement tosuch a direct factor in GIG'. Then M is permodular in G and [G/ M] is a finitechain, that is (1) holds, or G/M Cp, for some prime p and hence [G/M] 2- (N, <)in this case. Every subgroup of M that is invariant under P(M) is characteristicin M, hence normal in G, in particular permodular in G. Thus (3) holds.

Conversely, suppose that M is a proper permodular subgroup of G having oneof the properties (1)-(3). If (1) is satisfied, then IG : MI < x and hence GIMG is afinite group. By 5.1.3, it is a p-group or a P-group; in any case, G is not perfect.If (2) holds and x E X\ M, then [(x)/(x) n M] a- [(M, x>/M] and hence o(x, M)is infinite. By 6.2.12, M a A(M) a G and by 6.4.3, G/A(M) and A(M)/.M are metabe-lian. Thus Gl'4' < M < G and G is not perfect. Finally, suppose that (3) holds and, fora contradiction, that G is perfect. Then G" = G and by 6.2.19,

(4) M is permutable in G.

Since [G/M] 2t (N, <), it follows that

(5) G is not finitely generated modulo M.

Furthermore, there exists a subgroup M, of G such that M is maximal in M1. SinceM is permutable in G, the index IM1 : MI = p is a prime. For every proper subgroup

H of G containing M, [HIM] is a finite chain and hence

(6) H = M <x> for some x e G.

Therefore J H : MI is finite and by 5.1.3,

(7) HIM, is a finite p-group.

If M/MH were cyclic for every M < H < G, it would follow that G" < M. For, (5)and (6) show that any four elements a, b, c, d of G together with M generate a propersubgroup H = M<x> of G. It follows by 5.2.13, that H/MH = (M/MH)<xMH> ismetacyclic, and hence [[a,b],[c,d]] E H" < MH < M. Thus G" < M, a contradic-tion. Hence there exists a noncyclic factor group M/MH and therefore a normalsubgroup K of M such that

(8) M/ K is elementary abelian of order p2.

Let D = n K. If a E P(M), then K° is permodular in M and hence I M : K°I < cc.aCP(M)

Suppose first that K° is not normal in M and let N1, ..., NP+1 be the maximal sub-groups of M containing K°. Then none of the Ni is normal in M; for, N;:9 Mwould imply that K° = N; n N, g NN for all j i and hence K° g M. Therefore(K°)M = M. Since the lattice [M/K"] is directly indecomposable, 5.1.14 showsthat M/(K°)M is a P-group or K" is permutable in M. In the latter case, 6.2.10would imply that K° a Ni for all i and hence again K° a M, a contradiction. ThusM,'(K°)M is a P-group and, since K° is contained in exactly p + I maximal sub-groups of M, it follows that M/(K°)M E P(n, p) for some n e N. If K° a M, then, as aprojective image of M/K, M/K° E P(2, p). In both cases, M" < K" and xP(P-1) E K"for every x E M. This holds for all a E P(M) and hence M" < D and xP(p-1) E D forall x c M. Thus

(9) MID is metabelian of exponent dividing p(p - 1);

we remark that by Exercise 6.5.8, M/D is an elementary abelian p-group. Nowsuppose, for a contradiction, that M < Dc. Since I M/K I = p2, there exist x, y E Msuch that M = <K, x, y> and by 6.2.7 there exists a subgroup L > D of G, finite-ly generated modulo D, such that x,y c- DL. Write L = and H =<M, z 1, ... , By (5) and (6), H = M <g> < G for some g c G and it follows that

<D, x, y> < DL < DH = D"' <e> = D<9' < <D, g>.

Since D is invariant under P(M), our assumption (3) implies that D is permodular inG and hence [<D,g>/D] [<g>/<g> n D] is a distributive lattice. It follows that<D, x. y>/D is cyclic and hence <K, x, y>/K = M/K is cyclic, a contradiction. Thus

(10) M $ D°;

let Dc = N. Then N # G and by (9), MN/N a- M/M n N is metabelian of exponentdividing p(p - 1). Since G is perfect, MN # G. It follows that [G/MN] (1, <) andMN/N is a permutable subgroup of GIN of exponent dividing p(p - 1). For everyelement uN in some conjugate of MN/N, the index I(MN/N)<uN> : MN/NI divides

o(uN) and hence p(p - 1); on the other hand, by (7), IMN : MNI is a power of p.Thus u is contained in the atom of the infinite chain [G/MN] and, since uN wasarbitrary, it follows that (MN)' < G. But then again [G/(MN)c] (N, <) and, by1.2.3, G/(MN)G is abelian, a final contradiction. Thus G is not perfect.

As in the finite case, we can use Theorem 6.4.4 to characterize the final term Gr,of the (transfinitely extended) derived series of G. For, this subgroup is clearly per-fect, it contains every perfect subgroup of G, and hence it is the maximum perfectsubgroup of G. We only note the usual corollary; recall that a group G for whichG = 1 is called hypoabelian.

6.4.5 Corollary (Napolitani [1982]). If cp is a projectivity from the group G to agroup G, then (G :< )' = G a,. In particular, the classes of perfect and of hypoabeliangroups are invariant under projectivities.

Generalized soluble groups

We want to generalize our characterizations 5.3.5 and 5.3.7 of finite soluble andsupersoluble groups. There are several classes of infinite groups that are candidatesfor such a generalization. We remind the reader that a group G is called hyperabelian(hypercyclic) if every nontrivial epimorphic image of G contains a nontrivial abelian(cyclic) normal subgroup. Equivalent is (see Robinson [1972a], p. 14) that thereexists an ascending series of normal subgroups N. of G, y an ordinal number,such that No = 1, N. = G, Na+1/N1 is abelian (cyclic) for every ordinal x < y and

Np = U Na for limit ordinals fl. If y can be chosen finite, G is called soluble (super-2<P

soluble). We first characterize the class of hyperabeliah groups and for this we haveto study permodular subgroups with permodular subgroup lattice.

6.4.6 Lemma. If M is a nontrivial permodular subgroup of G with permodular sub-group lattice, then MG contains a nontrivial abelian normal subgroup of G.

Proof. First consider the case that MG is torsion-free. Let g E G and L = <M, g>.By 6.2.8, I ML : MI and I ML : M-11 are finite and hence so is l ML : M n M91. By 2.4.9,M and M9 are abelian and therefore M n M9 < Z(M u M9). It follows thatM u M9/Z(M u M9) is finite. A well-known theorem of Schur's (see Robinson [1982],p. 278) states that then (M u M9)' is finite and hence trivial since MG is torsion-free.Thus M u Mg is abelian for every g E G. This shows that M< Z(MG), and thatZ(MG) is a nontrivial abelian normal subgroup of G.

Now suppose that MG is not torsion-free. Then there exist elements of prime orderp, say, in MG. If any two of these commute, then A = is anelementary abelian characteristic subgroup of MG, and hence a nontrivial abeliannormal subgroup of G contained in MG. So suppose that u, v are elements of orderp of MG such that [u, v] : 1. Then u = a,... a, and v = a,+l ... a where ai e M9i,

gi e G; let L= <M, g 1, ... , and H = M u Ma l u . u Mg-. By 6.2.21, H is per-modular in G. Clearly,

(11) u,veH<ML

and we want to show that [u, v] E HG. For this let x E G such that o(x, H) is a primepower and let T = <H, x>. By 6.2.13, T/HT is nonabelian of order rs or a finiter-group for primes r, s. In the first case, H/HT is abelian and hence [u, v] E HT; in thesecond, H/HT is a core-free permutable subgroup of the r-group T/HT with cyclicsupplement <xHT>, and then (c) of 5.2.8 shows that [u, v] E HT. Thus in both cases,[u, v] e HT and, since HG is the intersection of these HT, it follows that

(12) 1 0 [u, v] E HG.

By 6.2.8, IML : MI < oc. Since H< ML, it follows that IH : MI, and hence H/MH,is finite. By 5.2.5, MH/MH is soluble. If Ho = MH and H; = MH(M9i u . U M9j )

for i = 1, ..., n, then MH = Ho < < H = H, Hi mod H and [Hi,,/Hi] is iso-morphic to an interval in [H, u <gi+1 >/Hi] [<gi+1 >/<gi+1 > n Hi]. Hence every[Hi,,/Hi] is distributive and, by 5.3.7, H/MH is supersoluble. By 6.4.3, M is soluble.Since H/MH, MH/MH and M. are soluble, we finally conclude that H is soluble.By (12), HG 1, so it contains a nontrivial abelian characteristic subgroup A.Since HG < MG, we see that A is an abelian normal subgroup of G contained in MG.

6.4.7 Theorem (Busetto [1980a], Zacher [1982b]). The group G is hyperabelian ifand only if it has an ascending series of permodular subgroups M,, of G, y anordinal number, such that MO = 1, My = G, [Ma+1/Ma] is permodular for every ordi-nal x < y and M. = U M, for limit ordinals /1.

a<fl

Proof. If G is hyperabelian, there exists an ascending series (M,),<,, of normal sub-groups M, of G such that Ma,,/M, is abelian for every a, and this series clearly hasthe desired properties. Conversely, suppose that is a series with the propertiesstated in the theorem and let N be a proper normal subgroup of G. Define 6 =min{a < -II M,,:4 Then M, < N for every e < b and hence 6 is not a limit ordinal.It follows that Mb_1 < N < MaN and thus MN/N is a nontrivial permodular sub-group of G/N with subgroup lattice isomorphic to [Mb/Ma n N], that is, with per-modular subgroup lattice. By 6.4.6 there exists a nontrivial abelian normal subgroupin GIN. Thus G is hyperabelian.

It is interesting to note that the property of Theorem 6.4.7 with finite y, whichseems to be the most natural lattice-theoretic approximation of solubility, in factdoes not characterize the class of soluble groups. Indeed, Stonehewer [1973] hasconstructed a nonsoluble group G that is the product of two metabelian subgroupsH and K which are permutable in G and have modular subgroup lattice. By 6.4.3,L(H) and L(K) are permodular and so I < H < G, H is permodular in G, [H/1] and[G/H] [K/H n K] are permodular. If we restrict ourselves to finitely generatedgroups, the situation is better.

6.4.8 Theorem (Previato [1978]). Let G be a finitely generated group. Then G issoluble if and only if there exist subgroups Mi of G such that I = MO < < M = G,Mi is permodular in G and [Mi+,/Mi] is permodular (i = 0,...,n - 1).

Proof. The members of the derived series of a soluble group of course have the givenproperty. So suppose, conversely, that G has a chain of subgroups Mi as describedin the theorem. We show by induction on the length n of this chain that G is soluble.For this let M, = M; by 6.4.3, M is soluble. Since G is finitely generated, 6.2.8yields that CMG : MI < oo and hence MG/MMG is a finite group. The existence of achain 1 = Mo < . < M = G of modular subgroups Mi with modular intervals[Mi+,/Mi] is inherited by subgroups and factor groups. Therefore MG/MMQ has sucha chain and hence, by 5.3.5, is soluble. Thus MG is soluble and, since the chain of theM; MG/MG (i = I__ , n) in GIMG has length n - I and G/MG is finitely generated,the induction assumption implies that G/MG is soluble. Thus G is soluble.

Since the cyclic groups are characterized lattice-theoretically, it is possible todecide in the subgroup lattice whether a group is finitely generated. Therefore Theo-rem 6.4.8 yields a lattice-theoretic characterization of the class of finitely generatedsoluble groups. A lattice-theoretic characterization of the class of all soluble groupsis not known. However, we shall prove in 6.6.4 that this class at least is invariantunder projectivities.

Recall that a group G is polycyclic if there exists a finite chain 1 = Go < < G = Gof subgroups G; such that Gi a G;+, and G;+, /Gi is cyclic (i = 0, ... , n - 1). It iswell-known (see Robinson [1982], p. 147) that the polycyclic groups are preciselythe hyperabelian or (finitely generated) soluble groups satisfying the maximal condi-tion for subgroups. Therefore 6.4.7 and 6.4.8 both yield lattice-theoretic character-izations of the class of polycyclic groups. We give another characterization which iscloser to the above definition.

6.4.9 Theorem. The group G is polycyclic if and only if there exist subgroups Mi of Gsuch that I = Mo < < M. = G, M; is permodular in Mi+ [Mi+,/Mi] is distribu-tive and satisfies the maximal condition (i = 0,...,n - 1).

Proof. A polycyclic group clearly has this property. So suppose, conversely, that Gis a group with such a chain of subgroups Mi. We use induction on the length n ofthis chain to prove that G is polycyclic. The induction assumption implies thatis polycyclic; furthermore, is permodular in G. Since satisfies themaximal condition, there exists a subgroup M of G containing and maximalwith respect to these two properties. Suppose, for a contradiction, that M is notnormal in G. Then by 6.2.13 there exists x c J(M) such that M MX and, if T =<M, x>, T/MT is nonabelian of order pq or a finite p-group, p and q primes. In bothcases, T/ MT and MT are polycyclic and hence so is T. In particular, M U MX ispolycyclic and, by 6.2.6, permodular in G. This contradicts the maximality of M.Thus M g G and, since [G/M] is distributive and satisfies the maximal condition,G/M is cyclic. It follows that G = M is polycyclic.

We remark that we did not use the deeper results of § 6.2 to prove Theorems6.4.7-6.4.9; note that we can use 6.2.6(d) instead of 6.2.21 in the proof of 6.4.6.

Supersoluble groups

In the following characterization of supersoluble groups, however, we have to studycyclic permodular subgroups and for this we need Theorem 6.2.17.

6.4.10 Lemma. If M is a finite cyclic permodular subgroup of the group G, thenMG/MG is hypercyclically embedded in G.

Proof. We may assume that M. = 1. By 6.2.17, M is permutable or P-embeddedin G. In the first case, 5.2.12 yields that MG is hypercentrally embedded in G.In the second case, G = S x T where S is a direct product of P-groups, MG =S x (M n T)G and M n T is permutable in G. Clearly, S is hypercyclically embeddedin G and so is (M n T)G, again by 5.2.12. Finally (c) of 5.2.1 shows that MG is hyper-cyclically embedded in G.

6.4.11 Theorem (Busetto [1980b], Zacher [1982b]). The group G is supersoluble ifand only if there exist subgroups M. of G such that I = Mo < < M = G, M;is permodular in G, [M;+1/M;] is distributive and satisfies the maximal condition(i=0,...,n-1).

Proof. In the first place, every supersoluble group has this property. Suppose, con-versely, that G is a group with such a chain of subgroups M.. We use induction onthe length n of this chain to prove that G is supersoluble. Let M = M1; then M iscyclic. By 2.1.7, L(G) satisfies the maximal condition. In particular, G is finitelygenerated and 6.2.18 shows that MG/MG is periodic. Therefore M/MG is a finitecyclic permodular subgroup of G/MG and, by 6.4.10, MG/MG is hypercyclicallyembedded in G. Since G satisfies the maximal condition, this implies that there existsa finite chain MG = No < - - - < N,,, = MG of normal subgroups Ni of G such thatNi,, /N; is cyclic (i = 0, ... , m - 1). Since MG is cyclic and G/MG, by induction, issupersoluble, it follows that G is supersoluble.

To characterize the class of hypercyclic groups, one also has to study infinitecyclic permodular subgroups.

6.4.12 Lemma. Let M be an infinite cyclic subgroup of G and let T(MG) be the torsionsubgroup of MG.

(a) If M is permutable in G, then T(MG) < Zx(G).(b) If M is permodular in G, then T(MG) is hypercyclically embedded in G.

The proof of 6.4.12 is complicated. We refer the reader to Busetto [1980b] wherehe proves (a) and instead of (b) the corresponding result for modular subgroups inTarski-free groups; this proof can be carried over literally to permodular subgroups

in arbitrary groups. Note that Exercise 6.3.5 shows that in general Ma Zx(G) if Mis an infinite cyclic permutable subgroup of G. In the same way as 6.4.7 follows from6.4.6, the results on cyclic permodular subgroups yield the following theorem.

6.4.13 Theorem (Busetto [1980b], Zacher [1982b]). The group G is hypercyclic ifand only if it has an ascending series (M,),<, of permodular subgroups M. of G, "l anordinal number, such that Mo = 1, M., = G, [M,+I/M,] is distributive and satisfies themaximal condition for every ordinal x < y and M. = U M, for limit ordinals fl.

2<a

Further topics

There are a number of interesting open problems. First of all, we do not know alattice-theoretic characterization of the class of soluble groups; this is perhaps themost exciting problem altogether on subgroup lattices of groups. Furthermore, wewould like to know which classes of groups are described by properties (b) and (c)of Theorem 5.3.5 if the word "modular" throughout is replaced by "permodular".Finally, there are many classes of generalized soluble groups of which it is not knownwhether they are invariant under projectivities; naturally, no lattice-theoretic char-acterizations of these classes are known. An important exception is the class ofSN-groups, that is, of groups possessing a series with abelian factors (see Robinson[1982], p. 365). As a special case of a general theorem on varieties of groups, Hickinand Phillips [1973] prove that G is an SN-group if and only if for every finitelygenerated subgroup K I of G, K' < K. Using Theorem 6.4.4, it is now easy to givea lattice-theoretic characterization of this class of groups.

We conclude this section with a lattice-theoretic characterization of a somewhatdifferent type of class of groups; for similar results see Exercises 3--5.

6.4.14 Theorem (de Giovanni and Franciosi [1983]). The group G is residuallysupersoluble if and only if for every nontrivial element x in G there exist subgroups Miof G such that x 0 Mo < ... < M = G, M. is permodular in G, [Mi+1 /Mi] is distribu-tive and satisfies the maximal condition (i = 0, ... , n - 1).

Proof. A residually supersoluble group of course has the given property. Forthe proof of the converse, we need some preliminaries. Let us call a chainM = M0 < < M = G of permodular subgroups M; of G such that [Mi+1 /Mi] isdistributive and satisfies the maximal condition (i = 0, ... , n - 1) a supersolublechain connecting M and G. Then for N a G,

(13) MN/N = MON/N < . . < MAN/N = GIN

is a supersoluble chain connecting MN/N and G/N. In particular, for N = MG, thechain (13) connects MG/MG with G/MG and 6.4.11 shows that

(14) G/Mc is supersoluble.

By 2.1.7, [G/M] satisfies the maximal condition and hence G is finitely generatedmodulo M. Thus 6.2.8 yields that IMG : MI < xc. Therefore if G/Mc is finite, then

IG : All < x and G/ MG is a finite group. By 5.2.5, MG/ MG is hypercyclically em-bedded in G and, considering (14), we conclude that

(15) G,/MG is supersoluble if G/MG is finite.

If G/MG is not finite, then G/MG has a torsion-free normal subgroup R/MG of finiteindex (see Robinson [1982], p. 148). Every x e R \ Al' has infinite order modulo Aland, by 6.2.12, M is permutable in G, M < A(M) < G and R < MGA(M) < A(M).Thus IG : A(M)( < x and hence M has only finitely many conjugates M9i, .., M9^.Then Al' = M5i ... M9^ and, by 6.3.2, MG/(M9,)M,, is finite and nilpotent for all

n

i = 1, ..., n. Since MG = n (M9i)M,,, it follows that MG/MG is a finite nilpotenti=I

group. Considering (14), we obtain in this case, and hence in general,

(16) G'MG is polycyclic.

Now suppose that G is a group with the property given in the theorem and letI x e G. We have to find a normal subgroup N of G such that x 0 N and G/N issupersoluble. By assumption there exist a permodular subgroup M of G and asupersoluble chain M = All < < Mn = G connecting M and G such that x 0 M.Then by (16), G/MG is polycyclic and a well-known theorem of Malcev's (see Rob-inson [1982], p. 148) says that M is the intersection of subgroups of finite index inG. Since x 0 M, there exists such a subgroup H of finite index such that x 0 H; letN = HG. Then GiN is finite, x 0 N and (13) is a supersoluble chain connectingMN/N and G/N. Since MN < H, we have (MN)G = N, and therefore (15) appliedto MN/N shows that GIN is supersoluble. Thus G is residually supersoluble.

Finally we remark that most of the results of this section differ slightly from thetheorems proved in the cited papers. For example, the classes of hyperabelian, hyper-cyclic, and hence also of supersoluble (that is hypercyclic with maximal condition)groups were characterized by Busetto [1980a], [1980b] using properties of modularsubgroups in certain Tarski-free groups. Zacher [l982b] realized that the main toolsin these papers can be proved literally the same way for permodular subgroups ofarbitrary groups using the corresponding properties of permodular subgroups; thisyielded Lemmas 6.4.6, 6.4.10 and 6.4.12. He then stated the characterizations of hyper-abelian and of hypercyclic groups that are more or less Theorems 6.4.7 and 6.4.13.

Exercises

1. Let p > 2 be a prime, G = <a, h, xIaP = h° = [a, h] = 1, a' = ab, hx = h> andAl = <a, hx">. Show that M per G and <a> is invariant under P(M) but notpermodular in G.

2. Use 6.4.10 and 6.4.12 to prove 6.4.13.3. (de Giovanni and Franciosi [1983]) Show that the group G is residually poly-

cyclic if and only if for every nontrivial element x e G there exist subgroups M; ofG such that x Mo < < Mn = G, M. is permodular in G, [M;+,/Mi] is per-modular and satisfies the maximal condition (i = 0, ... , n - 1).

6.5 Projective images of normal subgroups of infinite groups 321

4. Show that the group G is residually polycyclic if and only if for every nontrivialelement x e G there exist a permodular subgroup M and subgroups M; of G suchthat x 0 M = MO < < M = G, M; is permodular in Mi+1, [M;+1 /M;] is dis-tributive and satisfies the maximal condition (i = 0, ... , n - 1).

5. Find a lattice-theoretic characterization for the class of residually finitely gener-ated soluble groups (de Giovanni and Franciosi [1983]).

6.5 Projective images of normal subgroups of infinite groupsand index preserving projectivities

The main result of this section will be that N' and NG are normal subgroups of G ifN a G and cp is a projectivity from G to G; therefore cp induces a projectivity from

to G/NE;. This, in particular, will reduce all problems about projective imagesof subgroups of finite index to the corresponding problems for finite groups. We usethis fact to give some applications to index preserving projectivities. We show thata projectivity is index preserving if it preserves all the indices in cyclic subgroups.This enables us to generalize the results of §4.2 on singular projectivities of finitegroups to locally finite groups; for example, we prove that every projectivity of sucha group induces an index preserving projectivity in the second commutator sub-group. We prove some results on singular projectivities of groups with elements ofinfinite order and show that index preserving projectivities of arbitrary groups mapascending subgroups to ascending subgroups and therefore preserve the Gruenbergradical. Finally we give a lattice-theoretic characterization of the locally finite radi-cal of a group.

Criteria for normality

First of all we give a number of useful criteria for a projective image of a normalsubgroup to be normal. All these are immediate consequences of our results onpermodular subgroups in § 6.2.

6.5.1 Theorem (Zacher [1980]). If N a G and GIN is infinite cyclic, then N1, a G forevery projectivity tp from G to a group G.

Proof: By 6.2.1, N`° is permodular in G. Let x e G such that G = <N, x). ThenG=N' u <x> and <x> =(y) is cyclic. Since [(y)/(y) n N']^-[(x)/(x) n N]a-L(G/N), it follows that o(y,N") is infinite. By 6.2.3, y normalizes N`° and henceNIP aG.

In the sequel we frequently have to study nonnormal projective images of normalsubgroups. We collect the basic features of this situation.

6.5.2 Remark. Let cp be a projectivity from G to G, let N 9 G and suppose thatN° = N is not normal in G. Since N is permodular in G, 6.2.13 shows that

(1) Nc = n tN<N,v>iy e J(N)}.

In particular, there exists y E J(N) such that NY 0 N; let T, D, M < G such thatT = <N, y>, D = NT and M = N. As a projective image of N,

(2) M = N`°y`0 ' is permodular in G.

Furthermore, [T/N`] [T/N] ^-, [<y>/<y> n N] is a finite chain and henceNnNY <N<NuNl'.Thus

(3) D < N n M < N < NM < T and [TIN] [TIM] is a finite chain;

it follows that

(4) TIN is cyclic of prime power order.

Finally, (9) in 6.2.13 shows that

(5) T/D is a finite p-group

or nonabelian of order pq for primes p, q. In the second case let p > q and LID be aminimal subgroup of T/D different from N/D and M/D. Then D = N n M < M andD = N n L:9 L and hence D g M u L = T. Being a projective image of T/D, thegroup T/D belongs to P(2, p); since N _< T but N is not normal in T there exists aprime r such that

(6) IT/DI =pr,IN/DI =p>r,IT/DI =pq,N/DI =q<p.

We shall show in 6.5.6 that D < T. Therefore if (5) holds, then (p -' induces a projec-tivity from the nonabelian p-group T/D onto T/D and 2.2.6 implies that

(7) T/D and T/D are finite p-groups.

Note that for the moment only (5) or (6) holds in the situation we consider: as soonas 6.5.6 is proved, we shall know that (6) or (7) is satisfied.

The above discussion in particular yields (where again 6.5.6 is used for the lastassertion in (b)):

6.5.3 Theorem. Let cp be a projectivity from the group G to the group G, let N < Gand K < G such that K' = (N`°)6.

(a) if I G : NI = r is a prime, then either N" :9 6 or there exist primes p and q suchthat G/K`° and G/K are contained in P(2, p) and, more precisely,

I G/K I = pr, I N/K I = p > r, I G/K`° I = pq, I N`°/K° I = q 2. Then G/K' is a finite primarygroup; if N" is not normal in G, then G/K and G/K" are finite p-groups.

Proof. By assumption, G/N is cyclic of prime power order; let x e G such that G =<N, x> and let <x>" = <y>. If N" is not normal in G, then (N")'' N" and T = Gsatisfies the assumptions of 6.5.2. If IG : NI = r is a prime, then (5) cannot hold sincemaximal subgroups of finite p-groups are normal. Therefore (6) is satisfied and thatis the assertion in (a). Now suppose that IG/NI = p" and n >_ 2. Then (6) cannot holdand hence G/K" is a finite primary group. Clearly this is also true if N" a G. SinceGIN is a p-group, the last assertion in (b) follows from (7); for this we need 6.5.6.

In the following criterion we extend 5.4.9 (a) to infinite groups. Exercise 2, how-ever, shows that (b) and (c) of 5.4.9 do not hold for arbitrary groups. Since the groupG in this example is perfect, it shows in addition that normal subgroups of perfectgroups in general are not mapped to normal subgroups under projectivities (com-pare with Theorem 5.3.4 for finite groups).

6.5.4 Theorem. Let N be a normal subgroup of the group G and suppose that one ofthe following is satisfied:

(a) G/N has no nontrivial cyclic normal subgroup.(b) N is perfect.(c) N has no proper subgroup offinite index. _

Then N" a G for every projectivity co from G to a group G.

Proof. Suppose, for a contradiction, that N" is not normal in G and let y, T, M beas in 6.5.2. Then N n M < N < NM and [N/N n M] [NM/ M] is a finite chain.By 6.2.6, N n M is permodular in N and 6.2.5 yields that IN: N n M I is finite. Thiscontradicts (c) and also (b) because of 5.1.3. Furthermore, NM/N is a finite cyclicpermodular subgroup in G/N. By 6.4.10 there exists a nontrivial cyclic normal sub-group in G/N and this contradicts (a). Thus we get a contradiction if any of our threeassumptions hold. It follows that N" < G in all cases.

The normality of N° and Ni

We come to the main result of this section. We want to show that the preimages ofthe core and the normal closure of a projective image of a normal subgroup arenormal subgroups. The main step in the proof of this theorem is the following partialresult.

6.5.5 Lemma. Let N a G and G/N be finitely generated. If cp is a projectivity fromG to a group G and H" = (N")°, then H a G.

Proof. Let N" = N_and K = N. Then N is permodular in G and G is finitelygenerated modulo N. By 6.2.8, INc : NI < oc, hence [H/N] and therefore also

(8) 1 H : N I is finite.

We suppose, for a contradiction, that the lemma is false and choose a counter-example in which IH: NI is as small as possible. For every normal subgroup L of G

such that N < L < H, we have (LQ)G = H° and the minimality of I H : NJ implies thatL = N. Thus N is the maximal normal subgroup of G contained in H and since H isnot normal in G,

(9) N = HG < H.

If N is not permutable in G, then by 6.2.18, N is P-embedded in G. Thus G/K isperiodic, G/K = S/K x U/K with coprime groups S/K and U/K and S/K is a P-group whose nonnormal Sylow subgroup is N n S/K. By 1.6.4, L(G/K) L(S/K) xL(U/K) and [G/N] [S u N/N] x [U u N/N]. Now cp induces an isomorphismfrom L(G/N) onto this lattice and it follows from 1.6.5 that G/N = SN/N x UN/N.In particular, SN a G; but SN < H since 9!5; NG = H, and (9) implies that SN = N,a contradiction. Thus

(10) N is permutable in G.

Since N is not normal in G, there exists y c J(N) such that N" 0 N; let T = <N, y)and M = Ny as in 6.5.2 and let x e G such that <x> = <y)'. Then T = <N, x> andT/N is cyclic of order p" for some prime p and n e N. By 6.2.6, MN/N is permodularin G/N; furthermore, R< N u Ny < H and (9) imply that (MN)G = N. If MN is notpermutable in G, then by 6.2.18, MN is P-embedded in G; hence G/N is periodic andthere exist coprime groups S/N and U/N such that G/N = S/N x U/N and S/N is aP-group containing the p-group MN/N. Since N is subnormal but not normal in T,we see that [T/N] is a chain of length at least 2. Thus T/N is cyclic of order at leastp2 and must be contained in the P-group S/N, a contradiction. We conclude thatMN/N is a cyclic permutable subgroup of G/N. By 5.2.11, every subgroup of MN/Nis permutable in G/N. In particular,

(11) f)(T/N) = R/N is a permutable subgroup of order p of G/N contained inH/N.

By (9), R is not normal in G and (b) of 5.2.9 shows that we may choose z E G suchthat Rz -A R and o(zN) = p'° for some m E N. Let W = <N, x, z>, p be the projectivityinduced by p in W and Hg = (N')W. By 5.2.9, RG/N = R/N x (RG/N nZ(G/N)) and(9) implies that H n RG = R. Since x e W and R < MN, we have R < NW = Ho < Hand hence also R = Ho n RG. So if Ho < W, then R = Ho n RG < W, contradictingR : R. Thus Ho is not normal in W, that is, the lemma is false for the normalsubgroup N of W and the projectivity p. Therefore we may assume without loss ofgenerality that G = W, that is

(12) G = <N, x, z>.

Now let A/N be a cyclic q-subgroup of G/N for some prime q 0 p. Let A = <N, a>and <a> = ; then b e J(N). Therefore if N6 N, then, applying (11) to <N, a> =A in place of <N, x> = T, we obtain that f2(A/N) = R, IN is a permutable subgroupof order q of GIN contained in H/N. By 5.2.9, R, IN is centralized by the p-elementsxN and zN, and thus R, < G, contradicting (9). Thus Nb = N and hence

(13) N° -<A'*.

Write C/N = CG/N(R/N) and suppose next that A/N is a cyclic p-subgroup of GINsuch that A C. Then by (e) of 5.2.9, S2(A/N) < Z(G/N) and (9) implies thatA n H = N. Since H" g G, it follows that N" = A" n H" < A" in this case also.Finally, let B/N be any cyclic subgroup of GIN such that B $ C. Then by 5.2.9, B/Nis finite and its p'-part is contained in C/N. It follows that its p-part is not containedin C/N and therefore (13) holds for every Sylow subgroup A/N of B/N; thusN" < B". Now C/N G/N since RZ R, and hence G/N is generated by the cyclicsubgroups B/N not contained in C/N. It follows that N" < G and H = N < G, acontradiction.

6.5.6 Theorem (Busetto [1982]). Let N_be a normal subgroup of the group G, cp aprojectivity from G to a group G, H = NG and K = NG, that is H" the normal closureand K" the core of N" in G. Then H and K are normal subgroups of G.

Proof. We first show that H g G. So take g E G and let X be the set of all subgroupsF of G such that <N, g> < F and F/N is finitely generated. For F E I, let H(F) be thesubgroup of _F such that H(F)" = NF. By 6.5.5, H(F) < F; in particular, H(F)9 =H(F). Since NF < NE, we have H(F) < H; on the other hand, to every y E G thereexists x e G such that <x>" = <y> and hence N'' < NF for F = <N, g, x> E X. ThusH = <H(F)IF E 1>. It follows that H9 = H and since g was arbitrary, H < G.

We have proved the assertion of the theorem about the normal closure for arbi-trary normal subgroups and projectivities, of course. This result immediately impliesthe assertion for the core. For, if we apply it to the normal subgroup K" = NG of Gand the projectivity cp-1, it yields that (KG)" < G. Since K < N < G, we obtainK < KG < N, and hence NG= K9< (KG)" < N. But NZ; is the largest normal sub-group of G contained in N. It follows that (KG)" = NG = K" and hence K =KG<G.

Subgroups of finite index

An immediate consequence of Busetto's theorem and the Zacher-Rips Theorem isthe following result that was first proved by Rips in an unpublished manuscript.

6.5.7 Theorem (Rips). Let H be asubgroup of finite index of the group G and let cp bea projectivity from G to a group G. Then there exists a normal subgroup K of G suchthat K < H, K" < G and G/K and G/K" are finite groups.

Proof. Let N = HG and K < G such that K" = (N")z;. Then K < H, K" < G and, by6.5.6, K < G. Furthermore I G : NJ < oo and therefore by 6.1.7 I G : N"I < oc. It fol-lows that G/K" is finite and, as a projective image of this group, G/K is finite,too.

Index preserving projectivities

We give an application of Rips's theorem to a problem on index preserving projec-tivities. It is easy to see that a projectivity 9 between finite groups is index preserving

if it preserves indices in the cyclic subgroups. For, if this is the case, (p has to mapSylow p-subgroups to p-groups of the same order and 4.2.1 yields the assertion. Itwas unknown for some time whether this was also true for infinite groups.

6.5.8 Theorem (Zacher [1980]). A projectivity is index preserving if and only if itinduces in every cyclic subgroup an index preserving projectivity.

Proof. One implication is trivial. To prove the other, let (p be a projectivity fromG to G inducing in every cyclic subgroup an index preserving projectivity, andlet K < H < G. If I H : K I is infinite, then so is 1W': K`°I, by 6.1.7; so suppose thatI H : K I is finite. Then by 6.5.7 there exists a normal subgroup N of H such thatN < K, N`P a H11 and H/N and H`°/N`0 are finite. If A/N is a cyclic subgroup ofH/N and B/ N < A/N, then there exists a c H such that A/N = <aN> = <a> N/Nand hence B/N = (B n <a>)N/N. It follows that A`°/N`° = <a> PN`°/N`°, B`°/N`0 =)°N'/N`0 and hence

I A`0/N" : B`°/N`° I = I <a>": (B r <a> )`° I = I <a>: B n <a> I = I A/N : B/N I.

Thus the projectivity induced by cp in H/N induces index preserving projectivities inthe cyclic subgroups of H/N and hence is index preserving since H/N is finite. AsN < K < H, it follows that I H`° : K`° I = I H : K I, as desired.

Not much is known about singular projectivities between infinite torsion groups.For example, any two Tarski groups have isomorphic subgroup lattices and there-fore these groups possess singular projectivities. On the other hand, using Theorem6.5.8, it is not difficult to generalize the main results of §4.2 to locally finite groups;we do this for Corollary 4.2.9.

6.5.9 Theorem. Every projectivity of a locally finite group G induces an index preserv-ing projectivity in the second commutator subgroup G".

Proof. By 6.5.8, we have to show that for every x e G", I <x> "l = I <x>1. Now x is aproduct of commutators of elements of G' which in turn are products of commuta-tors of elements of G. Hence there exists a finitely generated subgroup F of G suchthat x c- V. Since G is locally finite, F is finite and, by 4.2.9, cp induces an indexpreserving projectivity in V. It follows that I (x> 11J = I <x>1, as desired.

Some results on singular projectivities of groups with elements of infinite ordercan be found in Zacher [1982a]. We prove one of these which will be used in the nextsection. For this we need two simple lemmas.

6.5.10 Lemma. Let (p be a projectivity from the cyclic group X to the group Y,let K < H < X such that I H : KI = p is a prime and suppose that I H" : K111 = q. 1fU < V < X such that I V : UI = p" (n e N), then I V": U`°I = q". (We say that (p mapsp-indices in X to q-indices in Y.)

Proof. There exists N < X such that I X : N I = p; let I Y : N" I = r. If S < X such thatIX : SI = p' (m c r\i), then XIS is cyclic of order p', and hence Y/S" is cyclic of order.s' for some prime s. Since S < N, we have that r divides the order of Y/S" and henceI Y : S"I = r'°. Now suppose that U < V < X such that I V : UI = p" and let IX: VI =

pk t where (p, t) = 1, k > 0. Then if R is the subgroup of index pk and T that of

index pk+" in X, it follows that T n V = U and TV = R. Therefore I V" : U"I =I VIP: T" n V"I = I R" : T"I = r". If we apply this to V = H and U = K, we get thatr = q, as desired.

6.5.11 Lemma. Let G = XP where X is infinite cyclic and IPI = p. If cp is a projec-tivity from G to a group G, then I P"I = p and cp maps p-indices in X to p-indices in X".

Proof: G/XG is finite since I G : X1 = p. Thus XG is infinite cyclic; let N be the sub-group of index pz in XG and H = XGP. We want to show that N" < H". If XG iscentralized by P, then H = XG x P has exactly p infinite cyclic subgroups inter-secting in the subgroup of index p in X. The image of this subgroup is centralizedby the images of these infinite cyclic groups and hence by H"; in particular,N" < Z(H'°). If XG is not centralized by P, then p = 2, X = XG and G is an infinitedihedral group. In this case, X is the only infinite cyclic subgroup of G and henceX" < G. In both cases, N" < H" and cp induces a projectivity from H/N to H"/N".Since H/N is neither cyclic nor elementary abelian, 2.2.6 shows that H"/N" is ap-group. It follows that IP"I = I P"N"/ N"I = p and, by 6.5.10, cp maps p-indices in Xto p-indices in X".

6.5.12 Theorem. Suppose that the group G has an abelian normal subgroup A contain-ing elements of infinite order and let qp be a projectivity from G to a group G.

(a) Then I HIP I =IHI for every subgroup H of G.(b) If there exists a subgroup of order p in G, p a prime, then I V11': U"I = p for all

U < V < A such that I V : U I = p (that is, (p maps p-indices in A to p-indices in A").

Proof. If IHI is infinite, then so is H"I; thus in the proof of (a), we may assume thatIHI is finite. If (a) holds for all subgroups of prime order in G, then cp induces indexpreserving projectivities in the Sylow subgroups of H, and 4.2.1 shows that IH"I _CHI. Therefore to prove (a), we may assume that

(14) CHI = p is a prime.

This is also the assumption in (b) and we now prove both assertions simultaneously.More precisely, we show that IH"I = p, that is (a) holds, and that

( 1 5 ) I V" : U" I = p if U < V < A, I V : U I = p and V is infinite cyclic.

For this we clearly may assume that G = AH. Let V be an infinite cyclic subgroupof A and write B = <V"Ih e H>. Then B is invariant under H and hence B a G. Asa finitely generated abelian group,

(16) B = B, x x B, with cyclic groups B. (r e N).

Since B is infinite, at least one of the B; is also infinite, say B,. If all the other B; arefinite, then I B : B, I is finite, hence so is I B : B, I for all h c- H and, since H = p, I B : Xis finite where X = n B;. It follows that X is an infinite cyclic group invariant

hcHunder H and 6.5.11, applied to XH, yields that I H11 = p and that cp maps p-indicesin X to p-indices in X". Since I B : XI is finite, X n V 0 1 and then 6.5.10 implies thatI V`° : U`°I = p. Thus (15) holds in this case.

Now suppose that B; is infinite for some i > 2, let N = <b°2Ib e B> and L = BH.Then N < L and L/N is a finite p-group that is neither cyclic nor elementary abelian.By 2.6.10, cp is induced by an isomorphism on B. In particular, I VP: U`°I = p, N`0 < B`°and I N : K I = I N": K" I for every subgroup K of N. By (a) of 6.5.3, N`° a (NH)".Thus M° < L`° and cp induces a projectivity from L/N to L`°/N4'. By 2.2.6, L`°/N`° is ap-group and hence I H4I = H4N`°/NI = p. This completes the proof of (a) and of(15).

We finally prove (b) for arbitrary V < A. Since I V : UI = p, we have V = U<g>for some g E A and I<g>: <g> n UI = IV: UI = p. Then V`° = U`0 u <g>4' and, by(a) or (15), I <g>4: <g>" n UPI = p. So if U`0 < VP, it follows that I VI: U4I =I <g>4': <g>4' n U4I = p. If U`° is not normal in V`° and K`° = (U`°),,., then 6.5.3(a)yields that V/K and V`0/K`0 are in P(2, s) where s = I V° : U" 1. Since V is abelian, V/Kis elementary abelian of order sz and it follows that s = I V : UI = p, as desired.

We mention without proof a much deeper result.

6.5.13 Theorem (Stonehewer and Zacher [1991b]). Let cp be a projectivity from G toG and suppose that M is a permodular subgroup of G having an ascending series withfactors locally finite or abelian. Let h(M) be the sum of the torsion free ranks of theabelian.factors of this series and assume that h(M) >_ 3 or h(M) = 2 and either L(M) ismodular or there exists g e G such that o(g, M) is infinite. Then cp is index preserving.

Ascendant subgroups and the Gruenberg radical

We finally generalize 5.4.11 to infinite groups and apply the result to the image of theGruenberg radical under a projectivity.

6.5.14 Lemma. Let cp be an index preserving projectivity from G to G.(a) if N < G, then M° is permutable in G. _(b) If N is ascendant in G, then N`° is ascendant in G.

Proof. (a) We have to show that M°<y> = < y> N`° for all y e G. For this purpose letx be an element of G such that <x)4' = <y> and put T = <N,x>. If TIN is infinite,then even N" < T`0 = <N4, y), by 6.5.1. If T/ N is finite, then by 6.5.7 there exists anormal subgroup K of T such that K < N, K`0 a T`° and T/ K and T4'/ K4' are finite.Since co induces an index preserving projectivity in T/ K, 5.4.11 implies that N'0/K10is permutable in T4/K`° and hence N4'<y> = N4K4<y> = M°<y>K4' = <y>K4N`° _<y> N`0.

(b) Let (N,),< be an ascending series such that No = N and N., = G. Then (a) and6.2.10 imply that (N,"),,,, is an ascending chain such that No = Nl*, N; is ascendantin N.",, for all x < y and N' = G. By transfinite induction, we obtain that N° isascendant in N; for all x < y. Thus N° is ascendant in G.

Recall that the Gruenberg radical of a group G is the join of all the cyclic ascen-dant subgroups of G. Now if (p is an index preserving projectivity from G to G, thencp-' is also index preserving and 6.5.14 shows that cp and (p -' map cyclic ascendantsubgroups to cyclic ascendant subgroups. It follows that the joins of all these sub-groups in G and G are mapped onto each other. Thus we have the following result.

6.5.15 Theorem (Zacher [1982a]). Every index preserving projectivity from G to Gmaps the Gruenberg radical of G to the Gruenberg radical of G.

The locally finite radical

We finally give a lattice-theoretic characterization of the locally finite radical GL%rthe join of all locally finite normal subgroups, of a group G. We also show that G'11is locally finite, a classical result in the theory of radicals (see Robinson [1972a], p.35). We need the following simple lemma.

6.5.16 Lemma. Let H and K be locally finite subgroups of a group G. If H is per-modular in G, then H u K is locally finite.

Proof. Let F be a finitely generated subgroup of H u K. Since every generator of Fis a product of finitely many elements of H and K, there exist finitely generatedsubgroups A of H and B of K such that F < A u B. Since H and K are locally finite,A and B are finite. By 6.2.6, M = H n (A u B) is permodular in A u B. ClearlyA< M< A u B, so A u B = M u B and, since M is modular in A u B, we see that[A u B/M] [B/B n M] is a finite lattice. By 6.2.5, JA u B : MI is finite. It followsthat M is finitely generated (see Robinson [1982], p. 36); therefore, as a subgroup ofH, it is finite. Thus A u B is finite and hence so is F.

6.5.17 Theorem (Stonehewer and Zacher [1994]). If G is a group, then GLI is locallyfinite and is the join of all locally finite permodular subgroups of G. In particular,(GL'Y)° = 6L-"for every projectivity (p from G to a group G.

Proof. Since every normal subgroup is permodular, GL's is contained in the join J ofall locally finite permodular subgroups of G. If F is a finitely generated subgroup ofJ, then F is contained in a join K of finitely many locally finite permodular sub-groups of G. By 6.2.21 and 6.5.16, the join of any two of these is permodular andlocally finite, and hence so is K. Thus F is finite and J is locally finite. It follows thatJ < GL'i and so GL's = J is locally finite. By 1.2.12 and 6.2.1, every projectivity mapslocally finite permodular subgroups to locally finite permodular subgroups. It fol-lows that (GL',)`' = GL't if cp is a projectivity from G to G.

Exercises

In Exercises 1, 3, 4, and 7 assume that rp is a projectivity from G to G. _1. (Zacher [I 982a]) If N < G and G/N is a finite perfect group, show that N° < G.2. (Busetto and Napolitani [1992]) Let G be the group constructed in Exercise

6.3.6, let i2(T) = (eo) and N = Ke1) x N, where <e1) is cyclic of order p. Let abe the automorphism of n(G) satisfying eo = eo, ei = eoe, and x° = x for allx e N,. Show that there exists an autoprojectivity cp of G induced by a on n(G)and by the identity on [G/S2(T)]. Note that N° is not normal in G although G isperfect and there exists no normal subgroup L of G such that L < N and NIL iscyclic.

3. Show that cp is index preserving if and only if for all U < V < G such that U is anormal subgroup of prime index in V, U' < V° and I VW : U1*J = I V : UI. _

4. If cp is index preserving and N g G, show that N' is w-ascendant in G, thatis, there exist subgroups Hi of G (i e NO) such that Ho = N", Hi < H;+1 forall i and U Hi = G. (Hint: Define H1= <g e G1o(g, NW) E P u {oc}) and H;+, _

iE:N<g cGIo(g, Hi) e P).)

5. Generalize Theorem 4.2.7 to locally finite groups.6. Let B be a normal subgroup of prime index in the group A and let (C1)iE, be a

family of normal subgroups of A such that Ci < B and A/Ci is a P-group for alli E I. If <Cili c- /) is a proper subgroup of B, show that A/n Ci is a P-group.

iEI

7. (Zacher [1980]) Let N 9 G such that GIN E P(n, p), p a prime, 2 < n < oc andlet K = (NP)d. Show that either N° 9 G or IN: K I = p and G/K e P(n + 1, p).

8. (Zacher [1982a]) Let N < G such that GIN is elementary abelian of order p",2 < n < oc. Show that G/ n Na is an elementary abelian p-group.

a E P(G)

9. (Zacher [1980]) Let N < G such that I G/N I is a prime. Show that G/ n Na isabelian or a P-group. ,IEP(G)

10. (Stonehewer and Zacher [1991b]) Let G be the semidirect product of a sub-group A of (Q, +) containing Z by an infinite cyclic group <g>, and suppose thatthe action of g on A is multiplication by m/n, (m, n) = 1. If p and q are primesdividing mn, show that there exists an autoprojectivity go of G and an elementx c A such that rp maps p-indices in <x> to q-indices in (x)4'.

6.6 The structure of NG/NG_ and TV-'ITV-6 andprojective images of soluble groups

We come now to the deeper results on the structure of NG/NU and NG/& for anormal subgroup N of G and a projectivity from G to G. The main theorem will bethat NG/NZ; and NG/N are soluble of derived length 4 and 5, respectively. As forfinite groups, we first prove that N is permutable or P-embedded in G. This willreduce our problem, and also many others, to the case that N is permutable in G.After further reductions we may assume that G and G are p-groups and LNG: NJ is

6.6 The structure of N',/N0- and N°,/N- and projective images of soluble groups 331

finite so that we can apply the results of §§ 5.5 and 6.3. As an important applicationof our methods we get that projective images of soluble groups are soluble andobtain a bound for the derived length of the image group.

The structure of N/N0- and N/N0-

We first study the structure of N/N0- and Here we reap the reward of ourefforts in § 5.5.

6.6.1 Theorem (Busetto and Napolitani [1991]). Let N be a normal subgroup of thegroup G and tp a project ivity.from G to a group G. Then N/N6- and N/N0- are subdirectproducts of finite primary groups with modular subgroup lattices; in particular, theyare metabelian. Furthermore, N/Nc is nilpotent of class at most 2 with commutatorsubgroup of exponent at most 2.

Proof. We shall prove all the assertions simultaneously. Since, by 6.5.6, rp induces aprojectivity from_ G/N to we may assume that & = I and, of course, N : I.Then by 6.5.2, N is the intersection of the N<N,Y> where y e J(N). If the statementsof the theorem hold for all these (N, y) in place of G, then they also hold for G;therefore we may finally assume that G = (N, y) for some y e J(N), and hence thatGIN is cyclic of prime power order. By 6.5.3, G and G then lie in P(2, p) or are finitep-groups for some prime p. In the first case, N and N are cyclic and, clearly, all theassertions of the theorem hold. In the second case, Hypothesis 5.5.1 is satisfied. By5.5.2 and 5.5.8, N and N are p-groups with modular subgroup lattices and I N' I < 2.In particular, N and N are metabelian.

It is not known whether N/NZ; is nilpotent (of class at most 2). In Example 5.5.5and Exercise 5.5.2, N0- = 1, c(N) = 2 and N' is cyclic of order 4 and p, respectively. Itis another interesting open problem whether N/N0- is an M-group and IN' I < 2 forarbitrary groups G.

Projective images of soluble groups

Another immediate consequence of the results in §§ 5.5 and 6.5 is the theorem onprojective images of soluble groups which we are now going to present. More gener-ally, we study projective images of normal subgroups with soluble factor group.Recall that for n e N, G(" is the n-th term of the derived series of G and G" =<g"Ig c- G); thus GZ is the smallest normal subgroup of G whose factor group is anelementary abelian 2-group. The basic result is the following lemma.

6.6.2 Lemma. Let N be a normal subgroup with abelian factor group of the group G.If co is a projectivity from G to a group G, then (G")2 and (N`'GZ)" are contained in N`0;in particular, 6(3) < N°.

Proof. We have to show that all the natural generators of (G")2 and (R62 )" arecontained in N. Since these generators are squares of products of iterated commuta-tors, or iterated commutators of products of elements of N with squares of elementsof G, we may assume that G is finitely generated modulo N. Then GIN is a finitelygenerated abelian group and hence a direct product of cyclic groups. It follows thatN is the intersection of normal subgroups of G_with cyclic factor groups of infiniteor prime power order. Let M be one of these. If M a G, then G/M is cyclic and henceG' < M. If M is not normal in G, then by 6.5.1, G/M is finite and 6.5.3 shows thatG/MZ; and G/MG- are either P-groups or finite p-groups for some prime p. In the firstcase, G" < M; in the second, Hypothesis 5.5.1 is satisfied for G/MZ; and By5.5.9, (G")2 and (NG2)" < (MG2)" are contained in MU < M. In any case, (G")2 and(NG2)" are contained in the intersection of all the M, that is, in N. Since G"1(6" )2 isabelian, it follows that G13j < (G")2 < N.

6.6.3 Theorem (Busetto and Napolitani [1991]). Let N be a normal subgroup of thegroup G such that GIN is soluble of derived length n E N. If go is a projectivity from Gto some group G, then 613ni < N.

Proof. We use induction on n, the case n = 1 being settled by Lemma 6.6.2. If n > 2,there exists a normal subgroup M of G such that N < M _<. G, G/M is abelianand M/N is soluble of derived length n - 1. By 6.6.2, 03) < M and, by induction,M,3n-31 < N; thus 613") < N, as desired.

6.6.4 Corollary. If G is a soluble group of derived length d(G) = n and cp is a projec-tivity from G to a group G, then G is soluble of derived length d(G) < 3n - 1.

Proof. We apply Theorem 6.6.3 to N = G("-" and get that &3n-3) < N. As a projec-tive image of an abelian group, N is metabelian by 2.4.21. It follows that G(3n-') = 1.

The above corollary improves an older result due to Yakovlev [1970]. Usingdifferent methods, he had proved that d(G) < 4n3 + 14n2 - 8n and thereby was thefirst to show that G is soluble. Even the new bound may not be best possible. For,since projective images of abelian groups are metabelian, an obvious conjecture isthat d(G) < 2d(G). Furthermore, the only examples known for which d(G) # d(G)are the projectivities between abelian and nonabelian groups, and Example 5.5.5in which d(G) = 2 and d(G) = 3. Therefore it is even possible that we have thesame phenomenon as for the Fitting length of finite groups (see 4.3.3), namely, thatonly for small derived length can it happen that d(G) : d(G), and d(G) = d(G) ford(G) > 4.

P-embedding

Our next aim is to show that N is either permutable or P-embedded in G. For thiswe first of all have to show that N/NN is periodic if N is not permutable in G. We

6.6 The structure of N°/NG- and N°/N0_ and projective images of soluble groups 333

prove a slightly more general result. Note that if N is permutable in G, N/Ni; neednot be periodic: in Example 6.2.9, N = (z) is infinite cyclic and core-free.

6.6.5 Lemma. Let N g G and let cp be a projectivity from G to G. If N is soluble andN' is not permutable in G, then N is periodic.

Proof. Assume, for a contradiction, that N is not periodic. Since N is soluble, thereexists m e N such that N'') = 1. Now N/N°j is periodic and N/N(') is not; hencethere exists i e N such that N/N'`-'> is periodic and N/N(') is not. We put A = N('-')and B = N(') = A'. Then since N a G,

(1) B < A < N, B < G, A a G, A/B is abelian, N/A is periodic and NB is notperiodic.

Thus there exists an element of infinite order in N/B and, by 6.2.12,

(2) B is permutable in G.

Since N/A is periodic, A/B contains an element of infinite order and, as an abeliangroup, is then generated by elements of infinite order. By 6.5.1,

(3) B < A.

Since N is not permutable in G, there exists y e G such that o(y, N) is a prime powerand N<y) zA <y)N. We may assume that G = <N, y); let K = NN. By 6.2.10, N isnot subnormal in G and then 6.5.3 shows that there exist primes p, q, r such that G/Kand G/K lie in P(2, p) and, more precisely,

(4) IG:NI=p>q=IN:KI,IG:NJ=r<p=IN:KI.Let S/K be one of the p + I minimal subgroups of G/K, and write IS: KI = s andS/K I = t, so that s = p or s = r and t = port = q. Then there exists g e G such that

S = K(g), and it follows that I(g): <g) n K I = IS: K I = s and I<g)4': <g)4' n K I =S:KI=t.By6.5.10,

(5) cp maps s-indices in (g) to t-indices in <g)4'.

Now suppose that o(g, B) is finite. Then since B = N(') < N' < K, we see that s =<g): <g) n KI divides I<g): <g) n BI = o(g,B), and therefore <g)/<g) n B con-

tains a subgroup X/<g) n B of order s. Then BX/BI = I X : X n BI = s and henceB a (BX)4' since B is maximal and ascendant, by (2) and 6.2.10, in (BX)4'. SinceB a A, cp induces a projectivity p from AX/B to (AX)4'/B and A/B is a nonperiodicabelian normal subgroup of AX/B. By 6.5.12, I(BX/B)°I = s and p maps s-indices inA_/B to s-indices in A/B. On the other hand, X < <g) and (5) imply that I(BX/B)°IB

=I

X/BI = I X/B n X I = J Y: (<g> n B)4'I = t and hence s = t. We have shown:

(6) If o(g, B) is finite, then s = t and the projectivity ' induced by cp in A/B mapss-indices in A/B to s-indices in A/B.

We now apply (5) and (6) to various subgroups of G/K. For S = N, (4) shows thats = p and t = q p. Therefore (6) implies that if N = K <u), then o(u, B) is infiniteand (5) yields that cp maps p-indices in to q-indices in <u)4'. Since G/A is periodic

and o(u) is infinite, n A = <a> : 1 and

(7) <aB> is an infinite cyclic subgroup of A/B in which 0 maps p-indices toq-indices.

Now let S/K be such that IS/KI = p and let S = K<v>. If IS/KI = p, then o(c,B) isinfinite since otherwise (6) and (7) would contradict each other. Again <v> n A =<h> : I and <bB> is an infinite cyclic subgroup of A/B. By (5), cp maps p-indices in<h> to p-indices in ". Then (7) implies that <aB> n <bB> = I and, by 2.6.10, 0 isinduced by an isomorphism. But this contradicts (7). Thus IS/KI : p, in particular

(8) r < p.

Finally, let T'K be a subgroup of order r of G/K that is different from S/K and letT = K <w>. Then I T/K I = q and by (5), cp maps r-indices in <w> to q-indices in<w>4'. If o(w, B) is finite, (6) implies that r = q and that 0 maps q-indices to q-indices.But then by (7), ii maps p-indices and q-indices in <aB> to q-indices, and this isimpossible. Therefore o(w, B) is infinite, again <w> n A = (c> 0 1 and <cB> is aninfinite cyclic subgroup of A/B in which 0 maps r-indices to q-indices. In (aB>, tlimaps p-indices to q-indices and since r p, it follows that (aB> n <cB> = 1. AgainBaer's theorem yields that 0 is induced by an isomorphism and this contradicts (7).

6.6.6 Theorem (Zacher [1982a], Napolitani and Zacher [1983]). Let N be a normalsubgroup of the group G, (p a projectivity from G to a group G, H = NG and K = NZ;,that is, H'' is the normal closure and K`0 the core of N ° in G. Then H and K are normalsubgroups of G. If N1* is not permutable in G, then G/K is periodic and there are primespi. qi and n; E N u ( x, } such that

G/K = (S, / K X S2/K x ) x T/K and G/K" = (Sr/K° x Sz /K" x ) x T`°/K'°

where the number of factors S1/K may be finite (nonzero) or infinite and for all i E N,(a) Si"M is a P-group of order p",qi, pi > qi and all the direct factors S,/K'

(j e N), T`°,, K`° are coprime,(b) S; / K E P(ni + 1, pi) and all the direct factors S;/K (j e N), T/K are coprime,(c) N/K = ((N n S, )/K x (N n S2)/K x ) x (N n T)11 K, I(N r Si)/K I = pi,

I(N n S1)`°/KPI = qi and (N n T)P is permutable in G,(d) H/K = x Sz/K x ) x (H n T)/K and (H n T)' is the normal closure

of (N n T`° in T`°.

Proof. That H and K are normal subgroups of G has been shown in 6.5.6. In theremainder of the proof we may therefore assume that K = I and that N is notpermutable in G. Then by 6.6.1, N is soluble and hence periodic, by 6.6.5; by 6.2.17,N is P-embedded in G. This means that G is periodic and a direct product G =(S1 X S2 x ) x T of coprime groups Si and T, where Si is a P-group of order p",q,Pi>gi,ni ENU,x},N=((NnN n T is permutable in G. By 1.6.6, G = (S1 X S2 x ) x T with coprime groups Siand T By 2.2.5, Si E P(ni + 1, p,) and IN n SiI = pi for all i since N a G. Thus (a)-(c)are satisfied and (d) follows from 2.2.2.

6.6 The structure of N' NG and N°/NiT and projective images of soluble groups 335

Like Theorem 5.4.7 for finite groups, the above result reduces questions aboutprojective images N of normal subgroups to the case where N is permutable in G.The difference from the situation for finite groups, however, is that there is noMaier-Schmid Theorem for permutable subgroups of infinite groups.

The structure of N°/N- and iv°/N6

We shall prove a number of deep structural restrictions for the groups N°/NU andN°; Ni;. One reason for this is that N°/N is generated by finite cyclic permutablesubgroups of the form '/N (x E G) and is therefore hypercyclically em-bedded in G (see Exercise 4). The main reason, however, is that the results of §§6.2and 6.3 reduce most of these problems to the case that N°/N is a finite p-group. Wemake this more precise.

6.6.7 Remark. Let N 9 G, cp a projectivity.from G to G, H = N°, K = NZ; and S =N" where H = Hl*. If we want to prove that

(a) S o G, or(b) H/N is metabelian, or(c) H°/K° is soluble of derived length at most 5,

we may assume without loss of generality that

(9) K = 1,

(10) N4' is permutable in G,

(11) IH : NJ and IH`°: N"I are finite,

(12) H and H`° are p-groups of finite exponent for some prime p, and

(13) H" is nilpotent.

Proof: First of all, all three statements deal with the structure of G/K or G;'K and,since cp induces a projectivity from G/K to G/K, clearly we may assume that K = 1.

Suppose next that N is not permutable in G. Then G = (S1 x S, x ) x T hasall the properties given in Theorem 6.6.6. In particular,

S = (S1 X S2 x ...) x (N n T)f''''

and, if (a) holds for the normal subgroup N n T of T with permutable image (N n T)4',then it also holds for N in G. Similarly, since the Si and S;° are metabelian, also (b)and (c) hold in general if they are true for N n T in T Thus we may assume that Nis permutable in G.

We show next that in all three cases we may assume that

(14) G/N is finitely generated.

In the proof of (a), we suppose, for a contradiction, that S is not normal in G. Thenthere exists g c S such that S9 S and, since 9 = <NYly E H), there exists y E H

such that ((N1')`° `)9 S. Since H = <N°Iz E G>, there exist z1, ..., z, c G such thaty E N" ... N"-. So if W = <N,g,_<z;>'P-'Ii = 1,...,r> and So = Nom, then (NY)'P< Soand ((NY)"-')9 So since y e NW and So < S. Thus So is not normal in W andwe may assume that G = W is finitely generated modulo N. In the proof of (b), wehave to show that H" < N, hence that for all xi e H, [[x1,x2], [x3,x4]] E N.Since <xi>'* = <yi> with y; e H = <N`Iz e G>, there exist z1, ..., z., E G such thatyi c- N=1 ... N== for i = 1, 4. So if W = <N, <zi)`° ' Ii = 1, ..., s> and (b)holds for normal subgroups with finitely generated factor groups, then[[x1,x2], [X3,x4]] E (NW)" < N and thus H" < N. For assertion (c), the situation issimilar; we just have to consider longer commutators of elements in H. Thus in allthree cases we may assume that (14) holds and then, of course, again that K = 1 forthis smaller group G.

It follows that G is finitely generated modulo N. By 6.2.18_and 6.3.3, I H : NI < xand H is nilpotent of finite exponent. Thus H = PI x . . x P, is a direct product offinitely many Sylow subgroups P. By 1.6.6, H = PI x . x P, with coprime groupsPi, and so N = (N n PI) x x (N n P,) where the N n Pi are characteristic sub-groups of N and hence normal subgroups of G. Now clearly H = HI x x H, andS = SI x x S, where H; = (N n Pi)° and Si = (N n Pi)"% Thus if one of our state-ments holds for every N n Pi, it holds for N. Thus we may assume that N is one ofthe N n Pi, that is, H is a nilpotent p-group of finite exponent for some prime p andIH: NJ < x. Then [H/N] and hence H/N is finite so that (11) and (13) hold. Fur-thermore H is locally finite and if H is not a p-group, 2.2.7 shows that H is abelian;thus S = N < G, H is metabelian and therefore (a)-(c) hold. Therefore we mayassume that H is a p-group; as a projective image of H, it clearly has finite exponent.

6.6.8 Lemma. Let N g G, cp a projectivity from G to G, H = N° and S = N" whereH = H`P. Then S g G and HIS is abelian.

Proof. By the above remark, we may assume in the proof of the first assertion that(9)-(13) are satisfied. Thus N is a p-group of finite exponent p" and we use inductionon n to show that S:9 G. Let M = fl(N). Then M a G and hence M is modular in6. Since H is nilpotent, M is ascendant in G (see Robinson [1982], p. 350) andso by 6.2.10, M is permutable in G. By 6.3.8 applied to the core-free permu-table p-subgroup N of G, 1V is elementary abelian and either k < NG = H orMc < N" = S. In the first case, S = N a G, as desired. So suppose that Mc < S andlet L < G such that L = Mo. By 6.5.6, L a G and cp induces a projectivity 0 fromGIL to GIL. Since L is elementary abelian, NL/L is a normal p-subgroup of exponentpn-1 of GIL and H/L is the normal closure of its image NL/L under IJi in GIL. SinceL < S a H, we have (NL/L)"!L = 91L and the induction assumption yields thatS/L g G/L. Thus S a G, as desired.

To prove that HIS is abelian, no reduction is necessary. We need only note thatS < H a G and S is modular in G since S a G; thus by 6.2.10,

(15) 9 is permutable in G.

6.6 The structure of Ns/N6- and N'/NG- and projective images of soluble groups 337

Now let x e H, g e G and write (x)" _ (x), (g)w _ (g) and (xg),* = (z). By 6.3.5applied to the permutable subgroup S of W = <9_t, g) = (S, x, z ), we have SW =SAX>SC9i = S<=>S<Z>. Since S a H and Y E H, it follows that S<x> = S and hence

SW = S<9> = S<=> < (g)S n (z>S = ((g)S)`° n ((xg)S),P.

It follows that (S<s>)v '/S < (g)S/S n <xg> S/S is centralized by gS and xgS, andhence by xS. Now SG = H implies that HIS is generated by these subgroups(S19>)1''/S and hence it follows that xS E Z(H/S). Since x e H was arbitrary, HIS isabelian.

Remark 6.6.7 implies that in the proofs of the next two theorems we may alsoassume that (9)-(13) are satisfied. We want to describe the basic structure of S/N inthis situation. Clearly, H = (NNXIx e G) and, since N < NNX < N(x), there existsy e (x) such that NNX = R(y). By (11), H : NJ < cc and hence there are finitelymany y, E H such that

(16) H = and N(y,) is permutable in G

since N (y,) is the product of two permutable subgroups. Then 6.3.5 implies that

(17) S = (N<Y,>1 i = 1,...,n).Now H/N and H_/NH are finite p-groups and hence all subgroups between N andH and between N and H are subnormal in G and G, respectively. Since cp and q0`map modular subgroups to modular subgroups, it follows from 6.2.10 that for allL E [H/N],

(18) L is permutable in G (in H) if and only if L is permutable in G (in H).

As a join of permutable subgroups, N<Y,> is clearly permutable in G and sinceN<r,> < N(YC),

(19) (N<,',>)`' '/N is a cyclic permutable subgroup of G/N (i = 1,...,n).

Of course, (17) and (19) show that S/N is generated by cyclic permutable subgroups.Since N(y,) is permutable in G and [(N, y;, yy)/N] is isomorphic to the subgrouplattice of the p-group (N, y,, yj)1* /N, it is clear that M = N and F = {yl,...,satisfy the assumptions of Lemma 6.3.10. It follows that every subgroup T of d suchthat N < T < N" = S is permutable in G and, by (18), this yields that

(20) every subgroup of S containing N is permutable in G.

In particular, S/N is an M-group and we finally want to show that it is an M*-group.So suppose, for a contradiction, that there is a quaternion group Q8 involved in S/N.Then by 2.3.8, S/N = S1/N X Sz/N where Sl/N Q8 and Sz/N is an elementaryabelian 2-group. By (17) and (19), S/N is generated by the cyclic subgroups(N<r,>)P '/N (i = 1, ... , n) and it follows that two of these, for i = 1, 2, say, have order4 and generate a quaternion group. However, by (16) and (18), the (N(y,))w /N,i = 1, 2, are cyclic permutable subgroups of the 2-group H/N. By 5.2.14, they gener-ate a metacyclic M-group containing this Q8 and therefore equal to this Q8. It

follows that N<''l > = N<yl ), a contradiction since all the conjugates of N in N<yl )must lie in the maximal subgroup of N<yl) containing N. Thus

(21) S/N is an M*-group.

We can now prove the announced result on the structure of H/N.

6.6.9 Theorem (Busetto andNapolitani [1990]). If N a G and cp is a projectivityfrom G to a group G, then N°/N is metabelian.

Proof. Again let H = N° and S = N". By 6.6.7, we may assume that (9)-(13) hold,so that (16)-(21) are also satisfied; let y; (i = I__ , n) be as defined in (16). In particu-lar, H is a p-group and H/N is finite. Therefore it is possible to use induction onIH : LI to prove the following result:

(22) If N < L< S and L g H, then H/L is metabelian.

The theorem follows on setting L = N. To prove (22), we first handle the case thatS/L is metacyclic. Since HIS is abelian, (22) clearly holds if S/L is abelian. Thus weassume that

(23) S/L is metacyclic but not cyclic.

By (17) and (19), S/L is generated by the cyclic permutable subgroups (N<Y,))c 'L/L(i = I__ , n) and, since S/L is metacyclic, there exist a, b e H such that <a>" = < y; ),<b)`° = <y,) and S/L = ((N<Y,>)'° L/L)((N<y,>)'° L/L). Since (N<a))° = N<y;) per G,we obtain N<a) per G, and hence <L,a)/L is permutable in H/L; similarly for<L, b)/L. Since S a G, finally,

(24) <L, a)/L and <L, b)/L are permutable subgroups of H/L and L<a>'L

We choose elements a, b e H satisfying (24) such that

(25) <L, a, b)/L has maximal order

and put R = <L, a, b). Then R/L is the product of two cyclic permutable subgroupsand hence, by 5.2.14, is a metacyclic M-group. If R/L _- Q8, then <L, a)/L and<L, b)/L have order 4 and by (24), S/L is contained in the subgroup of order 2 in thisQ8. But we assume that S/L is not cyclic, a contradiction. Thus

(26) R/L is a metacyclic M*-group.

We denote the natural epimorphism from H to H/L by v and want to show that forz c H,

(27) [S', z'] = 1 if <z") n R' = 1.

To see this, we study L Since 3:9 H and <L, z) n S < <L, z) n R = L, we seethat L`° = <L, z)" rSt g <L, z)°, and hence that L<z>' = L. By (18), L per H and6.3.5 implies that

L<a">' = L<a>"L<z>" = L<az>OL<z>' = L<a>° = L<az>®

6.6 The structure of NG/NG_ and NG/NG_ and projective images of soluble groups 339

Thus (L<°)")° L<a> n L<az> and this shows that (L<°>")`0 '/L iscentralized by aL and azL and hence by zL. The same holds for (L(6>°)I '/L and,since S/L is generated by these two groups, it is centralized by zL. This proves (27).

Since R' is metacyclic, there exists a cyclic normal subgroup T' of R' with cyclicfactor group R"/T" and we claim that

(28) T' n S' a H'.

For this we show that every cyclic permutable subgroup <z'> of H' normalizesT' -) S'; since, by (16), H" is generated by cyclic permutable subgroups, this willsuffice to prove (28). Now for <z'> there are three possibilities. If <z'> < R', itclearly normalizes T" n S' since T' n S' 9 R'. If (z'> n R' = 1, then by (27), z'centralizes S' and hence also T" n S". Thus it remains to consider the case that

(29) 1 <z'> n R' 0 <z">.

By 2.5.9, the M*-group R' is lattice-isomorphic to an abelian group in which theFrattini subgroup does not contain a maximal cyclic subgroup. It follows that thereexists c' e R' such that (z'> n R' < (c'> and c" 4)(R'). If <a', c'> and <b', c'>were both proper subgroups of R', then c" would be contained in the maximalsubgroup of R' containing <a'> and (b'>, respectively, and, since the intersection ofthese is V(R"), it would follow that c' E cb(R"). This is not the case and hence

(30) R' = (a', c'>,

say. From <z"> n R' < <c'> we obtain that

(31) (c", z'> n R' = (c'> (z'> n R' = <c"> (<z'> n R") = <c").

Since HIS is abelian and S < R < H, we have R a H and hence <c"> 9 <c",z").Thus <c',z'> is the product of two cyclic permutable subgroups and hence is ametacyclic M-group by 5.2.14. If this were a quaternion group, then o(c') =I (L, c> : LI = 4 and hence L1* a (L, c>", by (53) of § 5.5. Then by (24), 6.3.5 and (30),

S = L<°>"L1 = LR = L<°>'L<`>m = L<°>-.

Therefore S/L would be cyclic, contradicting our assumption (23). Thus

(32) W, z'> is a metacyclic M*-group.

Suppose, for a contradiction, that o(z') > o(c''). Then z' has maximal order in theM*-group <c', z'> and by 2.3.11 there exists t" E <c', z'> such that <c', z'> _<t', z'> and <t'> n <z'> = 1. Since 1 (z'> n R' < (c"), we have <c'> n <z'> 1

and hence <t'> n <c'> = 1. By (31), <t'> n R' = <t'> n <c', z"> n R" _<t"> n <c") = 1. It follows that <L, t> n S = L, hence L' = <L, t> n Sw 9 (L, t>and so L <'>' = L. Then <a",c",z'> = <R', z'> = <a',t",z""> and S a H togetherwith 6.3.5 imply that

9 = L<R,2>- = L<°>°L(Z>'L<`>m = L<°)°L(Z>"

and thus a, z satisfy (24). Now (a''> n <z') < <a"> n R' n <z'> < <a''> n <c'> andI <z''> I > I <c'> I; thus I <a", z'> I > I <a', c'> I = I R' l = I <L, a, b>/LI, which contradictsthe choice of a, b. Thus

(33) o(c') > o(z'').

Now c' has maximal order in the M*-group <c", z'> and there exists t' such that<c',z'> = <c'', t'> and <c'> n <t'> = 1. By (31), <t'> n R' < <t'> n <c'> = I and(27) shows that t" centralizes S'. Since S' n T' is normalized by c" e R', it followsthat <c', t'> < N,,,(S' n T'). In particular, z" normalizes S'' n T" and this proves(28).

By (28), S' n T' is a cyclic normal subgroup of H'. Since the automorphism groupof a cyclic group is abelian, H"/CH,(S" n T") is abelian and hence (H')' centralizesS' n T'. By 6.6.8, HIS is abelian and therefore (H")' < S". So, finally,T' n S' n (H')' = T' -) (H')' is a central subgroup of (H')' with cyclic factor group(H'')'/T" n (H')' T"(H")'/T". It follows that (H')' is abelian and thus H' ismetabelian. This proves (22) in the case that S/L is metacyclic.

For the general case, we use induction on I H : LI and assume that S/L is notmetacyclic. If S/L is abelian, then H/L is metabelian by 6.6.8. Thus we also assumethat S/L is not abelian. Then (16), (20) and (21) show that H/L and S/L satisfy thehypotheses of Lemma 6.3.9. Since S/L is not metacyclic but an M*-group, it isnot generated by two elements and 6.3.9 yields that it contains two different mini-mal normal subgroups L1/L and L2/L of H/L. By induction, H/Li is metabelian fori = 1, 2 and, since L, n L2 = L, it follows that H/L is metabelian. This completes theproof of (22) and of the theorem.

The main theorem

We come to our final result on the structure of N°/NU and No/NN. By 6.6.1 and 6.6.9,N/NG- and N°/N are metabelian so that N°/N_ is soluble of derived length at most4. Furthermore, since N°/N is metabelian, Theorem 6.6.3 yields that (N°)(6' < N,and hence N°/NG_ is soluble of derived length at most 6. However this can be im-proved.

6.6.10 Theorem (Busetto and Napolitani [1990]). Let N be a normal subgroup of thegroup G and cp a projectivity from G to a group G. Then N°/Ni7 and N'/NZ are solubleof derived length at most 4 and 5, respectively.

Proof. Again let H = N°and S = V. We have to show that H1" is contained in N;as a normal subgroup of G it will then be contained in &, and N°/NG_ will be solubleof derived length at most 5. By 6.6.7, we may assume that (9)-(13), and hence also(16)-(21), are satisfied. By (11), (12) and 6.6.9, H/N is a finite metabelian p-group; letN< M a Hsuch that HIM and M/N are abelian. Then if p > 2, Lemma 6.6.2shows that H" < M and M" < N, so that Hi4' < N.

Thus we can suppose that p = 2. Since HIS is abelian, we may assume that M < S.Let L/M = c2(S/ M) and TIN = (D(M/N) so that L/M and MIT are elementary abe-

6.6 The structure of N' /N0 and 9°/& and projective images of soluble groups 341

lian 2-groups. By 6.3.7, H/S is nilpotent of class at most 2; in particular, H" < S.Since HIM is abelian, (H")2 < M by 6.6.2. Therefore [H"M/M] 2t [H"/H" n k] isan interval in [H"/(H")2], and hence (H"M)' '/M is an elementary abelian subgroupof S/M. Thus (H"M) '/M < O(S/M) = L/M and H" < L.

Now M/N is abelian and therefore by 6.6.2, (NM2)" = T" < N. Finally, since LITis a 2-group of exponent at most 4, (53) of § 5.5 shows that T a L. By (21), S/N is anM*-group. Hence L/T and its projective image L/T are M*-groups of exponent atmost 4; by 2.3.7, LIT is abelian. It follows that H(5) < N.

Theorem 6.6.10 is the culmination of the work of four mathematicians from Padua:Busetto, Menegazzo, Napolitani and Zacher. The first step, however, was takenby Yakovlev [1970], who proved that under Hypothesis 5.5.1, d(G) < 24d(N) - 20.He used this to show that a projective image of an arbitrary soluble group Gis soluble and its derived length is at most 4n3 + 14n2 - 8n if d(G) = n. ThenMenegazzo [1978] realized that, in fact, N is abelian if Hypothesis 5.5.1 is satisfiedand p > 2; he deduced that for an arbitrary finite group of odd order, N/NG isabelian and therefore N/N6- is metabelian. The Zacher-Rips Theorem prepared theground for an investigation of projective images of normal subgroups in infinitegroups, and Zacher [1982a] showed that if it were possible to prove a similar, butperhaps somewhat weaker, result for p = 2-for example, that under Hypothesis5.5.1, d(N) is bounded by a constant c-then for arbitrary groups, N/N6- wouldbe soluble of derived length at most c. This was carried out by Busetto [1984]with c = 3, and he proved that N/N6- and N/N6- are soluble of derived length atmost 3 and 4, respectively. At the same time Busetto and Stonehewer [1985] pro-duced Example 5.5.5 showing that N/N6- in general need not be abelian. Meanwhile,Napolitani and Zacher [1983] had shown that N°/Nd and N°/N6- are soluble andthat d(N°/N) < 32. Busetto and Menegazzo [1985] proved Theorem 6.6.3 with 6ninstead of 3n, and improved Busetto's bound to d(N/Nd) < 3; their result alsoyielded better bounds for the derived length of N°/NZ and Finally Busettoand Napolitani [1990], [1991] obtained the results of §§5.5 and 6.6 in the formpresented here.

Elements of infinite order

We conclude this chapter with a remark that is perhaps somewhat surprising.Whereas for the proof of Theorem 6.6.10 we needed nearly all of our deep results ofChapters 5 and 6, the situation is much simpler if GIN contains an element of infiniteorder. In fact, N°/N6- and N°/& are abelian in this case and the proof of this is quiteelementary, using only 6.2.3, 6.2.11 and 6.3.5.

6.6.11 Theorem (Busetto and Napolitani [1990]). Let N be a normal subgroup of thegroup G and q a projectivity from G to a group G. If GIN contains an element ofinfinite order, then N°/N6- and N°/N6- are abelian.

Proof. Write H = N°, K = N-G and let g e G be such that o(gN) is infinite; by 6.2.12,N is permutable in G. By 6.5.1, N< >' = N and therefore 6.3.5 implies that for

x e G,

N<X.g>, = N<X>1R<g>m = N<Xg>wR<g>m = N<X>" = N<Xg>

It follows that (N<X>")4 ' < N(x) n N(xg) and (N<X>°)'P-'/N is centralized by xNand xgN and hence also by gN. Since H/N is generated by these (N<X>m)1'/N,x e G, the group H/N is centralized by gN. It follows that the group (H,g)/N =(H/N,gN) is generated by elements of infinite order and that these elements cen-tralize H/N. Thus H/N < Z((H,g)/N) and, as a cyclic extension of a central sub-group, (H, g)/N is abelian.

By 6.5.1, N a (H, g)`0 and p induces a projectivity from (H, g)/N to (H, g)`0/N.The first group is abelian and hence the second is an M-group with elements ofinfinite order. By 6.2.3, H/N is generated by finite cyclic subgroups of the formN<Y>/N, y e G; therefore it is contained in the torsion subgroup of (H, g)"/N, andhence is abelian by 2.4.8. Thus H' < N and since H' a G, H' < Nd = K, that is, H/Kis abelian.

If z e G, then a = cpzcp-' is an autoprojectivity of G. Since <H, g)/N is generatedby elements of infinite order, N° g (H, g)° and a induces a projectivity from(H, g)/N to (H, g)"/N". Thus (H, g)°/N° is an M-group with abelian torsion sub-group and therefore H°/N° is abelian. But H° = H since H g G and it follows thatH/ ft NP=`A ' = H/K is abelian.

_EG

There exist examples in which G/N is not periodic and N is not normal in G; seeExercise 6.

Exercises

In Exercises 1-5 let N a G, (p a projectivity from G to G, H = NG, K = Nd andS = NH. Exercises 3-6 are due to Busetto and Napolitani [1990].1. If G/N is a torsion group without nontrivial 2-elements and Gt") < N, show that

G'(2n) < N.2. If G is a finite group of odd order, show that H/K and H/K are soluble of derived

length at most 3 and 4, respectively.3. Let S, = S and S;+, = NS' (i E rkJ). Show that Si g G and S;/Si+1 < Z(H/S;+,) for

allieN.4. Show that H/N is hypercyclically embedded in G, and hypercentrally embedded

in G if N is permutable in G. (Hint: Use 6.4.10 and 5.2.12.)5. For every n e N, construct N, G, cp such that H/N is nilpotent of class n. (Hint:

By2.5.9,G=(a,b,claPr=bPrt'=cP'=[a,b]=1,a`=a'+° b`=b'+P) has anautoprojectivity mapping (a) to (c).) _

6. Let G = (a, b, cI a2 = b8 = [a, b] = [a, c] = 1, c'=c-'> and G = withthe same defining relations except that [a, b] = I is replaced by [a, b] = b4. Showthat there exists a projectivity from G to G mapping (a) to (a); note that (a) a G,G/(a) contains elements of infinite order and (a) is not normal in G (seeExercise 6.3.2).

Chapter 7

Classes of groups and their projectivities

In this final chapter on projectivities of groups we study certain interesting classesof groups and try to find out whether they are invariant under projectivities and,especially, which groups in these classes are determined by their subgroup lattices.Recall that a group G is said to be determined by its subgroup lattice if it is

isomorphic to every group G such that L(G) L(G); it is strongly determined byits subgroup lattice if every projectivity of G is induced by a group-isomorphism.Groups with this property we have met are the noncyclic elementary abelian 2-groups (see 2.2.5 and 2.6.7), the hamiltonian 2-groups (see 2.5.1), and the abeliangroups with two independent elements of infinite order (see 2.6.10). Not stronglydetermined but at least determined by their subgroup lattices are the following: theinfinite cyclic groups (see 1.2.6), the additive group of rational numbers (see 2.6.14),the generalized quaternion groups, certain dihedral groups, and the alternatinggroup A4 (see § 1.4). We shall show that many other groups have these properties.

In 1941 Sadovskii proved that every free group F of rank r > 2 is stronglydetermined by its subgroup lattice. We prove this, and, in fact, a more generalresult on 2-free groups, in § 7.1 using a lattice-theoretic description, due to Yakovlev[1974], of the subgroup lattice, the elements and the multiplication of F. This will,in addition, yield a characterization of those lattices which are isomorphic to thesubgroup lattice of an arbitrary group, a result interesting in itself. We mention,but do not prove, generalizations of Sadovskii's theorem to free products and freeproducts with amalgamation.

Another theorem of Sadovskii's is that every nonabelian torsion-free nilpotentgroup is strongly determined by its subgroup lattice. We use an unpublished manu-script of G.E. Wall's to prove this in § 7.2. In addition, we give a lattice-theoreticcharacterization, due to Kontorovic and Plotkin [1954], of the class of torsion-freenilpotent groups.

For mixed and periodic nilpotent groups the situation is not so good since thereare obvious examples of nilpotent and nonnilpotent groups with isomorphic sub-group lattices and of projectivities between nonisomorphic nilpotent groups. How-ever, in § 7.3 we give a lattice-theoretic characterization of the class of finitelygenerated nonperiodic nilpotent groups. There are also generalizations of Sadovskii'stheorem to mixed nilpotent groups. The main result in this direction, which is due toYakovlev [1988], is that a nilpotent group G of class n containing a free nilpotentgroup of class n as a subgroup is strongly determined by its subgroup lattice. How-ever, we shall not prove this but only note that the assumption cannot be weakenedto the hypothesis that G contains a torsion-free subgroup of class n or contains two

344 Classes of groups and their projectivities

independent elements of infinite order; this last assumption does not even imply thatG is determined by its subgroup lattice. On the other hand, we shall prove a theoremdue to Barnes and Wall [1964] stating that every normalizer preserving projectivitymaps a nonabelian almost free group of a nilpotent variety to an isomorphic group.This will yield as a corollary that every nilpotent group of class 2 and exponent p isdetermined by its subgroup lattice.

A nilpotent torsion group is the direct product of its primary components, andits subgroup lattice then is the direct product of the subgroup lattices of thesecomponents. Therefore in our study of subgroup lattices of periodic nilpotentgroups in § 7.4, we restrict our attention to nilpotent p-groups. We first use the ideasof Kontorovic and Plotkin to characterize the nilpotent groups among the p-groups.This yields that every projectivity maps nonabelian nilpotent p-groups to nilpotentp-groups, a result due to Yakovlev [1965]. Furthermore we obtain a lattice-theoreticcharacterization of the class of nilpotent primary groups and P-groups. Finally, westudy finite p-groups and prove that p-groups of maximal class are nearly alwaysmapped to p-groups of maximal class under projectivities, a result due to Caranti[1979].

The lattice-theoretic characterizations of §§ 5.3 and 6.4 show that certain classesof soluble and generalized soluble groups are invariant under projectivities. In § 7.5we prove in addition that the rank of a finite soluble group is preserved underprojectivities, and present some further results of Schmidt [1974], [1987] on forma-tions of finite soluble groups. Roughly speaking, we show that if I is defined locallyby classes j(p) which are invariant under projectivities, then I is also invariantunder projectivities and a-residual and a-projectors are mapped onto 1-residualand a-projectors, respectively, by every projectivity between finite soluble groups.At the other end of the spectrum of solubility, we prove that the class of radicalgroups is invariant under projectivities, a result due to Pekelis [1972b].

In the final three sections of this chapter we return to the question of whetherthe members of a given class of groups are determined or even strongly determinedby their subgroup lattices. In §7.6 we study direct products of groups and provea generalization of Suzuki's theorem that the direct product of two isomorphiccopies of a finite nonabelian simple group is determined by its subgroup lattice; weshow that this still holds for arbitrary perfect groups with trivial centre. We dealbriefly with the more general question whether the direct product of n isomorphicgroups (n > 2) is (strongly) determined by its subgroup lattice. There are results inthis direction for certain classes of groups due to Schmidt [1982] and Schenke[1987a], [1987b]. Similarly, we only sketch the results on projectivities of wreathproducts due to Menegazzo [1973] and to Arshinov and Sadovskii [1973]. Finallywe show that a finite perfect group with trivial centre has the property that the directproduct of all its pairwise nonisomorphic projective images is determined by itssubgroup lattice.

In § 7.7 we present a fruitful general method to prove that a given group generatedby involutions is determined by its subgroup lattice. This method is applied topermutation groups and we are able to show in this way that every triply transitivepermutation group of degree at least 4 which is generated by involutions is deter-mined by its subgroup lattice. The same holds for large classes of doubly transitive

7.1 Free groups 345

and even certain primitive groups. All these results are due to Schmidt [1975b],[1977b], [ 1980a].

Finally, in § 7.8 we note that the classification of finite simple groups yields thatevery finite nonabelian simple group is determined by its subgroup lattice. Unfortu-nately, a direct proof for this important result is not known; but the methods of § 7.7at least yield proofs for certain classes of finite simple groups. In contrast to this, itis not difficult to see that these groups in general are not strongly determined bytheir subgroup lattices. Here the alternating groups and also the 3-dimensionalprojective special linear groups yield examples; we study the groups of autoprojec-tivities of the alternating groups and also of the finite symmetric groups in somedetail. We finish this chapter by determining the finite lattice-simple groups, that is,the finite groups G having no nontrivial proper P(G)-invariant subgroup. Theseare precisely the cyclic groups of square-free order, the P-groups, and the directproducts of isomorphic nonabelian simple groups.

7.1 Free groups

We give necessary and sufficient conditions for a lattice L to be isomorphic to thesubgroup lattice of a free group F of rank r > 2. For this purpose we describe theelements of F and their multiplication in the lattice. And since we are able to identifythe normal subgroups of F, we also get a characterization of those lattices which areisomorphic to the subgroup lattice of an arbitrary group. Another immediate conse-quence will be that F and, more generally, every 2-free group, is strongly determinedby its subgroup lattice . First of all we need some definitions.

Cyclic elements and complexes

7.1.1 Definition. Let L = (L, <) = (L, n, u) be a complete lattice and 0 its leastelement. An element a c L is called cyclic if [a/O] is a distributive lattice withmaximal condition, the set of cyclic elements of L is denoted by C(L) or C, for short.For a, h E C and A, B C, we define a o h= {x E CJx u a= x u h= a u h} andAoB={xECJxEaoh, ac-A, hEB}. We also write aoB for AoB if A={a}consists of only one element.

All the a o h and A o B are subsets of C and we clearly have

(1) aoa=[a/O]andaoO={a};Indeed, every b c- [a/0] is contained in C and satisfies h u a = a = a u a. and theonly element x c L such that x u a = x u 0 = a, is x = a. By 1.2.5. the cyclicelements of the subgroup lattice L(G) of a group G are the cyclic subgroups of G and,if x,yEG,

(2) <x''y")E<x>o(vjfor all v,pEl= {+1,-1},

since <x''y"> u <x> = <x`y"> u <y> = <x> u <y> = <x,y>. It will be one of thefundamental properties of a free group F that for x, y E F with x :A1 1 0 y and<x> n <y) = 1, the set <x> o <y> precisely consists of the four elements <xy>,<x-' y), <xy-' > and <x-' y-`) (see 7.1.7). In general, however, <x> o <y> containsother cyclic subgroups; for example, if G = <x> x <y> is the elementary abeliangroup of order p2 for some prime p, every subgroup of order p of G different from<x> and <y> lies in <x> o <y>.

7.1.2 Definition. Let L and C = C(L) be as in 7.1.1, let n e F, ei e C (i = I__, n) andput E = (e,_., e a = (A1..... is called an a-com-plex with respect to E if for all i, j e { 1, ... , n} with i : j,

(3) A; is a 2-element subset of e; o a and

(4) aioaane;oej96 0forallaiEA;,aaeAj.

Let K(a, E) be the set of all a-complexes with respect to E. We call E _({e, }, ... , {e }) the 0-complex with respect to E and put K (0, E) = {f;}. Finally, acomplex is an a-complex for some a e C and .AY(E) is the set of all complexes withrespect to E, where two such complexes a = (A1..... and /3 = (B...... areconsidered equal if and only if Ai = B; for all i = 1, ..., n.

If there is no ambiguity, we omit the phrase "with respect to E" and speak ofcomplexes and a-complexes. Of course, K(a, E) may be empty for some a E C; forexample, this will certainly be so if a 0 0 and lei o al < 1 for some i. In subgrouplattices it is often possible to construct examples in the following way.

7.1.3 Example. Let G be a group and L = L(G) so that C is the set of cyclic sub-groups of G; take nontrivial elements gi (i = I...... n) of G, define ei = <gi> and putE = (e,,...,en) = (<gl>,...,<gn>)

(a) Let I a E G. By (2), ei o <a> = <gi> o <a> contains the (not necessarily dis-tinct) elements <gia>, <gi-' a>, <gia-' >, and <g ' a-') of C. Therefore if <gia> 4-<g; ' a> for all i e {1..... n}, then Ai = { <gia>, <g; ' a> } is a 2-element subset ofei o <a>; and for v, p e { + 1, - 1}, both <g,'a> o <gj a> and <g1> o <g1> contain theelement <gi",y,-"> of C, again by (2). It follows that <g;'a> o <gj a> n <g;> o <gj> 0 0if i j, and hence a = (A,,..., is an <a>-complex with respect to E.

(b) Suppose that G is torsion-free and g-'ag a-' for all a, g e G with a 0 1.(Note that in a free group, g-'ag = a-' implies a = I, see (18), so that the conditionis satisfied.) Then for q : 1, the element g-' a is different from the two generators gaand (ga)-` = a-'q-' of the group <ga> and it follows that <g'a> 4- <ga>. Thusthe assumption <gia> 0 <g; ' a> of (a) is satisfied, and a = (A...... with A; =I< gia>, <gi `a) } is an <a>-complex with respect to E. We denote it by x(a, E) andputr.(l,E)=E=({<g,)},...,{<gn>}).

Finally, we define a multiplication of complexes.

7.1.4 Definition. Let L, C, n, E, _Y(E) be as in 7.1.2 and let a = (A1..... and/3 = (B,,..., be complexes in .AY(E). We define the product a/3 of x and /3 to be the

7.1 Free groups 347

set of all complexes S = (D, , ... , D.) e W '(E) for which there exist a, h, d e C suchthat

(5) deaob,

(6) x c K (a, E), /3 E K (h, E), S e K (d, E), and

(7) all i,je{1,...,n}.

For subsets A, B of f(E), we define AB = U {af3Ia e A, /3 e B} and also write A(3 forAB if B = { /3} consists of only one element.

Clearly, x/3 and AB are (possibly empty) subsets of t -(E). We continue to studyExample 7.1.3 and show the following.

7.1.5 Lemma. Let G be a torsion free group and g-' xg : x-' for all x, g e G withx 1.Ifa,h,deGaresuch that ab=danda=x(a,E),/3=x(b,E), 5=x(d,E)arethe complexes constructed in 7.1.3, then S e af3.

Proof. Let x=(A,,...,A,1), /3=(B1 and S=(D,,...,Dr). By 7.1.3, xEK (<a>, E), /3 e K (<b), E), S e K (<d ), E) and by (2), <d> = <ab) E <a) o <h>. It re-mains to be shown that (7) holds. For a # 1, A; = { <g;a), <g,-'a> } and if a = 1,x = E so that A; = { <gi> } = { <gia) }. In both cases, (2) implies that <g;ag, ') =<g;db-' gj-.') is contained in <g;a) o <gi) c Ai o ej and in <g,d) o <g;h) c Di o B3.Thus Di o B; n Ai o ej : 0 and (7) holds.

We can now give a sufficient condition for a lattice to be isomorphic to thesubgroup lattice of a group.

7.1.6 Theorem (Yakovlev [1974]). Let L be a complete lattice in which every elementis the join of cyclic elements, and suppose that there exists a system E = (e,,...,en) ofelements e; E C = C(L) with the following properties.

(8) For every a e C such that a -A 0, IK(a,E)I = 2.

(9) If a E C, a = (A , , ... , E K (a, E), x' = (A', , ... , A;,) E K (a, E) and a x', thenA ' : A f o r

(10) If a,bEC,xeK(a,E),and (3EK(b,E)such that x=/3,then a=b.

(11) For all a, /3 e .K(E), the product af3 consists of a unique complex x

(12) For all x, /3. y E .flE), (a/3) y = a(f3'r).

(13) Let a e C and X C such that a< U X and let x c K(a, E). Thenthere exist finitely

//many elements hi e X and f3 i E K(hi, E) such that

x E ((...(f3,/32)(f'3...)f3m-1)flm

Then G = A (E) with the operation *: G x G -+ G given by (11), (a, f3) i- a * /3 forx, f3 e G, is a group whose subgroup lattice is isomorphic to L.

Proof. By (11) and (12), * is an associative binary operation on G and we show firstthat (G, *) is a group with identity element & = ({e, },..., For this let . E G, thatis, x = (A,,..., e K(a. E) for some a e C. We show that or c ae; it will follow from(11) that

(14) x*F=x.

If a = 0, then x = e and, since 0 o 0 = {O}, (5) and (6) show that EE can only containthe 0-complex r; by (11), F * E = r.. So suppose that a 0 0. Then by (1), a c a o 0 sothat (5) and (6) hold for /3 = P. and d = x. To prove (7), let i, j C { 1,..., n}. By (3), A; isa 2-element subset of e; o a s C and therefore contains an element c 0 0 of C. By(8) there exists a c-complex and hence ej o c 0 0. Since ee o c = c o eJ . c A; o e1, it

follows that A; o ej 96 0 and that (7) holds. Thus x E xE and this proves (14).Again let a 0 0 and x = (A,,..., e K(a, E). By (8) there exists a unique further

complex x'=(A',,...,A;,)in K(a,E).By (1)and (9),OEaoaand e;oA}nA;oej#0so that (5)-(7) hold for /f = oc' and S = E. Thus e c aa' and (11) yields that

(15) x*2'=E.

This shows that (G, *) is a group. By (10), for every x e G there exists a unique a e Csuch that x E K(a, E) and by (8), the map is G -i C defined by x' = a if x E K(a, E) is

surjective. Since L is a complete lattice, we may form arbitrary joins in L and hencethere exists the map cp: L(G) --> L defined by H° = U x' for H e L(G). We show that

aeHcp is an isomorphism, which will prove the theorem. First, to see that rp is surjective,let x c L and consider H = {x E G12' < x}. For x, /3 e H and S = of * /3, (5) and (6)show that b' E x' o /3'; then Definition 7.1.1 implies that S' < 7'u /3' < x and sod E H. By (15). (x-' )' = x' < x and hence x-' E H. Thus H is a subgroup of G. Byassumption, x is the join of the cyclic elements contained in x and, since i is surjec-tive, any such element is of the form x' where x e H. It follows that x = U a' = H°.Thus tp is surjective. 2eH

Now suppose that A, B < G such that A0 < B`1. If a E A, then x' < A" < B`° _U /3' and by (13) applied to a = x' and X = {(3'x/3 E B} there exist b,, ..., b, C X

fEBand /3i c K(b;, E) such that x Since b; E X, there exists /3 e B such that/3' = b; =/3; and(15)implies that /3;=/3or3;=/3-'.Inanycase, /3;eBforall i andhence x c B. Thus A < B. It follows that if A(P = B", then A < B and B < A andtherefore A = B. This shows that (p is injective and hence bijective. FurthermoreAll < B`° implies A < B, as we have just shown, and the reverse implication followstrivially from the definition of (p. By 1.1.2, cp is an isomorphism. D

Conversely we want to show that the subgroup lattice of a free group of rank atleast 2 satisfies the assumptions of Theorem 7.1.6. We handle a slightly larger classof groups since our method works in the more general situation as well.

2-free groups

A group is called 2-free if it is nonabelian and any two of its elements generate a freegroup. If G is a free group of rank r > 2, the Nielsen-Schreier Theorem says that

7.1 Free groups 349

every subgroup of G is free; in particular, G is 2-free. However, a 2-free group neednot even be locally free: an example can be found in Baumslag [1962].

We note some simple properties of the subgroup F = <a, h) generated by twoelements a, b of a 2-free group. By definition, F is free and has some rank r < 2, ofcourse. If r < 1, F is cyclic and, in particular, ah = ha. Since any two nontrivialsubgroups of an infinite cyclic group intersect nontrivially, it follows that r = 2 ifa 0 1 -A h and <a> n <h> = 1. Now suppose that r = 2. Then F is free on {a, b} (seeRobinson [1982], p. 160) and therefore every element in F can be written uniquelyin normal form with respect to the generators a, h. Therefore if a'b3 = h3a' for somei, j c- 7L, it follows that i = 0 or j = 0. In particular, <a> n <h> = 1. We have provedthe following three assertions.

Let G be a 2-free group and let a, b c- G.

(16) If <a> n <h) 0 1, then <a,b) is cyclic and ah = ha.

(17) If <a> n <h) = I and a :j4- 1 0 h, then F = <a, b) is free on {a, h}.

(18) If there exist i, j e 7L\{0} such that a'b' = b'a', then ab = ba.

Furthermore, let c e G and suppose that <a'c> n j4 1 0 <asc) n <h> for somer, s e 7L such that r 0 s. Then by (16), CG(b) contains a'c and asc and hence alsoa'c(asc)-' = a'-s. By (18), ah = ba and, using (17), we get the following.

(19) Let a 0 1 j4 b and <a> n <b) = 1. If c c- G and r c- Z such that <a'c) n <h) 0then <asc) n = I for every integer s 0 r.

Now we can prove the following simple, but basic fact.

7.1.7 Lemma. If G is a 2 free group and a, b c- G such that a I : b and (a) n (h) = 1,then

<a> o (h) = { <ab), (a-'b>, (ab-' ), (a-, b-,)}

and these four groups are distinct.

Proof. By (17), F = <a, b) is free on {a, b} and by (2), the four assigned groups liein <a> o <h>. If <x> E <a> o , then F = <a, h) = <a, x) = <h, x) contains x andtherefore x can be written in normal form with respect to {a, h}, that is, x =asi hPi ... a"-b°,, where n >- 1, x,, fl E 7L and xj, f, e 7L\{0} for i > 1 and ] < n. SinceF = <a, x>, there exist x, x' E 7L such that the normal form of y = a"xa" starts andends with nontrivial powers of h. If we write y = cwc-' where w e F is cyclicallyreduced and c c F is reduced, then y' = CW'c-' is reduced for every r E 7L\ {0}, andwe see that the normal form of y' starts and ends with nontrivial powers of b.It follows that an equation of the form b = a"yap ... a'1-Y 6, with 7, , Sm e 7L and

6j c- 7L\{0} for i > 1 and j < m can only hold for m = 1, 0 and ya, = h; on theother hand, h c- F = (a, x> = <a, a'xa'') _ <a, y) and hence b satisfies an equationof this form. Thus y = b" and hence x = a-zh"a-" where v e I = { + 1, - I } andx, x' e Z. In exactly the same way we see that x = bfla"h Q' where p e I and fl, E 77.

It follows that x = a"h" or x = b"a" where it, v E I. These 8 elements are clearlydistinct and generate the 4 assigned subgroups.

We need a technical lemma.

7.1.8 Lemma. Let G be a 2 -free group, k, 1, r, s, t c- 7L and a, h, c e G such thata l j4 hand<a>c = 1.

(a) (a'h'>o<a'bs)n<ak>o<a'> 0,then r=s.(b) If <ab'> o <b'> n <a> o <hi> 0 0, then <b'> = <b"'> or <h5> = <h-.+'>.

Proof. By (17), F = <a, h> is free on {a, h}.(a) If <akbr> n <a'bs> -A 1, then (16) implies that akb'a'bs = a'hsa' b' and, since

k j4 0 : 1, this equation can only hold if r = s. So suppose that <akbr> n <a'bs> = 1.Then by 7.1.7, (akb'> o <albs> consists of the four groups <akb'a'hs>, <b-'a'-kbs>,<akh'-sa-') and <h-'a-kb-'a-`>. By assumption, one of these groups lies in<ak> o <a'> and hence its generators are contained in <ak,a'> n-, <a>. In all cases,this implies that r = s.

(b) Let X E <ab'> o <b'> n <a> o . Since F is free on {a, b}, <a> n <b'> = 1.And if <ah'> n <hs> :j4- 1, then by (16), bs would centralize ab' and hence also a;therefore (18) would imply that ab = ba, a contradiction. Thus <ab'> n <b'> = 1.

If r = 0 = s, then by (1), X = <ab'> = <a> and hence t = 0 and <b'> = 1 =<b'+'>. Suppose next that r = 0 and s - 0. Then X = <a> and by 7.1.7 there existsigns µ, v c- 1 = { + 1, -1 } such that <a> = X = <(ab')''(bs)">. It follows that v = 1and t = -ps so that <b'> = <b'> = <b'+'>. Now suppose that r -A 0 and s = 0.Then there exist p, v c- I such that <ab'> = X = <a'b'o>. This implies v = 1and t = rµ, that is, <bs> = 1 = <b-°'+`J Finally, if r -A 0: s, there exist signsv, p, ., o) e I such that <(ab')'bs"> = <a"b',>. If v = - 1, we get that t = 0 andsy = rw so that <b5> = <hi> _ <b'+'>; if v = 1, it follows that t + s y = rcw andhence <bs> = <b-su> = <b-w'+'>. In all cases, <b5> = <h'+`> or <b5> = <b-'+'J asdesired.

We are now going to describe the product of two elements of a 2-free group in thesubgroup lattice.

7.1.9 Lemma. Let G he a 2 -free group and let u, v c G be such that u 1 -A v and n <v> = 1. Let 259 < m c N and for i = 1, ..., in, let k;, I. E l\{0} be such thatui = uk' and vi = v'' satisfy

(20) <ui> 0 <u3> and <vi> <vi> for all i 0j.

If a, h, c c- G are such that

(21) n <a> = n = n <c> = (v> n = 1

and i f f o r every pair (i, j) with i, j c { 1, ... , m} there exist signs i., It, v E I = { + 1, - 1 }satisfying

(22) <utc> o (vf b> n <ui a> o <vi> 0 0

then c = ab.

Proof. By (20), there can be at most one i for which ui'a = 1 with ) c I is satisfied;for this equation implies <u1> = <a>. The same holds for the equation u c = 1 and

7.1 Free groups 351

for the relations <ui"a) n <v> 0 1 and <u;"c) n <v> 0 1; the latter follows from (19)applied to u and v in place of a and b. Therefore we may assume without lossof generality that if one of these equations or relations holds for some ; E I andi E {1,...,m}, then i e {1,...,4}; that is,

(23) u a : 1 : u;"c and (u a) n <v) = 1 = n <v> for all i > 5 and i. E I.

In the sequel, we shall always assume that 5 < i < m. Then we may apply (19)with v, uic instead of a, b, and deduce that there is at most one l E 1L such that<v'b) n <uic) 0 1. By (20), there is at most one k = k1(i) e {1,...,m} such thatv' = vk for some u e I, that is, satisfying <v,"b> n (uic) 0 1. Similarly, there is atmost one k = k2(i) E { 1, ... , m} such that <vk b> n <u-.'c> A 1 for some p c- I. Thusif j is different from k1(i) and kz(i), then n (vj b) = 1 and, by (23), also<ui'a) n (vj ) = I for all ., p, v c I. This, in particular, holds for the triple (%, p, v) thatsatisfies (22) for the pair (i,j). By (23), furthermore, ui"c 0 1 ui'a and, since v 0 1and <v> n = 1, also vj 0 1 0 v; b. Therefore by 7.1.7,

 o (vj b> = { <(ui"c)'(vj b)1 > I x, fl e I j

and

<ui a> o (vi> = { ((ui a)1 v>I?, 8 E 11,

and (22) implies that there exist signs x, 6, co c I such that

(24) (ui c)'(v°b)' = ((ui'a)Yvis)aO

Thus if i c- { 5, ... , m} is fixed, then for every one of the at least 257 indicesjE{1,...,m}\{k1(i),k2(i)} there exists an 8-tuple ().,p,v,x,/1,y,6,w) of signs suchthat (24) holds. Since there are only 28 = 256 different such 8-tuples, there exist twodifferent indices j for which the 8-tuples in (24) coincide. That is, for the given i, thereexist an 8-tuple (;.,,u, v, x, Q, 6, (o) of signs and two indices j, k c { 1, ... , m} such thatj0k,

(25) (ui c)'(v; b)° = ((ui''a)7 vj )w and

(26) (ui c)'(v b)n = ((u, a)yvk)W.

We want to simplify these equations and first show that they can only hold if

(27) co = 1.

So suppose, for a contradiction, that w = - 1 and consider first the case that i3 = 1.Then (25) yields (u c)'v; b = vj-b(u, a)-' and hence vj (ui"c)'vj = (u;'a)-''b-1. From (26)we get the same equation with v; replaced by vk and it follows that

vf(ui C)2vj = Vk(ui C)2v ;

but this is impossible since by (23) and (17), F = (u c,v> is free on {u c,v} and

Vjb -A vk. Similarly for /3 = -1, we get the equation

vj a(ui a) ' vj = vk a(ui a) vk

and therefore the same contradiction. Thus (27) holds. Since w = 1, (25) implies

(28) (vjb)ev,-b = (u c)-'(ui a)'

and (26) yields the same equation with vj replaced by vk. It follows that

(29) (vj b)p vj'

= (vk b)fi vk'.

If b 1, then by (21) and (17), <b,v) is free on {b,v} and, since <vj> and <Vk > aredifferent subgroups of <v>, we obtain vj vk for all p c- Z\{0}. It follows that (29)can only hold for 1 and v; "_b

= vk°_b

= 1; thus (28) implies

(30) b-' = (u c)-'(u; a)'.

If b = 1, (29) yields vj P-' = vka_b and it follows that µ/i - 6 = 0; therefore onceagain (28) implies (30).

Thus we have shown that for every i e { 5, ... , m} there exists a quadruple (i., v, x, y)of signs such that (30) holds. Since there are only 16 such quadruples but m > 259,there exist two different indices i, h c- {5,... , m} for which these quadruples coincide;that is,

(31) (ui.c)'b-' = (u,'a)' and

(32) (uj c)'b-' = (una)'.

We study the four possibilities for the signs x and y. If x = y = 1, then (31) and (32)yield that uj = abc-' = un -". It follows from (20) that i - v = 0 and hence ab = c,as desired. In the other three cases, (31) and (32) yield equations of the form

(33) u°xun = yz = un xui"

where p, rl E I and x, y, z c- {a, b, c, a-', h-', c-' }: indeed, in the case x = 1we get ui "a-' ui cb-' = uh "a-' uh "; if y = 1 = - x, we have ui"c-' u,- = ab =uh`'c ' uh'; and for x = y = - 1, finally, uk b-' u; = ca-' = uh`b-' uN. If x --,4 I in(33), then by (21) and (17), <u,x> would be free on {u,x}; it would follow thatu° = ui°, contradicting (20). Thus x = 1 and hence u°+" = uf+", which again is onlypossible if p + r1 = 0. So (33) implies that yz = x = I and this in all three cases yieldsthe desired equation c = ab.

We can now prove, as stated above, that the subgroup lattice of a 2-free groupsatisfies the assumptions of Theorem 7.1.6. Of course, it is complete and any of itselements is the join of cyclic elements; so we have to find a system E with theproperties (8)-(13). For this we just have to take enough cyclic subgroups (u>, <v),<ui), <vi> as in Lemma 7.1.9. But first of all we give such a system a name.

7.1.10 Definition. A basic system of a complete lattice L is a family

E _ (e11,...,eim,e21,..., e2m,. ..,enl,...,enm)

7.1 Free groups 353

of elements ei, e C(L)\{D} satisfying

(34) e,J ekl for all i, k e { 1, ... , n} and j, I E { 1, ... , m} such that (i, j) (k,1),

(35) there exist el, ..., en e C(L) such that e; n ek = 0 for i 0 k and eij < e; for alliE{1,...,n},je{1,...,m},and

(36) n > 5, m > 259.

7.1.11 Theorem. Let G be a 2 -free group.(a) Then L(G) possesses basic systems.(b) Every basic system E of L(G) satisfies (8)-(13). The map o: G -+ .AY(E) mapping

every a E G to the complex x(a, E) defined in 7.1.3 is an isomorphism from G ontothe group (.,f (E), *) defined in 7.1.6. Furthermore, K(<a>, E) = {x(a, E),x(a-', E)} _{a°,(a-')°} for 1 0 a c G.

Proof (a) Let n, m c- F such that n >_ 5 and m >_ 259. Since G is not abelian, thereexist x, y c- G such that xy # yx and then F = (x, y> is a free group of rank 2. It iswell-known (see Robinson [1982], p. 157) that such a group contains a free sub-group H of infinite rank. So if H = <X> is free on X, we may choose n distinctelements g,..... gn E X, and then clearly <gi> n <gk> = 1 for i k. Every <g,>contains m distinct subgroups (g,,> 1 and the system of these <g1> is a basicsystem of L(G).

(b) Now let (g..>) be a basic system ofL(G) and let g; E G such that

(37) <g1 > < (g,) and (gi> n <gk) = 1 for all j e {1,...,m} and i,k e {1,...,n},i k;

let C be the set of cyclic elements of L(G). We have to prove that E satisfies (8)-(13)and we show first that (8) holds, that is, I K (<a>, E)I = 2 for every 1 # <a> E C. SinceG is torsion-free and, by (18), g-'ag a-' for all g c- G, 7.1.3 shows that K(<a>,E)contains the two complexes

(38) x = x(a, E) = (A 11, ... A.) and a' = x(a-' E) Anm)

where Aij = { (g,ja>, <g-'a> } and A;, = { <g,1a-' >, <gJ' a-') }. These are different.Indeed, by (37) there exists i e {1.....n} such that <a> n (g1> = I and, by (17),<a, g1> is free on {a, g1 } so that <g,1 a> <g,i a-'> for both ;! E I = { + 1, - 11; thusAil -0 A', and hence x # Y. To prove (8), we therefore have to show that there areno further <a>-complexes with respect to E.

So suppose that = (C11,..., Cnm) e K (<a>, E). By (37) there exists i e { 1, ... , n}such that <g1> n <a> = l; let j, I E { 1, ... , m} such that j 1. By (3) and 7.1.7, C,;is a 2-element subset of <gij> o <a> = {<g a")j%,u E I}. For <g"a"> E Cij and<g;ia`'') E C,l (i, p, v,w E I), (4) yields that <g a"> o <gnaw> n < > o (g'> :A 0 and(a) of 7.1.8 implies that µ = w. This shows that there exists p = µ(i) E I such thatC,j _ {<gij a"lil>,<g;J'a'">} for every j e {1,...,m}. We now fix an i e {I,...,n}such that <,g,> n <a> = 1, put p = u(i) and show that for all k e { 1, ... , n} andE {1,...,rn}

(39) Ckl = {( gkla'>,(gklla">}

This is clear if k = i. So let k : i and suppose first that <90 n <a> = 1. Thenµ(k) exists and (4) implies that <g;3a"(`)) o <gkla"(k,> n (g;3) ° <gk,) 0 for allj, l E { 1, ... , m}. It follows from 7.1.9 with u = g;, v = gk and a, b, c replaced by1, a"(k), a'"), that u(k) = u(i) = u and hence (39) holds for k. Now suppose that(90 n <a> A 1. By (16), (9k, a> = <h> is cyclic. Let <c> E Ck,. Then <c> E (gk,) o <a)and hence (c) < <9k1, a> < <gk,a> = <h). Thus there exist r, s, t e 1L such thatgk, = h', c = h' and a" = h'. By (4), again

<.gaa") ° <c) n <9a) ° <9ki) = <9nh') ° <h'> n <9u> ° <h') :?, 0

and (g;,) n <h> = 1 since (g;) n <gk) = I. Therefore the assumptions of 7.1.8(b)are satisfied with a, b replaced by g;,, h and hence <h5> = <h'+') or (h3> = (h-"' >that is, <c> = <gk,a, > or <c> = <gk,la"). Since ICk,I = 2, it follows that Ck, consistsof these two groups and (39) is also proved in this case. So (39) holds for all k, 1. Nowifp= 1,then 7=x, and if y 1, we get ;= 2'. Thus

(40) K(<a>, E) = {x, x'}

and (8) holds. Moreover, (2) shows that if i, k e { I__, n} and j, I E { I__, m}, then<g11a9ki1) is contained in <9i3) o <94ia-1) c <g;3) o Ak, and in <g;3a) ° <gk,)A;3 o <gk,). Thus <g;3) ° A'k, n A3 ° <gk,) : 0 and this proves (9).

Let 1 h e G and = x(b, E) = (B1 1 , ... , B..). Suppose that /3 = x, that is, B3 =A;3 for all i,j. Since n > 5 and <a> and , by (37), can intersect at most two of the(gk) nontrivially, there exist two indices, I and 2, say, such that <g1> n (a) =(91) n (b) = <92> n <a> = <92) n (b) = 1. If i, j e {1,...,m}, then <g11a) E Ailand <923b) E B2J = A23, and hence by (4), <911a) ° <923b) n <911) ° <923> 0.Now 7.1.9 applied with u = 91, v = g2, u; = gl; and v3 = g2j yields that a = h. Thuswe have shown:

(41) If a, b e G such that x(a, E) = x(b, E), then a = h.

So if <a>, E C, x e K((a), E) and /3 E K(<b), E) such that x = /3, then, by (40),x = x(a', E) and /3 = x(b", E) for some i.,,a e I and, by (41), a" = b" and hence (a). This proves (10).

By (40), Y(E) = {x(a, E)Ia E G} where we denote the trivial complex e by x(1, E).Thus if x, /3 E Y (E), there exist a, b e G such that x = x(a, E) and /3 = x(h, E). Weshow that

(42) x/3 = {x(ab, E)};

this will prove (11). By 7.1.5, x(ah, E) E x/3. So suppose that b = (D,,_.' E x/3

and let d e G such that 6 = x(d, E). Since n > 5 and any of the groups <a>, (b), (d )can intersect at most one of the <gk) nontrivially, there exist two indices, again I

( g , ( g ,

If we write x = (A11 , ... , and /3 = (B11,.. . , then by Definition 7.1.4,6 E x/3 implies that D1; a B23 n A 1; ° (g2J> e 0 for all i, j e { 1,._m),m}, and thismeans that for any such pair (Q) there exist signs y, v e I such that<g ;d) ° <g2Jb) n <g' ;a) o (923) 0. Thus the assumptions of Lemma 7.1.9 aresatisfied with u; = 91; and v3 = g21 and it follows that d = ab. This proves (42) andhence also (11).

7.1 Free groups 355

If we write x * /3 for the unique complex in x/3, the map *:.%((E) x 1'(E) -+ 1(E)x for a on andG - by a and

that a a an isomorphism from G ontothis group. In particular, * is associative, that is, (12) holds. Furthermore, by (40) and(38),

K(<a),E) = {x(a,E),x(a-',E)} = {a°,(a-1)°}

for I a e G. Finally, to prove (13), let <a> e C and X s C such that<a> , E). Then x = x(a A, E) for some i. e I and, sinceU X = <x I <x) e X ), there exist finitely many <b;) e X such that a" = b, ... b,. So ifwe write /3; = x(b;, E), then /3; E K ((be ), E) and

x=x(a'-,E)=(a')°=b; *...*b, =P,*...*ir

Thus (13) holds and this finishes the proof of the theorem.

Subgroup lattices of free groups

We can now characterize the subgroup lattices of free groups.

7.1.12 Theorem. Let r >_ 2 be a cardinal number. A group G is free of rank r if andonly if its subgroup lattice L has the following properties.

(43) For every c e C(L)\{O}, [c/O] is infinite.

(44) ifa,beC(L)suchthat auboC(L)andifdaob,thendna=dnb=0.(45) There exists a basic system E of L and a subset S of C(L) such that BSI = r,

U S = U L and that for every finite sequence b,, ... , b, of elements b; E S withb; : b,+, (i = 1, ... , s - 1) and a; e L with 0 -A a; < b; and x; e K (a;, E), thetrivial complex E is not contained in (...((2,x2)x3)...);.

Proof. Suppose first that G is free on X and JXJ = r. For c c C(L)\{O} there existsI g c G such that c = (g) and, since <g) is infinite cyclic, [c/O] ^ L((g)) isinfinite. Thus (43) holds. Let (a), <h> E C(L) such that (a) u 0 C(L) and sup-pose that (d) E <a> o (b). Then, by (16) and 7.1.7, <a> n = 1 and d = a)b"where i . , p c { + 1 , - I. If <d) n (a) o 1, then, again by (16), a would centralize dand hence also b; but this would contradict (17). Thus <d> n <a> = 1, similarly<d> n = I and so (44) holds. Finally, by 7.1.11 there exists a basic system E ofL; let a: G - 1(E) be the isomorphism given in 7.1.11(b) and let S = { <x)Ix E X}.Then BSI = r and U S = <xIx E X) = G = U L. Suppose that b;, a;, x; are as in (45).Then b. _ <x;) for some x; E X, a; = <y;) where 1 y; E (x;), and a; E K(a;, E) _{y (y; ')°}, by 7.1.11; thus there exists k; E 7L\{0} such that x; = (xi )°. As Esatisfies (11), the only complex contained in (... ((x, 22)x3)... ); is x, * *;; and

° ` ° ` = k.... xka 1 since G is free on X and x x forall i. Thus a, * * as 0 1° = E and (45) holds.

Conversely, suppose that G is a group whose subgroup lattice L satisfies (43)-(45)and let E and S be as in (45). By (43), G is torsion-free and (44) implies thatg-' xg x-' for all g, x e G such that x 1. Indeed, if x, g c G such thatg-' xg = x-' 1, then <x, g> would be nonabelian and hence, in particular,<x) u <g> 0 C(L) and g 0 1; by (2), <xg> e <x) o <g> but g-'xg = x-' would implyI * gZ = (xg)z E <xg> n <g>, contradicting (44). Thus G satisfies the assumptionsof 7.1.3(b) and 7.1.5. It follows that there exists the complex x(a, E) defined in 7.1.3and that x(ab, E) E x(a, E)x(b, E) for all a, b e G. A trivial induction yields that fora,,...,a.EG,

(46) x(a, ... as, E) e (... ((x(a,, E)x(a2, E))x(a3, E))...)x(a., E).

We want to prove that G is free on X where we obtain X by choosing from everysubgroup <x> E S one of its generators. For this we have to show (see Robinson[1982], p. 47) that every element of G can be written uniquely as a product of powersof the x; e X. Since U S = U L = G, the set X generates G. If there were an elementthat could be written in two different ways, there would exist s c N, x; E X(i = I__ , s) with x; # xi+, for all i and k; E 7L\{0} such that 1 = x; ... xs=. Ifb. = <xi> E S, ai = <x;`z> < <bi> and a, = x(xk-, E) E K(ai, E), then (46) implies thatE = x(1, E) = x(x;, ... xs=, E) E (... ((a, a2)a3)...)as and this contradicts (45). Thus Gis free onXandlXJ=r.

7.1.13 Theorem (Yakovlev [1974]). Let r > 2 be a cardinal number. The lattice L isisomorphic to the subgroup lattice of a free group of rank r if and only if L is complete,any of its elements is the join of cyclic elements and L has the properties (43)-(45)where the basic system E in addition satisfies (8)-(13).

Proof. If L has these properties, then by 7.1.6 there exists a group G such thatL L(G); and, since L(G) satisfies (43)-(45), G is free of rank r, by 7.1.12.

Conversely, if L = L(G) and G is free of rank r, then L is complete, any element ofL is the join of cyclic elements and, by 7.1.12, L has the properties (43)-(45). By (b)of 7.1.11, the basic system E in (45) satisfies (8)-(13).

Characterization of subgroup lattices

To characterize those lattices which are isomorphic to the subgroup lattice of anarbitrary group, we have to identify the normal subgroups in the subgroup lattice ofa free group. To do this, we describe conjugation by elements.

7.1.14 Definition. If L is a lattice and x, y c- C(L), we define

yTx=

7.1.15 Lemma. If G is a 2 -free group and a, b e G such that a # 1 j4 b and<a> n = 1, then T (a> = { <ba>, <b° `> }.

7.1 Free groups 357

Proof. By 7.1.7, <a> o = { <a"b"> 1µ, v c- I} where I = { + 1, -1 } and, since<a, b> is free on {a, b}, we have <a"b"> n <a> = 1 for all p, v c- 1. Thus, again by7.1.7,

(47) (<a> o ) -<a>= {<(a"b")°'a">Ip,v,i.,w c I}.

By (1), <a> -(a>= {<ak>Ik e Z} and therefore 7.1.7 and (1), in the case k = 0, yieldthat

(48) (<a> o <a>) o _ {<akb'>Ik c 71,1 E I}.

Now <(a"b')-' a"> = <b-"a-"+A> = <a"-' b"> and hence (47) and (48) show thatan element <c> E C(L(G)) is contained in (<a> o ) o <a> and not contained in(<a> o <a>) o if and only if

(49) <c> _ <a"b"a"> where p, i, v c 1.

By (48), <a> o ( o ) = { <a'bk> I, E I, k e 7l} and hence every element of(<a> o ( o )) o <a> is of the form <(a'bk)Iab> where ,, f, 6c I, kEZ. So ifp = ;! in (49), then <c2> = <a"b"a2µb"al> 0 (<a> o ( o )) o <c>; since, how-ever, <c2> E <c> o <c>, it follows that <c> o <c> (<a> o ( o )) o <a>. On theother hand, if p = then <c> = <a"b'a-"> = °" and, by (1),

<c> o <c> = [<c>l1] c {<a'b"a '>Ii c I,n c 7l} 9 (<a> o ( o )) o <a>.

This shows that <c> E T <a> if and only if u that is, if and only if <c> =<ba> or <c> = <b° 1>.

An immediate consequence of 7.1.15 is the following description of normal sub-groups in a 2-free group.

7.1.16 Lemma. Let G be a 2 -free group and let N < G. Then N g G if and only if T <a> c [N/1] for all a c G, b c N such that a I b and <a> n = 1.

Proof. If N g G and a E G, b c- N such that a 1 b and <a> n <b) = 1, then by7.1.15, T <a> (b°>, <b°-'>} [N/I]. Conversely, suppose that this conditionholds and let a c- G, b E N. If <a, b> is abelian, then clearly b° = b c- N. And if <a, b>is not abelian, then a 1 0 b and, by (16), <a> n = 1; the assumption and 7.1.15yield that <b°> E T <a> c [N/1] and hence b° E N. Thus N a G.

We can now give the desired characterization of subgroup lattices.

7.1.17 Theorem (Yakovlev [1974]). The lattice L is isomorphic to the subgroup lat-tice of some group if and only if there exist a lattice L* and an element d E L* with thefollowing properties.

(50) L* is a complete lattice in which every element is the join of cyclic elements;furthermore, L* satisfies (43)-(45) for some cardinal number r > 2 where forthe basic system E in addition (8)-(13) hold.

(51) If a, be C(L*)`,,{O} such that an b = 0 and b< d, then b T a c [d/O].

(52) L [I*/d] where 1* is the greatest element of L*.

Proof. If L L(G) for some group G, then there exist a cardinal number r > 2, afree group F of rank r, and a normal subgroup N of F such that G F/N. Thus, ifL* = L(F) and d = N, then L [I*/d] and (50) and (51) hold by 7.1.13 and 7.1.16.

Conversely, suppose that L* and d e L* satisfy (50)-(52). Then by 7.1.13 thereexists a free group F of rank r such that L(F) L*; let N be the image of d underthis isomorphism. By (51) and 7.1.16, N a F and (52) implies that L L(F/N).

Using similar ideas, Scoppola [1981], [1985] characterizes the subgroup latticesof torsion-free abelian groups of rank at least 2 and of abelian groups containingtwo independent elements of infinite order.

Projectivities of free groups

The results proved so far show at once that a free group is determined by its sub-group lattice. Indeed, if G is free of rank r > 2 and cp is a projectivity from G to agroup G, then L(G) inherits the properties (43)-(45) from L(G); by 7.1.12, G is free ofrank r and hence G G. It is also not difficult to show that cp is induced by anisomorphism from G onto G. We prove this, more generally, for 2-free groups.

7.1.18 Theorem (Yakovlev [1974]). Every projectivity of a 2 -free group is induced bya unique isomorphism.

Progf. Let G be a 2-free group and let cp be a projectivity from G to a group G.For u, v c G, there exist x, y e G such that <x> = and (y> = (v) and then(x, y> 0 = (x)1 u (y)`0 = (u, v>. Since G is 2-free, (x, y> is free and, by 7.1.12 or1.2.5, <u, v> is free. Thus G is 2-free.

By 7.1.11 there exists a basic system E = (ell , ... , en,,,) of L(G) and the mapa: G -- .X''(E) defined by x' = r(x, E) for x e G is an isomorphism from G onto(.K'(E), *). Then Definition 7.1.10 shows that E _ (e`1°, , ... , e) is a basic system ofL(G) and (b) of 7.1.11 implies that the map r: G _f(E) defined by u` = x(u, E) is anisomorphism from G onto (Jf''(E), *).

We extend cp in the usual way to subsets I of L(G), that is, define X' _{X`°IX e X}. Since cp is an isomorphism from L(G) onto L(G),

(53) (a o h)" = all o b`° for all a, b e C = C(L(G)).

For an a-complex x = (A,,,..., e if(E), we define x° = (A'1...., A" ). Then7.1.2 shows that A is a 2-element subset of (e;, o a)`° = eV o a* and that (4) holds forthe A? and e9?. Thus a° is an a`l-complex with respect to E, that is,

(54) 2 ° E K(a °, E) if x e K(a, E), a e C.

Thus p defines a map from -'''E) into 7f(E). The inverse map is defined in exactlythe same way, using cp-' instead of cp, so that p is bijective. If x, fi e Af(E), then

7.1 Free groups 359

6 = x * fi is the unique complex for which there exist a, b, d e C satisfying (5)-(7). By(53) and (54), the images xP, (3P, 6P and all, b", d`° also satisfy (5)-(7), that is, (a * fl )P =d° E aPf3P = {aP * 13P}. Thus (a * #)P = aP * 13P and p is an isomorphism. It followsthat µ = elpt ' is an isomorphism from G onto G. Now 7.1.11 and (54) show that forx e G and <x> ° = <u),

{C x 1}P = {Za (x-1)6 }Pi = K(<x>,E)Pr' = K(,E)L-' = {u,u-'}.

This implies that <x> = = <x>11, that is, cp and µ coincide on the cyclic andhence on all subgroups of G. Thus cp is induced by M. By 1.5.8 and 1.4.2, Pot G = 1and so µ is the unique isomorphism inducing cp.

As a corollary to Yakovlev's theorem we get a classical result of Sadovskii's.

7.1.19 Theorem (Sadovskii [1941]). Every nonabelian locally free group is stronglydetermined by its subgroup lattice.

Proof. Let G be a nonabelian locally free group and cp be a projectivity from G to agroup G. Let .q' be the set of all nonabelian finitely generated subgroups of G.Clearly, every element of G is contained in some X E 9; and if X, Y E So, thenZ = <X, Y> is finitely generated and hence lies in Y. Thus St is a local system ofsubgroups of G. Every X E .91 is a nonabelian free group and so, by 7.1.18, theprojectivity induced by cp in X is induced by a unique group-isomorphism. By 1.3.6,cp is induced by an isomorphism from G onto G.

Free products

It was shown by Sadovskii [1947] that every group decomposable into a nontrivialfree product is determined by its subgroup lattice; but he left open the questionwhether, in this case, every projectivity is induced by an isomorphism. This problemwas solved independently by Arshinov [1970a] and Holmes [1969] who proved thatevery nontrivial free product is strongly determined by its subgroup lattice. Holmes,in addition, studied free products with amalgamation. We give his main result with-out proof.

7.1.20 Theorem (Holmes [1969]). Let G be the free product of the groups A and Bwith amalgamated subgroup C = A n B where A :A1 C 0 Band suppose that IA : Cl > 2or IB : Cl > 2. Let cp be a projectivity from G to a group G.

(a) If CaG,then 6^fG.(b) If C < Z(G), then cp is induced by a unique group-isomorphism.

Another generalization of Sadovskii's theorem was given by Yakovlev. He studiedgroups with a set X of generators such that every projectivity of a subgroup of theform <x1,x2,x3> (x, E X, x1 # x2 -A x3 x1) is induced by a unique isomorphism.He was able to show that, under slight additional assumptions, such a group isdetermined by its subgroup lattice. As a corollary, he obtained the following result.

7.1.21 Theorem (Yakovlev [1986]). Assume that the group G has a presentation withgenerators x1, ..., x and a single defining relation w = 1. If w is cyclically reducedand contains at least three distinct generating elements, then G is determined by itssubgroup lattice.

7.2 Torsion-free nilpotent groups

Another important class of groups which are strongly determined by their subgrouplattices is the class of nonabelian torsion-free locally nilpotent groups. We prove thisresult, due to Sadovskii, using an unpublished manuscript of G.E. Wall's. First wegive a simple lattice-theoretic characterization for this class of groups.

Lattice-theoretic characterization

A group G is nilpotent if and only if it has a central series. And since a subgroup Hof G is central if and only if H u (.x> is abelian for every x c G, we can try to use theobvious description of cyclic and abelian groups to identify central subgroups andcentral series in L(G).

7.2.1 Definition. Let L be a complete lattice, 0 its least and I its greatest element,and let C(L) be the set of cyclic elements of L as defined in 7.1.1.

(a) The element a e L is modularly embedded in L if and only if [a v c/0] is amodular lattice for all c c C(L).

(b) A modular chain (of length t) in L is a finite chain 0 = ao < < a, = I suchthat is modularly embedded in [I/a;] for all i = 0, ..., t - 1.

7.2.2 Lemma. Let G be a torsion-free group and H < G. Then H is modularly em-bedded in L(G) if and only if H < Z(G).

Proof First suppose that H is modularly embedded in L(G) and let x e G. ThenL(H u (x>) -_ [H v (.x>, l] is modular and, since G is torsion-free, 2.4.9 shows thatH u <x> is abelian. Thus H < Z(G). Conversely, if H < Z(G) and X is a cyclicelement in L(G), then X is a cyclic group and hence H u X is abelian. Thus[H u X,`1] is modular and H is modularly embedded in G.

Of course, a group G is torsion-free if and only if [X/1] is infinite for everyX E L(G) such that X 0 1. Therefore to characterize the class of torsion-freenilpotent groups, we only have to tell when a torsion-free group is nilpotent. Forthis, let us call an element a of a complete lattice L isolated if a n c = 0 or a n c = cwhenever c e C(L). Then a subgroup H of a torsion-free group G is isolated in L(G)if and only if o(x, H) is infinite for every x e G\H, and a normal subgroup N of G isisolated in L(G) if and only if G/ N is torsion-free. It is well-known (see Robinson

7.2 Torsion-free nilpotent groups 361

[1982], p. 133) that for a torsion-free nilpotent group G and every n E N,

(1) is torsion-free;

thus these iterated centres are isolated elements of L(G).

7.2.3 Theorem (Kontorovic and Plotkin [1954]). Let G be a torsion free group.Then G is nilpotent (of class at most c) if and only if L(G) possesses a modular chain(qf length c) consisting of isolated elements.

Proof: If G is nilpotent, then the ascending central series of G is a modular chainin L(G) and, by (1), every ZJG) is isolated in L(G). Conversely, suppose that1 = Ho < . < H, = G is a modular chain in L(G) consisting of isolated elements H;.We use induction on c to show that G is nilpotent of class at most c. By 7.2.2,H = H, < Z(G). Thus if c = 1, G = H is abelian and we are done. So let c > 2 andassume that the assertion is true for c - 1. Since H is isolated in L(G), the group G/His torsion-free and 1 = H, /H < < HC/H = G/H is a modular chain of lengthc - I in L(G/H) consisting of isolated elements. By induction, G!H is nilpotent ofclass at most c - I and hence G is nilpotent of class at most c.

gyp-mappings

In the remainder of this section we show that every projectivity cp of a nonabeliantorsion-free nilpotent group G is induced by a group-isomorphism. Basic for theproof of this result is the notion of a (p-mapping, that is, an element map x: G --> Gsuch that

(2) H`0=H'={x'jxeH}forallH<G,and(3) xIA is an isomorphism for every abelian subgroup A of G.

By (2), G2 = G`0 = G so that x is surjective. And for x, y e G, (2) and (3) imply that x'generates <x>4'; so if x' = y', then <x>4' = <y>" and (3) yields that x = y. Thus

(4) every (p-mapping is bijective.

We use Theorem 2.6.10 to show that cp-mappings exist. First of all we note that cp isnormalizer preserving.

7.2.4 Lemma. Every projectivity of a torsion free locally nilpotent group is normal-izer preserving.

Proof. Let G be a torsion-free locally nilpotent group and let cp be a projectivityfrom G to a group G. By 5.6.5, we have to show that (H')`° = (H")' for every 2-generator subgroup H of G. Since G is locally nilpotent, H is nilpotent of class c, say.By 7.2.2, Z(H)`0 = Z(H11) and, since H/Z(H) is torsion-free, a trivial induction yieldsthat Z.(H)4 = Z;(H10) for i = 1, ..., c. In particular, H4' is nilpotent of class c (see alsoTheorem 7.2.3).

Now if H is abelian, then H`° is also abelian and (H')4' = I = (H°)'. If H is non-abelian, then H/Zc_2(H) is also nonabelian and so H/Zt_1(H) is not cyclic. On theother hand, H/Zc_1(H) is a homomorphic image of the abelian 2-generator groupH/H'. By the structure of finitely generated abelian groups there exist generators x,y of H/H' such that Zc_1(H)/H' = <x',y'> where i, j c Z. Since H/Z,(H) is torsion-free and not cyclic, it follows that H' = Zc_1(H). Similarly, (H4')' = Z,_1(H4') andhence (H')4' = (H`')', as required.

7.2.5 Lemma. Let G be a torsion free nilpotent group which is not locally cyclic andlet (p be a projectivity from G to a group G.

(a) There exist exactly two cp-mappings 21 and 22; for all x c G, x'2 = (x")-1.(b) If 1 H < G and 0 is the projectivity induced by cp in H, there exist exactly

two 0-mappings and these are a,IH and 0(2IH.

Proof. (a) By Zorn's Lemma, every abelian subgroup A : 1 of G is contained in amaximal abelian subgroup M of G. We claim that M contains two independentelements of infinite order. This is clear if M = G; for, in this case, by assumption, Mhas a finitely generated noncyclic subgroup. And if M G, then Z(G) < M andM/Z(G) is torsion-free; hence if I z e Z(G) and x c M\Z(G), then <x> n <z> = 1.By 2.6.10, (p is induced by an isomorphism on M and therefore also on A. Then 1.5.7shows that

(5) (p is induced on A by exactly two isomorphisms a and d and a' = (a6)-1 for allaEA.

Let (I: Z(G) -' Z(G)4' be one of these two isomorphisms for A = Z(G). For everyx e G, the subgroup A, = <x, Z(G) > is abelian and hence there is a unique isomor-phism ax: Ax -+ A4' inducing (p on Ax whose restriction to Z(G) is f3. Let x: G - G bedefined by x' = x'x for x e G. We want to show that a is a (p-mapping. For thissuppose that A and M are as before. Then Z(G) < M and by (5) there exists a uniqueisomorphism ;,: M - MP inducing cp on M such that ; IZ(G) If a E A, then xa =

and hence a' = a'. Thus 2IA = IIA is an isomorphism and (3) holds. The sameargument applied to A = A. shows that <x>' = <x)' = <x>" for x c G. It followsthat (2) is satisfied and x is a (p-mapping.

So if fl, and #2 are the isomorphisms inducing 9 on Z(G), we obtain two cp-mappings 21 and 22 such that 2i'Z(G) _ (3.. If x is any 9-mapping, then cp is inducedon Z(G) by the isomorphism 2l,(,) and it follows from (5) that 21z(a) = A for some i.Since cp is induced on A. by the isomorphisms ocIAx and aiIA, whose restrictions toZ(G) both are /3;, it follows that cIA. = 2;IAx and hence x' = x''. This holds for all.x E G and so x = x,. Thus 21 and x2 are the only (p-mappings and, since 21IAxand x2IAx are different isomorphisms inducing cp on A, again (5) yields that x'2 _(x2 )-1

(b) It is clear that the restriction of a (p-mapping to H is a /i-mapping and, by (a),)tl I II : x2IH. If H is abelian, then (3) implies that the -mappings are just the isomor-phisms inducing ,Li and, by 1.5.7, H has at most two such isomorphisms. If H isnonabelian, then by (a), there are exactly two 0-mappings. In both cases, it followsthat xlIH and 11 1H are all the 4-mappings.

Our aim is to show that one of the two cp-mappings constructed in the abovelemma is an isomorphism; the other will then be an anti-isomorphism, that is, willsatisfy (xy)' = y'x' for all x, y e G. We use induction on the class of G to prove thisand first settle the basis for this induction.

Nilpotent groups of class 2

We remind the reader that a torsion-free nilpotent group G is an R-group (seeKurosh [1956], p. 247) so that for all x, y e G:

(6) If x' = y' for some m c 7Z\{0}, then x = y.

(7) If there exist m, n e ZL\{0} such that xm'y" = y"x', then xy = yx.

Furthermore, if H = <x, y> has class 2, then <[x, y] > < H' < Z(H); and ifx'y'[x, y]k e Z(H), then y' E CH(x) and (7) implies that j = 0. Similarly i = 0 and itfollows that

(8) H' = < [x, y] > = Z(H).

7.2.6 Theorem (Pekelis and Sadovskii [1963]). Let G he atorsion-free nilpotentgroup of class 2 and let (p be a projectivity from G to a group G. Then one of the two(p-mappings (see 7.2.5) is an isomorphism, the other is an anti-isomorphism.

Proof. Let x be one of the two cp-mappings and let a, b e G. Then by (3),

(9) (ah)' = a'h' if ab = ba;

in particular,

(10) (a")' = (a')" for all n e Z.

Now suppose that ah -A ba and let H = <a, b>. By (8), H' = < [b, a] > = Z(H) and,since (p is normalizer preserving and induced by a,

(11) <[b,a]3> = Z(H)4 = Z(H4') = (H')' = <[b',a']>.

Since H/Z(H) is not cyclic, 2.6.10 yields that the projectivity induced by (P in H/Z(H)is induced by an isomorphism. On the other hand, (p is induced by x and it followsthat there exist E, 6 e { + 1, - 1) and zl E Z(H4') such that (ab)' = (a')`(b')azl. We putx = (a')' and y = (b')' so that H`° = <x,y> and Z(H`°) = (H`0)' _ <[y,x]>. If wefurther put c = [b, a] and z = [y,x], then by (11), <c'> = Z(HP) <z> and so

(12) c'=z" where ve{+1,-1},c=[b, a], z=[y,x];

in addition there exists r c 7L such that zl = z' and hence

(13) (ab)'=xyz'where rci,x=(a')`,y=(b')aande,6c{+1,-1}.By (3), (abc)' _ (ab)'c' and therefore

(14) (ba)' _ (abc)' = yxzr+v-1

For n e N, we put H. _ <a", b"> and compute (a"bn)a in two different ways. By(10), Hn = <(a")a,(b")"> _ <xn,yn>. By (10) and (13), ((ab)")a = ((ab)")" = (xY)"z'" _xny"z2 where z2 E Z(H"). Since anb" _ (ab)"c1 with c1 E Z(H), (9) implies that(a"b")a = ((ab)")acj = x"y"Z2C1 = X"y"Z3 where Z3 = Z2Ci e Z(H"). Now Z3 =y-"x-"(a"b")a E H. and hence z3 e Z(H") n Z(H,','). By (7), H. is not abelianand therefore Z(H") = (Hn )' = <[y",x"]>, by (11). It follows that there exists k" E 77such that z3 = [y", x"]k^ = [y x]"2k^ = z"2k^ Thus, finally,

(15) (a"b")' =x"Y"zn2k"

On the other hand, since (a"-'bn-')ba = a"-'b "a = a"b"[b",a] = a"b"c" andba(a"-' b' -') = ba"b"-' = a"b [b, a"] b"-' = anb"c", the elements ba and an'b"-'commute. Therefore (9) and (10) imply that

(16) (anbn)a(ca)" = (anbncn)a = (an-Ibn-Iba)a = (an-Ibn-I)a(ba)a.

Suppose first that v = 1 in (12), that is, c' = z. Then by induction we obtain that

(17) (a"b")a = x"y"z"' for all n E N.

Indeed, this is (13) if n = 1; and if (17) holds for n - 1, then (16) and (14) yield

(anbnT azn = (an-' bn-I)a(ba)a = x"-Iyn-1 Z(n-1)ryxZr = xnynZnZnrll 1 ,

as required. Now (15) and (17) imply that n2k" = nr and hence nk" = r for all n e N.For n > Irl, however, this can only hold if kn = 0 = r. Then (13) and (14) yield that

(18) (ab)' = xy and (ba)a = (abc)a = xyz = yx.

Now suppose that v = - I in (12), that is, c' = z-'. Then we similarly obtain that(19) (anbn)2 = xnyn.n(r+n-1).

Indeed, this again is (13) if n = 1; and if (19) holds for n - 1, then (16) and (14) yield

(anbnT 8z = (an-Ibn-I)2(ba)a = xn-Iyn-1Z(n-I)(r+n-2)yxZr-2

l = xnynZI,(n-111)(r+n-2)+r-2

and this implies (19). Now (15) and (19) imply that nkn = r + n - 1 and hencen(k" - 1) = r - I for all n E N; this yields r - 1 = 0. By (13) and (14),

(20) (ab)2 = xyz = yx and (ba)a = (abc)a = xy.

Now (18) and (20) show that either (ab)a = xy = (aa)`(ba)a or (ab)a = yx = (ba)a(a2Yand we want to prove that in both cases, e = S = 1. Since (18) and (20) aresymmetric in a and b, it suffices to show that e = 1. For this, we study(abc)a = ((ac)b)a and use our results with a, b replaced by ac, b. Then there existi., p E + 1, -1 } such that either (acb)a = ((ac)")Z(b2)" = (a'))(ba)"(ca)" or (acb)2 =(b2)"(a2)A(c2)" = (aa)'(ba)"[(ba)",(aa) ](ca)". Again suppose first that v = 1 in (12).Then by (18), (abc)a = xyz = (aa)`(b')aca and therefore

(a2)`(ha)aca = (aa)A(b")"(c')` or (aa)`(ba)°c2 = (a") (ba)"C(ba)",(aa)~J(ca)"

Since every element of H`° can be written uniquely in the form (a')'(b')`w with s, t E 77and w e Z(H`°), the first equation implies that e = i = 1 and the second that e = i,b = p and z = ca = [(b')a,(a')'](c')E = [y,x]zf = z'+E a contradiction. Thus e = 1in this case. And if v = - 1, (20) yields that (abc)' = xy = (a')E(b')a and hence

(a')E(b')a = (a')"(b')"(c')" or (aa)E(ba)s = (aa)z(b")"[(ba)µ (aa)i.](ca)d

This time the first equation leads to a contradiction and the second implies thate = i., S = p and 1 = [(b')b, (a')E] (c')E = [y, x] z-E = z' -E. Again it follows that c =1. Thus we have shown that e = S = 1 so that (18) and (20) yield the following result.

(21) Either (ab)' = a'b' and (ba)' = b'a' or (ab)2 = b'a' and (ba)' = a'b'.

By (9), this also holds if ab = ba and hence for all a, b e G. In the remainder of theproof we shall show that (21) and (9) imply that for all a, b E G, either (ab)' = a'b',that is, x is an isomorphism, or (ab)' = b'a', that is, x is an anti-isomorphism. By7.2.5, the other (p-mapping will then be an anti-isomorphism or an isomorphism,respectively. First of all, (21) implies that (aba)2 e{a'b'a',(a')2b',b'(a')2} and

{a2 b, bat }° = 1(a ')2 b', b'(a')2 }. By (4), x is bijective and it follows that

(22) (aba)' = a'b'a' for all a, b e G.

Now suppose, for a contradiction, that a is neither an isomorphism nor an anti-isomorphism. We then claim that there exist elements u, v, w E G such that

(23) (uv)' = u'v' # v'u', (uw)' = w'u' # Ow', and w'u'v' = vauaw'.

To prove this, we first show that the sets R = {a a G(ag)' = a'g' for all g E G} andS = {a e G(ag)' = g'a' for all g e G} are subgroups of G. For this let g E G and takea, b e R. Then (abg)' = a'(bg)' = a'b'g' = (ab)'g' and hence ab e R; furthermore,by (10), (ga-' )' = ((ag-' )')-' = (a'(g-' )')-' = g'(a-' )' and then (21) shows that(a-'g)' = (a-')'g° that is, a' E R. If a, b e S, then (abg)' = (bg)'a' = g'b'a' =g'(ab)' and hence ab e S; furthermore, (ga-')' = ((ag-')')-' = ((g')-'a')-' =(a-')'g' and then (a-' g)' = g'(a-' )', that is, a-' E S. Thus R and S are subgroups ofG. If R = G, x would be an isomorphism; if S = G, a would be an anti-isomorphism.Thus our assumption implies that R G : S and, since a group cannot be theset-theoretic union of two proper subgroups, there exists u E G such that u 0 R andu S. Let

V={gEGl(ug)'=u'g':g'u'} and W={gEGl(ug)'=g'u'#u'g'}.

Now V 0 since u S, and W : 0 since u R; further, an element q e G that doesnot lie in V or in W satisfies u'g' = g'u', hence (ug)' = (gu)' and therefore ug = gu.Thus G is the disjoint union of CG(u), V, and W. So if vw = wv for all v E V, w E W, wecould choose v c- V and investigate the three possibilities for uv. From uv E CG(u)it would follow that v c- CG(u), a contradiction. Also uv e V would imply thatu = uvv-' E <V> 5 CGW) and hence uw = wu, again a contradiction. It wouldfollow that uv c- W so that, by (22), (uuv)' = (uv)au' = u'v'U' = (uvu)'; since x is bijec-tive, this would imply that uv = vu, a final contradiction. Thus there exist v e V,

w c W such that vw # wv, and we claim that the three elements u, v, w satisfy (23).The first two equations are clear since v c- V and w e W. To prove the third,we compute (vuw)'. By (21), (v(uw))2 E {v'(uw)',(uw)2v'} _ {v'w'U',w'U'v'} and((VU)V%,)' E {(VU)2W2, w'(vu)'} = {v'U'w2, W'v'u2}. Since vw wv, we have v'w'u2w'v'u2, again by (21); similarly, v'w'u' # v'u2w' and w'u'v' # w'v'u'. It followsthat w'u'v' = (vuw)' = v2u'w2 and this proves (23).

We now take u, v, w e G satisfying (23) and study (uvuw)2. By (21) and (22),

((uvu)w)' E {U 2v'U'W', w'u'v'u'} and ((uv)(uw))' e {u'v2w'u2, W'u'u'v'}.

However, as before, u'v'u'w' 0- u2v'w'u' and w'u'v'u' w'u'u'v' and, by (23),u2v'u'w' = u'w'u'v' w'u2u'v2 and w'u'v'u' = v'u'w'u2 0 u2v'w2u'. This is acontradiction. Thus or is an isomorphism or an anti-isomorphism.

Finitely generated nilpotent groups

Before we carry out the induction announced above, we remind the reader of somefurther properties of a torsion-free nilpotent group G of class c > 2. As for regularp-groups, the commutator collecting process yields that for all primes p > c andiE N,

(24) GP' = <xP'lx e G> _ {xP'lx c- G} and

(25) K.(GP) < K;(G)P

If Li(G)/Ki(G) is the torsion subgroup of G/Ki(G), then

(26) G = L, (G) > L2(G) > . . . > LA(G) > LC+, (G) = I is a central series.

Indeed, if Z./L1(G) = Z(G/Li(G)), then by (1), G/Zi is torsion-free and, since[K1_1(G),G] = K;(G) < Li(G), it follows that Ki_1(G) < Zi and then Li_,(G) < Zi.Thus G = L 1(G) > . . . > Lc+, (G) = I is a central series and c(G) = c implies thatL;(G) > Li+1(G) for i = 1, ..., c.

In the sequel we shall mainly study torsion-free nilpotent groups which are finitelygenerated. It is well-known (see Robinson [1982], p. 132) that for such a group G,

(27) L(G) satisfies the maximal condition

and hence, in particular, every subgroup is finitely generated. Furthermore

(28) n G" = I for every infinite subset I of N.n E I

Indeed, every x c n G" is contained in n H" where H = <g c Gjg' = x for somenE! nEI

m e Z {0} >; by (7) and (27), H is a finitely generated torsion-free abelian group andhence n H" = 1.

nEI

Finally, if G = <x, y> is generated by two elements, then G/ZZ_, (G) is a torsion-free abelian group of rank 2 and hence also

(29) GIG' = <xG', yG'> is torsion-free of rank 2.

Moreover, G' = K2(G) = <[x, y], K3(G)> (see Huppert [1967], p. 258) and if thecyclic group G'/K3(G) were finite, it would follow that L3(G) > G' = L,(G), contra-dicting (26). Thus G/G' and G'/K3(G) are torsion-free and hence also

(30) G/K3(G) is torsion-free.

7.2.7 Theorem. Let G be a finitely generated torsion free nilpotent group, cp aprojectivity from G to a group G and x a cp-mapping. Then x is an isomorphism or ananti-isomorphism.

The proof of this theorem is long and technical; it will be given in the next threelemmas. We use induction on the class c of G. If G is abelian, then by (3), x is anisomorphism; if c = 2, Theorem 7.2.6 yields the assertion. So we assume in the sequelthat G, cp, x are as in the theorem, that

(31) c > 3

and that the assertion is true for groups of smaller class. By 7.2.4, cp is normalizerpreserving. Therefore 5.6.2 and a trivial induction yield that for all H < G and i E N,

(32) K;(H)l = K;(H`0) and hence L;(H)' = Li(HP).

Furthermore, if K < Z(H), then cp induces a projectivity 0 in H/K and a induces amap x*: H/K -+ H"/K" defined by (xK)a' = x1K`0 for x e H. Clearly, x* induces theprojectivity iii. And if K < A < H such that A/K is abelian, then c(A) < 2 and, by7.2.5, x'A is a p-mapping where p is the projectivity induced by cp in A. By 7.2.6 or(3), xIA is an isomorphism or an anti-isomorphism and hence the map induced by xor x* in the abelian group A/K is an isomorphism. Thus (3) holds for x* and 2* is a0-mapping. By (27), H/K is finitely generated so that our inductive hypothesis yieldsthe following.

(33) If H < G, K < Z(H) and H/K is torsion-free and of class c(H/K) < c, then 2induces an isomorphism or an anti-isomorphism in H/K.

For every H < G, we write H = H/L,(H) and apply (33) to this group. By (26),LJH) < Z(H) and H/L,(H) is torsion-free of class c(H) < c. Thus by (33),

(34) x induces an isomorphism or an anti-isomorphism in H = H,'LC(H).

7.2.8 Lemma. Let y, a e G, H = < y, a> and K = <y', (ysa)` > where r, s, t E Z. ff 2

induces an isomorphism in G but not in K, then [y, a] E L,(G), c(H) = 2 and )'I H is ananti-isomorphism.

Proof. By assumption and (34), x induces in G/L,(G) an isomorphism and inK/L,(K) an anti-isomorphism which is not an isomorphism. In particular, K is notcyclic and hence r : 0 # t. Since LJK) < L,(G) n K < K, we see that x induces inK/(L,(G) n K) an isomorphism and an anti-isomorphism. It follows that this groupis abelian and hence [y', (ysa)`] E LA(G). Since G/L,(G) is torsion-free, (7) then impliesthat [y, ysa] E LA(G) and so, finally, L y, a] E LA(G) < Z(G). Thus c(H) < 2. Since xdoes not induce an isomorphism in k, it follows that '7CIH is not an isomorphism. By(33), xIH is an anti-isomorphism and H is not abelian.

Let 1: G , G be defined by x' = x-' for x c G. By 7.2.5 and (34), one of the twocp-mappings a and ai induces an isomorphism in G, the other an anti-isomorphismwhich is not an isomorphism since c(G) = c - 1 > 2. We choose the map inducingan isomorphism and come to the main step in the proof of Theorem 7.2.7.

7.2.9 Lemma. Let a he the cp-mapping inducing an isomorphism in G' and suppose thatG = N <x> where N g G and aI N is an isomorphism or an anti-isomorphism. Then a isan isomorphism from G onto G.

Proof. Let t c 1 and a, b c N. Since x induces an isomorphism in G = G/L,(G), thereexist elements h,(a) and f(b, a) in L,(G) < Z(G) such that

(35) (x`a)x = (xa)`aah,(a)a and(36) (aa)lxb)' = (axb)2f(b, a)a.

We first show in several steps that the lemma will be proved if we can show thatf(l,a) = I for all a E N.

(a) If the assumptions of the lemma imply that f(l,a) = I for all a c N, thenf(h, a) = 1 for all a, b c- N.

Proof. Let b e N. Then G = N<xb> and the assumptions of the lemma also holdwith x replaced by xb. So if these assumptions imply that f(l, a) = I for all a E N,this will also be true for the function f *: N x N -* L,(G) defined via (36) for xbinstead of x. This means that (a2)(xb)' = (axb)f*(1 a)' = (axb)' and thus f(h,a) = 1.Since h e N was arbitrary, (a) holds.

(b) If f(b, a) = 1 for all a, b e N, then aIN is an isomorphism.

Proof. For a, b c N, (35) and (36) together with f(h,a) = f(l,a) = 1 and hl(b)' E Z(G)imply that

(axb)a = (aa))xb)' = (aa)x'b'h,(b)' = ((aa)x)bl = ((ax)a)b3

If alN is an isomorphism, we are done. And if IN is an anti-isomorphism, then(axb)a = (h-' axh)a = ha(ax)a(ha)-' = ((ax)a)'b'1 '. For arbitrary g, y e G,

(37) (gY)' = (ga)Y = (ga)(Y°) implies that gY = g.

Indeed, this assumption yields [(ya)z, ga] = 1 so that, by (7), [y', g'] = 1; thus g2 =(g')Y' = (gY)' and, since a is bijective, gY = g.

Now (37) applied with g = a' and y = h yields that [ax, h] = I and, since a andh are arbitrary elements of N, it follows that N is abelian. But then the anti-isomorphism IN is also an isomorphism and (b) holds.

(c) If f(l,a) = I for all a c N, then (a'')' = (a')'x')` for all a e N, t e Z.

Proof. By (36), (a')x' = (ax)' and hence ax = a'x''-' for all a e N. It follows thatax' = a2(x')12 ' for all t e 7L and this yields (c).

(d) If f(b, a) = 1 = h,(a) for all a, b e N and t e 71, then x is an isomorphism.

Proof. For a, b e N and s, t e 7L, (35), (b), and (c) imply that

(xsax'b)a = (xs+fax'b)a = (x2)s+((a`b)a = (xa)s+r(ax')aba

= (xa)s+`(aa)(x')^ba = (xs)sas(xs)`ba = (xsa)'(x`b)'.

Thus a is multiplicative and hence an isomorphism.

(e) Iff(b,a)=I for all a,bEN,then h,(a)=I for all aENandtE7L.

Proof. Let t e 7L and a E N. We put y = x' and study Uk = a''k-' e N anduk = (aa)(Y9)k '+ -+Y=+' fork e N. By induction on k, we get the well-known formulae

(38) (ya)k = ykuk and (y'a')k = (Y°)kUk.

Clearly, (b) and (c) together with (x`)' = (x')` imply that for all k e N,

(39) Uk = Vk

We choose a prime p > c, put H = <y, a> and K = <yp,(ya)p> = <yp, up>, and claimthat x induces an isomorphism in K. Indeed, otherwise 7.2.8 would imply thatc(H) = 2 and xIH is an anti-isomorphism. Using (c), we would get

(as)Y' = (ay)2 = (y-' ay)s = y2aa(Y2)-' =(a2)(Y,)-'

and, by (37), ay = a. But then H would be abelian, a contradiction. Thus Of inducesan isomorphism in k and hence

((ya)")2 = (Ypup)' _ (Y")aup = (Y2)"Up = (Y'a')p (mod LJK)4').

On the other hand, (35) and h,(a)' E Z(G) imply that

((Ya)p)s = ((x`a)s)p = ((xa)`aah,(a)a)p = (Ysas)p(h,(a)p)'.

It follows that h,(a)p E LJK). By (30), L3(K) = K3(K) and, since K < Gp, (25) yieldsthat LJK) < L3(K) = K3(K) < K3(Gp) < K3(G)p' < Gp3. By (24), h,(a)p = gp' forsome g e G and then (6) implies that h,(a) E GP'. This holds for all primes p > c; by(28), h,(a) = 1 and (e) holds.

Now (a), (d), and (e) yield that to prove the lemma, we only have to show thatf(l,a) = I for all a e N. To do this, however, we shall need some further steps. Weput f(1, a) = f(a) for a e N, first derive some general properties of this functionf : N -+ LA(G), then prove the desired result under additional assumptions on N; the

general case follows. First note that the defining equation (36) now reads

(40) (a2)x' = (ax)2f(a)2 for all a E N.

If xjN is an isomorphism, then ((ab)')x' = ((ab)x)'f(ab)' = (ax)'(bx)'f(ab)' for alla, b c- N and ((ab)')x' = (a')x (b')x' = (ax)'(bx)'f(a)'f(b)' since f(a)' E Z(6); if Of IN

is an anti-isomorphism, ((ab)')x' = ((ab)x)'f(ab)' = (bx)'(ax)'f(ab)' and ((ab)')x' =(b')x'(a')x' = (bx)¢(ax)¢.f(a)¢.f(b)'. In both cases, .f(ab)' = .f(a)¢.f(b)¢ = (f(a)f(b))8, by(3). Since a is bijective, it follows that

(41) f : N -+ Z(G) is a homomorphism.

Another consequence {o of (40), via a simple induction, is that

(42) (a')(x')' = (ax' )¢J (axt t }...+x+l )¢ for all a E N, t e N.

Indeed, if t = 1, this is (40); and if (42) holds for t - 1, then (40), L,(G)' < Z(6), and(41) yield that

((ax[I)¢f(a x' }...+s+1)2)x°(a')(x')t

=((a')`)'- I )x' =

_(ax')¢f(axI t)¢ f(ax'-1+...+s+1)¢

=(ax')¢f(ax'

'+-.-+x+I)¢

This proves (42) and, raising both sides of this equation to the s-th power, we obtain

(43) ((as)¢)(x')' = ((as)xI)¢(.f(ax'-1+...+x+1 )s)¢ for a e N, t e Nl, s E Z.

For every a E N, we define by induction ao = a and a;+i = [a;, x]. We put a', = f(a;)and claim that for all i > 0,

(44) v;: N Z(G) is a homomorphism.

This is clear for i = 0 since vo = f is a homomorphism by (41). Since a; = (ax);,

av"' = .f(a1+t) = .f(aT ' af) = .f(a1)-'.f((ax)j) = (av')-'(ax)v';

therefore if v; is a homomorphism, then since Nv' < Z(G), it follows that for alla,beN,

(ab)t..t = ((ab)v')-'((ab)x)v' = (av,)-'(ax)vi(bt,)-'(bx)v, = av,rtbv

Thus v;+1 is also a homomorphism and (44) holds.We study more closely the factor f(a' +- +x+l) appearing in (42) and first claim

that fort > 1,

(45) .f(ax`t+...+x+1)

= .f(ag)(`+i)

=o

For t = 1, both sides of this equation equal f(a); so suppose that (45) holds for t. Ifs e NJ, then ax'-' = [a,x]x'-' = La", x] = a-"+xs and hence ax' = ax=-'ai'-';

fur-

7.2 Torsion-free nilpotent groups

thermore (a, )i = ai+1 . Using (41), we therefore obtain

f(ax'+...+x+i) = f(axl-1+...+x+1)f(ax'-1+...+x+i)f(a)

= CII f(ai)(i+1))( f(ai+i)('+1)) f(a)1-o J `i=o l

f(ai)(i+1)+(i))f(ac)(`) _ fl f(ai)(!+i)

1=0 l i=o

371

since a, e Kc+, (G) = 1. This proves (45).

For 0 < i < c - 1, we write c! I i + 1) as a polynomial in t, that is, in the form

(46) c!t

_ Y_ ::,jtj where ;ij E Z.i+ j=,

Then we obtain from (45)

C-1 c c-1f(axt- I+...+x+1 ) C! = f(a1)Y-1

/1= f(al)"',"f

i=0 j=1 i=o

that isc

(47)f(ax'-1+...+x+1)c! _ fl gj(a)" where gj(a) = 11 f(ai)"".

j=1 i=0

By (46), x:01 = c! and -'c! so that

(48) g1(a) = f(ao)`f(a1) 112n'...

We can now prove the desired result in the crucial case.

(f) If G = <x,b) and N = <b)G, then f(a) = 1 for all a c- N and hence x is anisomorphism.

Proof. Let i c- 7L and a = bx'; let p > c be a prime and put K = <xP, aP). Then ainduces an isomorphism in k; indeed, otherwise, by 7.2.8, H = <x, a> = G would beof class 2 which contradicts (31). Thus ((aP)x")' _ ((aP)")(x')" (mod Lc(K)") and then

(43) shows that f(axo-I+...+x+1)P e Lc(K). By (47), fl f(ax°-1+...+x+1)dP E LJK).j=1

As in the proof of (e), Lc(K) < L3(K) = K3(K) < K3(GP) < GP' and it follows thatg, (a)Pzg2 (a)P3 ... E GP'. Thus g, (a) p2 e Go' and then (24) and (6) yield that g, (a) E GP.This holds for all p > c and therefore, by (28), g,(a) = 1, that is, g,(bx`) = 1. By (47)and (44), g, is a homomorphism from N to Z(G) and we have just shown that itskernel contains <bx'li e 7L) = " = N. Thus

(49) g, (a) = I for all a e N.

Now suppose, for a contradiction, that there exists a e N such that f(a) : 1. Thensince a; E K;+, (G) = I for i > c, there exists k e {0, ... , c - 1 } such that f(ak) : I andf(a1) = I for all i > k. By (48) with a replaced by ak,

gl(ak) = J (ak)c J (ak+l)-1112)c! = J(ak)c'

so that (49) implies f(ak)e! = 1. Since G is torsion-free, it follows that f(ak) = 1, thedesired contradiction. Thus f(a) = I for all a e N and, by (a), (d), and (e), of is anisomorphism. This proves (f).

We return to the general situation.

(g) Let b e N. Then (50) or (51) holds.

(50) (ab)' = (a2)(xb)' for all a e N.

(51) (a")' =(a' )uxb)=r' and [xb, a] E Z(G) for all a e N.

Proof. Let y = xb. For a E N, define H = (y, a> = (y, M> where M = <a>H a Hand let ,Li be the projectivity induced by tp in H. By 7.2.5, xdH is a 4i-mapping and,since M < N, xIM is an isomorphism or an anti-isomorphism. So if x induces anisomorphism in H, the assumptions of 7.2.9 are satisfied with G and x replaced by Hand 11H; by (f), xIH is an isomorphism and hence (ay)' _ (a')y'. And if x does notinduce an isomorphism in H, we apply 7.2.8 with K = H = (y, a>, that is, r = t = 1and s = 0; it follows that [y, a] E L,(G) < Z(G) and aIH is an anti-isomorphism, thatis, (ay)' = (y-1ay)' = y'a'(y')-1. Thus we obtain that either

(52) (a'')' = (a')y' or

(53) [y, a] E Z(G) and (a'')' = (a')ly')

Suppose, for a contradiction, that there exist u, v e N such that (uy)' = (u')y'(u')(y') ' and (vy)' = (v')1''"-' (v')y'. By (52) and (53) there exists e E { + 1, - 1 } suchthat (ay)' = (a')ly"' where a = uv. If aIN is an isomorphism, it follows that

(ua)y2(Ua)(Y')-' = (uy)a(vy)a = ((uv)y)3 = ((uv)')(Y.) = (u')(y)'(v a)ly')°.

For e = 1, we obtain (v')(y')-' = (v')y' and if e = - 1, (U2)y' _ (u')(y')-'; neither is thecase. Similarly, if xl,1, is an anti-isomorphism, we get the same contradiction. So wesee that the equations in (52) or (53) are satisfied simultaneously for all a e N. Thuseither (50) holds or every a e N satisfies the equation in (51). In the latter case, either[y, a] E Z(G) by (53), or (52) holds for a and then, by (37), [y, a] = 1 e Z(G). Thus(51) holds and this proves (g).

(h) For all a e N, f(a) = 1, that is, (ax)' = (a')x'

Proof. The assertion is that (50) holds for b = 1. Thus by (g), we may suppose that(51) holds for b = 1, that is

(54) (ax)' = (a')(x') for all a e N.

If [xb,d] E Z(G) for all b, d c N, then G = N<x> would imply that G' < Z(G) andthis would contradict (31). Thus there exist b, d e N such that

(55) [xb, d] 0 Z(G).

By (g), such an element b satisfies (50) and, since x induces an isomorphism in(xb)' - x'b' (mod Z(G)). It follows that

(56)(ab)a = (a')(xb)' = (aa)x'b' for all a E N.

So if ocl,, is an isomorphism, ((a')x')b' = (axb)a = (b-'axb)a = (b')-'(ax)aba = ((ax)a)b'

and hence (ax)' = (a')x', as desired. Therefore suppose, for a contradiction, that IN

is an anti-isomorphism. Then (56) and (54) imply that (a2)x'b' = (axb)a = (b-'axb)a =b2(ax)'(b')-' = for all a c N and hence (x')2 = (b')-2 (mod CG(N4')). By(7), G/Cj(N) is torsion-free and therefore by (6), x' _ (b')-' (mod CU(N")), that is,x'b' E Cj(N°). Now (56) yields (axb)' = a' and hence axb = a for all a e N; thiscontradicts (55). Thus (h) holds and Lemma 7.2.9 follows from (a), (d), (e), and (h).

The following lemma finishes the proof of Theorem 7.2.7.

7.2.10 Lemma. or is an isomorphism or an anti-isomorphism.

Proof. Let I be the set of all the subgroups H of G for which xIH is an isomorphismor an anti-isomorphism. Since G is finitely generated, L(G) satisfies the maximalcondition, by (27), and therefore X contains a maximal element N. We have to showthat N = G. So suppose, for a contradiction, that N # G. Then since G is nilpotent,there exists x c NG(N)\N; let H = <N, x> and (i be the projectivity induced by cp inH. Then H 0 X and hence, by (33), c(H) = c. By 7.2.5, xlH and atlH, where g' = g-'for all g c- G, are the two 0-mappings and (34) implies that xlH or ally induces in Han isomorphism. By 7.2.9, this map is then an isomorphism from H onto H°. Itfollows that H e 3:, a contradiction. Thus N = G and the lemma follows.

The main theorem

7.2.11 Theorem (Sadovskii [1964]). Every projectivity of a nonabelian torsion freelocally nilpotent group is induced by a unique isomorphism.

Proof. Let G be a nonabelian torsion-free locally nilpotent group and let cp be aprojectivity from G to a group G. By 1.5.8, cp is induced by at most one isomorphism.By 1.3.6, we may assume that G is finitely generated since the set of nonabelianfinitely generated subgroups is a local system of subgroups of G. Thus G is nilpotent.If c(G) = 2, the assertion follows from 7.2.6, and if c(G) > 3, then 7.2.5 and 7.2.7imply that there exists an isomorphism inducing cp.

We combine Theorem 7.2.11 and Sadovskii's Approximation Theorem to obtainthe following result.

7.2.12 Theorem. Let G he a nonabelian group and let .' he a family of normal sub-groups of G such that

(a) for every X, Y E .9' there exists Z E .90 with Z < X n Y,(b) n X = 1, and

XE V(c) GIN is torsion-free and locally nilpotent for every N E Y.

Then every projectivity of G is induced by a unique isomorphism.

Proof. Let (p be a projectivity from G to a group G. By (b) and (c), G is torsion-freeand hence 1.5.8 implies that (p is induced by at most one isomorphism. Let .% _(X E ,-VIG' X}. Then, clearly, (a) and (c) hold with 9 replaced by . And (b)implies that for I - g c G, there exist X, Y E. such that g X and G' 4 Y; by (a)there exists Z c Y such that Z < X n Y. Then Z E and g 0 Z; since G/Z is tor-sion-free, it follows that <g> n Z = 1. By 6.5.1 and 7.2.11, N° a G for every N Eand the projectivity induced by cp in GIN is induced by a unique isomorphism fromG/N onto G/N`0. By 1.3.8, (p is induced by an isomorphism from G to G.

It is well-known (see Robinson [1982], p. 159) that if F is a free group,n K.(F) = I and F/K;(F) is torsion-free. Thus Theorem 7.2.12 yields another proof

i E N

for the fact (see 7.1.18) that every projectivity of a nonabelian free group is inducedby a unique isomorphism. Another important class of groups which can be approxi-mated by torsion-free nilpotent groups is the class of free polynilpotent groups (seeSmelkin [1964]). So we have the following result.

7.2.13 Corollary (Sadovskii [1965a]). Every projectivity of a free polynilpotentgroup (in particular, of a free soluble group of given derived length) is induced by aunique isomorphism.

Exercises

1. Give a lattice-theoretic characterization of the n-th centre of a torsion-freenilpotent group.

2. (Scott [1957]) Let a be a bijective map from the group G to the group G such that(x),)' = x'y' or (xy)' = y'x' for all x, y c G. Show that x is an isomorphism or ananti-isomorphism. (Hint: First show that x satisfies (9) and (21).)

7.3 Mixed nilpotent groups

We would like to prove for mixed and periodic nilpotent groups analogues of theresults for torsion-free nilpotent groups. However, this is too much to expect sincethere are examples of nilpotent and nonnilpotent groups with isomorphic subgrouplattices and of projectivities between nonisomorphic nilpotent groups. For example,

7.3 Mixed nilpotent groups 375

in 6.2.9 we noted that the semidirect product G = T <z> of an abelian group T oftype px and an infinite cyclic group <z) with respect to the automorphism t -. t1+p

(t e T) is lattice-isomorphic to the direct product G = T x <z>. Clearly, G is abelianand IZ"(G)I = p" for all n c t\J so that G is not nilpotent.

Finitely generated groups

We want to show that the situation is better for finitely generated nonperiodicnilpotent groups. Such a group G has a finite torsion subgroup T(G), and if oneconstructs a central series in G containing T(G), it is well-known (see Robinson[1982], p. 133) that one can take infinite cyclic factor groups above T(G) and cyclicfactor groups of prime order below T(G). Thus L(G) has a cyclic modular chain inthe following sense.

7.3.1 Definition. A modular chain 0 = ao < 5 a" = I in a complete lattice Lis called cyclic if there exists k e {0,...,n - 1} such that [a,/a,_1] L(C2) fori = 1, ..., k and [ai/a,_1] c L(C,) for i = k + 1, ..., n; recall that C, is the cyclicgroup of order r.

7.3.2 Theorem. The group G is finitely generated non periodic nilpotent if and only ifL(G) has a cyclic modular chain.

Proof. Let I = Ho < .. < H" = G be a cyclic modular chain in L(G). We have toshow that G is finitely generated, nonperiodic, and nilpotent. For this we shall needthe following assertion which we prove by induction on i = 0, ..., n.

(*) If x c G is of finite order, then [<H,, x)/H,] is finite and distributive.

Indeed, for i = 0, [<H,,x)/H,] L(<x)) is finite and distributive. If [<H,,x)/H,]has this property, then <H,,x) is a cyclic element in [G,/H,] and, since Hi,, ismodularly embedded in this lattice, [<H,+1,x)/H,] is a modular lattice. By 2.1.5,[ <H,+1 , x)1 H,+1 ] = [H;+1 u <H,, x)/H1+1 ] is isomorphic to [<H,, x)1 <H,, x) n H;+1 ],and this is an interval in [<H,, x)/H,], hence is finite and distributive. Thus (*) holdsfor i=0,...,n.

We now prove the statements on G by induction on n, the length of the cyclicmodular chain in L(G); let k be as in Definition 7.3.1. If n = 1, then k = 0 and soL(G) L(Cx). By 1.2.6, G n C. is finitely generated nonperiodic nilpotent. So letn > I and suppose that the assertion is correct for groups with cyclic modular chainsof smaller lengths. If k = 0, then [H1+1/H,] L(C,) for all i and hence G is torsion-free. Indeed, if I x c- G had finite order, there would exist a maximal i e {0, ... , n}such that x H,; by (*), [<H,,x)/H,] would be a nontrivial finite interval in[H1+1/H1], which is impossible in L(Cj). Thus by 7.2.2, H1 < Z(G) and the inductionassumption implies that G/H1 is finitely generated, nonperiodic and nilpotent. SinceCx ^- H1 < Z(G), the group G also has this property.

Therefore let k > 0. Then H1 is cyclic of prime order p. Let x e G\Hn_1. If x hadfinite order, then by (*), [<Hn_1,x)/Hn_1] would be finite and nontrivial and this

again is impossible since [G/H,,_,] L(C.). Hence o(x) = x and L(<H,,x>) ismodular since H, is modularly embedded in L(G). By 2.4.11, H, :!9 <H,, x> andx induces a power automorphism a - a' in H, for which r - I (mod p). Thus xcentralizes H,. Since G is generated by the elements x e it follows thatH, < Z(G). By induction, G/H, is finitely generated, nonperiodic and nilpotent;since IH1 I = p, the group G also has this property.

An easy consequence of Theorem 7.3.2 is the following lattice-theoretic character-ization of the class of nonperiodic locally nilpotent groups.

7.3.3 Theorem. The group G is nonperiodic and locally nilpotent if and only if thereexists a cyclic element X in L(G) such that for every finite set of cyclic elementsX,, ..., X. in L(G), the lattice [X u X, u u has a cyclic modular chain.

Proof. If G is a nonperiodic locally nilpotent group, take x e G of infinite order andput X = <x>. Then for every finite set of cyclic elements X,, ..., X in L(G), H =<X, X, , ... , is a nonperiodic finitely generated and hence nilpotent subgroup ofG. By 7.3.2 there exists a cyclic modular chain in L(H) [H/I]. Conversely, supposethat G satisfies our condition, and take a finite subset S of G. If X,..... X. are thesubgroups generated by the elements of S and H = <X, X,, ... , then L(H) hasa cyclic modular chain and, by 7.3.2, H is nilpotent and nonperiodic. In particular,G is nonperiodic, <S> is nilpotent and thus G is locally nilpotent.

It follows from Theorems 7.3.2 and 7.3.3 that the classes of nonperiodic finitelygenerated nilpotent and of nonperiodic locally nilpotent groups are invariant underprojectivities, a result due to Pekelis [1961b]. Note that the order of succession ofthe cyclic intervals in Definition 7.3.1 is important. Indeed, the subgroup lattice ofthe direct product G of an infinite cyclic group and a nonabelian group of order pq, pand q primes, has a modular chain I = Ho < H, < Hz < H3 = G with [H1/Ho]L(C,) and [H;/Hi_,] L(C2) for i > 1; but G is not nilpotent.

Groups with torsion-free subgroups

In the light of Sadovskii's theorem on torsion-free nilpotent groups it is reasonableto expect that Baer's theorem on abelian groups with two independent elements ofinfinite order might generalize to nilpotent groups. However, the following exampleshows that this is not the case.

7.3.4 Example (Pekelis [ 1965] ). Let G = <a, b j [a, b] 3 = [[a, b], a] = [[a, b], h] = 1)'.If we put c = [h, a], then G' = <c> < Z(G) = <a3> x <b3> x <c)'; thus G is nil-potent of class 2 and possesses two independent elements a and h of infinite order.Every g e G can be written uniquely in the form g = zr where z e Z(G) andr e R = {a`h'IO < i,j < 2}. We define a bijective map a: G --+ G by g° = z'r' wherer° = r for all re {I,a,a2,b,b2,ab}, r° = rc2 for re {ab2,a2b,a2b2} and oiZ(c) isthe automorphism satisfying (a3)° = a3, (h3)° = b3, c° = c2. It is easy to check

that Q2 = I and that a satisfies the assumptions of Theorem 1.3.1; thus a induces anontrivial autoprojectivity cp in G. Since (a)° = (a), (b)° = (b), and(ab)° = (ab), we see that cp cannot be induced by an automorphism of G.

So it is clear that we need stronger assumptions for a nilpotent group G to obtainthat every projectivity of G is induced by an isomorphism. In this spirit, Pekelis[1967] proved that a nilpotent group of class 2 with a torsion-free subgroup of thesame class has this property. On the other hand, he gave an example showing thatthis assertion does not hold with 2 replaced by 3, and Arshinov [1971] producedsimilar examples for all c > 3 (see Exercise 2). Further results of this type, mainly forgroups with torsion-free direct factors, can also be found in Arshinov [1971]. Westate the most interesting general result in this direction without proof.

7.3.5 Theorem (Yakovlev [ 1988] ). Let n e N, n >_ 2.(a) If G is a nilpotent group of class n containing a subgroup isomorphic to a free

nilpotent group of class n, then G is strongly determined by its subgroup lattice.(b) If K is a torsion free nilpotent group of class n > 3 generated by 2 elements and

K is not a free nilpotent group of class n, then there exist a nilpotent group G of classn containing K as a subgroup and a projectivity of G which is not induced by anisomorphism.

We finally mention that for a nilpotent group G, the existence of two independentelements of infinite order does not imply that G is determined by its subgroup lattice.

7.3.6 Example (Arshinov [1971]). Let A be a torsion-free abelian group, p a primeand n E N such that 4 < n < p. For,,,,, y2 e {1.....p - 1 }, let B;= G"(y,) be the groupof order p" constructed in 5.6.7 and, finally, G = A x B1 and G = A x B2. By (b) of5.6.7 there exists a bijective map T: B1 B2 inducing a projectivity from B, to B2 andwhose restriction to every maximal subgroup of B1 is an isomorphism; let a: G - Gbe defined by (ab)° = ab' for a e A, b e B1. The reader may show that a satisfies theassumptions of Theorem 1.3.1 and therefore induces a projectivity from G to G. By(a) of 5.6.7, G G, in general.

Normalizer preserving projectivities

By 5.6.2, a normalizer preserving projectivity of a nilpotent group maps centralseries to central series. We study the projectivities induced in the central factors. Forthis we need the following result on multilinear mappings.

7.3.7 Lemma. Let A...... A,, L, M be (additive) abelian group, cp a projectivity fromL to M and suppose that f: (A,..... A,) -+ L and g: (A,..... A,) M are such that

(1) f and g are multilinear, that is, additive in every component,

(2) L = (f(x,,.. ,x,)Ix1 E A,), M = (g(x,,...,x,)Ix1 e A1), and

(3) (f(x1,...,x,))l = (g(xl,...,x,)) for all xi e A,.

Then there exists an isomorphism x from L onto M such that

(4) f(.xi,...,x,)2 = 9(x1...... ,) for all x; E Ai.

Proof: We use induction on t. For t = 1, f and g are epimorphisms from Al onto Land M, respectively, and by (3), Kerf = Kerg. Then the map x: L -+ M such that.f'(x)' = g(x) (x c A1) is well-defined and an isomorphism satisfying (4). So let t > 2,and assume that the lemma is correct for smaller numbers of abelian groups Ai. Firstnote that the assertion of the lemma is equivalent to the following.

(5) If C is a finite subset of A = (A1,..., A,,), then I f(x) = 0 if and only

if I g(x) = 0.XEC

XEC

Indeed, if x: L -* M is an isomorphism satisfying (4), then(EcY_ f(x)) _ Y g(x) for

XEC

every finite subset C of A; in particular, (5) holds. Conversely, suppose that (5) issatisfied. By (2), for every w c L there exists a finite subset C of A such thatw = Y f(x) and we put w° = g(x). Then (5) and (1) imply that x is well defined

XEC XEC

and injective; by (2), x is surjective and the additivity of x and (4) follow directly fromthe definition of x.

Now suppose that C is a finite subset of A. Every x c C has the formx = (x...... x,) where xi c Ai. For i = 1,...,t, we put A* = <xilx E C> 5 Ai;furthermore A* _ (A*....,.4*), f* = 9* = L* = <f*(a)lac A*>, M* _<g*(a)la c A*>, and consider the projectivity cq* induced by p in L*. Clearly, (1)-(3)hold for the starred groups and maps. So if we can prove (5) for these, then, since Cis a subset of A*,

I .f(x) = Op 1 f*(x) = 0G' 1 g*(x) = 0G' Y_ g(x) = 0,XEC XEC XEC XEC

as desired. Thus we may assume that all the Ai are finitely generated. Since f ismultilinear, it follows that L too is finitely generated and therefore a direct sum ofcyclic groups. Since every cyclic group is residually a cyclic primary group, thereexist subgroups Li of L (i c 1) such that L/Li is cyclic of prime power order forall i c I and n Li = 0. Let Mi = L;° and define.f : A L/ Li and gi: A - M/Mi by

IElfi(x) =.f(x) + Li and gi(x) = g(x) + Mi. Then again (1)-(3) hold for fi, gi, and theprojectivity'pi induced by (p in L/Li; in addition, since n Li = 0, also n Mi = 0.

IEC ,EC

Thus if (5) holds for cyclic groups of prime power order, then

Y f(x) = 0.<> Y f(x) c Li for all i.<> Y g(x) c Mi for all ie> Y g(x) = 0,XEC XEC SEC SEC

as desired. Therefore we may assume for the proof of the lemma that

(6) L is cyclic of order p°', p a prime, in c N.

As a projective image of L, M is cyclic of order q', q e P. For k e t\J and x =(x1,...,x,)EA, (kf(x)>0 = (f(kx1,x2....,x,)>' = (kg(x)>.Hence.f(x) and g(x) have the same order and (2) implies that p = q. We may there-fore assume that

(7) M = L.

Then q is the trivial autoprojectivity and (3) becomes

(8) (f(x)> = (g(x)> for all x e A.

Therefore there exists kx E Z such that (kx, p) = 1 and g(x) = kx f(x). Let rx denotethe order of f(x) in the group L. Then we claim that

(9) kx - k,, (mod ry) for all x, y e A such that rx > ry > 1.

This will prove the lemma. Indeed, by (2) there exists z E A such that <f(z)> = L and,if we put k = k_ then k - ky (modry) and hence kf(y) = kyf(y) = g(y) for all y E A.Thus the map x: L - L defined by wa = kw (w e L) is an automorphism of L sat-isfying f(y)° = g(y) for all y e A.

It remains to prove (9). For this take x = (x1....,x,) E A and y = (y...... y,) E Asuch that rx > ry > 1. Consider the maps a, r from A, to L defined by a' =J(x11...,xr_1,a) and a' = f(yl,..., y,-,, a) for a E A,. As f is multilinear, a and rare homomorphisms; and A; # 0: A;, since x' = f(x), y,° = f(y), and rx > ry > 1.Therefore pA° and pA, are subgroups of index p in A,° and A;, respectively, and itfollows that their preimages a'(pA') and r-'(pAr) are subgroups of index pin A,.Since a group cannot be the set-theoretic union of two proper subgroups, thereexists d E A, such that d 0 a-' (pA°) and d r-' (pA; ). Then d ° pA' and d' pA,so that (do> = A' and (d'> = A;. Let u = (x1.... ,x,_1,d) and v = (y1,..., yi_1,d).Then J(u) = d° and f(u) = d', so that A' = (f(u)) and A; = <f(v)>. Sincef(x) = x,° e A' = <.[(u)> and, similarly, f(y) = y' c (f(u)>, it follows that

(10)

We define a': A, - L and r': A, - L with respect to g in the same way as U, T for J;that is, a° = g(x....... ,_1,a) and a` = g(y,,...,y,_,,a) for all a c- A,. By (8), A,° _A is a subgroup L of L and we may apply the lemma in the case t = I to thehomomorphisms a and a' from A, to L. We obtain an isomorphism fl: L -i L sat-isfying u°° = a° for all a E A,. For a = x, and a = d, this yields f(x)fl = g(x) andf(u)' = g(u). Since automorphisms of L are multiplications with integers prime to p,there exists i e t\1 such that g(x) = if(x) and g(u) = if(u). It follows that k,f(x) =g(x) = if(x) and hence kX - i (mod rx); similarly, k = i (mod Since rx < r., finally,

(11) kx - k. (mod rx).

In the same way, applying the lemma in the case t = I to r and r', we obtain

(12) ky - k,, (modry).

Finally, for d e A, chosen above, there are multilinear mappings f andfrom (A A,-,) into L = <f(z,,...,zt_1,d)Izi E A,> defined by f(z,,...,zt_1) _f(z_., z,_1, d) and y(zl,..., zt_1) = g(z,,..., z,_ 1, d). By the induction assumption

there exits an automorphism h of L such that f(z,,...,z,_,)a = g(z,,...,zt_1) for allc A1. In particular,

f(u)' =.f(x,,...,x,-l,d)s = f(x,,...,x,-1)S = g(x,...... X,)

= g(x,,...,x,-l, d) = g(u)

and, similarly, .((r)" = g(v). Again there exists an integer j such that g(u) = jf(u) andg(v) = and it follows that k. = j (mod r.) and k,. - j (mod r,.). Thus k. = k,(mod r) where r = min(r., r,,). Now (10) and the assumption rx > ry, yield that r > r,.and then, by (11) and (12), kz = k = k,, = kv (mod ry). This proves (9) and thelemma.

We study properties of a fixed normalizer preserving projectivity between twogroups G and G. It is convenient for this purpose to represent G and G as factorgroups of an auxiliary group F. In the applications, F will be a suitable free orrelatively free group. _

So let G, G, F be groups, q a projectivity from G to G, i.: F - G and p: F -* Gepimorphisms. We say that cp is compatible with the pair (;.,P) if K'"° = K" for allK E L(F). More generally, we say that (p is compatible with the pair (2, p) on thesection P/Q of F if K"`' = K" whenever Q < K < P, that is, if cp induces a projectivityfrom P"/Q'- to P"/Q" which is compatible with the restrictions 2', p' of i., p to P/Q.

7.3.8 Remark. If co is compatible with the pair (2, p), then the map a: G - G definedby (x'')° = x" (x E F) is an isomorphism from G onto G inducing cp.

Proof. For every K < F, K" `O = K". Since cp is a projectivity, it follows that K" = 1if and only if K' = 1. Thus Ker i = Kerp and a is an isomorphism. For everyH < G, there exists K < F such that K" = H and then H`° = K"° = K" = H. Thuscp is induced by a.

7.3.9 Theorem. Let G, G, F he groups, 2: F -+ Gand p: F -+ G epimorphisms and cpa normalizer preserving projectivity from G to G. Let P,/Q,, ..., P,/Q, he centralsections of F, that is, [Pi, F] < Qi for i = 1, .... t, on each of which cp is compatiblewith (;.,,u). Finally, let

P = [P,..... P,] and Q= [Q1,P2....IP1][P,,Q2,P31 ... I P,]...[P1,...1P,-1,Q11.

Then P/Q is a central section of 'F and cp is compatible with (;.,,u) on P/Q.

Proof. The theorem is trivial for t = 1. And it suffices to prove the assertion fort = 2; the theorem for t sections will then follow by induction. Indeed, suppose thatt > 3 and the theorem holds for t - I sections. Then if P = [P...... P,-, ] and Q' =[Q1,P2....,P,-1][PI,Q2,P3,...,P,-1]...[P1....,P,_2,Q,-1], by induction, P/Q is acentral section of F on which cp is compatible with (;.,p); and the theorem fort = 2 applied to P/Q and P,/Q, yields the same for P*%Q* where P* = [P, P,] =

[P1 ,...,P,] = Pand

Q* = [Q, P] [P, Q,]

= [Q1,P2,...,P,]...[P1,...1 P,-2,Q,-1,P1][P1,...,P,-1,Q,] = Q,

that is, for P/Q.So suppose that t = 2, P = [P1,P2] and Q = [Q1,P2][P1,Q2]. By the Three

Subgroup Lemma (see (7) of § 1.5), [P, F] = [P1, P2, F] < IF, P1, P2] [P2, F, P, ] <[Q1, P2] [Q2, P1 ] = Q, that is, P/Q is a central section of F. Let u, v e P1 andx, y e P2. Then [u, x] E P and, since P/Q < Z(G/Q),

(13) [uu, x] _ [u, x]`[v,, x] - [u, x] [c, x] (mod Q).

Similarly,

(14) [u, xy] [u, x] [u, y] (mod Q).

If we choose v E Q1

and y E Q2, then [L, x] and [u, y] lie in Q so that [ut, x][u, x] _ [u, xy] (mod Q). It follows that the map (P1 /Q, , P2/Q2) -+ P/Q sending(uQ1,xQ2) to [u,x]Q is well-defined and, by (13) and (14), bilinear. Since i and it arehomomorphisms, the maps

f: (P1 /Q1,P2/Q2) - PA/QA; (uQ1,xQ2)~ [u A,xA]QA

and

9:(P1/Q1,P2/Q2)-'P"/Q"; (uQ1,xQ2)~[u",x"]Q"

are also bilinear and, as P = [P1,P2], their images generate P"/Q" and P";'Q", re-spectively. Now cp is compatible with (2,p) on P1/Q1 and P2/Q2; furthermore, cpis normalizer preserving and therefore, by 5.6.2, preserves mixed commutators. Itfollows that P"`0 = [P P2 ]"* = [Pi "0, Pz `0] = [Pi , P21 ] = P, similarly Q;-" = Q" and

<[u,x],Q>AW = <[<u,Q1>,<x,Q2>],Q>4 =

<[<u,Q1>",<x,Q2>"],Q"> _ <[u",x"],Q">

Therefore cp induces a projectivity ip from P"/Q" to P",/Q" satisfying

<J (uQ1,xQ2)>' = <[u",x"],Q">/Q" = <9(uQ1,xQ2)>

By 7.3.7 there exists an isomorphism a: P"/Q" P"/Q" such that J'(uQ1,xQ2)' =g(uQ1,xQ2) and hence ([u,x]Q)" = ([u,x]Q)" for all u e P1, x e P,. Since P =[Pl, P2], it follows that (wQ)" = (wQ)" for all w e P. So if Q < K < P, then(K"/Q")' = K"!Q" and (K"/Q")' = K"`0/Q" since x induces the projectivity Cp. ThusK" = K" and q is compatible with (;;,p) on P/Q.

A simple consequence of the above theorem is that a normalizer preservingprojectivity from G to G which is induced by an isomorphism on GIG', is induced byisomorphisms on every factor K;(G)/K;+1(G) of the lower central series of G (seeExercise 3). For certain nilpotent groups, we can say a little more.

Almost free groups of nilpotent varieties

We remind the reader of some basic properties of varieties of groups (see Robinson[1982], pp. 54-60). Let W be a nonempty set of words in the indeterminates xi(i e N). Then for every group G, the verbal subgroup of G determined by W is definedto be

(15) W(G) = <w(91,92.... )19i e G,w e W>.

The variety 23(W) determined by W is the class of all groups G such that W(G) = 1.Such a variety is called nilpotent (or abelian) if it consists of nilpotent (or abelian)groups. A group F e 23 is called 23 free on a subset X of F if for each map a from Xto a group G E 1J there exists a unique homomorphism /3: F -+ G extending a. If F isa free group on a set X c F, then

(16) F = F/W(F) is 23-free on X = {xW(F)Ix e X};

conversely, every 0-free group is of the form F/W(F) for some (absolutely) freegroup F.

If W = {xi'xz'x1x2}, then W(G) = G' for every group G,'Z3(W) is the variety ofabelian groups and a free abelian group F/F' is a direct product of infinite cyclicgroups (see Robinson [1982], p. 59). A verbal subgroup in this abelian group F/F'has the form (F'/F')" for some nonnegative integer n and therefore, if F is a free groupof an abelian variety,

(17) F is a direct product of isomorphic cyclic groups.

Finally, let 23 be a nilpotent variety and F a 0-free group; let c = c(F) be the classof F. Then a factor group

(18) G = F/M where M < K,(F)

is called an almost free group of 23.

7.3.10 Theorem (Barnes and Wall [1964]). Let 23 be a nilpotent variety, G a non-abelian almost free group of 0 and G e 13. If G is finite, assume further that G is notgenerated by two elements. If there exists a normalizer preserving projectivity from Gto G, then

G = F/M be as in (18) and let q be a normalizer preserving projectivityfrom G to G. Since M < K,(F), we have c(G) = c(F) = c and, as G is not abelian,c > 2. Thus KjF) < F' and hence G/G' F/F'. Let W be a set of words inx1, x2, ... such that 23 = B(W) and let W* = W v {xl'xz'x1x2}. Then F = F/W(F)for some free group F and, by (15), W*(F) = F'W(F). It follows that F/W*(F) =

F/(F'W(F)) F/F' ^- GIG'. Therefore GIG' is a free group of the abelian varietyO(W*) and so, by (17), is a direct product of isomorphic cyclic groups. Let cli be theprojectivity induced by (p in GIG'. It is well-known (see M. Hall [1961], p. 155) thatfor a subgroup H of a nilpotent group G,

(19) G = G'H implies that H = G.

In particular, G and GIG' have the same number of generators. Therefore if k is thenumber of isomorphic cyclic direct factors in the direct product GIG', then k > 2. IfGIG' is finite, then so is G (see Robinson [1982], p. 127) and the assumption of thetheorem yields k >_ 3. By 2.6.10 and 2.6.7 (or better Exercise 2.6.1), 0 is induced byan isomorphism a: GIG'- G/G'.

Let A : F - G = F/M and a: G --+ GIG' be the natural epimorphisms. Then;,au: F - 6/6' is an epimorphism. Now F is l3-free on some set X c F and for everyx E X, xZma = yG' where y e G. Since F is 8-free on X and G E 1l, there exists ahomomorphism p: F -+ G such that x" = y for all x E X. Thus f .1ma = f"G' for allf e F and hence G = G'F". Since G is nilpotent, (18) yields G = F", that is, p is anepimorphism. For F' < K <_ F, the definitions of 0, a, a, p imply that K''P/G' =(KAIG')" = (K'`*)"" = KAQe = K"/G'. Thus K-110 = K" and (p is compatible with ().,µ)on the section F/F' of F. We now apply 7.3.9 with P,/Q1 = F/F' for i = 1, ... , tand obtain that 9 is compatible with ().,,u) on K1(F)/K,+1(F) for t = 1, ... , c.Since Kc+1(F) = 1 < Ker.? = M < Kc(F), it follows that (Ker A)" = (Ker A)"' = 1,that is, Ker i! < Ker p. Suppose, for a contradiction, that Ker A < Ker p and takef e (Ker p)\(Ker A). Then f e K.(F)\Kt+1(F) for some i e { 1,..., c} and f a 0 Ki+1(F)'`since KerA < Kc(F). Since (p is compatible with (A, p) on K;(F)/Ki+1(F), it followsthat K.+1 (F)" = K,+1(F);"P < <f, Kj+1(F)> p = <f, K;+1(F)>" = K;+1 (F)", a con-tradiction. Thus Ker). = Ker p and G G.

7.3.11 Remark. Normally our aim is to prove that a given group (for example, analmost free group of a certain nilpotent variety) is determined by its subgroup lattice.In this regard, the above theorem contains three troublesome assumptions, namelythat

(a G is not generated by two elements if it is finite,(b) G E 'B, and(c) the projectivity (p from G to G is normalizer preserving.

However, we can say the following.(a) Barnes and Wall [1964] prove the theorem without the assumption that G is

not a finite 2-generator group. In this case, GIG' can be the direct product of twoisomorphic finite cyclic groups and cp need not be induced by an isomorphism onGIG'. After an easy reduction this leads to the case that G is a finite p-group gener-ated by two elements. If c(G) < 3, it is possible to write down generators and rela-tions for G and G and to see that G G. If c(G) > 4, then G/K4,(G) G/K4,(G) andone can show (see Exercise 4) that lp is induced by an isomorphism t on G/G'G"'where p' = Exp Kc(F). The argument then proceeds as in the general case using anisomorphism a: GIG' -+ G/G' which induces t.

(b) Sometimes G E0 follows from the fact that cp is normalizer preserving. Forexample, let W = {xr, 1X1, X2, ... , xc+1 ] } so that !O(W) = 23(p", c) is the variety of

nilpotent groups of class at most c and exponent dividing p" (p a prime). IfG C'I3(p",c) is nonabelian and (p is a normalizer preserving projectivity from G to G,then by 2.2.7, G is a p-group and so G E J3(p", c).

(c) Also the assumption that (p is normalizer preserving is not always needed. By5.6.6, for example, this is the case if Exp G = p. Therefore we get as a corollary toTheorem 7.3.10 that every nonabelian almost free group in B(p,c) is determined byits subgroup lattice. For c = 2, finally, we obtain the following result.

7.3.12 Theorem. Every nilpotent group of class 2 and exponent p (a prime) is deter-mined by its subgroup lattice.

Proof. Let c(G) = 2, Exp G = p and qp a projectivity from G to a group G. If G isgenerated by two elements, then I G I = p3 and G G; suppose therefore that G is notgenerated by two elements. Since Exp G = p, GIG' is a vector space over GF(p); let(g,G');E, be a basis of this vector space. By (16) there exists a group F which is13(p, 2)-free on a set { f I i E I} c F of the same cardinality. Since G E I3(p, 2), thereexists a homomorphism /3: F G such that g; for all i e I. By (19), /3 is anepimorphism. Thus (F')'1 = G' and the basis (f F');E, of F/F' is mapped to the basis(g;G);E, of G/G' by the epimorphism from F/F' onto G/G' induced by /i. It followsthat Ker /3 < F' and, since G is not abelian, Kerfl < F' = K2 (F). Thus G is a non-abelian almost free group in T(p, 2). By 2.2.7, G is a p-group of exponent p, of course.By 5.6.6, cp is normalizer preserving and hence c(G) = 2. Thus G E 23(p, 2) and all theassumptions of Theorem 7.3.10 are satisfied. It follows that G G.

Exercises

1. In Examples 7.3.4 and 7.3.6, verify that o satisfies the assumptions of Theorem1.3.1.

2. (Arshinov [1971]) Let p e P, 3 < n e N and

N = <a3) X ... X Can+1) x (C1) X ... X <Cn-2 )

where o(a1) = oo and o(c1) = p for all i. Let H = N<a2) and G = H<a1) besemidirect products with respect to the automorphisms a and /3, respectively,given by a3 = a3C1, a; = ai (i > 3), ca = cici+l (i = 1,...,n - 3), cn-2 = C-2 andof = a,a,+1 (i = 2,...,n), a#+1 = a"+1, cp = c; (i = 1,...,n - 2).(a) Show that G is nilpotent of class n and <a1,af> is a torsion-free nilpotent

subgroup of G of the same class.(b) For g E G, g = a i az b (r, s C 71, b E N), let g" =g if rs = 0 (mod p) and g °

9C,'-2 where r + s + t = 0 (mod p) if rs * 0 (mod p). Show that o : G -- G isbijective and induces an autoprojectivity lp of G.

(c) Show that (p is not induced by an isomorphism.

7.4 Periodic nilpotent groups 385

3. (Barnes and Wall [1964]) Let q be a normalizer preserving projectivity from G toG. If cp is induced by an isomorphism on GIG', show that qp is induced by isomor-phisms on Ki(G)/Ki+,(G) for all i >_ 1. (Hint: Represent G as a factor group of afree group and study the proof of Theorem 7.3.10.)

4. (Barnes and Wall [1964]) Let G be a finite p-group of class 3 generated by twoelements and suppose that G is Z-free for some variety 0; let p` = Exp K3(G). Ifcp is a normalizer preserving autoprojectivity of G, show that q is induced onG/G'GP' by an automorphism of G/G'G°`.

7.4 Periodic nilpotent groups

A nilpotent torsion group is the direct product of its primary components, and itssubgroup lattice is then the direct product of the subgroup lattices of these compo-nents. Therefore we have to study subgroup lattices and projectivities of primarynilpotent groups. We want to use the concepts of Kontorovic and Plotkin to give alattice-theoretic characterization of this class of groups where, of course, allowancemust be made for the P-groups. However, for every Tarski p-group G, L(G) ismodular and therefore has a modular chain 1 = Ho < H, = G of length 1; but G isnot nilpotent and is lattice-isomorphic to any other (nonprimary) Tarski group.Therefore we have to avoid the Tarski groups, and we do this just as in § 6.4. Recallthat a lattice L is permodular if it is modular and every interval of finite length in Lis finite.

Permodular chains

7.4.1 Definition. Let L be a complete lattice and let a, ao, ..., a, E L. Then a ispermodularly embedded in L if [a u c10] is permodular for all c E C(L), and0 = ao < < a, = I is a permodular chain in L if ai+1 is permodularly embeddedin [I/ai] for every i = 0, ..., t - 1.

We need some simple inheritance properties of these concepts.

7.4.2 Lemma. Let G be a p-group, p a prime, K a permutable subgroup of G,K < H < G and suppose that H is permodularly embedded in [G/K].

(a) Then H is permutable in G.(b) If C is a cyclic element in [G/K], there exists c c G such that C = K<c>.(c) If N < G, then NH is permodularly embedded in [G/NK].

Proof. (a) It suffices to prove the assertion in the special case that H = K<x> forsome x E G. Indeed, if x E H, then K<x> is permodularly embedded in [G/K]; and ifall these K <x> are permutable in G, then so is H = <K <x> Ix E H>. So suppose thatH = K <x>, let y c- G and put T = <H, y); we have to show that H < y> = < y> H.Since K per G, we have that [K<y>/K] ^ [<y)t/<y> n K] is a finite chain and hence

K(),> is cyclic in [G/K]. Since H is permodularly embedded in [G/K], it followsthat [K (y) u H/K] = [T/K] is a permodular lattice. In particular, H mod [T/K]and hence, by (c) of 2.1.6, H mod T. Then [T/H] L-t [ (y)/(y) n H] and [H/K] 2t[(x)/(x) n K] are finite chains. So [T/K] has finite length and, as a permodularlattice, is finite. By 6.2.5, IT: K I < x; thus T/KT is a finite p-group and H is asubnormal modular subgroup of T By 6.2.10, H per T; in particular, H(y> _(y) H, as desired.

(b) Since [C/K] satisfies the maximal condition, C is finitely generated modulo Kand hence, by 6.2.8, I Kc : K I < x. Furthermore, [C/Kc] is distributive, and so C/K`is cyclic. Since G is periodic, C/Kc is finite. Thus I C : K I < x and C/Kc is a finitep-group. Since [C/K] is distributive, there is exactly one maximal subgroup D of Ccontaining K and every element c e C\D satisfies C = K(c>.

(c) Since N and K are permutable in G, we have NK per G. So if C is a cyclic ele-ment in [G/NK], then by (b) there exists c c- G such that C = NK <c>. Since K per G,the element K <c> is cyclic in [G/K] and, as H is permodularly embedded inthis lattice, [H u K (c)/K] = [H u (c)/K] is permodular. Thus [NH u C/NK] =[H u <c> u NK/NK] [H u (c)/(H u (c)) n NK] is permodular and thereforeNH is permodularly embedded in [G/NK].

7.4.3 Corollary. Let G be a p-group and 1 = Ho < < H, = G be a permodularchain in L(G).

(a) Then H; is permutable in G for i = 0, ... , t.(b) If N a G, then 1 = NHo/ N < < NH,/N = G/N is a permodular chain in

L(G/N).

Proof. Clearly, Ho per G. If Hi per G, then H. = K and Hi,, = H satisfy the assump-tions of 7.4.2. Thus H;+1 per G and (a) holds. By (c) of 7.4.2, NH;,, is permodularlyembedded in [G/NH;] and this proves (b).

Nilpotent p-groups

7.4.4 Lemma. Let G be a p-group and suppose that H is permodularly embedded inL(G).

(a) If Exp H = p" (n E t\l), then H < Z2,,(G).(b) If Exp H = x, then H < Z(G).

Proof. (a) We use induction on n. Clearly, every subgroup of H is permodularlyembedded in L(G) and thus permutable in G by 7.4.2. So if M < H is of order p andX < G is cyclic, then I MX 1 2 and N = Z2(G). Then by 7.4.2, HN/Nis permodularly embedded in L(G/N) and Exp(HN/N) < p"-'. By induction,HN/N < Z2(n_1)(G/N) = Z2n(G)/N. Thus H < Z,"(G) and (a) holds.

(b) Let x c- G. Then <H, x> is a p-group with permodular subgroup lattice. If S isa finite subset of (H,x>, then by 2.1.5, L((S>) has finite length and hence is finite;

thus <S> is finite and (H, x> is locally finite. By 2.4.15, a nonabelian locally finitep-group with modular subgroup lattice has finite exponent. Thus (H, x> is abelianand H < Z(G).

The above lemma is the crucial step towards our lattice-theoretic characterizationof primary nilpotent groups. As an immediate consequence we get a characterizationof nilpotency in p-groups.

7.4.5 Theorem. Let G be a p-group, p a prime. Then G is nilpotent if and only if L(G)has a permodular chain.

Proof: If G is nilpotent, its ascending central series is a permodular chain in L(G).Conversely, let I = Ho < . . < H, = G be a permodular chain in L(G) and letN = HG. By 7.4.4, N< Z(G) or N< ZZ"(G) where p" = Exp H1. By 7.4.3,1 = H1 N/N < ... < H,N/N = GIN is a permodular chain of length t - 1 in L(G/N).By induction G/N, and hence also G, is nilpotent.

A second important consequence of Lemma 7.4.4 is that projective images ofnonabelian nilpotent p-groups are nilpotent. We also get an upper bound for theclass of such an image.

7.4.6 Theorem (Yakovlev [1965]). Let p be a prime and G a nilpotent _p-group whichis not elementary abelian. If G is a group such that L(G) L(G), then G is a nilpotentprimary group and

(a) c(G) = c(G) if Exp G = p or Exp G = x,(b) c(G)<, 2nc(G) if Exp G = p", 2 < n c- N.

Proof. Let co be a projectivity from G to G. As a nilpotent p-group, G is locally finite(see Robinson [1982], p. 147). If G is locally cyclic, then so is G and we are done; sosuppose that G is not locally cyclic. Then by 2.2.7, G is a_p-group. Put c = c(G) andHi = Zi(G)`° for i = 0, ..., c. Then I = Ho < < H, = G is a permodular chain inL(G) and, by 7.4.5, G is nilpotent. _

(a) If Exp G = p, then by 5.6.6, cp is normalizer preserving and hence Hi = Zi(G)for all i, that is, c(G) = c. So let Exp G = oo. Then we prove c(G) < c(G) by inductionon c; application of this result to cp-1 will yield the other inequality. It is well-known(see Robinson [1982], p. 133) that if the centre of a nilpotent group has finiteexponent, then so does the whole group. Therefore Exp Z(G) = x and, by 7.4.4,H1 < Z(G). Thus if c = 1, then G is abelian and the assertion holds; so let c > 1 andsuppose that the assertion is true for groups of smaller class. Clearly, cp induces aprojectivity from G/Z(G) to G/H1. Therefore if Exp(G/Z(G)) = x, the inductionassumption implies that c(G/H1) < c - I and hence c(G) < c. Finally, suppose thatExp(G/Z(G)) = p' < cc and let K = KE(G) be the last nontrivial term of thedescending central series of G. Since K`0 < H1 < Z(G), it follows that cp induces aprojectivity from G/K to G/K' and c(G/K) = c - 1. For x c- Kc_1(G) and y e G, wehave [x, y] E K < Z(G), and hence [x, y]P" _ [x P-, y] = 1, as xP"' E Z(G). Since K isgenerated by these commutators, it follows that Exp K < p'; thus Exp(G/K) = x.By induction, c(G/K") < c - 1 and hence c(G) < c.

(b) We use induction on c to show that a (nilpotent) p-group G of exponent pnwhose subgroup lattice possesses a permodular chain 1 = Ho < < H, = G oflength c satisfies c(G) < 2nc. By 7.4.4, Hl < Z2n(G) = N and, in particular, the asser-tion holds for c = 1. By 7.4.3, 1 = Hl N/N < . . < HCN/N = GIN is a permodularchain of length c - 1 in L(G/N) and, by induction, c(G/N) < 2n(c - 1). Thusc(G) < 2nc, as desired.

An arbitrary nilpotent torsion group G is the direct product of its primary compo-nents G. If q is a projectivity from G to a group G, then by 1.6.6, G = Dr GP with

P E P

coprime groups GP and, by 2.2.5 and 7.4.6, these GP are P-groups and nilpotentprimary groups. However, note that G need not be nilpotent even if all the Gp arenilpotent. And there are direct products of coprime finite p-groups which are notlattice-isomorphic to any nilpotent group.

7.4.7 Example. Let P = {pl,p2,...}._ (a) For every n e t\J, let G. be the abelian group of type (p," p."-') and letGn = <x,ylxP" = yP"-` = l,x'' = x'+P) where p = pn. Then G = Dr G. is abelian

_ _ nE N

and G = Dr Gn is lattice-isomorphic to G since every Gn is lattice-isomorphic to G.nE N

and the direct factors are coprime. Since c(Gn) = n, the group G is not nilpotent.(b) For every n e N, let Hn be a such that every projective image of Hn

has class at least n (by 7.4.12, we may take any pn-group of maximal class n + 1).Then every projectivity cp maps H = Dr H. to Dr H.", and this group is notnilpotent since c(Hf) > n. nE nE

Lattice-theoretic characterization

To give a lattice-theoretic characterization of the class of nilpotent primary groupswe combine our concept of permodular embedding with the obvious fact that everyelement of a group G has prime power order if and only if for every cyclic element Cof L(G), the lattice [C/1] is finite and directly indecomposable.

7.4.8 Lemma. Let G be a group such that

(1) [C/1] is finite and directly indecomposable for every cyclic element C of L(G)

and

(2) there exists a nontrivial subgroup of G which is permodularly embedded in L(G).

Then there is a prime p such that either(a) G is a p-group with Z(G) 0 1 or(b) G = PQ where P is a normal p-subgroup of G, IQ I is a prime dividing p - 1 and

Q operates fixed-point-freely on P. In this case, if G is not a P-group, then everypermodularly embedded element of L(G) is a normal elementary abelian p-subgroupof G.

Proof. By (1), G is a torsion group and

(3) every element of G has prime power order.

Let 1 : H < G be permodularly embedded in L(G) and take a minimal subgroup Mof H. Then

(4) IMI = p for some prime p.

Let x E G and let K be a nontrivial finite p-subgroup of H. Then K is permo-dularly embedded in L(G) and so [<K,x>/1] is a permodular lattice. Therefore[<K, x>/K] [<x>/<x> n K] is a finite chain, hence [<K,x>/1] has finite lengthand so is finite. Thus <K,x> is a finite group with modular subgroup lattice. By 2.4.4and (3),

(5) <K, x> is a finite p-group or a finite nonabelian P-group.

If G is a p-group, then M < Z2(G) by 7.4.4. Thus Z(G) 0 1 and (a) holds. So assumethat G is not a p-group and suppose first that there exist r-elements in G where r isa prime, r > p. We claim that G is a P-group in this case. To show this, let R be theset of r-elements in G, and take x, y E R such that x -A I -A y. By (5), <M, x> is a finiteP-group and so o(x) = r and <x>M = <x> since r > p. Furthermore, the P-group<M, x> is generated by conjugates of M and so x E MG. Since all conjugates of M arepermodularly embedded in L(G), <x> a MG. Similarly, <y> a MG and so <x> <y> iselementary abelian of order at most r2. It follows that xy = yx and xy-' E R. ThusR is an elementary abelian r-group and R < MG. By (3), CG(R) is an r-group, henceCG(R) = R. Since every subgroup of R is normal in MG, we see that M induces anontrivial power automorphism in R and so RM is a P-group. Furthermore, G/R isisomorphic to a subgroup of Aut R in which this power automorphism is central.Hence RM/R < Z(G/R) and, again by (3), G/R is a p-group. Since RM a G, we haveshown that

(6) MG = RM is a P-group.

Suppose, for a contradiction, that RM < G and take g E G\RM. Since G/R is ap-group, g is a p-element. If <M, g> were a finite nonabelian P-group, thenI<M,g>l = r' p for some k since no other primes are involved in G. But this wouldimply that g c- MG, a contradiction. Therefore, by (4), <M, g> is a finite p-groupoperating fixed-point-freely on R. Then M is the only minimal subgroup of <M, g>(see Huppert [1967], p. 502) and hence <M,g> _ <g> is cyclic. In particular,g c CG(M). Since G is generated by the elements g E G\RM, it follows thatM < Z(G), contradicting (3). Thus we have shown that G = RM is a P-group and(b) holds.

It remains to consider the case that H is a p-group where p is the largest primeinvolved in G. Let N = O°(G) = <x c- G I (o(x), p) = 1 >. Since G is not a p-group,N 1. Let K be a nontrivial cyclic subgroup of H and take 1 -A x c- G with(o(x),p) = 1. By (5), <K, x> is a nonabelian P-group of order p'q for some primeq < p. Thus

(7) 1 K I = p, o(x) is a prime and K g <K, x> < N.

Since x was arbitrary, K g N. In particular, M g N. So if P = CN,(M), then P g Nis a p-group and N/P is cyclic of order dividing p - 1. By (7), 1N : PI = q is a prime.Thus N = PQ where IQI = q and Q operates fixed-point-freely on P, by (3). Since Pis a characteristic subgroup of N, P a G. Every p-element of G/P normalizes NIP, acyclic group of order q < p, and thus centralizes N/P. But G/P does not containelements of composite order. This shows that every p-element of G is contained inP< N and that N = G. By (7), H is elementary abelian and normal in G. Thus (b)holds.

7.4.9 Lemma. Let G be a group such that

(8) every interval in L(G) is directly indecomposable and

(9) L(G) has a permodular chain.

Then there is a prime p such that either(a) G is a nilpotent p-group or(b) G = PQ where P is a normal abelian p-subgroup o f G, I Q I is a prime dividing

p - 1 and Q induces nontrivial power automorphisms in P.

Proof. We shall need the following simple fact.

(10) A finite group G of order p"q satisfying (8) has the structure given in (b) if pand q are primes such that q divides p - 1.

Indeed, if P E Sylp(G) and Q = <a> E Sylq(G), then P g G and P/cD(P) =N,/T(P) x x N,/D(P) where Ni a G and I N;/D(P)I = p for all i (see 4.1.7).If a induced different powers in N,/(D(P) and N2/D(P), say, then [G/H] ={G, N, H, N2 H, H} would be the direct product of two chains of length 1 whereH = N3 ... N,Q, a contradiction. Thus a induces the same power in every N;/Qt(P)and so G/(D(P) is a P-group. Hence if X is a proper subgroup of P, there existmaximal subgroups M, and M2 of G with X < M, n M2 and M, -A P -A M2. Byinduction, M, and M2 have the structure given in (b) and so X a <M1, M2 > = G.Thus every subgroup of P is normal in G, that is, Q induces power automorphismsin P, and P is abelian since p > 2. This proves (10).

To prove the lemma, we use induction on the length t of a permodular chain1 = Ho < H, < < H, = G in L(G). By (8), G is a torsion group and thereforesatisfies the assumptions of 7.4.8. If G is a p-group for some prime p, then by 7.4.5, Gis nilpotent and (a) holds; if G is a P-group, (b) is satisfied. So we may assume that Gis neither a p-group nor a P-group. Then by 7.4.8 there are primes p and q such thatq divides p - 1 and G = PQ where P is a normal p-subgroup of G, I Q I = q and Qoperates fixed-point-freely on P; furthermore, H, is an elementary abelian p-groupand normal in G. By induction, H, = P or P/H1 is abelian. In both cases, P ismetabelian and so locally finite. Therefore, if x and y are nontrivial elements of Pand Q = <a>, then

L = <x, x°, .. , x"-',y, y...... yp9-' >

is finite and invariant under Q. Hence LQ is a finite group of order p"q satisfying (8)and, by (10), L is abelian and <x>" = <x>. Thus xy = yx and since x and y arearbitrary, P is abelian. So <x> a G and Q induces power automorphisms in P.

We come to the desired characterization.

7.4.10 Theorem. The group G is a nilpotent p-group for some prime p or a P-group ifand only if it has the following properties.

(i) Every interval in L(G) is directly indecomposable.(ii) L(G) has a permodular chain.(iii) If S < T < G such that [S/1] is a chain of length 2 and S is maximal in T, then

[T/1] is modular or is isomorphic to the subgroup lattice of the dihedral group D8 oforder 8.

Proof. Let G be a nilpotent p-group. If [H/K] were a directly decomposable intervalin L(G), then, since G satisfies the normalizer condition, L(NH(K)/K) would bedirectly decomposable, in contradiction to 1.6.5. Thus (i) holds and (ii) follows from7.4.5. Finally, if S and T are as in (iii), then T is a group of order p3 with a cyclicsubgroup of order p2. Such a group has modular subgroup lattice or is isomorphicto D8. Hence (iii) is also satisfied. If G is a P-group, L(G) is isomorphic to thesubgroup lattice of a suitable elementary abelian group and therefore also satisfies(i)-(iii).

Conversely, let G be a group satisfying (i)-(iii). Then by 7.4.9, either G is anilpotent p-group or G = PQ where P is a normal abelian p-subgroup of G, IQ I is aprime dividing p - I and Q induces nontrivial power automorphisms in P. If Pcontained an element x of order p2, then S = <x> < <x>Q = T would satisfy theassumptions of (iii). But [T/1] is not modular since T = <Q, Q"). Also [T/1] is notisomorphic to L(D8) since p > 2. This contradiction shows that P is elementaryabelian and G is a P-group.

It should be mentioned that Theorem 7.4.10 is inspired by a lattice-theoreticcharacterization of finite p-groups and P-groups due to Suzuki [1951a]; seeExercises 3 and 4, also Suzuki [1956].

Finite p-groups

Apart from the results of §§ 2.5 and 2.6 on abelian groups, not much is known aboutthe problem: which finite p-groups are (strongly) determined by their subgrouplattices. So it seems better to ask which classes of p-groups are invariant underprojectivities; by 2.2.6, we may restrict our attention to projectivities between p-groups. Unfortunately, this question is not a very fruitful one either. It is rather easyto answer, in general. For example, the order, exponent, number of generators,Frattini length of a p-group are obviously invariant under these projectivities; on theother hand, properties depending on commutator calculus, like class or derivedlength, are not invariant, as the projectivities between p-groups with modular sub-

group lattices show. So there are only a few classes which are serious candidates forstudy. One of these, the class of metacyclic p-groups, was handled in 5.2.13. Webriefly look at regular p-groups and then consider groups of maximal class.

Recall that a finite p-group G is regular if for all x, y E G,

(11) xPyP = (xY)P fl dP

with suitable d; E <x, y>'. This class of groups is not invariant under index preservingprojectivities. Indeed, if we put m = p - 1 in Exercise 5.5.1, then c(G) = p - 1 andhence G is regular (see Huppert [1967], p. 322); in G, however, (bff_,)P = bPfo and,since Exp G' = p, this contradicts (11). In any case, since projectivities between p-groups satisfy and t (G)" = ?J (G) for all n e N, at least the powerstructure of regular p-groups is preserved under these projectivities (see Exercise 5).

p-groups of maximal class

Let G be a finite p-group, IGI = p", n > 2. Recall that G is said to be of maximal classif c(G) = n - 1. Then every factor of the upper and lower central series of G, exceptthe uppermost one, has order p and hence for i = 0, ..., n - 2,

(12) Z;(G) = is the unique normal subgroup of order p' of G.

We shall use the following characterization (see Huppert [1967], p. 375):

(13) G is of maximal class if and only if there exists s E G such that ICG(s)I = p2.

Let n > 4 and suppose that s e G such that ICG(s)I = p2. Then

(14) <Z(G),s> 0 Z2(G).

Indeed, since IZ2(G)I = p2, the equality <Z(G), s> = Z2(G) would imply that CG(s) _CG(Z2(G)) is a normal subgroup of index p in G; hence IGI = p3, a contradiction.Furthermore, we write Z = Z(G) and claim that

(15) ICGIz(sZ)I = p2.

Indeed, by (14), <Z2(G),s>/Z is a subgroup of order p2 of CG,z(sZ). So if (15) werefalse, there would exist H/Z < CG/z(sZ) containing <Z2(G), s>/Z as a maximal sub-group. Since <Z2(G), s>/Z is central in CGJz(sZ), it follows that H/Z is abelian andhence c(H) = 2. On the other hand, IHI = p4 and IC1,(s)I = p2; by (13), c(H) = 3.This contradiction proves (15).

In general, projectivities do not map p-groups of maximal class to p-groups ofmaximal class. For example, if p > 2, the nonabelian group of order p3 and exponentp2 has maximal class and is lattice-isomorphic to the abelian group of type (p2, p).However, the following two theorems show that the majority of groups in this classare invariant under projectivities.

7.4.11 Theorem (Caranti [1979]). Let G be a p-group of maximal class, IGI = p",n > 4 and lets E G such that ICG(s)I = p2 and o(s) = p. If cp is a projectivity from G toa group G, then 6 is a p-group of maximal class and Z;(G)I* = Z1(G) for all i.

Proof. By 2.2.6, G is a p-group. Write Z = Z(G) and S = <s>. Since I <Z, S> I = p2,the group <Z, S> is abelian. Suppose, for a contradiction, that ICZ;(S0)I > p3. Thenthere exists K< G such that <Z,S>0 < K4' < CU(S4) and IKI = p3. Since Z is theunique minimal normal subgroup of G, we have Z" a G, by 5.4.2. It follows that<Z4, S4') is central in K4' and hence K4° is abelian. But I CG(S)I = p2 implies that K isnot abelian and so K4' is of type (p2, p). Since I KZ2(G) : K1 < p, there exists H < Gsuch that KZ2(G) < H and IHI = p4. By (13), H is of maximal class. Furthermore,by assumption, ISI = p so that <Z, S> = S2(K) g H. Since H has only one normalsubgroup of order p2, it follows that <Z,S> = Z2(G), contradicting (14). ThusIC,(S,*)I = p2 and, by (13), 6 is of maximal class. For i = 1, ..., n - 2, Z;(G)/Zi_1(G)is the unique minimal normal subgroup of G/Zi_1(G) so that by 5.4.2 and a trivialinduction, Z;(G)4 < G. Since Z.(G) is the unique normal subgroup of order p` of G, itfollows that Z.(G)4 = Z-(G).

7.4.12 Theorem (Caranti [1979]). Let G be a p-group of maximal class, IGI = p",n > 4, and let cp be a projectivity from G to a group G. If G is not a p-group of maximalclass, then

(a) c(G) = c(G) - 1 and Z,(6) = Z;+1(G)" for all i >- 1,(b) Exp S2(G) = p and I G : S2(G)I = p,

(c) IGI < pp+1

Proof. Again by 2.2.6 and 5.4.2, IGI = p" and Z4 a G where Z = Z(G). _(a) First consider the case that IGI = p4. Suppose, for a contradiction, that Z;/Z`0

is abelian. Then G/Z" is of type (p2, p) and therefore contains two cyclic subgroupsAr/Z4' of order p2. Then A;/Z is cyclic, hence the A, are abelian and Al n A2 < Z(G).But this implies that IZ(G)I > p2, a contradiction. Thus G/Z4' is not abelian. Since Gis not of maximal class, it follows that c(G) = 2 and Z4' < Z(G). On the other hand,IZ(G/Z4)I = p and hence Z(G) = QD(G) = Z2(G)4. Thus (a) holds in this case. Nowsuppose that IGI >- p5 and let s e G such that ICG(s)I = p2. By 7.4.11, o(s) = p2and hence Z < <s>. So by (15), G/Z satisfies the assumptions of Theorem 7.4.11,and therefore c(G/Z4') = n - 2 and Zi+1(G)4/Z4' = Zi(G/Z4') for all i > 1. Sincec(G) < n - 1, we obtain Z4' < Z(G) and Z(G/Z4') = Z(G)/Z4'. Thus Z1(G/Z4')Zj(G)/Z4' for all i > 1 and (a) follows.

(b), (c) We prove both assertions simultaneously and consider a group G of mini-mal order for which (b) or (c) is false. Lets e G such that ICG(s)I = p2 and let Mbe amaximal subgroup of G containing s. By (13), M is of maximal class. By (a), Z(G) =Z2(G)4 < Z(M4) since Z2(G) < cb(G) < M; thus M" is not of maximal class. IfIMI >- p4, the minimality of G implies that (b) and (c) hold for M, that is, IMI < pP+1and S2(M) is a maximal subgroup of exponent p of M. Since S2(M) a G andI G : sl(M)I = p2, we have S2(M) = G' and hence

(16) Exp G' = p and IGI < pP+2,

If IMI < p4, then clearly IGI = p4 < pp+2. Let t e G\Z be such that o(t) = p andconsider H = CG(t). By 7.4.11, IHI = p3 and since <Z,t> < f (Z(H)), we concludethat H is abelian and I G: 12(H)I < p2. It follows that G' < S2(H) and hence (16) alsoholds in this case.

If IGI = pp", there exists a maximal subgroup G, of G such that i2(G,) = K3(G)(see Huppert [1967], p. 370 for p > 2; if p = 2, we can take the cyclic maximalsubgroup of G). But by (16), K2(G) = G' < fl(G,), a_contradiction. Thus IGI < pP+l.

Therefore by (a), c(G) = c(G) - 1 < p - 1 and so G is regular (see Huppert [1967],p. 322). Then Exp i2(G) = p and hence also Exp K2(G) = p. Since G is a counter-example to (b) or (c), it follows that IG : S2(G)i : p. By (16), G' < Q(G), and S2(G) = G'since o(s) = p2. By a well-known property of regular p-groups (see Huppert [1967],p. 327), IZ5(G)I = IG : Q(G)I and hence IZ5(G)I = IG : S2(G)I = p2. Thus there existsx e G such that x° 0 Z; let N = <x, G'>. Then S2(N) = G' and N is regularsince I NI < pP. Therefore IZ5(N)I = IN: S2(N)J = p and so <xP> = J(N) g G. Thiscontradicts (12).

Caranti [1979] proved that IGI < pP in 7.4.12(c) and he constructed all the p-groups of maximal class which are lattice-isomorphic to groups of smaller class.There are two series of such groups, one of which we present in Exercise 7. Wemention that the 2-groups of maximal class are the dihedral, generalized quaternion,and semidihedral groups (see Huppert [1967], p. 339), and all these groups are easilyseen to be determined by their subgroup lattices since they contain cyclic maximalsubgroups and differ in the number of minimal subgroups. So we could have as-sumed p > 2 in our investigations. Finally, for p > 2, two lattice-isomorphic butnonisomorphic groups of order p° and class 3 are the groups 9) and 10) in Exercise 6.

Exercises

1. If H is permodular in the group G, and C is a cyclic element in [G/H], show thatthere exists c c G such that C = <H, c>.

2. Let G be a nilpotent p-group such that Exp G = a,. Show that every projectivityof G is normalizer preserving.

3. (Suzuki [1951a]) Let G be a finite group such that L(G) is lower semimodular andall intervals in L(G) are directly indecomposable. If (D (G) = 1, show that G is aP-group. (Hint: By 5.3.11, G is supersoluble.)

4. (Suzuki [1951a]) Show that a finite group G has a lower semimodular subgrouplattice all of whose intervals are directly indecomposable if and only if G is eithera p-group or satisfies (b) of 7.4.9.

5. Let us call a finite p-group G an fl-group if for every section A of G and all n e N,is c AlaP" = 1}. (Thus every regular p-group is an fl-group.) Show that

if G is an Q-group and cp is a projectivity from G to a p-group G, then G is anfl-group.

6. It is well-known (see Huppert [1967], p. 346) that for p > 3 there are exactly 15pairwise nonisomorphic groups of order p°. Show that among these (in Huppert'snotation) just the following are lattice-isomorphic: 2) and 6), 3) and 7), 4) and 14),9) and 10), 13) and 15). Note that the groups of maximal class are 9), 10), 12),and 13).

7.5 Soluble groups 395

7. (Caranti [1979]) Let 3 < n < p and G = <s, s, , s2, ... , with defining rela-tions s° = s"_,, sp = I (1 < i < n - 1), [si,s] = s;+, (1 < i < n - 2) and [s;, s;] = I(1 < i, j < n - 1). Let G be the group obtained, replacing in these relations[sn_2,s] = s"_, by [sn_2,s] = 1. Show that IGI = IGI = p", c(G) = n - 1, c(G) _n - 2, and that G and G are lattice-isomorphic.

7.5 Soluble groups

We have seen that the class of soluble groups is invariant under projectivities. How-ever, there are only a few general theorems saying that the members of a certain classof soluble groups are determined (or even strongly determined) by their subgrouplattices. On the contrary, projectivities between nonisomorphic groups of this typeoccur in abundance. Examples can be found in 4.1.7, 4.2.12, 5.6.8, 5.6.9, and in theliterature cited in § 5.6. In addition, Yakovlev [1975] constructed a projectivitybetween two nonisomorphic nonabelian torsion-free soluble groups, and thusshowed that Sadovskii's theorem on torsion-free nilpotent groups does not general-ize to soluble groups. More accessible is the question as to which classes of solubleand generalized soluble groups are invariant under projectivities. We study thisproblem in two rather different situations. First and foremost we investigate forma-tions of finite soluble groups; and then at the end of this section we show that animportant class of generalized soluble groups, the class of radical groups, is invariantunder projectivities.

Finite soluble groups

We already know classes of finite soluble groups which are invariant under projecti-vities, and in some cases we have also given lattice-theoretic characterizations ofthese classes. For the convenience of the reader we give references in the followingtable. All groups considered are finite.

Class of Groups Reference Characterization

Soluble groups 4.3.4 5.3.5Groups of Fitting length k, k > 3 4.3.3 -Groups of p-length 1, 1 >- 2 4.3.8Supersoluble groups 5.3.8 5.3.7 and 5.3.9Supersoluble groups inducing automorphism 5.3.11 Groups with lower

groups of at most prime order in every semimodular subgroupchief factor lattice

The most important invariants of a finite soluble group G are the derived length,the Fitting length, and the rank of G. We have seen that derived length is notpreserved by projectivities, and we now show that the rank of G is preserved. Let us

recall that if 1 = Go < G1 < . < G = G is a chief series of G with I Gi : Gi_1 I = p;',pi E P and ri e N, i = I__ , n, then the rank r(G) of G is the unordered n-tuple(r...... of the dimensions of the chief factors of G. By the Jordan-Holder Theo-rem, r(G) is independent of the choice of the chief series. We write a(r) for the classof soluble groups of rank r = (r1, ... , and define Gk to be the class of all solublegroups whose chief factors have dimension at most k (k c- N).

7.5.1 Theorem (Schmidt [1972b]). If p is a projectivity from a finite soluble group Gto a group G, then r(G) = r(G). Thus all the classes Ca(r) and Sk are invariant underprojectivities.

Proof. We proceed by induction on IGI. Let N be a minimal normal subgroup of G.If NO is a minimal normal subgroup of G, then N and NO have the same dimension.By induction, r(61N0) = r(G/N) and hence r(G) = r(G). Now suppose that NO is nota minimal normal subgroup of G. Then by 5.4.4, N is cyclic and there exists a normalsubgroup S of G such that N < S, SO < G, and S and SO are hypercyclically em-bedded in G and G, respectively. Thus if m is the length of L(S), every chief series ofG through S has m of its factors below S and all of these have dimension 1. The sameholds for SO in G. By induction, r(G/S0) = r(G/S) and it follows that r(G) = r(G).

The above result generalizes Iwasawa's theorem that the class a1 of supersolublegroups is invariant under projectivities. It is not difficult also to extend the lattice-theoretic characterization 5.3.7 of G1 to the classes 3k (see Exercise 1).

Another interesting class of finite soluble groups is the class 1: of Sylow towergroups. Recall that G e 2 if and only if there exists a chain I = Go < - < Gr = G ofnormal subgroups G; of G such that Gi+1/Gi is a Sylow subgroup of G/Gi for i =0,...,r- 1.

7.5.2 Theorem (Schmidt [1972b]). The class Z of Sylow tower groups is invariantunder projectivities.

Proof. Let G e 2 and p be a projectivity from G to a group G. We use induction onI_GI to show that G E 1. If G = S x T where S is a P-group and (ISI, I TI) = 1, thenG = SO x TO, (ISO1,1 TO1) = 1, and SO is a P-group. Thus SO e 2 and, by induction,TO E 2; since (IS" I, I TOE) = 1, it follows that G e Z. Now suppose that G is notP-decomposable and let P I be a normal Sylow p-subgroup of G. By 4.3.2, PO _Oq(G) for some prime q. If PO e Sylq (G), then G e 2, by induction. And if PO is not aSylow q-subgroup of G, then cp-1 is singular at q and by 4.2.6 there exists a normalq-complement NO in G. By induction, NO E 2 and then also G e Z.

The types of the Sylow towers in G and G can of course differ. Indeed, the numberof primes dividing IGI can be twice the number of primes dividing IGI; for example,if G and G are direct products of coprime P-groups. But also if G is not P-decomposable, the number of primes involved in IGI and IGI can be different: Theo-rem 4.1.6 shows that the nonabelian group of order 63 and exponent 21 and the


dihedral group of order 42 are lattice-isomorphic. By Example 4.1.7(b) there arelattice-isomorphic groups of order 113.7 and 113. 19 with minimal normal sub-groups of order 113. This shows that the class of groups with a Sylow tower withrespect to the natural ordering of the primes is not invariant under projectivities.This contradicts Theorem 6.2 in Schmidt [1987]; a corrected version of this "theo-rem" is given in Exercise 2.

To obtain deeper results, we need some simple facts from the theory of formations.

Formations of finite soluble groups

A formation is a class R of soluble groups with the following properties:

(1) If NaGea,then GIN ea.

(2) If N1, N2 a G such that GIN; e a (i = 1, 2), then G/(N, n N2) e

A formation a is called saturated if G/D(G) E a implies G e a.It follows from (2) that if a is a formation, then in every finite soluble group G

there exists a unique smallest normal subgroup whose factor group lies in a; this iscalled the a-residual G. It is the intersection of all the normal subgroups N of Gsuch that GIN e a. Furthermore it is well-known that if R is saturated, then in everyfinite soluble group G there exists a canonical conjugacy class of subgroups S ofG, called a-projectors, defined by the condition that SN/N is a-maximal in GIN forevery normal subgroup N of G (see Robinson [1982], p. 272). We describe a con-struction for formations due to W. Gaschutz and give some examples.

For every prime p, let R(p) be a class of groups. Define a to be the class of allfinite soluble groups G such that G/CG(H/K) e R(p) for every chief factor H/K oforder divisible by p of G. It is clear that a satisfies (1). And since the chieffactors of G/(N, n N2) are G-isomorphic to those of GIN, or GIN2, it follows thatG/(N, n N2) e #1 if GIN, and GIN2 are in a. Thus a is a formation, called the forma-tion defined locally by the classes R(p). If all the 1(p) are formations, then a issaturated (see Robinson [1982], p. 269).

7.5.3 Examples. All groups considered are finite and soluble.(a) The class 2I of abelian groups clearly is a formation. Since G/D(G) e 2I for

every p-group G, the class 2I is not saturated.(b) If j(p) = { 1 } for all primes p, the formation defined locally by the R(p) is the

class of groups with central chief factors, that is, the class 91 of nilpotent groups.(c) Let I be a formation, R(p) = I for all primes p and let a be the formation

defined locally by the R(p). If G e a, then G/F(G) a I since F(G), the Fitting sub-group of G, is the intersection of the centralizers of the chief factors of G (see Rob-inson [1982], p. 129). Thus G e 911, the class of all groups with nilpotent I-residual.Conversely, if G e 913x, then G/F(G) a I and hence G e R. Thus R = 911. Of particu-lar interest in the theory of finite soluble groups are the class 9121 of nilpotent-by-abelian groups and the formations 915, s e 1, of groups of Fitting length atmost s.

(d) It is easily seen that the classes C3,,, k e N, are formations. Of these, only theformation G, of supersoluble groups is saturated: it is defined locally by the forma-tions 91P of abelian groups of exponent dividing p - 1 (see Huppert [1967], pp.712-715).

(e) For every prime p, let 3P be the class of groups consisting of the trivial groupand all the cyclic groups of prime order q dividing p - 1. Since 3P c 21p, theformation defined locally by these classes 3P consists of those supersoluble groupsinducing automorphism groups of at most prime order in their chief factors. By5.3.11, this is the class 971L of groups with lower semimodular subgroup lattice.

We come to our main result.

7.5.4 Theorem (Schmidt [1974], [1987]). For every prime p, let R(p) be a class offinite soluble groups such that

(3) 3,, c ti(p) and

(4) if X, Y are lattice-isomorphic irreducible subgroups of GL(n, p), n e N, andX E tar(p)\3p, then Y E R(p).

Then the formation I defined locally by the classes ti(p) is invariant underprojectivities.

Proof. Let G E fj and cp be a projectivity from G to a group G. We use induction onI G I to show that G e Suppose first that G is P-decomposable, that is, G = S x Twhere S is a P-group and (ISI, I TI) = 1. Then by 1.6.6, G = S4' x T"; furthermoreT G/S E It and hence by induction, 6/S" T" E . Therefore if H/K is a chieffactor of G of order a power of p such that K > S", then G/CG(H/K) e R(p). Also ifH 5 Se', then G/CG(H/K) e 3p c t` (p) since S' is a P-group. Thus G e R and we mayassume that G is not P-decomposable.

Now let N_ be a minimal normal subgroup of G. If N° is not a minimal normalsubgroup of G, then by 5.4.4 there exists a minimal subgroup M of Z(G) such thatMl' < Z_(G); let I M`0I = q. By induction, G/M° E a; also, G/CG-(M°) = I E j(q) andhence G E I. So suppose that N° is a minimal normal subgroup of G and letIN I = p". Then by induction, 6/N4' E c(J. If n > 2 or n = I and tp and p' areregular at p, then by 5.4.5, CG(N)4' = CU(N4) and q induces a projectivity fromX = G/CG(N) to Y = Since INS = p" = IN4j, both groups are irreduciblesubgroups of GL(n, p) and X e R(p) since G e 11. If X 0 3p, then by (4), Y =G/CC(N4') e c(p) and hence G e t`5. And if X E 3p, then IXI divides p - 1 and theirreducibility of X implies n = I (see Huppert [1967], p. 165). Since Y < GL(l, p)C,_, is lattice-isomorphic to X, the subgroup Y is cyclic of prime order dividingp - 1 (or Y = 1) and hence Y E 3p S R(p), by (3). Again it follows that G E I.

It remains to consider the case that N4 < G, INI = p and cp or (p-' is singular atp; let IN'I = q. If p-' is singular at q, then by 4.2.6 there exists a normal q-comple-ment with abelian factor group in G, since G is P-indecomposable. Then N4' is acentral chief factor so that 6/CU(NP) = 1 e (q) and G e . So suppose that tp-' isregular at q. Then p = q, cp is singular at p and by 4.2.6 there exists a normal

p-complement K in G such that K° g G_ and G/K'' is cyclic or a P-group. It followsthat G/CG-(N`') E 3P c J(p) and again G e It, as desired.

Theorem 7.5.4 and Example 7.5.3(c) show that if I is a formation of finite solublegroups satisfying (3) and (4) in place of I(p), then I = 91X is invariant underprojectivities. Now, obviously, 913` satisfies (3) and (4) and a trivial induction yieldsthat 9251 is invariant under projectivities for all s e N. We give examples of forma-tions X with this property. The formation 21 of abelian groups clearly satisfies (3). BySchur's Lemma, every irreducible subgroup of GL(n, p) has cyclic centre. And sincethe class of cyclic groups is invariant under projectivities, it follows that 21 satisfies(4). Thus the class 9121 is invariant under projectivities although 91 and 2I are not. By1.6.6 and 2.2.6, every projective image of a nilpotent group with cyclic centre isnilpotent and therefore 91 satisfies (4). Thus we get that for t >_ 2, the formations 91`are invariant under projectivities, a result already known from 4.3.3. Finally, by7.5.1, the formations Sk satisfy (3) and (4). So we have the following result.

7.5.5 Corollary. If X is a formation of finite soluble groups satisfying (3) and (4) inplace of I(p), then 9253 is invariant under projectiuities for every s e N. In particular,the saturated formations 9152t, 915Gk and 91' (s, k > 1; t > 2) are invariant underprojectivities.

Of course, the classes I(p) = 3P also satisfy (3) and (4) and are the smallest classeswith this property. In fact, the following holds and will be used later.

7.5.6 Remark. Let 10 { 1 } be a formation defined locally by classes a(p). If I isinvariant under projectivities, then 3P 9 I(p) for every prime p. Thus the formationJD1L of groups with lower semimodular subgroup lattice is the smallest nontrivialformation which is defined locally and is invariant under projectivities.

Proof. Let 1 G E a and p e P. Then there exists N a G such that GIN is cyclic ofprime order. Since GIN E a and a is invariant under projectivities, it follows thatCP E I; thus 1 E I(p). Furthermore, CP x CP E I and hence for every prime qdividing p - I, the nonabelian group of order pq lies in I. Therefore Cq E R(p) andhence 3P c I(p). Now 7.5.3(e) shows that 971L c I.

Residuals and projectors

We want to show now that every projectivity between finite soluble groups mapsa-residual to I(-residual and a-projectors to a-projectors if I is a formation whichis invariant under projectivities. In the case of the a-residual, we need an additionalassumption which will now be explained. For short, let us say that (G, H, K) is aP-hypercentral triple if H, K are normal subgroups of G such that K < H and

G/K = S, /K x . x Sr/K x T/K

where 0<relandforalli,je{1,...,r},(5) S;/K is a P-group,

(6) (ISf/K I,IS;/KI) = 1 = (IS1/KI,IT/KI) for i :j,

(7) H/K = S1/K x x Sr/K x (H n T)/K, and

(8) (H n T)/K is hypercentrally embedded in G.

A formation I is called P-hypercentrally closed if for every P-hypercentral triple(G, H, K), G/H e It implies that G/K e I. The formations studied above have thisproperty.

7.5.7 Lemma. If R is a formation defined locally by classes j(p) such that 3P c_ R(p)for every prime p, then tR is P-hypercentrally closed.

Proof. Let (G, H, K) be a P-hypercentral triple and suppose that G/H e R. We haveto show that G/K e R and may assume without loss of generality that K = 1. ThenG is soluble; let X/Y be a chief factor of G such that I X/YI is a power of the prime p.Then X/Y is G-isomorphic either to a chief factor of G/H, and so G/CG(X/Y) e R(p)since G/H e (1, or to a chief factor contained in H. By (7) and (8), in the latter case,X/Y is central or contained in some S.; in both cases, G/CG(X/Y) e 3P. Thus in anycase, G/CG(X/Y) E R(p) and G e R.

7.5.8 Theorem (Schmidt [1987]). Let R be a formation which is invariant underprojectivities and P-hypercentrally closed. If G is a finite soluble group and qP aprojectivity from G to a group G, then (G)'' = G.

Proof. Let Gj = N, H' = (Nc)o and K° = (N°)d. By 5.4.7, (G, H, K) is a P-hypercentral triple. Since G/N e t( and I is a formation, G/H e a and, since is

P-hypercentrally closed, G/K e R. But N is the smallest normal subgroup of G withfactor group in I. It follows that N = K and thus N° = K° a G. Since R is invariantunder projectivities, GIN" e R and therefore (G;j)' = N° > G. If we apply this re-sult to the soluble group G and q in place of G and cp, we obtain (G )" > Gj.Thus (G,)" = G,1, as desired.

7.5.9 Corollary. If (I is a formation defined locally by classes j(p) satisfying (3) and(4), then (G;,)° = G,1 for every projectivity cp from a soluble group G to a group G.

Proof. By 7.5.4 and 7.5.7, R is invariant under projectivities and P-hypercentrallyclosed. Now 7.5.8 yields the assertion.

Note that the above corollary improves Theorem 7.5.4 and applies to all theformations in Corollary 7.5.5. For a-projectors, the situation is even better. Theyexist in every soluble group if and only if It is saturated; and in this case, they behavewell.

7.5.10 Theorem (Schmidt [1987]). Let a be a saturated formation which is invariantunder projectivities. If S is an a-projector of the soluble group G and (p a projectivityfrom G to a group G, then S" is an a-projector of G.

Proof. If it = { 11, there is nothing to be shown; so suppose that a A { 11. It iswell-known (see Huppert [1967], p. 710) that a is defined locally by formations R(p).By 7.5.6, 3 -- R(p) for all primes p and, by 7.5.7, a is P-hypercentrally closed. Nowlet NO < G. We have to show that S"N"/N" is a-maximal in 6/N". Since S e a andii is invariant under projectivities, S" e I and hence SON"/N" S`°/S`0 n NO e a.Let M< G such that S"N" < Mw and M"/N" is a-maximal in 6/N". If we putH = Nc and K = No, then by 5.4.7, (G, H", K") is a P-hypercentral triple. SinceM"/N" E R and NO < M" n H" < Mw, we have M"H"/H" -_ M"/M" n H" e a.Also (M"H", H", K") is a P-hypercentral triple and a is P-hypercentrally closed;it follows that M"H"/K" e R. Thus MH/K E a. The choice of M implies thatS"K" <_ S"N" < M" and hence SK < MH. Since S is an a-projector, SK/K isa-maximal in G/K. It follows that SK = MH and then SON" > S"K" > M".Thus SON" = M" and SON"/N" is a-maximal in G;N". This shows that S" is ana-projector of G.

Lattice-theoretic characterizations of the formations studied in 7.5.5 are notknown; in fact, this is an interesting problem even for 9191 or 912. On the other handsuch characterizations are known for 91L and S1, and, more generally, S, (seeExercise 1). Then it is possible, under the usual assumptions on a, to give lattice-theoretic characterizations of the a-residual and of s-projectors (see Exercise 5).

A different approach is used by Hauck and Kurzweil [1990] to characterize theprefrattini subgroups discovered by Gaschiitz in 1962. They show that for a sub-group H of a finite soluble group G, all the subgroups K E [G/H] which are minimalwith respect to the property that the interval [G/K] is complemented are conjugatein G; furthermore they identify these groups as the H-prefrattini subgroups intro-duced by Kurzweil [1989]. For H = 1, one obtains the prefrattini subgroups, whichare therefore characterized in the subgroup lattice of G.

Further results on intervals in finite soluble groups are contained in Kurzweil[1985] and Heineken [1987].

Infinite soluble and generalized soluble groups

Some of the results of Chapters 4 and 5 have been generalized to infinite groups. By6.6.4, the class of soluble groups is invariant under projectivities, however a lattice-theoretic characterization of this class is not known. In § 6.4 we gave lattice-theoreticcharacterizations of the classes of hypoabelian, hyperabelian, finitely generatedsoluble, polycyclic, and supersoluble groups, and mentioned that the classes ofhypercyclic and of SN-groups can also be characterized.

Some rather special classes of infinite soluble groups were characterized byPlotkin [1957] and Pekelis [1961a]. Intervals in hyperabelian groups were studiedby Plaumann, Strambach and Zacher [1985], and Stonehewer and Zacher [1988].

We finish this section by showing that another important class of generalizedsoluble groups, the class of radical groups, is invariant under projectivities. Actually,we prove a somewhat stronger result on the iterated Hirsch-Plotkin radicals whichis a direct generalization of Theorem 4.3.3 on the iterated Fitting subgroups of finitegroups.

Radical groups

Recall that the Hirsch-Plotkin radical R(G) of a group G is the unique maximalnormal locally nilpotent subgroup of G. It is well-known (see Robinson [1982],p. 344) that

(9) R(G) contains all the ascendant locally nilpotent subgroups of G.

Furthermore, since R(G) is locally nilpotent, the elements of finite order in R(G) forma characteristic subgroup T = T(R(G)), the torsion subgroup of R(G). Clearly,

(10) R(G)/T is torsion-free and T is a direct product of p-groups.

If G is finite, R(G) = F(G) and so the following result generalizes Lemma 4.3.2(b).

7.5.11 Lemma. Let tp be a projectivity from G to G such that R(G)I° : R(G). ThenG is periodic and P-decomposable, that is, G = H x K where H E P(n, p) for somen c N u { x }, p c P, and H and K are coprime; furthermore, cp or (p-1 is singular at p.

Proof. We assume that the conclusions of the lemma do not hold and show thatthen

(11) R(G)'P < R(G);

by symmetry, it will follow that R(G)° = R(G), the desired contradiction.Let R = R(G) and suppose first that R is not periodic. Then by 7.3.3, R' is locally

nilpotent. Let T = T(R) be the torsion subgroup of R and let W = (x E GI o(x) = oc>.By (10), R/T is torsion-free. Since R is generated by the elements outside T, wehave R < W. By 6.5.1, T4 g W4' and cp induces a projectivity 0 from WIT to Wco/Tw.If R/T is abelian, then by 6.6.5, (R/T)O is permutable in (WIT)O. And if RIT isnonabelian, then by 7.2.11, ip is induced by an isomorphism on R/T and 6.6.6(c)shows that (R/T)" is permutable in (W/T)'°. In any case, it follows from 6.2.10 thatR' is ascendant in W°. Since W° = (x E GIo(x) = cc> a G, the subgroup R4' isascendant in G. By (9), R4' < R(G), as desired.

Now let R be periodic and suppose, for a contradiction, that (11) does not hold.By (10), R is a direct product of locally nilpotent p-groups and, by 1.6.6, R" is thedirect product of the images of these groups. Since R' R(G), there exists a p-component S of R such that S° 4 R(G). By (9), either SW is not locally nilpotent andhence by 2.2.7,

(12) S is an elementary abelian p-group and SW is a nonabelian P-group;

or SIO is not ascendant in G and hence, by 6.2.10, in particular,

(13) Sc is not permutable in G.

We show first that G is a torsion group. So suppose, for a contradiction, that thereexists an element x of infinite order in G. Then by 6.6.6 (or 6.2.12), Sc" is permutablein G and hence (12) is satisfied. Let Pc be the Sylow p-subgroup of Sc and L = S(x>.By 6.5.1, S1* a Lc and hence P10 g L' since P`° is characteristic in Sc"; again by 6.5.1,P :!g L. Thus cp induces a projectivity from L/P to Lc"/P`° and 6.5.11 applied to thisprojectivity yields that IS/PI = BSc"/P101. But IS/PI = p IS"/PI*I, a contradiction.Thus

(14) G is periodic.

We want to show next that (12) is satisfied. So assume that (13) holds and letH = SG, N = Sz;. Then by 6.6.6 applied to the normal p-subgroup S of G, GIN =H/N x K/N where H/N and K/N are coprime, H/N e P(n, p) for some n e N u { oc If,I S/N I = p and J Sc/N"I = q : p. Since we assume that the conclusions of the lemmado not hold, N # 1. By 2.2.7, S is either locally cyclic or elementary abelian. In thelatter case, (12) holds, as desired. So suppose that S is locally cyclic. Then S is finitesince it contains the maximal subgroup N. It follows that the Sylow p-subgroup ofH is a nilpotent normal subgroup of G; hence it is contained in R(G) and thereforeis S. Thus H is finite, cp is singular at p on H and, since S is cyclic of order at least p2,4.2.6 implies that there exists a normal p-complement in H. But S < H and henceH/N is a nonabelian group in P(n, p), a contradiction.

It remains to consider the case that (12) is satisfied. Then there exists M < S suchthat I M I = p and IM"I # p. We claim that

(15) M g G.

Since Mc is not permutable in the P-group S`°, and hence also not in G, it will followfrom 6.6.6 that G satisfies the conclusions of the lemma. This will contradict thechoice of G.

To prove (15), suppose, for a contradiction, that M is not normal in G and takex e G such that M" 0 M and o(x) = s is a prime power. Then

W =

is a finite group since all the Mx' are minimal subgroups of the abelian group S andtheir join is invariant under <x>. Clearly, cp induces a p-singular projectivity in W.Since M is a nonnormal and S n W > M is a normal p-subgroup of W, it is clear that(a) of 4.2.6 cannot be satisfied by this projectivity. Thus there exists a normal p-complement C in W with abelian factor group WIC. If x is a p-element, (S n W) <x>is a p-subgroup of W and hence is abelian; and if x is a p'-element, then x e C and[M, x] < S n C = 1. In both cases, M" = M, a contradiction. This proves (15) andthe lemma.

Recall that the upper Hirsch-Plotkin series of a group G is the ascending series1 = R0(G) < R, (G) < - - - in which Rz+1(G)/R,,(G) is the Hirsch-Plotkin radical of

G/R,(G) for every ordinal x. This series becomes stationary at a certain ordinal '/ andthe term R,(G) is then called the Hirsch-Plotkin hyperradical R(G) of G. Finally, G iscalled radical if R(G) = G. We come to our main result.

7.5.12 Theorem (Zacher [1982a]). If cp is a projectivity from the group G to a groupG, then for all n e N such that n > 2.

Proof. Suppose that the theorem is false. Then there exists a smallest integer n > 2for which there is a projectivity (p between two groups G and G such that R (G)'P 0-

If G is periodic and {S1,S2,...} is the set of P-subgroups which are coprimedirect factors of G, then G = (S1 X S2 x ) x K where_K is P-indecomposable.Since n > 2, (S1 X S2 x ) x R (K). By 1.6.6, G = (ST x SZ x ...) x KWand then Rn(G) = (S; x SZ x ) x R (K)" # R (K'P). Thuswe may assume without loss of generality that G = K is P-indecomposable if G isperiodic. Then 7.5.11 yields that

(16) R(G)" = R(G); write R(G) = R.

Let ip be the projectivity induced by cp in G/R. Since

R.-1(G/R)`D = (R,i(G)/R)`° = R,i(G)/Rw = R.-1(G/R`'),

the minimality of n implies that n = 2. And by 7.5.11 applied to (p, G/R is periodicand G/R = H/R x K/R where H/R and K/R are coprime, H/R E P(m, p) for somem e N u {oo}, p e P, and ip or 0 -1 is singular at p. Since the situation is symmetricin G and d, we may assume that ip is singular at p. Then if H/R is elementary abelian,there exist x, y c- H such that

(17) o(xR) = p = o(yR) and I<xR> I = p # q = I<yR>`°I;

and if H/R is not abelian, there exist x, y, z e H such that

(18) o(xR) = p : o(yR) = o(zR) and I<yR>0I = p q = I<xR> I = I<zR>#I

Now suppose first that R is periodic. Then in both cases we may choose x as ap-element and want to show that <x> g H. For this let w e H and consider W =<x, y, w>. Every finitely generated subgroup of R is a finitely generated nilpotenttorsion group and therefore is finite; thus R is locally finite and hence so is H (seeRobinson [1982], pp. 133 and 411). Thus W is finite and qp induces a p-singularprojectivity in W. By 4.2.6, either every p-subgroup of W is normal in W or W has anormal p-complement. In the first case, <x)' = <x>, as desired. In the second case,WR/R W/W r R also has a normal p-complement and hence is abelian. Therefore(17) holds and the Sylow p-subgroups of W are not cyclic. Then I<x>I = p = I<x)`°Iand, by 4.2.10, x e Z(W). So again <x> v = <x> and, since w e H was arbitrary, itfollows that <x> g H. Thus <x> is a subnormal nilpotent subgroup of G and, by (9),<x) < R. This contradicts (17) or (18).

It remains to consider the case that R is not periodic; let T = T(R). Then cpinduces a projectivity from G/T to 6/T" and since our final argument only concernsthis projectivity, we may assume that T = 1. Then R is torsion-free. Let u e H\R.

Since H/ R E P(m, p), we obtain u' E R for some prime r. If o(u) is finite, it followsthat o(u) = r. Take 1 : w e R and consider W = <w, w',..., w"'-` > (u>. Since R islocally nilpotent, <w, w",..., w"'-'> = W n R is a nontrivial nilpotent normal sub-group of W. Hence W has a nontrivial torsion-free abelian normal subgroup and by6.5.12,

(19) I`'I = II.

If o(u) is infinite and tp is induced by an isomorphism on R, then tp is indexpreserving on <u'> < R and hence, by 6.5.10, on also. But I <yR>0I : I <yR> I in(17) and (18). Therefore it follows from (19) that o(y) is infinite and that tp is notinduced by an isomorphism on R. By 7.2.11 and 2.6.10, R is abelian of rank 1. Nowif (17) holds, then by 6.5.12, there is no p-element in H and hence o(x) = o(y) = co.Furthermore, by 6.5.10, tp maps p-indices in <x> to p-indices in <x>`° and p-indicesin <y> to q-indices in <y>°. But this is impossible since <x> n (y> : 1. Similarly, if(18) holds and o(yR) = r = o(zR), then tp maps r-indices in <y> n (z> to p-indicesand q-indices in (<y> n <z>)`0, the same contradiction.

By an obvious transfinite induction, we get the result announced above.

7.5.13 Corollary (Pekelis [1972b], [1976]). If tp is a projectivity from the group Gto a group G, then R(G)4' = R(G). In particular, the class of radical groups is invariantunder projectivities.

Exercises

1. (Schmidt [1987]) For k e N, let us call a lattice L with least element 0 andgreatest element I poly-k-modular if there exist elements x, E L such that0 = xo < < X. = I, xi mod L and [xi+1/xi] is a modular lattice of length atmost k (i = 0, ... , r - 1). Show that a group G lies in Sk if and only if L(G) ispoly-k-modular.

2. For every ordering -< of the set P of primes, denote by I, the class of groupswith a Sylow tower with respect to -<; thus G e 2, if and only if there exists achain 1 = Go < < G, = G of normal subgroups Gi of G such that Gi/G1_1 is aSylow qi-subgroup of G/G1_1 for i = 1, ..., r and qr < q,_1 < - - -< q,. Show that2.< is not invariant under projectivities. [Hint: Suppose that 1 < is invariantunder projectivities and prove the following assertions.(a) IfgIp - 1, then q -< p.(b) If q -< p -< r, then oq(p) # or(p); here oq(p) is the order of p modulo q, that is,

the smallest natural number n such that p" = 1 (mod q).(c) There is no ordering of P satisfying (a) and (b). (Use a computer to produce a

table of all oq(p) for p, q < 250.)]3. (Schmidt [1973]) For p E P, let Up be the formation of abelian groups with

square-free exponent dividing p - 1 and let fl be the saturated formation de-fined locally by the Q..

(a) Show that UL c fl c a 1.(b) Show that C is the smallest nontrivial saturated formation which is invariant

under projectivities.4. Show that not every saturated formation containing UL is invariant under

projectivities.5. (Schmidt [1987]) Let I be a formation of finite soluble groups and 2 a class of

lattices such that for every group X, we have X e I if and only if L(X) e 2.Assume further that I is P-hypercentrally closed and that [I/x] e 2 for allx mod Le 2. Let G be a finite soluble group.(a) Show that G;, is the smallest modular subgroup M of G such that [G/M] e 2.(b) Show that a subgroup S of G is an a-projector of G if and only if for every

M mod G, we have that [S u M/M] e 2 and [T/M] 2 for all T:!:-, G suchthat SuM<T.

6. (Schmidt [1987]). Let k e F and G be a soluble group.(a) Show that the ak-residual of G is the smallest modular subgroup M of G with

[G/M] poly-k-modular.(b) Show that the subgroup S of G is an Gk-projector of G if and only if for every

M mod G, the interval [S u M/M] is poly-k-modular and [TIM] is not poly-k-modular for all T< G such that S u M < T

7.6 Direct products of groups

In 1951 Suzuki showed that the direct product of a finite simple group with itselfis determined by its subgroup lattice. We use his method to prove a generaliza-tion of this result and also study projectivities of direct products of groups moresystematically.

General properties

If G = Dr G., and cp is a projectivity from G to a group G, the first question to askAEA_

is whether G = Dr G. It is clear that this is false in general: every projectivity fromA A

a finite abelian to a nonabelian group is a counterexample. On the other hand, by1.6.6, the assertion is true in the important case where the GA are coprime. Otherpositive results follow from our general theorems on projective images of normalsubgroups. We prove an additional assertion of this type.

7.6.1 Lemma. Let G = H x K and suppose that X < K such that X n N(K) = 1where N(K) is the norm of K. If qp is a projectivity from G to a group G, then H° isnormalized by X`0.

Proof. Suppose, for a contradiction, that H° is not normalized by X° and takex e X" such that (H'*)" # H°. Then M = H4`P is a modular subgroup of G sat-

7.6 Direct products of groups 407

isfying M n K = H"'`°-' n K`°""-' = (H n K)c"c-' = 1. For every A < K, therefore,A = A u (M n K) = (A u M) n K a A u M since M mod G and K :!g G. Thus Mnormalizes A and it follows that (H u M) n K< N(K). Now H° u M° = H" u H""is contained in H° u X° and hence H u M < H x X. Since M H, we finallyobtain that 1 96 (H u M) n X < (H u M) n K < N(K), and this contradicts ourassumption on X.

In the following theorem we collect an obvious corollary of Lemma 7.6.1 andsome consequences of our results on projective images of normal subgroups.

7.6.2 Theorem. Let G = Dr GA with subgroups GA (A e A) and let cp be a projectivityA

from G to a group G. Then G = Dr GT if every GA has one of the following properties.AEn

(i) Z(GA) = 1.(ii) GA is perfect and finite.

(iii) GA is generated by elements of infinite order and order 2.

Proof. We have to show that Gx is normalized by G,`,° for every ).,,u e A; then all the

G° are normal subgroups of G and G = Dr GT. For this we clearly may assume thatAEA

i p and G = G. x G,,. Suppose first that 1. Then by 1.4.3, 1 and7.6.1 yields that G' is normalized by G,°. If G G/GA is a finite perfect group,then by 6.5.7 and (c) of 5.4.9, Gx < G. Finally, suppose that G, = <xlx e X> whereX G and o(x) E {2, oc } for every x e X. Let x e X and put H = <GA, x> = GA x <x>.If o(x) = oc, then GT < H° by 6.5.1. And if o(x) = 2 and Gw is not normal in Hg', thenby 6.5.3 there exists K < GA such that H/K and H`°/K° are contained in P(2, p) forsome prime p. But <x>K/K is a normal subgroup of order 2 in H/K. Thus p = 2 andH`°/K`° is elementary abelian of order 4. This contradiction shows that Gx < H".Since G,° = «x>"Ix E X>, it follows that GT is normalized by Gµ.

We show next that projectivities of direct products sometimes preserve indicesand normalizers.

7.6.3 Lemma. Let G = H x K be a finite group and p a prime dividing I HI and I KI.If G admits a p-singular projectivity, then G has elementary abelian Sylow p-subgroupsand a normal p-complement. Furthermore, in one of the two groups H or K, the Sylowp-subgroup is a direct factor.

Proof. Let rp be a p-singular projectivity from G to a group G, and let P e Sylo (G)such that I P" 10 I P I. Since p divides I H I and I K I, P is not cyclic and hence elemen-tary abelian, by 2.2.6.

Suppose, for a contradiction, that G does not possess a normal p-complement.Then by 4.2.6, G = S x T where (IS 1, I TI) = I and S is a P-group containing P asa proper normal subgroup. Let IS : PI = q and Q e Sylq(S). Then Q e Sylq(G) andIQ I = q. It follows that Q is contained in one of the two direct factors of G = H x K.So Q < H, say, and then P n K < Cp(Q) = 1 since S is a P-group. But P n K is aSylow p-subgroup of K, a contradiction. Thus G has a normal p-complement N.

Let Po be the Sylow p-subgroup of Pg'. Then Po is a maximal subgroup ofP and, by 4.2.10, PO < Z(G). If P n K < P0, then P n K < Z(K) and henceK = (P n K) x (N n K); and if P n K P0, then P = PO(P n K) < CG(H) and itfollows that H = (P n H) x (N n H).

It follows from 7.6.3, or simply from the fact that a nonabelian P-group is directlyindecomposable, that a projectivity cp of a direct product G = H x K of twoisomorphic finite groups H and K is index preserving if G° = H4 x K4'. Zacher[1981] proved that if G is infinite, it is at least true that (p induces index preservingprojectivities in the direct factors. We shall need not this but a similar result onnormalizers.

7.6.4 Lemma. Let G= H x K, H K and suppose that cp is a projectivity from G toa group G such that G = H4' x K4'. Then H4 K4' and the projectivity 0 induced bycp in H is normalizer preserving. In particular, (X')" = (X`°)' and Z(X)" = Z(X4) forevery subgroup X of H.

Proof. Clearly, (p maps diagonals with respect to H and K onto diagonals withrespect to 111' and K4'. By 1.6.2, H4 K4' and hence (p-1 satisfies the assumptions ofthe lemma. Therefore to prove that i is normalizer preserving, it suffices to showthat for every X < H, the inclusion NH(X)4' < NH.(X4') holds; application of thisresult to cp-' will yield the other inclusion.

So let X < H, N = NH(X), e: H -+ K be an isomorphism, M = N', and considerL = N x M < G. By 1.6.2, D = {xx'lx e N} is a diagonal in Land, since X :!g L, wesee that XD is a subgroup of L such that XD n N = X. Since G = H4' x K4', we haveL`° = N" x M4', and hence M4' centralizes X4'. Also, X4' = (XD)4' n N4 o (XD)4' sothat D4 normalizes X4'. Since L° = D4 u M4', it follows that X4 o L4'. In particular,NH(X)4' = N° < NHm(X4'), as desired. Thus 0 is normalizer preserving. By 5.6.4, 0 isalso centralizer preserving and, by 5.6.2 and 5.6.3, (X')4' = (X`°)' and Z(X)4' = Z(X4)for all X < H.

Note that if G is not the direct product of H4' and K", then (p in general neitherpreserves normalizers nor indices. This is shown by the obvious example of P-groups.

The direct product of two isomorphic groups

The basic tools in the study of projectivities of a direct product G = H x K of twoisomorphic groups H and K are the diagonals; these are the subgroups D of G suchthat D n H = 1 = D n K and DH = G = DK. By 1.6.2, these subgroups are in one-to-one correspondence with the isomorphisms from H to K, and hence also with theautomorphisms of H. It follows that a projectivity (p from G to a group G such thatG = H4' x K4' induces a bijective map i from Aut H to Aut H4'. We study this mapand want to show first that it induces an isomorphism i* from Aut H/Pot H ontoAut H4'/Pot H4'. This is particularly interesting if Z(H) = 1. For in this case, Pot H = 1

and Z(H`0) = I so that i = i* is an isomorphism from Aut H onto Aut Hl*. Hence(Inn H)' and Inn H° are normal subgroups of Aut H° isomorphic to H and H'°,respectively. We shall give an example to show that (Inn H)' Inn M, in general.However, our main result will be that (Inn H)' Inn H'°/(Inn H)' n Inn H'° is at leastabelian. This will imply that (Inn H)' = Inn H° and hence H H" if H has trivialcentre and is perfect.

So let G = H x K and E: H -+ K be a fixed isomorphism. By 1.6.2, for everyisomorphism 6 from H to K,

(1) D(6) = {xx'lx c H}

is a diagonal in G (with respect to H and K) and every diagonal has this form. Fora c Aut H, let 8 be the autoprojectivity of G induced by the automorphism i' of Gdefined by

(2) i1: xy-+ x°yfor xEH,yc- K.

Clearly, & is an autoprojectivity of G fixing H and every subgroup of K. In particu-lar, it has to map diagonals onto diagonals. Also

(3) D(TE)' = D(a-'TE)

since (xx")° = x°x" = x°(x°)°-"" for T E Aut H and x c H. We study images ofdiagonals under an autoprojectivity of G fixing H and every subgroup of K.

7.6.5 Lemma. Let be an autoprojectivity of G such that H'" = H and Y'" = Y forall Y K; let a, re AutH.

(a) If D' = D for some diagonal D, then X " = X for all X < D and all X < H.(b) If X'" = X for all X < H and D(ae) = D(TE), then aT-' e Pot H.(c) If D(E)S' = D(aE), then D(TE)'' = D(arxE) for some a c- Pot H.

Proof. (a) For every X < D, by Dedekind's law, X = XH n D = (H u (XH n K)) n D,and this subgroup is clearly invariant under 0. Similarly, if X < H, then X =XKnH=((XKnD)uK)nHand it follows that XO = X.

(b) Let a = UT-' and µ = &t/i'. By (3), D(aE)" =D(TE)4'-'

= D(aE) and then (a)implies that X" = X for all X < H. Since a = µ4i, it follows that X' = X for allX < H. Thus a = at-' e Pot H.

(c) This time, let p = 06. By (3), D(E)" = D(e) and hence again X" = X for allX < H. Then (b) implies that D(TE)" = D(Tae) where T(Ta)-1 = Tx"' T"' e Pot H.Thus x c- Pot H and, by (3), D(TE)O = D(TaE)a ' = D(araE).

Now suppose that cp is a projectivity from G to a group G_ such that G =H1* x K1*; put H° = H and K1* = K. Then D(E)'* is a diagonal in G with respect toH and k and hence D(E)'* = D(E') for some isomorphism e' from H onto K. Fora E Aut H, D(aE)' is a diagonal in G, and hence there exists a unique a' e Aut H suchthat D(aE)' = D(6 E'); define a' = a'. Then by 1.6.2, i is a bijective map from Aut Honto Aut H and we show that it induces an isomorphism in Aut H/Pot H.

7.6.6 Theorem (Suzuki [1951a]). Let G = H x K, H Kand let cp be a projectivityfrom G to a group 6 such that 6 = H" x Kl*; put H1* = H and K" = K. Let e be an

isomorphism from H onto K, D(E)" = D(e') and let i be the map from Aut H onto Aut Hdefined by

(4) D(ac)" = D(cr'E') for a E Aut H.

Then i induces an isomorphism i* from Aut H/Pot H onto Aut H/Pot H.

Proof. Let a, r e Aut H. If a = err-' E Pot H, then p = cp-' &cp is an autoprojectivityof G such that X'' = X for all subgroups X of H and K. Furthermore, by (3) and (4),

D(u'E')" = D(QE)"* = D(TE)`° = D(T'E').

Now (b) of 7.6.5 yields that a'(T')-' E Pot H. Conversely, E Pot H impliesQT-' E Pot H, since i-' is constructed from E' and cp-' in the same way as i from E andcp. Thus i induces a bijective map i* from Aut H/Pot H onto Aut H/Pot H. For

= (p-'&-'(p, D(E') ' = D(QE)I* = D(a'E'), by (3), and hence by (c) of 7.6.5 there existsQ E Pot H such that D(a'T'f E') = D(T'E')O = D(ATE)" = D((at)'E'). Thus (at)' =and i* is an isomorphism.

The above theorem is the essence of Suzuki's method. If H is a nonabelian simplegroup, then by 6.4.2 and 6.5.4, H" is also simple and G = H" x K4'. Hence theassumptions of Theorem 7.6.6 are satisfied and, furthermore, Pot H = 1 and Inn His the unique minimal normal subgroup of Aut H. It follows that i = t* is an isomor-phism and H (Inn H)' = Inn H4 H4'. Thus we get the theorem, proved by Suzuki[1951 a] in the finite case and by Zacher [1981] in general, that G is determined byits subgroup lattice if H is a nonabelian simple group. We want to prove a moregeneral result, and for this we have to study the map i more closely. We mention inpassing that i* is independent of the choice of e and that, in general, i is not anisomorphism. This can be seen, for example, if H is a cyclic group of prime orderp>5.

7.6.7 Lemma. Suppose that the hypotheses of Theorem 7.6.6 are satisfied and letX < H and a E Aut H. Then

(a) (Xe)w = (XI')`,(b) X' = X if and only if (X4')°' = X4', and(c) [X, x]`0 = [X`0, a'] if X. = X.

In particular, [H, a]' = [R, o('] and CH(af

Proof. (a) Clearly, (X x X`) n D(e) is a diagonal in X x X1 < G. Since G =H x k, it follows that (X x X`)4' = X4' x (X`)4' and hence ((X x X`) n D(E))4' _(X x (X`)'°) n D(E') is a diagonal in X'° x (X`)4'. It follows that (X`)4' = (X4')".

(b) Let us define D(X, a, E) = D(xE) n (X x X`). Then by (a) and (4),

(5) D(X, a, r)" = D(ae)4' n (X4' x (X`)4') = D(x'E') n (X4' x (X`0)`) = D(X4', a', E').

By 1.6.2, D(X, a, E) is a diagonal in X x X` if and only if the map x -i x°` (x E X) isan isomorphism from X onto X`. This is the case if and only if XS = X. Similarly,D(X4, a', E') is a diagonal in X4' x (X4')" if and only if (X`0)" = Xw. And finally, since

cp is a projectivity, D(X, x, c) is a diagonal in X x X` if and only if D(X, a, c)`0 =D(X`0,x',c') is a diagonal in X`° x (X`)4' = X`° x (X4')`. Summarizing, we obtain (b).

(c) Suppose that a, /i E Aut H such that X' = X = XB and let N a X. Then weclaim that

(6) D(X, a, c) < D(X, f3, c)N if and only if [X, a/i-' ] < N.

Indeed, if D(X, a, c) < D(X, fl, c)N, then for every x e X, there exist elements z e Xand a E N such that xx" = zzp`a = zazfl`. It follows that x = za, x' = za and hence[x, a fl-' ] = x-' x'p-' = a-' E N. Thus [X, af3-' ] < N. Conversely, if [X, a/3-' ] < N,then for every x e X, x' = (x [x, a f3-' ] )P = (xa)fl for some a e N and therefore xx" _xa(xa)a`a-' e D(X, f3, c)N. Thus (6) holds.

To prove (c), we apply (6) with N = [X, a] and /3 = idH, the trivial automorphism;note that N < X, by (6) of § 1.5. Thus by (6), D(X, a, c) < D(X, idH, c)N. By 7.6.4,N4 a X10 and, since cp is a projectivity, using (5) we get that D(X4,x',c') <D(X4,(idH)',c')N4'. Now (4) shows that (idH)' = idH and, by (b), (X4')" = X4'. Hence(6) for G = H x k yields that [X`°, a'] < N" _ [X, a]4'. If we apply this result to

9- and i-', we obtain that [X,2] < [X`0, a']4 and (c) follows. Now X = H andX = CH(a) satisfy the assumption of (c). It follows that [11,,i]" _ [Ti, a'] and[Cn(a)4, a'] = [CH(a), x]w = 1. Thus CH(")`P < CH(a') and, if we apply this result tocp - ' and i-', we get the other inclusion.

We come to our main result.

7.6.8 Theorem (Schmidt [1981]). Let G = H x K where H z- K and Z(H) = 1 andsuppose that cp is a projectivity from G to a group G. Put H4' = H and K`° = K, let uhe the natural isomorphism from H onto Inn H, that is, x=° = z -'xz for x, z e H;similarly r for H._

(a) Then G = H x k, H K, Z(H) = I and the map i constructed in 7.6.6 is anisomorphism from Aut H onto Aut H.

(b) If N is a characteristic subgroup of H or a normal subgroup of H contained inH', then N"N4't/N°' n Not is abelian. In particular, (Inn H)' Inn H/(Inn H)' n Inn H isabelian.

(c) The maximal perfect subgroup of H is isomorphic to the maximal perfect sub-group of H.

Proof. (a) By 7.6.2, G = H x K and then 7.6.4 implies that H -_ K and Z(H) = 1. Itfollows from 1.4.3 that Pot H = I = Pot H and, by 7.6.6, i = i* is an isomorphismfrom Aut H onto Aut H.

(b) Let X < H, E = X°' u X`°t, A = X°' n XP', and suppose that X°t a E andE. Then E/A = X°'/A x X`°t/A and we show that

(7) E/A is abelian.

For this let x e X' and x e X`0; let D < X such that D`0' = A. For z e H andfl eAutH,

z[X',6l = (Y-i(xzx-')'5 'X)p = (x1 xA)-'zx-1 x' = z[X'01,

Aut H

Figure 21

so that [x', #] = [x, #]T, a well-known formula. It implies that [x, x']T =[x", x'] E [XI", X°'] < A since X°' and xc" are normal subgroups of E. Therefore[x, oc'] E AT ' = D" for all x E X" and hence [X", a'] < D". Since x c- X°, we haveXz = X and, by (c) of 7.6.7, it follows that [X, x] < D. This holds for all x e X°; thusX' = [X, X°] < D. By 7.6.4, (X")' = (X')`° < D" and hence X`0/D'0 is abelian. So,finally, X`°`/D'0' = X"T/A is abelian. If we apply this result to c0-1 and t-1, we obtainthat X°/X"T' ' n X° is abelian. Thus X/X`°t n X°' = X°'/A is abelian and (7) holds.

Now if N is a characteristic subgroup of H, then by (b) of 7.6.7, N'' is characteristicin H and hence N°' and Nwt are normal subgroups of Aut H. By (7), N°'N'T/N°' n N"is abelian. In particular, (Inn H)' Inn H/(Inn H)'n Inn H is abelian and hence (H')°'and (H')T are contained in A = (Inn H)' n Inn H. Thus if N is a normal subgroupof H contained in H', then by 7.6.4, N" is a normal subgroup of H contained inH'. It follows that N°' and Nmt are normal subgroups of A > N°'N"T. By (7),N°'N`0T/N°' n Nwt is abelian.

(c) Let S be the maximal perfect subgroup of H. Then S is characteristic in H and,by 7.6.4 (or 6.4.5), Sc is the maximal perfect subgroup of H. It follows from (b) thatS°' = S°' n S"T = Sc,T and hence Sc" = S°'T ' S.

7.6.9 Corollary. If H is a perfect group with trivial centre and K Z H, then G =H x K is determined by its subgroup lattice.

Proof. By (a) of 7.6.8, a projective image G'° of G satisfies GI = HI x K'° whereH" K"' and, by (c),HP ^- H. Thus G" G.

The main idea in Theorem 7.6.8 is to try to prove that (Inn H)' = Inn H; sinceH Inn H and H Inn H, this implies that H H. We were able to do this whenH = H'; it can also be proved if this hypothesis is replaced by the assumption thatH is generated by involutions (see Zacher [1981]). We leave this as an exercise forthe reader, and remark that in 7.7.8 we shall prove H H in this situation using

quite different methods. We want to show now that 7.6.9 does not hold without theassumption that H is perfect so that in the situation of Theorem 7.6.8, in general,(Inn H)' : Inn H.

7.6.10 Example. Let q > 7 and p be primes such that ql p - 1, and let r E N suchthat r # I (mod p) and rQ - 1 (mod p). Recall that in 5.6.8 for every p e {2, ... , q - 1 },

we defined the Rottlander group

G"=<x,Y,alx"=Y°=a9=Cx, Y] = 1,a--lxa=xr a-1ya=yrl>

and proved that all the Gµ are lattice-isomorphic, but G G, if and only if p = v orpv = I (mod q). We claim that if p : q - 1 : v, then Gµ x Gµ and G, x G,, also haveisomorphic subgroup lattices.

Proof. Let H = <x1,x2,a> G K = <x3,x4,b> such that o(x1) = p, x;x, = xjx;for all i, j and xi = x'1, x2 = x2", x3 = x3, x4 = x4. Let G = H x K = NA whereN = <x1,x2ix3,x4> and A = <a,b>, and put Ni = <x,> for i = 1, ..., 4. We deter-mine the centralizers and eigenspaces in N of the subgroups of order q of A. Sincep 0 (mod q), we obtain CN(a) = N3 x N4, CN(b) = N1 x N2 and CN(ab") = 1 for0 < i. < q. Since p Ef= 1 (mod q), the other eigenspaces of a are N1 and N2, those of bare N3 and N4, and those of ab are N1 x N3 and N2 x N4. Since p 0 q - I, p2 * I(mod q) and there exists p' E {2, ... , q - 2} such that p 0 p' and pp' - 1 (mod q).Then the eigenspaces of ab" are N1, N2 x N3, N4 and those of ab," are N, x N4, N2,N3. Finally, if) i { 1, p, p' }, all the exponents r, r", r ;L' r' are pairwise incongruentmodulo p, and hence N,, N2, N3, N4 are the eigenspaces of ab". This eigen-space situation is more or less independent of p. Let H = <x1,x2ic> G,, K =<x3, x4+ d> where x` = x', x` = xr' X3 d = x'3, 4xd = x' 4°,

1 2 andand define G = H x K = NBwhere B = <(-,d>; let v' E {2__q - 2} such that vv' - I (mod p). Then the identityr on N and every projectivity T from A to B such that <a> = <c>, <h)` = <d>,<ab)' = <cd>, <ab">t = <cd`> and <abl">' = <cd,"> satisfy the assumptions of The-orem 4.1.6. Indeed, since qlp - 1, a subspace P of N is invariant under an elementof A or B if and only if it is a direct sum of subspaces of eigenspaces of this element.By 4.1.6, there exists a projectivity from G to G.

It is easy to see that for the projectivity constructed above, ((Inn H)')' = (Inn H)'in the notation of 7.6.8. It is an open problem whether this holds in general.

Direct products of isomorphic groups

The Fundamental Theorem of Projective Geometry states that every isomorphismof the lattice of subspaces of a vector space of dimension n > 3 onto the lattice ofsubspaces of another vector space is induced by a semilinear map between the twospaces. In particular, the ground fields are isomorphic. This theorem suggests thefollowing general question.

Given an algebraic structure X closed with respect to direct products, does thereexist an integer n = n(91) such that the n-fold direct product of every member of RI isdetermined by its lattice of substructures?

The P-groups show that the cyclic groups of prime order p > 2 are counter-examples to this conjecture for the algebraic structure "group"; and there are alsoless trivial examples (see Exercise 7). Therefore we have to make additional assump-tions on the projectivities considered. The most natural one is that they preserve thedirect decomposition. This leads to the following two problems.

7.6.11 Problem. Let X be a class of groups. Does there exist an integer n such thatwhenever X E X, G = X 1 x x X with X1 X for all i = 1, ..., n, and (p is aprojectivity from G to a group G satisfying G = X11 x x Xn , then

(A) X; X1 (and hence G G), or even(B) (p is induced by an isomorphism?

In particular, does there exist such an integer for the class X of all groups?

Clearly when n = 1 (A) or (B) is satisfied precisely when every group in X isdetermined or strongly determined by its subgroup lattice. So the two problems in7.6.11 can be considered as natural generalizations of the main problems studied inthis chapter. We want to discuss them briefly without giving proofs since most of theresults known are rather special and require long and technical computations.

Problem (A) is wide open. Example 7.6.10 shows that the Rottlander groups needat least three direct factors. But there is no example known in which n = 3 would notsuffice. So it might be true that, as in the Fundamental Theorem of ProjectiveGeometry, n = 3 will solve Problem (A) for the class X of all groups. This was provedby Baer for abelian groups (see 2.6.7 and 2.6.10, and note that the assumptions in7.6.11 imply that G is abelian). It is also proved in 7.6.9 for perfect groups with trivialcentre; for these groups and for abelian groups with elements of infinite order, evenn = 2 suffices. Another class of groups is handled by Schmidt [1982] who provesthat if X is a finite group with a proper elementary abelian normal Hall subgroupN such that Cx(N) = N, then n = 3 suffices. Note that the Rottlander groups arecovered by this theorem.

Problem (B) is solved for the class X of all groups. There are groups, even withtrivial centre, such that no integer n has the required property.

7.6.12 Example (Schmidt [ 1982] ). Let p > 2 be a prime and n c- N.(a) Let G = A1B1 x A2B2 be a finite group with elementary abelian normal p-

subgroups Ai, (p,lBil) = 1 = (IB1 l,IB2D) and CA;(Bi) = 1 for i = 1, 2. Let N = Al x A2,H = B1 x B,, take s c N such that 0 * s 0 1 (mod p), and define u: N - N by(xy), = xy.s for x c- A, y c- A2. Finally, for U < G, that is U = MK' where z e N,M < N, K < NH(M) (see 4.1.5), define U`° = MK` Then cp is an autoprojectivity ofG which induces the trivial autoprojectivity in N and in H and which is not inducedby an automorphism of G.

(b) Let q and r be different primes dividing p - 1 and let X be the direct productof a nonabelian group of order pq and a nonabelian group of order pr. If

G = X, x x X. where X; X for i = 1, ..., n, then there exists an autoprojec-tivity cp of G satisfying X? = X; for all i which is not induced by an automorphismof G. Clearly, I X I = p2 qr and Z(X) = 1.

Proof. (a) First note that N is an elementary abelian normal Hall subgroup ofG with complement H. Since (IB,I,1B21) = 1, every subgroup K of H has theform K = (K n B,) x (K n B2), and hence CN(K) = CN(K n B,) n CN(K n B2) =C,q,(K n B,) x CA2(K n B2). Since a fixes every subgroup of A, and A2, it followsthat CN(K)° = CN(K). If M < N is invariant under K, then by 4.1.3 and (1) of§1.5, M = [M, K] x C,(K) and [M, K] = [M, K n B2] [M, K rB, ]""2. Since[M, K n B2] < A2 and [M, K n Bt ] < A,, we obtain that [M, K]° = [M, K]. Thusif CN(K) < M, then M = [M, K] x CN(K) and hence M° = M.

Now let U; < G (i = 1,2), that is U; = M;K;i where zi e N, M. < N andK; < NH(Mi), and suppose that U1 < U2. Then by 4.1.5, K, < K2, M, < M2 andz, zZ 1 E CN(K, )M2. Therefore CN(K2) < CN(K1) and, since a is an isomorphism,

zi(zi)-1 = (z,zz1)Q E(CN(Kj)CN(K2)M2)' = CN(K1)CN(K2)M2 = CN(K,)M2.

Again by 4.1.5, M, K', < M2K2: and hence (p is well-defined and order preserving.The map constructed with respect to a-1 in the same way as (p with respect to a isthe inverse of gyp. Thus (p is a projectivity which clearly fixes every subgroup of N andH. An automorphism p of G inducing co therefore would have to fix H and to inducea power automorphism in N; on the other hand, it would satisfy H=' = (Hz)`* =(Hz)" = H=" and, since CN(H) = 1, it would follow from 4.1.1 that zQ = z" for allz c- N. But, since s * 1 (mod p), a is not a power automorphism.

(b) Let Xi = M;Q1 x L;R; where Mij = IL;I = p, JQil = q and JRij = r for alli = 1 , ..., n. Then A, = M, x . x M,,, A2 = L, x x L,,, B, = Q1 x x Q.and B2 = R, x x R satisfy the assumptions of (a). The autoprojectivity cpconstructed there fixes every subgroup of N and H, hence every Mi, Li, Qi, Ri andso, finally, all the Xi.

Similar examples of groups of order page with nontrivial centre are given bySchenke [1987b]; see Exercise 8. Although these examples destroy any hope for ageneral theorem, there are positive results for interesting classes of groups. Firstof all note that in Problem (A) every integer greater than n also has the desiredproperty; but this is not so clear for Problem (B). Baer's theorems mentioned aboveshow that (B) holds for every n > 2 if X consists of abelian groups with elements ofinfinite order, and for every n > 3 if X is the class of all abelian groups. In 7.7.6 weshall prove that (B) holds for n = 2 if 3 is the class of dihedral groups. AgainSchmidt [1982] proves that every n > 3 satisfies (B) for the class of all finite groupsX = NH with normal elementary abelian p-subgroup N and abelian q-group Hoperating faithfully on N, p and q being different primes. Also Schenke [1987a],[1987b] studies finite p-groups and proves that (B) holds for every

(8) n > 2 if X has class at most 4 and exponent p,

(9) n > p - 2 if X is metabelian of exponent p,

(10) n 3if Xis of class 2and p:t- 2.

In addition, for every prime p >_ 7, he constructs a group of class 5 and exponent pfor which n = 2 does not satisfy (B). And he shows that for every n E N there existsa prime p and a metabelian group X of exponent p such that n does not satisfy (B)for X. This shows that although the bound p - 2 in (9) might not be best possible,there is no integer n satisfying (B) for all metabelian groups of exponent p and allprimes p.

The methods of Schenke and Schmidt are quite different. In both cases, the resultsfollow from more general investigations. Schmidt considers under which conditionsa projectivity cp of a finite group G = NH with elementary abelian normal Hallsubgroup N and complement H is induced by an isomorphism. He shows that thisis true in a certain minimal situation and then examines how to extend isomor-phisms inducing cp on subgroups of the form NH;, Hi < H. In this way he is able toshow that cp is induced by an isomorphism if H is an abelian q-group with threeindependent elements of maximal order and CH(N) = 1. This implies the resultmentioned above.

Schenke's method is more closely related to the given problem. He uses Baer'sideas to study a projectivity cp from a direct product G = H x K of two p-groups Hand K of equal exponent to a p-group G = H° x K". His main tools are Baer maps;these are maps a: G -+ G for which there exist x E H, y e K with o(x) = o(y) = Exp Gsuch that for all aEHandbEK,

(11) a° = f(a; y, y°, (p), b" = f(b; x, x8, gyp) and (ab)2 = aab2

where f is the map defined in 2.6.2. In the situation of 7.6.11, Schenke shows thatevery Baer map x induces gyp, and that for every automorphism a of X;, a71Qa; is anautomorphism of X' if a; is the restriction of a to X;. Using this property, he is ableto handle the minimal situation that X is the nonabelian group of order p3 andexponent p. Then he uses induction to show that under the assumptions of (8)-(10)there is a Baer map which is an isomorphism.

Wreath products

A well-known construction which uses direct products of isomorphic groups is thewreath product of two groups A and B. To define this, let N be the set of functionsf : B -+ A satisfying f(b) = 1 for all but a finite number of elements b e B. With theobvious multiplication, f, f2(b) = f,(b)f2(b) (b E B), N is a group, and an action of Bon N is established as follows: if f e N and b e B, then f' is the function defined bythe equation

(12) fb(c) = f(cb-1) for c e B.

It is easy to check that, with this definition, B acts as a group of automorphisms onN. The semidirect product of N with B acting as above is the (standard restricted)wreath product of A with B and is denoted by A wr B; the normal subgroup N iscalled the base group of the wreath product. The following properties of A wr B =NB are easy to prove and are, in fact, well-known (see Suzuki [1982], pp. 268-279).

ForbeB,letA,= If ENJf(c)= I for all b cc B}. Then

(13) N = Dr Ab, A, n- A and (Ab)c = Abc for all b, c c B.beB

Conversely, this property characterizes the wreath product.

(14) If G = MC where M = Dr Mc and (Mc)° = Mc, for all c, d c C, then GM, wr C. CEC

As a consequence we get the following two properties which are basic for inductiveproofs.

(15) If X < Ab (b E B) and C < B, then <X, C> 2f X wr C.

(16) If C < B and T is a left transversal to C in B, then NC A wr C where A =Dr A,tET

We now briefly report the known results on projectivities of wreath productswithout giving proofs. So let G A wr B with nontrivial groups A, B and let cp be aprojectivity from G to a group G. First of all, the image Nwof the base group N is anormal subgroup of G (see Menegazzo [1973]). However, in general, G need not bea wreath product. For example, if I A I = 29 and I B I = 3, then I NJ = 293 and, since Bcentralizes the "diagonal" { (a, a, a) I a E A } of N but does not centralize N, it follows(see 4.1.7) that N = N, x N2 where IN, I = 29, 1 N21 = 292, N, = C,,,(B) and B oper-ates irreducibly on N2. Since 51292 - 1, a group C of order 5 can operate irreduciblyon N2 and, by 4.1.6, G = N, x N2 B is lattice-isomorphic to G* = N, x N2 C.Clearly, G* is not a wreath product. On the other hand, there is the followingpositive result.

7.6.13 Theorem (Menegazzo [1973]). Let G A wr B where A : 1 and 2 < J BI < cand let q be a projectivity from G to a group G.

(a) Then G = N4B4, N4 g G, N4' = Dr Ab and Ab A'* for all b, c c- B.bEB

(b) If A is locally finite and B4I = IB1, then N'° = Dr (A'*)b and hence GA" wr B`0. b E Bm

Here (b) is an immediate consequence of (a). Indeed, since A"< N`° andN`0 = (A, u B)w n N`° _ (Ai u B`0) n N'° _ (Af)B°, one only has to show thatAT n <(A")bI I b e B"> = 1. For this one may assume that A is finite, then (a) andthe assumption in (b) imply that I N4I = IAOIJBI = I AT j1B11, and this yields theassertion. To prove (a), one has to show that [A6 , Af] = I for all b, c c- B such thath c. Properties (15) and (16) reduce this to the case where A is cyclic of primepower order and CBI = q or 4, q a prime, q 96 2. Here Nw is abelian and, in general,G is even determined by its subgroup lattice (see Exercises 9 and 10). It is an openproblem to give precise conditions under which G G in 7.6.13. For torsion-freegroups, the situation is better.

7.6.14 Theorem (Arshinov and Sadovskii [1973]). The wreath product G = A wr Bof two nontrivial torsion free groups A and B is strongly determined by its subgrouplattice.

Sketch of proof. Let cp be a projectivity from G to a group Cr. First consider the casethat A and B are infinite cyclic. Then G is residually torsion-free nilpotent (seeSmelkin [1964]), and hence the family ." of all normal subgroups X of G withnonabelian torsion-free nilpotent factor group G/X satisfies the assumptions (18)and (19) of Sadovskii's Approximation Theorem in § 1.3. By 6.5.1 and 7.2.11, ifX E .", then X° a G and cp is induced by a unique isomorphism on G/X. Thus by1.3.8, cp is induced by an isomorphism from G to G. This proves Theorem 7.6.14 inthe special case that A ^- C, B.

If A and B are arbitrary torsion-free groups, then by 7.6.2 and 6.5.1,N4' = Dr Ab g G. Let b, c c- B. Since there are diagonals in Ab x A, there exist

IEB

diagonals in Ab x A° and hence all the Ab are isomorphic. By (15), if I * a c- A, and1 b c B, then <a, b> <a> wr is strongly determined by its subgroup lattice, aswe have just seen. By 1.5.8, Pot <a, b> = I and hence cp is induced on <a, b> by aunique isomorphism v(a, b): <a, b> <a, b>4'. So, if we fix a c A, we obtain a maprr: B -+ B4' defined by b° = b1"1" and satisfying " = <be> for all b e B. If we fixb c- B, we can define maps r,: Ac - A" (c c- B) by a'° = a"(°. b, for a c- A, and these canbe put together to a map r: N = Dr Ac - N' = Dr A". It is not difficult to show

CEB CEB

that a and 2 are independent of the choices of a c- Ac and b c B, and that they are, infact, isomorphisms satisfying <b-' ab>4' = <(ba)-l a`b°> for all a c N, b c B. Thusthere is an isomorphism co: G -+ G such that COIN = r and wIB = a, and it is possibleto show that the projectivity q is induced by co.

Direct products of lattice-isomorphic groups

Finally we study the following problem. If X is a group which has only finitelymany pairwise nonisomorphic projective images-call them X,,..., X.-isG = X1 x x X. determined by its subgroup lattice? The following exampleshows that this is not the case.

7.6.15 Example. Let p > 2 be a prime and X = X1 be the abelian group of type(p2, p). Then n = 2, X2 = <c, dl c°2 = d P = 1, [c, d] = c°> is the nonabelian group oforder p3 and exponent p2, and we claim that the group G = X1 X X2 is not deter-mined by its subgroup lattice.

Proof. Let A = X 1 x <c °, d > < G and a: G G be defined by x' = x 1 +° forall x e G. Since G' = <c"> < Z(G), it follows from (4) of § 1.5 that

) = xl+py1+p(xY)1+p = x1+pY1+p[y,x](2°1

for all x, y c G, that is,

(17) a is a power automorphism of G.

Therefore the situation is similar to the proof of Theorem 2.5.9: here IG : Al = p sothat s = 1 = m, but G is not abelian. Nevertheless we can construct a crossed

isomorphism of G in the same way as in that proof. We put r = 1 + p,,u = µ writex e G in the form x = ac") with a e A and 1 < i(x) < p, and define

(18) f(x) = a' for j e 7L such that µ(j) = i(x) (mod p).

We copy Baer's argument literally and obtain that f(x) is well-defined,

(19) A={xEGIf(x)=id}

and f : G End G satisfies (2)-(4) of § 2.5 and (d) of 2.5.8. By 2.5.2 and 2.5.8, there isa new operation o defined by

(20) x o y = xf(r) y

for x, y e G such that G* = (G, o) is a group and the identity is an f-isomorphisminducing a projectivity from G to G*. Clearly, i(c) = I and µ(1) = 1 so that f(c) = a.By (19) and (20), the inverse of an element of A is the same in G and G*. Let b be theinverse of c in G*, that is, I = b o c = bf(`)c. If a is an element of order p2 in X1, then

b o a o c o a-1 = (ba)f(c)ca-1 = bf<<iaf(Oca-1 = bf(`)ca"pa-1 = aP

and

b o d-1 o cod = (bd-')f (')cd = bf(<)(d-1)1+Pcd = bf(`)d-1cd = bf(`)ccP = cP

since o(d) = p. Thus the commutator subgroup of G* contains aP and cP and, sinceG' J = p, it follows that G* # G.

However, there is a positive result for a large class of finite groups.

7.6.16 Theorem (Schmidt [1980b]). Let X be a finite group such that Z(X) = 1 orX = X', and let G = X 1 x x X. where X,__ X. are all the pairwise nonisomor-phic projective images of X. Then G is determined by its subgroup lattice.

Proof. First note that, by 1.6.10, every finite group with infinitely many pairwisenonisomorphic projective images has a nontrivial cyclic direct factor. Thus X indeedhas only finitely many such images and G is well-defined. Let (p be a projectivityfrom G to a group G. We want to show that

(21) G = X; x . . x X .

If X is perfect, then by 5.3.4, every X. is perfect and (21) follows from 7.6.2. Nowsuppose that Z(X) = 1 and, for a contradiction, that X? is not normal in G for somei. Then there exists k e It__ n} such that Xr is not normalized by X11,. And, sinceXk is generated by its elements of prime power order, there exists a cyclic subgroupC of prime power order p' of Xk such that X,1' is not normalized by C". Clearly, pdivides IX,I since otherwise by 1.6.6, (Xi x C)" = X,? x C". If cp were singular at p,then by 7.6.3 applied to H = Xi and K = Dr X,, the group G would have a normal

Jmip-complement and, for some j e { 1, ... , n}, X, = P x Q where 1 0- P E Sylp (XX). Let

a be a projectivity from X3 to X. Then, again by 1.6.6, X = P° x Q° and Z(P°) = 1since Z(X) = 1. By 2.2.6, P° would be a nonabelian group in P(m, p) for some m. Butthe definition of G implies that X is isomorphic to a subgroup of G and hence wouldhave a normal p-complement. This contradiction shows that qp is not singular at p.

Let D = Q(C) be the minimal subgroup of C. By 4.3.5, X° < (Xi x D)4' andhence, in particular, D < C. Furthermore, by 7.6.1 applied to X, x Xk, we haveC n N(Xk) A 1 and hence D < N(Xk). Let r be a projectivity from Xk to X. By 1.4.3,N(X) = 1, and hence there exists a subgroup M of Xk such that M° = M but M` isnot normalized by DT. Again by 4.3.5, r is singular at p. Since C is cyclic of order atleast p2, 4.2.6 shows that there exists a normal p-complement N in Xk with cyclicfactor group such that NT a X. If P is a Sylow p-subgroup of MD containing D, thenP = (M n P)D; since P is cyclic and M n D = 1, it follows that P = D. Thus M =MD n N and then MT = (MT u DL) n N' :!g ML u D'. This final contradiction showsthat Xi g G for all i and (21) follows.

All the X7 are projective images of X. If two of them, Xr and X;, say, wereisomorphic, then by 1.6.2 there would exist diagonals in the group X5° x X7 =(Xi x XJP, and hence also in Xi x X;. It would follow that X, XX, contradictingour assumption. Therefore XT, ..., Xf are pairwise nonisomorphic projectiveimages of X and, since X possesses only n of these, it follows that the X"are isomorphic to the X1 in some arrangement. Thus G = X" x x X. 2 -X, x x X. = G, as required.

Exercises

1. (Lukacs and Palfy [1968]) If G = H x K, H K and L(G) is modular, showthat G is abelian. (Do not use the results of § 2.4.)

2. Show that the isomorphism t* in 7.6.6 is independent of the choice of C.3. Let X be a finite group such that Z(X) = 1 and let G = H x K where

H Aut X K. Show that G is determined by its subgroup lattice.4. Let G = H x K and define

R(H, K) = Ja c P(G)IH° = H and X° = X for all X <K},

S(H,K) = Ju e P(G)jX° = X for all X < H and all X < K}.

(a) Show that S(H, K) a R(H, K) and R(H, K)/S(H, K) is isomorphic to a sub-group of P(H) containing PA(H).

(b) If H K, show that R(H, K)/S(H, K) zt PA (H).5. (Zacher [1981]) If the group H in Theorem 7.6.8 is generated by involutions,

show that (Inn H)' = Inn H. Deduce that G G.6. Show that for the Rottlander groups G, we have L(Gi, x Gr,) L(GQ_, x G,_,)

if 2<p<q-2.In the next two exercises let 3 < n e N, p and q be different primes, X, _

N, H, where N, is an elementary abelian normal p-subgroup of X, and H, is aq-group. With the obvious notation, assume that X, = N;Hi are isomorphic to X,(i = 2,..., n) and let G = X, x x X°,N=N, x . x x . x H,,.

7.7 Groups generated by involutions 421

7. (Schmidt [1982]) Assume in addition that H1 is abelian of exponent q', r >_ 2and r > 3 in case q = 2, and that CHL(Nj) = c2(H1). Show that there exists asemidirect product G = NH lattice-isomorphic to G in which H is a nonabelianM-group. (Hint: Use 4.1.9.)

8. (Schenke [1987b]) Assume that 2 < qlp - 1, IN11 = p, H1 is the nonabeliangroup of order q3 and exponent q and JH1 : CH1(Nl)I = q. Show that G has anautoprojectivity cp which satisfies Xr = X, for all i and is not induced by anautomorphism. (Hint: Use 4.1.9 to construct an autoprojectivity of G which istrivial on G/CH(N) and induced on H1 by a suitable automorphism of H1.)

9. (Menegazzo [1973]) Let G = A wr B where A is cyclic of order p", IBS = q, p andq primes, n e N, and assume that cp is a projectivity from G onto a group G; letN be the base group of G.(a) If p q, show that NP N; if in addition IB"I = q, show that G G.

(b) If p=g02,show that G^--G.(c) If p = q = 2 and [D'*, Bl*] = 1 where D = {(a, a) I a E A} is the diagonal in

N, show that G _- G.10. (Menegazzo [ 1973]) (a) Show that G = A wr B is determined by its subgroup

lattice if A is cyclic of order 2" (n e N) and IBI = 4.(b) Show that C8 wr C2 is not determined by its subgroup lattice.

11. (Zacher [1981]) Show that for every group H there exists a group K such thatH x K is determined by its subgroup lattice. (Hint: Take a suitable free groupK.)

7.7 Groups generated by involutions

The following is a general idea to prove that a group G is determined by its sub-group lattice. It is clear that if cp is a projectivity from G to a group G, then Goperates on L(G) via the map r sending every x E G to the autoprojectivity r(x) of Ggiven by

(1) Hr(x) = H` X(,0 for H < G.

So one could try to prove that G is isomorphic to G by showing that G and Goperate in the same way on L(G). Unfortunately, this is much too difficult. Tosimplify matters, however, one could restrict oneself to a suitable subset A of L(G)and to a system D of generators of G and try to prove that for every d E D and<x) = <d>", the elements x and d operate in the same way on A. For this, A wouldhave to be a union of conjugacy classes of subgroups of G; otherwise G would notoperate on A. Furthermore, it seems that this can only work for elements d and xwhich are the unique generators of the subgroups they generate, that is, for involu-tions. We shall show in this section that this idea works and give some applications.The most interesting will be to show that certain multiply transitive permutationgroups and classical groups are determined by their subgroup lattices. In addition,an application to lattice-simple groups will be given in § 7.8.

2-singular projectivities

To prove these results, we first of all have to guarantee that I <d >"I = 2 if d E D is aninvolution. For any prime p and a projectivity cp between arbitrary groups G and G,let us say that q is p-regular if it maps every subgroup of order p of G to a subgroupof order p in G. Note that for finite groups this is equivalent to our definition ofp-regularity in §4.2. So we have to study 2-singular projectivities of groups gener-ated by involutions, and we start with groups generated by two involutions. Nowif G = <d, e) where o(d) = o(e) = 2 and o(de) = n E f01 u { oo }, then (de)" = dded =ed = (de)-' so that G is the semidirect product of a cyclic group <a> = <de> of ordern, and a cyclic group <d> of order 2 with respect to the automorphism given byad = a-1; such a group is called a dihedral group of order 2n. Note that (a'd)2 =a'a-' = 1 for all i e 1L so that every element in G\<a) has order 2.

7.7.1 Lemma. Let n E N u { oo }, G be a dihedral group of order 2n, that is, G = <d, e)where o(d) = o(e) = 2 and o(de) = n, and let cp be a projectivity from G to a group G;put <d) ° = <x> and <e> ° = <y>. _

(a) If n = oo, then cp is 2-regular, <de> ° = <xy) and G G.

(b) If n = p is a prime, then G E P(2, p). If q is 2-regular, then <de> = <xy).(c) If n is neither infinite nor a prime, then <de> is a cyclic normal subgroup of

order n in G and o(x) = o(y) = q for some prime q. If q = 2, then cp is 2-regular,<de)" = <xy) and G nf G. If q > 2, then either q does not divide n and x operatesfixed-point-freely on <de> or n = qn' where (n', q) = 1.

Proof. (a) If n = oo, <de> is the unique cyclic maximal subgroup of G. Since everyprojectivity maps cyclic subgroups to cyclic subgroups, <de> is the unique cyclicmaximal subgroup of G, and by 6.5.11, cp is 2-regular. In particular, o(x) = o(y) = 2,G = <x, y) _ <xy, x> is an infinite dihedral group and <de)'* = <xy).

(b) That G E P(2, p) follows from 2.2.5. If p is 2-regular, then <de> = <xy) since<de> and <xy) are the only minimal subgroups of G and G of order different from 2(if p > 2) or different from <d>, <e>, <x), <y) (if p = 2).

(c) Again, <de> is the unique cyclic maximal subgroup of G; hence <de> is acyclic normal subgroup of G and o(x) = 1 G/<de)'*l = o(y) is a prime q. Let I <de> l = m.We have to show that m = n.

First suppose that q = 2. Then G is generated by the involutions x and yand hence is a dihedral group with cyclic maximal subgroup <xy). It follows that<xy) = <de>". Since every element in G\<xy) has order 2, thenumber of minimalsubgroups of G not contained in <xy) is m. Thus m = n and G_ G. Clearly, q is2-regular. Indeed, if n is even, this follows from 2.2.6 since the Sylow 2-subgroups ofG are noncyclic. And if n is odd, the Sylow 2-subgroups of G are the complements to<de> in G; they are mapped to the complements to <xy) in G and hence to sub-groups of order 2.

Now let q > 2. If q does not divide m, then x operates fixed-point-freely on <de),*since every nontrivial fixed element z e <de> ' would yield a cyclic group <zx> ofcomposite order whose preimage <zx)"-' would be a dihedral group, a contradic-tion. Thus there exist exactly m complements to <de>" in G. The number of

complements to <de> in G is n and it follows that m = n. Finally, suppose that qdivides m and let Q" be the Sylow q-subgroup of <de> . Then Q"<x> is a noncyclicq-group whose preimage Q<d> is not a q-group. By 2.2.6, Q<d> E P(r,q) for somer E N and hence I Q I = q since Q is cyclic and Q g Q <d )'. Let R" be the complementto Q" in <de> . Then the dihedral group R <d> is mapped by (p to R" <x> and(IR"I,q) = 1. By (b) or the case of (c) just considered, IRI = IRI. It follows thatm = I R`° I I Q1 I = I R I q = n. Since I QP I = q, finally, n = qn' where (n', q) = 1.

The structure of finite groups with 2-singular projectivities is well-known. By4.2.6, such a group G is either P-decomposable, that is, G = S x T where S is aP-group of order 2p" for some prime p > 2, n e f\l, and (ISI, I TI) = 1, or G has anormal 2-complement K with cyclic factor group G/K such that K`0 a G. A similarresult for arbitrary groups is not known. Even the most moderate conjecture that agroup G generated by involutions with a 2-singular projectivity is either a P-groupor has a normal subgroup K such that I G : K I = 2, K`° a G and every element in Khas infinite or finite odd order has not been confirmed yet. An immediate conse-quence would be that all subgroups of order 2 of G are mapped onto subgroups ofthe same order if G is not a P-group. We can prove this much weaker result and itwill suffice for our applications.

7.7.2 Lemma. Let G be a group generated by involutions and cp a projectivity from Gto a group G. If there exist involutions a, b e G such that I <a>'I A I "I, then G is anonabelian P-group.

Proof. Let D be the set of involutions in G. If a E D, then I<a)'* I is a prime; let q bethe smallest prime occurring among the I <a)4I for a E D. We study the groups (a, b>and (a, b, c> where a, b, c e D and first of all note the following immediate conse-quence of 7.7.1.

(2) Let a, b e D such that I <a>4'I = q, I <b4'> I = p and q < p. Then <a, b> is adihedral group of order 2p and <a, b)4' is a nonabelian group of order pq.

Indeed, (a, b> is a dihedral group of order 2r for some r e N u { oc } and I <a>4I AI 4I. By 7.7.1, r is a prime and <a, b)4' E P(2, r). Since p and q divide the order of<a, b)'4', this group is nonabelian of order pq, r = p, and <a, b> is a dihedral group oforder 2p.

(3) Let a, b c- D such that I <a>4I = q, I 4I = p and q < p. If c c- D such thatc (a, b), then <a, b, c>4' is a P-group of order p2q, <a, b, c> E P(3, p) and<ab, ac, bc> is elementary abelian of order p2.

To prove this, we first show that H = <a, b, c> is finite. If I <c>4'I = r > q, then by(2), <a, b)4' and <a, c)'`0 are nonabelian of order pq and rq, respectively. It followsthat <a>" normalizes <b)4' and <c>"' and hence also <b)4' u <c)'4' = <b, c)4'. SinceI(b>'*I = p > q > 2, the projectivity 9 is 2-singular on the dihedral group <b, c).By 7.7.1, <b,c>4' is finite; hence so is <b,c)4'<a)'`° = H4' and H. Now suppose thatI <c>PI = q. Then by (2), <b)4' is normalized by <a>" and <c)4' and hence also by<a, c)'`°. If <a, c)4' were to contain an element of infinite order, it would follow from

6.5.11 that p = I <h)PI = I <b) I = 2, a contradiction. Therefore <a, c) is a finitedihedral group, and <b)'D<a,c)`D = H`0 and H are finite. Thus, in all cases, H =<a, b, c> is a finite group generated by involutions in which 9 induces a 2-singularprojectivity. If there existed a normal 2-complement K in H such that K`0 < H`°, thenq = I <a)°I = I H`0 : K111 = <b)10I = p, a contradiction. By 4.2.6, H is a P-group oforder 2r" for some prime r. Since c <a, b), we have n = 2; thus H E P(3, r) andtherefore also H" E P(3, r). Since p and q divide IH`°I and q < p, it follows that r = p,I H`°I = p2q and H E P(3, p). Now ab, ac, be are contained in the normal subgroup oforder p2 of H and, since H is generated by a and these three elements, it follows that<ab, ac, bc) is elementary abelian of order p2. This proves (3).

(4) Let b, c e D such that I <b)'°I = p > q, I <c)'PI > q and b c. Then <b, c)`° iselementary abelian of order p2.

To prove this, take an element a c D such that I <a)`0I = q. Then by (2), <a. b)`0 isnonabelian of order pq. Therefore is the unique minimal subgroup of orderdifferent from q in <a, b>' and hence c 0 <a, b). By (3), <a, h, c)`° is a P-group of orderp2q and <b, c)`0 = u <c> ° is the elementary abelian normal subgroup of orderp2 of <a, b, c)'.

To complete the proof of the lemma, we fix an involution b E D such that I ' =p > q and show that

(5) A = <abla c- D) is an elementary abelian p-group.

Since (ab)' = ba = (ab)-' for all a c D, it will follow that b inverts a system of genera-tors of the abelian group A and hence every element of A. Hence G = <aIa E D) _A will be a nonabelian P-group.

To prove (5), we show that all the generators ab (b a E D) of A have order p andany two of them commute. For this let a, c c D be such that a b 0 c, and supposefirst that <a>" or <c)4' has order q; let I <a)41 = q, say. Then by (2), <a, b) is adihedral group of order 2p and hence o(ab) = p. If c c <a, h), then <cb) = <ab) is thesubgroup of order p in <a, b), that is, cb has order p and commutes with ab. Ifc 0 <a, b), then by (3), <ab, ac, bc) is elementary abelian of order p2; again ab and cbhave order p and commute. Now suppose that I <a)4'I > q and I <c>4' > q. By (4),<b, a)`° and <b, c)`° are elementary abelian of order p2 and then so is <a, c>'P if a : c.Thus <a>", <b)`0, <c)4' are subgroups of order p in G centralizing each other andtherefore generating an elementary abelian subgroup of order p' or p2 of G. Itfollows that <a, b, c) is a P-group of order 2p2 or 2p and the elements ah and cb haveorder p and commute. This proves (5) and the lemma.

It is not difficult to generalize the above lemma to the situation that G is notgenerated by involutions; then G = S x T where S = <a E GIa2 = 1) is a non-abelian P-group and S and T are coprime (see Exercise 2). Conversely, if G is such agroup and S E P(n, p), there exist autoprojectivities cp of S, and hence of G, such thatI <a)4I = p : 2 = I 4I for certain involutions a, b c- S. We note an immediateconsequence of Lemma 7.7.2.

7.7.3 Corollary. If G is a group containing two commuting involutions, then everyprojectivity of G is 2-regular.

Proof. These two involutions generate a four group H and hence the subgroup Sgenerated by all the involutions in G cannot be a nonabelian P-group. Thus if cp is aprojectivity of G, then by 7.7.2 there exists a prime q such that I <a>" I = q for everyinvolution a e S. Since H' = 4, it follows that q = 2 and cp is 2-regular. Li

We want to compare the operation of an involution and its projective imageson the subgroup lattice. The next lemma takes care of the fixed points under thisoperation.

7.7.4 Lemma. Let G be a group generated by involutions, cp a projectivity from G toa group G and suppose that G is not a P-group.

(a) If a E G is an involution, <a> P = <x> and H < G such that H' = H, then(H"')x = H°.

(b) If N a G, then N`° a G.

Proof. Since G is generated by involutions, (b) of course follows from (a). To prove(a), suppose, for a contradiction, that (H"')" 0 H" and let T = H<a>. Then I T : HI =2 and by 6.5.3 there exist a prime p and a normal subgroup K of T contained in Hsuch that I T/K I = 2p, K`' a TW and T`°/KI* is a nonabelian group in P(2, p). By 7.7.2there exists a prime q such that I <d>111 = q for every involution d in G. So if U/K isa subgroup of order 2 of T/K, then U = (K<a>)' = K<a'> for some t c- T; henceU`° = Kl*<at>' and I UP,!KPI = I<a'> I = q. It follows that H`°/K" is the subgroup oforder p in T"/K`0, a contradiction. El

Involutions and conjugacy classes

We can now show that the idea described at the beginning of this section indeedworks. We present it in its most general form.

7.7.5 Theorem (Schmidt [1980a]). Let G be a group, D a set of involutions in G, A aset of subgroups of G and cp a projectivity from G to a group G with the followingproperties.

(6) G = <did c- D).

(7) A' =A(that is,H°EAfor all HeA,dED).

(8) If d c- D, H E A and <d> = <x>, then (H`°)"

(a) Then G/ n N0(H) is isomorphic to G/ n Nz;(H`°).H. A, HEe _

(b) Let n NC(H) = 1. 1J' G is finite or D G = D, then G is isomorphic to G.HcA

Proof. (a) For y c G and H < G, let W'') = HWY'W '. If d c D, H E A, and <x> = <d)10,

then H`(") = H° E A, by (8) and (7). Since G = <did c- D>, the group G is generated bythese elements x, and it follows that A9 = A for all g E G and Ar(Y) _ A for all y E G.Thus G and 6 operate on A via the homomorphisms v: G --+ Sym A given by

H'(Q' = H9 (g c G) and p: G - Sym A defined by H°(1) = H`('') (y e G) for H E A. By(8), v(d) = p(x) for every d e D and <x> = <d>". Since G is generated by theseinvolutions, it follows that G" = G. Thus G/Ker v z G" = G° G/Kerp; sinceKer v = n NG(H) and Ker p= n NeJH4 ), (a) follows.

IIEA HEA(b) By (a), G is isomorphic to G/ n Nd(H(O). If G is finite, it follows that G G

HEAsince the finite group G cannot admit a projectivity onto a_proper factor group.

Now let D° = D and suppose, for a contradiction, that G is not isomorphic to G;then Kerp 1. Ford c D and <x> = <d>", p(x) = v(d). Since visa monomorphism,it follows that o(p(x)) = 2, and hence o(x) = 2. Therefore, if G were a P-group, itwould follow that G G, a contradiction; thus G is not a P-group. Then also G isnot a P-group and is generated by involutions. By 7.7.2, cp and IP-' are 2-regular andby 7.7.4, K = (Kerp) ' a G. Since Z(G) = 1 and G = <did e D)', there exists d c- Dsuch that CK(d) K. Let u c- K such that ud u and let v = u-du. Then v,d =u ' ud = v ' and hence <v, d> is a dihedral group of order 2m where m = o(v) 1.

Since K a G, we have v c K and o(v) 2; indeed, otherwise <v>" would be a sub-group of order 2 of G normalizing every H' (H E A) and 7.7.4 applied to Ip-' wouldyield that v c n NG(H) = 1, a contradiction. Thus d and e = d° are two different

HEA

involutions in D (here we use that D° = D) and de E <v> < K. Let <d>" = <x>and <e>" = <y>. Since cp is 2-regular, 7.7.1 shows that <de>" = <xy> and hencexy e Kerp. For every H E A, also Hd c A and so by (8), H10 = (H4)YY = ((Hd)")y =(Hde)'', that is H = Hde. It follows that de c n NG(H) = I and hence d = e. Thiscontradiction shows that G G. HEA

To apply Theorem 7.7.5, we have to choose a set D of involutions and a suitableset A of subgroups of G satisfying (6)-(8). Here (6) and (7) have to be assumed, thatis, we have to assume that G is generated by (these) involutions and for A we have totake a union of conjugacy classes of subgroups of G, and then we have to try toprove that (8) holds. In view of Lemma 7.7.1 and our discussion on projective imagesof conjugacy classes in § 5.6, there are two situations in which this could perhaps bepossible, namely if

(i) A={<d>IdED}or(ii) A is a conjugacy class of maximal subgroups of G or, in the language of the

theory of permutation groups, G is a primitive (or multiply transitive) permutationgroup on a set S2 and A is the set of stabilizers of the points.

We start with the first possibility and need the following result which is interestingin its own right.

7.7.6 Lemma. Let n e t\J u { cc }. If S and T are dihedral groups of order 2n, thenG = S x T is strongly determined by its subgroup lattice.

Proof. Let cp be a projectivity from G to a group G. We have to show that cp isinduced by an isomorphism. By 7.7.3, cp is 2-regular and then 7.7.1 implies that ifd, e are involutions in G and <d>" = <x>, <e> _ <y>, then

(9) <de> = <xy> and o(de) = o(xy);

it follows that <d, e)" (d, e>. This holds in particular for S and T and by 7.6.2,

(10) G = S4' x T" where S° -- S and T° -- T; thus G 2w G.

We fix the following notation. Let S = <a, s) where o(a) = n, o(s) = 2 and a' = a-'Then s and as are involutions generating S. Therefore if <s>4' = and <as>'* =<cu> where c e S`°, then by (9), <a>4' = <c> and o(c) = o(a) = n; furthermore S`° =<c, u) and c" = c-'. Let Z. = Z be the ring of integers if n = x and Z. = 7Z/(n) ifn e N. Then every involution in S not contained in <a> can be written uniquely inthe form a's where i e Zn; similarly for the involutions in S4'\(c). Since <a>" = (c),for every i e Z. there exists a unique j e Z. such that <a's>'* = <ciu>. Thus we obtaina bijective map v: Z - Z. satisfying

(11) <a's>'* = <c''('1u) for all i e Zn.

Since <s>4' = and <as>'* = <cu>,

(12) v(0) = 0 and v(l) = 1.

We handle T and T4' in the same way. Let T = (b, t> where o(b) = n, o(t) = 2 andb` = h- '; let <t>' = <v> and <bt>'* = <dv). Then T`0 = <d, v), o(d) = n and d' _d-'. Let p: Zn -+ Z. be defined by

(13) <hir>' _ <d1'(i)v) for all j e Zn.

Then as above

(14) p(0) = O and p(1) = 1.

We want to show that v and p are the identity on Z. For this let i, j E Z,,. Then<a's)" = (c''""u) and (bit)" = <d°(i)v) and therefore by (9),

(15) <a'hist)" = = <c'(i)d°u'uv)

For i = j = 0, we obtain that <st)`0 = <uv) and then by (15) and (9),

(16) <a'hi)`° = <c'(')dv(i)).

Now (15) and (14) imply that <a'bst>'* = <c"(')duv) and (15), (12), and (14) yield that(ast)10 = <cuv). Again by (9),

<a'-'b)`' = <a'bstast)4' = <c'(')duvcuv) = <c'(')-'d).

On the other hand, by (16) and (14), <a'-'h)`° = <c""-"d) Thus <c''")-'d) =<c''"-"d) and, since o(c) = o(d) = n, it follows that v(i - 1) = v(i) - 1. This holdsfor all i e Zn. Since v(0) = 0, a trivial induction yields that v is the identity. In exactlythe same way it may be proved that p is the identity; thus we have shown that

(17) v(i) = i = p(i) for all i e Zn.

Now let a: G -+ G be the isomorphism satisfying a" = c, s'= u, b' = d, and t' = v.We want to show that H4' = H" for all H < G and it suffices to do this for cyclic H.So let H = <g), g = a'bis''t' where i, j e Z. and h, k e {0, 11. If h = k = I orh = k = 0, then <g)4' = (g)' by (15) and (17), or (16) and (17) respectively. So let

h = 1, k = 0 (the case h = 0, k = 1 is similar). Then g = a'sb' a <a's> x <ba> andg2 = b2j since as is an involution. So if <b2'> = <by>, then <g> = <a's> x <ba>. By(11) and (17), (a's>" = <c'u> _ <a's>' and, since <b'> ' = <bj>' as already shown, itfollows that <g>4' = (g>'. And if <b2j> <by>, then <g> is the unique subgroupdifferent from <a's> <b2j> and <by> contained properly between <a's> x <ba> and<b2j>. Since these four groups are mapped in the same way by q and a, it followsthat <g>4' = <g>'. Thus p is induced by a.

We can now give our first general application of Theorem 7.7.5.

7.7.7 Theorem (Schmidt [1980a]). Let F be a subgroup of the group G with thefollowing properties.

(18) Z(F) = 1 and

F is generated by a set D of involutions satisfying DF = D.

(20) Let d, e e D and S = <d, e>. If o(de) 0 {1, 2, 3, 4, 6, oo }, then there exists a sub-group T of CG(S) isomorphic to S or to an infinite dihedral group and satisfyingSnT=1.

If cp is a projectivity from G to a group G, then F`0 is isomorphic to F.

Proof. We verify the assumptions of Theorem 7.7.5 for F, the projectivity 0 from Fto F`° induced by gyp, D and A = {<d>Id e D}. By (19), (6) and (7) hold. We have toshow that (8) is satisfied; then, since n NF(H) = n CF(d) = Z(F) = 1 and DF = D,

HeAit will follow from 7.7.5 (b) that F P = F`0.

We show first that p is 2-regular. For this let d, e e D. If o(de) 0 { 1, 2, 3, 4, 6, oc },then by (20) there exist two commuting involutions in G and hence cp is 2-regular, by7.7.3. The same is true if o(de) = 2, 4, or 6. So assume that o(de) E {3, oo} for any twodifferent elements d, e e D. Since F = <did n D> and Z(F) = 1, we see that IDS >_ 2.So if the group Go generated by the involutions in G were a P-group, it would lie inP(m, 3) for some m c N u {oc} and, as F < Go, this would contradict (18). Thus Go isnot a P-group and by 7.7.2 there exists a prime q such that IH4I = q for everysubgroup H of order 2 of G. Suppose, for a contradiction, that q 0 2. Then by 7.7.1,o(de) = 3 and <d, e> is nonabelian of order 6. Therefore <d, e>" lies in P(2,3) andcontains three subgroups of order q 2. Thus q = 3, <d, e>10 is elementary abelian oforder 9 and so <d>4 and <e>4 are commuting subgroups of order 3 in G. This holdsfor every pair d, e of different elements in D and, since D generates F, it follows thatF`0 is an elementary abelian 3-group. But then F e P(n, 3) for some n, contradicting(18). Thus q = 2 and

(21) p is 2-regular.

To prove (8) fort , let d e D, H = <e> e A, <d>4' = <x>, and S = (d, e>. Suppose firstthat o(de) 0 { 1, 2,3,4,6, oo}. Then by (20) there exists T < CG(S) such that S n T = 1and T ^- S or T is an infinite dihedral group. If T a- S, then by 7.7.6, 9 is inducedby an isomorphism on <S, T> S x T. In particular,

(22) there exists an isomorphism a: S -+ S10 such that X' = X`° for all X < S.

We show that this is also true if T is an infinite dihedral group. Indeed, by 7.6.2and 7.7.1, (S x T)" = S" x T" and T" T Let N g T such that TIN S. Then qinduces a projectivity from <S, T )/N S x (TIN) onto <S, T >4'/N' and by 7.7.6there exists an isomorphism y: SN/N (SN)"/N" such that (V/N) = V"/N° for allN < V < SN. If /1: S SN/N and S: S"N"/N" S" are the natural isomorphisms,then x = flyd is an isomorphism from S onto S" satisfying Xa = XPy6 = (XN/N)'6 =((XN)"/N")6 = (X"N"/N")6 = X" for every X < S. Thus (22) also holds in this case,and it is clear that (22) implies (8). Indeed, since d e S and H < S, the equalities<x) = <d)" = <d> imply that x = d°. Then (HO)" = (H")x = (Ha )d- = (Hd)a =(H d)" = (Hd)'', as desired.

Now suppose that o(de) c { 1, 2, 3, 4, 6, oc }. If o(de) = 1, then H = <d > and so(Hd)'' = H'' = (Hi')'. If o(de) = 2, S and SO are elementary abelian of order 4 andhence again (Hd)'l = HO _ (HO )x. If o(de) = 4, then S and SO are dihedral groups oforder 8. Here H and Hd are the unique minimal subgroups of S which together with

S

<d>

o'de) = 2

d

5

o(de) = 3

Figure 22

d

<d>

S

o(de)-4

<d> generate S. Then HO and (Hd)O have the same property in SO with respect to <x>and hence are conjugate under x. Exactly the same argument applies if o(de) = 6;here 4 of the 6 involutions in S\<de) together with d generate subgroups of order 6,4, and 2 of S. Similarly, if o(de) = co, S and SO are infinite dihedral groups. Sinceevery element in S\<de) has the form (de)'d (i e Z), there are exactly two subgroupsof order 2 in S which together with <d> generate S, namely <(de)d) = Hd and<(de)-' d) = H. Their images have the same property in S'' with respect to <d >O =<x> and hence are conjugate under x. Thus (HO)' = (Hd)0, as desired. Finally, sup-pose that o(de) = 3. Then, since cp is 2-regular, SO is nonabelian of order 6 and HOand (Hd)'" are the unique subgroups of order 2 different from <d> O in SO. Thus(HO)x = (Hd)'' and (8) holds in all cases.

In general, we shall use Theorem 7.7.7 in the situation that F = G. However, therewill also be an application in 7.8.17 in which F < G and, of course, we have thefollowing general consequence.

7.7.8 Corollary. Let H be a group generated by involutions and with trivial centre. IfK H or K is an infinite dihedral group, then G = H x K is determined by its sub-group lattice.

Proof. Let cp be a projectivity from G to a group G. Then by 7.6.2 and7.7.3, G =H° x K`° and cp is 2-regular. We apply 7.7.7 with F = H to prove that H19 -_ H. Itwill also follow that K' -_ K if K ^-, H; and by 7.7.1, K`' K if K is an infinitedihedral group. So, in both cases, G H x K = G, as desired.

By assumption, Z(H) = 1. If H E P(n, 3) for some n, then by 2.2.5, H1* e P(n, 3) and,since cp is 2-regular, it follows that HP _- H. So we may assume that H 0 P(n, 3). Thus(18) holds for F = H and (19) is satisfied if we take for D the set of all involutions inH. Finally, S = <d, e) < H for all d, e e D. So if K H, there exists T !:-t:: K < CG(S)such that T S; and if K is an infinite dihedral group, we may choose T = K. Inboth cases, (20) is satisfied and by 7.7.7, H' = H.

Holmes [1971] has shown that for two lattice-isomorphic Rottlander groups H,and H2, the groups H1 x K and H2 x K are lattice-isomorphic if K is an infinitedihedral group.

Multiply transitive permutation groups

If G is a permutation group on a set Sl and a e fl, we write G. for the stabilizer of ain G and a° for the orbit of a under G, that is, G. _ {g E GIa9 = a} and ac ={a9I g e G}. Then it is well-known that

(23) GQ=G,gforallgeGandIa'I=IG:G,ISo if G is transitive on S2, the set A = {G,Ia E Sl} of stabilizers of the points of Sl is aclass of conjugate subgroups of G. Recall that G is said to be k-transitive on I(k E N,k < I(lI) if for any two ordered k-tuples (al,...,a,,) and (fl1,...,/fk) consistingof distinct elements of Q, there exists g e G such that z? = fl, for i = 1, ... , k. Clearly,for k > 2, every k-transitive group is (k - 1)-transitive, and the stabilizer G, of apoint a in a k-transitive group G is (k - 1)-transitive on 0\{a}.

Now let G be 2-transitive on S and assume that ISlI > 3. We put A = {G,Ia e 01and want to look out for the assumptions of Theorem 7.7.5. Clearly (6) has to beassumed. If G, < M < G, there exists x e M such that ax 4 a and, since G, is transi-tive on 0 \{a}, it follows that M is transitive on 0. Thus for g e G, there exists y e Msuch that a93' = a; hence gy c- G, < M and so g = (gy)y-1 e M. Therefore M = Gand this shows that

(24) A is a conjugacy class of maximal subgroups of G.

In particular, (7) holds. Moreover, since G, is transitive on b2\{a} and IK2I > 3, wehave G, # 1. Since G is faithful on n, it follows that NG(G,) = G. and

(25) n NG(H) = n G. = I.HEA aco

Furthermore, if S2 is finite, (23) shows that

(26) I G.: G.# I =I S2I -

1 = I G : G, I - 1 for o. 1 e Q.

If G E P(n, p) for some prime p and n c- Fi u {oo }, then, since every p-subgroup of Gis normal in G and G, has trivial core, it follows that IG,I = q is the smaller prime

involved in G and so IGI = pq. By (26), q = p - 1 and hence p = 3. Thus we haveshown:

(27) If G is a P-group, then 152 = 3 and G S3.

Now suppose that 101 >_ 4 and q is a projectivity from G to a group G; considerH = G. e A. If H4 g G, then, since H is a maximal subgroup of G, it would followthat IG: H°I is a prime and, by 6.5.3, G/Hc would be a P-group. But since HG = 1,this would contradict (27). Thus H41 is not normal in G and hence H`°. Sowe have the following:

(28) If IQI > 4, then H`° for all H e A and n NU(H`°) = 1.HED

Now 7.7.5 and (23)-(28) suggest the following procedure to prove that G is deter-mined by its subgroup lattice.

7.7.9 Remark. Let G be a doubly transitive permutation group on a set fl, IfIl > 4,and assume that G is generated by a set D of involutions. If we want to prove that Gis determined by its subgroup lattice and q is a given projectivity from G to a groupG, we put A = {G,ja a S2} and try to verify the assumptions of Theorem 7.7.5. Byassumption and (24), (6) and (7) are satisfied. So we have to prove (8); then 7.7.5(a)together with (25) and (28) will yield that G - G. To do this, we let d e D, H = G. e Aand <d>41 = <x). If d e H, then x E H`° and hence (H4)x = H41 _ (H")", as desired. So

we are left with the case that d 0 H, that is, a" _ /3 with a f3 e S2. Since d is aninvolution, it follows that f3" = a and hence by (23),

(29) G = Gp, Gf = G and G" = G.

By (27), G is not a P-group and then 7.7.4 implies that

(30) (Gp)x=Gp.

Finally, x 0 G = NG-(GQ) by (28) since d 0 G,; thus

(31) (G4)x :7-c G;.

Since H = G. and H" = Gp, we have to show that (G)' = G.; then (8) will hold.

If G is 3-transitive, this is quite easy since the stabilizers G. and GB are character-ized in the interval [G/G,p].

(32) Let G be triply transitive on S2, IQ > 4 and let a, Q e S2 such that or ft. Then[GIG,,,] has precisely 3 atoms, namely G Gp, and a subgroup X satisfyingIX:G,pj=2.

Proof. Since G. is 2-transitive on S2\{a}, G,p is a maximal subgroup of G. and,similarly, of G. Furthermore, since G is 2-transitive, there exists g e G such thata9 = /3 and Y9 = a. Then g 0 G,p, but gz a G,p and, by (23), Gas = (G, n Gp)9 =G; n GJ = Gp n G, = Gp. It follows that J<g, G,p>: G,pj = 2 and, since g 0 G. andg 0 Gp, <g, G,p> is a third atom in [G/Gp]. Therefore to prove (32) we have to show

that if X is an atom in [G/Gap] and Ga 0 X Gp, then X = <g, Gap>. Since Gap ismaximal in G. and Gp, we obtain X. = X n G. = Gap = X n Gp = Xp. If X weretransitive on S2, there would exist x c X such that ax = fi and hence by (23), G 'ft =XQ = Xp = G. It would follow that Xa = Gap a X and the transitivity of X wouldimply that Gap = 1. But since G is 3-transitive, Gap is transitive on fl\ lot, #I andIS2\ {a, /3} I >- 2, a contradiction. Therefore X is not transitive on S2. Since Gap < X istransitive on S2\{a,/3} and X. = Gap = Xp, it follows that ax = /3 and /fix = a forevery x c X \Gap. Then xg c Gap and hence x c- <g, Gap). Thus X = <g, G.,>, as de-sired.

We can now give our first application of Theorem 7.7.5 to multiply transitivepermutation groups.

7.7.10. Theorem. Every triply transitive permutation group of degree at least 4 whichis generated by involutions is determined by its subgroup lattice.

Proof. Let G be 3-transitive on the set fl, IS2I > 4, D the set of involutions in G andcp, A, d, x, a, fi as in Remark 7.7.9. We have to show that (G)x = GB. By (32) thereare precisely 3 subgroups of G containing G.", as a maximal subgroup, namely G.",G,", and <d, Gap>41. Since, by (30), G.", is normalized by x, these subgroups have to bepermuted by x. Now x c- <d, Gap>w and, by (31), (G, ',*)x : G". It follows that (G,',')' =G , as desired.

The most interesting groups satisfying the assumptions of Theorem 7.7.10 are thesymmetric and alternating groups. More generally, if for every set S2 and infinitecardinal A we define Sym(S2, A) = {g e Sym f1l I {a E S21a9 a} I < Al, then thesegroups too fall under Theorem 7.7.10. We collect these and some other examples inthe following corollary. For the finite symmetric and alternating groups, Exercises 9and 10 show some other methods to prove the desired result.

7.7.11 Corollary. The following groups are determined by their subgroup lattices:(a) Sym S2 and, more generally, Sym(S2,A) for every set S2 with IS2I > 4 and every

infinite cardinal A.(b) AItQ2, Ifl I > 4.(c) PGL(2, F) where F is a field, I F1 > 3.(d) The five Mathieu groups Mil, M12, M22, M23, M24.

Proof. (a) Every permutation can be decomposed into cycles, every finite cycle is aproduct of transpositions and

(..., -1,0, 1,...) = [(0, 1)(-1,2)(-2,3)...][(0,2)(-1,3)(-2,4)...].

Thus Sym(S2, A) is generated by involutions and, of course, 3-transitive on SZ ifE21 > 4.

(b) As shown in § 1.4, A4 is determined by its subgroup lattice. In addition for101 >- 5, A1tS2 is 3-transitive and generated by involutions since it is simple andcontains involutions.

(c) It is well-known that PGL(2, F) is sharply 3-transitive on the projective line(see Huppert [1967], p. 151 or (33) below). We leave it as an exercise for the readerto show that every element in this group is even the product of two involutions.

(d) The Mathieu group M. is 3-transitive of degree n and simple (see Huppert andBlackburn [1982b], pp. 297-303), and hence is generated by involutions.

We come now to doubly transitive permutation groups generated by involutions.Here we need additional properties to prove that such a group is determined by itssubgroup lattice. Some of these occur in the following groups.

7.7.12 Example. Let V be a vector space of dimension n >- 2 over the field F. Thenthe general linear group GL(V) operates on the set 0 of 1-dimensional subspaces ofV and the kernel of this operation is the group of scalar transformations of V. LetG = PSL(n, F) = SL(V)/Z where Z is the set of scalar transformations of determi-nant 1, and let a, /3, y be three distinct elements of S2. Then we claim that

(a) G is a doubly transitive permutation group on S2 and(b) G., fixes at most one further point of 0, except when n = 2 and IFJ = 3.

Furthermore, if n > 3, then(c) G°p has exactly 4 orbits on fl and(d) there exists an involution t E G such that t = (afJ)(y)... and G = <t"Ix E G).

Proof. Let 6, E E S2 such that S E and let a = <u), /I = <v>, S = <w), E = <z). Thenu, v and w, z are linearly independent and there exists a E SL(V) such that u° = w,v° = cz with suitable c E F. Then g = aZ satisfies a9 = b, I'9 = E and thus G is 2-transitive on n.

To prove the other assertions, suppose first that n = 2 and I FI > 4. Then thereexist u,vE V and C E F such that a = ,/3= (v>,y= (u+v>,and0 c2 1. LetaESL(V)be definedby u°=cu,v°=c-'v.Then g=aZEG°#and y90y;thus (b)holds in this case. We mention in passing that if 6 E 4\{a,/3}, then b = <au + v) forsome a c- F\ {0} and there exists a e GL(V) satisfying u° = au, v° = v so that 2° = a,

= /3, and y° = S; this shows that PGL(V)°, is transitive on S2\lot, /#} and hence

(33) PGL(V) is triply transitive on f2 if n = 2.

Now assume that n > 3 and define Fa, _ {S E S2\{a,/3}I6 < a + /3} and &, ={b E Q\{a,Q}1S $ a + Q}. If y, 6 E F.., we may again choose u, v E V such that a =<u), /i = <v>, y = , 6 = <au + v> for some a c- F\ {0}; also there exists w E V

such that u, v, w are linearly independent. Then there are a, z c- SL(V) such thatu°=au,v°=v,w°=a-'w and ut=v,vt=u,wt=-w,rz=1.And if y,6EO°,,then a = , /I = <v), y = <w>, 6 = <z> where u, v, w and u, v, z are linearly inde-pendent. Then there are a, r a SL(V) satisfying u° = u, v° = v, w° = cz for suitablec e F, ut = v, vt = u, wt = -w, and t2 = 1. In both cases, g = aZ E G°, and satisfiesy9 = S, whereas t = xZ is an involution in G such that a` = /3, fl' = a, and y` = y. Itfollows that {x}, {Q}, IF,,,, and A,# are the orbits of G°, on 0 and since J0°,I > 1 thisalso implies (b). Finally, since G is simple (see Robinson [1982], p. 72), the conju-gates of t = (x/3)(y)... generate G.

We now show that if G is a finite 2-transitive group generated by involutions, then(b) of 7.7.12 yields that G is determined by its subgroup lattice.

7.7.13 Theorem (Schmidt [1975b]). Let G be a doubly transitive permutation groupon a finite set Q, IS21 = n > 4, such that the stabilizer Gap of two distinct pointsa, Q E Q fixes at most one further point. If G is generated by involutions, then G isdetermined by its subgroup lattice.

Proof. Let cp be a projectivity from G to a group G, D the set of involutions in G andA = {Gala e (2}. We first prove that

(34) (A")9 = A" for all g e G.

For this we have to show that HLc9 = H`°9`°-' e A for all H E A and g e G. Since thestabilizers of the points are maximal subgroups, their images generate G, and henceit suffices to show that HT(9) E A for all g e G,','(2 E S2). Clearly, Ga E A and fora $ e S2, we have G1(9) _ (Gp n Ga)`(9) = GpT(9) n G. If G were P-decomposable, itwould be a P-group since it is generated by involutions; but this contradicts (27).Thus G is not P-decomposable and, by 5.6.10, r(g) is index preserving. Thus

G(9): Gt(9) n Ga 1 =1 G(e): G=as) I= I G: G I= n - 1a a a as a a

by (26). This shows that the orbit of a under G,'(s) has length n - 1, and hence thereexists a unique point y c- Q not contained in this orbit. It follows that G'(9) < G. andthen G;(9) = G. since both groups are maximal subgroups of G. Thus A'(') = A and(34) holds.

To prove the theorem, we have to show that (G, 'P)' = G where d, x, a, $ are as inRemark 7.7.9. By (30) and (31), (G p)x = G,',', and (G.11)' G.". Since (A")x = A", itfollows that x permutes the Gy" (y c- S2) containing G. If Gap does not fix a third pointof fl, then G. and Gp are the unique elements of A containing Gap, and therefore(G,',*)' = G'. So suppose that there exists y e S2 such that a 0 y # 3 and Gap fixes y.Then by assumption, Ga, Gp, and G., are the unique elements of A containing Gap. By(29), Gp = Gap, and hence d operates on {Ga, Gp, G,,}. Since G = Gp and G° = GG, itfollows that G° = G. Then (24) implies that d c- GY and so x c- G'. Therefore (GP)' =Gf and hence again (G,")' = Gp, as desired. El

By 7.7.12, the groups PSL(n, q), n > 2, q a prime power, q > 4 if n = 2, satisfy theassumptions of Theorem 7.7.13. In fact, it is known that in every finite simple 2-transitive permutation group the stabilizers of two points fix at most one furtherpoint. Thus the unitary groups PSU(3, q2), q > 3 a prime power, the symplecticgroups Sp(2m, 2), m > 3, the Suzuki groups, the Ree groups, the Higman-Sims groupand Conway's group (.3) also fall under our theorem.

Futhermore note that the proof of (34) did not use our additional assumption onGap; nor did it really require that G be generated by involutions. Thus, in general, ifG is a finite doubly transitive permutation group of degree at least 4 and cp is aprojectivity from G to a group G, then G operates on A and it is possible to showthat G is 2-transitive on A. A consequence of this for soluble doubly transitivegroups is given in Exercise 5.

We now show that for a possibly infinite 2-transitive permutation group G, prop-erties (c) and (d) of 7.7.12 yield that G is determined by its subgroup lattice.

7.7.14 Theorem (Schmidt [1980a]). A doubly transitive permutation group G on a setQ, ICII >_ 4, is determined by its subgroup lattice if it has the following three properties:

(35) G is generated by involutions having fixed points on f2.

(36) If a, /3, y are three distinct elements of fl, there exists s e G such thats = (a)(fly)... .

(37) For or, /3 E 0 such that a 54- /3, Ga, has exactly 4 orbits on fl.

Proof. Let G be a group satisfying (35)-(37), D the set of involutions in G havingfixed points on f2 and cp, A, d, x, a, /3 as in Remark 7.7.9. We have to show that(G, )" = G"_ By (31), (GQ )" # Ga ; let F = G"9_'. Since d E D, there exists y E 0 suchthat d = (af)(y)... and by (36) there exists s e G such that s = (a)(/3y).... Let S =<s,Ga,), T = Ga,Y and W = <T,s,d). Then TS = T = T°, s2 E T, sd = (a/3y)... andhence (sd)3 E T; it follows that WIT is a dihedral group of order 6. Therefore d anddS are involutions such that W = <T,d,ds) and again by 7.7.4, T" a W". Thus cpinduces a projectivity from WIT onto W"/TI*, and hence

(38) W°/TI* is either elementary abelian of order 9 or dihedral of order 6.

Suppose, for a contradiction, that I W"/T`°I = 9. Then (<T, s)")" = <T, s>', and henceby (30), x normalizes (T, s, Ga,)° = <s, G8,)"' = S". Since s = (a) (fly)... and S"", ' _S, we have Ga, < S < G. and so S n S° < G. n GQ = G. Therefore S is not normalin <S, d) and, by 6.5.3, <S, d)/S(s.e> E P(2, p) for some prime p. It follows that Gaa =S n Sd a S, and hence by (23), GaQ = (Gafl)S = G.Y. Thus Gaff fixes y and therefore, by(37), has the four orbits {a}, { /3}, {y}, and S2\ {a, /3, y} = I' on (2. The orbits of F >_ Gaflon Q are joins of these orbits of Ga,. Since yd = y, we have x e G' and so

F9 = (F n G,)1* = F" n G"' = (GQ )" n (Gv)" = (Ga,)" = (G s)" = Gas,

by (30). Thus FY = GaR and, since Ga, < S < F, it follows that IyFI = IF: GaQI > 4.Thus yF n F Q and hence r s yF. Also /3 E yF since s = (a)(/3y)... E S < F. NextF G8, so S2\ {a} is not an orbit of F and, finally, y' = 0. Let g e F such that ya = a.Then GQ, = F9 = FF = F n Q. > S. However this is impossible since Ga, has 4 orbitson fl whereas S has at most 3 since it contains Ga, and s = (/3, y)... .

This contradiction shows that W"IT° is dihedral of order 6. Again by 7.7.4,<T, sd>" a W"' and hence o(x) = I< T, d)": T`°I = 2. The involution xT° permutes thetwo subgroups of order 2 of W"/T"' not containing xT". Thus (<T, s)")" = (<T, s)d)cOand then S = <T, s, Ga,) and (30) imply that (S`°)" = (S° )c. Therefore if S = Ga, we aredone; so suppose that S < G. By 7.7.2, cp is 2-regular, hence G is generated byinvolutions and not a P-group. Since (G,1,0)' = F"' and x is an involution, (G, ','r-) F4)"= G, n F° and then 7.7.4 yields that (Ga n F)" = Ga n F. It follows that Ga n F <G. n GQ = Ga,; on the other hand, by (30), Ga# < Ga n F and thus G., = Ga n F = Fa.Now let {a}, {f3}, r, and F2 be the orbits of G., on fl, and let y e F1, say. Since

s° = (f3)(ay)... E Sd = S`0X`°- < F, we see that F, s aF. If aF = {a} v I,1, then F. =G2 is transitive on aF\{a} and hence F is 2-transitive on or'; but as F. < S° < F, thiscontradicts (24). If /3 e aF, there exists g e F such that a9 = f3 and then G'9', = Fa =FF > S°. This is impossible since GQ, has 4 orbits on S2 whereas S° has at most 3.Since aF is a join of orbits of G2 , the only remaining possibility is that aF ={a} u F, u F2. Then fl\{aF} = { f3} is invariant under F, that is F = G,. Thus (G,,,')X= G.10, as desired.

7.7.15 Corollary. If n > 3 and F is any field, then PSL(n, F) is determined by itssubgroup lattice.

Proof. This follows from 7.7.12 and 7.7.14.

If F is a field in which - 1 is not a square, it is easy to see that PSL(2, F) does notcontain involutions with fixed points on Q. Therefore these groups do not satisfy theassumptions of Theorem 7.7.14. However it can be shown, using 7.7.5 directly, thatPSL(2,F) also is determined by its subgroup lattice if BFI > 3; see Schmidt [1980a]where this result and 7.7.15 are proved for arbitrary skew-fields F. Yakovlev [1986]studies the groups EL (R) generated by all transvections of dimension n over anassociative ring R with identity. He shows that such a group is determined by itssubgroup lattice if n > 4 and the additive group of R either contains an element ofinfinite order or is generated by elements of prime order.

Classical groups

The methods of this section can also be applied to the other classical groups, that isthe symplectic, orthogonal, and unitary groups over fields. We only state the resultsand give some brief comments on the proofs; for details see Schmidt [1980a].

7.7.16 Theorem. Let V be a finite-dimensional nonsingular symplectic, orthogonal, orunitary vector space over the field F. In any of the following cases, G is determined byits subgroup lattice.

(A) G = PSp(V), V symplectic, dim V > 12 if char F # 2, dim V > 4 if char F = 2.(B) G = PS1(V), V orthogonal, ind V > 12 if char F : 2, ind V > 6 if (char F = 2

or) - 1 is a square in F.(C) G = PSU(V), V unitary, ind V > 6 if char F # 2, ind V > 2 if char F = 2.

Sketch of proof. In all cases, G = G'/Z where G is the group of all isometries of Vand Z = Z(G). First let char F zA 2 and recall that the index of V is the dimension ofa maximal isotropic subspace of V. Thus our assumptions imply the existence ofhyperbolic planes H in V, that are subspaces H = <u, v> such that (u, u) = 0 = (v, v)and (u, v) = 1. For every such H, we let aH be the linear map on V satisfying x°N =-X for x e H and x°N = x for x c H'. Since V = H +Q H1, a, is an isometry of V andallZ is a well-defined involution in G. We let be the set of all hyperbolic planes inV, D = {arZIH E -5} and verify the assumptions of Theorem 7.7.7 for Gin place of

F. Since G is simple, G = (did e D>; so (18) and (19) are satisfied. If d = aHZ ande = aKZ are elements in D and S = <d, a>, then S is trivial on H' n K' and, sinceind V is big enough, it is possible to find subspaces X, Y of V and an isometry p ofV such that H + K <_ X, X 1 Y and X° = Y. Then it is easy to see that T = S°satisfies (20). So by 7.7.7, G is determined by its subgroup lattice. If char F = 2, wecan work with transvections, that is to say, maps a of the form x° = x + A(x, v)v forcertain ;. E F and v e V satisfying (v, v) = 0. Here aZ is an involution in G and a fixeselementwise the hyperplane <v>'. That is the reason why our method yields betterresults in this case.

Every finite classical simple group is determined by its subgroup lattice; see 7.8.1and 7.8.2.

Finite Coxeter groups

We finish this section with a result of quite a different flavour. Recall that a Coxetergroup is a group G generated by a set S of involutions such that the relations

(39) (st)m(s,') = 1 for (s, t) E I

are defining relations for G; here, for s, t e S, m(s, t) is the order of st and I is the setof pairs (s, t) for which m(s, t) is finite. The number BSI is called the rank of G; soCoxeter groups of rank 2 are just the dihedral groups.

7.7.17 Theorem (Uzawa [1986]). Every finite Coxeter group G of rank r > 3 is de-termined by its subgroup lattice.

Proof. Let rp be a projectivity from G to a group G and let S be as above. It iswell-known (see Suzuki [1982], p. 359) that every Coxeter group of rank r > 3contains two commuting involutions (even in S). By 7.7.3, cp is 2-regular and then7.7.1 shows that o(st) = o(xy) if s, t e S, <s>" = (x> and (t)" = (y>. Thus G isgenerated by a set of involutions satisfying the relations (39). Since these are definingrelations for G, there exists an epimorphism from G onto G. But a finite groupcannot be lattice-isomorphic to a proper factor group and it follows that G ^ G.

Exercises

1. (Uzawa [1986]) Let G be the dihedral group of order 2n and n = p;1 ... p'' theprime factorization with p, < . . < p,. Show that G is determined by its sub-group lattice if and only if one of the following holds:(a) n is even,(b) n is odd, e, > 1 and gcd(p, - 1,...,p, - 1) = 2, or(c) n is odd, e , = 1 and gcd(p,, p2 - 1, ... , p, - 1) = 1

2. (Schmidt [1980a]) Let G be a group and cp a projectivity from G to a group G.If there exist involutions a, b e G such that I (a)"I I <b)"I, show that G =S x T where S is a nonabelian P-group generated by involutions and S and Tare coprime.

3. (Schmidt [1980a]) Let G be a group, D a set of involutions and A a set ofsubgroups of G with the following properties.(a) G=<dldED)andG0P(n,3)forallnEN u{cc}.(b) AG = A and n NG(H) = 1.

HcA(c)If a E D and H E A, then K° = K for all K e A\{H, Ha} such that

HnH"<K<<H,H°).Show that if there exists a projectivity cp from G to a group G such that(A")G = AO, then G G.

4. (Schmidt [1975b]) Let G be a finite doubly transitive permutation group on theset 0, 1012!! 4, and let cp be a projectivity from G to a group G. Show that G canbe regarded as a doubly transitive permutation group on A = {Gala e S2}.

5. (Schmidt [1975b]) Show that for p' >- 4, the group I'(p') of all semilinear map-pings

(a OO,aEAutGF(p'))

on GF(p') is determined by its subgroup lattice. (Hint: Use Exercise 4 andHuppert's theorem on soluble doubly transitive permutation groups; seeHuppert and Blackburn [1982b], p. 379.)

6. (Schmidt [I975b]) Let p and r be primes, p 1 (modr), r > 2, S2 = GF(p')and S = F(p'). Let T= {x - x + bib e i2}, M = {x -* axIO a e S2} and A ={x -+ xaia e Aut f2}.(a) Show that S contains precisely r + 1 subgroups of index r, namely TM, a

subgroup containing A and r - 1 other subgroups, all 2-transitive on fl.(b) Show that any two of these r - I remaining subgroups are lattice-

isomorphic but not isomorphic.7. (Schmidt [1980a]) Show that a doubly transitive permutation group G on a set

fl, 101 >- 4, is determined by its subgroup lattice if it has the following twoproperties.(a) G is generated by involutions having fixed points on ft(b) If a, /3, y are three distinct elements of S2, there exists s e G such that

s = (a) (#y)... and G. = <s, Gae).

8. (Holmes [1971]) Show that if H1 and H2 are lattice-isomorphic Rottlandergroups and K is an infinite dihedral group, then H1 x K and H2 x K are lattice-isomorphic. (Hint: Use 1.6.1.)

9. (a) Show that the symmetric group S is the only group of order n! whichcontains a core-free subgroup of index n. Use 4.2.6 and 5.4.11 to conclude thatS. is determined by its subgroup lattice if n > 4.(b) Give a similar proof for A,,, n it 5.

10. Show that if D is the set of transpositions in the symmetric group S and d, e e D,then o(de) E { 1, 2,3 1. Use Theorem 7.7.7 to conclude that S is determined by itssubgroup lattice if n >- 4.

7.8 Finite simple and lattice-simple groups 439

7.8 Finite simple and lattice-simple groups

Many problems in group theory were solved by the classification of finite simplegroups. Among these is one of the most interesting problems on subgroup lattices ofgroups.

7.8.1 Theorem. Every nonabelian finite simple group is determined by its subgrouplattice.

Proof. Let G be a nonabelian finite simple group and let cp be a projectivity from Gto a group G. By 5.3.2 and 4.2.8, G is a simple group of the same order as G. Usingthe classification of finite simple groups, Kimmerle, Lyons, Sandling and Teague[1990] have shown that the only pairs of nonisomorphic finite simple groups of thesame order are PSL(3,4) and PSL(4,2) and, in the Lie notation, andfor some n > 3 and some odd q. In the same paper they show that PSL(3,4)and PSL(4, 2) have different numbers of elements of order 5, and that and

have different numbers of involutions if n > 3 and q is odd. Since cp is indexpreserving, this cannot happen for G and G. It follows that G G. (Note that allthe exceptional groups mentioned above are classical simple groups, so that themethods of § 7.7 show that they are determined by their subgroup lattices; seeRemark 7.8.2 below.)

Elementary proofs

Unfortunately, a direct proof of Theorem 7.8.1 is not known. But more elementaryproofs are available for certain classes of finite simple groups. For doubly transitivesimple groups, in particular the alternating and the projective special linear groups,Theorem 7.7.13 yields the desired result. But the methods of § 7.7 also work underweaker assumptions than 2-transitivity and therefore it is possible to handle all theclassical finite simple groups in this way. We only sketch this briefly; for details seeSchmidt [1977b].

7.8.2 Remark. (a) A primitive permutation group G on a finite set S2 is determinedby its subgroup lattice if it has the following four properties.

(i) G is generated by involutions.(ii) The stabilizer G, of a point a E S2 has precisely two orbits on 0\{7} (that is, G

is a rank 3 group) and these have different lengths; let A, be the larger orbit.(iii) G = <G,,I a e 12, P e A,>.(iv) The stabilizer of two different points fixes at most one further point.

The proof of this theorem uses 7.7.5 and elementary properties of rank 3 groups.(b) Let V be a nonsingular symplectic, orthogonal, or unitary vector space of

dimension n over the field GF(q) with q elements, and assume that one of the follow-ing holds:

(A) V symplectic, n > 4, q 0 3, G = PSp(V) = PSp(n,q).(B) V orthogonal, n > 7, q odd, G = PQ(V) = Pfl(n, q).

(C) V orthogonal, n = 2m > 8, q even, G = PS2(V) = Pc(n, q).(D) V unitary, n > 4, q a square, G = PSU(V) = PSU(u, q).

In any of these cases, G is a primitive permutation group on the setf) = { <v> 10 v e V, (v, v) = 0}, respectively 0 = { <v> 10 : v e V, f (v) = 0} in case(C) where f is the quadratic form associated to V. By Witt's theorem, the stabilizerG. of a point a = <v> E SZ has the orbits Al = { <w> e S2I(v, w) = 0} andA2 = { (w) I(v, w) = 1 }, and G also satisfies the other assumptions in (a).

(c) The classical simple groups not included in (A)-(D) are isomorphic either togroups on the list or to groups PSL(2, q) and are therefore covered by (a) or Theo-rem 7.7.13. The only exceptions are the groups PSp(2n, 3) in which the stabilizer oftwo points fixes two further points; they are handled in Exercise 1. Thus all theclassical finite simple groups are covered by Theorem 7.7.5 and its consequences.

An immediate consequence of Theorem 7.7.7 is that every finite simple groupgenerated by a conjugacy class D of {3,4}-transpositions, that is, satisfying o(de) < 4for all d, e e D, is determined by its subgroup lattice. This covers all the Chevalleyand Steinberg groups over GF(2) and the three Fischer groups.

Finally, there is a short general argument for arbitrary groups of Lie type, which,however, uses a deep theorem of Tits'.

7.8.3 Theorem (Li [1983]). Every finite simple group G of Lie type of rank n > 2 isdetermined by its subgroup lattice.

Proof. Let cp be a projectivity from G to a group G. Then G operates on L(G) via themap T, sending every x e G to the autoprojectivity T(x) of G defined by H`(") =Hl" I" " for H < G. Let p be the characteristic of the ground field and B the normal-izer of a Sylow p-subgroup of G; recall that the parabolic subgroups of G are thesubgroups containing some conjugate of B, and that the set A of all parabolicsubgroups together with the dual of the inclusion relation is the building of G. By4.2.8, every autoprojectivity of G is index preserving and hence maps Sylow p-subgroups to Sylow p-subgroups. Since the normalizer of a Sylow p-subgroup P isthe largest subgroup of G containing P but no other Sylow p-subgroup, T(x) per-mutes the normalizers of the Sylow p-subgroups and therefore maps parabolicsubgroups to parabolic subgroups. So T(x) induces an automorphism o(x) of thebuilding A of G. If x e (B9)`°, then T(x) fixes every subgroup containing B9, and hencea(x) is type preserving. Since G is generated by the (B9)0, we conclude that o is ahomomorphism from G into the group of all type preserving automorphisms of A.By a well-known theorem of Tits [1974], this group contains a normal subgroupG* G with soluble factor group. Since G is simple, it follows that G is isomorphicto a subgroup of G. But G and G are lattice-isomorphic finite groups; thus G G, asdesired.

Unfortunately, all these partial results cannot prove 7.8.1. No simple direct proofof this theorem, without using the classification of finite simple groups, is known; itwould be highly desirable to have such a proof even if it used the Feit-ThompsonTheorem.

The second main problem, whether or not every nonabelian finite simple group isstrongly determined by its subgroup lattice, is much easier to solve. This is equiva-lent to asking the question if, for every nonabelian finite simple group G,

(1) P(G) = PA(G).

Indeed, (1) is satisfied for every group G which is strongly determined by its sub-group lattice. Conversely, suppose that (1) holds for every nonabelian finite simplegroup G. Then, since every projective image G of G is simple, 1.4.3 and (1) wouldimply that

Aut G PA(G) = P(G) ^- P(G) = PA(G) Aut G;

and, since a nonabelian simple group is isomorphic to the unique minimal normalsubgroup of its automorphism group, it would follow that G ^- G. So we wouldobtain, without using 7.8.1 or the classification of finite simple groups, that G isdetermined by its subgroup lattice, and (1) then says that every projectivity of G isinduced by an isomorphism. We mention in passing that the weaker property

(2) PI (G) a P(G),

for every nonabelian finite simple group G, would suffice to give an elementary proofof Theorem 7.8.1 (see Exercise 2). Metelli [1971] proved that all the minimal simplegroups that is, nonabelian finite simple groups all of whose proper subgroups aresoluble, satisfy (1) and therefore also (2). However, we are now going to show thatneither property holds for arbitrary finite simple groups. For this, surprisinglyenough, it suffices to study the alternating groups. But for completeness sake we willalso treat the finite symmetric groups and give some brief comments on infinitesymmetric and alternating groups.

Symmetric and alternating groups

In the sequel let n >_ 4, S2 = { 1, ... , n} and G = Sym S2 = S. or G = Alt D = A.; write

G; = {g E Gli9 = i} and A = {G;Ii E S2}

so that G. is the stabilizer of the point i e S2 in G and A is the set of all stabilizers. Wewant to investigate the group P(G) of autoprojectivities of G. The idea is to showthat (except when n = 6) P(G) operates on A, and then to study this operation. Forthis purpose we have to establish some basic properties of A.

7.8.4 Lemma. If M < G such that I G : M = n, there exists x E Aut G such thatM.

Proof. Let { Mx,,..., Mx } with x = I be the set of cosets of M in G and considerthe permutation representation a: G -+ S. of G on this set: g' maps i e S2 to j c S2 ifand only if Mx;g = Mx1. Since A. is simple for n > 5, n Mx = 1; this also holds for

xeG

n = 4 since I M I = 3 or 6 in that case. So a is a monomorphism and, since G is theunique subgroup of its order in Sn, it follows that G' = G, that is, x E Aut G. Nowg e M if and only if Mg = M and this holds if and only if n9' = n. Thus M' = Gn, asdesired.

Let D be the set of 3-cycles in S. Since (12n) 0 and An_, is a maximalsubgroup of An, a trivial induction yields that

(3) An = <(123),(124),...,(12n)>.

A product of two 3-cycles has one of the following four types

(4) (123)(124) = (14)(23), (123)(142) = (234), (123)(145) = (12345), (123)(456).

Again a simple induction yields the following result.

(5) Ifl <m<-n-3and x...... that o(xixj)=2for i0j,then M. . . . . . . . X , , > has at least n - m - 2 fixed points on 0.

Indeed, this is obvious for m = 1; so suppose that m > 2 and <x...... x,n_, > has atleast n - (m -1) - 2 = n - m - 1 fixed points. Since o(x,x,n) = 2, we see from (4)that two of the points moved by x,n are also moved by x,. It follows that<x,,..., has at least (n - m - 1) - 1 fixed points.

7.8.5 Lemma. If n 0 6, G has only one conjugacy class of subgroups of index n(namely A).

Proof. Assume first that G = A. and, since the assertion is obvious for n = 4, thatn > 5. We show that

(6) D' = D for all a e Aut G.

To see this, first note that by (4), D' is a conjugacy class of elements of order 3 of Gsuch that o(de) < 5 for all d, e e D. Suppose, for a contradiction, that D' 0 D. Thenthere exists x e D'\D. Since o(x) = 3 and x 0 D, the cycle decomposition of xcontains at least two 3-cycles. If x has a fixed point, so that we may assume x =(123)(456)(7)..., then for g = (34)(67) E G, y = x9 = (124)(357)(6)... E D' and xy =(1473256) ... has order divisible by 7, a contradiction. On the other hand, if x hasno fixed points, then n 6 implies that x has the form (123) (456) (789) ... , say;once again, if g = (34)(67), then y = x9 = (124)(357)(689)... E D' and now xy =(147932586)... has order divisible by 9. This contradiction shows that D' = D.

Now let M be a subgroup of index n in G. By 7.8.4 there exists a e Aut G such thatG = M and, by (3), G = An_, _ <(123),(124),. .. , (12 n -1)>. If xi = (12 i + 2)', thenby (6) and (4), xi E D and o(xix;) = o((12i+2)(12j+2)) = 2 for i 0 j. Thus by (5), thegroup M = G' = <x 1, ... , xi_3 > has at least n - (n - 3) - 2 = 1 fixed points on 0.So M < Gi for some i and then M = Gi since both groups have the same order. Thisproves the lemma for G = An.

If G = S. and M <- G such that I G : MI = n, then I An: M n Anl = n or n/2. Sincethe simple group A. does not admit a homomorphism into the symmetric group Skfork < n, it follows that IAn: M n AnI = n. So M n A. = Gi n A. for some i, as we have

just shown. Now Ana G, so M j4 G. implies that M n A = Gi n A o M u G. = G,a contradiction. Thus M = Gi, as desired.

7.8.6 Lemma. If n = 6, G has exactly two conjugacy classes F and A of subgroups ofindex 6; forMEVandNEA,IM:MnNJ=6.

Proof. Let S = S5 and T = <(12345),(1243)> < S. Then ITI = 20 and henceIS: TI = 6. Thus the permutation representation lI of S on the cosets of T is ahomomorphism from S into G = S6 with Ker/I = n Tx = 1. It follows that

xeSSO S = S5, and so I G : SO 1 = 6. If { Tx1,... , Tx6 } are the cosets of Tin S, then thestabilizer of i in SO is SO n G. _ (Tx')fl, and hence SO n G. T Since Gi n GG S4 fori : j, this shows that SO 0 A = {Gili e S2}. Clearly, SO n A6 is a subgroup of index 6in A6 which intersects Gi in a subgroup of order 10. So there exist at least twoconjugacy classes of subgroups of index 6, both in S6 and in A6.

Now let G = S6 or A6 and M, N< G such that I G : MI = I G : N J = 6. We claimthat

(7) 1M: M n NI = 6 if M and N are not conjugate.

To see this, consider the operation of M on the set A of cosets of N in G. Thestabilizer of the coset N in M n N and hence I M: M n N I 5. Also IM: M n NJ = 5 would imply that Mhas one orbit of length 5 and therefore also one fixed point on A. Thus M < Nx forsome x e G and, since both groups have the same order, it would follow thatM = Nx, contradicting our assumption. So IM: M n NI = 6 and (7) holds.

Now let M1, ..., Mk be representatives of the conjugacy classes of subgroups ofindex 6 in G; thus k > 2 as we have seen. Since 311MiI, there exists x, e Mi such thato(xi) = 3. If xi were conjugate to x; for some i 0 j, then x9 = x; E M9 n M; for someg e G; but by (7), I M9 n M;I is not divisible by 3, a contradiction. Since G has onlytwo conjugacy classes of elements of order 3, with representatives (123) and(123)(456), we obtain that k = 2. The second assertion of 7.8.6 follows from (7).

7.8.7 Remark. By 4.2.8, every autoprojectivity a of G is index preserving. So 7.8.5shows that for n : 6, a induces a permutation Q of A. The map a -+ 6 is clearly ahomomorphism from P(G) into Sym A S. with kernel

K = {a e P(G) I G' = Gi for all i e 01.

By the homomorphism theorem, P(G)/K is isomorphic to a subgroup of Sn.

To study the kernel K, we need the following simple result.

7.8.8 Lemma. Let n, k, q be natural numbers such that kQ - 1 (mod n) and k' * I(mod n) for 1 < r < q.

(a) If x = (1, ... , n) is a cycle of length n and s: S2 -a S2 is defined by i 5 = ik (mod n)foriES2={I__ nl, then seSn,o(s)=gandxs=xk.

(b) If k = n - I in (a), then s e A. if and only if n - 1 (mod 4) or n - 2 (mod 4).(c) If y e S. and T E Aut<y>, there exists t c S. such that o(t) = o(r) and y` = yT.

Proof. (a) Clearly, i" - ik' (mod n); since k' - 1 (mod n), it follows that s9 is theidentity on 0. Therefore s is a permutation and, since k' # 1 (mod n) for 1 < r < q,we have o(s) = q and is-' - ik9-' (mod n). Thus is-'"s - (ik9-' + 1)k - i + k - ik(mod n) and hence s-' xs = xk.

(b) If k = n - 1, then s interchanges i and n - i for all i = 1, ... , n - 1 and hences is a product of (n - 1)/2 or (n - 2)/2 transpositions according as n is odd or even.These numbers are even if and only if n - 1 (mod 4) or n - 2 (mod 4), respectively.

(c) Let Y = Y1...Yd be the cycle decomposition of y and suppose that y' = yk =Yi .. Yd, k E N. If yi is a cycle of length m and q is the smallest natural numbersatisfying k9 - 1 (mod m), then qI o(r). By (a) there exists si E Sn, operating only on them points moved by yi, such that o(si) = q and yj' = y'. Then t = sl ... sd E S satisfiesy` = yk = y` and o(t) = o(r).

7.8.9 Lemma. If G = S. and a E P(G) such that (G1)° = Gi for all i e S2, then a = 1.

Proof. We have to show that <g)° = <g> for every g E G and use induction on thenumber v(g) of points moved by g to prove this first for involutions g. If g is atransposition (j), then <g> = n Gk and hence <g>° = <g>. So suppose that

i#k#jv(g) > 2. Then the cycle decomposition of g yields involutions gi such thatv(gi) < v(g) and g = 9192 = 9291. By the induction assumption, <gi)° = <gi> andhence <g1,g2>" = (g1,g2>; thus the third subgroup <g> of order 2 of the four-group <g1, 92 > is invariant under a.

Now let g E G such that o(g) > 2. By 7.8.8 there exists s E G such that s2 = 1 andg' = g-'. Then H = <g, s> is generated by involutions and hence H° = H. Sinceevery element in H\<g> has order 2, each maximal subgroup of H different from <g>is generated by involutions. Thus a fixes all these maximal subgroups and hence also(g)

7.8.10 Theorem (Suzuki [1951a], Zacher [1965]). Let n e N, n >- 4.(a) If n : 6, z S..(b) P(S6) = PA(S6), PI(S6) S61 I P(S6) : PI(S6)I = 2.(c) Every projectivity of S. is induced by a unique isomorphism.

Proof. (a) By 7.8.7 and 7.8.9, is isomorphic to a subgroup of S. On the otherhand, since 1, it follows from 1.4.3 that S,,. Thus P(S°) = S..

(b) By 7.8.6, G = S6 has two conjugacy classes F and A of subgroups of index 6.If M, N E A and a e P(G), then M n N ^ S4 and, since a is index preserving,M' n N° = (M n N)° has order 24. By 7.8.6, M° and N° are conjugate. It followsthat A° = A or A° = F, and hence I P(G) : LI < 2 if L = IT e P(G)jAt = Al. By 7.8.9,L operates faithfully on A and so is isomorphic to a subgroup of S6. On the otherhand, S6 ^J PI(G) < L and hence L = PI(G) S6. By 7.8.4, P1(G) < PA(G) and so,finally, PA(G) = P(G) and I P(G) : PI(G)I = 2.

(c) If cp is a projectivity from G = S. to a group G, then by 7.7.11, G G. By (a) or(b), (p is induced by an isomorphism a which is unique, by 1.4.2 and 1.4.3.

Note that by 1.4.3, Aut S,,; thus 7.8.10(a) implies that Aut S S. forn # 6. It is well-known (see Huppert [1967], p. 199) that A6 PSL(2,9). ThereforeS6 is a subgroup of index 2 of Aut A6 n-, PI'L(2, 9), and hence this group operatesas a group of automorphisms on S6. So 7.8.10(b) implies that P(S6) = PA(S6)Aut S6 PTL(2, 9).

It is quite obvious that we could handle the alternating groups in the same way asthe symmetric groups if we could prove the analogue of Lemma 7.8.9. However thisis only possible for certain integers n.

7.8.11 Lemma. Let n >_ 5, G = A. and a E P(G) such that (Gi)° = Gi for all i e n. Ifx c G such that o(x) = p' (p c P, r c J) and <x>° -t- <x>, then

(a) p = 3,(b) x fixes at most one point of S2,(c) x is a cycle (of length 3'),(d) r is odd.

Proof. (a) We show first in several steps that most cyclic subgroups of G areinvariant under a. So let g E G.

(8) If g is a 3-cycle, then <g>° _ <g>.

Indeed, if g = (i, j, k), then <g> is the intersection of all the G. with m E Sl\{i, j, k}.

(9) If g2 = 1, then <g>° = <g>.

As in the proof of 7.8.9, we use induction on v(g) to prove this. If v(g) = 0, then g = 1,while v(g) = 2 is impossible since G = A. So let v(g) = 4 and g = (12)(34), say. Sincen >_ 5, the subgroups H = n Gi and K = ((125)> are invariant under a. Clearly,

i>5H ^-, A5 and <(125),(12)(34)> = NH(K) is invariant under a since it is the uniquesubgroup of order 6 between H and K. The same argument applies to the group<(345),(12)(34)>, and it follows that a also fixes the intersection <(12)(34)> of thesetwo groups. Now let v(g) > 4. Then the cycle decomposition of g yields involutionsgi E G such that 4 < v(gi) < v(g) and g = 9192 = 9291. By induction, <gi>° = <gi>and hence <91,92>° = <g1,g2>; therefore the third subgroup (g> of order 2 of thefour-group <g1,g2> is invariant under a.

(10) If there exists an involution s e G such that gS = g-1, then <g>° = <g>.

Indeed, <g, s> and all maximal subgroups of <g, s> different from <g> are generatedby involutions; thus a fixes all these groups and hence also <g>.

(11) If g' = 1, then <9>° = <g>.

Again we use induction on v(g) to prove this. By (8), (11) holds if v(g) = 3. If v(g) = 6,then g = (123)(456), say, and s = (12)(45) E G satisfies gS = g-1; thus (10) impliesthat <g>° = <g>. Finally, if v(g) > 6, the cycle decomposition of g yields elementsgi E G of order 3 with disjoint supports such that g = 919293. By induction, <g1 >,<9293>, <9192>, <93> are invariant under a. Since <g1,g2g3> and <g192,93>are different elementary abelian groups of order 9 containing g, it follows that

<g> = (<91 > v <9293>) n (<9192> v <g3>) is invariant under a.

(12) If o(g) = 2", then <gy' = <g>.

Let k > 2 and g = g1... g,, be the cycle decomposition of g. Since every cycle of evenlength is an odd permutation, m is even. By 7.8.8 there exist involutions s; E Soperating on the points moved by gi such that g'! = g.-'; let s = s, ... If one of thegi is a transposition, then either s E G or sgi e G, and g' = g-' = g'gi since gi central-izes g; by (10), <g>c = <g>. If none of the gi is a transposition, then v(gi) = 0 (mod 4)and hence, by (b) of 7.8.8, si G for i = 1, ..., m. Since m is even, it follows that s E Gand <g>6 = <g>, again by (10).

(13) If o(g) = p' and p > 3, then <g>a = <g>.

We use induction on o(g) to prove this. So suppose that (13) holds for elements ofsmaller order than g and let r be an automorphism of order p - I of <g>. By 7.8.8there exists t c- S such that o(t) = p - I and g' = gr. Since p > 3 and IS.: GI = 2, wehave <t> n G 1; let s e <t> n G be of prime order q, say. Then <g, s> and allmaximal subgroups of <g, s> different from <g> are generated by subgroups of orderq. By (9), (11), or induction, all these groups are invariant under a and then so is <g>.This proves (13); clearly, (12) and (13) imply (a).

(b) If g c- G fixes two points i and j of Q, and s E S. is the involution satisfyinggs = g-' constructed in 7.8.8, then s or (Q )s is an involution in G inverting g. Thus(10) implies that <g>d = <g> and this proves (b).

(c) We show first that the cycle decomposition of x contains at most two cycles.To see this, suppose that g c- G such that o(g) = 3' and g = 819293 where the gi : Ihave disjoint supports. Then o(g) is the least common multiple of the o(gi)and so 3' = o(g) = o(g2) > o(gi) (i = 1, 3), say. Let S = <g1g2> v <g3> andT = <91> v <9293). If y c S n T, then y = (g192)i93 = 9',(9293)' for i, j, k, m E N.Since the product of the <gi> is direct, it follows that i = m (mod 3') and g3 = 93 = 93Thus y = (8,8283)', E <g> and so S n T = <g>. The 3-elements g1, 93, 9192, 9293 allhave more than one fixed point on fl; therefore by (b), a fixes S and T and hence <g>.

Now let g = 8192 with disjoint cycles g1 of length 3' and g2 of length 3'. If r = k,the involutions si inverting gi constructed in 7.8.8 are either both odd or both even;in any case, s = s1s2 E G and by (10), <g>° = <g>. So let r > k. Consider the semi-direct product H = N of an elementary abelian group N = <a, > x x <a3, ,of order 331-' and a cyclic group of order 3r-1 with respect to the operationa, = ai+, 3'-' - 1), ail-Ib = a, (note that H C3 wr Car- i; see (14) of § 7.6).Then c = a, b satisfies c3' = a,... air-i and hence o(c) = Y. If M is a complementto S2(<c>) in N, then M<c> = H and M n <c> = 1; furthermore, Ml = 1 sinceZ(H) = S2(<c>). Thus H operates faithfully on the 3' cosets of M in H and c is a cycleof length 3'. Since all cycles of length 3' in S are conjugate, there exists a subgroupS C3 wr C31-1 of G containing g, and having the same support as g,. Let Abe the base group of S and T = A<g>. Then S = A<g1>; and Tn <g1,g2> =<g>(A n <g1,g2>) = <g> since A is elementary abelian and intersects <g2> triviallyso that An <g1,g2> = S2(<g>). Since r > k, we have T n <g2> = <g> n <g2> = 1and T <g2> = <A, g1, g2) = S x <g2>. Thus T is isomorphic to S and hence is gen-erated by elements of order at most 3'-'. Since n > 3r + 3k, every such element either

has two fixed points on 0- as do g, and g2-or has three cycles in its cycledecomposition. By (b) and the first part of the proof of (c), a fixes T and <g g2>,and hence also T n <g g2> _ <g>. Thus x is a cycle.

(d) If r were even, then 3' - I (mod 4) and hence by 7.8.8 there exists an involutions c- G inverting the cycle x of length 3'. By (10), <x> = <x>, a contradiction. Thus ris odd.

7.8.12 Theorem (Schmidt [1977a]). If n > 5, n : 6 and not of the form n = 3'or n = 3' + 1 where r is odd, then PA(A,,) S,,; furthermore P(A6) =PA(A6) PFL(2,9). Every projectivity of these alternating groups is induced by aunique isomorphism.

Proof. Let G = A,, and K = {a c- P(G) I G° = G; for all i c- 0}. If K : 1, there wouldexist a c K and x e G of prime power order such that <x>a # <x>. By 7.8.11, xwould then be a cycle of length 3', r odd, with at most one fixed point, and it wouldfollow that n = 3' or n = 3' + 1, a contradiction. Thus K = 1 and by 7.8.7, P(G) isisomorphic to a subgroup of S,,. On the other hand, S. is a subgroup of Aut A.PA(AJ < P(G), and it follows that P(G) = PA(G) S,,. The proofs of the otherstatements are similar to the corresponding proofs for S. in 7.8.10 and are left to thereader.

Using Sadovskii's Local Theorem, Corollary 1.3.7, we can extend the results of7.8.10 and 7.8.12 to infinite sets.

7.8.13 Corollary. If X is an infinite set, then the alternating group AR X and thegroup of finitary permutations Sym(X, N,) are strongly determined by their subgrouplattices.

Zacher [1977] has shown that for infinite X, every normal subgroup of Sym X isstrongly determined by its subgroup lattice; see Exercise 7.

We now treat the cases n = 3' and n = 3' + 1, r odd.

7.8.14 Lemma. Let n = 3' or n = 3' + I (r > 2) and suppose that X is a cyclic sub-group of order 3' in G = 5,,. If X < H < G, then 1.

Proof. Suppose that the lemma is false and choose a minimal counterexampleH < G; write H; for the stabilizer of the point i c- S in H. Then H = QX where Q 1

is a normal 3'-subgroup of H. If q is a prime dividing IQ I, then by 4.1.3 there existsan X-invariant Sylow q-subgroup of Q, and the minimality of H implies that Q is aq-group. If n = 3', then X is transitive on S and hence so is H. Therefore I H : Hil = 3'and so Q < H; for all i e 0. It follows that Q = 1, a contradiction. Thus n = 3' + 1.If H is not transitive on S2, then H has orbits of lengths 3' and 1 and, as above, Q = 1.So H is transitive on S2 and then 3' + 1 = n = q' for some m since X is containedin the stabilizer H. of some point i c Q. It follows that q = 2 and m < 2, since3' + 1 ° 2 or 4 (mod 8). Thus r < 1, contradicting our assumption r > 2.

7.8.15 Theorem. Let n = 3' or n = 3' + 1, r odd, r > 3 and let G = A. Then PA(G)and P1(G) are not normal in P(G).

Proof. Let x = (12 ... 3') be a cycle of length 3' in G and X = <x>. If y e Cs,(X)and I' = 1, then ii' = IX'- 'y = 1YX'-' = Ix'-' = i for all i e { 1,..., 3'}, and hence y = 1.Since X < Cs,(X), this shows that CC,(X) operates regularly on { 1, ... , 3r } and so hasorder 3'. Thus Cs,(X) = X. Therefore Ns.(X)/X and NG(X)/X are isomorphic tosubgroups of Aut X, and this group is cyclic of order 2.3r-1 (see Huppert [1967], p.84). By 7.8.8, Ns,(X)/X Aut X; since r is odd, 3' - 3 (mod 4) and therefore theinvolution s satisfying x' = x-1 constructed in 7.8.8 is not contained in G. SoNG(X)/X is cyclic of order 3r-1. Let H be the subgroup of order 3r+1 between X andNG(X). Then X is a cyclic subgroup of index 3 in H, and hence the structure of H iswell-known. By 2.3.4 and 2.3.5, for example, H is an M-group and I H : Q'._1(H) J = 3.So H has 3 cyclic subgroups of order 3'; let A be the set of these cyclic maximalsubgroups of H. Then we claim the following.

(14) If YeAand Y<M<G,then H<M.To prove this, suppose, for a contradiction, that H -4 M. If Y = <y>, then y is a cycleof length 3' since n < 3' + 1. So Y is conjugate to X in S. and hence NA(Y)/YNG(X)/X is cyclic of order 3r-1. Thus H is the unique subgroup of NG(Y) in which Yis a maximal subgroup. Since H M, it follows that Y = NM(Y) and hence Y is aSylow 3-subgroup of M contained in the centre of its normalizer. By a theorem ofBurnside (see Robinson [1982], p. 280) there exists a normal 3-complement Q in M.By 7.8.14, Q = I and hence Y = M, a contradiction. Now we can construct auto-projectivities of G which are not induced by automorphisms.

(15) If r is a permutation of A, then a: L(G) -- L(G) defined by Y' = Y` for Y e Aand Y° = Y for Y 0 A is an autoprojectivity of G.

Of course, a is bijective. Let U < V< G. If U, V 0 A, then U° = U < V = V. IfV e A, then U° = U < b(V) = D(H) < V°; and if U E A, then by (14), H < V andhence U°< H < V = V°. Thus in all cases, U < V implies that U° < V°. The sameholds for a-1 and so a is a projectivity.

An automorphism x inducing this projectivity a has to fix every involution in G.Since G is generated by involutions, it follows that x = 1; but this is impossible ifr 0 1. Thus a is not induced by an automorphism and hence P(G) :?-4 PA(G). But wecan improve on this. Since G is simple, there exists g c G such that H9 H. Let it bethe autoprojectivity of G induced by the inner automorphism induced by g; son c P1(G). Then X°"-' 0 A and hence X°-"n-Iff" = X°-'"-"N = X°-' X if we chooser c- Sym A such that X` X. Thus a ' n-1 Qn 0 1. On the other hand, if U < G suchthat I UI = 2, then U '" °" = U"-" = U, since a fixes all subgroups of order 2. Soagain an automorphism x of G inducing the projectivity a-1 n-1 Qn would have to fixevery involution in G and would therefore be the identity, a contradiction. Thusa-1 n-1 air 0 PA(G) and, since it E PI(G) < PA(G), this shows that PI(G) and PA(G)are not normal in P(G).

The following theorem gives more information about the structure of P(A") in theexceptional cases; the proof is left to the reader (see Exercises 4-6).

7.8.16 Theorem (Schmidt [1977a]). Let n = 3' or n = 3' + 1, r odd, r > 3 and letG = A,,. Write K = {a e P(G)IG7 = G; for all i c- S1}, and let L be the set of allv e P(G) fixing every noncyclic subgroup of G. Then K and L are normal subgroups ofP(G) and

(a) P(G)/K S.,(b) K/L is isomorphic to a subgroup of a direct product of groups isomorphic to S3

(it is not known whether K L),(c) L is the direct product of (3' - 1)!/2.3' (if n = 3') or (3' - 1)!(3' + 1)/2.3' (if

n = 3r + 1) groups isomorphic to S3.

Simple groups of Lie type

A large number of simple Chevalley groups have been shown to be strongly deter-mined by their subgroup lattices. This was done for the simple groups PSL(2, q) byMetelli [1968], [1969], and for PSL(3, 3) and the Suzuki groups in Metelli [1971].For a Chevalley group G of rank at least 2, P(G) operates as a group of auto-morphisms on the building A of G, as was shown in the proof of Theorem 7.8.3; andif K is the kernel of this operation, it follows from Tits' theorem that P(G)/KAut G. So it remains to study this kernel K. It was proved by Volklein [1986a] thatK = I for the groups of type B,, C,, D21, 2D21, 3D4, G2, F4 if p > 3 and of type E7, E8if p > 7. Similar results have been obtained by Costantini [1989b] for simple alge-braic groups over the algebraic closure of the prime field GF(p). These groups arestrongly determined by their subgroup lattices for all odd primes with the soleexception of the groups of type A2. These groups are in fact also exceptional in thefinite case. Volklein [1986a], [1986b] and Costantini [1990] show that the groupsPSL(3, q) in general are not strongly determined by their subgroup lattices;PSL(3, 17) is the smallest example of this kind (see Costantini [1989a]).

Lattice-simple groups

A group G is called lattice-simple if for every subgroup H of G such that 1 < H < G,there exists a e P(G) such That H° H; that is, G does not contain any nontrivialproper P(G)-invariant subgroup. Obvious examples of such groups are the directproducts of isomorphic simple groups: these groups are even characteristicallysimple (see Robinson [1982], p. 84). In particular, every elementary abelian groupand therefore, by 2.2.3, every P-group is lattice-simple. By 1.2.7, the subgroup latticeof a cyclic group of square-free order is a direct product of chains of length one, andhence such groups are lattice-simple. We now show that, conversely, every finitelattice-simple group is one of the examples mentioned above.

7.8.17 Theorem (Suzuki [1951a], Schmidt [1978]). The finite group G is lattice-simple if and only if it has one of the following properties.

(i) G is a P-group.(ii) G is cyclic ofsquare free order.

(iii) G is a direct product of isomorphic (nonabelian) simple groups.

Proof. It remains to be shown that every finite lattice-simple group G has one of theproperties (i)-(iii). To see this, consider the set 97l of all modular subgroups of primeorder in G and let L = <MIM E 972). Clearly, L is P(G}invariant and hence L = Gor L = 1.

Suppose first that L = G. If there exists M E 972 which is not permutable in G, thenby 5.1.9, G = MG x K where M' is a P-group and (IMGI, IKI) = 1. By 4.2.4, M' isP(G)-invariant. It follows that G = M' and (i) holds. So suppose that every M E 972is permutable in G. Then by 5.2.9, L = <MGI M E 9J1> is the product of nilpotentnormal subgroups, and hence G = L is nilpotent. Since 'D(G) is P(G)-invariant,(D(G) = 1. Therefore if G is a p-group, (i) holds. If G is not a p-group and P is anontrivial Sylow p-subgroup of G, there exists a e P(G) such that P° 0 P. Since P isthe unique Sylow p-subgroup of G, it follows that P° is not a p-group. But P° isnilpotent and hence by 2.2.6, P is cyclic. So G is cyclic and of square-free order since(D(G) = 1. Thus (ii) holds.

Now suppose that L = 1, that is, G has no modular subgroup of prime order. LetN be a minimal normal subgroup of G. Then N is not cyclic and hence by 5.4.4, N°is a minimal normal subgroup of G for every a E P(G). Clearly, M = <N°l a E P(G)>is P(G)-invariant and hence M = G; it follows that G = N x N°' x x N forcertain a; E P(G). As a minimal normal subgroup, N is a direct product of isomor-phic simple groups; since it is a direct factor of G, it follows that N is simple. By 7.8.1,all the N are isomorphic to N and so G satisfies (iii). However, we need not use theclassification of finite simple groups here; using the Feit-Thompson Theorem only,we can verify the assumptions of Theorem 7.7.7 for N and a = a, (i = 1, ... , r) inplace of F and cp. Indeed, since N is simple, Z(N) = 1 and N is generated by involu-tions. Let D be the set of all involutions in N, let d, e c- D and S = <d, e>. By 4.2.8, ais index preserving since G is a direct product of simple groups. So 7.7.1 shows thatS° ti S and, clearly, S° < No < CG(S). Thus T = S° satisfies (20) of § 7.7. By Theorem7.7.7, N° N and G satisfies (iii).

Exercises

1. (Schmidt [1977b]) Show that for n> 2, the group G = PSp(2n, 3) =Sp(2n, 3)/Z(Sp(2n, 3)) satisfies the assumptions of Theorem 7.7.5 if one takesD = {aZ(Sp(2n, 3))1a2 = 1 # a E Sp(V)} and A = {Gala E S2} in the notation of7.8.2(b).

2. Let G and G be nonabelian finite simple groups such that P1(G) 9 P(G) andP1(G) a P(G). If L(G) L(G), show that G ti G (without using 7.8.1, of course).

In the next three exercises let n = Y or n = 3' + 1, r odd, r >_ 3 and G = A,,.3. Show that CP(G)(PI (G)) : 1 (compare with the well-known result that

CA,,,X(Inn X) = 1 if Z(X) = 1).4. Let S E Sy13 (G) and write K(S) for the set of all a e P(S) fixing every subgroup of

exponent at most 3'-' of S. Let a E K(S) and X < S such that X° # X.(a) Show that X is metacyclic of exponent 3'.(b) If X is not cyclic, show that there exist K(S)-invariant subgroups U, V, W of S

such that U < V < W, U <X < W, and IW/UI =9.

(Hint: Note that S is an iterated wreath product of cyclic groups of order 3; seeRobinson [1982], p. 41.)

5. If H < G and a e K (defined in 7.8.7) such that H° 0 H, show that H is a meta-cyclic 3-group of exponent Y.

6. Prove Theorem 7.8.16.7. (Zacher [1977]) Let X be an infinite set and G a nontrivial normal subgroup of

Sym X. Show that every projectivity of G is induced by a unique isomorphism.(Hint: Consult Scott [1964], pp. 305-315 for the basic properties of Sym X. Inparticular, note that G = Alt X or G = Sym(X, A) for some infinite cardinal A sothat by 7.7.11, G is determined by its subgroup lattice.)

Chapter 8

Dualities of subgroup lattices

We say that a group G has a dual if there exists a duality from its subgroup latticeonto the subgroup lattice of a group G, that is, a bijective map 6: L(G) -* L(G) suchthat for all H, K E L(G), H < K if and only if Kb < H. Not every group has a dual,for instance the quaternion group has none, and it is an interesting problem todetermine all groups with duals_

In 1939 Baer proved that every group with dual is a torsion group and that anabelian group has a dual if and only if it is a torsion group whose primary compo-nents are finite. In 1951 Suzuki determined the nilpotent and the finite solublegroups with duals. It turned out that the latter are precisely the projective images offinite abelian groups. In 1961 Zacher showed that every finite group with dual issoluble and determined the infinite soluble groups with duals. Again these are pro-jective images of abelian groups; as torsion groups, they are locally finite. In 1971Zacher proved conversely that every locally finite group with dual is soluble. Thusthe structure of these groups is known and we present it in this chapter.

It should be mentioned that all these results will be proved using elementarymethods, except the last one, that every locally finite group with dual is soluble. Thisproof uses the Feit-Thompson Theorem. Finally, the Tarski groups are examples ofgroups with duals which are not locally finite and the structure of an arbitrary groupwith dual is not known.

8.1 Abelian groups with duals

In this section we determine the abelian groups with duals and also give somegeneral results on dualities between subgroup lattices. First of all, we have thefollowing "dual" to Theorem 1.1.2.

8.1.1 Lemma. Let 6 be a bijective map of a lattice L to a lattice L. Then the followingproperties are equivalent.

(a) For all x, y e L, x < y if and only if xa > ya(b) (x n y)a = xb u yb for all x, y e L.(c) (xuy)a=x'3nybforallx, y E L.

Furthermore, if 6 satisfies (a) and S is a subset of L such that n S exists, then U Sbexists and (n S)a =U Sa; similarly (U S)° = n Sb if U S exists.

8.1 Abelian groups with duals 453

Proof. Let x, y e L. By 1.1.1, the three statements x < y, x n y = x and x u y = y areequivalent. Therefore if (b) is satisfied, x < y if and only if x° = (x n y)° = x° u y°,and this holds if and only if x° > y°. Thus (b) (and similarly (c)) implies (a). Con-versely, if S satisfies (a) and S is a subset of L such that z = n S exists, then z° > s°for every s e S. Hence z° is an upper bound of S°. If w is any upper bound of S°, thenw > s° and so w°-' < s for all s e S. It follows that w°-' < n S = z and hence w > z°.This shows that z° is the least upper bound of S° and that (n S)° =U S°. In particu-lar (b) holds. The corresponding result for U S follows similarly.

A bijective map S: L L with one and therefore all of the properties (a)-(c)of 8.1.1 is called a duality from L to L. For brevity, if G and G are groups andb: L(G) L(G) is a duality, we also call b a duality from G_onto G; and we say thatG has a dual if there exists a duality from G to some group G. It is clear that not everygroup has a dual; for instance the quaternion group and a quasicyclic group havenone, since a group G with a maximal subgroup containing every proper subgroupof G is cyclic. It is also not difficult to see that the infinite cyclic group has no dual.We prove a more general result.

8.1.2 Theorem (Baer [1939b]). If a group G has a dual, then G is a torsion group.

Proof. Suppose, for a contradiction, that z is an element of infinite order in G. LetZ = <z> and Z; = <z2' > for i c- N. Then Z > Z, > Z2 > .. is a descending chain ofsubgroups of G such that n Z, = 1. By 8.1.1, Z° < Z; < ZZ < . is an ascending

ic rIchain of subgroups of G whose join is 1° = G. Consider the subgroup W = <z3> ofZ. Since Z covers W, we see that Z° is a maximal subgroup of W° and thereforeW° = <Z°, w> for some w e W°\Z°. Since G = <Z°li E > and the join of an ascend-ing chain of subgroups is the set-theoretical union of these subgroups, there existsk c- J such that w e Zk°. It follows that W° = <Z°, w> < <Z°, Zk) = Z,° and henceW > Zk. But z2k 0 <z3> = W, a contradiction.

Finite abelian groups

We want to show next that every finite abelian group A has a dual. For this we haveto consider the character group of A.

8.1.3 Lemma. Let A be a finite abelian group, written multiplicatively, and let A* =Hom(A, C*) be the set of all homomorphisms from A into the multiplicative group C*of complex numbers. For a, T E A*, define c o T E A* by x°°` = x°x` (x c A).

(a) Then (A*, o ) is a group isomorphic to A; it is called the character group of A.(b) If 1 x c- A, there exists a e A* such that x° 1.

Proof. (a) It is trivial to verify that (A*, o) is an abelian group. As a finite abeliangroup, A = <a, > x x <an> for certain a, E A. Suppose that o(a;) = n, and let r, bea primitive n.-th root of unity in C. For every y = air ... a map

454 Dualities of subgroup lattices

XY: A C* by

(a ... ax")X' = E1 IY1. . . E."Y" (xi E 77).

The reader may easily verify that XY E A* and that the map x: A -+ A* sending y e Ato xY is a homomorphism. If a E A*, then (a°)"' = (a°')° = 1° = 1 for all i = 1, ..., n.Therefore a° is an n.-th root of unity in C and there exists y; E 77 such that a' = &Y-.It follows that

(ai' ... Ei'x' (aj ...

where y =ail ... E A. Thus X is surjective. And if z = a71... E Ker x,, then1 = a!- = E;' for i = 1, ... , n. Since E; is a primitive n;-th root of unity, it follows thatn; divides z; and hence a;' = 1. Thus z = 1 and x is injective. So, finally, x is anisomorphism between A and A*.

(b) Let x = ai' ... a Then there exists i e { 1, ... , n} such that a; ' 1 and, sinceo(a;) = o(s1), it follows that a = Xa, satisfies x° = E; 1.

8.1.4 Theorem (Baer [1937]). A finite abelian group A is self-dual, that is, there existsa duality from A onto A.

Proof. We show first that there exists a duality from A onto its character group A*.For X < A and S< A*, we define

X1={aEA*Ix°=IforallxEX} and ST={aEAla°=l for allaES}.

Of course, X1 and ST are subgroups of A* and A, respectively, so thatl: L(A) -- L(A*) and T: L(A*) -+ L(A) are well-defined maps. If x e X and a E X1,then x° = 1. This shows that x E (X1)T and hence X < (X1)T. If z e A\X, thenzX 0 1 in AIX and by (b) of 8.1.3 there exists T E (A/X)* such that (zX)` 0 1. So ifa: A -+ AIX is the natural epimorphism, then a = aT E A* satisfies x' = xa` = P = 1for all x e X and z° = (zX)` 0 1. Thus a E X1 and hence z (X1)T. This shows that(X 1)T = X and that 1 is injective; for, X; = XZ implies X, = (X; )T = (X . )T = X2.Since A and A* are isomorphic, !L(A)I = IL(A*)I and it follows that 1 is bijective.Clearly, X < Y if and only if X1 > Y1. Thus 1 is a duality from A onto A*. And ifp is the projectivity induced by an isomorphism from A* to A, it is obvious thatb = 1p is a duality from A onto A.

Basic properties of dualities

To determine the infinite abelian groups with duals we need some general propertiesof dualities between subgroup lattices. These, of course, will also be used in the studyof arbitrary locally finite groups with duals. The first two results are rather obvious.

8.1.5 Lemma. (a) Suppose that Lo, ..., L. are lattices and that for i = 1, ..., n themapping Q;: L;_, - L. is a duality or an isomorphism. Then p = a,... Q" is an isomor-

phism (respectively a duality) from Lo onto L if the number of dualities among the o;is even (respectively odd).

(b) Let G and G be groups and S a duality from G onto G. If a E P(G), then6`66 e P(G); if r e P(G), then STb-' e P(G). In particular, if x c- G, then the mapv: L(G) -+ L(G) given by H" = ((Ha-')" )a for H < G is an autoprojectivity of G;similarly, for u c- G, the map p: L(G) -+ L(G) given by K" = ((K')') a_' for K < G is anautoprojectivity of G. We call v the autoprojectivity of G induced by x c- G (via S) andusually write v = S-'x(; similarly, p = Sub-' is the autoprojectivity of G induced byuEG(via 6).

Proof. (a) As a product of bijective maps, p is bijective. Every duality among the aiinverts the order relation. So it will finally be preserved if the number of dualitiesamong the a, is even, and it will be inverted if this number is odd.

(b) Application of (a) to S-'a8 and STS-' yields the first statement of (b). Then ifwe apply this to the autoprojectivities of G and G induced by the inner automor-phisms given by x and u, respectively, we get the second statement.

8.1.6 Lemma. Let S be a duality from G onto G, 1 < H < K < G and suppose thatH a K and Kb a Ha. Then b induces a duality from K/H onto H8/K'. In particular, ifN _a G, then S induces a duality from GIN onto Na.

Proof. Clearly, the restriction /3 of S to [K/H] is a duality from [K/H] onto[Ha/Ko]. If a: L(K/H) - [K/H] and y: [Ha/Ka] -+ L(Ha/Ka) are the natural isomor-phisms, then by 8.1.5, a/3y is a duality from K/H onto Ha/K6 and (X/H)"PY = Xa/Kafor H < X < K. This proves the first statement of the lemma. If H = N g G = K,then Ka = 1 a Ha and so S induces a duality from GIN onto Na.

The following result is an important special case of the above situation.

8.1.7 Lemma. Let S be a duality from G onto G and let X < G. If X is invariant underP(G), then Xa is invariant under P(G) and S induces dualities in X and in G/X.

Proof. If T E P(G), then by 8.1.5, STS -' E P(G). Since X is invariant under P(G), wehave X = Xa`a-' and then Xa = (X°)T. Thus Xa is invariant under P(G). In particular,X a G and Xa g G. By 8.1.6, S induces dualities from G/X onto Xa and from X onto

61V.

We finally reduce our main problem to the study of groups with directly indecom-posable subgroup lattices.

8.1.8 Lemma. Let (GA)AEA be a family of coprime torsion groups and let G = Dr GA._ AEA

(a) If there exists a duality S from G onto a group G, then G = Dr Ga where

GA

zEA= U Gµ and the Gz are coprime torsion groups. In particular, S induces a duality

µ#zfrom G/GA GA onto Gs, that is, every GA has a dual.

(b) Conversely, if every GA is self-dual, then so is G.

Proof. (a) By 8.1.1, U Gx = n Gx = la = G and U d'n GA =ZeA AEA #Z

a

n Gµ v G, _ (Gz v Gz)° = G° = 1. Since GA and Gz are coprime and G =µ#A

Gx x Gz, it follows that G. is the unique complement to G, in G. If x E G6, then dxdis an autoprojectivity of G fixing Gx and hence also GA. It follows that (G°)x = G.Thus Gz a G and G = Dr G. If two of these groups, Gz and Gµ, say, were to contain

AEA _elements of the same prime order p, there would exist a subgroup P of order p of Gneither contained in GA nor in Gz. But then P°-' would be a maximal subgroup ofG neither containing Gx nor Gz and this would contradict 1.6.4. Thus the Gz arecoprime.

(b) Suppose that (L1)AEA is a family of lattices, L = Dr Lx and, for every ). E A, 8Zzen

is an autoduality of LA. For f e L define f a E L by f a(A) = f (2)a& (A e A). Since all the8,i are bijective, so is b: L -. L. Further f < g with f, g e L, if and only if f(A) < g(A)for all A E A; moreover this is the case if and only if f(A)" >_ g(2)ax for all A E A, thatis, f a > gb. Thus S is an autoduality of L. Now by 1.6.4, L(G) Dr L(G,A). Since

zenevery L(GA) has an autoduality, so does Dr L(GA) and then also L(G).

AEA

We remark that part (a) of Lemma 8.1.8 follows from a more general property ofdualities between lattices (see Exercise 2) together with 1.6.4 and 1.6.5. It is an openproblem whether (b) holds under the weaker assumption that every Gx has a dual.

Infinite abelian groups

By 8.1.8 we need only study abelian p-groups; the crucial case is handled by

8.1.9 Lemma. An infinite elementary abelian p-group has no dual.

Proof. Let P be an infinite elementary abelian p-group and let B = {aAJ2 e Al be abasis of P; write r = I A I. Then P = Dr <a,> and every element of P is a product

IEnof finitely many different elements of B* = E A, 0 < k < p - 1}. Hence thereexists an injective map from P into the set of finite subsets of B* and it follows thatIPI <- I kl IPIBI = J BI = r. In particular, P has at most r cyclic subgroups; on the otherhand, P contains all the <a'> (A E A) and therefore P has precisely r minimal sub-groups. Every maximal subgroup M of P determines exactly p - 1 nontrivial homo-morphisms of P into a cyclic group Z of order p having M as their kernel. Thus thenumber of maximal subgroups of P is IHom(P, Z)I. Since B is a basis of P, every mapfrom B into Z can be extended to a unique homomorphism from P to Z. ThereforeI Hom(P, Z)I = plBI = 2' and we have shown that

(1) P has r minimal and 2' maximal subgroups.

Now suppose that b is a duality from an elementary abelian p-group G onto a group6. If 1 j4 u e G, then by 8.1.6, 6 induces a duality from G/'-' onto . Thus

G/<u)a-' is an elementary abelian p-group with distributive subgroup lattice andhence is cyclic of order p. So

(2) every nontrivial element of G is of prime order.

Let 1 v E G such that <v> and put W = <u)a-' n <v)a-'. By (2), '-' and<v)a-' are two different maximal subgroups of G, and hence G/W is elementaryabelian of order p2. Again by 8.1.6, b induces a duality from G/W onto Ws = <u, v)and, since G/W is self-dual, <u, v> is a projective image of G/W. By 2.2.5,

(3) <u, v) E P(2, p) if 1 0 u, v e e; and <u) <v>.

Let H = { x E G I o(x) is a power of p}. If x, y E H, then by (2), either I <x, y) l divides por x 1 y and <x> <y>. In this case, (3) implies that <x, y) is elementary abe-lian of order p2 since a nonabelian group in P(2, p) has only one subgroup of orderp. In any case, we see that o(x) divides p, xy-' e H and xy = yx. Thus

(4) H is an elementary abelian p-subgroup of G.

Clearly, H a G. By 8.1.6, 6 induces a duality from Ha-' onto the p'-group 61H, and(3) applied to this duality yields that G/H is cyclic of prime order.

Now suppose, for a contradiction, that G is infinite. Then H too is infinite and 6induces a duality bo from the infinite elementary abelian p-group P = G/Ha-' ontoH. Let r be the number of minimal subgroups of P. Then by (1), P has 2' maximalsubgroups. These are mapped by So onto the minimal subgroups of H. Therefore,again by (1), H has 2Z" maximal subgroups and these are mapped by So' onto theminimal subgroups of P. It follows that r = 22', a contradiction.

It follows from 8.1.9 that every abelian p-group with dual is finite. Indeed, weprove a much more general result.

8.1.10 Lemma. Every locally finite p-group with dual is finite.

Proof. Let G be a locally finite p-group and S a duality from G to some group G.Since finite p-groups are nilpotent, G is locally nilpotent. Therefore every maximalsubgroup of G is normal in G (see Robinson [1982], p. 345) and hence G/ b(G) iselementary abelian. By 8.1.6, G/(D(G) has a dual and therefore is finite. Thus G =H.V(G) for some finitely generated subgroup H of G. Since G is locally finite, H isfinite. If H < G, then H6 > I and by 8.1.2 there would exist a subgroup M6 of primeorder contained in Ha. Then M would be a maximal subgroup of G containing Hand it would follow that G = HD(G) < M, a contradiction. Thus G = H is finite.

We can now prove our final result on abelian groups with duals.

8.1.11 Theorem (Baer [1939b]). An abelian group G has a dual if and only if G is atorsion group and every primary component of G is finite. In this case, G is evenself-dual and every dual of G is lattice-isomorphic to G.

Proof. Suppose first that G has a dual. Then by 8.1.2, G is a torsion group andtherefore is the direct product of its primary components. By 8.1.8, every such com-ponent has a dual and hence is finite, by 8.1.10. Conversely, let G = Dr Gp with finite

pEPabelian p-groups Gp. By 8.1.4, every Gp is self-dual and then so is G, by 8.1.8. Now8.1.5 shows that if b is a duality from G to a group G and 9 is an autoduality of G,then 9S is a projectivity from G to G.

Exercises

1. (Baer [1937]) In the notation of the proof to Theorem 8.1.4, show that for everyX < A, X A*/X1 -_ A/Xa and AIX X' Xa.

2. Let (L; )A E s be a family of complete lattices; write IA, OA for the greatest and leastelements of LA, respectively. Let L = Dr LA and, for all ). E A, define gA E L by

.LEA

gA(p) = I. for p i) and gA(a) = OA. If b is a duality from L onto a lattice L, and0 is the least element of L, show that the map cp sending every x E L to thefunction x`° defined by xw(A.) = x n gx (A, E A) is an isomorphism from L ontoDr [gz/0].AEA

8.2 The main theorem

By 8.1.5, every projective image of a group with dual has a dual. Therefore theresults of § 8.1 and Chapter 2 yield that the following groups have duals.

8.2.1 Theorem. Let G = Dr GA with finite coprime groups GA, and suppose that everyAEA

GA either is a P-group or a nonhamiltonian p-group with modular subgroup lattice.Then G is self-dual.

Proof. By 2.2.3 or 2.5.9, GA is lattice-isomorphic to a finite abelian group. By 8.1.4,every GA is self-dual and then so is G, by 8.1.8.

The aim of this chapter is to prove that, conversely, every locally finite group Gwith dual has the above structure. We state this result in a form which also describesthe image of G under the duality.

8.2.2 Main Theorem (Suzuki, Zacher). Let G be a locally finite group and S a dualityfrom G to a group G. Then there exist subgroups GA of G with the following properties.

(a) G = Dr GA and the GA are coprime.AEA

(b) G = Dr Ga where GA = U G,, and the Gz are coprime.*AACA

(c) For every ). e A, there are primes p, q (not necessarily distinct) such that

8.2 The main theorem 459

(1) Gx is a cyclic p-group, G° is a cyclic q-group; or

(2) G;, and G$ are in P(n, p) for some n e N; or

(3) GA and Gx are finite nonhamiltonian p-groups with modular subgrouplattices.

As mentioned in the introduction of this chapter, Suzuki [1951a] proved thistheorem for finite soluble groups and Zacher [1961a] extended it to arbitrary solu-ble groups; later Zacher [1961b], [1966], [1971] showed that, respectively, everyfinite, locally soluble, and, finally, locally finite group with dual is soluble. We shallprove directly in § 8.3 that an arbitrary soluble group with dual has the structuregiven in 8.2.2, and then in §§ 8.4 and 8.5 we shall study finite and locally finite groups.In addition, we shall prove in § 8.5 the following result on arbitrary groups withdual. Recall that GL' is the largest locally finite normal subgroup of the group G; letG' (respectively G°) be the join of all finite (respectively soluble) normal subgroupsof G and write G;j (respectively G;) for the intersection of all subgroups of finiteindex (respectively normal subgroups with soluble factor group) of G.

8.2.3 Theorem (Stonehewer and Zacher [1994]). Let 6 be a duality from the group Gto the group G. Then

(a) G" = G;j = Ga is a perfect group and has no proper subgroup of finite index,(b) G' = G5,(c) (G")a = G' and (G')a = G",(d) Z(G") = G" n G' and hence G"G' /Z(G") = G"/Z(G") x G'/Z(G"), and(e) G"/Z(G") and G'/Z(G") are coprime.

It follows from 8.1.6 that 6 induces dualities from GIG" onto G' and from G' ontoG/G", so that these groups have the structure given in 8.2.2. Furthermore 6 inducesa duality between the perfect groups G"/Z(G") and G"G'/G' G"/Z(G"). The struc-ture of these groups is not known: every Tarski group is self-dual and, using atheorem of Obraztsov [1990], it is easy to construct many other infinite simplegroups with duals.

Consequences of the main theorem

In the remainder of this section we give some corollaries to the Main Theorem andhandle some special cases. These corollaries will follow once 8.2.2 has been proved,but of course they will also hold in the inductive situations of this proof for dualitieswhich are not counterexamples and therefore satisfy 8.2.2.

First of all, if we are only interested in the structure of a group with dual, we havethe following more elegant version of the statements in 8.2.1 and 8.2.2.

8.2.4 Theorem. The locally finite group G has a dual if and only if G is a directproduct of finite coprime groups G. such that every Gx is either a P-group or a non-hamiltonian p-group with modular subgroup lattice.


By 2.4.5, every finite group with modular subgroup lattice is metabelian. There-fore all the G. in 8.2.4 are metabelian and then so is G. By 2.2.3 and 2.5.9, everyG. is lattice-isomorphic to an abelian p-group where p is a prime dividing GAI.

Thus these abelian groups are coprime and, by 1.6.4, their direct product is lattice-isomorphic to G. Using 8.1.11, we get the following result.

8.2.5 Corollary. Every locally finite group G with dual is metabelian and lattice-isomorphic to an abelian group. Thus G is self-dual and every dual of G is lattice-isomorphic to G.

Finally, every GA is supersoluble and a nonnormal Sylow subgroup of G,; hasprime order. So we get the following result.

8.2.6 Corollary. Let G be a finite group with dual. Then G is supersoluble and everynonnormal Sylow subgroup of G has prime order.

Special cases

In the remainder of this section let G be a group and 6 a duality from G to a group6. We start with the proof of the Main Theorem and handle some special cases.

8.2.7 Lemma. If G is abelian, d satisfies 8.2.2.

Proof. By 8.1.11, G = Dr GA with finite p-groups GA. By 8.1.8, G = Dr d' whereE A A E A

G; = U G,,, the images G are coprime torsion groups, and d induces a duality fromu#

G!G;, GA onto Ga; again by 8.1.11, Gx is lattice-isomorphic to G.. So (a) and (b) of8.2.2 hold and, by 2.2.6, (1) or (2) is satisfied, or 16'1 I = G;, I. In this case, G. and Gaclearly are nonhamiltonian p-groups with modular subgroup lattices for some primep. Thus S satisfies 8.2.2.

8.2.8 Lemma. If G is a P-group, then G and G lie in P(n, p) for some n E N, p e P.

Proof: By 2.2.3 there exists a projectivity cp from G to an elementary abelian groupG*. Then (P `8 is a duality from G* to G and, by 8.1.9, G* is finite. Thus G* E P(_ n, p)for some n e N, p e P. By 8.1.11, G is lattice-isomorphic to G*; by 2.2.5, G and G_ liein P(n, p).

8.2.9 Lemma. If G is a locally finite p-group, b satisfies 8.2.2.

Proof: By 8.1.10, G is finite and we use induction on IGI to show that 6 has therequired properties. By 8.2.7, we may assume that G is not abelian; in particular, Gis neither elementary abelian nor cyclic and hence

(4) ((G) y-L 1, 1 G : 1(G)I = pt, m > 2.

8.2 The main theorem 461

By 8.1.6, d induces a duality from G/D(G) onto D(G)6 and. 8.2.8 shows that(D(G)' E P(m, p). Suppose, for a contradiction, that (D(G)' is not a p-group. Then'D(G)' is a nonabelian P-group of order p'°-'q, p > q e P. If J(D(G)j > p, there existsa minimal normal subgroup N of G such that 1 < N < (D(G). Since GIN is neithercyclic nor elementary abelian, the induction assumption implies that G/N and Naare p-groups; but '1(G)6 < Na is not a p-group, a contradiction. Thus J(D(G)j = p andIGI = pm+'

By 8.1.7, (D(G)' < G. Let P be the Sylow p-subgroup of (D(G)a. Then P is character-istic in (D(G)' and hence normal in G; let Ma be a minimal normal subgroup of Gcontained in P. Since L(G) satisfies the Jordan-Dedekind chain condition, so doesL(G) and hence, by 5.3.9, G is supersoluble. Thus IMbI = p and I M I = p'. By 8.1.6, ijinduces a duality from M onto 6/Ma which, by induction, satisfies 8.2.2. Since(D(G)'/M' is a normal subgroup of order p n-2q of G/Ma, the group 6/Mb is not ap-group, nor does it lie in P(m, p). Thus M and G/Ma are cyclic of prime power order.This implies that m = 2 so that I G I = p3 and 161 = pq2.

The Sylow q-subgroups of G are cyclic and intersect trivially since (D(G)' containsnonnormal subgroups of order q. Thus 1(G) = 1 and hence G is generated by mini-mal subgroups. Since p > q >_ 2, it follows that G has exponent p; but M is cyclic oforder p2. This contradiction shows that

(5) D(G)' is a p-group.

It follows that every minimal subgroup of G has order p, and Sylow's theoremimplies that G is a p-group. By 2.1.8, L(G) is lower semimodular and so its imageL(G) under the duality S' is upper semimodular. Thus L(G) is lower and uppersemimodular and hence modular, by 2.1.10. If G were hamiltonian, then by 2.3.12,G = Q x A where Q QB and d induces a duality from G/A Q8 onto A'. Then Abhas only one maximal subgroup, hence is cyclic, and L(A°) is a chain; but this is notthe case. Thus Gis a nonhamiltonian p-group with modular subgroup lattice. Thesame is true for G, so (3) holds and d satisfies 8.2.2.

Finally, we have the following rather obvious general remark.

8.2.10 Lemma. Let G = A x B with coprime groups A and B. If d induces dualities inG/A and in G 'B which satisfy 8.2.2, then S satisfies 8.2.2.

Proof. By 8.1.8, G = Aa x B° where Ab and Ba are coprime. By assumption, G/A B

and Aa are direct products of groups satisfying (1), (2), or (3). The same holds forG/B A and Bb and then also for G = A x B and G = AS x B.

Exercises

1. (Suzuki [1951a]) Show that a nilpotent group G has a dual if and only if G is atorsion group and every primary component of G is a finite M*-group.

8.3 Soluble groups with duals

The aim of this section is to prove the following special case of our Main Theorem.

8.3.1 Theorem (Suzuki [1951 a], Zacher [1961a]). Every duality S of a soluble groupG to a group G satisfies 8.2.2.

To prove this we suppose the theorem is false and choose a "minimal counter-example" in the following way: if there exists a finite soluble group with a dualityviolating 8.2.2, we take a group G of minimal order with this property; and if thereis no such finite group, we take a counterexample G with minimal derived lengthd(G). In both cases, we study a duality 6 from G to a group G for which 8.2.2 doesnot hold. Then b has the following properties:

(1) S does not satisfy 8.2.2,

(2) if G is finite, every duality 9 of a finite soluble group X such that I X < I G Isatisfies 8.2.2, and

(3) if G is infinite, every duality 9 of a soluble group X such that IXI is finite ord(X) < d(G) satisfies 8.2.2.

Throughout this section we assume that S: L(G) -+ L(G) is such a minimal counter-example; let d(G) = n. The special cases settled in § 8.2 yield the following results.

8.3.2 Lemma. G is a nonabelian locally finite group. If G is finite, L(G) is directlyindecomposable.

Proof By 8.1.2, G is a soluble torsion group and hence locally finite (see Robinson[1982], p. 147). By 8.2.7, G is not abelian. Suppose, for a contradiction, that G isfinite and L(G) is directly decomposable. Then by 1.6.5, G = A x B where A : I V- Band (IAI,IBI) = 1. By (2), the dualities induced by b in G/A and GIB satisfy 8.2.2and hence so does S, by 8.2.10. But this contradicts (1). Thus L(G) is directlyindecomposable.

We prove another general remark.

8.3.3 Lemma. If G is finite, G is supersoluble.

Proof. Suppose, for a contradiction, that G is not supersoluble. If N is a minimalnormal subgroup of G, then by 8.1.6, 6 induces a duality in GIN. By (2), this dualitysatisfies 8.2.2 and hence, by 8.2.6, GIN is supersoluble. Since G is not supersoluble, itfollows that N is the unique minimal normal subgroup of G, N is not cyclic and, byHuppert's theorem (see Robinson [1982], p. 268), N D(G). Since N is not cyclic,5.4.4 implies that N° is a minimal normal subgroup of G for every a E P(G). ThusN° = N for every a E P(G) and, by 8.1.7, Nb g G. Since N (D(G), there exists amaximal subgroup M of G such that N M. Then N n M a NM = G and henceN n M = 1. Therefore 6 = Na u Ma = N°M° and, since Mb is a minimal subgroup

8.3 Soluble groups with duals 463

of G, it follows that Na has prime index in G. Thus N is cyclic of prime order, acontradiction.

Since d(G) = n, we have G(") = 1 and G("-') 1. Consider first the case that G("-')is a p-group for some prime p.

8.3.4 Lemma. If 0"-'1 is a p-group, every autoprojectivity of G is regular at p.

Proof. Suppose that this is false. Then there exist o E P(G) and a E G such thatI<a>I = p and I <a>61 p. Since G'"-') = (G"n-2))' :e 1, there exist x, y E G(n-2) suchthat [x, y] :e- 1. Let H be a finite subgroup of G containing a, x, y. Then a inducesa projectivity from H onto H" which is singular at p since a c H. If a satisfies (b)of 4.2.6, H contains a normal p-complement N with abelian factor group. Thus[x, y] E H'< N and so [x, y] E N n G("-') = 1, since N is a p'-group and G("-'is a p-group. This contradiction shows that (a) of 4.2.6 holds, that is, H = S x Twhere S is a P-group containing a Sylow p-subgroup of H and (ISI, I TI) = 1. Since1 :e- [x, y] E H n G("-'), H n G("-') is a nontrivial normal p-subgroup of H and hencep is the larger prime dividing ISI. Thus

(4) H = S x T,(ISI,ITI)=1,SEP(m,p)for some mEN.

In particular, since G is locally finite, this holds for H = <a, x, y>. Let <a, x, y> _S o x T o where ISOI = p'kq, p > q, and let b e So such that o(b) = q and ab = a' (r e 7/);note that o(a) = p and hence <a> a So. Now let P be the set of all p-elements and Kthe set of all {p,q}'-elements of G, and let w be an arbitrary q-element. For c, d e Pand u, v e K, consider H = <a, x, y, c, d, u, v, w>. Since G is locally finite, H is a finitesubgroup of G and hence H = S x T as in (4). It follows that a, c, d c- S and socd-' E P, cd = dc, and o(c) < p; thus P is an elementary abelian p-group. Further-more, a E S implies b 0 CG(S); therefore q is the smaller prime dividing ISI and b c- S.Thus cb = c' and, since c was an arbitrary element of P, this shows that P is aP-group. Also w e S < P, which implies that P is the set of all {p, q}-elementsof G. Finally, u, v e T and so uv-' E T s K; hence the set K of all {p,q}'-elements isa subgroup of G. It follows that G = P x K.

By 8.1.6, S induces dualities 6, in G/K P and S2 in G/P(b>. By 8.2.8, G/Kis finite and Sl satisfies 8.2.2. Since Gl"-') < P, we have d(G/P) < n and so (2)or (3) implies that S2 satisfies 8.2.2. But then by 8.2.10, S also satisfies 8.2.2, acontradiction.

8.3.5 Lemma. If G4"-" is a p-group and P E Sylp(G), then P!9 G and G is finite.

Proof. Suppose, for a contradiction, that P is not normal in G. By (2) or (3), G/O"-')has the structure given in 8.2.2 and, since PIG('-') is a nonnormal Sylow p-subgroupof G/G("-'), it follows that G/G("-') = S/G("-1) x T/G("-1) where SIG('-') andT/G") are coprime and S/G("-') is a P-group of order rmp, r > p. Then G("-') is theintersection and S the join of all the Sylow p-subgroups of G. If a E P(G), then by8.3.4, a and a-' are regular at p and hence a maps Sylow p-subgroups of G to Sylowp-subgroups. Thus G("-') and S are invariant under P(G) and therefore, by 8.1.7, 6

induces dualities b, in G("-" and SZ in S. By 8.1.10, 0`1 is finite and so S is a finite{ p, r}-group with indecomposable subgroup lattice. If S < G, then by (2) or (3),6, satisfies 8.2.2 and hence S is a P-group. But G("-" is a normal p-subgroup of Sand since p < r, it follows that G("-" = 1, a contradiction. Thus G = S is a finite{p, r}-group. By 8.3.3, G is supersoluble and, since r > p, R < G if R e Sylr(G) (seeRobinson [1982], p. 145). By 8.3.2, n > 2 and therefore G' >_ G(`). Since GIG("-')is a P-group of order rmp, it follows that G' = R x G("-') is abelian and hence1 = G" < G("-" < G'. This contradiction shows that

(5) P<G.

Again by 8.3.4, 8.1.7 and 8.1.10, P is invariant under P(G); therefore it has a dual andso is finite. Suppose, for a contradiction, that G is not finite. Then G/P is infinite andG'"-" < P. By (3), G/P is an infinite direct product of finite coprime groups satisfying(1)-(3) of 8.2.2. If there exists a prime q p and a q-element x 0 CG(P), then let K bethe set of all {p,q}'-elements and y be any q-element of G; if there is no such prime,let K be the set of p'-elements in G and put x = 1 = y. In any case, if u, v e K, thereexists a finite join H/P of components of the above decomposition ofG/P such thatu, v, x, y e H. Since Ha is characteristic in Pa < G, we obtain Ha < G and hence 6induces a duality in the finite subgroup H of G. By (3), this duality satisfies 8.2.2 andit follows that H = Q x L where Q = <P, x> and L are coprime, and Q is a P-groupif x 1. Thus uv-' e L c K and hence K is a subgroup of G. Now if x: 1, then Qcontains y and hence is the set of {p,q}-elements of G; thus G = Q x K. If x = 1,then P = Q and G = P x K = Q x K in this case too. By (3), the dualities inducedby 6 in G/K and in G/Q satisfy 8.2.2; for, G/K Q is finite and d(G/Q) < n sinceG'"-" < P. By 8.2.10, S satisfies 8.2.2, a contradiction. Thus G is finite.

8.3.6 Lemma. G'"-') is not a p-group.

Proof. Suppose that G("-') is a p-group. Then by 8.3.5, G is finite and has a normalSylow p-subgroup P. By 8.3.4, P is invariant under P(G) and so by 8.1.7, Pa < G.Finally, by 8.2.9, G is not a p-group. Thus

(6) P<G,Pa<G,P0G.Since G is soluble, there exists a complement K to P in G. Thus P n K = 1 and henceG = P6Ka. So if X < P, Dedekind's law implies that Xa = Pa (Xa n Kb) and henceX = P n (X u K) < X u K since P < G. By 8.3.2, L(G) is not directly decomposableand hence K CG(P). Thus

(7) P has a complement K inducing nontrivial power automorphisms in P.

We show next that P is elementary abelian. So suppose, for a contradiction, that(D(P) 1 and let N be a minimal normal subgroup of P contained in'i(P). By (7),N < G and, by (2), the duality induced by 6 in GIN satisfies 8.2.2. Now P:5 Gimplies that 1(P) < cD(G) (see Robinson [1982], p. 131). Therefore if L(G/N) weredirectly decomposable, then so would L(G/O(G)) be, and then also L(G), by 1.6.9; butthis contradicts Lemma 8.3.2. Thus L(G/N) is directly indecomposable and, sinceP/N is a nontrivial proper normal Sylow p-subgroup of GIN, it follows that GIN is

a P-group of order pm'q, p > q, m e N. Thus IKI = q and ING(K)I = q or pq. Hencethere are I G : NG(K)I = pni+1 or p' conjugates to K, that is, complements to P in G.Since I G : PI = q, clearly I P61 is a prime s. And since P is not elementary abelian and8.2.2 holds for the duality induced in P, the group G/Pa is a t-group for some primet. By 8.2.9, G is not a primary group; for, in this case 6-1, and hence also b, wouldsatisfy 8.2.2. Thus s r t and the complements to Pa in G are the Sylow t-subgroupsof G. Their number is 16: N6(K°)I = 1 or s. Since S maps complements of P tocomplements of Pa, it follows that pn` or prn+1 is equal to 1 or s. This is only possibleif m = 1 and s = p = I G: NG(K)I. Thus IGI = p2q, NG(K) = NK and P is cyclic sinceD(P) 1. But then K centralizes N and hence P; so G = P x K, a contradiction.Thus V(P) = 1, that is

(8) P is elementary abelian.

We show next that K is an abelian q-group for some prime q. First of all, if there aretwo different minimal normal subgroups N, and N2 of G, then the dualities inducedin GIN; satisfy 8.2.2 and hence, by 8.2.5, G" < N, n N2 = 1. So n = 2 and G' =

G("-' < P; this implies that K G/P is abelian. By (7) and (8), every minimal sub-group of P is normal in G. Thus K is abelian if IPI > p2. If IPI = p and CK(P): 1,then CK(P) contains a minimal normal subgroup of G different from P and so K isabelian; and if CK(P) = 1, then K is isomorphic to a subgroup of Aut P and hence iseven cyclic. Thus in all cases,

(9) K is abelian.

Now K = Q, x . x Q, is the direct product of its Sylow subgroups Q; and G/ P =S, /P x x S,/P where Si = PQ;. Suppose, for a contradiction, that r > 2. By 8.1.8,P° = Tb x x T,b where T, = U SJ and the are coprime. So Ps = Ss x

jmiwhere Si' is a characteristic subgroup of Pb 9 G and hence Ss a G. Thus S induces aduality from Si onto G/Sf.. By (2), Si = P x Q; or S; is a P-group. In the latter case, Pis a normal p-subgroup of S, and hence S, e P(k, p) for some k e N. Also, by 8.2.8,6/Ss e P(k, p) and, as Ps/Sa is a nontrivial normal subgroup of this P-group, p dividesI Pb/Sa l = I T81. Since these Ti' are coprime, this can happen for at most one i. Thusthere exists j e { 1, ... , r} such that S; = P x Qj and, since K is abelian, it follows thatQ; <_ Z(G). But Q, is a Sylow subgroup of G, hence L(G) is directly decomposableand this contradicts 8.3.2. Thus

(10) K is a q-group for some prime q.

Suppose, for a contradiction, that IPI > p2 and let N < P such that INI = p. By (7)and 1.5.4, K induces nontrivial universal power automorphisms in P. So K does notcentralize P/N and, by (2), G/N is a P-group. Thus I K I = q and G = PK is a P-group. By 8.2.8, 6 satisfies 8.2.2, a contradiction. Thus

(i1) IPI = p.

Now suppose that CK(P) 0 1. Since K is abelian, CK(P) < Z(G); let N < CK(P) suchthat INI = q. Then GIN has a dual and, by (7), K/N is not normal in GIN. By (2) and8.2.8, GIN is a P-group of order pq and Na e P(2, p). Since Pa n Na a N6, we have

I P6 n N°1 = p. So S induces a duality from the q-group G/P K of order q2 onto P6,a group whose order is divisible by p > q. It follows that G/P and Pa are cyclic ofprime power order. By 8.2.9, G is not a p-group. It follows that I K6 0 p, Na isnonabelian and is the unique maximal subgroup of G containing K6. Thus NN(K6) =Ka and Ks has p2 conjugates in G. These are mapped by S-' to complements of P inG; but P has only p complements. This contradiction shows that

(12) CK(P) = I.

Now K is isomorphic to a subgroup of Aut P and hence is cyclic. Then G/P and Pbare also cyclic so that Mb < G if M6 is the minimal subgroup of P6. By (1) and 8.2.8,G P(2, p) and hence M contains P properly, has a dual and has nonnormal Sylowq-subgroups, by (12). Thus by (2), M is a P-group of order pq and Z;/Ma E P(2, p).Since Pb < G, we obtain I Pa : _M61 = p and hence I Pa I = p2. Again by 8.2.9, G is nota p-group. So I K61 p and G/M° is nonabelian. Now P has p complements, thesame holds for Pa and this implies that Ms < Ni;(K6). It follows that K° centralizesMa and hence also P6; but then G/ M6 is abelian, a final contradiction.

It remains to study the case that G("-') is not a primary group.

8.3.7 Lemma. (a) n = 2.(b) For every prime r, let S, be the r-component and T, the r'-component of G', so

that G' = Sr x Tr. Then either

(13) G/Tr is a direct product of coprime finite primary groups, or

(14) G,', = X,/T, x Y,/T, where Xr/T, e P(m, r) for some m e N, Y/T, is a directproduct of coprime finite primary groups and X,/T, and Y/T, are coprime; inthis case, G/G' is an r'-group.

(c) Every primary subgroup of G is finite.

Proof. Let r be a prime. Since G("-') is an abelian torsion group, G("-" = R x Kwhere R is the r-component and K the r'-component of G("-'). Then the duality boinduced by S in G/K satisfies (2) and (3) and (G/K)("-') = G(`" /K is an r-group; by8.3.6, So cannot satisfy (1), that is, 8.2.2 holds for 60. Thus G/K is a direct product ofcoprime finite P-groups and p-groups. In particular, G/K has finite Sylow r-sub-groups and is metabelian, by 8.2.5. Since K is an r'-group, it follows that G has finiteSylow r-subgroups and G" is an r'-group. This holds for every prime r and thereforeevery primary subgroup of G is finite and G" = 1. Thus (a) and (c) hold.

Now G("-')/K = G'/K is an r-group and therefore lies in a component X/K of theabove direct decomposition of G/K. The other components are then abelian andhence primary groups. If X/K also is a primary group, (13) holds. If X/K is notprimary, it is a P-group with nontrivial normal r-subgroup G'/K and hence lies inP(m, r) for some m e N. Then (X/K)' = G'/K is the Sylow r-subgroup of G/K so thatG/G' is an r'-group. Thus (14) holds.

8.3.8 Lemma. Let p > q be primes, P e Sylp(G), Q E Sylq(G) and suppose that[P, Q] 1. Then G = PQ x Y where PQ is a finite P-group and Y a normal { p, q}-complement of G.

Proof. Let r be a prime and let T, be as in 8.3.7. If r zA p, then PT/T, and QTr/Tr liein different components of the decomposition of G/Tr; for, either these componentsare primary groups or they lie in P(m, r) and r 0 p > q. It follows that [P, Q] < T,.If also [P, Q] < TP, we would obtain [P, Q] < n Tr = 1, a contradiction; thus

rEP[P, Q] TP. It follows that (14) holds for r = p and PQTP/TP = XP/TP e P(m,p) forsome m e N; let Y = YP. Then G'/TP = PTP/TP and hence

(15) P<G'andIQG':G'I = q.

Thus G/G' is not a q'-group and so 8.3.7 yields that (13) holds for r = q. So QT,,/T, isa direct factor in G/Tq containing the q-group G'/T,. But then by (15), G'/T" =(G/Tq)' = (QT/T,)' = (QG'/Tq)' is of index q in the finite q-group QG'/Tq and thisimplies that QG'/Tq is cyclic of order q. It follows that Tq = G', that is, G' is aq'-group, and IQI = q. Then TP is a {p,q}'-group and, since P is characteristic inG', P:9 G and PQ is a { p, q}-group. So PQ n TP = 1 and PQ PQTP/TP = XP/TP.Also PQ n Y = I since Y is a {p, q}'-group and PQ u Y = PQTPY = XPYP = G.Thus

(16) PQ is a finite P-group and Y is a normal {p,q}-complement of G.

We have to show that PQ g G; then G = PQ x Y, as desired. So suppose, for acontradiction, that PQ is not normal in G. Then since P a G, there exist a prime rand a Sylow r-subgroup R of G such that p : r 0 q and [Q, R] 1. By (15) and (16),R < G', RQ is a finite P-group and there exists a normal {r, q}-complement Z in G.Then PQR is a { p, q, r}-group and Y n Z a { p, q, r}'-group so that PQR n (Y n Z) = 1.

Furthermore, G/Y n Z is finite. So by (2) or (3), if Y n Z 0 1, the duality inducedin G/Y n Z would satisfy 8.2.2. But PQR G/Y n Z is not directly decomposable, acontradiction. Thus Y n Z = 1 and G = PQR is finite. If N is a minimal subgroup ofP, then N is centralized by PR < G' and normalized by Q so that N 9 G. By (2), theduality induced in G/N satisfies 8.2.2 and it follows that N = P. Thus IPI = p, simi-larly I R I = r and so I G I = pqr. Then PR is the unique cyclic subgroup of compositeorder of G and hence is invariant under P(G). By 8.1.7, (PR)' = Pan Ra a G.Now Pa e P(2, r) since G/P e P(2, r), and so I Ps n Rb I = r; but R° E P(2, p) sinceG/R E P(2, p), and hence I Pa n R'J = p. This contradiction shows that PQ 9 G andG=PQ x Y.

Proof of Theorem 8.3.1. By 8.3.7, every Sylow subgroup of G is finite. If any twoSylow subgroups of different order centralize each other, G is the direct product ofits Sylow subgroups. And if there exist Sylow subgroups of different order not cen-tralizing each other, then by 8.3.8, the join of any such pair is a finite P-group and adirect factor of G. In any case, G = Dr GA with coprime finite groups GA which are

AEA

p-groups or P-groups. By 8.1.8, G = Dr Gx where GA = U Gµ and the Gz are co-AEA M#A

prime. For every . e A, 6 induces a duality SA from G/GA GA onto G. By 8.2.8and 8.2.9, these aA satisfy 8.2.2. It follows that 6 satisfies 8.2.2, a final contradiction.

Exercises

1. Try to prove Theorem 8.3.1 for finite groups in the shortest possible way.

8.4 Finite groups with duals

The aim of this section is to prove that every finite group G with dual is soluble.Together with Theorem 8.3.1 this will imply that every duality 6 of G satisfies 8.2.2.In the crucial case of this proof, G will be simple so that 8.1.6-8.1.8 will not be ofmuch help. However, it is possible to find certain proper sections of G which haveduals and to which the induction assumption may therefore be applied. For thispurpose, we introduce the following two concepts both of which are generalizationsof P(G)-invariance.

P(G)-connected subgroups

Let H and X be subgroups of an arbitrary group G. We say that X is P(G)-connectedto H if H° = H implies X° = X for a e P(G).

8.4.1 Remark. (a) If H = G (or H = 1), every a c- P(G) satisfies H° = H and there-fore X is P(G)-connected to G if and only if X is invariant under P(G).

(b) Conversely, if X < H and X is invariant under P(H), then X is P(G)-connected to H; for, every u c- P(G) satisfying H° = H induces an autoprojectivityin H and hence fixes X. Thus subgroups like fi(H) or Q(H) are P(G)-connected to H.

(c) Finally, note that if X is P(G)-connected to H. then NG(H) < NG(X). For, everyg c- NG(H) induces an autoprojectivity in G fixing H and therefore also X; thusX9 = X and g e NG(X).

Since we are studying dualities of groups, we are not only interested in the casethat X < H, but also in the situation that H < X or even that H and X are incompa-rable. This is shown by the following simple result and, in particular, its corollary.

8.4.2 Lemma. Let 6 be a duality from the group G to the group G and let H, X < G.If X is P(G)-connected to H, then X' is P(G)-connected to Ha; in particular,Ha < NN(X°).

Proof. Let T E P(G) such that (Ha)` = Ha. By 8.1.5, bib-' e P(G) and H"'-' H.Since X is P(G)-connected to H, it follows that X6`a-' = X and hence (XO)` = Xa.Thus Xa is P(6)-connected to Ha. By (c) of 8.4.1, Ha < N-(X°).

This lemma will mainly be used in the following two situations.

8.4 Finite groups with duals 469

8.4.3 Corollary. Let 6 be a duality from G to G, let H, X< G and suppose that X isP(G)-connected to H.

(a) If H g X, then Xs a Ha and 6 induces a duality from X/H onto Ha/X'.(b) If H n X = 1, then Xa g G and 6 induces a duality from X onto G/Xa.

Proof. If H g X, then Xa < Ha and hence Xa < Ha, by 8.4.2. And if H n X = 1, thenNG-(X°) > Ha u Xa = (H n X)a = G, that is, Xa g G. By 8.1.6, b induces dualities inX/H and X, respectively.

P(G)-normality

Let H and K be subgroups of a group G such that H < K. We say that H isP(G)-normal in K if H° g K° for all a e P(G).

Of course, H :!g K if H is P(G)-normal in K; but P(G)-normality strongly dependson the group G (see Exercise 2). It is important that if G has a dual, then K/H iscontained in a factor L/H with dual.

8.4.4 Lemma. Let G be a group, H < K < G and suppose that H is P(G)-normal inK. If 6 is a duality from G to a group G, and La is the core of Ka in Ha, thenH g L = U Ka"a-' and hence 6 induces a duality from L/H onto Ha/La.

uUH°

Proof. Clearly, Ha >_ Ka > La so that H < K < L. By 8.1.1, L = n (K0)" _C

-

H,

U Kaua Now 8.1.5 shows that (5u(' e P(G) for every u e Ha, and thereforeu EH°

H = Ha"a-' < Kaua-' since H is P(G)-normal in K. Thus H:9 L and, by 8.1.6, 6induces a duality from L/H onto Ha/La.

The main theorem

In the remainder of this section we prove the result announced above.

8.4.5 Theorem (Zacher [1961b]). Every finite group with dual is soluble.

To prove this, we suppose the theorem is false and choose a counterexample G ofminimal order; let b be a duality from G to a group G. Then if X is a group such thatX I < I and X has a dual, it follows that X is soluble and, by 8.3.1, that

(1) every duality of X satisfies 8.2.2.

Of course, G is not soluble; indeed we show first that our inductive hypothesisimplies that G and G are in fact simple.

8.4.6 Lemma. G and 6 are nonabelian simple groups.

Proof. We suppose the lemma is false and show that there exists a subgroup Rof G such that 1 < R < G and R° = R for all a E P(G). By 8.1.7, S induces dualitiesin R and in G/R; then by induction, R and G/R, and therefore G, are soluble, acontradiction.

If G is not simple, there exists N < G such that 1 < N < G. By 8.1.6, GIN has adual and therefore, by induction, it is soluble. Thus if R = Ga is the soluble residualof G, then R < N < G and, of course, R 1 since G is not soluble. By 5.3.6, R° = Rfor all a e P(G). _

Now suppose that G is not simple. Then there exists N < G such that 1 < N < Gand Na g G. Again by 8.1.6, S induces a duality from N onto G/Na; by (1), N andGIN' are soluble. Let R < G be such that Ra = G. Then Ra < Na < G; if Rb = 1, thegroup G is soluble and hence by 8.3.1 applied to S', G is soluble, a contradiction.Thus 1 < R < G and, again by 5.3.6, Ra is invariant under P(G). By 8.1.7, R° = R forall a e P(G).

In the sequel let q be the smallest prime dividing IGI and let Q E Sylq(G). Of course,the Feit-Thompson Theorem implies that q = 2; however, we do not need this. Infact, the proof of Theorem 8.4.5 does not use any of the deeper results proved in thecourse of the classification of finite simple groups. We shall only need some transferarguments culminating in Griin's Second Theorem. So, for example, the additionalassumption in the following lemma, that Q is not a generalized quaternion group, issuperfluous by a well-known theorem of Brauer and Suzuki. However, it will be noproblem to get rid of this assumption later on.

8.4.7 Lemma. If Q is not a generalized quaternion group, there exists Q1 E Sylq(G)such that Q, 0 Q and Q n Q1 : 1.

Proof. Suppose, for a contradiction, that Q n Q1 = 1 for all Q1 E Sylq(G) such thatQ1 Q. Since G is simple, 1 < Q < G and Q is not cyclic (see Robinson [1982],p. 280). By assumption, Q is not a generalized quaternion group and it follows (seeRobinson [1982], p. 138) that Q has at least two minimal subgroups. Let H be oneof these and let a E P(G) such that H° = H. By 4.2.8, a is index preserving and henceQ° E Sylq(G). So if Q° Q, then 1 :?6 H < Q n Q°, a contradiction. Thus Q° = Q, thatis Q is P(G)-connected to H. By 8.4.2, Qb < Ha. So if K is another minimal subgroupof Q, it follows that Qa _< HS u Ka = (H n K)s = 1a = G. But this contradicts thesimplicity of G.

The above lemma suggests the study of maximal intersections of different Sylowq-subgroups of G. For this we shall need the following technical result.

8.4.8 Lemma. Suppose that N < K < G, INI = q, N < K and K/N is a P-group oforder p"q where q < p E P. Then n = 1.

Proof. Let P E Sylq(K). Then P operates on the cyclic group N of order q and sinceq < p, we can assert that P centralizes N. Thus PN = P x N. Let H < P such thatCHI = p. Then H is characteristic in H x N g K and hence H :!g K. So if S E Sylq(K),

then H is the unique subgroup of order p in HS. By 4.2.8, every autoprojectivity r ofG is index preserving. It follows that P° is an elementary abelian p-group and thatH° is the unique subgroup of order p in (HS)°. Thus H° a P° u S° = K°, that is, His P(G)-normal in K. By 8.4.4 there exists L < G such that K < L, H g L and binduces a duality in L/H; by (1), this duality satisfies 8.2.2. Since q2 divides the orderof L/H, this group has a normal Sylow q-subgroup, by 8.2.6. Then K/NH also has anormal Sylow q-subgroup and since K/N is a P-group of order p"q, it follows thatIK : NHI = q. Thus H = P and n = 1, as desired.

We come to the main step in the proof of Theorem 8.4.5. Recall that q is thesmallest prime dividing IGI.

8.4.9. Lemma. Let D 0 1 be a maximal intersection of two different Sylow q-subgroups of G. Then IDI = q, I NG(D) I = pq2 where q NQ;(D) > D and thechoice of D implies that Q. = Ti. Thus Si < Qi and hence S; = NQ;(D) and D = S1 n S2.This shows that NG(D) has nonnormal Sylow q-subgroups and D = Oq(NG(D)).Let r E P(G) such that D° = D. By 8.4.6 and 4.2.8, a is index preserving. Thus D =D° = Oq(NG(D)°) a NG(D)° and hence NG(D)° < NG(D). Since G is finite, this impliesthat NG(D)° = NG(D), that is, NG(D) is P(G)-connected to D. By 8.4.3, S induces aduality in NG(D)/D and since NG(D)/D has nonnormal Sylow q-subgroups, it followsfrom (1) that

(2) NG(D)/D = A/D x BID where (IA/DI,IB/DI) = 1 and AID is a P-group of or-der p"q,p>q,nER.

We have to study A and show first that (D(A) = 1 or 4)(A) = D; in fact, we prove amore general result that will be applied several times in the sequel.

(3) Let H < D and D < K < A such that pq divides I K: DI. If H is P(G)-normalin K, then H=IorH=D.

To prove this, suppose that 1 < H < D. By 8.4.4 there exists a subgroup L of G suchthat K < L, H g L and L/H has a dual. Since pq divides I K : DI, K/D has nonnormalSylow q-subgroups and then so does L/H. Therefore by (1) and 8.2.6, IL/HI is notdivisible by q2. It follows that H = D, as desired. Now D(A)' = D(A°) g A° for alla e P(G). Thus sD(A) is P(G)-normal in A and, since A/D is a P-group, '(A) < D. Sowe may apply (3) with H = D(A) and K = A to obtain that

(4) '(A) = 1 or cb(A) = D.

We want to show that IDI = q and, as a first step, we prove this in a special case.

(5) Let P E Sylq(A). If C,(D) 1, then IDI = q.

To see this, put C = C,.(D) and K = CS where S E Sylq(A). Since CD g A, it followsthat K = CS = CDS is a subgroup of A containing D; also pq divides IK : DI since

Figure 23

C 1. Let a E P(G). If H D, then HC = H x C and hence H' a (HC)', by 1.6.7.So if H° 9 S°, it follows that H° a C° u S° = K°. We now apply this argument threetimes. First, H ='b(S) < D and, clearly, H° = (D(S°) < S. Thus H° < K° and so His P(G)-normal in K. By (3),'b(S) = 1 or t(S) = D. If'b(S) = D, then S is cyclic; henceS° is cyclic and every subgroup H of D satisfies H° < S. If I(S) = 1, then S iselementary abelian; since a is index preserving, S° too is elementary abelian andagain H° a S° for all H < D. In both cases, it follows that every subgroup H of D isP(G)-normal in K. By (3), H = 1 or H = D, that is IDI = q, as desired.

(6) IDI = q.

Suppose that this is false and let P E Sylp(A), as before. The Frattini argument ap-plied to the normal subgroup PD of A yields that A = DNA(P). So if I(A) = D, itwould follow that A = NA(P) and hence [P, D] q2.

Since q is the smallest prime dividing IGI, a minimal subgroup of D which is normalin A is centralized by A. Therefore if every minimal subgroup of D were normal inA, then C,(D) = P and, again by (5), IDI = p, a contradiction. Thus there exist mini-mal subgroups of D which are not normal in A. Since G is simple, Da can be normalin the image of at most one of these minimal subgroups, that is, there exists

(6b) H < D such that IHI = q, H A and D° H6.

Let La = n (D6)" be the core of Da in H6. Then by 8.1.1, L = U D'"'-' and, by" E H° 11 E HI

(6a) and (6b), H < D < L. Since every autoprojectivity a of G is index preserving, D°is elementary abelian and hence H° < D° for every a c P(G). By 8.4.4, H 9 L and binduces a duality from L/H onto Ha/La; by (1), this duality satisfies 8.2.2. Since everyautoprojectivity bub-' of G (u E H6) is index preserving, L is a join of q-groups; inparticular, L(L/H) is directly indecomposable. Furthermore, since D is elementaryabelian, Da/La is the intersection of maximal subgroups of Ha/La, and Da 4 Ha im-

plies that Ha/La is not a primary group. It follows that

(6c) Ha/LO is a P-group of order rms, r > s,

and L/H e P(m + 1, r); since L is generated by q-groups,

(6d) L/H is an elementary abelian q-group (and q = r) or a nonabelian P-groupof order rmq, r > q.

Now H < D < L. So if L/H is abelian, L< NG(D) and then IL: DI = q, by (2). And ifL/H is not abelian, 8.4.8 applied with N = H and K = L yields that I L : HI = rq. Inboth cases,

(6e) D is a maximal subgroup of L.

We finally show that

(6f) Da is a p-group

where p is the prime appearing in (2). For this we choose x e A such that HX 0 H,and consider the autoprojectivity r = 6-'xb of G induced by x. Then Ha and HOT =HxO are different maximal subgroups of G and hence G = Ha u HOT. Furthermore,DOT = DXO = DO since x e A < NG(D). And finally, by (6e), LO is a maximal subgroupof DO; also ADO/LOI = t e P since LO a DO. Because r is index preserving, 4.3.5 yieldsthat LOT -< DOT = Da and ADO/LaTI = t. So if N = 0'(D°), then N = OT(LO) = Ot(LOT).Thus N is characteristic in LO -< Ha and hence N < Ha. Moreover, since Ha/La is aP-group, Ha is the join of subgroups in which La is maximal and so 4.3.5 implies thatLOT < HOT. Therefore N g HOT also and hence N < HO u HOT = G. But G is simpleand it follows that N = 1, that is Da is a t-group. Since AID is a P-group, AO is theintersection of maximal subgroups of DO and so AO < Da. By 8.1.6, b induces aduality from A/D onto D6/A6 and, by 8.2.8, D6/A" E P(n + 1, p). Thus t = p and (6f)holds.

It follows from (6f) that p divides MHO/LOS. Therefore r > p > q in (6c) and henceL/H cannot be an elementary abelian q-group. By (6d), L/H is a nonabelian P-groupof order r'q and so D is not normal in L. By (6e), D is a maximal subgroup of L and,by (2), D is also maximal in S if S E Sylq(A). Now S : L since D < S; let M = L u S.

Then LO and SO are different maximal subgroups of the p-group Da and henceDa/(La n Sa) = DO/MO is elementary abelian of order p2. So £ _ [MID] \, {M, D, S}is an antichain on which S operates by conjugation. Since 1 2CI = p, the q-group Smust fix some X e X and then X g S u X = M. It follows that D = X n Y :!g Y forevery Y e 1 such that Y X, and, since these Y generate M, we get that D a M. Inparticular, D g L and this contradiction finally proves (6).

Now we may apply Lemma 8.4.8 with N = D and K = A and obtain that I A/ D _pq. Thus

(7) J A I = pqz.

It remains to be shown that NG(D) = A; then (6), (7), and (2) contain all theassertions of the lemma. Since IDS = q is the smallest prime dividing IGI, we haveD < Z(NG(D)); in addition, D is a normal Hall subgroup of B and so the Schur-Zassenhaus Theorem implies that B = D x H where (I D 1, 1 H 1) = 1. Then H is char-

acteristic in B :!g NG(D) and hence H g NG(D); thus NG(D) = A x H and (I A I,1 HI) = 1.By 1.6.6, NG(D)" = A° x H° for every a e P(G), that is, H is P(G)-normal in NG(D).By 8.4.4 there exists L< G such that NG(D) < L, H g L and L/H has a dual. SinceA NG(D)/H < L/H, q2 divides IL/HI. So if H 1, (1) and 8.2.6 would imply thatL/H has a normal Sylow q-subgroup; but (2) shows that A has nonnormal Sylowq-subgroups, a contradiction. Thus H = 1, that is NG(D) = A, as desired.

Recall that q is the smallest prime dividing IGI and Q e Sylq(G). Lemma 8.4.9implies that Q is small and q = 2; so we can finally work with involutions.

8.4.10 Lemma. IQI < q2; hence q = 2, Q is elementary abelian of order 4 andI NG(Q)lCG(Q)I = 3.

Proof. Suppose, for a contradiction, that IQI > q3. We show first that for everygE G,

(8) Z(Q) < Q9 implies Q9 = Q.

For, if Z(Q) < Q9 and Q9 Q, there would exist a maximal intersection D of twodifferent Sylow q-subgroups of G such that D > Q n Q9 > Z(Q). By 8.4.9, IDI = qand I NG(D)I = pq2 for some prime p. So it would follow that D = Z(Q) and henceQ < NG(D), impossible since IQI >_ q3. Thus (8) holds and a similar argument yieldsthat

(9) Z(Q) < Q.

Indeed this is clear if Q is a generalized quaternion group. If Q is not of this type,then by 8.4.7 there exists a maximal intersection D : 1 of Q and another Sylowq-subgroup Q 1 . Again by 8.4.9, I D I = q and I NG(D)I = pq2 for some prime p. SinceZ(Q) < NG(D), it follows that IZ(Q)I < q2 and hence Z(Q) < Q.

Now Z(Q) is a characteristic subgroup of Q, and hence NG(Q) < NG(Z(Q)). Ifg e NG(Z(Q)), then Z(Q) = Z(Q)9 < Q9 and so by (8), Q9 = Q. Thus g e NG(Q), that is,NG(Q) = NG(Z(Q)). Let a E P(G) such that Z(Q)" = Z(Q). Since a is index preserving,Q" is a Sylow q-subgroup of G containing Z(Q) and hence by (8), Q° = Q. It followsthat Q is the unique Sylow q-subgroup of NG(Q)° and so NG(Q)" < NG(Q). Since G isfinite, this implies that NG(Q)° = NG(Q). Thus we have shown that

(10) X = NG(Q) = NG(Z(Q)) is P(G)-connected to Z(Q).

By 8.4.3, 6 induces a duality in X/Z(Q) and by (1), this duality satisfies 8.2.2. ThusX / Z(Q) is a direct product of coprime P-groups and p-groups. By (9), Z(Q) < Q < X.Since q is the smallest prime dividing IGI, the direct factor of the decompositionof X/Z(Q) containing Q/Z(Q) is either a q-group or a nonabelian P-group of orderp"q, p > q, n e N. In both cases, X = NG(Z(Q)) has a normal subgroup of index q.Now (8) shows that G is q-normal, that is, Z(Q)9 < Q implies Z(Q)9 = Z(Q); for,Z(Q)9 < Q implies Z(Q) < Q9-', hence Q9 = Q and then Z(Q)9 = Z(Q9) = Z(Q).So Grun's Second Theorem (see Robinson [1982], p. 285) yields that G has a nor-mal subgroup of index q; but by 8.4.6, G is simple. This contradiction shows thatQI < q2. It is a well-known consequence of Burnside's criterion for p-nilpotence (see

Robinson [1982], p. 280) that now q = 2, Q is elementary abelian of order 4 andI NG(Q)ICG(Q)I = 3.

8.4.11 Lemma. If t e G is an involution, CG(t) g G.

Proof. By 8.4.7 there exists a nontrivial intersection D = Q n Q1 of Q and anotherSylow 2-subgroup Q1 of G. By 8.4.10, IQI = 4 so that D has order 2 and is a maximalintersection of two different Sylow 2-subgroups; in addition, ING(Q)/CG(Q)I = 3 im-plies that all involutions in G are conjugate. So we may assume that D = <t); putC = CG(t) = NG(D). Then by 8.4.9, C/D is nonabelian of order 2p for some primep > 2; let P E Sylp(C). Then PD = P x D < C so that P < C, and hence P is theunique subgroup of order p of C.

We want to show that C is P(G)-connected both to D and to P; by 8.4.2, this willimply that C° < D° u P° = (D n P)° = G, as desired. Clearly, C is the join of all theSylow 2-subgroups of G containing D. So if a c- P(G) such that D° = D, a is indexpreserving and hence permutes these Sylow 2-subgroups. Thus C° = C, that is, C isP(G)-connected to D.

Now let a e P(G) such that P° = P. Since a is index preserving, (DP)' is a cyclicgroup of order 2p containing P. We show that t is the unique involution in Gcentralizing P. This will imply that D° = <t) = D and, as we have just shown, it willfollow that C° = C, as desired. So suppose, for a contradiction, that s e G is aninvolution such that t s E CG(P). Then <s, t> is a dihedral group centralizing P;hence <s, t) n P = 1. Furthermore, since C/D is nonabelian, P Z(C); hence s 0 Cand <s, t) is not a 2-group. Now P is the unique subgroup of order p of C andtherefore P is P(G)-normal in C = CG(t). Then P is the unique subgroup of order pof CG(s) since CG(s) is conjugate to CG(T); thus it is P(G)-normal in CG(s). HenceP is P(G)-normal in CG(t) U CG(s) and by 8.4.4 there exists L < G such thatCG(t) U CG(s) < L and L/P has a dual. Since IC/PI = 4, (1) and 8.2.6 imply that L/Phas a normal Sylow 2-subgroup; but <s, t> PIP _- (s, t> does not. This contradictionshows that t is the unique involution of G centralizing P. Thus D° = D and C° = C,as desired.

Proof of Theorem 8.4.5. By 8.4.6, a minimal counterexample G to Theorem 8.4.5is simple and, by 8.4.10, I G I is even. If t e G is an involution, it follows that1 < CG(t) < G. By 8.4.11, CG(t)° 9 G; but by 8.4.6, G is simple. This contradictionproves the theorem.

Exercises

1. Give examples of subgroups H, X of a group G such that X is P(G)-connected toH and(a) 1 < H < X <G,or(b) H$XandX H, or(c) X < H and X is not invariant under P(H).

2. Find groups H:,--- K < G < G* such that(a) H is P(G)-normal in K but H is not P(G*)-normal in K,(b) H is not P(G)-normal in K but H is P(G*)-normal in K.

3. (Calcaterra [1987]) Let A be a group acting on a finite group G and suppose that(A, G) satisfies the double centralizer property (or DCP), that is, C0CA(H) = H forany subgroup H of G and CACG(B) = B for any subgroup B of A.(a) If A = A, x x A with coprime groups A;, show that G = G1 x x G

with coprime groups G. such that (A;, G;) satisfies the DCP for all i.(b) If A is nontrivial and L(A) is directly indecomposable, show that either

(i) A I= q and I G I= p where p and q are primes such that q I p- 1, or(ii) A and G lie in P(n, p) for some prime p and integer n.

8.5 Locally finite groups with duals

The aim of this section is to show that every locally finite group with dual is soluble;this together with Theorem 8.3.1 will complete the proof of the Main Theorem 8.2.2of this chapter. For this purpose we first of all have to prove Theorem 8.2.3(a),without using 8.2.2, of course.

Subgroups of finite index

8.5.1 Theorem. If G is a group with dual, then G" = G; = G a where Gj is the intersec-tion of all subgroups of finite index in G and G: is the intersection of all normalsubgroups of G with soluble factor group.

Proof. First of all, if H is a subgroup of finite index in G, then G/Hc is a finite group;by 8.1.6, G/Hc has a dual and hence is soluble, by 8.4.5. Now if N is an arbitrarynormal subgroup of G with soluble factor group, then 8.3.1 and 8.2.5 yield that GINis metabelian. Thus G" < N; in particular, G" < HG < H. It follows that G" < Gaand G" < Ga. Conversely, since G/G" is soluble, obviously G < G"; furthermore, theduality induced in G/G" satisfies 8.2.2 and hence G/G" is aVdirect product of finitegroups. Therefore G" is the intersection of normal subgroups of finite index in G andsoGj<G". Thus G" = G:= Gj.

8.5.2 Corollary. If G is a group with dual, G" is a perfect group with dual and has noproper subgroup of finite index.

Proof: By 8.5.1, G" = G,,j = Ga. Since Ga_ G and GIG"' is soluble, we see that(G")' = G> G a = G"; thus G" is perfect. The Zacher-Rips Theorem 6.1.7 showsthat G; is invariant under P(G). By 8.1.7, G" = G, has a dual and then 8.5.1 appliedto this group yields that (G")j = (G")" = G". Thus G" has no proper subgroup offinite index.

8.5 Locally finite groups with duals 477

The locally soluble radical S(G)

For a locally finite group G, define S(G) to be the join of all locally soluble normalsubgroups of G.

Suppose that N/S(G) is a locally soluble normal subgroup of G/S(G) and let H bea finitely generated subgroup of N. Since G is locally finite, H is a finite group.Therefore H/H n S(G) ^- HS(G)/S(G) is soluble since N/S(G) is locally soluble. Andthe finite subgroup H n S(G) of S(G) is contained in the join of a finite number oflocally soluble normal subgroups N,, ..., Nk of G. If K is a finitely generated sub-group of N, N2, then again K is finite and therefore K n N, is soluble; and sinceN,N,/N, NZ/N, n NZ is locally soluble, KN1/N1 K/K n N, is soluble. Thus Kis soluble and so N, N2 is locally soluble. An obvious induction yields that N, ... Nkis locally soluble and hence H n S(G) is soluble. So, finally, H is soluble and thisshows that N is locally soluble. This holds in particular for N = S(G), that is

(1) S(G) is locally soluble;

moreover, we have shown that N < S(G), that is

(2) S(G/S(G)) = 1.

Now suppose that a E P(G). By 6.6.10 and (2), S(G)' a G. By 6.6.4 and 1.2.11, theclass of locally soluble groups is invariant under projectivities. Thus by (1), S(G)' isa locally soluble normal subgroup of G and hence S(G)° < S(G). This result appliedto u-' yields the other inclusion. Thus S(G)' = S(G) and hence

(3) S(G) is invariant under P(G).

By 8.1.2, a locally soluble group with dual is locally finite. Thus our next result is apartial case of the main theorem of this section.

8.5.3 Theorem (Zacher [1966]). Every locally soluble group with dual is soluble.

Proof. Suppose that the theorem is false and consider a counterexample G. Then by8.5.2, G" 0 1 is a perfect locally soluble group with dual and we may assume thatG = G" is perfect.

Since G is locally soluble, every chief factor H/K of G is abelian (see Robinson[1982], p. 367) and we show that H/K is even central. For this we may assume thatK = I since G/K is also a perfect locally soluble group with dual. Then H is anormal p-subgroup of G for some prime p and hence lies in the intersection OP(G) ofall the Sylow p-subgroups of G. By 6.5.9, every autoprojectivity of G is index preserv-ing and therefore OP(G) is invariant under P(G). By 8.1.7 and 8.1.10, OP(G) has a dualand is finite. Thus H is finite and, since G/CG(H) is isomorphic to a group of auto-morphisms of H, it follows that G/CG(H) is finite. But by 8.5.2, G has no propersubgroup of finite index; so CG(H) = G, that is, H/K is central.

Thus every chief factor of G is central and this implies that every finitely gener-ated, hence finite, subgroup of G has a central series. Thus G is locally nilpotent andhence it is the direct product of its primary components (see Robinson [1982],p. 342). By 8.1.6 and 8.1.10, these are finite p-groups. Since G is perfect, it followsthat G = 1, a contradiction.

8.5.4 Corollary. If G is a locally finite group and b a duality from G to a group G, thenS(G) is a soluble group with dual and S(G)° is perfect.

Proof. By (1), (3), and 8.1.7, S(G) is a locally soluble group with dual, and 8.5.3implies that S(G) is soluble. Suppose, for a contradiction, that S(G)° is not perfect.Since S induces a duality from G/S(G) onto S(G)° and S(G/S(G)) = 1, we may assumethat S(G) = 1. Then G' < G and hence there exists a subgroup H of prime orderof G such that G' < H° < G. If 6 E P(G), then b-'ub E P(G) and therefore HO _(H°)° 'a' is a projective image of the normal subgroup H° of prime index in G.By 6.5.3, this projective image contains G"; thus Ha° > G" for all a e P(G).NowL= U H° is clearly invariant under P(G) and, by 8.1.1, L° = n Ha° >_ G"; by

ae P(G) _ a EP(G)

8.1.7, 6 induces a duality from L onto the soluble group G/L°. But then 8.3.1 impliesthat L is soluble, a contradiction since S(G) = 1.

The main theorem

We come to the proof of the result announced above.

8.5.5 Theorem (Zacher [1971]). Every locally finite group with dual is soluble.

Again we need several steps to prove this. In contrast to the finite case, however,we shall use the Feit-Thompson Theorem at several points in this proof. First of all,we need the following fact (see Robinson [1982], p. 415):

(4) every infinite locally finite group has an infinite abelian subgroup.

No proof of this is known which avoids using the Feit-Thompson Theorem. An-other frequently used result is:

(5) if G is locally finite and S < G is such that o(s) 0 2 for all s e S, then S is locallysoluble;

indeed, every finitely generated subgroup of S is a finite group of odd order andhence soluble.

In the sequel, let G be a locally finite group and S a duality from G to a groupWe start with the following special case of 8.5.5.

8.5.6 Lemma. If G has a finite Sylow 2-subgroup, G is soluble.

Proof. We suppose that this is false and choose a counterexample b: L(G) - L(G)for which the order of the Sylow 2-subgroups of G is minimal; recall that these Sylow2-subgroups are conjugate since one of them is finite (see Robinson [1982], p. 413).By 8.4.5, G is infinite. By 8.5.2 and 8.1.6, S induces a duality in G"/S(G") and theSylow 2-subgroups of this group cannot have larger order than those of G. Thus wemay assume that G = G" is perfect and, by (2), that S(G) = 1; it follows from 8.5.4that G is perfect. Thus

(6) G and 6 are perfect.

For c c- G of prime order, we define Xc = Q(CG(c)) = <x E CG(C)10(X) C- P>. Leta E P(G) such that <c> = <c). By 6.5.9, a is index preserving. If x E CG(c) such thato(x) E P, then <c, x> is either elementary abelian of order p or p2 or cyclic of orderpq for primes p and q. The same then holds for <c, x)°, and therefore <x)° centralizes<c)° = <c>. Thus <x> < X, and, since these elements x generate X, it follows thatX, < Xc. This result applied to a-' yields the other inclusion. Thus

(7) Xc is P(G)-connected to <c> and hence X,/<c) has a dual,

by 8.4.3. Now suppose, for a contradiction, that every element of G has prime powerorder. By (4), G has an infinite abelian subgroup A which then has to be a p-groupfor some prime p. Moreover, if c E A has order p, then CG(c) is also a p-group andhence is a locally finite p-group with dual. By 8.1.10, Xc is finite and, sinceQ(A) < Xc, it follows that A has finite p-rank. Thus A is a direct product of finitelymany cyclic and quasicyclic groups (see Robinson [1982], p. 107) and, since A isinfinite, there exists a quasicyclic subgroup K of A; let H < K such that I H I = p. By1.2.3, every projective image of K is locally cyclic (in fact, quasicyclic, by Exercises1.2.1 and 1.2.2). Thus H is P(G)-normal in K and, by 8.4.4, L/H has a dual whereL = U Ka"a-'. Since H = Ha"a-' is centralized by Kb"a-' for every u c- Ha, we have

YEHs

L< CG(H) and hence L is a p-group. Again by 8.1.10, L/H is finite, a contradictionsince H < K < L. Thus not every element of G has prime power order and hencethere exists c c- G such that

(8) o(c) = q c- P and Xc is not a q-group.

Let q be the smallest prime such that there exists c c- G satisfying (8). If q = 2, thenXc/<c) has Sylow 2-subgroups of smaller order than G; the choice of G implies thatX,/<c) is soluble. If q j4 2, then XX does not contain elements of order 2; by (5) and(7), Xc/<c) is a locally soluble group with dual and hence is soluble, by 8.5.3. In bothcases, by 8.3.1, Xc/<c) is a direct product of finite primary groups and P-groups.Since q is the smallest prime involved in X, it follows that X/<c) has a normalq-complement N/<c). Let Q be the set of q'-elements of Xc. If x, y c- Q, then (c, x, y)is a finite subgroup of N in which <c> is a central Hall subgroup; thus <c, x, y) =<c) x (Q n <c, x, y)) and hence xy -' E Q. So Q is a subgroup of N and N = <c> x Q.

If a E P(G) such that <c> = <c>, then by (7), X' = Xc and hence Q° = Q, since ais index preserving. So Q is P(G)-connected to <c) and, by 8.4.3, Qb < G and Sinduces a duality from Q onto G/Qb. The choice of q implies that the q'-group Q doesnot contain elements of order 2 and, by (5), Q is locally soluble. By 8.5.3 and 8.3.1, Qand G/Qa are soluble and, since G is perfect, Q = 1. But then X, is a q-group,contradicting (8).

8.5.7 Lemma. Let H < G such that H is not locally soluble. Then every primarysubgroup of CG(H) is finite.

Proof. We suppose the lemma is false and consider a counterexample H < G; letC = CG(H). Since H is not locally soluble, there exists a finitely generated subgroupof H which is not soluble; since G is locally finite, this subgroup is finite and its

soluble residual K is a nontrivial finite perfect subgroup of H. Let B be a maximalnormal subgroup of K. Then K/B is simple and, by 5.4.9, B is P(G)-normal in K. By8.4.4, A/B has a dual where A = U K°u°-' is perfect since, by 5.3.4, it is a join of

uEBo

perfect groups. On the other hand, B is a finite normal subgroup of A, and soA/BCA(B) is a finite group with dual and hence is soluble, by 8.4.5. Thus A = BLA(B)and so K = B(CA(B) n K)=BCK(B). Then CK(B)/(B n CK(B)) ^ K/B and B n CK(B)=Z(CK(B)). So if F is the soluble residual of CK(B), then F is a finite perfect subgroupof H such that F/Z(F) is simple. Since CG(H) < CG(F), the group F is a counter-example to the lemma and so we may assume that H is a finite perfect group andH/Z(H) is simple.

It follows that Z(H) is P(G)-normal in HC. For, since H is perfect and the factorgroup HC/C H/Z(H) is simple, 6.5.4 implies that H° a (HC)' and C' :!9 (HC)' forevery a e P(G); hence Z(H)° = H° n C° a (HC)'. Again by 8.4.4 there exists W< Gsuch that HC < W, Z(H) a W and W/Z(H) has a dual. In this group, H/Z(H) is afinite simple subgroup whose centralizer contains C/Z(H); therefore, since Z(H)is finite, this centralizer contains an infinite primary subgroup. Thus H/Z(H) is acounterexample to the lemma and so we finally may assume that

(9) H is a finite simple group.

It follows that CG(H)° = C0(H°) for every a e P(G). For, if X is a p-subgroup ofC0(H), then, since H = 0°(H), 1.6.8 implies that [H°, X°] = 1. Since CG(H) is gener-ated by primary subgroups, we obtain that CG(H)° < C0(H°); and if we apply thisresult to a-1 and the simple group H°, we get the other inclusion. This shows, inparticular, that C is P(G)-connected to H and, since H n C = 1, 8.4.3 yields that

(10) C°aG.

Let D = U H°u° and put M = C u D and N = C n D. As we have just shown,u E C6

H°ud_'

is centralized by C°ub-' = C for all u e G. Thus D is centralized by C; soN< Z(M) and M/N = C/N x D/N. Similarly, it is clear that D° = n (H°)u is in-

U C,

variant under Cs and so by (10), M° = Co n D° 9 C°D° = N° and the factor groupN°/M° = C°/M° x D°/M°. By 8.1.6, b induces a duality from M/N onto N°/M° andwe claim that

(11) C°/M° and D°/M° are coprime.

To prove this, suppose, for a contradiction, that both groups contain a subgroupof order p e P. The join of these is elementary abelian of order p2 and thereforecontains a subgroup T°/M° of order p which is neither contained in C°/M° norin D°/M°. Thus T° n C° = M° = T° n D° and hence T u C = M = T u D. Since[C, D] < N and D g M, it follows that T n D is normalized by C and by T andhence by C u T = M. In particular, T n D 9 D and [D/T n D] is dual isomorphic to[T° u D°/D°] nt [T°/M°]. So D/T n D is cyclic of prime order, a contradiction sinceD = U H°ub ' is generated by simple groups and hence is perfect. Thus (11) holds.

v E Co

Now (11) and 8.1.8 imply that C/N and DIN are coprime. Since H n C = 1, thefinite simple group H is isomorphic to a subgroup of D/N and the Feit-Thompson

M

Figure 24

Theorem implies that D/N has elements of order 2. Thus C/N has none and so, by(5), it is locally soluble; since N < Z(C), we conclude that C is locally soluble. Now 6induces a duality from C onto 6/0 ° and 8.5.3 implies that C is soluble. By 8.3.1,every primary subgroup of C is finite, a final contradiction.

We come to the last step in the proof of Theorems 8.5.5 and 8.2.2.

8.5.8 Lemma. IfG01,G0G'.

Proof. Suppose, for a contradiction, that G = G'. By 8.5.4, S(G) is soluble and, sinceS(G/S(G)) = 1, we may assume that S(G) = 1. So again by 8.5.4, we have that

(12) G and G are perfect.

By 8.5.6, every Sylow 2-subgroup of G is infinite. Let S n Sy12(G) and suppose, for acontradiction, that S has a subgroup of the form A x B where A and B are infiniteelementary abelian or quasicyclic groups; let H = U Baxa By 6.5.9, every auto-

x6Ad

projectivity v of G is index preserving. So (A x B)° is either elementary abelian ora locally finite 2-group with modular subgroup lattice and infinite exponent; by2.4.15, (A x B)° is abelian in this case also. It follows that if x n A', then A = Aaxa-'

centralizes Baxa-' and hence A < CG(H). In particular, A a AH and, since Hb =n (B°)x is invariant under A°, (AH)b = Ab n Hs a A°. By 8.1.6, 6 induces a duality

xeAdfrom AH/A onto Ab/(AH)a. Now AH/A contains the infinite 2-group AB/A andhence, by 8.5.3 and 8.3.1, cannot be locally soluble; it follows that H is not locallysoluble. But CG(H) contains the infinite 2-group A, and this contradicts Lemma8.5.7. The contradiction first of all shows that every abelian subgroup of S has finite

rank and therefore (see Robinson [1982], p. 107) S satisfies the minimal conditionfor abelian subgroups. So S is a Cernikov group (see Robinson [1982], p. 408), thatis, contains a subgroup E of finite index isomorphic to a direct product of quasicyclicgroups. Since S is infinite and does not contain subgroups isomorphic to the directproduct of two quasicyclic groups, it follows that E is quasicyclic. Thus

(13) S has a quasicyclic subgroup E of finite index.

If M is any group containing a quasicyclic subgroup E of finite index, then for everyquasicyclic subgroup F of M, the subgroup EM n F has finite index in F, and henceEM n F = F. Thus F 5 E and then F = E; so E is the unique quasicyclic subgroupof M. In particular, E < S, and we want to show that S = NG(E). First of all, ifa e P(G) and X is a cyclic subgroup of NG(E), then E is of finite index in M = EXand hence is the unique quasicyclic subgroup of M. Then E' is the unique qua-sicyclic subgroup of M°, that is, X° < NG(E°). It follows that NG(E)° < NG(E°), andthis result for u-1 and E° yields the other inclusion. Thus

(14) NG(E)° = NG(E°) for all a e P(G).

In particular, NG(E) is P(G)-connected to E. By 8.4.3, 6 induces a duality in NG(E)/E.This group has a finite Sylow 2-subgroup S/E and hence is soluble, by 8.5.6. Then8.3.1 implies that NG(E)/E is a direct product of P-groups and p-groups, and there-fore possesses a normal 2-complement K/E. Since Aut E is a 2-group, K < CG(E)and, since G is locally finite, it follows that the set Q of 2'-elements of K is a subgroupof K and K = E x Q. By (14), every a e P(G) satisfying E° = E fixes NG(E); thereforeit fixes Q since a is index preserving. Thus Q is P(G)-connected to E and, by 8.4.3,Qb < G. By (5), Q is locally soluble; hence Q and G/Qb are soluble, by 8.5.3 and 8.3.1.Since G is perfect, it follows that Q = 1. So K = E and

(15) NG(E) = S.

Now let D be the subgroup of order 4 of E, and suppose, for a contradiction, thatNG(D) is a 2-group. If a E P(G) satisfies D' = D, then D is a characteristic subgroupof E° < S°, and hence D g S. It follows that S° < NG(D) = S and, similarly, S°-' < Sso that S° = S. Thus S is P(G)-connected to D and, by 8.4.3, S/D has a dual; butthis contradicts 8.1.10 since S/D is infinite. So we see that NG(D) is not a 2-group,that is, there exists a subgroup H of prime order p > 2 normalizing and hencecentralizing D. Let X = H<x e CG(H)lo(x) = 4> and consider a e P(G) such thatH° = H. By 1.6.7 and since a is index preserving, a maps cyclic subgroups of order4 of CG(H) to cyclic subgroups of order 4 of CG(H), and it follows that X° = X. Thus

(16) X is P(G)-connected to H

and, by 8.4.3, S induces a duality in X/H. Clearly, D < X. Let T be a Sylow 2-subgroup of X containing D, and let Sl e Sy12(G) such that T< S. By (13) and (15)there exists a quasicyclic subgroup E1 of finite index in Sl, and NG(E1) = S. SoH:4 CG(EI), but H 5 CG(X) < CG(T) and hence E1 n T < E1. Since E1 is quasi-cyclic, it follows that E1 n T is finite. Furthermore, T/E1 n T is isomorphic to asubgroup of S1/E1 and hence is also finite. Thus T is finite and, by 8.5.6, X/H issoluble. Since Exp T > 4, TH/H cannot be contained in a P-group. Thus 8.3.1 im-

plies that the Sylow 2-subgroup TH/H is a direct factor of X/H. Since H < Z(X), wehave TH = T x H and hence T :!g X. By (16), every a E P(G) satisfying H° = H fixesX and hence T. So T is P(G)-connected to Hand, by 8.4.3, Tag G. Thus S inducesa duality from T onto G/T° and, by 8.4.5, G/T° is soluble. Since G is perfect, itfollows that T = 1; but D < T, a final contradiction.

Proof of Theorems 8.5.5 and 8.2.2. Let G be a locally finite group with dual. Then by8.5.2, G" is a perfect locally finite group with dual, and so by 8.5.8, G" = 1. Thus G issoluble and this proves Theorem 8.5.5. The Main Theorem 8.2.2 follows from 8.5.5and 8.3.1.

Arbitrary groups with duals

In the remainder of this section we give the proof of Theorem 8.2.3. Since Theorem8.5.1 and Corollary 8.5.2 are just part (a) of this theorem, it remains to prove (b)-(e).Of these, (b)-(d) are immediate consequences of 8.2.2 and (a).

Proof of Theorem 8.2.3. Let 6 be a duality from the group G to the group G. Then binduces a duality from G/G" onto (G")°. Therefore (G")° has the structure given in8.2.2 and hence is the join of finite soluble characteristic subgroups. By 8.5.1, G" =Ga and the Zacher-Rips Theorem shows that this group is invariant under P(G). By8.1.7, (G")° g G and hence every characteristic subgroup of (G")° is normal in G. Itfollows that (G")° < 611 < GLa and (G")° < Ga < GLa. Conversely, by 6.5.17, GLa isinvariant under P(G); again by 8.1.7, N = (GLa)° g G and S induces a duality fromG/N onto GLa. Since 6L11 is locally finite, 8.2.2 implies that GIN is metabelian; thusN > G" and GLa = N° < (G")°. It follows that (G")° = GLa = Ga = Ga. If we applythis result to 6-', we obtain (G")° = GLa = Ga = Ga and so (Ga)° = G". Thus (b)and (c) hold.

By (c), b induces a duality from G" n Ga onto G/GaG" and 8.2.2 implies thatG" n G' is a direct product of coprime finite groups Hi. It follows that Hi a G andG'/CG..(Hi) is finite since it is isomorphic to a subgroup of Aut Hi. But G" has noproper subgroup of finite index, so Hi <_ Z(G"). Thus G" n Ga < Z(G"). The reverseinclusion is obvious, so Z(G") = G" n G' and (d) holds.

Suppose, for a contradiction, that G'/Z(G") and Ga/Z(G") are not coprime, and letp be a prime such that both groups contain a subgroup of order p. Since Ga/ Z(G")has the structure given in 8.2.2, there exists a normal subgroup N of Ga such thatZ(G") < N < G' and Ga/N is cyclic of order p or nonabelian of order pq, p _> q E P.Then S induces a duality from H = G"Ga/N = G"N/N x Ga/N onto N°/G" n Ga,and H" = G"N/N, by (a). Moreover, if Ha = K/N, then K is locally finite as anextension of a locally finite group by a locally finite group (see Robinson [1982],p. 411); hence K < (G"Ga)a < Ga and so Ha = Ga/N. Thus H is a counterexample to(e) and we may assume that G = H; that is,

(17) G = G" x F where F = Ga has order p or pq and G" has a subgroup A oforder p.

Let B be the normal subgroup of order p of F and consider T = A x B. Clearly T iselementary abelian of order p2; therefore it contains a third subgroup C of order p.If Ba a G, it follows that Ta = Aa n Bb _= Ca n Bb < Aa u Ca = G. Thus S induces aduality from T onto G/Ta and hence G/Ta is metabelian. Therefore 6"< Ta; how-ever, by (c), Fa = G". It follows that Fa < T° and hence T < F, a contradictionsince A 4 F. _

Thus Ba is not normal in G. Since Fa = G" a G, it follows that B # F and soIF = pq, p > q. Now b induces a duality from F onto G/Fa and so G/Fa e P(2, p).Hence there exists a subgroup Q of order q of F such that Qa < G; let M = A x F =TQ. Since p > 2, there exist two different subgroups C and D of T which arenot normalized by Q. It follows that C u Q = D u Q = TQ = M and hence Ma =Ca n Qa = Da n Qa < Ca u Da = G. Again this implies that Z;/Ma is metabelian;hence Fa = G" < Ma and so M < F, a final contradiction.

Exercises

1. (Stonehewer and Zacher [1994]) Let G be a group with dual, p a prime and P thep-component of Z(G").(a) If X is a p-subgroup of G", show that either X < P or P < X.(b) If G"/Z(G") has a nontrivial p-element, show that P is cyclic.

2. Let G be a p-group with dual and p a prime.(a) Show that Z(G") = 1.(b) If G is self-dual, show that either G is finite or G is perfect and has trivial

centre.3. (Foguel [1992]) If G is a locally finite group such that (Aut G, G) satisfies the

double centralizer property (see Exercise 8.4.3), show that G is trivial, cyclic oforder 3, or the symmetric group on 3 letters.

Chapter 9

Further lattices

In this final chapter we study briefly the other four lattices connected with a groupwhich were introduced in § 1.1 together with L(G): the lattice 91(G) of normal sub-groups, the lattice R(G) of composition subgroups, the centralizer lattice (E(G), andthe coset lattice 91(G). In each case we wish to determine the structure of groupsfor which the given lattice has special properties and to find out how far a group isdetermined by this lattice. Since 91(G), R(G), and 1(G) may be rather small comparedwith L(G), the latter problem in these cases can only be tackled under rather strongassumptions on the groups involved, and the results are usually much weaker thanthose for subgroup lattices.

First we describe the groups G for which 91(G) is directly decomposable, give acriterion for a finite group to have a distributive lattice of normal subgroups, andshow that 91(G) is complemented if and only if G is a direct product of simple groups.Then we show that isomorphisms between lattices of normal subgroups map abeliangroups to abelian groups, with the obvious exceptions of groups of rank 1. In theremainder of §9.1 we study isomorphisms and dualities between 91(G) and 92(G),where we assume that G is a finite nilpotent and G a finite soluble group. The mainresult here is due to Heineken [1965] and states that if in addition G has onlynoncyclic Sylow subgroups, and c(G) < 2 or G' is nilpotent, then G is nilpotent andI G I = I G I. This theorem has been generalized to many classes of infinite groups.

In the study of R(G) in § 9.2 we restrict our attention to finite groups althoughsome of the results hold more generally for groups with a composition series. Wedetermine when R(G) is directly decomposable, show that R(G) is modular if andonly if every subnormal section of order p3 has modular subgroup lattice, and thatR(G) is distributive if and only if every subnormal section of order p2 is cyclic.Moreover R(G) is complemented if and only if 91(G) is complemented. Our mainresult on isomorphisms cp: R(G) -+ R(G) is due to Pazderski [1972] and says that ifp > 2 is a prime and OP(G) is not cyclic, then OP(G)w = OP(G); similarly OP(G)4' =OP(G) if G/OP(G) is not cyclic. Both assertions also hold for p = 2 under theadditional assumption that G, or in the second case G and G, are soluble; theseassumptions are needed since R(Q8) a- R(PI'L(2, q2)) for 2 < q e P. Finally we studydualities between lattices of subnormal subgroups of finite soluble groups.

In §9.3 we show that every group with distributive centralizer lattice is abelian,and characterize groups G for which (E(G) is directly decomposable. The structureof groups with modular centralizer lattice is not known. However we present thecharacterization, due to Schmidt [1970c], of finite groups G for which l(G) is (mod-ular) of length 2 and use it to show that the groups SL(2, 2") are the only nonabelian

486 Further lattices

finite simple groups with modular centralizer lattice, a result due to Antonov [1987].Finally we determine those finite groups G for which E(G) = L(G) or, more gener-ally, in which every subgroup containing the centre is a centralizer.

The study of coset lattices of groups was initiated by Curzio [1953]. Since theatoms of 93(G) are the sets {g} with g c- G and the interval [G/{g}] in 9R(G) isisomorphic to L(G), the coset lattice determines rather strongly the structure of G.In particular, every isomorphism a: 91(G) -+ 91(G) can be considered as a bijectivemap from G to G preserving cosets; such a map will be called an R-isomorphism.Using our results on projectivities of abelian groups of torsion-free rank r > 2 andon torsion-free locally nilpotent groups, we prove in § 9.4 that every R-isomorphismof such a group is an isomorphism or an antiisomorphism. These results are due toLoiko who also showed that the situation is different for mixed abelian groups oftorsion-free rank 1. Gaschiitz and Schmidt [1981] introduced the concept of anaffinity; this is an R-isomorphism a: G -. G satisfying (xH)° = x°H° and (Hx)° _H°x° for all x e G, H < G. In general even such a map need not be an isomorphism,and there exist affinities between nonisomorphic groups. But we show that everyaffinity of G is an isomorphism if G is generated by elements of infinite order andinvolutions, or if Z(G) = 1, or if G is a finite perfect group.

9.1 Lattices of normal subgroups

We have seen in 2.1.4 that the lattice of normal subgroups is a modular sublatticeof the subgroup lattice of the group G. Since %(G) can be much smaller than L(G),rather different groups may have isomorphic lattices of normal subgroups. We givesome examples of this phenomenon.

9.1.1 Examples. Let G be a group.(a) %(G) is a chain of length one if and only if G is simple.(b) If G is abelian, %(G) = L(G). In particular, 9l(Cp.,) is a chain of length n for

every prime p and n E N.(c) For 3 < n 4, A. is a simple normal subgroup of index 2 in S and hence

91(S,,) is a chain of length 2. Similarly, for every odd prime power q > 5, Z(SL(2, q))is a normal subgroup of order 2 with simple factor group in SL(2, q) and hence also9l(SL(2, q)) is a chain of length 2. Finally, since A4 and Klein's four group are theonly normal subgroups of S4, it follows that 91(S4) is a chain of length 3.

(d) For every odd prime p >_ 5, 91(PI'L(2, pz)) 91(Aut A6) ^- 91(D8) 91(Q8) _L(Q8) since all these groups have a unique minimal normal subgroup with factorgroup elementary abelian of order 4.

We introduce a concept which is quite useful in the study of 91(G).

G-isomorphic normal subgroups

Let A and B be normal subgroups of G. A map a: A -- B is a G-isomorphism if itis an isomorphism satisfying (ag)a = (a')9 for all a e A, g e G; A and B are called

9.1 Lattices of normal subgroups 487

G-isomorphic if there exists a G-isomorphism between them. We note some simpleproperties of this concept.

9.1.2 Lemma. Let A and B be G-isomorphic normal subgroups of G.(a) Then CG(A) = CG(B).(b) If A<RaG,B<SaGand RnS=1,then A x B<Z(R x S).(c) If A n B = 1, then A x B is abelian.

Proof. (a) Let a: A -+ B be a G-isomorphism. If c e CG(A) and b = a' E B, then b` =(a')` _ (a`)' = a' = b. Thus c e Co(B) and hence C0(A) < CG(B). Since (b9)'-' =((a')9)'-' = a9 = ((a')' ')9 = (b'-')9, also a-' is a G-isomorphism. It follows thatCA(B) < CG(A) and hence that CG(A) = CG(B), as desired.

(b) Since A n S = 1, we have S < CG(A); similarly, R < CG(B). By (a), CG(A) _CA(B) and hence A x B is centralized by R x S.

(c) Apply (b) with A = R and B = S.

We now show that it is possible to decide in 91(G) whether or not two normalsubgroups A and B with trivial intersection are G-isomorphic.

9.1.3 Lemma. Let A, B g G such that A n B = 1.(a) If a: A -+ B is a G-isomorphism, then C = {aa'la c A} is a normal subgroup of G

satisfying A x B = A x C = B x C.(b) Conversely, if there exists a normal subgroup C of G such that A x B =

A x C = B x C, then A and B are G-isomorphic and abelian.

Proof. (a) By 1.6.2, C is a diagonal in A x B and we only have to show that C a G;then A x B = A x C = B x C. So let g c G. Since a is a G-isomorphism, (aa')9 =a9(a9)' E C for all a e A. Thus C9 < C and hence C a_ G.

(b) By assumption, C is a diagonal in A x B and so, by 1.6.2, C = {aa'ja e Alfor some isomorphism a: A -+ B. Let a e A, g e G. Since C a G, we have (aa')9 =a9(a')9 e C and hence (a')9 = (a9)'. Thus a is a G-isomorphism. By 9.1.2, A and B areabelian.

If we apply 9.1.3 to minimal normal subgroups, we obtain information aboutcertain intervals in 91(G). We add an index N to distinguish in our notation betweenthese and intervals in L(G); thus if S, T e 91(G) such that T:5 S, then

[S/T]N = {X E 91(G)IT < X < S}.

Furthermore, recall that M is the (modular) lattice of length 2 with n atoms.

9.1.4 Lemma. Let A and B be different minimal normal subgroups of G.(a) If A and B are G-isomorphic, they are abelian and [A x B/1]N M for some

cardinal n >_ 3.(b) If A and B are not G-isomorphic, then [A x B/1]N M2.

Proof. Since A and B are different minimal normal subgroups, A n B = 1. As 91(G)is modular, [A x B/1]N has length 2 and therefore is some M where n 2. By 9.1.3,A and B are G-isomorphic if and only if n > 3.

Groups with special lattices of normal subgroups

We now study groups whose lattices of normal subgroups have certain special pro-perties. First we use 1.6.1 to characterize those groups G for which 91(G) is directlydecomposable.

9.1.5 Theorem (Curzio ([1964b]). The lattice 91(G) of normal subgroups of the groupG is directly decomposable if and only if G = H x K where H # 1 0 K and any twonontrivial central factors X/Y of H and SIT of K are coprime (recall that X/Y is acentral factor of H if Y < X < H and [X, H] < Y).

Proof. Suppose first that G = H x K has this property. We claim that for everyU E 92(G),

(1) U=(UnH) x (UnK).Indeed, since U a G and [H, K] = 1, we obtain [UK n H, H] < [UK, H] < U n Hand [UH n K, K] < U n K. Thus if (1) did not hold, then by 1.6.1, UK n H/U n Hand UH n K/U n K would be isomorphic nontrivial central factors in H and Krespectively. But this implies that there exist central factors of the same prime orderp in H and K, which contradicts our assumption. Thus (1) holds. Now we de-fine maps cp: 92(G) - 91(H) x 91(K) and i/i: 91(H) x 91(K) 9t(G) by

U''=(UnH,UnK) and (V,W)'k=V X W

for U E 9t(G), V E 91(H), W E 91(K). Then (1) shows that U° = U and since, trivi-ally, (V, W)0' = ((V x W) n H, (V x W) n K) = (V, W), it follows that cp and 0 areinverse maps. Thus cp is bijective and hence is an isomorphism by 1.1.2. Thus 91(G)is directly decomposable.

Suppose conversely that 92(G) = L1 x L2 where the Li are nontrivial lattices.If Ol, 11 and 02, 12 are least and greatest elements of L1 and L2, respectively,then H = (I1, 02) and K = (01,12) are normal subgroups of G such that H n K =(01, 02) = 1 and H u K = (11,12) = G. Thus G = H x K. Suppose, for a contradic-tion, that there are nontrivial central factors in H and K which are not coprime.Since an infinite cyclic group has factors of every prime order, this implies thatthere are central factors X/Y of H and SIT of K of the same prime order p. Then[X x S, G] = [X, H] [S, K] < Y x T so that (X x S)/(Y x T) is a central elemen-tary abelian subgroup of order p2 in Gi(Y x T). Thus [(X x S)/(Y x T)]N ti MD+1;but [(X x S)/(Y x T)]N [X/Y]N x [S/T]N M2, a contradiction.

Note that 1.6.5 and 9.1.5 imply that if L(G) is directly decomposable for a groupG, then so is 91(G). However, there are groups with 9t(G) decomposable and L(G)indecomposable; see Exercise 1 or simply take G = S3 X C3.

The structure of groups with distributive lattice of normal subgroups is notknown. For finite groups, we have the following characterization.

9.1.6 Theorem (Pazderski [1987]). The following properties of a finite group G areequivalent.

(a) 91(G) is distributive.(b) In every factor group GIN, any two G/N-isomorphic minimal normal subgroups

coincide.(c) In every factor group GIN, any two GIN-isomorphic normal subgroups coincide.(d) The socle of every factor group GIN is a direct product of pairwise non-GIN-

isomorphic minimal normal subgroups.

Proof. Suppose first that 91(G) is distributive. If A/JV and BIN were different G/N-isomorphic minimal normal subgroups of GIN, then by 9.1.4, [AB/N]N M forsome n >_ 3, impossible in a distributive lattice. Thus (a) implies (b).

Now suppose that (b) holds and that L1/N and L2/N are two G/N-isomorphicnormal subgroups of GIN; let a: L1/N -+ L2/N be a G/N-isomorphism. If L, = N,then L2 = N and L1 = L2. So suppose that L1 > N and let M1/N < L, IN be aminimal normal subgroup of GIN. Since a is a G/N-isomorphism, (M1/N)° = M2/Nis a minimal normal subgroup of GIN which is G/N-isomorphic to M1/N. By (b),M, = M2 = M. Now a induces a G/M-isomorphism a': L1/M - L2/M and an obvi-ous induction yields that L1 = L2. Thus (b) implies (c).

Now suppose that (c) holds. It is well-known that a modular lattice is distributiveif and only if it does not contain a sublattice isomorphic to M3 (see Gratzer [1978],p. 59). Thus if 91(G) were not distributive, there would exist distinct normal sub-groups A, B, C of G such that AnB=AnC=BnC=Nand AuB=ABC=B u C. By 9.1.3, A/N and B/N would be G/N-isomorphic, which contradicts ourassumption (c). Thus (c) implies (a).

Finally, it is well-known that the socle of a group is a direct product of minimalnormal subgroups and that any two such decompositions are G-isomorphic (seeRobinson [1982], pp. 80-84). So (b) and (d) are equivalent.

9.1.7 Remark. There are a number of recent results on groups with distributivelattice of normal subgroups; we can only mention the most interesting of these. Forthe proofs and for further results we refer the reader to the literature cited.

(a) Let G be a finite group. Pazderski [1987] proves that if 91(G) is distributive,then every factor group of G has a faithful irreducible character. Longobardi andMaj [1986a] show that the converse is true for groups with nilpotent commutatorsubgroup, but does not hold in general. Their example has supersoluble commuta-tor subgroup.

(b) Again Longobardi and Maj [1986a] show that there are finite soluble groupsof arbitrarily large derived length with distributive lattice of normal subgroups.

(c) (Pazderski [1987]) If G is a finite soluble group with 91(G) distributive, thenAut G is soluble of derived length at most 2k where k is the socle length of G.

(d) (Maj [1984]) The following properties of an infinite supersoluble group G areequivalent.

(i) 91(G) is distributive.(ii) Every finite factor group of G has distributive lattice of normal subgroups.(iii) G = G' (x) where G' is finite of odd order, 92(G/Z(G)) is distributive, and

Z(G/N) is cyclic for every N a G.(e) Longobardi and Maj [1986b] determine the neutral elements in the lattice of

normal subgroups of a finite supersoluble group and prove that if G is an infinitesupersoluble group, then N is neutral in 9t(G) if and only if NT/T is neutral in9t(G/T) for every finite factor group G/T of G.

We mention a result of a different character: every finite distributive lattice isisomorphic to the lattice of normal subgroups of a wreath product of finite non-abelian simple groups (Silcock [1977]) and of a finite soluble group (Palfy [1986]).

Straightforward applications of Zorn's Lemma yield the following result.

9.1.8 Theorem. Let G be a group. Then 91(G) is complemented if and only if G is adirect product of simple groups.

Proof. If G = Dr G, with simple groups G,1 and N a G, Zorn's Lemma impliesAEn

that there exists K c 91(G) maximal with the property that N n K = 1. If NK G,there would exist i e A such that Gz $ NK. Since G, is simple, it would follow thatNK n G, = 1 and hence NKG, = N x K x G,1. So KG, :!g G and N n KG = 1,contradicting the maximality of K. Thus NK = G and K is a complement to N in92(G).

Conversely, suppose that 92(G) is complemented and G 0 1. Then we claim thatG contains a simple normal subgroup. Indeed, if 1 g e G, Zorn's Lemma impliesthat there exists M E 92(G) maximal with the property that g M. Thus M G; letN be a complement to M in 92(G). Then G = N x M and if N were not simple, therewould exist N1 c 92(G) such that 1 < N, < N. Let K be a complement to N, in 91(G)and let N2=NnK.Then N= NnN1K=N,(NnK)=N1N2so that N2is acom-plement to N1 in N. Thus G = N1 x N2 x M and the maximality of M implies thatg c N1M n N2M = M, a contradiction. Thus N is simple. Now Zorn's Lemma im-plies that there exists a maximal direct product S of simple normal subgroups of G.If S 0 G, then G = S x T for some T 1 and, since 91(T) must be complemented,there exists a simple normal subgroup N of T, as we have shown. But this contra-dicts the maximality of S. Thus S = G, as desired.

Isomorphisms between lattices of normal subgroups

In the remainder of this section we study isomorphisms and dualities between thelattices of normal subgroups of two groups G and G. As usual, we want to show thatG and G must be to some extent similar. The examples in 9.1.1 show that this canonly be done under rather restrictive assumptions on G and G. We shall study thesituation where one of the two groups is abelian or nilpotent. First of all we presentsome simple general properties.

9.1.9 Lemma. Let cp: 91(G) 91(G) be an isomorphism.(a) If A a G, then 91(G/A) f-- 91(G/A4').(b) If A and B are G-isomorphic normal subgroups of G such that A n B = 1, then

A" and B1* are 6-isomorphic and abelian.(c) If NA g G (i! E A) such that N = (NAIL e A> = Dr NA, then N`P = Dr N,"'. If, in

AEA AEAaddition, all the NA are G-isomorphic and IAI >_ 2, then N',' is abelian.

Proof. (a) This is obvious since 91(G/A) fit [G/A]N.(b) By (a) of 9.1.3 there exists a normal subgroup C of G which is a diagonal in

A x B. Then C° has the same property in Ac' x B`° and now (b) of 9.1.3 yields thatAc' and B4' are G-isomorphic and abelian.

(c) Since N = Dr NA, we have that 1 = 1`° _ (NA n (N, j 9 µ e A))`° _AEA

Nx n <N''0µ I 0 t e A> and hence N4' = Dr N, ,P. If NA and Nµ are G-isomorphic andAEA

i 96 µ, then by (b), N' and N, are abelian. Thus Nc is abelian if IAI >_ 2.

It follows immediately from 9.1.9 that if G is a torsion-free abelian group of rankro(G) > 2, then G is abelian. So cp is a projectivity from G to G and by 2.6.10, G G.

We want to show that this result holds for a larger class of abelian groups.

Images of abelian groups

We start with torsion groups; clearly, by 9.1.5, we may restrict our attention top-groups. By 9.1.1, we have to assume that G is not cyclic and the crucial case ishandled in the following lemma.

9.1.10 Lemma. Let G = R x S be abelian, R Co., (1 < n < oo) and B < S such thatCBI = p. If (p: 91(G) - 91(G) is an isomorphism, then G = R`° x S`° and R" R.

Proof. Clearly G = R`° x S`°; so we have to show that R`° R. Let A be the sub-group of order p of R. Then A and B are G-isomorphic; hence by 9.1.9, A" and B`°are G-isomorphic and 9.1.2 (b) shows that A`° x B`° < Z(G). It follows that

L(A4' x B`°) _ [A`° x B`°/l], ^- [A x B/1]N = L(A x B)

and hence A`° x Bc is elementary abelian of order p2. Thus

(2) IA4I = p and A" < Z(G).

It suffices to show that R`° is abelian. Indeed, since G = R`° x Sc, it will then followthat L(R`°) = [R4'/l]N [R/1], = L(R) is a chain of length n. And since I A`°I = p,Exercise 1.2.2 (or Theorem 1.2.3) will imply that R4 Cp. R, as desired.

For n c- N, we prove the lemma by induction on n. The induction assumptionapplied to G/A = R/A x SA/A yields that R`°/A`° is cyclic of order pn-'. SinceAc' < Z(R4'), it follows that R`° is abelian.

So let n = x and A. be the subgroup of order p' of R. Then (2) applied to theisomorphism from 91(G/A;) to 92(G/A;°) induced by co yields that Ai±1/A? < Z(G/A;°)and I AT 1 /A;° I = p. Thus

(3) IA" p' for allkEN.

Now suppose, for a contradiction, that Z(R°) < Rw. Since Z(RIO) a G, it follows thatZ(R`°) = A? for some j e N. Choose a e Aj+1 \A? and define the map a: R`° -+ R`° byx° = [x, a] where x e RIP. Since AJ+1/A? < Z(G/Ar), we have x° = [x, a] E A° _Z(R11) and hence by (1) of § 1.5, (xy)° = [xy, a] _ [x, a] [y, a] = x°y° for all x, y c- R.Thus a is a homomorphism from R° into Z(R"); let K be its kernel. Then thehomomorphism theorem implies that R`°/K - (R")° < A? is finite. Since a 0 Z(R`°),we have (R`0)6 0 1 and hence K < R". Every proper normal subgroup of R ° is

normal in G and therefore of the form Ak for some k e N. By (3), K is finite and henceR' is finite. But this is impossible since R° has an infinite chain of subgroups A;°.Thus R4' is abelian, as desired.

9.1.11 Theorem (Curzio [1965]). Let G be an abelian p-group which is not locallycyclic. If cp: 91(G) 91(G) is an isomorphism, then G is abelian (and isomorphic to G).

Proof. We use induction on n to show that the groups S2n(G)`° are abelian for alln e N. Since G = U S2n(G), it follows that G = U S2n(G)" is abelian. So cp is a

MEN MEN

projectivity from G to G and by 2.6.8, G ^- G.By assumption, S21(G) is not cyclic and hence 9.1.9(c) yields that S21(G)`° is abe-

lian. Now suppose that n > 2 and that Qn_1(G)`° is abelian. It is well-known (seeRobinson [1982], p. 105) that S2n(G) is a direct product of cyclic groups and henceQM(G) = Rn x S. where R. is a direct product of k cyclic groups of order p"(0< k< x) and S. < S2n_1(G). Then S,"< SZn_1(G)`° is abelian and n,,(G)`° = R x S.".If k > 2, then 9.1.9(c) implies that R1 is abelian; therefore so is S2n(G)`°. If k = 0,we are done. So suppose that k = 1. Then G/S2n_1(G) has a unique subgroup of orderp and hence is isomorphic to C,, (1 < m < x). If m < x, then G = Stn+rn_1(G) =R x S where R ^- Cp"+,,-, and 1 S < S2n_1(G). Again S° < S2n_1(G)11 is abelian andby 9.1.10, RIO R; thus G = R4' x S( is abelian. So, finally, assume that m = x.Then G/S2n_1(G) C, . and the map r: G -+ G defined by x` = x""-' is an endomor-phism satisfying KerT = S2n_1(G), and hence GT G/KerT C°<. Since a divisiblesubgroup of an abelian group is a direct factor (see Robinson [1982], p. 93), itfollows that G = G` x T where 1 -A T< S2n_1(G). Again T`0 < S2n_1(G)`° is abelianand by 9.1.10, (Gt)r and G = (GT)`° x T' are abelian.

We can now also prove the desired result for abelian groups of torsion-free rankat least 2.

9.1.12 Theorem (Brandl [1986]). Let G be an abelian group which contains two ele-ments a and b of infinite order such that <a> n = 1. If cp: 91(G) -+ 91(G) is anisomorphism, then 6 is abelian (and isomorphic to G).

Proof. Let ro(G) be the torsion-free rank of G. By assumption, ro(G) >_ 2 and thedefinition of ro(G) (or Zorn's Lemma) implies that there exists a free abelian groupF < G of rank ro(F) = ro(G) such that G/F is a torsion group. For p e P, let Sr/F bethe p'-component of G/F. If F° = {xPIx e F}, then G/FP is also a torsion group; letTP/FP be the p'-component of G/FP. Since F/FP is a p-group and TP/FP a p'-group,F n TP = FP. So TPF/F TP/FP is a p'-group and hence TP < SP. Since FTP/TP-F/FP and ro(F) > 2, we see that G/TP is a p-group which is not locally cyclic. By9.1.11, G/ p`° is abelian and hence G' < p" < S. Now F = n S, and it follows thatG'<ns;=(nsv)w=Fm. PEP

vPEP PEP

With F, every F" _ {x"jx e F} (n c N) is free abelian of rank ro(G) and G/F" is atorsion group. So we may apply the above result with F" in place of F and obtainthat G' < (F")". Thus G' < n (F")' = (n F")' = 1 and G is abelian. Now (p is a

n"nEN

projectivity from G to G and by 2.6.10, G ^- G.

As an application of 9.1.12, we get the following result for arbitrary groups Gand G.

9.1.13 Corollary. Let cp: 91(G) - 91(G) be an isomorphism. If ro(G/G') >_ 2, thenG/G' G/G' and (G')m = G'.

Proof. Since q induces an isomorphism from 9t(G/G') onto 9.1.12 im-plies that G/G' ^- G/(G')`°. In particular, G' < (G')' and ro(G/G') >_ 2. So we mayapply 9.1.12 to the isomorphism induced by cp-1 in 91(G/G') and obtain thatG/(G')4'-' is abelian. Thus G'< (G')"' and hence (G')' < G'. It follows that (G')' = G',as desired.

The case ro(G) = 1 has not been settled yet. For example, it is not known whethera group whose lattice of normal subgroups is isomorphic to that of an infinite cyclicgroup is abelian (and hence infinite cyclic). Under the additional assumption that Gis soluble, however, we get the desired result.

9.1.14 Theorem (Curzio [1965]). Let G_be abelian of torsion free rank ro(G) = 1. Ifcp: 9t(G) - 9t(G) is an isomorphism and G is soluble, then G is abelian and ro(G) = 1.

Proof. Assume first that G is torsion-free. Then it is well-known (see Robinson[1982], p. 112) that G is isomorphic to a subgroup of (0, +) and hence is countable;it follows that there exist infinite cyclic groups N such that N, < N2 < and G =XU N. Let N = <x>. Since N,1' is soluble, there exists a characteristic abelian sub-

group A of NNP such that A 1. Then A a_ G and A"' = (x") for some n EFor every prime p not dividing n, Ni = <x") u <x"); hence N`P = A u <x">' andNT/(xP)m A/A n <x"> is abelian. Thus (Nil)' < n (x">' _ (n <x")), = 1

PYR Pk"and hence Ni'D is abelian. Since G = U Nil' and N1° < NN < , it follows that G is

i=1

abelian. Then q is a projectivity from G to 6 and therefore ro(G) = ro(G) = 1.

Now suppose that G is an arbitrary abelian group with ro(G) = 1 and let T =T(G) be the torsion subgroup of G. If X e G\T, then o(x) = oe and hence (x> n T = 1.It follows that (T x <x))O = T° x <x>', that is, T' is centralized by <x>". SinceG = U <x)', the group G is generated by these <x> and hence T° < Z(6). By the

xeG Tcase already settled, G/T" is torsion-free abelian of rank 1. Therefore if x, y e G,<x, y>/<x, y> n T <x, y> T/T is cyclic and, since <x, y> n T< Z(6), it follows that<x, y> is abelian. Thus G is abelian, cp is a projectivity from G to G, and ro(G) =ro(G) = 1.

Images of nilpotent groups

We now study the situation when G is a finite nilpotent group. Example 9.1.1(d)shows that here we must assume not only that the Sylow subgroups of G are notcyclic but also that the image group G is soluble. Then, of course, we can only hopeto prove that G is nilpotent and not that G G. However, the following exampleshows that even this is not possible.

9.1.15 Example. Let N = <x, y> be elementary abelian of order 9, and D the sub-group of Aut N generated by the automorphisms a and /3 of N defined by x° = y,y'=xandxQ=x-1,yv=y. Then o(a)=o($)=2and o(a/J)=4since x'#=yandy''# = x-'. Thus D is a dihedral group of order 8 and we consider the semidirectproduct G = ND. Then N is the unique minimal normal subgroup of G and G/N D.

The dihedral group G of order 16 has a unique minimal normal subgroup Z(G)of order 2 with factor group G/Z(G) D. It follows that 91(G) - R(G), but G is notnilpotent.

In the above example, G' = NZ(D) is supersoluble and c(G) = 3. However, underslightly stronger assumptions, we can prove the desired result. For this we need asimple lemma.

9.1.16 Lemma. Let G be a finite p-group of class 2. If G has only one minimal normalsubgroup N, then GIN has at least 2 minimal normal subgroups.

Proof. Since G is not abelian, G/Z(G) is not cyclic; so there exist two differentminimal subgroups X/Z(G) and Y/Z(G) of G/Z(G). Let X/Z(G) = <xZ(G)> andg e G. Then x° e Z(G) and 1 = [x", g] = [x, g]P since G' < Z(G). Therefore [x, g] e N,the unique minimal subgroup of Z(G), and, since this holds for all g e G, wesee that xN e Z(G/N). Thus X/N = <x, Z(G)>/N < Z(G/N) and, similarly,YIN < Z(G/N). So Z(G/N) is not cyclic since it contains two different subgroups ofthe same order. It follows that G/N has at least 2 minimal normal subgroups.

9.1.17 Theorem (Heineken [1965]). Let G and G be finite groups such that91(G) 91(G) and assume that G is nilpotent with all its nontrivial Sylow subgroupsnoncyclic. If

(a) G' is nilpotent or(b) c(G) < 2 and G is soluble,

then G is nilpotent and IGI = I G!.

Proof. We use induction on IGI. Let cp: 91(G) 91(G) be an isomorphism. Since Gis nilpotent, G = Gl x x G, with nontrivial Sylow subgroups G; and by 9.1.9,G = G4' x ... x G;. Clearly, all the G; and G;° satisfy the assumptions of the theoremand therefore we may assume that G is a p-group.

If G is abelian, then by 9.1.11, G is isomorphic to G and we are done. So let G' 0 1and N be a minimal normal subgroup of G. Then N < Z(G) and, since G is notabelian, G/N is not cyclic. By induction,

(4) G/N'* is a p-group which is not cyclic

since GIN is not cyclic. Suppose, for a contradiction, that G is not a p-group. Thensince G is soluble,

(5) N'' is an elementary abelian q-group for some prime q p.

If M were a minimal normal subgroup of G such that M 0 N, then by (4) 6/M", andhence also N9M"/M9 N`°, would be a p-group, a contradiction. Thus

(6) N is the unique minimal normal subgroup of G.

By Schur's Lemma, every irreducible subgroup of GL(m, q) has cyclic centre andhence

(7) Z(G/CG(N`°)) is cyclic.

We show that this implies that N" < CU(N"). Indeed, if (b) holds, then by 9.1.16, G/Nhas two different minimal normal subgroups. Then G/N' has the same property and,since it is a p-group, this implies that its centre is not cyclic. By (7), N1* QjN`°).Now suppose that (a) holds. Then we claim that G' < ;(N1*). Indeed, this is clear ifN" n G' = 1; and if N10 n G' 1, then N4 r) Z(G') : 1 by a well-known property ofthe nilpotent group G'. Since N' is a minimal normal subgroup of G, it follows thatN' < Z(G'). Thus G' < QJNI'); hence G/CU(N") is abelian and therefore cyclic, by(7). Since GIN" is not cyclic, N° 0 Q(N`°).

Thus in both cases, N'' < CU(N`°) and hence by (4) and (5), ;(N9') = N° x P forsome nontrivial p-group P. Since P is a characteristic subgroup of Ci;(N'0) g G, itfollows that P 9 G and hence P"-' a G, contrary to (6). This contradiction showsthat G is a p-group and then IGI = p" = IGI where n is the length of 91(G).

Theorem_ 9.1.17(a) had been proved by Curzio [1962] under the stronger assump-tion that G is supersoluble. If c(G) = 2 in 9.1.17, however, even the assumption thatG is soluble is rather strong. Using 9.1.16, it is not difficult to show that Example9.1.1(d) is "worst possible" in this situation; that is, if G is a p-group of class 2 andis not a p-group, then G has a unique minimal normal subgroup N'' and I G1N4'I =IG/NI (see Exercise 11). Much deeper is the result of Longobardi and Maj [1987]that if G and G are finite p-groups with 91(G) - 91(G) and c(G) = 2, then c(G) < 3.An example in which c(G) = 3 is given in Exercise 12.

9.1.18 Remark. Heineken's theorem has been generalized to many classes of infinitegroups. We mention some of these results; for the proofs and for further results inthis direction we refer the reader to the literature cited. So let G and G be groupssuch that 91(G) 91(G) and assume that G is not locally cyclic.

(a) (Franchetta and Tuccillo [1976]) Let G be a hypercentral p-group with thefollowing property.

(8) If G' 0 1, there exists N a G such that N < G' and I G'/NI = p.

If G' is hypercentral, then G is a hypercentral p-group such that IGI = IGIThe additional assumption (8), in fact, is needed; indeed, if G = N and G =

M<v> with N C2x, M C,, (p > 2), u and v involutions inverting N and M,respectively, then G and G have isomorphic lattices of normal subgroups. Here G isa hypercentral 2-group and G' = N has no subgroup of index 2. _

(b) (de Giovanni and Franciosi [1985]) If G is a nilpotent p-group and G' ishypercentral, then G is a hypercentral p-group such that IGI = IGI. _

(c) (Longobardi and Maj [1987]) If G is a nilpotent p-group of class 2 and G issoluble, then G is a nilpotent p-group of class at most 3._ (d) (de Giovanni and Franciosi [1985]) Let G be a torsion-free nilpotent group. IfG is hypercyclic or if G is finitely generated and G' hypercentral, then G is torsion-free nilpotent and c(G) = c(G).

(e) (Brandl [1986]) If G is a finitely generated torsion-free nilpotent group of class2, then so is G.

Dualities

Results similar to those for isomorphisms hold for dualities between lattices ofnormal subgroups. Again we consider only finite nilpotent groups G, but now weassume that G is supersoluble; recall that then (see Robinson [1982], pp. 145-146)

(9) G has nilpotent commutator subgroup and normal subgroups of order r andof index s where r is the largest and s the smallest prime dividing IGI.

9.1.19 Lemma. Let G be a noncyclic finite p-group, G a finite supersoluble group and6: 91(G) -+ 91(G) a duality.

(a) If Z(G) is not cyclic, IGI =IGI(b) If Z(G) is cyclic, then G' is a p-group and E;/G' a cyclic q-group, p > q E P.

Proof. Since G is not cyclic, G/D(G) is elementary abelian of order p" where n >_ 2.So if M = Q>(G)a, then IM/1]N is dual and hence also isomorphic to [G/(D(G)]NL(G/O(G)). By 9.1.4, M_= M, x x M. with G-isomorphic minimal normal sub-groups M. of G. Since G is supersoluble, I M; I = r for some prime r and an elementg e G induces the same power in every M,. Thus every subgroup of M is normal inG and so [M/1]N = L(M). It follows that r = p and I M I = p"

(a) Now if Z(G) is not cyclic, then the product S of all the minimal normal sub-groups of G is an elementary abelian central subgroup of order p'° of G where m >_ 2.


Therefore [G/S6]N [S/1]N = L(S) and by 9.1.4, as above, G/S' is elementary abelianof order p'. Every maximal normal subgroup of G contains Sb and therefore hasindex p in G, every minimal normal subgroup of G is contained in M and thereforehas order p. By (9), G is a p-group and then IGI = p' = IGI where k is the lengthof 91(G).

(b) Since Z(G) is cyclic, G has a unique minimal normal subgroup N and Na is theunique maximal normal subgroup of G. Thus 6/6' is an abelian group with only onemaximal subgroup and hence is a cyclic q-group for some prime q. By (9), G' isnilpotent and hence is a p-group since every minimal normal subgroup of G hasorder p. If p = q, then G is cyclic since it has only one maximal normal subgroup.But M is elementary abelian of order p", n > 2, a contradiction. Thus p 0 q and thenby (9), p > q.

Note that if G is a nonabelian group of order p3 and G a P-group of order pzq,then 91(G) is dual to 91(G). Thus case (b) in Lemma 9.1.19 actually occurs.

9.1.20 Theorem (Curzio [1962]). Let G be a finite nilpotent group all of whose non-trivial Sylow subgroups have noncyclic centre. If G is_a finite supersoluble group and6: 91(G) -+ 91(G) is a duality, then G is nilpotent and IGI = IGI.

Proof. Since G is nilpotent, G = G, x x G, with nontrivial Sylow subgroups G.Then, as in 8.1.8, G = G; x x G; where G; = U G. So b induces a duality from

#i91(G/G;) 91(G;) onto 91(G°) and by (a) of 9.1.19, I GbI = I G;I. Thus G is nilpotent andIGI = IGI.

Theorem 9.1.20 has also been generalized to infinite groups; see Franciosi [1981]and de Giovanni and Franciosi [1986].

Exercises

1. (Curzio [1964b]) (a) Let G be an infinite supersoluble group. If 91(G) is directlydecomposable, show that G = H x K where H is infinite and K is a finitegroup of odd order.

(b) Construct an infinite supersoluble group for which 91(G) is directly decom-posable. (Note that by 1.6.5 and (a), L(G) cannot be directly decomposable.)

2. Let G be a finite group.(a) (Curzio [1957b]) If any two normal subgroups of G have different order,

show that 91(G) is distributive.(b) Give an example of a group G with 91(G) distributive having two normal

subgroups of the same order.3. (Curzio [1964b]) Let G be a supersoluble group. Show that 92(G) is a chain if

and only if G is either a cyclic p-group or a group of order p"q' with cyclic Sylowsubgroups and trivial centre (p and q primes, p : q).

4. (Longobardi and Maj [1986a]) Let G be a finite group and A, B 9 G such that(JAI, IBI) = 1. If 91(G/A) and %(G/B) are distributive, show that 92(G) is distribu-tive.

5. (Longobardi and Maj [1986a]) Let G = P<x> be a finite supersoluble group, Pa normal Sylow p-subgroup and P n <x> = 1. Show that 91(G) is distributive ifand only if <a>" = " for all a, b e P for which there exists r e 77 such thatax= a' andbx=b'.

6. (Maj [1984]) If G is an infinite supersoluble group, show that 91(G) is distribu-tive if and only if 91(GIN) is distributive for every finite factor group GIN of G.(Hint: First show that this condition implies that G = G' <x> where G' is finite ofodd order.)

7. Show that 91(G) is a Boolean algebra (that is, it is distributive and comple-mented) if and only if G is a direct product of simple groups such that no twoabelian factors are isomorphic.

8. (Zitarosa [1952]) If G is a finite nilpotent group, show that N is a neutralelement in 91(G) if and only if N is the unique normal subgroup of its order in G.

9. (Longobardi and Maj [1986b]) Let G be a finite group with normal Sylowp-subgroup P. Show that P is a neutral element of 91(G) and that N e 91(G) isneutral in 91(G) if and only if N n P and NP/P are neutral elements in 92(G) and91(G/P), respectively.

10. (Longobardi and Maj [1986b]) If G is an infinite supersoluble group and N aneutral element in 9I(G), show that either N or GIN is finite.

It. (Brandl [1986]) Let G and G be finite groups, G a p-group of class 2 and supposethat 91(G) 91(G). _(a) If Z(G) is not cyclic, show that IGI = IGI.(b) If Z(G) is cyclic and N is the (unique) minimal normal subgroup of G, show

that I G/NIQ I = I G/N I.12. (Heineken [1965]) Let p > 2, G = <x,y,zIx"2 = y°Z = z"2 = EX, Z] = [y,z] = 1,

[x, y] = z> and let G be the group with the same defining relations except that[x, z] = 1 is replaced by [x, z] = z°. Show that 92(G) ^- 91(G), c(G) = 2 andc(G) = 3.

9.2 Lattices of subnormal subgroups

We have seen in § 1.1 that the set R(G) of all composition subgroups of a group G isa sublattice of L(G) and that if G has a composition series, the composition sub-groups coincide with the subnormal subgroups of G. In this section we are going tostudy the lattice R(G), but mainly for finite groups. Since in this case the term"subnormal subgroup" is more common, we shall usually speak of the lattice ofsubnormal subgroups of G. Some of the results also hold in the more general situa-tion where G has a composition series, but in most cases it is not known whathappens for arbitrary groups.

General properties of subnormal subgroups

Recall that S :9!9 G if and only if there exist subgroups S; of G (i = 0, ... , n) such that

(1) S=So9S1 g...aS=G;

9.2 Lattices of subnormal subgroups 499

the smallest integer n for which there exists such a chain (1) is called the defect of Sin G. If H<G and (1) holds, then

S n H :g:9 H. In particular, S !9:9 H if S g g G and S< H< G. It followsthat

(2) S :!5i H if S, H E R(G) and H covers S in R(G).

It is obvious that Sag H :9!9 G implies S g g G and that if N g G and N< S< G,then S :9:9 G if and only if S/N :9!9 GIN. So we have the following basic propertyof the lattice of subnormal subgroups of a group G with a composition series.

(3) If N < H < G, H !9:9 G and N a H, then [H/N]K =- R(H/N).

Here, for arbitrary S, T ER(G) such that T < S, we write

[S/T]K = {X ga GET < X < S}

for the interval in R(G).Now let G be finite, it a set of primes and S satisfy (1). Then 0"(S;) is a char-

acteristic subgroup of Si g Si,, and hence a normal 7r-subgroup of Si,,. Thus0.(S;)< On(Si+1) and so, by induction,

(4) 0.(S) :!g On(G).

In particular, S < O"(G) if S is a 7t-subgroup of G. In just the same way we see thatif S is soluble, then S < Ga, the soluble radical of G. It follows that SG < Ga and sowe have:

(5) If S is a soluble subnormal subgroup of G, then SG is soluble.

Dually, suppose that N a G and GIN is a rr-group. Then O"(G) < N and N/O"(G) isa rr-group; so O"(N) <_ O"(G). On the other hand, O"(N) is a characteristic subgroupof N a G and hence a normal subgroup of G with factor group a rr-group. ThusO"(G) < 0"(N) and then 0"(N) = O"(G). This generalizes to a subnormal subgroupS whose index in G is a rr-number, that is, is divisible only by primes in it.

(6) If S a g G and I G: SI is a rr-number, then 0"(S) = O"(G).

Indeed, for the S. in (1), O"(S) = 0"(S,) = = O"(S") = O"(G). Clearly, (4) impliesthe following property which we, in fact, proved in 6.2.15 for arbitrary groups G.

(7) If S is a subnormal 7r-subgroup of G, then SG is a n-group.

It follows that

(8) [S, T] = 1 if S and T are coprime subnormal subgroups of G.

Indeed, [S, T] < S' n TG = 1 since SG is a it-group and TG a ir'-group for a suitableset it of primes. Less trivial is the following result of Wielandt.

(9) Let S, T :9:9 G such that S n T = 1. If S is a nonabelian simple group, then[S,T]=1.

Proof. Suppose that (9) is wrong and choose a counterexample S, T for which thedefect d of S in J = <S, T> is minimal; let S = So a a Sd = J. If d < 1, then S a Jand [S, T] g S. Since [S, T] 0 1 and S is simple, it follows that S = [S, T], sothat S < T' and V = J. Therefore T = J by subnormality of T, and this impliesS = S n T = 1, a contradiction. Thus d > 2 so that S S' for some t E T. SinceS n S' !9!9 S and S is simple, S n S' = 1. Now S` < Sd_1 and the minimality of dyields [S, S'] = 1. So for any x, y e S, by (1) of § 1.5,

I = [x, Y'] = [x,Y[Y, t]] = [x, [y, t]] [x, y][r.r[;

since [S, T] a J, it follows that [x, y] e [S, T]. Thus S' < [S, T]; but S = S' since Sis nonabelian simple and so S < [S, T]. As before this leads to T' = J and thecontradiction S = 1.

Again let S be a nonabelian simple subnormal subgroup of G. By (9), any twodifferent conjugates of S centralize each other and hence S" g SG = <S91g e G> forall x e G. By Zorn's Lemma, SG is the direct product of some of these S". Thus wehave shown:

(10) If S !9!9 G and S is a nonabelian simple group, then SG is a direct product ofconjugates of S.

It is easy to see that SG is also a minimal normal subgroup of G. Note that neitherthis nor any of properties (9) and (10) in general holds if IS! is a prime; this is shownby any nonnormal subgroup S of order p of a finite p-group G. We discuss somefurther examples.

9.2.1 Examples. Let G be a finite group.(a) 1(G) is a chain of length one if and only if G is simple. (Note that for p e P,

R(Cp< x Cp) is also a chain of length 1.)(b) If G is nilpotent, every subgroup of G is subnormal and hence R(G) = L(G). In

particular, R(Cph) is a chain of length n for every prime p and n E N.(c) The unique maximal normal subgroup of S4 is A4 and that of A4 is V4. So we

get the lattices S1(G) in Figure 25 for G = Q8, A4, and S4. Now let G = Q8C3 be thesemidirect product of Q8 by a subgroup of order 3 of its automorphism group. This

R(08) R(A4) R(S,) R(Q8C3)

Figure 25

C3 permutes transitively the subgroups of order 4 of Q8 and hence G has no normalsubgroup of index 2. Thus Q8 is the unique maximal normal subgroup of G andR(G) consists of G and the subgroups of Q8.

(d) Finally, as in 9.1.1 we see that for 3 < n 4 and every odd prime power q > 5,R(S.) R(SL(2, q)) is a chain of length 2; furthermore, for every odd prime p 5,

R(PFL(2, p2)) R(Aut A6) R(Q8) = L(Q8).

Groups with special lattices of subnormal subgroups

We now study groups whose lattices of composition subgroups have certain specialproperties. First we use 1.6.1 to characterize those groups G with composition seriesfor which R(G) is directly decomposable; note the similarity to Theorem 9.1.5.

9.2.2 Theorem (Tamaschke [1961]). Let G be a group with a composition series.Then R(G) is directly decomposable if and only if G = H x K where H 1 K andno abelian composition factor of H is isomorphic to a composition factor of K.

Proof. Suppose first that G = H x K has this property. We claim that for everyUeR(G),

(11) U=(UnH)x(UnK).Indeed, let U = U a Ui_1 a a Uo = G and suppose, for a contradiction, that(11) does not hold for U. Since (11) is trivially satisfied for Uo = G, there existsa smallest integer k such that (11) does not hold for Uk. We may suppose that U =Uk so that U a U1i_1 = S x T where S = Uk_1 n H and T = U1i_1 n K. By 1.6.1,UT n S/U n S and US n T/U n T are nontrivial isomorphic factors in S < H andT< K, respectively. Since U a S x T, we have [UT n S, S] < [UT, S] < U n S, soUT n S/U n S is abelian, as a central factor in S. Since U and S x T are subnormalsubgroups of G, it follows that UT n S:!9!9 H and US n T a a K. Thus a composi-tion factor of UT n S/U n S is an abelian composition factor of H which is iso-morphic to a composition factor of K, a contradiction. This proves (11). Now asin the proof of 9.1.5, cp: R(G) -+ R(H) x R(K) defined by U`° = (U n H, U n K) forU e R(G) is an isomorphism.

Suppose conversely that R(G) = L1 x L2 where the L. are nontrivial lattices. If01, , and 02, 12 are least and greatest elements of L1 and L2, respectively, thenH = (11, 02) and K = (01, I2) are subnormal subgroups of G such that H n K =(01, 02) = 1 and H u K = (11,12) = G. Since H is the unique complement of Kin R(G), every (inner) automorphism of G fixing K also fixes H. It follows thatK < NG(H) and so H a G. Similarly, K g G and hence G = H x K. If X/Y and SITwere composition factors of the same prime order p in H and K, respectively, thenas in the proof of Theorem 9.1.5, (X x S)/(Y x T) would be an elementary abelianfactor of order p2 in G and, by (3), [(X x S)/(Y x T)]K !-- Mp+1, a contradiction.Thus H and K have no isomorphic abelian composition factors.

Next we study finite groups with modular and distributive lattices of subnormalsubgroups. Here we have two similar results.

9.2.3 Theorem (Zappa [1956b]). The following properties of the finite group G areequivalent.

(a) R(G) is modular.(b) If T :g Sag G and SIT is a p-group, p a prime, then L(S/T) is modular.(c) If T :g S as G and IS/TI = p3, p a prime, then L(S/T) is modular.

Proof. If R(G) is modular and T!9 S :9!9 G such that SIT is a p-group, then by (3),L(S/T) = R(SIT) [SIT]K'S modular. Thus (a) implies (b), while it is trivial that (b)implies (c).

Now suppose that G satisfies (c). By 2.1.8, R(G) is lower semimodular. We showthat R(G) is also upper semimodular; then by 2.1.10, R(G) is modular and (c) implies(a). For this, since R(G) is finite, it suffices to show that if H, K E R(G) such that Hand K cover H n K, then H u K covers H (see Gratzer [1978], p. 173). By (2),H n K a H u K and by (3), we may assume that G = H u K and H n K = 1. ThenH and K are minimal subnormal subgroups of G, hence simple groups, and we claimthat HK = KH. Indeed, this follows from (9) if H or K is nonabelian, and from (8) ifH I and I K I are different primes. And if I H I_ I K I = p e P, then by (4), H u K is a

p-group. Our assumption implies that every section of order p3 of H U K has modu-lar subgroup lattice. By 2.3.3, H u K is an M-group and by 2.3.2, HK = KH, asdesired. Now if X E R(G) such that H< X < HK, then X = H(X n K) and, since Kis simple, X n K = I or X n K = K. Thus X = H or X = HK and so HK covers Hin R(G).

9.2.4 Theorem (Zacher [1957]). The following properties of the finite group G areequivalent.

(a) R(G) is distributive.(b) If T g S 99 G and S/T is a p-group, p a prime, then SIT is cyclic.(c) If T :Q S:9 -_9 G and IS/TI = p2, p a prime, then SIT is cyclic.

Proof. If R(G) is distributive and T:9 S 99 G such that SIT is a p-group, then by(3), L(S/T) = R(SIT) [S/T]K is distributive and hence S/T is cyclic by 1.2.4. Thus(a) implies (b) and again (b) clearly implies (c).

Now suppose that G satisfies (c). If K a H a a G such that I H/K I = p3, then everysubgroup of order p2 of H/K is cyclic and hence, by 2.3.3, L(H/K) is modular; by9.2.3, R(G) is modular. Suppose, for a contradiction, that R(G) is not distributive.Then since R(G) is finite, there exists a 5-element sublattice {A, B, C, N, M} of R(G)such that AnB=AnC=BnC=N,AuB=AuC=BuC=M andA,B,Call cover N (see Birkhoff [1948], p. 134). Then M covers A, B, C and hence by(2), A, B, C, and N are normal subgroups of M. Thus M/N = A/N x B/N =A/N x C/N = B/N x C/N with simple groups A/N, B/N, C/N. By 9.1.3 applied toM/N, A/N B/N is abelian and hence M/N is elementary abelian of order p2,contradicting our assumption. Thus R(G) is distributive and (c) implies (a).

9.2.5 Corollary (Zappa [1956a]). Let G be a finite soluble group. Then R(G) is dis-tributive if and only if every Sylow subgroup of G is cyclic.

Proof. If R(G) is distributive, then by 9.2.4, every chief factor of G is cyclic, that is, Gis supersoluble. So G has a Sylow tower (see Robinson [1982], p. 145) and again9.2.4 implies that every Sylow subgroup of G is cyclic. Conversely, any group withcyclic Sylow subgroups clearly satisfies (b) of 9.2.4.

Another characterization of finite groups with distributive lattice of subnormalsubgroups is given in Exercise 2. Using some deeper properties of the permutizerPH(K) = (X IX < H, XK = KX> of two subnormal subgroups H and K, it is possi-ble to generalize 9.2.3 and 9.2.4 to infinite groups. We give this result without proof.

9.2.6 Theorem (Napolitani [1970]). Let G be a group and SN(G) the partially or-dered set of all subnormal subgroups of G.

(a) SN(G) is a modular sublattice of L(G) if and only if for all S, T< G such thatT :!g Sag G and IS/TI = p3, p a prime, L(S/T) is modular.

(b) SN(G) is a distributive sublattice of L(G) if and only if for all S, T< G such thatT :g Sag G and IS/TI = p2, p a prime, S/T is cyclic.

Our next result together with Theorem 9.1.8 shows that for a finite group, R(G) iscomplemented if and only if 91(G) is complemented; and in this case, R(G) = 91(G).

9.2.7 Theorem (Curzio [1958]). The following properties of the finite group G areequivalent.

(a) R(G) is complemented.(b) Every minimal normal subgroup of G is a direct factor of G.(c) G is a direct product of simple groups.(d) R(G) is relatively complemented.

Proof. Suppose first that R(G) is complemented and let N be a minimal normalsubgroup of G. Then N E R(G) and hence there exists M e R(G) such thatG = N u M = NM and N n M = 1. Since M is a proper subnormal subgroup of G,there exists L a G such that M < L < G. By Dedekind's law, L = (N n L)M. SinceN is a minimal normal subgroup of G, N n L = N or N n L = 1. In the firstcase, L = NM = G, a contradiction. Hence N n L = 1 and so M = L g G andG = N x M. Thus (a) implies (b).

We use induction on IGI to show that (b) implies (c). Let N be a minimal normalsubgroup of G and let M < G such that G = N x M. If R is a minimal normalsubgroup of M, then since M is a direct factor, R is a minimal normal subgroup ofG. By assumption there exists S < G such that G = R x S; it follows that M =R x (M n S). Thus M satisfies (b) and therefore, by induction, it is a direct productof simple groups. Then G = N x M is also such a direct product and (c) holds.

Now if G satisfies (c), then by 9.1.8, 91(G) is complemented. Thus every normalsubgroup N of G is a direct factor and hence every normal subgroup of N is normalin G. It follows that every subnormal subgroup is normal, that is, 51(G) = 91(G).Then R(G) is modular and complemented and hence relatively complemented. Thus(c) implies (d), while it is clear that (d) implies (a).

We finally mention that Pazderski [1972] determined the finite soluble groupshaving a unique minimal subnormal subgroup (see Exercise 4) and that Tamaschke[1961] characterized the neutral elements of R(G) for a group G with a compositionseries. He shows that N E 52(G) is neutral in R(G) if and only if for every S E 51(G), noabelian composition factor between N and N n S is isomorphic to a compositionfactor between S and N n S.

Isomorphisms between lattices of subnormal subgroups

We are now going to study isomorphisms between lattices of subnormal subgroupsof two groups G and G. Here we restrict our attention to finite groups and, in fact,most often even to finite soluble groups. First we want to show that under suitableassumptions, suggested by the examples in 9.2.1, such an isomorphism maps p-groups to p-groups and, more generally, Op(G) to Oo(G) and O°(G) to O°(G). For thiswe need the following simple lemma.

9.2.8 Lemma. Let G and G be finite groups and (p: 51(G) -+ R(G) an isomorphism. IfN, H E R(G) such that N < H and H/N is a noncyclic elementary abelian p-group,then Nw 9 H' and H`0/NI* H/N.

Proof. Since H/N is not cyclic, there exists D E R(G) such that N < D < H andIH/DI=p2;letM,ER(G)such that D<M.<H(i= 1,2,3)andM1 :A M2:94 M3 :0 MI -

By (2), M;° < H`0; hence D4' = M; n MZ < H`° and H`0/D4' = Mr /D`0 x Mf/D4' for alli j. By 9.1.3, H4'/D`0 is abelian and then L(H'/D`0) ^- [H4!D4']K [H/D]K - L(H/D).By 2.2.5, H4'/D`0 is elementary abelian of order p2. Since N' is the intersection of allsubgroups D4' such that N < D < H and H/DI = p2, it follows that N10 < H4' andH10/N10 is an elementary abelian p-group. But then I H`0/N4 I = p" = I H/N I where n isthe length of [H4'/N']K - [H/N]K and hence H4'/N4 - H/N.

9.2.9 Theorem (Curzio [1957a], Zacher [1962]). Let G be a finite p-group andq : R(G) -+ R(G) an isomorphism. _

(a) If G has at least 2 minimal subgroups, then G =IGI _(b) If G is generalized quaternion and G is a finite soluble group, then G G.

Proof. (a) We use induction on I G I. Since G has 2 minimal subgroups, G is not cyclicand hence G/(D(G) is an elementary abelian p-group of order p" for some n >_ 2. By9.2.8, IG: b(G)4I = p". So if D(G) = 1, we are done. And if I(G) :A 1, we chooseA, B < G of order p such that A < Z(G) and A : B. Then AB is elementary abelianof order p2, hence AB ; G and so there exists a maximal subgroup M of G contain-ing AB. By induction, M101 = M and, since 16: M1* 1 = p, it follows that IGI =IGI

(b) Let I G I = 2" and N= Z(G). Then Nw is the unique minimal subnormal sub-group of G. Hence N`° < G and IN4I = p, a prime, since G is a finite soluble group.Now G/N is a dihedral group and so by (a), JO/N`01 = I G/NI = 2"-'. Since IN4I = p,the group G/C;(N4') is cyclic; on the other hand, R(G/N4') is not a chain and henceG/N`0 is not cyclic. Thus N4' < CG(N`0). So if p > 2, then Q(N4') = N`P x S for some

nontrivial 2-group S; but then S would contain a minimal subnormal subgroup ofG, a contradiction. It follows that p = 2 and 161 = I G I. Then L(G) = R(G), so G hasonly one minimal subgroup and hence is generalized quaternion of order 2". ThusGAG.

We want to generalize Theorem 9.2.9 to an assertion on the images of OP(G) andOP(G). For this we need the following lemma.

9.2.10 Lemma. Let G and G he finite groups, cp: R(G) -+ R(G) an isomorphism and ita set of primes. If G is soluble, then OjG)° and O' (G)" are characteristic subgroupsof G.

Proof. (a) Let N = On(G). We use induction on I G I to show that N° is characteristicin G. We may assume that N A 1 and claim that there exists

(12) L g G such that 1 -A L < N and LW is characteristic in G.

By induction, it will follow that NW/LW is characteristic in G/LW and then N° ischaracteristic in G, as desired. To prove (12), consider a minimal normal subgroupM of G contained in N. If MW is characteristic in G, we are done. So assume that M'is not characteristic in G. Since M is the join of minimal subnormal subgroups, thesame holds for MW and hence there exist a minimal subnormal subgroup R' < M"and a e Aut G such that (R4)a RW. Since G is soluble, I RWI = I(RW)aI = p, a prime.Let L< G such that L° = OP(G). By (4), R° and (R")a are different minimal sub-groups of Li" and hence by 9.2.9, L is a p-group. Therefore I R I = p and p e it sinceR < M < N. Again by (4), L < OP(G) < N. Hence OP(G) has more than one minimalsubgroup; so 9.2.9 and (4) imply that OP(G)° is a p-group and OP(G)Y < OP(G) = Lu'.It follows that L = OP(G) a G and L satisfies (12).

(b) Now let N = 0"(G). There are subgroups L of G such that

(13) N < L as G and LIP is characteristic in G;

for example, L = G satisfies (13). Let L be minimal with this property and suppose,for a contradiction, that N < L. Since N10 g g L", there exists a maximal subnor-mal subgroup KW of L" containing NW. By (2), K4' a LO and, since G is soluble,ELI': KWI = p, a prime. The minimality of L implies that K° is not characteristic in G;thus there exists a e Aut G such that (KW)a -* K". For every such a, (K")8 is a normalsubgroup of index p in (LW)a = LW; hence M" = n_ (KW)a is characteristic in G

ac AutGand LW/MW is a noncyclic elementary abelian p-group. By 9.2.8, M a L and L/M isan elementary abelian p-group. Thus I L : K I = p and p e it since N < K. It followsthat I G : MI = I G : LI IL: MI is a rr-number and hence by (6), N = OR(G) = O(M) < M;but this contradicts the minimality of L. Thus N = L and N" is a characteristicsubgroup of G.

We can now prove the desired result on OP(G)" and OP(G)"'. We clearly have toassume that OP(G) and G/OP(G) are not cyclic; but the examples in (d) of 9.2.1 showthat for p = 2, an additional assumption is needed.

9.2.11 Theorem (Pazderski [1972] ). Let G and G be finite groups, p: R(G) -+ R(G) anisomorphism and p a prime.

(a) Assume that OP(G) is not cyclic and if p = 2, in addition, that G is soluble. ThenOP(G)4' = Op(G). _

(b) Assume that G/OP(G) is not cyclic and if p = 2, in addition, that G and G aresoluble. Then OP(G)4' = OP(G).

Proof. (a) Let N = OP(G). By 9.2.9, N° is a p-group and so N4' < OP(G), by (4).If N4' < OP(G), take H such that N4' < H< OP(G) and I H : N"I = p. Since N4' isnot cyclic, neither is H and hence 'D(H) < N4' < H. By 9.2.8, H4' '/D(H)4-' is anelementary abelian p-group. So H4-' is a subnormal p-subgroup of G and by (4),H4 < OP(G) = N, a contradiction. Thus N4' = OP(G), as desired. _

(b) We use induction on IGI to show that OP(G)4' > OP(G). Then 6/OP(G) is notcyclic and the above assertion for cp-t yields the other inclusion. So let N = OP(G),DIN = (D(G/N) and assume that the assertion is true for groups of smaller order.Since GIN is not cyclic, neither is G/D and so by 9.2.8, D4 < G and G/D4' is anelementary abelian p-group. Thus D4' > OP(G) and we are done if D = N. So we mayassume that G/N is not elementary abelian.

If M is a maximal subgroup of G containing N, then 16: M4I = p and by (6),OP(M) = OP(G) = N and OP(MP) = OP(G). If M/N is not cyclic, the induction as-sumption yields that N4' = OP(M)4' > OP(M10) = OP(G), as desired. So we may as-sume that G/N has only cyclic maximal subgroups. Since it is not elementary abelianof order p2, it then has only one minimal subgroup and is isomorphic to Q8. Thusp = 2, hence G is soluble, by assumption, and so N4' a G, by 9.2.10. Now 9.2.9 yieldsthat GIN" -_ Q8 and hence N4' > 02(G), as desired.

Our second main result on isomorphisms between lattices of subnormal sub-groups is that under suitable assumptions they are "index preserving".

9.2.12 Theorem (Heineken [1965]). Let G and G be finite soluble groups,cp: R(G) - R(G) an isomorphism and p a prime. If OP'(G)/OP'P(G) is not cyclic, thenHl*: K4I = p for every composition factor H/K of order p of G.

Proof. We use induction on I G I and first note that for any normal subgroup N andsubgroups Y< X of a group,

(14) 1X:Yj=IXN:YNIIXnN:YnNI.Indeed, if we write T = X n YN = Y(X n N), then IX : TI _ IX : X n YNI _IX N : YNI and I T : YJ _ I Y(X n N) : YI = IX n N : Y n NI, and (14) follows.

Now let K a H << G such that JH : KI = p. Since G is soluble and K4' is a maxi-mal subnormal subgroup of H4',

(15) 1 H4': K4' is a prime.

If G is a p-group, then OP'(G) = G and OP'P(G) = 1 so that G is not cyclic. By 9.2.9,G is a p-group and IH4' : K"I = p. So we may assume that G is not a p-group; writeR = OP'(G), S = OP(R), P = OP(G), and Q = Op-(G). By 9.2.10, RIO, S4', P4', Q4' are

normal subgroups of G. By (14), I HR: KR I H n R : K n R I= I H: K I= p and, sinceG/R is a p'-group, it follows that I H n R : K n R I = p. If R G, the inductionassumption applies to R and so p = j(H n R)W : (K n R)1*I = I H4 r) RP : K° n R"j;then (14) shows that p divides I H": K"I and (15) implies that I H4' : K4I = p, asdesired. So we may assume that

(16) G = R = OP'(G).

Suppose that there exists T < S such that 1 < T a G, T4' < G, and HT # KT. Thenby (14), JHT: KTI = p and we may apply the induction assumption to G/T. Weobtain that I H4T4 : K°T4' = p and then again IH'P: KWI = p, by (14) and (15). So wemay assume that

(17) HT = KT for all T<Ssuch that I <T<GandT"< G.

Since G/S is a p-group and Q a p'-group, Q< Sand I H n Q: K n Q I is prime to p.So (14) implies that HQ # KQ and, since Q4 g G, it follows from (17) that

(18) Q = 1.

This implies that P is the Fitting subgroup of G and so CG(P) < P, by a well-knownproperty of soluble groups (see Robinson [1982], p. 144). If P were cyclic, thenGICG(P), as a subgroup of Aut P, would be abelian. Thus G/P is abelian and (16)implies that G/P is a p-group. But then G is a p-group, contrary to our assumption.Thus P is not cyclic and by 9.2.9,

(19) P° is a p-group.

Since G is not a p-group, S # 1 and by (18), every minimal normal subgroup of G isa p-group; thus T = P n S 1. Of course, T a G and T" = P4' n S4' < G. By (17),HT = KT and then (14) shows that I H n T : K n Tj = p. Since T4' < P° is a p-group, I H4 n T4': Mr) T °J = p and again (14) and (15) yield that JH`° : K"I = p.

9.2.13 Corollary. Let G and G be finite soluble groups, q : R(G) -* R(G) an isomor-phism and assume that for every prime p dividing IGI, the group OP'(G)/OP'P(G) is notcyclic. Then IH4I = IHI for every H E R(G).

Proof. By 9.2.12, cp maps every composition factor of G onto a composition factorof the same order in G. It follows that IH4I = I H I for every H ER(G).

Exercise 6 shows that it is not possible to replace the assumption on the factorgroup OP'(G)/OP'P(G) in Theorem 9.2.12 by the assumption that the Sylow p-subgroups of G are not cyclic.

Dualities

We have seen in Chapter 8 that for a group G to have a duality of L(G) is a ratherrestrictive property. The same holds for R(G), at least if G has sufficient subnormalsubgroups. We want to show that if G is a finite soluble group of odd order with a

duality 6: R(G) -+ R(G) and G is finite soluble, then G is supersoluble. For this weneed some preliminaries.

9.2.14 Lemma. If R(G) has finite length and b: R(G) --+.q(C) is a duality, then R(G) ismodular.

Proof. By 2.1.8, R(G) and R(G) are lower semimodular. It follows that R(G) is uppersemimodular and hence modular, by 2.1.10.

By (3), we have the following inheritance properties (compare with 8.1.6).

9.2.15 Lemma. Let G and G be finite groups and 6: R(G) R(G) a duality.(a) If N g G, then b induces a duality from R(GIN) to R(N").(b) If N° a G, then 6 induces a duality from R(N) to R(6/N°).

More interesting are the following two results on maximal elementary abeliansections.

9.2.16 Lemma. Let G and G be finite soluble groups and 6: R(G) -+ R(G) a duality.Suppose that p is a prime and N is the join of all subnormal subgroups of order p of G.Then N is an elementary abelian normal p-subgroup of G and N° is a characteristicsubgroup of G. If N is not cyclic, 6IN° _- N.

Proof. This is clear if there are no subnormal subgroups of order p in G, that is ifN = 1; so let N # 1. Then N° is the intersection of maximal subnormal subgroups ofG and hence N° G, by (2). Thus b induces a duality from R(N) onto R(G/N°). By(4), every subnormal subgroup of order p of G is contained in Op(G) and therefore Nis a normal p-subgroup of G. So R(N) = L(N) and by 9.2.14, L(N) is modular. SinceN is generated by subgroups of order p, it follows from 2.3.5 that N is elementaryabelian. Thus R(N) is self-dual and so R(6/N°) R(N). By 9.2.8, 6IN° N if N isnot cyclic; in any case, since G is soluble, G/N° is an elementary abelian q-group oforder q" for some prime q (with q = p if n > 1). Suppose, for a contradiction, that thereexists a normal subgroup M° of index q in G such that N° M°. Then GIN'(-) M°is elementary abelian of order q"+', b induces a duality from R(NM) ontoR(6/N° n M°), and hence NM is elementary abelian of order q"+'. Since INS = p, itfollows that q = p and then M < N, by definition of N. Thus N° < M°, a contradic-tion. So we see that there is no such M°, that is, N° is the intersection of all normalsubgroups of index q in G. In particular, N° is a characteristic subgroup of G and allassertions of the lemma hold.

9.2.17 Lemma. Suppose that N a G such that N is elementary abelian of order p",GIN is elementary abelian of order q', p and q primes, CG(N) = N and Z(G) = 1. If Gis a finite soluble group and S: R(G) -- R(G) a duality, then either I G(= pq or G A4

andG^ Q8.

Proof. In the first place p 0 q since Z(G) = 1. If INI = p, then GICG(N) is cyclic andhence I G I = pq. So suppose that n >_ 2. By 9.2.16, N° a 6 and 6/N° is elementary

abelian of order p". Let Q e Sylq(G). Then by 4.1.3, N = [N, Q] CN(Q) = [N, Q] sinceCN(Q) < Z(G) = 1. Thus N< G' and so every maximal normal subgroup of G con-tains N. It follows that

(20) every minimal subnormal subgroup of G is contained in Nb.

Now 9.2.16 applied to the duality induced in R(GIN) yields that Nb is elementaryabelian of order q`° if m > 2; and if m = 1, IN'I is a prime since G is soluble. Sup-pose, for a contradiction, that G is not a p-group. Then (I NbI, I G/N' I) = I and soCU(N') = N8 x P for some p-group P; by (20), P = 1, that is, CU(N°) = N°. Since61N° is not cyclic, it follows that IN°I is not a prime; so m > 2 and IN°I = q'°. As in4.1.7, by Maschke's Theorem, N is a completely reducible GF(p)Q-module, that is,N = N, x x N, with irreducible Q-modules Ni. By Schur's Lemma, Q/CQ(Ni) iscyclic and, since Q is elementary abelian, IQ: CQ(Ni)I < q for all i. Now CG(N) = Nand it follows that

qm=IQI=IQ:CQ(N)I= Q: n CQ(Ni)=1

< qr < q"

since I N I = p" and so r < n. Thus m < n and equality implies that I Ni I = p andIQ: CQ(Ni)I = q for all i, so thatgjp - 1. Since CG(N°) = Nb, the same argumentapplies to G = N°S for S e Sylq(G) and yields that n < m and equality implies thatplq - 1. But since m < n and n < m, we have equality and so get the contradictionthat qlp - 1 and pIq - 1. Thus G is a p-group. It follows that m = 1 and by (20), Nbis the unique minimal (subnormal) subgroup of G. Since 6/N° is elementary abelian,G Qg and so INI = 4; then IG : NI = 3 since CG(N) = N, and G ^- A4, as desired.

We can now prove the result announced above.

9.2.18 Theorem (Zacher [1962]). Let G and G be finite soluble groups and letS: R(G) R(G) be a duality. If I GI is odd, then G is supersoluble.

Proof. We use induction on I GI. By 9.2.15, S induces a duality in R(G/ I (G)). IfD(G): 1, then by induction, G/D(G) is supersoluble and then so is G (see Robinson[1982], p. 268). So let d>(G) = 1. Then F(G) is a direct product of minimal normalsubgroups of G (see Robinson [1982], p. 131). If N, and N2 are two of these, thenagain by induction, G/Ni is supersoluble (i = 1, 2) and then so is GIN, n NZ = G.Thus we may assume that N = F(G) is the unique minimal normal subgroup of G;let I NI = p". By 9.2.16, N is the join of all subnormal subgroups of order p of G andNb is a characteristic subgroup of G. If N = G, we are done. So let N G. Then forsome prime q, the join M/N of all subnormal subgroups of order q of G/N is non-trivial; 9.2.16 applied to the duality induced by S in R(GIN) yields that M/N iselementary abelian of order q'° for some m e N and Mb is a characteristic subgroupof N°. So Mb < G and 6 induces a duality from R(M) onto R(G/Mb) which satisfiesthe assumptions of 9.2.17. Indeed, N = F(G) and a well-known property of solublegroups (see Robinson [1982], p. 144) yields that N = CG(N) = CM(N). It follows that

Z(M) < N; since Z(M) a G and N is a minimal normal subgroup of G, we haveZ(M) = 1. Now I G I is odd and so 9.2.17 implies that I N = p. By induction, GIN issupersoluble and so, finally, G is supersoluble.

The assumption that I G I is odd is needed in Theorem 9.2.18. In fact, in (c) of 9.2.1,we saw that A4 and Q8 have dual lattices of subnormal subgroups and that R(Q8C3)is self-dual; clearly, A4 and Q8C3 are not supersoluble. On the other hand, these arethe only bad examples.

9.2.19 Theorem (Zacher [1962]). Let G and G be finite soluble groups and letd: R(G) - R(G) be a duality. Then G = H x K where (I HI, I K I) = 1, H is supersolubleand K is isomorphic to A4i Q8C3, or 1.

We sketch a proof of this theorem in Exercise 8 and mention that it has beenextended to infinite groups by Franchetta [1978].

X-subnormal subgroups

We finally report on an interesting generalization of the lattice of subnormal sub-groups. Let I be a class of finite groups which is closed under extensions, epimor-phic images, and subgroups. A subgroup H of G is called X-normal in G if H a G orG/Hc e 1; then, of course, H is called 1-subnormal in G if there exist Hi < G suchthat H=Ho<H1 Hi is X-normal in H1+1fori=0,...,n-1.

9.2.20 Theorem (Kegel [1978]). For every finite group G, the set Rr(G) of all 1-subnormal subgroups of G is a sublattice of L(G).

Clearly, R3(G) contains R(G) for every class 1; furthermore A.T(G) = R(G) ifI =

{I

}and R}(G) = L(G) if I is the class of all groups. For every set it of primes,

we may choose the class I of rt-groups. In this way we obtain infinitely manyfunctors R.., assigning to every finite group G a sublattice R.T(G) of L(G) containingR(G).

Exercises

1. If S is a nonabelian simple subnormal subgroup of G, show that So is a minimalnormal subgroup of G.

2. (Zacher [1957]) If G is a finite group, show that R(G) is distributive if and only iffor all S, T :!g G such that T < S and SIT is a p-group for some prime p, SIT iscyclic. (Hint: First show that if G has this property and H e S1(G) is soluble, thenevery Sylow subgroup of H is cyclic and H a G.)

3. Let G be a finite soluble group. Show that R(G) is a chain if and only if either Gis a cyclic p-group or G = PQ where P a- G is a cyclic p-group, Q a cyclic q-group,p and q different primes, and CG(P) = P.

9.3 Centralizer lattices 511

4. (Pazderski [1972]) Let G be a finite soluble group. Show that G has only oneminimal subnormal subgroup if and only if it has one of the following properties.(a) G = PQ where P a G is a cyclic p-group, p a prime, IQI divides p - 1, and

CG(P) = P.(b) G is a generalized quaternion group.(c) G = Q8C3 as defined in Example 9.2.1.(d) G has a normal subgroup N Q8 such that G/N S3 and CG(N) = Z(N).

5. Determine R(G) for the groups G in Exercise 4(d). Note that there are two differ-ent such groups, a semidirect product and a non-splitting extension of Q8 by S3-

6. (Heineken [1965]) Show that there are groups G and G with R(G) R(G) sat-isfying l G/G' l= 2, 16101 = 3, and G' G' A x B where J A I= 21819, I B I=31819; here a cyclic group of order 19 operates irreducibly on elementary abeliangroups of order 218 and 318, respectively.

7. (Curzio [1957d]) Let G be a finite supersoluble group without nontrivial cyclicSylow subgroups. If R(G) is self-dual, show that G is nilpotent.

8. (Zacher [1962]) Let G and G be finite soluble groups and let S: R(G) - R(G) bea duality.(a) If 02(G) is cyclic, show that G is supersoluble.(b) If G is not supersoluble and IGI = p''20 for some prime p, show that G A,

orG^_-Q8C3.(c) Prove Theorem 9.2.19. (Hint: Study 02(G)H where H is a Hall {2,31'-sub-

group of G.)

9.3 Centralizer lattices

In this section we study the centralizer lattice E(G) which we defined in § 1.1. Againour main aim will be to determine the structure of groups whose centralizer latticeshave special properties and to find out which groups are determined by their central-izer lattices (modulo abelian direct factors). But since E(G) is only a meet-sublatticeof L(G), there is the additional question: under which conditions is CM(G) a sublatticeof L(G) or even 5(G) = L(G)?

Throughout this section we shall write C(X) for CG(X) if X < G and there is noambiguity. We also write Z for the centre Z(G) of G and CC(X) for C(C(X)) and usean index C to denote intervals in CE(G). Thus if H, K E CM(G) and K < H, then

[H/K]c = {X E E(G)IK < X < H}.

Basic properties of centralizers

It is well-known and easy to prove that for every set X of subgroups of G,

(1) n C(X) = C U X I and U C(X) < CI ( X I.xE.r XE. / XEz \XE[ /

By 1.1.6, it follows from (1) that CM(G) = {C(X)IX E L(G)} is a complete meet-sublattice of L(G). We write A and v for the operations of this lattice and have thatfor H, K EE(G),

(2) HAK=HnKand HuK<HvK=n{XE(Y (G)IH<XandK<X}.The least element of CM(G) is Z(G) = C0(G) and the largest element is G = Ca(l). Forall subgroups X, Y of G, we clearly have that X < Y implies C(Y) < C(X) and thatX < CC(X). This yields the following important property of E(G).

9.3.1 Theorem. The map CI L(a): (((G) CM(G) is an involutory duality, that is, for allH, K E L(G),

(3) H < K if and only if C(K) < C(H) and

(4) CC(H) = H.

It follows that

(5) C(H A K) = C(H) v C(K) and C(H v K) = C(H) A C(K).

Proof. Let H, K E (((G) and take X< G such that H = CG(X). Then X < CC(X)and H < CC(H); and the first inclusion implies that C(CC(X)) < C(X), that is,CC(H) < H. Thus H = CC(H) and (4) holds. It follows that the map CI,,a) is bijec-tive; indeed, it is surjective since H = C(C(H)), and it is injective since C(H) = C(K)implies that H = CC(H) = CC(K) = K. Now if H< K, then clearly C(K) < C(H),and C(K) < C(H) yields H = CC(H) < CC(K) = K. Thus (3) holds and CI a(G) is aduality of CM(G). Finally, (5) follows from 8.1.1; in fact, since E(G) is a complete lattice,8.1.1 even yields the infinite analogue of (5).

We give some simple examples of centralizer lattices.

9.3.2 Examples. Let G be a group, Z = Z(G), p and q primes such that qj p - 1, andk,ncN.

(a) G is abelian if and only if G = Z, that is, if and only if C(G) is the trivial latticecontaining only one element.

(b) On the other hand, if G is nonabelian of order pq, then every subgroup of G isa centralizer so that CE(G) = L(G). Similarly, if G is a p-group and IG : ZI = p2, thenevery subgroup of G lying properly between G and Z is abelian and hence is its owncentralizer; thus E (G) = [G/Z].

(c) We write G E 7R(k) if s(Q) is (modular) of length 2 and has k atoms. Thus allthe groups considered in (b) lie in 9R(p + 1). More generally, we claim that if G is afinite nonabelian group with an abelian normal subgroup N of prime index, thenG E TI(IN : ZI + 1). Indeed, since G is nonabelian and G/N is cyclic, N Z andhence N = CG(N) E CE(G). Let IG : NJ = p and A = U-(G)\ {Z, N, G}. If H E A, thenH = CG(X) for some X < G such that X $ N. Therefore Ca(N n H) > NX = G andso N n H < Z. On the other hand, every centralizer contains Z and so N n H = Z.It follows that H N, hence NH = G and III: ZI = I G : NJ = p. Thus all the H E Ahave the same order and A u {N} is an antichain, that is, (l:(G) is of length 2. If

x E G \, N, then <x> Z < Cc(<x>) E A and hence I <x> Z : Z I = p. It follows that everyelement in G/Z not contained in N/Z has order p and, since there are (p - 1)IN/ZIsuch elements, Al I= IN: ZI. Thus G E 9JI(IN : ZI + 1) and this proves our assertion.

As special cases we obtain that A4 E M(5) and that G e 9JI(p" + 1) if G is a non-abelian P-group of order p"q. Here, if n >- 2, there are subgroups of order q in Ggenerating a subgroup of order pq and hence E (G) is not a sublattice of L(G).Another special case is the dihedral group D2" of order 2n whose centre is nontrivial(and of order 2) if and only if n is even. Thus D4k E 9J2(k + 1) for all k >- 2 andD2kE9Jl(k+ 1) ifkisodd, k> 3.

(d) Let G = SL(2, 2"), n > 2. Then G has a partition E consisting of the Sylow2-subgroups of order 2" and the maximal cyclic subgroups of order 2" + 1 and2" - I (see 3.5.1). We claim that ((G) consists of Z = 1, G, and the groups in E.Indeed, every component X of E is abelian and so X < CG(x) for every 1 x E X. IfX < CG(x) and y e CG(x)\X, then by 3.5.2, x and y have the same prime order p.Then <x, y> is elementary abelian of order p2 and contained in a Sylow p-subgroupP. Since the Sylow subgroups for odd primes are cyclic (they are contained in com-ponents of E), p = 2 and P e E. But then x E P A X implies P = X and so y E X, acontradiction. Thus X = CG(x) for every 1 x e X. By (1), every centralizer is anintersection of element centralizers and since E is a partition, the assertion follows.Now it is well-known (see Huppert [1967], pp. 191-194) that the normalizers of thecomponents of E have orders 2"(2" - 1), 2(2" + 1), 2(2" - 1), respectively, and, sinceevery component contains a Sylow subgroup of G, all components of the same orderare conjugate. Since IGI = (2" + 1)2"(2" - 1), it follows that

IEI =(2"+ 1)+q(2"- 1)2"+Z(2"+ I)2"=22n+2"+ 1.

Thus SL(2, 2") E 9JI(22n + 2" + 1). So we see that SL(2, 2") and the dihedral group oforder 4(22n + 2"), that is, a nonabelian simple and a metacyclic group, have isomor-phic centralizer lattices.

(e) Let G = S4. Then G has no element of order 6 and hence every proper central-izer in G is either a 2-group or a 3-group. It follows easily that CE(G) consists of 1, G,and the centralizer lattices of the Sylow subgroups of G. Since G has 4 Sylow 3-

Figure 26

subgroups of order 3 and 3 Sylow 2-subgroups isomorphic to D. intersecting in theKlein four-group, we obtain the lattice ((G) displayed in Figure 26.

If H is a subgroup of G, the only connection between (E(H) and E(G) in general isthat every element K in E(H) is of the form K = CH(X) = CG(X) n H for someX < H and hence is the intersection of H with an element of Cr(G). Since the intersec-tion of two centralizers is a centralizer, it follows that K e E(G) if H E tr(G). Thus

(6) E (H) -_ [H/Z(H)]e if H e E(G).

The following two simple remarks are used quite often. First, if x e G, then

(7) CG(x) = Cc(<x>) EE(G),

that is, every element centralizer belongs to E(G). The other is that every abeliansubgroup of G is contained in a maximal abelian subgroup of G and these alsobelong to CSE(G).

(8) If A is a maximal abelian subgroup of G, then A = C(A) E CE(G).

Indeed, A < C(A) since A is abelian. And if x e C(A), then B = <A, x> is an abeliansubgroup of G containing A and hence equal to A by maximality of A. Thus x e Aand A = C(A) E L(G). Conversely, if H E tr(G) satisfies H = C(H), then H is clearly amaximal abelian subgroup of G. Thus the maximal abelian subgroups are preciselythe fixed points of the duality C in,Theorem 9.3.1.

We come to another important general property of (r(G).

9.3.3 Lemma. Let H E CY(G). Then CY(H v C(H)) = [H v C(H)/H A C(H)]e.

Proof. For short, let F = H v C(H) and D = H A C(H). Then by 9.3.1,

C(F) = C(H v C(H)) = C(H) A CC(H) = C(H) A H = D

and hence C(D) = CC(F) = F. Since D < F, it follows that D = Z(F). Now ifX E C1(F), then X > Z(F) = D and, by (6), X EE(G); thus X E [F/D]c. Conversely,if X c [F/D],, then D < X < F and hence by (3), D = C(F) < C(X) < C(D) = F.Thus C(X)= CF(X), and if we apply this to C(X) in place of X, we obtain thatX = CC(X) = CF(C(X)) E Cr(F).

Let us call an atom of CV-(G) a minimal centralizer and an antiatom a maximalcentralizer of G. If N is a minimal centralizer of G and x E N\Z(G), thenx E CG(x) n N E Cr(G) so that Z(G) < CG(x) n N < N. The minimality of N impliesthat CG(x) n N = N and hence N < CG(x). Since this clearly also holds for x e Z(G),we have shown that

(9) every minimal centralizer is abelian.

If M is a maximal centralizer, then by (4), Z(G) C(M) and hence there exists anelement x (of prime power order if G is finite) in C(M)\Z(G). Then M < CG(x) < Gand so M = CG(x). Thus

(10) every maximal centralizer is of the form M = CG(x) where x can be chosen ofprime power order if G is finite.

But the main property of a maximal centralizer is the following consequence of 9.3.3and (9).

9.3.4 Lemma. If M is a maximal centralizer of G, then QM) < M and Ch(M)[M/C(M)]C.

Proof. Since CIc(G) is a duality, C(M) is a minimal centralizer of G and hence isabelian, by (9). It follows that QM) < CC(M) = M. Now 9.3.3 yields that L(M) =[M/C(M)]C.

Finally, suppose that G is nonabelian and (1(G) satisfies the maximal condition.Then maximal centralizers exist and, by (8), every maximal abelian and hence

(11) every abelian subgroup of G is contained in a maximal centralizer.

In particular, every element of G is contained in a maximal centralizer and since Gcannot be the set-theoretic union of two proper subgroups, it follows that

(12) G has at least 3 maximal centralizers.

Distributive centralizer lattices

There are no distributive centralizer lattices except the trivial one, a fact that can beproved quite simply.

9.3.5 Theorem. Let G be a group. Then (1(G) is distributive if and only if G is abelian.

Proof. If G is abelian, then I1(G)I = 1 and therefore (1(G) is distributive. Supposeconversely that (1(G) is distributive and let a, b e G. Then CG(b) A CG(ab) _CG(b) n CG(ab) < CG(a) and hence the distributive law implies that

CG(a) = CG(a) v (CG(b) A CG(ab)) = (CG(a) v CG(b)) A (CG(a) v CG(ab)).

Now b e CG(b) < CG(a) v CG(b) and b = a-lab e CG(a) u CG(ab) < CG(a) v CG(ab),by (2). Therefore b is contained in the right hand side of the above equation and sob c CG(a). Thus ab = ba and G is abelian.

Direct products

It is easy to see that, in contrast to subgroup lattices, the centralizer lattice of anarbitrary direct product of groups is the direct product of the centralizer lattices ofthe direct factors. We restrict ourselves to a finite number of direct factors and provea more general result. For this we need a name for the set of all commutators of

elements of a group, which, as is well-known, in general is different from the commu-tator subgroup. So let us write Com G = { [x, y] Ix, y e G}.

9.3.6 Theorem. If G = HK such that [H, K] = 1 and Com H n Com K = 1, thenE(G) 1r(H) x C;(K). In particular, if G = H x K, then t(G) E(H) x E(K).

Proof. We show first that for every subset M of G,

(13) C0(M) = CH(M)CK(M).

For this let c E CG(M) and x e M. Since G = HK, there exist elements c,, x, E Hand c2, x2 E K such that c = c, c2 and x = x, x2. Since [H, K] = 1, we have1 = [c, x] = [c,, x, ] [c2, x2] and hence I= [x2, c2] E Com H n Com K = 1.So c, centralizes x, and hence also x; this holds for all x e M so that c, E CH(M).Similarly, c2 E CK(M) and we see that CG(M) < CH(M)CK(M). Since the otherinclusion is trivial, (13) follows.

Now let X = CH(R) E !(H) and Y = CK(S) E (_E (K) where R < H and S < K. ThenRS is a subgroup of G since [H, K] = 1, and, by (13), CG(RS) = CH(RS)CK(RS) =CH(R)CK(S) = X Y. This shows that there is a well-defined map 0: &(H) x &(K) - E(G)satisfying (X, Y)'' = X Y. Since K n H < Z(H) < X,

(14) XKnH=X(KnH)=X.So if A e t(H), B E C!(K), and (A, B)'" _ (X, Y)O, then X = XK n H = X YK n H =ABK n H = AK n H = A; similarly, Y = B and >G is injective. For T< G = HK,Dedekind's law implies that TK = (TK n H)K and TH = H(TH n K). So by (13),

CG(T) = CH(T)CK(T) = CH(TK)CK(TH) = CH(TK n H)CK(TH n K)

and this shows that t,1i is surjective. Finally, (A, B) < (X, Y) clearly implies (A, B)'' _AB < X Y = (X, Y)''; conversely, if (A, B)'" < (X, Y)'', then AB :!5; X Y < XK and by(14), A < X K n H = X. Similarly, B < Y and so (A, B) < (X, Y). By 1.1.2, 0 is anisomorphism.

Exercise 3 shows that the assumption Com H n Com K = 1 is needed in theabove theorem. In fact, we are now going to show that for groups whose centralizerlattice has finite length, the converse of 9.3.6 almost holds. For this we need thefollowing auxiliary result.

9.3.7 Lemma. Assume that G = HK where H = CG(K) and K = CG(H). If A E CS(H)and B e CE(K), then AB E (E(G).

Proof. By (6), A, B e f(G) and, by (2), AB < A v B. Let x e A v B and write x =x, x2 where x, E H, x2 E K. Then by 9.3.1, x e A v B< A v K = CC(A) v C(H) =C(C(A) A H) = C(CH(A)). Therefore every c e CH(A) satisfies 1 [c, x] = [c, x, x2] _[c, x, ] since [c, x2] E [H, K] = 1. Thus x, E CH(CH(A)) = A. Similarly, x2 e B andso x = x, x2 E AB. It follows that AB = A v B e E(G).

9.3.8 Theorem. Let G be a group such that E(G) has finite length. Assume that CM(G)is the direct product of two lattices L, and L2, write 0, 11 and 02, 12 for the least andgreatest elements of L, and L2, respectively, and let H = (I1, 02) and K = (01 ,12).Then

(a) CG(H) = K and CG(K) = H,(b) Com H n Com K = 1, and(c) E(H) = [H/Z(G)lc L1, (E(K) = [K/Z(G)lc L2, and hence C1(HK) ('(G).

Proof. (a) If IL1 1 = 1, then H = Z(G), K = G, and (a)-(c) hold trivially. So we mayassume that I L 1 I I I L2 and show first by induction on the length of CM(G) that

(15) [H, K] = 1.

By (12), G contains at least 3 maximal centralizers. These have the form (1 x) or(Y, 12) with antiatoms x of L2 and y of L1i hence at least one of the two lattices hasmore than one antiatom. So let x1, x2 be two different antiatoms of L2, say, andlet M = (I,, x) where x = x1 or x = x2. Then by 9.3.4, C(M) < M and E(M) _[M/C(M)]c. So C(M) = (w,z) where w e L1, z e L2 and z < x; it follows that

[M/C(M)lc = [(11,x)/(w, z)] = {(s, t)Iw < s < I, z 5 t < x} = [11/w] x [x/z].

Since [M/C(M)lc is a proper interval in E(G), by induction, (15) holds for M. So ifHo = (11, z) and Ko = (w, x), then [Ho, KO] = 1. Now H < Ho and (01,x) < K0;thus (01,x) < C(H). If we apply this with x = x1 and x = x2, it follows thatK = (01,12) _ (01,x1) v (01,x2) < C(H) and (15) holds.

By (15), C(H) > K = (01,12) and hence C(H) = (u,12) for some u c- L1; similarlyC(K) = (11, v) with v e L2. Since G = (11,12) = H v K, Theorem 9.3.1 yields thatZ(G) = C(H) A C(K) = (u, v). It follows that u = 01 and v = 02; thus C(H) _(01,12) = K and C(K) = H.

(b) Let a E H and b e K. Every X E 2(G) = L 1 x L2 satisfies X= (X A H) V (X A K)and so CG(ab) = CH(ab) v CK(ab) = CH(a) v CK(b) since [H, K] = 1. By 9.3.7applied to the group HK there exists T:!!-, HK such that CH(a)CK(b) = CHK(T).Then CH(a) and CK(b) are contained in CG(T); it follows that

CHK(ab) = CG(ab) n HK = (CH(a) v CK(b)) n HK < CG(T) n HK

and hence CHK(ab) < CHK(T) = CH(a)CK(b). Since the other inclusion is trivial, wehave shown that

(16) CHK(ab) = CH(a)CK(b) for all a e H, b E K.

Now suppose, for a contradiction, that Com H n Com K 0 1. Then there exista, c E H and b, d e K such that 1 0 [a, c] = [d, b]. Since [H, K] = 1, this impliesthat 1 = [a, c] [b, d] = [ab, cd] and hence cd e CHK(ab) = CH(a)CK(b), by (16). Sincedc-K, it follows that c E CH(a)K n H = CH(a)(K n H) = CH(a), contradicting[a, c] 0 1. Thus Com H n Com K = 1.

(c) Of course, [H/Z]c = { (x, 02 )I X E L 1 } ^ L 1 and we show that 5(H) = [H/Z]c.

Indeed, if X e (1(H), then by (6), X e (1(G) and so Z < X < H; thus 1(H) c [H/Z]e.Conversely, if X e [H/Z],, then CG(X) CG(H) = K and hence CG(X) = (y,12) _Y v K for some y c L1 and Y = (y, 02) < H. By 9.3.1,

X = CG(CG(X)) = CG(Y v K) = CG(Y) A H = CH(Y) E 1(H).

Thus 1(H) = [H/Z]c L,, similarly (1(K) L2 and, by 9.3.6,1(HK) - L1 x LZ =(s (G).

If we combine Theorems 9.3.6 and 9.3.8, we obtain a result that was proved byAntonov [1987] for finite groups with modular centralizer lattices.

9.3.9 Corollary. Let G be a group such that (1(G) has finite length. Then 1(G) is adirect product of two lattices L1 and L2 if and only if G contains subgroups Hand K such that 1(HK) 1(G), [H, K] = 1, Com H n Com K = 1, (1(H) L1 and(1(K) L2.

It is not known whether 9.3.8 and 9.3.9 are true without the assumption that 1(G)has finite length. Exercise 4 shows that HK may be a proper subgroup of G inTheorem 9.3.8, although, of course, G is the join of H and K in the centralizer lattice.

Modular centralizer lattices

There are many groups with modular centralizer lattice and so there are only a fewgeneral results about these groups and lattices. One of them is the simple fact thattheir length is even if it is finite. After proving this we shall study finite groups withcentralizer lattice of length 2 and, as an application, determine all finite simplegroups with modular centralizer lattice.

9.3.10 Lemma. If L(G) is modular and of finite length 1, then I is even.

Proof. Recall from 2.1.10 that (1(G) satisfies the Jordan-Dedekind chain condition,and consider a maximal abelian subgroup A of G. By (8), A = C(A) e (1(G) and hencethere exists a maximal chain Z(G) = A0 < A, < < A. = A < Ai+1 < < A, = Gin (1(G) containing A. Then Z=A0 < < A,=A and Z=C(A,) < ... < A

are two maximal chains from Z to A and therefore have the same length. Thusn = I - n, that is, l = 2n is even.

Let us write G e 9J1 if (1(G) is (modular) of length 2. So 1771 is the union of the classes931(k), k e N u {ox}, defined in 9.3.2(c). We shall also say that G is an 971-group ifG E 971, trusting that the reader will not confuse these with the M-groups studied inChapter 2. If G is an 971-group, then every H e (1(G)\{G, Z} is a minimal centralizerand hence is abelian, by (9). Conversely, if G is nonabelian and every proper central-izer of G is abelian, then every H e (E(G)\ { G, Z} is contained in a maximal abeliansubgroup A of G. If H < A, it would follow that A = C(A) < C(H) and so C(H)

would not be abelian, a contradiction. Therefore H = A and since two differentmaximal abelian subgroups cannot be contained in each other, it follows that G E 93t.Thus

(17) G e 931 if and only if G is nonabelian and every proper centralizer of G isabelian.

Now if H < G e 931, then every proper centralizer of H is contained in a propercentralizer of G and hence is abelian. So (17) implies that

(18) every subgroup of an 931-group is an 991-group or abelian.

The main property of 971-groups, however, is the following. Recall from § 3.5 that apartition of a group G is a set E of nontrivial subgroups of G such that everynonidentity element of G is contained in a unique subgroup X e E.

9.3.11 Lemma. If G is an 931-group, then the set E of all M/Z(G) where M is amaximal abelian subgroup of G is a normal partition of G/Z(G); we call E the central-izer partition of G/Z(G).

Proof. Let Z = Z(G). If x e G, then x e <x> < M for some maximal abelian sub-group M of G; thus xZ e M/Z. And if xZ e M, /Z n M2/Z for two different maximalabelian subgroups M, and M2 of G, then x e M, n M2 = M, A M2 = Z and hencexZ = I in G/Z. This shows that E is a partition. If M is a maximal abelian subgroupof G and g e G, then Mg is a maximal abelian subgroup of G and so Y. is normal.

Using the results of § 3.5 on partitions of finite groups, we obtain the followingcharacterization of finite 991-groups.

9.3.12 Theorem (Schmidt [1970c]). A finite group G is an 931-group (and lies in 931(k))if and only if it is a group of one of the following types:

(i) G = A x P where A is abelian and P is a p-group in 931(k), p a prime; here k= 1(mod p).

(ii) G is nonabelian and has an abelian normal subgroup N of prime index; herek = IN:Z(G)I+1.

(iii) G/Z(G) is a Frobenius group with Frobenius kernel F/Z(G) and Frobeniuscomplement K/Z(G) where F and K are abelian; here k = IF: Z(G)I + 1.

(iv) G/Z(G) is a Frobenius group with Frobenius kernel F/Z(G) and Frobeniuscomplement K/Z(G) where K is abelian, Z(F) = Z(G), F/Z(G) is a p-group, p a prime,and F is an 931-group; here k = IF: Z(G)j + m if F E 931(m).

(v) G/Z(G) S4 and V is not abelian if V/Z(G) is the Klein four-group; here k = 13.(vi) G/Z(G) PSL(2, p") or PGL(2, p"), G' SL(2, p"), p a prime, p" > 3; here k =

pen+p"+1.(vii) G/Z(G) ^ PSL(2, 9) or PGL(2, 9), G' is isomorphic to the representation group

of PSL(2,9) in the sense of Schur; here k = 121.

Proof. We show first that every 972-group has one of the properties (i)-(vii); then weprove that, conversely, every such group indeed lies in 931(k) where k is as stated. Soassume that G is an 971-group, let Z = Z(G) and let E be the centralizer partition ofG%Z. By 3.5.10 and 3.5.11, we know the possible structure of G/Z and, since thepartition E is normal, 3.5.7 and 3.5.9 show in addition that if G/Z is a Frobeniusgroup, then the Frobenius complements are components of E. Also, if G/Z S,,then E consists of the maximal cyclic subgroups of G/Z. So we have to consider thefollowing six possibilities for G/Z.

(a) G/Z is a p-group for some prime p. Let A be the p'-component of Z andP e Sylp(G). Then G = A x P, A is abelian and, by 9.3.6, E(P) CM(G). Thus G is oftype (i).

(b) G/Z has a normal subgroup N/Z e Y. of prime index. Since N/Z e E, the groupN is abelian and so (ii) holds.

(c) G/Z is a Frobenius group and the Frobenius complements are components of E.Let F/ Z be the Frobenius kernel and K/Z a Frobenius complement of G/Z. Then Kis abelian since K/Z e E. If F too is abelian, then (iii) holds. Thus we assume that Fis not abelian. By (18), F is an 931-group and hence Z(F) = A, n A2 for some maximalabelian subgroups A; of F. If M, is a maximal abelian subgroup of G containing A;(i = 1, 2), it follows that M, M2 and Z(F) = A, n A2 < M, n M2 = Z. Since Z < F,the other inclusion is trivial and so Z(F) = Z. As a Frobenius kernel, F/Z is nil-potent (see 3.5.1) and hence is a p-group for some prime p, by 3.5.3. Thus (iv) holds.

(d) G/Z S, and E consists of the maximal cyclic subgroups of S,. Since V/ Z is notcyclic, it is not contained in a component of E and so V is not contained in amaximal abelian subgroup of G. Thus V is not abelian and (v) holds.

(e) G/Z a- PSL(2, p") or PGL(2, p"), p a prime, p" > 4. Let H = G'Z. Since H/ZPSL(2, p") is simple, H'Z = H and hence H/H' is central in G/H' of index at most 2.Thus G/H' is abelian, that is, G' = H' = (G'Z)' = G". Therefore G' r) Z < (G')' n Z(G')and, by a well-known theorem of Schur's (see Huppert [1967], p. 629), G' n Zis isomorphic to a subgroup of the Schur multiplier M of G'/G' n Z H/ZPSL(2, p"). Now this Schur multiplier and the representation group of PSL(2, p") arewell-known (see Huppert [1967], p. 646). If p = 2 and p" 0 4, then M = I and soG' PSL(2, 2") = SL(2, 2"), as desired. If p > 2 and p" 0 9, then I MI = 2 and soG' PSL(2, p") or SL(2, p"). Let p" 0 5 and suppose, for a contradiction, that G' 2fPSL(2, p"). Then p" + 1 or p" - 1 has 4 as a proper factor and G' contains a dihedralgroup D of this order (see Huppert [1967], p. 213). Then Z(D) 0 1 and CG(Z(D)) is anonabelian proper centralizer in G, a contradiction. Thus G' SL(2. p"), as desired.We are left with the exceptional cases p" = 4, 5, 9. Here PGL(2,4) = PSL(2,4)PSL(2, 5) has representation group SL(2, 5), so that if G/Z at PSL(2, 5), then (vi)holds with p" = 4 or p" = 5. Now let G/Z PGL(2, 5) and suppose, for a contradic-tion, that G' PSL(2, 5). Then G r) Z = I and if A e Sy12(G'), S e Sy12(G) and A < S,then A = G n S a S and so A n Z(S) 0 1. Thus S < Co(A n Z(S)) < G and, sinceS is nonabelian, this is impossible. Thus G' ti SL(2, 5) and (vi) holds. Finally, ifp" = 9, then I M I = 6 and, as shown above, G' PSL(2,9). So G' n Z I and weclaim that IG' n ZI # 3. Indeed, if IG' n ZI = 3, a Sylow 2-subgroup S of G' wouldbe dihedral of order 8 and, since Z(S) 4 Z, there would exist a nonabelian central-izer in G. Thus IG' n ZI = 2 or 6 and so (vi) or (vii) holds.

(f) G/Z Sz(q), q a power of 2. The group Sz(q) has nonabelian Sylow 2-subgroups which, by 3.5.3, have to be contained in components of E. Since allcomponents of Y. are abelian, this case cannot occur.

Now assume that, conversely, G has one of the properties (i)-(vii). If G satisfies (i),then by 9.3.6, L(G) L(P); so G e M(k) where k = 1 (mod p), by 3.5.10. If G is of type(ii), then G E M(IN : ZI + 1), by (c) of 9.3.2. If G satisfies (iii) or (iv), then, since K/Zoperates fixed-point-freely on F/Z, every proper centralizer of G is contained in F orin a conjugate of K. So if F and K are abelian, (1(G) = {Z,F,G} u {K9Ig e C} andtherefore G e W (k) where k= I G: NG(K) I+ 1= I G: K I+ I =I F: Z I+ 1. And ifF E 9JI(m) and Z(F) = Z, then (1(G) consists of Z, G, the maximal abelian subgroupsof F, and the conjugates of K; thus G E TI(k) where k = IF: ZI + m.

Now let G be of type (v) and M a maximal centralizer of G. We claim that M/Z isa maximal cyclic subgroup of G/Z; since S4 has 13 of these, it will follow thatG E M(13). By (10), M = CG(x) for some p-element x e G\Z and, since M/Z central-izes xZ, we have that M/Z is a p-group. Therefore we are done if p = 3. So let p = 2,S/Z be a Sylow 2-subgroup of G/Z containing M/Z, and D/Z be the cyclic subgroupof order 4 of S/Z. As a cyclic extension of a central subgroup, D is abelian. Since Vis nonabelian, Z(V) = Z and hence Z(S) = Z. So I M : ZI 5 4 and Z < Z(M) impliesthat M is abelian. If D = M, we are done. And if D 0 M, then D n M < Z(S) = Zand hence M/Z is a maximal cyclic subgroup of order 2 of G/Z, as desired.

We leave it as an exercise for the reader to show that every group of type (vi) and(vii) is an 9JI-group. In the first case, the centralizer partition E of G/Z is described in3.5.1(e); it has the Sylow p-subgroups of G/Z as components. As in Example 9.3.2(c)for SL(2,2") one shows that jEI = pen + p" + 1. In the second case, the centralizerpartition of G/Z consists of the maximal cyclic subgroups of G/Z, of which there are34+4(32+ 1) = 121.

As particular cases of the above theorem we obtain that GL(2, p") and SL(2, p") liein 9JI(p2n + p" + 1) for p" > 3. Also note that the two representation groups of S4satisfy (v); one of them is GL(2, 3) (see Huppert [1967], p. 653).

Our theorem does not say much in the case where G is a p-group in M. Here ifx E Z2(G)\Z, then G'< CG(Z2(G)) < CG(x) < G and so N = CG(x) is a normal maxi-mal abelian subgroup of G with abelian factor group; by 3.5.3, every element inG/Z\,N/Z has order p. If IG: NI = p, then G is a p-group with an abelian subgroupof index p and so its structure is well-known. If IG: NI > p2, Rocke [1975] showsthat c(G) 2 andr > 0, he constructs a p-group G E 931 of class p having a normal maximal abeliansubgroup N of index p" such that IN: G'ZI = p'

Since SL(2, p") is not simple for p > 2, Theorem 9.3.12 yields that the only finitenonabelian simple `J-groups are the groups SL(2, 2"). We shall now show that thesegroups are the only finite nonabelian simple groups with modular centralizer lattice.For this we need the following two results which are perhaps of independent interest.The first one implies that every finite group in 9JI(p" + 1) is soluble. We mention inpassing another strange result of this kind: If k is even, every finite group in 9JI(k) issoluble. This follows immediately from Theorem 9.3.12 and the simple fact thatpen + p" + 1 is odd for every prime p and n e N.

9.3.13 Lemma. Let p be a prime, n e Ni, and G a finite group.(a) If G e 9R(p" + 1), then either G/Z(G) is a p-group or G has a maximal abelian

subgroup A such that A/Z(G) is a p-group and B/Z(G) is a p'-group for every othermaximal abelian subgroup B of G.

(b) If E(G) is isomorphic to the lattice of subspaces of a vector space over GF(p"),then G e 9R(p" + 1) or G/Z(G) is a p-group.

Proof. (a) Let G e 931(p" + 1). Since 12, q2m + q' = gm(gm + 1), and 120 are notprime powers, G is not of type (v), (vi), or (vii). Suppose, for a contradiction, that Gis of type (iv). Then F/Z is a q-group in 93(m) for some q e P, m e N and so p" + 1=IF: ZI + m = 1 (mod q), by (i) of 9.3.12. Thus p = q, let IF: ZI = p'. Since m is thenumber of components of a partition of F/Z, clearly m < p' and so m = pst + 1where s < r and (t, p) = 1. It follows that p" = p' + pst = ps(p'-s + t); but p'-s + t is

not a power of p. This contradiction shows that G is of type (i), (ii), or (iii). Now if Gis of type (i), then G/Z is a q-group and p" + I = 1 (mod q) implies that p = q. If Gis of type (ii), then IN: ZI = p" and IG : NI = q e P. So if q = p, G/Z is a p-group; andif q : p, every maximal abelian subgroup B 0 N of G satisfies IB/ZI = I G/NI = q sothat B/Z is a p'-group. Similarly, if G is of type (iii), IF : ZI = p" and the Frobeniuscomplements have order prime to IF/ZI. This proves (a).

(b) If H E E(G), then by (6), C&(H) c [H/Z(H)]c. Suppose that this interval haslength 2. Then E(H) has length 2, that is, H is an 931-group. Further [H/Z(H)]c isisomorphic to the lattice of subspaces of a 2-dimensional vector space over GF(p")and therefore has p" + 1 atoms. If X is such an atom and x e X \ Z(H), there existsa maximal abelian subgroup A of H such that x e A. Then x e X n A and, since Xand A are atoms in [H/Z(H)]c and x 0 Z(H), it follows that X = A E E(H). Thus wehave shown:

(19) If H E CM(G) such that [H/Z(H)]c has length 2, then 3(H) = [H/Z(H)]c andHE9R(p"+ 1).

By 9.3.10, !(G) has even length 1. If I= 2, then (19) shows that G E 931(p" + 1),as desired. So we may assume that l >_ 4 and we have to show first that A/Z isa p-group for every minimal centralizer A of G. For this we need the followingassertion.

(20) Let A, B be atoms of CS(G) such that [A, B] = 1. If A/Z is a p-group, then sois B/Z.

To show this we may, of course, assume that A 0- B. Since A and B are atoms, C(A)and C(B) are antiatoms of CM(G). And since any two maximal subspaces of a vectorspace have a common complement, there is an atom D of 3(G) such thatD C(A) and D C(B). It follows that H = A v B v D is not abelian. However,since [A, B] = 1, (9) yields that AB is abelian and hence is contained in a maximalabelian subgroup X of H. By (8) and (6), X e CE(G) and so A v B< X; since[H/Z(G)]c has length at most 3, it follows that A v B = X. Now H = X v D impliesthat Z(H) = H A C(H) = H A C(X) A C(D) = CH(X) A C(D) = X A C(D). SinceC(D) is an antiatom of E(G) and E(G) is modular, X n C(D) is maximal in X and soZ(H) is an atom of CE(G). Thus [H/Z(H)]c has length 2 and, by (19), H E 9R(p" + 1).

By (a), H/Z(H) is a p-group or has a maximal abelian subgroup N such that N/Z(H)is a p-group and Y/Z(H) is a p'-group for any other maximal abelian subgroup Y ofH. In the latter case, since AZ(H)/Z(H) A/A n Z(H) = A/Z is a nontrivial p-groupcontained in X/Z(H), it follows that X = N. Thus in both cases X/Z(H) is a p-group;hence B/Z is a p-group since B/Z BZ(H)/Z(H) < X/Z(H). This proves (20).

(21) If A is an atom of K(G), then A/Z is a p-group.

Indeed, let B be an atom of E(G) such that B C(A) and let H = A v B. Since((H) c [H/Z]c and this interval has length 2, we must have Z(H) = Z and soH E TI(p" + 1), by (19). Now (a) implies that H contains an atom X of CM(G) such thatX/Z is a p-group. Since 12! 4, we have H 0 G and so C(H) contains an atom Y ofLE(G). Then [X, Y] = 1 = [Y, A] and, applying (20) twice, we obtain that Y/Z andA/Z are p-groups. This proves (21).

Finally we use induction on the length l of ((G) to show that G/Z is a p-group (ifI > 4); this will complete the proof of the lemma. Since every element x e G lies in amaximal abelian subgroup of G, it suffices to show that A/Z is a p-group for everymaximal abelian subgroup A of G. For this let M be a maximal centralizer contain-ing A. Then Z < C(M) = Z(M) < A < M < G. So if I = 4, [M/Z(M)]c has length 2and by (19), M E 9JI(p" + 1). Since [A/Z]c is isomorphic to the lattice of subspacesof a 2-dimensional vector space, there exists an atom X Z(M) of L(G) such thatX 5 A. By (21), X/Z and Z(M)/Z are p-groups and X/Z XZ(M)/Z(M) < A/Z(M);it follows from (a) that A/Z(M) is a p-group. Thus A/Z is a p-group, as desired. IfI > 4, then by 9.3.10, 1 > 6 and so (QM) = [M/Z(M)]c has length at least 4. Byinduction and (21), M/Z(M) and Z(M)/Z are p-groups and then so is A/Z.

9.3.14 Theorem. (Antonov [1987]). If G is a finite nonabelian simple group withmodular centralizer lattice, then G SL(2, 2")for some n > 2.

Proof. Suppose, for a contradiction, that the length I of K(G) is at least 4. Sincethe join of all atoms of (1;(G) is clearly a normal subgroup of G and G is simple, thisjoin-in L(G) and in E(G)-equals G. By a well-known property of modular lattices(see Crawley and Dilworth [1973], p. 30), L(G) is complemented.

Now suppose, for a contradiction, that &(G) is directly decomposable. Then thereexist nontrivial lattices Li (i = 1, 2) with least elements Oi and greatest elements Iisuch that L(G) = L, x L2 and L, is directly indecomposable. If H = (11,02) andK = (01,12), then by 9.3.8, [H, K] = 1 and E(H) = [H/Z(G)]c ^- L1. For g e G,H9 E C!(G); hence H9 = (x,y) with x e L1, y e L2 and

((H9) _ ([H/Z(G)]c)9 = [H9/Z(G)]c [x/0i] x [Y/02].

On the other hand, E(H9) L(H) is directly indecomposable and it follows thatx = 0, or y = 02, that is, H9 < K or H9 = H. In any case, H9 < HK for all g e Gand, since G is simple, it follows that HK = G. But K centralizes H, a contradiction.

Thus L(G) is a finite, directly indecomposable, complemented, modular lattice oflength at least 4 and it is well-known (see Crawley and Dilworth [1973], Chapter 13)that such a lattice is isomorphic to the lattice of subspaces of a vector space over

some finite field GF(p"). But now 9.3.13(b) yields that G is a p-group, a contradic-tion. It follows that 1 < 3 and so l = 2, by 9.3.10. Thus G is an 912-group and, by9.3.12, G SL(2, 2") for some n > 2.

The structure of arbitrary finite groups with modular centralizer lattices is notknown; only very special classes of such groups have been investigated. By 9.3.13,every finite group whose centralizer lattice is isomorphic to the lattice of subspacesof a vector space of dimension at least 3 over GF(p") is a p-group. Antonov [1987]shows that IA : ZI < p' for every minimal centralizer A of such a group and deter-mines the group if I A : ZI = p" for all such A.

Jiirgensen [1976] studies finite groups with modular centralizer lattices of length4. She shows that such a group is soluble if it has trivial centre and the intersectionN of all the maximal centralizers of G is different from Z; Exercise 9 gives anexample of such a group. If N = Z, then either G/Z is a p-group or (((G) is decom-posable into a direct product of two lattices of length 2; in this case 9.3.8 and 9.3.12give the structure of G.

Finally, Vasileva [1977] studies 9112-groups, that is, non-912-groups whose maxi-mal centralizers are either abelian or in 991. These centralizer lattices, of course, neednot be modular, an example is C!(S4). Her main result is that the finite simple 912-groups are precisely the groups PSL(2, q) (q odd, q > 5), Sz(22n+1) (n > 1), PSL(3, q)(q > 2), PSU(3,q) (q > 2), the Janko group J,, the alternating group A7, and theMathieu group M,,.

Characterization by centralizer lattices

By 9.3.6, ((G) (QG x A) for every abelian group A. So at best it is possible tocharacterize a group modulo abelian direct factors or in a certain class of groups, forexample, among the groups with trivial centre. Vasileva's theorem and Theorem9.3.12 (or 9.3.14) yield such characterizations. We give three other results of this typewithout proof; all these proofs are quite technical and use rather special propertiesof the groups studied.

9.3.15 Theorem. Let G be a finite group, p a prime, n e N.(a) (Schmidt [1970d]) If L(G) ^- CS(J,), then G = A x J, where A is abelian.(b) (Schmidt [1972a]) If (V-(G) (((PSL(2, p")) and Z(G) = 1, then G PSL(2, p").(c) (Vasileva [1976]) If 11(G) ^_ p")) and Z(G) = 1, then G PSL(3, p").

11(G) as a sublattice of L(G)

Finally we turn to the question of when L(G) is a sublattice of L(G). This is certainlythe case if 1(G) = L(G), that is, if every subgroup of G is a centralizer. The finitegroups with this property were determined by Gaschiitz in 1955. More generally, westudy finite groups G for which 1(G) = [G/Z(G)], that is, in which every subgroupcontaining the centre is a centralizer. This property is inherited by subgroups.

9.3.16 Lemma. If (-f (G) = [G/Z(G)] and H < G, then .(H) = [H/Z(H)].

Proof. By 9.3.1, the centralizer map is a duality of [G/Z]. In particular, for allX, Y > Z, we have C(X n Y) = C(X) u C(Y). Using this, 9.3.1, and Dedekind's law,we get for Z(H) < A < H,

CH(CH(A)) = C(C(A) n H) n H = C((C(A) n H)Z) n H = C(C(A) n HZ) n H

= (CC(A) u C(H)) n H = (CC(AZ) u C(H)) n H

= AC(H) n H = A(C(H) n H) = AZ(H) = A.

Thus A e E(H), as desired.

9.3.17 Theorem. The finite group G satisfies CM(G) = [G/Z(G)] if and only ifG = G1 x x G, x A where (IG1I,IG;I) = 1 = (IGiI,JAI) for i 0 j, A is abelian,and every G. is either a { p, q}-group with IGi/Z(Gi)I = pq or a p-group satisfying(((Gi) = [Gi/Z(Gr)], p and q primes.

Proof. Suppose that G has this structure. Then G/Z G1/Z(G1) x x G,/Z(G,).If IGi/Z(Gj)I = pq and Z(Gi) < X < Gi, then X is a maximal abelian subgroup of G.and hence X E E(G1). Thus !(G1) = [Gi/Z(G1)] for all i; by 9.3.6 and 1.6.4,

1(G) ((G1) x ... x E(Gr) = [G1/Z(G1)] x ... x [Gr/Z(Gr)] [G/Z].

Since E(G) is a subset of [G/Z], it follows that E(G) = [G/Z].Suppose conversely that E(G) = [G/Z]. Then by 9.3.1, G/Z has a dual and hence,

by 8.2.3, is a direct product of coprime p-groups and P-groups. Now, in general, ifG/Z = X/Z x Y/Z where X/Z is a n-group and Y/Z is a n'-group, then G = H x Kwith n-group H and n'-group K satisfying H/Z(H) X/Z. Indeed, 0,.(Z) is anormal Hall subgroup in X and hence has a complement H in X, by the Schur-Zassenhaus Theorem. Then X = H x 0,.(Z) and H a G since H is characteristicin X < G. Similarly, Y = O,(Z) x K where K is a normal n'-subgroup of G. Since0,,(Z)< H and 0,.(Z)< K, it follows that G = H x K. Now Z(H) = H n Z and soH/Z(H) -- X/Z.

If we apply this to the given direct decomposition of G/Z, we see that G =G1 x x G, x A where A is abelian, (IGiI,IG;I) = 1 = (IGiI,IAI) for i j, andGi/Z(GL) is a p-group or a P-group. By 9.3.16, E(G1) _ [Gi/Z(G3)] for all i. If Gi/Z(G3)is a P-group of order p"q, p > q, then G,/Z(G1) has a unique maximal normal sub-group (of order p"). Since the centralizer duality maps normal subgroups to normalsubgroups, it follows that G]/Z(G1) has only one minimal normal subgroup. Butevery subgroup of order p of G;/Z(G;) is normal, so n = 1.

If we add the condition Z(G) = 1 in the above theorem, we get the structure offinite groups in which every subgroup is a centralizer.

9.3.18 Corollary (Gaschutz [1955]). Let G be a finite group. Then S.(G) = L(G) ifand only if G = G1 x - - - x G, where ((Gil,JG;j) = 1 for all i j and every G; is non-abelian of order pq, p and q primes.

Theorem 9.3.17 characterizes the finite groups G with E(G) = [G/Z(G)] modulothe p-groups with this property. The structure of these groups is also known; we giveit without proof.

9.3.19 Theorem (Reuther [1977], Cheng [1982]). The following properties of thefinite p-group G are equivalent:

(a) (_V(G) = [G/Z(G)],(b) [G/Z(G)] is modular and G' is cyclic,(c) G' = <a> is cyclic and [<a>, G] < <a4>.

Independently, and more generally, Antonov [1980] determined all groups G forwhich G/Z(G) is locally finite and CM(G) = [G/Z(G)].

We return to the more general question of when E(G) is a sublattice of L(G). By(1), this is clearly the case if

(22) C(X n Y) = C(X) u C(Y) for all X, Y< G.

However, this condition is much stronger. Reuther [1975] shows that a finite groupG satisfies (22) if and only if G = G1 x . . . x G, where (IGig, J G,I) = 1 for all i O j andthe G. are abelian, isomorphic to Q8, or have order pq", p > q, and cyclic Sylowsubgroups. Clearly (22) implies that the map C2: L(G) -+ L(G) sending every X < Gto CC(X) is a homomorphism. Antonov [1980] studies the much weaker propertythat CZ is a homomorphism from L(G) to CM(G); since

C2(X U Y) = C(C(X) n C(Y)) = C2(X) V C2(Y)

holds in general, this amounts to the condition that

(23) C2(X n Y) = CZ(X) A C2(Y) for all X, Y < G.

Nevertheless he is able to determine all groups G satisfying (23) for which G/Z(G) islocally finite. For finite groups, he obtains the same structure theorem as Reutherso that (22) and (23) are equivalent for these groups. Both Reuther and Antonovalso study groups for which the equations in (22) and (23) only hold for allX, Y E [G/Z(G)].

We finish this section by mentioning a result of quite a different nature.

9.3.20 Theorem (Antonov [1991]). If (E(G) is a sublattice of L(G) for a finite groupG, then L(G) is modular.

Exercises

1. Show that every abelian centralizer is the intersection of maximal abelian sub-groups.

9.4 Coset lattices 527

2. If (G;,)AEA is a family of groups, show that L I Dr GA) Dr !(G;).\AEA AeA

3. (Schmidt [1970c]). Let G1 = H, x K, with H, ^- Q8 K1 and let G = G1/Dwhere D is the diagonal in the direct product Z(H1) x Z(K 1); let H = H1 D/Dand K = K1D/D. Show that H Q8 K, G = HK and [H,K] = 1, butL(G) * L(H) x L(K).

4. (Jurgensen [1976]) Let P = <a> x be elementary abelian of order 25 andlet c, d e Aut P be defined by a` = a-1, b` = b, and xd = x2 for all x e P. LetQ = <c, d> < Aut P and G = PQ be the semidirect product of P and Q. Finally,let e = cd 2, H = <a, c> and K = <b, a>. Show that H K D10, H = CG(K),K = Cc, (H), I G : HK I = 2, and L(G) E(H) x E, (K).

5. (Antonov [1987]) If G is a finite group and H < G such that IL(H)I = IL(G)I,show that L(H) L(G).

6. (Vasileva [1976]) If G is a finite group, H < K < G and L(H) L(G), show thatL(K) L(G).

7. (Antonov [1987]) Let G be a finite p-group such that L(G) is isomorphic to thelattice of subspaces of a vector space over GF(p"). If I A : Z(G)I = p" for everyminimal centralizer A of G, show that L(G) is a sublattice of L(G).

8. (Jurgensen [1976]) Suppose that G is a finite group such that L(G) is modular oflength 4 and the intersection N of all the maximal centralizers of G is differentfrom Z(G).(a) Show that [G/N]c has length 2 and GIN has a normal partition.(b) If Z(G) = 1, show that G is soluble.

9. (Jurgensen [1976]) Let

M = <a,b,cIa3 = b3 = c3 = 1,[a,b] = c,[a,c] = [b,c] = 1>

be the nonabelian group of order 27 and exponent 3 and let x E Aut M such thatas = a2, ba = b, ca = c2. Show that the semidirect product G = M<a) hascentralizer lattice modular of length 4.

10. If G is a finite group such that L(G) L(S4), show that G/Z(G) S4.

11. (Reuther [1975]) If G is a finite nonabelian p-group satisfying (22), show thatG ." Q8_

9.4 Coset lattices

In this final section we study the coset lattice 91(G) which was defined in § 1.1.Our first main problem, to determine the groups whose coset lattices have specialproperties, is not a very fruitful one: almost any interesting lattice property issatisfied by nearly all or by very few coset lattices, or else it holds in 91(G) if and onlyif it holds in L(G). More interesting is the study of isomorphisms between cosetlattices. Since the atoms of 91(G) are the one-element subsets of G and for any suchatom {g}, the interval [G/{g}] in 93(G) is isomorphic to L(G), two groups withisomorphic coset lattices have the same order and isomorphic subgroup lattices.

Nevertheless there are nonisomorphic groups with isomorphic coset lattices. Anisomorphism o: 93(G) -+ fl(G) can be regarded as a bijective map from G to Gpreserving cosets. We consider under which conditions a is an isomorphism from Gto G.

Basic properties of 9i(G)

Recall that ER(G) is the set of all right cosets of subgroups of G together with theempty set 0. With the set-theoretic inclusion as relation, 93(G) is a complete meet-sublattice of the lattice of all subsets of G; we write A and v for the intersection andjoin in 9(G). If H, K G and a, b E G, then Ha n Kb = 0 or

(1) Ha A Kb =Ha nKb =HxnKx=(HnK)xfor every xEHa nKb.

Clearly, Ha and Kb are contained in <H, K, ba-' >a. Conversely, if Mx is a cosetcontaining Ha and Kb, then a c- Mx implies Ha c Mx = Ma and hence H < M;similarly K < M and ba-' E M since a, b e Mx. Thus <H, K, ba-' >a is the smallestcoset containing Ha and Kb, that is,

(2) Ha v Kb = <H, K, ba-' >a.

Since aH = (aHa-' )a, every left coset is a right coset and conversely; thus fl(G) is theset of all (right or left) cosets of G. We shall need the following characterization ofcosets.

(3) A subset X of G is an element of IN(G) if and only if xy-' z e X for allx, y,zEX.

Indeed, for h; e H, we have hta(h2a)-'h3a = h1h2'h3a e Ha. Conversely, supposethat X 0 satisfies the condition, let S = {xy ' jx, y E X} and x, y, u, v e X. Then(xy-')(uv-')-' = xy-'vu-' E S since xy-'v e X, and so S is a subgroup of G. Fur-thermore xy-'v E X implies Sv c X and x = xv-'v E Sv yields X c Sv; thus X = Svis a coset.

The least element of 93(G) is the empty set 0 and the greatest element is the cosetG. The atoms in the lattice of all subsets of G are the one-element subsets of G andthese are cosets of the trivial subgroup 1. Therefore

(4) the atoms of tR(G) are the sets {g} where g c G.

Since the subgroups of G are precisely the cosets containing 1, we have that

(5) [G/1] = L(G).

Here, and throughout this section, intervals are taken in 31(G). Note that if T is asubgroup of G, an interval [S/T] in 9i(G) only contains subgroups of G so that wedo not need to distinguish between intervals in fl(G) and L(G).

For g e G and X, Y c G, it is obvious that X c Y if and only if Xg c Yg; and ifX = Ha is a right coset, so is Xg = Hag. Therefore the map

(6) pg: `R(G) given by X°9 = Xg for 0 # X E fl(G) is an isomorphism.

Since 9R(G) is the set of all left cosets in G, the map

(7) A9:'R(G) -+ 'R(G) given by Xz9 = g-1X for 0 0 X E SR(G) is an isomorphism.

Now 1°9 = {g} yields [G/{g}] [G/1] and hence

(8) [G/{g}] L(G) for every atom {g} of tR(G).

Finally, note that for every subgroup H of G, clearly,

(9) [H/O] = HR(H).

Groups with special coset lattices

As mentioned above, coset lattices do not usually have nice lattice properties. Forexample, if I < H < G and a e G\H, then {0, {a}, Ha, <H, a>, H} is a nonmodularsublattice of 9R(G). Thus a group with modular coset lattice is cyclic of at most primeorder. With a little more effort, we can prove the following.

9.4.1 Theorem (Curzio [1957c], Loiko [1962]). The following properties of a groupG are equivalent.

(a) ¶R(G) is modular.(b) 91(G) is lower semimodular.(c) I G I is a prime or 1.

Proof. By 2.1.5, every modular lattice is lower semimodular and so (a) implies(b). Now let SR(G) be lower semimodular and suppose, for a contradiction, that1 < H < G; let a e G\H. By Zorn's Lemma there exists a subgroup M maximal withrespect to the properties H < M < <H, a> and a 0 M (see the proof of Lemma 3.1.1).Clearly, M is a maximal subgroup of (H, a> and, by (2), (H, a> = M v Ma. SoM v Ma covers M in 94(G), but M A Ma = 0 c {a} c Ma. As 93(G) is lower semi-modular, this is a contradiction. It follows that G is cyclic of at most prime order.Finally, if IGJ is a prime p, then tR(G) is of length 2 and has p atoms. So 9 (G) ismodular.

Since a lattice of length 2 is distributive if and only if it has at most 2 atoms, weobtain from 9.4.1 the following.

9.4.2 Corollary. 'R(G) is distributive if and only if IGI < 2.

9.4.3 Theorem (Curzio [1957c]). ER(G) is directly decomposable if and only ifG1 = 2.

Proof. If G = { 1, g}, then ER(G) = {0, { 1 }, { g}, G} is the direct product of two chainsof length 1. Assume conversely that ¶R(G) = L1 x L2 with nontrivial lattices L.. LetOi and 1i be the least and greatest elements of L1, respectively, and let (11, 02) = Huand (01,12) = Kv where H, K < G and u, v c- G. Every atom { g} of 'R(G) is of the

form (x,02) or (Ol,y) with x e L1, y e L2 and hence is contained in Hu or Kv. So ifI e Hu, say, we see that G is the set-theoretic union of H and Kv. It follows thatK H and Hv E- Kv, that is, H = K is a subgroup of index 2 in G. But thenH = (11, 02) is maximal in G = (11,12) and hence (01, 02) = 0 is also maximal in(01,12) = Hv. Thus Hv is an atom of 91(G) and this yields H = 1 and I GI = 2.

Exceptions to the statement made at the beginning of this subsection are thefollowing two results.

9.4.4 Theorem (Curzio [1957c]). If every proper subgroup of G is contained in amaximal subgroup, then 91(G) is complemented. In particular, if G is finite or cyclic,91(G) is complemented.

Proof. Let X e 91(G) such that 0 c X c G. Then there exist H < G and a e G suchthat X = Ha. By assumption there exists a maximal subgroup M of G containing H.For every b e G\Ma, we have ba-' 0 M and hence by (2), Ha v Mb = G. Clearly,Ha A Mb 9 Ma n Mb = 0 and so Mb is a complement to X = Ha in 91(G).

9.4.5 Theorem (Curzio [1957c]). If G is a finite group, then 91(G) is relativelycomplemented if and only if L(G) is relatively complemented.

Proof. If 91(G) is relatively complemented, then so is its interval L(G). Supposeconversely that L(G) is relatively complemented and let [X/Y] be a nontrivial inter-val in 91(G). If Y # 0, then Y = bK for some b e G, K < G and the automorphism/!b of 91(G) maps [X/Y] to [b-'X/K]. This is an interval in L(G) and hence iscomplemented. Now let Y = 0. Then X = aH for some a e G, H < G and ;,, maps[X/Y] to [H/QS] = 91(H). Since H is finite, 91(H) is complemented, by 9.4.4. Thusevery interval of 91(G) is complemented and so 91(G) is relatively complemented.

Groups whose coset lattices are self-dual, or upper semimodular, or satisfy theJordan-Dedekind chain condition are treated in Exercises 1-3. Similar results as for91(G) were proved by d'Andrea [1969] for the lattice 91N(G) of all cosets of G withrespect to normal subgroups of G (see Exercise 4).

91-isomorphisms

A bijective map a: G - G is called an IX-isomorphism if for every subset X of G, wehave X E 91(G) if and only if X° E 91(G); a is called normed if l° = 1. Clearly, every91-isomorphism induces an isomorphism from 91(G) onto 91(G). Conversely, anisomorphism 0: 91(G) --+ 9R(6) maps every atom {g} of 91(G) to an atom {g} of 91(G)and the map g - g is an 91-isomorphism from G to G inducing 0. Therefore itis an equivalent problem to study isomorphisms between coset lattices or 91-isomorphisms between groups; we prefer to do the latter. So in the sequel weconsider the maps p9 and 1.9 defined in (6) and (7) as element maps; these are 91-

isomorphisms. The product of two 9i-isomorphisms clearly is an 91-isomorphism.The maps p and i sending every g e G to p9 and ate, respectively, are isomorphismsfrom G to subgroups Rt G and Lt G of the group RP(G) of all 9R-automorphisms ofG; to obtain this, we have to multiply by g-1 in (7). We call Rt G (and Lt G) the groupof right (and left) translations of G and have that

(10) RP(G) = RP(G)1 Rt G = RP(G)1 Lt G

if RP(G)1 is the group of all normed 91-automorphisms of G. _More generally, if a: G G is any 9t-isomorphism, then r = ai. 1,: G -+ G satisfies

1` = (1°) 1 1° = I and hence is a normed 9I-isomorphism. So we have shown that

(11) 92(G) and 9R(G) are isomorphic if and only if there exists a normed 9R-isomorphism from G to G.

Therefore it suffices to study normed 93-isomorphisms. Every isomorphism or anti-isomorphism a is a normed 9I-isomorphism since in this case (Hx)° = Haxa or x°H°is a coset if H < G and x e G. Conversely, we wish to find criteria for a normed93-isomorphism a: G -+ G to be an isomorphism or an antiisomorphism. We start bygiving some basic properties of a. A subset X of G is a subgroup of G if and only ifX e 9I(G) and 1 e X. Since a is a normed 91-isomorphism, this holds if and only ifX° e 9t(G) and I = 1° e X°, that is, if and only if X° < G. Thus

(12) a induces a projectivity & from G to G.

In particular, a maps cyclic subgroups to cyclic subgroups and hence satisfies

(13) <g>° _ <g°> for every g e G;

this also follows from the fact that <g> is the smallest coset containing 1 and g.Clearly, (13) implies that o(g) = o(g°) and so <g> ^- <g>°. We have shown:

(14) Every cyclic group is determined by its coset lattice.

This is one of the few cases in which (12) and I GI = 161 yield that G and G areisomorphic. If o(g) is infinite, we can even show that a induces an isomorphism inW.

9.4.6 Lemma (Loiko [1961]). If a: G -+ G is a normed 9i-isomorphism and u e G isan element of infinite order, then (u°)k = (uk)° for all k e Z.

Proof. Put u° = v. Then by (13), ° = <v> and we show by induction on k that(uk)° = vk for k e N; since uk and u-k are the only generators of (uk>, it will followthat (u-k)° = v-k. For k = 0 and k = 1, the assertion is clear. So let k >_ 1 and assumethat (u')° = v' for i = 0, ..., k. By (6) and (7), r = a ),k is an 91-isomorphismsatisfying It = v-k(uk)° = 1, by our induction assumption. By (12), r induces aprojectivity and, since ` = v-k° = <v>, it follows that uT = v or u= = v-1. Onthe other hand, the definition of r yields u` = v-k(uk+1 )° and hence (uk+1 )° = vk+1or vk-1. Since a is bijective and vk-1 = (uk-1)a we obtain that (uk+1)° = Vk+1 asdesired.

Further lattices

The above Icmiu. is fundamental in the study of 9I-isomorphisms. As a firstconsequent, wr note the following.

9.4.7 Lemma. If o: G -- G is a normed'.R-isomorphism and a, b e G such that o(b) _o(ah) = ac, then (aba)° = a°ba°.

Proof. Let r = pb-, u p,.. Then r is a normed 93-isomorphism since 1 t =(b-')°b° = 1, by 9.4.6. Furthermore (ab)t = (abb-')°b° = a°b° and, since o(ab) = oc,again 9.4.6 yields (aba)°b° = (abab)' = ((ab)')2 = a°b°a°b°. Thus (aba)° = a°b°a°.

It is to be expected that if the projectivity 8 in (12) is induced by an isomorphism,then a has some interesting properties. We show that if G has enough elements ofinfinite order, then a is in fact an isomorphism or an antiisomorphism in thissituation. For this we need the following result.

9.4.8 Lemma. Let G be a group which is torsion free or has an element of infiniteorder in its centre. If co: G G is a normed 9l-automorphism such that H' = H for allsubgroups H of G, then either x' = x for all x E G or x`° = x-' for all x E G.

Proof. Suppose first that G is torsion-free and let 1 0 x c- G. Then x and x-' are theonly generators of <x> and, since <x>°' = <x>, it follows that x' = x or x' = x-'.Suppose, for a contradiction, that a, b e G such that a 0 1 0 b, a' = a and b' = b-'.Then by 9.4.7, (aba)' = a°'b°'a' = ab-'a; on the other hand, (aba)' = aba or(aba)-'. In the first case, we obtain b = b-', a contradiction since o(b) = cc. Soab-' a = a-' b-' a-' and hence b-' a2b = a-2. This holds for every nontrivial elementb e G which is inverted by co. By 9.4.6, (b2)' = b-2 and so b_2a2b2 = a-2; butb-'a2b = a-2 implies b-'a 2b2 = a2. This contradiction shows that co maps everyelement of G to the same power.

Now suppose that Z(G) contains an element z of infinite order. Again x' = x orx-' for all x E G of infinite order; let z°' = z' where e E { + 1, -1}. We claim thatx°' = x` for all x e G. If <x> n <z> 0 1, then I ; x' c- (z> for some i c- 77; by 9.4.6,(x`)' = (x')' and then x' = x'. If (x> n (z> = I and o(x) = cc, then (x, z><x> x <z> is torsion-free and hence x`° = x', as we have just shown. Thus x' = x`for all x c G of infinite order. Now suppose that o(x) is finite. Since z e Z(G), x(z> isa coset containing a unique element of finite order, namely x. By (13), to preservesthe order of an element and hence (x<z>)' is a coset in which x' is the uniqueelement of finite order. Now (x(z>)`° contains (xz)' = x`z` and (xz2)` = x`(z`)2 andtherefore, by (3), it contains xez`(x`(zt)2)-'x`z` = x`. Thus x°' = x`, as desired.

9.4.9 Theorem. Let G be a group which is torsion free or has an element of infiniteorder in its centre. If a: G - G is a normed 9R-isomorphism such that the projectivity& induced by a is induced by an isomorphism, then a is an isomorphism or an anti-isomorphism.

Proof. Let a be an isomorphism inducing 6. Then co = ace' is a normed 'R-automorphism of G satisfying H' = H for all H < G. By 9.4.8, co is an isomorphismor an antiisomorphism and then so is a = coot.

We have shown in a number of cases that every projectivity of a given group isinduced by an isomorphism. So Theorem 9.4.9 covers some classes of groups whichwere handled directly, without using the results on projectivities, by Loiko.

9.4.10 Corollary. Let a: G G be a normed 9i'-isomorphism.(a) (Loiko [1961], [1963]) If G is torsion free abelian or mixed abelian of torsion-

free rank r > 2, then a is an isomorphism.(b) If G is a locally free group, then a is an isomorphism or an antiisomorphism.(c) (Loiko [1965c], Bruno [1970]) If G is a locally nilpotent torsion free group or

a free polynilpotent group, then a is an isomorphism or an antiisomorphism.

Proof. In all cases, G is torsion-free or has an element of infinite order in its centre.To apply Theorem 9.4.9, we have to show that 6 is induced by an isomorphism. By2.6.10, 7.1.19, 7.2.11, and 7.2.13, this holds in all cases except possibly when G istorsion-free abelian of rank 1. In this case, if a, b e G, then <a, b> is cyclic and hence(ab)° = a°b" by 9.4.6; thus a is an isomorphism.

Mixed abelian groups of torsion-free rank 1 do not fall under Theorem 9.4.9 sincein general they are not even determined by their subgroup lattices: we have shownin 2.5.15 that every M-group with elements of infinite order is lattice-isomorphic tosuch an abelian group. For coset lattices, the situation is not so simple; it wasclarified by Loiko in a special case. We give his result without proof.

9.4.11 Theorem (Loiko [1965a]). Let G be a mixed abelian group of torsion freerank 1 which splits over its torsion subgroup T(G). There exists an 91-isomorphismfrom G to the group G if and only if T(G) T(G), G/T(G) G/T(G), G = T(G)Hfor some torsion free subgroup H, and for every element z e G of infinite order andevery Sylow p-subgroup Tp of T(G) there exists an integer i(z, p) with the followingproperties:

(i) z-laz = a3-2i(z.°) for all a e Tp,(ii) i(z, p) = 1 if Exp T, = oc,(iii) 2(i(z, p) - 1)2 = 0 (mod p") and i(zk, p) 1 + (i(z, p) - 1)k (mod p") for all

kEJifExpT,=p".

It follows from this theorem that there are mixed abelian groups G of torsion-freerank 1 and nonabelian groups G_such that (i) 91(G) 9I(G), and, on the other hand,(ii) L(G) L(G) and 'R(G) 93(6) (see Exercises 6 and 7). For finite abelian groups,the situation is different. Here Theorem 2.5.9 shows that every locally finite non-hamiltonian p-group G with modular subgroup lattice is ¶R-isomorphic to a suitableabelian group G (see the proof of 9.4.12). Since Baer proved this theorem in 1944,these were the first examples of nonisomorphic groups with isomorphic coset latticesthat were discovered. The 91-isomorphisms a: G - 6 constructed in 2.5.9 satisfy

(Hx)° = H°x° for all H < G and x e G; so every right coset of H is mapped to a rightcoset of H°. However, this is not true for left cosets. Indeed, since x° E (xH)° and Gis abelian, it would follow that x°H° = (xH)° = (Hx)° = H°x°; so H°9 G for allH < G, a contradiction if G is not abelian. Similar phenomena occur in all the otherexamples mentioned above and we use the remainder of this section to study 91-isomorphisms for which this does not happen. Since they preserve the affin structureof G, they will be called affinities.

Affinities

An 9i-isomorphism a: G -+ G is called an affinity if

(15) (Hx)° = H°x° and (xH)° = x°H° for all H < G and x e G.

Note that a bijective map a satisfying (15) in general is not an 9R-isomorphism. It israther obvious that a satisfies 1° = 1 and maps subgroups of G to subgroups of G,but G may have more subgroups. For example, any normed bijective map from acyclic to an elementary abelian group of order 4 satisfies (15), and this is by no meansthe only example (see Exercise 9). So the extra condition in the following criterion isneeded.

9.4.12 Lemma. A bijective map a: G - G is an affinity if and only if a induces aprojectivity from G to G and satisfies (15).

Proof. If a is an affinity, then {1°} = (<1> l)° = <1>°1° = {(1°)2} and hence 1° = 1.By (12), a induces a projectivity. Conversely, suppose that a induces a projectivityand satisfies (15); let 0 X gi G. If X E 91(G), then X = Hx for some H < G, x E Gand our assumptions on a imply that H° < G and X° = H°x° E 91(G). If X° E 9R(G),there exist K < G and Y E G such that X° = Ky = H°x° = (Hx)° for some H < G,x c- G; so X = Hx e 91(G). Thus a is an 91-isomorphism satisfying (15).

Recall from (4) of § 1.3 the associativity identities for the amorphy a of a mapa: G -- G. We defined a(x, y) = (y°)-1(x°)-1(xy)° and obtained that

(16) a(x, y)`"a(xy, z) = a(y, z)a(x, yz) for all x, y, z E G.

We show next that affinities are characterized by the behaviour of their amorphies.

9.4.13 Theorem. The bijective map a: G G with amorphy a is an affinity if andonly if

(a) a induces a projectivity from G to G and(b) a(x,y) E <x> n <y>° for all x, y c- G.

If a is an affinity, then (xy)° = x°y° for all x, y e G such that o(x) or o(y) is infinite.

Proof. If a is an affinity, then 9.4.12 shows that (a) holds and that for x, y E G, wehave (xy)° E (x(y>)° = x°<y>°; hence a(x,y) e <y>° since <y>° is a subgroup of G.Similarly, (xy)° E <x>°y° and so a(x, y)(Y°)-' e <x>°. Since <y>° is cyclic, a(x, y)°Y°)-' _a(x, y) and (b) holds.

Now assume that a satisfies (a) and (b) and let x c- G, H < G. For h E H, we have(xh)" = x"h"a(x, h) E x"H" since a(x, h) E <h>" 5 H" < G. Hence

(17) (xH)" c x"H" for all x c- G, H < G.

If u c- G is of infinite order and U = <u)', then U" is infinite cyclic. Using inductionon I U : H 1, we show that (xH)" = x"H" for all x E U, H < U. This is true forU : HI = 1; so assume it to be true for subgroups of smaller index than I U : HI = n

and let U = x1 H u u x"H with x = x1 be the coset decomposition of U withrespect to H. Then

U" =(x1H)"u ...u (x"H)" x'H"u ...u x"H"

and therefore I U': H"I < n. The induction assumption implies that every subgroupof index <n of U" has a preimage different from H and so I U" : H"I = n. It followsthat the x°H" are distinct and that (x,H)" = x'H" for all i. In particular, (xH)" _x"H", as desired. By 9.4.12, a induces an affinity in U and so by 9.4.6,

(18) (u")k = (ilk)" for all k E Z.

Now assume that a(x,y) 0 1 for some elements x, y E G such that o(x) or o(y) isinfinite. Since a(x, y) E <x>" n (y>', it follows that o(x) = o(y) = oo and a(x, y) =(x")' = (y")f for r, s E Z. By (18), x' = y' and, applying the associativity identities (16)to x, x-', x'y, we get

a(x X-')(x,Y)°a(X1-r xrY) = a(X-',x'Y)a(x,Y).

Again by (18), a(x, x-') = I and so (y")' = a(x, y) E <x'y>" = <ys+1 >" = <(Y")s+1

This contradiction shows that a(x, y) = 1 and proves that

(19) (xy)" = x"y" for all x, y e G such that o(x) or o(y) is infinite.

By 9.4.12, it remains to be shown that a satisfies (15). So let x E G, H < G and h E H.If o(h) is infinite, then by (19), x"h" = (xh)" E (xH)"; if o(h) is finite, then I(x<h>)"I =a(h) = I x"<h>" I and so x"h" E x" <h>" = (x <h>)", by (17). Thus x"H" c (xH)" andso (xH)" = x"H". Since (hx)" = h"x"a(h, x) = h"a(h, x)x" E H"x", we obtain in thesame way (Hx)" = H"x".

The above theorem enables us to construct affinities which are not isomorphismsand even affinities between nonisomorphic groups.

9.4.14 Examples. (a) Let G be a group and let N and M be subgroups of G suchthat 1 <N< M < G and N< <x> for all x E G\M. For 1 od E N, the mapa=Qd:G-+Gdefined by x"=xfor xEMand x"=xdfor xEG\M is an affinity;it is an automorphism if and only if I G: MI = 2 and dz = 1.

Note that if G = x V is a p-group such that o(u) = p" > p'° = Exp V andm > 1, then N = <u"°-') = t$ _1(G) and M = <a""-"> x B = satisfy theabove assumptions.

(b) Let 2<pEP,andlet H=<u,v,wju"=v"=w"=1,[v,u]=w=w°=w")be the nonabelian group of order p3 and exponent p. Denote by a the automorphism

of order p of H given by u2 = uv-', v° = v. Let s be a quadratic nonresidue modulopandG=H<g>,G=H<g>where gP=w,gP=ws,andh9=ha=ha forallhEH.Then G and G are nonisomorphic groups of order p° (see Huppert [1967], p. 347)and the map a: G - G defined by (g'x)° = g'x for x c- H, i c- {0,...,p - l} is anaffinity.

Proof. (a) Of course, a is bijective. For H < G, we have H° = H if H < M andH° < HN = H if H -t M. Thus H° = H in any case. Now N < Z(G) since G isgenerated by the elements in G\M and these elements centralize N. It followsthat a(x, y) = 1 if one of x and y is contained in M. If .x, y E G\M, thena(x, y) = d- y-' d-' x-' (xy)° = d-2 if xy e M and a(x, y) = d' if xy 0 M. In anycase, a(x, y) E N < <x> r <y> = <x>° r <y>°. By 9.4.13, a is an affinity. And a is anautomorphism if and only if a(x, y) = 1 for all x, y c- G\M, and this is the case if andonly ifd2 = 1 and xy EM for all x, y G\\M.

(b) Clearly, a is bijective and an easy computation shows that for x = g'h, y = g'k(h, k E H, i, j E {0, ... , p - 11), (xy)° = x°y° if i + j - p. Now H = S2(G) = f2(G) and <w> = i(G) = U(G) and hence we see that(xy)° E (x°, y°> and a(x, y) E (x)° r <y)°. Since a-' has the corresponding pro-perties, it follows from 1.3.1 that a induces a projectivity, and then from 9.4.13 thata is an affinity.

It is another immediate consequence of Theorem 9.4.13 that affinities map abeliangroups onto abelian groups.

9.4.15 Theorem. If a: G -- G is an affinity and H < G is abelian, then H° is abelianand Z(G)° = Z(G).

Proof. Let x, y e G such that xy = yx. If o(x) or o(y) is infinite, then by 9.4.13,x°y° = (xy)° = (yx)° = y°x°. So assume that o(x) and o(y) are finite. Then W =<x, y> is a finite abelian group, hence W = <w,,..., w,) with <wr> n <wj> = I andso, by (b) of 9.4.13, a(wi, ww) = 1 for i 0 j if a is the amorphy of Q. Since a inducesa projectivity, W° = <wi,...,w, > and w'wj = (w;ww)° _ (www,)° = wJ'w;° for i 0 j.Thus W° is abelian and xy = yx. This shows that every abelian subgroup H of G ismapped to an abelian subgroup of G and that Z(G)° < Z(G). Since a-' is also anaffinity, we get the other inclusion.

If a: G G is any map, an element x c- G is called right regular for a if (_xg)° _x°g° or, equivalently, a(x, g) = I for all g e G; Reg'(a) is the set of right regularelements for a. Left regular elements and Reg`(a) are defined correspondingly.

9.4.16 Lemma. If a: G -- G is an affinity, then Reg'(a) and Reg'(a) are subgroups of G.

Proof. Since 1° = 1, 1 E Reg'(a). If x, y E Reg'(a), then (xyg)' = x°(yg)° = x°y°g° =(xy)°g° for all g e G and hence xy E Reg'(a). If o(x) is finite, this also implies thatx-' E Regr(a), and if o(x) is infinite, x-' E Reg'(a) by 9.4.13. Thus Regr(a) is a sub-group of G and the proof for Reg'(a) is similar.

We come to our first main result on regular elements. For this we need thefollowing lemma.

9.4.17 Lemma. If a: G - G is an affinity, then t: G - G defined by gt = ((g-')')-'for g e G is also an affinity.

Proof. Of course, t induces the same projectivity as a. For H< G and x e G, wehave

(Hx)t = ((x-'H)`)-1 = ((x-' )f HQ)-' = Ha((x-1)a)-l = H`xt.

Similarly, (xH)t = xtHT and by 9.4.12, t is an affinity.

9.4.18 Theorem (Schmidt [1984a]). Let G be a group, let Reg'G be the set of allelements of G which are right regular for every affinity from G to any group G anddefine Reg' G similarly. Then Reg' G = Reg' G =: Reg G is a characteristic subgroupof G containing all elements of infinite order and all involutions of G.

Proof. Let a: G -- G be an affinity. By 9.4.16, Reg'G and Reg'G are subgroups of G.Let X E Reg' G. If t is defined as in 9.4.17, then x-' E Reg'(t) and so for g E G,

((gx)°)-' = (x-ig-i)t = (x-')t(g-1)t = (xa)-'(ga)-' = (gaxa)-',

that is, (gx)a = g°xa. Thus X E Reg'(a) and, since a was arbitrary, x e Reg' G. SoRegr G < Reg' G and the other inclusion is proved similarly.

Now let x c- Reg G and of e Aut G. For every g c- G there exists y e G such thatg = ya and, since as is an affinity, (x'g)° = (xay')° = (xy)Q' = xaayaa = (xa)'g'. Thusx' E Reg'(a) and, since a was arbitrary, it follows that xa E Reg' G = Reg G. So(Reg G)a < Reg G and Reg G is a characteristic subgroup of G.

By 9.4.13, Reg G contains every element of infinite order. So, finally, let x c- G bean involution and H = 11, x}. Then for g e G, Hg = { g, xg} and by (15),

{ga,(xg)a} = (Hg)° = Haga = {g°,xaga}.

It follows that (xg)' = xaga and so x e Reg' G.

9.4.19 Corollary. If G is generated by involutions and elements of infinite order, everyaffinity of G is an isomorphism.

Semiaffinities

We call an 9R-isomorphism a: G -- G a left affinity if it satisfies (xH)' = x°H' and aright affinity if it satisfies (Hx)' = H' x' for all H < G and x e G. It is interestingto note that these semiaffinities have similar properties to those just established foraffinities. Schmidt [1983] shows that for a left affinity, every element of infinite order

and every involution is left regular, but also gives examples in which certain suchelements are not right regular. Nevertheless we obtain as a corollary that for groupsgenerated by elements of infinite order, every left affinity is an isomorphism, a resultdue to Loiko [1972]. Similar assertions hold for right affinities since for every rightaffinity a, the map z defined in 9.4.17 is a left affinity and conversely.

The structure of Am G and G/Reg G

We come to our main result on the amorphy of an affinity.

9.4.20 Theorem (Gaschiitz and Schmidt [1981], Schmidt [1984a]). If v: G -> G isan affinity and a the amorphy of a, then a(x, y) E Z(G) for all x, y e G.

Proof. Suppose, for a contradiction, that there exist x, y e G such that a(x, y) Z(G);by 9.4.13, x and y have finite order, and we can choose them so that o(x) is as smallas possible. If x = xlx2 with o(xl) and o(x2) less than o(x), then the associativityidentities (16) with x1, x2, y would imply

a(X,Y) = (a(x1,x2)(v1)-'a(X2,Y)a(XI ,X2Y) E Z(G),

a contradiction. Thus o(x) = p", p a prime, n e N.Let z e G such that a(x, y) does not commute with z°. Then a(x, y) 0 <z'>. Since

o(x°) = o(x) = p", the subgroups of <x°) form a chain and by 9.4.13,

<a(z,x)) <_ <X°) n <z°) < <a(X,Y)) < <x°) n <Y°).

Hence a(z, x)y' = a(z, x) = a(x, y)kp with k e Z and the associativity identities forz, x, y yield

(20) 1 = a(X, Y)' -kpa(Z, xY)a(zX, Y)-1.

Since a(x,y) and a(zx,y) are elements of <y°), they commute with y° and (20)implies that [a(z, xy), y°] = 1. In addition, a(z, xy) commutes with (xy)° =x°y°a(x y) = x°(y°)S (s E 71), and hence also with x°. Since a(x, y) E <x°), equation(20) yields [a(zx, y), x°] = 1. Now a(zx, y) also commutes with (zx)° = z"(x°)` (t E Z)and hence with z°. As a(z,xy) E <z°), again (20) implies that a(x,y)l-kp commuteswith z°. Since a(x, y) E <x°) is a p-element, <a(x, y)1-kp) = <a(x, y)) and so a(x, y)commutes with z°, the desired contradiction.

The above theorem is fundamental in the study of affinities. If a: G - G is anaffinity and N :!g G, then x°N° = (xN)° = (Nx)° = N°x° for all x e G. So N° a Gand 9.4.13 shows that the map

(21) ir: G/N -. 6/N°, Nx i- N°x° is an affinity.

And if N = Z(G), Theorems 9.4.15 and 9.4.20 show that this map is an isomorphism.

9.4.21 Corollary. Every affinity o : G - G induces an isomorphism from G/Z(G) ontoG/Z(G).

Another immediate consequence of 9.4.20 is that the associativity identitiesbecome

(22) a(x, y)a(xy, z) = a(y, z)a(x, yz) for all x, y, z c- G

if a is the amorphy of an affinity o : G G. To study the situation inside G, weintroduce Am(a) = <a(x, y)'-'Ix, y e G) and define the amorphy Am G of G to be thesubgroup generated by all the Am(o) where a ranges over the affinities from G togroups G.

9.4.22 Theorem. If G is a group, then Am G is a characteristic subgroup of Gcontained in Z(G).

Proof. By 9.4.20 and 9.4.15, Am G < Z(G). Let a E Aut G and let a: G - G be anaffinity with amorphy a. Then T = or-'a is an affinity and its amorphy b satisfies

x°Y°a(x,Y) _ (xY)Q = (x2Y2)2= x°y°b(x°,Y°)

for all x, y e G. So (a(x, y)'-')* _ (b(x°`, y8))`-' e Am G.

In the remainder of this section we want to show that G/Reg G is 7t-closed forevery set it of primes; this will yield another criterion for a group to have theproperty that every affinity is an isomorphism.

9.4.23 Lemma. Let a: G - G be an affinity, a the amorphy of a, and let u, v e Gsuch that uv = vu and (u) n (v) = 1. Then a(uv, x) = a(u, x)a(v, x) and a(x, uv) _a(x, u)a(x, v) for all x e G.

Proof. By (22), a(u, v)a(uv, x) = a(v, x)a(u, vx) and by 9.4,13, a(u, v) e (u)° n (v)° = 1.Hence

(23) a(uv, x) = a(v, x)a(u, vx).

Interchanging u and v, we get a(uv, x) = a(u, x)a(v, ux). Since

a(v, x), a(v, ux) E <v)° n Z(G) and a(u, vx), a(u, x) E (u)° n Z(G)

and the product of these two groups is direct, it follows that a(u, vx) = a(u, x). By(23), a(uv, x) = a(u, x)a(v, x), as desired. The other equation is proved in just the sameway applying (22) to x, u, v in place of u, v, x.

By a trivial induction, 9.4.23 yields that if x, y e G are elements of finite order andx = fl xp, y = fl yp are their primary decompositions, then

PCP PEP

(24) a(x, Y) = fl a(xp, Y) = fl a(x, y) = fl a(xp, Yp)PEP pEP PEP

and a(x,, vP) is the p-primary component of a(x, y); for the proof of the last equationnote that a(x, yP) _ fl a(xq, yp) = a(xp, yP) since <xq>a n <yp> = 1 for q 0 p.

qc PRecall that a group G is n-closed for a set it of primes if the product of any two

rr-elements of G is a rr-element.

9.4.24 Theorem (Schmidt [1984a]). If G is a group, then G/Reg G is 7r-closed forevery set it of primes. In particular, if G/Reg G is finite, it is nilpotent.

Proof. Suppose that x(Reg G) and y(Reg G) are nontrivial 7r-elements in G/Reg G.Then by 9.4.18, o(x) and o(y) are finite and so we may choose x and y as 7r-elements.If o(xy) is infinite, then xy e Reg G. So we may assume that o(xy) is finite; letxy = uv = vu where u is a n-element and v is a n'-element. We have to show that forevery affinity a: G G with amorphy a and every z c- G, we have a(v, z) = 1; thenv c- Reg'G = Reg G and x(Reg G)y(Reg G) = u(Reg G) is a n-element. By 9.4.18 and(24), it suffices to do this for z e G of prime power order p".

If p e it, then 9.4.13 implies that a(v, z) e <v> n <z)° = 1. So let p 0 it. Then a(x, y),a(y, z), a(x, yz) all are ir-elements in Z(G), whereas a(xy, z) is a n -element; hence (22)implies that a(xy, z) = 1. Since a(u, z) e c n <z> = 1, it follows from 9.4.23 thata(v, z) = a(uv, z) = a(xy, z) = 1, as desired. Thus G/Reg G is n-closed for every set itof primes and a finite group with this property is nilpotent.

9.4.25 Corollary. If G is a finite perfect group, then every affinity of G is anisomorphism.

Much more can be said about Am G and G/Reg G. First of all, Am G is locallycyclic. Furthermore it is possible to describe Am G and Reg G within G; note that forthe definition of these subgroups one has to consider every affinity from G to anygroup G. Call a subgroup S of G a p-collecting subgroup of G if it contains everyelement of infinite order and every p'-element of G and satisfies K = n <x> o 1;

xEG Scall K the p-collector of S. It is easy to see that the intersection TP(G) of all p-collecting subgroups of G is a p-collecting subgroup of G, and it is possible to showthat for p > 5, Reg G < TP(G) and TP(G)/Reg G is the p'-component of G/Reg G.Dually, the join of all p-collectors in G is not in general a p-collector, but it is thep-component of Am G. If p = 2 or 3, the same results hold modulo a few exceptionalcases. For details and proofs see Schmidt [1984a].

Exercises

1. (Curzio [1957c]) If 93(G) is a self-dual lattice, show that IGI is a prime or 1.2. (Loiko [1962]) Let G be a finite group. Show that 91(G) is upper semimodular if

and only if G is a p-group.3. (Loiko [1962]) Show that 91(G) satisfies the Jordan-Dedekind chain condition if

and only if L(G) does.

4. (d'Andrea [1969]) Let SRN(G) be the lattice consisting of all cosets of G withrespect to normal subgroups and the empty set.(a) Show that SRN(G) is modular if and only if G is simple or G = 1; show that

9RN(G) is distributive if and only if I G I < 2.(b) Show that ¶RN(G) is directly decomposable if and only if I G I = 2.(c) Show that 'RN(G) is complemented if and only if every proper normal

subgroup of G is contained in a maximal normal subgroup of G.5. (Loiko [1972]) If a: G - G is an 'R-isomorphism, show that (a<a-' b))° =

a"((a°)-'ba> and ((ba-' )a)" = <b°(a°)-' )a".6. (Loiko [1965a]) Let p > 2, G = (a) x <w> and G = (z) where o(a) =

o(b) = p2, o(w) = o(z) = cc and z-'bz = b'-2P. Show that the map a: G , Gdefined by (a'ws)° = b'"+SP)zs is an SR-isomorphism. Find H < G and x E G suchthat (Hx)" 0 H°x°.

7. (Loiko [1965a]) Let p > 2, G = <a> x <w> and G = <b) <z> where o(a) _o(b) = p3, o(w) = o(z) = oo and z-`bz = b'+2P. Show that ER(G) %(G); notethat by 2.5.15, L(G) n-- L(6).

8. (Loiko [1965a]) Let G = <a> x <w> where o(a) = 2", n > 2, and o(w) = x. De-fine a: G G by (a`w')° = c(i, k)a`wk where c(i, k) = a2 -1 if i and k are odd andc(i, k) = I in all other cases. Show that o. is a normed 931-automorphism but noautomorphism of G.

9. If G is a cyclic and G an arbitrary group of order p", show that there exists abijective map a: G -+ G satisfying (15).

10. (Schmidt [1983]) Let a: G G be bijective, a the amorphy of a and assume thatv induces a projectivity from G to G. Show that

(i) (xH)°=x°H° for allH<GandxeG

implies

(ii) a(x, y) E <y> for all x, y e G,

and this implies

(iii) (xy)° = x"y° for all x, y e G such that o(x) = o(y) = cc.

11. Use Exercise 10 to prove Loiko's theorem saying that for a group generated byelements of infinite order, every left affinity is an isomorphism.

12. Let p > 2 be a prime, G = <g) a cyclic group of order p", n >_ 1, and let H =<gP>. Show that the map a: G --+ G defined by (gx)° = g2x and (g2x)" = gx forall x c- H and y° = y for all y e G\(gH u g2H) is an affinity of G.

13. (Gaschiitz and Schmidt [1981]) Let G be a group.(a) If x, y e G such that <x> n (y) = 1 = <xy) n (y), show that y e Reg G.(b) If x and y are p-elements of G such that x 0 Reg G and y Reg G, show that

14. (Schmidt [1984a]) Show that Am G is locally cyclic.

Bibliography

IN. Abramovskii1967. Groups whose lattice of subgroups is a lattice with relative complements (Rus-

sian). Algebra i Logika 6, 5-8.A.B. d'Andrea

1969. Sul reticolo SN(G). Ricerche Mat. 18, 128-140.V.A. Antonov

1980. Groups of Gaschutz type and groups close to them. Math. Notes 27, 405-414.1987. Finite groups with a modular lattice of centralizers. Algebra and Logic 26, 403-

422.1991. The finite groups in which the lattice of centralizers is a sublattice of the subgroup

lattice. Soviet Math. (Iz. VUZ) 35, no. 3, 4-12.M.N. Arshinov

1970a. Projection of a free product and group isomorphisms. Siberian Math. J. 11, 8-14.1970b. Projectivities of pure supersoluble groups (Russian). Dokl. Akad. Nauk BSSR 14,

984-985.1971. Lattice isomorphisms of certain classes of infinite groups. Dissertation Moscow.

M.N. Arshinov and L.E. Sadovskii1972. Some lattice-theoretical properties of groups and semi-groups. Russian Math. Sur-

veys 27, no. 6, 149-191.1973. The lattice definability of wreath products (Russian). Trudy Moskov Mat. Obsc. 29,

207-213.R. Baer

1934. Der Kern, eine charakteristische Untergruppe, Compositio Math. 1, 254-283.1937. Dualism in abelian groups. Bull. Amer. Math. Soc. 43, 121-124.1939a. The significance of the system of subgroups for the structure of the group. Amer.

J. Math. 61, 1-44.1939b. Duality and commutativity of groups. Duke Math. Journal 5, 824-838.1943. A theory of crossed characters. Trans. Amer. Math. Soc. 54, 103-170.1944. Crossed isomorphisms. Amer. J. Math. 66, 341-404.1952. Linear algebra and projective geometry. Academic Press, New York.1956. Norm and hypernorm. Publ. Math. Debrecen 4, 347-350.1961 a. Partitionen endlicher Gruppen. Math. Z. 75, 333-372.1961b. Einfache Partitionen endlicher Gruppen mit nichttrivialer Fittingscher Unter-

gruppe. Archiv Math. 12, 81-89.1964. Irreducible groups of automorphisms of abelian groups. Pacific J. Math. 14, 385-

406.D.W. Barnes

1962. Lattice embeddings of prime power groups. J. Austral. Math. Soc. 2, 17-34.1964. Projectivities in finite groups. J. Austral. Math. Soc. 4, 308-326.

D.W. Barnes and G.E. Wall1964. On normaliser preserving lattice isomorphisms between nilpotent groups. J. Aus-

tral. Math. Soc. 4, 454-469.

Bibliography 543

P.P. Barysovec1977. Finite nonabelian groups with complemented nonabelian subgroups. Ukrainian

Math. J. 29, 541-544.1981a. Finite nonnilpotent groups in which all the nonabelian subgroups are comple-

mented. Ukrainian Math. J. 33, 117-122.1981b. On a class of finite groups with complemented nonabelian subgroups. Ukrainian

Math. J. 33, 225-229.G. Baumslag

1962. On generalized free products. Math. Z. 78, 423-438.H. Bechtell

1965. Elementary groups. Trans. Amer. Math. Soc. 114, 355-362.1969. A note on permutable complements. Amer. Math. Monthly 76, 799-800.

T.R. Berger and F. Gross1982. A universal example of a core-free permutable subgroup. Rocky Mountain J.

Math. 12, 345-365.Ja.G. Berkovic

1966. Finite groups satisfying weakened Jordan-Dedekind conditions concerning chains(Russian). Finite Groups, 14-23. Nauka i Tehnika, Minsk.

M. Bianchi and M.C. Tamburini Bellani1977. Sugli IM-gruppi finiti e i loro duali. Istit. Lombardo Accad. Sci. Lett. Rend. A 111,

429-436.G. Birkhoff

1967. Lattice Theory. Amer. Math. Soc. Coll. Publ. XXV, 3rd Ed., Providence.N. Bourbaki

1968. Elements of mathematics: Theory of sets. Hermann, Paris.R.H. Bradway, F. Gross and W.R. Scott

1971. The nilpotence class of core-free quasinormal subgroups. Rocky Mountain J.Math. 1, 375-382.

R. Brandl1986. On groups with certain lattices of normal subgroups. Archiv Math. 47, 6-11.1988. The Dilworth number of subgroup lattices. Archiv Math. 50, 502-510.1989. Groups whose lattices of normal subgroups are distributive. Glasgow Math. J. 31,

183-188.B. Bruno

1970. Un teorema di approssimazione per gli S-isomorfismi. Boll. Un. Mat. Ital. (4) 3,225-229.

1972. Coppie n-distributive ed u-distributive net reticolo dei sottogruppi di un grup-po. Ist. Veneto Sci. Lett. Arti Atti Cl. Sci. Mat. Natur. 131, 147-153.

G. Busetto1979. Propriety di immersione dei sottogruppi quasinormali periodici. Matematiche 34,

243-249.1980a. Una caratterizzazione reticolare dei gruppi iperabeliani. Atti Accad. Naz. Lincei

Rend. Cl. Sci. Fis. Mat. Natur. (8) 68, 95-98.1980b. Propriety di immersione dei sottogruppi modulari localmente ciclici nei gruppi.

Rend. Sem. Mat. Univ. Padova 63, 269-284.1982. Sottogruppi normali e proiettivity. Rend. Sem. Mat. Univ. Padova 67, 105-

110.1984. Normal subgroups and projectivities. J. Algebra 88, 52-62.1985. On the embedding of core-free projective images of normal subgroups. Ann. Mat.

Pura Appl. (4) 142, 91-103.

544 Bibliography

G. Busetto and F. Menegazzo1985. Groups with soluble factor groups and projectivities. Rend. Sem. Mat. Univ.

Padova 73, 249-260.G. Busetto and F. Napolitani

1990. On projectivities of groups. Rend. Circ. Mat. Palermo (2) Suppl. 23. 45-55.1991. On the behaviour of normal subgroups under projectivities. J. Algebra 141, 93

105.

1992. On some questions concerning permutable subgroups of finitely generated groupsand projectivities of perfect groups. Rend. Sem. Mat. Univ. Padova 87, 303-305.

G. Busetto and S.E. Stonehewer1985. A nonabelian normal subgroup with a core-free projective image. J. Algebra 97,

329-346.R.A. Calcaterra

1987. Group actions with inverting centralizers. Archiv Math. 49, 465-469.A. Caranti

1979. Proiettivita dei p-gruppi di classe massimale. Rend. Sem. Mat. Univ. Padova 61.393-404.

M. Cardin1984. Su una classe di gruppi con molti sottogruppi quasinormali. Rend. Sem. Mat.

Univ. Padova 72, 157-161.S.N. Cernikov

1954. Groups with systems of complemented subgroups. Amer. Math. Soc. Transl. (2) 17(1961), 117-152.

1967. Groups with given properties of systems of infinite subgroups. Ukrainian Math. J.19, 715-731.

1980. Groups with complemented infinite abelian subgroups. Math. Notes 28, 788-792.

N.V. Cernikova1956. Groups with complemented subgroups. Amer. Math. Soc. Transl. (2) 17, (1961),

153-172.Y. Cheng

1982. On finite p-groups with cyclic commutator subgroup. Archiv Math. 39, 295-298.C. Christensen

1964. Complementation in groups. Math. Z. 84, 52-69.C.D.H. Cooper

1968. Power automorphisms of a group. Math. Z. 107, 335-356.M. Costantini

1989a. Sul gruppo delle autoproiettivita di PSL(3, q). Riv. Mat. Pura Appl. 4, 79-88.1989b. On the lattice automorphisms of certain algebraic groups. Dissertation Warwick.1990. On lattice automorphisms of the special linear group. Atti Accad. Naz. Lincei

Rend. Cl. Sci. Fis. Mat. Natur. (8) 83, 33-38.P. Crawley and R.P. Dilworth

1973. Algebraic theory of lattices. Prentice-Hall, Englewood Cliffs.M. Curzio

1953. Su di un particolare isomorfismo di struttura. Ricerche Mat. 2, 288-300.1955. Gli automorfismi del reticolo dei laterali dei sottogruppi d'un gruppo. Ricerche

Mat. 4, 3-14.1957a. Sul reticolo dei sottogruppi di composizione di alcuni gruppi finiti. Boll. Un. Mat.

Ital. (3) 12, 284-289.

Bibliography 545

1957b. Alcune osservazioni sul reticolo dei sottogruppi d'un gruppo finito. Ricerche Mat.6,96-110.

1957c. Propriety di struttura delle classi laterali relative ad un gruppo. Matematiche(Catania) 12, 51-57.

1957d. Sui gruppi supersolubili per cui it reticolo dei sottogruppi di composizione 6autoduale. Matematiche (Catania) 12, 74-79.

1957e. Sui gruppi tp-isomorfi ad un gruppo speciale finito. Rend. Accad. Sci. Fis. Mat.Napoli (4) 24, 70-75.

1958. Sui sottogruppi di composizione dei gruppi finiti. Ricerche Mat. 7, 265-280.1960. Alcuni criteri di finitezza per i gruppi a condizione massimale o minimale.

Ricerche Mat. 9, 248-254.1962. Certains criteres de nilpotence pour les groupes finis. Bull. Sci. Math. (2) 86, 74-80.1964a. Elementi u-quasi-distributivi in alcuni reticoli di gruppi. Ricerche Mat. 13, 70-79.1964b. Sui gruppi per cui sono riducibili alcuni reticoli di sottogruppi notevoli. Mate-

matiche (Catania) 19, 1-10.1965. Una caratterizzazione reticolare dei gruppi abeliani. Rend. Mat. e Appl. (5) 24,

1-10.M. Curzio and S. Rao

1981. Sui fattori principali degli IM-gruppi. Ricerche Mat. 30, 279-295.G. Daues and H. Heineken

1975. Dualiti ten and Gruppen der Ordnung p6. Geometriae Dedicata 4, 215-220.F. De Giovanni and S. Franciosi

1981. Debole complementazione in teoria dei gruppi. Ricerche Mat. 30, 35-56.1982. Sui gruppi sottomodulari infiniti. Rend. Circ. Mat. Palermo (2) 31, 257-266.1983. Una caratterizzazione reticolare dei gruppi residualmente supersolubili. Boll. Un.

Mat. Ital. A(6) 2, 355-360.1985. Isomorfismi tra reticoli di sottogruppi normali di gruppi nilpotenti senza torsione.

Ann. Univ. Ferrara Sez. VII (N.S.) 91, 91-98.1986. Alcuni isomorfismi duali tra reticoli di sottogruppi normali. Boll. Un. Mat. Ital. A

(6) 5, 185-192.1987. Sui gruppi nei quasi ogni sottogruppo finito a dotato di complemento. Mate-

matiche 38, 201-220.WE. Deskins

1963. On quasinormal subgroups of finite groups. Math. Z. 82, 125-132.R.P. Dilworth

(See P. Crawley and R.P. Dilworth)L. Di Martino and M.C. Tamburini Bellani

1980. Do finite simple groups always contain subgroups which are not intersection ofmaximal subgroups? Istit. Lombardo Accad. Sci. Lett. Rend. A 114, 65-72.

1981. On the solvability of finite IM-groups. Istit. Lombardo Accad. Sci. Lett. Rend. A115, 235-242.

N.T. Dinerstein1968. Finiteness conditions in groups with systems of complemented subgroups. Math.

Z. 106, 321-326.M. Emaldi

1969. Sui gruppi risolubili complementati. Rend. Sem. Mat. Univ. Padova 42, 123-128.1970. Una nota sui gruppi risolubili complementati. Boll. Un. Mat. Ital. (4) 3, 858-862.1971. Sui gruppi relativamente complementati. Rend. Sem. Mat. Univ. Padova 46, 15-18.1978. Una caratterizzazione reticolare dei gruppi completamente fattorizzabili. Rend.

Sem. Mat. Univ. Padova 59, 17-18.

546 Bibliography

1985a. Sui gruppi completamente fattorizzabili. Rend. Accad. Sci. Fis. Mat. Napoli (4) 52,35-39.

1985b. Sui gruppi risolubili con molti sottogruppi massimali sensitivi. Matematiche(Catania) 40, 3-9.

M. Emaldi and G. Zacher1965. I gruppi risolubili relativamente complementati. Ricerche Mat. 14, 3-8.

W. Feit1983. An interval in the subgroup lattice of a finite group which is isomorphic to M,.

Algebra Universalis 17, 220-221.T. Foguel

1992. Galois-theoretical groups. J. Algebra 150, 321-323.A. Fort

1975. Gruppi finiti debolmente modulari. Rend. Sem. Mat. Univ. Padova 53, 269-290.

1978. Gruppi finiti debolmente supersolubili. Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis.Mat. Natur. (8) 64, 270-272.

1983. I gruppi finiti nei quali tutti i sottogruppi massimi sono duali di Dedekind. Rend.Accad. Naz. Sci. XL Mem. Mat. (5) 7, 1-5.

A. Franchetta1978. Sui gruppi con la struttura subnormale dotata di duale. Ricerche Mat. 27, 333-

341.A. Franchetta and F. Tuccillo

1976. Su una classe di gruppi ipercentrali. Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis.Mat. Natur. (8) 59, 232-237.

S. Franciosi1981. Alcuni omomorfismi duali tra reticoli di sottogruppi e reticoli di sottogruppi nor-

mali. Ricerche Mat. 30, 179-193.(See also F. De Giovanni and S. Franciosi)

L. Fuchs1960. Abelian groups. Pergamon Press, New York.

W. Gaschiitz1955. Gruppen, deren samtliche Untergruppen Zentralisatoren rind. Archiv Math. 6,

5-8.W. Gaschutz and R. Schmidt

1981. Affinitaten endlicher Gruppen. J. Reine Angew. Math. 326, 187-197.E. Gasparini and C. Metelli

1984. On projectivities of abelian groups of torsionfree rank one. Boll. Un. Mat. Ital. A(6) 3, 363-371.

L. Gerhards1970. Uber die Struktur bizyklischer Gruppen. J. Reine Angew. Math. 241, 180-199.

Yu.M. Gorcakov1960. Primarily factorizable groups. Soviet Math. Dokl. 1, 1001-1002.

D. Gorenstein1982. Finite simple groups: An introduction to their classification. Plenum Press, New

York and London.G. Gratzer

1978. General lattice theory. Birkhauser Verlag, Basel and Stuttgart.F. Gross

1971. p-subgroups of core-free quasinormal subgroups. Rocky Mountain J. Math. 1,541-550.

Bibliography 547

1975a. Subnormal core-free, quasinormal subgroups are solvable. Bull. London Math.Soc. 7, 93-95.

1975b. p-subgroups of core-free quasinormal subgroups. II. Rocky Mountain J. Math. 5,349-359.

1982. Infinite permutable subgroups. Rocky Mountain J. Math. 12, 333-343.(See also T.R. Berger and F. Gross; R.H. Bradway, F. Gross and W.R. Scott)

P. Hall1937. Complemented groups. Journal London Math. Soc. 12, 201-204.

M. Hall1961. On the lattice of normal subgroups of a group. Proc. Sympos. Pure Math., Vol. II,

American Math. Soc., Providence, R.1., 185-194.P. Hauck and H. Kurzweil

1990. A lattice-theoretic characterization of pre-Frattini subgroups. Manuscripta Math.66, 295-301.

H. Heineken1965. Uber die Charakterisierung von Gruppen durch gewisse Untergruppenverbiinde.

J. Reine Angew. Math. 220, 30-36.1987. A remark on subgroup lattices of finite soluble groups. Rend. Sem. Mat. Univ.

Padova 77, 135-147.(See also G. Daues and H. Heineken)

K.K. Hickin and R.E. Phillips1973. On classes of groups defined by systems of subgroups. Archiv Math. 24, 346-

350.D.G. Higman

1951. Lattice homomorphisms induced by group homomorphisms. Proc. Amer. Math.Soc. 2, 467-478.

C.S. Holmes1969. Projectivities of free products. Rend. Sem. Mat. Univ. Padova 42, 341-387.1971. Direct products with isomorphic lattices. Rend. Sem. Mat. Univ. Padova 45, 71-

80.1976. Groups of order p3q with identical subgroup structures. Atti Accad. Sci. Ist. Bolo-

gna Cl. Sci. Fis. Rend. (13) 3, 113-123.1984. Generalized Rottlander, Honda, Yff groups. Houston J. Math. 10, 405-414.1988. Automorphisms of the lattice of subgroups of 7Lp.n x 7L,, . Archiv Math. 51, 491-

495.1989. Nearly normal projectivities. Rend. Sem. Mat. Univ. Padova 82, 151-156.

K. Honda1952. On finite groups whose Sylow subgroups are all cyclic. Comm. Math. Univ. Sancti

Pauli 1, 5-39.D.R. Hughes and J.G. Thompson

1959. The Hp problem and the structure of Ho groups. Pacific J. Math. 9, 1097-1102.B. Huppert

1960. Gruppen mit modularer Sylow-Gruppe. Math. Z. 75, 140-153.1961. Zur Sylowstruktur autlosbarer Gruppen. Archiv Math. 12, 161-169.1967. Endliche Gruppen I. Springer-Verlag, Berlin Heidelberg New York.

B. Huppert and N. Blackburn1982a. Finite groups 11. Springer-Verlag, Berlin Heidelberg New York.1982b. Finite groups III. Springer -Verlag, Berlin Heidelberg New York.

N. Ito1951. Note on (LM)-groups of finite orders. Kodei Math. Sem. Rep. 1-6.

548 Bibliography

N. Ito and J. Szep1962. Uber die Quasinormalteiler von endlichen Gruppen. Acta Sci. Math. (Szeged) 23,

168-170.S.G. Ivanov

1969. Standard and dually standard elements of the lattice of subgroups of a group.Algebra and Logic 8, 253-256.

1985. L-homomorphisms of locally soluble torsion-free groups (Russian). Mat. Zametki37, 627-635.

K. Iwasawa1941. Uber die endlichen Gruppen and die Verbande ihrer Untergruppen. J. Fac. Sci.

Imp. Univ. Tokyo. Sect. 1.4,171-199.1943. On the structure of infinite M-groups. Jap. J. Math. 18, 709-728.

A.W. Jones1945. The lattice isomorphisms of certain finite groups. Duke Math. J. 12, 541-560.

R. Jiirgensen1976.

O.H. Kegel

Gruppen mit modularem Zentralisatorverband der Dimension 4. DissertationKiel.

1961. Die Nilpotenz der HP Gruppen. Math. Z. 75, 373-376.1978. Untergruppenverbande endlicher Gruppen, die den Subnormalteilerverband echt

enthalten. Archiv Math. 30, 225-228.E.I. Khukhro

1993. Nilpotent groups and their automorphisms. Walter de Gruyter, Berlin New York.W. Kimmerle, R. Lyons, R. Sandling, and D.N. Teague

1990. Composition factors from the group ring and Artin's theorem on orders of simplegroups. Proc. London Math. Soc. (3) 60, 89-122.

R. Kochendbrfer1969. Soluble groups with complemented subnormal subgroups. J. Austral. Math. Soc.

10, 465-468.P. Kohler

1983. M, as an interval in a subgroup lattice. Algebra Universalis 17, 263-266.P.G. Kontorovic and B.I. Plotkin

1954. Lattices with an additive basis (Russian). Mat. Sbornik 35 (77), 187-192.A.G. Kurosh

1955. The theory of groups 1. Chelsea Publ. Comp., New York.1956. The theory of groups II. Chelsea Publ. Comp., New York.

H. Kurzweil1985. Endliche Gruppen mit vielen Untergruppen. J. Reine Angew. Math. 356, 140-

160.

1989. Die Praefrattinigruppe im Intervall eines Untergruppenverbandes. Archiv Math.53, 235-244.(See also P. Hauck and H. Kurzweil)

J.C. Lennox1981. A note on quasinormal subgroups of finitely generated groups. J. London Math.

Soc. (2) 24, 127-128.J.C. Lennox and S.E. Stonehewer

1987. Subnormal subgroups of groups. Clarendon Press, Oxford.S. Li

1983. On the subgroup lattice characterization of finite simple groups of Lie type. Chi-nese Ann. Math. Ser. B 4, 165-169.

Bibliography 549

W. Lindenberg1970. Uber die Struktur zerfallender bizyklischer p-Gruppen. J. Reine Angew. Math.

241, 118-146.N.V. Loiko

1961. The S-isomorphism of torsion-free Abelian groups (Russian). Ural. Gos. Univ.Mat. Zap. 3, 67-71.

1962. Lattice properties of cosets of a group (Russian). Sibirsk. Mat. Z. 3, 79-86.1963. S-isomorphisms of mixed abelian groups of rank r 2 (Russian). Ural. Gos. Univ.

Mat. Zap. 4, 57-59.1965a. S-isomorphisms of mixed abelian groups of rank r = 1 (Russian). Sibirsk. Mat. Z.

6, 1053-1067.1965b. S-isomorphisms of metabelian torsion-free groups (Russian). Ural. Gos. Univ. Mat.

Zap. 5, 61-66.1965c. Natural S-isomorphisms of torsion-free groups (Russian). Mat. Zap. Krasnoyarsk

1,46-53.1972. On S-isomorphisms of groups generated by elements of infinite order. Siberian

Math. J. 13, 151-154.P. Longobardi

1982. Gruppi finiti a fattoriali modulari. Note Mat. 2, 73-100.P. Longobardi and M. Maj

1976. Su di un teorema di Heineken. Riv. Mat. Univ. Parma (4) 2, 315-320.1985. Su una classe di gruppi proiettivamente invariante. Ricerche Mat. 34, 325-

332.1986a. Finite groups with nilpotent commutator subgroup, having a distributive lattice of

normal subgroups. J. Algebra 101, 251-261.1986b. Elementi neutri nel reticolo dei sottogruppi normali di un gruppo supersolubile.

Boll. Un. Mat. Ital. A (6) 5, 121-128.1987. On the nilpotence of groups with a certain lattice of normal subgroups. Group

Theory (Bressanone 1986), 107-114, Lecture Notes in Math. 1281, Springer-Verlag,Berlin New York.

A. Lucchini1986. Sui gruppi it cui gruppo delle autoproiettivita a transitivo sugli atomi. Rend. Sem.

Mat. Univ. Padova 75, 275-294.1988. On p-groups whose L-automorphism group is transitive on the atoms. Rend. Sem.

Mat. Univ. Padova 80, 45-53.1991. On subgroups of index p'. Comm. Algebra 19, 2851-2864.

E. Lukacs and P.P. Palfy1986. Modularity of the subgroup lattice of a direct square. Archiv Math. 46, 18-19.

R. Lyons(See W. Kimmerle, R. Lyons, R. Sandling, and D.N. Teague)

K. Mahdavi and J. Poland1992. On lattice-isomorphic abelian groups. Archiv Math. 58, 220-230.1993. On lattice isomorphic of mixed abelian groups. Archiv Math. 60, 327-329.

R. Maier and P. Schmid1973. The embedding of quasinormal subgroups in finite groups. Math. Z. 131, 269-272.

M. Mainardis1991. Projektivitaten zwischen abelschen and nichtabelschen Gruppen. Archiv Math.

57, 332-338.1992. A simple proof of Baer's and Sato's theorems on lattice-isomorphisms between

groups. Atti. Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (9) 3, 173-176.

550 Bibliography

M. Maj1984. Gruppi infiniti supersolubili con it reticolo dei sottogruppi normali distributivo.

Rend. Accad. Sci. Fis. Mat. Napoli (4) 51, 15-20.(See also P. Longobardi and M. Maj)

A. Mann1978. Conjugacy classes in finite groups. Israel J. Math. 31, 78-84.

C. Marchionna Tibiletti1972. Su alcuni reticoli legati ad un gruppo. 1st. Lombardo Accad. Sci. Lett. Rend. A 106,

470-504.E. Martino

1973. Elementi neutri nel reticolo dei sottogruppi d'un gruppo. Rend. Sem. Mat. Univ.Padova 50, 345-354.

F. Menegazzo1970a. Sui gruppi relativamente complementati. Rend. Sem. Mat. Univ. Padova 43, 209-

214.1970b. Gruppi nei quali ogni sottogruppo e intersezione di sottogruppi massimali. Atti

Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur (8) 48, 559-562.1971. Dual-Dedekind subgroups in finite groups. Rend. Sem. Mat. Univ. Padova 45,

99-111.1973. Isomorfismi reticolari e prodotti intrecciati. Rend. Sem. Mat. Univ. Padova 50,

331-344.1974. On index preserving projectivities of finite groups. Rend. Sem. Mat. Univ. Padova

52, 227-242.1978. Normal subgroups and projectivities of finite groups. Rend. Sem. Mat. Univ.

Padova 59, 11-15.(See also G. Busetto and F. Menegazzo)

C. Metelli1968. Sugli isomorfismi reticolari del gruppo proiettivo lineare speciale PSL(2,p). 1st.

Veneto Sci. Lett. Arti Atti Cl. Sci. Mat. Natur. 127, 73-78.1969. Sugli isomorfismi reticolari di PSL(2, pf). Atti Accad. Naz. Lincei Rend. Cl. Sci.

Fis. Mat. Natur. (8) 47, 446-452.1971. I gruppi semplici minimali sono individuati reticolarmente in senso stretto. Rend.

Sem. Mat. Univ. Padova 45, 367-378.(See also E. Gasparini and C. Metelli)

M.D. Miller1975. On the lattice of normal subgroups of a direct product. Pacific J. Math. 60, 153-

158.

L.S. Mozarovskaja1978. Some properties of finite supersoluble groups. Ukrainian Math. J. 30, 191-192.

K. Nakamura1966. Uber den Quasinormalteiler der regularen p-Gruppe von der Klasse 2. Nagoya

Math. J. 26, 61-67.1968. Uber einige Beispiele der Quasinormalteiler einer p-Gruppe. Nagoya Math. J. 31,

97-103.1970. Beziehungen zwischen den Strukturen von Normalteiler and Quasinormalteiler.

Osaka J. Math. 7, 321-322.1979. Charakteristische Untergruppen von Quasinormalteilern. Archiv Math. 32, 513-515.

F. Napolitani1965. Elementi u-quasi distributivi nel reticolo dei sottogruppi di un gruppo. Ricerche

Mat. 14, 93-101.

Bibliography 551

1967a. Sui p-gruppi modulari finiti. Rend. Sem. Mat. Univ. Padova 39, 296-303.1967b. Sui gruppi risolubili complementati. Rend. Sem. Mat. Univ. Padova 38, 118-120.1968. Elementi r'-quasi distributivi ed u.c.r.-elementi del reticolo dei sottogruppi di un

gruppo finito. Ricerche Mat. 17, 95-108.1970. Modularity e distributivity nell'insieme dei sottogruppi subnormali. Rend. Sem.

Mat. Univ. Padova 43, 215-220.1971. Gruppi finiti minimali non modulari. Rend. Sem. Mat. Univ. Padova 45, 229-

248.1973. Submodularity nei gruppi finiti. Rend. Sem. Mat. Univ. Padova 50, 355-363.1982. Isomorfismi reticolari e gruppi perfetti. Rend. Sem. Mat. Univ. Padova 67, 181-

184.

F. Napolitani and G. Zacher1983. Uber das Verhalten der Normalteiler unter Projektivitaten. Math. Z. 183, 371-

380.(See also G. Busetto and F. Napolitani)

B.H. Neumann1954. Groups covered by permutable subsets. J. London Math. Soc. 29, 236-248.

W. Niegel1973a. Die Untergruppenverbande von Gruppen, deren Ordnungen hochstens drei Prim-

faktoren enthalten. J. Reine Angew. Math. 258, 1-22.1973b. Die Untergruppenstruktur der Gruppen mit quadratfreier Ordnung. J. Reine

Angew. Math. 259, 48-64.V.N. Obraztsov

1990. An imbedding theorem for groups and its corollaries. Math. USSR-Sb. 66, 541-553.

A. Yu. Olshanskii1979. Infinite groups with cyclic subgroups. Soviet Math. Dokl. 20, 343-346.1991. Geometry of defining relations in groups. Kluwer Academic Publishers, Dordrecht.

0. Ore1937. Structures and group theory I. Duke Math. J. 3, 149-173.1938. Structures and group theory II. Duke Math. J. 4, 247-269.

U. Ostendorf1991. Projektivitatstypen torsionsfreier abelscher Gruppen vom Rang 1. Rend. Sem.

Mat. Univ. Padova 86, 183-191.P.P. Palfy

1986. On partial ordering of chief factors in solvable groups. Manuscripta Math. 55,219-232.

1988. On Feit's examples of intervals in subgroup lattices. J. Algebra 116, 471-479.P.P. Pylfy and P. Pudlyk

1980. Congruence lattices of finite algebras and intervals in subgroup lattices of finitegroups. Algebra Universalis 11, 22-27.(See also E. Lukycs and P.P. Pylfy)

K.J. Paulsen1975. Uber Untergruppenverbande von Gruppen mit Hallschen Normalteilern. Disser-

tation Kiel.G. Pazderski

1972. Uber Gruppen mit isomorphen Subnormalteilerverbanden. Publ. Math. Debrecen19, 1-19.

1987. On groups for which the lattice of normal subgroups is distributive. BeitrdgeAlgebra Geom. 24, 185-200.

552 Bibliography

A.S. Pekelis1961a. Lattice isomorphisms of solvable groups. Soviet Math. Dokl. 1, 845-847.1961 b. Lattice isomorphisms of locally nilpotent nonperiodic groups (Russian). Ural. Gos.

Univ. Mat. Zap. 3, 72-76.1963. Lattice isomorphisms of locally nilpotent torsion-free groups (Russian). Uspehi

Mat. Nauk 18, 187-190.1965. Lattice isomorphisms of mixed nilpotent groups (Russian). Sibirsk. Mat. Z. 6,

1315-1321.1967. Structural isomorphisms of mixed metabelian groups. Siberian Math. J. 8, 625-

630.1972a. The lattice of subgroups of a nilpotent nonperiodic group. Math. Notes 11, 240-

243.1972b. Lattice isomorphisms of radical groups. Soviet Math. Dokl. 13, 649-653.1976. Lattice characterization of radicals in a group (Russian). Latviisk. Mat. Ezegodnik

Vyp. 18, 128-142.AS. Pekelis and L.E. Sadovskii

1963. Projections of a metabelian torsion-free group. Soviet Math. Dokl. 4, 918-920.R.E. Phillips

(See K.K. Hickin and R.E. Phillips)P. Plaumann, K. Strambach and G. Zacher

1985. Gruppen mit geometrischen Abschnitten im Untergruppenverband. MonatshefteMath. 99, 117-145.

B.I. Plotkin1954. Lattice isomorphisms of soluble R-groups (Russian). Doklady Akad. Nauk SSSR

(N.S.) 95, 1141-1144.1957. Radical and semi-simple groups (Russian). Trudy Moskov. Mat. Obsc. 6, 299-

336.(See also P.G. Kontorovic and B.I. Plotkin)

J. Poland1985. On verifying lattice isomorphisms between groups. Archiv Math. 44, 309-3 10.

(See also K. Mahdavi and J. Poland)S. Pontier

1966a. Sur les groupes (demi-groupes) dont le treillis des sous-groupes (sous-demi-groupes) est geometrique. C.R. Acad. Sci. Paris Ser. A-B 262, A1435-A1437.

1966b. Demi-groupes dont le treillis des sous-demi-groupes est geometrique. Groupesdont le treillis des sous-groupes est geometrique. Proprietes annexes. Publ. Dep.Math. (Lyon) 3, fasc. 4, 1-66.

E. Previato1975. Una caratterizzazione dei sottogruppi di Dedekind di un gruppo finito. Atti

Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 59, 643-650.1976. Sui sottogruppi di Dedekind nei gruppi infiniti. Atti Accad. Naz. Lincei Rend. Cl.

Sci. Fis. Mat. Natur. (8) 60, 388-394.1978. Una caratterizzazione reticolare dei gruppi risolubili finitamente generati. Istit.

Veneto Sci. Lett. Arti Atti Cl. Sci. Mat. Natur. 136, 7-11.1982. Some families of simple groups whose lattices are complemented. Boll. Un. Mat.

Ital. B (6) 1, 1003-1014.P. Pudlak and J. Tuma

1980. Every finite lattice can be embedded in a finite partition lattice. Algebra Univer-salis 10, 74-95.(See also P.P. Palfy and P. Pudlak)

Bibliography 553

S. Rao(See M. Curzio and S. Rao)

M. Reuther1975. Endliche Gruppen, in denen das Erzeugnis zweier Zentralisatoren wieder ein

Zentralisator ist. Dissertation Kiel.1977. Endliche Gruppen, in denen alle das Zentrum enthaltenden Untergruppen

Zentralisatoren sind. Archiv Math. 29, 45-54.D.J.S. Robinson

1964. Groups in which normality is a transitive relation. Proc. Camb. Phil. Soc. 60,21-38.

1965. Joins of subnormal subgroups. Illinois J. Math. 9, 144-168.1968. A note on finite groups in which normality is transitive. Proc. Amer. Math. Soc. 19,

933-937.1972a. Finiteness conditions and generalized soluble groups, Vol. 1. Springer-Verlag,

Berlin.1972b. Finiteness conditions and generalized soluble groups, Vol. 2. Springer-Verlag,

Berlin.1982. A course in the theory of groups. GTM 80, Springer-Verlag, New York Heidelberg

Berlin.D. M. Rocke

1975. p-groups with abelian centralizers. Proc. London Math. Soc. (3) 30, 55-75.J.S. Rose

1965. Abnormal depth and hypereccentric length in finite soluble groups. Math. Z. 90,29-40.

A. Rottlander1928. Nachweis der Existenz nichtisomorpher Gruppen von gleicher Situation der Un-

tergruppen. Math. Z. 28, 641-653.R. Rudolph

1982. Ein Untergruppensatz fur modulare Gruppen. Monatsh. Math. 94, 149-153.L.E. Sadovskii

1941. Uber die Strukturenisomorphismen von Freigruppen. Doklady 32, 171-174.1944. Structural isomorphisms of free groups and of free products (Russian). Mat.

Sbornik 14 (56), 155-173.1947. On structural isomorphisms of free products of groups (Russian). Mat. Sbornik 21

(63), 63-82.1957a. The lattice of subgroups of a nilpotent torsion-free group (Russian). Uspehi Mat.

Nauk (N.S.) 12, 201-204.1957b. Structure isomorphisms of a free metabelian group (Russian). Mat. Sbornik 42 (84),

445-460.1964. On the definability by lattices of a locally nilpotent torsion-free group. Soviet

Math. Dokl. 5, 294-297.1965a. An approximation theorem and lattice isomorphisms. Soviet Math. Dokl. 6, 424-

427.1965b. Projectivities and isomorphisms of nilpotent groups (Russian). Izv. Akad. Nauk

SSSR Ser. Mat. 29, 171-208.1968. Some lattice-theoretical problems in the theory of groups. Russian Math. Surveys

23, no. 3, 125-156.(See also M.N. Arshinov and L.E. Sadovskii; A.S. Pekelis and L.E. Sadovskii)

R. Sandling(See W. Kimmerle, R. Lyons, R. Sandling, and D.N. Teague)

554 Bibliography

S. Sato1949. On groups and the lattices of subgroups. Osaka Math. J. 1, 135-149.1951. Note on lattice-isomorphisms between Abelian groups and non-Abelian groups.

Osaka Math. J. 3, 215-220.1952. On the lattice homomorphisms of infinite groups I. Osaka Math. J. 4, 229-234.

M. Schenke1986. On weakly Dedekind subgroups. Ricerche Mat. 35, 113-123.1987a. Analoga des Fundamentalsatzes der projektiven Geometrie in der Gruppen-

theorie I. Rend. Sem. Mat. Univ. Padova 77, 255-303.1987b. Analoga des Fundamentalsatzes der projektiven Geometrie in der Gruppen-

theorie II. Rend. Sem. Mat. Univ. Padova 78, 175-225.E. Schenkman

1960. On the norm of a group. Illinois J. Math. 4, 150-152.P. Schmid

(See R. Maier and P. Schmid)R. Schmidt

1968. Eine verbandstheoretische Charakterisierung der auflosbaren and der uberaufli s-baren endlichen Gruppen. Archiv Math. 19, 449-452.

1969. Modulare Untergruppen endlicher Gruppen. Illinois J. Math. 13, 358-377.1970a. Modular subgroups of finite groups II. Illinois J. Math. 14, 344-362.1970b. Endliche Gruppen mit vielen modularen Untergruppen. Abh. Math. Sem. Univ.

Hamburg 34, 115-125.1970c. Zentralisatorverbande endlicher Gruppen. Rend. Sem. Math. Univ. Padova 44,

97-131.1970d. Charakterisierung der einfachen Gruppe der Ordnung 175 560 durch ihren Zen-

tralisatorverband. Math. Z. 116, 299-306.1972a. A characterization of PSL(2, p") by its centralizer lattice. J. Algebra 21, 280-29 1.1972b. Verbandsisomorphismen endlicher autlosbarer Gruppen. Archiv Math. 23, 449-

458.1974. Lattice isomorphisms and saturated formations of finite soluble groups. Proceed-

ings of the Second International Conference on the Theory of Groups (AustralianNat. Univ., Canberra, 1973), 605-610. Lecture Notes in Math., Vol. 372, Springer,Berlin.

1975a. Normal subgroups and lattice isomorphisms of finite groups. Proc. London Math.Soc. (3) 30, 287-300.

1975b. Untergruppenverbande zweifach transitiver Permutationsgruppen. Math. Z. 144,161-168.

1977a. Verbandsautomorphismen der alternierenden Gruppen. Math. Z. 154, 71-78.1977b. Untergruppenverbande endlicher einfacher Gruppen. Geometriae Dedicata 6, 275-

290.1978. Untergruppenverbande direkter Produkte von Gruppen. Archiv Math. 30, 229-235.1980a. Untergruppenverbande involutorisch erzeugter Gruppen. Rend. Sem. Mat. Univ.

Padova 63, 95-126.1980b. Projektivitaten direkter Produkte endlicher Gruppen. Archiv Math. 35, 79-84.1981. Der Untergruppenverband des direkten Produktes zweier isomorpher Gruppen.

J. Algebra 73, 264-272.1982. Untergruppenverbande endlicher Gruppen mit elementarabelschen Hallschen

Normalteilern. J. Reine Angew. Math. 334, 116-140.1983. Semiaffinitaten von Gruppen. Math. Z. 182, 291-302.1984a. Affinities of groups. Rend. Sem. Mat. Univ. Padova 72, 163-190.

Bibliography 555

1984b. Verbandstheoretische Charakterisierungen der Endlichkeit des Indexes einer Un-tergruppe in einer Gruppe. Archiv Math. 42, 492-495.

1986. Gruppen mit modularem Untergruppenverband. Archiv Math. 46, 118-124.1987. Untergruppenverbande endlicher aufl6sbarer Gruppen. Group Theory (Bressanone

1986), 130-150, Lecture Notes in Math. 1281, Springer-Verlag, Berlin New York.(See also W. Gaschiitz and R. Schmidt)

C.M. Scoppola1981. Sul reticolo dei sottogruppi di un gruppo abeliano senza torsione di rango diverso

da 1: una caratterizzazione reticolare. Rend. Sem. Mat. Univ. Padova 65, 205-221.1985. Una caratterizzazione reticolare del reticolo dei sottogruppi di un gruppo abe-

liano contenente due elementi aperiodici indipendenti. Rend. Sem. Mat. Univ.Padova 73, 191-207.

W.R. Scott1957. Half-homomorphisms of groups. Proc. Amer. Math. Soc. 8, 1141-1144.1964. Group theory. Prentice-Hall, Englewood Cliffs.

(See also R.H. Bradway, F. Gross and W.R. Scott)G.M. Seitz and C.R.B. Wright

1969. On finite groups whose Sylow subgroups are modular or quaternion-free. J. Alge-bra 13, 374-381.

M.I. Sergeev1964. Completely FN-factorizable groups. Soviet Math. Dokl. 5,461-464.

H.L. Silcock1977. Generalized wreath products and the lattice of normal subgroups of a group.

Algebra Universalis 7, 361-372.A.L. Smelkin

1964. Free polynilpotent groups. Soviet Math. Dokl. 4, 950-953.A.E. Spencer

1971. Self dual finite groups. 1st. Veneto Sci. Lett. Arti Atti Cl. Sci. Mat. Natur. 130,385-391.

R.F. Spring1963. Lattice isomorphisms of finite non-abelian groups of exponent p. Proc. Amer.

Math. Soc. 14, 407-413.S.E. Stonehewer

1972. Permutable subgroups of infinite groups. Math. Z. 125, 1-16.1973. Permutable subgroups of some finite p-groups. J. Austral. Math. Soc. 16, 90-

97.1974. Permutable subgroups of some finite permutation groups. Proc. London Math.

Soc. (3) 28, 222-236.1976a. Modular subgroups of infinite groups. Symposia Mathematica, Vol. X VII (Con-

vegno sui Gruppi Infiniti, INDAM, Roma, 1973), 207-214. Academic Press, Lon-don.

1976b. Modular subgroup structure in infinite groups. Proc. London Math. Soc. (3) 32,63-100.

S.E. Stonehewer and G. Zacher1988. On groups with an interval isomorphic to a projective geometry. J. Pure Appl.

Algebra 53, 187-196.1990a. Lattice homomorphisms of groups and dual standard subgroups. Rend. Circ. Mat.

Palermo (2) Suppl. 23, 289-297.1990b. Dual-standard subgroups in nonperiodic locally soluble groups. Atti. Accad. Naz.

Lincei Rend. Cl. Sci. Fis. Mat. Natur. (9) 1, 101-104.

556 Bibliography

1991a. Dual-standard subgroups of finite and locally finite groups. Manuscripta Math.70, 115-132.

1991b. Lattice homomorphisms of nonperiodic groups. J. Algebra 138, 48-90.1994. Dualities of groups. To appear.

(See also G. Busetto and S.E. Stonehewer; J.C. Lennox and S.E. Stonehewer)K. Strambach

(See P. Plaumann, K. Strambach and G. Zacher)M. Suzuki

1951a. On the lattice of subgroups of finite groups. Trans. Amer. Math. Soc. 70, 345-371.1951b. On the L-homomorphisms of finite groups. Trans. Amer. Math. Soc. 70, 372-386.1956. Structure of a group and the structure of its lattice of subgroups. Springer-Verlag,

Berlin Heidelberg New York.1961. On a finite group with a partition. Archiv Math. 12, 241-254.1982. Group theory I. Springer-Verlag, Berlin Heidelberg New York.1986. Group theory II. Springer-Verlag, Berlin Heidelberg New York.

Ja.P. Sysak1977. Finite elementarily factorizable groups. Ukrainian Math. J. 29, 51-58.

J. Szep(See N. Ito and J. Szep)

0. Tamaschke1961. Gruppen mit reduziblem Subnormalteilerverband. Math. Z. 75, 211-214.

M.C. Tamburini Bellani(See M. Bianchi and M.C. Tamburini Bellani; L.Di Martino and M.C. TamburiniBellani)

D.N. Teague

(See W. Kimmerle, R. Lyons, R. Sandling, and D.N. Teague)J.G. Thompson

1967. An example of core-free quasinormal subgroups of p-groups. Math. Z. 96, 226-227.(See also D.R. Hughes and J.G. Thompson)

J. Tits1974. Buildings of spherical type and finite BN-pairs. Lecture Notes in Math. 386,

Springer-Verlag, Berlin Heidelberg New York.M.V. Tsybanev

1973. On abelian normal subgroups all of whose subgroups are m-complemented. Alge-bra and Logic 13, 41-45.

1980. Solvable F-factorizable groups. Siberian Math. J. 21, 298-302.F. Tuccillo

(See A. Franchetta and F. Tuccillo)J. Tuma

1989. Intervals in subgroup lattices of infinite groups. J. Algebra 125, 367-399.(See also P. Pudlak and J. Tuma)

T. Uzawa1986. Finite Coxeter groups and their subgroup lattices. J. Algebra 101, 82-94.

A.V. Vasileva1976. A characterization of the group PSL3(q) by its centralizer lattice. Algebra and

Logic 15, 320-335.1977. Centralizer lattices of finite simple groups. Siberian Math. J. 18, 188-196.

Ph. Vassiliou1967. Eine Bemerkung fiber die von dem Zentralisator bestimmte Abbildung. Prakt.

Akad. Athenon 42, 286-289.

Bibliography 557

P. Venzke1972. Finite groups with many maximal sensitive subgroups. J. Algebra 22, 297-308.1975. Finite groups whose maximal subgroups are modular. Atti Accad. Naz. Lincei

Rend. Cl. Sci. Fis. Mat. Natur. (8) 58, 828-832.H. Volklein

1986a. On the lattice automorphisms of the finite Chevalley groups. Indag. Math. 48,213-228.

1986b. On the lattice automorphisms of SL(n,q) and PSL(n,q). Rend. Sem. Mat. Univ.Padova 76, 207-217.

R.W.v.d. Waall1973. On modular p-groups. J. Algebra 25, 125-127.1975. On modular p-groups II. Indag. Math. 37, 79-82.

G.E. Wall1965. On Hughes' HD problem. Proc. Internat. Conf. Theory of Groups (Canberra 1965),

357-362. Gordon and Breach, New York.(See also D.W. Barnes and G.E. Wall)

P.M. Whitman1946. Lattices, equivalence relations, and subgroups. Bull. Amer. Math. Soc. 52, 507-

522.G.M. Whitson

1977. A lattice theoretic generalization of normal subgroups. J. Algebra 49, 387-410.1978. Finite groups whose subgroup, composition subgroup, or normal subgroup lattice

is an ortholattice. Algebra Universalis 8, 123-127.H. Wielandt

1939. Eine Verallgemeinerung der invarianten Untergruppen. Math. Z. 45, 209-244.C.R.B. Wright

(See G.M. Seitz and C.R.B. Wright)B.V. Yakovlev

1965. On the lattice isomorphisms of periodic nilpotent groups (Russian). Ural. Gos.Univ. Mat. Zap. 5, 106-116.

1970. Lattice isomorphisms of solvable groups. Algebra and Logic 9, 210-222.1974. Conditions under which a lattice is isomorphic to the lattice of subgroups of a

group. Algebra and Logic 13, 400-412.1975. An example of solvable torsion-free groups which are lattice isomorphic but not

isomorphic. Algebra and Logic 14, 282-301.1986. Lattice definability of matrix and certain other groups. Algebra and Logic 25,

696-744.1988. Strict lattice determination of mixed nilpotent groups. Algebra and Logic 27, 376-

384.P. Yff

1967. Groups with identical subgroup structures. Math. Z. 99, 178-181.G. Zacher

1952. Determinazione dei gruppi d'ordine finito relativamente complementati. Rend.Accad. Sci. Fis. Mat. Napoli (4) 19, 200-206.

1953. Caratterizzazione dei gruppi risolubili d'ordine finito complementati. Rend. Sem.Mat. Univ. Padova 22, 113-122.

1955a. Sugli elementi modulari in un p-gruppo. Rend. Sem. Mat. Univ. Padova 24, 165-182.

1955b. Un criterio di non semplicita di un gruppo finito. Rend. Sem. Mat. Univ. Padova24, 215-219.

558 Bibliography

1956. Sugli elementi modulari di un gruppo finito. Rend. Sem. Mat. Univ. Padova 26,70-84.

1957. Sui gruppi finiti per cui it reticolo dei sottogruppi di composizione a distributivo.Rend. Sem. Mat. Univ. Padova 27, 75-79.

1960. On lattice dual-homomorphisms between finite groups. Rend. Sem. Mat. Univ.Padova 30, 65-75.

1961 a. I gruppi risolubili con duale. Rend. Sem. Mat. Univ. Padova 31, 104-113.1961b. Caratterizzazione dei gruppi immagini omomorfe duali di un gruppo finito. Rend.

Sem. Mat. Univ. Padova 31, 412-422.1962. Sui gruppi risolubili finiti col reticolo dei sottogruppi di composizione dotato di

duale. Rend. Sem. Mat. Univ. Padova 32, 325-327.1965. Sul reticolo dei sottogruppi di un gruppo di permutazione. Cedam, Padova.1966. Sui gruppi localmente finiti con duale. Rend. Sem. Mat. Univ. Padova 36, 223-

242.1971. Determination of locally finite groups with duals. J. Algebra 18, 426-

431.1977. Sul reticolo dei sottogruppi del gruppo simmetrico. Rend. Istit. Mat. Univ. Trieste

9,122-126.1978. La classe dei gruppi iperciclici a proiettivamente invariante. Rend. Sem. Mat. Univ.

Padova 59, 263-268.1980. Una caratterizzazione reticolare della finitezza dell' indice di un sottogruppo in

un gruppo. Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur (8) 69, 317-323.

1981. Sul reticolo dei sottogruppi del quadrato cartesiano di un gruppo semplice. Rend.Sem. Mat. Univ. Padova 64, 235-241.

1982a. Sulle immagini dei sottogruppi normali nelle proiettivita. Rend. Sem. Mat. Univ.Padova 67, 39-74.

1982b. Una relazione di normality sul reticolo dei sottogruppi di un gruppo. Ann. Mat.Pura Appl. (4) 131, 57-73.

1984. Union-omomorfismi tra gruppi. Rend. Accad. Naz. Sci. XL Mem. Mat. (5) 8, 247-260.

1985. Sottogruppi normali e r-omomorfismi completi tra gruppi. Ann. Mat. Pura Appl.(4) 139, 83-106.

1988. Conoscenze attuali relative ad alcuni aspetti del reticolo dei sottogruppi di ungruppo. Rend. Circ. Mat. Palermo (2) Suppl. 19, 39-60.(See also M. Emaldi and G. Zacher; F. Napolitani and G. Zacher; P. Plaumann,K. Strambach and G. Zacher; S.E. Stonehewer and G. Zacher)

G. Zappa

1949. Sulla condizione perche un omomorfismo ordinario sia anche un omomorfismostrutturale. Giorn. Mat. Battaglini (4) 78, 182-192.

1951a. Sulla condizione perche un emitropismo inferiore tipico tra due gruppi sia unomotropismo. Giorn. Mat. Battaglini (4) 80, 80-101.

1951b. Sulla risolubilita dei gruppi finiti in isomorfismo reticolare con un gruppo risolu-bile. Giorn. Mat. Battaglini (4) 80, 213-225.

1951c. Determinazione degli elementi neutri net reticolo dei sottogruppi di un gruppofinito. Rend. Accad. Sci. Fis. Mat. Napoli (4) 18, 22-28.

1956a. Sui gruppi finiti risolubili per cui it reticolo dei sottogruppi di composizione edistributivo. Boll. Un. Mat. Ital. (3) 11, 150-157.

1956b. Sui gruppi finiti per cui it reticolo dei sottogruppi di composizione a modulare.Boll. Un. Mat. Ital. (3) 11, 315-318.

Bibliography 559

H. Zassenhaus1958. The theory of groups. Vandenhoeck and Ruprecht, Gottingen.

1. Zimmermann1989. Submodular subgroups in finite groups. Math. Z. 202, 545-557.

A. Zitarosa1952. Sugli elementi neutri del reticolo dei sottogruppi normali di un gruppo speciale

finito. Ricerche Mat. 1, 249-254.

Index of Names

Abramovskii, IN. 129 Feit, W. 10

d'Andrea, A.B. 530, 541Antonov, V.A. 518, 523, 524, 526, 527

Foguel, T. 484Fort, A. 87, 232, 234

Arshinov, M.N. 344, 359, 377, 384, 417 Franchetta, A. 496, 510F i i 125 133 135S 319 320 321ranc os , , ,. , , , ,

Baer, R. 13, 14, 25, 34, 41, 50, 51, 89, 93, 496, 49794, 101, 103, 104, 106, 115, 121, 146, 148, Fuchs, L. 110152, 156, 414, 416, 452, 453, 454, 457, 458

Barnes, D.W. 272, 274, 275, 279, 344, 382, Gaschiitz, W. 23, 397, 486, 524, 526, 538,383 385 541,

Barysovec, P.P. 125 Gasparini, E. 108, 113Baumslag, G. 349 Gorcakov, Yu.M. 125B ll H 125 126 134h E 34Gec te , . , , oursat, .

Berger, T.R. 220 Gross, F. 220, 222, 227, 304, 310, 311B k i 232 234J Gov ,er c, a. .

Bi hi 131M H ll P 114 122 126anc , .

B d R H 220, . , ,a

H k P 401way,ra . . auc , .

Brandl, R. 40, 492, 496, 498 Heineken, H. 277, 401, 485, 494, 498, 506,Bruno, B. 16,533 511

Busetto, G. 65, 200, 225, 227, 253, 254, Hickin, K.K. 319264, 270, 271, 282, 303, 304, 305, 310, 311, Higman, D.G. 142316, 318, 319, 320, 325, 33 0, 331, 332, 338, Holmes, C.S. 104, 200, 276, 277, 359, 430,

342340 341 438, ,

277H d K 200 276on a, . , ,

Calcaterra, R.A. 476 Hughes, D.R. 147Caranti, A. 277, 344, 3 92, 393, 394, 395 Huppert, B. 34, 187Cardin, M. 310, 311Cernikov, S.N. 125 Ito, N. 222, 233, 243Cernikova, N.V. 123, 124, 126 Ivanov, S.G. 136, 144Cheng, Y. 526 Iwasawa, K. 41, 55. 65, 71, 75, 79, 86, 199,Christensen, C. 121 228, 231C C D H 1 29 30 34ooper, . . . , , ,

C t ti i M 449 A W 65 87Jos an n , .

Curzio, M. 10, 121, 134, 137, 486, 488,ones, . . ,

Jiirgensen, R. 524, 527492 493 495 497 504503 511 529 530, , ,

540

, ,, , , ,

Kegel, O.H. 148, 510

Daues, G. 277Khukhro, E.I. 148Kimmerle, W. 439

De Giovanni, F. 125, 133, 135, 319, 320, Kochendorfer, R. 125321, 496, 497 Kontorovic, P.G. 343, 361, 385

Deskins, W.E.Di Martino, L.Dinerstein, N.T.

222131, 134

121

Kurosh, A.G. 43Kurzweil, H. 401

L J C 304Emaldi, M. 118. 120, 121, 123, 126, 129,

ennox, . .

Li, S. 440130, 134 Loiko, N.V. 529, 531, 533, 538, 540, 541

562 Index of Names

Longobardi, P. 87, 489, 490, 496,Lukacs, E. 420

497, 498 Schenkman, E. 30Schmid, P. 218

Lyons, R. 439 Schmidt, R. 23, 41, 82, 157, 175, 180, 184,199, 200, 202, 209, 215, 216, 219, 229, 230,

Mahdavi, K. 112 232, 236, 239, 241, 243, 277, 278, 287, 288,Maier, R. 218 344, 345, 396, 397, 398, 400, 401, 405, 406,Mainardis, M. 77 , 96, 97, 98, 100 411, 414, 415, 416, 419, 421, 425, 428, 434,Maj, M. 87, 489, 490, 496, 497, 498 435, 436, 438, 439, 447, 449, 450, 485, 486,Martino, E. 143 519, 524, 527, 537, 538, 540, 541Menegazzo, F. 65, 114, 129, 130, 131, 132,

134, 157, 187, 195, 196, 197, 200, 250, 253,254, 271, 282, 341, 344, 417, 421

Metelli, C. 108, 113, 441, 449Mozarovskaja, L.S. 126

Nakamura, K. 222, 227Napolitani, F. 65, 68, 87, 120, 137, 139,

143, 145, 200, 226, 254, 264, 270, 282, 303,305, 310, 311, 315, 330, 331, 332, 334, 338,340, 341, 342, 503

Neumann, B.H. 283Niegel, W. 277

Obraztsov, V.N. 459Olshanskii, A.Yu. 82, 311Ore, O. 1, 12, 16, 43, 222Ostendorf, U. 111

Scoppola, C.M. 358Scott, W.R. 220, 374Seitz, G.M. 67, 187, 197Sergeev, M.I. 126Silcock, H.L. 490Smelkin, A.L. 374, 418Spring, R.F. 273Stonehewer, S.E. 43, 144, 145, 209, 220,

253, 254, 293, 297, 302, 311, 312, 316, 328,329, 330, 341, 401, 459, 484

Strambach, K. 401Suzuki, M. 1, 25, 37, 39, 41, 52, 65, 115,

142, 144, 146, 152, 157, 172, 181, 185,200, 228, 229, 230, 2.33, 239, 391, 394,406, 409, 410, 444, 449, 452, 458, 459,461, 462

Sysak, Ja.P. 125Szep, J. 222, 243

Palfy, P.P. 10, 420, 490Paulsen, K.-J. 160, 164, 176, 186, 187, 197,

198Pazderski, G. 485, 489, 504, 506, 511Pekelis, A.S. 344, 363, 376, 377, 401, 405Phillips, R.E. 319Plaumann, P. 401Plotkin, B.I. 343, 361, 385, 401Poland, J. 19, 112Previato, E. 116, 120, 204, 209, 215,

317Pudlak, P. 9

Rao, S. 134Reuther, M. 526, 527Rips, E. 281, 286, 325Robinson, D.J.S. 33Rocke, D.M. 521Rose, J.S. 40Rottlander, A. 200, 275, 276Rudolph, R. 83

Sadovskii, L.E. 1, 21, 22, 343, 344, 359,360, 363, 373, 374, 417

Sandling, R. 439Sato, S. 41, 65, 96, 99, 142, 233Schenke, M. 48, 344, 415, 416, 421

Tamaschke, O. 501, 504Tamburini Bellani, M.C. 131, 134Teague, D.N. 439Thompson, J.G. 147, 227Tits, J. 440Tsybanev, M.V. 126Tuccillo, F. 496Tuma, J. 9, 10

Uzawa, T. 437

Vasileva, A.V. 524, 527Venzke, P. 133, 234, 235Volklein, H. 449

v.d.Waall, R.W. 67Wall, G.E. 148, 272, 274, 275, 279, 343,

344, 360, 382, 383, 385Whitman, P.M. 9Whitson, G.M. 216Wielandt, H. 6, 499Wright, C.R.B. 67, 187, 197

Yakovlev, B.V. 282, 332, 341, 343, 344,347, 356, 357, 358, 360, 377, 387, 395,436

Yff, P. 200,276,277

Index of Names 563

Zacher, G. 114, 118, 119, 128, 129, 144, 444,447,451,452,458,459,462,469,477,145, 200, 226, 243, 281, 282, 286, 287, 289, 478, 484, 502, 504, 509, 510, 511293,294,297,300,301,302,310,312,313, Zappa, G. 115, 141, 142, 143, 144, 157,316,318,319,320, 321, 326, 328, 329, 330, 181,502334, 341, 401, 404, 408, 410, 412, 420, 421, Zitarosa, A. 498

Index of Subjects

Aabelian group(s) 43, 86, 94-113, 491-494,

536character group of 453elementary 25, 49-52, 131of torsion-free rank 1 107-113,493,533of torsion-free rank > 2 106, 358, 492,

533periodic 94, 96, 103-105, 397, 492power automorphisms of 32projectivities between 103, 104,

110-113projectivities of 94, 96, 99, 106R-isomorphisms of 533with duals 452-458

abelian Sylow subgroup 185-187, 197abelian variety 382absorption identities 2admissible subgroup of a partition 147affinity 534-541algebraic lattice 9

almost free group of a variety 382alternating group 10, 26, 115-116, 432,

441-449amorphy of a group 539-541amorphy of a map 17,534,538-541antiatom 3

antichain 3

antiisomorphism 90, 363, 374, 532ascendant subgroup 293, 328ascending chain condition 4, 46associativity identities 17, 534, 539atom 3

automorphismof group 24of lattice 4power 24, 29-34

autoprojectivity 23-29

BBaer map 416basic system 352bound

greatest lower 2least upper 2

lower/upper 2

Ccentre 273, 536

of a finite perfect group 229of a p-group 243of a Sylow subgroup 195-197of a torsion-free group 360ZJG) 183tcen ra zerli

duality 512lattice 8, 511-527maximal 514minimal 514partition 519preserving projectivity 272-273

C-group 121-126chain 3, 16

modular 360, 375permodular 385-391supersoluble 319

chain condition 4ascending 4, 46descending 4Jordan-Dedekind 47, 231--234, 540

character group 453characterization

lattice-theoretic 14

of subgroup lattices 357classical group 436, 439class of groups invariant under

projectivities 14classical groups 436. 439class with lattice-theoretic

characterization 14finite nilpotent-by-abelian groups 399finite soluble groups of Fitting length

k 180, 399finite soluble groups of rank r 396formation 398groups of Lie type 440hypoabelian groups 315locally 2"-groups 15

P-groups 51radical groups 405soluble groups 332Sylow tower groups 396

class of groups with lattice-theoretic

Index of Subjects 565

characterization 14 core 199cyclic groups 13 coset lattice 8, 527-541finite groups 15 cover

finitely generated nonperiodic nilpotent elements of a lattice 3

groups 375 irreducibly a group 283finitely generated soluble groups 317finite soluble groups 229, 230finite soluble groups with chief factors of

dimension at most k 405free groups 356hyperabelian groups 316hypercyclic groups 319locally cyclic groups 12locally finite groups 15-16nilpotent p-group or P-group 391nonperiodic locally nilpotent

groups 376perfect groups 229, 313P-groups 70, 394polycyclic groups 317residually finite groups 287residually finitely generated soluble

groups 321residually polycyclic groups 320residually supersoluble groups 319simple groups 228, 312SN-groups 319supersoluble groups 230, 231, 318torsion-free nilpotent groups 361

commutators 29set of 516

commutator subgroup 29, 272second 174, 326, 476

compact element 9comparable elements 3

compatible projectivity 380complement 115

permutable 121relative 127sectional (or S-) 122

complemented lattice 115of cosets 530of normal subgroups 490of subgroups 115-135of composition subgroups 503

completelattice 2

meet-sublattice 7

complex 346component of a partition 145composition series 6composition subgroup 6,498-511conjugacy class 278-280, 425conjugacy preserving projectivity 272coprime groups 36

Coxeter group 437crossed isomorphism 90-96cyclic

element 345group 12-16,28

L, (G) 18

modular chain 375permutable subgroup 223-225, 307-

311, 318

Ddecomposable subgroup lattice 36-39Dedekind element 43Dedekind subgroup 289

defect of a subnormal subgroup 125, 499D-embedded subgroup 289derived length 332descending chain condition 4determined by subgroup lattice vii, 343

alternating group 432classical group 436, 439dihedral group 422, 437direct product of two isomorphic

groups 412,429finite Coxeter group 437finite simple group 432, 434, 439infinite cyclic group 13

nilpotent group of class 2 and exponentp 384

projective general/special lineargroup 156, 432, 434, 436

quaternion group 28strongly 343Suzuki group 156, 434symmetric group 432, 438

diagonal 35, 408-411dihedral group 27, 422, 426, 437

generalized 149directly decomposable

centralizer lattice 515-518, 527coset lattice 529lattice 8lattice of composition subgroups 501lattice of normal subgroups 488subgroup lattice 36-39

direct product ofgroups 34-40isomorphic groups 408-416, 426, 429lattice-isomorphic groups 418-420lattices 8

566 Index of Subjects

distributivecentralizer lattice 515coset lattice 529element 136lattice 11

lattice of compositionsubgroups 502-503, 510

lattice of normal subgroups 489, 497-498

pair of elements 16, 40subgroup lattice 12-13

double centralizer property 476, 484doubly transitive permutation group 434,

435, 438duality 453

abelian groups with dual 454, 457between lattices of composition

subgroups 507-511between lattices of normal

subgroups 496centralizer 512groups with dual 452-484; 458, 459

dually modular element 87

Eelementary abelian group 25, 49-52, 131element map 17,20,101-107

Baer map 416derived from a projectivity 101-103inducing a projectivity 17, 20cp-mapping 361

element of a groupheight of 107inducing the power k in a subgroup 257left regular 536order modulo a subgroup 206right regular 536type of 108

element(s) of a latticecompact 9comparable 3

cyclic 345Dedekind 43dually modular 87greatest 3

isolated 360join-distributive 136-137, 141-142join-quasi-distributive 136-137least 3

maximal 3

meet-distributive 136, 138, 143-145meet-quasi-distributive 136, 138,

143-145minimal 3modular 43-46, 144

modularly embedded 360neutral 135,142-145,498permodularly embedded 385polymodular 234semimodular 49, 216standard 136-137, 141uniquely complemented 136, 138-141

embeddinghypercentral 217hypercyclic 217modular 360of lattices in subgroup lattices 9permodular 385

exponent of a group 178extended Tarski group 82, 88

Ffinite group 15Fitting

length 180, 399series 118, 180subgroup 116,178-181

formation of finite solublegroups 397-401, 405-406

.F-projector 397, 401Frattini subgroup 38, 115free group 345-359, 374, 533

of a variety 3822-free group 348-359

free polynilpotent group 374, 533free product of groups 359.-residual 397, 400Frobenius group 145, 519

GG-isomorphism 486greatest element 3

greatest lower bound 2group(s)

abelian 43, 86, 94-113, 397, 452-458,491-494, 533, 536

alternating 10, 26, 115-116, 432,441-449

amorphy of 539-541C- 121-126classical 436, 439coprime 36covered irreducibly by cosets (subgroups)

283Coxeter 437cyclic 12-16, 28determined by its subgroup lattice vii,

343dihedral 27, 422, 426, 437elementary abelian 25, 49-52, 131


finite 15finitely generated modulo subgroup

292free 345-359, 374, 533free polynilpotent 374, 533Frobenius 145, 519generalized dihedral 149generated by involutions 421-451, 537hamiltonian 55, 60, 68, 89hyperabelian 315, 316hypercyclic 123, 315, 319hypoabelian 315IM- 130-135K- 115-124lattice-isomorphic 4lattice-simple 449locally cyclic 12locally finite 15-16locally free 359, 533locally nilpotent 373, 376, 533locally X- 15M- 54-100;55,71,75,79,82902- 518-523M*- 55, 58-69metacyclic 225nilpotent 47, 54, 360-395, 494-498,

533nilpotent-by-abelian 397of all autoprojectivities 23-29of all power automorphisms 24, 29-34of Lie type 440,449i2- 394p- 52-69, 79, 94, 220-225, 243-274,

279, 385-395, 460, 504P- 49-54P*- 69, 89P-decomposable 170perfect 186, 219, 229, 243, 313, 330, 540permutation 115, 430-449polycyclic 317projective general/special linear 116,

120, 146, 152, 156, 177, 432-436, 519,523-524

Prefer 16p-soluble 183, 196quaternion 28radical 402-405RC- 129-130residually finite 287residually finitely generated soluble 321residually polycyclic 320residually supersoluble 319RK- 127-135Rottlander 162, 276, 413, 430SC- 122-124

simple 116, 156, 228, 312, 432, 434,439-441,449,523

SK- 126SN- 319soluble 118-121, 130-132, 181, 229,

315-321, 332, 395-406, 462-468,506-511

strongly determined by its subgrouplattice 343

supersoluble 122, 230-234, 318Suzuki 116, 146, 152, 156, 434Sylow tower 396, 405symmetric 115, 149, 177, 432, 438,

441-445, 447, 451, 513, 527T-, T*- 127Tarski 82, 882-free 348-359with dual 452-484; 458, 459with identical subgroup

structures 275-277with normal Hall subgroup 158-166with partition 145-156; 152

Gruenberg radical 329

HHall subgroup 158-166hamiltonian group 55, 60, 68, 89Hasse diagram 5

height of element in abelian group 107Hirsch-Plotkin

hyperradical 404, 405radical 402series 403, 404

homomorphism of lattice 4Hughes subgroup 147hyperabelian group 315, 316hypercentral embedding 217hypercentre 183, 243hypercyclic

embedding 217group 315, 319K-group 123

hypoabelian group 315

Iidentical subgroup structures 275-277IM-group 130-135index preserving projectivity 166-178,

325-329, 408induce a power in a subgroup 257inner automorphism 24interval 5, 10, 401invariant under projectivities 14

(see class of groups invariant underprojectivities)

568

involution 421-451, 537isolated element 360isomorphism

between lattices of compositionsubgroups 504-507

between lattices of normalsubgroups 490-496

crossed 90-96of lattice 4R- 530

Iwasawa triple 60-69

Index of Subjects

Jjoin-distributive element 136-137,

141-142join-distributive pair 16join-quasi-distributive element 136-137Jordan-Dedekind chain condition 47,

231-234,540

KK-group 115-124

Llattice 2

algebraic 9centralizer 8, 511-527complemented 115complete 2coset 8, 527-541directly decomposable 8distributive 11

lower semimodular 47-48modular 41-48of composition subgroups 6, 47, 498-

511of cosets with respect to normal

subgroups 541of divisors of n 11

of normal subgroups 5, 43, 486-498of subnormal subgroups 7,47,498-511partition 10permodular 312poly-k-modular 405polymodular 234relatively complemented 127sectionally complemented 126semimodular 46-48, 233subgroup 3

upper semimodular 46-48weakly join-complemented 133weakly meet-complemented 135weakly modular 87

lattice-isomorphic groups 4lattice-simple group 449

lattice-theoretic characterization 14(see class of groups with lattice-theoretic

characterization)least element 3

least upper bound 2

left affinity 537length of a poset 3, 39locally cyclic group 12

locally finite group 15-16M-group 78-82of finite exponent 178p-group 53, 79, 94, 273RK-group 130SC-group 123with dual 458, 478

locally finite radical 329, 459locally free group 359, 533locally nilpotent group 373, 376, 533locally P-embedded subgroup 302locally soluble radical 477locally I-group 15

local system of subgroups 21lower semimodular lattice 47-48, 233,

394, 399, 529

Mmap

amorphy of 17, 534, 538-541Baer 416

element 17, 20, 101-107normed 530rp- 361

maximal centralizer 514maximal condition 4, 46maximal element 3

maximal sensitive subgroup 133, 234meet-distributive element 136, 138,

143-145meet-distributive pair 16, 40meet-quasi-distributive element 136, 138,

143-145meet-sublattice 7metacyclic p-group 225M-group 54-100;55,71,75,79,82971-group 518-523M*-group 55, 58-69minimal centralizer 514minimal condition 4minimal element 3

minimal normal subgroup 235-238modular

centralizer lattice 518-524, 527coset lattice 529dually 87element 43--46, 144


lattice 41-48lattice of composition subgroups 502-

503subgroup 43,200-220,232subgroup lattice 54-88weakly 87

modular chain 360cyclic 375

modularly embedded element 360

Nneutral element 135, 142-145, 498nilpotent-by-abelian group 397nilpotent group 360-395

finite 47, 54, 494-498of class 2 and exponent p 384periodic 387, 391torsion-free 361, 373, 533

nilpotent residual 181

nilpotent variety 382-385normal closure 199, 207

of a permutable subgroup 304normal Hall subgroup 158-166normality relation 302normalizer preserving projectivity 272-

275, 377-385, 408normal partition 146normal subgroup 5, 43, 486-498

minimal 235-238projective image of 239, 321-323, 334,

340d'-normal 510

normed map 530norm of a group 24, 30

0Q-group 394OP(G) 181, 506OP(G) 178, 506o(x, H) 206

Pp-adic unit 32-33pair

join-distributive 16meet-distributive 16, 40

partially ordered set 1

partition lattice 10partition of a group 145-156, 519p-collecting subgroup 540p-collector 540P-decomposition 170P-embedded subgroup 297perfect group 186, 219, 229, 243, 313, 330.

540

permodular chain 385-391permodular lattice 312permodularly embedded element 385permodular subgroup 288-303permutable complement 121

permutable subgroup 43, 201, 202, 216-227, 293, 296, 303-311, 318

permutably complemented subgroup121

permutation group 430-449doubly transitive 434, 435, 438primitive 115, 439triply transitive 432

P(G)-connected subgroups 468P(G)-normal subgroup 469p-group

finite 52-69, 220-223, 243-272, 274,279, 391-395, 504

locally finite 53, 79, 94, 273, 460metacyclic 225nilpotent 387, 391of maximal class 392-395regular 392with modular subgroup lattice 54-69,

79, 94P-group 49-54P*-group 69, 89p -mapping 361P-hypercentrally closed formation 400P-hypercentral triple 399p-length 183p-normal 195polycyclic group 317poly-k-modular lattice 405polymodular element 234polymodular lattice 234poset 1

length 3, 39width 3, 39-40

power automorphism 24, 29-34of abelian group 32of M*-group 68of nonperiodic group 33universal 32

prefrattini subgroup 401p-regular projectivity 167-178, 422-

424primitive permutation group 115, 439product

direct 8, 34-40, 406-421, 426, 429free 359wreath 416-418

projective general/special linear group116, 120, 146, 152, 156, 177, 432-436,519,523-524


projectivity 4between abelian groups 103, 104, 110-

113between finite groups with normal Hall

subgroups 160centralizer preserving 272-273compatible with a pair of epimorphisms

380conjugacy preserving 272

index preserving 166-178, 325-329,

408

induced by a crossed isomorphism 93induced by a group-isomorphism 20-

24induced by an element map 17, 20induced in a factor group 22induced in a subgroup 15invariant under 14normalizer preserving 272-275, 377-

385, 408of abelian groups 94, 96, 99, 106of M-groups 89, 94, 99p-regular (or regular at p) 167-178,

422-424p-singular (or singular at p) 167-178,

422-424singular 166-178

projector of a formation 397, 401Priifer group 16p-singular projectivity 167-178, 422-424p-soluble group 183, 196

Qquasinormal subgroup 43quaternion group 28

Rradical

Gruenberg 329Hirsch-Plotkin 402

locally finite 329, 459locally soluble 477soluble 230, 234

radical group 402-405rank of soluble group 396RC-group 129-130regular at p 167-178, 422-424

regular element 536regular p-group 392relative complement 127relatively complemented lattice 127residual

nilpotent 18192k- 240of a formation 397, 400

soluble 230, 234supersoluble 234

residually finite group 287residually finitely generated soluble

group 321residually polycyclic group 320residually supersoluble group 319right affinity 537R-isomorphism 530RK-group 127-135Rottlander group 162, 276, 413, 430

Ssaturated formation 397SC-group 122-124Schmidt structure 297sectional complement (or S-complement)

122sectionally complemented lattice 126semiaffinity 537semimodular element 49, 216semimodular lattice 46-48, 233, 394, 399,

529, 540serial subgroup 293simple group 116, 156, 228, 312, 432, 434,

439-441, 449, 523singular at p 167-178, 422-424singular projectivity 166-178SK-group 126SN-group 319soluble group(s) 181, 229, 317, 332,

395-406, 506-511derived length of 332Fitting length of 180, 399formation of 397-401, 405-406generalized 315-321IM-group 132K-group 118-121rank of 396RK-group 130with dual 462-468

soluble radical 230, 234soluble residual 230, 234standard element 136-137, 141strongly determined by subgroup lattice

343abelian group of torsion-free rank > 2

106alternating group A,,, n : 3' or 3' + 1,

r odd 447direct product of isomorphic dihedral

groups 426free group of rank > 2 358free polynilpotent group 374hamiltonian 2-group 89

Index of Subjects

nonabelian locally free group 359nonabelian torsion-free locally nilpotent

group 373simple group of Lie type (# A 2) 449symmetric group 444, 447, 4512-free group 358wreath product of torsion-free groups

417strongly embedded subgroup 153subgroup(s)

admissible (of a partition) 147ascendant 293, 328commutator 29, 272composition 6, 498-511cyclic 15, 18Dedekind 289D-embedded 289diagonal 35, 408-411Fitting 116, 178-181Frattini 38, 115G-isomorphic 487Hall 158-166Hughes 147locally P-embedded 302local system of 21maximal sensitive 133, 234modular 43, 200-220, 232of finite index 282-288, 325, 476p-collecting 540P-embedded 297permodular 288-303permutable 43,201,202,216-227,293,

296, 303-311, 318permutably complemented 121P(G)-connected 468P(G)-normal 469prefrattini 401quasinormal 43serial 293strongly embedded 153subnormal 6, 240-243, 498-511Sylow 166, 185-187, 195-198verbal 382

subgroup lattice(s) 3

chain 16characterization of 357complemented 115-126determined by vii, 343directly decomposable 36-39distributive 12-13IM- 130-135Jordan-Dedekind chain condition 231-

234lower semimodular 47, 233, 394, 399modular 54-88; 55, 71, 75, 79, 82

571

permodular 312relatively complemented 127-135sectionally complemented 126strongly determined by 343upper semimodular 233weakly join-complemented 133weakly meet-complemented 135weakly modular 87

sublattice 5

meet- 7

subnormal subgroup 6, 240-243, 498-511

defect of 125, 499projective image of 240-243:2"- 510

supersolublechain 319group 230-234, 318K-group 122residual 234

Suzuki group 116, 146, 152, 156, 434Sylow subgroup 166

abelian 185-187, 197--198centre of 195-197

Sylow tower group 396, 405symmetric group 115, 149, 177, 432, 438,

441-445, 447, 451, 513, 527S4 149, 513, 527

TTarski group 82, 88T-group 127T*-group 127tripleIwasawa 60-69P-hypercentral 399

triply transitive permutation group 432type of abelian group 104, 108type of element in abelian group 108

Uuniquely complemented element 136,

138-141unit (p-adic) 32-33universal power automorphism 32upper Hirsch-Plotkin series 403, 404upper semimodular lattice 46-48, 233,

540

Vvariety 382-385verbal subgroup 382

Wweakly join-complemented lattice 133


weakly meet-complemented lattice 135 Xweakly modular lattice 87 I-normal subgroup 510width of a poset 3, 39-40 I-subnormal subgroup 510wreath product 416-418

ISSN 0935-6572ISBN 3-11-011213-2

Subgroup Lattices of Groups - Department of Mathematics · Subgroup Lattices of Groups by Roland...

Documents

Transcript of Subgroup Lattices of Groups - Department of Mathematics · Subgroup Lattices of Groups by Roland...