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LEFT DISTRIBUTIVE LEFT QUASIGROUPS

DAVID STANOVSKY

PhD Thesis

Charles University in Prague

July 2004

1

2 David Stanovsky: Left distributive left quasigroups

Acknowledgement

I wish to thank to everybody I learned anything from during past twenty sixyears.

Above all, to my advisor, Jaroslav Jezek, for his support and insightful advices.To my inofficial coadvisor, Ralph McKenzie, for numerous inspirative conversa-

tions and for financial and other support during my stay at Vanderbilt Universityin Nashville.

To Vera Trnkova, Ales Drapal, Tomas Kepka, Premysl Jedlicka and other peoplefrom the Department of Algebra at Charles University in Prague, for their interestand useful advices.

To people from the Department of Mathematics at Vanderbilt University inNashville, who made my stay in the United States nice and fruitful.

To Anna Romanowska and Barbara Roszkowska from the Technical Universityin Warsaw, for providing useful materials and for an invitation to visit Warsaw.

And, especially, to my parents for their support during long years of my studies.

A partial financial support of the Grant Agency of the Czech Republic undergrants 201/02/0594 and 201/02/0148 is greatfully acknowledged.

David Stanovsky: Left distributive left quasigroups 3

Contents

Acknowledgement 2

Contents 3

I. Introduction 5

II. Preliminaries 91. Basic facts 92. Normal form of terms 123. Definable sets are left ideals 134. Quasigroups and loops 14

III. Left symmetric left distributive idempotent groupoids 161. Normal form of terms 172. Exponent 193. Representation by Bruck loops 204. Applications 235. Cycles 236. Group of displacements 257. Left ideal decomposition and BL-sum 268. Cores 27

IV. On loops isotopic to left distributive quasigroups 29

V. Equational theory of group conjugation 311. The variety generated by conjugation 312. The equational theory of conjugation 343. On the role of idempotency 36

VI. Subdirectly irreducible non-idempotent left distributive leftquasigroups 381. Description 382. Examples 42

VII. On varieties of left distributive left idempotent groupoids 451. Varieties satisfying xn+1 ≈ x 452. Varieties satisfying xm+n ≈ xm 47

VIII. Miscellaneous remarks and open problems 491. Miscellaneous remarks 492. Open problems 50

Appendix: Homomorphic images of subdirectly irreducible algebras 521. The necessary condition 522. Groupoids 543. Algebras with rich signature 584. Monounary algebras 62


5. Unary algebras 646. Particular varieties 65

References 66


I. Introduction

Selfdistributive groupoids arise on borders of several mathematical disciplines,such as algebra, geometry and set theory. An interested reader is encouraged to lookat the recent monography [11] of P. Dehornoy. We consider a particular subclass:left distributive left quasigroups, known also as racks, wracks, quandles, automor-phic sets, pseudo-symmetric sets, crystals, etc. (some of the names refer to theidempotent case, some of them don’t).

A groupoid (i.e. a set equipped with a binary operation) is called left distributive,if it satisfies the identity

x(yz) ≈ (xy)(xz).(LD)

It is called left quasigroup, if

for every a, b there is a unique c with ac = b;(LQ)

such c is denoted a\b. A groupoid is called idempotent, if it satisfies

xx ≈ x.(I)

There is a very natural construction of LDI left quasigroups: on a group G,define a new operation by a ∗ b = aba−1. It turns out that the groupoid G(∗),called the conjugation groupoid of G, is an LDI left quasigroup.

In the present thesis we study several problems regarding left distributive leftquasigroups. We do not build a compact theory, the chapters are rather inde-pendent. We are concerned with two types of problems, quite different in theirnature: “idempotent problems” and “non-idempotent problems”. Problems in gen-eral (non-idempotent) case are usually reduced to the idempotent case, which isusually more complicated. For instance, we find all non-idempotent subdirectly ir-reducible left distributive left quasigroups, but we have no idea how the idempotentones look like. We describe part of the lattice of subvarieties, modulo the lattice ofidempotent subvarieties (which is probably very complicated). A short overview ofresults follows.

Chapter II contains notation, terminology and basic properties of left distribu-tive left quasigroups used throughout the thesis. A necessary introduction to quasi-groups and loops is also there. Except for these preliminaries, the other chaptersare intended to be more or less self-contained.

Chapters III, IV and V deal with “idempotent problems”. In Chapter III, westudy the variety of left symmetric left distributive idempotent groupoids (LDI leftquasigroups satisfying the identity x(xy) ≈ y) in full detail. Several structuralresults are obtained, mainly for quasigroups. The main result establishes a corre-spondence between LSLDI groupoids of odd exponent and the well known class ofBruck loops of odd exponent.


In Chapter IV, we show a solution to the eighth Belousov’s problem from hisbook [1]. We find a smallest example of a left distributive quasigroup such that itis isotopic to a loop which is not a Bol loop.

The equational theory of conjugation groupoids is investigated in Chapter V. Weshow that it coincides with the equational theory of left distributive left quasigroups(and some other classes), it contains the well known class of groupoid modes and,mainly, we are concerned about finding a basis of the equational theory. We cannotfind it, but we conjecture, that the theory is not finitely based. The results of thischapter were published in [53].

Chapter VI and VII deal with “non-idempotent problems”. Subdirectly irre-ducible non-imdepotent left distributive left quasigroups are described in ChapterVI. We generalize the results of the paper [18]. In Chapter VII we study varieties ofleft distributive left idempotent groupoids of finite exponent. LDLI groupoids arenot necessarily left quasigroups, however, LD left quasigroups are the most impor-tant examples of LDLI groupoids and the results of this chapter are well applicableto them.

Finally, Chapter VIII contains miscellaneous remarks and open problems.The Appendix consists of a completely independent topic: homomorphic images

of subdirectly irreducible algebras. The main result was published in [52], the wholematerial appeared in a contest thesis (SVOC 2001).

The thesis consists of original achievements of the author. Results of otherauthors are explicitly quoted. Parts of the thesis appeared in author’s Master’sthesis, author’s thesis for the SVOC contest and in papers [52], [53] and [54].

Historical remarks. The first explicit allusion to selfdistributivity is, perhaps, inthe work of C. S. Peirce [39] from 1880. He discusses various forms of a distributivelaw and he pleads also that selfdistributivity, ”which has hitherto escaped notice,is not without interest.” Another early work, which mentions an example of a non-associative distributive structure, is [51] of E. Schroder (1887). The first knownarticle fully pursued to selfdistributive structures is [8] of C. Burstin and W. Mayerfrom 1929. They investigated two-sided distributive idempotent quasigroups. Thefirst articles dealing with non-idempotent distributive groupoids were [49] of J.Ruedin (1966) in the two-sided case and [26] of T. Kepka (1981) in the one-sidedcase. A comprehensive survey for study of two-sided selfdistributive systems is [21],regarding one-sided distributivity we recommend a recent book [11] of P. Dehornoy.There are several rather new papers that develop and summarize an algebraic theoryof left distributive left quasigroups, mainly from a geometrical point of view: e.g.[13] of R. Fenn and C. Rourke, [50] of H. Ryder or [4] of E. Brieskorn.

The knot quandle. The main reason that led to the development of the theory ofLDI left quasigroups was the discovery of the so called knot quandle (independentlyby D. Joyce [22] and S. V. Matveev [33] in 80’s), an LDI left quasigroup assignedto a knot, which turns out to be a full knot invariant. The construction works asfollows.

A knot is a subspace of a (3-dimensional) sphere S3 homeomorphic to a circle.Two knots K1 and K2 are equivalent, if there is an (auto)homeomorphism of thesphere such that its restriction to K1 is a homeomorphism of K1 and K2. One of themost important problems in the knot theory is, given two knots, to decide, whether


they are equivalent. For this reason, invariants (with respect to the equivalence) ofknots are searched.

Usually only tame knots are studied (tame means equivalent to a close polygonalcurve). A classical invariant of tame knots is the fundamental group. It is definedfor a knot K to be the fundamental group of the topological space S3 r K. It isa full invariant, i.e. two tame knots are equivalent, iff their fundamental groupsare isomorphic. The problem is that fundamental groups are usually not very wellcomputable, so one would like to look for “more simple” invariants. The knotquandle is a possibility.

Consider a regular projection of a tame knot, i.e. a mapping to a (2-dimensional)plane such that there are only finitely many crossings and none of them is a three-fold crossing. This projection divides the knot to arcs (by an arc we mean a segmentfrom one underpass through some overpasses to the next underpass); denote the setof arcs A. Choose an orientation of the knot and define the knot quandle to be anLDI left quasigroup generated by A and presented by relations (1) ab = c for everyunderpass, where “a coming under b yields c” and where b is going over a in theleft-right direction (in a sense of the orientation of the knot), and (2) a\b = c in thesituation, where b is going over a in the right-left direction. The involutory knotquandle is an LSLDI groupoid (i.e. an LDI left quasigroup satisfying x\y ≈ xy)defined analogously regardless the orientation. Examples:

(i) a

�b c

trefoil

The trefoil quandle is presented by

〈a, b, c; ab = c, bc = a, ca = b〉 .

One can easily compute that the involutory trefoil quandle is the core ofthe cyclic group Z3. (The core of an abelian group G is the groupoid G(∗)defined by a ∗ b = 2a− b.)

(ii) c

�a d b

figure-eight

The the figure-eight quandle is presented by

〈a, b, c, d; ab = d, b\c = a, cd = b, d\a = c〉 .


One can easily compute that the involutory figure-eight quandle is the coreof the cyclic group Z5 (recall that it is generated by arcs, not the set ofarcs).

Joyce and Matveev proved that two tame knots are equivalent, iff their knotquandles are isomorphic. (Thus, in particular, the knot quandle does not dependon the chosen regular projection.) The proof is based on investigation of the roleof conjugation in fundamental groups and it is rather difficult. Involutory knotquandles are also invariant, however, there are two non-equivalent tame knots withisomorphic involutory knot quandles; on the other hand, known examples are rathercomplicated.


II. Preliminaries

In the present chapter, we summarize basic properties of left distributive leftquasigroups used throughout the thesis. We assume only very basic preliminaryknowledge of universal algebra. An introduction to the theory can be found forinstance in the textbooks [7] or [34]. Our notation and terminology is rather stan-dard.

We denote FV(X) the free groupoid over a set X in a variety V. We considerthe standard representation of free groupoids by terms modulo equivalence withinthe theory. The letters x, y, z usually refer to term variables.

A subgroupoid generated by a set X is denoted 〈X〉.

1. Basic facts

In the present thesis, we consider mostly binary algebras; they are called shortlygroupoids. The binary operation is denoted implicitly multiplicatively, unless othernotation is introduced (such as ◦ or ∗). In this section, we summarize some termi-nology regarding groupoids.

First, we list some of the most often used groupoid identities:

(LD) x(yz) ≈ (xy)(xz) left distributivity(RD) (zy)x ≈ (zx)(yx) right distributivity(M) (xy)(uv) ≈ (xu)(yv) mediality (also entropy)

(n-LS) x(x(· · · (x︸︷︷︸n

y) ≈ y left n-symmetry

(LI) (xx)y ≈ xy left idempotency(I) xx ≈ x idempotency

Left and right distributive groupoids are called distributive (D). Note that medialidempotent groupoids are distributive. Left symmetry refers implicitly to left 2-symmetry.

Let G be a groupoid and a ∈ G. The left translation by a in G is a mappingLa : G → G, x 7→ ax; the right translation is defined dually and denoted Ra. Foroperations other than ·, for instance ∗, we use the notation L∗a, R∗a.

It is often useful to translate identities into the language of translations. Agroupoid G is left (right) distributive, iff all left (right) translations are endomor-phisms. It is left n-symmetric, iff Ln

a = id for every a ∈ G. And it is left idempotent,iff La = Laa for every a ∈ G. Also, note that a groupoid G is medial, iff the mappingG×G→ G, (a, b) 7→ ab, is a homomorphism.

We say that a groupoid G is left cancellative, if all left translations are injective,and left divisible, if all left translations are onto. It is called a left quasigroup, ifall left translations are permutations; or equivalently, if the equation ax = b has aunique solution x for any a, b ∈ G. Such x is usually denoted a\b. Left quasigroupsdo not form a groupoid variety (i.e. they cannot be axiomatized by identities in


the language of multiplication). However, they are axiomatized by the identitiesx(x\y) ≈ y and x\(xy) ≈ y in the language {·,\}.

For a left quasigroup G, we define the left multiplication group LMlt(G), to be asubgroup of the symmetric group over G generated by all La, a ∈ G.

Left n-symmetric groupoids are indeed left quasigroups with a\b = a(a(. . . (a︸︷︷︸n−1

b))).

On the other hand, any finite left quasigroup is left n-symmetric for some n.

1.1. Lemma. (1) A groupoid is a left distributive left quasigroup, iff all its lefttranslations are automorphisms.

(2) If G is a left distributive left quasigroup, then Lϕ(a) = ϕLaϕ−1 for every

a ∈ G and every automorphism ϕ of G. In particular, Lab = LaLbL−1a for

every a, b ∈ G(3) The mapping λ : a 7→ La is a homomorphism of an LD left quasigroup G

into the conjugation groupoid of the left multiplication group of G.

Proof. (1) follows immediately from the definition.(2) Since ϕ(ab) = ϕ(a)ϕ(b) for every a, b ∈ G, we have ϕLa = Lϕ(a)ϕ and thus

also Lϕ(a) = ϕLaϕ−1 for every a ∈ G.

(3) It follows from (2) that λ(ab) = Lab = LaLbL−1a = λ(a) ∗ λ(b). �

1.2. Lemma. Let G be a left distributive left quasigroup. Then(1) G is left idempotent;(2) if G is right cancellative, it is idempotent;(3) if G is right distributive, it is idempotent.

Proof. (1) xy ≈ x(x(x\y)) ≈LD (xx)(x(x\y)) ≈ (xx)y.(2) x(xx) ≈LD (xx)(xx) and use right cancellativity.(3) (xx)(xx) ≈RD (xx)x and use left cancellativity. �

There is a recent important structural result on non-idempotent LDLI groupoids,due to P. Jedlicka [17]. Let ipG denote the smallest equivalence on a groupoid Gcontaining the set {(a, aa) : a ∈ G}.

1.3. Lemma. (P. Jedlicka) Let G be an LDLI groupoid. Then (a, b) ∈ ipG iff thereare k, l such that ak = bl, the equivalence ipG is a congruence, G/ipG is idempotentand ipG is the smallest congruence such that the corresponding factor is idempotent.Moreover, for any (a, b) ∈ ipG, ac = bc holds for every c ∈ G.

Proof. See [17]. �

Consequently, any block of ipG is a subalgebra of G and it is term equivalent to aconnected monounary algebra. We will thus use the terminology of unary algebras(and directed graphs) for ipG-blocks.

By a circle of length n we mean a groupoid isomorphic to the groupoid Cn,defined on the set {0, . . . , n−1} by ab = b+1 mod n. An infinite path is a groupoidisomorphic to the groupoid C∞ of integers with the operation ab = b+ 1. It is easyto see that any ip-block of an LD left quasigroup is either a circle, or an infinitepath. And any ip-block of an n-LSLD groupoid is a circle of length k for some k|n.

Example. Let G be a group and put a ∗ b = aba−1 for every a, b ∈ G. Then G(∗)is an LDI left quasigroup, called the conjugation groupoid of G. Moreover, if A is


a subset of G such that an = 1 for every a ∈ A, the subgroupoid 〈A〉 of G(∗) isn-LSLDI.

Example. Let G be a group, e a central element in G and put a ∗e b = aba−1e forevery a, b ∈ G. Then G(∗e) is an LD left quasigroup and it is idempotent, iff e = 1.All ip-blocks of G(∗e) are cycles of length equal to the order of e. Again, if A is asubset of G such that an = 1 for every a ∈ A and en = 1, the subgroupoid 〈A〉 ofG(∗e) is n-LSLD.

Example. Let G be a group, e a central involution in G and put a ∗e b = ab−1aefor every a, b ∈ G. Then G(∗e) is an LSLD groupoid. It is idempotent, iff e = 1; insuch a case it is called the core of G. Cores play quite important role in the theoryof LSLDI groupoids. See more in Chapter III.

There are also other constructions of LSLD groupoids from groups. We refer tothe paper [54].

Example. Let R be a commutative ring with a unit, M a module over R andk ∈ R such that the mapping x 7→ kx is a permutation of M . For all a, b ∈M , puta ◦k b = (1− k)a+ kb. Then M(◦k) is a medial idempotent left quasigroup.

More examples of LD left quasigroups (mostly from geometry) can be found, forinstance, in the paper [13].

Example. There are two isomorphism types of left distributive left quasigroupson a 2-element set (one of them is idempotent, the other is not):

a b

a a bb a b

a b

a b ab b a

There are six isomorphism types of left distributive left quasigroups on a 3-element set (three of them idempotent, three not):

a b c

a a b cb a b cc a b c

a b c

a a c bb a b cc a b c

a b c

a a c bb c b ac b a c

a b c

a b c ab b c ac b c a

a b c

a a b cb a c bc a c b

a b c

a a c bb a c bc a c b

The third idempotent example, the core of the cyclic group Z3, is a commutativemedial quasigroup and it forms the smallest Steiner triple system. It appears as anexample of a non-associative distributive structure already in the paper [51] from1887.

Example. According to D. Joyce [23], finite simple LDI left quasigroups can bedescribed in terms of triples (G,C,m), where G is a finite simple group, C a con-jugacy class in G and m a positive integer. It follows from Lemma 1.3 that cycles


of prime length are the only simple non-idempotent LD left quasigroups (see also[17]).

2. Normal form of terms

In order to decrease the number of parentheses in terms, we assume implicitlythat letters in terms are right associated, i.e. xyz = x · yz = x(yz). We definethe n-th power by xn = x · · ·xx︸︷︷︸

n

. (Note that other possible definitions of powers

do not really make sense in left idempotent groupoids: one can check by an easyinduction that any unary term t(x) is LI-equivalent to the term xn, where n is thedepth of the rightmost variable in t.) Also, x[n]y will stand for x · · ·x︸︷︷︸

n

y = Lnx(y).

There should be no confusion, because it follows from left idempotency by an easyinduction that xky ≈ xy for any k. For instance, the left n-symmetry takes theform x[n]y ≈ y in this notation.

It is important to observe that in any LD left quasigroup

xy · z ≈ xy · x(x\z) ≈LD xy(x\z).

In case of left n-symmetry we get

xy · z ≈ xyx[n−1]z

and similarly by induction

(x1x2 · · ·xm) · z ≈ x1x2 · · ·xm−1xmx[n−1]m−1 · · ·x

[n−1]2 x

[n−1]1 z.

Consequently, using left n-symmetry and left distributivity, one can write any termover an alphabet X in the form x1x2 · · ·xn−1xm, for some m and x1, . . . , xm ∈ X.The following theorem shows a useful normal form of terms in the variety of n-LSLDI and n-LSLD groupoids. It is a generalization of a similar result of theauthor obtained for 2-LSLD(I) groupoids in [54].

2.1. Theorem. Let n ≥ 2 and X be a non-empty set. For every x ∈ X, let Cx bea cyclic group of order n generated by x and denote G the free product of all Cx,x ∈ X.

(1) The free n-LSLDI groupoid over X is isomorphic to the subgroupoid 〈X〉 ofthe conjugation groupoid of the group G. For every term over the alphabetX there is a unique n-LSLDI-equivalent term of the form

x[k1]1 x

[k2]2 · · ·x[km−1]

m−1 xm,

where x1, . . . , xm ∈ X, xi 6= xi+1 and ki ∈ {1, . . . , n − 1} for every i =1, . . . ,m− 1.

(2) The free n-LSLD groupoid over X is isomorphic to the subgroupoid 〈X〉 ofthe groupoid H(∗(1,e)), where H denotes the direct product G×Ce and Ce isa cyclic group of order n generated by e. For every term over the alphabetX there is a unique n-LSLD-equivalent term of the form

x[k1]1 x

[k2]2 · · ·x[km−1]

m−1 xkmm ,

where x1, . . . , xm ∈ X, xi 6= xi+1 for every i = 1, . . . ,m − 1 and ki ∈{1, . . . , n− 1}.


Proof. We have seen that any term is n-LSLD-equivalent to a term x1x2 · · ·xm.Collect equal neighbours and use left n-symmetry to decrease the exponents bellown (resp. idempotency in (1) for the rightmost variable).

(1) Denote F the subgroupoid of the conjugation groupoid of G generated by X.Clearly, F is left n-symmetric, because it is generated by elements of order n. Weshow that different terms in the described form represent different elements in F .Let x1, . . . , xm ∈ X such that xi 6= xi+1, i = 1, . . . ,m− 1, and ki ∈ {1, . . . , n− 1}.The word

x[k1]1 ∗ x[k2]

2 ∗ . . . x[km−1]m−1 ∗ xm = xk1

1 xk22 . . . x

km−1m−1 xmx

−km−1m−1 . . . x−k2

2 x−k11

is irreducible in the free product, so such words are pairwise different.(2) Proceed similarly. Since

x[k1]1 ∗ x[k2]

2 ∗ . . . x[km−1]m−1 ∗ xkm

m = xk11 x

k22 . . . x

km−1m−1 xmx

−km−1m−1 . . . x−k2

2 x−k11 e−1+

∑i ki ,

different terms represent different elements in the groupoid. �

Remark. The free n-LSLD groupoid over X is isomorphic to the product of thefree n-LSLDI groupoid over X and the cycle of length n. It can be checked directlyfrom the representation proven above or one can apply Theorem VII.1.3.

Remark. According to [12], the free groupoid over X in the variety generatedby LDI left quasigroups is isomorphic to the subgroupoid 〈X〉 of the conjugationgroupoid of the free group over X (see also Chapter V). Consequently, in the lan-guage {·, \}, there is a normal form for terms in the variety of LD(I) left quasigroupssimilar to that from Theorem 2.1 (in fact, one can use the representation of termsby words in free groups).

3. Definable sets are left ideals

Let G be a groupoid. A non-empty subset I ⊆ G is called a left ideal, if GI ⊆ I.In other words, if ab ∈ I for every a ∈ G, b ∈ I. Clearly, if G is a left quasigroupand I ⊂ G is a left ideal, then Gr I is also a left ideal.

A subset S ⊆ G is called definable inG if there exists a formula Φ (in the languageΛ consisting of the multiplication symbol) with a single free variable such thatS = {a ∈ G : Φ(a)}. A relation α on G is called definable, if there exists a formula Φ(in the language Λ) with two free variables such that α = {(a, b) ∈ G×G : Φ(a, b)}.A relation α is called right stable, if (a, b) ∈ α implies (ac, bc) ∈ α for every c.

A variant of the following proposition can be found in Jerabek, Kepka, Stanovsky[18] for 2-LSLD groupoids.

3.1. Theorem. Let G be a left distributive left quasigroup. Then(1) every definable subset in G is either empty, or a left ideal;(2) every definable right stable equivalence on G is a congruence of G.

Proof. Let S = {a ∈ G : Φ(a)} for a formula Φ(x). We need to prove that Φ(a)implies Φ(ba) for every a, b ∈ G.

Let Φ(x) be a formula of the form (Q1x1)(Q2x2) . . . (Qnxn)Ψ(x, x1, . . . , xn),where the letters Q1, . . . , Qn stand for quantifiers and Ψ is a formula with n + 1free variables without quantifiers. Since all left translations are permutations, it isenough to prove the following statement: for all a, a1, . . . , an, b ∈ G

Ψ(a, a1, . . . , an) implies Ψ(ba, ba1, . . . , ban).(†)


By induction. Assume Ψ is an atomic formula of the form t ≈ s, t, s terms.If t(a, a1, . . . , an) ≈ s(a, a1, . . . , an), then t(ba, ba1, . . . , ban) = bt(a, a1, . . . , an) =bs(a, a1, . . . , an) = s(ba, ba1, . . . , ban) (the first and third equalities follow from thefact that Lb is an automorphism) and thus (†) holds.

Assume Ψ is of the form Ψ1 ◦ Ψ2, where ◦ ∈ {∨,∧}. It is clear that if Ψ1,Ψ2

satisfy (†), then so does Ψ.Finally, assume that Ψ is of the form ¬Ψ1 and Ψ1 satisfies (†). Suppose that

Ψ(a, a1, . . . , an) is true and Ψ(ba, ba1, . . . , ban) is not. Then Ψ1(ba, ba1, . . . , ban) istrue and so is Ψ1(L−1

b (ba), L−1b (ba1), . . . , L−1

b (ban)) ≡ Ψ1(a, a1, . . . , an), since L−1b

is an automorphism of G. Hence, a contradiction with Ψ = ¬Ψ1.For right stable equivalences proceed analogously. �

It follows from Lemma 1.3 that every equivalence bellow ipG is right stable,hence one can use Theorem 3.1 in such a situation.

Remark. Note that Theorem 3.1 holds also for sets and equivalences definable inthe language {·,∈ J1, . . . ,∈ Jn}, where J1, . . . , Jn are left ideals and ∈ J1, . . . ,∈ Jn

the corresponding unary relational symbols. (We have no use for this more generalstatement.)

4. Quasigroups and loops

In this section, we summarize some terminology and basic facts regarding quasi-groups and loops. We will use them mainly in Chapters III and IV.

A quasigroup is a groupoid, where all left and right translations are permutations(i.e. which is both left and, dually, right quasigroup). A loop is a quasigroup witha unit element 1 (i.e. x · 1 ≈ 1 · x ≈ x holds). An alternative definition says thatquasigroups are groupoids with left and right cancellation, it means ab = ac⇒ b = cand ba = ca ⇒ b = c holds for any a, b, c, and left and right division, it means forevery a, b the equations ax = b and ya = b have a solution. By cancellativity, thesolutions are unique and they are denoted a\b and b/a. Quasigroups and loops canbe, indeed, axiomatized by indentities in the language {·, /, \}. An introduction tothe theory of quasigroups and loops can be found, for instance, in the book [40].

We say that quasigroups Q1 and Q2 are isotopic, if there are bijective mappingsα, β, γ : Q1 → Q2 such that α(ab) = β(a)γ(b) for every a, b ∈ Q1. It means,Q2 is obtained from Q1 by permuting rows, columns and renaming entries in themultiplication table.

A left quasigroup Q is said to have the left inverse property (LIP), if for everya ∈ L there is a−1 ∈ Q such that a\b = a−1b for every b ∈ Q. The RIP is defineddually.

Loops may be regarded as non-associative groups. Weaker forms of associativ-ity are extensively studied. A loop is called Moufang, if the following equivalentidentities hold:

xyxz ≈ (xy · x)z, (zx · y)x ≈ z(xyx), xy · zx ≈ (xyz)x ≈ x(yz · x).

Commutative Moufang loops are usually denoted additively. They can be based(relatively to the axioms of loops) by the identity

(x+ x) + (y + z) ≈ (x+ y) + (x+ z).

A loop is called left Bol, ifxyxz ≈ (xyx)z


holds (the word left is usually dropped). A loop is Moufang, iff it is left and rightBol. Bol loops have LIP, thus Moufang loops have both LIP and RIP. Moufangloops are diassociative, i.e. any 2-generated subloop is a group (hence all terms intwo variables are associative). Bol loops are power-associative, i.e. any 1-generatedsubloop is a cyclic group. Consequently, powers are well-defined and we can speakabout the order of an element and about the exponent of a Bol loop. Also, the Bolidentity implies xxy ≈ x2y (left reflexivity). It is well-known that any loop isotopicto a Bol loop is a Bol loop. More information about Moufang and Bol loops canbe found for instance in [9], [29] or [1].

Later we will need the following two lemmas (they can be found e.g. in [29]).

4.1. Lemma. Let L be a left quasigroup with a unit satisfying LIP and the Bolidentity. Then L is a Bol loop, a\b = a−1b and a/b = b−1(ba · b−1) for everya, b ∈ L.

Proof. Since a(a−1b) = b, it is the unique solution of ax = b. To prove rightcancellation, suppose ac = bc, compute ca = cacc−1 =Bol (cac)c−1 = (cbc)c−1 =Bol

cbcc−1 = cb and use left cancellation. Finally, we need to check that (b−1(ba ·b−1))b = a. Indeed, by the Bol identity, the left side is equal to b−1(ba)b−1b =b−1ba = a. �

A loop with LIP is said to have the automorphic inverse property (AIP), if itsatisfies (xy)−1 ≈ x−1y−1.

4.2. Lemma. Let L be Bol loop. Then the following properties of L are equivalent.(1) AIP.(2) (xy)2 ≈ xy2x.(3) L2

ab = LaL2bLa for every a, b ∈ L.

Proof. (1) ⇒ (2). xy ≈ xy2xx−1y−1 ≈AIP xy2x(xy)−1 ≈Bol (xy2x)(xy)−1. Henceby 4.1 xy2x ≈ xy/(xy)−1 ≈ (xy)((xy)−1(xy) · (xy)) ≈ (xy)2.

(2) ⇒ (3). To show (ab)2c = abbac for every a, b, c ∈ L, compute (ab)2c =(2)

(ab2a)c =Bol ab2ac =LR abbac.

(3) ⇒ (1). For any a, b ∈ L we have ab = ab2aa−1b−1 = LaL2bLa(a−1b−1) =(3)

L2ab(a

−1b−1) = (ab)2(a−1b−1). Left cancellation by ab yields (ab)(a−1b−1) = 1,hence AIP holds. �

Bol loops satisfying the properties of the previous lemma are called Bruck loops(also known as K-loops or gyrocommutative gyrogroups)1. A Bruck loop is Moufang,iff it is commutative. Indeed, we derive commutativity from (2) by diassociativity.On the other hand, commutative Moufang loops obviously satisfy AIP.

There are two recent books on Bruck loops: H. Kiechle’s [29] and A. A. Ungar’s[59]. The first one is written from a purely algebraic point of view, whereas thesecond one is motivated by relativistic physics: the key example (pointed out byUngar) which led to the fast growth of the theory of Bruck loops is the set of alladmissible velocities in relativistic physics with the Einstein addition. A remarkablepaper [16] of G. Glauberman deals with Bruck loops of odd order (called B-loopsthere). For historical notes see e.g. the appendix of [29].

1The name is chosen in honor of R. H. Bruck, who first explicitly used the notion in his book[5]. It seems to be the oldest name for this structure, see for instance [44].


III. Left symmetric left distributiveidempotent groupoids

In the present chapter we consider left 2-symmetric left distributive idempotentleft quasigroups; we call them shortly LSLDI groupoids. In literature, they appearunder a bunch of other names, such as kei, symetric sets, symmetric groupoids orinvolutory quandles. The first paper on LSLDI groupoids seems to be [56] of M.Takasaki (1943), however, most results come from 70’s and 80’s.2

Our aim is to continue investigations of this interesting variety, mainly in direc-tion of representation of LSLDI groupoids by other structures. We also summarizesome of the older results and shortly reprove some of them. We start with a normalform of terms in LSLDI and LSMI (M refers to mediality) and define the exponent.Then we proceed with the main theorem, a representation of LSLDI groupoids ofodd exponent by Bruck loops of odd exponent. In the next section, we investigatecycles, an important structural feature of LSLDI groupoids. Further, we present aconstruction, called BL-sum, of LSLDI groupoids from Bol loops. Finally, the lastsection is devoted to cores.

The origin of the theory of LSLDI groupoids lies in the book [32] of OttmarLoos (1969). He studied so called symmetric spaces and he found that they can bedescribed equivalently as differentiable manifolds equipped with a smooth LSLDIoperation satisfying the following condition: for any point a, there is a neighborhoodU of a such that a is the only fixed point of the left translation La on U . A basicexample of a symmetric space comes as follows: on any manifold with metric, onecan define a product x ◦ y to be the image of y by the symmetry through x. Forinstance, on the real line x ◦ y = 2x − y, on the n-sphere x ◦ y = 2 〈x, y〉x − y.Actually, the 3-sphere gives a reason why the fixed point condition is defined locally— the point symmetric to the south pole through the north pole is the south pole.A different example is the set of all n×n positive definite symmetric matrices overthe reals with the operation A ◦ B = AB−1A. Grassman manifolds and Jordanalgebras are also symmetric spaces. And, of course, any LSLDI groupoid can beconsidered as a discrete symmetric space. Motivated by the above natural example,Nobuo Nobusawa [37] and others started to investigate LSLDI groupoids from apurely algebraic point of view in 70’s. The most remarkable references are [37],[24], [41], [42], [22], [23]. The results are surveyd in [55].

0.1. Lemma. Let G be a groupoid. Then

(1) G is LSLD, iff every left translation is either the identity, or an involutoryautomorphism of G;

(2) if G is LSLD, then Lab = LaLbLa for every a, b ∈ G.

2Actually, the content of the rather long Takasaki’s paper is a big puzzle — it is in Japanese.It seems that it contains some basic properties and many examples. The journal is available in

the library of the Mathematical Institute of the Polish Academy of Sciences in Warsaw.


(3) if G is LSLD, the mapping λ : a 7→ La is a homomorphism of G into thecore of the left multiplication group of G.

Proof. Easy (cf. II.1.1). �

A groupoid is called effective, if its left translations are pairwise different. Aneffective LSLDI groupoid thus can be embedded into the core a group. The kernelof the homomorphism λ was called a central congruence by R. S. Pierce and it wasextensively studied in his papers [41], [42].

Let L be a Bol loop. We define the core of L, denoted by Core(L), to be thegroupoid L(∗) with a∗b = ab−1a. It is easy to check that cores of groups are LSLDI.It is not that easy to check that cores of Bol loops are also LSLDI (a different proofcan be found in [1], for instance).

0.2. Lemma. The core of a Bol loop is LSLDI.

Proof. For idempotency, a ∗ a = aa−1a = a. For left symmetry, a ∗ a ∗ b =a(ab−1a)−1a = a(ab−1a)−1ab−1aa−1b =Bol a(ab−1a)−1(ab−1a)a−1b = aa−1b = b.For left distributivity,

a ∗ (b ∗ c) = a(bc−1b)−1a = a(bc−1b)−1aa−1bc−1bb−1cb−1a =Bol

a(bc−1b)−1(bc−1b)b−1cb−1a = ab−1cb−1a

and similarly we get

(a ∗ b) ∗ (a ∗ c) = (ab−1a)(ac−1a)−1(ab−1a) =

ab−1a(ac−1a)−1(ab−1a)(ab−1a)−1ac−1aa−1ca−1(ab−1a) = ab−1cb−1a.

�

Cores will be discussed in the last section in detail.

1. Normal form of terms

We recall from the introduction that LSLD groupoids satisfy

xy · z ≈LS xy · xxz ≈LD xyxz

and similarly by induction also

(x1x2 · · ·xn) · z ≈ x1x2 · · ·xn · · ·x2x1z.

Consequently, using left symmetry and left distributivity, one can write any termover X in the form x1x2 · · ·xn−1xn, xi ∈ X. Note that mediality takes the form

xyzt ≈ zyxt.

Using these observations, one can derive a useful normal form of terms in LSLDI.(The first part of the following theorem is a particular case of Theorem 2.1.)

1.1. Theorem. Let X be a non-empty set.(1) The free LSLDI groupoid over X is isomorphic to the subgroupoid 〈X〉 of

the core of the free group over X and any term over X is LSLDI-equivalentto a unique term of the form

x1x2 · · ·xn−1xn, x1, . . . , xn ∈ X and xi 6= xi+1, i = 1, . . . , n− 1.(†)


(2) The free LSMI groupoid over X is isomorphic to the subgroupoid 〈X〉 ofthe core of the free abelian group over X and any term over X is LSMI-equivalent to a unique term of the form

x1x2 · · ·xn−1x, x1, . . . , xn−1, x ∈ X, {x1, x3, . . . } ∩ {x2, x4, . . . } = ∅,x1 4 x3 4 . . . , x2 4 x4 4 . . . ,(‡)x /∈ {xn−1, xn−3, . . . },

where (X,4) is a fixed linear order.

Proof. We have seen that any term over X can be written in LSLDI in the formx1x2 · · ·xn. If xi = xi+1, we use left symmetry to cancel them out. Using mediality,we can reorder the variables to the form (‡).

Let F (G resp.) be the subgroupoid 〈X〉 of the core of free (abelian, resp.) groupover X. We show that different terms in the form (†) ((‡) resp.) represent differentelements in F (G resp.). For x1, . . . , xn ∈ X with xi 6= xi+1, i = 1, . . . , n − 1, putε = (−1)n. The word

x1 ∗ x2 ∗ · · · ∗ xn−1 ∗ xn = x1x−12 · · ·xε

n−1x−εn xε

n−1 · · ·x−12 x1

is irreducible in the free group, so such words are pairwise different. Similarly, inthe free abelian group (denoted additively)

x1 ∗ x2 ∗ · · · ∗ xn−1 ∗ x = 2x1 + 2x3 + · · · − 2x2 − 2x4 − · · ·+ εx.

When {x1, x3, . . . }∩{x2, x4, . . . } = ∅ and x /∈ {xn−1, xn−3, . . . }, the value uniquelydetermines the term up to the order of x1, x3, . . . and x2, x4, . . . �

A groupoid G is called n-medial, if every n-generated subgroupoid of G is me-dial. Clearly, 4-medial groupoids are medial. It follows from 1.1 that every LSLDIgroupoid is 2-medial, since FLSLDI(2) = FLSMI(2). Jezek, Kepka, Nemec proved in[21], section IV.2, that an idempotent groupoid is (left and right) distributive, iffit is 3-medial. The backward implication is indeed easy: zx · yx ≈ zy · xx ≈ zy · x.Using 1.1, one can prove rather easily the forward implication for left symmetricgroupoids: using right distributivity in the form xyzx ≈ zyx, one can transformany term in three variables into its medial normal form (the details are omitted).

The following two lemmas will be used later.

1.2. Lemma. Let G be an LSLD groupoid. Then G is medial, iff LaLbLcLd =LcLdLaLb for every a, b, c, d ∈ G.

Proof. Using the normal form, it is easy to see that abcdx = cbadx = cdabx forevery a, b, c, d, x ∈ G in LSMI groupoids. On the other hand, if the latter is true,(xy)uv ≈LSLD xyxuv ≈ xuxyv ≈LSLD (xu)yv. �

1.3. Lemma. Let G be a LSMI quasigroup. Then LaLbLc = L(a/b)bc for all a, b, c ∈G.

Proof. Using several times Lemmas 0.1 and 1.2, we get

LaLbLc = L(a/b)bLbLc =(0.1) La/bLbLa/bLbLc = La/bLbLa/bLcLcLbLc =(1.2)

La/bLbLcLbLa/bLcLc =(0.1) La/bLbcLa/b = L(a/b)bc.

�


2. Exponent

We define a sequence of binary terms wn by w1(x, y) = x and wn+1(x, y) =xwn(y, x) for every n ≥ 1. In other words,

wn(x, y) is xy · · · yx︸︷︷︸n

for n odd and xy · · ·xy︸︷︷︸n

for n even.

It is easy to see that FLSLDI(x, y) = FLSMI(x, y) = {wn(x, y), wn(y, x) : n ∈ N}and they are isomorphic to the core of the additive group of integers. Underthe isomorphism ϕ determined by x 7→ 1, y 7→ 0 one has ϕ(wn(x, y)) = n andϕ(wn(y, x)) = 1− n.

2.1. Lemma. Let wm(x, y) ≈ y and wn(x, y) ≈ y hold in a variety V of LSLDIgroupoids. Then wk(x, y) ≈ y holds in V, where k is the greatest common divisorof m,n.

Proof. Suppose m < n. For m,n of the same parity we have

y ≈ wn(x, y) = xy · · ·xywm(x, y) ≈ xy · · ·xyy = wm−n(x, y).

Otherwise,

y ≈ wn(x, y) = xy · · · yxwm(y, x) ≈ xy · · · yxx = wm−n(x, y).

Proceed further by the Euclid’s algorithm. �

By the exponent of an LSLDI groupoid G we mean the least positive integer nfor which wn(x, y) ≈ y holds in G, if such exists, and ∞ otherwise. It is denotedexp(G). The exponent of a variety V of LSLDI groupoids is defined to be the leastn such that all its elements have exponent at most n (in other words, the least nsuch that wn(x, y) ≈ y holds in V, if such exists).

2.2. Proposition. (1) Every locally finite variety of LSLDI groupoids has afinite exponent.

(2) Every finite LSLDI groupoid has a finite exponent.

Proof. (1) Since FV(x, y) is finite, there are m,n such that wm(x, y) ≈ wn(x, y)or wm(x, y) ≈ wn(y, x) holds in the variety. Using left multiplication and leftsymmetry, we can derive wk(x, y) ≈ y for certain k. Take the smallest then.

(2) The variety generated by a finite groupoid is indeed locally finite, hence boththe variety and the groupoid have a finite exponent. �

2.3. Proposition. Let G be an LSLDI groupoid of finite exponent. Then the fol-lowing statements are equivalent.

(1) G is a quasigroup;(2) G is right cancellative;(3) G has odd exponent.

Proof. (1) ⇒ (2) is trivial.(2) ⇒ (3) Suppose G is an LSLDI quasigroup of exponent 2n. Then w2n(x, y) ≈

y and there is a, b ∈ G such that wn(a, b) 6= b. However, wn(x, y)y ≈ w2n(x, y) ≈y ≈ yy, hence, by right cancellativity, wn(x, y) ≈ y. A contradiction with minimal-ity of the exponent.

(3) ⇒ (1) Let n be the exponent of G and a, b ∈ G. Then w(n+1)/2(a, b)b =wn+1(a, b) = awn(b, a) = aa = a, so w(n+1)/2(a, b) is a solution of the equation xb =


a. Note that w(n+1)/2(xy, y) ≈ wn+1(x, y) ≈ xwn(y, x) ≈ xx ≈ x. Consequently,whenever ac = bc, we have a = w(n+1)/2(ac, c) = w(n+1)/2(bc, c) = b. �

2.4. Proposition. Let G be a finite LSLDI quasigroup. Then |G| is odd.

Proof. Fix some e ∈ G and assume all sets Ga = {a, ea}, a ∈ G. Clearly, |Ge| = 1,|Ga| = 2 for all a 6= e (since G has odd exponent, ea 6= a) and for every a, beither Ga = Gb or Ga, Gb are disjoint. Consequently, G is a disjoint union of one1-element and several 2-element sets, thus it has an odd number of elements. �

The converse is not true, groupoids satisfying xy ≈ y may have arbitrary numberof elements, though they have exponent 2. However, a finite Bruck loop L has oddexponent, iff |L| is odd (see [16], cf. next section).

3. Representation by Bruck loops

Let L be a Bol loop. Put for all a, b ∈ La ? b = a2b−1 = a · ab−1.

The resulting groupoid L(?) is denoted DG(L).

3.1. Lemma. Let L be a Bruck loop. Then DG(L) ' Core(L) via a 7→ a2.

Proof. We need to verify that (a ? b)2 = (a2b−1)2 = a2b−2a2 = a2 ∗ b2 for everya, b ∈ L. However, this follows directly from 4.2(3). �

Let G be an LSLDI quasigroup of an odd exponent n. Choose e ∈ G and putfor all a, b ∈ G

a ◦ b = (LaLe)(n+1)/2(b) = ae · · · ea︸︷︷︸n

eb.

The resulting algebra G(◦, e) is denoted BLe(G). The definition of the operation◦ appears, for instance, in the paper [46] of B. Roszkowska, where it is used toestablish a correspondence of LSMI quasigroups of odd exponent and abelian groupsof odd exponent. We extend the corresponence to LSLDI quasigroups and Bruckloops.

3.2. Theorem. Let G be an LSLDI quasigroup of an odd exponent n and e ∈ G,let L be a Bruck loop of an odd exponent n. Then

(1) BLe(G) is a Bruck loop of exponent n;(2) DG(L) is an LSLDI quasigroup of exponent n;(3) DG(BLe(G)) = G and BL1(DG(L)) = L.

Proof. (1) Put m = (n + 1)/2. Note that wn(x, y) ≈ y implies that (LaLe)n = idfor any a, e. Indeed, (LaLe)n(b) = wn(a, e)eb = eeb = b. Now, we are ready tocheck the axioms of Bruck loops. Let a, b, c ∈ G.

• e ◦ a = (LeLe)m(a) = a.• a ◦ e = (LaLe)m(e) = awn(e, a) = aa = a.• a ◦ (ea) = (LaLe)m(ea) = (LaLe)m−1(aeea) = (LaLe)m−1(a) = wn(a, e) =e.

• (ea) ◦ a = (LeaLe)m(a) = (LeLaLeLe)m(a) = (LeLa)m(a) = ewn(a, e) =ee = e.

• (ea) ◦ (a ◦ b) = (LeaLe)m(LaLe)m(b) = (LeLa)m(LaLe)m(b) =(LeLa)m(LeLa)−m(b) = b.


Consequently, a−1 = ea.

• the Bol identity: (a ◦ b ◦ a) ◦ c = (La◦b◦aLe)m(c) =

((LaLe)m(LbLe)mLa(LeLb)m(LeLa)mLe)m(c) =

(LaLe)m(LbLe)m[La(LeLb)m(LeLa)mLe(LaLe)m(LbLe)m]m−1La(LeLb)m(LeLa)mLe(c)

and we need to prove that this is equal to

a ◦ b ◦ a ◦ c = (LaLe)m(LbLe)m(LaLe)m(c).

Hence, it remains to show that the mapping

ϕ = [La(LeLb)m(LeLa)mLe(LaLe)m(LbLe)m]m−1La(LeLb)mLe

is the identity. Indeed, we have

La(LeLb)m(LeLa)mLe(LaLe)m(LbLe)m = La(LeLb)m(LeLa)n+1Le(LbLe)m =

La(LeLb)mLeLaLe(LbLe)m = (La(LeLb)mLe)2

and so

ϕ = (LaLe(LbLe)m)n−1LaLe(LbLe)m = (LaLe(LbLe)m)n =

Le(LeLaLe(LbLe)mLe)nLe = Le(La◦eLb◦e◦b◦...)nLe = L2e = id.

• AIP: a−1 ◦ b−1 = (LeaLe)m(eb) = (LeLa)m(eb) = e(LaLe)m(b) = e(a ◦ b) =(a ◦ b)−1.

It follows from 4.1 that G(◦) is a Bruck loop. We claim that k-th power of ain BLe(G), denoted by ak, is wk(a, e). Proceed by induction: a = w1(a, e) andak = a ◦ ak−1 = (LaLe)m(wk−1(a, e)) = wn+k(a, e) = wk(a, e) for k ≥ 2.

(2) According to Lemmas 3.1 and 0.2, DG(L) ' Core(L) is LSLDI. We showthat wk(a, b) = b in DG(L), iff (ab−1)k = 1 in L. For k, we have

wk(a, b) = a ? b ? · · · ? b ? a = a2b−2 . . . a2b−2a = a2b−2 . . . aab−2a =Bruck

a2b−2 . . . a2b−2a(ab−1)2 =Bol a2b−2 . . . (a2b−2a)(ab−1)2 =Bruck

a2b−2 . . . a(ab−1)4 = · · · = a(ab−1)k−1.

In order to get b, one needs k divisible by the order of ab−1.(3) The first statement follows from a ◦ (a ◦ b−1) = (LaLe)m(LaLe)m(eb) =

(LaLe)n+1(eb) = (LaLe)n(aeeb) = ab. For the second, a ? 1 ? · · · ? 1 ? a ? 1 ? b =a21−2 · · · 1−2a21−2b = a2 · · · a2︸︷︷︸

(n+1)/2

b = an+1b = ab. �

3.3. Corollary. Let n be odd. The variety of pointed LSLDI groupoids satisfyingwn(x, y) ≈ y is term equivalent to the variety of Bruck loops satisfying xn ≈ 1. (Apointed groupoid is an algebra of type (2, 0), without any additional condition onthe constant.)

3.4. Corollary. (B. Roszkowska [46]) In the notation of 3.2, if G is medial, thenBLe(G) is an abelian group, and if L is an abelian group, then DG(L) = Core(L)is medial. Consequently, LSMI quasigroups of finite exponent are cores of abeliangroups of the same exponent.


Proof. For the first assertion, it is enough to prove asociativity. Write (a ◦ b) ◦ c as

(La◦bLe)m(c) = ((LaLe)mLb(LeLa)mLe)m(c) = ((LaLe)mLbLe(LaLe)m)m(c),

use 1.2 to get

((LaLe)m(LaLe)mLbLe)m(c) = ((LaLe)nLaLeLbLe)m(c) = (LaLeLbLe)m(c)

and, again by 1.2, conclude that (LaLe)m(LbLe)m(c) = a ◦ (b ◦ c). Commutativityfollows from AIP. For the second statement, observe that a∗b∗c∗d = 2a−2b+2c−d =c ∗ b ∗ a ∗ d for any a, b, c, d ∈ L. �

3.5. Corollary. In the notation of 3.2, if G is distributive, then BLe(G) is a com-mutative Moufang loop, and if L is a commutative Moufang loop, then DG(L) =Core(L) is distributive. Consequently, LSDI quasigroups of finite exponent are coresof commutative Moufang loops of the same exponent.

Proof. For the first part, it is enough to prove that BLe(G) is commutative. Ac-cording to Jezek, Kepka, Nemec [21], distributive groupoids are trimedial, i.e. every3-generated subgroupoid of a distributive groupoid is medial. Since the proof ofcommutativity of BLe(H) for H medial involves only 3-generated subgroupoids (a◦bis in the subgroupoid generated by a, b, e), we get commutativity in the distributivecase too.

To prove that cores of commutative Moufang loops are distributive, use forinstance the Belousov’s theorem ([1], theorem 9.7) saying that the core of a Moufangloop is distributive, iff the loop satisfies xy2x ≈ yx2y. This is certainly satisfiedin the commutative case. The direct proof is also easy: (x ∗ z) ∗ (y ∗ z) ≈ 2(2x −z) − (2y − z) ≈ 2(2x − z) + (z − 2y) ≈CML (2x − z + z) + ((2x − z) − 2y) ≈2x+ ((−2y) + (2x− z)) ≈Bol (2x− 2y+ 2x)− z ≈ 2(2x− y)− z ≈ (x ∗ y) ∗ z usingdiassociativity and commutativity several times. �

A particularly interesting case is the exponent 3. The identity w3(x, y) ≈ y isindeed equivalent to commutativity (and also to right symmetry) and commutativeLSLDI groupoids are traditionally called distributive symmetric quasigroups. Theycorrespond to commutative Moufang loops of exponent 3 via the operation a ◦ b =aeaeb = eab (see Belousov [1]).

The operation ◦ can be defined equivalently in the following way:

a ◦ b = R−1e (a)Le(b),

where Re is the right translation by e. Consequently, the loop BLe(G) is isotopicto G and the above theorems say that every LSLDI (LSMI, LSDI) quasigroupis isotopic to a Bruck (abelian, commutative Moufang) loop (group). Also, wecan define the operation ◦ for any LSLDI quasigroup (not necessarily with infiniteexponent) in this way. It is proven independently in [1], [36], [30] and maybealso elsewhere that Q(◦) is a Bruck loop for any LSLDI quasigroup Q. However,none of the authors observed that the operation ◦ can be defined polynomially (forfinite exponent) and that there is a there-and-back correspondence as described inTheorem 3.2.


4. Applications

Example. Let L be a Bol loop of an odd exponent n. Then Core(L) has exponentn and the operation of BL1(Core(L)) is a ◦ b = a(n+1)/2ba(n+1)/2 =

√ab√a. Hence

L(◦) is a Bruck loop. In particular, every Bol loop is isotopic to a Bruck loop.

Theorem 3.2 puts together the results of G. Glauberman [16], Kano, Nagao,Nobusawa [24] and V. M. Galkin [14], [15]. They proved that both Bruck loops ofodd order (Glauberman) and finite LSLDI quasigroups (K.N.N. and independentlyGalkin) are solvable, possess the Lagrange property and satisfy a Sylow theorem.(A Bruck loop is said to be solvable, if it can be constructed from the trivial loop by achain of extensions by abelian groups. An LSLDI quasigroup is said to be solvable, ifit can be constructed from the trivial quasigroup by a chain of extensions by medialLSLDI quasigroups (note that the definitions of solvability are in accordance withthe correspondence 3.2). A groupoid G is said to have the Lagrange property, if |H|divides |G| for every subgroupoid H of G. A groupoid G satisfies a Sylow theorem,if for every prime divisor p of G and every subgroupoid H with |H| = pk thereis a subgroupoid K of G containing H such that |K| is a maximal prime powerdivisor of |G|, i.e. |K| = pl so that pl+1 does not divide |G|.) The approach of thefirst two papers is rather similar. K.N.N. use the natural embedding of an LSLDIquasigroup G into the core of the symmetric group over G, while Glaubermanembeds a Bruck loop L into the group under the operation

√ab√a (cf. the previous

example). Then they translate properties of groups into those of the embeddedstructures. Galkin’s approach is more general. He proves that an LSLDI quasigroupG (not necessarily finite) is solvable, iff for every ϕ in the commutant of the leftmultiplication group of G and for every a ∈ G the mapping Laϕ has a uniquefixpoint. An example of a non-solvable (infinite) LSLDI quasigroup is provided:on the set {(x1, x2, x3) ∈ R3 : x2

1 + x22 − x2

3 = −1, x3 > 0} take the operationx ◦ y = 2〈x, y〉x− y, where 〈x, y〉 = −x1y1 − x2y2 + x3y3.

5. Cycles

In this section we study cycles, introduced by N. Nobusawa in [37]. They seemto be an important structural feature of LSLDI groupoids. We gather partiallyfrom the paper [37], but our approach is inovative.

A sequence w1(a, e), w2(a, e), w3(a, e) . . . is called a cycle generated by a overe. If no confusion may happen, we denote the elements a1, a2, a3, . . . . Let orde(a)denote the order of the cycle, that is the least n such that wn(a, e) = e, if suchexists, or ∞ otherwise. A cycle is said to be k-periodic, if wi(a, e) = wi+k(a, e) forevery i.

Let G be an LSLDI groupoid and a, e ∈ G. Put

n = orde(a) and m = ord(LaLe)

(i.e. m is the order of LaLe in the symmetric group over G). For m finite, wehave a2m = e, because w2m(a, e) = (LaLe)m(e) = id(e) = e. Hence n ≤ 2m.On the other hand, for n finite, (LaLe)n(t) = wn(a, e)et = eet = t for any t, so(LaLe)n = id and m divides n. Consequently,(C1) m = ∞, iff n = ∞. If m,n are finite, n = m or n = 2m.

Let us discuss periodicity. Since wk+2m(a, e) = (LaLe)m(wk(a, e)) = wk(a, e), wesee that


(C2) any cycle is 2m-periodic.Moreover, even powers are n-periodic: a2k+n = w2k+n(a, e) = (LaLe)k(wn(a, e)) =(LaLe)k(e) = w2k(a, e) = a2k. How about odd powers?(C3) a cycle is n-periodic, iff an+1 = a.

Indeed, if w(2k+1)+n(a, e) = w2k+1(a, e), we get by left symmetry wn+1(a, e) =a. On the other hand, when an+1 = a, compute a(2k+1)+n = w(2k+1)+n(a, e) =(LaLe)k(wn+1(a, e)) = w2k+1(a, e) = a2k+1.

We call a cycle generated by a over e regular, if orde(a) = orda(e). A regularcycle is n-periodic, because an+1 = wn+1(a, e) = awn(e, a) = aa = a. On the otherhand, if wn+1(a, e) = a, then by cancellation wn(e, a) = a, so n′ = orda(e) ≤ n.Analogously, if wn′+1(e, a) = e, then n ≤ n′. So(C4) a cycle generated by a over e is regular, iff the cycle generated by a over e

is n-periodic and the cycle generated by e over a is n′-periodic.These observations help us to classify cycles.

(1) n is odd. Then m = n and the cycle is regular and thus n-periodic. Indeed,ord(LaLe) = ord(LeLa) and thus m = n = n′.

(2) n is even, m = n/2. Then the cycle n-periodic by (C2).(3) n is even, m = n.

Finally, we prove for every a ∈ G that(C5) ak = al implies m|(k − l).(C6) ak = e, iff n|k.Since (LaLe)kLe = Lak = Lal = (LaLe)lLe (using 0.1(2)), we get (LaLe)k−l =

id. Therefore m divides k − l and this finishes the proof of (C5).For (C6), assume ak = e with k not divisible by n. Due to (C5), m|k, so by (C1)

n = 2m and 2m does not divide k. Then k = lm for l odd and ak = wlm(a, e) =(LaLe)mw(l−2)m(a, e) = w(l−2)m(a, e) = · · · = wm(a, e) 6= e, contradiction.

For the other direction, assume n|k. If n is odd, the statement follows fromn-periodicity. If n is even, we prove by induction that aln = e for every l. Thecase l = 1 follows from the definition and whenever aln = e, we have a(l+1)n =(LaLe)n/2(aln) = (LaLe)n/2(e) = wn(a, e) = e.

5.1. Lemma. Let G be an effective LSLDI groupoid. Then for any a, e ∈ Gorde(a) = ord(LaLe). Consequently, all cycles are regular.

Proof. Let m = ord(LaLe). Then id = (LaLe)m = LamLe, so Lam = Le andeffectivity implies am = e. By (C1), orde(a) = m. Regularity follows from the factthat ord(LaLe) = ord(LeLa). �

5.2. Lemma. Let G be an LSLDI groupoid. Then exp(G) is equal to the leastcommon multiple of all orde(a), a, e ∈ G.

Proof. If there are a, e with orde(a) = ∞, then G has, clearly, infinite exponent. Ifall of them are finite, every orde(a) must divide exp(G) by (C6), because aexp(G) =wexp(G)(a, e) = e. On the other hand, if orde(a) divides k for all a, e ∈ G, we haveby (C6) wk(a, e) = e for all a, e. Consequently wk(x, y) ≈ y and k is a multiple ofexp(G). �

Consequently, an LSLDI groupoid of finite exponent is a quasigroup, iff it hasan odd exponent, iff all its cycles have odd length.


5.3. Proposition. (N. Nobusawa [37]) An effective LSMI groupoid of a finite ex-ponent is a quasigroup.

Proof. Suppose there is a cycle generated by a over e of even length n. By 5.1,orde(a) = ord(LaLe) = n, thus (LaLe)n/2 6= id. Choose c such that (LaLe)n/2(c) 6=c. Look at the cycle generated by aec over c. Using 1.2, we see that

LaecLc = LaLeLcLeLaLc = LaLeLaLcLcLe = LaLeLaLe.

Consequently, ordc(aec) = ord(LaecLc) = ord((LaLe)2) = n/2. On the other hand,observe that wk(aec, c) = (LaLe)k(c) for any k. Indeed, for k odd

wk(aec, c) = (LaecLc)(k−1)/2(aec) = (LaLe)k−1(aec) = (LaLe)k(c)

and for k evenwk(aec, c) = (LaecLe)k/2(c) = (LaLe)k(c).

Consequently, c = wn/2(aec, c) = (LaLe)n/2(c), contradiction. �

The assumptions of the proposition are necessary: the core of the symmetricgroup S3 is effective non-medial, but not a quasigroup; and the core of the additivegroup of integers is medial and effective, but not a quasigroup.

In the last proposition, we focus on the Lagrange property of LSLDI quasigroups.We present a short proof of two particular cases (the proof in general case is quitecomplicated, see remarks in the previous section).

5.4. Proposition. (1) Let G be a finite LSLDI quasigroup and a, e ∈ G. Thenorde(a) divides |G|.

(2) Let G be a finite LSMI quasigroup and H its subquasigroup. Then |H|divides |G|.

Proof. Let H be a subquasigroup of G. Obviously, all right cosets Ha, a ∈ G,have the same number of elements. It is enough to prove for every a, b ∈ G thateither Ha = Hb or Ha,Hb are disjoint. Assume that there are h, k ∈ H so thatha = kb. So let la ∈ Ha and we prove that la ∈ Hb. Indeed, a = hkb and thusla = lhkb. If H is 2-generated (by a, e), then write l, h, k in the generators andobtain lhkb = wm(a, e)wn(a, e)wp(a, e)b = wq(a, e)b for certain m,n, p, q. If G ismedial, use Lemma 1.3 and get la = ((l/h)hk)b for (l/h)hk ∈ H. �

6. Group of displacements

In this section, we mostly follow ideas of N. Nobusawa from [37]. The group ofdisplacements of an LSLDI groupoid G is the subgroup 〈LaLb : a, b ∈ G〉 of the leftmultiplication group over G. It is denoted D(G).

The structure of D(G) determines certain properties of G. For instance, Lemma1.2 says that G is medial, iff D(G) is abelian. If G has exponent n, then D(G) hasexponent n or n/2, and vice versa. H. Nagao showed in [35] that D(G) is the leastnormal subgroup of LMlt(G) in any simple LSLDI groupoid (see also VIII.1.3).

Note that the mapping a 7→ LeLa is a homomorphism of an LSLDI groupoid Ginto the core of D(G), for any e ∈ G. If G is effective, it is an embedding. Theaim of the rest of this section is to show that under mediality this mapping is anisomorphism.

6.1. Lemma. Let G be an LSLDI groupoid with D(G) = {LeLa : a ∈ G}. Then Gis medial.


Proof. The permutation ϕ : x 7→ x−1 is an automorphism of D(G), because(LeLa)−1 = LaLe and thus ϕ is in fact a restriction of the inner automorphismx 7→ LexLe of the symmetric group over G. Consequently, D(G) is abelian and Gis medial. �

6.2. Lemma. Let G be an LSMI quasigroup and e ∈ G. Then D(G) = {LeLa : a ∈G}.

Proof. According to Lemma 1.3, every generator LaLb ∈ D(G) can be written asLeLeLaLb = LeL(e/a)ab, its inverse LbLa too, the identity is LeLe and a composi-tion (LaLb)(LcLd) is equal to a generator L(a/b)bcLd. Hence 〈LaLb : a, b ∈ G〉 ={LeLa : a ∈ G}. �

6.3. Corollary. (N. Nobusawa [37]) Let G be an LSLDI quasigroup. Then G ismedial, iff D(G) = {LeLa : a ∈ G} for a fixed e ∈ G. Moreover, if the equivalentconditions are satisfied, G is isomorphic to the core of D(G).

Proof. Use the preceding two lemmas. The mapping a 7→ LeLa is an isomorphismof G and the core of D(G). �

This is a completely different approach to the result of Corrollary 3.4.Yet another approach is to use the classical result K. Toyoda [57] about represen-

tations of medial quasigroups. It says that Q is a medial quasigroup, iff there is anabelian group Q(+), two commuting automorphisms ϕ,ψ of Q(+) and an elemente ∈ Q such that ab = ϕ(a) + ψ(b) + e for every a, b ∈ Q. From left symmetry weget ϕ(a) +ϕψ(a) +ψ2(b) +ψ(e) + e = b. For a = b = 0 we obtain ψ(e) + e = 0, andthereafter b = 0 yields ϕ(a)+ϕψ(a) = 0 and thus ϕ−1(ϕ(a)+ϕψ(a)) = a+ψ(a) = 0for every a ∈ G. Consequently, ψ(a) = −a. In particular, we obtained the followingtheorem.

6.4. Theorem. Let Q be a left symmetric medial quasigroup. Then there is anabelian group Q(+), an automorphism ϕ of Q(+) and e ∈ Q such that ab = ϕ(a)−b+ e for every a, b ∈ Q.

The theorem includes also the non-idempotent case. Every left distributive quasi-group is idempotent — by left distributivity x · xx ≈ xx · xx and use right cancel-lation — but medial quasigroups need not to be! In the idempotent case, we haveaa = ϕ(a)− a+ e = a, for a = 0 we obtain e = 0 and finally ϕ(a) = 2a, as before.

7. Left ideal decomposition and BL-sum

The following construction is a folklore. It can be done generally for LD leftquasigroups, but we wish to avoid here technical difficulties.

Let G be an LSLD groupoid. Put

(a, b) ∈ δG iff there is n ≥ 0 and c1, . . . , cn ∈ G such that a = c1 . . . cnb.

It is easy to see that δG is an equivalence and its blocks are minimal left idealsof G. Moreover, it is a congruence: whenever a = c1 . . . cnb and d ∈ G, we getda = dc1 . . . cnb = (dc1) . . . (dcn)db and ad = (c1 . . . cnb)d = c1 . . . cnbcn . . . c1d =c1 . . . cn(bcn) . . . (bc1)bd, for every d ∈ G. And δG is the smallest congruence suchthat the corresponding factor satisfies the identity xy ≈ y. Indeed, if G/α doesso, we have (ab, b) ∈ α for every a, b ∈ G and thus α contains δG: whenever


a = c1 . . . cnb, we have (c1 . . . cnb, c2 . . . cnb) ∈ α, (c2 . . . cnb, c3 . . . cnb) ∈ α, . . . ,(cnb, b) ∈ α and so (a, b) ∈ α.

For instance, note that whenever orda(b) is odd, then w(n+1)/2(a, b)a = b, andthus a, b fall into the same minimal left ideal.

The decomposition to minimal left ideals plays a key role in the structure ofLSMI groupoids. B. Roszkowska [47] proved that one can define on every minimalleft ideal a structure of an abelian group so that the left ideal becomes the core ofthe group. Moreover, any LSMI groupoid of finite exponent is isomorphic to a socalled AG-sum of cores of abelian groups. We introduce a BL-sum, a generalizedconstruction of LSLDI groupoids from a family of Bol loops.

7.1. Proposition. Let Li(·, ei), i ∈ I be Bol loops and hij : Li → Lj be mappingssuch that hii is the identity for every i ∈ I and

hjk(hij(a) ∗ b) = hik(a) ∗ hjk(b) for every i, j, k ∈ I,(†)where ∗ is the core operation (thus, in particular, hij is a homomorphism of coresof Li and Lj). Let G be a disjoint union of all Li, i ∈ I and define a multiplicationon G by

a~ b = hij(a) ∗ b for every a ∈ Li, b ∈ Lj ,

Then G(~) is an LSLDI groupoid, each Li is a left ideal of G and ~ is the coreoperation on each Li.

Proof. The latter two statements are obvious. Left symmetry is also easy: a~a~b =hij(a) ∗ hij(a) ∗ b = b for every a ∈ Li, b ∈ Lj , because cores are left symmetric.We prove left distributivity, using, again, the fact that cores of Bol loops are leftdistributive. For every a ∈ Li, b ∈ Lj and c ∈ Lk, a~b~c = hik(a)∗hjk(b)∗c =LD

(hik(a) ∗ hjk(b)) ∗ (hik(a) ∗ c) and (a~ b) ~ (a~ c) = hjk(hij(a) ∗ b) ∗ (hik(a) ∗ c),while the two statements are equal due to the identity (†). �

The groupoid G defined in the proposition is called the BL-sum of Bol loops Li,i ∈ I. If Li’s are abelian groups, then G is exactly Roszkowska’s AG-sum and it ismedial. If Li’s are commutative Moufang loops, then G is distributive.

In constrast to the result of B. Roszkowska on representation of LSMI groupoidsby AG-sums, there is a 15-element LSLDI groupoid which is not a BL-sum of Bolloops. Let G be the conjugation class of involutions in the alternating group A5

(clearly, G has 15 elements) with the core operation. According to D. Joyce [23],G is simple, hence it contains no proper left ideal. Consequently, if G is a BL-sum,then it is isomorphic to the core of a Bol loop L. However, since G is simple, Lmust be also simple. But there is no simple Bol loop with 15 elements.

8. Cores

Let Q be a quasigroup. For every a, b ∈ Q, put

a ∗ b = a(b\a).

The groupoid Q(∗) is called the core of Q and denoted Core(Q). The definition isdue to R. H. Bruck. He used cores in his investigations of isotopy classes of loops.He proved in [5] that isotopic Moufang loops have isomorphic cores.

According to 0.2, cores of Bol loops are LSLDI. Nevertheless, there is a 6-elementnon-Bol LIP loop with left distributive core and there is a 10-element non-Bol LIPloop such that its core satisfies the indentity xy ≈ y (and thus is indeed LSLDI).


It is an open problem (number 10 in the Belousov’s book [1]) to describe all loopswith (left symmetric) left distributive core.

8.1. Proposition. Let G be a group.(1) The core of G is medial, iff G is 2-nilpotent (i.e. G satisfies the identity

[x, y]z ≈ z[x, y], where [x, y] denotes the commutator).(2) The core of G is (left and right) distributive, iff G satisfies the identity

xy2x ≈ yx2y.

The first assertion if due to B. Roszkowska (unpublished). The second statementcan be found in the book of V. D. Belousov [1] (p. 159), in fact, more generally forMoufang loops.

8.2. Proposition. Let L be a Bol loop.(1) The core of L is effective, iff there do not exist a, b ∈ L with Rab = L−1

a L−1b .

(2) The core of L is a quasigroup, iff x 7→ x2 is a permutation of L.

Proof. (1) We show that L∗a = L∗b iff Rba−1 = L−1b La. We have L∗a = L∗b , iff

axa = bxb for every x, iff (*) (axa)a−1 = (bxb)a−1 for every x. Using the Bolidentity, the left side is equal to axaa−1 = ax and the right side is equal to bxba−1,hence the equality (*) is equivalent to x = a−1bxba−1 for every x. The last equalitycan be rewritten as L−1

a LbRba−1 = id and we are done.(2) When L(∗) is a quasigroup, for any a there is a unique b with a = b ∗ 1 = b2.

Hence the latter is true. For the other direction, let√a detote the unique b with

b2 = a. We need to find a solution of x ∗ a = b for any a, b ∈ L. The equal-ity (**) xa−1x = b holds, iff (xa−1x)

√a−1 = b

√a−1, iff

√a−1(xa−1x)

√a−1 =√

a−1b√a−1. Using the Bol identity, the left side is equal to

√a−1xa−1x

√a−1 =√

a−1x(√a−1)2x

√a−1 =

√a−1x

√a−1

√a−1x

√a−1 = (

√a−1x

√a−1)2 (again by

Bol), hence the equality (**) is equivalent to√a−1x

√a−1 =

√√a−1b

√a−1. Thus

x =√a√√

a−1b√a−1

√a is the only possible solution of x ∗ a = b. �

8.3. Proposition. Let G be a group.(1) The core of G is effective, iff the center of G contains no involution.(2) For G finite, the core of G is a quasigroup, iff |G| is odd, iff G contains no

involution.

Proof. (1) For a, b ∈ G, Rab = L−1a L−1

b iff (*) xab = a−1b−1x for every x ∈ G. First,suppose the conditions are met. For x = 1 we get ab = a−1b−1 and thus a2b2 = 1,for x = b we get bab = a−1 and thus (ab)2 = 1. Consequently, a2b2 = ababand ab = ba. We show that the involution ab is in the center: for any x ∈ G,xab = a−1b−1x = (ba)−1x = (ab)−1x = abx. For the other direction, suppose thereis an involution a in the center of G. It means ax = xa for every x ∈ G and thusRa = La. Since a = a−1, we have La = L−1

a and the condition (*) is met for b = 1.(2) is clear. �

The statement 8.3(1) appears in the paper [58] of N. Umaya and 8.3(2) can befound in the paper [24] of Kano, Nagao, Nobusawa. The equivalence 8.2(2) and itsproof is taken from the book [1] of V. D. Belousov.


IV. On loops isotopic toleft distributive quasigroups

We present a solution to the eighth Belousov’s problem from his book [1]: thereis a left distributive quasigroup such that it is isotopic to a loop which is not a Bolloop. This is the example:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 140 0 2 4 1 6 3 8 5 10 7 9 12 11 14 131 3 1 0 11 5 4 12 14 9 8 13 7 2 6 102 1 12 2 6 0 13 3 10 11 14 7 4 5 9 83 5 0 7 3 11 12 10 2 14 13 6 1 9 8 44 2 8 11 14 4 9 0 13 1 5 12 3 6 10 75 7 9 12 0 8 5 13 11 4 1 14 10 3 2 66 4 13 10 7 12 14 6 3 0 11 2 8 1 5 97 9 11 6 10 14 0 2 7 13 12 3 5 8 4 18 6 5 14 13 9 1 11 12 8 4 0 2 10 7 39 10 4 13 12 1 8 14 0 5 9 11 6 7 3 210 8 14 3 2 13 11 7 6 12 0 10 9 4 1 511 14 3 8 4 2 7 9 1 6 10 5 11 13 12 012 13 6 1 5 10 2 4 9 7 3 8 14 12 0 1113 11 10 5 9 7 6 1 8 3 2 4 0 14 13 1214 12 7 9 8 3 10 5 4 2 6 1 13 0 11 14

The classical results of Toyoda [57] and Belousov [1] say that medial quasigroupsare isotopic to abelian groups and (left and right) distributive quasigroups areisotopic to commutative Moufang loops. Moreover, on a distributive quasigroup Qone can define an operation by

a ◦ b = (a/e)(e\b),

where e is a fixed element of Q, and the resulting structure Q(◦) is a commutativeMoufang loop. It is proven independently in [1], [36], [30], in Theorem III.3.2 (forfinite exponent) and maybe also elsewhere that Q(◦) is a Bol loop (with AIP) alsowhenever Q is a left distributive left symmetric quasigroup. One could expect ageneral result, that Q(◦) is a Bol loop for all left distributive quasigroups. Theabove example shows that this idea is incorrect.

For any idempotent quasigroup Q, Q(◦) is a loop, where e is the identity element.Indeed, the mappings a 7→ a/e and b 7→ e\b are permutations (and thus Q(◦)is an isotop of Q) and also a ◦ e = (a/e)e = a and e ◦ b = e(e\b) = b, sincex/x ≈ x\x ≈ x in idempotent quasigroups (LD quasigroups are idempotent). Sinceany loop isotopic to a Bol loop is again a Bol loop (see Belousov [1], for instance),it is enough to search for an LD quasigroup Q such that Q(◦) does not satisfy theBol identity. The above example was found by an automatic model builder SEM


[61] and it is the smallest left distributive quasigroup which is not isotopic to a Bolloop. Also, this example is the smallest such that it does not satisfy the identity

(e/(e\x))e ≈ (e(x/e))\e,which says a simple fact that the solutions of equations a ◦ x = e and x ◦ a = e areidentical. (They indeed must be in Bol loops, because of power associativity.)

Let us note that the eighth Belousov’s problem was solved by V. I. Onoı [38].However, as far as I know, this is the smallest known example and, in fact, nosmaller exists. The negative solution rises a natural question (the ninth Belousov’sproblem): describe those loops to which left distributive quasigroups are isotopic.There are several attempts; let us mention for instance [2] and [3].


V. Equational theory of group conjugation

In this chapter, we study the identities satisfied by group conjugation. Strictlyspeaking, for a group G we put a ∗ b = aba−1 for all a, b ∈ G and we ask, whichidentities hold in all conjugation groupoids G(∗), G a group. Particularly, doesthere exist a finite base of identities they satisfy?

It is easy to see that every G(∗) is an LDI left quasigroup and it turns outthat LDI left quasigroups satisfy the same identities as all G(∗) do (it means, theygenerate the same variety). However, it wasn’t clear, whether they satisfy someadditional identities, not following from LDI. In 1999, D. Larue found in [31] aninfinite independent set of such identities of the form

(xy · y)(x · z) ≈ (xy)(yx · z)(yxy · xy · y)(yx · z) ≈ (yxy · xy)(yyx · z)

(yyxy · yxy · xy · y)(yyx · z) ≈ (yyxy · yxy · xy)(yyyx · z)etc.

We use this result to get a conjecture which implies that the identities of conju-gation are not finitely based. We also provide a couple of properties of the varietygenerated by all G(∗).

1. The variety generated by conjugation

Let FG(X) denote a free group over a set X and F(X) the subgroupoid ofFG(X)(∗) generated by the set X.

1.1. Proposition. (A. Drapal, T. Kepka, M. Musılek)

(1) F(X) is free over X in the variety generated by all G(∗), G a group.(2) If |X| ≤ ℵ0, then F(X) can be embedded into F(2).(3) The groupoid F(2) is right cancellative.


1.2. Lemma. (T. Kepka) Every LDI left cancellative groupoid can be embeddedinto an LDI left quasigroup.


1.3. Lemma. (D. Larue) Every identity satisfied by LDI left cancellative groupoidsis satisfied by LDI left divisible groupoids.


An LDI left quasigroup with pairwise distinct left translations can be embeddedinto the conjugation groupoid of its left multiplication group by mapping an elementa onto its left translation La (see II.1.1). However, the LDI left quasigroup


0 1 2

0 0 1 21 0 1 22 1 0 2

cannot be embedded into any G(∗), since x ∗ y = y iff y ∗ x = x for all x, y ∈ G,while 0 · 2 = 2 and 2 · 0 = 1. Still, the following holds.

1.4. Lemma. Every LDI left quasigroup A(·) is a homomorphic image of a sub-groupoid of FG(A)(∗).Proof. Put B = {wxw−1 : x ∈ A,w ∈ FG(A)}. Clearly, B(∗) is a subgroupoid ofFG(A)(∗).

For every a, b ∈ A, let a\b denote the (unique) c ∈ A such that a · c = b.Now, we define by induction a mapping f : B → A. For x ∈ A put f(x) = x. If

f(w) is already defined, put f(xwx−1) = x·f(w) and f(x−1wx) = x\f(w). Clearly,f is a well-defined surjective mapping.

To prove f(r ∗ s) = f(r) · f(s) for all r, s ∈ B, we use induction on the length ofr. For every s ∈ B and x ∈ A we have f(x ∗ s) = f(xsx−1) = x · f(s) = f(x) · f(s)directly from the definition of f . Now, let r = xwx−1 and f(w ∗ u) = f(w) · f(u)for every u ∈ B. Then f(r ∗ s) = f(xwx−1sxw−1x−1) = x · (f(w) · (x\f(s))) =(x · f(w)) · (x · (x\f(s))) = (x · f(w)) · f(s) = f(xwx−1) · f(s) = f(r) · f(s). Finally,let r = x−1wx and f(w ∗ u) = f(w) · f(u) for every u ∈ B. Then x · f(r ∗ s) =x·f(x−1wxsx−1w−1x) = x·(x\(f(w)·(x·f(s)))) = f(w)·(x·f(s)) = (x·(x\f(w)))·(x · f(s)) = x · ((x\f(w)) · f(s)) = x · (f(r) · f(s)), and we obtain the desired usingleft cancellativity of A. �

The following theorem is an easy consequence of the preceeding lemmas.

1.5. Theorem. The following classes of groupoids generate the same variety:(1) all G(∗), G a group;(2) the subgroupoid 〈x, y〉 of FG(x, y)(∗);(3) LDI left quasigroups;(4) LDI left cancellative groupoids;(5) LDI left divisible groupoids;(6) LDI (both left and right) cancellative groupoids.

Let us denote this variety Conj.

Remark. It follows from Kepka’s lemma 1.2 that LDI left quasigroups generate thequasivariety of LDI left cancellative groupoids (given by LDI and the quasiidentityxy ≈ xz → y ≈ z). However, the quasivariety generated by LDI left divisiblegroupoids is strictly bigger, since there are LDI left divisibles which are not leftcancellative. On the other hand, the quasivariety generated by all conjugationgroupoids is strictly smaller, since it satisfies the quasiidentity xy ≈ y → yx ≈ x.

Let R be a commutative ring with a unit, M a module over R and k ∈ R. Forall x, y ∈ M , put x ◦k y = (1 − k)x + ky. It is easy to see that M(◦k) is an MIgroupoid. It is left cancellative (left divisible, resp.), iff the mapping x 7→ kx isinjective (surjective, resp.) on M .

1.6. Theorem. The variety of medial idempotent groupoids is generated by each ofthe following sets of groupoids:


(1) the set of all Q(◦k), k ∈ Z (where Q denotes the field of rational numbers);(2) the set containing a single groupoid R(◦k), where R is a field extension of

the field Q and k ∈ R is not algebraic over Q.

Proof. Denote M the assumed generating set. Let an identity t(x1, . . . , xn) ≈s(x1, . . . , xn) hold in M. We prove that MI ` t ≈ s.

Let m denote the greater of the depths of t, s. Let t′ (s′, resp.) be a termsuch that the corresponding binary tree is complete of depth m and I ` t ≈ t′

(I ` s ≈ s′, resp.). E.g., we may choose (xy · (uv · u))′ to be (xy · xy)(uv · uu) or(xx · yy)(uv · uu). Clearly, M � t′ ≈ s′ and it is enough to prove that MI ` t′ ≈ s′.

We say that a certain occurence of a variable in a term has property Di, ifwe reach it in the corresponding complete binary tree by turning i-times rightand (m − i)-times left. Let ξij (ζij , resp.) denote the number of variables xj

having the property Di in the term t′ (s′, resp.). It follows that t′(a1, . . . , an) =∑mi=0 k

i(1−k)m−i(∑n

j=1 ξijaj) and s′(a1, . . . , an) =∑m

i=0 ki(1−k)m−i(

∑nj=1 ζijaj)

in R(◦k) for every field R and k, a1, . . . , an ∈ RWe show that ξij = ζij . Put δij = ξij − ζij . From M � t′ ≈ s′ it follows

that∑m

i=0

∑nj=1 k

i(1 − k)m−iaj δij = 0 for every assumed k and a1, . . . , an. Fixl ∈ {1, . . . , n} and put al = 1 and aj = 0 for j 6= l. In the first case, choosingk = 2, . . . ,m + 2, we obtain m + 1 linear equations for δ0l, . . . , δml. It is easy tosee that the only solution in Q is δ0l = · · · = δml = 0. In the second case, since nopolynomial can have k as its root, we get all δil equal 0 directly.

We have proven that t′ and s′ may differ only by permuting occurences of vari-ables satisfying certain Di. So it is enough to prove the following. If u, v are twoterms such that their corresponding binary trees are complete of depth m and theydifer just by transposing two occurences of the variables x, y satisfying Di for somei, then MI ` u ≈ v.

By induction on m. For m ≤ 2 it is clear. If m > 2, write u = u1u2 · u3u4 andv = v1v2 · v3v4. Without loss of generality, suppose x occurs first.

(a) Both x, y occur in the left (right, resp.) subterm of u. Use inductionassumption on the left (right, resp.) subterm.

(b) x occurs in u1, y occurs in u3. By mediality, u ≈ u1u3 · u2u4, by inductionassumption on u1u3 we get u ≈ v1v3 ≈ v2v4 and again mediality yieldsMI ` u ≈ v.

(c) x occurs in u1, y occurs in u4. Let w be the term obtained from u1u2

by transposition of the occurence of x and the rightmost variable with Di.Certainly, then this x is in the right subterm of w. Now, use inductionassumption to get MI ` u1u2 ≈ w, then proceed as in (b) on the termw · u3u4 to get w′ ≈ v3v4 (transposing that x and y) and finally, useinduction assumption on w′ to get MI ` w′ ≈ v1v2.

(d) x occurs in u2, y occurs in u3. Proceed as in (c). We must take special careon the case when x is the rightmost variable in u2 and y is the leftmostvariable in u3.

�

1.7. Corollary. The variety of medial idempotent groupoids is a subvariety of thevariety Conj.

Proof. R(◦k) is an LDI left quasigroup for every field R and k ∈ Rr {0}. �


It follows from Proposition 1 that Conj is a subvariety of the variety generated byLDI right cancellative groupoids. We show that the inclusion is proper (and thusanswer a question of A. Drapal, T. Kepka, M. Musılek from [12]). The followinggroupoid is an LDI right quasigroup which does not satisfy the first Larue’s identityfrom the list in the introduction: ((0 · 1) · 1)(0 · 2) = 4 6= 3 = (0 · 1)((1 · 0) · 2).

0 1 2 3 4 5 6 7 8

0 0 2 6 6 6 0 2 0 21 3 1 5 5 5 3 1 3 12 6 7 2 4 3 8 0 1 53 8 5 4 3 2 1 7 6 04 1 0 3 2 4 6 5 8 75 5 3 1 1 1 5 3 5 36 2 6 0 0 0 2 6 2 67 7 4 8 8 8 7 4 7 48 4 8 7 7 7 4 8 4 8

The groupoid was found by a computer search using an automated model builderSEM by J. Zhang and H. Zhang [61] and it is a smallest such example. However,the following problem remains unsolved.

Problem. (A. Drapal, T. Kepka, M. Musılek; D. Larue) Do LDI right cancella-tive groupoids generate the variety of LDI groupoids? In particular, are free LDIgroupoids right cancellative? How about LDI right divisible groupoids?

2. The equational theory of conjugation

Let A be an LDI groupoid and c ∈ A such that cx 6= c for all x ∈ A r {c}.We denote A[c] the following groupoid: the universe is A ∪ {c′} for c′ /∈ A and theoperation of A[c] is an extension of the operation on A defined by c′x = cx for allx ∈ A, xc′ = xc for all x ∈ Ar {c} and c′c′ = cc′ = c′. It is easy to check that A[c]

is an LDI groupoid.

A[c] c′ c x

c′ c′ c cxc c′ c cxx xc xc x

In further text, let R always be an extension of the field Q containing all squareroots of positive integers. For k, c ∈ R, we denote R[c]

k the groupoid R(◦k)[c].

2.1. Theorem. (D. Larue) There are (logically) independent identities εn, n ∈ N,satisfied in Conj but not in the variety of LDI groupoids. Moreover, there existkn ∈ R such that for every c ∈ R the groupoid R

[c]kn

satisfies all the identities εi,i ∈ N r {n}, but not the identity εn.


Remark. The identities εn can be taken e.g. from the list in the introductionomitting every (2`)2-th one, ` ∈ N. In such a case one can choose kn = 1 + 1√

`,

whenever εn is `-th in the list.


2.2. Conjecture. (D. Larue) Let c ∈ R. Let an identity ε hold in Conj. ThenR

[c]k 6� ε for at most finitely many k ∈ R.

The theorem and validity of the conjecture directly imply that Conj is not finitelybased — if there were a finite base B, then R

[c]k � B for all but finitely many k,

hence R[c]k � εn for all n and all but finitely many k, so we get a contradiction with

the theorem.

Below, x1, x2, . . . always denote variables. By n-ary term t we mean a groupoidterm in variables x1, . . . , xn. Without loss of generality, suppose xn is the rightmostvariable of t. Denote t1 the left subterm of t, t2 the left subterm of the right subtermof t and so forth. Hence, t is a product t1(t2(. . . (tmxn))) for some m.

Let ϕ : A[c] → A be the mapping defined by ϕ(x) = x for all x ∈ A and ϕ(c′) = c.It is easy to check that ϕ is an onto homomorphism.

2.3. Lemma. Let t, s be n-ary terms such that A � t ≈ s. Then A[c] 6� t ≈ s, iffthere are a1, . . . , an ∈ A[c] such that one of the following conditions takes place:

(1) t(a1, . . . , an) = c′ and s(a1, . . . , an) = c;(2) t(a1, . . . , an) = c and s(a1, . . . , an) = c′.

Proof. Take a1, . . . , an ∈ A[c] such that u = t(a1, . . . , an) 6= s(a1, . . . , an) = v. Byassumption, ϕ(u) = ϕ(v), hence {u, v} = {c, c′}. �

First of all, observe that for a, b ∈ A[c] we have ab = c′, iff b = c′ and a ∈{c, c′}. Thus, for n-ary terms t = t1(t2(. . . (tmxn))) and s = s1(s2(. . . (slxn))),the condition (1) is equivalent to an = c′ and ti(a1, . . . , an) ∈ {c, c′} for all i andsi(a1, . . . , an) /∈ {c, c′} for at least one i. Let bi = ϕ(ai), i = 1, . . . , n. If (1)holds, then bn = c and ti(b1, . . . , bn) = c for all i and si(a1, . . . , an) 6= c for at leastone i. On the other hand, given such b1, . . . , bn ∈ A, then a1 = b1, . . . , an−1 =bn−1, an = c′ satisfy the condition (1). Consequently, (1) is equivalent to existenceof b1, . . . , bn−1 ∈ A such that

(i) ti(b1, . . . , bn−1, c) = c for all i;(ii) si(b1, . . . , bn−1, c) 6= c for at least one i.

Note that for every n-ary term r there are polynomials ρ1, . . . , ρn over the fieldR so that in R(◦k)

r(a1, . . . , an) =n∑

j=1

ρj(k)aj .

2.4. Lemma. Let t = t1(t2(. . . (tmxn))), s = s1(s2(. . . (slxn))) be n-ary terms suchthat R(◦k) � t ≈ s. Let τi1, . . . , τin be the polynomials corresponding to the termsti and σi1, . . . , σin be the polynomials corresponding to the terms si. Then thefollowing conditions are equivalent:

(1) There are a1, . . . , an ∈ R[c]k so that t(a1, . . . , an) = c′ and s(a1, . . . , an) = c.

(2) There are b1, . . . , bn−1 ∈ R so that ti(b1, . . . , bn−1, c) = c for all i andsi(b1, . . . , bn−1, c) 6= c for at least one i.


(3)

rank

τ11(k) . . . τ1,n−1(k). . .

τm1(k) . . . τm,n−1(k)

< rank

τ11(k) . . . τ1,n−1(k)

. . .τm1(k) . . . τm,n−1(k)σ11(k) . . . σ1,n−1(k)

. . .σl1(k) . . . σl,n−1(k)

Proof. The equivalence of (1) and (2) was proved above (in general). Now, (2) isequivalent to the fact that the system of m linear equations over the field R τ11(k) . . . τ1,n−1(k)

. . .τm1(k) . . . τm,n−1(k)

∣∣∣∣∣∣(1− τ1n(k))c

. . .(1− τmn(k))c

has a solution b1, . . . , bn−1 ∈ R, but the system of l linear equation over the field Rσ11(k) . . . σ1,n−1(k)

. . .σl1(k) . . . σl,n−1(k)

∣∣∣∣∣∣(1− σ1n(k))c

. . .(1− σln(k))c

has not the solution b1, . . . , bn−1. This gives the equivalence of (2) and (3). �

2.5. Corollary. Under the same assumptions, define matrices T (k) = (τij(k)),

S(k) = (σij(k)) and U(k) =(

T (k)S(k)

). Then the following conditions are equivalent:

(1) R[c]k � t ≈ s;

(2) rank(T (k)) = rank(S(k)) = rank(U(k)).

So, we got an equivalent form of Larue’s conjecture.

2.6. Conjecture. Let an identity t ≈ s hold in the variety Conj. Then

rank(T (k)) < rank(U(k))

for at most finitely many k ∈ R.

3. On the role of idempotency

It is interesting that the equational theories of LD groupoids and LD left can-cellative groupoids coincide (P. Dehornoy [10]), but the equational theory of LDIgroupoids is strictly smaller than that of LDI left cancellative groupoids. On theother hand, the equational theory of LD left cancellative groupoids is strictly smallerthen that of LD left divisibles, but the equational theories of LDI left cancellativeand LDI left divisible groupoids coincide. (P. Dehornoy proved even more — freeLD groupoids are left cancellative.)

Consider the first Larue’s identity ε1 (xy · y)(xz) ≈ (xy)(yx · z) (found indepen-dently in [31] and [12]).

3.1. Proposition. (1) ε1 is satisfied in every LD left divisible groupoid.(2) ε1 is satisfied in every LDI left cancellative groupoid.(3) There is an LD left cancellative groupoid which does not satisfy ε1.


Proof. (1) Let A be an LD left divisible groupoid and a, b, c ∈ A. Take d ∈ A suchthat c = bd. Then (ab · b)(ac) = (ab · b)(ab · ad) = (ab)(b · ad) = (ab)(ba · bd) =(ab)(ba · c).

(2) b((ab ·b)(ac)) = (b(ab ·b))(b ·ac) = ((b ·ab)(bb))(ba ·bc) = ((b ·ab)b)((ba ·b)(ba ·c)) = (b · ab)(b · (ba · c)) = b((ab)(ba · c)) and use left cancellativity. Alternatively,verify the identity for conjugation groupoids.

(3) Since the equational theories of LD groupoids and LD left cancellative grou-poids coincide, it is enough to check that R[c]

k 6� ε1 for every field R, c ∈ R andk ∈ { 1

2 , 2}. However, it is not easy to find a particular example of an LD leftcancellative groupoid which does not satisfy ε1. Here it is: take a group G, anelement e ∈ G and an endomorphism f of G such that ef2(x) = f2(x)e for all x ∈ Gand ef(e)e = f(e)ef(e). Put x ◦ y = xf(y)ef(x−1) for all x, y ∈ G. Then G(◦) isan LD groupoid (see [28], IV.2.2). Moreover, G(◦) is left cancellative (left divisible,resp.) iff f is injective (surjective, resp.). Now, take for G the symmetric group onω, e = (0 1) and f the endomorphism given by the transformation i 7→ i+ 1 of ω.Then G(◦) is an LD left cancellative groupoid and choosing a = (0 1 2), b = (0 2 1)and c = (0 1), one can verify G(◦) 6� ε1. �


VI. Subdirectly irreducible non-idempotentleft distributive left quasigroups

It is well-known that a groupoid G is subdirectly irreducible, iff the intersection ofall non-trivial congruences is non-trivial. The intersection is called the monolith ofGand denoted µG. The aim of the present chapter is to describe the structure of non-idempotent subdirectly irreducible left distributive left quasigroups. We generalizemost of the results from the paper [18] of Jerabek, Kepka and the author, wherenon-idempotent subdirectly irreducible LSLD groupoids were described.

Unfortunately, we have no idea how idempotent subdirectly irreducible LD leftquasigroups look like. For our considerations, it is essential that a non-idempotentelement is present. The task in the idempotent case is probably far more compli-cated. Even the charaterization of subdirectly irreducible LSMI groupoids, madeby B. Roszkowska in [48], is rather complicated, though they are quite few. Sim-ple LDI left quasigroups were characterized by D. Joyce [23] (see an example inChapter II).

Let Aut(G) denote the automorphism group of a groupoid G and let Autn(G) ={ϕ ∈ Aut(G) : ϕn = id}. It is easy to check that Autn(G) is a left n-symmetricsubgroupoid of the conjugation groupoid of Aut(G).

Note that a subgroupoid of a left quasigroup is not necessarily left quasigroup(it is indeed left cancellative, but not necessarily left divisible). We will thus usethe notion of left subquasigroup. Subgroupoids of finite left quasigroups (or, moregeneraly, of left n-symmetric groupoids) are indeed left subquasigroups.

1. Description

Let G be an LD left quasigroup. We denote IdG the left ideal of idempotentelements and KG = Gr IdG the left ideal of non-idempotent elements of G.

Let I be a left ideal of G. We put (a, b) ∈ ρI , iff either a = b, or a, b ∈ I and(a, b) ∈ ipG. Clearly, ρI is a congruence.

1.1. Lemma. Let G be a non-idempotent subdirectly irreducible LD left quasigroup.Then KG contains no proper left ideal. In particular, it contains no definable propersubset.

Proof. If I is a proper left ideal, then ρI and ρKGrI are two non-trivial congruenceswith trivial intersection. The rest follows from II.3.1: definable subsets are leftideals. �

Let γk be an equivalence generated by all pairs (a, ak+1), a ∈ G. According toII.3.1, γk ⊆ ipG is a congruence (which glues together every k-th element in eachcycle). Moreover, if l|k, then γk ⊆ γl. Note that the congruence lattice of Cn

consists of (pairwise different) congruences γk, k|n.

1.2. Proposition. Let G be a non-idempotent subdirectly irreducible LD left quasi-group. Then there is a prime p and a positive integer r such that


(1) each non-trivial (i.e. with more then one element) block of ipG is a circleof length pr;

(2) the monolith of G is γpr−1 .Moreover, if G is n-LSLD, then pr|n.

Proof. (1) First, assume that all non-trivial ipG-blocks are infinite. Then γk 6= γl

for every k 6= l, and so there is an infinite decreasing sequence γ2 ⊃ γ4 ⊃ · · · ⊃ γ2k ⊃. . . with trivial intersection. Hence G is not subdirectly ireducible, a contradiction.

So, let n be the least number such that there is a non-trivial ipG-block whichis a cycle of length n. Then, according to II.3.1, Kn = {a ∈ G : an+1 = a} is a(non-empty) left ideal and thus Kn = KG. It means that all non-trivial blocks arecycles of length n. If n = kl for k, l relatively prime, then γk and γl are non-trivialcongruences with trivial intersection, a contradiction. Hence n is a prime power.

(2) Clearly, ipG = γ1 ⊃ γp ⊃ · · · ⊃ γpr−1 ⊃ γpr = idG, so µG ⊆ γpr−1 . Assumethat it is a proper subcongruence. Then there is an ipG-block B such that µG isidentical on B and thus the set I = {a ∈ G : (a, b) ∈ µG for some b 6= a} is aproper subset of KG. However, it is easy to see that I is a left ideal, so we have acontradiction with 1.1. �

An LD left quasigroup satisfiying the condition (1) of Proposition 1.2 is said tohave type pr.

1.3. Lemma. Let K be an idempotent-free LD left quasigroup and I be a leftsubquasigroup of the conjugation groupoid of Aut(K). Let G be a disjoint union ofI and K. Then the following conditions are equivalent.

(1) The operations of I and K can be extended onto G so that G becomes anLD left quasigroup with ϕ · u = ϕ(u) for all ϕ ∈ I, u ∈ K.

(2) LKu ϕ(LK

u )−1, (LKu )−1ϕLK

u ∈ I for all ϕ ∈ I, u ∈ K (where LKu denotes the

left translation of u in K).If the conditions are satisfied, the operation on G is uniquely determined and u ·ϕ =LK

u ϕ(LKu )−1 for all ϕ ∈ I, u ∈ K.

Moreover, G is n-LSLD, iff K is n-LSLD and ϕn = id for every ϕ ∈ I.

Proof. For every u ∈ K, we need to extend the left translation LKu of u in K to

a left translation LGu of u in G. From left distributivity we get for every u, v ∈ K

and ϕ ∈ I the identity u(ϕv) = (uϕ) · (uv); it can be rewritten as LKu ϕ = LG

u (ϕ)LKu

and thus we must define LGu (ϕ) = LK

u ϕ(LKu )−1. Of course, this is possible, iff

LKu ϕ(LK

u )−1 ∈ I.Now, it is straightforward to check that the resulting groupoid is left distribu-

tive. The left translation by an element ϕ ∈ I is a permutation on both I (byassumptions) and K (since it acts as the automorphism ϕ). Moreover, (LG

ϕ )n = idiff ϕn = id. The left translation by an element u ∈ K is a permutation on K (byassumptions) and it is a permutation on I, iff LK

u ϕ(LKu )−1 has a preimage for every

ϕ ∈ I, i.e. iff (LKu )−1ϕLK

u ∈ I. Moreover, (LGu )n = id iff (LK

u )n = id. �

The groupoid G from Lemma 1.3 will be denoted by I tK.

I tK ψ vϕ ϕψϕ−1 ϕ(v)u LK

u ψ(LKu )−1 uv


The groupoid Aut(K)tK will be called the full extension of K and denoted byFull(K). The groupoid Autn(K) tK will be called the full n-extension of K anddenoted by Fulln(K).

Now, we are ready to prove the main theorem, describing non-idempotent sub-directly irreducible LD left quasigroups.

1.4. Theorem. Let G be a non-idempotent subdirectly irreducible LD left quasi-group. Then G embeds into Full(KG) via an injective homomorphism Φ definedby

Φ(u) = u for u ∈ KG and Φ(a) = La|KGfor a ∈ IdG.

Moreover, if G is n-LSLD, then it embeds into Fulln(KG).

Proof. First, we show that the mapping Φ is a homomorphism. Let a, b ∈ G, weprove that Φ(ab) = Φ(a)Φ(b). The case a, b ∈ KG is trivial. For a, b idempotent, itfollows from left distributivity that LabLa = LaLb and thus Lab = LaLbL

−1a . For a

idempotent and b ∈ KG, we have Φ(a) = La|KG, Φ(b) = b, thus both sides are equal

to ab. Finally, for a ∈ KG and b idempotent, Φ(ab) = Lab|KGand the right side is

aLb|KG= La|KG

Lb|KGL−1

a |KG= Lab|KG

according to left distributivity again.Put (a, b) ∈ α iff either a = b, or a, b are idempotent and au = bu for every

u ∈ KG. Obviously, α is an equivalence and we show that α is a congruence.Indeed, if (a, b) ∈ α, then ca · u = LcLaL

−1c (u) = LcLbL

−1c (u) = cb · u for every

u ∈ KG and thus (ca, cb) ∈ α. Similarly, ac·u = LaLcL−1a (u) = LbLcL

−1b (u) = bc·u

for every u ∈ KG, hence (ac, bc) ∈ α for every c ∈ IdG. If c ∈ KG, then ac = bcand thus (ac, bc) ∈ α too. Since the intersection α and ipG is trivial, α is necessarytrivial. It means that La|KG

6= Lb|KGfor all idempotent elements a, b with a 6= b.

The last claim is obvious. �

1.5. Lemma. Let G and H be non-idempotent LD left quasigroups, H ≤ G andKH = KG. If H is subdirectly irreducible, then G is so.

Proof. According to Theorem 1.4, we may assume that G and H are subgroupoidsof Full(KG). Let KG be of type pr. It means that γpr−1 is the monolith of H and weneed to prove that it is also the monolith of G. Let α be a non-trivial congruenceof G. Clearly, α∩H2 is a congruence of H, hence, if we prove that it is non-trivial,we get γpr−1 ⊆ α and we are done.

Assume that (a, b) ∈ α for some a ∈ GrH. Then a is idempotent and if b is so,we have La|KG

6= Lb|KGby Theorem 1.4, hence there is u ∈ KG such that au 6= bu

and thus (au, bu) ∈ α ∩H2. So assume that b ∈ KG. If a 6= Lb|KG, there is again

u ∈ KG such that au 6= bu and thus (au, bu) ∈ α ∩ H2. So let a = Lb|KG. Then

ba = a and bb 6= b. Hence (ba, bb) = (a, bb) ∈ α and (b, bb) ∈ α ∩H2. �

1.6. Lemma. Let K be an idempotent-free LD left quasigroup of type pr and I be aleft subquasigroup of the conjugation groupoid of Aut(K) such that every non-trivialI-invariant congruence of K contains γpr−1 (I-invariant means that (a, b) ∈ αimplies (ϕ(a), ϕ(b)) ∈ α for every ϕ ∈ I). Then I tK is subdirectly irreducible.

Proof. Assume that α is a non-trivial congruence of I t K. Indeed, α ∩ K2 isan I-invariant congruence of K (I-invariancy is obtained by left multiplication byelements of I), hence it is enough to prove that α is non-trivial. First, if thereare ϕ,ψ ∈ I with (ϕ,ψ) ∈ α, then there is u ∈ K such that ϕ(u) 6= ψ(u) andthus (ϕ(u), ψ(u)) ∈ α ∩ K2. So assume that (ϕ, u) ∈ α for ϕ ∈ I and u ∈ K. If


ϕ 6= Lu|K , there is again v ∈ K such that ϕ(v) 6= uv and thus (ϕ(v), uv) ∈ α∩K2.So assume ϕ = Lu|K . Then uϕ = ϕ and uu 6= u and so (u, uu) ∈ α ∩K2. �

We define mappings ok by ok(a) = ak. Since (xy)k ≈LD xyk ≈LI xkyk, ok is an

automorphism of every LD left quasigroup. Moreover, it commutes with any otherautomorphism ϕ, since ϕ(ak) = ϕ(a)k for every a. Consequently, each ok forms aone-element left ideal in Full(K) for any K.

We put Aut−(K) = Aut(K) r {ok : k ∈ N} and Full−(K) = Aut−(K) tK and,similarly, also Aut−n (K) and Full−n (K).

The following proposition establishes conditions under which an idempotent-freeLD left quasigroup possesses a subdirectly irreducible extension (by idempotentelements).

1.7. Proposition. (A) Let K be an idempotent-free LD left quasigroup of type pr.The following statements are equivalent:

(1) There is a subdirectly irreducible LD left quasigroup G with KG = K.(2) Full(K) is subdirectly irreducible.(3) Full−(K) is subdirectly irreducible.

The three statements are implied by(4) Every non-trivial Aut(K)-invariant congruence of K contains γpr−1 .

(B) Let K be an idempotent-free n-LSLD groupoid of type pr. The following state-ments are equivalent:

(1) There is a subdirectly irreducible n-LSLD groupoid G with KG = K.(2) Fulln(K) is subdirectly irreducible.(3) Full−n (K) is subdirectly irreducible.(4) Every non-trivial Autn(K)-invariant congruence of K contains γpr−1 .

Proof. The implications (2) ⇒ (1) and (3) ⇒ (1) are trivial, (1) ⇒ (2) follows fromTheorem 1.4 and Lemma 1.5 and (4) ⇒ (2) follows from Lemma 1.6.

To prove (2) ⇒ (3), assume that α is a non-trivial conguence of Full−(K). It isenough to prove that α∪ id is a congruence of Full(K) — in such a case α containsγpr−1 and we are done. Let (a, b) ∈ α. We need to check that (oka, okb) ∈ α ∪ idand (aok, bok) ∈ α∪ id for every k. In the letter case, aok = ok = bok, since {ok} isa left ideal, hence we are done. In the former case, observe that okc = ck for everyc (for c is idempotent, okc = c because ok commutes with any automorphism, andfor c non-idempotent by definition) and thus (a, b) ∈ α implies (ak, bk) ∈ α. In thecase (B) proceed similarly.

Finally, we prove (2) ⇒ (4). Assume that α is a non-trivial Autn(K)-invariantcongruence of K. Put

β = α ∪ {(ϕ,ψ) ∈ Autn(K)2 : (ϕ(u), ψ(v)) ∈ α for every (u, v) ∈ α}.

We prove that β is a congruence of Fulln(K) — then (2) yields that γpr−1 ⊆ α.Clearly, β is an equivalence. First, let (u, v) ∈ α. Then (uw, vw), (wu,wv) ∈ βfor every w ∈ K, because α is a congruence, (ϕu, ϕv) ∈ β, because α is Autn(K)-invariant and (uϕ, vϕ) = (LuϕL

−1u , LvϕL

−1v ) = (LuϕL

n−1u , LvϕL

n−1v ) ∈ β, be-

cause (x, y) ∈ α clearly implies (uϕ(u · · ·ux), vϕ(v · · · vy)) ∈ α. Now, let (ϕ,ψ) ∈β ∩ Autn(K). Then (ϕu, ψu) ∈ α follows immediately from the definition of β,(ϕρϕ−1, ψρψ−1) = (ϕρϕn−1, ψρψn−1) ∈ β follows from the definition (apply the


condition n-times) and from Autn(K)-invariancy of α, (ρϕρ−1, ρψρ−1) ∈ β simi-larly and (uϕ, uψ) is just a particular case for ρ = Lu. �

2. Examples

Let k, ` be positive integers and C(k, `) = k × C` (where C` denotes the cycleof length `). We denote P (k, `) the group of all permutations π of the set C(k, `)such that π(i, a) = (j, b) implies π(i, a2) = (j, b2) (where x2 refers to the succesorof x in the cycle). Let Pn(k, `) = {π ∈ P (k, `) : πn = id}.

2.1. Proposition. Let G be a non-idempotent subdirectly irreducible LD left quasi-group of type pr with k non-trivial ipG-blocks. Then

|G| ≤ kpr + |P (k, pr)| = kpr + k!(pr)k.

Moreover, if G is n-LSLD, then |G| ≤ kpr + |Pn(k, pr)|.

Proof. It follows from the embedding established in Theorem 1.4. Indeed, everyautomorphism of KG satisfies the condition from the definition of P (k, `). �

The following example shows that the upper bound on the number of idempotentelements is best possible. For every k and pr, we construct a subdirectly irreducibleLD left quasigroup G of type pr with k non-trivial ipG-blocks such that |G| =kpr + |P (k, pr)|. The bound is also best possible in case of n-LSLD groupoids,provided that k ≥ n (and, indeed, pr|n — otherwise, there is no such n-LSLDgroupoid).

Example. Let K = C(k, pr) and put (i, a) · (j, b) = (j, b2) for every i, j ∈ k anda, b ∈ Cpr . It is easy to see that K is an LD left quasigroup with Aut(K) = P (k, pr)and Autn(K) = Pn(k, pr). We prove that Full(K) and Fulln(K) are subdirectlyirreducible using Proposition 1.7. Let α be a non-trivial Autn(K)-invariant congru-ence of K. Let ((i, a), (j, b)) ∈ α. If i 6= j, we use the permutation π that fixes allcycles except for the j-th one, and it shifts the j-th cycle by one. Autn-invariancyyields that ((i, a), (j, b2)) ∈ α and thus ((j, b), (j, b2)) ∈ α. Hence we may assumethat ((i, a), (i, b)) ∈ α for some i, a, b. Then, clearly, α contains also the restrictionof γpr−1 to the i-th cycle. Hence we may use for every j 6= i a permutation thatsends the i-th cycle onto the j-th one (there is such a permutation in Pn(k, pr) forany k ≥ n) and we get γpr−1 ⊆ α.

The bound is not necessarily reached in the case k < n. For example, we provethat there is no subdirectly irreducible 3-LSLD groupoidG (of type 3) with two non-trivial ip-blocks and |P3(2, 3)| = 6 idempotent elements. Indeed, every permutationfrom P3(2, 3) sends elements of the first cycle onto the first cycle and elements ofthe second one onto the second one. Since the only two-element 3-LSLDI groupoidis

a ba a bb a b

both KG and Full3(KG) contain two proper left ideals: namely, each of the two ipG-blocks. Hence, according to Lemma 1.1, Full3(KG) is not subdirectly irreducible.

In the rest of the section, we characterize all small subdirectly irreducible LDleft quasigroups. We proceed by the number of ip-blocks.


One ip-block. There are 2pr

subdirectly irreducible LD left quasigroups of type pr

with one non-trivial ip-block. The ip-block is isomorphic to Cpr . Since Cpr is sub-directly irreducible, any idempotent extension is subdirectly irreducible. Clearly,Aut(Cpr ) is the cyclic group of order pr generated by the mapping x 7→ x2. Inparticular, it is abelian, hence every idempotent element of Full(Cpr ) forms a one-element left ideal. Consequently, I tCpr is a subdirectly irreducible LD left quasi-group for every I ⊆ Aut(Cpr ).

Two ip-blocks. Let G be a subdirectly irreducible LD left quasigroup with twoip-blocks. According to the structure of the two-element LD left quasigroup (seeabove) and due to the fact that the only automorphisms of cycles are rotations, KG

must consist of two cycles of length pr (let us denote them B1, B2 with elements0, . . . , pr − 1 and 0, . . . , pr − 1) with the following multiplication table:

KG c b

a c2 dj+1

b ck+1 d2

Now, if j 6= k, then the set B1 is definable by {y : (∃x) xy = yj+1} in G, acontradiction. Hence j = k and it is easy to see that Aut(KG) = P (2, pr). UsingLemma 1.6, one can check the following facts. If k 6= 1, then I tK is subdirectlyirreducible, iff I contains a permutation switching B1 and B2. If k = 1, then I tKis subdirectly irreducible, iff I contains a permutation switching B1 and B2 anda permutation which rotates the cycles by a different angle each. In the case ofpr = 2, the situation is quite simple. The conjugation groupoid Aut−(KG) lookslike this:

Aut−(KG) a b c d e fa, b a b d c f ec, d b a c d f ee, f b a d c e f

where a = (0 1), b = (0 1), c = (0 0)(1 1), d = (0 1)(1 0), e = (0 0 1 1) andf = (0 1 10). Clearly, no one-element extension of KG is subdirectly irreducible,hence we may extend only by one of the subgroupoids {a, b}, {c, d}, {e, f} or theircombinations. If j = 1, the first and the second one do not yield a subdirectlyirreducible I tKG, for j = 0 only the first one doesn’t (no switching permutation).And any combination does the job. Consequently, there are 4·5 = 20 subdirectly ir-reducible extensions in the case j = 1 and 4·6 = 24 in the case j = 0 (multiplicationby 4 because of additional extensions by o1, o2, both or none).

Three ip-blocks. All subdirectly irreducible 2-LSLD groupoids with at mostthree non-trivial ip-blocks were described in [18]. The table shows the numberof isomorphism classes of subdirectly irreducible 2-LSLD groupoids with a givennumber of non-idempotent (2, 4 and 6) and idempotent elements.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 201 2 10 0 1 2 3 4 20 0 0 0 0 0 4 8 4 8 16 8 6 12 6 4 8 4 2 4 2


Summary. In the following table, we summarize the number of subdirectly ir-reducible LD left quasigroups (2-LSLD groupoids, resp.) with a given number ofnon-idempotent elements. The structure is described as the number of cycles of agiven type. The blank spaces haven’t been computed.

|KG| structure |IG| ≤ # 2-LSLD1 0 02 1× C2 2 4 43 1× C3 3 8 04 1× C4 4 16 0

2× C2 8 44 125 1× C5 5 32 06 1× C6 6 64 0

2× C3 18 03× C2 48 96

7 1× C7 7 128 08 1× C8 8 256 0

2× C4 32 04× C2 384


VII. On varieties of left distributiveleft idempotent groupoids

The purpose of this chapter is to continue recent investigations of P. Jedlicka [17]on LDLI groupoids. We apply his result to compute the lattice of subvarieties ofLDLI groupoids satisfying xn+1 ≈ x for some n modulo the lattice of subvarieties ofLDI groupoids (see Theorem 1.4). A very particular case, the lattice of subvarietiesof left symmetric left distributive groupoids, was computed in a similar way by T.Kepka in [27]. This is, in my opinion, an appropriate generalization. Note that theresult can be applied also to the varieties of n-LSLD groupoids.

In the second part, we show some properties of subvarieties satisfying indetitiesxm+n ≈ xm.

1. Varieties satisfying xn+1 ≈ x

We say that a variety V of groupoids has exponent n, if n is the least positiveinteger such that xn+1 ≈ x holds in V. (Of course, such n does not necessarilyexist, however, many important varieties, for instance n-LSLD groupoids, havefinite exponent.)

Recall that Cn denotes the groupoid on the set {0, . . . , n−1} with the operationab = b+ 1 mod n.

1.1. Lemma. Let a variety V have exponent n. Then n is the greatest positiveinteger such that Cn ∈ V.

Proof. It follows from the fact that Ck satisfies xl+1 ≈ x, iff k divides l. Note thatthe presence of Ck in a variety depends only on the one-variable identities of thevariety, because Ck satisfies an identity, iff it satisfies its substitution instance withall variables unified. �

1.2. Lemma. Let G be an LDLI groupoid with xn+1 ≈ x. Then

(1) any block of ipG is isomorphic to Ck for some k|n;(2) Cn is a homomorphic image of G, if and only if G is isomorphic to Cn ×

(G/ipG).

Proof. (1) follows from Lemma II.1.3. For (2), choose a projection g : G→ Cn andput f(x) = (g(x), x/ipG). Then f : G→ Cn × (G/ipG) is a homomorphism. Sincethere is a homomorphism Ck → Cl iff l|k, every block of ipG is isomorphic to Cn

(g restricted to a block of ipG is a homomorphism). Hence g is bijective on everyblock of ipG, because rotations are the only endomorphisms of Cn and thus f is anisomorphism. The other implication is clear. �

Let FV(X) denote the free groupoid over X in a variety V. Let I denote thevariety of idempotent groupoids.


1.3. Theorem. Let V be a subvariety of LDLI groupoids and assume V has exponentn. Then FV(X) is isomorphic to Cn × FV∩I(X). Consequently, the variety V isgenerated by (V ∩ I) ∪ {Cn}.

Proof. Since Cn ∈ V, it is a homomorphic image of FV(X). Hence, by Lemma 1.2,FV(X) ' Cn×H, where H = FV(X)/ipFV(X). It is easy to see that H ' FV∩I(X),because ip is the smallest idempotent congruence. �

1.4. Theorem. Let L denote the lattice of subvarieties of LDI groupoids, K itssublattice of varieties containing right zero bands (i.e. groupoids satisfying xy ≈ y)and N the lattice of positive integer divisors of n. The lattice of subvarieties ofthe variety of LDLI groupoids satisfying xn+1 ≈ x is isomorphic to the lattice(L× {1}) ∪ (K × (N r {1})) (regarded as a subposet of L×N), sending a varietyV of exponent m to the pair Φ(V) = (V ∩ I,m).

Proof. First, we check that the mapping Φ is well-defined: the exponent m of asubvariety V is clearly a divisor of n and since V contains Cm, it contains a rightzero band (Cm×Cm)/ip and thus it contains the whole variety of right zero bands.Next, Φ is injective: if V1 and V2 are distinct varieties of exponent m, then V1 ∩ Iand V2 ∩ I are distinct, because Vi is generated by (Vi ∩ I) ∪ {Cm}, i = 1, 2.The mapping Φ is onto, a pair (m,W) is the image of the variety generated byW ∪ {Cm}. Indeed, let G be an idempotent groupoid in the variety generatedby W ∪ {Cm} and we show that G ∈ W. By Birkhoff’s HSP theorem, there areH ∈ W, K ≤ H ×Ck

m (for some k) and an onto homomorphism ϕ : K → G. SinceipK is the smallest idempotent congruence and G is idempotent, there is an ontohomomorphism ψ : K/ipK → G. Further, K/ipK ≤ (K × Ck

m)/ip ' K × (Ckm/ip).

However, Ckm/ip is a right zero band and thus it is in W. Consequently, G is

a homomorphic image of a subgroupoid of a groupoid from W, thus it is in W.Finally, Φ clearly preserves the order and it follows from Theorem 1.3 that alsoΦ−1 preserves the order. Consequently, Φ is a lattice isomorphism. �

Example. B. Roszkowska proved in [45] that the lattice of subvarieties of leftsymmetric medial idempotent (LSMI) groupoids (those where x(xy) ≈ y, xy ·uv ≈xu·yv and xx ≈ x hold) is isomorphic to the lattice of integers ordered by divisibilitywith a top element added. A number n corresponds to a variety based on wn(x, y) ≈y (relatively to LSMI), where

wn(x, y) = x(y(x(y(. . . ))))︸︷︷︸n

.

Note that right zero bands satisfy wn(x, y) ≈ y iff n is even. Thus, using Theorem1.4, it is easy to describe bases of all proper subvarieties of left symmetric leftdistributive medial groupoids (relatively to LSLDM):

(1) xx ≈ x;(2) wn(x, y) ≈ y and xx ≈ x, for every n;(3) wn(x, y) ≈ y, for every n even.

(Note that mediality and idempotency imply left distributivity, however, non-idempotent medial groupoids are not necessarily left distributive.)

Example. J. P lonka [43] investigated idempotent groupoids satisfying

x(x(. . . (x︸︷︷︸n

y)) ≈ y, x(yz) ≈ y(xz) and xz ≈ (yz)z.


He called them n-cyclic groupoids. It is easy to see that they are LDLI. P lonkaproved that the only non-trivial subvarieties of n-cyclic groupoids are m-cyclicgroupoids for m|n. One can use Theorem 1.4 to describe the subvarieties of non-idempotent n-cyclic groupoids. These are indeed varieties generated by idempotentm-cyclic groupoids and the groupoid Ck, wherem, k|n; hence there are exactly q2+1subvarieties, where q is the number of divisors of n.

2. Varieties satisfying xm+n ≈ xm

Theorems 1.3 and 1.4 cannot be generalized to varieties satisfying an identityxm+n ≈ xm for m > 1. For instance, in an LDLI groupoid G with x3 ≈ x2,every ipG-block is a constant groupoid, i.e. ab = cd for all ipG-congruent elementsa, b, c, d (the corresponding unary algebra is a loop with several “tails” of length 1).However, this variety is not generated by LDI and constant groupoids. Indeed, bothLDI and constant groupoids satisfy the identity xy ≈ x(yy), while the groupoid

a b c da, b b b c dc, d a b d d

does not, though it is LDLI with x3 ≈ x2. We thus present a weaker result for suchvarieties.

For a variety V, let Vm,n denote the subvariety of V given by the additionalidentity xm+n ≈ xm.

It is well known that a groupoid G is subdirectly irreducible, iff it possesses thesmallest non-trivial congruence, and that any variety is generated by its subdirectlyirreducible members.

2.1. Lemma. Let G be a subdirectly irreducible LDLI groupoid. If Cn is a subalgebraof G for some n ≥ 2, then G contains no fork (i.e. elements a 6= b with a2 = b2)and, vice versa, if G contains a fork, then Cn is not a subalgebra of G, for anyn ≥ 2.

Proof. Put α = {(a, b) ∈ G × G : a2 = b2}. It is clear that α is an equivalence,which glues each fork. It is a congruence, because whenever a2 = b2, we getat = a2t = b2t = bt and (ta)2 = ta2 = tb2 = (tb)2.

Put (a, b) ∈ β iff there are k, l such that ak = b and bl = a. By a similarargument, it is easy to see that β is a congruence, which glues each circle.

Clearly, α ∩ β = idG, hence either α = idG or β = idG or both, and thus eitherG contains no fork, or no circle with two or more elements, or both. �

Remark. One can prove that a subdirectly irreducible LDLI groupoid G eithercontains a fork, or there is a prime p and a natural number k such that all ipG-blocks are circles of length either 1, or pk (in Proposition 1.2, it is enough to assumethat G is a subdirectly irreducible LDLI groupoid without forks; almost no changeof the proof is necessary). On the other hand, there seems to be no uniformityin the former case. In the following example the ip-blocks have different length oftails and, moreover, one contains a ‘pure fork’ (such that b 6= b2 = c2 6= c), whilethe other don’t. The smallest non-trivial congruence has the only non-trivial block


{a2, a3}.a a2 a3 b c b2 d

a, a2, a3 a2 a3 a3 a2 a2 a3 db, c, b2 = c2 a3 a3 a3 b2 b2 b2 d

d a3 a3 a3 a2 a3 a3 d

2.2. Lemma. Let V be a subvariety of LDLI groupoids and m,n be positive integers.Then Vm,n is the join of the varieties Vm,1 and V1,n.

Proof. Let G be an LDLI groupoid with xm+n ≈ xm. Then ipG-blocks consistof a circle of length k, where k|n, and possibly some “tails” of length at most m(precisely, for any element a out of the circle, am lies on the circle). Now, assumethat G is subdirectly irreducible. It follows from the previous lemma that either allthe circles are of length one, or there are no tails (because whenever a tail joins acircle, there is a fork). Hence, G satisfies either xm+1 ≈ xm (tails of length at mostm only), or xn+1 ≈ x (circles only). Now, the claim follows from the fact that anyvariety is generated by its subdirectly irreducible members. �

2.3. Theorem. Let V be a subvariety of LDLI groupoids and let k, l,m, n be positiveintegers. Then

Vk,l ∨ Vm,n = VLCM(k,m),max(l,n)

andVk,l ∧ Vm,n = VGCD(k,m),min(l,n).

Proof. For the first equality, use the previous lemma and compute Vk,l ∨ Vm,n =Vk,1∨Vm,1∨V1,l∨V1,n = VLCM(k,m),1∨V1,max(l,n) = VLCM(k,m),max(l,n). The secondclaim is rather clear. �

Let V be a variety of LDLI groupoids such that it satisfies no identity xm+n ≈ xm.We show that the mapping (m,n) 7→ Vm,n is injective on N × N. The identityxz ≈ yz implies an identity t ≈ s iff the depth of the rightmost variable in t equals tothe depth of the rightmost variable in s and the two variables are identical. Indeed,all identities of V have the latter property — otherwise t(x, . . . , x) ≈ s(x, . . . , x)were a non-trivial identity in one variable. Hence V contains all groupoids withxz ≈ yz, i.e., in fact, unary algebras. It is easy to see that for any m1, n1 and m2, n2

(so that (m1, n1) 6= (m2, n2)) there is a unary algebra such that it satisfies exactlyone of the identities xm1+n1 ≈ xm1 , xm2+n2 ≈ xm2 . Hence, Vm1,n1 6= Vm2,n2 .Moreover, V1,1 is always non-trivial.

However, not every subvariety of LDLI groupoids is equal to Vm,n for someV,m, n. For instance, assume the variety C of constant groupoids (satisfying theidentity xy ≈ uv); clearly, x3 ≈ x2 holds in C. Suppose there is a variety V suchthat V2,1 = C. Then V1,1 is a non-trivial idempotent subvariety of C. However,there is no non-trivial idempotent groupoid in C, a contradiction.


VIII. Miscellaneous remarksand open problems

1. Miscellaneous remarks

In the present section, we collect a couple of interesting properties, for which,however, we have no use.

1.1. Proposition. Let G be an LD left quasigroup. Then G can be embedded intoan LD left quasigroup H such that for every integer n and for every a ∈ H, thereis b ∈ H with Ln

a = Lb. In particular, H has the LIP.

Proof. Let H = G × Z and put (a, k) · (b, l) = (a[k]b, l). Clearly, a 7→ (a, 1) is anembedding of G into H. The groupoid H is a left quasigroup, since a left translationby L(a,k) acts on each G×{l} as the permutation Lk

a. It is left distributive, (a, k) ·((b, l) · (c,m)) = (a, k) · (b[l]c,m) = (a[k]b[l]c,m) =LD ((a[k]b)[l]c,m) = (a[k]b, l) ·(a[k]c,m) = ((a, k)·(b, l))·((a, k)·(c,m)). And for every integer n and (a, k) ∈ H, weclaim that L(a, k)n = L(a,kn). Indeed, (a, k)[n]·(b, l) = (a[kn]b, l) = (a, kn)·(b, l). �

There is a connection between congruences of an LD left quasigroup and nor-mal subgroups of its left multiplication group. The following two statementsare straightforward generalizations of observations of H. Nagao [35] for LSLDgroupoids.

1.2. Proposition. Let G be an LD left quasigroup.(1) For a congruence ρ of G, put Nρ = 〈LaL

−1b : (a, b) ∈ ρ〉. Then Nρ is a

normal subgroup of LMlt(G).(2) For a normal subgroup N of LMlt(G), put (a, b) ∈ ρN , iff there is ϕ ∈ N

such that ϕ(a) = b. Then ρN is a congruence of G.

Proof. (1) It is enough to show that LcLaL−1b L−1

c ∈ Nρ for any (a, b) ∈ ρ andc ∈ G. Due to II.1.1(2), the expression is equal to LcaL

−1cb and indeed (ca, cb) ∈ ρ.

(2) Clearly, ρN is an equivalence on G. Let (a, b) ∈ ρN and c ∈ G. We havebc = ϕ(a)c = ϕ(a)(a\ac) = Lϕ(a)L

−1a (ac) = ϕLaϕ

−1L−1a (ac), and so (ac, bc) ∈ ρN ,

because ϕ ∈ N and, by normality of N , also Laϕ−1L−1

a ∈ N . Further, cb =cϕ(a) = ϕ(ϕ−1(c)a) = ϕ(ϕ−1(c)(c\ca)) = ϕLϕ−1(c)L

−1c (ca) = LcϕL

−1c (ca), and so

(ca, cb) ∈ ρN , because LcϕL−1c ∈ N . �

1.3. Corollary. Let G be a simple LD left quasigroup. Then D(G) = 〈LaL−1b :

a, b ∈ G〉 is the least normal subgroup of LMlt(G).

Proof. It follows from II.1.1(2) that D(G) is a normal subgroup of LMlt(G). Now,let N be a non-trivial normal subgroup of LMlt(G). Then ρN is a non-identicalcongruence of G, and thus ρN = G × G. Hence, for every a, b ∈ G, there existsϕ ∈ N such that b = ϕ(a). Consequently, LaL

−1b = LaL

−1ϕ(a) = LaϕL

−1a ϕ−1 ∈ N ,

because N normal and thus both ϕ−1 and LaϕL−1a lie in N . �


Recall that a groupoid is called effective, if its left translations are pairwisedifferent.

1.4. Proposition. The left multiplication group of an effective LD left quasigroupis centerless. Consequently, an effective LD left quasigroup can be embedded intothe conjugation groupoid of a centerless group.

Proof. First, note that La\b = L−1a LbLa. Indeed, L−1

a (bac) = L−1a (b)L−1

a (ac) =(a\b)c, because L−1

a is an automorphism. Now, assume that an element ϕ =Lε1

a1. . . Lεn

an, εi ∈ {±1}, of the left multiplication group is in its center. Then ϕ

commutes with every Lb, hence Lε1a1. . . Lεn

anLbL

−εnan

. . . L−ε1a1

= Lb for every b andthus La1◦1···◦n−1an◦nb = Lb, where ◦i stands for multiplication iff εi = 1 and for leftdivision otherwise. Now, effectivity says that a1 ◦1 · · · ◦n−1 an ◦n b = b for every band thus Lε1

a1. . . Lεn

anis the identity.

For the second claim, use Lemma II.1.1(3). �

2. Open problems

This is a list of my favourite open problems regarding the topic of the thesis.Not all of them were attempted hardly, so it may happen that some of them haveeasy answers.

1. Is the equational theory of group conjugation finitely based? Find a (nice) base.(See Chapter V.)

2. B. Roszkowska described in [47] finite LSMI groupoids. They can be decomposedto minimal left ideals and a structure of an abelian group can be defined on eachof the left ideals (the construction is called AG-sum).

(a) Does something similar work for LSLDI groupoids? The straightforwardgeneralization, BL-sum, does not work (see III.7.1 and remarks bellow), but maybesomething more complicated does. Note that the description of LSMI groupoidsof odd exponent is fully generalized in Theorem III.3.2 to LSLDI. Try at least forLSDI groupoids.

(b) Does something similar work for n-LSMI groupoids or even MI left quasi-groups? (Cf. with Toyoda’s theorem.)

3. Can one find a good description of non-idempotent LSM groupoids? Note thatthey are not necessarily left distributive or left idempotent.

In particlar, describe free LSM groupoids. Even the structure of the free mono-generated one seems to be unknown and complicated. In contrast to the free mono-generated LSLDM groupoid, which has two elements, the free monogenerated LSMgroupoid is infinite. For instance, Z(�) and Z(~) are infinite monogenerated LSMgroupoids, where x � y = x − y and x ~ y = −x − y. In Z(�), the terms x, xx,(xx)x, ((xx)x)x, etc. represent pairwise distinct elements.

4. B. Roszkowska described in [45] the lattice of subvarieties of LSMI groupoids(see Chapter VII for details). In particular, she proved that the lattice is countableand that any set of identities is LSMI-equivalent to a single identity in two variables.

I suspect that none of the statements is true for the variety of LSLDI groupoids.In particular, find a non finitely based subvariety of LSLDI groupoids.

Are the statements true for the varieties of n-LSMI groupoids?


5. A couple of problems regarding cores.(a) The tenth Belousov’s problem: classify loops with (left symmetric) left dis-

tributive core. (See the last section of Chapter III.)(b) Is the variety of LSDI groupoids generated by cores of commutative Moufang

loops? It follows from Theorem III.1.1 that the variety of LSLDI groupoids isgenerated by cores of groups (and thus also by cores of Bol loops) and the varietyof LSMI groupoids is generated by cores of abelian groups.

(c) (JDH Smith) Describe the quasivariety generated by cores of groups (abeliangroups, Bol loops, etc.).

6. The ninth Belousov’s problem: which loops are isotopic to left distributivequasigroups? (See Chapter IV.)

7. There is a conjecture related to Chapter V.

Conjecture. Let w ∈ FConj(X). Then there are two words u, v ∈ FConj(X), eachof them shorter than w, such that w = u ∗ v.

It implies two interesting facts. First, there is an algorithm to decide, whether agiven free group word over X belongs to the free groupoid FConj(X). Next, there isa simple direct proof of the fact that every LDI left divisible groupoid (and hencealso every LDI left quasigroup) is a homomorphic image of a conjugation groupoid.


APPENDIX:Homomorphic images of

subdirectly irreducible algebras

We investigate the following problem: Which algebras are homomorphic imagesof subdirectly irreducible algebras? Which of them are isomorphic to a factoralgebraof a subdirectly irreducible algebra over its monolith?

For instance, it is proven in [25] that every groupoid with an absorbing elementis a factorgroupoid of a subdirectly irreducible groupoid over its monolith. On theother hand, it is easy to prove that the additive semigroup of positive integers isnot a factorgroupoid of a subdirectly irreducible groupoid. In fact, it is not hard toprove that a homomorhic image of a subdirectly irreducible algebra must have non-empty intersection of its ideals. Our main result is that this condition is sufficientfor algebras with at least one at least binary operation. Moreover, for such analgebra A we find a subdirectly irreducible B such that the A is the factor of Bover its monolith and if A is finite, B is also finite.

The first section contains a proof of the necessary condition. In next two sections,we prove the converse for groupoids and for algebras with rich enough signature. Inthe fourth section, we answer the questions for monounary algebras. However, thecase of (multi)unary algebras remains open. We do not know an answer to eitherof the two questions. The last section contains a couple of examples.

The main result was published in [52], the whole material appeared in a contestthesis (SVOC 2001). The main result was achieved independently by J. Jezekand T. Kepka in [20]; their construction is completely different and a little bit lesseffective (in the sense of the size of the constructed subdirectly irreducible algebra).

1. The necessary condition

By a signature Σ we mean a mapping ar : Σ → ω, assigning to each symbol itsarity. An algebra of a signature Σ is a non-empty set A equipped with operationsFA : Aar(F ) → A for each F ∈ Σ.

A non-empty subset I of A is said to be an ideal of A, if FA(x1, . . . , xn) ∈ Iwhenever for every symbol F ∈ Σ of arity n ≥ 1 and for every x1, . . . , xn ∈ A withxi ∈ I for at least one i. We denote by Int(A) the intersection of all ideals of A.The element o ∈ A is called absorbing, if {o} is an ideal of A.

A signature is called rich, if it contains at least one at least binary symbol.

1.1. Lemma. Let A be an algebra. Then Int(A) is either empty or an ideal ofA. If the latter is true, then Int(A) is the smallest ideal of A. Moreover, if A hasrich signature, then A possesses an absorbing element o, if and only if Int(A) is aone-element set; then Int(A) = {o}.

Proof. The intersection of a family of ideals of A is either empty or an ideal. Therest is clear. �


1.2. Lemma. Let ϕ be a projective homomorphism of an algebra A onto an al-gebra B. If J is an ideal of B, then the inverse image ϕ−1(J) is an ideal of A.Consequently, Int(A) ⊆ ϕ−1(Int(B)). In particular, if Int(A) 6= ∅, then Int(B) 6= ∅.Proof. Let F ∈ Σ of arity n ≥ 1 and x1, . . . , xn ∈ A such that xi ∈ ϕ−1(J)for at least one i. Then ϕ(FA(x1, . . . , xn)) = FB(ϕ(x1), . . . , ϕ(xn)) ∈ J , becauseϕ(xi) ∈ J for at least one i, and thus FA(x1, . . . , xn) ∈ ϕ−1(J). The rest is anobvious consequence. �

It is well-known that an algebra A is subdirectly irreducible, iff the intersectionof all non-trivial congruences is non-trivial. The intersection is called the monolithof A and denoted µA.

We put ρX = (X ×X) ∪ idA for a subset X ⊆ A. Clearly, ρX is a congruence,whenever X is an ideal.

1.3. Lemma. Let A be a subdirectly irreducible algebra with a rich signature. ThenI = Int(A) 6= ∅. Moreover, if |I| ≥ 2, then µA ⊆ ρI = (I × I) ∪ idA.

Proof. If A contains an absorbing element o, then Int(A) = {o}. Now, assume thatA has no absorbing element. If I is an ideal of A, then |I| ≥ 2 and thus ρI is anon-trivial congruence of A and µA ⊆ ρI . Consequently, if (u, v) ∈ µA, u 6= v, then(u, v) ∈ ρI and u, v ∈ I. Thus u, v ∈ Int(A). �

1.4. Corollary. Let A be a homomorphic image of a subdirectly irreducible algebrawith rich signature. Then Int(A) 6= ∅.Example. Let G = N(+) is the additive semigroup of non-negative integers. ThenInt(G) = ∅.1.5. Lemma. Let A be an algebra with a rich signature. Then the intersection ofa non-empty finite set of ideals of A is again an ideal of A. Consequently, if theset of ideals of G is finite, then Int(G) 6= ∅.Proof. The intersection of two ideals is non-empty and thus an ideal. �

Example. If A is a finite algebra with a rich signature, then Int(A) 6= ∅.An algebra with signature consisting of κ unary symbol is called κ-unary (or

just unary; for κ = 1 monounary). A subset of a unary algebra is an ideal, iff itis a subalgebra. Therefore, the intersection of two ideals is not always non-emptyand Lemma 1.3 does not hold. In particular, not every finite unary algebra is ahomomorphic image of a subdirectly irreducible one.

Let Int∗(A) denote the intersection of all at least two-element subalgebras of A.

1.6. Lemma. Let A be a subdirectly irreducible unary algebra. Then I = Int∗(A)contains at least two elements and µA ⊆ ρI = (I × I) ∪ idA.

Proof. Let I be a subalgebra of A with |I| ≥ 2. Then ρI is a non-trivial congruenceof A and so µA ⊆ ρI . Consequently, if (u, v) ∈ µA, u 6= v, then (u, v) ∈ ρI andu, v ∈ I. Thus u, v ∈ Int∗(A). �

1.7. Corollary. Let A be a homomorphic image of a subdirectly irreducible unaryalgebra. Then Int∗(A) 6= ∅.

A subalgebra C of a unary algebra A is called a component of A, if it is a minimalsubalgebra satisfying (FA)−1(C) = C for every F ∈ Σ. A unary algebra is said tobe connected, if it contains only one component.


1.8. Lemma. Let A be a homomorphic image of a subdirectly irreducible unaryalgebra. Then A is either connected, or A contains two components and one ofthem has one element.

Proof. Assume that A is not connected. If it contains one-element components onlyand |A| ≥ 3, then clearly Int∗(A) = ∅. If not, choose a component C with |C| ≥ 2.If |ArC| ≥ 2, then C and ArC are two disjoint at least two-element subalgebrasof A and Int∗(A) = ∅, a contradiction. Hence |Ar C| = 1. �

Observe that constants involve neither congruences, nor ideals of an algebra.Therefore, to avoid technical difficulties, we assume without loss of generality thatsignatures contain no constant symbols.

2. Groupoids

An algebra with signature consisting of one binary symbol is called groupoid.Note that an ideal of a groupoid G is a non-empty subset I satisfying GI ∪ IG ⊆ I.

In this chapter, we prove that a groupoid G is a homomorphic image of a sub-directly irreducible groupoid (over its monolith), if and only if Int(G) 6= ∅.

Construction. Let n ≥ 3 be an odd number. We define a groupoid Zn(∗) on theset {0, 1, . . . , n− 1} in the following way:

• 0 ∗m = 0 for every 0 ≤ m ≤ n− 1;• k ∗ 0 = 0 for every odd 1 ≤ k ≤ n− 2;• l ∗ 0 = 1 for every even 2 ≤ l ≤ n− 1;• 1 ∗m = m for every 1 ≤ m ≤ n− 1;• m ∗m = m for every 2 ≤ m ≤ n− 1;• k ∗ l = k + 1 for all 2 ≤ k ≤ n− 2, 1 ≤ l ≤ n− 1, k 6= l;• (n− 1) ∗ l = 0 for every 1 ≤ l ≤ n− 2.

It is easy to check that Zn(∗) is a simple idempotent groupoid and that no righttranslation of this groupoid is a permutation. Moreover, 0 is a left absorbing elementand 1 is a left neutral element.

Zn (odd) 0 1 2 3 4 . . . n− 10 0 0 0 0 0 01 0 1 2 3 4 n− 12 1 3 2 3 3 33 0 4 4 3 4 44 1 5 5 5 4 5. . . . . .n− 1 1 0 0 0 0 n− 1

Construction. Let n ≥ 4 be an even number. We define a groupoid Zn(∗) on theset {0, 1, . . . , n− 1} in the following way:

• 0 ∗m = 0 for every 0 ≤ m ≤ n− 1;• k ∗ 0 = 1 for every odd 1 ≤ k ≤ n− 1;• l ∗ 0 = 0 for every even 2 ≤ l ≤ n− 2;• 1 ∗m = m for every 1 ≤ m ≤ n− 1;• m ∗m = m for every 2 ≤ m ≤ n− 1;• k ∗ l = k + 1 for all 2 ≤ k ≤ n− 2, 1 ≤ l ≤ n− 1, k 6= l;• (n− 1) ∗ l = 0 for every 1 ≤ l ≤ n− 1.


Again, Zn(∗) is a simple idempotent groupoid where no right translation is a per-mutation. The element 0 is left absorbing.

Zn (even) 0 1 2 3 4 . . . n− 10 0 0 0 0 0 01 1 1 2 3 4 n− 12 0 3 2 3 3 33 1 4 4 3 4 44 0 5 5 5 4 5. . . . . .n− 1 1 0 0 0 0 n− 1

For n = 2 we define the groupoid Z2(∗) in the following way:

Z2 0 10 1 11 0 1

Construction. Let κ be an infinite cardinal number. We define a groupoid Zκ(∗)on the set κ in the following way:

• 0 ∗ α = 0 for every 0 ≤ α < κ;• (β + k) ∗ 0 = 1 for all limit ordinals β < κ and finite even numbers k ≥ 0,β + k 6= 0;

• (β + l) ∗ 0 = 0 for all limit ordinals β < κ and finite odd numbers l ≥ 1;• 1 ∗ α = α for every 1 ≤ α < κ;• α ∗ α = α for every 2 ≤ α < κ;• α ∗ β = α+ 1 for all 2 ≤ α < κ and 1 ≤ β < κ.

Again, Zκ(∗) is a simple idempotent groupoid where no right translation is a per-mutation. The element 0 is left absorbing and the element 1 is left neutral.

Zκ 0 1 2 3 4 . . .0 0 0 0 0 01 0 1 2 3 42 1 3 2 3 33 0 4 4 3 44 1 5 5 5 4. . . . . .

2.1. Theorem. The following conditions are equivalent for a groupoid H:(i) H is a homomorphic image of a subdirectly irreducible groupoid.(ii) H is isomorphic to G/µG for a subdirectly irreducible groupoid G.(iii) The intersection of all ideals in H is non-empty.

Proof. In a view of 1.1 and 1.4, it is enough to show that (iii) implies (ii). Hence,let I = Int(H), K = H r I and κ = max(|I|, |K|).

If κ = 1, then |H| ≤ 2. If |H| = 1, then G can be chosen to be any simplegroupoid. If |H| = 2, then |I| = 1, i.e. I = {o}, where o is an absorbing element ofH = {o, a}. We take G = H ∪ {z}, z /∈ H, and we put u ◦ v = uv for all u, v ∈ H,a ◦ z = z, o ◦ z = z ◦ o = z ◦ a = z ◦ z = o. Clearly, G(◦) is a subdirectly irreduciblegroupoid and H ' G/µG.

Now, assume that κ ≥ 2 and K 6= ∅.


Claim 1. There exist permutations πa,u ∈ κ! for all a ∈ I, u ∈ K, such that thefollowing two conditions are satisfied:

(A) πa,u 6= πa,v for all a ∈ I, u, v ∈ K, u 6= v;(B) for all a, b ∈ I, a 6= b, and all 0 ≤ α < κ, there exists some u ∈ K with

πa,u(α) 6= πb,u(α).

Proof. (a) Let |I| = κ. Choose w ∈ K, a bijection ξ : I → κ and a quasigroup Q(�)defined on κ. Put πa,w(α) = α � ξ(a) for all a ∈ I and α < κ. Now it is easy to findthe remaining permutations πa,u, a ∈ I, u ∈ K r {w}.

(b) Let |I| < κ, 4 ≤ κ. Choose a permutation ρ ∈ κ! without fix points,an injective mapping ξ : I → K and permutations πu ∈ κ!, u ∈ K such thatρπu 6= πv 6= πu for all u, v ∈ K, u 6= v. Now, define πa,u = πu for all a ∈ I, u ∈ Ksuch that ξ(a) 6= u and πb,ξ(b) = ρπξ(b) for every b ∈ I.

(c) Let |I| < κ ≤ 3. This case is easy. �

Put G = (I × κ) ∪K and define an operation ◦ on G in the following way:• u ◦ v = uv for all u, v ∈ K such that uv ∈ K;• u ◦ v = (uv, 0) for all u, v ∈ K such that uv ∈ I;• (a, α) ◦ (b, β) = (ab, α ∗ β) for all a, b ∈ I, 0 ≤ α, β < κ (the operation ∗

defined above);• u ◦ (a, α) = (ua, πa,u(α)) for all a ∈ I, u ∈ K, 0 ≤ α < κ;• (a, α) ◦ u = (au, πa,u(α)) for all a ∈ I, u ∈ K, 0 ≤ α < κ.

For a ∈ I, let Ia = {a} × κ ⊆ I × κ.

G(◦) (b, β) v(a, α) (ab, α ∗ β) (av, πa,v(α))u (ub, πb,u(β)) uv

Claim 2. The groupoid G(◦) is subdirectly irreducible and

µG(◦) =⋃a∈I

(Ia × Ia) ∪ idG.

Proof. It is clear that µ =⋃

a∈I(Ia × Ia) ∪ idG is a non-trivial cogruence of Gand we have to show that µ ⊆ ν for any non-trivial congruence of G(◦). For thispurpose, put J = {a ∈ I : Ia × Ia ⊆ ν}. If κ = 2, then obviously JI ∪ IJ ⊆ J .If κ ≥ 3, a ∈ J , b ∈ I and 0 ≤ α < κ, then ((a, 0), (a, α)) ∈ ν, and therefore((ab, 0), (ab, α)) = ((a, 0) ◦ (b, α), (a, α) ◦ (b, α)) ∈ ν. From this ab ∈ J and, quitesimilarly, ba ∈ J . Thus JI ∪ IJ ⊆ J . If a ∈ J , u ∈ K and 0 ≤ α, β < κ, then((au, πa,u(α)), (au, πa,u(β))) = ((a, α)◦u, (a, β)◦u) ∈ ν. Since πa,u is a permutation,we get au ∈ J . Quite similarly, ua ∈ J , and we conclude that J is an ideal of G,J = I and µG ⊆ ν, provided that J 6= ∅. Consequently, it remains to show that Jis non-empty. This will be done in next five steps.

(1) Assume that ((a, α), (a, β)) ∈ ν for some a ∈ I and 0 ≤ α < β < κ.If α = 0 and β = 1, then, using the right translation by (a, γ) for all 0 ≤ γ < κ, weget aa ∈ J .If α∗0 6= β∗0, then using the right translation by (a, 0), we get ((aa, 0), (aa, 1)) ∈ ν,and hence aa · aa ∈ J .Finally, if α ∗ 0 = β ∗ 0 and 2 ≤ β, then using the right translation by (a, α) forα 6= 0 and by (a, 1) for α = 0, we get ((aa, α), (aa, β⊕)) ∈ ν (here β⊕ = β + 1 for κ


infinite or κ finite and β ≤ κ−2, and β⊕ = 0 for κ finite and β = κ−1). Accordingto the preceding part of the proof, we have (aa · aa)(aa · aa) ∈ J .

(2) Assume that ((a, α), (b, β)) ∈ ν for some a, b ∈ I, a 6= b, 0 ≤ α < β < κ andtake arbitrary c ∈ I.First, suppose 2 ≤ α. Then, applying the right translations by (c, 1) and (c, α),we get ((ac, α⊕), (bc, β⊕)) ∈ ν and ((bc, β⊕), (ac, α)) ∈ ν. So ((ac, α⊕), (ac, α)) ∈ νand our result follows from (1).If α = 1, then, applying the right translations by (c, γ) and (c, 1), where 2 ≤ γ 6= β,we get ((ac, γ), (bc, β⊕)) ∈ ν and ((bc, β⊕), (ac, 1)) ∈ ν. Thus ((ac, γ), (ac, 1)) ∈ νand (1) applies again.If α = 0 and 2 ≤ β, then, using the right translations by (c, γ) and (c, β), where1 ≤ γ 6= β, we get ((ac, 0), (bc, β⊕)) ∈ ν and ((ac, 0), (bc, β)) ∈ ν. That is((bc, β⊕), (bc, β)) ∈ ν and (1) takes place.If α = 0, β = 1 and 3 ≤ κ, then, because of the right translations by (c, 2)and (c, 1), we get ((ac, 0), (bc, 2)) ∈ ν and ((ac, 0), (bc, 1)) ∈ ν. It follows that((bc, 2), (bc, 1)) ∈ ν and (1) finishes the job.Finally, if α = 0, β = 1 and κ = 2, then, because of the right translations by (c, 1)and (c, 0), we get ((ac, 1), (bc, 1)) ∈ ν and ((ac, 1), (bc, 0)) ∈ ν. So ((bc, 1), (bc, 0)) ∈ν and (1) works.

(3) Assume that ((a, α), (b, α)) ∈ ν for some a, b ∈ I, a 6= b and 0 ≤ α < κ.Then, by (B) (see Claim 1), there is u ∈ K such that β = πa,u(α) 6= πb,u(α) = γ.Thus, using the right translation by u, we get ((au, β), (bu, γ)) ∈ ν. Now, either (1)or (2) can be used.

(4) Assume that ((a, α), u) ∈ ν for some a ∈ I, u ∈ K and 0 ≤ α < κ.If ba 6= bu for some b ∈ I, then, using the left translation by (b, 0), we get ((ba, 0 ∗α), (bu, πb,u(0))) and either (2) or (3) can be used.If κ ≥ 3 and ca = cu for some c ∈ I, then, using the facts that πc,u is a permutationof κ, but no right translation of Zκ(∗) is a permutation, we find 0 ≤ β < κ suchthat β ∗ α 6= πc,u(β) = γ. We apply the left translation by (c, β) and we get((ca, β ∗ α), (cu, γ)) ∈ ν. Thus (1) takes place.The case κ = 2 is clear.

(5) Assume that (u, v) ∈ ν for some u, v ∈ K, u 6= v, and take arbitrary a ∈ I.By (A) (see Claim 1), there is 0 ≤ α < κ such that β = πa,u(α) 6= πa,v(α) = γ.Now, applying the right translation by (a, α), we get ((ua, β), (va, γ)) ∈ ν. Thus atleast one of (1) and (2) can be used. �

Claim 3. G(◦)/µG(◦) ' H.

Proof. Easy to see. �

Now, we will discuss the case K = ∅, i.e. H is an ideal-free goupoid and I =IntH = H. Put G = H × κ and define an operation ◦ on H by (a, α) ◦ (a, β) =(aa, β ∗ α) and (a, α) ◦ (b, β) = (ab, α ∗ β) for all a, b ∈ H, a 6= b, 0 ≤ α, β < κ. Fora ∈ H, let Ha = {a} × κ.

G(◦) (a, β) (b, δ)(a, α) (aa, β ∗ α) (ab, α ∗ δ)(b, γ) (ba, γ ∗ β) (bb, δ ∗ β)


Claim 4. The groupoid G(◦) is subdirectly irreducible and

µG(◦) =⋃

a∈H

(Ha ×Ha).

Proof. We have to show that µG =⋃

a∈H(Ha × Ha) ⊆ ν for any non-trivial con-gruence ν of G(◦). Proceeding similarly as in the proof of Claim 2, it is sufficientto check that Ha ×Ha ⊆ ν for at least one a ∈ H. This will be done in the nextthree steps.

(1) Assume that ((a, α), (a, β)) ∈ ν for some a ∈ H and 0 ≤ α < β < κ. Now,using left translations instead of the right ones, we can proceed similarly as in Claim2 (1).

(2) Assume that ((a, α), (b, β)) ∈ ν for some a, b ∈ H, a 6= b, and 0 ≤ α < β < κ.If κ ≥ 3, then we can proceed similarly as in Claim 2 (2); we have to choosea 6= c 6= b. The case κ = 2 is clear.

(3) Assume that ((a, α), (b, α)) ∈ ν for some a, b ∈ H, a 6= b and 0 ≤ α < κ.There is 0 ≤ β < κ such that α ∗ β 6= β ∗ α, and hence, using the right translationby (a, β), we get ((aa, β ∗α), (ba, α∗β)) ∈ ν. Now, either (1) or (2) takes place. �

Claim 5. G(◦)/µG(◦) ' H.


Now, the proof of Theorem 2.1 is completed. �

2.2. Corollary. Let H be a finite groupoid, |H| = n and | Int(H)| = m. Then thereexists a finite subdirectly irreducible groupoid G such that G/µG ' H. Moreover,G can be chosen so that

(1) |G| = 2 if n = 1;(2) |G| = 3 if n = 2;(3) |G| = m2 + (n−m) if n ≥ 3 and n ≤ 2m;(4) |G| = (m+ 1)(n−m) if n ≥ 3 and n > 2m.

Proof. Clear from 1.5 and from the construction above. �

3. Algebras with rich signature

Theorems 2.1 and 2.2 can be easily generalized to all algebras with at least one atleast binary operation (i.e. with a rich signature). The proof of the main theoremis rather similar to that for groupoids.

3.1. Theorem. The following conditions are equivalent for an algebra B with arich signature:

(i) B is a homomorphic image of a subdirectly irreducible algebra.(ii) B is isomorphic to A/µA for a subdirectly irreducible algebra A.(iii) The intersection of all ideals in B is non-empty.

Proof. In a view of 1.1 and 1.4, it is enough to show that (iii) implies (ii). Hence,let I = Int(B), K = B r I and κ = max(|I|, |K|).

If κ = 1, then |B| ≤ 2. If |B| = 1, then A can be chosen to be any simple algebra.If |B| = 2, then |I| = 1, i.e. I = {o}, where o is an absorbing element of B = {o, a}.We take A = B ∪ {z}, z /∈ B, and for every symbol F ∈ Σ of arity n we putFA(x1, . . . , xn) = FB(x1, . . . , xn) for all x1, . . . , xn ∈ B, FA(a, z, x1, . . . , xn−2) = z


for all x1, . . . , xn−2 ∈ A and FA(x1, . . . , xn) = o for all x1, . . . , xn ∈ A such that FA

is not defined yet. Clearly, A is a subdirectly irreducible algebra and B ' A/µA.Now, assume that κ ≥ 2 and K 6= ∅. Again, put A = (I × κ) ∪ K and define

operations on A in the following way (for every symbol F ∈ Σ of arity n):• FA(u1, . . . , un) = FB(u1, . . . , un) = v for all u1, . . . , un ∈ K such thatv ∈ K;

• FA(u1, . . . , un) = (FB(u1, . . . , un), 0) for all u1, . . . , un ∈ K not satisfyingthe previous condition;

• if F is unary, then FA((a, α)) = (FB(a), α) for every a ∈ I, 0 ≤ α < κ;• if F is at least binary, then

FA(u1, . . . , uk, (a, α), uk+1, . . . , un−1) = (FB(u1, . . . , uk, a, uk+1, . . . , un−1), πa,u1(α))

for all a ∈ I, u1, . . . , un−1 ∈ K, 0 ≤ α < κ, k = 0, . . . , n− 1 (the permuta-tions πa,u are those from the Claim 1 of 2.1);

• if F is at least binary, then

FA(x1, . . . , xn) = (FB(ξx1, . . . , ξxn), α1 ∗ (α2 � · · · � αk))

for all x1, . . . , xn ∈ A such that xi1 , . . . , xik∈ I, k ≥ 2, denoting xij =

(aj , αj), j = 1, . . . , k, and ξ : A→ B, ξ|K = idK , ξ(a, α) = a for all a ∈ I,0 ≤ α < κ; � is some group operation on κ with a neutral element ε, and∗ is the operation on κ defined above.

For a ∈ I, let Ia = {a} × κ ⊆ I × κ.Note that in the case of groupoids the construction yields the same subdirectly

irreducible algebra as in the proof of Theorem 2.1.For the purpose of this proof, let us establish the following terminology. Let

F ∈ Σ be an n-ary symbol, ν a congruence on an algebra X, (y, z) ∈ ν. Using ofthe k-translation by x1, . . . , xn−1 on (y, z) means the conclusion that

(FX(x1, . . . , xk−1, y, xk, . . . , xn−1), FX(x1, . . . , xk−1, z, xk, . . . , xn−1)) ∈ ν.

The left (right) translation means the 1-translation (n-translation).Claim 1. The algebra A is subdirectly irreducible and

µA =⋃a∈I

(Ia × Ia) ∪ idA.

Proof. It is clear that µ =⋃

a∈I(Ia × Ia) ∪ idA is a non-trivial cogruence of Aand we show that µ ⊆ ν for any non-trivial congruence of A. For this purpose,put J = {a ∈ I : Ia × Ia ⊆ ν}. Let F ∈ Σ be a symbol of arity n, a ∈ J andx1, . . . , xn−1 ∈ B. We want to show that w = FB(x1, . . . , xk−1, a, xk, . . . , xn−1) ∈J . This is trivial for F unary, so assume F is at least binary.

(a) Let xi ∈ K for all i = 1, . . . , n − 1. Then, using the k-translation byx1, . . . , xn on all pairs ((a, α), (a, β)) gives ((w, πa,x1(α)), (w, πa,x1(β))) ∈ ν,0 ≤ α < β < κ. Since πa,x1 is a permutation, we get w ∈ J .

(b) Let x1, . . . , xk−1 ∈ K, but there is xi ∈ I for at least one i ≥ k. Put yαi = xi

for xi ∈ K, 1 ≤ i ≤ n− 1, and yαi = (xi, α) for i being the least such that

xi ∈ I and yαi = (xi, ε) in the other cases. Then using the k-translation

by yα1 , . . . , y

αn on ((a, 0), (a, 1)) gives ((w, 0), (w,α)) ∈ ν. Applying for all

1 ≤ α < κ (if κ = 2, then for α = 0), we get w ∈ J .


(c) Let xi ∈ I for at least one i < k. Put yi = xi for xi ∈ K, 1 ≤ i ≤ n−1, andyi = (xi, 1) for i being the least such that xi ∈ I and yi = (xi, ε) in the othercases. Then using the k-translation by y1, . . . , yn on all pairs ((a, α), (a, β))gives ((w,α), (w, β)) ∈ ν, 0 ≤ α < β < κ, (with the exception of the casesκ = 2 or κ even and α = 0; they must be held separately). Hence, we getw ∈ J .

Hence, we conclude that J is either empty, or an ideal of A, J = I and µA ⊆ ν.Consequently, it remains to show that J is non-empty. This will be done in nextfive steps. Choose F ∈ Σ at least binary.

(1) Assume that ((a, α), (a, β)) ∈ ν for some a ∈ I and 0 ≤ α < β < κ.If α = 0 and β = 1, then, using the right translation by (a, γ), (a, ε), . . . , (a, ε) forall 0 ≤ γ < κ, we get w = FB(a, . . . , a) ∈ J .Next, if α ∗ 0 6= β ∗ 0, then, using the right translation by (a, 0), (a, ε), . . . , (a, ε), weget ((w, 0), (w, 1)) ∈ ν, and hence FB(w, . . . , w) ∈ J .Last, if α ∗ 0 = β ∗ 0 and 2 ≤ β, then, using for α 6= 0 the right translation by(a, α), (a, ε), . . . , (a, ε) and for α = 0 the right translation by (a, 1), (a, ε), . . . , (a, ε),we get ((w,α), (w, β⊕)) ∈ ν (here β⊕ = β+1 for κ infinite or κ finite and β ≤ κ−2,and β⊕ = 0 for κ finite and β = κ − 1). According to the preceding part of theproof, we find an element of J .

(2) Assume that ((a, α), (b, β)) ∈ ν for some a, b ∈ I, a 6= b, 0 ≤ α < β < κ andtake arbitrary c ∈ I. Put x = FB(a, c, . . . , c), y = FB(b, c, . . . , c).First, suppose 2 ≤ α. Then, applying the right translations by (c, 1), (c, ε), . . . , (c, ε)and (c, α), (c, ε), . . . , (c, ε), we get ((x, α⊕), (y, β⊕)) ∈ ν and ((y, β⊕), (x, α)) ∈ ν.So ((x, α⊕), (x, α)) ∈ ν and (1) can be used.If α = 1, then, applying the right translations by (c, γ), (c, ε), . . . , (c, ε) and by(c, 1),(c, ε),. . . , (c, ε), where 2 ≤ γ 6= β, we get that ((x, γ), (y, β⊕)) ∈ ν and also((y, β⊕), (x, 1)) ∈ ν. Thus (x, γ), (x, 1)) ∈ ν and (1) applies again.If α = 0 and 2 ≤ β, then, using the right translations by (c, γ), (c, ε), . . . , (c, ε)and (c, β), (c, ε), . . . , (c, ε), where 1 ≤ γ 6= β, we get ((x, 0), (y, β⊕)) ∈ ν and((x, 0), (y, β)) ∈ ν. That is ((y, β⊕), (y, β)) ∈ ν and (1) takes place.If α = 0, β = 1 and 3 ≤ κ, then, using the right translations by (c, 2), (c, ε), . . . , (c, ε)and (c, 1), (c, ε), . . . , (c, ε), we get ((x, 0), (y, 2)) ∈ ν and ((x, 0), (y, 1)) ∈ ν. Itfollows that ((y, 2), (y, 1)) ∈ ν and (1) applies.If α = 0, β = 1 and κ = 2, then, using the right translations by (c, 1), (c, ε), . . . , (c, ε)and (c, 0), (c, ε), . . . , (c, ε), we get ((x, 1), (y, 1)) ∈ ν and ((x, 1), (y, 0)) ∈ ν. So wehave ((y, 1), (y, 0)) ∈ ν and (1) works.

(3) Assume that ((a, α), (b, α)) ∈ ν for some a, b ∈ I, a 6= b and 0 ≤ α < κ. Then,by (B) (see Claim 1 of 2.1), there is u ∈ K such that β = πa,u(α) 6= πb,u(α) = γ.Thus, using the right translation by u, . . . , u, we get

((FB(a, u, . . . , u), β), (FB(b, u, . . . , u), γ)) ∈ ν.

Now, either (1) or (2) can be used.(4) Assume that ((a, α), u) ∈ ν for some a ∈ I, u ∈ K and 0 ≤ α < κ.

If FB(b, u, . . . , u, a) 6= FB(b, u, . . . , u, u) for some b ∈ I, then, using the left trans-lation by (b, 0), u, . . . , u, we get

((FB(b, u, . . . , u, a), 0 ∗ α), (FB(b, u, u, . . . , u), πb,u(0))) ∈ ν

and either (2) or (3) can be used.


If κ ≥ 3 and FB(c, u, . . . , u, a) = FB(c, u, . . . , u, u) for some c ∈ I, then, using thefacts that πc,u is a permutation of κ, but no right translation of Zκ(∗) is a permuta-tion, we find 0 ≤ β < κ such that β ∗α 6= πc,u(β) = γ. We apply the left translationby (c, β), u, . . . , u and we get ((FB(c, u, . . . , u, a), β ∗α), (FB(c, u, u, . . . , u), γ)) ∈ ν.Thus (1) takes place.The case κ = 2 is clear.

(5) Assume that (u, v) ∈ ν for some u, v ∈ K, u 6= v, and take arbitrary a ∈ I.By (A) (see Claim 1 of 2.1), there is 0 ≤ α < κ such that β = πa,u(α) 6= πa,v(α) = γ.Now, applying the right translation by (a, α), u, . . . , u, we get

((FB(u, a, u, . . . , u), β), (FB(v, a, u, . . . , u), γ)) ∈ ν.Thus at least one of (1) and (2) can be used. �

Claim 2. A/µA ' B.


Now, we will discuss the case K = ∅, i.e. B is an ideal-free algebra and I =IntB = B. Put A = B × κ and define operations on B for every symbol F ∈ Σ ofarity n by

• if F is unary, then FA((a, α)) = (FB(a), α) for every a ∈ B, 0 ≤ α < κ;• if F is at least binary, then

FA((a1, α1), . . . , (an, αn)) = (FB(a1, . . . , an), α1 ∗ (α2 � · · · � αn))

for all (a1, α1), . . . , (an, αn) ∈ A such that ai 6= aj for some i, j = 1, . . . , n;• if F is at least binary, then

FA((a, α1), . . . , (a, αn)) = (FB(a, . . . , a), (α2 � · · · � αn) ∗ α1)

for all a ∈ B, 0 ≤ α1, . . . , αn < κ.For a ∈ B, let Ba = {a} × κ.

Claim 3. The algebra A is subdirectly irreducible and

µA =⋃

a∈B

(Ba ×Ba).

Proof. We have to show that µA =⋃

a∈B(Ba × Ba) ⊆ ν for any non-trivial con-gruence ν of A. Proceeding similarly as in the proof of Claim 1, it is sufficient tocheck that Ba ×Ba ⊆ ν for at least one a ∈ B. This will be done in the next threesteps.

(1) Assume that ((a, α), (a, β)) ∈ ν for some a ∈ B and 0 ≤ α < β < κ. Now,using the left translations instead of the right ones, we can proceed similarly as inClaim 1 (1).

(2) Assume that ((a, α), (b, β)) ∈ ν for some a, b ∈ B, a 6= b, and 0 ≤ α < β < κ.If κ ≥ 3, then we can proceed similarly as in Claim 2 (2); we have to choosea 6= c 6= b. The case κ = 2 is clear.

(3) Assume that ((a, α), (b, α)) ∈ ν for some a, b ∈ B, a 6= b and 0 ≤ α < κ.There is 0 ≤ β < κ such that α∗β 6= β∗α, and hence, using the right translation by(a, β), (a, ε), . . . , (a, ε), we get ((FB(a, a, . . . , a), β ∗α), (FB(b, a, . . . , a), α ∗ β)) ∈ ν.Now, either (1) or (2) takes place. �

Claim 4. A/µA ' B.



Now, the proof of Theorem 3.1 is completed. �

3.2. Corollary. Let B be a finite algebra with a rich signature, |B| = n and| Int(B)| = m. Then there exists a finite subdirectly irreducible algebra A suchthat A/µA ' B. Moreover, A can be chosen so that

(1) |A| = 2 if n = 1;(2) |A| = 3 if n = 2;(3) |A| = m2 + (n−m) if n ≥ 3 and n ≤ 2m;(4) |A| = (m+ 1)(n−m) if n ≥ 3 and n > 2m.

Proof. Clear from 1.5 and from the construction above. �

4. Monounary algebras

An algebra with signature consisting of one unary symbol is called monounary.The operation is denoted F . We find all subdirectly irreducible monounary alge-bras, compute their congruence lattices and characterize their homomorphic images.

4.1. Lemma. Let A be a subdirectly irreducible monounary algebra. Then everyfinitely generated subalgebra of A is finite.

Proof. Let B be an infinite subalgebra of A generated by a finite set X. For everyn let Bn denote the subalgebra of A generated by {(FA)n(a) : a ∈ X}. ThenB = B0 ⊃ B1 ⊃ B2 ⊃ . . . and

⋂nBn = ∅. Consequently,

⋂n ρBn = idA and A is

not subdirectly irreducible, a contradiction. �

4.2. Lemma. Let A be a subdirectly irreducible monounary algebra. Then there donot exist a, b ∈ A, a 6= b, satisfying FA(a) = FA(b) /∈ {a, b}.

Proof. Assume there are such a, b ∈ A. Put X = {a, b} and let Y be the subalgebragenerated by a and Z the subalgebra generated by b. Then ρX , ρY and ρZ arecongruences, |X ∩ Y ∩Z| ≤ 1 and thus ρX ∩ ρY ∩ ρZ = idA. Hence a contradictionwith subdirect irreducibility of A. �

We define the following monounary algebras.• the cycle of length n: Cn = {0, . . . , n− 1}, FCn(k) = k+ 1 mod n for all k;• the path of length n: Pn = {0, . . . , n − 1}, FPn(k) = k − 1 for all k 6= 0,FPn(0) = 0;

• the infinite path: Pω = {0, 1, 2, . . . }, FPω (k) = k−1 for k 6= 0, FPω (0) = 0.

��Cn Pn Pω

4.3. Corollary. Let A be a subdirectly irreducible monounary algebra. Then everycomponent is isomorphic either to a cycle or to a path.

4.4. Lemma. For every n(1) the congruence lattice of Cn is isomorphic to the lattice of divisors of n;


(2) the congruence lattice of Pn is a linear order on n elements;(3) the congruence lattice of Pω is isomorphic to the linear order (ω,≤).

Proof. (1) For k|n put νk = {(a, b) ∈ Cn×Cn : k|(a−b)}. It is clearly a congruenceand n

k ↔ νk is a lattice isomorphism.(2),(3) For every k, let Xk denote the subalgebra of Pk generated by k. Then

k ↔ ρXkis a lattice isomorphism. �

Let A, B be two unary algebras. We denote A+B the disjoint union (categoricalcoproduct) of A and B.

�Cn + C1

4.5. Lemma. A monounary algebra A + C1 is subdirectly irreducible, if and onlyif A is isomorphic to Cn, where n is a prime power or n = 1.

Proof. C1 + C1 is obviously subdirectly irreducible. Now, assume |A| ≥ 2. IfA + C1 is subdirectly irreducible, then, due to 4.3, A is isomorphic to a cycle orto a path. First, if A = Cn, ν is a congruence on Cn + C1 and (k, x) ∈ ν, wherek ∈ Cn and x ∈ C1, then ν = (Cn + C1)× (Cn + C1); thus Cn + C1 is subdirectlyirreducible, if and only if Cn is so. On the other hand, for arbitrary 2 ≤ n ≤ ω isInt∗(Pn + C1) = {0} and thus, by 1.6, Pn + C1 is not subdirectly irreducible. �

4.6. Theorem. The following are (up to an isomorphism) all subdirectly irreduciblemonounary algebras:

(1) Cn, where n is a prime power;(2) Cn + C1, where n is a prime power or n = 1;(3) Pn for 2 ≤ n ≤ ω.

The only simple monounary algebras are P2, C1 + C1 and Cp, where p is a primenumber.

Proof. Let A be a subdirectly irreducible monounary algebra. Due to 1.8, A iseither connected, or it has two components and one of them is isomorphic to C1.In the first case, by 4.3 and 4.4, A is isomorphic either to Cn, n a prime power, orto Pn, where 2 ≤ n ≤ ω. In the latter case, use 4.5. �

4.7. Corollary. The congruence lattice of a subdirectly irreducible monounary al-gebra is a linear order.

Proof. This is a direct consequence of 4.4, the proof of 4.5 and 4.6. �

4.8. Theorem. The following conditions are equivalent for a monounary algebraA:

(i) A is a subdirectly irreducible or |A| = 1.(ii) A is a homomorphic image of a subdirectly irreducible monounary algebra.(iii) A is isomorphic to B/µB for a subdirectly irreducible monounary algebra

B.


Proof. An easy computation based on 4.4, 4.5 and 4.6. �

The characterization of finite subdirectly irreducible monounary algebras can befound in the paper [60] of M. Yoeli. His approach is different: he used in his searchthe original definition of subdirect irreducibility (saying that an algebra is s.i., if itis not a proper subdirect product). Our proof is much shorter.

5. Unary algebras

We provide a couple of examples which show that the problem for unary algebrasis quite different. First, the following example, due to Jezek, Kepka [19], shows aunary algebra which is a homomorphic image of a subdirectly irreducible one, butwhich is not a factoralgebra of any subdirectly irreducible unary algebra over itsmonolith.

Example. Let A be a κ-unary algebra on the set {a, b, c, d} with all operationsequal to a, b 7→ c and c, d 7→ d.

�Assume that A = B/µB . Clearly, ρc∪d is a congruence of B. If |c| = 1, thenρa∪b is a congruence of B and ρa∪b ∩ ρc∪d = idB , a contradiction. Hence |c| ≥ 2.Thus also |d| ≥ 2, otherwise there are congruences ρc ∩ ρX = idB , where X is thesubalgebra of B generated by some x ∈ c. However, then µB ⊆ ρd and we get acontradiction with |c| ≥ 2.

Now, assume an algebra B on the set {a, b, c, c′, d, d′} with one operation defined

by a 7→ c 7→ d 7→ d, b 7→ c′ 7→ d′ 7→ d and all other oprations defined by a 7→ c 7→d 7→ d′, b 7→ c′ 7→ d′ 7→ d′. It is easy to check that B is subdirectly irreducible andA is a homomorphic image of B via a 7→ a, b 7→ b, c, c′ 7→ c and d, d′ 7→ d.

Another important difference is that there is a bound on the size of subdirectlyirreducible unary algebras (a thus, indeed, their homomorphic images): it is well-known that subdirectly irreducible κ-unary algebras have at most 2κ·ℵ0 elements.We have no idea whether every κ-unary algebra A with non-empty Int∗(A) and|A| ≤ 2κ·ℵ0 is a homomorphic image of a subdirectly irreducible one. The followingtwo examples deal with “the simplest case”.

Example. Let κ be infinite. Let A be a κ-unary algebra on the set 2κ ∪ {•} withall operations constant onto •. Let B be a κ-unary algebra on the set 2κ ∪ {0, 1}defined in the following way: 0, 1 are fixed by each operation and let the result ofthe α-th operation on f ∈ 2κ be f(α). It is easy to check that B is subdirectlyirreducible, µB = {0, 1}2 ∪ idB and B/µB ' A.

Example. Let κ be finite. Let A be a κ-unary algebra on the set 2ℵ0 ∪ {•} withall operations constant onto •. We show that there is no subdirectly irreducible Bsuch that B/µB ' A. (We do not know, whether there is an s.i. B such that A is


a homomorphic image of B.) Assume there is such B. Then, of course, the onlynon-trivial block of µB corresponds to • and it is simple and thus countable (itis well-known that simple κ-unary algebras have at most κ · ℵ0 elements). Hence,there are uncountably many points, but only countably many results of operations.So there are two points a 6= b such that all operations on a and b coincide. Thusν = {a, b}2 ∪ id is a congruence of B and µB ∩ ν = idB , a contradiction.

6. Particular varieties

Once the initial problem is solved, another natural question appears: given analgebra with non-empty intersection of ideals from a variety V, is it a homomorphicimage (over the monolith) of a subdirectly irreducible algebra from the variety V?

For instance, our construction yields for an idempotent algebra an idempotentsubdirectly irreducible one (except for the case κ = 2, which can be easily solvedseparately). Actually, it is easy to check that it preserves any identity of the formt(x) ≈ x, where t is a unary term. And if the algebra contains no proper ideal, thenthe construction preserves any identity in one variable. However, it seems that noother interesting identities are kept.

Of course, the question has negative answer in many cases, such as in the va-riety of distributive lattices, which contains only one (two-element) subdirectlyirreducible algebra. On the other hand, in several familiar varieties the answeris known to be positive. S. Bulman-Fleming, E. Hotzel and J. Wang [6] answerthe question affirmatively for the variety of semigroups. However, there is a finitesemigroup (e.g. any right zero band) such that it is not isomorphic to the factorof a finite subdirectly irreducible semigroup over its monolith. Every group andevery lattice (indeed, all of them have no non-trivial ideals, in our sense) is alsoa homomorphic image of a subdirectly irreducible one and for a finite one the s.i.can be constructed finite. For lattices, an easy construction was found by RalphFreese (unpublished). For groups, the solution was pointed out by Ralph McKenziein private communication (hint: use a wreath product).

What about other familiar varieties? For instance, commutative groupoids?


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