The Mathieu groups - Luctr.maths.lu.se/media11/thesis/2009/leo-taslaman-MATM01.pdfInthe 19thcentury...

The Mathieu groups

Leo Taslaman

May 26, 2009

Abstract

In the 19th century E. Mathieu discovered and studied ve multiply transitivepermutation groups. The groups are called the Mathieu groups and it turnedout that all ve are simple. In this thesis these remarkable groups areconstructed, with special focus on the largest Mathieu group M24. Allmaximal subgroups of M24 will be described and classied. It is also shownthat the other Mathieu groups are subgroups of M24. Finally the simplicityof the ve Mathieu groups is proved.

Acknowledgements

I wish to express my deepest respect and most sincere gratitude to mysupervisor Professor Arne Meurman, for guidance, editorial support andengagement in my questions and ideas. I also wish to thank him for thechoice of subject, which suited me perfectly.

Leo Taslaman, May 2009

1

Contents

Introduction 4

Notation 5

1 Preliminaries 6

1.1 Basic denitions and lemmas . . . . . . . . . . . . . . . . . . . 61.2 Multiply transitive groups . . . . . . . . . . . . . . . . . . . . 81.3 The nite projective groups . . . . . . . . . . . . . . . . . . . 12

2 The Steiner System S(5, 8, 24) and the Golay code 15

2.1 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Basic machinery . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 MOG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Uniqueness of S(5, 8, 24) and C . . . . . . . . . . . . . . . . . 24

3 The Mathieu group M24 31

3.1 Eight maximal subgroups . . . . . . . . . . . . . . . . . . . . 323.1.1 The projective group L2(23) . . . . . . . . . . . . . . . 323.1.2 The Octad Stabilizer . . . . . . . . . . . . . . . . . . . 333.1.3 The Duum Stabilizer . . . . . . . . . . . . . . . . . . . 333.1.4 The Sextet Stabilizer . . . . . . . . . . . . . . . . . . . 343.1.5 The Trio Stabilizer . . . . . . . . . . . . . . . . . . . . 35

3.2 The involutions in M24 . . . . . . . . . . . . . . . . . . . . . . 373.3 A further maximal subgroup . . . . . . . . . . . . . . . . . . . 423.4 Proof of maximality . . . . . . . . . . . . . . . . . . . . . . . . 453.5 Generators of M24 . . . . . . . . . . . . . . . . . . . . . . . . . 58

4 Simplicity of the Mathieu groups 59

2

5 Uniqueness of the small Mathieu groups 62

A GAP and permutation groups 68

3

Introduction

The classication theorem of nite simple groups states that except for 26groups, all nite simple groups are isomorphic to a cyclic group of primeorder, an alternating group of degree at least ve or a simple group of Lietype. The 26 exceptional groups are called sporadic. The ve Mathieu groupswere the rst sporadic groups found. They were discovered and studied byEmile Mathieu in the 19th century (see [11], [12] and [13]). It took almostcentury before the next sporadic group was found. Mathieu found the groupswhen he investigated multiple transitivity and did not know that the groupswere simple. This was rst shown in the 1930's by Ernst Witt, when he didresearched on the Steiner System S(5, 8, 24) (see [14]). It turned out thatthe largest Mathieu group M24 could be dened as the automorphism groupof this system.

In Chapter 1 we will introduce the concept of multiply transitive groupsand theory that is necessary for the main part of this thesis. In Chapter 2 weuse Curtis [1] approach to discuss the Steiner system S(5, 8, 24), the Golaycode and the link in between. We will here introduce the powerful tool MOG(Miracle Octad Generator) and nd the group M24. Chapter 3 is dedicatedto the Mathieu groupM24. We will go through all maximal subgroups and onthe way nd the other four Mathieu groupsM11,M12,M22 andM23. A majorpart of this chapter discuss a proof that simultaniously shows the maximalityof claimed maximal groups and that there are no further maximal groups.In Chapter 4 we show simplicity of the Mathieu groups and in Chapter 5 wegive a new uniqueness proof of the small Mathieu groups, i.e M11 and M12.

4

Notation

1 the identity element

〈1〉 the trivial group

Ga the isotropy subgroup of G xing a

Gρ the conjugate ρGρ−1

A.B a group G such that ACG and G/A ∼= B

Fq the nite eld of order q

pk the elementary abelian group of order pk, where p is a prime

Ln(q) the projective linear group PSLn(Fq)

MX−1 the isotropy subgroup of the Mathieu group MX

Zm cyclic group of order m

Sn the symmetric group of degree n

An the alternating group of degree n

C the Golay code

H the extended (8,4) Hamming code

A ⊆ B A is a subset of B

A ⊂ B A is a proper subset of B

H ≤ G H is a subgroup of G

H < G H is a proper subgroup of G

5

1. Preliminaries

1.1 Basic denitions and lemmas

Lemma 1.1. If H ≤ G then NG[H]/CG[H] is isomorphic to a subgroup ofAut(H).

Proof. The kernel of the homomorphism

φ : NG[H] −→ Aut(H)

∈ ∈

g 7→ (h 7→ ghg−1)

is CG[H]. So the lemma follows from the rst isomorphism theorem.

Lemma 1.2. Let G be a group and N a normal subgroup of G. If P is aSylow p-subgroup of N then G = NG[P ]N .

Proof. Sylow's second theorem and the fact that N is normal implies that∀g ∈ G ∃n ∈ N such that P g = P n. Thus gn−1 ∈ NG[P ], implyingg ∈ NG[P ]N .

Lemma 1.3. If ρ and σ are two elements in a permutation group then σρσ−1

and ρ has the same cycle shapes. Further if (i1, i2, ..., ik) is a cycle in ρ then(σ(i1), σ(i2), ..., σ(ik)) is a cycle in σρσ−1.

Proof. If ρ = ρ1ρ2 . . . ρk, where ρi are disjoint cycles, then

σρσ−1 = σρ1σ−1σρ2σ

−1 . . . σρkσ−1.

So it is enough to prove the lemma for one arbitrary cycle. Let x = (i1, i2, . . . , in)and σ(im) = jm for m = 1, 2, . . . , n. Then it is straight forward to check thatσxσ−1 = (j1, j2, . . . , jn).

6

Remark 1.4. If σ centralizes ρ the above lemma implies that σ permutes thexed points of ρ.

We say that a permutation group is 1/2-transitive if all orbits havethe same length and transitive if there is only one orbit. If all isotropysubgroups are trivial we say that the group is semiregular. A semiregulartransitive group is called regular.

If a group G is transitive on a set A then there is a one to one correspondencebetween the cosets of an isotropy subgroup Ga and the elements in A, namelyhGa ↔ b if and only if h(a) = b. Thus we have the shown the for this textvery important lemma.

Lemma 1.5. If G is transitive on A and a ∈ A then |A| = [G : Ga].

Corollary 1.6. If G acts regularly on a set A then |A| = |G|.

If G is a transitive group acting on a set A we say that a subset B ⊆ A,|A| > |B| > 1, is a set of imprimitivity if ∀g ∈ G gB ∩ B = B orgB ∩ B = ∅. If we remove the condition on the size of B we say that B isa block. Clearly a set of imprimitivity yields a natural partition of A intosubsets of size |B|. Thus |B|

∣∣|A|. If no set of imprimitivity exist we say thatG is primitive. If |A| is a prime G is obviously primitive.

Lemma 1.7. If G is a transitive permutation group and N C G then N is1/2-transitive. If in addition N is nontrivial and G is primitive then N istransitive.

Proof. Let S be an orbit of N of minimal length. Then ∀g ∈ G we haveN(gS) = (Ng)S = (gN)S = g(NS) = gS so gS is a union of orbits of N .Since S is of minimal length and |S| = |gS| we see that gS is exactly oneorbit. Now G is transitive so every orbit of N has the same length (havingthe form gS). Thus N is 1/2 transitive. Finally two dierent orbits aredisjoint so S form a set of imprimitivity for G. So if G is primitive then wehave two cases. Either |S| = 1 implying N is trivial, or |S| = |A| implyingN is transitive.

Theorem 1.8. Let G be transitive on a set A. If and a ∈ A then G isprimitive if and only if Ga is a maximal subgroup.

7

Proof. If G is not primitive there is a set of imprimitivity B containing a.We want to show that Ga < GB < G. As G is transitive G 6= GB. Since Gis transitive ∃g ∈ G such that ga ∈ B \ a. Thus g ∈ GB \Ga.Conversely suppose there exist H such that Ga < H < G. We want to showthat Ha = B is a set of imprimitivity. Since Ga < H we have |B| > 1. If|A| = |B| then H is transitive and by Lemma 1.5 |A| = [G : Ga] = [H : Ga].Contradiction. Finally suppose that gB ∩ B 6= ∅ and b ∈ gB ∩ B, then∃h1, h2 ∈ H such that h1a = gh2a = b and hence h−1

1 gh2 ∈ Ga. This yieldsg ∈ H and gB ∩B = B.

1.2 Multiply transitive groups

A permutation group G on set A is called k-fold transitive if for every twoordered subsets of A, (a1, a2, . . . , ak), ai 6= aj and (b1, b2, . . . , bk), bi 6= bj,there is a permutation ρ ∈ G such that ρ(ai) = bi for i = 1, . . . , k. If inaddition the identity is the only permutation xing k elements we say thatG is sharply k-fold transitive. Note that if G is sharply k-fold transitiveand σ is another element satisfying σ(ai) = bi for i = 1, . . . , k, then ρσ−1

xing k elements and thus equals the identity. Hence a permutation group issharply transitive if and only if for each pair of ordered sets (a1, a2, . . . , ak)and (b1, b2, . . . , bk) there is exactly one permutation in G satisfying ai 7→ bi.If G is a sharply k-fold transitive on a set A, |A| = n, and ai, bi ∈ A fori = 1, . . . , n, then there is a one to one correspondence between ordered sets(b1, b2, . . . , bk) and permutations in G, namely:

(b1, b2, . . . , bk)↔(a1 a2 . . . ak . . .

b1 b2 . . . bk . . .

).

Thus |G| = n(n− 1) · · · (n− k + 1).

Example 1.9. It is obvious that Sn is sharply n-fold transitive. Since (n−1)-fold transitivity implies n-fold transitivity we see that An is(n− 2)-fold transitive. There is exactly two permutations in Sn xing n− 2elements and only one of them is even. Thus An is sharply (n − 2)-foldtransitive. Alternatively we can see this by considering the order of An.

N

Lemma 1.10. Let G be a transitive permutation group on a set A where|A| ≥ 2. Then G is k-fold transitive if and only if Ga is (k− 1)-transitive onA \ a.

8

Proof. If G is k-fold transitive Ga obviously (k − 1)-fold transitive. Forthe other direction pick any two ordered subsets of A, a1, a2, . . . , ak andb1, b2, . . . , bk. Since G is transitive we can nd ρ ∈ G such that ρ(a1) = b1.Now use the (k − 1)-transitivity and take σ ∈ Gb1 such that σρ(ai) = bi. Itfollows that σρ ∈ G satises ai 7→ bi for i = 1, . . . , k.

Lemma 1.11. If G is doubly transitive on a set A then G is primitive.

Proof. Assume B ⊂ A is a set of imprimitivity and pick a, b ∈ B and c ∈A \ B. Since G is doubly transitive we can nd g ∈ G such that g(a) = aand g(b) = c, contradicting that B is a set of imprimitivity.

Remark 1.12. Note that together with Lemma 1.8 this says that doublytransitivity implies that the isotropy groups are all maximal subgroup.

So far the only multiply transitive groups we have seen is the symmetricand alternating groups. We consider these groups trivial and start the workof ruling out the possibility for a nontrivial group to be more than sharply5-fold transitive.

Lemma 1.13. Let G be a k-fold transitive group and H a subgroup xing kpoints. If P is a Sylow p-subgroup of H that xes m ≥ k points then NG[P ]is k-fold transitive on the m points xed by P .

Proof. Let a1, a2, . . . , ak be the points xed H and b1, b2, . . . , bk k arbitrarypoints xed by P . We want to show that there is an element in NG[P ]mapping ai 7→ bi for i = 1, . . . , k. Since G is k-fold transitive there is atleast one such element g ∈ G. So we have g(ai) = bi for i = 1, . . . , k. Now Pxes b1, b2, . . . , bk and so P g must x a1, a2, . . . , ak and thus be another Sylowp-subgroup of H. By the second Sylow theorem ∃h ∈ H such that P = P gh.Finally gh ∈ NG(P ) and gh(ai) = g(ai) = bi for i = 1, . . . , k.

We need the following lemma to achieve contradictions later on.

Lemma 1.14. If G is a sharply k-fold transitive group of degree n then wecannot have (k = 4 and n = 10) or (k = 6 and n = 13).

Proof. We deal with the two very similar cases separately.

Case k = 4 and n = 10We know that |G| = 10 · 9 · 8 · 7 and so there is Sylow 7-subgroup P . Since

9

P must have order seven it is generated by a 7-cycle, say g = (1 2 3 4 5 6 7).Further G is 3-transitive and P is a Sylow 7-subgroup xing the points 8,9and 10. Thus Lemma 1.13 implies that NG[P ] is 3-transitive on 8, 9, 10 andwe have a restriction homomorphism ϕ : NG[P ] −→ S3. From Lemma 1.1 weknow that NG[P ]/CG[P ] ∼= H ≤ Aut(P ) ∼= Z6. As S3 is not abelian ϕ[CG[P ]]cannot be trivial and so since it is a normal subgroup of S3 it is isomorphic toeither A3 or S3. In either case 3

∣∣|CG[P ]| and we can nd h ∈ CG[P ] of order3. But then gh has order 21 and (gh)7 6= 1 xes seven points. Contradiction.

Case k = 6 and n = 13Now |G| = 13 · 12 · 11 · 10 · 9 · 8 so G has a Sylow 5-subgroup P of order ve.Hence P = 〈g〉 for some g. If g has only one 5-cycle then it xes eight points.Contradiction. So g consist of two 5-cycles, say g = (1 2 3 4 5)(6 7 8 9 10).With the same argument as in the previous case CG[P ] contains an elementh of order 3. Further the element gh has order 15 and must consist of 3-cycles and 5-cycles. Since (gh)6 = g we see that gh has two 5-cycles and one3-cycle. But then (gh)5 6= 1 xes ten points. Contradiction.

Lemma 1.15. The involutions in S4 that acts without xed points formtogether with the identity a regular normal subgroup.

Proof. It straight forward to check thatH = 1, (12)(34), (13)(24), (14)(23)forms a regular subgroup. Since conjugation preserves cycle shapes and Hcontains all elements of cycle shape 22 the normality follows.

We now show that there are at most two choices for the degree of a sharplyk-fold transitive group, for k ≥ 4.

Theorem 1.16. If G is a nontrivial sharply k-fold transitive group of degreen, where k ≥ 4, then (k = 4 and n = 11) or (k = 5 and n = 12).

Proof. Suppose G is sharply 4-transitive. If we consider the order G we seethat the degree cannot be four, ve or six since it then would be isomorphicto S4, S5 and A6 respectively. Assume G is of degree seven. Then |G| = 7!/6and [S7 : G] = 6. Since S7 permutes the left cosets of G transitively wehave a homomorphism ϕ : S7 −→ S6. Further ϕ cannot be injective so thekernel is nontrivial. Since the only nontrivial proper normal subgroup of S7

is A7 we must have Ker(ϕ) = A7. A contradiction since S7/A7∼= 2 cannot

be transitive on six elements. Thus n ≥ 8.

10

Next we want to show that any two nonidentity involutions are conjugates.Since such an involution xes at most three points it has at least two 2-cycles.Let x = (a b)(c d) . . . and y = (e f)(g h) . . . . Since G is 4-transitivity we

can nd z =

(a b c d . . .

e f g h . . .

)and so zxz−1 and y agree on four points. Thus

zxz−1 = y.

Let g = (1)(2)(3 4) . . . and h = (1 2)(3)(4) . . . and suppose g, h ∈ G. Sinceg2 and h2 xes four points we have g2 = h2 = 1. Further gh and hg agreeon four points so gh = hg. Let k = gh = (1 2)(3 4) . . . . Now g might haveanother xed point, say 7. To avoid writing two cases we will always whenwriting (7) be open for the possibility that this cycle do not occur. Sinceh centralize g Remark 1.4 implies that also h contains (7). Now since bothg and k are nonidentity involutions they are conjugates and have the samecycle shape. So k xes say 5 and 6. Using Remark 1.4 again, we get that gand h both contains (5 6). Hence

g = (1)(2)(3 4)(5 6)(7)

h = (1 2)(3)(4)(5 6)(7)

k = (1 2)(3 4)(5)(6)(7)

on 1, 2, . . . , 7. These generate an elementary abelian subgroup 〈g, h, k〉 =H of order four. Assume x ∈ CG[H] and x 6= 1. As above x must permuteall xed points to g, h and k each, and thus looks like

x = (1 2)σ1(3 4)σ2(5 6)σ3(7) . . .

where σ1, σ2, σ3 = 0 or 1. Since x xes at most three points at least two ofσ1, σ2 and σ3 equals one. If exactly two of them equals one we have x ∈ Hand if all three equals one xg 6= 1 xes four points. Contradiction. HenceH = CG[H] and by Lemma 1.1 we have |NG[H]/CG[H]| ≤ |Aut(H)|. Now His elementary abelian of order four so Aut(H) ∼= GL2(2) and |Aut(H)| = 6.Thus |NG[H]| ≤ 24. Now n ≥ 8 implies that H has an orbit O disjoint from1, 2, 3, 4, 5, 6, 7. Since no nonidentity element in H can have more xedpoints, H acts regularly on O. Thus Corollary 1.6 implies that |O| = 4. AsG is sharply 4-transitive there exist a subgroup W ∼= S4 that permutes theelements in O. By Lemma 1.15 H CW . Since |W | = 24 and NG[H] ≤ 24 soNG[H] = W . Let y ∈ NG[H] \ H be an involution. If H has another orbit

11

P then Lemma 1.15 implies that y has two xed points in both O and P .Contradiction. Thus n = 10 or 11 depending on whether (7) occurs or not.But Lemma 1.14 forbid the case k = 4 and n = 10. Thus n = 11.

Let k ≥ 5 and suppose Ga is sharply (k − 1)-fold transitive of degree n− 1.If Ga is one of the trivial cases Ga has order (n− 1)! or (n− 1)!/2, implyingthat G has order n! or n!/2 and thus is also trivial. Assume G is nontrivial ofdegree n. If k = 5 thenGa is sharply 4-fold transitive and of degree n−1 = 11.Thus n = 12. If k = 6 then Ga is 5-transitive of degree n − 1 = 12 fromwhat we just proved. This yields k = 6 and n = 13 contradicting Lemma1.14. Finally if k > 6 we get hold of case k = 6 by looking at the pointwisestabilizer of k − 6 points.

1.3 The nite projective groups

The general linear group GLn(q) is dened as the set of invertible n × n-matrices with entries in Fq. The special linear group SLn(q) ≤ GLn(q) isdened as the subgroup of matrices with determinant one. Let V be the setof n × 1 vectors over Fq. The projective space Pn−1 is dened as the set ofone-dimensional subspaces of V . We call P1 the projective line. If v ∈ V wewrite the respective element in Pn−1 as 〈v〉. Clearly |Pn−1| = (qn−1)/(q−1).Since GLn(q) permutes the elements in Pn−1 there is an homomorphism

ϕ : GLn(q) −→ SPn−1 .

Lemma 1.17. The kernel Ker(ϕ) consists of the matrices of the form cI,c ∈ F×q .

Proof. Obviously cI ∈ Ker(ϕ) for c ∈ F×q . For the other direction letA ∈ Ker(ϕ). Let e1, e2, . . . , en be the natural basis in Fnq . The calculation

Aei = aiei, for some ai ∈ F×q

shows that the ith column in A equals aiei. As this holds ∀i, A is diagonal.Finally

aiei + ajej = A(ei + ej) = c(ei + ej)

shows that c = ai = aj for all i and j. Thus A = cI.

12

If we in GLn(q) and SLn(q) factor out respective kernel of ϕ we get twoprojective groups. As permutation groups these act faithfully on Pn−1. Thelatter group

Ln(q) = SLn(q)/(SLn(q) ∩Ker(ϕ)),

plays an important role in this thesis.

Lemma 1.18. Let q be a prime power. Then

(1) |GLn(q)| = (qn − 1)(qn − q) · · · (qn − qn−1)

(2) |SLn(q)| = |GLn(q)|/(q − 1)

(3) |Ln(q)| = |SLn(q)|/(n, q − 1).

Proof. Let A ∈ GLn(q). The rst column can be chosen in qn − 1 ways.Suppose we already have chosen i columns. Since the columns in A arelinearly independent we can chose the (i+ 1)th column in qn− qi ways. Thisyields (1). Further SLn(q) is the kernel of the determinant homomorphism.As the determinant clearly is onto (2) follows. For (3) we notice that SLn ∩Ker(ϕ) consist of all matrices cI where cn = 1. Since F×q form a cyclic group

the number of such c's is the number of solutions to x(n,q−1) = xan+b(q−1) = 1in F×q . Thus |SLn(q) ∩Ker(ϕ)| = (n, q − 1).

Theorem 1.19. Let n ≥ 2. As a permutation group on Pn−1, Ln(q) is a2-transitive.

Proof. Let e1, e2, . . . , en be the natural basis in Fnq and 〈e1〉, 〈e2〉, . . . , 〈en〉the respective elements in PL. Pick 〈x1〉, 〈x2〉 ∈ PL, 〈x1〉 6= 〈x2〉. Clearlythe representatives x1, x2 ∈ Fnq are linearly independent, so we can extendthem to a basis x1, x2, . . . , xn. Choose A ∈ GLn(q) such that Aei = xi fori = 1, . . . , n. Scale the rst column in A with 1

det(A)and call the new matrix

B. Clearly ϕ(B) ∈ Ln(q) have the action 〈e1〉 7→ 〈x1〉 and 〈e2〉 7→ 〈x2〉.

If n = 2 then |P1| = q + 1. We see that P1 consist 〈(0, 1)T 〉 together with

〈(x, 1)T 〉, x ∈ Fq. Let(a b

c d

)∈ SLn(q). Since

a b

c d

x

1

=

ax+ b

cx+ d

and 〈(ax+b, cx+d)T 〉 = 〈(ax+ b

cx+ d, 1)T 〉

13

if cx+ d 6= 0 and 〈(1, 0)T 〉 otherwise,

we see that it is convenient to rename 〈(1, 0)T 〉 =∞ and 〈(x, 1)T 〉 = x. Withthe convention 1/0 =∞, L2(q) consist of all Möbius transformations

x 7→ ax+ b

cx+ d, ad− bc = 1, a, b, c, d ∈ Fq.

Lemma 1.20. The group L2(q) is generated by α : x 7→ x + 1 andγ : x 7→ − 1

x.

Proof. Dene βa : x 7→ a2x ∀a ∈ Fq. If c = 0 then d = a−1 and

ax+ b

cx+ d= a2x+ ab = αabβa,

otherwiseax+ b

cx+ d=a

c− 1

c2x+ cd= αa/cγαcdβc.

Finally 1 1

0 1

1 0

a− 1 1

1 −a−1

0 1

1 0

a− a2 1

=

a 0

0 a−1

,

were the matrices represent α, γαa−1γ, α−a−1, γαa−a

2γ and βa respectively,

shows α and γ generate every βa. Hence 〈α, γ〉 = L2(q).

14

2. The Steiner System S(5, 8, 24)and the Golay code

2.1 Construction

We will in this section use Curtis [1] approach and construct the SteinerSystem S(5, 8, 24) and the Golay code simultaneously.

Denition 2.1. A Steiner System S(i, j, k) is a family of j-element subsetsof a k-element set, Ω, with the property that any i elements in Ω lie in exactlyone of them.

If a Steiner System S(i, j, k) exist it is of size

(k

i

)/(ji

). It can be seen

as the number of ways we can choose i points divided by the number of wayseach j-set can be represented by i points. This rules out a lot of cases sincethe quotient has to be an integer to make sense. For example the SteinerSystems S(2, 4, 8) and S(4, 7, 14) cannot exist.

The equality

(24

5

)/(8

5

)= 759 shows that there is a chance for the Steiner

System S(5, 8, 24) to exist and the following theorem shows that this is thecase. In the remaining part of this thesis we assume that

Ω = ∞, 0, 1, 2, 3, ..., 22.

Why we use these numbers will become more clear later when we come tothe projective subgroup.

Theorem 2.2. There exists a Steiner System S(5, 8, 24).

15

Method of Proof. We will think of the power set P (Ω) as a 24 dimensionalvector space over F2 in the following way. Each element in Ω represent acoordinate in the vector space. If a member of P (Ω) contains an elementthe corresponding coordinate is one, otherwise zero. Further the symmetricdierence is used as addition. In this vector space we shall nd a linear code,C , were all non-empty code words have weight at least eight. Clearly no pairof code words has ve points in common since their sum then would haveweight less than eight. Moreover C will contain 759 codewords with weightexactly eight and so form a Steiner System S(5, 8, 24).

Proof. Let Λ be an eight element set and consider the vector space P (Λ). Topleasure the eye we will in an obvious manner use a matrix notation on thevectors. Dene the two three dimensional subspaces

P =

0 0

0 0

0 0

0 0

0

,

0 0

0 0

1 1

1 1

A

,

0 0

1 1

0 0

1 1

B

,

0 0

1 1

1 1

0 0

C

,

0 1

0 1

0 1

0 1

D

,

0 1

0 1

1 0

1 0

E

,

0 1

1 0

0 1

1 0

F

,

0 1

1 0

1 0

0 1

G

L =

0 0

0 0

0 0

0 0

0

,

0 1

1 1

0 1

0 0

a

,

0 1

0 0

1 1

0 1

b

,

0 1

0 1

0 0

1 1

c

,

0 1

1 0

1 0

1 0

d

,

0 0

1 0

0 1

1 1

e

,

0 0

1 1

1 0

0 1

f

,

0 0

0 1

1 1

1 0

g

Note that for every element l ∈ L\0 there are exactly three nonzero elementsin P that cuts l evenly. We now extend the dimension by taking three copiesof Λ and glue them together and identify

Ω =Λ1 Λ2 Λ3

We will call Λ1,Λ2 and Λ3 bricks. To describe the sets in P (Ω) we usenotation [R, S, T ]t, where R+ t is the intersection with Λ1, S+ t with Λ2 andT + t with Λ3. If t = 0 we simply write [R, S, T ]. We now dene C as theimage of

φ : F2 × F2 × F2 × P × P × L −→ P (Ω)

16

(σ1, σ2, σ3, X, Y, t) 7→ [X + σ1Λ1, Y + σ2Λ2, X + Y + σ3Λ1]t

The space above to the left has dimension 12 and φ is obviously a linearmap, so the dimension of C is 12-dim(Ker(φ)) by the dimension theorem.Let x = (σ1, σ2, σ3, X, Y, t) ∈ Ker(φ). Since the image of x is zero theintersections with the three bricks all has to be zero and so the sum of theintersections, seen as members of P (Λ), has to be zero. This implies that tis zero and we have Ker(φ) = 0. Thus C has dimension 12.

To investigate the sizes of sets in C we let X, Y ∈ P \ 0, X 6= Y andt ∈ L \ 0. Up to permutation of the three bricks and complementationmodulo Λi there are eight cases. In the Table 1 we see how each case cutsthe three bricks.

Case Shape Λ1 Λ2 Λ3

1 [0, 0, 0] 0/8 0/8 0/8

2 [X, X, 0] 4/4 4/4 0/8

3 [X, Y, X + Y ] 4/4 4/4 4/4

4 [0, 0, 0]t 4/4 4/4 4/4

5 [X, X, 0]t |X ∩ t| ≡ 0 (mod 2) 4/4 4/4 4/4

6 [X, X, 0]t |X ∩ t| ≡ 1 (mod 2) 2/6 2/6 4/4

7 [X, Y, X + Y ]t |X ∩ t| ≡ |Y ∩ t| ≡ 0 (mod 2) 4/4 4/4 4/4

8 [X, Y, X + Y ]t |X ∩ t| ≡ |Y ∩ t| ≡ 1 (mod 2) 2/6 2/6 4/4

Table 1

The notation m/n in column Λi means that the case cut Λi in m or n points,depending on the value of σi in the function φ above. It is worth notingthat m/n gives the same information as n/m. The only nontrivial cases are7 and 8 and especially their cut with Λ3. In both cases X+Y ∈ P imply that|(X + Y ) ∩ t| equals 1,2 or 3. On the other hand|(X + Y ) ∩ t| = |(X ∩ t) + (Y ∩ t)| is easily seen even, so it must be 2.Further we have |X + Y + t| = |X + Y |+ |t| − 2|(X + Y ) ∩ t| = 4.

From Table 1 we see that C contains sets of size 0, 8, 12, 16 and 24. We callthe subset of octads (i.e 8-sets) C8 and calculate its cardinality.

17

case shape number of elements

1 [0, 0, 0] 3

2 [X, X, 0] 7 · 3 · 2 · 2 = 84

6 [X, X, 0]t 7 · 4 · 3 · 2 = 168

8 [X, Y, X + Y ]t 7 · 4 · 3 · 3 · 2 = 504

Table 2

Folding up this gives 759 octads.

Denition 2.3. The linear code C described in Theorem 2.2 is called The(extended) Golay code.

The code C contains 212 = 4096 codewords. From the construction wesee that they have weight 0, 8, 12, 16 and 24. From the equality

1 + 759 + 2576 + 759 + 1 = 4096

we see that the number dodecads (i.e codewords of weight twelve) is 2576and so C has weight enumerator

x24 + 759x16y8 + 2576x12y12 + 759x8y16 + y24. (2.1)

We will later see that the Golay code is the unique linear code with thisweight enumerator.

2.2 Basic machinery

In this section we will introduce some useful lemmas for later use. Untilwe have proved uniqueness of the Steiner System S(5, 8, 24) we denote anarbitrary such system C8.

Lemma 2.4. If A,B ∈ C8 and |A ∩B| = 4 then A+B ∈ C8.

Proof. Let A = a1, a2, a3, a4, a5, a6, a7, a8, B = a1, a2, a3, a4, b1, b2, b3, b4where a5, a6, a7, a8 ∩ b1, b2, b3, b4 = ∅. Assume A + B is not an octad.The octad X containing a5, a6, a7, a8, b1 cannot contain a further a and sofor the intersection with B to be even it has to contain a further b, say b2.Similarly the octad Y containing a5, a6, a7, a8, b3 must contain a further b,

18

but this b cannot be b1 or b2 without equal X and implying X = Y = A+B.Assume b4 ∈ Y . The octad containing a5, a6, a7, b1, b3 must contain a furthera. If it contains a8 it will have ve points in common with X and Y and oncemore X = Y = A+ B, so assume it contains a1. But now it must contain afurther b and equals X and Y . Contradiction.

Corollary 2.5. For each tetrad, i.e a set of size 4, there exist an uniquepartition of Ω into six tetrads such that the union of any two is an octad.

Proof. Let S be a tetrad. Then each point not in S denes another tetrad Tsuch that S+T ∈ C8. Assume T and T ′ are two dierent such tetrads. Thenthey are disjoint since otherwise S + T and S + T ′ would have ve pointsin common. Let T1 and T2 be to dierent tetrads in the partition. Then|(S + T1) ∩ (S + T2)| = 4 and by Lemma 2.4 the union is an octad.

Such a partition will be called a sextet.

Lemma 2.6. An octad cut the tetrads of a sextet 42.04, 31.15 or 24.02.

Proof. Octads intersect each other in 0, 2, 4 or 8 elements. We divide theproblem in three cases. The largest cut with a tetrad is of size

4. For the octad to cut all other octads found in the sextet properly, ithas to cut one more tetrad in four points and so be disjoint from theremaining. Thus the cut is 42.04.

3. Again due to the intersection of the octads found in the sextet, theintersections with the other ve tetrads are all of size one and we havethe cut 31.15.

2. Since the cut is even the cuts with every tetrad has to be even. Thusthe cut is 24.02.

Lemma 2.7. If x1, x2, ..., x8 is an octad, then the number of octads intersectingx1, x2, ..., xi in x1, x2, ..., xj is the (j + 1)th entry in the (i+ 1)th line inTable 3.

19

759506 253

330 176 77210 120 56 21

130 80 40 16 578 52 28 12 4 1

46 32 20 8 4 0 130 16 16 4 4 0 0 1

30 0 16 0 4 0 0 0 1

Table 3

Proof. The number of octads containing x1, . . . , xi is(

24− i5− i

)/(8− i5− i

)if

i < 5 and 1 otherwise. Thus the rightmost entries in the table are established.If we see the table as a tree the remaining entries can be lled with the rule:

parent−right child = left child.

For instance the middle entry in the third row is 253− 77 = 176.

Lemma 2.8. A linear code with weight enumerator (2.1) has a purely octadbasis.

Proof. The only nontrivial part to prove is that all dodecads are sums ofoctads. Let F1 be an octad and a, b ∈ F1. By Table 3 there are 16 octadscutting F1 in exactly a and b. Thus∣∣∣∣Fi, Fj ∣∣∣ Fi, Fj ∈ C8, |F1 ∩ F2| = 2

∣∣∣∣ = 759 ·(

8

2

)· 16 · 1

2.

It is clear that all members of the above set form dodecads, but somedodecads might be represented multiple times. Since each ve points ina dodecad determines a sixth point the number of ways a dodecad can bepresented as a sum Fi + Fj is (

12

5

)· 1

6· 1

2,

were we divide by two since the complement to the six points denes thesame sum. Thus the number of unique dodecads that is the sum of two

20

octads is759 ·

(82

)· 16 · 1

2(125

)· 1

6· 1

2

= 2576,

i.e all dodecads.

Remark 2.9. Note that this says that each ve points in a dodecad dene anoctad O1 that cuts the dodecad in 6 points. Further the remaining six pointsare in another octad O2 and the dodecad is the sum of O1 and O2.

Remark 2.10. As

(12

5

)· 1

6· 1

2=

(12

2

)each two points outside a dodecad D

induce a partition of D into two sets of size six such that each half togetherwith the two points form an octad.

Remark 2.11. Since every set of ve points in a dodecad determines a uniquesixth point each dodecad determines a Steiner System S(5, 6, 12).

2.3 MOG

Curtis has invented a tool, MOG (Miracle Octad Generator), for visualizingall octads constructed in Theorem 2.2. Figure 2.1 is a version of it generatedby a simple C++ program. To nd an octad (distinct from the three bricks)we start by picking one of the 35 pictures. In this picture we choose black orwhite in the heavy brick (i.e the left brick) and a x suit in the square (i.ethe two rightmost bricks). Finally we are allowed to permute the bricks.

Example 2.12. The octad in MOG[3,2] with black and heart and the tworst bricks interchanged is

x x

x x

x x

x x

= 4, 13, 10, 2, 22, 19, 12, 5.

N

MOG is constructed using the fact that each octad cuts one of the bricksin exactly four points. This would lead to 70 sextets if we draw them modulopermutations of the three bricks. But if two octads O1 and O2 cut the samebrick in two disjoint sets of four points then they must belong to the samesextet. So the 35 sextets in Figure 2.1 describes them all. Two importantobservations are the following lemmas.

21

Where the underlying table of numbers is:

∞ 14 17 11 22 19

0 8 4 13 1 9

3 20 16 7 12 5

15 18 10 2 21 6

Figure 2.1: An equivalent to Curtis MOG. The sextet row i and column jwill be denoted MOG[i, j].

Lemma 2.13. Each tetrad in the square cuts the rows with the same parityand the columns with the same parity.

Proof. For the rows we consider an octad O that cuts Λi in four points andlet S be the sextet corresponding to MOG[2,2] with Λi as the heavy brick.The octad O cuts one of the tetrads in the heavy brick in one, two or fourpoints. If the cut is of size four it is trivial. If the cut is of size two Lemma2.6 implies that O cuts two of the square tetrads in 2 points each and are

22

disjoint from the remaining. Similarly if the cut is of size one Lemma 2.6implies that O cuts each square tetrad in exactly one point. Thus the paritycondition is satised for rows. For the columns it follows analogously usingMOG[4,5].

Lemma 2.14. Interchanging the two bricks in the square preserves the sextet.

Proof. By Corollary 2.5 the tetrad in the heavy brick denes an unique sextet.Suppose Λ1 is the heavy brick (analogously for Λ2 and Λ3). Then we simplyobserve that if [X, Y, Z]t is an octad found in the sextet dened by X + t sois [X,Z, Y ]t.

In the remaining part of this thesis MOG will be used frequently formainly two tasks:

• to nd the octad containing ve given points.

• to check whether or not an element is in C .

Example 2.15. To nd the octad containing 1, 10, 13, 16, 17 we start bydrawing them in the MOG table and observe that Λ2 is the heavy brick.

x

x x

x

x

−→x x x

x x x

x

x

= 1, 10, 13, 16, 17, 9,∞, 14.

Further MOG[3,1] help us to ll in remaining points. N

Example 2.16. A bit trickier is to nd the octad containing ∞, 16, 2, 12, 5.As before we start drawing the points in a table.

x

x x x

x

It is clear that Λ1 cannot be the heavy brick since 16, 2, 12, 5 do not satisfythe parity condition. If Λ3 is the heavy brick MOG[7,1] force us to includethe point 8 to satisfy the parity condition. But then the heavy bricks do notmatch. Thus Λ2 is the heavy brick and using the parity condition we haveto include 14. Finally we nd our octad using MOG[3,5].

x x x

x

x x x

x

= ∞, 14, 11, 13, 16, 2, 12, 5.

23

N

Example 2.17. To see that

0, 8, 20, 18, 4, 13, 16, 2, 1, 9, 12, 6 = x x x x x x

x x x

x x x

/∈ C ,

we simply observe that the octad containing 0, 8, 20, 18, 13 cut the set inve points. N

Example 2.18. By Lemma 2.8 D = 0, 8, 20, 18, 4, 11, 16, 2, 1, 9, 12, 6 forma dodecad if and only if it is the symmetric dierence of two octads. Orequivalent D is an dodecad if and only if for an octad O that cuts Din six points the sum D + O is an octad. From MOG[7,1] we see thatO = 0, 8, 20, 18, 4, 11, 21, 5 is such an octad, but

D +O = 16, 2, 1, 9, 12, 6, 21, 5 = x x

x x x

x x x

is obviously not an octad. HenceD /∈ C . N

Example 2.19. To see if a set is a dodecad it is often easier to go back tothe construction. For instance we see directly that

x x x

x x x

x x x

x x x

= [F, F, F ] ∈ C .

N

2.4 Uniqueness of S(5, 8, 24) and C

We have already found the Golay Code and its associated Steiner SystemS(5, 8, 24). What we mean by uniqueness of C is: For an arbitrary linearcode C ′, over Ω′, with the same weight enumerator as C , there is a bijectionρ : Ω −→ Ω′ such that code words are mapped to code words. Uniquenessfor S(5, 8, 24) is dened analogously. Since it is not relevant how we namethe points in Ω′ we might as well assume Ω′ = Ω and we characterize ρ as apermutation in SΩ.

24

Lemma 2.20. ρ preserves all octads if and only if ρ preserves all the sextets.

Proof. Assume all sextets are preserved by ρ. Since each octad can be foundas the union of two tetrads in a sextet the image of each octad is obviouslyan octad. For the other direction assume there exist a sextet S such thatρ(S) is not a sextet. Then there exist two tetrads T1, T2 ⊂ S such thatρ(T1) ∪ ρ(T2) /∈ C ′8, but T1 ∪ T2 ∈ C8.

Lemma 2.21. If every octad cutting one specic octad O in four points areknown, then all octads follows by symmetric dierencing.

Proof. Let a, b, c be three arbitrary distinct points in O. From row four inTable 3 we see that there are 21 octads containing a, b, c. For the intersectionto be even any two of these must intersect in four points. By Lemma 2.4 the

symmetric dierence is an octad disjoint from a, b, c. There are

(21

2

)= 210

pairs among these 21 octads. If A,B and C,D are two such pairs andA + B = C + D then two points from A \ a, b, c has to be in C or D,say C. So A and C has ve points in common and A = C, implying B = D.Hence all 210 pairs are unique. Again using row four in Table 3 we see thatthese are all octads disjoint from a, b, c. This shows that all octads disjointfrom some a, b, c are known. I.e all octads, since every octad (not O) has todier by at least 3 points.

Theorem 2.22. The Steiner System C8 is unique.

Method of Proof. We will show that if ρ is predened on seven points wecan in one and only one way dene ρ on the remaining 17 points such thatseven sextets similar to seven sextets from MOG exist. To nd these sevensextets we use MOG and the fact that sextets has to be mapped to sextets.By choosing suitable octads from these sextets and taking the symmetricdierence we will obtain all sextets that cut a specic octad 42. Finally theresult follows from Lemma 2.21.

Proof. Let C ′8 be an arbitrary Steiner System S(5, 8, 24), (x1, x2, x3, x4, x5, x6)an ordered set of points from O ∈ C ′8 and x7 ∈ Ω \ O. The idea is to showthat there exists exactly one permutation ρ satisfying ρ[C8] = C ′8 and

25

ρ(∞) = x1 ρ(0) = x2 ρ(3) = x3

ρ(15) = x4 ρ(14) = x5 ρ(8) = x6

ρ(17) = x7

For the rest of the proof we are going to describe ρ with a matrix R in thefollowing way. If

M =

∞ 14 17 11 22 19

0 8 4 13 1 9

3 20 16 7 12 5

15 18 10 2 21 6

we dene R[i, j] = ρ(M [i, j]). So we start with

R =

x1 x5 x7

x2 x6

x3

x4

and we wish to ll in the blanks in R in such a way that the octads arepreserved. From MOG[4,5] we see that the columns of M form a sextetassociated with C8 and hence due to Lemma 2.20 the columns of R mustform a sextet associated with C ′8. So in our rst attempt to dene ρ we llR with the remaining 17 points such that the tetrads in the sextet denedby x1, x2, x3, x4 are columns. In what order does not matter at the moment.We write this sextet

S1 =

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

where the corresponding entries in R, to a specic number, form a tetrad.We will by Oiab denote octad formed by tetrads a and b from Si. Note thatS1 implies that the two rst columns of R form the octad O. Now MOG[2,2]implies that, for the theorem to hold, we must require

S2 =

2 1 3 3 3 3

1 2 4 4 4 4

1 2 5 5 5 5

1 2 6 6 6 6

to be a sextet.

26

To achieve this we observe that the octad containing x2, x3, x4, x5, x7 cutsS1 3.15 so we can, if necessary, permute the 17 points to obtain S2 withoutdestroying S1. Similarly, MOG[7,1] implies that we must require

S3 =

1 1 3 4 5 6

2 2 4 3 6 5

1 2 5 6 3 4

1 2 6 5 4 3

to be a sextet.

To see that can be achieved we observe that the octad containing x1, x3, x4, x5, x7

cuts both S1 and S2 3.15. By permuting the rows in the lower right 3 × 4-submatrix of R, we can start writing the sextet dened by x1, x3, x4, x5 (i.ethe sextet we want to force to be S3) as

1 1 3

2 2 3

1 2 3

1 2 3

−→1 1 3 4

2 2 4 3

1 2 3 4

1 2 4 3

Where the 4-tetrad must be since O156 +O234 cuts the sextet 42.04. Now thesecond column together with one of the four last imply that there should beone 5 and one 6 in each row. Using that O145 + O235 cuts the sextet 24.02

we see that the sextet actually is S3. It is easy to convince oneself that thegroup of rearrangements of the 17 un-named points that preserves S1, S2 andS3 is generated by

α =

p p p p p pp p p p p pp p p p p pp p p p p p-

?

@@@R@@

@@R

, β =

p p p p p pp p p p p pp p p p p pp p p p p p@@ and γ =

p p p p p pp p p p p pp p p p p pp p p p p pwhere dots denote xed points and lines cycles. Similarly MOG[3,2] requirethat

S4 =

1 1 3 3 5 5

1 1 3 3 5 5

2 2 4 4 6 6

2 2 4 4 6 6

has to be a sextet.

The octad X containing x1, x2, x5, x6, x7 cuts S1 24.02 and by using α wemight assume it cuts the rst four columns. X also cuts S2 and S3 24.02

so it has to be the upper left 2 × 4-submatrix of R. As X cuts the sextetdened by x1, x2, x5, x6 42.04 the tetrads 1 and 3 above are settled. Now

27

O112, O134, O234 all cutting this sextet 42.04. Hence S4 is a sextet. As aboveMOG[6,1] require that

S5 =

1 1 3 4 5 6

1 2 5 6 3 4

1 2 6 5 4 3

2 2 4 3 6 5

has to be a sextet.

The octad Y containing x1, x2, x3, x5, x7 cannot contain any further pointsfrom O113, O213 or O313 since it then would have ve points in common witha dierent octad. Using that Y cuts S1, S2 and S3 3.15 this leaves only twocases:

x x x

x x

x x

x

or

x x x

x x

x x

x

,

where the corresponding entries in R form the octad. By possibly using βto permute we might assume that the latter describes Y . Notice that βpreserves S4. Both Y and O112 cuts the sextet dened by x1, x2, x3, x5 42.04

so the tetrads 1,2 and 3 in S5 are settled. To establish the other tetrads weuse the fact that Oi1j, i = 1..4 and j = 3..6, all cuts the sextet 31.15. Thisveries that S5 is a sextet. Using the same technique, with the added casei = 5, on the sextet dened by x1, x2, x4, x5 we easily nd that

S6 =

1 1 3 4 5 6

1 2 6 5 4 3

2 2 4 3 6 5

1 2 5 6 3 4

is a sextet

Notice that S6 is the image of MOG[7,3]. For our last sextet, MOG[3,2]require

S7 =

1 1 3 3 5 5

2 2 4 4 6 6

1 1 3 3 5 5

2 2 4 4 6 6

to be a sextet.

To see this we look at the octad dened by

x x x x

x x x.

28

Since this octad cuts S1 24.02 it has one more element in the second column.This implies that it also cuts S2 and S6 24.02. Using this we get three cases:

x x x x

x

x x x,

x x x x

x x x xor

x x x x

x x x

x

.

The rst case fails to cut S4 properly. The other two are equivalent under γso we might assume that the middle case describes our octad. Notice that γpreserves all previous sextets. Since it cut S1 24.02, the tetrads 1,2,3,4 aboveare established. Finally, the fact that O234 and O256 both cut the sextet 42.04

implies that tetrads 5 and 6 are established and S7 is a sextet. Since 〈α, β, γ〉was the full group of permutations preserving S1, S2 and S3 and we used allgenerators in the construction of the other sextets, it is clear that no furthernonidentity permutations exist that preserves all seven sextets.

To obtain the image sextets of 28 remaining sextets fromMOG, we repetitivelytake symmetric dierences of already known octads that overlap in fourpoints. By Lemma 2.4 the symmetric dierence is an octad. If we lookhow this new octad cuts already found sextets it is an easy task to write outthe sextet. To start, using O114 and O414, we get

x x

x x

x x

x x

+

x x

x x

x x

x x

=

x x

x x

x x

x x

.

O112 cuts the sextet dened by the four points in the left brick 42.04. Furtherthe cut with O413 and O414 is 24.02. Thus the sextet contains tetrads

1 2 3 4 ? ?

1 2 3 4 ? ?

2 1 4 3 ? ?

2 1 4 3 ? ?

. Now O535 cuts the sextet 42.04. Thus

S8 =

1 2 3 4 5 6

1 2 3 4 5 6

2 1 4 3 6 5

2 1 4 3 6 5

.

29

Next using O613 and O714 gives

x x x

x x

x

x x

+

x x x x

x x x x=

x

x x

x x x

x x

.

As above O112 cuts the sextet dened by the four points in the left brick42.04. Further both O735 and O846 cut the sextet 42.04. Thus

S9 =

1 1 5 3 6 4

2 1 6 4 5 3

2 2 3 5 4 6

2 1 4 6 3 5

By using new sextets when found we can continue this tedious process andwill after a while end up with all 35 sextets dened by four points in O112.Hence we have found all octads cutting O112 in four points. The uniquenessnow follows from Lemma 2.21.

Theorem 2.23. The Golay code is unique.

Proof. Since each linear code with weight enumerator 2.1 denes a Steinersystem S(5, 8, 24) and has a purely octad basis the assertion follows from theuniqueness of S(5, 8, 24).

Corollary 2.24. The permutations of Ω that preserves C8 form a 5-foldtransitive permutation group of order 244, 823, 040.

Proof. The permutations that preserves C8 form obviously a group. To pickan element in this group we have, from the uniqueness theorem, 24 · 23 · 22 ·21·20 choices to pick x1, x2, x3, x4, x5. For x6 we have three choices and for x7

we have 16 choices. Putting it all together we have 244, 823, 040 choices.

Denition 2.25. The 5-fold transitive group preserving C8 is called M24.

Note that since C has a purely octad basis it is also preserved by M24.

30

3. The Mathieu group M24

A common problem that will haunt us for the rest of this thesis is whetheror not x ∈ S24 is in M24. The proof of Theorem 2.22 gave us one solution,namely to check if x maps Si, i = 1..7, to sextets. This is however anextremely simple and tedious calculation so to increase our quality of lifeGAP (see Appendix) is a good option.

Lemma 3.1. In M24 the only element in M24 xing seven points not in anoctad is the identity.

Proof. We will later see that the octad stabilizer has form 24.A8 where A8 isthe restriction of the action to the octad being stabilized and 24 is action onthe remaining 16 points when the octad is stabilized pointwise. Further allnontrivial permutations in the 24 has cycle shape 18.24, i.e xed point freeon the points outside the octad.

Suppose ρ ∈M24 xes x1, x2, x3, x4, x5, x6, x7 and the seven points are not inan octad. If x1, x2, x3, x4, x5, x6 is in an octad O1, then ρ is in the stabilizerof O1. Now the restriction of ρ to its action on O1 is an even permutationso ρ must x O1 pointwise. Since the 24 is acting xed point free and ρxes a point in Ω \ O, ρ must be the identity. Let O2 be the octad denedby x1, x2, x3, x4, x5. If x6 /∈ O2 we know from Theorem 2.22 and the octadstabilizer that ρ is either the identity or has a cycle of length three, i.e thethree remaining points in O2. Assume the case is the latter and call thepoints in the three cycle y1, y2, y3. Then ρ

3 xes O2 pointwise and a point inΩ \ O2 so o(ρ) = 3. Thus ρ has cycle shape 16+3k.36−k for some 1 ≤ k ≤ 5.Assume x6, z1, z2, z3 xed points to ρ outside O2. The octad Q dened byx5, x6, z1, z2, z3 contains either one further point from O2 being xed by ρ,or three further points from O2. If it contains three further points they haveto be y1, y2, y3 since ρ otherwise would stabilize another octad pointwise and

31

have bad cycle shape. If y1, y2, y3 ∈ Q then ρ stabilizes the octad O2 + Qpointwise and has bad cycle shape. Hence Q contains exactly one furtherpoint from O2 xed by ρ. But then ρ xes six points in Q (i.e Q being xedpointwise) and has bad cycle shape. Contradiction.

The following corollary will be used more frequently than the lemma itself.

Corollary 3.2. Only the identity xes nine points.

3.1 Eight maximal subgroups

The goal in this section is to nd and characterize eight natural subgroupsof M24. We will later see that they all are maximal subgroups.

The subgroups of M24 xing one, two an three points are called M23, M22

and M21 respectively.

Denition 3.3. The groups M24, M23 and M22 are Mathieu groups.

We will later see that all four groups above are simple. The group M21

is not a sporadic group thou. In fact M21∼= PSL3(4), a group of Lie type,

but the proof of this is out of the scope for this thesis. From the 5-foldtransitivity of M24 it follows that the stabilizers of a monad (one elementset), dyad (two element set) and triad (three element set) are M23, M22.2and M21.S3 respectively.

3.1.1 The projective group L2(23)

Since 23∣∣|M24| there is an element α ∈M23 of order 23. The numbers in the

MOG table are adapted to this element. We have α : i 7→ i+ 1 or

α = (∞)(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22).

We dene another element γ : i 7→ −1/i, i.e

γ = (0∞)(1 22)(2 11)(3 15)(4 17)(5 9)(6 19)(7 13)(8 20)(10 16)(12 21)(14 18).

To see that γ ∈ M24 we could check its action on sextets Si, i = 1, . . . , 7,and see that the images are indeed sextets. Alternatively we wait until weknow more about the involutions in M24. Finally Lemma 1.20 implies that〈α, γ〉 ∼= L2(23). By Lemma 1.20 L2(23) is doubly transitive on Ω.

32

3.1.2 The Octad Stabilizer

Since M24 clearly is transitive on octads all octad stabilizers are conjugates,so it is enough to look at one of them. Let G be the stabilizer of Λ1 and letϕ be the restriction map to Λ1. From the proof of Theorem 2.22 we knowthat the ϕ(G) is 6-fold transitive. Thus

G = K.

S8

A8

where K is the kernel of ϕ. By Lemma 1.5 |G| = |M24|/759. Further thepermutations

p p p p p pp p p p p pp p p p p pp p p p p p ,


p p p p p pp p p p p pp p p p p pp p p p p p and

p p p p p pp p p p p pp p p p p pp p p p p p

are all in M24. They generate an elementary abelian subgroup in K of order16. Finally 16 · |A8| = |G|. Thus K = 24 and thus G ∼= 24.A8.

3.1.3 The Duum Stabilizer

The complement of a dodecad is itself a dodecad. Such a pair of complementarydodecads is called a duum. To see the structure of the duum stabilizer weneed the order of the dodecad stabilizer and so we prove

Lemma 3.4. M24 is transitive on dodecads.

Proof. The stabilizer to Λ1 has a subgroup H xing top two points as a pair.By the last row in Table 3 there is exactly 16 octads that cuts Λ1 in thesetwo points. One such octad is

x x x x x

x

x

x

It is readily checked that no element in the 24 xing every point in Λ1 xesthis octad. Thus the 24 is transitive over the 16 octads cutting Λ1 in the toptwo points. Finally if D1 and D2 are two dodecads Remark 2.9 implies that

33

D1 = O1 + O2 and D2 = O3 + O4 for some octads O1, O2, O3 and O4. Takeg, h ∈ M24 such that g(O1) = Λ1 and h(O3) = Λ1. Then we can nd k ∈ Hsuch that kg(O2) = h(O4) and so h−1kg(D1) = D2.

Lemma 3.5. The dodecad stabilizer is sharply 5-fold transitive on its twelvepoints.

Proof. Let D be a dodecad and pick two subsets X = x1, x2, x3, x4, x5 andY = y1, y2, y3, y4, y5 of D. These sets dene octads Ox and Oy. FromRemark 2.9 we know that D = Ox + Rx = Oy + Ry for some octads Rx

and Ry. Pick g ∈ M24 such that g(Ox) = Oy. Since the stabilizer H of Oy

acts as A8 on Oy we can nd h ∈ H such that hg(Ox ∩ Rx) = Oy ∩ Ry.If we let K be the kernel of the restriction map of H to Oy we know fromthe proof of Lemma 3.4 that K is transitive on the octads cutting Oy inOy ∩ Ry. Especially we can choose k ∈ K such that khg(Rx) = Ry. Thenkhg stabilize D and map X to Y . Finally the A8 xing Oy contains a S6

acting on Oy \Ry and so there must exist a permutation mapping xi to yi fori = 1, 2, 3, 4, 5. From Lemma 1.5 we know that the stabilizer of a dodecad hasorder |M24|/2576 = 12·11·10·9·8, so it must be sharply 5-fold transitive.

Thus we have found the Mathieu group M12 and its isotropy group, i.ethe Mathieu group M11. Finally since M24 is transitive on dodecads, theduum stabilizer is isomorphic to M12.2.

3.1.4 The Sextet Stabilizer

Clearly M24 is transitive on sextets, so it is enough to investigate one sextetstabilizer. We choose the stabilizer G to the sextet MOG[4,5] where the

columns are tetrads. There are

(24

4

)· 1

6= 1771 sextets. By Lemma 1.5

G has order |M24|/1771 = 26 · 3 · 6!. Let ϕ : G −→ S6 be the restrictionmap to the action on the tetrads. Clearly ϕ[G] is doubly transitive. As the

image of

p p p p p pp p p p p pp p p p p pp p p p p p@@ is a transposition, ϕ[G] contains all transpositions

and ϕ[G] ∼= S6. We see that the kernel has order 26·3. Further the involutions

34



and


all xing two tetrads pointwise and preserving the others. Again due to thedoubly transitivity on the tetrads, each pair of of tetrads has three such

involutions. Let X be the set containing these

(6

2

)· 3 = 45 involutions.

It is clear that all elements in 〈X〉 commute. Since 〈X〉 ≤ Ker(ϕ) and〈X〉 do not contains an element of order three we have 〈X〉 = 26. Finally

g =

p p p p p pp p p p p pp p p p p pp p p p p p666666 ∈ Ker(ϕ) has order three and normalize all

elements in X. Thus g normalize 〈X〉 and Ker(ϕ) ∼= 26.3.Hence G ∼= (26.3).S6.

3.1.5 The Trio Stabilizer

A partition of Ω into three octads is called a trio.

Lemma 3.6. The Mathieu group M24 is transitive on trios.

Proof. From Theorem 2.22 we know that we can nd a permutation ρ ∈M24

with cycle shape 5.3 on an octad O that xes a point in Ω \ O. Since ρ15

xes nine points we know that ρ15 = 1 and so ρ has shape 5.3.1.1i.3j.5k.15l

for some i, j, k, l. If l = 0 then ρ5 6= 1 or ρ3 6= 1 xes at least nine points.Contradiction. Hence ρ has shape 3.5.1.15. By Table 3 there are 30 octadsdisjoint from O. Now ρ has two orbits of length 15 on these 30 octads. Oneorbits consists of those octads containing the xed point and the other theremaining. So the stabilizer of O is transitive on the 15 trios containing O.Finally let T1 and T2 be two trios containing octads O1 and O2 respectively.Pick σ ∈M24 such that σ(O1) = O2 then σ(T1) is a trio containing O2 so wecan nd π the stabilizer to O2 such that πσ(T1) = T2.

By studying the MOG table we see that there are 759 · 5 = 3795 trios.Let G be the stabilizer to the MOG trio, i.e the three bricks. Then dueto the above lemma |G| = |M24|/3795 = 26 · 3 · 2 · 168. If S is a sextet

35

such that the tetrads can be paired to form a trio T , then we say that S is

a renement of T . As each sextet is a renement of

(6

2

)= 15 trios, the

number of renements to a xed trio is (1771·15)/3795 = 7. Clearly G has topermute the renements of the MOG trio. We observe that the renements tothe MOG trio is the nonzero elements in the point space P from Theorem 2.2(the corresponding sextets are MOG[2,3], MOG[3,2], MOG[4,1], MOG[4,5],MOG[5,4], MOG[6,3] and MOG[7,2]). Since these generate an elementaryabelian group of order 8, using symmetric dierencing as multiplication, themaximum action G can have on the renements is

Aut(23) ∼= GL3(2) ∼= L3(2) ∼= L2(7).

The last isomorphism will be visible inside the Octern group later. Besidesthe action on the renements G contains

p p p p p pp p p p p pp p p p p pp p p p p p and


and so the full S3 bodily permuting the bricks. These actions do not interfere.Hence we can identify the restriction map to G's action on bricks and therenements with

ϕ : G −→ S3 × L2(7).

Further the three elements

a =

p p p p p pp p p p p pp p p p p pp p p p p p666666, b =


and c =

p p p p p pp p p p p pp p p p p pp p p p p p@@are in G and stabilize the bricks. As o(ϕ(a)) = 3, o(ϕ(abc)) = 7 ando(ϕ(acb)) = 4 we have ϕ[〈a, b, c〉] ≤ 〈1〉 × L2(7) and |ϕ[〈a, b, c〉]| = 84 or168. To eliminate the rst case we need the following lemma.

Lemma 3.7. The projective group L2(7) is simple.

Proof. On the projective line L2(7) is genererated by

ρ = (∞)(0 1 2 3 4 5 6) and σ = (∞ 0)(1 6)(2 3)(4 5)

36

Lemma 1.20 together with Lemma 1.11 implies that L2(7) is primitive. AssumeN C L2(7) is proper and non trivial. By Lemma 4.1 N is transitive. Thus8∣∣|N | and N contains all Sylow 2-subgroups of L2(7). Suppose |N | = 8.Then as ρiσρ−i ∈ N for i = 1..7 all have order 2, i.e N is elementary abelian.But o(ρσρ−1σ) = 4, contradiction. Thus |N | = 8 · 3 or 8 · 7. If 3

∣∣|N | thenN contains all Sylow 3-subgroups. Hence ρσ = (∞ 1 0)(2 4 6) ∈ N . Butthen (ρσ)σ = ρ ∈ N and N = G. Contradiction. Similarly if 7

∣∣|N |, thenN contains all Sylow 7-subgroups and ρ ∈ N . Thus ρσ ∈ N and againN = G.

This shows that ϕ[〈a, b, c〉] = 〈1〉 ×L2(7) and ϕ[G] ∼= S3×L2(7). Finally|Ker(ϕ)| = |G|/|ϕ[G]| = 26. Clearly Ker(ϕ) contains the subgroup xing therst brick pointwise and preserving the bricks. By conjugating this subgroupwith the S3 bodily permuting the bricks we see that Ker(ϕ) contains







It's readily checked that these generate an elementary abelian group of order26. Thus

G ∼= 26.(S3 × L2(7)).

Remark 3.8. Galois proved that the group L2(p) does not have a nontrivialpermutation representation on fewer than p + 1 symbols for p > 11 ([9],p.268). This shows that the above representation of L2(7) on seven symbolsis very special.

3.2 The involutions in M24

We have already seen one type of involution. I.e involutions xing an octadpointwise. We will call this an involution of type one. Since the subgroupxing an octad pointwise has order 16 and is transitive on the 16 pointsoutside the octad, there is one and only one involution xing an octad F andinterchanging i, j /∈ F . Call this involution Fi,j.

37

Lemma 3.9. Let F be an octad and let i, j, k, l ∈ Ω \ F be four distinctpoints. Then Fi,j = Fk,j if and only if i, j, k, l is a special tetrad of F + Ω.

Proof. If i, j, k, l is a special tetrad of Ω \ F , ∃x1, x2, x3, x4 ∈ F suchthat O = i, j, k, l, x1, x2, x3, x4 is an octad. Clearly Fi,j stabilize O. ThusFi,j = Fk,l. For the other direction we notice that i, j together with threepoints y1, y2, y3 ∈ F denes a fourth point in y4 ∈ F plus two pointsk, l ∈ Ω \ F , such that they all lie in the same octad. As this octad beingstabilized Fi,j = Fj,k. Obviously i, j, k, l is a special tetrad of Ω\F . We can

choose y1, y2, y3, y4 in1

4

(8

3

)= 14 ways. Modulo F each choice denes two

points in Ω\F . Since no two octads intersect in more than four points everytwo dierent choices modulo F denes disjoint pairs in Ω \ F . All togetherthis gives seven pairs plus the points i, j, i.e all of Ω \ F .

Knowing this we can easily construct any involution Fi,j. The method is

(i) Pick x1, x2, x3, x4 ∈ F

(ii) Let a, b be in the octad dened by k, l, x1, x2, x3, x4, where the pair k, lforms a known 2-cycle in Fi,j. As the octad is being stabilized we have(a b) ∈ Fi,j.

(iii) If the action on some point is unknown go to (i) and pick four newpoints.

Example 3.10. Suppose we want to nd (Λ3)∞,0 =

p pp pp pp p .

The unknown action on the octad

x x

x x

x x

x x

is forced to be

a 2-cycle. Thus (Λ3)∞,0 =

p pp pp pp p . Similarly

x x x x

x x x x and

x x x x

x x x x

imply (Λ3)∞,0 =

p pp pp pp p . Four octads later we end

38

up with (Λ3)∞,0 =

p p p p p pp p p p p pp p p p p pp p p p p p .N

Multiplying (Λ3)∞,0 and (Λ1)4,16 gives

σ =

p p p p p pp p p p p pp p p p p pp p p p p p ×p p p p p pp p p p p pp p p p p pp p p p p p

=


an involution that acts without xed points. We notice that this involutionxes the sextet S1 tetradwise. Further the involution xes the dodecad

M = [A,B,C]Λ =

x x x x x x

x x

x x

x x

cutting across the tetrads 26. We will call this an involution of type two.

Theorem 3.11. Let S be a sextet and D a dodecad cutting the tetrads of S26. Then there is an involution in M24 xing the tetrads of S and D. Furtherall involutions of type two are conjugates.

Proof. In this proof we will by M12 mean the stabilizer of D. Since M24 istransitive on dodecads ∃ρ ∈ M24 such ρ(M) = D. Clearly D cuts ρ(S1) 26.Thus ρσρ−1 xing D and the tetrads of ρ(S1). Were done if we can show thatM12 xing D is transitive on the sextets cutting D 26. Since then ∃α ∈M12

such that αρσρ−1α−1 has the desired properties. Consider a sextet denedby four points in D. Since D cannot contain an octad this sextet cuts D 4.24.Hence we have found two sextets containg a tetrad cutting D in two points.Our goal is to show that M12 only has two orbits on such tetrads. Clearlyone of this orbits contains all sextets cutting D 26. Note that we alreadyknow that M12 has at least two orbits. Let a, b, c, d and s, t, u, v be twotetrad such that a, b, s, t ∈ Ω \D and c, d, u, v ∈ D. Since M12 is 5-transitiveon Ω \D there ∃β ∈M12 such that

β(s) = a and β(t) = b.

By Remark 2.10 a, b induce a partion of D into two halves X and Y . AsM12 is 5-transitive the subgroup G xing X acts as S6 on X and xing a, b

39

as a pair. By symmetry G also acts as S6 on Y . Consider g ∈ G with cycleshape 1.2.3 on X. No matter what cycle shape g has on Y g6 xes at leastnine points. By Corollary 3.2 o(g) = 6. Thus the possible cycle shapes for gY are

1.2.3, 12.22, 13.3, 14.2, 16, 23, 32 and 6

Case 1, 3, 4, 5, 6 and 7 implies that g3 6= 1 is xing seven points not inan octad, contradicting Lemma 3.1. Similary case 2 yields g2 6= 1 is xing8 points not in an octad. Hence 〈g〉 acts transitively on Y . Further thesubgroupM10.2 xing a, b as a pair clearly preservs the partion X and Y as apair. SinceM12 is sharply 5-transitive [M10.2 : G] = 2. As G already containsall actions on X the sharply 5-transitivity implies that there ∃γ ∈M10.2 \Gswapping X and Y . This shows that two tetrads that cut D in two pointsare in the same orbit if and only if they split accross X and Y in the sameway. If they do there are two cases

• If β(s), β(t) ∈ A and c, d ∈ B for A,B = X or Y , then by possiblyusing γ we can map so they all belong to the same half. Finally we usethe S6 xes that half to map β(s) 7→ c and β(t) 7→ d.

• If the pairs β(s), β(t) and c, d both split across X and Y , we can usethe S6 stabilizing one half to map β(s) 7→ c, say. As we can x c andstay transitive on the other half we map β(t) 7→ d.

Remark 3.12. As the S6 acted dierently (by consideration of the cycleshapes) on X and Y we have shown that S6 is not unique up to relabelingof the points in its action on six points.

Theorem 3.13. Every involution in M24 is of type one or two.

Proof. Let π ∈ M24 be an involution. As the number of octads is odd, πxes at least one octad. Suppose π xes Λ1, i.e the left MOG brick, and letG = 24.A8 be the stabilizer of Λ1. Since G acts as A8 on Λ1 π xes zero,four or eight points in Λ. If eight points are xed π is an involution of type1. Suppose π xes zero points of Λ1. Up to conjugation by G we can assume

that π has the action

p pp pp pp p on Λ1. Multiplying with ρ =

p p p p p pp p p p p pp p p p p pp p p p p p40

gives πρ = α ∈ 24. Clearly ρ and π commute. Since π = αρ = ρα weare looking for α ∈ 24 commuting with ρ. It is readily checked that theonly non trivial elements in the 24 commuting with ρ are the seven elementsxing each brick. If we let A, D and E be the sextet dened by correspondingelement in the space P from from Theorem 2.2, π has one of the followingactions:p p p p p pp p p p p pp p p p p pp p p p p p ×

p p p p p pp p p p p pp p p p p pp p p p p p =

p p p p p pp p p p p pp p p p p pp p p p p pp p p p p pp p p p p pp p p p p pp p p p p p ×p p p p p pp p p p p pp p p p p pp p p p p p =

p p p p p pp p p p p pp p p p p pp p p p p p@@

@@

= A[E,F,C]


@@ @@

@@@@

=


@@

= A[D,C,G]


=


= D[E,F,C]


=


= D[E,C,F ]


@@

@@ =

p p p p p pp p p p p pp p p p p pp p p p p pAAA

AAA

@@ = E[D,B,F ]


AAA

AAAAAA

AAA =

p p p p p pp p p p p pp p p p p pp p p p p pAAA

AAA

@@ = E[D,F,B]

If π xes four points of Λ1 we use the same idea as above. Up to conjugation

by G, π has the action

p pp pp pp p on Λ1. Multiplying by

41

σ =

p p p p p pp p p p p pp p p p p pp p p p p p@@ gives πσ = β ∈ 24. Clearly the only elements in

the 24 commuting with σ are the three elements xing each tetrad. Weend up with three cases. So up to conjugation π has one of the followingactions:

p p p p p pp p p p p pp p p p p pp p p p p p@@ ×p p p p p pp p p p p pp p p p p pp p p p p p =

p p p p p pp p p p p pp p p p p pp p p p p p@@p p p p p pp p p p p pp p p p p pp p p p p p@@ ×p p p p p pp p p p p pp p p p p pp p p p p p =


p p p p p pp p p p p pp p p p p pp p p p p p@@ ×p p p p p pp p p p p pp p p p p pp p p p p p

@@ @@

@@@@

=


Clearly all three are of type one.

3.3 A further maximal subgroup

All subgroups so far have been rather visible from our construction. Thissubgroup however is not. The subgroup was found later and did not occuron the complete list of maximal subgroups of M24 rst published [10]. Thesubgroup is transitive and is called the Octern group. It can be denedas the stabilizer of a special purely dodecad four dimensional subspace C .The quotation marks is because we allow ∅ and Ω to be in the subspace.Consider the matrix

∞ 14 3 15 20 0 8 18

17 16 10 7 4 13 2 11

22 9 6 19 12 21 5 1

42

Let ρ be the element xing the rows and rotating the last seven columns. Itcan be veried that ρ ∈M24. We now name the columns in the above matrix∞, 0, 1, . . . , 6. Since we only will consider elements preserving these columnsthere is a well dened restriction map θ to the action on the columns. Weuse the notation θ(x) = x. For instance ρ = (0 1 2 3 4 5 6). In a natural waythe subsets of ∞, 0, . . . , 6 corresponds to subsets of Ω. It turns out that

D1 = ∞, 0, 1, 3, D2 = ∞, 0, 2, 6 and D3 = ∞, 1, 2, 4

are dodecads. Together with 1 = ∞, 0, . . . , 6 these generate 14 dodecads.We want to show that these are all dodecads that is union of four columns.

Totally there are

(8

4

)= 70 possibilities. Assume there is a further such

dodecad D4. Then 1, D1, D2, D3 and D4 generate 30 such dodecads. But itturns out that none of

F1 = ∞, 0, 1, 2, F2 = ∞, 0, 1, 4, F3 = ∞, 0, 1, 5,

F4 = ∞, 0, 1, 6, F5 = ∞, 0, 2, 4 and F6 = 0, 1, 2, 3are dodecads. Its readily checked that the set

ρiFj∣∣ i = 0, . . . , 6 , j = 1, . . . , 6

contains 42 elements that is not dodecads. Since 70 − 42 < 30 a furtherdodecad D4 cannot exist. Dene H to be the linear code spanned by1, D1, D2, D3 and dene the Octern group G = M24 ∩ Aut(H ). If we take1, D1, D2, D3 as a basis for H we can express θ[G] as the subgroup of GL4(2)consisting of matrices

g =

1 a b c

0

0 Ag

0

a, b, c ∈ 0, 1 and Ag ∈ L3(2)

We now dene the restriction map

ϕ : θ[Aut(H )] −→ L3(2)

∈ ∈

g 7→ Ag

43

and get θ[G]/Ker(ϕG) ≤ 23.L3(2), where ϕG is the restriction of ϕ to θ[G].Note that Ker(ϕ) is an elementary abelian group, containing exactly the eightelements satisfying Di 7→ Di or Di 7→ Di + 1 for i = 1, 2, 3. Our objectivesare:

(i) Show that |θ[G]| = |L3(2)|.

(ii) Show that Ker(θ) is trivial.

(iii) Show that Ker(ϕG) is trivial.

The rst step might look like magic, but it is not. GAP has been used (seeAppendix).

(i) It turns out that

σ = (02)(16)(318)(419)(513)(720)(821)(917)(1011)(1215)(14∞)(1622) ∈ G.

Note ρ and σ together implies that G is transitive. Now θ[G] contains

ρ = (0 1 2 3 4 5 6) and σ = (∞ 0)(1 6)(2 3)(4 5).

By Lemma 1.20 these generate an L2(7). Finally |L2(7)| = |L3(2)|.

(ii) Assume α ∈ M24 xes ∞, 0, . . . , 6 and consider the octads denedby ∞ and two points from 0. That is

x x x x

x x

x x

,x x

x x

x x x x

andx x

x x x

x x x

Clearly α has to permute these octads among each other. On the other handthey do not cut the same columns. Thus α xes all three octads and has thefollowing xed points

α =

But by Corollary 3.2 only the identity can x nine points. Hence no permutationbut the identity can x all columns.

(iii) The element β = (∞ 2)(0 6)(1 6)(3 5) xes H . Further

β(D1) = D1 + 1, β(D2) = D2 and β(D3) = D3.

44

So β ∈ Ker(ϕ). Assume β ∈ ϕG[G]. Then

γ = ρ2σρ−2β = (0 5)(3 6) ∈ ϕG(G).

Since γ2 x all columns γ2 = 1. Thus γ is an involution xing exactly 8 points,i.e xing two columns and two points outside these columns, pointwise. Butno two columns are in an octad so seven points not in an octad being xed,contradicting Lemma 3.1. Thus σiβσ−i /∈ ϕG[G] for i = 0, . . . , 6 implyingthat Ker(ϕG) is trivial.

Adding it all together we get G ∼= L3(2) ∼= L2(7).

Besides the exceptional isomorphism between the projective groups abovethere is a further remarkable bonus with this construction. If we write thegenerator matrix to H

1 1 1 0 1 0 0 0

1 1 0 0 1 0 0 1

1 0 1 1 0 1 0 0

1 1 1 1 1 1 1 1

we see that H is self dual. Hence the above matrix is also the check matrix.But the top left submatrix is recognized as check matrix to the [7, 3] Hammingcode when the last column is a parity check column. Thus H is the extendedHamming code and we have shown the following theorem.

Theorem 3.14. The automorphism group of the extended Hamming code isisomorphic to 23.L2(7).

3.4 Proof of maximality

Even though some cases are proved in a whole new way in the followingproof, the main idea comes from Curtis article [2].

Let A ⊂ Ω of size n ≤ 12. We say that A is of type Sn if it contains oris contained in an octad. Otherwise we say that A is of type Un if it is

45

contained in a dodecad. If A is neither of type Sn nor Un we say that Ais of type Tn. It turns out that all types denes orbits of M24. Using thatthe octad stabilizer is 6-fold transitive on the octad being stabilized we seedirectly that Sn denes orbits for n ≤ 8. That the remaining types denesorbits might not be that obvious, but most types are irrelevant for this proof.We need the following thou.

Lemma 3.15. The type U8 denes an orbit of M24.

Proof. Suppose X and Y be two sets of type U8 and let

X = D1+a, b, c, d, a, b, c, d ∈ D1 and Y = D2+e, f, g, h, e, f, g, h ∈ D2.

Since M24 is transitive on dodecads exists a mapping D1 7→ D2. Furtherthe stabilizer of D2 is 5-fold transitive on D2. Thus ∃g ∈ M24 such thatg(X) = Y .

Lemma 3.16. Let X ⊆ Ω. Then X is congruent an unique monad, dyad,triad or sextet (i.e any tetrad of the sextet) or ∅ modulo C .

Proof. If |X| ≥ 5 the symmetric dierence with an octad having 5 points inX is always a smaller set. Clearly |X| is xed modulo C if |X| ≤ 4. Finallysuppose X = Λ +C ≡ Λ′+D, where Λ and Λ′ are monads, dyads or sextetsand C,D ∈ C . Then Λ + Λ′ ∈ C and the result follows.

Lemma 3.17. Let G be a permutation group of composite order actingprimitively on a set A. If ni is the largest orbit length of Ga acting on Aand nj is any other orbit length such that nj > 1, then (ni, nj) 6= 1.

Proof. The proof is out of the scope of this thesis but can be found in [4] p.48.

Lemma 3.18. Let G be a permutation group of composite order actingprimitively on an even set A. Then Ga cannot have an orbit of length 1or 2 on A \ a.

Proof. Since G is primitive Lemma 1.8 implies that Ga is maximal. SupposeGa xes a further point b, in other words Ga = Gb. Then since G is transitive∃π ∈ G such that π(a) = b. Clearly π ∈ NG[Ga] \Ga. As Ga is maximal thisyields Ga C G. Thus Ga = Gx ∀x ∈ A, i.e G is regular. But Ga is maximalso this can only be true if G is of prime order. Contradiction.

46

Suppose Ga has an orbit b, c. Then from the restriction mapϕ : Ga −→ Sym(b, c) we see that [Ga : Gab] = 2, and by symmetry[Gb : Gab] = 2. Thus NG[Gab] contains Ga and Gb. Since Ga is maximalNG[Gab] = G. Consider the factor group G/Gab. Clearly Ga/Gab is containedin some 2-group P < G/Gab. The inverse image of P is strictly between Ga

and G, contradicting that Ga is maximal.

Lemma 3.19. Any subgroup of M24 acting primitively of degree 4, 6, 8 or12 in its action on blocks in Ω, is doubly transitive on those blocks.

Proof. Let s be the length of an orbit in the stabilizer of a block. By Lemma3.18 s ≥ 3 and by Lemma 3.17 s has a factor in common with the largestorbit length. This rules out all cases.

Lemma 3.20. If G ≤M24 is primitive then G ∼= L2(23) or G ∼= M24.

Proof. The proof is out of the scope for this thesis but can be found in [5] p.30.

Before we start with the big theorem we need some elements from M24.As usual we leave the task to show that the elements actually are in M24 tothe interested reader. We start by dening three matrices D, H and L.

∞ 0 1 2 3 4

1 ∞ 14 17 11 19 22

2 0 18 4 2 6 1

3 3 8 16 13 9 12

4 15 20 10 7 5 21 D

∞ 0 1 2 3 4 5 6

1 ∞ 14 3 15 20 0 8 18

2 17 11 16 10 7 4 13 2

3 22 19 12 21 5 1 9 6H

∞ 0 1 2 3 4 5 6 7 8 9 10

1 ∞ 15 7 14 5 10 20 17 11 22 21 19

2 0 3 6 12 1 2 4 8 16 9 18 13 L

Not that the rows in H are the MOG trio and the rows in L form a duum.We use the notation [x × y]A for the element acting with x on the columnsand y on the rows of matrix A. Similarly [G×H]A denotes the group actingwith G on the columns and H on the rows of matrix A. The names of therows and columns are adapted to the following groups:

[L2(5)× A4]D, [L2(7)× S3]H and [L2(11)× 1]L.

They all turn out to be subgroups of M24. Further the element

a b c e g i k l j h f d

a b d f h j l k i g e cL

∈M24,

where elements corresponding to the same letter form transpositions.

47

Theorem 3.21. The following nine groups describes all maximal subgroupsof M24.

Description Group Order Factorized order

Monad stabilizer M23 10,200,960 27 · 32 · 5 · 7 · 11 · 23

Dyad stabilizer M22.2 887,040 28 · 32 · 5 · 7 · 11

Octad stabilizer 24.A8 322,560 210 · 32 · 5 · 7

Triad stabilizer M21.S3 241,920 28 · 33 · 5 · 7

Duum stabilizer M12.2 190,080 27 · 32 · 5 · 11

Sextet stabilizer (26.3).S6 138,240 210 · 33 · 5

Trio stabilizer 26.(S3 × L2(7)) 64,512 210 · 32 · 7

Projective group L2(23) 6,072 23 · 3 · 11 · 23

Octern group L2(7) 168 23 · 3 · 7

Method of Proof. We will see that there are no containment among the abovenine groups. Then we show that all proper subgroups of M24 are containedin one of the above groups.

Proof. By only considering the orders and the fact that a group xing aset cannot have a subgroup moving this set, we see that the only possiblecontainments among the nine groups are that a Trio group contains an Octerngroup. Assume the elements ρ and σ in the Octern group xes a trio. Sinceρ has shape 13.73 it xes least one octad O. Clearly O is the union of axed point and a 7-cycle. But then another octad being xed, i.e all threeoctads in the trio being xed. Thus the trio is a partion of three pairs(xed point, 7-cycle), related to ρ. We get six possible trios but σ xesneither one of them. Contradiction.

If G ∈M24 is intransitive let Λ denote one of the orbits. Either G is containedin the octad or duum stabilizer or by Lemma 3.16 Λ = X + C where X isan unique monad, dyad, triad or sextet and C ∈ C . But then G xes X andthus is contained in correspondning stabilizer.

48

If G is primitive Lemma 3.20 implies that G ∼= L2(23) or G ∼= M24. IfG is imprimitive we divide six cases, namely the possible block sizes.

Blocks of size 12. Clearly the two blocks are congruent modulo C . ByLemma 3.16 they are congruent to an unique monad, dyad, triad, sextet or∅ modulo C . Note that for G to be transitive the only possible options are∅ and a sextet. Thus G is contained in corresponding stablizer.

Blocks of size 8. Suppose Ω = X + Y + Z, |X| = |Y | = |Z| = 8. IfX (and thus Y and Z) is of type S8 G is in the trio stabilizer. If X is of typeT8 it is congruent to an unique dyad modulo C . Thus X, Y and Z denes aset of three dyads being stabilized, contradicting that G is transitive. Finallysuppose X is of type U8. Then X = T+D for some tetrad T and the dodecadD containing X. Clearly T and X are disjoint. Since the cut of a dodecadand an octad is 2, 4 or 6, the sextet containing T has further tetrads cuttingD in 0 or 2 points. Thus there are exactly two tetrads disjoint from X, theseform an octad OX . Similarly Y and Z denes octads OY and OZ . Clearly Gpreserves these three octads. As G is transitive we have OX +OY +OZ = Ω,i.e OX , OY , OZ form a trio. Hence G is in the trio stabilizer.

Blocks of size 6. Suppose the blocks are X, Y , Z and T . By Lemma3.19 G is doubly transitive on the blocks. If the blocks are of type S6, theoctad containing X has at least one point in Y , say, and zero points in Z.As G is doubly transitive on the blocks we can x X, and thus the octadcontaining X, and map Y to Z. Contradiction. Hence all blocks are of typeU6. Clearly the sum of any two blocks are congruent an even set modulo C .As G is transitive on Ω the sum cannot be a dyad. If the sum is congruent ∅we get X ≡ Y ≡ Z ≡ T mod C . Since each block are congruent to the samesextet G is in the sextet stabilizer. If the sum is congruent a sextet there arethree sextets S1 = X + Y , S2 = X + Z and S3 = Y + Z, with the propertyS1 + S2 = S3. By Lemma 2.6 an octad cuts a sextet 42.04, 24.02 and 3.15.Using this and the fact that all concerned tetrads must cut each other in 0,1 or 2 points, it is easy to show that there are only two ways two sextets canintersect such that the sum is a further sextet. By permuting the rows and

49

columns we get the following intersection matrices

2 2 0 0 0 0

2 2 0 0 0 0

0 0 2 2 0 0

0 0 2 2 0 0

0 0 0 0 2 2

0 0 0 0 2 2

and

2 0 0 0 1 1

0 2 0 0 1 1

0 0 2 0 1 1

0 0 0 2 1 1

1 1 1 1 0 0

1 1 1 1 0 0

.

where the rows correspond to S1, say, and columns to S2. In the rst caseS1 and S2 are renements to a unique trio. Thus G is in the trio stabilizer.In the second case let S1 be the columns in following matrix (MOG is notnecessarily the underlying table) and S2 the written out numbers.

1 2 3 4 1 1

1 2 3 4 2 2

5 5 5 5 3 3

6 6 6 6 4 4

We note that S3 has to cut S1 in the same way. Summing the tetrads andpossibly swapping the two top entries in the 4 rst rows (S1 and S2 and arepreserved under this action) we may assume that

S3 =

5 5 5 5 1 1

6 6 6 6 2 2

1 2 3 4 3 3

1 2 3 4 4 4

.

Now each sextet has a unique octad that cuts the other two sextets 24.02.Written out in the same table as above,

A A A A C C

A A A A C C

B B B B C C

B B B B C C

,

we see that they form a trio. Thus G is in the trio stabilizer.

Blocks of size 4. By Lemma 3.19 G is doubly transitive on the blocks.Thus 5

∣∣|G| and one of the Sylow 5-subgroups in M24 is in G. Since allSylow 5-subgroups are conjugates we may assume that the element of order5 in G, xing one block, is ρ = [(∞)(0 1 2 3 4) × I]D. Clearly the group

50

H = 〈[I × A4]D, [(∞)(0)(1 2 4 3) × (∞ 0)]D〉 normalize ρ. Hence we canconjugate with H without budge the rst column. Using ρ we see that up toconjugation by H the possible block structures are

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5D

(i)

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4D

(ii)

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 4 5 ¶ 2 3 D(iii)

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 5 ¶ 2 3 4D

(iv)

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 3 4 5 ¶ 2 D(v)

0 ¶ 2 3 4 5

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 2 3 4 5 ¶ D(vi)

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 4 5 ¶ 2 3

0 3 4 5 ¶ 2 D(vii)

0 ¶ 2 3 4 5

0 5 ¶ 2 3 4

0 3 4 5 ¶ 2

0 4 5 ¶ 2 3 D(viii)

If two blocks are of type S8 thenG is in the sextet stabilizer, due to the doubletransitivity on the blocks. If two blocks are of type T8 they are congruent to

a dyad modulo C . Totally this gives

(6

2

)= 15 dyads. Suppose a ∈ Ω is in

exactly n of the 15 dyads. Since conjugation by g ∈ G, such that g(a) = bpreserves the blocks, b is also in exactly n dyads. As this holds for ∀b ∈ Ωthe 15 dyads contains 24n elements, counted with repitition. Contradiction.Clearly the sum of block 0 and ¶ is of type S8 in case (i) and of type T8 incase (ii). Suppose two blocks are of type U8. We want to show that thatthere is a unique pairing of the points in the set such that the union of anythree pairs is of type S6. As both S6 and U8 denes orbits it is enough to

investigate one set of type U8. Consider the set

x x

x x

x x

x x

. Since a set

of type S6 with 5 points in Λ1 has to be a subset of Λ1, we immediately seethat the two points not in Λ1 form a pair. By gradually removing two pointsto see if the remaining form a set of type S6, we easily eliminate all cases but

4 4

1 1

2 2

3 3

. For G to be doubly transitive on the blocks this pairing has

to cut the blocks in the same way. For case (iii) we have that

0 1

0 1

0 1

0 1D

induces the pairing

2 3

1 2

1 4

3 4D

51

while1 2

1 2

1 2

1 2D

induces the pairing

4 1

3 2

4 2

3 1D

,

a contradiction. We got similar contradictions in the cases (iv)-(vii). Forcase (viii) we write down the sextets corresponding to each block. The x:esdenotes the blocks dening the sextets.

x 1 2 3 4 5

x 1 2 3 4 5

x 1 2 3 4 5

x 1 2 3 4 5

2 x 4 2 1 1

5 5 x 5 2 x

3 4 3 x 5 1

1 5 3 2 3 4

4 3 x 5 3 4

1 1 1 2 5 5

2 2 1 2 x x

3 4 5 x 3 4

2 5 5 x 2 4

1 x 2 1 2 3

3 5 1 4 4 3

5 4 3 1 x x

3 3 4 4 2 x

1 2 2 1 x 1

5 x x 5 5 2

4 3 4 3 5 1

4 3 4 2 x 2

5 1 4 x 5 5

1 1 3 2 5 1

2 x x 3 3 4

The tetrads that are not named x are called non-special. We now create thegraph (Ω, E), where (a, b) ∈ E if and only if a and b occur together in exactlyone non-special tetrad. We draw the graph:

11

19

22

14

17

9

12

8

16

13

∞

3

3

3

3

3

1

4

6

18

2

10

5

20

7

21

0

15

15

15

15

15

We see that the graph consists of two disjoint subgraphs containing thedodecads

x x x x x x

x x x x x xand x x x x x x

x x x x x x

respectively.

52

As the blocks are preserved by G, so is the graph. Hence G is in the duumstabilizer.

Blocks of size 3. By Lemma 3.19 G is doubly transitive on the blocks.Thus 7

∣∣|G| and as in the preivious case we may assume that the element oforder 7 in G, xing one block, is ρ = [(∞)(0 1 2 3 4 5 6) × I]H . The groupH = 〈[I×S3]H , [(∞)(0)(124)(365)×I]H〉 normalizes ρ and up to conjugationby H the possible block structures are

0 ¶ 2 3 4 5 6 7

0 ¶ 2 3 4 5 6 7

0 ¶ 2 3 4 5 6 7H

(i)

0 ¶ 2 3 4 5 6 7

0 ¶ 2 3 4 5 6 7

0 7 ¶ 2 3 4 5 6H

(ii)

0 ¶ 2 3 4 5 6 7

0 ¶ 2 3 4 5 6 7

0 5 6 7 ¶ 2 3 4H

(iii)

0 ¶ 2 3 4 5 6 7

0 7 ¶ 2 3 4 5 6

0 6 7 ¶ 2 3 4 5H

(iv)

0 ¶ 2 3 4 5 6 7

0 7 ¶ 2 3 4 5 6

0 5 6 7 ¶ 2 3 4H

(v)

0 ¶ 2 3 4 5 6 7

0 7 ¶ 2 3 4 5 6

0 3 4 5 6 7 ¶ 2H

(vi)

For case (i) we write the columns/blocks in MOG.

x x x

∞

x x x

0

x x x

1

x x x

2

x x x

3

x x x

4

x x x

5

x x x

6

Clearly any pair of blocks satisfy the parity condition in the square and thusis congruent to a sextet visualized in MOG. We write down seven of the thepairs

∞+ 0 ≡x

x

x

x

, ∞+ 1 ≡x

x

x

x

, ∞+ 2 ≡x

x

x

x

, ∞+ 3 ≡x x

x x

,

∞+ 4 ≡x

x

x

x

, ∞+ 5 ≡ x x

x xand ∞+ 6 ≡ x x

x x

.

53

These are renements of the MOG trio. It turns out that all further pairsalso are congruent to one of the above sextets. It is easy to convince oneselfthat the MOG trio is unique trio having all seven sextets as renements. Asthe seven sexets are preserved, as a set, G is in the trio stabilizer.

In case (ii) the sum of block 0 and ¶ is of type U6 while the sum of block¶ and 2 is of type S6, contradicting that G is doubly transitive on the blocks.

Similarly in case (iii) the sum of block 0 and ¶ is of type S6 while thesum of block 1 and 3 is of type U6.

For case (iv) we write the blocks in MOG,

0 ¶ 0 7 0 6

5 6 4 5 3 4

2 4 ¶ 3 7 2

3 7 2 6 ¶ 5

.

The sum of block 0 and ¶ is congruent to the sextet

6 3 4 3 6 2

1 5 2 3 6 4

5 6 5 4 4 1

3 1 5 2 2 1

and the

sum of block 2 and 3 is congruent to

5 5 4 3 2 5

4 3 1 1 1 6

2 2 4 3 2 5

3 4 6 6 6 1

. These two sextets do

not cut the blocks in the same way, contradicting that G is doubles transitiveon the blocks.

Suppose the case is (v). Let G be the maximal group preserving theblocks. We write out the blocks in MOG,

0 4 0 3 0 ¶

¶ 2 7 ¶ 5 6

5 7 4 6 2 4

6 3 5 2 3 7

.

Now any pair of blocks is of type U6 and is congruent to a sextet moduloC . Let B denote the set of blocks and let b1, b2 ∈ B. Suppose b1 + b2

∼= S(mod C ). If x ∈ b1 + b2 and O is the octad dened by b1 + b2 + x, clearlyx is the unique element in b1 + b2 that is in the tetrad b1 + b2 + O ∈ S. Wesay that a pair x, y is special if x and y lies in dierent blocks. Each specialpair x, y ∈ Ω denes a sextet S such that x ∈ Ti and y ∈ Tj for some tetradsTi, Tj ∈ S, Ti 6= Tj. Thus there is a well dened function f on special pairs

f(x, y) =∣∣∣ b ∈ B

∣∣|b ∩ Ti| = |b ∩ Tj| = 1 ∣∣∣, where x ∈ Ti, y ∈ Tj.

We now dene a relation ∼ on Ω by

54

x ∼ x, ∀x ∈ Ω,

x ∼ y if f(x, y) ≥ 2,

in other cases x 6∼ y. If x, y is a speical pair lying in tetrads Ti and Tj and

g ∈ G we have

f(g(x), g(y)) =∣∣∣ b ∈ B

∣∣|b ∩ gTi| = |b ∩ gTj| = 1 ∣∣∣

=∣∣∣ b ∈ B

∣∣|gg−1b ∩ gTi| = |gg−1b ∩ gTj| = 1 ∣∣∣

=∣∣∣ b ∈ B

∣∣|g(g−1b ∩ Ti)| = |g(g−1b ∩ Tj)| = 1 ∣∣∣

=∣∣∣ b ∈ B

∣∣|g−1b ∩ Ti| = |g−1b ∩ Tj| = 1 ∣∣∣

=∣∣∣ b ∈ B

∣∣|b ∩ Ti| = |b ∩ Tj| = 1 ∣∣∣ = f(x, y).

Hence G preserves the relation ∼. We will to show that ∼ induce a partitionof Ω into three sets which will turn out to be the MOG trio. We start bynoting that ρ preserves the MOG trio and having a 7-cycle in each brick. We

see that the element σ =


@@

preserves B. Clearly 〈ρ, σ〉 is doubly

transitive on the rst brick. Further the element

α = (∞ 17 22)(0 13 19)(1 8 11)(2 21 3)(4 9 14)(5 20 7)(6 15 16)(10 12 18) ∈M24

preserves B and is transitive on the bricks. From the sextet,

4 5 6 6 3 2

5 4 1 1 3 2

2 2 4 5 6 1

3 3 4 5 1 6

dened by the points ∞ and 0,

we see that the blocks 0 and ¶ both the cut the tetrads 4 (containing∞) and5 (containing 0). Thus f(∞, 0) = 2 and∞ ∼ 0. Since G is doubly transitiveon the rst brick and preserves ∼, all points in the rst brick are related. Letx ∈ Λ1 and y ∈ Λ2 and suppose x ∼ y. Pick g ∈ 〈ρ, σ〉 such that g(x) =∞.As ∞ and 17 are in the same block ∞ 17. Thus g(y) ∈ Λ2 + 17.

55

Multiplication with ρi for a suitable i implies that ∞ ∼ 11. But by lookingat the sextet

S =

1 1 5 6 2 4

2 4 3 1 5 5

2 4 1 3 6 6

3 3 5 6 4 2

dened by the points ∞ and 11,

wee see that no block cuts the tetrads 1 (containing ∞) and 6 (containing11) in one point each. Contradiction. Thus no elements in Λ1 are related toan element in Λ2. Further multiplication with α shows that all element inone brick are related and no two element from dierent bricks are related.Since G ≤ G, G stabilize the MOG trio.

Finally, we constructed the Octern group as the maximal subgroup ofM24 preserving the block structure in case (vi).

Blocks of size 2. By Lemma 3.19 G is doubly transitive on the blocks.Clearly 11

∣∣|G|. As before we may assume that the element of order 11 isρ = [(∞)(0 1 2 3 4 5 6 7 8 9 10)× I]L. The group

H = 〈[(∞)(0)(1 3 9 5 4)(2 6 7 10 8)× I]L,a b c e g i k l j h f d

a b d f h j l k i g e c

〉

normalizes ρ. Up to conjugation of H the possible block structures are

0 ¶ 2 3 4 5 6 7 8 9 X Y

0 ¶ 2 3 4 5 6 7 8 9 X YL

(i)

0 ¶ 2 3 4 5 6 7 8 9 X Y

0 Y ¶ 2 3 4 5 6 7 8 9 XL

(ii)

0 ¶ 2 3 4 5 6 7 8 9 X Y

0 X Y ¶ 2 3 4 5 6 7 8 9L

(iii)

For case (i) let G be the maximal subgroup preserving the blocks, andG∞2 the subgroup stabilizing the blocks ∞ and 2. We write the sextetcorresponding to ∞+ 2.

1 5 6 1 2 5 3 4 6 6 2 5

1 4 2 1 3 4 3 2 3 4 6 5L

Let X be the octad containing 1's and 5's. Then ∀g ∈ G∞2 we haveX + gX ∈ C implying that gX = X or |X + gX| = 8. Note that g preserves

56

both the blocks and the sextet above. If gX = X then g2 = 1. If g isan involution of type one, tetrads 1, 3 and 5 has to be xed. On tetradsthere are two possibilities for g, plainly g = 1 or g = (2 6). In the rst casewe see that columns 3, 6, 7, 8 has to be xed. But 3 + 6 + 7 + 8 /∈ C , acontradicion. In the second case no column containing a 2 or a 6 can containa xed point. Thus there must be an octad in the sum of the remainingcolumns. I.e ∞ + 0 + 2 + 4 + 5 + 10 contains an octad. There are only twosuch octads,

x x x

x x

x

x x

and

x

x x

x x x

x x

.

Both contain the points 0, 12, 15, 19 which spread over ∞, 0, 2, 10. Thus 4and 5 cannot contain xed points. Especially in column 4 tetrad 4 and 5 hasto be swapped. Contradiction. If g is an involution of type two, we end upwith two cases.

g = (∞ 0)(1 16)(2 3)(4 20)(5 11)(6 7)(8 22)(9 17)(10 15)(12 14)(13 19)(18 21)

or

g = (∞ 0)(1 16)(2 3)(4 20)(5 11)(6 18)(7 21)(8 22)(9 17)(10 15)(12 14)(13 19),

but neither is in M24. Thus g = 1. Suppose |X + gX| = 8 and letX ′ = X+∞+2. If gX ′ = hX ′ for some h ∈ G∞2 then h

−1gX ′ = X ′ implying

that g = h. Hence if g′ ∈ G∞2 and g 6= g′ then gX ′ and g′X ′ are disjoint sets.

Thus |G∞2| ≤|Ω +∞+ 2||X ′|

= 5, implying that |G| ≤ 12 · 11 · 5 = |L2(11)|.

But G contains [L2(11)× I]L so they must be isomorphic. Hence G is in thedodecad stabilizer and so is G.

For doubly transitivity to hold each two blocks must be contained in thesame number of sets of type S6 which are union of blocks. In case (ii) theve tetrads that together with block 0 and ¶ form an octad are:

1, 11, 14, 16, 4, 13, 20, 21, 2, 8, 17, 22, 10, 12, 17, 17 and 3, 5, 19, 9.

There are exactly 4 blocks that are contained in one of the above sets. Thusthe two blocks 0 and ¶ are contained in 4 sets of type S6 which are unionsof blocks. The ve tetrads that together with block ¶ and 2 form an octadare:

∞, 0, 10, 18, 11, 13, 16, 20, 5, 8, 9, 17, 1, 4, 14, 21 and 2, 3, 19, 22.

57

There are only 3 blocks that are contained in one of these sets. Hence case(ii) cannot occur. Similar reasoning rules out case (iii).

3.5 Generators of M24

Let α and γ, i.e the usual generators to L2(23), be two of the generators.Since L2(23) is maximal and 5 - |L2(23)| any δ ∈ M24 such that o(δ) = 5works as a nal generator to M24. It turns out that M24 is generated by

α = (∞)(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22)

γ = (0 ∞)(1 22)(2 11)(3 15)(4 17)(5 9)(6 19)(7 13)(8 20)(10 16)(12 21)(14 18)

δ = (∞)(0)(1)(2 16 9 6 8)(3 12 13 18 4)(5)(7 17 10 11 22)(14 19 21 20 15).

58

4. Simplicity of the Mathieu

groups

Lemma 4.1. If G is a transitive permutation group on a set A and N CGthen N is 1/2-transitive. If in addition N is nontrivial and G is primitivethen N is transitive.

Proof. Let S be an orbit of N of minimal length. Then ∀g ∈ G we haveN(gS) = (Ng)S = (gN)S = g(NS) = gS so gS is a union of orbits of N .Since S is of minimal length and |S| = |gS| we see that gS is exactly oneorbit. Now G is transitive so every orbit of N has the same length (havingthe form gS). Thus N is 1/2 transitive. Finally two dierent orbits aredisjoint so S form a set of imprimitivity for G. Thus if G is primitive wehave two cases, either |S| = 1 implying N is trivial, or |S| = |A| implying Nis transitive.

We will handle the small and large Mathieu groups separately. Themethods on the two cases are very similar thou. The main idea, taken from[7], is to use induction and the main dierence is the base cases. The keylemma used is the following.

Lemma 4.2. Let G be a primitive permutation group on a set A. If Ga issimple then G is simple or every proper and nontrivial normal subgroup ofG is regular.

Proof. Assume 〈1〉 < N C G. By Lemma 4.1 N is transitive. Since Ga

is simple and (Na ∩ Ga) C Ga we must have Na = 〈1〉 or Na = Ga. IfNa = 〈1〉 then N is a regular normal subgroup and If Na = Ga then|A| = [G : Ga] = [N : Ga] implying that G = N .

59

Suppose N is a regular normal subgroup of a 2-transitive group G. Ifn,m ∈ N ,

n(a) = b and m(a) = c,

then ∃g ∈ G such that gmg−1(a) = b. Since gmg−1n−1 ∈ N xes b it followsthat all nonidentity elements in N are conjugates. Hence all nonidentityelements in N have the same cycle shape. As N is regular this cycle shapeis pn for some prime p.

Lemma 4.3. Let G be a 2-transitive permutation group on a set A. If Gcontains a regular normal subgroup N then |A| = |N | = pn for some primep.

Proof. Using Lemma 1.11 we see that G is primitive and by Lemma 4.1 N istransitive. Finally Corollary 1.6 implies that |A| = |N | and the result followsfrom the above discussion.

Theorem 4.4. The Mathieu group M11 is simple.

Proof. Assume N C M11 is proper and nontrivial. Since M11 is doublytransitive it is primitive. By by Lemma 4.1 N is transitive and so 11

∣∣|N |. LetP be a Sylow 11-subgroup of N and s11 the number of Sylow 11-subgroupsin M11. From Sylow's third theorem we have that

s11 =11 · 10 · 9 · 8|NM11 [P ]|

and s11 ≡ 1 (mod 11)

implying that s11 = 144 and |NM11 [P ]| = 55. Now since P < N CG and allSylow 11-subgroups are conjugates, N contains all 144 Sylow 11-subgroups.Using Sylow's third theorem again we obtain |N | = 144 ·NN [P ] = 11 ·2 ·9 ·8.Note that NN [P ] contains exactly eleven elements since it contains P andN and is proper. Further [N : Na] = 11 yields N ∩ M10 6= 〈1〉, whereM10 < M11 is the isotropy group xing a. Now N ∩M10 CM10. As M10

is doubly transitive and Lemma 1.11 implies that M10 is primitive. FurtherLemma 4.1 implies that N ∩ M10 is transitive on ten points. This yields10∣∣|N ∩M10| implying 10

∣∣|N |. Contradiction.

Theorem 4.5. The Mathieu group M12 is simple.

Proof. Since M11 is simple Lemma 4.2 implies that M12 is simple or containsa regular normal subgroup. But 12 is not a prime power so by Lemma 1.16M12 cannot have a regular normal subgroup.

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Theorem 4.6. The subgroup M21 is simple.

Proof. We rst note that |M21| = 26 · 32 · 5 · 7. Assume N CM21 is nontrivialand proper. Since M21 is 2-transitive Lemma 1.11 and Lemma 4.1 impliesthat N is transitive. Hence 7

∣∣|N |. Let 〈g〉 be a Sylow 7-subgroup of N . ByLemma 1.2 M21 = NM21 [〈g〉]N . Let H = NM21 [〈g〉].

We now want to show that |H| = 21. By Corollary 3.2 g can only have cycleshape 13.73. Thus H is imprimitive with these 7-cycles as blocks. Assume5∣∣|H|. Then ∃x ∈ H such that o(x) = 5. Dene the restriction map to thethree blocks ϕ : H −→ S3. By considering the order we get x ∈ Ker(ϕ).But then x xes at least two points in each block plus the three points xedby M21, i.e nine points, contradicting Corollary 3.2. Further assume 2

∣∣|H|.Then H contains an involution y. By Theorem 3.13 there are only two typesof involutions so y xes exactly eight points, i.e the three points xed byM21

plus ve more. For y to not have more than eight xed points this leavestwo cases. Either all ve points are in one block or one block contains threepoints and two other one point. In either case ygy−1 /∈ 〈g〉. Contradiction.Thus |H| = 21 or 63. By Sylow's third theorem the latter is impossible.

Since N is proper we cannot have H = N and so H ∩ N = 〈g〉. By thesecond isomorphism theorem we have

M21/N = HN/N ∼= H/(N ∩H) = H/〈g〉 ∼= Z3.

Hence |N | = 26 · 3 · 5 · 7 and by Lemma 1.5 |Na| = 26 · 5. Now Na CGa. Let〈h〉 be a Sylow 5-subgroup of Na. Then Lemma 1.2 gives Ga = NGa [〈h〉]Na.By consideration of the order 3

∣∣|NGa [〈h〉]| and so ∃n ∈ NGa [〈h〉] such thato(n) = 3. As Ga xes a sextet and one tetrad pointwise in this sextet,there is a well dened restriction map to the action on the remaining tetradsθ : Ga −→ S5. Clearly θ(h) is 5-cycle. If n ∈ Ker(θ) then n xes at leastone points each of the ve tetrads, i.e at least nine points, contradictingCorollary 3.2. Thus θ(n) is a 3-cycle. But in S5 no 3-cycle normalize a Sylow5-subgroup.

Theorem 4.7. The Mathieu groups M22, M23 and M24 are simple.

Proof. SinceM21 i simple Lemma 4.2 implies thatM22 is simple or contains aregular normal subgroup. As 22 is not a prime power Lemma 4.3 implies that

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M22 is simple. Similarly M23 is simple or contain a regular normal subgroupor order 23. But M23 is 3-transitive so no subgroup of order 23 is normal.HenceM23 is simple. FinallyM24 is simple since 24 is not a prime power.

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5. Uniqueness of the small

Mathieu groups

The goal is to show that up to isomorphism (i.e up to relabeling of the points)M11 is the unique nontrivial sharply 4-fold transitive group. Then using thisto prove uniqueness for M12. As a bonus we will get an explicit constructionof the small Mathieu groups.

Lemma 5.1. All permutations of order two in M11 are conjugates and haveexactly three xed points. Further let g and h be two such permutations thatcommute. If (a b) is a cycle in g then one of the following holds:

• h contains (a b)

• h contains (a)(b)

• h contains (a c)(b d) where g contains (c d)

Proof. In the proof of Theorem 1.16 we saw that all such involutions wereconjugates. Because of the 4-transitivity ∃x ∈ M11 such that x contains(1 2)(3)(4). Further x2 xes four points and so x2 = 1. Hence x is of ordertwo. Since the degree of M11 is odd x has an odd number (less than four) ofxed points and so x xes exactly three points. Since all other permutationsof order two are conjugates to x all they xes three points.

If h(a) = a then h(b) = hg(a) = gh(a) = g(a) = b and we got the rstcase. If h(a) = b then h(b) = hg(a) = gh(a) = g(b) = a and we gotthe second case. Finally let h(a) = c, where c is neither a nor b. Thenh(b) = hg(a) = gh(a) = g(c) = d for some d not equal to a, b or c. This isthe third case.

Theorem 5.2. The Mathieu group M11 is unique up to isomorphism.

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Note. Even if the following proof is possible to do by hand I want to pointout that it has been created using GAP (see Appendix).

Proof. Through this proof we will be using Lemma 5.1 frequently and so referto it as the lemma. Up to isomorphism we can assume that the elementcontaining (1 2)(3 4) in M11 is

x1 = (1 2)(3 4)(5 6)(7 8).

Due to the sharply 4-fold transitivity there is an unique element x2 containing(1 2)(3)(4). Since x1x2 and x2x1 agree on 1, 2, 3, 4 they are identical so x1

and x2 commute. Since both x1 and x2 contains the cycle (1 2) they cannothave another common 2-cycle. Using the lemma on (5 6) and the fact thatx2 already xes two points we can assume that up to isomorphism

x2 = (1 2)(5 7)(6 8)(9 10) implying x1x2 = (3 4)(5 8)(6 7)(9 10).

Let x3 be the unique element containing (1 3)(2 4). As above x1 and x3

commute. Using the lemma on (5 : 6) and the fact that all three pairs of2-cycles in 5, 6, 7, 8 are already present implies that x3 contains exactlyone of (5 6) and (7 8). These are equivalent up to isomorphism so we canassume x3 contains (5 : 6). Now x2x3 contains (1 3 2 4) and have order foursince (x2x3)4 = 1. If x2x3 do not contain (9 10) is also contains a 3-cycle, acontradiction. Hence

x3 = (1 3)(2 4)(5 6)(9 10) implying x1x3 = (1 4)(2 3)(7 8)(9 10).

Let x4 be the unique element containing (1 3)(2)(4). As above x3 and x4

commute so by lemma 5.1 x4 contains two 2-cycles from 5, 6, 9, 10. Since x3

and x4 already agree on 2 points x4 contains (5 : 9)(6 : 10) or (5 : 10)(6 : 9).These are equivalent up to isomorphism so we assume x4 contains the rst.Now x4 contains a 2-cycle from 7, 8, 11. If this 2-cycle contains 11 thenx1x4 contains a 4-cycle and a 3-cycle, a contradiction. Thus

x4 = (1 3)(5 9)(6 10)(7 8) implying x3x4 = (2 4)(5 10)(6 9)(7 8),

x4x2x−14 = (2 3)(5 6)(7 10)(8 9) and x4x1x2x

−14 = (1 4)(5 6)(7 9)(8 10).

Let x5 be the unique element containing (7 8)(9 11). By the lemma the xedpoints are 1, 2, 10, 3, 4, 10 or 5, 6, 10. Now all permutations having a

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2-cycles in 1, 2, 3, 4 are already present so by the lemma there are only fourcases for x5 namely:

(1 5)(2 6)(7 8)(9 11), (1 6)(2 5)(7 8)(9 11),

(3 5)(4 6)(7 8)(9 11) and (3 6)(4 6)(7 8)(9 11)

Case one and three multiplied by x4 yields a 4-cycle and a 3-cycle. Contradiction.Further if g = (1 3)(2 4)(7 8) then 〈x1, x2, x3, x4〉g = 〈x1, x1x2, x3, x4〉 =〈x1, x2, x3, x4〉 so case two and four are equivalent under isomorphism so wecan assume that

x5 = (1 6)(2 5)(7 8)(9 11).

We have now shown that 〈x1, x2, x3, x4, x5〉 ≤M11 is unique up to isomorphism.Finally we want to show equality. The idea is to use Lagrange theorem topush up the order. We nd the elements:

y1 = x5x2x3 = (1 3 5 7 2 4 6 8)(9 11)

y2 = x1x2x4x5 = (1 9 11 8 5 2 10 7 6 4 3)

y3 = x2x4x5 = (1 9 11 7 5)(2 10 8 6 3)

a1 = x4x2x1 = (1 3 4)(5 7 10)(6 8 9)

s = x5x4x−15 = (1 10)(2 11)(3 6)(7 8)

a2 = sa1s−1 = (1 5 8)(3 7 9)(4 10 6).

Now o(y1) = 8, o(y2) = 11, o(y3) = 5 and |〈a1, a2〉| = 9 yields

|〈x1, x2, x3, x4, x5〉| ≥ 11 · 5 · 9 · 8 =|M11|

2.

As M11 is simple it cannot contain a subgroup of index two. Hence

M11 = 〈x1, x2, x3, x4, x5〉.

Lemma 5.3. Let G be a sharply doubly transitive group of degree nine. ThenG = H oGa where H ∼= Z3×Z3. Further Ga acts transitively on H \ 1 byconjugation.

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Proof. Suppose G is acting on 1, 2, . . . 9 and let g ∈ G be an element oforder three. Let Xi = g ∈ G|g(1) = i. Then G = X1 ∪ · · · ∪X9. Since Gis sharply doubly transitive |Xi| = 8 for i = 1, . . . , 9. For all i 6= 1 the sevenisotropy subgroups not xing 1 and i is transitive and each containing anelement in Xi. Because of the doubly transitivity we can conjugate g withsome αi such that gαi ∈ Xi. Hence are there exactly eight element of orderthree and none of order nine. These eight element together with the identityform a normal Sylow 3-subgroup H ∼= Z3×Z3. Finally pick two nonidentityelements k ∈ Ga and h ∈ H and assume hk = h ⇐⇒ hk = kh. Supposeh(a) = b. Then k(b) = kh(a) = hk(a) = h(a) = b, a contradiction since kcannot x two points. Hence Ga acts transitively on H by conjugation.

Theorem 5.4. The Mathieu group M12 is unique up to isomorphism.

Method of Proof. Let H be the sharply 4-fold transitive group constructedabove and embed it in S12. Further let G < S12 be an arbitrary sharply 5-foldtransitive group such that G12 = H. We will show that there exist an uniqueelement σ ∈ G \H having ve special properties. Moreover we will see thatσ do not depend on the choice of G. Finally as H is maximal in G it followsthat H〈σ〉 = G. The uniqueness of G now follows from the uniqueness of Hand σ.

Proof. From above we have H = 〈x1, x2, x3, x4, x5〉 where

x1 = (1 2)(3 4)(5 6)(7 8)

x2 = (1 2)(5 7)(6 8)(9 10)

x3 = (1 3)(2 4)(5 6)(9 10)

x4 = (1 3)(5 9)(6 10)(7 8)

x5 = (1 6)(2 5)(7 8)(9 11)

Since |H| = 11 · 10 · 9 · 8 and |G| = 12 · 11 · 10 · 9 · 8 we see that the Sylow3-subgroups in H and G have order 9 and 27 respectively. Our aim is to ndσ in a Sylow 3-subgroup. Let A = 〈a1, a2〉 be the subgroup of order nine inTheorem 5.2. Clearly A < H2,11.

Next we want to show that there is an element σ ∈ G satisfying properties(1)-(5):

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(1) σ(12) 6= 12,

(2) σ(10) = 10,

(3) o(σ) = 3,

(4) σa1 = a1σ,

(5) σa2 = a1a2σ.

Let P be a Sylow 3-subgroup of G such that A < P . If we consider theorbits of the point 10 we get

A(1) = 1, 3, 4, 5, 6, 7, 8, 9, 10 ⊆ P (1) ⊆ 1, 2, . . . , 12 =⇒ 9 ≤ |P (10)| ≤ 12.

But |P (10)| = [P : P10]∣∣|P | = 27 so |P (10)| = 9. Thus |P10| = 3 and we can

nd σ, o(σ) = 3, such that P10 = 〈σ〉. It is clear that σ satises properties(2) and (3) above. Since no nonidentity element in A xes the point 10 wehave |〈σ〉 ∩ A| = 1 and |A〈σ〉| = |P |. Hence P = A〈σ〉. If σ(12) = 12 thisimplies that P ≤ H, a contradiction since |P | - |H|. So σ satises property(1) above. Next we want to investigate Z(P ). Since P is a 3-group we knowthat |Z(P )| = 3, 9 or 27. Assume the latter, i.e P is abelian. Then ∀i ∈ A(1)we can nd a ∈ A such that a(10) = i. Hence ∀i ∈ A(1) we have

σi = σa(10) = aσ(10) = a(10) = i.

A contradiction since the only element in G xing ve points is the identity.Hence P is non abelian. If |Z(P )| = 9 then P/Z(P ) is cyclic, implying Pis abelian. Contradiction. Thus |Z(P )| = 3. Suppose |Z(P ) ∩ A| = 1.Then since |AZ(P )| = |P | and both A and Z(P ) are normal subgroups weget P ∼= A × Z(P ), i.e again abelian. Contradiction. Thus Z(P ) < Aand Z(P ) = 〈a〉 for some a ∈ A. By Lemma 5.3 the group H2,11,10 actingtransitively on A by conjugation and xing A as a set. So we can ndh ∈ H2,11,10 such that hah−1 = a1. We set

P = P h, σ = hσh−1.

Now Z(P ) = Z(P )h = 〈hah−1〉 = 〈a1〉. Clearly o(σ) = 3, σ(10) = 10 andσ(12) 6= 12. Further the same arguments that implied P = A〈a〉 yieldsP = A〈a〉. Hence σ also satisfy properties (1), (2) and (3) above. So if weswap the names P, σ and P , σ we have achieved

Z(P ) = 〈a1〉.

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Since a1 ∈ Z(P ) we have a1σ = σa1 and σ satises property (4). To obtainproperty (5) consider the commutator subgroup of P . As P is non abelianwe have [P, P ] 6= 〈1〉. Further P/Z(P ) is abelian (prime square order) yields[P, P ] ≤ Z(P ). Thus [P, P ] = Z(P ). In particular

[σ, a2] = 1, a1 or a21

If [σ, a2] = 1 then σ, a1 and a2 all would commute. As P = A〈σ〉 this yieldsthat P is abelian. Contradiction. Thus [σ, a2] = a1 or [σ, a2] = a2

1. Now σ−1

also satises properties (1)-(4), so by replacing σ with σ−1 if necessary wemight assume

[σ, a2] = a1 ⇐⇒ σa2 = a1a2σ

I.e property (5). Knowing that σ satises properties (1)-(5) we now explicitnd the action on each point in 1, . . . , 12.

σ(5) = σa1(10) = a1σ(10) = a1(10) = 5

σ(7) = σa1(5) = a1σ(5) = a1(5) = 7

σ(6) = σa2(10) = a1a2σ(10) = a1a2(10) = a1(6) = 8

σ(8) = σa1(6) = a1σ(6) = a1(8) = 9

σ(9) = σa1(8) = a1σ(8) = a1(9) = 6

σ(1) = σa2(8) = a1a2σ(8) = a1a2(9) = a1(3) = 4

σ(3) = σa1(1) = a1σ(1) = a1(4) = 1

σ(4) = σa1(3) = a1σ(3) = a1(1) = 3

It remains to nd the action on 2, 11, 12. Since σ already xes three points2, 11, 12 is in a three cycle. Hence

σ = (1 4 3)(6 8 9)(2 11 12) or σ = (1 4 3)(6 8 9)(2 12 11)

Consider the second case. Then since H < G is a maximal subgroupH〈σ〉 = G. But the element

x5x3σ = (1 5)(2 9)(4 6 7 8 10 11 12)

has order 14 - |G|. Contradiction. Thus σ = (1 4 3)(6 8 9)(2 11 12) is the onlypossibility.

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A. GAP and permutation groups

GAP (Groups, Algorithms, Programming) is a system for computationaldiscrete algebra. It is free and can be found at http://www.gap-system.org.GAP is a very good tool for tedious calculations. Even if it is a computerprogram the outcome often is veriable by hand. The web page contains adetailed manual, but to get started here is short list of for this thesis usefulcommands.

gap> x:=(1,2)(3,4,5);

Denes the permutation x = (1 2)(3 4 5). All permutations in GAP acts ona set of positive integers.

gap> S:=[1,2,5];

Formally this denes a list, but if the numbers are written in ascending orderwe can see this as the denition S = 1, 2, 5.

gap> OnPoints(5,x);

Determine x(5).

gap> OnSets(S,x);

Assuming S is an ordered list this command returns the image of S under xas an ordered list.

gap> G:=Group(x,y,z);

Denes the Group G=〈x, y, z〉.

gap> MathieuGroup(24);

ReturnsM24 with the same generators used in thesis thesis. The non innityelements x ∈ Ω are relabeled as x+ 1 and ∞ as 24.

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http://www.gap-system.org

gap> Order(G);

Determine the order of the group G.

gap> x in S;

If S is any set this command returns true if x ∈ S and false otherwise.

gap> S:=Filter(T,x->P(x));

Denes S as the set of elements in T satisfying P , where P is a predicate onT .

Example A.1. Suppose we want to nd σ in the Octern group. We look foran involution in M24 having the action (1 2)(3 8)(4 5)(6 7) on the columns in

∞ 14 3 15 20 0 8 18

17 16 10 7 4 13 2 11

22 9 6 19 12 21 5 1

Since GAP is relatively slow we start by ltering out all involutions.

gap> I:=Filtered(MathieuGroup(24),x->x*x=());

gap> P:=function(x)

> if OnSets([18,23,24],x)=[10,15,17] and

> OnSets([4,7,11],x)=[2,12,19] and

> OnSets([8,16,20],x)=[5,13,21] and

> OnSets([1,14,22],x)=[3,6,9] then

> return true;

> else

> return false;

> fi;

> end;;

gap> Filtered(I,x->P(x));

[ (1,3)(2,7)(4,19)(5,20)(6,14)(8,21)(9,22)(10,18)(11,12)

(13,16)(15,24)(17,23) ]

N

Remark A.2. It is worth noting that GAP, unlike us, uses left to rightmultiplication. To show that this is irrelevant let · denote left to right

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multiplication and ∗ right to left multiplication. Let G be the set of allpermutations in the group GAP calculated on. Then ϕ : (G, ·) −→ (G, ∗)dened by ϕ(x) = x−1 is obviously a bijection. Further

ϕ(x · y) = (x · y)−1 = (y ∗ x)−1 = x−1 ∗ y−1 = ϕ(x) ∗ ϕ(y).

So ϕ is an isomorphism.

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Bibliography

[1] R.T. Curtis: A new combinatorial approach to M24, Cambridge (1974).

[2] R.T. Curtis: The maximal subgroups of M24, Cambridge (1974).

[3] D.S. Passman: Permutation Groups, New York, W.A. Benjamin, Inc.(1968).

[4] H. Wielandt: Finite permutation groups, New York, Academic Press(1964).

[5] C. Choi: On Subgroups of M24. II: The Maximal Subgroups of M24,Transactions of the American Mathematical Society, Vol. 167, 29-47(May, 1972).

[6] J.H. Conway, N.J.A Sloane: Sphere Packings, Lattices and Groupssecond edition, New York Berlin Heidelberg, Springer-Verlag (1993).

[7] H. Cuypers: The Mathieu groups and designs (1994), Lecture notes,http://www.win.tue.nl/ hansc/eidmamathieu.pdf

(online May, 2009).

[8] A. Wool: Sharply transitive permutation groups (2006),http://www.geocities.com/assafwool/TransGrp.pdf

(online May, 2009).

[9] J.H. Conway, N.J.A Sloane: Sphere Packings, Lattices and Groupssecond edition, New York Berlin Heidelberg, Springer-Verlag (1993).

[10] J.A. Todd: A representation of the Mathieu group M24 as a collineationgroup, Ann. Mat. Pura Appl., 71, 199-238 (1966).

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[11] E. Mathieu: Mémoire sur le nombre de valeurs que peut acquérir unefonction quand on y permut ses variable de toutes les formes possiblesCrelle, J. 5 9-42 (1860).

[12] E. Mathieu: Mémoire sur l'etude de fonctions de plusieurs quantitiés,sur la maniére des formes et des substitutions qui les laissent invariables.Crelle, J. 6 241-323 (1861).

[13] E. Mathieu: Sur la fonction cinq fois transitive des 24 quantitiés Crelle,J. 18 25-46 (1873).

[14] E. Witt: Die 5-fach transitiven Gruppen von Mathieu Hamburg, Abh.Math. Sem. (1938).

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