3-local identifications of somefinite simple groups

by

SARAH ASTILL

A thesis submitted toThe University of Birmingham

for the degree ofMaster of Philosophy (Sc, Qual)

School of MathematicsThe University of Birmingham

Abstract

In this thesis we characterize the groups PSL3(3), M12, G2(3) and PSp4(3) by their 3-local

structure. We identify each group as a finite faithful completion of a certain type of rank

2-amalgam known as a weak BN -pair.

Acknowledgements

I am grateful for the help, support and encouragement first and foremost from my super-

visor Professor Chris Parker. Thanks also go to Dr Rachel Fowler, Dr Murray Clelland,

Dr Rebecca Waldecker and James Sprittles. Finally, I gratefully acknowledge the financial

support of the EPSRC.

Contents

Introduction 1

1 Some Group Theory 6

1.1 Some General Group Theoretic Results . . . . . . . . . . . . . . . . 6

1.2 Extraspecial Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Fixed-Point-Free Actions . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Strongly Embedded Subgroups . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Automorphisms of Abelian Groups . . . . . . . . . . . . . . . . . . . 19

2 Some Representation Theory 23

2.1 Endomorphism Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Natural GL2(3)-Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.4 Sym(4)-Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.5 Some Modular Character Theory . . . . . . . . . . . . . . . . . . . . 37

2.6 Natural SL2(2n)-Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3 Amalgams and the Coset Graph 49

4 Background Results 53

5 An Amalgam of Type PSL3(3) 61

5.1 Some Structure of the Amalgam (A,B,C) . . . . . . . . . . . . . . . 63

5.2 A Presentation for the Amalgam . . . . . . . . . . . . . . . . . . . . . 65

6 An Amalgam of Type G2(3) 75

6.1 The Amalgam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.2 The Coset Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.3 Controlling the 3-Local Structure . . . . . . . . . . . . . . . . . . . . 89

6.4 Controlling the Centralizer of an Involution . . . . . . . . . . . . . . 96

7 An Amalgam of Type PSp4(3) 101

7.1 The Amalgam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.2 The Coset Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

7.3 Controlling the 3 and 2-Local Structure . . . . . . . . . . . . . . . . 109

8 Higman’s Theorem on Fixed-Point-Free Elements of Order Three 119

Appendix 137

Introduction

A p-local subgroup (p a prime) of a group is the normalizer of a non-trivial p-subgroup.

To identify a group p-locally is to determine the isomorphism type of a group given some

information about its p-local structure. Suppose we have a finite group G with a Sylow

p-subgroup S such that S has subgroups A and B which are normalized by NG(S) then

the intersection of NG(A) and NG(B) contains NG(S). The Amalgam Method aims to

characterize G given the subgroup structure associated to NG(A) and NG(B).

A rank two amalgam of finite groups, A = (P1, P2, C, φ1, φ2), is a collection of three

finite groups, P1, P2 and C, and two injective homomorphisms, φ1 : C → P1 and φ2 : C →

P2. A completion, (G,ψ1, ψ2), of the amalgam is a group, G, such that ψ1 : P1 → G and

ψ2 : P2 → G are homomorphisms with G = 〈ψ1(P1), ψ2(P2)〉 and such that φ1ψ1 = φ2ψ2.

In [7], Delgado and Stellmacher first defined weak BN -pairs of rank two. They described

a situation when a rank two amalgam (P1, P2, C) satisfied some local conditions notably

that Op(Pi) (i = 1, 2 and p a fixed prime) admitted an action from Op′(Pi)/Op(Pi) ∼=

PSL2(pni), SL2(p

ni),U3(pni), SU3(p

ni), Sz(2ni) or Dih(10) (and p = 2) or Ree(3ni),Ree(3)′

(and p = 3). They classified the possible types of amalgam and described each one. In

[23], Parker and Rowley examined the BN -pairs for which the characteristic prime, p,

is odd. They characterized finite faithful completions satisfying some p-local conditions

under a K-proper hypothesis. In this thesis we are interested in the weak BN -pairs

(P1, P2, C) such that, for i = 1 or 2, O3(Pi) admits an action from SL2(3) of which there

1

are four types. One of these, the exceptional weak BN -pair of type F3, has recently been

studied by Rachel Fowler to characterize the Thompson sporadic simple group [10]. Her

characterization relies on a K-proper hypothesis to recognize a section of the Thompson

group which is isomorphic to G2(3). In this thesis we will describe the remaining three

weak BN -pairs which have types PSL3(3), G2(3) and PSp4(3). Furthermore, we will

characterize the groups PSL3(3) and M12; G2(3); and PSp4(3) as finite faithful completions

of the respective amalgams. Notably, the characterization of G2(3) will be used in [11] to

strengthen the characterization of the Thompson sporadic simple group thereby removing

any reliance on the K-proper hypothesis.

We will examine the amalgams by studying the local group theoretic structure of the

groups which constitute the amalgam. Therefore we begin in Chapter 1 with some well

known group theoretic results which will be referred to throughout this thesis. In Chapter

2 we discuss some representation theory. In particular, we describe the SL2(3)-modules

which will appear within the group theoretic structure of the groups in the amalgams. In

Chapter 3 we will formally define an amalgam and a completion of an amalgam. We will

also associate a certain graph, called the coset graph, to a completion of an amalgam. The

coset graphs associated to G2(3) and PSp4(3) are Moufang Polygons (these are defined

and classified in [29]) and so in Chapter 4 we define such graphs and state the classification

theorem of Tits and Weiss. Also in Chapter 4, we present a character theoretic result due

to Feit and Thompson ([9]) which describes a finite group containing a self-centralizing

element of order three. We also discuss a theorem of the same nature by Smith and

Tyrer ([27]). Their theorem describes groups in which the automizer of an abelian Sylow

p-subgroup (odd prime p) has order just two. We will use both results several times in

the group recognition theorems in later chapters.

In Chapter 5 we begin to prove the main theorems of this thesis. The main hypothesis

and theorem in Chapter 5 are as follows.

2

Hypothesis A. Let G be a finite group with subgroups A ∼= B such that C := A ∩

B contains a Sylow 3-subgroup of both A and B and such that no non-trivial normal

subgroup of C is normal in both A and B. Suppose further that

(i) G = 〈A,B〉;

(ii) A/O3(A) ∼= B/O3(B) ∼= GL2(3);

(iii) O3(A) and O3(B) are natural modules with respect to the actions of A/O3(A) and

B/O3(B) respectively; and

(iv) for S ∈ Syl3(C), NG(Z(S)) = NA(Z(S)) = NB(Z(S)).

Theorem A. If G satisfies Hypothesis A, then G ∼= M12 or PSL3(3).

We may refer to the amalgam (A,B,C) as an amalgam of type PSL3(3). This amalgam is

particularly interesting as it completes to PSL3(3) and also to the sporadic simple group

M12 under the same 3-local assumption, namely condition (iv) of the hypothesis.

The main theorem and hypothesis in Chapter 6 are:

Hypothesis B. Let G be a finite group with non-conjugate subgroups A1 and A2 such

that A12 := A1 ∩ A2 contains a Sylow 3-subgroup of both A1 and A2, and such that no

non-trivial normal subgroup of A12 is normal in both A1 and A2. Suppose further that,

for i = 1, 2,

(i) |O3(Ai)| = 35;

(ii) Ai/O3(Ai) ∼= GL2(3);

(iii) O3(O3(Ai))/Z(Ai) and Z(O3(Ai))/O3(Ai)′ are natural modules with respect to the

action of Ai/O3(Ai); and

(iv) NG(O3(Ai)′) = Ai.

3

Theorem B. Let G be a group satisfying Hypothesis B. Then G ∼= G2(3).

In some sense this is the strongest amalgam theoretic result we prove since we begin

with a group containing a completion of the amalgam, as opposed to Theorem A which

characterizes just the completion itself.

The methods successfully used in Chapter 6 are extended in Chapter 7 where we

describe an amalgam of type PSp4(3). The hypothesis we assume is strong and so the

theorem is less successful than other characterizations of PSp4(3) (see for example [24]

and [16]).

Hypothesis C. Let A1 and A2 be subgroups of a finite group G = 〈A1, A2〉 such that

A12 := A1∩A2 contains a Sylow 3-subgroup of both A1 and A2 and such that no non-trivial

normal subgroup of A12 is normal in both A1 and A2. Suppose further that

(i) A1/O3(A1) ∼= Sym(4);

(ii) A2/O3(A2) ∼= SL2(3);

(iii) O3(A1) is elementary abelian of order 33 and is a faithful A1/O3(A1)-module;

(iv) O3(A2) is extraspecial of order 33 and O3(A2)/Z(O3(A2)), is a natural A2/O3(A2)-

module;

(v) NG(O3(A1)) = A1 and NG(Z(O3(A2))) = A2; and

(vi) O3(A1) normalizes no non-trivial 3′-subgroup of G.

Theorem C. Let G be a group satisfying Hypothesis C. Then G ∼= PSp4(3).

Notably condition (vi) of the hypothesis makes this theorem less successful than Theo-

rem B. However, it is possible to construct groups which have shape 26 : PSp4(3) and

4

56 : PSp4(3) which satisfy the 3-local condition (v). Certainly condition (vi) in the hy-

pothesis rules out such groups however a preferable alternative hypothesis might be to

assume O3′(G) = 1. At present it is unclear whether this alternative hypothesis would be

sufficient. The uncertainty about such groups does however raise an interesting question:

for which finite groups G does there exist 1 6= N E G such that G/N ∼= PSp4(3)? Higman

answered a similar question about SL2(2n) in [17] and so Chapter 8 together with much

of the representation theory in Chapter 2 is devoted to understanding Higman’s result.

An open question is whether the methods from Chapter 8 can be extended to PSp4(3).

We include also a short appendix which contains some elementary number theoretic

proofs concerning orders of fields. The appendix also contains a proof related to the coset

graph in Chapter 6.

Before we begin we note that all groups will be finite and all vector spaces will be

finite dimensional. We use Atlas [6] notation for groups and group extensions with the

exception of the alternating, symmetric and dihedral groups which are labeled Alt(n),

Sym(n) and Dih(n) respectively.

5

Chapter 1

Some Group Theory

1.1 Some General Group Theoretic Results

Lemma 1.1. Let p be a prime and let A be a p-group acting on a non-trivial p-group V .

Then CV (A) 6= 1 and [V,A] < V . In particular if V is an FA-module where F is a field

of characteristic p then CV (A) is a non-trivial submodule and [V,A] is a proper submodule

of V .

Proof. If we form the semidirect product S := V : A then S is a p-group with normal sub-

group V . Thus, S has non-trivial centre and Z(S)∩V 6= 1. Hence CV (A) 6= 1. Now since

S is a finite p-group, S is nilpotent and so for some integer n, 1 = [S, S, n] = [S, S, . . . , S]

(n’th term in lower central series). So suppose [V,A] = V then V = [V,A,A, . . . , A] =

[V,A, n] 6 [S, S, n] = 1 which is a contradiction.

A group, G, is said to act equivalently on two sets Ω1 and Ω2 if there is a bijection

α : Ω1 → Ω2 such that for ω ∈ Ω1, g ∈ G, α(ω · g) = α(ω) · g.

Lemma 1.2. Suppose G is a group with subgroups A and B which normalize each other.

Suppose further that both are acted on by another group X. Then the action of X on

6

AB/A is equivalent to the action of X on B/(A ∩ B). In particular X commutes with

AB/A if and only if X commutes with B/(A ∩B).

Proof. Let α be the natural isomorphism α : AB/A → B/(A ∩ B) then for x ∈ X and

Ab ∈ AB/A,

α(Abx) = α(A(bx)) = (A ∩B)bx = ((A ∩B)b)x = α(Ab)x

and so the action is equivalent.

Lemma 1.3. Let G be a group and let a, b, c ∈ G. The following identities hold.

(i) [a, bc] = [a, c][a, b]c.

(ii) [ab, c] = [a, c]b[b, c].

(iii) If p is a prime and if G is a p-group then [[a, b], c][[b, c], a][[c, a], b] ∈ [G,G,G,G].

Proof. The first two identities are easy to check the third is Lemma 5.6.1 (iv) in [13,

p209].

Definition 1.4. Let A be a group acting on a group G. The action of A on G is coprime

if |A| and |G| are coprime.

Theorem 1.5. (Coprime Action) Suppose A is a group acting on the group G and sup-

pose the action of A on G is coprime. The following hold.

(i) G = [G,A]CG(A) and if G is abelian G = [G,A]× CG(A).

(ii) [G,A] = [G,A,A].

(iii) CG/N(A) = CG(A)N/N for any A-invariant N E G.

7

(iv) If A is an elementary abelian p-group of order at least p2 then G = 〈CG(a)|a ∈

A#〉 = 〈CG(A1)|[A : A1] = p〉.

Proof. For (i) and (ii) see [20, 8.2.7 p187]. For (iii) see [20, 8.2.2 p184]. For (iv) see [20,

8.3.4 p193].

A group G is said to have a normal p-complement if Op′(G) = Op(G) or equivalently if

G has a normal subgroup N such that G is the semidirect product of N with some Sylow

p-subgroup.

Theorem 1.6. (Burnside) Let P be a Sylow p-subgroup of a group G. Suppose NG(P ) =

CG(P ). Then G has a normal p-complement.

Proof. See [20, p169].

Lemma 1.7 (Frattini Argument). Let K be a normal subgroup of a group G and let

P be a Sylow p-subgroup of K (for some prime p) then G = KNG(P ).

Let K be a normal subgroup of a group G and let R 6 G. Then G = NG(R)K if and

only if RG = rg|r ∈ R, g ∈ G = rk|r ∈ R, k ∈ K = RK.

Proof. The first result is a special case of the second since if P is a Sylow p-subgroup

of K E G then for any g ∈ G, P g 6 K implies P g = P k for some k ∈ K by Sylow’s

Theorem. So we prove the second result.

Suppose RG = RK . For any g ∈ G there exists k ∈ K such that Rg = Rk. Therefore

gk−1 ∈ NG(R) and so g ∈ NG(R)k. Hence G = NG(R)K. Now suppose G = NG(R)K

and suppose RK ⊂ RG. Then there exists g ∈ G such that Rg * RK . However g = nk

where n ∈ NG(R) and k ∈ K and so Rg = Rnk = Rk ⊆ RK which is a contradiction.

Theorem 1.8. (Gaschutz) Let p be a prime and V an abelian normal p-subgroup of a

group G. Let V 6 P ∈ Sylp(G). Then G splits over V if and only if P splits over V .

8

Proof. See [1, 10.4 p31].

Lemma 1.9. (Burnside) Let G be a group and let P be a Sylow p-subgroup of G. Suppose

A1 and A2 are normal subsets of P which are conjugate in G. Then they are already

conjugate in NG(P ).

Proof. See [20, 7.1.5 p167].

Definition 1.10. Let G be a group, p be a prime and S ∈ Sylp(G). Let A(S) be the

set of abelian subgroups of S of order pr where pr is as large as possible. The Thompson

subgroup of S is J(S) = 〈A|A ∈ A(S)〉.

Lemma 1.11. Let G be a group, p be a prime and S ∈ Sylp(G). Suppose J(S) = Z(J(S))

and suppose a, b ∈ J(S) are conjugate in G. Then a and b are conjugate in NG(J(S)).

Proof. Suppose ag = b for some g ∈ G. Notice first that it follows immediately from the

definition of the Thompson subgroup that J(S)g = J(Sg). Now J(S), J(Sg) 6 CG(b).

Let P,Q ∈ Sylp(CG(b)) such that J(S) 6 P and J(Sg) 6 Q. Again, by the definition of

the Thompson subgroup, it is clear that J(S) 6 P implies J(S) = J(P ) and similarly

J(Sg) = J(Q). By Sylow’s Theorem, there exists x ∈ CG(b) such that Qx = P and so

J(S) = J(P ) = J(Q)x = J(S)gx. Thus gx ∈ NG(J(S)) and agx = bx = b as required.

Definition 1.12. The Fitting subgroup of a group G is the group, F (G), generated by

all nilpotent normal subgroups in G

Theorem 1.13. Let G be a group and F (G) the Fitting subgroup. Then the following

hold.

(i) F (G) is the unique maximal nilpotent normal subgroup of G.

(ii) If G is soluble then CG(F (G)) 6 F (G).

Proof. See [13, Thm6.1.2, Thm6.1.3 p218].

9

1.2 Extraspecial Groups

Definition 1.14. Let p be a prime and let E be a p-group. If E ′ = Z(E) = Φ(E) is

cyclic of order p then E is said to be extraspecial.

Lemma 1.15. Let E be an extraspecial p-group. Exactly one of the following holds.

(i) p = 2 and E is a central product of n copies of Dih(8).

(ii) p = 2 and E is a central product of n− 1 copies of Dih(8) and one copy of Q8.

(iii) p 6= 2 and E has exponent p.

(iv) p 6= 2 and E has exponent p2.

We denote such groups as E ∼= 21+2n+ , 21+2n

− , p1+2n+ and p1+2n

− respectively.

Proof. See [8, Thm 20.5].

It is well known that Dih(8)∗Dih(8) ∼= Q8 ∗Q8 so the description of extraspecial 2-groups

given here is not unique.

Theorem 1.16. Suppose E is an extraspecial p-group of order p2n+1.

(i) If p = 2 and E ∼= 21+2n+ then Out(E) ∼= O+

2n(2).

(ii) If p = 2 and E ∼= 21+2n− then Out(E) ∼= O−

2n(2).

(iii) If p is odd and E ∼= p1+2n+ then Out(E) ∼= Sp2n(p)Cp−1.

(iv) If p is odd and E ∼= p1+2n− then Out(E) ∼= p2n−1

+ Sp2n−2(p)Cp−1.

Proof. See [15].

10

Corollary 1.17. Let E be an extraspecial 2-group of order 25. If E admits a group of

automorphisms of order nine then E ∼= 21+4+ .

Proof. This is clear since 9 does not divide |O−4 (2)| = 120.

Lemma 1.18. The extraspecial 2-group 21+4+ contains exactly two subgroups isomorphic

to Q8 and they commute.

Proof. Firstly 21+4+

∼= Dih(8) ∗ Dih(8) ∼= Q8 ∗ Q8 so there are at least two subgroups

isomorphic to Q8 and we need to prove there cannot be more. We let Q1 and Q2 be the

two commuting quaternion subgroups and count elements of order four in Q := 〈Q1, Q2〉.

Clearly we see six elements of order four in Q1 and six more in Q2. So suppose qi ∈ Qi

for i = 1, 2 and q1q2 is another element of order four. If q1 has order two then it is central

in Q1 and then q1q2 ∈ Q2. Therefore q1, and in the same way q2, is an element of order

four and so both q1 and q2 square to the same element in z ∈ Z(Q) ∼= C2. However, this

means (q1q2)2 = q2

1q22 = z2 = 1 which is a contradiction. Therefore there are exactly 12

elements of order four in Q. So if Q3 6 Q is a subgroup isomorphic to Q8 then Q3 must

contain an element of order four from Q1 and an element of order four from Q2. However

these two elements commute so Q3 cannot be isomorphic to Q8.

1.3 Fixed-Point-Free Actions

An automorphism of a group G is said to be fixed-point-free if it leaves only the identity

of G fixed. The following famous theorem was proved by Thompson and says that groups

admitting such an action from an automorphism of prime order are nilpotent.

Theorem 1.19. (Thompson) If G is a group which admits a fixed-point-free automor-

phism of prime order then G is nilpotent.

Proof. See [13, Thm 2.1 p337].

11

Lemma 1.20. Let φ be a fixed-point-free automorphism of order two of a group G. Then

φ inverts every element of G and G is abelian.

Proof. Define

θ : G −→ G

g 7−→ g−1φ(g).

Then θ is injective since x−1φ(x) = y−1φ(y) if and only if yx−1 = φ(y)φ(x)−1 = φ(yx−1)

if and only if yx−1 = 1 if and only if x = y. Also θ is onto since G is finite. So for each

g ∈ G there is some h ∈ G such that g = h−1φ(h) therefore φ(g) = φ(h−1)h = g−1. So φ

inverts every element of G. Let x, y ∈ G. Then

x−1y−1 = φ(x)φ(y) = φ(xy) = (xy)−1 = y−1x−1

and so xy = yx and G is abelian.

The following lemma and its proof generalizes a well known result which dates back

to Burnside the proof of which can be found in [17, Thm 8.1]. We will use the notation

a ≡ b mod H when Ha = Hb where a, b ∈ G > H.

Lemma 1.21. Suppose that p is a prime and Q is a p-group. Furthermore, suppose that

z ∈ Aut(Q) has order three and CQ(z) = 〈t〉 where t ∈ Z(Q) has prime order. Then Q

is nilpotent of class at most three. In particular [Q,Q,Q] 6 〈t〉 and for any a, b ∈ Q,

[a, bz] ≡ [az, b] ≡ [a, b]z2

mod 〈t〉.

Proof. Let Q = Q1 > Q2 > Q3 > . . . be the lower central series of Q. Then for each i,

〈t〉Qi/Qi+1 is a central subgroup of Q/Qi+1 which is acted on by z. Moreover, z can only

centralize the subgroup 〈t〉Qi+1/Qi+1.

For any a ∈ Q, aazaz2is a fixed point modulo Q2 so

aazaz2 ≡ 1 mod 〈t〉Q2.

12

It follows from the commutator relations (Lemma 1.3) and from Q2/Q3 being abelian that

[a, b][az, b][az2, b] ≡ [aazaz2

, b] ≡ 1 mod Q3 and [a, b][a, bz][a, bz2] ≡ [a, bbzbz

2] ≡ 1 mod Q3

for any a, b ∈ Q. Therefore

[a, b] ≡ [b, az][b, az2

] ≡ [bz, a][bz2

, a] mod Q3.

Rearranging gives

[a, bz][b, az] ≡ [az2

, b][bz2

, a] mod Q3.

Therefore [a, bz][b, az]Q3 is a z-invariant element of Q2/Q3 and so must be in 〈t〉Q3/Q3.

Hence

[a, bz] ≡ [az, b] mod 〈t〉Q3. (1.1)

Now we repeat these arguments modulo Q4. For a, b ∈ Q the element [a, b][a, b]z[a, b]z2

is

a fixed point modulo Q3 and so

[a, b][a, b]z[a, b]z2 ≡ 1 mod 〈t〉Q3.

Using the commutator relation again and since Q3/Q4 is abelian, for a, b, c ∈ Q,

[[a, b], c][[a, b]z, c][[a, b]z2

, c] ≡ [[a, b][a, b]z[a, b]z2

, c] ≡ 1 mod Q4.

Also,

[[a, b], c][[a, b], cz][[a, b], cz2

] ≡ [[a, b], cczcz2

] ≡ 1 mod Q4.

Therefore

[[a, b], c] ≡ [c, [a, b]z2

][c, [a, b]z] ≡ [cz, [a, b]][cz2

, [a, b]] mod Q4.

13

Rearranging gives

[c, [a, b]z2

][[a, b], cz2

] ≡ [cz, [a, b]][[a, b]z, c] mod Q4.

The corresponding element modulo Q4 is z-invariant and therefore must be in 〈t〉Q4/Q4

and so

[[a, b]z2

, c] ≡ [[a, b], cz2

] mod 〈t〉Q4,

and in a similar way

[[a, b]z2

, c] ≡ [[a, b], cz2

] ≡ [[a, b], c]z ≡ [[az, b], c] ≡ [[a, bz], c] mod 〈t〉Q4. (1.2)

Now Lemma 1.3 (iii) gives

[[a, b], c][[b, c], a][[c, a], b] ≡ 1 mod Q4 (1.3)

and

[[az, b], c][[b, c], az][[c, az], b] ≡ 1 mod Q4. (1.4)

Conjugating (1.3) by z and applying (1.2) gives

[[az, b], c][[b, c], az2

][[c, az], b] ≡ 1 mod 〈t〉Q4

which together with (1.4) gives

[[b, c], az2

][[c, az], b] ≡ [[b, c], az][[c, az], b] mod 〈t〉Q4.

Hence [[b, c], az2] ≡ [[b, c], az] mod 〈t〉Q4. Finally, using (1.2) we get

[[b, c], a]z ≡ [[b, c], az2

] ≡ [[b, c], az] ≡ [[b, c], a]z2

mod 〈t〉Q4

14

is a fixed point mod Q4 which gives [[b, c], a] ∈ 〈t〉Q4 for any a, b, c ∈ Q, however this

implies that Q3 6 〈t〉Q4 6 〈t〉Q3. Since 〈t〉 has prime order, either 〈t〉Q3 = 〈t〉Q4 and

then Q3 = Q4 (so Q has nilpotence class two), else, Q3 = 〈t〉Q4. In this second case we

get Q5 = [Q4, Q] = [〈t〉Q3, Q] = [Q3, Q] = Q4 and so Q4 = 1 and Q3 = 〈t〉. In either

case Equation 1.1 becomes [a, bz] ≡ [az, b] mod 〈t〉 for any a, b ∈ Q and applying this

congruence again replacing a and b with az2and bz

2gives [a, bz] ≡ [a, b]z

2mod 〈t〉.

Corollary 1.22. Let Q be a group and let z ∈ Aut(Q) have order three and CQ(z) = 〈t〉

where t ∈ Z(Q) has prime order. Then Q is nilpotent of class at most three.

Proof. By Theorem 1.19, Q/〈t〉 is nilpotent and so is a direct product of its Sylow p-

subgroups. Since 〈t〉 is central of prime order, Q is nilpotent and a direct product of its

Sylow subgroups. Moreover each Sylow subgroup is nilpotent of class less or equal to

three and so Q is nilpotent of class less or equal to three.

Corollary 1.23. Suppose Q is a 2-group with z ∈ Aut(Q) of order 3 and CQ(z) =

〈t〉 ∼= C2. If Q = 〈X, Y 〉 with X ∼= Y ∼= Q8 and Z(X) = Z(Y ) = 〈t〉 then [x, yz] ≡

[xz, y] mod 〈t〉 for all x ∈ X and all y ∈ Y .

In fact such a group Q has order at most 27 and there exist two isomorphism classes

of groups of this order admitting such an automorphism. They can be constructed in

Magma (see [4]) by deriving possible relations from the congruence [a, bz] ≡ [az, b] mod 〈t〉

for all a, b ∈ Q.

Lemma 1.24. (Burnside) Suppose that Q is a group and suppose that z ∈ Aut(Q) has

order three and acts fixed-point-freely on Q\1. Then Q is nilpotent of class at most two.

Proof. Apply Corollary 1.22 with t = 1.

Lemma 1.25. Suppose S is a 2-group and A,B are fours groups with A 6= B and S =

〈A,B〉. If z ∈ Aut(S) has order 3, A and B are z-invariant and CS(z) = 1, then either

S is elementary abelian of order 24 or special of order 26 with |Z(S)| = 22.

15

Proof. By Lemma 1.24, S has nilpotence class at most two. If S is abelian then S has

order 24 so assume not then S has class two. Thus commutators are central giving us

that S ′ = 〈[a, b]|a ∈ A, b ∈ B〉 and for every a ∈ A, b ∈ B,

1 = [[a, b], b] = (baba)b(abab)b = (baba)(baba) = [b, a]2,

and so S ′ is elementary abelian. Since both A and B are z-invariant, we can write A\1 =

a0, a1, a2 where a1 = az0 and a2 = az2

0 and similarly we label B\1 = b0, b1, b2. Thus,

for i, j ∈ Z3, [ai, bj][ai+1, bj+1][ai+2, bj+2] ∈ CS(z) = 1. And so

[ai, bj][ai+1, bj+1] = [ai+2, bj+2]

= [aiai+1, bjbj+1]

= [ai, bj][ai, bj+1][ai+1, bj][ai+1, bj+1].

Hence for every i, j ∈ Z3,

[ai, bj+1] = [ai+1, bj],

and so S ′ = 〈[a1, b1], [a1, b2], [a1, b0]〉. But now [a1, b1][a1, b2] = [a1, b0] and so S ′ has

order at most 22. Since S ′ is z-invariant, |S ′| = 22. Now A and B do not commute

and so A ∩ S ′ 6 A ∩ Z(S) = 1 and similarly B ∩ S ′ = 1 (the intersections could not

have order two because they are z-invariant). Therefore AS ′/S ′ ∼= BS ′/S ′ ∼= 2 × 2

and so 〈A,B〉S′ = ( A

S′ )(BS′ ) ∼= 2 × 2 × 2 × 2 and so S has order 26. Now suppose S ′ <

Z(S). Since S 6= Z(S) and Z(S) is z-invariant, Z(S) must have order 24. However,

Z(S) ∩ A = 1 so S = Z(S)A but this is a contradiction since S ′ = [Z(S)A,Z(S)A] =

[Z(S), Z(S)][A,Z(S)][A,A][Z(S), A] = 1 (see Lemma 1.3).

16

1.4 Strongly Embedded Subgroups

Definition 1.26. Let G > E be groups of even order. We say E is a strongly embedded

subgroup of G if E ∩ Eg has odd order for every g ∈ G\E.

We immediately see that a strongly embedded subgroup is necessarily self-normalizing.

The following arguments originate from Bender (see [3]) and can, in particular, be

found in Aschbacher’s Sporadic Groups [2, Lemma 7.6]. Let G and E be groups of even

order with E a strongly embedded subgroup of G. Let I be the set of involutions in G

and let Ω be the set of conjugates of E in G. We consider the action of I on Ω.

Lemma 1.27. Each i ∈ I fixes precisely one element of Ω.

Proof. Let A ∈ Ω then A is a conjugate of E and so has even order. Therefore there

exists some u ∈ I ∩A and so Au = A. Suppose Bu = B for some B ∈ Ω\A. Since B is

strongly embedded, B is self-normalizing and so u ∈ B. Therefore u ∈ B ∩ A. However

A ∩ B has odd and so u ∈ I ∩ A can fix no other element of Ω. Also, since u is an

involution, u must swap pairs in Ω and so Ω has odd order. Now, since Ω has odd order,

any involution acts on Ω and fixes at least one element. So suppose v ∈ I and Cv = C

for some C ∈ Ω. As we have seen, C is self normalizing and so v ∈ I ∩C and then v fixes

precisely one element of Ω.

Lemma 1.28. Let E 6= Eu ∈ Ω. Any involution swapping E and Eu lies in the coset

u(E ∩ Eu).

Proof. Let w be such an involution. Then Ew = Eu and so uw ∈ E. Also, (uw)−1 =

wu = (uw)u ∈ Eu and so uw ∈ E ∩ Eu. Therefore w ∈ u(E ∩ Eu).

Lemma 1.29. Let E 6= Eu ∈ Ω and set K = E ∩ Eu. The number of involutions in I

that swap E and Eu is equal to |J | where J = k ∈ K|ku = k−1.

17

Proof. Any such involution, v say, which swaps E and Eu lies in the coset u(E∩Eu) = uK.

So let v = uk for k ∈ K. Then 1 = v2 = kuk and so ku = k−1 and then k ∈ J

and v = uk ∈ uJ . Of course any element of uJ is an involution and swaps E and

Eu. Therefore uJ is the set of all involutions in uK and so |J | = |uJ | is the number of

involutions swapping E and Eu.

Theorem 1.30. Let G be a group and E a strongly embedded subgroup of G. If for some

involution t ∈ E, CG(t) = CE(t) intersects trivially with every distinct conjugate of E,

then G contains an abelian subgroup K such that E = KCG(t) and K ∩ CG(t) = 1.

Proof. Set n = [G : E]. Count the triples (v, A,B) such that v ∈ I and A,B ∈ Ω

with Av = B. Let T be the set of such triples. By Lemma 1.27, each involution fixes a

unique element of Ω and permutes n−12

pairs so |T | = |I| (n−1)2

> [G : CG(t)] (n−1)2

= [E :

CE(t)]n(n−1)2

.

Now we choose an involution u ∈ I such that |I ∩ 〈u〉(E ∩ Eu)| is maximal. Equiva-

lently we have chosen two conjugates of E which are permuted by a maximal number of

involutions. As before set K = E ∩ Eu and J = k ∈ K|ku = k−1.

Now |T | =∑

A,B∈ΩMAB where MAB is the number of involutions swapping A and B.

So MAB 6 ME,Eu = |J |. Thus |T | 6(

n2

)ME,Eu = n(n−1)

2|J |. So [E : CE(t)] 6 |J |. Now

CG(t) = CE(t) intersects trivially with every distinct conjugate of E and so CG(t)∩K = 1.

Therefore every element of K lies in a unique coset of CG(t). There are [E : CE(t)] cosets

of CE(t) in E and so this number is greater than |CE(t)k|k ∈ K| = |K|. Hence

|J | 6 |K| 6 [E : CG(t)]. So we have that

|J |n(n− 1)

26 |K|n(n− 1)

26 [E : CG(t)]

n(n− 1)

26 |T | = |I|(n− 1)

26n(n− 1)

2|J |.

Hence J = K and |I| = n[E : CG(t)] = [G : CG(t)]. Furthermore, since u inverts every

element of J = K, K is an abelian subgroup by Lemma 1.20 and from the group orders

18

we get E = KCG(t).

1.5 Automorphisms of Abelian Groups

LetA be an additive abelian group that is isomorphic to a direct product ofm cyclic groups

of order q where q = pn for some prime p. Consider the ring of endomorphisms of A,

End(A). This is the set of group homomorphisms from A to itself with ring multiplication

given by composition of maps and addition given by (θ + ϕ)(a) := θ(a) + ϕ(a) for all

homomorphisms θ and ϕ and all a ∈ A. We will identify this endomorphism ring with

a ring of matrices and then determine which matrices define automorphisms of A. A

reference for this section is [25]. From now on we will view A as a direct sum of m copies

of the integers modulo q. So any a ∈ A has the form a = (a1, a2, . . . , am) where each ai

represents an equivalence class of integers modulo q and ai an integer representative of the

class. Let π be the natural group homomorphism π : Zm → A given by π(a1, a2, . . . , am) 7→

(a1, a2, . . . , am). Consider the map

ψq : Mm(Z) −→ End(A) = End(Zmq )

M 7→ ψq(M),

where ψq(M) : (a1, a2, . . . , am) 7→ π((a1, a2, . . . , am)M).

Lemma 1.31. The map ψq is a surjective ring homomorphism.

Proof. We first show each ψq(M) is a well defined endomorphism ofA. Suppose (a1 . . . , am)

and (b1, . . . , bm) satisfy π(a1 . . . , am) = π(b1, . . . , bm). Then q|ai − bi for each 1 6 i 6 m.

19

Thus

ψq(M)(a1 . . . , am)− ψq(M)(b1 . . . , bm) = π((a1 . . . , am)M)− π((b1, . . . , bm)M)

= π(Σaimi1, . . . ,Σaimim)− π(Σbimi1, . . . ,Σbimim)

= π(Σ(ai − bi)mi1, . . . ,Σ(ai − bi)mim)

= 0A.

So ψq(M) is well defined and also clearly a group homomorphism of A as π is linear and

matrix multiplication is distributive. So ψq is a map into End(A) and it remains to prove

ψq is a surjective homomorphism. Again it is clear that ψq is a ring homomorphism as π is

linear and matrix multiplication is distributive and associative. Now any endomorphism

of A is determined by where it maps each wi := (0, . . . , 1, . . . 0), the vector whose only

non-zero entry is a 1 in the i’th position. So suppose θ ∈ End(A) and θ(wi) = (a1, . . . , am).

Then consider a matrix M = (mij) where mij = aj for each j 6 m. Do this for each

i 6 m and it follows that θ = ψq(M). Thus ψq is a surjective ring homomorphism.

Lemma 1.32. The kernel of ψq is the set K := M = (mij) ∈Mn(Z)| q|mij ∀i, j.

Proof. It is clear that for all k ∈ K and each wi := (0, . . . , 1, . . . 0) defined as before,

ψq(k)(wi) = 0A. Since A is generated by the wi’s, K ⊆ Ker(ψq). However if we suppose

that some matrix, M = (Mij), is in the kernel then ψq(M)(wi) = 0A. Therefore q divides

mij for each j. We do this for each i to see that q divides each entry in M and so M ∈ K.

Lemma 1.33. θ = ψq(M) ∈ End(A) is an automorphism of A if and only if det(M) is

prime to p.

Proof. Suppose M ∈ Mm(Z) has determinant prime to p. Consider the cofactor matrix,

N , of M . This is the matrix which satisfies MN t = det(M)I. Since M has integer entries,

so does N . Since det(M) is prime to p, we can find s ∈ Z such that sdet(M) ≡ 1 mod q.

Set ψ−1M := ψ(sN) and we see that ψM ψsNt = ψsMNt = ψsdet(M)I = 1End(A) and so ψM

is invertible.

20

Conversely suppose M ∈Mm(Z) with ψM an invertible endomorphism of A. Suppose

ψM−1 = ψN for some N ∈Mm(Z). Then

ψ(MN − I) = ψ(MN)− ψ(I) = ψMψN − 1End(A) = 0End(A).

Thus MN − I ∈ Ker(ψ) and so q divides every entry of MN − I and so the entries of

MN are equal to the entries of I modulo p. Thus

1 ≡ det(MN) ≡ det(M)det(N) mod p.

Hence det(M) is prime to p.

Lemma 1.34. Set GLm(Zq) to be the subset of Mm(Zq) containing matrices with deter-

minant prime to p. Then GLm(Zq) ∼= Aut(A) and has order p(n−1)m2|GLm(p)|.

Proof. The set GLm(Zq) is a group under matrix multiplication. Consider the map,

Ψ : GLm(Zq) −→ Aut(A),

where each (aij) ∈ GLm(Zq) Ψq((aij)). This map is a well defined homomorphism as Ψq

is and non-zero since for each such (aij), the matrix (aij) is not in Ker(Ψq). It is also

injective since if (aij), (bij) ∈ GLm(Zq) with Ψq((aij)) = Ψq((bij)) then aij − bij ∈ Ker(Ψq)

and then q|(aij − bij) for each i, j and so (aij) = (bij). Now, for any automorphism of

A we can find M = (mij) ∈ Mn(Z) such that M has determinant prime to p. Therefore

(mij) ∈ GLm(Zq) and so Ψ is surjective.

Consider the map Θ : GLm(Zq) → GLm(Zp) where for each M ∈ GLm(Zq) we restrict

the matrix entries modulo p. Then Θ(M) has non-zero determinant modulo p and so is

an element of GLm(p). Clearly Θ is a surjective group homomorphism. The kernel is the

set of matrices (mij)|mii ≡ 1 mod p ∀i and mij ≡ 0 mod p ∀i 6= j. So there are pn−1

21

choices for each entry and so p(n−1)m2elements in the kernel. Thus GLm(Zq) ∼= Aut(A)

has order p(n−1)m2|GLm(p)|.

We have seen that if A = Zmq then elements of Aut(A) can be viewed as m by m

matrices with elements in Zq which have determinant prime to p. Notice GLm(Zq)

acts on the vector space V := A/Ωn−1(A) ∼= Zmp and so there is an homomorphism

GLm(Zq) → GLm(Zp) = Aut(Zmp ). Moreover this homomorphism simply restricts matrix

entries modulo p.

Lemma 1.35. Let p be a prime and set q = pn. Let G be a group acting on an abelian

group A ∼= Zmq . Then there are homomorphisms

α : G −→ GLm(Zq),

β : G −→ GLm(Zp)

for which entries of α(g) are congruent modulo p to entries of β(g) for each g ∈ G. Fur-

thermore β is a representation affording the module A/Ωn−1(A) and if β1 : G→ GLm(Zp)

is an equivalent representation then there is a homomorphism α1 : G → GLm(Zq) for

which entries of α1(g) are congruent modulo p to entries of β1(g) for each g ∈ G.

Proof. Since G acts on A, there is a homomorphism, α, from G to Aut(A) ∼= GLm(Zq)

as described previously. Moreover we have seen that restricting matrix entries modulo p

describes the action of G on the quotient group A/Ωn−1(A) ∼= Zmp and so we have the

required representation β. So suppose β1 : G→ GLm(Zp) is an equivalent representation

then there is a matrix M ∈ GLm(Zp) such that for each g ∈ G, β1(g) = M−1β(g)M .

Since M is a matrix whose elements are integers modulo p and whose determinant is

not a multiple of p, we can view M as a matrix in GLm(Zq) simply by viewing each

integer modulo p as the same integer modulo q. Now it makes sense to consider the

homomorphism α1 : G → GLm(Zq) where α1(g) = M−1α(g)M and so entries of α1 are

congruent modulo p to entries of β1(g) as required.

22

Chapter 2

Some Representation Theory

In this chapter we discuss endomorphism rings, tensor products, some modular represen-

tation theory and natural modules. The results on natural GL2(3) and SL2(3)-modules

will be used throughout Chapters 5, 6 and 7. The natural SL2(2n)-modules will appear

again in Chapter 8 as will much of the work on tensor products and modular character

theory.

Theorem 2.1. (Maschke) Let G be a group and let F be a field whose characteristic does

not divide |G|. Then every FG-module is completely reducible.

Proof. See [19, 1.9 p4].

Lemma 2.2. Let p be a prime and let K = GF(pn). If H is a group with H ∼= K× then

any irreducible KH-module has dimension one.

Proof. Suppose V is an irreducible KH-module with representation ρ : H → GL(V ). Let

〈h〉 = H then h has order pn − 1 and so any eigenvalue for ρ(h) is a pn − 1’th root of

unity and so will lie in K. Thus we can find a corresponding eigenvector for ρ(h). The

eigenspace spanned by the eigenvector is 1-dimensional and H invariant and therefore a

KH-submodule and so V must have dimension one.

23

2.1 Endomorphism Rings

Definition 2.3. Let G be a group, F a field and let V and W be FG-modules. Define

HomFG(V,W ) to be the ring of all FG-homomorphisms from V to W . The endomorphism

ring of V is the ring EndFG(V ) := HomFG(V, V ).

If V is an FG-module and E := EndFG(V ) then V can be viewed as a vector space

over E by defining scalar multiplication as e · v := e(v) for e ∈ E, v ∈ V . Thus V is also

an EG-module.

Lemma 2.4. (Schur) If V is an irreducible FG-module then EndFG(V ) is a division ring.

If, furthermore, F is a finite field then EndFG(V ) is a field.

Proof. Let 0 6= φ ∈ EndFG(V ). Set W = φ(V ). Since φ is non-zero, W 6= 0. Hence for

all v ∈ V and g ∈ G

(φ(v))g = φ(vg) ∈ W

and so W is an FG-submodule. Since V is irreducible and W 6= 0, V = W . So

φ is onto and by the rank-nullity result φ is a non-singular linear transform on V and

thus invertible. Clearly φ−1 is a linear transform and vg = φ(φ−1(v))g = φ(φ−1(v)g)

and so φ−1(vg) = φ−1(v)g and φ−1 ∈ EndFG(V ). A quick check verifies EndFG(V ) is a

ring with usual addition and composition of functions and so it is a division ring. Now

Wedderburn’s Theorem says that a finite division ring is a field. So if F is a finite field

then HomFG(V, V ) is finite and hence a field.

Lemma 2.5. If V is a faithful irreducible FG-module where F is a finite field and G an

abelian group then EndFG(V ) is a field of order |V |.

Proof. By Schur’s Lemma, E := EndFG(V ) is a field. Since elements of F define endo-

morphisms of V which commute with the action of G, we see an embedding of F in E.

24

Also, as G is abelian, G embeds in the multiplicative group of E. We can view E as an

FG-module by defining scalar multiplication and an action of G simply by composition

of maps. We then define an FG-module homomorphism,

Φ : E −→ V

α 7−→ α(v),

where v is some fixed non-zero vector in V . This map is non-zero and so the kernel is an

ideal and so must be trivial. It follows that this map is a bijection and so |E| = |V |.

Lemma 2.6. Let G be a group and H 6 G. Let F be a finite field of characteristic p and

let V be an irreducible FG-module. Suppose χ : G → C is the character afforded by V

and finally suppose that p does not divide |H|. Then dim(CV (H)) = 〈χ|H , 1|H 〉.

Proof. By Maschke’s Theorem, V decomposes into irreducible H-modules, V = V1⊕ . . .⊕

Vn, and χ|H decomposes into irreducible characters for H, χ|H = χ1 + . . .+χn. The inner

product in H, 〈χ|H , 1|H 〉 gives us the number of times the trivial character appears in

the decomposition and similarly the number of 1-dimensional modules on which H acts

trivially. Hence dim(CV (H)) = 〈χ|H , 1|H 〉.

2.2 Tensor Products

See, for example, [1] for more explanation and proofs of results. Let G be a group, K a

field and F an extension field of K.

Definition 2.7. Let U and V be KG-modules. A K-balanced map θ : U × V → A,

where A is an abelian group, is a map which satisfies

kθT (u, v) = θT (ku, v) = θT (u, kv),

25

θ(u+ u′, v) = θ(u, v) + θ(u′, v)

and

θ(u, v + v′) = θ(u, v) + θ(u, v′)

for k ∈ K, u, u′ ∈ U , v, v′ ∈ V . An abelian group T is a tensor product of U and V over

K if there is a surjective K-balanced map θT : U × V → T such that if S is an abelian

group and θS : U×V → S is a surjective K-balanced map then there is a unique K-linear

homomorphism α : T → S such that αθT = θS.

We denote a tensor product of U and V by U ⊗K V (or U ⊗V if there is no ambiguity for

the field). The tensor product exists and is unique up to isomorphism and is a KG-module

when we define

(u⊗ v) · g := u · g ⊗ v · g

for u ∈ U, v ∈ V, g ∈ G.

Suppose ui|1 6 i 6 n is a basis for U overK and vj|1 6 j 6 m is a basis for V over

K. Then U⊗KV is a vector space overK with basis ui⊗Kvj|1 6 i 6 n, 1 6 j 6 m. Now

consider the field extension F/K. Since F is a vector space over K, we can form the tensor

product F ⊗K V . This gives a vector space over F when we define f1 · (f⊗v) := (f1f⊗v).

Notice that 1F ⊗ vi|1 6 i 6 n is an F -basis for F ⊗ V and so the dimension of this

space over F is equal to the dimension of V over K.

An irreducibleKG-module, V , is called absolutely irreducible if the FG-module F⊗KV

is irreducible for each extension field F of K.

Lemma 2.8. Let V be an irreducible KG-module. Then V is absolutely irreducible if and

only if EndKG(V ) ∼= K.

Proof. See [1, 25.8].

26

Lemma 2.9. Let U and V be KG-modules. Then the FG-modules (U⊗KF )⊗F (V ⊗KF )

and (U ⊗K V )⊗K F are FG-module isomorphic.

Proof. Consider the map

Θ : (U ⊗K F )× (V ⊗K F ) −→ (U ⊗K V )⊗K F

(u⊗ f, v ⊗ e) 7−→ (u⊗ v)⊗ fe.

We claim Θ is an F -balanced map. Let f, f1, f2, e ∈ F , u, u1, u2 ∈ U , v ∈ V and observe

Θ(u1 + u2 ⊗ f, v ⊗ e) = (u1 + u2 ⊗ v)⊗ fe

= (u1 ⊗ v)⊗ fe+ (u2 ⊗ v)⊗ fe

= Θ(u1 ⊗ f, v ⊗ e) + Θ(u2 ⊗ f, v ⊗ e)

Θ(u⊗ f1 + f2, v ⊗ e) = (u⊗ v)⊗ f1e+ f2e

= (u⊗ v)⊗ f1e+ (u⊗ v)⊗ f2e

= Θ(u⊗ f1, v ⊗ e) + Θ(u⊗ f2, v ⊗ e)

and linearity in the second variable holds in the same way. Also

f1Θ(u⊗ fv ⊗ e) = (u⊗ v)⊗ f1fe = Θ(u⊗ f1f, v ⊗ e) = Θ(u⊗ f, v ⊗ f1e)

= Θ(f1(u⊗ f), v ⊗ e) = Θ(u⊗ f, f1(v ⊗ e)).

So Θ is an F -balanced map. Also if g ∈ G then

Θ(u⊗f, v⊗e)g = (u⊗v)g⊗fe = (ug⊗vg)⊗fe = Θ((u⊗f)g, (v⊗e)g) = Θ((u⊗f, v⊗e)g).

Now, by the definition of the tensor product, there is an F -linear map α : (U ⊗K F )⊗F

(V ⊗K F ) → (U ⊗K V ) ⊗K F such that αθT = Θ where θT is the usual tensor product

map θT : (U ⊗K F ) × (V ⊗K F ) → (U ⊗K F ) ⊗F (V ⊗K F ). It is easy to check θT

27

commutes with the action of G and so we get that α is FG-linear. Since θT is surjective

by definition and Θ is also clearly surjective, α is surjective. Finally since the dimension

of (U ⊗K F )⊗ F (V ⊗K F ) over F equals the dimension of (U ⊗K V )⊗K F over F , α is

an FG-isomorphism.

Given a KG-module, V , and corresponding representation, ρ : G → GL(V ), we can

define a representation ρF : G → GL(F ⊗K V ) where for g ∈ G, ρF : g 7→ 1 ⊗ ρ(g) and

for f ⊗ v ∈ F ⊗K V , 1⊗ ρ(g) : f ⊗ v 7→ f ⊗ ρ(g)(v).

Lemma 2.10. Let F be an extension field of K and suppose θ : G → GLn(K) is a

representation with corresponding module V . Let ι be the inclusion map ι : GLn(K) →

GLn(F ). Then the representation θF : G→ GLn(F ) corresponding to the module F ⊗K V

is equivalent to the representation ιθ.

Proof. Fix a K-basis v1, . . . , vn for V and write each g ∈ G as a matrix θ(g) = (aij)

where θ(g)(vi) := ai1v1 + . . . + ainvn and each aij is in K 6 F . The module F ⊗K V

affords the representation θF : G → GL(F ⊗K V ) where g ∈ G maps to the linear

transformation 1 ⊗ θ(g) : f ⊗ v 7→ f ⊗ θ(g)(v). Choose a basis for F ⊗K V to be

1 ⊗ v1, . . . , 1 ⊗ vn. Then, with respect to this basis, the matrices representing G via

θF : G→ GL(F ⊗K V ) are exactly the matrices representing G via θ since θF (g)(1⊗vi) =

1⊗ θ(g)(vi) = 1⊗ ai1v1 + . . .+ ainvn = 1⊗ ai1v1 + . . .+ 1⊗ ainvn.

If V is a vector space over a field F and φ1, . . . , φn are F -linear transformations of V ,

we say φ1, . . . , φn are simultaneously diagonalizable if there is a basis for V , v1, . . . , vm,

such that for each 1 6 i 6 n and each 1 6 j 6 m, vj is an eigenvector for φi.

Lemma 2.11. Let V be a vector space over a finite field K = GF(p) and let φ1, . . . , φn

be commuting K-linear transformations of V with order prime to p. Furthermore let

F be a field extension of K that contains all eigenvalues for φ1, . . . , φn. Then the linear

transformations of F⊗K V , 1⊗K φ1, 1⊗K . . . , 1⊗K φn, can be simultaneously diagonalized.

28

Proof. Each φi has order, ri say, prime to p and so has eigenvalues which are all ri’th

roots of unity and so there exists some extension field, F > K, which contains every

eigenvalue for every φi. Let V F = F ⊗K V and let φFi := 1 ⊗K φi : f ⊗ v 7→ f ⊗

φi(v) be the corresponding linear transformations. Notice that the linear transformations,

φF1 , . . . , φ

Fn , all commute. Find the eigenspaces for some φ := φF

i . Let ε1, . . . , εm ∈ F

be the eigenvalues for φ and Eig(ε1), . . . ,Eig(εm) the corresponding eigenspaces in V F .

Suppose 1 6 k 6 m and v ∈ Eig(εk), then, φφFj (v) = φF

j φ(v) = φFj (εv) = εφF

j (v). So each

φFj preserves each eigenspace for φ. Now φ is diagonalizable with respect to a basis of

eigenvectors. We now show we can choose a basis for V F with respect to which every φFj

is diagonalizable. If we can find a basis for each eigenspace with respect to which each φFj

is diagonalizable then we are done. So we proceed by induction on the dimension of the

eigenspace. If Eig(εk) has dimension one then, as it is preserved by each φFj , each φF

j is

scalar. So suppose Eig(εk) has dimension n+1 and that each φFj is diagonalizable on any

eigenspace of lesser dimension. Suppose some φFj is not diagonalizable on Eig(εk). Then

the vector space Eig(εk) decomposes into eigenspaces for φFj each of which is preserved by

each linear transformation, φF1 , . . . , φ

Fn and by induction there is a basis for each for which

each φF1 , . . . , φ

Fn is diagonalizable. However this gives us a basis for Eig(εk) on which φj

is diagonalizable contrary to assumption.

In the following corollary K is a field extension of L and is therefore a vector space

over L. We use the notation GL(L,K) for the group of invertible L-linear transformations

of K.

Corollary 2.12. Let p be a prime and let L = GF(p) and K = GF(pn) for some n > 1.

Each k ∈ K× defines a linear transformation of K as a vector space over L and there is

an embedding φ : K× → GLn(L) → GLn(K). Furthermore there exists M ∈ GLn(K)

such that φ(k)M = diag(k) for all k ∈ K×.

Proof. It is clear thatK is an n-dimensional vector space over L and each non-zero, k ∈ K,

29

defines a linear transformation, λk : a 7→ ka (for each a ∈ K). Thus there is an embedding

θ : K× → GLn(L) ∼= GL(L,K) and since L 6 K, we see the inclusion map ι : GLn(L) →

GLn(K) thus φ := ιθ : K× → GLn(K). Now, the linear transformations λk|k ∈ K×

commute pairwise and are all scalar on K so by Lemma 2.11, the corresponding linear

transformations of K⊗LK (as an n-dimensional vector space over K), 1⊗L λk|k ∈ K×

are simultaneously diagonalizable. Therefore we can assume there is an image of K× in

GL(K,K⊗LK), via θK say, in which θK(k) is scalar for each k ∈ K×. Finally, by Lemma

2.10, the representation φ is equivalent to θK . Therefore, there exists M ∈ GLn(K) ∼=

GL(K,K ⊗L K) such that φ(k)M = diag(k) for each k ∈ K×.

2.3 Natural GL2(3)-Modules

In later chapters we will consider amalgams. These will be subgroups inside a larger group

and the structure of the subgroups will be described via GL2(3) and SL2(3)-modules. In

the following section set G ∼= GL2(3), G > H ∼= SL2(3) and F = GF(3).

Lemma 2.13. The following are true in G ∼= GL2(3).

(i) Any subgroup of G isomorphic to 2 × 2 is conjugate to the subgroup of diagonal

matrices.

(ii) Q8∼= O2(G) 6 H.

(iii) H ∼= SL2(3) is the unique subgroup of G at index two.

Proof. Let 2× 2 ∼= T 6 G. Consider the eigenvalues for each t ∈ T#. Each t ∈ T satisfies

the polynomial X2 = 1 and so the possible eigenvalues are 1 and −1. If t has only the

eigenvalue 1 then t is the identity. If t has only the eigenvalue −1 then t must be the

central scalar matrix diag(−1,−1). If t has eigenvalues 1 and −1 then t is conjugate to

30

the matrix diag(−1, 1). If T contains the central involution then we are done since we

only need to diagonalize one more element of T . So suppose not then we must have t ∈ T

with eigenvalues 1 and −1. Let u and v be eigenvectors for t such that t(u) = u and

t(v) = −v. Let s ∈ T\〈t〉 then s(u) = s(t(u)) = t(s(u)) so s(u) is in the eigenspace 〈u〉

similarly s(v) is in the eigenspace 〈v〉 and it follows that s must be diagonal with respect

to the basis u, v and so we can simultaneously diagonalize T . Therefore (i) holds.

Notice that

K := 〈

(0 −1

1 0

),

(1 1

1 −1

)〉

is a normal subgroup of G isomorphic to Q8 and that this subgroup is contained in H.

Furthermore G cannot have a larger normal 2-subgroup since we can find two distinct

Sylow 2-subgroups of order 24 by including the element

(1 −1

0 −1

)or

(0 −1

−1 0

)so (ii)

holds.

Now, H is a normal subgroup of G at index two so suppose K is another subgroup of

G at index two. Then H ∩K E G has at index four in G and has order 322. If H ∩K

contains a normal Sylow 3-subgroup, S say, then S char H ∩K E G implies S E G which

is a contradiction since G does not have a normal Sylow 3-subgroup. In fact G contains

four Sylow 3-subgroups namely,

⟨ 1 0

1 1

⟩ ,

⟨ 1 1

0 1

⟩ ,

⟨ 0 1

−1 −1

⟩ ,

⟨ −1 1

−1 0

⟩

which pairwise intersect trivially. Each is normalized in G by a group of order 322 and so

by Sylow’s Theorem, G has exactly four Sylow 3-subgroups. Also, by Sylow’s Theorem,

H∩K contains four Sylow 3-subgroups and so H∩K contains 104 elements of order three.

Therefore H ∩K can contain only three elements of order two and so necessarily has a

normal subgroup of order four, F say. However, H contains a Sylow 2-subgroup which is

quaternion of order eight and so F must be cyclic of order four and therefore admits no

31

automorphism of order three. This implies that CH∩K(F ) = NH∩K(F ) = H ∩K and so

F commutes with each Sylow 3-subgroup but then each Sylow 3-subgroup in H ∩ K is

normal in H ∩K which is a contradiction.

Definition 2.14. Let V be a 2-dimensional vector space over GF(3). We identify G

with GL(V ) and H with SL(V ) and call V the natural G-module and natural H-module

respectively.

Lemma 2.15. Let V be a natural G-module or a natural H-module. The following hold.

(i) For S ∈ Syl3(G), CV (S) is a subspace of V of dimension one.

(ii) Both G and H act transitively on the subspaces of V of dimension one.

(iii) Both G and H act transitively on the non-zero vectors in V .

(iv) The action of G (and H) on V is faithful.

Proof. (i) Let S ∈ Syl3(G) then S is conjugate to the group of strictly lower triangular

matrices. It is easy to check that this group of matrices fix precisely the 1-space 〈(1, 0)〉.

Thus CV (S) has dimension one.

(ii) There are four subspaces of V with dimension one. The stabilizer in G of the

space 〈(1, 0)〉 is the subgroup of lower triangular matrices which has index four in G.

Similarly the stabilizer in H of 〈(1, 0)〉 is the subgroup of lower triangular matrices with

determinant 1 which has index four in H. Therefore both G and H are transitive on the

1-spaces in V .

(iii) We repeat the same arguments to see that the stabilizer in G (and in H) of a

non-zero vector in V has index eight. Therefore both G and H are transitive on the eight

non-zero vectors in V .

(iv) This is just by definition since we identified GL2(3) with the group of automor-

phisms GL(V ).

32

The following lemma describes a situation which will occur many times when we are

dealing with groups involving natural modules.

Lemma 2.16. Suppose S is a 3-group with subgroups P 6= Q 6 S of index three. Suppose

further that t is an involution in Aut(S) which leaves P and Q invariant. Then t acts

on S/P (and S/Q) and either centralizes or inverts S/P . If t centralizes S/P then there

exists x ∈ Q\P such that t centralizes x. If t inverts S/P then there exists x ∈ Q\P such

that t inverts x.

Proof. It is clear that t acts on S/P such that for any element Py ∈ S/P , Pyt := P (yt).

Pick y ∈ Q\P then the coset Py is either inverted or centralized by t. Suppose first that

Py is centralized by t then t describes a permutation of the set Py. Furthermore elements

py ∈ Py are in Q if and only if p ∈ Q ∩ P and t preserves Q so preserves those elements.

Since t is an involution and Py∩Q is a set of odd order, |P ∩Q|, t must fix some element,

x say. Thus x is in Q\P and is centralized by t. Now suppose that Py is inverted by t

then [y, t] /∈ P . Also [y, t]t = (y−1yt)t = (y−1)ty = [y, t]−1 and so [y, t] is inverted by t.

Thus [y, t] ∈ Q\P and is inverted by t.

2.4 Sym(4)-Modules

Lemma 2.17. Suppose that X ∼= Sym(4) and that V is a faithful 3-dimensional GF(3)X-

module. Let q1, q2, q3 = O2(X)#. The following hold.

(i) There is a set of distinct 1-dimensional subspaces B := 〈v1〉, 〈v2〉, 〈v3〉 where 〈vi〉 =

CV (qi).

(ii) X has orbits of lengths 3, 4 and 6 on the 1-dimensional subspaces of V with repre-

sentatives

Type 1A - 〈v1〉,

33

Type 1B - 〈v1 + v2 + v3〉 and

Type 1C - 〈v1 + v2〉

respectively.

(iii) X has orbits of lengths 3, 4, and 6 on the 2-dimensional subspaces of V with repre-

sentatives

Type 2A - 〈v1, v2〉,

Type 2B - 〈v1 + v2, v2 + v3〉 and

Type 2C - 〈v1, v1 + v2 + v3〉

respectively.

(iv) If x ∈ X has order three then the subspace CV (x) has dimension one and lies in the

orbit of 1-spaces with representative 〈v1 + v2 + v3〉.

(v) For i 6= j, vi is negated by qj. For i, j, k = 1, 2, 3, qk negates vi + vj.

(vi) |CX(vi)| = 4, |CX(vi + vj)| = 2 for i 6= j.

(vii) If x ∈ X\O2(X) has order two then x centralizes a 1-space of type 1C.

Proof. Set Q = O2(X) then Q# = q1, q2, q3. By coprime action of Q on V ,

V = 〈CV (qi)|i = 1, 2, 3〉.

Each CV (qi) is a non-trivial subspace of V and as X acts transitively on Q#, X acts

transitively on the three subspaces CV (q1), CV (q2) and CV (q3). Thus each has the same

dimension. SinceX acts faithfully on V , each subspace is proper. Suppose each CV (qi) has

dimension two. Consider CV (q1)∩CV (q2). This must be non-trivial. Moreover any vector

fixed by q1 and q2 must be fixed by Q and so this intersection is X-invariant. However

34

this is a contradiction since X cannot act faithfully on a 1-dimensional subspace. Choose

vectors vi such that 〈vi〉 = CV (qi) then

V = 〈v1〉 ⊕ 〈v2〉 ⊕ 〈v3〉.

Clearly 〈vi〉 | 1 6 i 6 3 forms an orbit of length 3 on the 1-dimensional subspaces

of V . We also have that the subspaces 〈v1 ± v2 ± v3〉 form an orbit of length 4 on V and

the subspaces 〈vi ± vj〉, where i 6= j form an orbit of length 6.

Clearly the subspaces 〈vi, vj〉 for i 6= j give an orbit of length 3 on the 2-dimensional

subspaces of V . The subspaces 〈vi ± vj, vj ± vk〉 form an orbit of length 4. Finally the

subspaces 〈vi, vi + vj − vk〉, for i j and k distinct form an orbit of length 6. We note that

for each choice of i we have two choices for k.

We see that any element of order three cycles the set Q# and so cycles the spaces

〈v1〉, 〈v2〉, 〈v3〉. But if x ∈ X cycles 〈v1〉, 〈v2〉, 〈v3〉, it must also cycle the spaces of type

〈v1 + v2〉 of which there are six. Since spaces of type 〈v1 + v2 + v3〉 lie in an orbit of length

four, x necessarily fixes at least one of them. However if it fixes them all it acts trivially

on V which is impossible.

The space 〈vi〉 is preserved by NX(qi). So for j 6= i, qj fixes this one space. However

we have seen that CV (qi)∩CV (qj) is trivial thus qj negates vi. Now if we let k be distinct

from both i and j then qk negates vi and vj and thus the vector vi + vj.

We consider the action of X on the set of vectors. We have seen that vi lies in an orbit

with its inverse and with the non-trivial elements of the other 1-spaces of type 1A. Thus

vi lies in an orbit of length six and so |CX(vi)| = 4. Also vi + vj lies in an orbit with its

inverse and the non-trivial elements of the other 1-spaces of type 1C. Thus the orbit has

length 12 and so |CX(vi + vj)| = 2. The element of order two in X centralizing vi + vj

cannot lie in O2(X) by (iv) and so must be in the other conjugacy class of involutions.

35

Since all involutions outside of O2(X) lie in this conjugacy class, each fixes some conjugate

of vi + vj.

Lemma 2.18. Suppose that X ∼= Sym(4) and that V is a faithful 3-dimensional GF(3)X-

module. Then the subspaces of order 32 have the following forms.

(i) Type 2A - These contains two 1-dimensional subspaces conjugate to 〈v1 + v2〉 and

two conjugate to 〈v1〉. There are three such subspaces.

(ii) Type 2B - These contain one 1-dimensional subspace conjugate to 〈v1+v2+v3〉 three

1-dimensional subspaces conjugate to 〈v1 + v2〉. There are four such subspaces.

(iii) Type 2C - These contain two 1-dimensional subspaces conjugate to 〈v1 + v2 + v3〉,

one conjugate to 〈v1 + v2〉 and one conjugate to 〈v1〉. There are six such subspaces.

Proof. Let Y = 〈v1, v2〉. Then Y contains the 1-dimensional subspaces 〈v1〉, 〈v2〉, 〈v1 +v2〉

and 〈v1 − v2〉. Hence Y is of type 2A.

Now let W = 〈v1+v2, v2−v3〉. Then W contains the 1-dimensional subspaces 〈v1+v2〉,

〈v2 − v3〉, 〈v1 + v3〉 and 〈v1 − v2 − v3〉. Hence W is of type 2B.

Finally let U = 〈v1, v1 + v2 + v3〉. Then U contains the 1-dimensional subspaces 〈v1〉,

〈v1 + v2 + v3〉, 〈v2 + v3〉 and 〈v1 − v2 − v3〉. Hence U is of type 2C.

The number of each type of subgroup follows from Lemma 2.17 (iii).

There are two isomorphism types of 3-dimensional Sym(4)-modules and we give an

example of both. Firstly, let

A = 〈(123), (456), (789), (147)(258)(369), (14)(2536)〉.

36

Then A has shape 33 : Sym(4) and A 6 Alt(9). Secondly, let

B = 〈(123), (456), (789), (147)(258)(369), (14)(2536)(78)〉,

then B also has shape 33 : Sym(4). We see that B 6 Sym(9) but B Alt(9). No-

tice in both groups that we can apply Lemma 2.17 by setting v1 = (123), v2 = (456)

and v3 = (789). Notice also that in B the element (789) is inverted by the element

(14)(2536)(78) however one can check that CA((789)) = NA((789)). Also, by Lemma

2.17 (vi), |CA((789))| = |CB((789))| = 4 however in A, (789) commutes with an element

of order four, (14)(2536), and again one can check that in B, (789) commutes with an

elementary abelian group of order four. We will use Lemma 2.17 and Lemma 2.18 in

Chapter 7 where it will become clear we have a Sym(4)-module of the second type.

2.5 Some Modular Character Theory

We introduce a small amount of modular character theory which we will require in Chapter

8. Results will be stated without proof. The main reference for this chapter has been

Chapter 15 of [19].

Let A be the ring of algebraic integers and let M be a maximal ideal of A containing

Ap where p is a fixed prime. Set F to be the field A/M which necessarily has characteristic

p and set ∗ : A → F to be the natural ring homomorphism. Notice that we may have

some choice over M and hence over F but our choice makes no difference to the following

results. Finally, set U = ε ∈ A|εn = 1 for n ∈ Z with (n, p) = 1. The following hold.

(i) ∗∣∣U

: U → F× is a group isomorphism.

(ii) F is an algebraically closed field.

37

Let G be a group and let ρ : G → GLn(F ) be a representation of G affording the

character χ. Let Ω be the set of elements of G with order prime to p. We define the

Brauer character of G to be the map φ : Ω → C as follows. For each x ∈ Ω, ρ(x) has

eigenvalues ε1, . . . , εn ∈ F× as F is algebraically closed. Since ∗∣∣U

is a group isomorphism

onto F×, we can find a unique ui ∈ U such that ∗ : ui 7→ εi. Thus we can define

φ(x) = Σui. A Brauer character is said to be irreducible if the representation affording it

is irreducible. Each Brauer character is clearly constant on conjugacy classes of G in Ω

and so is an example of a class function defined on the elements of Ω.

Lemma 2.19. Suppose U and V are FG-modules affording characters χ1 and χ2 respec-

tively. Then U ⊗F V affords the character χ1χ2 defined by χ1χ2(g) = χ1(g)χ2(g) for each

g ∈ G.

Proof. If u ∈ U is a λ-eigenvector and v ∈ V is a µ-eigenvector then u⊗ v ∈ U ⊗ V is an

eigenvector and the eigenvalue is λµ. Suppose ρ1 : G→ GL(U) and ρ2 : G→ GL(V ) are

the corresponding representations. Let g ∈ G and let λ1, . . . , λn be the eigenvalues for

ρ1(g) and let µ1, . . . , µm be the eigenvalues for ρ2(g). Then λiµj|1 6 i 6 n, 1 6 j 6 m

are the eigenvalues for ρ(g) where ρ : G → GL(U ⊗ V ). Now, by the definition of the

Brauer character,

χ1(g) =n∑

i=1

ai and χ2(g) =m∑

j=1

bj

where ∗(ai) = λi and ∗(bj) = µj. The Brauer character afforded by U ⊗ V is χ(g) =∑ni=1

∑mj=1 cij where ∗(cij) = λiµk = ∗(ai) ∗ (bj) = ∗(aibj) since ∗ is a group homomor-

phism. Therefore

χ(g) =n∑

i=1

m∑j=1

aibj =n∑

i=1

ai

m∑j=1

bj = χ1(g)χ2(g).

Theorem 2.20. The irreducible Brauer characters give a basis for the space of C-valued

38

class functions defined on Ω. The number of irreducible Brauer characters of G is equal

to the number of conjugacy classes of elements of order prime to p.

Proof. This theorem is Theorem 15.10 in [19].

Lemma 2.21. Let χ be an ordinary character of G and let χ be the restriction of χ to

the elements in Ω. Then χ is a Brauer character of G.

Proof. [19, 15.6 p265]

Theorem 2.22. Let χ be an ordinary irreducible character of G such that |G|/χ(1) is

prime to p. Then χ is an irreducible Brauer character of G.

This theorem is essentially Theorem 15.29 in [19] which proves that the set χ, χ is a

block.

Let V be an n-dimensional module for G over some finite field, K, of characteristic p.

Then consider the module F ⊗K V . This is an n-dimensional module for G over F when

we define (f ⊗ v) · g = f ⊗ (v · g) for f ∈ F , v ∈ V and g ∈ G. A basis for F ⊗K V is

1 ⊗K vi|1 6 i 6 n over F where vi|1 6 i 6 n is a basis for V over K. By Lemma

2.8, if V is irreducible and EndKG(V ) = K then V ⊗K F is irreducible. So we have some

tools for classifying possible modules for G over finite fields.

2.6 Natural SL2(2n)-Modules

In all that follows let X ∼= SL2(2n) where n ≥ 2, T ∈ Syl2(X) and set L := GF(2) and

K := GF(2n). We will calculate the Brauer character table for X and so we must choose

an algebraically closed field, F , of characteristic two in which to calculate.

Lemma 2.23. The following hold in X ∼= SL2(2n).

39

(i) There are at most 2n conjugacy classes of elements of odd order.

(ii) Let S, T ∈ Syl2(X). Then S∩T = 1 and S acts transitively on the set Syl2(X)\S.

(iii) For S, T ∈ Syl2(X), NX(S)∩NX(T ) has order 2n−1 and fixes precisely two elements

in Syl2(X).

(iv) X acts sharply 3-transitively on Syl2(X).

(v) For S, T ∈ Syl2(X) and t ∈ T#, X = 〈S, T 〉 = 〈S, t〉.

Proof. The characteristic polynomial for a non-identity element, A, ofX is x2−tr(A)x+1.

Since conjugate elements have the same characteristic polynomial and values of the trace

lie in K, we have 2n possibilities. So together with the identity we have at most 2n + 1

conjugacy classes in X. As one of these classes contains elements of order 2, we have at

most 2n classes of elements of odd order.

Every Sylow 2-subgroup has order 2n and is normalized by a subgroup of X conjugate

to the Borel subgroup of lower triangular matrices. This has order 2n(2n−1) and so there

are 2n +1 Sylow 2-subgroups in X. We consider the action of X on the set Ω := Syl2(X).

Let S ∈ Ω. It is clear that no non-trivial element of S fixes any other element of Ω else

we would have a larger 2-group and since |Ω\S| = 2n = |S|, we see S is transitive on

Ω\S. Thus X is 2-transitive on Syl2(X). Consider two elements of Ω namely, U , the

upper triangular and, L, the lower triangular matrices. These intersect trivially and so

any pair of Sylow 2-subgroups intersect trivially. Notice that both are normalized by the

subgroup of diagonal matrices, D. In fact D = NX(U)∩NX(L) for the intersection can be

no larger with out having even order. Suppose D normalizes a third element of Ω, P say.

Find x ∈ U such that Lx = P . Then D,Dx 6 NX(Lx) ∩NX(U). But then D = Dx and

so x ∈ NU(D) = 1. Thus D fixes exactly 2 elements of Ω and so is transitive on Ω\U,L.

So now we have shown that a 2-point stabilizer is transitive on the remaining 2n−1 points

and that the stabilizer of three points is trivial. Hence X is sharply 3-transitive on Ω.

40

Now for any distinct pair of Sylow 2-subgroups, S and T , and any t ∈ T#, the group

Y := 〈S, t〉 contains Sylow subgroups of order 2n and also an element from each Sylow

2-subgroup. As they all intersect trivially it contains all Sylow subgroups. However, the

subgroup of X generated by all Sylow 2-subgroups must be normal in X but X is simple

and so it must be all of X.

Definition 2.24. Let R be the polynomial ring in two commuting indeterminates, x and

y, with coefficients in K. Then R is a KX-module when we set

x ·

(a b

c d

)= ax+ by

and

y ·

(a b

c d

)= cx+ dy.

Let R(1) = f ∈ R|degf = 1 ∪ 0. Then R(1) is a 2-dimensional submodule which we

will call the natural module over K.

Lemma 2.25. Let V be a natural module for X over K. The following hold.

(i) V is irreducible.

(ii) X is transitive on non-zero vectors of V .

(iii) X is 2-transitive on the 1-subspaces of V .

(iv) Elements of X with odd order act fixed-point-freely on V .

Proof. The stabilizer of a 1-space is clearly the normalizer of a Sylow 2-subgroup, i.e.

conjugate to a group of triangular matrices, which again is transitive on all remaining

1-spaces. It is then clear that V is irreducible since there can be no X-invariant subspace.

Moreover, the group of triangular matrices is transitive on the non-zero vectors in the

41

1-space and so X is transitive on non-zero vectors. Finally, the stabilizer of a non-zero

vector is a Sylow 2-subgroup and so any non-identity element of X with odd order fixes

no non-zero vector in V .

Lemma 2.26. There are at least n non-isomorphic, 2-dimensional, irreducible modules

for X over K, namely the natural module over K and its n algebraic conjugates.

Proof. If φ : X → GL(V ) is the representation of X affording a natural module, V , over

K and σ ∈ Aut(K) then, after choosing a basis for V and writing X as matrices, we

can define another representation φσ : X → GL(V ) simply by applying σ to the matrix

entries. The corresponding module we label V σ and call it a Galois twist of V . It is clear

that no choice of basis elements allow the matrices to be written purely by elements of L.

See [1, 26.3] for a proof that this means V and V σ cannot be isomorphic as KX-modules.

Since K = GF(2n) has n automorphisms, namely k 7→ k2i0 ≤ i ≤ n−1, we see n distinct

irreducible modules for X over K.

Let V1, . . . , Vn be these 2-dimensional irreducible KX-modules.

Theorem 2.27. Let V be an irreducible KX-module and let K ′ > K be a field extension.

Then the K ′X-module K ′ ⊗K V is irreducible.

Proof. Recall that an irreducibleKX-module, V is called absolutely irreducible ifK ′⊗KV

is an irreducible K ′X-module for each extension K ′ > K. The splitting field for X is the

field, S, with the property that every irreducible SX-module is absolutely irreducible.

Remark 2.8.8 in [14] says that the splitting field for X is K. Thus every irreducible

KX-module is absolutely irreducible.

Corollary 2.28. Let V be an irreducible KX-module then EndKX(V ) ∼= K.

Proof. By Theorem 2.27, V is absolutely irreducible and so by Lemma 2.8, EndKX(V ) ∼=

K.

42

Let Mi = F ⊗ Vi for 1 ≤ i ≤ n. Recall these are 2-dimensional modules over the

algebraically closed field F and by Theorem 2.27, each Mi is irreducible.

Lemma 2.29. Let x ∈ X have odd order and let χi be the Brauer character afforded by

the FX-module Mi. Then χi(x) = ε2i−1+ ε−2i−1

where ε is some fixed m’th root of unity

and m and the order of x both divide 2n − 1 or both divide 2n + 1.

Proof. Any element, x ∈ X, of odd order either normalizes a Sylow 2-subgroup and so

has order dividing 2n− 1 or is fixed-point-free on the Sylow 2-subgroups and so has order

dividing 2n+1. Let ρ1 be the natural K-representation affording the KX-module V1. The

possible eigenvalues for ρ1(x) in the algebraically closed extension field F are therefore

field elements, α and α−1, satisfying the polynomial X2n−1 − 1 or X2n+1 − 1. By Lemma

2.10, the Brauer character afforded by the module M1 for x is ε+ε−1 where ε is some fixed

m’th root of unity in C and m divides either 2n− 1 or 2n + 1. The entries of the matrices

representing the action of X on each module Vi are altered by a Galois automorphism. So,

by Lemma 2.10, the Brauer character afforded by the module Mi for x is ε2i−1+ ε−2i−1

.

Lemma 2.30. There are 2n irreducible modules for X over F . They are the 2n possible

tensor products⊗

i∈I Mi where I ⊆ 1, 2, . . . , n.

Proof. We begin by identifying an irreducible character for X in characteristic zero of

degree 2n. Consider the action of X on its Sylow 2-subgroups. Let χ be the corresponding

class function and set χ1 = χ− I where I is the trivial character. It follows from Lemma

2.23 that any element of X which normalizes a Sylow 2-subgroup either has order two and

fixes precisely one Sylow 2-subgroup or lies outside of it and fixes exactly two of them.

Thus the class function χ gives values 1 for elements of order two, 2 for elements whose

order divides 2n− 1 and 0 for elements whose order divides 2n + 1. So χ1 gives respective

values 0, 1, −1. We see that this class function is an irreducible character by taking its

43

inner product. This gives

〈χ1, χ1〉 = 22n

2n(2n−1)(2n+1)+ 0

2n + a(

12n−1

)+ b(

(−1)2

2n+1

)

= 122n−1

(2n + a(2n + 1) + b(2n − 1)) ,

where a is the number of conjugacy classes of elements whose order divides 2n − 1 and b

is the number of classes of elements whose order divides 2n + 1. Thus a+ b = 2n − 1 and

it follows that 〈χ1, χ1〉 is strictly less than 2 and so must be 1 and so χ1 is irreducible.

Now by Theorem 2.22, the restriction of χ1 to the elements of x with odd order is an

irreducible Brauer character, χ1. Clearly χ(1) = |Syl2(X)| = 2n + 1 and so χ1(1) = 2n.

Therefore we have found an irreducible Brauer character for X in characteristic two of

degree 2n.

We now begin to identify all irreducible representations of X over F . Consider the

Brauer character for some element, x, of odd order afforded by the module M1 ⊗F . . .⊗F

Mn. By Lemma 2.29 and Lemma 2.19, this will give values

n∏i=1

(ε2i−1

+ ε−2i−1

) =ε2n − ε−2n

ε− ε−1,

where ε is some fixed m’th root of unity and m divides 2n − 1 or 2n + 1. So if x has

order dividing 2n − 1 then the character value is 1 otherwise it is −1. Now we have

a direct agreement between the character induced by the module M1 ⊗ . . . ⊗ Mn and

the character χ1. This means that M1 ⊗ . . . ⊗Mn is an irreducible module. It follows

immediately that every module Mr1 ⊗ . . . ⊗Mrkwhere 1 6 r1 < r2 < . . . < rk 6 n is

irreducible since any proper submodule of Mr1⊗ . . .⊗Mrkwould give a proper submodule

of M1 ⊗ . . . ⊗Mn. If M is any X-module then the module M ⊗M is not irreducible

since it has a proper submodule 〈m⊗ n− n⊗m|m,n ∈M〉. So suppose Mr1 ⊗ . . .⊗Mrk

and Ms1 ⊗ . . . ⊗Mskare isomorphic. Then their tensor product would be reducible but

44

then there is some repeated natural module Mi. Continuing in this way we would see

Mr1 ⊗ . . . ⊗ Mrk= Ms1 ⊗ . . . ⊗ Msk

. Thus we have identified 2n distinct irreducible

modules for X over F . We have already seen that there are at most 2n conjugacy classes

of elements of odd order and so by Theorem 2.20, there are exactly 2n classes and 2n

irreducible modules.

Lemma 2.31. There are exactly 2n irreducible modules for X over K. They are all

possible tensor products (over K)⊗

i∈I Vi where I ⊆ 1, 2, . . . , n.

Proof. For any set I ⊆ 1, . . . , n, (⊗

i∈I,K Vi) ⊗K F ∼= ⊗i∈I,FMi by repeated us of

Lemma 2.9. Suppose (⊗

i∈I,K Vi) had a proper KX-submodule W . Then W ⊗K F would

be a proper FX-submodule of the irreducible module ⊗i∈I,FMi and so must be trivial.

Thus each (⊗

i∈I,K Vi) is an irreducible KX-module. If for I, J ⊆ 1, . . . , n,⊗

i∈I,K Vi

was KX-isomorphic to⊗

j∈J,K Vj then⊗

i∈I,F Mi∼=⊗

j∈J,F Mj and so by Lemma 2.30,

I = J . Therefore we have at least 2n irreducible KX-module. Suppose we could find

one more, U say. Then by Lemma 2.27, U ⊗K F is irreducible and we would have

U ⊗K F ∼=⊗

i∈I,F Mi∼= (⊗

i∈I,K Vi)⊗K F and so U ∼=⊗

i∈I,K Vi. Thus there are exactly

2n irreducible KX-modules.

The previous lemma is also a famous theorem by Robert Steinberg.

Theorem 2.32. (Steinberg) There are exactly 2n irreducible modules for X over K. They

are all possible tensor products (over K)⊗

i∈I Vi where I ⊆ 1, 2, . . . , n.

Proof. See [28, 13.1].

Lemma 2.33. (i) The only irreducible modules for X over F which admit an element

of order three acting fixed-point-freely are the modules M1, . . . ,Mn.

(ii) The only irreducible modules for X over K which admit an element of order three

acting fixed-point-freely are V1, . . . , Vn.

45

Proof. We consider the possible Brauer character values for some element g of order three

acting fixed-point-freely. Let χi be the character corresponding to Mi. Then χi(g) =

ω + ω2 = −1 where ω is a primitive cube root of unity. Thus if χ is the character value

corresponding to some irreducible FX-module, M = Mr1 ⊗ . . . ⊗Mrk, then χ(g) = 1 or

−1. By Lemma 2.6,

CM(〈g〉) = 〈χ|〈g〉 , 1|〈g〉〉 =1

3(χ(1) + χ(g) + χ(g2)).

So g acts fixed-point-freely on M if and only if χ(1) = dim(M) = 2 if and only if M = Mi

for some 1 6 i 6 n.

Suppose now that V is some irreducible KX-module admitting g fixed-point-freely.

By Theorem 2.27, F ⊗K V is an irreducible FX-module. Observe that g acts fixed-point-

freely on F ⊗K V since (f ⊗ v) · g = f ⊗ v · g = f ⊗ v if and only if v · g = v for f ∈ F

and v ∈ V . So now F ⊗K V must have dimension two and so V must be KX-isomorphic

to Vi for some 1 6 i 6 n. Finally, by Lemma 2.25, X acts fixed-point-freely on V1 and

similarly on each algebraic conjugate.

Definition 2.34. A natural module for X over L is an irreducible 2n-dimensional mod-

ule, V , such that EndLX(V ) ∼= GF(2n).

Lemma 2.35. (i) Let V be a natural module for X over L. Then V can be viewed as

a 2-dimensional vector space over K and as such is KX-isomorphic to an algebraic

conjugate of the natural module over K.

(ii) Let U be an algebraic conjugate of the natural module for X over K. Then U can

be viewed as a 2n-dimensional vector space over L and as such is LX-isomorphic

to a natural module over K.

(iii) The natural module over L is unique up to LX-isomorphism.

46

Proof. Let V be a natural module over L. Then V admits an action by its endomorphism

ring and so is a vector space over K of dimension two. Let U represent the vector space

V over K. Suppose U is not irreducible then U contains a KX-invariant proper subspace

U0 say. However this would mean that U0 was LX-invariant and so also a submodule of

V , the vector space over L, which contradicts that V is irreducible by definition. So U is

irreducible of dimension two and therefore must be the natural module over L or one of

its algebraic conjugates.

Now let U be a natural module for X over K then U is also a vector space over L. Let

V represent this restriction module then V is a vector space over L and V is irreducible

as an X-module since X is transitive on non-zero vectors of U by Lemma 2.25. We

can define an action of K on V and so K is embedded in E := EndLX(V ). By Schur’s

Lemma, E is a field. However, since all elements of E commute with K, it follows that

E = EndKX(V ) = EndKX(U) = K by Lemma 2.28. So by definition V is a natural

LX-module and so the restriction of any of the irreducible 2-dimensional modules over K

gives a natural module over L.

Let V be the natural module overK and V σ a Galois twist of V . Consider the following

map

Φ : V → V σ

(k1, k2) 7→ (k1σ, k2

σ).

Then Φ is an LX-module isomorphism since for all v = (k1, k2) ∈ V and every matrix in

47

the corresponding representation of X,

Φ

(k1, k2)

a b

c d

= ((k1a+ k2c)σ, (k1b+ k2d)

σ)

= (kσ1 , k

σ2 )

aσ bσ

cσ dσ

= Φ ((k1, k2)) ·

a b

c d

.

Also because σ is trivial on L we have Φ(lv) = lΦ(v) for all l ∈ L. Thus V and V σ are

isomorphic as LX-modules. In particular it will follow that their restrictions over L are

isomorphic as LX-modules.

48

Chapter 3

Amalgams and the Coset Graph

An amalgam of finite groups is an ordered collection of three finite groups and two injective

homomorphisms A = (A,B,C, φ1, φ2), such that φ1 : C → A and φ2 : C → B.

We say that two amalgams A1 = (A1, B1, C1, φ1, φ2) and A2 = (A2, B2, C2, θ1, θ2) are

isomorphic if there exist isomorphisms α : A1 → A2, β : B1 → B2 and γ : C1 → C2 such

that αφ1 = θ1γ and βφ2 = θ2γ. We say that A is a simple amalgam if, whenever K 6 B,

with φ1(K) E A1 and φ2(K) E A2, then K = 1.

A completion of an amalgam, A = (A,B,C, φ1, φ2), is a collection triple (G,ψ1, ψ2)

where G is a group such that ψ1 : A → G and ψ2 : B → G are homomorphisms with

G = 〈ψ1(A), ψ2(B)〉 and such that ψ1φ1 = ψ2φ2. That is that the following diagram

commutes:

A

@@

@R

C G

@@

@R

B

φ1 ψ1

φ2 ψ2

49

A completion, G, is faithful if ψ1 and ψ2 are injective. In this case we often suppress

use of monomorphisms and identify A, B and C with their images in G. In fact, when

we describe particular amalgams in this thesis we will often assume we have a group G

with subgroups A, B and C := A ∩B such that G = 〈A,B〉. Therefore we are implicitly

assuming that (G, I1, I2) is a completion of the amalgam (A,B,C, i1, i2) where i1 and i2

are the inclusion maps i1 : C → A and i2 : C → B and I1, I2 are the inclusion maps

I1 : A → G and I2 : B → G.

A completion (F,Ψ1,Ψ2) ofA is called universal if given any other completion (G,ψ1, ψ2)

there is a unique homomorphism π : F → G such that Ψiπ = ψi for i = 1, 2. This im-

plies that any completion of A is isomorphic to a quotient of the universal completion

of A and that the universal completion is itself unique up to isomorphism. Isomorphic

amalgams have the same groups as completions so to recognize possible completions we

need to recognize the possible isomorphism classes of the amalgam. A fundamental result

which has become known as the Goldschmidt Lemma (see [12, 2.7]) gives us a method for

calculating the number of isomorphism classes of a given type of amalgam.

Lemma 3.1. (Goldschmidt Lemma) Let (X, Y, Z) be an amalgam of finite groups and

set X∗ = NAut(X)(Z)/CAut(X)(Z), Y ∗ = NAut(Y )(Z)/CAut(Y )(Z) then there is a one-to-one

correspondence between isomorphism classes of type (X, Y, Z) and (X∗, Y ∗)-double cosets

in Aut(Z).

Let A = A(A1, A2, B) be a simple amalgam and let G be a faithful completion of A.

We identify A1, A2 and B with their images in G and regard φi as the inclusion map of

B into Ai. We also suppose that B = A1 ∩A2. The coset graph of the amalgam A is the

graph Γ = Γ(G,A1, A2, B) that has vertex set

V (Γ) = Aig | g ∈ G, i ∈ 1, 2,

50

and edge set

E(Γ) = Aig, Ajh | Aig ∩ Ajh 6= ∅, i 6= j.

Since G is a simple amalgam, G acts faithfully by right multiplication on Γ and this action

preserves the edge set E(Γ).

For γ ∈ V (Γ), let Gγ = StabG(γ) and let Γ(γ) = δ ∈ V (Γ) | γ, δ ∈ E(Γ) be the

set of vertices adjacent to γ in Γ. For γ, δ ∈ E(Γ), let Gγδ = StabG(γ, δ) = Gγ ∩Gδ.

The following results are well known and easy to prove.

Lemma 3.2. Let G be a faithful completion of a simple amalgam of groups (A1, A2, B)

and let Γ be the coset graph.

(i) G acts faithfully on Γ.

(ii) G has two orbits on V (Γ) and is transitive on E(Γ).

(iii) For γ ∈ V (Γ), Gγ is G-conjugate to either A1 or A2.

(iv) For γ, δ ∈ E(Γ), Gγδ is G-conjugate to B.

(v) For γ ∈ V (Γ), Gγ acts transitively on Γ(γ). In particular, |Γ(γ)| = [Gγ : Gγδ] for

any δ ∈ Γ(γ).

(vi) The graph Γ is connected.

Proof. See [22, Lemma 4.1, Lemmas 4.3 and 4.5].

Parts (ii) and (iii) of Lemma 3.2 imply that the amalgam A = A(A1, A2, B) is isomorphic

to the amalgam A′ = A′(Gγ, Gδ, Gγδ) for γ, δ ∈ E(Γ).

Definition 3.3. Let (α1, α2, . . . , αn+1) and (β1, β2, . . . , βn+1) be any two paths of n + 1

vertices in Γ where α1 and β1 lie in the same G-orbit. If there exists g ∈ G such that

ai · g = bi for each 1 6 i 6 n = 1 we say that G acts locally n-arc transitively on Γ.

51

Lemma 3.4. G acts locally n-arc transitively if and only if the stabilizer of any path of

m vertices, (α1, . . . , αm), is transitive on Γ(αm)\αm−1 for each 1 6 m 6 n.

Proof. Suppose G acts locally n-arc transitively on Γ. Let 1 6 m 6 n and let (α1, . . . , αm)

be a path of m vertices in Γ. Suppose β1, β2 ∈ Γ(αm)\αm−1 then by n-arc transitivity,

there is some g ∈ G that maps the path (α1, . . . , αm, β1) to the path (α1, . . . , αm, β2) and

so β1g = β2.

Now suppose that (α1, α2, . . . , αn+1) and (β1, β2, . . . , βn+1) are two paths of n + 1

vertices in Γ with α1 and β1 lying in the same G-orbit. By edge-transitivity, there exists

g1 ∈ G such that α1, α2g1 = β1, β2. Now α3g1 ∈ Γ(β2)\β1 so, by assumption,

there exists g2 ∈ Gβ1,β2 such that α3g1g2 = β3 and then α1, α2, α3g1g2 = β1, β2, β3.

Continuing in this way proves G acts locally n-arc transitively on Γ.

52

Chapter 4

Background Results

In this chapter we include several known results which will be used in this thesis. We

discuss theorems of Feit and Thompson, Smith and Tyrer, and Hayden which are all

character theoretic as well as a result from representation theory by McLaughlin. In

contrast, we also discuss a graph theoretic theorem of Tits and Weiss.

The following result of Feit and Thompson (see [9] and [17]) characterizes a finite

group containing a self-centralizing element of order three. The proof determines some of

the character theory of a minimal counter example.

Theorem 4.1. (Feit-Thompson) Let G be a finite group containing a subgroup, X, of

order three such that CG(X) = X. Then one of the following hold.

(i) G contains a nilpotent normal subgroup, N , such that G/N is isomorphic to Sym(3)

or C3.

(ii) G contains a nilpotent normal 2-subgroup, N , such that G/N ∼= Alt(5).

(iii) G is isomorphic to PSL2(7) ∼= PSL3(2).

When applying Theorem 4.1 we will often need the following.

53

Lemma 4.2. (i) Neither Alt(5) or PSL2(7) ∼= PSL3(2) admit an outer automorphism

of order three.

(ii) If G is a group with a central involution, t, such that G/〈t〉 ∼= PSL2(7) then G ∼=

SL2(7).

Proof. It is well known (see [6]) that Alt(5) and PSL2(7) both have outer automorphism

groups of order two. It is also well known that the Schur multiplier of PSL2(7) has order

two so if G is a group satisfying G/N ∼= PSL2(7) where N is a central subgroup of G then

N has order two and G ∼= SL2(7).

The following result of Smith and Tyrer is of a similar nature to Theorem 4.1. It

describes a group containing a p-subgroup (odd prime p) with restricted normalizer. The

situation it describes is one which will appear several times in this thesis and so the

theorem is one of the most well used here. We will use it when we have a group with an

elementary abelian p-subgroup with restricted normalizer to say that the group has either

a proper normal subgroup at index a power of p or the group is p-soluble of length one.

First we define p-soluble.

Definition 4.3. A groupG is p-soluble if every composition factor ofG is either a p-group

or a p′-group.

Consider the following series

1 E Op′(G) E Op′,p(G) E Op′,p,p′(G) E . . .

where Op′,p(G) is the preimage in G of Op(G/Op′(G)) and Op′,p,p′(G) is the preimage in

G of Op′(G/Op′,p(G)) and so on. This series defines a minimal factorization of G into p

and p′ factors (minimal in the sense that the number of factors is as few as possible). We

call this the lower p-series for G.

54

Definition 4.4. A p-soluble group G has length n if there are n factors in the lower

p-series for G which are p-groups. In particular, G is p-soluble of length one if G =

Op′,p,p′(G). Alternatively, G is p-soluble of length one if for any Sylow p-subgroup, S, of

G, Op′(G)S E G.

Theorem 4.5. (Smith-Tyrer) Let G be a finite group and let P be a Sylow p-subgroup of

G for an odd prime p. Suppose P is abelian and [NG(P ) : CG(P )] = 2.

(i) If G is perfect then P is cyclic.

(ii) If P is non-cyclic, then Op(G) < G or G is p-soluble of length one.

Proof. See [27].

This theorem is used many times in Chapters 6 and 7. It is particularly well suited to

the problems encountered in Chapter 6 since each time it is used we can rule out one of

the possibilities given by the theorem. However the group structures which we encounter

in Chapter 7 satisfy both conclusions of the Smith-Tyrer theorem and so the theorem is

less useful in those cases. It is easy to construct such groups which are both p-soluble of

length one and have a proper p-residue. For example, consider the following subgroup of

Sym(10).

G := 〈(1, 2, 3), (4, 5, 6), (7, 8, 9, 10), (1, 2)〉.

Notice first that G has a normal Sylow 3-subgroup for which the centralizer of the Sylow 3-

subgroup has index two in G. Notice also that O3′(G) = 〈(7, 8, 9, 10〉 and then O3′,3(G) =

〈(7, 8, 9, 10), (1, 2, 3), (4, 5, 6)〉 has index two in G and so G is 3-soluble of length one.

However, O3(G) = 〈(7, 8, 9, 10), (1, 2), (1, 2, 3)〉 6= G and so G satisfies both conclusions of

Theorem 4.5.

Corollary 4.6. Let G be a group and let S be a Sylow p-subgroup of G (odd prime p).

Suppose S is elementary abelian of order at least p3 and |NG(S)/CG(S)| = 2. Finally, for

some x ∈ S of order p, suppose x ∈ Z(G). Then

55

(i) G is either p-soluble of length one or Op(G) < G; and

(ii) if G is not p-soluble of length one then Op(G) has index at least p2 in G and does

not contain x.

Proof. We apply the theorem of Smith and Tyrer to G to see that either G is p-soluble of

length 1 or Op(G) < G. We can apply these arguments again to G := G/〈x〉 to see that

G is either p-soluble of length one or Op(G) < G. So suppose G is p-soluble of length one.

Let N := Op′(G) and consider its preimage in G, N say. Then 〈x〉 is a Sylow p-subgroup

of N and so N splits over 〈x〉 and therefore N = M〈x〉 say. It follows that M is a normal

subgroup of G of order prime to p. Now MS = NS is normal in G since NS is normal in

G because G is p-soluble of length one. Thus we have that G is p-soluble of length one.

So we see that G being p-soluble of length one forces G to have the same property.

We now suppose that G does not have this property then neither does G and so both

have normal subgroups of index a proper power of p. By Gaschutz’s Theorem, there is a

complement to 〈x〉 in G, L say, which is necessarily normal in G. Thus Op(G) 6 L and

〈x〉 /∈ Op(G). So consider now the preimage in G of Op(G). This preimage gives a normal

subgroup of index at least p which contains x and so is distinct from L. Thus Op(G) is

contained in the intersection of the preimage of Op(G) and L and so G/Op(G) has order

at least p2 and does not contain x.

The following theorem is of the same nature as the previous theorems and is also

proved using characters. It can be found in [16, 3.3 p545].

Theorem 4.7. (Hayden) Let G be a finite group and let T ∈ Syl3(G) be elementary

abelian of order nine. Assume that

(i) NG(T )/CG(T ) ∼= 2× 2;

(ii) CG(T ) = T ; and

56

(iii) CG(t) 6 NG(T ) for each t ∈ T#.

Then G = NG(T ).

The following theorem of Jack McLaughlin comes from the paper Some Groups Gen-

erated by Transvections [21] and will be used in the proof of Theorem B. It can be used

to describe a situation in which an elementary abelian group is normalized by a group

generated by elements which each act trivially on a maximal subgroup.

Theorem 4.8. (McLaughlin) Let V be a vector space over GF(p) for odd prime p. Sup-

pose H is a subgroup of GL(V ) such that H = 〈h1, . . . , hr〉 where [V : CV (hi)] = p for

each 1 6 i 6 r. Furthermore, suppose that H contains no non-trivial normal p-subgroup.

Then for some n > 1,

V = V0 ⊕ V1 ⊕ . . .⊕ Vn,

where each Vi is a G-module and

H = H1 × . . .×Hn

where each Hj∼= SL(Vj) or Hj

∼= Sp(Vj). Finally for i 6= j, Hj acts trivially on Vi.

Proof. See [21].

We demonstrate an application of this theorem by introducing a scenario which will occur

inside the amalgam of type G2(3).

Lemma 4.9. Let M be a group and let V := O3(M) be elementary abelian and self-

centralizing in G of order 34. Let M = M/V and let R ∈ Syl3(M) have order 32 and

let S, T be non-conjugate subgroups of R of order three and suppose [V : CV (S)] = [V :

CV (T )] = 3. Then 〈SM〉 ∼= SL2(3), 〈TM〉 ∼= SL2(3) and [〈SM〉, 〈TM〉] = 1.

57

Proof. Let H = 〈SM, T

M〉 6 M . Since V is self-centralizing in M , H is isomorphic to

a subgroup of the group of automorphisms of V , Aut(V ) ∼= GL(V ). Since V = O3(M),

by correspondence, M contains no normal 3-subgroup. Thus the intersection of all its

Sylow 3-subgroups is trivial. Clearly H is a subgroup of M containing every Sylow 3-

subgroup of M and therefore H can have no normal 3-subgroup either. Thus we can

apply Theorem 4.8 to decompose both V and H. Therefore there is an integer n such

that V = V0 ⊕ V1 ⊕ . . . ⊕ Vn and H = H1 × . . . × Hn. We have several possibilities to

consider.

(i) n = 1. Then V = V0 ⊕ V1 where V1 has order 3, 32, 33 or 34 and so H ∼= 1, SL2(3),

SL3(3), SL4(3) or Sp4(3). However H has Sylow 3-subgroups of order 32 and so

n 6= 1.

(ii) n = 2. Then V = V0 ⊕ V1 ⊕ V2 where V1 has order 3 or 32 or 33 and V2 has order 3

or 32 and so H ∼= SL2(3)× 1, SL2(3)× SL2(3) or SL3(3)× 1. The only possibility is

H ∼= SL2(3)× SL2(3) again because of the order of the Sylow 3-subgroup.

(iii) n = 3. Then V = V0⊕V1⊕V2⊕V3 where V1 has order 3 or 32 and V2, V3 both have

order 3. In this case H ∼= SL2(3)× 1× 1 or 1× 1× 1. Neither of which is possible.

(iv) n = 4. Then V = V1⊕V2⊕V3⊕V4 where each has order 3 and so H ∼= 1× 1× 1× 1

which cannot happen.

Therefore V can be decomposed as V = V1 ⊕ V2 where Vi is a 2-dimensional H-module

for i = 1, 2 and H = SL(V1) × SL(V2). We consider the elements, x, of order three in

SL(V1)× SL(V2). There are three types.

(i) x = a where a ∈ SL(V1)#. These act trivially on V2 and fix a 1-subspace of V1 and

are all conjugate in SL(V1).

58

(ii) x = ab where a ∈ SL(V1)# and b ∈ SL(V2)

#. These act non-trivially on both V1 and

V2 but fix a 1-subspace in both.

(iii) x = b where b ∈ SL(V2))#. These act trivially on V1 and fix a 1-subspace of V2 and

are all conjugate in SL(V2).

Recall that S, T 6 H fix 3-dimensional subspaces of V and so have type (i) or (iii).

However S and T are non-conjugate and so, without loss of generality, we must have

〈SM〉 = SL(V1) and 〈TM〉 = SL(V2) and H = 〈SM〉 × 〈TM〉.

We now introduce a graph called a Moufang polygon. Such a graph is bipartite and

contains circuits of length 2n for some fixed integer n as well as satisfying the so-called

Moufang condition. Theorem B and Theorem C in this thesis rely on the classification of

Moufang polygons which was completed by Jacques Tits and Richard Weiss in 2002 (see

[29]).

Definition 4.10. The diameter of a graph, ∆, is n where n is the largest possible integer

for which there exists vertices α and β in ∆ with d(α, β) = n. The girth of ∆ is the

smallest integer m where ∆ contains a circuit of length m.

Definition 4.11. A generalized n-gon is a bipartite graph which has girth n and diam-

eter 2n. In particular, a generalized hexagon is a generalized 6-gon and a generalized

quadrangle is a generalized 4-gon.

Definition 4.12. Let ∆ be a generalized n-gon in which each vertex has valency three

or more. Let Θ = (α0, . . . , αn) be a path of length n which is contained in a cir-

cuit of length 2n. We set UΘ to be the stabilizer in Aut(∆) of the set of vertices

Γ(α1),Γ(α2), . . . ,Γ(αn−1) and call UΘ the root group corresponding to Θ. If UΘ is

transitive on Γ(αn)\αn−1. Then ∆ is said to satisfy the Moufang condition.

59

Let ∆ be a generalized n-gon then ∆ is a bipartite graph so we consider the subgroup of

the automorphism group of ∆ which preserves the partition. We call this group Auto(∆).

Now let Θ = (α0, α1, . . . , αn) be a path of n + 1 vertices contained in an n-cycle and

consider the root group UΘ. The subgroup of Auto(∆) generated by all root groups is a

normal subgroup of Auto(∆) which we call G†.

Theorem 4.13. (Tits/Weiss) Let ∆ be a finite generalized quadrangle or hexagon satis-

fying the Moufang condition. Then ∆ is uniquely defined by its order. In particular, if

∆ is a Moufang quadrangle with 80 vertices then G† ∼= U4(2) ∼= PSp4(3) and if ∆ is a

Moufang hexagon with 728 vertices then G† ∼= G2(3).

Proof. See section 17 and 34 in [29]. In particular pages 378-379.

60

Chapter 5

An Amalgam of Type PSL3(3)

In [23] amalgams of type PSL3(pa) are classified when a > 1 and when p is an odd prime.

The methods used to identify completions for a > 1 do not extend to the smaller cases

of type PSL3(p). In this chapter we describe an amalgam of type PSL3(3). In PSL3(3)

there are two maximal parabolic subgroups which intersect at the normalizer of a Sylow

3-subgroup. The amalgam consists of these two parabolic subgroups and we will assume

we have a completion of this amalgam which satisfies a 3-local condition. Under such

conditions the amalgam completes to PSL3(3) and also to the sporadic simple group M12.

We can describe the amalgam in each group as follows. Consider the action of a group

G1∼= PSL3(3) on a vector space of dimension three over a field of order three. Let

A1 := StabG1(〈(1, 0, 0)〉) and B1 := StabG1(〈(1, 0, 0), (0, 1, 0)〉). Now consider the action

of G2∼= M12 on a set of twelve points Ω := 1, 2, . . . , 12. Let A2 = StabG2(1, 2, 3) and

B2 = StabG2(1, 2, 34, 5, 67, 8, 910, 11, 12). A result of Goldschmidt can be used

to prove that the two amalgams (A1, B1, A1 ∩B1) and (A2, B2, A2 ∩B2) are isomorphic.

The two groups, PSL3(3) and M12, share an isomorphic amalgam and an isomorphic

centralizer of a 3-central element but differ in their centralizer of an involution. We will

apply the theorem of Feit and Thompson (Theorem 4.1) to the involution centralizer in

61

an abstract finite completion of the amalgam which satisfies a 3-local condition. This,

together with a result of Burnside about fixed-point-free automorphisms of order three

(Lemma 1.24), will allow us to restrict the structure of the completion. Furthermore, the

Burnside lemma we use will give us a particular group relation. The techniques which will

be used in later chapters are not suitable in this amalgam since the subgroups are small.

Instead we use a computer algebra package. We will create the subgroups which constitute

the amalgam in Magma and then make the universal completion as the amalgamated

free product. Factoring the universal completion by a relation forced by Lemma 1.21 and

using the coset enumeration package will then confirm the only possible completions are

M12 and PSL3(3).

Hypothesis A. Let G be a finite group with subgroups A ∼= B such that C := A ∩

B contains a Sylow 3-subgroup of both A and B and such that no non-trivial normal

subgroup of C is normal in both A and B. Suppose further that

(i) G = 〈A,B〉;

(ii) A/O3(A) ∼= B/O3(B) ∼= GL2(3);

(iii) O3(A) and O3(B) are natural modules with respect to the actions of A/O3(A) and

B/O3(B) respectively; and

(iv) for S ∈ Syl3(C), NG(Z(S)) = NA(Z(S)) = NB(Z(S)).

Our main theorem in this chapter is the following.

Theorem A. If G satisfies Hypothesis A, then G ∼= M12 or PSL3(3).

62

5.1 Some Structure of the Amalgam (A,B,C)

We assume Hypothesis A and define some further notation. Set QA := O3(A) and QB :=

O3(B).

Lemma 5.1. The following hold.

(i) S := QAQB is a normal Sylow 3-subgroup of C and S ∼= 31+2+ .

(ii) QA ∩QB = Z(S).

(iii) C = NA(S) = NB(S) = NA(Z(S)) = NB(Z(S)) = NG(Z(S)) has index four in A

and in B.

(iv) CG(Z(S)) has index two in C.

Proof. (i) If QA = QB then A and B would share a normal subgroup which is not possible.

Also, Sylow 3-subgroups of A and B have order 33 and C contains one of them so C

contains QA and QB which necessarily normalize each other. Thus S := QAQB ∈ Syl3(C)

is normal in C. Now choose some x ∈ QB\QA then 1 6= QAx ∈ A/QA∼= GL2(3). Since QA

is a natural A/QA-module, QAx acts non-trivially on QA. In particular, S is non-abelian.

If Z(S) has index three in S then we could choose any x ∈ S\Z(S) which would necessarily

commute with 〈x, Z(S)〉 = S and S would be abelian. Therefore Z(S) must have order

three. It is then immediate that Z(S) = S ′. Furthermore S/Z(S) = ( QA

Z(S))( QB

Z(S)) is

elementary abelian so S is extraspecial. Now since QA and QB have exponent three, it

follows that S ∼= 31+2+ .

(ii) Since QA and QB are not equal and have index three in S, QA ∩ QB has order

three and is a normal subgroup of S. It is therefore central in S and so QA ∩QB = Z(S).

(iii) The normalizer of a Sylow 3-subgroup in A is the preimage in A/QA∼= GL2(3)

of the normalizer of a Sylow 3-subgroup which has index four. Therefore the normalizer

63

of a Sylow 3-subgroup in A (and in B) has index four. We have NA(S) 6 NA(Z(S)) =

NB(Z(S)) 6 B and NB(S) 6 NB(Z(S)) = NA(Z(S)) 6 A and so C 6 NA(S), NB(S) 6

A ∩B = C which means C = NA(S) +NB(S) has index four in A and in B.

(iv) We have seen that C = NA(S) = NB(S) 6 NG(Z(S)). Since NG(Z(S)) =

NA(Z(S)) = NB(Z(S)) 6 A ∩ B, C = NG(Z(S)). Choose t ∈ A such that 〈tQA〉 =

Z(A/QA). Then t normalizes S and so lies in C. Also t inverts the natural module QA

and so t inverts Z(S) which is not central in C. Therefore CC(Z(S)) has index two in

C.

Lemma 5.2. If t ∈ A is an involution such that tQA ∈ Z(A/QA) then CA(t) ∼= GL2(3)

and there is x ∈ CA(t) of order three such that 〈x〉 = Z(P ) for some P ∈ Syl3(B).

Proof. Let t ∈ A be an involution such that tQA ∈ Z(A/QA) ∼= Z2. Then tQA commutes

with each Sylow 3-subgroup of A/QA and in particular commutes with QAQB/QA∼=

QB/(QA∩QB). By Lemma 2.16, there is some x ∈ QB\QA that commutes with t. Notice

that A has at most 27 Sylow 2-subgroups of A and since t is in the centre of a Sylow

2-subgroup, at most 27 A-conjugates of t. However, t commutes with an element of order

three, x ∈ QB\QA, and so there can be at most nine A-conjugates of t in A. Moreover t

inverts QA and so the set tQA = tq|q ∈ QA has order nine and therefore 〈t〉QA = 〈t〉A. By

a Frattini argument (Lemma 1.7), A = NA(〈t〉)QA = CA(t)QA and so by an isomorphism

theorem,

CA(t) ∼=CA(t)

CA(t) ∩QA

∼=CA(t)QA

QA

=A

QA

∼= GL2(3).

Finally, since x ∈ Q#B and QB is a natural B/QB-module, Lemma 2.15 (ii) implies that

〈x〉 is conjugate to Z(S) in B. Thus there exists P ∈ Syl3(B) such that 〈x〉 = Z(P ).

Lemma 5.3. Let u ∈ C be an involution then CS(u) has order three. In particular, if u

is an involution in G normalizing a Sylow 3-subgroup P of G then |CP (u)| = 3.

64

Proof. We can find a complement, T ∼= 2 × 2, to S in C by Sylow’s Theorem and

TQA/QA 6 A/QA∼= GL2(3) and TQB/QB 6 B/QB

∼= GL2(3). By Lemma 2.13, any

subgroup of GL2(3) isomorphic to 2×2 contains the central element. So choose t ∈ T such

that tQA ∈ Z(A/QA) and choose s ∈ T such that sQB ∈ Z(B/QB). We have seen that t

inverts each element of QA and centralizes some element x ∈ QB\QA. Similarly s inverts

QB and centralizes some element y ∈ QA\QB. In particular, t 6= s and therefore T = 〈t, s〉.

Now we see that st centralizes the group QA∩QB = Z(S) and inverts the groups 〈x〉 and

〈y〉. So suppose CS(st) had order nine. Then S = CS(st)〈x〉 = CS(st)〈y〉 and therefore

x = zy for some z ∈ CS(st). However this implies y−1z−1 = x−1 = xst = zstyst = zy−1

and so [y, z] = z2. Therefore z ∈ S ′ = Z(S) (Lemma 5.1). However this implies that

z2 = [y, z] = 1 and so z = 1 and x = y which is not possible. Hence each element of T#

centralizes just an element of order three in S.

Lemma 5.4. Let t ∈ A be as in Lemma 5.2 then CG(t)/〈t〉 contains a self-centralizing

element of order three.

Proof. We have seen that there is an element of order three, x ∈ CA(t), such that 〈x〉 is

the centre of some Sylow 3-subgroup in B. By Lemma 5.1 (iv), CG(x) has order 332 and

is contained in B. More precisely, CG(x) = P 〈t〉 for some P ∈ Syl3(B). Now Lemma

5.3 implies that CP (t) has order three and so CG(x) ∩ CG(t) = 〈x, t〉. Therefore x〈t〉 is a

self-centralizing element of order three in CG(t)/〈t〉.

5.2 A Presentation for the Amalgam

Lemma 5.5. A ∼= B ∼= AGL2(3)

Proof. If V is a 2-dimensional vector space over GF(3) then AGL2(3) is defined to be the

set of affine transformations φ + w where w ∈ V and φ ∈ GL(V ) and for each v ∈ V ,

φ+ w : v 7→ φ(v) + w. There are 9|GL(V )| = 2433 such transformations.

65

Notice that by Lemma 5.2, we can choose an involution t ∈ A such that CA(t) ∼=

GL2(3) and so D := CA(t) is a complement to QA in A. We will define a map, Θ, from

A to AGL2(3) as follows. Each a ∈ A can be written uniquely as a = bc where b ∈ QA

and c ∈ D. Now, for each v ∈ QA∼= V , define Θ(bc) : v 7→ b(vc−1

). We check that Θ is a

homomorphism. Let bc and b1c1 be in A such that b, b1 ∈ QA and c, c1 ∈ D. There exists

b2 ∈ QA such that b2 = cb1c−1 and so bcb1c1 = bb2cc1. Therefore Θ(bcb1c1) = Θ(bb2cc1)

and for v ∈ V ,

Θ(bb2cc1)(v) = bb2(vc−11 c−1

) = bbc−1

1 (vc−11 c−1

) = b((b1vc−11 )c−1

) = Θ(bc)(Θ(b1c1)(v)).

So Θ is a homomorphism. Suppose bc ∈ Ker(Θ) then for each v ∈ QA, bvc−1= v and

so bv2c−1

= v2. However bv2c−1

= b(vc−1)2 = b(b−1v)2 = b−1v2 and it follows that b = 1.

Therefore vc = v for all v ∈ QA. However QA is a natural A/QA-module so only the

identity acts trivially (see Lemma 2.15 (iv)). Therefore the kernel of this map is trivial

and since |A| = |AGL2(3)|, the map is an isomorphism.

There are two faithful representations of AGL2(3) in PSL3(3) which we shall see.

Consider an amalgam of type (X,Y, Z) where X ∼= Y ∼= AGL2(3) and Z ∼= NX(S)

where S ∈ Syl3(X). BothX and Y have an elementary abelian normal 3-subgroup, O3(X)

and O3(Y ) respectively, of order 9. Suppose φ1 and φ2 define embeddings of Z into X and

Y respectively. The image of Z in X contains O3(X) and the image of Z in Y contains

O3(Y ). If H is some completion of (X, Y, Z) we can ask if the corresponding images in H

share the same 3-radical subgroup. We easily see instances of both possibilities and so can

immediately identify at least two isomorphism classes of this amalgam. Firstly suppose

X = Y . Such an amalgam cannot be isomorphic to the following amalgam of subgroups

66

in H ∼= PSL3(3). Let H act on the vector space of dimension three over GF(3) and set

X = StabH(〈(1, 0, 0)〉)

=

α β 0

γ δ 0

ε ζ η

∣∣α, β, γ, δ, ε, ζ ∈ GF(3), η = (αδ − βγ)−1 ∈ GF(3)\0

and

Y = StabH(〈(1, 0, 0), (0, 1, 0)〉)

=

η 0 0

ε α β

ζ γ δ

∣∣α, β, γ, δ, ε, ζ ∈ GF(3), η = (αδ − βγ)−1 ∈ GF(3)\0

.

Let A represent the isomorphism class of this amalgam (X,Y, Z).

Lemma 5.6. Let X ∼= AGL2(3) and Z = NX(S) where S ∈ Syl3(X). Then the following

hold.

(i) Both Z and X have trivial centres.

(ii) CX(Z) = 1 and NX(Z) = Z.

(iii) Z/S ∼= 2× 2.

(iv) CZ(S) = Z(S)

Proof. The representation of AGL2(3) in PSL3(3) makes checking most of these results

trivial when we identify Z with the group

α 0 0

δ β 0

ε ζ γ

∣∣α, β, γ, δ, ε, ζ ∈ GF(3), αβγ = 1

.To show NX(Z) = Z we observe that S E Z so Z 6 NX(Z) 6 NX(S) = Z.

Lemma 5.7. Let X ∼= AGL2(3) and Z ∼= NX(S) where S ∈ Syl3(X).

67

(i) Inn(Z) ∼= Z and |Aut(Z)/Inn(Z)| = 2.

(ii) Aut(X) = Inn(X) ∼= X.

Proof. (i) By Lemma 5.6, Z has trivial centre so we identify Z ∼= Inn(Z) with its image

in Aut(Z). Let Aut(Z) ≥ R > Z be such that R/Z has prime order q. Suppose first that

CR(S) > CZ(S) = Z(S) then CR(S)/CZ(S) has order q. Since Z is soluble and R/Z is

cyclic of prime order, R is soluble. The Fitting subgroup of R contains all nilpotent normal

subgroups of R so S 6 F (R). By Lemma 1.13, CR(F (R)) 6 F (R) so F (R) > CR(S).

Since F (R) is nilpotent, it is a direct product of its Sylow subgroups so as no element

of order two in Z commutes with S, we have F (R) ∩ Z = S. Therefore it follows that

|F (R)| = 33q. If q 6= 3 then R contains a normal subgroup, N say, of order q and then

[N,Z] 6 N ∩ Z = 1 but then N would act trivially on Z, thus q = 3. Let T be a Sylow

2-subgroup of Z then T acts on the normal subgroup CR(S). Moreover, this action is

coprime so CR(S) = [CR(S), T ]CCR(S)(T ). However [CR(S), T ] 6 CR(S) ∩ Z = CZ(S)

and so 1 6= CCR(S)(T ). This means that some x ∈ R\Z is trivial on 〈T, S〉 = Z which

is a contradiction. So suppose instead that CR(S) = CZ(S). Then R/CR(S) embeds

into Aut(S) and in particular R/S embeds into Out(S) ∼= GL2(3) (Lemma 1.16). Now

2× 2 ∼= Z/S E R/S and any subgroup of GL2(3) isomorphic to 2× 2 is conjugate to the

group of diagonal matrices (see Lemma 2.13). So we identify Z/S with the subgroup of

diagonal matrices in GL2(3) which sits at index two in its normalizer in GL2(3). However,

this implies that q = 2. Therefore the only prime dividing Aut(Z)/Inn(Z) is two. So we

repeat the same argument since 2 × 2 ∼= Z/S E Aut(Z)/S and so |Aut(Z)/Inn(Z)| 6

2. Finally we see that Aut(Z)/Inn(Z) has order exactly two with the following outer

automorphism.

α :

α 0 0

γ αβ 0

δ ε β

7→

β 0 0

εα αβ 0

γε − αβδ γβ α

.

(ii) The centre of X is trivial so we immediately identify X ∼= Inn(X) with its image

68

inside Aut(X). Let Q := O3(X) ∼= 32 and let R be such that Aut(X) ≥ R > X and R/X

is cyclic of prime order q. By an isomorphism theorem,

Aut(Q) ∼= GL2(3) ∼=X

Q=

X

CX(Q)=

X

X ∩ CR(Q)∼=XCR(Q)

CR(Q)6

R

CR(Q)→ Aut(Q).

So R/CR(Q) ∼= GL2(3) and CR(Q)/Q must have order q. Now CR(Q) E R and has order

32q. Suppose q 6= 3 then R contains a normal subgroup, N say, of order q. But then

[N,X] 6 N ∩ X = 1 and N acts trivially on X which is a contradiction. Hence q = 3

and CR(Q) is a group of order 27. Now by a Frattini argument, R = XNR(S) and by an

isomorphism theorem.

NR(S)

Z=

NR(S)

NR(S) ∩X∼=R

X∼= C3.

In particular, NR(S) normalizes Z and so NR(S) = NR(Z). By part (i), |Aut(Z)|3 = 33

and so |NR(Z)/CR(Z)|3 6 33 and it follows that 3 divides CR(Z) and so R = X : CR(Z).

Now CR(Z) 6 CR(t) where t =

−1 0 0

0 −1 0

0 0 1

and CX(t) ∼= GL2(3). Therefore CR(Z)

normalizes O2(CX(t)) ∼= Q8 and so by coprime action we get

O2(CX(t)) = CO2(CX(t))(CR(Z))[O2(CX(t)), CR(Z)].

However CR(Z) 6 CR(Q) E R so [O2(CX(t)), CR(Z)] 6 O2(CX(t)) ∩ CR(Q) = 1. Thus

CR(Z) commutes with 〈Z,O2(CX(t))〉 = X which is a contradiction.

Lemma 5.8. There are exactly two isomorphism classes of amalgams of type (X, Y, Z)

with X ∼= Y ∼= AGL2(3) and Z ∼= NX(S) where S ∈ Syl3(X).

Proof. We will use the Goldschmidt Lemma (3.1), Lemma 5.6 and Lemma 5.7. Firstly,

X∗ = NAut(X)(Z)/CAut(X)(Z) ∼= NX(Z)/CX(Z) ∼= NX(Z) ∼= Z

and similarly Y ∗ ∼= Z giving us exactly two (X∗, Y ∗)-double cosets in Aut(C). Now the

69

Goldschmidt Lemma implies that there are exactly two isomorphism classes of amalgams

of type (X, Y, Z).

Corollary 5.9. Suppose A1 = (X1, Y1, Z1) is an amalgam with X1∼= Y1

∼= AGL2(3) and

Z1∼= NX(S) where S ∈ Syl3(X1). If O3(X1) 6= O3(Y1) then A1 is isomorphic to A.

Proof. We have seen there are exactly two isomorphism types of this amalgam and we

have seen examples of both. The first can be seen when X1 = Y1 and the second can be

seen in PSL3(3) which can, in particular, be characterized by the fact that the 3-radical

subgroups in the two groups constituting the amalgam are distinct.

Lemma 5.10. (A,B,C = A ∩B) is isomorphic to the amalgam A.

Proof. Lemma 5.5 gives A ∼= B ∼= AGL2(3) and Lemma 5.1 gives that C = NA(S) for

some S ∈ Syl3(A). If O3(A) = O3(B) then A and B would share a normal subgroup and

so O3(A) 6= O3(B) and so the amalgam (A,B,C) is isomorphic to A.

We can now identify our amalgam (A,B,C) with the amalgam A of matrices seen in

PSL3(3). In the following lemma we give a presentation for the universal completion of

the amalgam A.

Lemma 5.11. Let

RC = a3 = b3 = [a, b]3 = [a, [a, b]] = [b, [a, b]] = t2 = u2 = [t, u] = 1,

au = a, bu = b−1, at = a−1, bt = b,

RA = RC ∪ p4 = q4 = 1, p2 = q2 = t, [a, b]p = a, [a, b]q = [b, a]a,

pu = p−1, pb = pq, qb = p,

and

RB = RC ∪ r4 = s4 = 1, r2 = s2 = u, [a, b]r = b−1, [a, b]s = b−1[b, a],

rt = r−1, ra = rs, sa = r.

70

Then C ∼= 〈a, b, t, u|RC〉, A ∼= 〈a, b, p, q, t, u|RA〉, B ∼= 〈a, b, r, s, t, u|RB〉 and F :=

〈a, b, p, q, r, s, t, u|RA,RB〉 is the universal completion of A.

Proof. We make the following identifications of matrices in C to elements in the presen-

tation.

a ∼

1 0 0

1 1 0

0 0 1

, b ∼

1 0 0

0 1 0

0 1 1

, t ∼

1 0 0

0 −1 0

0 0 −1

, u ∼

−1 0 0

0 −1 0

0 0 1

.

Therefore 〈a, b, t, u|RC〉 is a group with a quotient group isomorphic to C since the el-

ements of C satisfy the relations. Coset enumeration using Magma ([4]) ensures it is

isomorphic to C. Similarly if we identify

p ∼

1 0 0

0 0 1

0 −1 0

, q ∼

1 0 0

0 −1 1

0 1 1

then coset enumeration gives 〈a, b, p, q, t, u|RA〉 ∼= AGL2(3) and again if we identify

r ∼

0 −1 0

1 0 1

0 −1 0

, s ∼

−1 1 0

1 1 0

0 0 1

then 〈a, b, r, s, t, u|RB〉 ∼= AGL2(3). By Corollary 5.9, the amalgam

(〈a, b, p, q, t, u|RA〉, 〈a, b, r, s, t, u|RB〉, 〈a, b, t, u|RC〉)

is isomorphic to A and so F := 〈a, b, p, q, r, s, t, u|RA,RB〉 is the universal completion of

the amalgam (see [26, Thm 1 p3] for justification that F is universal).

Recall Lemma 5.2 and Lemma 5.4 where we found an involution in A whose centralizer

in A was isomorphic to GL2(3). Notice that with the identifications we made in Lemma

5.11, the involution t satisfies the properties described in Lemma 5.2. For the remainder

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of this chapter we identify A and B with the finitely presented groups and in particular

we identify t ∈ A = 〈a, b, p, q, t, u|RA〉 with the involution in Lemmas 5.2 and 5.4. We

also identify G with some quotient of F .

Lemma 5.12. Let G′ be any faithful completion of A then CG′(t) contains two (not neces-

sarily distinct) subgroups isomorphic to Q8 both of which are normalized but not centralized

by the group 〈b, u〉 ∼= Sym(3).

Proof. It is clear from the presentation that P := 〈p, q〉 and 〈r, s〉 are quaternion groups

of order eight. Notice that upr = t follows from the relations pu = p−1 and rt = r−1. So

R := 〈r, s〉pr is a quaternion group with central element t. It follows from the relations of

〈a, b〉 as an extraspecial group of order 27 that [a, b]−1 = [a−1, b] and so we notice that apr =

[a, b]p2r = [a, b]tr = [a−1, b]r = ([a, b]−1)r = b. Therefore prb = apr and so 〈rprb, sprb〉 =

〈rs, r〉pr is normalized by 〈b〉. Also pru = pur = up−1r = utpr and so 〈rpru, spru〉 =

〈r−1, sr〉pr is normalized by 〈u〉 also. We point out that any faithful completion of the

amalgam contains injective images of the groups constituting the amalgam. Therefore

the groups 〈p, q〉, 〈r, s〉 and 〈b, u〉 will always have the isomorphism types claimed.

We continue to set P = 〈p, q〉 and R = 〈x, y〉, where x = rpr and y = spr.

Lemma 5.13. CG(t) contains a nilpotent normal subgroup, N and P,R 6 N . In partic-

ular P and R generate a 2-group.

Proof. By Lemma 5.4, we can apply Theorem 4.1 to CG(t)/〈t〉. Suppose CG(t)/〈t〉 ∼=

PSL2(7) then by Lemma 4.2, CG(t) ∼= SL2(7) which only contains one element of order

two. This is a contradiction since we know that t commutes with another involution

u ∈ C. Therefore CG(t)/〈t〉 contains a nilpotent normal subgroup of order prime to 3.

Let N be the preimage in CG(t) of this normal subgroup then 〈t〉 ≤ N E CG(t) and

CG(t)/N ∼= Alt(5) or Sym(3) or C3. By Lemma 5.12, CG(t) contains the two groups

P ∼= R ∼= Q8 which are normalized in CG(t) by 〈b, u〉 ∼= Sym(3). We consider P ∩ N .

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Notice first that D := P 〈b, u〉 ∼= GL2(3) so O3′(D) = O2(D) = P ∼= Q8. Also N ∩D is a

normal 3′ subgroup and so N ∩D = N ∩ P . Suppose P ∩N = 〈t〉 then

∣∣∣∣CG(t)

N

∣∣∣∣ > ∣∣∣∣DNN∣∣∣∣ =

∣∣∣∣ D

N ∩D

∣∣∣∣ =243

2= 233

which is not possible. So suppose |P ∩ N | = 4. The group of order three, 〈b〉, in D

acts regularly on P/〈t〉 and normalizes N and so we have a contradiction since P =

〈(P ∩ N)〈b〉〉 6 N . Thus P 6 N . We can argue in the same way with R since R is

normalized but not centralized by 〈b, u〉. Thus P,R 6 N . Since N is a nilpotent group

and P and R are 2-groups, 〈P,R〉 is a 2-group.

We have that G is a faithful completion of the amalgam A so is a quotient of the

universal completion F . Furthermore it contains the 2-group generated by P and R and

so 〈P,R〉 satisfies the congruences described in Corollary 1.23. Therefore, for any α ∈ P

and any β ∈ R, [αb, β][βb, α] ∈ 〈t〉. We make the following four sets of relations:

(i) R1 = [yb, q][qb, y] = 1, [pqb, x][xb, pq] = 1;

(ii) R2 = [yb, q][qb, y] = t, [pqb, x][xb, pq] = 1;

(iii) R3 = [yb, q][qb, y] = 1, [pqb, x][xb, pq] = t;

(iv) R4 = [yb, q][qb, y] = t, [pqb, x][xb, pq] = t.

Lemma 5.14. |G| divides 95040 or |G| divides 5616.

Proof. G is a quotient of one of the following groups

F1 = 〈a, b, p, q, r, s, t, u|RA,RB,R1〉,

F2 = 〈a, b, p, q, r, s, t, u|RA,RB,R2〉,

F3 = 〈a, b, p, q, r, s, t, u|RA,RB,R3〉,

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F4 = 〈a, b, p, q, r, s, t, u|RA,RB,R4〉.

Coset enumeration, using Magma ([4]), gives respective group orders of 95040, 1, 5616,

1.

A final corollary proves Theorem A

Corollary 5.15. G ∼= M12 or PSL3(3).

Proof. Both M12 and PSL3(3) satisfy Hypothesis A and so are completions of the amalgam

A. Therefore each must be a quotient of one of the four groups presented in the previous

lemma. However the orders of the two non-trivial groups give the orders of M12 and

PSL3(3). It follows that F1∼= M12 and F3

∼= PSL3(3).

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Chapter 6

An Amalgam of Type G2(3)

In Local Characteristic p Completions of Weak BN-Pairs [23] a completion of an amalgam

of type G2(pa) (odd prime p) is proven to be isomorphic to G2(p

a) under a K-proper

hypothesis and some strong p-local conditions. An amalgam of type G2(3) is an amalgam

of groups comprising two maximal 3-local subgroups in G2(3) which intersect at the

normalizer of a Sylow 3-subgroup. Here we will prove that a group, G, which contains

a faithful completion, G, of an amalgam of type G2(3) and satisfies a 3-local condition

must be isomorphic to G2(3). We will describe the structure of this amalgam as in [23]

and analyse the conjugacy classes of elements of order three which we see. This allows

us to fully determine the centralizer of an involution. We consider the coset graph given

by a completion of the amalgam and this will turn out to be a generalized 6-gon with

728 vertices. Thus we can appeal to the classification of generalized polygons by Tits and

Weiss to recognize the uniqueness of this graph and force the completion of the amalgam

to be isomorphic to G2(3). We are left to rule out the possibility of a strict containment

G > G. However knowledge of the full centralizer of an involution gives us that G is

strongly embedded in G and a contradiction follows.

Limiting the structure of the centralizers of 3-elements and proving that exactly 5

75

conjugacy classes exist allows us to limit the structure of many 3-local subgroups. Thus

a modular character theoretic result of Stephen Smith and Peter Tyrer proves useful and

we use it on three occasions (see Theorem 4.5).

The hypothesis we will assume is:

Hypothesis B. Let G be a finite group with non-conjugate subgroups A1, A2 such that

A12 := A1 ∩ A2 contains a Sylow 3-subgroup of both A1 and A2 and such that no non-

trivial normal subgroup of A12 is normal in both A1 and A2. Suppose further that, for

i = 1, 2,

(i) |O3(Ai)| = 35;

(ii) Ai/O3(Ai) ∼= GL2(3);

(iii) O3(Ai)/Z(O3(Ai)) and Z(O3(Ai))/O3(Ai)′ are natural modules with respect to the

action of Ai/O3(Ai);

(iv) NG(O3(Ai)′) = Ai.

The main theorem of this chapter is:

Theorem B. Let G be a group satisfying Hypothesis B. Then G ∼= G2(3).

6.1 The Amalgam

We assume Hypothesis B and define some further notation. Let G := 〈A1, A2〉 6 G then

G is a faithful completion of the simple amalgam (A1, A2, A12). We can thus introduce the

coset graph Γ = Γ(G,A1, A2, A12) and by naming the vertices α = A1, β = A2 ∈ V (Γ) we

get Gα := A1 = StabG(A1), Gβ := A2 = StabG(A2) and Gαβ := A12 = StabG(A1, A2).

Let γ ∈ α, β and set

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(i) Qγ = O3(Gγ) - this has order 35 and is normal in Gγ;

(ii) Zγ = Z(Qγ) - this is also normal in Gγ and has order 33 since Qγ/Zγ is a natural

module so has order 32;

(iii) Yγ = Q′γ - this has order three and is normalized by Gγ;

and choose Sαβ ∈ Syl3(Gαβ). We also introduce the following notation for the correspond-

ing intersections Qαβ := Qα ∩Qβ, Zαβ := Zα ∩ Zβ.

We now begin to build up an understanding of the structure of the subgroups in the

amalgam (Gα, Gβ, Gαβ).

Lemma 6.1. The following hold.

(i) Sαβ = QαQβ E Gαβ.

(ii) Zαβ = Z(Sαβ) = YαYβ has order 32.

(iii) For γ, δ = α, β, Zγ 6 Qδ.

(iv) ZαZβ = Qαβ is elementary abelian of order 34.

(v) Qγ has exponent 3, for γ ∈ α, β.

These results are from Lemma 6.5 in [23]. In the following proof we use Lemma 2.15 several

times. Lemma 2.15 gives us information on the natural GL2(3)-module and in particular

that elements of order three in Gγ/Qγ act non-trivially on the natural module and that

Gγ/Qγ is transitive on vectors and 1-subspaces of the natural module (γ ∈ α, β).

Proof. (i) We have that Qα 6= Qβ as the amalgam is simple and so QαQβ ∈ Syl3(Gαβ).

Moreover Qα E Gαβ and Qβ E Gαβ and so QαQβ is normal in Gαβ and therefore is the

unique Sylow 3-subgroup of Gαβ. Thus Sαβ = QαQβ.

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(ii) Notice first that both Yα and Yβ are central in Sαβ. Also Zα is central in Qα and

Zβ is central in Qβ and so Zαβ is central in QαQβ = Sαβ. Suppose Z(Sαβ) Qα then

Sαβ = QαZ(Sαβ). However this would mean that Sαβ/Qα acts trivially on the natural

modules Qα/Zα and Zα/Yα which is not possible by Lemma 2.15 and so Z(Sαβ) 6 Z(Qα)

and similarly Z(Sαβ) 6 Z(Qβ). Hence Z(Sαβ) = Zαβ. Also, since the amalgam is simple,

Zα 6= Zβ and Yα 6= Yβ. Thus YαYβ 6 Z(Sαβ) Zα, Zβ and so we get the order 32.

(iii) Let γ, δ = α, β and suppose that Zγ Qδ. Then

Zγ > Zγ ∩Qδ > Zγ ∩ Zδ = Z(Sγδ)

by (ii). Hence Zγ ∩Qδ = Z(Sγδ) follows by considering group orders. So,

[Zγ, Qδ] 6 Zγ ∩Qδ = Z(Sγδ) 6 Zδ.

Therefore, Sγδ/Qδ = ZγQδ/Qδ acts trivially on the natural module Qδ/Zδ which is not

possible by Lemma 2.15. Hence Zγ 6 Qδ.

(iv) We have seen that Qα 6= Qβ and that both subgroups have index three in Sαβ.

Thus Qαβ has index 32 in Sαβ and so has order 34. We have also seen that Zα 6= Zβ and

that Zα ∩ Zβ has order 32. So by an isomorphism theorem, C3∼= Zβ/Zαβ

∼= ZαZβ/Zα

and so ZαZβ has order 34. However Zα 6 Qα ∩ Qβ and Zβ 6 Qβ ∩ Qα by (iii) and so

ZαZβ = Qαβ. Now let γ, δ = α, β and consider the cosets in Zγ/Yγ. Every element

in the cosets Zγδ/Yγ has order dividing three and Gγ/Qγ is transitive on the non-identity

cosets of Zγ/Yγ by Lemma 2.15 so every element in each coset of Zγ/Yγ has order dividing

three. Hence Zα and Zβ are elementary abelian. Furthermore Zα 6 Qβ and Zβ = Z(Qβ)

by (iii) so [Zα, Zβ] = 1. Thus ZαZβ is elementary abelian.

(v) We have seen that ZαZβ = Qα ∩ Qβ is elementary abelian. Let γ, δ = α, β.

The elements of the cosets of ZγZδ/Zγ have order dividing three. We have that Gγ/Qγ∼=

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GL2(3) acts transitively on the non-trivial elements of Qγ/Zγ and therefore the elements

of every coset of Zγ in Qγ have order dividing three. Hence Qγ has exponent three.

Lemma 6.2. (i) If u ∈ Qα\Qβ and v ∈ Qβ\Qα, [v, u, u] 6= 1 6= [v, u, v].

(ii) If z ∈ Sαβ has order three, then z ∈ Qα ∪Qβ.

(iii) Zαβ S ′αβ Qαβ and S ′

αβ 6= Zα, Zβ.

(iv) The cube of any element of order nine lies in the set ab|a ∈ Y #α , b ∈ Y

#β .

Proof. (i) Let u ∈ Qα\Qβ and v ∈ Qβ\Qα We show that [v, u, u] 6= 1, the proof for

[v, u, v] is similar. Suppose [v, u] ∈ Zα. Since u ∈ Qα and v /∈ Qα, we consider the action

of Qαv ∈ Gα/Qα# on Zαu ∈ Qα/Zα

#. Since [v, u] ∈ Zα, Qαv fixes Zαu. By Lemma

2.15, Qαv centralizes only the subspace ZαZβ/Zα = Qαβ/Zα. Thus u ∈ Qαβ which is

a contradiction since u /∈ Qβ. Now suppose [v, u, u] = 1. Then CQαβ(u) > 〈Zα, [v, u]〉

and |〈Zα, [v, u]〉| = 34. Therefore u commutes with Qαβ. So consider the action of

Qβu ∈ Gβ/Qβ# on Zβ/Yβ. Then u centralizes Qαβ > Zβ so Qβu acts trivially on Zβ/Yβ

which is not possible and therefore [v, u, u] 6= 1.

(ii) Let z ∈ Sαβ\(Qα ∪Qβ). Since Sαβ = QαQβ, we can write z = xy for x ∈ Qα\Qβ

and y ∈ Qβ\Qα. Both Qα and Qβ have exponent three so x and y both have order three

and hence z has order three or nine (no element of Sαβ can order greater than nine, if

w ∈ Sαβ had order 27 then Sαβ/Qα = 〈w〉Qα/Qα∼= 〈w〉/(〈w〉 ∩ Qα) and then 〈w〉 ∩ Qα

would contain elements of order nine). Suppose z has order three. We will perform a

calculation under this assumption for which we must first observe that S ′αβ 6 Qαβ since

Qαβ = Qα∩Qβ is normal in Sαβ and has index nine. Furthermore [S ′αβ, Qα] 6 [Qαβ, Qα] 6

Q′α = Yα and similarly [S ′

αβ, Qβ] 6 Yβ. Thus commutators of the form [y, x, x] and [y, x, y]

are central in Sαβ. Also notice that

y[y, x][y, x, y] = [y, x]y (∗)

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and

x[y, x][y, x, x] = [y, x]x (∗∗).

We do the following calculation.

1 = xyxyxy

= x2y([y, x]y)xy

= x2y(y[y, x][y, x, y])xy by (∗)

= x2y2([y, x]x)y[y, x, y]

= x2y2(x[y, x][y, x, x])y[y, x, y] by (∗∗)

= x2y2x([y, x]y)[y, x, x][y, x, y]

= x2y2x(y[y, x][y, x, y])[y, x, x][y, x, y] by (∗)

= [x, y][y, x][y, x, x][y, x, y]2

= [y, x, x][y, x, y]2.

So [y, x, x] = [y, x, y]. However [y, x, x] ∈ Yα, [y, x, y] ∈ Yβ and Yα ∩ Yβ = 1. So [y, x, x] =

[y, x, y] = 1 and this contradicts part (i). Hence z has order 32.

(iii) Let u ∈ Qα\Qβ and v ∈ Qβ\Qα then [u, v, u] 6= 1 and [u, v, v] 6= 1. In particular,

the derived subgroup, S ′αβ is not central in Qα or Qβ and so is not contained in Zα or Zβ

but it does contain Zαβ = YαYβ = Q′αQ

′β. Thus

S ′αβ = [Qαβ〈u, v〉, Qαβ〈u, v〉] 6 [Qαβ, Qαβ][Qα, Qα][Qβ, Qβ]〈[u, v]〉 = YαYβ〈[u, v]〉

has order 33. It follows that S ′αβ lies strictly between Qαβ and Zαβ but is not equal to Zα

or Zβ.

(iv) Let z = xy be an element of order nine as before (where x ∈ Qα\Qβ and y ∈

Qβ\Qα) then (xy)3 = [y, x, x][y, x, y]2. We have seen that S ′αβ 6 Qαβ and so [y, x, x] ∈

[S ′αβ, Qα] 6 Q′

α = Yα and [y, x, y] ∈ [S ′αβ, Qβ] 6 Q′

β = Yβ. Finally [y, x, x] 6= 1 6= [y, x, y]

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and so the result holds.

6.2 The Coset Graph

Recall that Γ = Γ(G,Gα, Gβ, Gαβ) is the coset graph of the amalgam where G = 〈Gα, Gβ〉.

Lemma 6.3. The coset graph Γ has valency four.

Proof. We have seen that the valency of the vertex α is given by [Gα : Gαβ] and the

valency of β by [Gβ : Gαβ]. We have seen also that Sαβ E Gαβ 6 NGα(Sαβ), NGβ(Sαβ). It

follows from the structure of Gα/Qα∼= Gβ/Qβ

∼= GL2(3) that both Gα and Gβ have four

Sylow 3-subgroups and so Gαβ has index at least four in both Gα and Gβ. By Lemma 6.2,

the normalizer of Sαβ preserves Qα ∪ Qβ as this is the set of all elements of order three

in Sαβ together with the identity. In particular this implies that Sαβ contains no other

subgroup isomorphic to Qα∼= Qβ. Therefore NGα(Sαβ) normalizes Qα and so must also

normalize Qβ. Thus NGα(Sαβ) 6 NG(Qβ) = Gβ (NG(Qα) = Gα and NG(Qβ) = Gβ follow

from Hypothesis B). Similarly NGβ(Sαβ) 6 NG(Qα) = Gα and so NGα(Sαβ) 6 Gβ ∩ Gα

and NGβ(Sαβ) 6 Gα ∩Gβ. Therefore Gαβ has index exactly four in both Gα and Gβ and

since G has two orbits on V (Γ) with representatives α and β, we see that every vertex in

Γ has valency four.

We extend our earlier notation to all groups Gλ where λ ∈ V (Γ).

Lemma 6.4. Let (λ, µ, ν) be a path of three distinct vertices in Γ. Then

(i) Sµν = QµQν;

(ii) Qµ = QλµQµν = Sλµ ∩ Sµν;

(iii) Qµν = ZµZν;

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(iv) Zµ = Qλµ ∩Qµν = ZλµZµν;

(v) Zµν = YµYν; and

(vi) Yµ = Zλµ ∩ Zµν.

Proof. (i) Since G is edge transitive, this is just Lemma 6.1 (i).

(ii) Suppose Qλµ = Qµν . Notice that Gµ = 〈Gλµ, Gµν〉. Lemma 6.2 implies that Qαβ is

characteristic in Sαβ and thus normal in Gαβ. So if Qλµ = Qµν then this group is normal

in Gµ. However this is a contradiction since Qµν/Zµ is a 1-space of Qµ/Zµ and is not

preserved by Gµ. Thus Qλµ 6= Qµν and both groups commute modulo Zµ and therefore

normalize each other. Hence Qµ = QλµQµν . Also Qµ 6 Sλµ ∩ Sµν = QλQµ ∩ QµQν and

Qλµ 6= Qµν so we have Qµ = Sλµ ∩ Sµν .

(iii) This is just Lemma 6.1 (iv).

(iv) Since Qµ = QλµQµν , Qλµ∩Qµν has order 33. Since Qλµ and Qµν are both abelian

groups, their intersection commutes with Qµ. Thus Zµ = Qλµ ∩Qµν . Suppose Zλµ = Zµν

then Z(Sλµ) = Z(Sµν) is normalized by 〈Gλµ, Gµν〉 = Gµ which leads to a contradiction

as before. Thus Zµ = ZλµZµν .

(v) This is just Lemma 6.1 (ii).

(vi) We have seen that Zλµ 6= Zµν and so this follows immediately.

Lemma 6.5. Let (α, β, γ, δ) be a path of vertices in Γ. Then

Z3∼= Sαβ/Qα

∼= Qβ/Qαβ∼= Qβγ/Zβ

∼= Zγ/Zβγ∼= Zγδ/Yγ

∼= Yδ.

Proof. This follows from repeated use of Lemma 6.4 and an isomorphism theorem.

Lemma 6.6. Given any fours group T 6 Gαβ such that Gαβ = SαβT and we can choose

involutions tα, tβ ∈ T such that Qαtα ∈ Z(Gα/Qα) and Qβtβ ∈ Z(Gβ/Qβ).

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Proof. Notice that for γ ∈ α, β, Gγ/Qγ∼= GL2(3) and GL2(3) has a central involution.

Notice also that this central involution normalizes every Sylow 3-subgroup in GL2(3).

Also, in GL2(3) every Sylow 3-subgroup is normalized by an elementary abelian group

of order four. Choose such a fours group T 6 Gγ such that Gαβ = SαβT . Of course T

contains three involutions and so we choose one of these, tγ, such that Qγtγ 6 Z(Gγ/Qγ).

This will of course hold in any choice of complement, T , such that Gαβ = SαβT .

Of course we can do a similar thing for any pair of adjacent vertices, γ, δ in Γ. Notice

however that any choice of tγ and tδ relies on having first chosen a fours group.

Lemma 6.7. Let γ, δ be adjacent vertices in Γ. The following hold.

(i) tγ inverts all non-trivial elements in Qγ/Qγδ, Qγδ/Zγ, Zγ/Zγδ, Zγδ/Yγ, Sγδ/Qδ,

Qγδ/Zδ, Zδ/Zγδ, and Yδ.

(ii) tγ centralizes Sγδ/Qγ, Yγ, Qδ/Qγδ and Zγδ/Yδ.

(iii) Given any choice of complement T ′ such that Gγδ = SγδT′, tγ is the unique element

in T ′ centralizing Yγ. Furthermore, tγ 6= tδ and T ′ = 〈tγ, tδ〉.

(iv) Gγ = CGγ (Yγ) 〈tδ〉.

Proof. We show the results hold for γ, δ = α, β and then the lemma follows by edge

transitivity.

Having chosen tα ∈ T 6 Gαβ as in Lemma 6.6 we consider the action of Qαtα on Qα.

Since Qα/Zα and Zα/Yα are natural Gα/Qα-modules, Qαtα acts on these spaces as the

matrix Diag(−1,−1) in GL2(3). So Qαtα inverts non-trivial elements. Now Yα E Gα and

CGα(Yα) has index at most two in Gα. Also Gα/Qα contains elements of order four which

square to Qαtα. So suppose tα doesn’t commute with Yα then 〈tα〉 is a complement to

CGα(Yα) in Gα. Then by correspondence Gα/Qα = (CGα(Yα)/Qα) : (〈tα〉Qα/Qα) however

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〈tα〉Qα/Qα = Z(Gα/Qα) and so we have a direct product Gα/Qα = CGα(Yα)/Qα ×

〈tα〉Qα/Qα which is not possible.

Consider the following natural isomorphisms.

(i) Sαβ/Qα = QαQβ/Qα∼= Qβ/Qαβ;

(ii) Qα/Qαβ∼= QαQβ/Qβ = Sαβ/Qβ;

(iii) Qαβ/Zα = ZαZβ/Zα∼= Zβ/Zαβ;

(iv) Zα/Zαβ∼= ZαZβ/Zβ = Qαβ/Zβ;

(v) Zαβ/Yα = YαYβ/Yα∼= Yβ/(Yα ∩ Yβ) = Yβ;

(vi) Yα = Yα/(Yα ∩ Yβ) ∼= YαYβ/Yβ = Zαβ/Yβ.

By Lemma 1.2 we see equivalent actions of Qαtα (and of Qβtβ) on naturally isomorphic

groups. This is to say Qαtα inverts Qα/Qαβ∼= Sαβ/Qβ, Qαβ/Zα

∼= Zβ/Zαβ, Zα/Zαβ∼=

Qαβ/Zβ and Zαβ/Yα∼= Yβ and centralizes Sαβ/Qα

∼= Qβ/Qαβ and Yα∼= Zαβ/Yβ. So (i)

and (ii) hold.

Recall tα was chosen to lie inside a fours group T 6 Gαβ and such that Qαtα ∈

Z(Gα/Qα) and tβ was chosen such that Qβtβ ∈ Z(Gβ/Qβ). Notice that tβ must centralize

Yβ whereas we know tα inverts Yβ and so tα 6= tβ and therefore T = 〈tα, tβ〉. Hence

Gβ = CGβ(Yβ)〈tα〉 and in the same way tβ must invert Yα and so Gα = CGα(Yα)〈tβ〉.

Thus (iii) and (iv) hold.

For the remainder of this chapter we will be interested in the involutions tα ∈ Gα and

tβ ∈ Gβ so we will assume that we have made a fixed choice 〈tα, tβ〉 = T 6 Gαβ.

Lemma 6.8. T fixes a circuit in Γ.

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Proof. Consider the action of tα on the graph Γ. We chose tα inside Gαβ so it certainly

fixes the edge α, β in Γ. We have seen also that each vertex in Γ has valency four. Thus

tα acts on the set Γ(β)\α which has size three. By the orbit-stabilizer theorem tα fixes

an element of this set. Let γ be the fixed vertex. We repeat the argument to see tα fixing

some other neighbour of γ, δ say. By assumption G is a finite group and so Γ has a finite

number of vertices and so tα fixes some cycle in Γ, (α, β, γ, δ, . . . , α). Of course tβ also

fixes some cycle (a, β, . . . , α) and if ε is any vertex fixed by tα then ε · tβ = ε · tαtβ = ε · tβtα

and so tβ preserves the subgraph of Γ fixed by tα. In particular this means we can choose

the circuit (α, β, γ, δ, . . . , α) to be fixed by T .

Lemma 6.9. Let Θ = (α, β, γ, δ, . . . , α) be a path in Γ fixed by T . Then T is a comple-

ment to Sλµ in Gλµ for each pair of adjacent vertices λ, µ in Θ furthermore, for each

λ ∈ Θ if we choose tλ ∈ T as in Lemma 6.6 then tα = tν for each vertex ν such that

d(α, ν) is a multiple of three.

Proof. Since T fixes the path Θ then T is a subgroup of each edge stabilizer Gλµ and

is therefore a complement as claimed. We have seen how to choose tα ∈ T such that

tα ∈ Z(Gα/Qα). We have seen also that T = 〈tα, tβ〉 and that tα is the only choice of

involution from T which centralizes Yα. We choose tδ in T uniquely such that it centralizes

Yδ. By Lemma 6.5, we get the natural isomorphism Sαβ/Qα∼= Yδ and by Lemma 1.2, we

see that tα centralizes Yδ and so tα = tδ. This argument holds for any vertex at distance

three from α and so for any vertex at distance a multiple of three.

We illustrate the way tα acts in Figure 6.1 which shows how tα acts equivalently on

naturally isomorphic groups. If tα is acting trivially on a section we label with 1 and if it

is inverting a section we label with −1.

Given a path (α, α+ 1, α+ 2, . . .) of vertices in Γ fixed by T we use Lemma 6.9 to see

tα = tα+3n for all n ∈ Z.

85

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Figure 6.1: Some partial subgroup structure inside G and the action of tα on sections.

86

Lemma 6.10. The elements of T# are G-conjugate.

Proof. Let T 6 S ∈ Syl2(Gβ) then SQβ/Qβ∼= S. So we consider the images of tαQβ and

tβQβ = tα+1Qβ in Gβ/Qβ∼= GL2(3). Thus we identify

tβQβ ∼

(−1 0

0 −1

), tαQβ ∼

(−1 0

0 1

).

Consider x ∼(

0 1

−1 0

). Then 〈txα〉Qβ/Qβ = 〈tαtβ〉Qβ/Qβ = 〈tα+2〉Qβ/Qβ. So by Sylow’s

Theorem, tα is conjugate to tα+2. By the same arguments tβ is conjugate to tα+3 = tα.

Lemma 6.11. G is locally 7-arc transitive on Γ.

Proof. See [7, p73 and p98] or Lemma A.4.

Lemma 6.12. For each vertex γ ∈ Γ we can choose an involution tγ ∈ Gγ such that

CGγ (tγ) has shape 3 : GL2(3).

Proof. Define tγ ∈ Gγ, as in Lemma 6.6, to be an involution such that tγQγ ∈ Z(Gγ/Qγ).

Then tγ is in the centre of a Sylow 2-subgroup of Gγ and also in some complement to Sγδ

in Gγδ for each δ ∈ Γ(γ). We have seen that the element tγ inverts the quotients Qγ/Zγ

and Zγ/Yγ and centralizes Sγδ/Qγ and Yγ. By Lemma 2.16, CSγδ(tγ) = 〈Yγ, b〉 for some

b ∈ Sγδ\Qγ has order 32 and is elementary abelian and CQγ (tγ) = Yγ. Thus CGγ (tγ) has

order 2432 and by an isomorphism theorem,

CGγ (tγ)/Yγ = CGγ (tγ)/(CGγ (tγ) ∩Qγ) ∼= CGγ (tγ)Qγ/Qγ 6 Gγ/Qγ∼= GL2(3).

So it follows from the order that CGγ (tγ)/Yγ∼= GL2(3) and so CGγ (tγ) ∼ 3.GL2(3).

Finally, since the Sylow 3-subgroup is elementary abelian and therefore splits over Yγ,

CGγ (tγ) splits over Yγ by Theorem 1.8.

Lemma 6.13. O2(CGγ (tγ))∼= Q8 is transitive on Γ(γ).

87

Proof. By Lemma 6.12, GL2(3) ∼= CGα(tα)/Yα. Since Z3∼= Yα E CGα(tα), the centralizer

of Yα in CGα(tα) has index at most two and since tβ inverts Yα the index is exactly

two. By Lemma 2.13, GL2(3) has a unique index two subgroup and so it follows that

CCGα (tα)(Yα)/Yα∼= SL2(3). Also by Lemma 2.13, O2(CGα(tα)/Yα) ∼= Q8 is contained

in CCGα (tα)(Yα)/Yα. Let L be the preimage in CGα(tα) of O2(CGα(tα)/Yα) then L ∼=

3 × Q8 and so it follows that CGα(tα) has a normal subgroup isomorphic to Q8 and by

correspondence it is the largest normal 2-subgroup. Now, this group of order eight acts on

Γ(α). If any element of order four fixes a vertex in Γ(α) then we have an edge stabilizer

containing elements of order four which cannot happen. Thus the action is transitive.

Lemma 6.14. Let (α, α + 1, α + 2, α + 3) be a path of vertices in Γ fixed by tα. Let

X := O2(CGα(tα)) and Y := O2(CGα+3(tα+3)). Then X 6= Y and if [X, Y ] = 1 then Γ is

a Moufang hexagon with 728 vertices. In particular, if [X, Y ] = 1 then |G| = 4245696.

Proof. Suppose X = Y . Let γ ∈ Γ(α). Since X fixes α and α + 3 and is transitive on

Γ(α), we can find x ∈ X such that γ · x has distance two from α + 3. But G preserves

distance so α and α + 3 lie on a circuit of length six. However this is not possible since

G acts 7-arc transitively on Γ.

So suppose [X, Y ] = 1. Since X 6= Y , we can choose p ∈ Y \X and consider the vertex

α ·p 6= α. This vertex has distance three from a+3 and therefore distance six from α and

so we will name it α+6 = α ·p. Let q ∈ X then (α+6) ·q = α ·pq = α ·qp = α ·p = α+6.

ThusX fixes α and α+6 and moves α+3 and so arguing as before we see that α, α+3, α+6

lie on a circuit of length 12. However G = 〈Gα, Gβ〉 acts 7-arc transitively on Γ and so

every vertex is contained in a 12-cycle. Now 7-arc transitivity ensures every vertex is in

a 12-cycle and so no vertex can be more than six vertices away from any other (so Γ has

diameter 6 and girth 12 so is a generalized hexagon). It follows from 7-arc transitivity

that the number of vertices at distance 0− 6 from α are 1, 4, 12, 36, 108, 324, 243 and so Γ

has 728 vertices. Since |Gα| = 2436 = 11664 and there are 728/2 = 364 images of α under

88

G, we get |G| = 11664× 364 = 4245696 as required. Finally let (α0, α1, . . . , α6) be a path

of seven vertices in Γ. Since G acts 7-arc transitively on Γ, the stabilizer of the path is

transitive on Γ(α6)\α5 and three divides the order of the path stabilizer. Therefore an

element of order three must necessarily fix⋃

16i65 Γ(αi) and so Γ satisfies the Moufang

condition.

Theorem 6.15. Let α, α + 1, α + 2, α + 3 be a path in Γ fixed by tα. If O2(CGα(tα))

commutes with O2(CGα+3(tα)) then G ∼= G2(3).

Proof. By Theorem 4.13 and Lemma 6.14 the coset graph Γ is a Moufang hexagon with 728

vertices and is uniquely defined. Notice that the stabilizer of a path of length six contains a

group of order three conjugate to Yγ for γ ∈ α, β. So the group 〈Yγ|γ ∈ V (Γ)〉 6 G†∩G

(where G† is defined in 4.13). To see that 〈Yγ|γ ∈ V (Γ)〉 = G we recall Lemma 6.4

which can be used to show that each Sylow 3-subgroup is contained in 〈Yγ|γ ∈ V (Γ)〉.

We have also seen that both Gα and Gβ are generated by their Sylow 3-subgroups. Thus

G = 〈Gα, Gβ〉 6 〈Yγ|γ ∈ V (Γ)〉 6 G. Thus G 6 G† ∼= G2(3). Now it follows from the

order of G that G = G† ∼= G2(3).

6.3 Controlling the 3-Local Structure

We have our group G containing a faithful completion of an amalgam (A1, A2, A12) =

(Gα, Gβ, Gαβ) of type G2(3). First we will find all conjugacy classes of elements of order

three. In particular we will identify an element of order three which acts fixed-point-freely

on a section of the centralizer of an involution and apply Lemma 1.25. We will use this to

force two subgroups to commute and then invoke Theorem 6.15 to recognize G2(3) as the

completion contained in G. Finally we will show that this containment cannot be proper.

First we require further notation.

89

For γ ∈ α, β, set Xγ = [Zγ, tγ] and choose elements yγ such that 〈yγ〉 = Yγ. As in

Lemmas 6.6 and 6.7 we fix T = 〈tα, tβ〉. Set tαβ := tαtβ.

We saw in Lemma 6.2 that the derived subgroup of Sαβ lies strictly between Qαβ and

Zαβ but is not equal to Zα or Zβ. SinceQαβ is elementary abelian, there are four subgroups

in Sαβ lying strictly between Qαβ and Zαβ three of which are normalized by Gαβ, Zα, Zβ

and S ′αβ. Thus the fourth is necessarily normalized by Gαβ too. Let A and B be the

two subgroups distinct from Zα and Zβ and we will not need to distinguish which is the

derived subgroup of Sαβ. We will however choose an element from each. Now by Lemma

6.7, both tα and tβ invert non-trivial elements in Qαβ/Zαβ and both A/Zαβ and B/Zαβ

are 1-dimensional subspaces so are inverted also. Thus tαβ centralizes them. In particular

this means we can choose a ∈ A\Zαβ and b ∈ B\Zαβ such that [a, tαβ] = [b, tαβ] = 1 (see

Lemma 2.16).

Lemma 6.16. Let γ, δ = α, β. Then Lγ := CG(yγ), has index two in Gγ, [Gγδ :

Lγ ∩Gδ] = 2, Lγ ∩ Lδ = Sγδ and Lγ/Qγ∼= SL2(3).

Proof. For each vertex δ we choose tδ as in Lemma 6.6. By Lemma 6.7, CG(yγ) = Lγ

has index two in NG(Yγ) = Gγ. Suppose Gγδ 6 Lγ then Sγδ char Gγδ E Lγ and so

Sγδ char Lγ. However, Lγ is normal in Gγ and so we have Sγδ E Gγ. This contradicts

that Qγ = O3(Gγ). Thus Gγδ Lγ and so LγGγδ = Gγ. By an isomorphism theorem,

Gγδ/(Lγ ∩ Gγδ) ∼= LγGγδ/Lγ = Gγ/Lγ and so Lγ ∩ Gγδ has index two in Gγδ. In the

same way Lδ ∩ Gγδ has index two in Gγδ. By Lemma 6.7, tγ /∈ Lδ but tγ ∈ Lγ ∩ Gγδ so

Lγ ∩Gγδ 6= Lδ ∩Gγδ. However Lγ ∩ Lδ 6 Gγ ∩Gδ = Gγδ so Lγ ∩ Lδ = Lγ ∩ Lδ ∩Gγδ has

index four in Gγδ. Thus Lγ ∩ Lδ = Sγδ. Finally, Lγ/Qγ has index 2 in Gγ/Qγ∼= GL2(3)

and by Lemma 2.13 (iii), GL2(3) has a unique subgroup at index two and therefore

Lγ/Qγ∼= SL2(3).

Lemma 6.17. Xγ = [Zγ, tγ] has order 32, is a complement to Yγ in Zγ and Xγ = 1 ∪

ygδ |g ∈ Gγ for γ, δ = α, β.

90

Proof. By coprime action on an abelian group, Zα = CZα(tα)×Xα. Recall from Lemma

6.7 that tα centralizes Yα and inverts Zα/Yα and Yβ. So Yα 6 CZα(tα). However if

this containment is strict then CZα(tα)/Yα is a proper subspace of Zα/Yα that tα acts

trivially on which is not possible. So CZα(tα) = Yα. Moreover tα inverts Yβ and so

yβ = [yβ, tβ] ∈ Xα. Now Zα is central in Qα, and Qα〈tα〉 E Gα by correspondence so we

see that [Zα, tα] = [Zα, Qα〈tα〉] EGα. Therefore ygδ |g ∈ Gγ ⊆ Xα. Lemma 6.16 implies

that CGα(Yβ) = Gα ∩ Lβ has index eight in Gα and so we see eight conjugates of yβ in

Xα = 1 ∪ ygβ|g ∈ Gα. A similar argument holds for Xβ.

Lemma 6.18. Zαβ contains three different G-conjugacy classes of elements of order three

and NG(Zαβ) = NG(Sαβ) = Gαβ.

Proof. Recall tα inverts yβ and tβ inverts yα so T = 〈tα, tβ〉 is transitive on the set

Ω := yαyβ, y2αyβ, yαy

2β, y

2αy

2β. Recall also that by Lemma 6.2 (iv), Sαβ contains elements

of order nine and the cube of any such element lies in Ω. Since Zαβ = Z(Sαβ), we can

apply Burnside’s Lemma 1.9 to see that elements of Zαβ are conjugate in G only if they

are conjugate in NG(Sαβ). However any element of NG(Sαβ) maps elements of order nine

to elements of order nine so no element of Ω can be conjugate to yα or yβ in G. Also

we have that yα and yβ are both G-conjugate to their inverses but not to each other

since by Hypothesis B, Gα is not conjugate to Gβ. Hence there are three G-conjugacy

classes of elements of order three in Zαβ. In particular, any element of G normalizing Zαβ

normalizes Yα and Yβ and so lies in Gαβ. Thus NG(Zαβ) = Gαβ and since Zαβ = Z(Sαβ),

NG(Sαβ) = Gαβ.

Recall we chose a ∈ A\Zαβ and b ∈ A\Zαβ such that [a, tαβ] = [b, tαβ] = 1.

Lemma 6.19. CSαβ(a) = CSαβ

(b) = Qαβ, |CGαβ(a)| = |CGαβ

(b)| = 342 and |NGαβ(〈a〉)| =

|NGαβ(〈b〉)| = 3422.

91

Proof. We show the results for a and similar arguments work for b. By Lemma 6.7,

tα inverts the quotient Qαβ/Zαβ and so inverts the 1-dimensional subspace A/Zαβ. By

Lemma 2.16, we can choose a′ ∈ A\Zαβ such that tα inverts a′. Consider CSαβ(a′) > Qαβ

and suppose this containment is strict. If Qα 6 CSαβ(a′) then a′ ∈ Z(Qα) ∩ A = Zαβ

which would contradict our choice of a′. So Qα CSαβ(a′) and similarly Qβ CSαβ

(a′).

So choose x ∈ CSαβ(a′)\Qαβ. By Lemma 6.2, x has order nine. Since tα inverts a′,

tα normalizes CSαβ(a′) so xtα ∈ CSαβ

(a′). Now Qαx ∈ Sαβ/Qα is centralized by tα so

xtαx−1 ∈ Qα ∩ CSαβ(a′) = Qαβ. Also, Qβx ∈ Sαβ/Qβ which is inverted by tα (by Lemma

6.7) so xtαx ∈ Qβ ∩ CSαβ(a′) = Qαβ. Therefore (xtαx−1)−1xtαx = x2 ∈ Qαβ and it follows

that x ∈ Qαβ which gives a contradiction. Thus CSαβ(a′) = Qαβ which has index nine

in Sαβ. Now we observe nine conjugates of a′ in A and since a′ is inverted by tα we

see a further nine. Since A\Zαβ is a set of order 18, each element is a conjugate of a′.

Thus CSαβ(a) = Qαβ. Finally since a is centralized by an involution and a′ is inverted

by an involution and the two elements are conjugate, it follows that |CGαβ(a)| = 342 and

|NGαβ(〈a〉)| = 3422.

Lemma 6.20. CG(Qαβ) = Qαβ, NG(Qαβ) = Gαβ.

Proof. Firstly by Lemmas 6.16 and 6.19,

CG(Qαβ) 6 CG(yα) ∩ CG(yβ) ∩ CG(a) = Lα ∩ Lβ ∩ CG(a) = Sαβ ∩ CG(a) = Qαβ,

and since Qαβ is abelian, CG(Qαβ) = Qαβ. Now consider O3(NG(Qαβ)). We have

NG(Qαβ) > Gαβ and since Sαβ is a Sylow 3-subgroup of G, Sαβ ∈ Syl3(NG(Qαβ)) and so

Qαβ 6 O3(NG(Qαβ)) 6 Sαβ. Suppose O3(NG(Qαβ)) = Qα, then NG(Qαβ) 6 NG(Qα) = Gα

and so NG(Qαβ) = NGα(Qαβ). However Qαβ/Zα is 1-subspace of the natural Gα/Qα-

module, Qα/Zα, and the stabilizer of a 1-space has order 12. But then NG(Qαβ) =

NGα(Qαβ) = Gαβ as claimed. Similarly if O3(NG(Qαβ)) = Qβ. So suppose instead that

O3(NG(Qαβ)) = Qαβ. In this case we can apply a result which followed from a theo-

92

rem of McLaughlin, namely Lemma 4.9. We apply it with M = NG(Qαβ), V = Qαβ,

S = Qα/Qαβ and T = Qβ/Qαβ. This implies that 〈QNG(Qαβ)α 〉/Qαβ

∼= SL2(3) and

〈QNG(Qαβ)

β 〉/Qαβ∼= SL2(3) intersect trivially and commute. This means that 〈QNG(Qαβ)

β 〉

normalizes Qα and then 〈QNG(Qαβ)

β 〉 6 Gα ∩ NG(Qαβ) = Gαβ. However, this implies

〈QNG(Qαβ)

β 〉/Qαβ 6 Gαβ/Qαβ∼= 32 : (2 × 2), a contradiction. Therefore we get that

O3(NG(Qαβ)) = Sαβ and NG(Qαβ) 6 NG(Sαβ) = Gαβ by Lemma 6.18.

Lemma 6.21. Qαβ ∈ Syl3(CG(a)) ∩ Syl3(CG(b)) and a is not G-conjugate to b.

Proof. We prove Qαβ ∈ Syl3(C) where C := CG(a) and the same proof will work for

CG(b). Suppose S ∈ Syl3(C) such that Qαβ < S then there exists some S0 6 S such that

Qαβ C S0. Then Qαβ < S0 6 NG(Qαβ) = Gαβ and S0 is a 3-group. Thus Qαβ < S0 6

Sαβ ∩ C = Qαβ by Lemma 6.19. So Qαβ ∈ Syl3(C).

Suppose now that there is g ∈ G such that ag = b. Then Qαβ, Qgαβ ∈ Syl3(CG(b)) so

there exists some h ∈ CG(b) such that Qghαβ = Qαβ. But then gh ∈ Gαβ and agh = b.

However Gαβ normalizes A and B and so agh = b ∈ A ∩ B = Zαβ which contradicts our

choice of a and b.

Lemma 6.22. Let d ∈ a, b then CG(d) 6 Gαβ. In particular |CG(d)| = |CGαβ(d)| = 342

and |NG(〈d〉)| = |NGαβ(〈d〉)| = 3422.

Proof. We prove C := CG(a) 6 Gαβ and the same proof will work for CG(b). Let R :=

O3′(C) and assume R 6= 1. We see that R is normalized by Xα = 1 ∪ yβGα as a

and Xα commute. By coprime action, R := 〈CR(y)|y ∈ Zα\1〉 = 〈CR(ygβ)|g ∈ Gα〉 by

Lemma 6.17. Consider CG(yβ)g ∩R = Lβg ∩R for any g ∈ Gα. Since Lβ is a 2, 3-group

and R has order prime to 3, this is a 2-subgroup of Lβg. Recall Lβ/Qβ

∼= SL2(3) and

so any 2-subgroup of Lβ is isomorphic to a subgroup of Q8. Now R E C and Qαβ 6 C

so Qαβ normalizes R. Also Qαβ 6 Qα = Qgα 6 Lβ

g so Qαβ normalizes Lgβ too. Thus

Qαβ normalizes Lgβ ∩R. But this is isomorphic to a subgroup of Q8 and so any subgroup

93

normalizing it centralizes the unique involution. However, this contradicts Lemma 6.20

which says CG(Qαβ) = Qαβ. Hence R = 1.

Now, because of Lemmas 6.19 and 6.20, we can apply a corollary to the theorem of

Smith and Tyrer, Corollary 4.6, to C withQαβ as its Sylow 3-subgroup. If C were 3-soluble

of length one then, since O3′(C) = 1, we would have that Qαβ E C 6 NG(Qαβ) = Gαβ

and we would be done. So let us assume not. Let N := O3(C) then C/N has order at

least 32. Recall that a was chosen to commute with the involution tαβ and that tαβ = tαtβ

inverts both Yα and Yβ. Thus tαβ ∈ N and so yα = [yα, tα], yβ = [yβ, tα] ∈ [Qαβ, tαβ] ⊆ N

and so Zαβ ∈ Syl3(N). By Lemma 6.18, NG(Zαβ) = Gαβ so NN(Zαβ) 6 Gαβ ∩C and so it

follows from Lemma 6.19 that 〈Zαβ, tαβ〉 = NN(Zαβ). Thus we apply Theorem 4.5 to see

that N is either 3-soluble of length one or O3(N) 6= N . If N has a normal 3′-subgroup

then so does C. So if N is 3-soluble of length one then Zαβ E N and so N = 〈Zαβ, tαβ〉

and then |C| = 342 and we would be done. So suppose O3(N) 6= N . Arguing as before,

tαβ ∈ O3(N) and Zαβ normalizes O3(N) and so it follows that Zαβ = [Zαβ, tαβ] 6 O3(N)

which contradicts that O3(N) 6= N . Hence C 6 Gαβ has order 342.

The remaining results now follow immediately from Lemma 6.19.

Lemma 6.23. There are five conjugacy classes of elements of order 3 in G and for γ ∈

α, β, Zγ contains only elements from the conjugacy classes with representatives yα, yβ

and yαyβ.

Proof. We have seen at least five conjugacy classes of elements of order three in G (with

representatives yα, yβ, yαyβ, a and b) and that any element of order three in a Sylow

3-subgroup, Sαβ, lies in Qα ∪Qβ. Thus it remains to prove that each of the 404 elements

of order three in Sαβ is in one of the known conjugacy classes.

First we look for conjugates of yα in Sαβ. By Lemma 6.16, CG(yα) ∩ Gβ = Lα ∩ Gβ

has index 8 in Gβ so we see 8 conjugates yα in Zβ. Of these 8, 6 lie outside Yα but inside

94

Qα. Since Zβ is normal in Gαβ but not in Gα, the normalizer in Gα of Zβ has index 4 in

Gα. Thus we see 4 conjugates of each of the 6 we saw already and so we have at least

4× 6 + 2 = 26 conjugates of yα in Sαβ and similarly at least 26 conjugates of yβ.

In particular, at this point we have counted all elements of order three in Zα. We

counted 16 conjugates of y, 2 conjugates of yα and 8 of yβ.

Now we count conjugates of yαyβ =: y. In Lemma 6.18 we saw four conjugates of y

and so it follows that CGαβ(y) = Sαβ. We saw also that NG(Zαβ) = Gαβ. If y commutes

with some 2-element, u, in Gα then u normalizes Zαβ and so must be in Gαβ. However,

no such element in Gαβ commutes with y and so no 2-element in Gα (or in Gβ by a similar

argument) commutes with y. Thus we see 16 conjugates of y in Zα and 16 in Zβ of which

4 are shared in Zαβ. Of the 16 in Zα, 12 lie outside Zαβ and since Zα is not normal in

Gβ, each has four conjugates lying inside Qβ giving 64 conjugates in Qβ and similarly 64

in Qβ of which the 28 inside Zα ∪ Zβ are shared giving 100 in total.

Finally we count conjugates of a and conjugates of b. We have seen that CG(a) = CG(b)

has order 342 and is contained in Gαβ. Therefore we see 72 conjugates in Gα and 72 in

Gβ of which the 18 in Qαβ are shared. So we have 126 conjugates of a and 126 of b.

Thus we have counted 26 + 26 + 100 + 126 + 126 = 404 elements of order three and so

can conclude that G has exactly five such conjugacy classes.

Notice that all conjugation we observed in this proof occurred within Gα ∪ Gβ. So if

any two elements of order three in Sαβ are G-conjugate then they are 〈Gα, Gβ〉-conjugate.

Thus we have the following corollary.

Corollary 6.24. Let x and y be elements of order three in G = 〈Gα, Gβ〉. Suppose x and

y are G-conjugate then there exists g ∈ G such that xg = y.

Proof. Since x and y are elements of order three in G and G is transitive on its Sylow

3-subgroups, we can assume x and y lie in the same Sylow 3-subgroup of G. So let us

95

assume x, y ∈ Sαβ. As x and y are elements of order three, x, y ∈ Qα ∪ Qβ. In Lemma

6.23 we counted all such elements of order three and observed that all conjugacy occurred

within 〈Gα, Gβ〉 and so we can find g ∈ G such that xg = y.

Lemma 6.25. NG(〈a, b〉) 6 NG(〈a〉) ∩ NG(〈b〉) 6 Gαβ and Qαβ is the unique Sylow 3-

subgroup of NG(〈a, b〉).

Proof. By Lemma 6.21, a is not conjugate to b so consider the set Ω := ab, a2b2, a2b, ab2 ⊆

〈a, b〉. Each element in Ω lies in Qαβ but not in A or B and so must lie in Zα or

Zβ. However by Lemma 6.23, no element of Ω can be conjugate to either a or b.

Thus any element normalizing 〈a, b〉 must normalize 〈a〉 and 〈b〉. By Lemma 6.22,

NG(〈a〉) 6 Gαβ and NG(〈b〉) 6 Gαβ and so NG(〈a, b〉) 6 Gαβ. By Lemma 6.19 and

Lemma 6.22, Qαβ 6 NG(〈a〉) 6 Gαβ and NG(〈a〉) has order 3422. Therefore Qαβ is a

Sylow 3-subgroup of NG(〈a〉) 6 Gαβ and is necessarily normal.

6.4 Controlling the Centralizer of an Involution

Lemma 6.26. Let S ∈ Syl3(CG(tαβ)). Then S has order 32 and |NCG(tαβ)(S)/CCG(tαβ)(S)| 6

2.

Proof. Consider CQαβ(tαβ). By our choice of a and b, 〈a, b〉 6 CQαβ

(tαβ). By Lemma

6.7, tαβ = tαtβ inverts every element in Zαβ = YαYβ. Therefore CQαβ(tαβ) has order nine.

Suppose 〈a, b〉 is not a Sylow 3-subgroup of CG(tαβ) and let S ∈ Syl3(CG(tαβ)) and suppose

〈a, b〉 < S then 〈a, b〉 C NS(〈a, b〉) 6 Qαβ ∩ CG(tαβ) by Lemma 6.25. However CQαβ(tαβ)

has order nine so 〈a, b〉 is a Sylow 3-subgroup. Furthermore if we set S = 〈a, b〉 then

by Lemma 6.25, NCG(tαβ)(S) 6 NCG(tαβ)(〈a〉) and since 〈a〉 is a group of order three it is

immediate that |NCG(tαβ)(〈a〉)/CCG(tαβ)(〈a〉)| 6 2 and so |NCG(tαβ)(S)/CCG(tαβ)(S)| 6 2.

Recall that tα, tβ and tαβ are all conjugate in G by Lemma 6.10. So for any conjugate

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of tα, the centralizer in G contains an element from each of the four conjugacy classes of

elements of order three with representatives yα, yβ, a, and b. From now on set t := tα

and H := CG(t). Let (α, β = α + 1, α + 2, . . .) be a path in Γ fixed by 〈tα, tβ〉. We

have that tα = tα+3 and so CGα(t) ∼= CGα+3(t). Let Pα = O2(CGα(t)) ∼= Q8 and Pα+3 =

O2(CGα+3(t))∼= Q8.

Lemma 6.27. V := 〈yα, yα+3〉 is a Sylow 3-subgroup of H.

Proof. First observe that since t = tα = tα+3, Yα, Yα+3 6 H. We use Lemma 6.4 to see

that YαYα+1 = Zα,α+1 and Yα+1Yα+2 = Zα+1,α+2 and Yα+2Yα+3 = Zα+2,α+3. Also we see

that Zα,α+1Zα+1,α+2 = Zα+1 and Zα+1,α+2Zα+2,α+3 = Zα+2. Finally we see by Lemma 6.4,

that Zα+1Zα+2 = Qα+1,α+2 and so YαYα+1Yα+2Yα+3 = Qα+1,α+2. In particular Yα 6= Yα+3

since Qα+1,α+2 has order 34. Also YαYα+3 = 〈yα, yα+3〉 = V has order 32 and by Lemma

6.26 is a Sylow 3-subgroup of H.

Recall that the involution tβ ∈ H inverts Yα (by Lemma 6.7) and since tβ = tα+4

where α+ 4 ∈ Γ(α+ 3), tβ inverts Yα+3 also.

Lemma 6.28. O3′(H) = 〈Pα, Pα+3〉 is a 2-group and O3′(H)/〈t〉 admits a fixed-point-free

automorphism of order three.

Proof. By Lemma 6.26, we have |NH(V )/CH(V )| 6 2 but since tβ inverts V , the index

is exactly 2. Since V is abelian, we can apply the theorem of Smith-Tyrer, Theorem

4.5, to H. Clearly tβ ∈ O3(H) and tβ inverts yα and yα+3 and so yα = [yα, tβ], yα+3 =

[yα+3, tβ] ∈ O3(H). Therefore O3(H) contains a Sylow 3-subgroup and so O3(H) = H.

Thus H must be 3-soluble of length one. Let R = O3′(H). Choose x ∈ S such that x

is a conjugate of a. Then CG(x) has order 342 and so, as t ∈ CH(x), |CH(x)| 6 18. In

particular, x acts fixed-point-freely on R := R/〈t〉. By Corollary 1.22, R is nilpotent.

Let p be any prime dividing |R| and let P be a Sylow p-subgroup of R. By coprime

action, P = 〈CP (v)|v ∈ V #〉. Since the centralizer of every non-trivial element of V is a

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2, 3-group, P is trivial unless p = 2. Thus R = 〈CR(v)|v ∈ V #〉 is a 2-group. We have

V = 〈yα, yα+3〉 so we consider CR(v) for each v ∈ V #. Firstly CR(yα) 6 Lα ∩ CGα(t) and

Lα ∩CGα(t) has order at most 2332. Also, CR(yα) is a 2-group so has order at most eight

and so CR(yα) 6 Pα and similarly CR(yα+3) 6 Pα+3. The remaining elements in V are

conjugates of a and b and so centralize just 〈t〉 in R. Therefore R 6 〈Pα, Pα+3〉. Finally,

since H is 3-soluble of length 1, Pα normalizes RV . Also, V normalizes Pα so by coprime

action, Pα = [Pα, V ]CPα(V ) = [Pα, V ]〈tα〉 6 RV and then Pα 6 R. Similarly Pα+3 6 R.

Thus R = 〈Pα, Pα+3〉 is a 2-group.

Lemma 6.29. PαPα+3 = O3′(H) ∼= 21+4+ and 21+4

+ : 32 : 2 ∼ CG(t) 6 〈Gα, Gβ〉.

Proof. Applying Lemma 1.25 to R = O3′(H)/〈t〉 = 〈Pα, Pα+3〉/〈t〉 forces R to be special

of order 26 or elementary abelian of order 24. Suppose the former then R has order 27

and has a characteristic subgroup, Z say, of order 23. Recall V = 〈yα, yα+3〉 is a Sylow

3-subgroup of H and so normalizes Z. However it is easy to check that the automorphism

group of any group of order eight does not contain a group of order nine. Thus CV (Z)

must be non-trivial and so some element of order three in V = 〈yα, yα+3〉 commutes with

a group of order eight, Z. The only possibilities are yα, yα+3 and their inverses. Hence

Z 6 Lα or Z 6 Lα+3 and so Z = Pα or Z = Pα+3 and then R = 〈Pα, Pα+3〉 has order

less than 27. So we must have that R is elementary abelian. Since R contains quaternion

subgroups, R is non-abelian and since Z(R) is normalized by an element of order three,

it must have order two or eight. However, the order cannot be eight since no abelian

group of order eight has an automorphism of order three. Therefore R is extraspecial and

by Corollary 1.17 and Lemma 1.18, R is the central product of Pα and Pα+3. We have

seen already using the theorem of Smith and Tyrer that H is 3-soluble of length 1 and so

R E RV E H. By a Frattini argument H = RV NH(V ) = R〈V, tβ〉. Thus H 6 〈Gα, Gβ〉.

The next result now follows immediately from Theorem 6.15.

98

Lemma 6.30. G contains a subgroup isomorphic to G2(3).

We are left with the possibility that G > G ∼= G2(3) however we know enough about

the structure of G now to rule this out. The following argument relies on the result Lemma

6.23 which implies that G has five conjugacy classes of elements of order three, moreover

if we look only at the elements of order three in G (or any conjugate of G in G) then we

see the same conjugacy classes. So if some element of order three, x say, lies in G and in

some conjugate G1 then CG(x) ∼= CG1(x).

Lemma 6.31. G ∼= G2(3).

Proof. SupposeG2(3) ∼= G = 〈Gα, Gβ〉 < G. By a Frattini argument, NG(G) = GαβG = G

so G is self-normalizing. Suppose Gg is a distinct conjugate of G for some g ∈ G. Suppose

further that |G∩Gg| is even. Since G2(3) contains just one conjugacy class of involutions

(see, for example, [6]), we can, without loss of generality, assume t ∈ G ∩ Gg. But then

t = sg for some s ∈ G. Since t and s are involutions in G, they are conjugate in G,

say sgh = th = s for some h ∈ G. However this implies gh ∈ CG(s) = CG(th) 6 G and

then g ∈ G which is a contradiction. Thus G is a strongly embedded subgroup of G. Let

x ∈ CG(t)\1. If x has order two then x is a conjugate of t and so x cannot lie in any

other conjugate of G. If x has order three then x is a conjugate of yα, yβ, a or b. In any

case CG(x) 6 G. Suppose x lies in some other conjugate of G, x ∈ G ∩ Gh say. Thus

x = yh for some y ∈ G. By Corollary 6.24, there is some g ∈ G such that yhg = xg = y.

Hence hg ∈ CG(y) = CG(xg) 6 G and so h ∈ G. Hence CG(t) intersects trivially with

any conjugate of G. Thus we can apply Lemma 1.30 to get that G contains an abelian

subgroup, K say, of order 34.7.13. In particular G contains an abelian subgroup, A say,

of order 34. Without loss of generality we assume A 6 Sαβ. If A 6 Qα and A 6 Qβ

then A = Qαβ and [Qαβ, K] = 1 which contradicts Lemma 6.20. So we may assume that

A 66 Qα. Then |A ∩Qα| = 33 and so A ∩Xα > 1. But then A contains a conjugate of yβ

99

and so K is contained in some subgroup of G conjugate to Gβ which is a contradiction.

So G can have no such abelian subgroup and hence G = G ∼= G2(3).

100

Chapter 7

An Amalgam of Type PSp4(3)

A Sylow 3-subgroup of PSp4(3) is isomorphic to a Sylow 3-subgroup of Sym(9) which

has order 34, centre of order three and for which the Thompson subgroup is elementary

abelian subgroup of order 33. Given a Sylow 3-subgroup in PSp4(3), S say, with Thompson

subgroup Q, the amalgam of type PSp4(3) comprises of the group A, the normalizer in

PSp4(3) of Q, together with B, the centralizer in PSp4(3) of a 3-central element of order

three. Furthermore these two 3-local subgroups intersect at C, the normalizer in PSp4(3)

of S. Of the three amalgams described in this thesis, this is the only one in which the two

groups, A and B, comprising the amalgam are not isomorphic. Also, in the amalgams of

type PSL3(3) and G2(3) we observed a natural GL2(3) action whereas, in the amalgam

of type PSp4(3), the elementary abelian subgroup Q is a faithful Sym(4)-module and

we see an SL2(3) action in B. We do however proceed in the same spirit as Chapter 6

since the task of recognizing PSp4(3) as a completion of the amalgam, in some respects,

seems to be similar. The centralizer of an involution, which we will determine some of

the structure of, admits an action from an elementary abelian subgroup of order nine and

the theorem of Smith-Tyrer seems well suited once again. However, things do not run as

smoothly as they did in G2(3). The problem appears to be much harder and consequently

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we will assume more in our hypothesis. We will assume that Q normalizes no subgroup of

the completion of the amalgam with order prime to 3. This is a hypothesis employed in

several other characterizations of PSp4(3) to be found in the literature (see for example

[24] and [16]). Therefore the following theorem is certainly weaker than its predecessors.

It is a necessary condition though since groups with the structure 26 : PSp4(3) and

56 : PSp4(3) can be constructed with Magma and furthermore these constructions satisfy

the 3-local conditions imposed in Hypothesis C. This does however raise several interesting

questions. Firstly, if Qα normalizes some 3′-subgroup does this force there to be a normal

3′-subgroup as in the constructions 26 : PSp4(3) and 56 : PSp4(3)? Secondly, given

the appropriate 3-local conditions, what do the modules for PSp4(3) look like and is it

possible that we get a normal 3′-subgroup which is not elementary abelian? To answer

these question further works needs to be done perhaps involving signalizer functors. The

second question is reminiscent of a theorem of Graham Higman about a group G for which

G/O2(G) ∼= SL2(2n). We will discuss and prove his result in Chapter 8.

The hypothesis and theorem for this chapter are as follows.

Hypothesis C. Let A1 and A2 be subgroups of a finite group G = 〈A1, A2〉 such that

A12 := A1 ∩ A2 contains a Sylow 3-subgroup of both A1 and A2. Suppose further that

(i) no non-trivial normal subgroup of A12 is normal in A1 and A2;

(ii) A1/O3(A1) ∼= Sym(4);

(iii) A2/O3(A2) ∼= SL2(3);

(iv) O3(A1) is elementary abelian of order 33 and is a faithful A1/O3(A1)-module;

(v) O3(A2) is extraspecial of order 33 and O3(A2)/Z(O3(A2)), is a natural A2/O3(A2)-

module;

(vi) NG(O3(A1)) = A1 and NG(Z(O3(A2))) = A2;

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(vii) O3(A1) normalizes no non-trivial 3′-subgroup of G.

Theorem C. Let G be a group satisfying Hypothesis C. Then G ∼= PSp4(3).

7.1 The Amalgam

Associate the amalgam (A1, A2, A12) to a coset graph, Γ. Let α and β be adjacent vertices

in Γ and so set A1 = Gα, A2 = Gβ.

Let Sαβ ∈ Syl3(Gαβ), Qαβ = Qα ∩Qβ, Zβ = Z(Qβ).

Lemma 7.1. The following hold in G.

(i) O3(Gαβ) = Sαβ = QαQβ ∈ Syl3(Gαβ).

(ii) Qαβ has order nine and Qβ∼= 31+2

+ .

(iii) Zβ = Z(Sαβ), Qα = J(Sαβ) are characteristic subgroups of Sαβ. CG(Qα) = Qα.

(iv) Sαβ ∈ Syl3(G).

(v) Gβ = NG(Zβ) = CG(Zβ).

(vi) NG(Sαβ) = Gαβ and Gαβ has index 4 in both Gα and Gβ.

(vii) If z ∈ Sαβ has order three then z ∈ Qα ∪Qβ.

(viii) Qβ is characteristic in Sαβ and S ′αβ = Qαβ.

Proof. (i) We have that Gα and Gβ intersect at a group containing a Sylow 3-subgroup

of G which has order 34. Also Qα E Gα and Qβ E Gβ. Thus the product QαQβ is normal

in Gαβ. If Qα = Qβ then the amalgam would not be simple and so QαQβ is a Sylow

3-subgroup of Gαβ.

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(ii) Since Qα and Qβ have index three in Sαβ, their intersection Qαβ has index nine and

so has order nine. In particular, the extra special group, Qβ has an elementary abelian

subgroup of order nine. Since Qβ/Zβ is a natural Gβ/Qβ-module, Lemma 2.15 implies

that Gβ is transitive on the subgroups of Qβ with order nine. Thus Qβ is extra special of

+-type.

(iii) Suppose Z(Sαβ) Qβ then Sαβ = Z(Sαβ)Qβ. But then Sαβ/Qβ∼= Z(Sαβ) is a

Sylow 3-subgroup of Gβ/Qβ but acts trivially on the natural module Qβ/Zβ. This is a

contradiction (see Lemma 2.15) and so Z(Sαβ) 6 Qβ and so must be equal to Zβ.

Suppose Sαβ contains another elementary abelian subgroup, A, of order 33. Then

Qα and A intersect at a group containing a subgroup of order 32 which commutes with

QαA = Sαβ. However this contradicts Z(Sαβ) = Zβ. Thus Qα is the Thompson subgroup

of Sαβ and in particular is a characteristic subgroup of Sαβ.

(iv) Since Gα/Qα acts faithfully on Qα, no element commutes with all of Qα. Also,

since CG(Qα) 6 NG(Qα) = Gα, we see that CG(Qα) = Qα.

(v) Suppose Sαβ < S ∈ Syl3(G). Then Sαβ C R for some R 6 S. However, this means

Zβ C R 6 NG(Zβ) = Gβ which is a contradiction as Sαβ ∈ Syl3(Gβ). Thus Sαβ ∈ Syl3(G).

(vi) Suppose that CG(Zβ) < Gβ then the index must be 2. However this would give

an index 2 subgroup of Gβ/Qβ∼= SL2(3) which is not possible.

(vii) The normalizer of a Sylow 3-subgroup in both Gα/Qα∼= Sym(4) and Gβ/Qβ

∼=

SL2(3) has index 4. Thus by correspondence the same holds in Gα and Gβ. Now Sαβ E

Gαβ and so Gαβ has index at least 4 in both Gα and Gβ. Since Z(Sαβ) = Zβ, NG(Sαβ) 6

NG(Zβ) = Gβ. Thus NGα(Sαβ) 6 Gαβ and it follows that Gαβ has exactly index 4 in Gα

and so order 342. But then Gαβ must have index 4 in Gβ also and NG(Sαβ) = Gαβ.

Lemma 7.2. (i) If z ∈ Sαβ has order three then z ∈ Qα ∪Qβ.

(ii) Qβ is characteristic in Sαβ and S ′αβ = Qαβ.

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Proof. (i) We have that Sαβ = QαQβ so any element in Sαβ\(Qα ∪ Qβ) has the form yx

where y ∈ Qβ\Qα and x ∈ Qα\Qβ. Notice that S ′αβ 6 Qα and so [x, y, x] = 1 as Qα is

abelian. Suppose [y, x, y] = 1, then the element Qαy ∈ Gα/Qα acting on Qα fixes the

1-space 〈[y, x]〉. By Lemma 2.17, Qαy can fix just a 1-space but it already fixes the space

Zβ and so we must have Zβ = 〈[y, x]〉.

We also have that Qβx ∈ Gβ/Qβ acts on Qβ/Zβ and since the action of Gβ/Qβ on

this module is natural, we see that Qβx fixes precisely the 1-subspace Qαβ/Zβ. However,

y /∈ Qαβ and Qβx fixes the space Zβy since [y, x] ∈ Zβ. This is a contradiction. Thus

[y, x, y] 6= 1 and [y, x] /∈ Zβ .

Now suppose yx as order three. We use the relations x[x, y][x, y, x] = [x, y]x and

y[x, y][x, y, y] = [x, y]y together with [x, y, x] = 1 in the following calculations. Also we

use that x ∈ Qα and so commutes with all commutators. Hence,

1 = yxyxyx

= y2x[x, y]xyx

= y2x2[x, y][x, y, x]yx

= y2x2[x, y]yx

= y2x2y[x, y][x, y, y]x

= y2x2y[x, y]x[x, y, y]

= y2x2yx[x, y][x, y, x][x, y, y]

= y2x2yx[x, y][x, y, y]

= [y, x][x, y][x, y, y]

= [x, y, y].

Which gives a contradiction. Thus yx has order nine.

105

(ii) If Sαβ contained another extraspecial subgroup of exponent three then we could

choose some non-central element of order three inside it that would not be in Qα ∪ Qβ

and that is not possible. Thus Qβ is a characteristic subgroup. Now Qαβ = Qα ∩ Qβ is

a normal subgroup of Sαβ and so S ′αβ 6 Qαβ. Suppose the containment was strict then

Sαβ would have order three and would be a central subgroup of Sαβ which would mean

S ′αβ = Zβ. However we have seen already that with x and y as before, [x, y] Zβ. Thus

Zβ 6= S ′αβ and so S ′

αβ = Qαβ.

7.2 The Coset Graph

Lemma 7.3. The coset graph Γ has valency four.

Proof. We have seen that the valency of the vertex α is given by [Gα : Gαβ] and the

valency of β by [Gβ : Gαβ] both of which are four by Lemma 7.1. Since G has two orbits

on V (Γ) with representatives α and β, we see that every vertex in Γ has valency four.

Lemma 7.4. (i) There is an involution tβ ∈ Gβ such that Gαβ = Sαβ〈tβ〉 and Qβtβ ∈

Z(Gβ/Qβ).

(ii) tβ fixes a circuit Θ = (α, β, . . . , α) in Γ.

(iii) For each vertex γ ∈ Θ lying in the same G-orbit as β, Qγtβ ∈ Z(Gγ/Qγ).

Proof. Notice that the centre of Gβ/Qβ∼= SL2(3) has order two and that the central

involution normalizes each Sylow 3-subgroup in Gβ/Qβ. So choose an element of order

two, tβ, as a coset representative such that Qβ〈tβ〉/Qβ = Z(Gβ/Qβ). Since tβ ∈ Gαβ =

NGβ(Sαβ), tβ fixes the edge α, β in Γ and since every vertex has valency four it follows

that tβ fixes a circuit in Γ. In particular, this means for any vertex γ ∈ Θ lying in the

same G-orbit as β, tβ ∈ Gγ and furthermore tβ normalizes a Sylow 3-subgroup of Gγ and

so Qγtβ ∈ Z(Gγ/Qγ) as before.

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Lemma 7.5. G is locally 5-arc transitive on Γ.

Proof. See [7, p73 and p98].

Lemma 7.6. CQβ(tβ) = Zβ, 3× 3 ∼= CSαβ

(tβ) 6 Qα and CGβ(tβ) ∼= 3× SL2(3).

Proof. By the choice of tβ, we have that Qβtβ ∈ Z(Gβ/Qβ) and so tβ is in the centre of a

Sylow 2-subgroup of Gβ. Also tβ centralizes Zβ and Sαβ/Qβ. Clearly CQβ(tβ) = Zβ else

we have a non-trivial subgroup of Qβ/Zβ centralized by tβ which contradicts that this

group is a natural Qβ/Zβ-module. By Lemma 2.16, we can choose some x ∈ Qα\Qβ such

that tβ commutes with x. Hence CSαβ(tβ) = 〈x, Zβ〉 6 Qα is elementary abelian of order

nine. Consider the set tGβ

β = tgβ|g ∈ Gβ. Since Gβ has at most 34 Sylow 2-subgroups

and tβ is in the centre of one of them, this set has order at most 34. However tβ commutes

with a group of order nine, 〈Zβ, x〉 so it follows this set has order at most nine and so

〈tβ〉Gβ = 〈tβ〉Qβ . Now by a Frattini argument (Lemma 1.7), Gβ = CGβ(tβ)Qβ and by an

isomorphism theorem,

CGβ(tβ)/Zβ = CGβ

(tβ)/(CGβ(tβ) ∩Qβ) ∼= CGβ

(tβ)Qβ/Qβ = Gβ/Qβ∼= SL2(3).

The element 〈x〉 gives a complement to Zβ in CSαβ(tβ) and so by Gaschutz’s Theorem

(Theorem 1.8), CGβ(tβ) splits over Zβ and so CGβ

(tβ) ∼= 3× SL2(3).

At this point we recognize that CGβ(tβ) has a normal subgroup, X say, isomorphic to

SL2(3). Therefore Q8∼= O2(X) = O2(CGβ

(tβ)). Also, if γ is a vertex fixed by tβ at

distance two from β thenQγtβ ∈ Z(Gγ/Qγ) by Lemma 7.4 and so it follows that CGγ (tβ) ∼=

3× SL2(3) and so Q8∼= O2(CGγ (tβ)) also.

Lemma 7.7. Let (γ, α, β) be a path of vertices in Γ fixed by tβ. Let X := O2(CGβ(tβ))

and Y := O2(CGγ (tβ)). Then X 6= Y and if [X, Y ] = 1 then Γ is a Moufang quadrangle

with 80 vertices. In particular, if [X, Y ] = 1 then |G| = 25920.

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Proof. Suppose X = Y . Each x ∈ X fixes β and γ. Notice, edge stabilizers have order

342 so do not contain elements of order four so if x ∈ X has order four then x does not fix

α. So there are elements in X which fix β and γ but move α. Therefore Γ must contain

a four cycle however this is impossible since G acts 5-arc transitively on Γ.

Since X 6= Y , we can choose p ∈ Y \X and consider the vertex β · p 6= β. This

vertex has distance two from γ and therefore distance four from β. Let q ∈ X then

β · pq = β · qp = β · p. Thus X fixes β and β · p and moves γ and so arguing as before we

see that β, γ, β ·p lie on a circuit of length 8. However, G = 〈Gα, Gβ〉 acts 5-arc transitively

on Γ and so every vertex is contained in an 8-cycle and no vertex can be more than four

vertices away from any other (so Γ has diameter 4 and girth 8). The number of vertices at

distance 0− 4 from α are 1, 4, 12, 36, 27 and so Γ has 80 vertices. Since |Gα| = 2334 = 648

and there are 80/2 = 40 images of α under G, we get |G| = 40× 648 = 25920 as required.

Finally let (α0, α1, α2, α3, α4) be a path of five vertices in Γ. Since G acts 5-arc transitively

on Γ, the stabilizer of the path is transitive on Γ(α4)\α3 and so we must have three

dividing the stabilizer of the path. Therefore an element of order three must necessarily

fix⋃

16i63 Γ(αi) and so Γ satisfies the Moufang condition.

Theorem 7.8. Let γ, α, β be a path in Γ fixed by tβ. If O2(CGβ(tβ)) commutes with

O2(CGγ (tβ)) then G ∼= PSp4(3).

Proof. By Theorem 4.13 and Lemma 7.7, the coset graph Γ is a Moufang quadrangle with

80 vertices and is uniquely defined. Notice that the stabilizer of the path of length three,

γ, α, β, δ, contains Qαβ. Thus the group G† (as defined in 4.13) contains all conjugates of

Qαβ. Notice that QαβQβδ = Qβ so Qβ 6 G† and similarly Qα 6 G†. It now follows that

G† contains every Sylow 3-subgroup of G and since Gα and Gβ are both generated by

their Sylow 3-subgroups, G 6 G†. Now it follows immediately from the order of G and

Theorem 4.13 that G = G† ∼= PSp4(3).

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7.3 Controlling the 3 and 2-Local Structure

By Hypothesis C, Qα is a 3-dimensional faithful Sym(4)-module for Gα/Qα so we will

use Lemmas 2.17 and 2.18 and we will use the terminology introduced by referring to the

proper, non-trivial subgroups of Qα as being of type 1A, 1B, 1C, 2A, 2B or 2C. Find a,

b and c in Qα as in Lemma 2.17 where a ∼ v1, b ∼ v2 and c ∼ v3 give the subspaces fixed

by the non-trivial elements in O2(Gα/Qα). Then Qα = 〈a, b, c〉. Notice that the only type

of 1-space centralized by an element of order three in Gα/Qα has type 1B therefore Zβ

has type 1B. Choose a, b and c in such a way that Zβ = 〈abc〉.

Lemma 7.9. Qαβ = S ′αβ has type 2B.

Proof. The group Gα has four Sylow 3-subgroups which intersect at O3(Gα) = Qα. Now,

Gα is transitive on its Sylow 3-subgroups and thus on their corresponding derived sub-

groups which each have order 32. However, by Lemma 2.17, there is no orbit of subgroups

of order nine in Qα of length one. Therefore the derived subgroups must be distinct and

have type 2B.

Lemma 7.10. Sαβ has four G-conjugacy classes of elements of order three with represen-

tatives a, ab, abc, a2b2c2. In particular abc, a2b2c, a2bc2, ab2c2 and a2b2c2, abc2, ab2c, a2bc

are in disjoint conjugacy classes.

Proof. A lemma of Burnside (1.9) says that abc ∈ Zβ is conjugate to its inverse in G only

if it is conjugate to it in NG(Sαβ) = Gαβ 6 CG(Zβ). Thus abc and any other 3-central

element is not conjugate to its inverse. By Lemma 2.17, elements such as a and ab are

conjugate to their inverses and so cannot be conjugate to elements of type abc.

Subgroups of type 2B contain one subgroup of type 1B and three of type 1C. Now

consider S ′αβ = Qαβ. This group of order 32 has type 2B. Since Qβ/Zβ is a natural

Gβ/Qβ-module, Gβ is transitive on the 1-spaces and so Gβ is transitive on subgroups of

109

order 32 contained in Qβ. By Lemma 7.2, every element of order three in Sαβ lies in

Qα∪Qβ and so any element of order three in Sαβ lies in a subgroup of type 1A, 1B or 1C.

Finally, since Qα = J(Sαβ), Lemma 1.11 proves that no conjugate of ab can be conjugate

to any conjugate of a in G else they would already be conjugate in Gα. Thus there are

exactly four conjugacy classes of elements of order three in G.

In particular, notice that by Lemma 2.17 (v), abc is conjugate to a2b2c, a2bc2 and

ab2c2.

Lemma 7.11. 〈a, abc〉 = CSαβ(tβ) 6 Qα has order nine and has type 2C, CGα(tβ) ∼

3× 3 : 2× 2 and 〈a, abc, tβ〉 = CGαβ(tβ) ∼ 3× 3× 2.

Proof. In Lemma 7.6 we saw that 3× 3 ∼= CSαβ(tβ) 6 Qα. Choose some element d ∈ Qα

such that 〈Zβ, d〉 = CSαβ(tβ). Notice that Qαtβ ∈ (Gα/Qα) \ O2(Gα/Qα) (we see this is

true since Qαtβ normalizes a Sylow 3-subgroup of Gα/Qα). Now by Lemma 2.17 (vi),

Qαtβ centralizes a subgroup of Qα of type 1C and so we can choose d so that 〈d〉 has type

1C. By Lemma 2.18, 〈Zβ, d〉 has type 2B or 2C. It is easily checked that there are four

groups of order nine in Qα containing Zβ of which three have type 2C and one has type

2B. We have seen that the subgroup of type 2B is S ′αβ = Qαβ however CQβ

(tβ) = Zβ by

Lemma 7.6 and so 〈Zβ, d〉 must have type 2C. In particular this means CSαβ(tβ) has a

subgroup of type 1A and so we may as well assume 〈a〉 6 CSαβ(tβ) and then d ∈ 〈bc〉.

Now we see that

QαCGα(tβ)

Qα

6 CGα/Qα(Qαtβ) ∼= 2× 2

since Qαtβ is an involution in Sym(4) which normalizes a Sylow 3-subgroup and so com-

mutes with just a group of order four. In fact, we have equality in the previous expression

since Gα has Sylow 2-subgroups which are dihedral of order eight and so tβ at least

110

commutes with another involution. Now, by an isomorphism theorem,

2× 2 ∼=QαCGα(tβ)

Qα

∼=CGα(tβ)

CQα(tβ)

and we have seen already that CQα(tβ) ∼= 3× 3.

It is immediate that CGαβ(tβ) has order 18 and is abelian.

We have now that CSαβ(tβ) is a subgroup of type 2C and therefore contains subgroups

of type 1A, 1B and 1C. Thus, this group of order nine contains an element from each of

the four conjugacy classes of elements of order three in G. We therefore aim to restrict the

structure of the normalizer in G of this subgroup and then use the Theorem of Smith and

Tyrer once again. First we must restrict the structure of some of the 3-local subgroups

in G.

We continue to set 〈Zβ, d〉 = CSαβ(tβ) where d ∈ Qα\Qβ is chosen so that 〈d〉 has type

1C.

Lemma 7.12. 〈Qα, tβ〉 = CGβ(d) = CGα(d) = CGαβ

(d) ∼ 33.2.

Proof. By Lemma 2.17 (iv), CGα/Qα(d) has order two. Also, as d is not central in Sαβ, it

commutes with exactly Qα here. So Qαtβ ∈ CGα(d)/Qα 6 CGα/Qα(d) ∼ 2 which implies

that CGα(d) = 〈Qα, tβ〉 has order 332.

We chose d such that d ∈ Qα\Qβ and so Qβ〈d〉/Qβ is a Sylow 3-subgroup of Gβ/Qβ.

Therefore CGβ/Qβ(Qβd) ∼ 3× 2. By an isomorphism theorem,

CGβ(d)Qβ/Qβ

∼= CGβ(d)/CQβ

(d) = CGβ(d)/Qαβ.

Also, it is easily seen that CGβ(d)Qβ/Qβ 6 CGβ/Qβ

(d) = CGβ/Qβ(Qβd). Together this

gives us

CGβ(d)/Qαβ

∼= CGβ(d)Qβ/Qβ 6 CGβ/Qβ

(Qβd) ∼ 3× 2.

111

Hence CGβ(d) ∼ 33.2 and so the result follows.

Lemma 7.13. Qα ∈ Syl3(CG(d)) and CG(d) is either 3-soluble of length one or has a

normal subgroup at index nine.

Proof. Set C := CG(d) and notice that Qα 6 C. Now NC(Qα) = NG(Qα)∩C = Gα∩C =

CGα(d) ∼ 33.2. So suppose Qα is not a Sylow 3-subgroup of C then Qα < P ∈ Syl3(C)

and then Qα C NP (Qα) which is a contradiction since |NC(Qα)| = 233. By Lemma 7.1

(iv), CG(Qα) = Qα and so |NC(Qα)/CC(Qα)| = 2. Therefore C satisfies Corollary 4.6

and so C is either 3-soluble of length one or O3(C) has index at least nine. Suppose

|C/O3(C)| = 33 then O3(C) = O3′(C) and C would be 3-soluble of length one so we are

done.

Now we make full use of our strong assumption in Hypothesis C that Qα does not nor-

malize any non-trivial 3′-group in G.

Lemma 7.14. CG(d) 6 Gαβ, NG(d) 6 Gα.

Proof. Again we set C := CG(d) and we consider the two possibilities given by Lemma

7.13. First we suppose C is 3-soluble of length one. Since Qα does not normalize any

non-trivial subgroup of G of order prime to three, O3′(C) = 1. Thus C has a normal

Sylow 3-subgroup and so C = NC(Qα) = Gα ∩ C = 〈Qα, tβ〉 6 Gαβ and then C 6 Gαβ

and NG(d) 6 NG(Qα) 6 Gα as claimed. So suppose C is not 3-soluble of length one then

C has a normal subgroup, N say, at index nine. Let x be an element of order three in

N and consider CN(x). By Burnside’s normal p-complement Theorem (1.6), CN(x) has

a normal 3-complement, M say. However M is normalized by Qα and has order prime

to 3 and so must be trivial. Therefore x is self-centralizing in N and so we can apply a

theorem of Feit and Thompson (4.1) to N . Since O3′(C) = 1, O3′(N) = 1. Therefore by

Theorem 4.1, N ∼= Alt(5) or N ∼= Sym(3) or N ∼= PSL3(2). Notice N admits an action

from Zβ and CN(Zβ) = N ∩ Gβ 6 C ∩ Gβ ∼ 322. Hence CN(Zβ) = 〈x, tβ〉 can have

112

order just six. Suppose N ∼= Alt(5) or N ∼= PSL3(2). By Lemma 4.2, N does not admit

an outer automorphism of order three. Therefore 〈x, Zβ〉 ∩ CC(N) is non trivial. Clearly

neither x nor Zβ commutes with N so choose zβ ∈ Z#β such that xzβ ∈ CC(N). Then

1 = [xzβ, tβ] = txzβ

β tβ as tβ ∈ N . Therefore tβ = tz−1β

β = txβ which contradicts that x is self

centralizing in N . Thus N ∼= Sym(3) and so |C| = 6.32 which forces C 6 Gαβ. Finally,

by Lemma 2.17 (v), an element of order two in Gα inverts d. Thus NG(〈d〉) 6 Gα.

Corollary 7.15. Let d′ ∈ Qα such that 〈d′〉 is a subgroup of type 1C. Then 33 : 2 ∼

CG(d′) 6 Gα.

Proof. This is now immediate since all such elements in Qα are conjugate in Gα.

We now know the centralizers in G of elements of order three from three of the four

conjugacy classes. So we will consider CG(a) and by our choice of a we have tβ ∈ CG(a).

Lemma 7.16. CGα(a) ∼ 33 : (2× 2) and Qα ∈ Syl3(CG(a)). In particular NCG(a)(Qα) ∼

33 : (2× 2).

Proof. By Lemma 2.17, CGα/Qα(a) has order four. Also Qαtβ ∈ CGα/Qα(a). Notice that

this forces CGα/Qα(a) to be elementary abelian (since Qαtβ normalizes an element of order

three in Gα/Qα∼= Sym(4) and so no element of order four squares to Qαtβ). Now, we

clearly have CGα(a)/Qα 6 CGα/Qα(a), however, equality follows since a ∈ Qα is abelian.

Thus CGα(a)/Qα∼= 2× 2 and so CGα(a) ∼ 33 : (2× 2).

We have NCG(a)(Qα) = NG(Qα) ∩ CG(a) = CGα(a) ∼ 33 : (2 × 2). So if Qα were not

a Sylow 3-subgroup of CG(a) then we would have Qα < T for some T ∈ Syl3(CG(a)) and

then Qα C NT (Qα) 6 NCG(a)(Qα) ∼ 33 : (2 × 2). This is a contradiction and so Qα is a

Sylow 3-subgroup of CG(a) with normalizer as described.

We will apply a theorem of Hayden (Theorem 4.7) to the group CG(a) = CG(a)/〈a〉 so

we must first work to show the hypothesis of the theorem holds.

113

Lemma 7.17. Set T := Qα 6 CG(a) =: A then A and T satisfy the following.

(i) NA(T )/CA(T ) ∼= 2× 2.

(ii) CA(T ) = T .

(iii) CA(t) 6 NA(T ) for each t ∈ T#.

In particular, A = NA(T ).

Proof. By Lemma 7.16, T = Qα is a Sylow 3-subgroup of A = CG(a) and it is clear that

NA(T ) = NCG(a)(Qα)/〈a〉. We now choose some coset representatives for the elements of

T and consider their centralizers in A. We choose 〈ab, abc〉 = T and so we must consider

the groups (i) CA(ab), (ii) CA(abc), (iii) CA(a2b2c) and (iv) CA(c). We will show each is

contained in CGα(a).

(i) Suppose x ∈ CA(ab). Then [ab, x] = 1 implies [ab, x] ∈ 〈a〉 and then (ab)x ∈

ab, a2b, b. Lemma 7.10 implies that ab cannot be conjugate to b in G but is conjugate

to a2b in Gα. So there exists y ∈ Gα such that (ab)y = a2b. So suppose (ab)x = a2b

then (ab)xy−1= ab. Therefore xy−1 ∈ CG(ab) 6 Gα by Corollary 7.15 and so x ∈ Gα. If

instead (ab)x = ab then x ∈ CG(ab) 6 Gα so again x ∈ Gα.

(ii) Suppose x ∈ CA(abc). Then [abc, x] = 1 implies [abc, x] ∈ 〈a〉. Therefore (abc)x ∈

abc, a2bc, bc. By Lemma 7.10, the only possibility is (abc)x = abc. Therefore x ∈

CG(abc) = Gβ. So x ∈ CG(abc) ∩ CG(a) 6 CG(bc) 6 Gα by Corollary 7.15.

(iii) Suppose x ∈ CA(a2b2c). Then [a2bc2, x] = 1 implies [a2b2c, x] ∈ 〈a〉. Therefore

(a2b2c)x ∈ a2b2c, b2c, ab2c. Lemma 7.10 implies that (a2b2c)x = a2b2c and so x ∈

CG(a) ∩ CG(a2b2c) 6 CG(b2c) 6 Gα by Corollary 7.15.

(iv) Finally we suppose x ∈ CA(c). Then [c, x] = 1 implies [c, x] ∈ 〈a〉. Therefore

cx ∈ c, ac, a2c. Once again by Lemma 7.10, we have cx = c and so x ∈ CG(a)∩CG(c) 6

CG(ac) 6 Gα by Corollary 7.15.

114

So we have shown that CA(t) 6 CGα(a) = NCG(a)(Qα) = NA(T ) for each t ∈ T#. Our

calculations also show that if x ∈ CA(T ) then x ∈ CG(〈abc, c, a〉) = CG(Qα) = Qα by

Lemma 7.2. Therefore CA(T ) = T and NA(T )/CA(T ) ∼= 2 × 2. Thus we have shown A

satisfies the hypothesis of Theorem 4.7 and so A = NA(T ) = NCG(a)(Qα)/〈a〉.

Corollary 7.18. 33 : (2× 2) ∼ CG(e) 6 Gα for each e ∈ Qα such that 〈e〉 has type 1A.

Proof. By Lemma 7.17, we have that CG(a)/〈a〉 = NCG(a)(Qα)/〈a〉 = (CG(a) ∩ Gα)/〈a〉

and so CG(a) 6 Gα. If e ∈ Qα and 〈e〉 has type 1A then e is a conjugate of a by an

element of Gα and so the result holds.

We have now calculated the centralizer in G of every conjugacy class of element of order

three so the following result is immediate.

Corollary 7.19. If x ∈ G has order three then CG(x) is a 2, 3-group. In particular if

x ∈ Sαβ then CG(x) 6 Gα or CG(x) 6 Gβ.

We now work to restrict the structure of the group H := CG(tβ).

Lemma 7.20. S := CSαβ(tβ) is a Sylow 3-subgroup of H and |NH(S)/CH(S)| = 2.

Proof. By Lemma 7.11, S has type 2C and so contains elements of order three from each

of the four G-conjugacy classes of elements of order three. In particular the only conjugate

of d in S is d−1 so for any x ∈ NH(S)\S, x ∈ NG(〈d〉) 6 Gα. Therefore, by Lemma 7.11,

S 6 NH(S) 6 Gα ∩ H = CGα(tβ) ∼ 32 : 22 has a normal 3-subgroup of order nine. So

NH(S) = CGα(tβ) has order 2232. In particular, this means S ∈ Syl3(H) since if not then

S < T ∈ Syl3(H) and then S C NT (S) 6 NH(S) which contradicts that |NH(S)|3 = 32.

Now, CH(S) 6 CH(Zβ)∩CH(d) = CGβ(d) ∼ 33.2 by Lemma 7.12 and so CH(S) has order

at most 232. In fact this is the exact order as S commutes with itself and with tβ. Hence

|NH(T )/CH(T )| = 2.

115

Recall Lemma 7.4 implies that the involution tβ fixes a circuit in Γ, (γ, α, β, . . .)

and Qγtβ ∈ Z(Gγ/Qγ). Thus CGγ (tβ) ∼= 3 × SL2(3). Furthermore Zγ 6 CGγ (tβ) and

〈Zβ, Zγ〉 6 Qα is abelian. So 〈Zβ, Zγ〉 6 CSαβ(tβ) = S and therefore S = 〈Zβ, Zδ〉.

Lemma 7.21. 1 6= O3′(H) is nilpotent and contains A := O2(CGβ(tβ)) and B := O2(CGγ (tβ)).

Proof. By Lemma 7.20, |NH(S)/CH(S)| = 2 and so we can once again apply the theorem

of Smith and Tyrer (Theorem 4.5) to see that either H is 3-soluble of length one or

O3(H) 6= H. If H is 3-soluble of length one then X := O3′(H)S is a normal subgroup of

H. Then a Frattini argument implies that H = XNH(S). It is clear that tβ ∈ O3′(H) and

so 〈tβ, S〉 6 X ∩NH(S) and so by an isomorphism theorem, H/X ∼= NH(S)/(X ∩NH(S))

has order at most two. Therefore X is a normal subgroup of H at index at most two.

If H = X then H has a normal 3-complement and clearly A,B 6 O3′(H). So we may

as well assume X has index two in H. Suppose A X. Then by an isomorphism

theorem, 2 = |H/X| = |AX/X| = |A/(A ∩ X)| and so A ∩ X has order four. However

3 × SL2(3) ∼= CGβ(tβ) 6 H and so X > 〈(A ∩ X)CGβ

(tβ)〉 = A which is a contradiction.

Therefore A 6 X and similarly B 6 X and then clearly both quaternion groups are

contained in O3′(H).

So we are left to consider the possibility that Y := O3(H) 6= H. By Lemma 7.11 and

Lemma 7.14, CH(d) 6 Gαβ ∩H ∼= 3× 3× 2. So CH(d) = 〈S, tβ〉.

Since we know |NH(S)/CH(S)| = 2, there exists an element of even order x ∈

NH(S)\CH(S). If x ∈ CH(d) then x commutes with S. Recall, S has type 2C and

the only conjugate of d in S is d−1 (see Lemma 2.18 and Lemma 7.10). Thus x must

invert d as d−1 is the only other conjugate of d in S. Since x has even order, x ∈ Y and

therefore Y 3 dxd−1 = dxdx = d2x and so d ∈ Y . Now CY (d) = 〈S, tβ〉 ∩ Y has order 6

so if we set H := H/〈tβ〉 then d is a self-centralizing element of order three in Y . Hence

we can apply Theorem 4.1 to Y and we have three cases to consider.

Firstly, suppose Y ∼= PSL3(2). Then by Lemma 4.2, S ∩ CG(Y ) is non-trivial else we

116

would have an outer automorphism of PSL3(2) of order three which is not possible. Thus

for some s ∈ S#, and each y〈tβ〉 ∈ Y , y〈tβ〉s = y〈tβ〉. Since the coset y〈tβ〉 has order

two and s ∈ S has order three, s must centralize each element of the coset. In particular

we could choose y〈tβ〉 to have order seven. Then y has order seven and commutes with

s ∈ S# and this contradicts Corollary 7.19.

Secondly, suppose Y has a normal 2-subgroup, M , such that Y/M ∼= Y /M ∼= Alt(5).

Then by Lemma 4.2, S ∩ CG(Y/M) is non-trivial else we would have an outer automor-

phism of Alt(5) of order three which is not possible. Thus for some s ∈ S#, and each

yM ∈ Y/M , yM s = yM . Since the coset yM has order a power of 2 and s ∈ S has order

three, s must centralize each element of the coset. In particular we could choose yM to

have order five. Then y has order a multiple of five but commutes with s ∈ S# and this

contradicts Corollary 7.19.

So we are left only with the third case given by Theorem 4.1. Therefore Y has a

normal 3′-subgroup M such that Y/M ∼= Y /M ∼= Sym(3) or C3. In particular this means

that M : S is a subgroup of H at index at most two and so is normal. It follows that

H must be 3-soluble of length one and so by the previous argument we see that H has a

normal 3′-subgroup containing the necessary quaternion subgroups.

It remains only to prove that in each case O3′(H) is nilpotent. First notice that

tβ ∈ O3′(H) and CH(d) ∩ O3′(H) = 〈tβ〉 so we can apply Corollary 1.22 to O3′(H) to see

that O3′(H) is nilpotent.

Lemma 7.22. X := 〈O2(CGβ(tβ)), O2(CGγ (tβ))〉 is an extraspecial 2-group and has shape

21+4+ . In particular, [O2(CGβ

(tβ)), O2(CGγ (tβ))] = 1.

Proof. Since O3′(H) is nilpotent, it is a direct product of its Sylow subgroups and so the

two 2-groups A := O2(CGβ(tβ)) and B := O2(CGγ (tβ)) generate a 2-group. We apply

Lemma 1.25 to X/〈tβ〉 = 〈A,B〉/〈tβ〉 to get that X/〈tβ〉 is either elementary abelian

117

of order 24 or special of order 26 with centre of order four. Suppose the latter then,

by correspondence, X has a normal subgroup of order eight, Z say, containing tβ. By

coprime action, Z = 〈CZ(s)|s ∈ S#〉 so we consider the conjugacy class of each s ∈ S#.

If s = d (or d−1) then CZ(s) = 〈tβ〉 since we already saw that d acts fixed-point-freely on

O3′(H)/〈tβ〉. Consider CZ(s) 6 X ∩Gβ = A for s ∈ Z#β . Notice

CZ(s) 6 X ∩ CG(s) = X ∩Gβ 6 CGβ(tβ) ∼= 3× SL2(3)

and CZ(s) will be a normal 2-subgroup of CGβ(tβ) and so CZ(s) 6 A. If A 6 Z then

A = Z E X which is a contradiction since X = 〈A,B〉 has order 27. Suppose |A∩Z| = 4

then A∩Z is normalized by S and consequently centralized by S which is a contradiction

since CZ(d) = 〈tβ〉. Thus A ∩ Z = CZ(s) = 〈tβ〉. Similarly if s ∈ Z#γ then CZ(s) =

B ∩ Z = 〈tβ〉. So we must have Z = 〈CZ(s)|s ∈ S#〉 = CZ(a) has order eight. However

this is a contradiction since eight does not divide the order of CG(a) by 7.18. Hence

X/〈tβ〉 must be elementary abelian. Finally, since X contains non-abelian quaternion

groups, X is non-abelian and tβ ∈ Z(X) admits an automorphism of order three which

centralizes precisely 〈tβ〉 so Z(X) has order two or eight. However, the order cannot be

eight since no abelian group of order eight has an automorphism of order three. Thus X

is extraspecial of order 25. Now Corollary 1.17 together with Lemma 1.18 concludes the

result.

We can now use Theorem 7.8 to prove Theorem C.

Corollary 7.23. G ∼= PSp4(3).

118

Chapter 8

Higman’s Theorem on

Fixed-Point-Free Elements of Order

Three

Theorem 8.2 of Higman’s notes, Odd Characterisations of Finite Simple Groups [17], is

a powerful result about an action of SL2(2n) in which an element of order three acts

fixed-point-freely on a 2-group. The theorem is as follows.

Theorem 8.1. Let G be a finite group with a normal 2-subgroup, Q, such that X :=

G/Q ∼= SL2(2n) for some n > 2 and some element of order three in G acts fixed-point-

freely on Q. Then Q is elementary abelian and a direct product of natural modules.

Higman’s proof is sketchy and incomplete so we give a complete proof here. We will

use new methods of Chermak and Meierfrankenfeld found in [5] to give conditions under

which SL2(2n) modules can be decomposed and then finally we use Higman’s arguments to

prove that the subgroup Q is elementary abelian and thus an SL2(2n) module. In all that

follows let X ∼= SL2(2n) where n ≥ 2, T ∈ Syl2(X) and set L := GF(2) and K := GF(2n).

119

We must recall much of Chapter 2. In particular the definitions of the natural modules

over K and that these are the only irreducible modules over K for SL2(2n) which admit

an element of order three acting fixed-point-freely. We recall the definition of the natural

module over L also. We continue notation from Chapter 2 where we set V1, . . . , Vn to be the

irreducible 2-dimensional KX-modules. We also fixed an algebraically closed extension

field F of K over which we could define Brauer characters and we set Mi = F ⊗K Vi.

We proved these were the only 2-dimensional X-modules over an F . Recall also that by

Lemma 2.33, these were the only irreducible FX-modules admitting an element of order

three in X acting fixed-point-freely.

Lemma 8.2. Let V1, . . . , Vn be the n algebraic conjugates of the natural KX-module.

Then HomKX(Vi ⊗K Vj, Vk) = 0 for all i, j, k 6 n.

Proof. Firstly if i 6= j then we have seen in Lemma 2.31 that Vi ⊗ Vj is irreducible so the

only KH-module homomorphism Vi⊗K Vj → Vk is the zero map. So we consider the case

i = j. Suppose α : Vi ⊗K Vi → Vk is a KX-homomorphism. Since Vk is irreducible, this

map is either a zero map or surjective so we will assume α is surjective. Let U = Ker(α)

then U is a 2-dimensional submodule of V := Vi⊗KVi. Set W := 〈m⊗n+n⊗m|m,n ∈ Vi〉

then W is a KX-submodule of V . Let v1, v2 be a basis for Vi and write all elements of

X as matrices with respect to this basis. Any element of Vi can be written v = av1 + bv2

for a, b ∈ K. So for a, b, c, d ∈ K, a general element of W has the form

(av1 + bv2)⊗ (cv1 + dv2) + (cv1 + dv2)⊗ (av1 + bv2) = (ad+ bc)v1 ⊗ v2 + (ad+ bc)v2 ⊗ v1.

Thus W has dimension 1 and is spanned by the element 〈v1 ⊗ v2 + v2 ⊗ v1〉. If W U

then 1 6= WU/U V/U ∼= Vk where the isomorphism is an isomorphism of KX-modules.

This is not possible as Vk is irreducible. So W 6 U . Consider the 3-dimensional KX-

module V/W . This contains a 1-dimensional submodule U/W . The only 1-dimensional

KX-module is the trivial module (see Lemma 2.31) and so X acts trivially on U/W . A

120

basis for V := V/W is v1 ⊗ v1, v2 ⊗ v2, v1 ⊗ v2. Consider the element x :=

(0 1

1 1

)in X (represented with respect to the module Vi and the basis chosen) and let v :=

av1 ⊗ v1 + bv2 ⊗ v2 + cv1 ⊗ v2 ∈ V for a, b, c ∈ K. Now v1 · x = v2 and v2 · x = v1 + v2 and

so

v · x = bv1 ⊗ v1 + (a+ b+ c)v2 ⊗ v2 + cv1 ⊗ v2.

So if v were contained in U then it would be fixed by x and so a = b = c. Since K > L,

there is some ω ∈ K such that ω 6= 0, 1. So consider y :=

(ω 0

0 ω−1

)∈ X. Then

v1 · y = ωv1 and v2 · y = ω−1v2. So w := v1 ⊗ v1 + v2 ⊗ v2 + v1 ⊗ v2 ∈ U must be

centralized by y. However

w · y = ω2v1 ⊗ v1 + ω−2v2 ⊗ v2 + v1 ⊗ v2 6= w.

Thus there can be no such 0 6= α ∈ HomKX(Vi, Vi, Vk).

Corollary 8.3. Let U, V,W be non-trivial KX-modules admitting an element of order

three acting fixed-point-freely then HomKX(U ⊗K V,W ) = 0.

Proof. Suppose we have a surjective map, ρ : U ⊗ V → W . Every irreducible KX-

submodule of U , V and W is KX-isomorphic to Vi for some 1 6 i 6 n by Lemma 2.33.

Let W1 be an irreducible submodule of W and consider the preimage in U ⊗ V of W1.

There must exist submodules U1 6 U and V1 6 V such that ρ : U1 ⊗ V1 → W1. Suppose

U1 is not irreducible. For each proper submodule U2 of U1, ρ maps U2⊗V1 to a submodule

of W1. Since W1 is irreducible, this image is either W1 or 0. Since ρ∣∣U1⊗V1

is non-zero

we must be able to choose an irreducible submodule, U2 of U1 such that ρ : U2⊗V1 → W1.

We repeat these arguments in the case that V1 is not irreducible to find an irreducible

submodule V2 6 V1 such that ρ must map U2 ⊗ V2 to W1 where each of U2, V2,W1 is

KX-isomorphic to a natural KX-module. However, by Lemma 8.2, this is not possible

and so we have a contradiction.

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In the following lemma we will refer to some known number theoretic results which

can be found in the appendix.

Lemma 8.4. (Chermak)[5, 4.1] If V is an LX-module satisfying [V, T, T ] = 0 and such

that CV (X) = 0 then [V,X] is a direct sum of natural modules for X.

Proof. Set B = NX(T ) and let H ∼= Cq−1 be a complement to T in B. Choose an element

s ∈ X that normalizes H but is not in B. Now, T is a 2-group and V is a module

over a field of characteristic two so CV (T ) 6= 0 and is B-invariant. In particular |H| is

coprime to |CV (T )| so CV (T ) is completely reducible as a sum of irreducible LH-modules

by Maschke’s Theorem. Let U be one of these irreducible modules. Set E := EndLH(U).

By Schur’s Lemma, E is a finite field of order 2m for some m. Moreover, as H is abelian,

there is a homomorphic image of H in the multiplicative group E×. Let H be this image

and let r = |H|. Since U is an irreducible H-module, it is also an irreducible H-module.

Now H 6 E× and so r divides |E×| = 2m − 1. Also, r divides |H| = 2n − 1. Let d be the

greatest common divisor or 2n − 1 and 2m − 1 and let g be the greatest common divisor

of n and m. Then d = 2g − 1 by Lemma A.2 and r divides 2g − 1. Since g divides m,

E contains a subfield, E0 say, of order 2g (See Lemma A.3). Also, since r divides 2g − 1

and multiplicative groups of fields are cyclic, it follows that H 6 E×0 . We can view U

as a vector space over E and also as a vector space over E0. Since E0 contains H, any

1-dimensional E0-subspace of U is H-invariant so U has dimension 1 over E0. It follows

that |U | = |E0| 6 2n.

For any C ⊆ X set UC := 〈uc|u ∈ U, c ∈ C〉 and for D 6 X set [UC , D] := 〈[uc, d] =

ucd − uc|u ∈ U, c ∈ C, d ∈ D〉. Now consider the LX-module W := UX . By the BN-

property of X, we have X = B ∪BsB and so

W = UB + UBsB = U + U sB = U + U sHT = U + U sT

because s normalizes H. Now for any u ∈ U and any t ∈ T , ust = us+[us, t] ∈ U s+[U s, T ]

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so U sT 6 U s + [U s, T ]. Also, U s 6 U sT and [U s, T ] 6 U sT . Thus W = U + U s + [U s, T ].

Since the action of T is quadratic on V and U 6 CV (T ), U + [U s, T ] 6 CW (T ). Suppose

[U s, T ] = 0, then U commutes with 〈T, T s−1〉 = X (by Lemma 2.23) which contradicts

that CV (X) = 0. Hence W/CW (T ) ∼= U s has order |U |.

Now, [U s, T s] = 0, so U s 6 CW (T s) and therefore W = CW (T ) + CW (T s). Consider

CW (T ) ∩ CW (T s). This space commutes with 〈T, T s〉 = X and so must be trivial. Now

by an isomorphism theorem,

U s ∼=W

CW (T )=CW (T ) + CW (T s)

CW (T )∼=

CW (T s)

CW (T ) ∩ CW (T s)∼= CW (T s).

Since W is X-invariant, and therefore s-invariant, CW (T s) = CW (T )s and so CW (T ) has

order |U | and therefore |W | = |CW (T )||U s| = |U |2 and W = U ⊕ U s.

Let 0 6= v ∈ U s. If [v, t] = 0 for any 1 6= t ∈ T then v ∈ CV (T s)∩CV (t) = CV (〈T s, t〉) =

CV (X) = 0 (by Lemma 2.23). Hence CT (v) = 1 and so the set [v, t] = v · t − v|t ∈

T ⊆ [v, T ] has order |T | = 2n. Since T acts quadratically on V , [v, T, T ] = 0 and so

[v, T ] 6 CW (T ) = U . Hence |T | 6 |U |. But we already saw that |U | 6 2n and therefore

|U | = |T | = 2n and so |W | = 22n. Also, CT (v) = 1 implies that T acts regularly on

non-zero vectors in U s. Similarly T s acts regularly on non-zero vectors in U . So suppose

W1 is a proper X-invariant submodule of W . If u ∈ U# is in W1 then all of U is in W1

and then so is U s and so W = W1. Similarly if us in W1 then so is all of U s and again all

of U . So suppose u1, u2 ∈ U# and w = u1 +us2 ∈ W1. Then for t ∈ T#, wt = u1 +ust

2 ∈ W

and then w − wt = us2 − ust

2 ∈ U ∩W1 = 0 and so us2 = ust

2 which contradicts that T

acts regularly on U s. Hence W is an irreducible LX-module.

By Lemma 2.5, EndLH(U) is a field of order |U | and so EndLH(U) ∼= K. Notice that

any element of EndLX(W ) preserves CW (T ) = U and so any LX-endomorphism of W

restricts to give an endomorphism of U . Also, any LH-endomorphism of U , α say, extends

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to an endomorphism of W = U⊕U s, α : u1+us2 7→ α(u1)+α(u2)

s. Thus EndLX(W ) ∼= K.

Also, W has dimension 2n over L and is irreducible and therefore is a natural module

over L.

Now [V,X] 6 [V, T ] + [V, T s] 6 CV (T ) + CV (T s) and CV (T ) is completely reducible

as a product of irreducible H-modules say CV (T ) = U1 ⊕ . . .⊕ Ur. Thus

CV (T ) + CV (T s) = (U1 ⊕ U s1 )⊕ . . .⊕ (Ur ⊕ U s

r )

gives a direct sum of natural modules for X and so [V,X] is a submodule of this decom-

posed module and so also decomposes.

Lemma 8.5. The only irreducible module for X over L which admits an element of order

three acting fixed-point-freely is the natural module.

Proof. Suppose V is an irreducible LX-module admitting an element of order three acting

fixed-point-freely. Set E := EndLX(V ). Then V is an irreducible EX-module and satisfies

E = EndEX(V ). Recall that F is the algebraically closed field of characteristic two over

which we defined Brauer characters. By Lemma 2.8, the FX-module F⊗EV is irreducible

and also admits an element of order three acting fixed-point-freely. By Lemma 2.33,

F ⊗E V has dimension two over F and so V has dimension two over E. Now, [V, T ] is a

proper E-subspace of V and so has dimension one and therefore [V, T, T ] is trivial. So by

Lemma 8.4, [V,X] (as an LX-module) is a direct sum of natural LX-modules. However,

some x ∈ X acts fixed-point-freely on V and so [V,X] > [V, x] = [V, x] ⊕ CV (x) = V

by coprime action. Thus, V is a direct sum of natural LX-modules. Finally, since V is

irreducible, V must be a natural LX-module.

Lemma 8.6. Let K/L be a field extension and let B be a group. Let U1 and U2 be KB-

modules such that EndLB(Ui) ∼= K for i = 1, 2. Suppose that there exists an LB-module

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isomorphism α : U1 → U2. Then there exists β ∈ Gal(K/L) such that for every u ∈ U1

and every k ∈ K, α(ku) = β(k)α(u).

Proof. Let λi : K → EndLB(Ui) be the ring isomorphism where λi(k) is the LB-homomorphism

induced by multiplication by k. Clearly λi(k+ j) = λi(k) + λi(j) and λi(kj) = λi(k)λi(j)

for j, k ∈ K. For each k ∈ K, αλ1(k)α−1 is an LB-endomorphism of U2 and so applying

λ−12 to it gives an element of K, λ−1

2 (αλ1kα−1). Consider the map β : K → K given by

β(k) = λ−12 (αλ1(k)α

−1). We check this is a ring homomorphism. Let k, j ∈ K then

β(k + j) = λ−12 (αλ1(k + j)α−1)

= λ−12 (α(λ1(k) + λ1(j))α

−1)

= λ−12 ((αλ1(k)α

−1) + (λ1(j)α−1))

= λ−12 (αλ1(k)α

−1) + λ−12 (αλ1(j)α

−1)

= β(k) + β(j).

Also,

β(kj) = λ−12 (αλ1(kj)α

−1)

= λ−12 (αλ1(k)λ1(j)α

−1)

= λ−12 (αλ1(k)α

−1)λ−12 (αλ1(j)α

−1)

= β(k)β(j).

So β is a non-zero field homomorphism and thus an isomorphism. For l ∈ L, αλ1(l)α−1

is just multiplication of U2 by l as α and α−1 are L-linear maps. So αλ1(l)α−1 = λ2(l)

and therefore β(l) = λ−12 (λ2(l)) = l. Thus β is an automorphism of K fixing L. Now

for each u ∈ U1 we can find v ∈ U2 such that u = α−1(v). Thus α(ku) = α(kα−1(v)) =

(αλ1(k)α−1)(v) = k1α(u) where k1 ∈ K satisfies λ2(k1) = αλ1(k)α

−1 and therefore k1 =

λ−12 (αλ1(k)α

−1) = β(k) as required.

Lemma 8.7. Let T ∈ Syl2(X) and let Cp−1∼= H ≤ NX(T ) be a complement to T in

NX(T ). Then T is a KH-module. Let V,W be natural KX-modules and σ a Frobenius

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map such that V σ and W are isomorphic as KX-modules. Let V1 = CV (T ) and W1 =

CW (T ).

(i) If α : T → HomK(V1,W1) is an LH-module isomorphism then there exists β ∈

Gal(K/L) such that for 0 6= k ∈ K, β(k2) = k−1σ(k).

(ii) If α : T → HomK(V1,W/W1) is an LH-module isomorphism then there exists β ∈

Gal(K/L) such that for 0 6= k ∈ K, β(k2) = k−1σ(k−1).

(iii) If α : T → HomK(V/V1,W1) is an LH-module isomorphism then there exists β ∈

Gal(K/L) such that for 0 6= k ∈ K, β(k2) = kσ(k).

(iv) If α : T → HomK(V/V1,W/W1) is an LH-module isomorphism then there exists

β ∈ Gal(K/L) such that for 0 6= k ∈ K, β(k2) = kσ(k−1).

Proof. Identify X with the 2-dimensional representation affording the module V . Choose

T to be the upper triangular matrices and H to be the diagonal elements. Fix the

parameterizations K× → H of H by k 7→ hk and K → T of T by l 7→ tl where

hk =

(k−1 0

0 k

), tl =

(1 l

0 1

).

So T is a KH-module when we define ktl := tkl and tl · hk := h−1k tlhk = tk2l for every

k ∈ K,hk ∈ H, tl ∈ T . In particular tl ·hk = k2 ·tl. By Lemma 2.5, we see EndLH(T ) ∼= K.

Now given our choice of representation we can identify both V and W σ−1with the

vector space (a, b)|a, b ∈ K. For (a, b) ∼ v ∈ V and hk ∈ H we have

v · hk ∼(

a b

)( k−1 0

0 k

)=(

k−1a kb

)and for (a, b) ∼ w ∈ W = V σ

w · hk ∼(

a b

)( σ(k−1) 0

0 σ(k)

)=(

σ(k−1)a σ(k)b).

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Thus we identify V1 and W σ−1

1 with (0, b)|b ∈ K and V/V1 and (W/W1)σ−1

with (a, 0)+

〈(0, 1)〉|a ∈ K. So for each hk ∈ H,

(i) v · hk = kv for all v ∈ V1,

(ii) w · hk = σ(k)w for all w ∈ W1,

(iii) v · hk = k−1v for all v ∈ V/V1,

(iv) w · hk = σ(k−1)w for all w ∈ W/W1.

Consider the K-vector space MAB := HomK(A,B) where A = V1 or V/V1 and B = W1

or W/W1. Since A and B are 1-dimensional K-modules and elements of MAB commute

with K, each φ ∈ MAB is determined by where it maps one non-zero element of A.

Therefore, M has order |B| = 2n. Also, MAB is aKH-module when we define (f ·hk)(a) :=

f(a ·h−1k ) ·hk for f ∈MAB, k ∈ K, hk ∈ H and a ∈ A. Now K embeds into EndLX(MAB)

and MAB is a vector space over EndLX(MAB). However, |MAB| = |K| so MAB has

dimension one over K and therefore we must have that K ∼= EndLX(MAB).

We are given the LH-module isomorphism α and so we can apply Lemma 8.6 with

U1 = T , U2 = MAB, to see that there exists β ∈ Gal(K/L) such that β(k2) ·α(t) = α(k2 ·t)

for every t ∈ T and every k ∈ K. Now α(k2 · t) = α(t · hk) = α(t) · hk as α is an LH-

module homomorphism. The action we defined of H on MAB means for any a ∈ A,

(α(t) · hk)(a) = α(t)(a · h−1k ) · hk. We have four possibilities:

(i) If A = V1, B = W1 then α(t)(a · h−1k ) · hk = σ(k)α(t)(k−1a) = k−1σ(k)α(t). Thus

β(k2) = k−1σ(k).

(ii) If A = V1, B = W/W1 then α(t)(a · h−1k ) · hk = σ(k−1)α(t)(k−1a) = k−1σ(k−1)α(t).

Thus β(k2) = k−1σ(k−1).

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(iii) If A = V/V1, B = W1 then α(t)(a · h−1k ) · hk = σ(k)α(t)(ka) = kσ(k)α(t). Thus

β(k2) = kσ(k).

(iv) If A = V/V1, B = W/W1 then α(t)(a · h−1k ) · hk = σ(k−1)α(t)(ka) = kσ(k−1)α(t).

Thus β(k2) = kσ(k−1).

Lemma 8.8. (Meierfrankenfeld)[5, 4.2] Let V be an LX-module with a submodule W

such that V/W and W are natural modules over L. Then V is completely reducible.

Proof. By Lemma 2.25, any non-identity element of X with odd order, g say, acts fixed-

point-freely on W and on V/W . Therefore g acts fixed-point-freely on V and so CV (X) 6

CV (g) = 0 and by coprime action, V = [V, g] ⊕ CV (g) = [V, g] 6 [V,X] 6 V . Thus if T

acts quadratically on V then V is completely reducible by Lemma 8.4 and we are done, so

assume not. Set V = V ⊗L K so V is a module over K with dimK(V ) = dimL(V ). Every

irreducible KX-submodule of V is a natural module over K or a Galois conjugate by

Lemma 2.33. Assume for a contradiction that there is an indecomposable KX-invariant

section V0 of V of dimension four. Again we can assume that T does not act quadratically

on V0.

Let W0 be a 2-dimensional submodule of V0 and let σ ∈ Aut(K) such that (V0/W0)σ

and W0 are isomorphic as KX-modules. Find 0 6 s < n such that σ raises elements of k

to the power 2s. Set U = CW0(T ) then U is a 1-dimensional subspace of W0 as is CW0(t)

for each t ∈ T# so U = CW0(t) also. Let R be the preimage in V0 of CV0/W0(T ) then R has

dimension three and [R, T ] 6 W0. For each t ∈ T , [R, t, t] = 0 and so [R, t] 6 CW0(t) = U .

Thus [R, T ] = U .

Since W0 is 2-dimensional, by Lemma 1.1, we get [W0, T, T ] = 0 and since U =

CW0(T ) is 1-dimensional, we get U = [W0, T ]. Similarly R/W0 = [V0/W0, T ] and so

R = W0 +[V0, T ]. Since H is transitive on the non-trivial elements of T , H is transitive on

the spaces [V0, t] for t ∈ T# as [v, th] = v·(th)−v = (v·h−1)·t·h−(v·h−1)·h ∈ [V0, t]·h. Now

128

if [V0, t, T ] = 0 for all t ∈ T then [V0, T, T ] = 0. So for some t ∈ T# (and thus all t ∈ T#)

we have [V0, t, T ] 6= 0. The space [V0, t] is invariant under T so using that [V0, t] 6 R

and [R, T ] = U we have 0 6= [V0, t, T ] 6 [V0, t] ∩ [R, T ] = [V0, t] ∩ U . Thus U 6 [V0, t] for

each t ∈ T . If [V0, t] has dimension one then [V0, t] = U but then [V0, t, T ] = 0 and so

this containment is strict. Since [V0, t, t] = 0, [V0, t] 6= [V0, T ] else [V0, T, t] = 0 for every

t ∈ T and again we would have [V0, T, T ] = 0. So we have U [V0, t] [V0, T ] = R.

Also, U 6 CV0(T ) 6 CV0(t) and [V0, t] 6 CV0(t) R. It follows from the dimensions that

CV0(t) = [V0, t] and when we observe that [V0, t] 6= CV0(T ) (else [V0, t, T ] = 0) we get

U = CV0(T ) [V0, t] = CV0(t) [V0, T ] = R.

Now R/U is a 2-dimensional KH-module so by Maschke’s Theorem it is completely

reducible. By Lemma 2.2, all irreducible submodules are 1-dimensional. Therefore, we

can find some 2-dimensional H-invariant submodule D 6 R with D ∩ W0 = U and

D + W0 = R and since CV0(T ) = U , [D,T ] 6= 0. We see immediately that [D,T ] = U

since [D,T ] 6 [R, Y ] = U . Consider the map,

α : T −→ HomK(D/U,U)

t 7−→ λt,

where λt : d+U 7→ [d, t]. This map is well defined as [U, T ] = 0. Also, since α is non-zero

and both T and HomK(D/U,U) are 1-dimensional, α is an isomorphism. By Lemma 8.7,

we see T is a KH-module with H-action t · h = h−1th and HomK(D/U,U) is a KH-

module with H-action described by (f · h)(x) = f(x · h−1) · h for f ∈ HomK(D/U,U),

129

h ∈ H and x ∈ D/U . Thus, for t ∈ T , h ∈ H, and d+ U ∈ D/U we get

(λt · h)(d+ U) = λt(d+ U · h−1) · h

= λt(d · h−1 + U) · h

= [d · h−1, t] · h

= (d · h−1 · t− d · h−1) · h

= d · h−1 · t · h− d · h−1 · h

= [d, t · h]

= λt·h(d+ U).

Therefore α(t · h) = λt·h = λt · h = α(t) · h and so α is an LH-module isomorphism.

We have D/U = D/W0 ∩ D ∼= W0 + D/W0 = R/W0 as KH-modules so we can

apply Lemma 8.7 (i) to show there exists some β ∈ Gal(K/L) = Aut(K) such that for

each non-zero k ∈ K, β(k2) = k−1σ(k). We find 0 6 b < n such that β(k) = k2b. Thus

(k2)2b= k−1k2 and therefore 2b+1 ≡ 2s−1 mod 2n−1. Since b, s < n, 2b+1−2s+1 = 2n−1

and then it follows that s = 1 and b = n− 1.

Consider the map,

γ : T −→ HomK(V0/R,R/D)

t 7−→ µt,

where µt : v + R 7→ [v, t] + D. We proceed as before. Firstly µt is well defined as

[R, T ] = U 6 D. Also, γ is non-zero as R = [V0, T ] and both T and HomK(V0/R,R/D)

are 1-dimensional so γ is an isomorphism. As before T and HomK(V0/R,R/D) are KH-

130

modules. Thus for t ∈ T , h ∈ H, and v +R ∈ V0/R we get

(λt · h)(v +R) = λt(v +R · h−1) · h

= λt(v · h−1 +R) · h

= [v · h−1, t] +D · h

= (v · h−1 · t− v · h−1 +D) · h

= v · h−1 · t · h− v · h−1 · h+D

= [v, t · h] +D

= λt·h(v +R).

Thus γ(t · h) = λt·h = λt · h = γ(t) · h and so γ is an LH-module isomorphism.

Now R/D = D + W0/D ∼= W0/(W0 ∩ D) = W0/U as KH-modules so we can apply

Lemma 8.7 (iv) to show there exists some δ ∈ Gal(K/L) such that for each non-zero

k ∈ K, δ(k2) = kσ(k−1). So, as before, we have some 0 6 d < n such that δ(k) = k2d

and then 2d+1 ≡ −2s + 1 mod 2n − 1. Therefore 2n − 1 divides 2d+1 + 2s − 1 = 2d+1 + 1

as s = 1. It follows that 2n − 1 = 2d+1 + 1 and then d = 0 and n = 2. So suppose

X ∼= SL2(4). Then T is normalized by an element of order three. Moreover, the element

of order three, x say, acts fixed-point-freely on T and fixed-point-freely on V0. Let vt be

some element of the split semidirect product V0T that is fixed by x. Then (vt)x = vt and

so v−1vx = tt−x ∈ V0 ∩ T = 1. Thus vx = v and tx = t and so CV0T (x) = CV0(x)CT (x).

This means x acts fixed-point-freely on the split semidirect product V0T . Thus by a lemma

of Burnside, 1.24, V0T has nilpotence class at most two. This gives a contradiction since

now 1 = [V0T, V0T, V0T ] > [V0, T, T ] and so T acts quadratically on V0.

Corollary 8.9. Let V be an LX-module such that an element of order three in X acts

fixed-point-freely on V . Then V is completely reducible as a product of natural modules

for L.

Proof. By Lemma 8.5, the only irreducible submodules of V are natural modules over L.

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Suppose V is not completely reducible then there must be some submodule of dimension

4n which cannot be reduced and satisfies Lemma 8.8 which is a contradiction.

Lemma 8.10. Let G be a finite group with a normal 2-subgroup, Q, such that X :=

G/Q ∼= SL2(2n) for some n > 2 and some element of order three in G acts fixed-point-

freely on Q. Suppose G is minimal subject to Q being non-elementary abelian then Q is

abelian.

Proof. Suppose Q is non-abelian then by minimality, Φ(Q) = Q′ = Z(Q) = Ω1(Q). Set

Q = Q/Φ(Q) then Q and Q′ are LX-modules (where vector addition is group multipli-

cation and the action of X is by conjugation) admitting an element of order three acting

fixed-point-freely. Construct the KX-modules A := Q ⊗L K and B := Q′ ⊗L K then

these also admit a fixed-point-free element of order three.

We define a surjective map,

φ : A× A → B

(p⊗L k1, q ⊗L k2) 7→ [p, q]⊗L k1k2,

We need to check that φ is K-balanced (see Definition 2.7). Let p, q, r ∈ Q, k, k1, k2 ∈

K. Then

φ((pq ⊗ k1, r ⊗ k2)) = [pq, r]⊗ k1k2

= [p, r][q, r]⊗ k1k2

= [p, r]⊗ k1k2 + [q, r]⊗ k1k2

= φ((p⊗ k1, r ⊗ k2)) + φ((q ⊗ k1, r ⊗ k2))

132

and

φ((p⊗ k + k1, q ⊗ k2)) = [p, q]⊗ (k + k1)k2

= [p, q]⊗ kk2 + k1k2

= [p, q]⊗ kk2 + [p, q]⊗ k1k2

= φ((p⊗ k, q ⊗ k2)) + φ((p⊗ k1, q ⊗ k2))

and finally

kφ((p⊗ k1, q ⊗ k2)) = k[p, q]⊗ k1k2

= [p, q]⊗ kk1k2

= φ((p⊗ kk1, q ⊗ k2))

= φ((p⊗ k1, q ⊗ kk2)).

The map is bilinear in the second coordinate in exactly the same way and so φ is K-

balanced. By the definition of A⊗KA, there must be aK-homomorphism α : A⊗KA→ B

such that αθT = φ where θT : A×A→ A⊗K A is the map θT : (p, q) 7→ p⊗K q. We aim

to show the action of X on these modules commutes with α. So let x ∈ X, k ∈ K and

p, q ∈ Q. Then

φ((p⊗ k1, q ⊗ k2)) · x = [px, qx]⊗ k1k2 = φ(px ⊗ k1, qx ⊗ k2) = φ((p⊗ k1, q ⊗ k2) · x)

and

θT ((p⊗k1, q⊗k2))·x = (p⊗k1)⊗K (q⊗k2)·x = (px⊗k1)⊗K (qx⊗k2) = θT ((p⊗k1, q⊗k2)·x).

So φ and θT commute with the action of X and since αθT = φ, α commutes with the

action of X also. So α is a KX-homomorphism. However, by Corollary 8.3, such a map

must be the zero map and so Q must be abelian.

Theorem 8.11. Let G be a finite group with a normal 2-subgroup, Q, such that X :=

G/Q ∼= SL2(2n) for some n > 2 and some element of order three in G acts fixed-point-

freely on Q. Then Q is elementary abelian and a direct product of natural modules.

133

Proof. If Q were elementary abelian and hence an LX-module then by Corollary 8.9, we

would be done. Thus all rests on proving Q is elementary abelian.

Let G be a minimal counter example to the theorem. Then Q is not elementary

abelian but Q is abelian by Lemma 8.10 and so contains elements of order four. By

minimality, Q/Ω1(Q) is elementary abelian and therefore Q has exponent four. Let W be

the characteristic subgroup of Q generated by all elements of order four. Then W/Ω1(W )

is an LX-module and so is decomposable into a direct product of natural modules. Let W1

be the preimage in W of some irreducible submodule of W/Ω1(W ). Then the irreducible

submodule has dimension 2n and soW1 is a direct product of 2n cyclic groups of order four.

Since X acts on W1, there is a homomorphism Φ : X → Aut(W1) ∼= Aut(Z4 × . . .× Z4).

In Section 1.5 we investigated such groups and saw that each element of X can be viewed

as a 2n× 2n matrix with entries in Z4 and with odd determinant. We also saw how each

of these matrices acts on the natural module W1/Ω1(W1) and that the corresponding

representation simply restricts matrix entries modulo 2. By Lemma 1.35, we have the

map

α : X −→ GL2n(Z4)

and the representation affording the natural LX-module

β : X −→ GL2n(2)

where for each x ∈ X, the entries in α(x) are congruent modulo 2 to the entries in β(x).

We saw in Lemma 2.35 that given any algebraic conjugate of the natural KX-module,

we can view it as a module over L and as such it is the natural LX-module. We represent

elements of X as 2 × 2 matrices with entries in K. We then obtain a natural LX-

representation as follows. The field K is a vector space of dimension n over L and each

a ∈ K defines an L-linear transformation, namely, for each a ∈ K, λa : k 7→ ak for each

134

k ∈ K. So if we fix a basis for K over L we can make the n-dimensional matrix with

entries in L representing λa. We will call this matrix [a] for each a ∈ K. Notice that

the map K× → GLn(2) is a homomorphism and in particular this means for a, b ∈ K×

[a][b] = [b][a]. Now consider the map

β1 : X −→ GL2n(2)

such that β1 :

(a b

c d

)7→

([a] [b]

[c] [d]

). Then β1 is a natural LX-representation of X.

The natural LX-representation is unique up to equivalence. Lemma 1.35 implies that

there is a homomorphism α1 : X → GL2n(Z4) such that for each x ∈ X entries of α1(x)

are congruent modulo 2 to entries of β1(x). Consider the images of the matrices

(1 a

0 1

)for each a ∈ K. It follows that such a matrix is represented via α1 as(

I + 2A1 [a] + 2A2

0 + 2A3 I + 2A4

)

where A1, A2, A3, A4 are n× n matrices with 0, 1 entries. This matrix has order two and

so we can simplify to get (I + 2A1 [a] + 2A2

0 −I + 2[a]−1A1[a]

).

We can now choose non-zero elements a, b, c ∈ K such that a + b + c = 0 thus we follow

the same process for a, b and c and we see(I + 2A1 [a] + 2A2

0 −I + 2[a]−1A1[a]

)(I + 2B1 [b] + 2B2

0 −I + 2[b]−1B1[b]

)(I + 2C1 [c] + 2C2

0 −I + 2[c]−1C1[c]

)

is equal to the identity matrix. This gives that A1 + B1 + C1 is the zero matrix and

[a]−1A1[a] + [b]−1B1[b] + [c]−1C1[c] is the identity matrix. Now consider Corollary 2.12.

We can view [a], [b] and [c] as matrices over K and so there exists M ∈ GLn(K) such that

[a]M = diag(a), [b]M = diag(b) and [c]M = diag(c). Since A1, B1 and C1 are in Mn(2), it

135

makes sense to conjugate each by M and so we have the following.

I = IM = ([a]−1)MAM1 [a]M + ([b]−1)MBM

1 [b]M + ([c]−1)MCM1 [c]M = AM

1 +BM1 +CM

1 = 0M

which is a contradiction. Therefore, there is no minimal counter example and so Q is

elementary abelian and a direct product of natural modules over L.

136

Appendix

Lemma A. 1. Let p be a prime and let a and b be natural numbers. Then pa− 1 divides

pb − 1 if and only if a divides b.

Proof. If a divides b then we can write pb − 1 = (pa − 1)(pb−a + pb−2a + . . .+ pa + 1) and

so pa− 1 divides pb− 1. So assume that pa− 1 divides pb− 1. Choose an integer k so that

ka ≤ b < (k + 1)a. Then pa − 1 divides pka − 1 and so pa − 1 divides pb − 1− (pka − 1) =

pka(pb−ka−1). Since pa−1 is coprime to p, pa−1 divides pb−ka−1. But pb−ka−1 < pa−1

by the choice of k. So pb−ka − 1 = 0 and this means that b = ka as required.

Lemma A. 2. Let p be a prime and let m and n be natural numbers with greatest common

divisor g. Then pg − 1 is the greatest common divisor of pn − 1 and pm − 1 for any prime

p.

Proof. Since g divides m and n, pg−1 is a common divisor of both pn−1 and pm−1. Let

us assume m > n and m is not a multiple of n or else the result holds trivially. We will use

the Euclidean algorithm with pn−1 and pm−1 to find their greatest common divisor. Find

the integer k such that 0 < m− kn < n. Notice that pm − 1 = (pn − 1)pm−n + (pm−n − 1)

and also pm−n−1 = (pn−1)pm−2n +(pm−2n−1). So substituting the second into the first

gives

pm − 1 = (pn − 1)(pm−n + pm−2n) + (pm−2n − 1)

137

and we continue in this way k times to get

pm − 1 = (pn − 1)(pm−n + pm−2n + . . .+ pm−kn) + (pm−kn − 1).

This gives the first step in the Euclidean algorithm and so we have GCD(pm−1, pn−1) =

GCD(pn − 1, pm−kn − 1). We continue in this way until the process terminates and we

have that the greatest common divisor of pm − 1 and pn − 1 has the form pl − 1 for some

positive integer l. But now we have that l divides both n and m and so l divides g and

furthermore pl − 1 divides pg − 1. Thus pg − 1 is the greatest common divisor of pm − 1

and pn − 1.

Lemma A. 3. Let p be a prime and let F be a field of order pn for some natural number

n. Then any subfield of F has order pm where m divides n. Furthermore if m divides n

then there is a subfield of F of order 2m.

Proof. Suppose K is a subfield of F then K is a field of characteristic p so has order pm

for some m 6 n. Also K× is a subgroup of F×. So pm − 1 divides pn − 1. However this

can only occur when m divides n.

So suppose now that some natural number m divides n. The group F× is cyclic of

order pn−1. Since pm−1 divides pn−1, there is a unique subgroup H say of F× of order

pm − 1. We claim K := H, 0F is a subfield of F of order pm. We need to check K is

closed under addition. So let a, b ∈ K. If either a or b is zero it is clear that a + b ∈ K

so we will assume a, b ∈ H then both a and b have order dividing pm − 1 so apm= a and

bpm

= b. Notice that for any natural number r,

(a+ b)r = apm

+ rar−1b+r(r − 1)

2ar−2b2 + . . . rabr−1 + br.

So if r = pm then (a+ b)pm= apm

+ bpm

= a+ b. Thus either (a+ b)pm−1 = 1 or a+ b = 0.

If (a + b)pm−1 = 1 then a + b lies in the unique subgroup, H, of F× of order pm − 1. In

138

either case a + b ∈ K so K is closed under addition. Now we show K contains additive

inverses. We have seen that K is closed under addition so fix some non-zero k ∈ K. Then

define a map τ : K → K such that for l ∈ K, τ : l 7→ l + k. This map is clearly injective

and since K is a finite set, it must be surjective. In particular this means that there exists

some l ∈ K such that l + k = 0 and so K has additive inverses. All other axioms follow

trivially so K is a subfield of F .

Lemma A. 4. Let G be a faithful completion of an amalgam of type G2(3) and Γ the

coset graph. Then G is locally 7-arc transitive on Γ.

Proof. Let (α1, α2, α3, α4, α5, α6, α7) be a path of seven vertices. By Lemma 3.4, it

is necessary to prove that the stabilizer of the path (α1, α2, . . . , αi) is transitive on

Γ(αi)\αi−1 for each i 6 7. By the Orbit-Stabilizer Theorem, it is enough to show

|Gα1,α2,...,αi: Gα1,α2,...,αi+1

| = 3 for each i 6 7. We apply Lemma 6.4 and Lemma 6.1

several times. Firstly i = 1 is just edge transitivity which is always true in a coset

graph. When i = 2, Gα1α2 ∩ Gα2α3 has Sylow 3-subgroup Sα1α2 ∩ Sα2α3 = Qα2 and so

3 divides |Gα1,α2 : Gα1,α2,α3|. When i = 3, Gα1,α2,α3 ∩ Gα2,α3,α4 has Sylow 3-subgroup,

Qα2 ∩ Qα3 = Qα2α3 , with order 34 and so 3 divides |Gα1,α2,α3 : Gα1,α2,α3,α4|. When i = 4,

Gα1,α2,α3,α4 ∩ Gα2,α3,α4,α5 has Sylow 3-subgroup, Qα2α3 ∩ Qα3α4 = Zα3 , with order 33 and

so 3 divides |Gα1,α2,α3,α4 : Gα1,α2,α3,α4,α5|. When i = 5, Gα1,α2,α3,α4,α5 ∩ Gα2,α3,α4,α5,α6 has

Sylow 3-subgroup, Zα3 ∩ Zα4 = Zα3α4 , with order 32 and so 3 divides |Gα1,α2,α3,α4,α5 :

Gα1,α2,α3,α4,α5,α6|. When i = 6 Gα1,α2,α3,α4,α5,α6 ∩ Gα2,α3,α4,α5,α6,α7 has Sylow 3-subgroup,

Zα3α4 ∩ Zα4α5 = Yα4 , with order 3 and so 3 divides |Gα1,α2,α3,α4,α5,α6 : Gα1,α2,α3,α4,α5,α6,α7|.

When i = 7 Gα1,α2,α3,α4,α5,α6,α7 ∩Gα2,α3,α4,α5,α6,α7,α8 has Sylow 3-subgroup, Yα4 ∩ Yα5 = 1,

and so 3 divides |Gα1,α2,α3,α4,α5,α6,α7 : Gα1,α2,α3,α4,α5,α6,α7,α8|. Hence G acts locally 7-arc

transitive on Γ.

139

Bibliography

[1] M. Aschbacher. Finite group theory, volume 10 of Cambridge Studies in Advanced

Mathematics. Cambridge University Press, Cambridge, second edition, 2000.

[2] Michael Aschbacher. Sporadic groups, volume 104 of Cambridge Tracts in Mathe-

matics. Cambridge University Press, Cambridge, 1994.

[3] Helmut Bender. Transitive Gruppen gerader Ordnung, in denen jede Involution genau

einen Punkt festlaßt. J. Algebra, 17:527–554, 1971.

[4] Wieb Bosma, John Cannon, and Catherine Playoust. The Magma algebra system.

I. The user language. J. Symbolic Comput., 24(3-4):235–265, 1997. Computational

algebra and number theory (London, 1993).

[5] Andrew Chermak. A class of p-local finite groups of local characteristic p. Unpub-

lished, 2007.

[6] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson. Atlas of

finite groups. Clarendon Press, Oxford, 1985.

[7] A. Delgado, D. Goldschmidt, and B. Stellmacher. Groups and graphs: new results

and methods, volume 6 of DMV Seminar. Birkhauser Verlag, Basel, 1985. With a

preface by the authors and Bernd Fischer.

140

[8] Klaus Doerk and Trevor Hawkes. Finite soluble groups, volume 4 of de Gruyter

Expositions in Mathematics. Walter de Gruyter & Co., Berlin, 1992.

[9] Walter Feit and John G. Thompson. Finite groups which contain a self-centralizing

subgroup of order 3. Nagoya Math. J., 21:185–197, 1962.

[10] Rachel Fowler. A 3-local characterization of the Thompson sporadic simple group.

PhD thesis, 2006.

[11] Rachel Fowler and Christopher Parker. A 3-local characterization of the Thompson

sporadic simple group. In preparation.

[12] David M. Goldschmidt. Automorphisms of trivalent graphs. Ann. of Math. (2),

111(2):377–406, 1980.

[13] Daniel Gorenstein. Finite groups. Chelsea Publishing Co., New York, second edition,

1980.

[14] Daniel Gorenstein, Richard Lyons, and Ronald Solomon. The classification of the

finite simple groups. Number 3., volume 40.3 of Mathematical Surveys and Mono-

graphs. American Mathematical Society, Providence, RI, 1998.

[15] Robert L. Griess, Jr. Automorphisms of extra special groups and nonvanishing degree

2 cohomology. Pacific J. Math., 48:403–422, 1973.

[16] John L. Hayden. A characterization of the finite simple group PSp4(3). Canad. J.

Math., 25:539–553, 1973.

[17] Graham Higman. Odd Characterisations of Finite Simple Groups, Lecture Notes of

University of Michigan. Ann Arbor. 1968.

[18] Christopher Hillar and Darren Rhea. Automorphisms of finite abelian groups. Un-

published.

141

[19] I. Martin Isaacs. Character theory of finite groups. Academic Press [Harcourt Brace

Jovanovich Publishers], New York, 1976. Pure and Applied Mathematics, No. 69.

[20] Hans Kurzweil and Bernd Stellmacher. The theory of finite groups. Universitext.

Springer-Verlag, New York, 2004. An introduction, Translated from the 1998 German

original.

[21] Jack McLaughlin. Some groups generated by transvections. Arch. Math. (Basel),

18:364–368, 1967.

[22] Christopher Parker and Peter Rowley. Symplectic amalgams. Springer Monographs

in Mathematics. Springer-Verlag London Ltd., London, 2002.

[23] Christopher Parker and Peter Rowley. Local characteristic p completions of weak

BN -pairs. Proc. London Math. Soc. (3), 93(2):325–394, 2006.

[24] A. R. Prince. A characterization of the simple groups PSp(4, 3) and PSp(6, 2). J.

Algebra, 45(2):306–320, 1977.

[25] Arthur Ranum. The group of classes of congruent matrices with application to the

group of isomorphisms of any abelian group. Trans. Amer. Math. Soc., 8(1):71–91,

1907.

[26] Jean-Pierre Serre. Trees. Springer-Verlag, Berlin, 1980. Translated from the French

by John Stillwell.

[27] Stephen D. Smith and A. P. Tyrer. On finite groups with a certain Sylow normalizer.

I, II. J. Algebra, 26:343–365; ibid. 26 (1973), 366–367, 1973.

[28] Robert Steinberg. Endomorphisms of linear algebraic groups. Memoirs of the Ameri-

can Mathematical Society, No. 80. American Mathematical Society, Providence, R.I.,

1968.

142

[29] Jacques Tits and Richard M. Weiss. Moufang polygons. Springer Monographs in

Mathematics. Springer-Verlag, Berlin, 2002.

143