The Vertex Primitive and Vertex Bi-Primitive S-Arc Regular Graphs

Home , Conway group, Hall subgroup, Quaternion group

THE VERTEX PRIMITIVE AND VERTEX BI-PRIMITIVE S-ARC REGULAR GRAPHS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of the Ohio State University

Liang Niu, B.S., M.S.

*****

The Ohio State University 2008

Dissertation Committee: Approved by Professor Akos´ Seress, Advisor

Professor Ronald Solomon Advisor Professor Michael Davis Graduate Program in Mathematics

ABSTRACT

A complete classiﬁcation is given of vertex primitive and vertex bi-primitive s-arc regular graphs with s ≥ 3. In particular, it is shown that the Petersen graph and

Coxeter graph are the only vertex primitive 3-arc regular graphs, and that vertex bi-primitive 3-arc regular graphs consist of the complete bipartite graph K3,3, the standard double covers of the Petersen graph and Coxeter graph, and three graphs admitting PGL(2, 11), PGL(2, 13) and PΓL(2, 27), respectively.

ii To My Family

iii ACKNOWLEDGMENTS

First and foremost, I would like to thank my advisor Akos´ Seress for allowing me to work on this thesis. I have appreciated the many conversations we have had, both mathematical and otherwise, and your consistently warm and kind demeanor. You have always been quick to answer questions. It has been much appreciated.

I would also like to thank my other committee members, Professor Ronald Solomon and Professor Michael Davis, for agreeing to serve on my committee. Professor

Solomon gave me great help when I was working on this paper, and I would like to say that you were one of the very best professors I have had. You have been very giving of your time and gave great inspiration to me.

I would also like to thank my former advisor, Professor Cai Heng Li. Although you have not been my advisor for a long time, you still gave me great advice on this paper. I highly appreciate your help.

iv VITA

2003-Present ...... Graduate Teaching Associate, The Ohio State University

1999-2002 ...... Graduate Student, Peking University

2002 ...... M.S. in Mathematics, Peking University

1997 ...... B.S. in Mathematics, Hubei University

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Algebraic Graph theory

v TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita...... v

List of Figures ...... viii

CHAPTER PAGE

1 Introduction and main results ...... 1

1.1 Basic background ...... 1 1.2 Main results ...... 5

2 Vertex stabilizers of s-arc regular graphs ...... 8

2.1 Theorem for the vertex stabilizers of s-arc regular graphs 8 2.2 The vertex stabilizers of s-arc regular graphs: s = 3 . . . 9 2.3 The vertex stabilizers of s-arc regular graphs: s ≥ 4 . . . 31 2.4 A more general result ...... 33

3 A key lemma ...... 37

3.1 The key lemma ...... 37 3.2 Some useful facts ...... 38 3.3 Proof of the key lemma, in the case p=2 ...... 43 3.4 Proof of the key lemma, in the case p odd ...... 45

4 Proof of the main theorems ...... 56

4.1 Basic background on coset graphs ...... 56 4.2 Proof of Theorem 1.2.1 and Corollary 1.2.2 ...... 58

vi 4.3 Proof of Theorem 1.2.3 and Corollary 1.2.4 ...... 63

Bibliography ...... 77

vii LIST OF FIGURES

FIGURE PAGE

1.1 The Petersen Graph and Coxeter Graph ...... 3

viii CHAPTER 1

INTRODUCTION AND MAIN RESULTS

1.1 Basic background

Denote by Γ an undirected connected graph with vertex set V and edge set E. For a vertex α ∈ V , denote by Γ (α) the neighborhood of α, that is the set of vertices adjacent to α in Γ . For a positive integer s, an s-arc of Γ is an (s + 1)-tuple

(α0, α1, . . . , αs) of vertices such that αi ∈ Γ (αi−1) for 1 ≤ i ≤ s and αi−1 6= αi+1 for 1 ≤ i ≤ s − 1. A regular graph is a graph where each vertex has the same number of adjacent vertices, i.e. every vertex has the same valency. If Γ is regular of valency at least 3 and G ≤ Aut(Γ ) is transitive on the set of s-arcs of Γ , then Γ is called

(G, s)-arc transitive. If G is regular on the set of s-arcs, then Γ is called (G, s)-arc regular. (Recall that a permutation group G is regular on a set Ω if G is transitive and |G| = |Ω|.) A graph Γ is called s-arc regular if it is (Aut(Γ ), s)-arc regular.

The class of s-arc regular graphs is closely connected to some important classes of combinatorial objects, such as regular Mobius maps [16], near-polygonal graphs [15], and half-transitive graphs [18].

The purpose of this paper is to complete the classiﬁcation of vertex primitive and vertex bi-primitive s-arc regular graphs with s ≥ 3. Recall that a graph Γ is called

1 vertex primitive if Aut(Γ ) is a primitive permutation group on the vertex set V . For a bipartite graph Γ with bi-parts ∆1 and ∆2, G ≤ Aut(Γ ) is called bi-primitive on

+ + the vertex set V of Γ if G is primitive on the bi-parts ∆i, i = 1, 2, where G is the setwise stabilizer of ∆1 (also ∆2) in G. A graph Γ is called vertex bi-primitive if Aut(Γ ) is bi-primitive on the vertex set V of Γ .

Let Γ be a connected (G, s)-arc regular graph with vertex set V such that G is primitive or bi-primitive on V . If s ≥ 4, then (G, Γ ) can be read out from the main theorem of [14], which gives the classiﬁcation of vertex primitive or vertex bi-primitive s-arc transitive graphs when s ≥ 4. If s = 2 and G is primitive on V then (G, Γ ) is classiﬁed in [6]. In this paper, we classify (G, Γ ) for s = 3.

There are examples of (G, 3)-arc regular graphs with G primitive on the vertex set of the graph, such as the Petersen graph and Coxeter graph. First let us look at the Petersen graph, a regular graph of valency 3 with 10 vertices. The graph is ∼ ∼ the upper graph of Figure 1.1. It is well known that Aut(Γ ) = PGL(2, 5) = S5, and G = Aut(Γ ) is regular on the the set of 3-arcs. Further, G is primitive on the vertex ∼ set, since the vertex stabilizer Gα = D12 is maximal in G. Therefore the Petersen graph is a (G, 3)-arc regular graph with G = Aut(Γ ) primitive on the vertex set of the graph.

Next let us look at the Coxeter graph, a regular graph of valency 3 with 28 vertices.

The graph is the lower graph of Figure 1.1. We will prove later that Aut(Γ ) ∼= PGL(2, 7) and G = Aut(Γ ) is regular on the the set of 3-arcs. Further, we will ∼ see that G is primitive on the vertex set, since the vertex stabilizer Gα = D12 is

Figure 1.1: The Petersen Graph and Coxeter Graph

3 maximal in G. Therefore the Coxeter graph is another (G, 3)-arc regular graph with

G = Aut(Γ ) primitive on the vertex set of the graph.

In Corollary 1.2.2, we will see that the Petersen graph and the Coxeter graph are the only vertex primitive 3-arc regular graphs.

A permutation group G ≤ Sym(Ω) is called sharply 2-transitive if it is 2-transitive on Ω and the identity is the only element of G with more than one ﬁxed point. All sharply 2-transitive groups are listed below, see Theorem 9.4, chapter XII of [2]:

Theorem 1.1.1. Assume that G is a sharply 2-transitive permutation group. Then

d d G = Zp:Xω is aﬃne, |Xω| = p − 1, and one of the following holds:

d (I) Xω has a cyclic normal subgroup which is irreducible on Zp, and

d Xω ≤ ΓL(1, p ) = Zpd−1.Zd;

d d (II) Xω is solvable but has no irreducible cyclic normal subgroup on Zp, and (Xω, p ) is listed in the following table:

d Xω p SL(2, 3) 52

2 G48 7

2 SL(2, 3) × Z5 11

2 G48 × Z11 23

where G48 = SL(2, 3).Z2 = Q8.S3 is the group deﬁned in Deﬁnition 8.4, chapter XII of [2].

4 d (III) Xω is not solvable, and (Xω, p ) is listed in the following table:

d Xω p SL(2, 5) 112

2 SL(2, 5) × Z7 29

2 SL(2, 5) × Z29 59

1.2 Main results

For the primitive case,we have:

Theorem 1.2.1. Let Γ be a connected (G, 3)-arc regular graph such that G is primitive on the vertex set. Then either

(i) Γ is the Petersen graph and G ∼= PGL(2, 5), or

(ii) Γ is the Coxeter graph and G ∼= PGL(2, 7).

And we have the corollary:

Corollary 1.2.2. A connected graph Γ is vertex primitive 3-arc regular if and only if either

(i) Γ is the Petersen graph and Aut(Γ ) ∼= PGL(2, 5), or

(ii) Γ is the Coxeter graph and Aut(Γ ) ∼= PGL(2, 7).

The above corollary shows that the Petersen graph and the Coxeter graph are the only vertex primitive 3-arc regular graphs. However, by [6, 14], there are inﬁnitely many vertex primitive 2-arc regular graphs and 4-arc regular graphs.

5 For the bi-primitive case, we have:

Theorem 1.2.3. Let Γ be a connected (G, 3)-arc regular graph such that G is bi- primitive on the vertex set. Then one of the following holds:

(i) Γ = Kpd,pd , a complete bipartite graph, for some prime p;

(ii) Γ is the standard double cover of the Petersen graph or Coxeter graph, and ∼ G = PGL(2, 5) × Z2 or PGL(2, 7) × Z2, respectively;

(iii) Γ is one of the following three graphs:

∼ ∼ (a) Γ is cubic, G = PGL(2, 11), and D12 = Gα < G;

∼ ∼ (b) Γ is cubic, G = PGL(2, 13), and D12 = Gα < G;

∼ ∼ (c) Γ has valency 4, G = PΓL(2, 27), and Z3 × A4 = Gα < G.

And we have the corollary:

Corollary 1.2.4. A connected graph Γ is vertex bi-primitive 3-arc regular if and only if one of the following holds:

∼ (i) Γ = K3,3, and Aut(Γ ) = S3 o Z2;

(ii) Γ is the standard double cover of the Petersen graph or the Coxeter graph, and ∼ Aut(Γ ) = PGL(2, 5) × Z2 or PGL(2, 7) × Z2, respectively;

(iii) Γ is one of the following three graphs:

∼ ∼ (a) Γ is cubic, Aut(Γ ) = PGL(2, 11), and D12 = (Aut(Γ ))α;

6 ∼ ∼ (b) Γ is cubic, Aut(Γ ) = PGL(2, 13), and D12 = (Aut(Γ ))α;

∼ ∼ (c) Γ has valency 4, Aut(Γ ) = PΓL(2, 27), and Z3 × A4 = (Aut(Γ ))α.

By the above corollary, there are only six vertex bi-primitive 3-arc regular graphs.

In chapter 2, we prove a theorem for the structure of the vertex stabilizer Gα for a (G, s)-arc regular graph for all s. Then in chapter 3, we give an explicit list of almost simple groups which have a maximal subgroup with the form of the vertex stabilizer of a (G, 3)-arc regular graph. In chapter 4, we prove the above two theorems and their corollaries.

7 CHAPTER 2

VERTEX STABILIZERS OF S-ARC REGULAR GRAPHS

2.1 Theorem for the vertex stabilizers of s-arc regular graphs

Let Γ be a graph with vertex set V , and let G ≤ Aut(Γ ). For a vertex α ∈ V and its

Γ(α) neighborhood Γ (α), we denote by Gα the permutation group induced on Γ (α) by [1] Γ(α) ∼ [1] Gα, and by Gα the kernel of Gα acting on Γ (α). Then Gα = Gα/Gα . The vertex stabilizers of s-arc regular graphs have simple structure, given as follows:

Theorem 2.1.1. Let Γ be a (G, s)-arc regular graph of valency k, where s ≥ 1 and

G ≤ Aut(Γ ). Let (α, β) be a arc of Γ . Then one of the following holds:

(i) s = 1 and Gα is faithful and regular on Γ (α);

(ii) s = 2 and Gα is faithful and sharply 2-transitive on Γ (α);

n [1] Γ(α) Γ(α) (iii) s = 3 and k = p is a prime power. Moreover, Gα = Gα .Gα , where Gα

[1] contains a sharply 2-transitive subgroup K and Gα is isomorphic to a normal

∼ n subgroup of Kβ. The Sylow p-subgroup P = Zp of Gα is normal in Gα, and either:

8 Γ(α) [1] (a) Gα = K is a sharply 2-transitive group on Γ (α) and Gα = Gα × (P o [1] Gβ ), or

Γ(α) n Γ(α) (b) Gα ≤ AΓL(1, p ) and K < Gα is a sharply 2-transitive group of type

[1] pn−1 Γ(α) n n I. In this case, the order of Gα is e and the order of Gα is p (p −1)e, where e is a non-trivial common divisor of pn − 1 and n.

(iv) k = 3, and (s, Gα) = (4, S4) or (5, S4 × Z2).

Parts (i) and (ii) are clear, and (iv) follows easily from known theorems; let us consider (iii) ﬁrst.

2.2 The vertex stabilizers of s-arc regular graphs: s = 3

In this section, we will prove Theorem 2.1.1 (iii), when s = 3. The proof consists of a series of lemmas.

To start, recall a primitive permutation group G is of type HA if soc(G) is abelian,

∼ d and so soc(G) = Zp and G ≤ AGL(d, p), where p is a prime and d ≥ 1; G is of type AS if soc(G) = T is a non-abelian simple group, and T ≤ G ≤ Aut(T ). Let Γ be a

(G, 3)-arc regular graph, we then claim:

Γ(α) Lemma 2.2.1. The action of Gα on Γ (α) is 2-transitive of type HA.

Proof. Since Γ is (G, 3)-arc regular, G is transitive on the set of 2-arcs of Γ . In

Γ(α) Γ(α) particular, Gα is 2-transitive on Γ (α). Hence Gα is either of type AS or type HA, by Theorem 4.1B of [5]. Since G is regular on the set of 3-arcs of Γ , the order

2 Γ(α) of Gα is the number of 3-arcs of Γ starting from α, k(k − 1) . So the order of Gα

9 2 Γ(α) divides k(k − 1) . Also it is clear that the order of Gα is divisible by k(k − 1), since

Γ(α) it is 2-transitive on Γ (α). We prove that these two conditions on the order of Gα

Γ(α) restrict the Gα to HA type.

Γ(α) Suppose B = Gα is of type AS. By Table 7.4 of [3](or Section 7.7 of [5]), the list of 2-transitive permutation groups of type AS, B and Ω = Γ (α) can only be one of the following:

• An ≤ B(n ≥ 5), and |Ω| = n;

n! n! 2 Now the order of An is 2 . But 2 | n(n − 1) implies (n − 2)! | 2(n − 1), so n = 2, 3, or 4, which is not possible since n ≥ 5.

• PSL(d, q) ≤ B, q = pm is a prime power and d ≥ 2 with (d, q) 6= (2, 2) or (2, 3),

qd−1 and |Ω| = q−1 ; d(d−1) 2 Qd i q i=2(q −1) The order of PSL(d, q) is e , where e = GCD(d, q − 1), where d(d−1) 2 Qd i d q i=2(q −1) q −1 GCD stands for the greatest common divisor. Hence e | ( q−1 ) · qd−1 2 qd−1 qd−1 2 2 ( q−1 − 1) . Now ( q−1 , q) = 1, and ( q−1 − 1) = q · r, where (q, r) = 1, so d(d−1) q(q2−1) 2 2 ≤ 2 and d = 2. Now we have e | (q + 1)q , i.e. q − 1 | q · e, thus q − 1 | e. Notice e | q − 1, so q − 1 = e. By e = GCD(2, q − 1), e = 1 or 2, hence

q = 2 or 3, which is a contradiction to (d, q) 6= (2, 2) or (2, 3).

• B = Sp(2m, 2), where m ≥ 3, and |Ω| = 2m−1(2m + 1) or 2m−1(2m − 1);

m2 Qm 2i Now Sp(2m, 2) has order 2 i=1(2 − 1), so we have

m 2 Y 2m (22i − 1) | 2m−1(2m + 1)(22m−1 + 2m−1 − 1)2 i=1

10 or m 2 Y 2m (22i − 1) | 2m−1(2m − 1)(22m−1 − 2m−1 − 1)2. i=1 Both of these conditions imply m2 ≤ m − 1, which is imposible when m ≥ 3.

• PSU(3, q) ≤ B, where q = pm > 2, and |Ω| = q3 + 1;

3 (q2−1)(q3+1) Now the order of PSU(3, q) is q d , where d = GCD(3, q + 1), so 3 (q2−1)(q3+1) 3 3 2 q d | (q + 1)(q ) , which is impossible.

• Sz(q) ≤ B, where Sz(q) is a Suzuki group with q = 22m+1 > 2, and |Ω| = q2 +1;

Now the order of Sz(q) is (q2 + 1)q2(q − 1), so (q2 + 1)q2(q − 1) | (q2 + 1)q4,

which implies q − 1 | q2, which is impossible for q > 2.

• Ree(q) ≤ B, where Ree(q) is a Ree group with q = 32m+1 > 3, and |Ω| = q3 +1;

Now the oder of R(q) is (q3 + 1)q3(q − 1), so (q3 + 1)q3(q − 1) | q6(q3 + 1), which

implies q − 1 | q3, which is impossible.

• B = PSL(2, 11), |Ω| = 11

The order of PSL(2, 11) is 660, which does not divide 11 · 102 = 1100.

• B is one of Mathieu groups M11, M12, M22, M23, or M24, or B = M22.Z2;

Here we just inspect the order of B and |Ω|, and we see that no one satisﬁes

the |B| | |Ω|(|Ω| − 1)2:

11 B Order |Ω|

4 2 M11 2 · 3 · 5 · 11 11

4 2 M11 2 · 3 · 5 · 11 12

6 3 M12 2 · 3 · 5 · 11 12

7 2 M22 2 · 3 · 5 · 7 · 11 22

8 2 M22.Z2 2 · 3 · 5 · 7 · 11 22

7 2 M23 2 · 3 · 5 · 7 · 11 · 23 23

10 3 M24 2 · 3 · 5 · 7 · 11 · 23 24

• B = A7, and |Ω| = 15;

7 7! 2 The order of A7 is 2 !, but 2 - 15 · 14 .

• PSL(2, 8) ≤ B, and |Ω| = 28;

(82−1)(82−8) 3 2 The order of PSL(2, 8) is 8−1 = 504 = 2 · 3 · 7, which does not divide 28 · 272.

• B is the Higman-Sims group HS, and |Ω| = 176;

The order of HS is 29 ·32 ·53 ·7·11 = 44, 352, 000, which does not divide 176·1752.

• B is the Conway group Co3, and |Ω| = 276;

10 7 3 The order of Co3 is 2 · 3 · 5 · 7 · 11 · 23 = 495, 766, 656, 000, which does not divide 276 · 2752.

This completes the proof of the lemma.

12 Γ(α) n Now Gα on Γ (α) is of HA type, hence the valency k = p , for some prime p.

Γ(α) n n [1] pn−1 Clearly the order of Gα is p (p − 1)e and the order of Gα is e , where e is a n Γ(α) ∼ n divisor of p − 1. Let P = soc(Gα ). Then P = Zp and the following lemma is easy to prove:

∼ Γ(α) Γ(α) Lemma 2.2.2. Let (α, β, σ) be a 2-arc of Γ . Then Gαβσ = Gαβσ , Gαβσ is a regular Γ(α) Γ(α) permutation group on Γ (α)\{β} and K = P oGαβσ ≤ Gα is a sharply 2-transitive [1] ∼ [1] ∼ [1] Γ(α) Γ(α) permutation group on Γ (α). Further, Gα = Gβ = (Gβ ) ≤ Gαβσ = Kβ, and

[1] Γ(α) Γ(α) (Gβ ) ¡ Gαβ .

Proof. All the statements above come from the fact that Γ is (G, 3)-arc regular. To

[1] Γ(α) Γ(α) [1] see(Gβ ) ¡ Gαβ , notice that Gβ ¡ Gαβ.

Γ(α) Now we can get further information about Gα , as the following lemma shows:

Γ(α) Γ(α) Lemma 2.2.3. For Gα = P o Gαβ and K as above, only one of the following can happen:

Γ(α) n (i) Gα ≤ AΓL(1, p ), and K is a sharply 2-transitive group of type I;

Γ(α) n 2 2 2 2 (ii) SL(2, 3) ¡ Gαβ , and p ∈ {5 , 7 , 11 , 23 }; or

Γ(α) n 2 2 2 2 (iii) SL(2, 5) ¡ Gαβ , and p ∈ {11 , 19 , 29 , 59 }.

Proof. We check the Table 7.3 of aﬃne 2-transitive groups of [3]. (or Remark 7.5, chapter XII of [2].) Notice that in this table, we need to add one more case in the row with SL(2, 5) ¡ H, where q = 92 (see [17, p.513]):

13 Γ(α) n d • SL(d, q) ≤ Gαβ ≤ ΓL(d, q), and p = q ;

If d ≥ 2, then the order of SL(d, q) is divisible by q, hence divisible by p. But

Γ(α) n 2 the order of Gαβ divides (p − 1) , clearly a contradiction. So d = 1 and

Γ(α) n Γ(α) n Gαβ ≤ ΓL(1, p ). Therefore we get Gα ≤ AΓL(1, p ). This implies that Γ(α) Kβ = Gαβσ , deﬁned in Lemma 2.2.2 is solvable. Hence K = P o Kβ is either of type I or of type II. Suppose it is one of the four exceptional groups in type

n 2 2 2 2 n II. Then SL(2, 3) ≤ Kβ and p ∈ {5 , 7 , 11 , 29 }. Since Kβ ≤ ΓL(1, p ), we

know that Kβ has a unique subgroup of order 3. But SL(2, 3) = Q8 o Z3 ≤ Kβ, a contradiction. Therefore K is of type I.

Γ(α) n 2d • Sp(2d, q) ¡ Gαβ , where d ≥ 2, and p = q ;

d2 Qd 2i Now the order of Sp(2d, q) is q i=1(q − 1), which is divisible by q, hence by

Γ(α) n 2 p. But the order of Gαβ divides (p − 1) , clearly a contradiction.

Γ(α) n 6 • G2(q) ¡ Gαβ , and p = q with p = 2;

6 6 2 Now the order of G2(q) is q (q − 1)(q − 1), which is divisible by q, hence by

Γ(α) n 2 p. But the order of Gαβ divides (p − 1) , clearly a contradiction.

Γ(α) n 4 • E5 ¡ Gαβ , where E5 is an extraspecial group of order 32, and p = 3 ;

Γ(α) Now Gαβ /E5 ≤ S5, as in the remark before Table 7.3 of [3]. Since K is a

4 4 sharply 2-transitive group of degree 3 , it is of type I. Thus Kβ ≤ ΓL(1, 3 )

is of order 80 and Kβ has a normal cyclic subgroup A of order 20. We know

that for an element a of order 5 in A, its imagea ¯ in S5 is still of order 5, and

CS5 (¯a) = ha¯i. Thus C Γ(α) (a) ≤ E5 o hai and A ≤ E5 o hai, so a commutes Gαβ

14 with an element of order 4 in E5. Now a induces a linear transformation in

∼ 4 E5/Z(E5) = Z2 by conjugation and ﬁxes a non zero vector. Since the order of the point stabilizer GL(4, 2) is 3(24 − 2)(24 − 22)(24 − 23), coprime to 5,

4 a induces a trivial permutation on Z2. We claim that a ∈ C Γ(α) (E5). To Gαβ a ∼ see this, notice ﬁrst that t = t, where Z2 = hti = Z(E5), because Z(E5) is

a a characteristic subgroup of E5. So if x = xt for some x ∈ E5, then we get x = xa5 = t5x = tx, which is a contradiction. By Remark 7.5, chapter XII of [2],

Γ(α) Γ(α) Γ(α) C Γ(α) (E5) = Z(Gαβ ), thus a ∈ Z(Gαβ ) and Gαβ = C Γ(α) (a) = E5 × hai. Gαβ Gαβ 4 ∼ Γ(α) Now Kβ ≤ ΓL(1, 3 ) centralizes a, thus Kβ = Z80. Therefore Gαβ has an Γ(α) ∼ element of order 8, which is not possible since Gαβ = E5 o hai = E5 o Z5.

Γ(α) n 4 • Gαβ = A6, and p = 2 ;

4 2 2 This is not possible since the order of A6 does not divide (2 − 1) = 15 .

Γ(α) n 4 • Gαβ = A7, and p = 2 ;

4 2 2 This is not possible since the order of A7 does not divide (2 − 1) = 15 .

Γ(α) n 6 • Gαβ = PSU(3, 3), and p = 2 ;

This is not possible since the order of PSU(3, 3) is 33(32 −1)(33 +1), which does

not divide (26 − 1)2.

Γ(α) n 6 • Gαβ = SL(2, 13), and p = 3 ;

(132−1)(132−13) This is not possible since the order of SL(2, 13) is 12 , which does not divide (36 − 1)2.

15 Γ(α) n 2 2 2 2 2 • SL(2, 5) ¡ Gαβ , and p ∈ {9 , 11 , 19 , 29 , 59 };

Notice that pn 6= 92, because in this case (pn − 1)2 = 80 · 80 = 6400 is not

divisible by the order of SL(2, 5), which is 120.

Γ(α) n 2 2 2 2 • SL(2, 3) ¡ Gαβ , and p ∈ {5 , 7 , 11 , 23 }.

Hence the lemma is proved.

Now let us look at the (ii) and (iii) in Lemma 2.2.3. We claim that in both cases,

Γ(α) K must be a sharply 2-transitive group of type II or III and K = Gα . Thus pn 6= 192 by Theorem 1.1.1.

Before we prove this, let us state three results that will be used in the proof:

∼ Lemma 2.2.4. The normal subgroups of SL(2, 3) = Q8 o Z3 are {1},Z(SL(2, 3))

(which is isomorphic to Z2), Q8 and SL(2, 3). The normal subgroups of SL(2, 5) are

{1}, Z(SL(2, 5))(which is isomorphic to Z2), and SL(2, 5).

∼ Proof. First let us consider SL(2, 3). Notice that Z(Q8) = Z(SL(2, 3)) = Z2 is the unique subgroup of order 2 in SL(2, 3). Now consider the canonical surjective ∼ homomorphism SL(2, 3) to PSL(2, 3) = A4, with kernel Z = Z(SL(2, 3)). For any non-trivial normal subgroup N of SL(2, 3), either N ∩Z = 1 or Z ≤ N. If N ∩Z = 1, ∼ then N is isomorphic to a non-trivial normal subgroup of PSL(2, 3) = A4. Since a non-trivial normal subgroup of A4 is either the Sylow 2-subgroup of A4 (which is

2 isomorphic to Z2) or A4, we have 2 | |N|. Thus Z ≤ N because SL(2, 3) has a unique involution, which is a contradiction to N ∩ Z = 1. Therefore we have Z ≤ N. Now there is a one to one correspondence between the non-trivial normal subgroups of

16 ∼ SL(2, 3) and the normal subgroups of A4. Clearly N can only be Z = Z(SL(2, 3)) =

Z2, Q8, or SL(2, 3). ∼ For SL(2, 5), we know that its composition factors are Z2 and PSL(2, 5) = A5. ∼ By the Jordan − Holder¨ Theorem, If 1 6= N ¡ SL(2, 5), then either N = Z2,A5, or ∼ ∼ SL(2, 5). If Z2 = N¡SL(2, 5), then N ≤ Z(SL(2, 5)) = Z2, therefore N = Z(SL(2, 5)). ∼ ∼ If N = SL(2, 5), then N = SL(2, 5). If N = A5, then SL(2, 5)/N is abelian and [SL(2, 5), SL(2, 5)] ≤ N, but [SL(2, 5), SL(2, 5)] = SL(2, 5) since SL(2, 5) is a perfect group. Thus we get a contradiction.

Lemma 2.2.5. Let A be a subgroup of GL(2, 7) or GL(2, 23) such that A can be ∼ written as SL(2, 3).Z2 and has a unique involution, then A = G48, where G48 is the group deﬁned in Deﬁnition 8.4, chapter XII of [2].

Proof. Consider the case when A ≤ GL(2, 7). By the Atlas [4] we see that Z(SL(2, 3)) ≤

Z(GL(2, 7)). It is then clear that Z(GL(2, 7))∩A = Z(SL(2, 3)). Also A/Z(SL(2, 3)) ∼=

S4. By Lemma 8.5, chapter XII in [2], all such A are isomorphic to a uniquely deﬁned group in Deﬁnition 8.4, chapter XII of [2]. The same argument works for the case when A ≤ GL(2, 23).

∼ Lemma 2.2.6. PSL(2, 59) contains subgroups isomorphic to A5. If A5 = H ≤

PSL(2, 59), then NPGL(2,59)(H) = H.

Proof. By the Atlas [4], there is a maximal subgroup in PSL(2, 59) which is isomorphic ∼ to A5. Let A5 = H ≤ PSL(2, 59). Then NPSL(2,59)(H) = H, since H is maximal and

PSL(2, 59) has no nontrivial normal subgroups. Therefore either NPGL(2,59)(H) = H,

17 ∼ or NPGL(2,59)(H) = H.Z2. Suppose NPGL(2,59)(H) = H.Z2. We claim that H.Z2 = S5. ∼ To see this, ﬁrst notice that either H.Z2 = H × Z2 or H.Z2 = Aut(H) = S5. If

3 H.Z2 = H × Z2, then the Sylow 2-subgroup of PGL(2, 59) is Z2, which is not possible since the Sylow 2-subgroup of PGL(2, 59) is dihedral by Lemma 1.2(d), chapter 10 ∼ of [12]. Therefore H.Z2 = S5. Now we can ﬁnd a involution t ∈ H.Z2 \ H, such ∼ that CH.Z2 (t) = Z2 × S3. On the other hand, by Lemma 1.3(b), chapter 10 of [12], ∼ 2 CPGL(2,59)(t) = D2·58. Then we have 3 | 2 · 58 = 2 · 29, a contradiction. Thus

NPGL(2,59)(H) = H.

Now we are ready to prove the following lemma:

Γ(α) Lemma 2.2.7. In lemma 2.2.3 case (ii), K = Gα is one of the four exceptional

Γ(α) sharply 2-transitive group of type II; in case (iii), K = Gα is one of the three exceptional sharply 2-transitive group of type III. In particular, pn 6= 192 in case (iii).

Proof. let us look at the groups in Lemma 2.2.3 one by one:

Γ(α) n 2 • SL(2, 3) ¡ Gαβ , p = 5 .

Now SL(2, 3) ≤ GL(2, 5). Since SL(2, 3) = Q8 o Z3 is generated by order 3 ele-

ments, we have SL(2, 3) ≤ SL(2, 5). We claim that NSL(2,5)(SL(2, 3)) = SL(2, 3). To see this, ﬁrst notice that Z(SL(2, 3)) = Z(SL(2, 5)), otherwise PSL(2, 5) =

A5 will contain a subgroup isomorphic to SL(2, 3), which is not possible by

the order of these two groups. Now since NA5 (A4) = A4, NSL(2,5)(SL(2, 3)) = SL(2, 3). Consider SL(2, 3) ≤ SL(2, 5) ¡ GL(2, 5). Next we claim that for all

g ∈ GL(2, 5), there is a f ∈ SL(2, 5) such that SL(2, 3)g = SL(2, 3)f . To

18 see this, notice that Q8 ≤ SL(2, 3) is also a Sylow 2-subgroup of SL(2, 5),

g f thus for the g mentioned above, there is f ∈ SL(2, 3), such that Q8 = Q8 .

Now NSL(2,5)(Q8) = SL(2, 3), otherwise Q8 ¡ SL(2, 5), which is not possi-

g g f ble by Lemma 2.2.4. Therefore SL(2, 3) = NSL(2,5)(Q8) = NSL(2,5)(Q8 ) =

f SL(2, 3) . By Frattini’s argument, GL(2, 5) = SL(2, 5)NGL(2,5)(SL(2, 3)). Thus ∼ ∼ NGL(2,5)(SL(2, 3))/SL(2, 3) = GL(2, 5)/SL(2, 5) = Z4, and NGL(2,5)(SL(2, 3)) = Γ(α) Γ(α) SL(2, 3).Z4. Hence Gαβ = SL(2, 3), SL(2, 3).Z2 or SL(2, 3).Z4. If Gαβ = [1] Γ(α) SL(2, 3).Z2, then Gα is of order 12 and isomorphic to a subgroup A of Gαβ =

SL(2, 3).Z2 by Lemma 2.2.2. Thus A ∩ SL(2, 3) ¡ SL(2, 3) has order 12 or 6, Γ(α) which is not possible by Lemma 2.2.4. If Gαβ = SL(2, 3).Z4, then the order [1] of Gα is 6, and SL(2, 3) will have a normal subgroup of order 3 or 6, again

Γ(α) Γ(α) not possible by Lemma 2.2.4. Thus Gαβ = SL(2, 3) and Gα is a sharply Γ(α) 2-transitive group. Finally, we show that Gα is of type II. Suppose on the

2 contrary it is of type I. Then SL(2, 3) is a subgroup of ΓL(1, 5 ) = Z24.Z2. Then

we will have Z12 ¡ SL(2, 3), which is not possible by Lemma 2.2.4.

Γ(α) n 2 • SL(2, 3) ¡ Gαβ , p = 7 . ∼ Now SL(2, 3) ≤ GL(2, 7). It is clear that Z2 = Z(SL(2, 3)) ≤ Z(GL(2, 7)), otherwise PGL(2, 7) would contain a subgroup isomorphic to SL(2, 3), which

is not possible by the Atlas [4]. Thus the image of SL(2, 3) in PGL(2, 7) is

a A4. Checking the Atlas again, we see that NPGL(2,7)(A4) = S4. Therefore

0 0 NGL(2,7)(SL(2, 3)) ≤ N , where N is the pre-image of S4 in GL(2, 7). Notice

0 that the order of N is 24 · 6 and SL(2, 3) × Z3 ≤ NGL(2,7)(SL(2, 3)), so we can

19 0 Γ(α) 0 write N = (SL(2, 3)) × Z3).Z2. Now Gαβ ≤ NGL(2,7)(SL(2, 3)) ≤ N . Since Γ(α) Γ(α) Gα is 2-transitive, the order of Gαβ is divisible by 48. Therefore for the

0 SL(2, 3) here, NGL(2,7)(SL(2, 3)) = N = (SL(2, 3)) × Z3).Z2. Now if the order Γ(α) Γ(α) of Gαβ is 48, then Gα is a sharply 2-transitive group and it is of type II by the Γ(α) same argument as above. For later discussion, notice that Gαβ = SL(2, 3).Z2, Γ(α) as a Frobenius complement, has an unique involution. Thus Gαβ = G48 by Lemma 2.2.5.

Γ(α) Γ(α) If the order of Gαβ is 48 · 3, then Gαβ = (SL(2, 3) × Z3).Z2. It contains the Γ(α) [1] subgroup G48 = SL(2, 3).Z2, so Gαβ = G48 × Z3. Now Gα has order 16 so, Γ(α) by Lemma 2.2.2, Gαβ has a normal subgroup of order 16. Notice that G48

can be written as (Q8.S3), as in Deﬁnition 8.4, chapter XII of [2]. Therefore

G48 = Q8.S3 has a normal subgroup of order 16. But a Sylow 2-subgroup of

Q8.S3 is not normal, so we get a contradiction in this case.

Γ(α) n 2 • SL(2, 3) ¡ Gαβ , p = 11

Now SL(2, 3) ≤ GL(2, 11). By the same argument as in the case pn = 52,

SL(2, 3) ≤ SL(2, 11) and Z(SL(2, 3)) = Z(SL(2, 11)). Consider the image

of SL(2, 3) in PSL(2, 11), which is isomorphic to A4. Since NPSL(2,11)(A4) =

A4 by the Atlas [4], we obtain NSL(2,11)(SL(2, 3)) = SL(2, 3). Notice that

2 NPSL(2,11)(Z2) = A4 and NSL(2,11)(Q8) = SL(2, 3) by the Atlas [4]. By Frattini’s

n 2 argument as in the case p = 5 , NGL(2,11)(SL(2, 3)) = SL(2, 3).Z10. Notice that

SL(2, 3)×Z5 ≤ NGL(2,11)(SL(2, 3)) and NGL(2,11)(SL(2, 3)) = (SL(2, 3)×Z5).Z2. Γ(α) Now Gαβ has order divisible by 120 and is a subgroup of NGL(2,11)(SL(2, 3)),

20 Γ(α) Γ(α) thus Gαβ = SL(2, 3) × Z5 or Gαβ = (SL(2, 3) × Z5).Z2. In the ﬁrst case, it is Γ(α) a sharply 2-transitive group. Clearly Gα is of type II, by the same argument

Γ(α) as in the previous cases. In the second case, Gαβ contains a normal subgroup [1] A isomorphic to Gα , whose order is 60. Therefore A ∩ (SL(2, 3) × Z5) is of order 30 or 60. If it is 30, then A ∩ SL(2, 3) is normal in SL(2, 3) with order

6; if it is 60, then A ∩ SL(2, 3) is normal in SL(2, 3) with order 12. Neither of

these cases are possible by Lemma 2.2.4.

Γ(α) n 2 • SL(2, 3) ¡ Gαβ , p = 23

Now SL(2, 3) ≤ GL(2, 23). By the same argument as before, SL(2, 3) ≤

SL(2, 23) and Z(SL(2, 3)) = Z(SL(2, 23)). By the Atlas [4], NPGL(2,23)(A4) ≤

0 PSL(2, 23) and NPSL(2,23)(A4) = S4, hence NPGL(2,23)(A4) = S4. Let N be the

0 0 pre-image of S4 in GL(2, 23), then NGL(2,23)(SL(2, 3)) ≤ N and |N | = 24 · 22.

Let NSL(2,23)(SL(2, 3)) = SL(2, 3).Z2. Notice that NSL(2,23)(SL(2, 3)) × Z11 ≤

0 0 NGL(2,23)(SL(2, 3)) ≤ N , so NGL(2,23)(SL(2, 23)) = N = NSL(2,23)(SL(2, 3)) ×

Γ(α) Z11. Thus Gα = NSL(2,23)(SL(2, 3)) × Z11 is a sharply 2-transitive group of type II, by the same argument as before. For later discussion, notice that

NSL(2,23)(SL(2, 3)) = SL(2, 3).Z2, as a subgroup of a Frobenius complement, has

a unique involution. So, by Lemma 2.2.5, NSL(2,23)(SL(2, 3)) = G48.

Γ(α) n 2 • SL(2, 5) ¡ Gαβ , p = 11

Now SL(2, 5) ≤ GL(2, 11). Since SL(2, 5) is generated by order 3 elements,

we have SL(2, 5) ≤ SL(2, 11). By the Atlas [4], PSL(2, 11) has no subgroup

of order 120. Hence Z(SL(2, 5)) = Z(SL(2, 11)). By the Atlas [4] again,

21 we see that NPGL(2,11)(PSL(2, 5)) = NPSL(2,11)(PSL(2, 5)) = A5. Therefore

NGL(2,11)(SL(2, 5)) has order at most 60 · 10. Noticing that SL(2, 5) × Z5 ≤

NGL(2,11)(SL(2, 5)), we have NGL(2,11)(SL(2, 5)) = SL(2, 5) × Z5. Now either Γ(α) Γ(α) Gαβ = SL(2, 5) or Gαβ = SL(2, 5) × Z5. If the ﬁrst case happens, then Γ(α) clearly Gα is a sharply 2-transitive group of type III. If the second case hap- Γ(α) [1] ∼ [1] pens, then Gαβ contains the normal subgroup Gβ = Gα , of order 24. This implies that SL(2, 5) has a normal subgroup of order 24, which is not possible

by Lemma 2.2.4.

Γ(α) n 2 • SL(2, 5) ¡ Gαβ , p = 19

Now SL(2, 5) ≤ GL(2, 19). Since SL(2, 5) is also generated by order 5 elements,

SL(2, 5) ≤ SL(2, 19). By the same argument as before, NGL(2,19)(SL(2, 5)) =

Γ(α) 2 SL(2, 5) × Z9. Since the order of Gαβ is divisible by 360 = 19 − 1, either Γ(α) Γ(α) Gαβ = SL(2, 5) × Z3 or Gαβ = SL(2, 5) × Z9. The ﬁrst case can not happen since we do not have non-solvable sharply 2-transitive groups of degree 192.

Γ(α) If the second case happens, then Gαβ has a subgroup Kβ of order 360 and

2 Kβ ≤ ΓL(1, 19 ). Therefore Kβ has a cyclic subgroup of order 180, and this

Γ(α) cyclic subgroup is also in Gαβ = SL(2, 5) × Z9. Then SL(2, 5) has a element of ∼ order 20 and PSL(2, 5) = A5 has an element of order 10, which is not possible.

Γ(α) n 2 • SL(2, 5) ¡ Gαβ , p = 29

Now SL(2, 5) ≤ GL(2, 29). Since SL(2, 5) is generated by order 3 elements,

SL(2, 5) ≤ SL(2, 29). Following the same argument as before, we have that

NGL(2,29)(SL(2, 5)) = SL(2, 5) ◦ Z28, the central product of SL(2, 5) and Z28

22 ∼ (which means their intersection is Z(SL(2, 5)) = Z2), which is of order 120 · 14. Γ(α) Γ(α) Γ(α) Since 28·30 divides Gαβ , either Gαβ = SL(2, 5)×Z7 or Gαβ = SL(2, 5)◦Z28. Γ(α) If the ﬁrst case happens, then clearly Gα is a sharply 2-transitive group of

Γ(α) type III. If the second case happens, then Gαβ contains the normal subgroup [1] ∼ [1] [1] Gβ = Gα , whose order is 420. Now Gβ ∩ SL(2, 5) is a normal subgroup of

420 420 420 420 SL(2, 5), possibly with order 1 = 420, 2 = 210, 7 = 60 or 14 = 30. None of them is possible, either because the order is too large or because of Lemma

2.2.4.

Γ(α) n 2 • SL(2, 5) ¡ Gαβ , p = 59

Now SL(2, 5) ≤ GL(2, 59). Since SL(2, 5) is generated by order 3 elements,

we have SL(2, 5) ≤ SL(2, 59). Following the same argument as before, we will

get Z(SL(2, 5)) = Z(SL(2, 59)). Consider the image of SL(2, 5) in PGL(2, 59),

which is a subgroup of PSL(2, 59) and isomorphic to A5. By Lemma 2.2.6,

we see that the order of NGL(2,59)(SL(2, 5)) is at most 60 · 58. Noticing that

SL(2, 5)×Z29 ≤ NGL(2,59)(SL(2, 5)), we have NGL(2,59)(SL(2, 5)) = SL(2, 5)×Z29. Γ(α) Now Gα = NGL(2,59)(SL(2, 5)) is a sharply 2-transitive group of type III.

Hence the lemma is proved.

Now by Lemma 2.2.3 and Lemma 2.2.7, we get the following conclusion:

Γ(α) Γ(α) Γ(α) Γ(α) Lemma 2.2.8. Let Gα = P o Gαβ and K = P o Gαβσ ≤ Gα as before. Then Γ(α) Gαβ and K can be only one of the following:

23 Γ(α) Γ(α) n (i) Gα = P o Gαβ ≤ AΓL(1, p ), and K is a sharply 2-transitive group of type I;

Γ(α) n 2 2 2 2 (ii) K = Gα is a sharply 2-transitive of type II, and p ∈ {5 , 7 , 11 , 23 }; or

Γ(α) n 2 2 2 (iii) K = Gα is a sharply 2-transitive of type III, and p ∈ {11 , 29 , 59 }.

Next let us prepare some lemmas for the next result that a Sylow p-subgroup of

Gα, still denoted as P , is normal in Gα. Recall the following fact, as it will be used in the lemma next to it:

Lemma 2.2.9. An automorphism of a group G which commutes with every inner automorphism must act trivially on G/Z(G) and on [G, G].

This can be found in a lot of group theory books. We include a proof:

Proof. Suppose σ ∈ Aut(G) is such a automorphism. Then we have σaˆ =aσ ˆ , for all aˆ ∈ Inn(G), wherea ˆ ∈ Inn(G) is induced by a ∈ G. Therefore xa = xaˆ = xσ−1aσˆ = xσ(a), for all x, a ∈ G. Thus a−1 · σ(a) ∈ Z(G) for all a ∈ G and σ acts trivially on

G/Z(G).

Now for all a, b ∈ G,[a, b]σ = (a−1b−1ab)σ = (a−1aˆb)σ = (aσ)−1b−1aσb = (ac)−1b−1acb =

−1 −1 σ a b ab = [a, b], where a = ac and c ∈ Z(G). So σ acts trivially on [G, G].

We need the following lemma to prove that P ¡Gα in cases (ii) and (iii) of Lemma 2.2.8:

Lemma 2.2.10. In cases (ii) and (iii) of lemma 2.2.8, the structure Kβ, for the group K deﬁned in Lemma 2.2.2, is described in the following table:

24 n Kβ Aut(Kβ) |Aut(Kβ)| p

2 SL(2, 3) S4 24 5

2 G48 S4 × Z2 48 7

2 SL(2, 3) × Z5 S4 × Z4 96 11

2 G48 × Z11 (S4 × Z2) × Z10 480 23

2 SL(2, 5) S5 120 11

2 SL(2, 5) × Z7 S5 × Z6 720 29

2 SL(2, 5) × Z29 S5 × Z28 3360 59 where G48 is the group deﬁned in Deﬁnition 8.4, chapter XII of [2].

Proof. Suppose we know that Aut(SL(2, 3)) = S4, Aut(SL(2, 5)) = S5, and Aut(G48) =

S4 × Z2. Then for the other Kβ’s, we can see that Aut(Kβ) is just a direct product of Aut(Zm−1)(m ∈ {5, 11, 7, 29}) and one of the automorphism groups above. For example, for Kβ = SL(2, 3) × Z5, we have Aut(Kβ) = Aut(SL(2, 3)) × Aut(Z5) =

S4 × Z4. To see this, notice that SL(2, 3) and Z5 are characteristic subgroups of Kβ, since SL(2, 3) is generated by elements of order 3 in Kβ and Z5 is the normal Sylow 5- subgroup of Kβ. Thus we just need to show that Aut(SL(2, 3)) = S4, Aut(SL(2, 5)) =

S5, and Aut(G48) = S4 × Z2.

To see that Aut(SL(2, 3)) = S4, notice that Aut(PSL(2, 3)) = Aut(A4) = S4. Now consider the surjective homomorphism from SL(2, 3) to PSL(2, 3) = SL(2, 3)/Z(PSL(2, 3)).

Any a ∈ Aut(SL(2, 3)) will induce an automorphism on PSL(2, 3) = A4 (since ∼ Z(G) = Z2 is ﬁxed under a). We claim that if the induced automorphism is trivial, then a ﬁxes all the order 3 elements of SL(2, 3). To see this, suppose g ∈ SL(2, 3) is

25 an element of order 3. Then ga = g or gz, where hzi = Z(SL(2, 3)). Now (g2)a = g2, thus ga = g. Since SL(2, 3) is generated by the set of all the order 3 elements, we claim a = 1. Thus Aut(SL(2, 3)) is isomorphic to a subgroup of Aut(A4) = S4. Consider the conjugation action of GL(2, 3) on SL(2, 3). The kernel of this ac- ∼ ∼ tion is CGL(2,3)(SL(2, 3)) = Z(GL(2, 3)) = Z2. Thus S4 = GL(2, 3)/Z(GL(2, 3)) ≤

Aut(SL(2, 3)) and Aut(SL(2, 3)) = S4. The same argument also works in the case of Aut(SL(2, 5)). Now Aut(PSL(2, 5)) = ∼ Aut(A5) = S5, CGL(2,5)(SL(2, 5)) = Z(GL(2, 5)) = Z4, and GL(2, 5)/Z(GL(2, 5)) = ∼ PGL(2, 5) = S5, implying Aut(SL(2, 5)) = S5.

For Aut(G48), notice that G48 = SL(2, 3).Z2 = Q8.S3, as in Deﬁnition 8.4, chapter ∼ ∼ XII of [2]. Thus we have Z(G48) = Z(SL(2, 3)) = Z2 and Inn(G48) = G48/Z(G48) =

S4. Also we claim that CG48 (SL(2, 3)) = Z(G48). To see this, notice that Z(G48) ≤

CG48 (SL(2, 3)), SL(2, 3)/Z(G48) = A4 and CS4 (A4) = 1.

Next consider the action of Aut(G48) on Inn(G48) by conjugation. Suppose the kernel is C. Now we have a homomorphism δ from Aut(G48) to Aut(Inn(G48)) =

Aut(S4) = S4. Also we know that the image of Inn(G48) is exatly S4, because ∼ Z(Inn(G48)) = Z(S4) = 1. So δ is surjective. Therefore Aut(G48) = Inn(G48) × C since Inn(G48) ∩ C = 1. For any c ∈ C, c acts trivially on [G48,G48] and c acts trivially on G48/Z(G48) by Lemma 2.2.9. We claim that [G48,G48] = SL(2, 3). To see this, notice that Q8 = [SL(2, 3), SL(2, 3)] ≤ [G48,G48], implying that [G48,G48] is either

Q8 or SL(2, 3). Since G48/[G48,G48] is abelian, we obtain [G48,G48] = SL(2, 3). The

c group G48 is generated by SL(2, 3) and another element t. We have t = t or tz, where

Z(G48) = hzi. So there are at most 2 possibilities for c and |C| = 1 or 2. We claim that

26 ∼ the order of C is 2, thus Aut(G48) = S4 × Z2. To see this, we deﬁne an injection c be-

c c tween G48 and G48 by g = g for all g ∈ SL(2, 3), and g = gz for all g ∈ G48 \SL(2, 3).

c c c Then it is clear that (ab) = a b for all a, b ∈ G48. Also c∈ / Inn(G48), since if it is,

then c is induced by some element in CG48 (SL(2, 3)) = Z(G48), because c acts trivially ∼ ∼ on SL(2, 3). Thus c = 1, which is not possible. So Aut(G48) = S4 × hci = S4 × Z2.

Γ(α) To show that P ¡ Gα when Gα is in case (i) of Lemma 2.2.8, we need the following three lemmas:

Lemma 2.2.11. The subgroups of a generalized quaternion group are either cyclic or generalized quaternion.

Proof. For Q8, we know it is true that a subgroup of it is either cyclic (1, Z2, Z4), or quaternion (itself). For a generalized quaternion group Q with order 2n ≥ 16, we know that Q = ha, b|a2n−1 = 1, a2n−2 = b2, b−1ab = a−1i, where n ≥ 4. Clearly hai ¡ Q. Then consider any subgroup Q1 of Q. If |Q1| = 2, then Q1 = Z(Q), since the order 2 element is unique in Q. If |Q1| = 4, then we see that Q1 has to be cyclic, again by the fact that the order 2 element of Q is unique. If |Q1| ≥ 8, then we can see that either Q1 ≤ hai (and so Q1 is cyclic) or Q1/(Q1 ∩ hai) = Z2. Since any b0 ∈ Q \ hai has order 4 and acts the same way on hai as b in the representation of

Q, we see that in the case Q1/(Q1 ∩ hai) = Z2, Q1 is another generalized quaternion group.

[1] Lemma 2.2.12. In case (i) of Lemma 2.2.8, Gα is meta-cyclic and each Sylow

[1] subgroup of Gα is either cyclic or a generalized quaternion group.

27 [1] ∼ [1] Γ(α) Γ(α) n Proof. We have Gα = (Gβ ) ≤ Kβ ≤ Gαβ ≤ ΓL(1, p ) by Lemma 2.2.2, and

n [1] ΓL(1, p ) ia meta-cyclic. Since a subgroup of a meta-cyclic group is meta-cyclic, Gα is meta-cyclic. Also Kβ is a Frobenius complement , thus each Sylow subgroup of

Kβ is either cyclic or a generalized quaternion group, by the structure theorem of ﬁnite Frobenius groups in [5] section 3.4. By Lemma 2.2.11, we see that each Sylow

[1] subgroup of Gα is either cyclic or a generalized quaternion group.

Recall that a meta-cyclic group is super-solvable. Also for any super-solvable

t1 t2 tm group G of order p1 p2 ...pm with primes p1 < p2 < ... < pm, there exists a unique

Qm tj subgroup Gi with order j=m−i+1 pj , for all i = 1, 2, ..., m, as mentioned in Corollary 10.5.2 of [13]. For a group G, a Hall subgroup H is a subgroup whose order is relatively prime to its index. If π is a set of primes such that any prime that divides the order of H is in π, and any prime that divides the index of H is not in π, then H is called a Hall π-subgroup of G. By Hall’s Theorem, a solvable group contains a Hall

π-subgroup, for any subset π of the prime divisors of |G|. Thus all the above Gi’s are

Hall-subgroups, and {1} = G0 ¡ G1 ¡ ... ¡ Gm = G. Recall that for a prime p, a p0-group is a group with order relative prime to p.

Lemma 2.2.13. Suppose we have a p-group P acting faithfully on a p’-group P 0. If

0 0 0 0 there exists a normal series {1} = P0 ¡P1 ¡...¡Pm = P which is invariant under P ,

0 0 0 0 0 0 then P is isomorphc to a subgroup of Aut(P1/P0) × Aut(P2/P1) × ... × Aut(Pm/Pm−1).

0 0 0 0 Proof. There is a natural homomorphism c from P into Aut(P1/P0) × Aut(P2/P1) ×

0 0 0 0 ...×Aut(Pm/Pm−1), induced by the conjugation action of P on each quotient Pi /Pi−1,

28 for i = 1, 2, ..., m. By the proof of Theorem 3.2, section 5.3 of [9], the kernel of this action is 1. Thus c is injective.

Now we claim:

∼ n Lemma 2.2.14. The Sylow p-subgroup of Gα is normal in Gα, P = Zp and P [1] commutes with Gα .

Γ(α) [1] Proof. First suppose that Gα is as in case (i) of Lemma 2.2.8. Let the order of Gα

pn−1 Qm tj be e = j=1 pj , where p1 < p2 < ... < pm are primes. By Lemma 2.2.12 and the discussion after that lemma, we can ﬁnd a normal series {1} = G0 ¡ G1 ¡ ... ¡ Gm =

[1] [1] Qm tj Gα , such that Gi is the unique subgroup of Gα of order j=m−i+1 pj and Gi+1/Gi is either cyclic or a generalized quaternion group. If pi 6= 2, then Gi+1/Gi is cyclic.

If p1 = 2 and Gm/Gm−1 is generalized quaternion of order greater than 8, then there

[1] is a unique subgroup G 1 of index 2 in Gα . Now G 1 /Gm−1 is cyclic and we m− 2 m− 2 can add G 1 into the normal series to make all the quotients cyclic. If p1 = 2 m− 2 ∼ [1] and Gm/Gm−1 = Q8, then there is a unique subgroup G 1 of index 4 in Gα , and m− 2 ∼ 2 Gm/G 1 = . m− 2 Z2

Take a Sylow p-subgroup P0 of Gα, and consider the conjugation action in-

[1] duced by P0 on Gα . If the action is faithful then, by Lemma 2.2.13, P0 is isomorphic to a subgroup of Aut(G1/G0) × Aut(G2/G1) × ... × Aut(Gm/Gm−1) (or up to

[1] pn−1 Aut(Gm/G 1 )). But the order of the latter group is less than |Gm| = |Gα | = m− 2 e ∼ (note that in the case when Aut(Gm/G 1 ) is not cyclic, Aut(Gm/G 1 ) = S3 and m− 2 m− 2 pn−1 n |Aut(G 1 /Gm−1)| · |Aut(Gm/G 1 )| = 6 ≤ |Gm/Gm−1| = 8). Since < p , m− 2 m− 2 e the order of P0, we get a contradiction. So there is some x ∈ P0 \{1}, such that

29 [1] Γ(α) Γ(α) x ∈ C = CGα (Gα ). Now C ¡ Gα and 1 6= C ¡ Gα . Since the unique minimal

Γ(α) n Γ(α) n normal subgroup of Gα is isomorphic to Zp , the order of C is divisible by p . Thus the order of C is also divisible by pn and C contains a Sylow p-subgroup P

[1] Γ(α) n Γ(α) [1] of Gα. We see that (P × Gα ) = Zp ¡ Gα , thus P × Gα ¡ Gα. This implies [1] P ¡ Gα, since P is the unique Sylow p-subgroup of P × Gα .

Γ(α) Next suppose that Gα is as in cases (ii) and (iii) of Lemma 2.2.8. Take a

n Sylow p-subgroup P of Gα of order p and consider the conjugation action of P on [1] ∼ [1] [1] Gα = Kβ. From the table in Lemma 2.2.10, p - |Aut(Gα )|, so P acts on Gα trivially.

[1] [1] Γ(α) Therefore we have a subgroup P × Gα in Gα, and (P × Gα ) is the unique Sylow

Γ(α) [1] p-subgroup of Gα by its order. Thus P × Gα ¡ Gα. Now as the unique Sylow

[1] p-subgroup of P × Gα , P is a normal subgroup of Gα.

Proof of Theorem 2.1.1 (iii):

n [1] Γ(α) Γ(α) By Lemma 2.2.1, k = p . Clearly Gα = Gα .Gα . By Lemma 2.2.2, Gα

[1] contains a sharply 2-transitive group K and Gα is isomorphic to a normal subgroup

∼ n of Kβ. The Sylow p-subgroup P = Zp is normal in Gα, by Lemma 2.2.14. Γ(α) Suppose Gα is a sharply 2-transitive group on Γ (α) and consider Gαβ. We see

[1] [1] [1] [1] [1] [1] that Gα ¡Gαβ and Gβ ¡Gαβ. since Gα ∩Gβ = {1}, we know that Gα ×Gβ ≤ Gαβ.

[1] [1] n 2 [1] [1] Now the order of Gα × Gβ is (p − 1) , thus Gα × Gβ = Gαβ. Noting that [1] P ¡ Gα and P commutes with Gα by Lemma 2.2.14, we see that Gα = P o Gαβ = [1] [1] [1] [1] P o (Gα × Gβ ) = Gα × (P o Gβ ). Thus we get (iii)(a) of Theorem 2.1.1. Γ(α) Suppose Gα is not a sharply 2-transitive group on Γ (α). Now by Lemma 2.2.8,

(Γ(α)) n Γ(α) we know that Gα ≤ AΓL(1, p ) and Gα contains a sharply 2-transitive group K

30 [1] pn−1 Γ(α) n n of type I. It is clear that the order of Gα is e and the order of Gα is p (p −1)e, where e is a non-trivial common divisor of pn − 1 and n. Thus we get (iii)(b) of

Theorem 2.1.1.

2.3 The vertex stabilizers of s-arc regular graphs: s ≥ 4

For Theorem 2.1.1 (iv), we have the following proof. Recall that a graph Γ is called

(G, s)-transitive if Γ is (G, s)-arc transitive but not (G, s + 1)-arc transitive, where

G ≤ Aut(Γ ); and Γ is called s-transitive if Γ is (Aut(Γ ), s)- transitive. Further, for a group G and a subgroup H ≤ G, we denote the index of H in G by |G : H|.

Proof of Theorem 2.1.1 (iv):

Let Γ be (G, s)-arc regular for some s ≥ 4, then Γ is (G, s)-transitive. By [22],

Γ(α) e we know that Gα £ PSL(2, q), where q = p with p prime and k = q + 1. We obtain

2 where A ≤ Gα is of order 3 ; or s = 7 and A o GL(2, 3) ≤ Gα, where A ≤ Gα is of order 35. Therefore |PGL(2, 3)| = 24 divides 4·3s−1, which is not possible. Thus q = 2 and k = 3, and by Tutte’s well-known theorem, (s, Gα) = (4, S4), or (5, S4 × Z2).

Now by the above result, we check [14] and get the following results:

Theorem 2.3.1. A connected graph Γ is vertex primitive 4-arc regular if and only if it is isomorphic to a cubic graph Cos(G, H, HgH), where G = Aut(Γ ) = PSL(2, p), ∼ p is a prime and p = ±1(mod 16), H = S4. A connected graph Γ is vertex primitive

31 5-arc regular if and only if it is isomorphic to a cubic graph Cos(G, H, HgH), where ∼ G = Aut(Γ ) = Aut(PSL(3, 3)) = PSL(3, 3).Z2, H = S4 × Z2. Further, such a graph Γ is uniquely determined by its order V (Γ ).

Theorem 2.3.2. A connected graph Γ is vertex bi-primitive 4-arc regular if and only if either it is:

(i) a standard double cover of a connected vertex primitive 4-arc regular graph;

(ii) isomorphic to a cubic graph Cos(G, H, HgH), where G = Aut(Γ ) = PGL(2, p), ∼ p is a prime and p = ±1(mod 8), H = S4;

(iii) isomorphic to a cubic graph Cos(G, H, HgH), where G = Aut(Γ ) = PGL(2, 7) = ∼ Aut(PSL(3, 2)), H = S4.

Further, such a graph Γ is uniquely determined by its order V (Γ ).

Theorem 2.3.3. A connected graph Γ is vertex bi-primitive 5-arc regular if and only if either it is:

(i) a standard double cover of a connected vertex primitive 5-arc regular graph, or

(ii) isomorphic to a cubic graph Cos(G, H, HgH), where G = Aut(Γ ) = PΓL(2, 9), ∼ H = S4 × Z2. (This is Tutte’s 8-cage, or the Levi graph)

Further, such a graph Γ is uniquely determined by its order V (Γ ).

For coset graphs mentioned in the above theorems, see the introduction before

Lemma 4.1.1.

32 2.4 A more general result

By Theorem 2.1.1 (iii), we know that for a (G, 3)-arc regular graph Γ , the valency is

n k = p , where p is a prime, and the Sylow p-subgroup P of Gα is normal in Gα with

∼ n [1] Γ(α) P = Zp . From the previous discussion, we also know that Gα = Gα .Gα , where [1] Gα is isomorphic to a normal subgroup of a point stabilizer of a sharply 2-transitive

n Γ(α) n group of degree p , and Gα is a 2-transitive group with socle Zp . In this section we prove a more general result.

Lemma 2.4.1. Suppose a group G has a normal subgroup H and G/H ∼= K, where H is isomorphic to a normal subgroup of a point stabilizer of a sharply 2-transitive

n n group of degree p and K is a 2-transitive group with socle Zp . Then there exists a ∼ n normal subgroup P of G with P = Zp .

When H is isomorphic to a normal subgroup of a point stabilizer of a type I

n pn−1 sharply 2-transitive group of degree p , H is meta-cyclic with order e . By the same argument as when we show that P ¡ Gα in the previous section, we have

∼ n that CG(H)H ¡ G contains the pre-image of soc(K) = Zp , under the canonical homomorphism φ from G to K. Since CG(H) ¡ CG(H)H and |CG(H)H : H| is not

∼ n divisible by p, CG(H) contains a subgroup P = Zp such that φ(P ) = soc(K). Then P ¡ G since P × H ¡ G.

When H is isomorphic to a normal subgroup of a point stabilizer of a type II or

III sharply 2-transitive group of degree pn, we check the point stabilizer of the seven sharply 2-transitive groups of type II or III one by one, and list all the possibilities for normal subgroups H in each of them. We shall prove that the order of Aut(H) is

33 not divisible by the corresponding p for all cases. Thus CG(H) contains a subgroup

∼ n P = Zp such that φ(P ) = soc(K). Then P ¡ G since P × H ¡ G. Now let us look at the possibilities of H, according to the point stabilizer in the seven sharply 2-transitive groups of type II or III.

Lemma 2.4.2. If H ¡ SL(2, 3), where SL(2, 3) is a point stabilizer of a sharply 2-

n 2 ∼ transitive group of type II of degree p = 5 , then H = 1, Z(SL(2, 3)) = Z2, Q8, or SL(2, 3).

Proof. By Lemma 2.2.4.

Lemma 2.4.3. If H ¡ G48, where G48 = SL(2, 3).Z2 is a point stabilizer of a sharply n 2 ∼ 2-transitive group of type II of degree p = 7 , then H = 1, Z(SL(2, 3)) = Z2, Q8,

SL(2, 3) or G48.

Proof. It is clear that all subgroups listed here are normal in G48. ∼ If H ≤ SL(2, 3), then by Lemma 2.4.2, we have H = 1, Z(SL(2, 3)) = Z2,Q8, or SL(2, 3).

Suppose H is not in SL(2, 3). Then |H : H ∩ SL(2, 3)| = 2 and H ∩ SL(2, 3) = 1, ∼ Z(SL(2, 3)) = Z2,Q8, or SL(2, 3). We claim that H ∩ SL(2, 3) 6= 1, because otherwise

G48 has more than one involution, which is a contradiction to the fact that the

Frobenius complement has at most one involution. If H ∩ SL(2, 3) = Z2 or Q8, then hQ8,Hi = Q8H ¡ G48. Therefore G48 has a normal Sylow 2-subgroup, clearly ∼ impossible since G48 = Q8.S3, as mentioned in Theorem 1.1.1. So H ∩ SL(2, 3) =

SL(2, 3) and H = G48.

34 Lemma 2.4.4. If H ¡ SL(2, 3) × Z5, where SL(2, 3) × Z5 is a point stabilizer of a n 2 ∼ sharply 2-transitive group of type II of degree p = 11 , then H = 1, Z(SL(2, 3)) = Z2, ∼ Q8, SL(2, 3), Z5, Z(SL(2, 3)) × Z5 = Z10, Q8 × Z5, or SL(2, 3) × Z5.

Proof. It is clear that all subgroups listed here are normal in SL(2, 3) × Z5. If the order of H is not divisible by 5, then H ¡SL(2, 3) since SL(2, 3)¡SL(2, 3)×

Z5. Therefore H is one of the ﬁrst four groups in the statement of the lemma by

Lemma 2.4.2. If the order of H is divisible by 5, then Z5 ¡ H since Z5 is the normal

Sylow 5-subgroup of SL(2, 3) × Z5. Now H = (H ∩ SL(2, 3)) × Z5 and H ∩ SL(2, 3) ¡

SL(2, 3), thus we get the remaining four groups in the statement of the lemma .

Lemma 2.4.5. If H ¡ G48 × Z11, where G48 × Z11 is a point stabilizer of a sharply n 2 ∼ 2-transitive group of type II of degree p = 23 , then H = 1, Z(SL(2, 3)) = Z2, Q8, ∼ SL(2, 3), G48, Z11, Z(SL(2, 3)) × Z11 = Z22, Q8 × Z11, SL(2, 3) × Z11, or G48 × Z11.

Proof. The same argument as in the proof of Lemma 2.4.3 and Lemma 2.4.4 works in this case as well.

Lemma 2.4.6. If H ¡ SL(2, 5), where SL(2, 5) is a point stabilizer of a sharply 2-

n 2 ∼ transitive group of type III of degree p = 11 , then H = 1, Z(SL(2, 5)) = Z2, or SL(2, 5).

Proof. By Lemma 2.2.4.

Lemma 2.4.7. If H ¡ SL(2, 5) × Z7, where SL(2, 5) × Z7 is a point stabilizer of a sharply 2-transitive group of type III of degree pn = 292, then H = 1, Z(SL(2, 5)) ∼= ∼ Z2, SL(2, 5), Z7, Z(SL(2, 5)) × Z7 = Z14, or SL(2, 5) × Z7.

35 Proof. By the same argument as in the proof of Lemma 2.4.4, using Lemma 2.4.6.

Lemma 2.4.8. If H ¡ SL(2, 5) × Z29, where SL(2, 5) × Z29 is a point stabilizer of a sharply 2-transitive group of type III of degree pn = 592, then H = 1, Z(SL(2, 5)) ∼= ∼ Z2, SL(2, 5), Z29, Z(SL(2, 5)) × Z29 = Z58, or SL(2, 5) × Z29.

Proof. By the same argument as in the proof of Lemma 2.4.4, using Lemma 2.4.6.

Lemma 2.4.9. For all H described in Lemma 2.4.2 to Lemma 2.4.8, the order of

Aut(H) is not divisible by the corresponding p.

Proof. For a group G = A × Zq, where q is a prime and the order of A is not divisible by q, Aut(G) = Aut(A) × Aut(Zq) = Aut(A) × Zq−1, since A and Zq are both characteristic subgroups of G. Also notice that Aut(Q8) = S4 is a well known fact which can be found in [23, p.148]. These observations, combined with Lemma 2.2.10, prove this lemma.

36 CHAPTER 3

A KEY LEMMA

In this chapter, we determine the almost simple groups G which have a maximal subgroup isomorphic to the vertex stabilizer of a 3-arc regular graph.

3.1 The key lemma

First let us state our key lemma.

Lemma 3.1.1. If G is an almost simple group with a maximal subgroup M which has a structure as in Theorem 2.1.1(iii), then G and M can only be one of the following:

(i) G = PGL(2, 5), M = Z2 × (Z3 o Z2) = D12;

(ii) G = PGL(2, 7), M = Z2 × (Z3 o Z2) = D12;

(iii) G = PSL(2, 11), M = Z2 × (Z3 o Z2) = D12;

(iv) G = PSL(2, 13), M = Z2 × (Z3 o Z2) = D12;

2 (v) G = PΣL(2, 27), M = Z3 × (Z2 o Z3) = Z3 × A4,

and in each case, such M exists and is unique under conjugation in G.

37 Before we show that the ﬁve cases in Lemma 3.1.1 are the only cases, let us ﬁrst note that for each case of Lemma 3.1.1, M exists and is unique under conjugation:

Lemma 3.1.2. In each case of Lemma 3.1.1, M exists. If M ∼= M 0 ≤ G, then M and M 0 are conjugate in G. Thus M is unique under conjugation in G.

Proof. By the Atlas [4]. Note that in the case G = PΣL(2, 27), if we let f be the ﬁeld ∼ automorphism of order three in G, then we can assume that M = hfi × CK (f) = ∼ 1 ∼ Z3 × A4, where K = PSL(2, 27). This is true because CK (f) = PSL(2, 27 3 ) = A4, by Lemma 1.4(a), chapter 10 of [12].

Next let us show: For an almost simple group G, which has a maximal subgroup M

0 ∼ n with a structure like in Theorem 2.1.1(iii), i.e., M = H.(P o H ), where P = Zp ¡ M and P oH0 contains a sharply 2-transitive group K on pn points; H is isomorphic to a

n n 2 normal subgroup of Kβ, where Kβ is a point stabilizer of K; M is of order p (p −1) ; then G and M can only be one of the ﬁve cases in Lemma 3.1.1.

In the following discussion of this chapter, L = soc(G), unless further speciﬁed.

Also all the other notation are the same as in Lemma 3.1.1 and the above statement.

3.2 Some useful facts

To start, let us state some useful facts:

Lemma 3.2.1. L M, therefore hM,Li = LM = G.

Proof. Suppose L ≤ M; then M is not solvable. By Theorem 2.1.1 (iii), this can only happen when P o H0 ≤ M is a non-solvable sharply 2-transitive group. There

38 are only three non-solvable sharply 2-transitive groups (with pn = 112, 292, or 592) ∼ and for all such cases the only non-abelian composition factor is PSL(2, 5) = A5. So ∼ for M, the only non-abelian composition factor is also PSL(2, 5) = A5. Since L ¡ M, ∼ L = A5. Therefore M ≤ Aut(A5) = S5, and |M : L| = 1 or 2. Then we get a contradiction here, since Zp is another composition factor of M, and p > 2 in all three cases. So we have L M. Then LM = G, since M is maximal in G.

n ∼ Lemma 3.2.2. NG(P ) = M, CG(P ) = H × P , and Zp = P ≤ M is a Sylow p-subgroup of G.

Proof. Since P ¡ M, we have M ≤ NG(P ). Notice that M is maximal in G, and so

NG(P ) = G or M. If P ¡ G, then by the fact that L is the unique minimal normal subgroup of G, we get L ≤ P , a contradiction. Thus NG(P ) = M.

To see that CG(P ) = H × P , ﬁrst notice that CG(P ) ≤ NG(P ) = M, thus

CG(P ) = CM (P ). Now H × P ≤ CM (P ) by Lemma 2.2.14. To see that CM (P ) ≤

0 H × P , consider the image of CM (P ) in P o H under the canonical homomorphism.

We know that this image is contained in the centralizer of P in P o H0, which is P

0 since P o H is a 2-transitive group of HA Type. Thus CM (P ) is contained in the pre-image of P , i.e., H × P . Thus H × P = CM (P ) = CG(P ). Suppose that P is not a Sylow p-subgroup of G. Then there is a Sylow p-subgroup

P 0 of G, such that P < P 0, by Sylow’s Theorem. We know that in a p-group, any proper subgroup is strictly contained in its normalizer. Therefore P < NP 0 (P ). On

0 0 the other hand, NP 0 (P ) = NG(P ) ∩ P = M ∩ P = P . Thus we get a contradiction.

So P is a Sylow p-subgroup of G.

39 For the next two lemmas, notice that if N ¡ M, then either N ≤ H or P ≤ N, by the structure of M.

Lemma 3.2.3. For the above P , M and G, we have that P ≤ [M,M] ≤ [G, G] .

Proof. Notice that [M,M]¡M, so either [M,M] ≤ H or P ≤ [M,M]. If [M,M] ≤ H,

0 then G/H = P o H is abelian, which is not possible. Thus P ≤ [M,M] ≤ [G, G].

Lemma 3.2.4. For the above P and L, we have P ≤ L .

Proof. Notice that L∩M ¡M, thus either L∩M ≤ H or P ≤ L∩M. If L∩M ≤ H, then P ∩ L = {1}. Now hL, P i = L o P ≤ G, and P is a Sylow p-subgroup of G by

Lemma 3.2.2, thus p does not divide the order of L. Consider (L o P )/L ≤ Out(L). ∼ We know that P = (L o P )/L ≤ [Out(L), Out(L)], by Lemma 3.2.3. Now let us look at the families of ﬁnite non-abelian simple groups one by one, to show that this is not possible.

If L is either an alternating group or one of the 26 sporadic simple groups, then the order of Out(L) is 1 or 2 or 4 (when L = A6). Therefore Out(L) is abelian and ∼ [Out(L), Out(L)] = 1, which is a contradiction to P = (L o P )/L ≤ [Out(L), Out(L)]. If L is a ﬁnite simple group of Lie type, then by Theorem 2.5.12(b) of [11], we see that Out(L) is a split extension of Outdiag(L) and ΦLΓL. By Deﬁnition 2.5.10 of this ∼ book, ΦLΓL is abelian unless L = D4(q), where q is a prime power. ∼ If L D4(q), then P = (LoP )/L ≤ [Out(L), Out(L)] ≤ Outdiag(L). By Theorem 2.5.12(c) of [11], the prime divisors of Outdiag(L) are all prime divisors of L and we get a contradiction.

40 ∼ ∼ Now suppose that L = D4(q). By Theorem 2.5.12(e) of [11], Out(L)/Outdiag(L) = ∼ ∼ ΦL ×ΓL, where ΦL = Aut(Fq) and ΓL = S3. Taking a Sylow 3-subgroup A in ΓL, we see ∼ ∼ that Out(L)/(Outdiag(L) o A) = ΦL × Z2, which is abelian. Thus P = (L o P )/L ≤

[Out(L), Out(L)] ≤ Outdiag(L) o A. Since the order of D4(q) is divisible by 3, p 6= 3. Therefore the order of Outdiag(L) is divisible by p, which is again a contradiction to the fact that the prime divisors of Outdiag(L) are all prime divisors of L.

Thus P ≤ L.

The next result is clear:

Lemma 3.2.5. For an odd prime power pn, if n = 2ts and s is odd, then 2t+1 | pn −1.

Proof. By mathematical induction on t.

∼ n Lemma 3.2.6. If P = Zp and p is odd, then n = 1.

Proof. Suppose that p is odd and n ≥ 2. Since e divides both n and pn −1, by Lemma

pn−1 3.2.5 we have 2 | e . Therefore H has a involution z. By Theorem 2.1.1 (iii), we see [1] that H = Gα is isomorphic to a subgroup of a Frobenius complement of even order. By Theorem 3.4(a) of [5], we know that H has a unique involution. This implies hzi ¡ M, since H ¡ M. Thus M ≤ CG(z). Since M is a maximal subgroup of an almost simple group G, we obtain CG(z) = M. Next, since P ≤ L by Lemma 3.2.4,

P ≤ CL(z) = M ∩ L ≤ M. We also know that P ¡ M, therefore P ≤ O20 (CL(z)). ∼ By [11] Theorem 7.7.1, we see that O20 (CL(z)) is cyclic unless L = PSL(3, 4), and in

2 this case CL(z) = Z3 o Q8. If O20 (CL(z)) is cyclic then n can not be greater than ∼ 2 one, a contradiction. Suppose that L = PSL(3, 4) and L ∩ M = CL(z) = Z3 o Q8.

41 ∼ 2 2 Then P = Z3. Notice that the order of M is 9 · 8 now, thus M/L ∩ M is of order ∼ ∼ 8. On the other hand, M/L ∩ M = LM/L ≤ Out(L), and Out(L) = Z2 × S3 by the

Atlas [4]. Therefore we get a contradiction and n = 1.

By the Theorem 2.1.1, M = Zp−1×(ZpoZp−1) when n = 1. Thus by Lemma 3.2.6, we see that either p = 2, or p is odd and M = Zp−1 × (Zp o Zp−1). To treat these two cases, we need the following two results regarding ﬁnite non-abelian simple groups:

Theorem 3.2.7. (see [9, Theorem 16.6]) Let L be a ﬁnite simple group whose Sylow

2-subgroup is abelian. Then L is isomorphic to PSL(2, q) with q = 3 or 5 (mod 8),

m e PSL(2, 2 ), J1, or Ree(q) with q = 3 .

Recall that the p-rank for a ﬁnite group is the maximum dimension of an elemen- tary abelian p-subgroup, considered as a vector space over GF(p).

Theorem 3.2.8. (see [10, Theorem 1.86]) Let L be a ﬁnite simple group of 2-rank 2.

Then L is isomorphic to PSL(2, q) (q ≥ 5, q odd), PSL(3, q) or PSU(3, q) (q odd),

PSU(3, 4), A7, or M11.

we will use Theorem 3.2.7 to treat the case when p = 2, and in this case we will see that we can only get (v) of Lemma 3.1.1. We will use Theorem 3.2.8 to treat the case when p is odd, and in this case we can only get (i)-(iv) of Lemma 3.1.1. Both statements above will be proved by a series of lemmas.

42 3.3 Proof of the key lemma, in the case p=2

In this section, we assume p = 2.

3 Lemma 3.3.1. If p = 2 then (G, M) In Lemma 3.1.1 can only be (PΣL(2, 3 ), Z3 ×

A4).

Proof.

n By Lemma 3.2.4, we know that P = Z2 ≤ L and by Lemma 3.2.2 P is a Sylow 2-subgroup of G, hence P is a Sylow 2-subgroup of L. Now L is a ﬁnite non abelian simple group containing an abelian Sylow 2-subgroup so, by Theorem 3.2.7, L is

m isomorphic to PSL(2, q)( q = 3 or 5 mod 8), PSL(2, 2 ), J1, or a Ree group.

Suppose that L is the Janko group J1. Then by the Atlas [4] we know that

Aut(J1) = J1, so G = J1. Again by the Atlas, we ﬁnd that no maximal subgroup of

n n 2 J1 has order of the form 2 (2 − 1) . So L cannot be J1. Suppose that L = Ree(q), where q = 32m+1 with m ≥ 1 is a Ree group. Then

|L| = q3(q3 + 1)(q − 1). By 32m+1 = 3 mod 4, we see that q − 1 = 2 mod 4, and by

32m+1 = 3 mod 8, we see that q3 + 1 = 4 mod 8. Therefore 8 | |L| and 16 - |L|. Since ∼ 3 3 P is a Sylow 2-subgroup of L, P = Z2. Thus M = Z7 × (Z2 o Z7). By Theorem 6.5.5

3 2 (f) of [11], we know that NL(P ) = P.F21, so |F21| = 21 | |NG(P )| = |M| = 2 · 7 , which is a contradiction. So L cannot be a Ree group.

Suppose that L = PSL(2, 2m). Then the Sylow 2-subgroup P of L is isomorphic

m m m 2 to Z2 . Hence n = m and M should have order 2 (2 − 1) . The full automorphism ∼ m m m m m m group Aut(L) = PSL(2, 2 ).Zm has order 2 (2 + 1)(2 − 1)m. Thus, 2 (2 +

43 1)(2m − 1)m is divisible by 2m(2m − 1)2. From this we get (2m + 1)m − (2m − 1)m is divisible by 2m − 1, i.e., 2m is divisible by 2m − 1, which is not possible for m ≥ 2.

Now consider L = PSL(2, q), where q = 3 or 5 mod 8. By Lemma 1.2(c), Chapter

∼ 2 10 of [12], we know that a Sylow 2-subgroup of L is dihedral. So P = Z2 and

2 M = Z3 × (Z2 o Z3). By Lemma 1.3(f), Chapter 10 of [12], we can choose a 4-subgroup U of L, such ∼ that NL(U) = A4 or S4. Since |P | = 4 and P ≤ L by Lemma 3.2.4, we have P = U by Sylow’s Theorem. Without loss of generality, we can assume that P = U, since all

Sylow subgroups are conjugate in L. Now we know that NL(P ) = A4 or S4. On the other hand, from Lemma 3.2.2 we have NG(P ) = M, so either A4 ≤ M, or S4 ≤ M.

Since 24 - |M| = 36, we must have NL(P ) = L ∩ NG(P ) = L ∩ M = A4.

Next, also by Lemma 1.3(f), Chapter 10 of [12], we know that CAut(L)(P ) = P ×F ,

∼ ∼ 2 where F = Aut(GF(q)). By Lemma 3.2.2, CG(P ) = H × P = Z3 × Z2, hence ∼ Z3 = H ≤ F . Suppose H = hfi, where f ∈ F , has order 3. Then by Lemma 1.4(a), ∼ 1 Chapter 10 of [12], we know that CL(f) = PSL(2, q 3 ).

What is CL(f)? Let us consider CG(f) ﬁrst. Clearly M ≤ CG(f). Since M is maximal in G, either CG(f) = M or G. Clearly CG(f) 6= G, thus CG(f) = M.

Therefore, CL(f) = L ∩ M, which is A4, from above. ∼ 1 1 Now A4 = PSL(2, q 3 ), so q 3 = 3, and q = 27. So we have L = PSL(2, 27) and

2 M = Z3 × (Z2 o Z3). ∼ Now |M/L ∩ M| = 3, so M/L ∩ M = Z3. By Lemma 3.2.1, G = LM, thus ∼ ∼ ∼ G/L = LM/L = M/L ∩ M = Z3. Also f ∈ G \ L is a ﬁeld automorphism of order 3, so G = L o hfi = PΣL(2, 27)

44 Hence we get case (v) of Lemma 3.1.1, when p = 2.

3.4 Proof of the key lemma, in the case p odd

In this section, we assume that p is odd. Remember that now M = Zp−1 ×(Zp oZp−1). First let us show that the 2-rank of L is 2. We need the following lemmas:

∼ Lemma 3.4.1. There is a unique involution z in H = Zp−1. For this z, CG(z) = M.

Proof. Clear.

Lemma 3.4.2. If S is a 2-group, and there is an involution z ∈ S such that CS(z) =

∼ 2 T = hzi × hti = Z2, then S has a cyclic subgroup of index 2.

Proof.

We prove this by induction on |S|.

If S = T , then the statement is clear. ∼ ∼ If |S| = 8, then S = Q8 or S = D8. In either case, S has a cyclic subgroup of order 4.

Suppose that |S| ≥ 16. Notice ﬁrst that Z(S) = hti. This is true because

Z(S) ≤ CG(z) = T = hzi × hti, and z∈ / Z(S) since T < S. On the other hand, Z(S) is not trivial since S is a 2-group. Hence |Z(S)| = 2 and we can assume that

Z(S) = hti, without loss of generality.

Next consider the quotient group S¯ = S/Z(S). We know thatz ¯2 = 1,¯ wherez ¯

¯ ¯ ∼ 2 is the image of z in S. Now we will show that C = CS¯(¯z) = Z2, therefore by the inductive hypothesis S¯ has a cyclic group of index 2, denoted by L¯.

45 ¯ ∼ 2 ¯ ¯ To show that C = Z2, ﬁrst we prove that |C| = 4. Let a ∈ S such thata ¯ ∈ C.

a a a Then z = z or z = zt, implies that a ∈ N = NS(T ) (because t = t). Hence C¯ ≤ N¯, where N¯ is the image of N in S¯. Consider the conjugation action of N on T .

The orbit of z is a subset of {z, t, zt} and the point stabilizer for z is N ∩ CG(z) = T . Hence either |N : T | = 1, 2 or 3. Clearly T 6= N, since in the 2-group S, the normalizer of T is strictly bigger than T . Also |N : T | 6= 3, since both of them are

2-groups. So |N : T | = 2, |N| = 8, and N¯ = N/hti has order 4. Thus C¯ ≤ N¯ is of order 2 or 4. ¯ ¯ ¯ ¯ Suppose that |C| = 2, then Z(S) ≤ CS¯(¯z) = hz¯i, and Z(S) 6= 1, because S ¯ ¯ is a 2-group, so Z(S) = hz¯i. We then have S = CS¯(¯z) = hz¯i, therefore |S| = 4, contradicting |S| ≥ 16.

So C¯ has order 4 and C¯ = N¯. ¯ ∼ ¯ If C = Z4, then the pre-image of C, which is N, is either Z8 or hti × Z4. In either case, N is abelian and N ≤ CS(z) = T , a contradiction. ¯ ∼ 2 So C = Z2. Now consider the pre-image of L¯, denoted by L. Then L is either cyclic of index 2 in S, or L = hti×L0, where L¯ ∼= L0 ≤ L is a cyclic group of index 4 in S. Suppose the

0 second case happens and let L1 = ht1i be the unique subgroup of L of order 2. Then

0 L1 is characteristic in L because for the generator g of L and for an automorphism σ of L, either gσ ∈ L0, or gσ = tg0, where g0 ∈ L0 and o(g) = o(g0) = |L0| ≥ 4. In both

σ cases, we get t1 = t1. Now L ¡ S by |S : L| = 2, hence L1 ¡ S and t1 ∈ Z(S) = hti, a contradiction.

So L is a cyclic subgroup of S of index 2 and the lemma is proved.

46 Now we can prove:

Lemma 3.4.3. G has 2-rank 2.

Proof.

0 Let M = Zp−1 × (Zp o Zp−1) = H × (P o H ) and T = C × D, where C is the Sylow 2-subgroup of H and D is the Sylow 2-subgroup of H0. They have the same

m ∼ ∼ order 2 for some m ≥ 1. Clearly T is a Sylow 2-subgroup of M and C = D = Z2m , so T has 2-rank 2.

First consider the case when m ≥ 2.

Suppose T ≤ S, where S is a Sylow 2-subgroup of G. Either T = S or T < S.

If T = S, then S has 2-rank 2. If T < S, then T < T1 = NS(T ). We claim that

|T1 : T | = 2.

To see this, consider the conjugation action of T1 on T . Let the orbit containing z, the unique involution of C, be O. Let z0 ∈ D be the unique involution of D, then

O ⊂ {z, z0, zz0}, since they are the only involutions in T . Now in this action the point stabilizer of z is CS(z) = S ∩ M = T , by Lemma 3.4.1. So |T1 : T | = |O| = 1, 2 or 3. But T < T1 implies |T1 : T |= 6 1. Noting that T , T1 are 2-groups, we obtain

|T1 : T | = 2.

Next we claim that T1 = S. Suppose that T1 6= S, and so T1 < NS(T1). Then there

g 0 0 0 2m−1 exists g ∈ NS(T1) \ T1 such that T = T < T1, but T 6= T . Thus |T ∩ T | = 2 by

0 0 0 0 hT,T i = TT = T1. The group T ∩ C is not trivial since m ≥ 2, so T ∩ C contains

0 g ∼ 0 the unique involution z of C. Now T = T = T is abelian, hence T ≤ CG(z). This

47 0 implies T1 = hT,T i ≤ CG(z). By Lemma 3.4.1, M = CG(z). We get a contradiction, so T1 = S. The facts that T ¡S and |S : T | = 2 imply that if S has a subgroup E isomorphic

3 ∼ 2 to Z2, then |E : E ∩T | = 1 or 2. Since T has 2-rank 2, E∩T = Z2. But we know that

2 0 the unique subgroup in T isomorphic to Z2 is hzi×hz i, so z ∈ E and E ≤ CG(z) = M, which is a contradiction. Thus S has 2-rank 2. By Sylow’s Theorem, G has 2-rank 2.

0 ∼ 2 0 Next, consider the case m = 1. Now T = hzi × hz i = Z2, where z, z are as above. Consider a Sylow 2-subgroup S of G with T ≤ S. Since CG(z) = M, we have

CS(z) = S ∩ M = T . If S = T , then clearly S has 2-rank 2, hence G has 2-rank 2.

Now suppose that T < S. Since CS(z) = T , by Lemma 3.4.2 we know that S has a cyclic subgroup of index 2. Thus the 2-rank of S is either 1 or 2. Since T ≤ S has

2-rank 2, S has 2-rank 2 and G has 2-rank 2.

Furthermore, we claim that in the case m = 1 S can only be dihedral or semi- dihedral. To see this, ﬁrst notice that S is not abelian (because CS(z) = T < S) and has a cyclic subgroup of index 2. Then by Theorem 4.4, chapter 5 of [9], S is either

2l−1 2 y dihedral, semi-dihedral, generalized quaternion, or Ml(2) = hx, y|x = y = 1, x =

1+2l−2 x i(l ≥ 4). Suppose that S = Ml(2). Then by Theorem 4.3(i)(b), chapter 5 of

l−2 ∼ 2 [9], Z(S) is cyclic of order 2 ≥ 4, but Z(S) ≤ CS(z) = Z2, which is a contradiction. If S is a generalized quaternion group, then S has a unique involution, contradicting

2 ∼ to Z2 = T ≤ S. Therefore S can only be dihedral or semi-dihedral.

Now G has 2-rank 2, so L = soc(G), as a subgroup of G, has 2-rank at most 2.

48 We know that a non-abelian simple group with 2-rank ≤ 2 can not have 2-rank 0 or 1, see [10, p.1]. So L has 2-rank 2. Now we can quote Theorem 3.2.8 to get the following lemma:

Lemma 3.4.4. In Lemma 3.1.1, if p is odd then L is isomorphic to PSL(2, q) (q odd, q ≥ 5), PSL(3, q) or PSU(3, q) (q odd), PSU(3, 4), A7, or M11.

We will see that except for L ∼= PSL(2, q)(q odd, q ≥ 5), the other cases can not happen here. ∼ Let us ﬁrst exclude the three cases when L = PSU(3, 4), A7, or M11, using the ∼ Atlas [4]. Remember that G has a maximal subgroup M = Zp−1 × (Zp o Zp−1) whose order is p(p − 1)2, where p is an odd prime. Notice that M is solvable.

Lemma 3.4.5. In Lemma 3.1.1, if p is odd then L = soc(G) is not isomorphic to

PSU(3, 4), A7, or M11.

Proof.

We look at the three cases one by one:

• L ∼= PSU(3, 4)

By the Atlas [4], Out(L) = 4, so G = L, L : 2, or L : 4.

2+4 2 If G = L, then the maximal subgroups of G are 5 × A5, 2 : 15, 5 :S3, and

13 : 3 = F39. The ﬁrst one of these is not solvable, and the last three have orders 26 ·3·5, 2·3·52, and 3·13, respectively. None of these has form p(p−1)2.

If G = L : 2, then the maximal subgroups of G are PSU(3, 4), D10 × A5,

2+4 2 2 : (3 × D10), 5 :D12, and 13 : 6 = F78. The ﬁrst two are not solvable and

49 the last three have orders 27 · 3 · 5, 22 · 3 · 52, and 2 · 3 · 13, respectively. None

of these has form p(p − 1)2.

If G = L : 4, then the maximal subgroups of G are PSU(3, 4) : 2, D10 × A5.2,

2+4 2 2 : (3×D10).2, 5 : (4×S3), and 13 : 12 = F156. The ﬁrst two are not solvable and the last three have orders 28 · 3 · 5, 23 · 3 · 52, and 22 · 3 · 13, respectively.

None of these has form p(p − 1)2.

∼ • L = A7

By the Atlas [4], Out(L) = 2, so G = L or S7.

If G = L, then the maximal subgroups of G are A6, PSL(2, 7), PSL(2, 7), S5,

and (A4 × 3) : 2. The ﬁrst four are not solvable and the last one has order 23 · 32, which is not of the form p(p − 1)2.

If G = S7, then the maximal subgroups of G are A7,S6,S5 × 2, S4 × S3, and 7 : 6. The ﬁrst three are not solvable and the last two have orders 24 · 32 and

2 · 3 · 7, respectively. None of these has form p(p − 1)2.

∼ • L = M11

By the Atlas [4], Out(L) = 1, so G = L.

Now the maximal subgroups are M10 = A6.23, PSL(2, 11), S5, M9 : 2, and 2S4. The ﬁrst three are not solvable and the last two have orders 24 · 32 and 24 · 3,

respectively. None of these has form p(p − 1)2.

So L is not isomorphic to any of the above three groups.

50 Next we show that L is not isomorphic to PSL(3, q) or PSU(3, q) with q odd.

We need the following lemmas to show that L cannot be isomorphic to PSL(3, q) or PSU(3, q), q is odd:

Lemma 3.4.6. L is not isomorphic to PSL(3, 3) or PSU(3, 3).

Proof. We look at the two cases separately:

• L ∼= PSL(3, 3)

By the Atlas [4], Out(L) = 2, so G = L or L : 2.

2 2 If G = L, then the maximal subgroups of G are 3 : 2S4, 3 : 2S4, 13 : 3, and

4 3 4 3 3 S4. They have orders 2 · 3 , 2 · 3 , 3 · 13, and 2 · 3, respectively. None of these has form p(p − 1)2.

If G = L : 2, then G has a maximal subgroup S4 × 2, and so the 2-rank of G is at least 3, which is a contradiction to Lemma 3.4.3.

• L ∼= PSU(3, 3)

By the Atlas [4], Out(L) = 2, so G = L or L : 2.

1+2 If G = L, then the maximal subgroups of G are PSL(2, 7), 3 : 8, 4.S4, and

2 4 :S3. The ﬁrst one of these is not solvable and the last three have orders 23 · 33, 25 · 3, and 25 · 3, respectively. None of these has form p(p − 1)2.

If G = L : 2, then the maximal subgroups of G are PSU(3, 3), PSL(2, 7) : 2,

1+2 2 3 : 8 : 2, 4.S4 : 2, and 4 :D12. The ﬁrst two are not solvable and the last three have orders 24 · 33, 26 · 3, and 26 · 3, respectively. None of these has form

p(p − 1)2.

51 So L is not isomorphic to PSL(3, 3) or PSU(3, 3).

∼ Lemma 3.4.7. Let L = PSL(3, q), with q odd. For any involution z ∈ L, CL(z) has a composition factor isomorphic to PSL(2, q). The same conclusion holds for

L ∼= PSU(3, q), with q odd.

Proof. The proof of this lemma can be found in Lemma 4.1 (a), chapter 10 of [12].

Lemma 3.4.8. In Lemma 3.1.1, when p is odd, L = soc(G) is not isomorphic to

PSL(3, q) or PSU(3, q), with q odd.

Proof.

All the notations that we use here are as in Lemma 3.4.3.

Suppose L = soc(G) is isomorphic to PSL(3, q) or PSU(3, q), with q odd. By

0 ∼ Lemma 3.4.6, we have q ≥ 5. Now M = H × (P o H ) = Zp−1 × (Zp o Zp−1).

Let z be the unique involution in H. We have M = CG(z) by Lemma 3.4.1, and so

CL(z) = L∩M is solvable. If z ∈ L then, by Lemma 3.4.7, CL(z) = CG(z)∩L = M∩L has a composition factor isomorphic to PSL(2, q). This is not possible since PSL(2, q) is not solvable when q ≥ 5. If z∈ / L, then by Proposition 4.9.1 and Table 4.5.1 of [11], we see that z is either conjugate to a ﬁeld automorphism, a graph-ﬁeld automorphism, or a graph automorphism in Aut(L). Again by the above proposition and table, CL(z) is not solvable when q ≥ 5. Thus we get a contradiction again.

Next, we show that if L ∼= PSL(2, q), with q > 3 odd, then we get cases (i)-(iv) of Lemma 3.1.1. We need the following lemma:

52 m Lemma 3.4.9. If L = PSL(2, q), q = p1 > 3 is odd, and L ∩ M is a dihedral group, then the order of L ∩ M is either 2p or 4p, and p = 3 or 5.

∼ Proof. By Lemma 3.2.4, we know that Zp = P ≤ L ∩ M. Thus L ∩ M = P o (L ∩ 0 ∼ 0 ∼ (H × H )), where H = H = Zp−1. Now we claim that (L ∩ H) × P ≤ L ∩ M is the base group (the unique cyclic normal subgroup of index 2) of L ∩ M. To see this, notice that the base group of L∩M is contained in CG(P ), which is H ×P by Lemma 3.2.2. Thus the base group of L ∩ M is contained in L ∩ (H × P ) = (L ∩ H) × P .

Considering L ∩ H ≤ Z(L ∩ M), we see that L ∩ H is of order at most 2 since L ∩ M is dihedral. Therefore the order of L ∩ M is either 2p (when |L ∩ H| = 1) or 4p

(when |L ∩ H| = 2). If the order of L ∩ M is 2p, then L ∩ (H × H0) is of order 2. As

L ∩ H = 1, we have that either L ∩ M = P o htt0i or L ∩ M = P o ht0i, where t ∈ H is the unique involution of H and t0 ∈ H0 is the unique involution of H0. In either

∼ 0 0 case it is easy to see that M/L ∩ M = p−1 × p−1 (notice that H × H = hxi × hxx i, Z Z 2 where H = hxi and H0 = hx0i, and tt0 ∈ hxx0i). If the order of L ∩ M is 4p, then

0 2 0 0 L ∩ (H × H ) is of order 4, and is isomorphic to Z2. Hence L ∩ (H × H ) = hti × ht i and M/L ∩ M ∼= p−1 × p−1 . Z 2 Z 2

From the above, we see that p−1 × p−1 is isomorphic to a subgroup of L/L∩M ≤ Z 2 Z 2 ∼ Out(L). We know that Out(L) = Zm × Z2, by Lemma 1.2(a), chapter 10 in [12]. If

p−1 × p−1 ≤ m × 2 then p−1 × p−1 has a cyclic subgroup of index at most 2, Z 2 Z 2 Z Z Z 2 Z 2 p−1 implying that 2 ≤ 2. Hence p = 3 or 5.

Now we prove:

53 ∼ m Lemma 3.4.10. In Lemma 3.1.1, if p is odd and L = PSL(2, q), q = p1 > 3 odd, then (G, M) can only be one of the cases (i)-(iv) listed in the lemma.

Proof. We claim that either L∩M is maximal in L, or G = PGL(2, 7) and M = D12. The pairs (G, M) with M maximal in G but L ∩ M not maximal in L are listed in

Table 1 of [8]. All cases in this table, with the exception of G = PGL(2, 7), M = D12, are eliminated because |M| is not of the form p(p − 1)2 for any primes p.

Assume now that L ∩ M is maximal in L. The maximal subgroups of L are grouped into eight cases in Theorem 2.2 of [8]. The non-solvable maximal subgroups ∼ are eliminated because L∩M ≤ M is solvable. The cases L∩M = A4 = PSL(2, 3) or ∼ S4 = PGL(2, 3) are impossible, because P ≤ L ∩ M by Lemma 3.2.4 and this implies ∼ p = 3. However, if p = 3 then M = D12 and L ∩ M cannot be A4 or S4.

∼ m ∼ m Suppose now that L ∩ M = p q−1 . In this case, soc(L ∩ M) = p . Using Z 1 o Z 2 Z 1 ∼ m Lemma 3.2.4 again, we see that P ≤ L ∩ M. Hence P = Zp1 , implying p = p1

p(p−1) 2 and m = 1, and |Out(L)| = 2. Thus |M| ≤ 2 · |L ∩ M| = 2 · 2 < p(p − 1) , a contradiction.

The only remaining case is that L ∩ M is dihedral of order q − 1 and q ≥ 13, or of order q + 1 and q 6= 7, 9. By Lemma 3.4.9, we have p = 3 or 5. By the same lemma, if p = 5 then |L ∩ M| = 10 or 20, implying q = 19. Then M/L ∩ M is of order 4 and ∼ M/L ∩ M ≤ Out(L) = Z2, clearly a contradiction. Thus p = 3 and M = D12. So we have 3 cases here: q = 13, L ∩ M = D12; q = 5, L ∩ M = D6; q = 11, L ∩ M = D12. It is clear that G = L when q = 11, 13, and G = PGL(2, 5) when q = 5. So we get case (i)-(iv) in Lemma 3.1.1.

54 Now combining Lemma 3.3.1, Lemma 3.4.4, Lemma 3.4.5, Lemma 3.4.8, Lemma 3.4.10, and Lemma 3.1.2, we proved Lemma 3.1.1.

55 CHAPTER 4

PROOF OF THE MAIN THEOREMS

In this chapter, we prove Theorem 1.2.1 and Theorem 1.2.3, together with their corollaries.

4.1 Basic background on coset graphs

We ﬁrst introduce coset graphs. Recall that for a subgroup H of a group G, the core of H in G is the largest normal subgroup of G contained in H. If the core of H in G is {1}, then we say that H is core-free in G. For a core-free subgroup H of G, G has a faithful transitive permutation representation on the set of right cosets {Hx|x ∈ G} by right multiplication. A coset graph

Γ = Cos(G, H, HgH), where g ∈ G, and g2 ∈ H is deﬁned as the graph whose vertex set is the set of all right cosets of H in G, and

Hx is adjacent to Hy if and only if yx−1 ∈ HgH.

We have the following lemmas for coset graphs, which are easy to prove: (see

Theorem 2.1 of [7])

56 Lemma 4.1.1. A connected (undirected) graph Γ is G-arc transitive if and only if

Γ = Cos(G, H, HgH), where hH, gi = G, H is core-free in G, and g2 ∈ H.

Lemma 4.1.2. A coset graph Γ = Cos(G, H, HgH) is (G, 2)-arc transitive if and only if H acts 2-transitively on the set of cosets {(H ∩ Hg)x|x ∈ H}.

The above two lemmas are useful when we construct G-arc transitive graphs.

Let Γ = Cos(G, H, HgH) be a coset graph and let L = H ∩ Hg. We claim that

2 we can always ﬁnd a 2-element x ∈ NG(L) \ L such that x ∈ L, Cos(G, H, HgH) = Cos(G, H, HxH) and L = H ∩ Hg = H ∩ Hx. To see this, notice that by g2 ∈ H and

2 −1 2 g 2 g g g = g g g ∈ H , g ∈ H ∩ H = L. It is clear that L = L, so g ∈ NG(L). Now hL, gi/L has order 2. Thus we can ﬁnd a 2-element x in hL, gi\L ≤ NG(L)\L. Since Lg = Lx, HgH = HxH. Therefore Cos(G, H, HgH) = Cos(G, H, HxH). Clearly x2 is in L. Also it is clear that L = H ∩ Hg = H ∩ Hx.

Now label the vertices corresponding to H and Hg to be α and β, respectively.

g Notice that the valency of Γ is |Gα : Gα ∩ Gβ| = |H : H ∩ H |. We have the following lemma by the above discussion:

Lemma 4.1.3. Suppose we have a coset graph Γ = Cos(G, H, HgH). Let L = Gαβ =

g Gα ∩ Gβ = H ∩ H . Then g can be chosen in a way such that g is a 2-element in

2 NG(L) \ L and g ∈ L.

This lemma will be very useful when we try to determine the number of non- isomorphic graphs for given G and Gα = H.

Notice that in this chapter we use L to denote Gαβ, which is diﬀerent from the notation in the previous chapter.

57 4.2 Proof of Theorem 1.2.1 and Corollary 1.2.2

Now let us look at Theorem 1.2.1 ﬁrst. Notice that by Lemma 4.1.1 and Lemma 4.1.3 we can always assume the graph Γ that we are concerned in this paper, i.e., a (G, 3)- arc regular graph, is a coset graph Cos(G, H, HgH), where g is 2-element in NG(L)\L, L = H ∩ Hg , and g2 ∈ L.

The following lemma is well known:

Lemma 4.2.1. The Petersen graph is a connected (G, 3)-arc regular graph for G =

PGL(2, 5), acting primitively on the vertex set. The Coxeter graph is a connected

(G, 3)-arc regular graph for G = PGL(2, 7), acting primitively on the vertex set.

We need the following lemma to prove that the two graphs above are the only

(G, 3)-arc regular graphs such that G acts primitively on the vertex set.

Lemma 4.2.2. In Lemma 3.1.1, let L ≤ M be a Sylow 2-subgroup of M for the ﬁrst four cases and be a Sylow 3-subgroup of M in the last case. Then L is unique in M ∼ ∼ under conjugation. Furthermore, NG(L) = D8 in cases (i) and (ii); NG(L) = A4 in

∼ 2 cases (iii) and (iv); and NG(L) = Z3 o Z3 in case (v).

Proof. Clearly L is unique in M under conjugation, since L is a Sylow subgroup of

M. We show that the normalizer of L in G is as stated by considering the ﬁve cases one by one.

2 ∼ ∼ ∼ 2 In case (i), we have G = PGL(2, 5) and Z = L ≤ M = D12. Since L = Z2 is in a Sylow-2-subgroup of G, |NG(L)| is divisible by 8. By the Atlas [4], NG(L) ≤ S4, and so NG(L) is isomorphic either to D8 or S4. If NG(L) = S4, then we claim that

58 L ≤ PSL(2, 5). To see this, ﬁrst notice that the order of L∩PSL(2, 5) can only be 2 or

4, since G/PSL(2, 5) is of order 2. Also notice that L∩PSL(2, 5)¡S4 ∩PSL(2, 5) = A4.

Since A4 has no normal subgroup of order 2, L ∩ PSL(2, 5) is of order 4, implying ∼ L ≤ PSL(2, 5). Also the unique Sylow 3-subgroup H = Z3 of M is in PSL(2, 5), because G/PSL(2, 5) has order 2. Thus hL, Hi = M ≤ PSL(2, 5), a contradiction to ∼ the fact that M is maximal in G. Therefore NG(L) = D8.

2 ∼ ∼ In case (ii), we have G = PGL(2, 7) and Z = L ≤ M = D12. By the same argument as above, L cannot be contained in PSL(2, 7). Thus NG(L) ≤ D16, by the ∼ ∼ Atlas [4]. Since L = Z2 × Z2, by a simple calculation we obtain NG(L) = D8. 2 ∼ ∼ In case (iii), we have G = PSL(2, 11) and Z2 = L ≤ M = D12. By the Atlas 0 ∼ 2 [4], there are subgroups of G isomorphic to A4, so there are subgroups L = Z2 with

0 0 NG(L ) ≥ A4. Since L, L are both Sylow 2-subgroups of G, NG(L) ≥ A4. Also by ∼ the Atlas [4], the only maximal subgroups of G containing a subgroup N = A4 are

2 isomorphic to A5 and NA5 (Z2) = A4, so NG(L) = A4. 2 ∼ ∼ In case (iv), we have G = PSL(2, 13) and Z2 = L ≤ M = D12. By the same argument as in case (iii), NG(L) ≥ A4. Since A4 is maximal in G, NG(L) = A4.

2 ∼ ∼ In case(v), we have G = PΣL(2, 27) and Z3 = L ≤ M = Z3×A4. By Lemma 3.1.2, we can assume M = CG(f), where f is the ﬁeld automorphism of order 3 in G. Let

L = hfi × hbi, where hfi, hbi are all isomorphic to Z3 and b ∈ PSL(2, 27). It is clear that NG(L) = NPSL(2,27)(L) o hfi, since f∈ / PSL(2, 27) and |G : PSL(2, 27)| = 3.

Any g ∈ NPSL(2,27)(L) normalizes L ∩ PSL(2, 27) = hbi, therefore g ∈ NPSL(2,27)(hbi). By the Atlas [4], we see that b and b2 are in diﬀerent conjugacy classes of PSL(2, 27), so NPSL(2,27)(hbi) = CPSL(2,27)(hbi). This implies g ∈ CPSL(2,27)(hbi). By the Atlas

59 ∼ 3 [4] again, CPSL(2,27)(hbi) = Z3. Summarizing, we obtain NPSL(2,27)(L) ≤ P3, where ∼ 3 P3 = Z3 is the (unique) Sylow 3-subgroup of PSL(2, 27) containing hbi.

Finally, we determine which g ∈ P3 belong to NPSL(2,27)(L). Consider the isomorphism induced by the conjugation of g on hfi × hbi. Since hbi is ﬁxed in this ∼ isomorphism, g induces a isomorphism on the quotient group hfi × hbi/hbi = Z3. This has to be a identity isomorphism since g3 = 1. Therefore f g ∈ hbif.

The elements of P3 can be identiﬁed with the transformations gβ : x → x + β on GF(27), for some β ∈ GF(27), and f can be identiﬁed with the Frobenius

3 f automorphism f : x → x of GF(27). If b = gγ : x → x + γ then b = b : x → (x9 +γ)3 = x+γ3, implying γ ∈ GF(3). The elements of hbif are the transformations

3 g x → x + δ, for δ ∈ GF(3). Now g = gβ ∈ NPSL(2,27)(L) if and only if f : x → (x − β)3 + β = x3 + β − β3 = x + δ for some δ ∈ GF(3). Consider the group homomorphism

ϕ : GF(27) → GF(27) where

β 7→ β − β3, for all β ∈ GF(27).

This is a group homomorphism under addition with kernel GF(3). Notice that

3 3 x −x+1 is irreducible over GF(3), so there exists a β0 ∈ GF(27) such that β0−β0 = 1. ∼ Therefore 1 ∈ Image(ϕ), and Z3 = GF(3) ≤ Image(ϕ). Thus the pre-image of GF(3)

2 2 under ϕ is isomorphic to Z3. As a result, NPSL(2,27)(L) = Z3, hence in the bigger ∼ 2 group G = PΣL(2, 27), NG(L) = NPSL(2,27)(L) o hfi = Z3 o Z3.

60 Now with Lemma 4.2.2, we prove the following lemma. For this lemma, recall that for a group G and a subset S of G with S−1 = S, a (undirected) Cayley graph

Cay(G, S) is deﬁned to be the graph Γ with vertex set V (Γ ) = G such that

x is connected to y if and only if yx−1 ∈ S.

Also recall that a primitive permutation group G is of type TW if soc(G) is non- abelian and regular on Ω.

Lemma 4.2.3. If Γ is a (G, 3)-arc regular graph such that G is primitive on the vertex set, then (Γ ,G) can only be either:

(i) Γ is the Petersen graph, and G ∼= PGL(2, 5), or

(ii) Γ is the Coxeter graph, and G ∼= PGL(2, 7).

Proof. Let Γ be a (G, 3)-arc regular graph such that G is primitive on the vertex set, and let M = Gα be a vertex stabilizer. It is clear that G has a 2-transitive sub-orbit. By the result in [21], G can only be of HA, TW, or AS. However, G cannot be of type

TW, because if it is, then Gα has no nontrivial solvable normal subgroup by Theorem

∼ n 4.7B of [5]. But we have P = Zp ¡ M by Theorem 2.1.1(iii), a contradiction. We claim that G cannot be of type HA type. Otherwise Γ is a Cayley graph by Lemma

16.3 of [1], and the elements of Gα act on Γ (α) as group automorphisms. Therefore Γ cannot be (G, 3)-arc transitive, since a 3-arc (y, 1, x, yx) is ﬁxed by the vertex stabilizer Gα, where α = 1. Thus, G can only be of type AS and G can be only one of the ﬁve cases listed in Lemma 3.1.1.

61 ∼ In case (i), we have G = PGL(2, 5) and Gα = M = D12. Now the valency of Γ is

2 ∼ 3 and for β ∈ Γ (α), Z2 = Gαβ ≤ Gα. So we can write Γ = Cos(G, M, MgM), where g ∼ 2 g is a 2-element in NG(L) \ L, L = Gαβ = M ∩ M = Z2 × Z2, g ∈ L. We claim that this graph has to be the Petersen graph.

To see this, ﬁrst notice that by Lemma 4.1.1 and Lemma 4.1.3, the Petersen graph

0 0 0 0 0 0g0 ∼ 0 can be written as Γ = Cos(G, M , Mg M), where L = M ∩ M = Z2 × Z2, g is a

0 0 02 0 2-element in NG(L ) \ L and g ∈ L . Notice that M is unique under conjugation in G by Lemma 3.1.1(i), also L is unique under conjugation in M by Lemma 4.2.2. We

0 0 0 can assume M = M and L = L . As a result, we can assume that g ∈ NG(L) \ L.

∼ ∼ 0 By Lemma 4.2.2, NG(L) = D8, hence NG(L)/L = Z2. Since g, g ∈ NG(L) \ L, Lg = Lg0 and MgM = Mg0M. Now Cos(G, M, MgM) = Cos(G, M, Mg0M) is the

Petersen graph.

In case (ii), the same argument as above shows that Γ is the Coxeter graph.

Cases (iii)-(v) can not happen here. To see this, notice that NG(L)/L has odd order for each of these three cases by Lemma 4.2.2. Thus NG(L)\L has no 2-element and so the appropriate coset graph does not exist.

Therefore we have shown that a primitive (G, 3)-arc regular graph can only be the Petersen graph or the Coxeter graph.

By Lemma 4.2.1 and Lemma 4.2.3, we get Theorem 1.2.1.

To prove Corollary 1.2.2, we need the following lemma. Recall a graph Γ is called s-transitive if Aut(Γ ) is transitive on the set of s-arcs but not on the set of (s+1)-arcs.

Lemma 4.2.4. Let Γ be a connected s-transitive graph of valency 3 with s ≥ 3.

62 Then s = 3, 4 or 5 and Γ is s-regular. Let A = Aut(Γ ) and let α be a vertex of Γ . Then we have one of the following: (s, Aα) = (3, D12), (s, Aα) = (4, S4) or

(s, Aα) = (5, S4 × Z2).

The proof of lemma can be found in [1], chapter 18 and we will see that this lemma is also useful in the proof of corollary 1.2.4.

Now let us prove corollary 1.2.2:

Proof of Corollary 1.2.2: It is well known that the full automorphism group of the Petersen graph is PGL(2, 5). Let A = Aut(Γ ) be the full automorphism group of the Coxeter graph Γ . We shall prove that A = PGL(2, 7). From Theorem1.2.1, we know that PGL(2, 7) ∼= G ≤ A. Suppose that A 6= G, then Γ is either 4-transitive or 5-transitive. Since D12 ≤ Aα, by Lemma 4.2.4 we know that Γ is 5-transitive and |A : G| = 4. Consider the action of A on the cosets of G. The kernel of this action must contain H = soc(G) because H is simple and has no faithful permutation representation on four points. Hence H ¡ A. Since A is primitive, A ≤ Aut(H) = G by the O’Nan-scott theorem, a contradiction.

4.3 Proof of Theorem 1.2.3 and Corollary 1.2.4

In the following discussion, let Γ be a connected bipartite graph with bi-parts ∆1 and ∆2, G ≤ Aut(Γ ) such that Γ is (G, 3)-arc regular, and G is bi-primitive on the vertex set. Recall that G is bi-primitive on the vertex set of Γ if G is transitive on

+ + V (Γ ) and G is primitive on ∆i, where G is the setwise stabilizer of ∆1 (and also of ∆2) in G.

63 We start with the following lemma:

Lemma 4.3.1. For each pair (G, Gα) listed in Theorem 1.2.3(iii), there exist a connected bipartite graph Γ such that G ≤ Aut(Γ ); Γ is a (G, 3)-arc regular graph; G is bi-primitive on the vertex set; and Gα is a vertex stabilizer of Γ in G.

Proof. First consider G = PGL(2, 11). Now PSL(2, 11) ≤ G. By Lemma 3.1.1(iii), ∼ we know that PSL(2, 11) has a maximal subgroup M = D12, which is unique under conjugation. We want to construct a connected (G, 3)-arc regular graph, which satisﬁes our requirements.

Consider the Sylow 2-subgroup L of M, which is unique in M under conjugation

2 by Lemma 4.2.2. We need an 2-element g ∈ NG(L)\L , such that g ∈ L, hM, gi = G (for the connectivity), and Γ = Cos(G, M, MgM) is a cubic (G, 3)-arc regular graph with G bi-primitive on Γ . Actually we just need to show that Γ is (G, 3)-arc transitive and G is bi-primitive on Γ , since then Γ has to be (G, 3)-arc regular by the order of

G. ∼ By Lemma 4.2.2, NPSL(2,11)(L) = A4. By the Atlas[4], the only maximal subgroups of G containing A4 are S4 and PSL(2, 11), so A4 ≤ NG(L) ≤ S4. we claim that ∼ NG(L) = S4.

There exists a Sylow 2-subgroup L1 of G with order 8 such that L ≤ L1. Since

0 0 |L : L| = 2, we have L ¡ L . Since A4 ≤ NG(L) and L1 ≤ NG(L), we obtain ∼ NG(L) = S4.

Next, we determine NG(M). By the Atlas [4], the group G contains maximal ∼ subgroups H = D24, and such subgroups contain index 2, and therefore normal,

64 ∼ subgroups H1 = D12, H1 ≤ PSL(2, 11). From the maximality of H, we obtain

NG(H1) = H. Again by the Atlas [4], there is a unique conjugacy class of subgroups ∼ ∼ isomorphic to D12 in PSL(2, 11), so H1 = M and NG(M) = D24. ∼ We claim that NG(L) ∩ NG(M) = D8. Let L1 be a Sylow 2-subgroup of NG(M), containing L. We have L ¡ L1 because |L1 : L| = 2, so L1 ≤ NG(L). This implies ∼ D8 = L1 = NG(L)∩NG(M), since |L1| = 8 and NG(L)∩NG(M) is a proper subgroup of the group NG(M) of order 24. ∼ The group NG(L) = S4 has four involutions in NG(L) \ L1, let g be any of these four involutions. We claim that Γ = Cos(G, M, MgM) is a graph which satisﬁes our requirements.

It is clear that M is core-free and g2 = 1 ∈ L. To see that Γ is connected,

g we need to show that hM, gi = G. Since g∈ / NG(M), we have M 6= M, implies M < hM,M gi ≤ PSL(2, 11), Now, as M is maximal in PSL(2, 11), we can deduce that hM,M gi = PSL(2, 11), and so hM, gi ≥ hM,M gi = PSL(2, 11). We also have ∼ g∈ / PSL(2, 11), because NG(L) ∩ PSL(2, 11) = A4. Hence PSL(2, 11) < hM, gi ≤ G, implying hM, gi = G.

To see that Γ is cubic, we have to show |M : M ∩ M g| = 3. It is clear that

g ∼ 2 g M ∩M ≥ L = Z2, since g ∈ NG(L). Also M ∩M is a proper subgroup of M, because

g g g∈ / NG(M). Therefore M ∩ M = L, and the valency of Γ is |M : M ∩ M | = 3. To see that Γ is (G, 3)-arc transitive hence (G, 3)-arc regular, ﬁrst notice that

M acts 2-transitively on the set of the cosets {(M ∩ M g)x : x ∈ M}. Hence, by

Lemma 4.1.2, Γ is (G, 2)-arc transitive. Suppose Γ is not (G, 3)-transitive and con- ∼ sider Gαβγ, the stabilizer of a 2-arc. We have Gαβγ = Z2 by the order of G, so there

65 exists a ∈ Gαβγ with a 6= 1. This a must ﬁx the neighborhood of γ, otherwise Γ is (G, 3)-arc transitive. Since Γ is (G, 2)-arc transitive, any element of G which ﬁxes a

2-arc will ﬁx all the neighborhood of the end point of that 2-arc. Hence, by the connectivity of Γ , a ﬁxes all vertices of Γ , contradicting a 6= 1. Therefore Γ is (G, 3)-arc transitive.

To see that Γ is a bipartite graph, notice that g ∈ G \ PSL(2, 11), and |G :

PSL(2, 11)| = 2. We claim that Γ is a bipartite graph with bi-parts ∆1 = {Mx : x ∈

PSL(2, 11)} and ∆2 = {Mx : x ∈ PSL(2, 11)g}. Indeed, if Mx, My ∈ ∆i for some i ∈ {1, 2}, then xy−1 ∈ PSL(2, 11) and so xy−1 ∈/ MgM and {Mx, My} is not an

+ edge. Also it is clear that G = G∆1 = PSL(2, 11).

+ + To see that G is bi-primitive on Γ , notice that Gα = (G )α = M is maximal in G . So we conclude that Γ is a connected cubic (G, 3)-arc regular with G ∼= PGL(2, 11) ∼ acting bi-primitively on Γ , and the vertex stabilizer Gα = D12.

For later use, we mention that no matter which of the four involutions of NG(L) \

L1 do we pick as g, we get the same graph. The reason is because for any pair

0 l 0 l g, g of these involutions, there exists l ∈ L1 such that g = g and, for this l, L = L and M l = M. Hence Cos(G, M l,M lglM l) = Cos(G, M, Mg0M) and obviously

Cos(G, M, MgM) ∼= Cos(G, M l,M lglM l). ∼ For G = PGL(2, 13) and the maximal subgroup M = D12 of PSL(2, 13) ≤ G, the same argument shows that there exists a unique connected, cubic (G, 3)-arc regular ∼ graph Γ with G acting bi-primitively on V (Γ ), and the vertex stabilizer Gα = M =

D12. Now consider G = PΓL(2, 27). By Lemma 3.1.1(v), we know that PΣL(2, 27) ≤ G

66 ∼ has a maximal subgroup M = Z3 o A4. We want to construct a connected (G, 3)-arc ∼ regular graph of valency 4, such that G acts bi-primitively on V (Γ ) and Gα = Z3 ×A4.

By the description of the maximal subgroup Z3 × A4 in the Atlas [4] (or the proof of Lemma 3.1.2), we can assume that M = CPΣL(2,27)(f) = hfi × CK (f) = hfi × A4, where f is a ﬁeld automorphism of order 3 in PΣL(2, 27) and K = PSL(2, 27). Also notice that M is core-free in G since soc(G) = PSL(2, 27).

2 ∼ 2 Let Z3 = L ≤ M. We need some g ∈ NG(L) \ L such that g ∈ L, hM, gi = G, and Γ = Cos(G, M, MgM) is (G, 3)-arc-regular with G acting bi-primitively on V (Γ ).

Again we just need to show that Γ is (G, 3)-arc transitive, because then Γ has to be

(G, 3)-arc regular by the order of G.

Let us ﬁrst determine NG(L). From the proof of Lemma 4.2.2, we know that

∼ 2 ∼ NPΣL(2,27)(L) = L o A = Z3 o Z3, where L = hfi × hbi = Z3 × Z3, b ∈ CK (f) and ∼ A = Z3 is a subgroup of the Sylow 3-subgroup P3 of PSL(2, 27) containing b. We can write GL(2, 27), as a matrix group, in a basis such that P3 corresponds to the lower triangular matrices with 1’s on the diagonal, and f corresponds to the transformation of matrices raising each entry to the third power. Let σ ∈ PGL(2, 27) be deﬁned by corresponding matrix

  −1 0     . 0 1

Clearly σ is an involution and normalizes L, and σ∈ / NPΣL(2,27)(L). It is easy to see that σ also normalizes NPΣL(2,27)(L) because f and σ commute. Therefore

67 ∼ NG(L) ≥ hNPΣL(2,27)(L), σi = (LoA)ohσi = LoD6. Since |NG(L): NPΣL(2,27)(L)| = ∼ |G : PΣL(2, 27)| = 2, NG(L) = (L o A) o hσi = L o D6.

Next, we determine NG(M). We claim that NG(M) = CG(f) = hfi × (CK (f) o hσi) = hfi × (A4 o Z2).

To see that CG(f) = hfi × (CK (f) o hσi), ﬁrst notice that σ ∈ CG(f) implies hCK (f), σi ≤ CG(f). By K ¡ G, we have CK (f) ¡ CG(f). Since σ∈ / PΣL(2, 27), and so σ∈ / K, we have that hCK (f), σi = CK (f) o hσi. Thus hfi × (CK (f) o hσi) ≤

CG(f). Since |CG(f): CK (f)| ≤ |G : K| = 6, we obtain CG(f) = hfi × (CK (f) o hσi).

To see that NG(M) = CG(f), notice that M = CPΣL(2,27)(f) ¡ CG(f) since

PΣL(2, 27) ¡ G. Thus CG(f) ≤ NG(M). Then by |NG(M): M| = |NG(M):

NPΣL(2,27)(M)| ≤ |G : PΣL(2, 27)| = 2, we have NG(M) = CG(f).

Next consider NG(M)∩NG(L). We claim that NG(M)∩NG(L) = Lohσi. To see

2 3 this, notice that L o hσi ≤ NG(M) ∩ NG(L). Since |NG(M)| = 3 · 2 and |NG(L)| =

3 2 3 ·2, we know that |NG(M)∩NG(L)| divides 3 ·2. Thus NG(M)∩NG(L) = Lohσi.

Let g be one of the two involutions in NG(L) \ NG(M). We claim that Γ = Cos(G, M, MgM) is a graph which satisﬁes our requirements.

Clearly M is core-free in G and g2 = 1 ∈ L. We claim that hM, gi = G, and so Γ

g is connected. Indeed, since g∈ / NG(M), we see that M 6= M. Since M is maximal in PΣL(2, 27), this implies that hM,M gi = PΣL(2, 27). Also, since g∈ / PΣL(2, 27), we obtain hM, gi ≥ hM,M g, gi PΣL(2, 27), implying hM, gi = G. To see that Γ has valency 4, we have to show that |M : M ∩ M g| = 4. We claim that M ∩ M g = L. To see this, ﬁrst notice that L ≤ M ∩ M g. If L 6= M ∩ M g,

68 then by |L| = 9 and |M| = 36, either M ∩ M g = M or |M : M ∩ M g| = 2. Since

g g g∈ / NG(M), we can not have M ∩ M = M and so |M : M ∩ M | = 2. This implies that M ∩ M g ¡ M. Now, as g normalizes M ∩ M g, we have M ∩ M g ¡ hM, gi = G, which is not possible since soc(PΓL(2, 27)) = PSL(2, 27). Therefore L = M ∩ M g and Γ has valency |M : L| = 4.

To see that Γ is (G, 3)-arc transitive hence (G, 3)-arc regular, ﬁrst notice that M acts 2-transitively on the set of the cosets {(M ∩ M g)x : x ∈ M}. Therefore, Γ is

(G, 2)-arc transitive by Lemma 4.1.2. Suppose that Γ is not (G, 3)-arc transitive. ∼ Consider Gαβγ, the stabilizer of a 2-arc. By the order of G, we know that Gαβγ = Z3.

Leta ∈ Gαβγ, a 6= 1. Then a induces a permutation s on the 3-element set. The order of s divides 3 and s can not act transitively, otherwise Γ is (G, 3)-arc transitive. This implies that s = 1. Since Γ is (G, 2)-arc transitive, any element of G which ﬁxes a

2-arc will ﬁx the entire neighborhood of the endpoint of this 2-arc. Hence, by the connectivity of Γ , a ﬁxes V (Γ ) pointwise. This fact contradicts to a 6= 1, implying that Γ is (G, 3)-arc transitive.

By the same argument as in the discussion for the case G = PGL(2, 11), we can see that Γ is a bipartite graph and G is bi-primitive on Γ .

Also, by the same argument as in the case G = PGL(2, 11), we obtain isomorphic graphs no matter which of the two involutions in NG(L) \ NG(M) do we choose as g, because these two involutions are conjugates under NG(L) ∩ NG(M).

For the next lemma, recall that for a (G, s)-arc transitive graph Σ = Cos(G, H, HgH),

69 and for F = G × hzi, where z is an involution, Γ = Cos(F, H, HgzH)) is an (F, s)-arc transitive bipartite graph. This graph Γ is called the standard double cover of Σ.

Lemma 4.3.2. All the graphs listed in Theorem 1.2.3 exist and are connected (G, 3)- arc regular graphs for the corresponding G. Also, G is bi-primitive on the vertex set.

Each of the three graphs in Theorem 1.2.3(iii) is the unique (G, 3)-arc regular graph for the corresponding G and Gα.

Proof.

It is clear that the graphs in Theorem 1.2.3 (i) and (ii) exist, and they are (G, 3)- arc regular graphs for the corresponding G by the explanation before this lemma.

Also it is clear that G is bi-primitive on the vertex set.

To see that they are connected, notice that it is obvious for case (i).

For case (ii), ﬁrst consider the standard double cover of the Petersen graph. To show that Γ is connected, we need that hH, gzi = G = PGL(2, 5) × hzi, where

∼ g 2 + H = D12 and g ∈ PGL(2, 5) \ H such that H ∩ H = Z2. Consider hH, gzi ∩ G , + ∼ + + where G = G∆1 = PGL(2, 5). We see that hH, gzi ∩ G = H or G , since H is maximal in G+. If it is H, then H = hH, gzi ∩ G+ ¡ hH, gzi by G+ ¡ G. This implies that H = Hgz = Hg, which is not possible. Hence hH, gzi ∩ G+ = G+. Using that gz∈ / G+, we obtain hH, gzi = G. The same argument shows that the standard double cover of the Coxeter graph is connected.

For case (iii), we proved in Lemma 4.3.1 that such graphs Γ exist. Now we need to show that Γ is the unique bi-primitive (G, 3)-arc regular for the corresponding G and Gα.

70 Suppose that Γ = Cos(G, M, MgM) and Γ 0 = Cos(G, M 0,M 0g0M 0) are such graphs, with M ∼= M 0. In all the three cases, there is a unique conjugacy class of maximal subgroups in G+ isomorphic to M, so we can assume M = M 0. This also implies that we can assume L = M ∩ M g = L0 = M 0 ∩ M 0g0 and then, as noted in the

0 proof of Lemma 4.3.1, the graphs Γ and Γ are isomorphic.

The following lemma ﬁnishes the proof of Theorem 1.2.3. Recall that for a graph

Γ and N ¡ G ≤ Aut(Γ ), where G is transitive on the vertex set V , the normal

N N quotient ΓN of Γ is deﬁned as the graph with vertex set B = {x : x ∈ V }, and x is adjacent to yN if and only if some vertex u ∈ xN is adjacent in Γ to some vertex v ∈ yN . Also recall that a primitive permutation group G is of type PA if Ω = ∆m,

m B = T ≤ G ≤ M o Sm ≤ Sym(∆) o Sm, where M is a primitive permutation group on ∆ of type AS with socle T , and G acts transitively on the m simple factors of B

(m ≥ 2).

Lemma 4.3.3. Let Γ be a connected (G, 3)-arc regular graph such that G acts bi- primitively on V (Γ ). Then Γ can be only one of the graphs in Theorem 1.2.3.

Proof.

+ Let ∆i, i = 1, 2, be the bi-parts of Γ and let G∆1 = G∆2 = G be the set-wise stabilizer of ∆1 and ∆2.

+ + If G does not act faithfully on ∆1, then the point-wise stabilizer K1 of ∆1 in G

+ is nontrivial. Now on one hand, G acts primitively on ∆2 and K1 is a nontrivial

+ normal subgroup of G , we obtain that K1 is transitive on ∆2. On the other hand, for any α ∈ ∆1 we have K1 ≤ Gα, so Γ (α) is ﬁxed set-wise by K1. This implies

71 d Γ (α) = ∆2 for all α ∈ ∆1, so Γ = K∆1,∆2 . By Lemma 2.2.1, |∆1| = p for some prime p.

+ Now suppose that G acts faithfully on ∆1. By Theorem 2.3 of [20], we know that (G+)∆1 ∼= (G+)∆2 is of type HA, AS, PA, or TW. (Notice that in that paper, “bi-quasiprimitive” means “faithful” and “bi-quasiprimitive”.)

+ ∆1 + If (G ) is of type TW, then for any α ∈ ∆1, Gα has no solvable nontrivial

n + normal subgroups by Theorem 4.7B of [5]. Howerver, Zp ¡Gα by Theorem 2.1.1 (iii), for some prime p and integer n. This is a contradiction and hence (G+)∆1 cannot be of type TW.

If (G+)∆1 is of type HA, then by Theorem 1.3(c) of [20], there exists γ ∈ Aut(Γ ), such that hG, γi contains a normal regular subgroup R, then Γ is a Cayley graph by

Lemma 16.3 of [1], and elements of Gα act on Γ (α) as group automorphisms. By the same argument as in the proof of the HA case in Lemma 4.2.3, we have that Γ cannot be (G, 3)-arc transitive, which is a contradiction.

We claim that G+ cannot be of type PA either. Suppose on the contrary that

G+ is of type PA. We have soc(G+) ∼= T m, for some nonabelian simple group T

+ m and integer m ≥ 2. Let B = soc(G ) and consider the point stabilizer Bα = Tλ , where α = (λ, λ, ..., λ)(m λ’s here). We know that Bα 6= 1. We claim that 1 6=

Γ(α) Γ(α) Γ(α) Γ(α) (Bα) ¡ (Gα) . To see this, ﬁrst notice that (Bα) ¡ (Gα) because B ¡ G.

Γ(α) g g Suppose that (Bα) = 1. Then for any g ∈ G and β = α , we have Bα = g g Γ(β) g Γ(αg) ∼ Γ(α) (B ∩ Gα) = B ∩ Gαg = B ∩ Gβ = Bβ. Therefore (Bβ) = (Bα) = Bα = 1. This implies that any element in B which ﬁxes a point will ﬁx the neighborhood of that point. Hence by the connectivity of the Γ we get Bα = 1, which is a contradiction

72 Γ(α) Γ(α) to Bα 6= 1. Since (Bα) is a nontrivial normal subgroup of (Gα) and by

Γ(α) n ∼ Γ(α) Γ(α) Lemma 2.2.1 the group (Gα) is of type HA, Zp = soc((Gα) ) ¡ Bα . Thus n ∼ n the order of Bα is divisible by p . By Lemma 2.2.14, the Sylow p-subgroup P = Zp of Gα is unique in Gα, thus P ≤ Bα. We claim that all non-trivial elements of P are

Γ(α) conjugate in Gα. To see this, notice that by Theorem 2.1.1 (iii), Gα has a sharply

n ∼ Γ(α) 2-transitive subgroup K with Zp = P ¡ K. Therefore all non-trivial elements in

Γ(α) P are conjugate in K. Since P is the unique Sylow p-subgroup of Gα, all non- trivial elements in P are conjugate in Gα. Now let Pi be the Sylow p-subgroup of the

∼ m i-th factor of Bα = Tλ , for i = 1, 2, ..., m. We have P = P1 × P2 × ... × Pm. Take an element x 6= 1 from the ﬁrst factor P1 and an element y 6= 1 from the second factor

P2. We claim that x ∈ P and xy ∈ P are not conjugate in Gα. To see this, notice that G+ is a primitive permutation group of type PA, so G+ ≤ M o N, where M is a primitive group with socle T and N is a transitive group on {1, 2, ..., m}. Thus x is

+ + not conjugate to xy in (G )α = Gα, which is a contradiction. As a result, G cannot be of type PA.

So far, we have shown that G+ is of type AS. Let T ¡ G+ ≤ Aut(T ), where T is a non-abelian simple group. Notice that T is a characteristic subgroup of G+ and

G+ ¡ G, we have T ¡ G.

+ If soc(G) 6= T , then CG(T ) 6= 1. By |G : G | = 2 and CG+ (T ) = 1, the only ∼ + possibility is that C = CG(T ) = Z2. Since C ¡ G, G = G × C. Let Σ = ΓC , the quotient of Γ induced by C. Then Γ is the standard double cover of Σ. The graph

Σ is connected, since Γ is connected. Furthermore, G+ ∼= G/C ≤ Aut(Σ), G+ is primitive on V (Σ). By [19] Theorem 4.1, Σ is (G+, 3)-arc transitive (actually, this

73 fact can be shown directly in our content). Furthermore, Σ is (G+, 3)-arc regular by the order of G+. So Γ is the standard double cover of Σ, a primitive (G+, 3)-arc regular graph. By Theorem 1.2.1, there are two possibilities for Σ and Γ is one of the two graphs in Theorem 1.2.3(ii).

Finally, suppose that soc(G) = T and so T ≤ G+ ≤ G ≤ Aut(T ). By Lemma 3.1.1, we have only ﬁve possibilities for G+.

Cases (i) and (ii) of Lemma 3.1.1 are not possible here, because in these two cases

G+ = Aut(T ) cannot have an extension G+ G ≤ Aut(T ). So G+ can only be one of the cases (iii)-(v) of Lemma 3.1.1.

+ By |G : G | = 2, we get the following three possibilities for G and Gα:

∼ ∼ (1) G = PGL(2, 11), Gα = D12, or

∼ ∼ (2) G = PGL(2, 13), Gα = D12, or

∼ ∼ (3) G = PΓL(2, 27), Gα = Z3 o A4

By Lemma 4.3.2, Γ has to be one of the three graphs in Theorem 1.2.3(iii).

The combination of Lemma 4.3.2 and Lemma 4.3.3 proves Theorem 1.2.3.

To prove Corollary 1.2.4, we need the following lemma:

Lemma 4.3.4. Let Γ be the graph of valency 4 as in (iii) of Theorem 1.2.3, G =

PΓL(2, 27) ≤ Aut(Γ ), and A = Aut(Γ ). Then |A : G| is at most 2.

+ Proof. Let ∆i, i = 1, 2 be the bi-parts of Γ , let G = G∆1 = G∆2 be the set-wise

+ stabilizer of ∆1 in G, A∆1 = A∆2 = A be the set-wise stabilizer of ∆1 in A. Note

+ that |∆1| = |G : Z3 × A4| = 819.

74 Consider PSL(2, 27) ∼= soc(G+) ¡ G+ ≤ A+. Since G+ is a primitive permuta-

+ + tion group on ∆1, we see that A is also primitive on ∆1. Also A acts faithfully on ∆1, otherwise Γ would be a complete bipartite graph by the proof of Lemma 4.3.3. From the list of primitive groups of [5], we see that the primitive groups of

3 3 degree |∆1| = 819 are almost simple with socle PSL(2, 27), D4(2 ), or A819. In our

+ 3 3 3 3 3 3 3 3 case, soc(A ) cannot be D4(2 ), since neither D4(2 ) or Aut( D4(2 )) = D4(2 ).Z3

+ + contains G = PΣL(2, 27) as a subgroup by the Atlas [4]. soc(A ) cannot be A819

+ + either, because Γ is of valency four and |∆1| = 819. Thus soc(A ) = soc(G ) and

+ ∼ A ≤ Aut(PSL(2, 27)) = PΓL(2, 27) = G, implying that |A : G| is at most 2.

Now let us prove Corollary 1.2.4:

Proof of Corollary 1.2.4:

Let A = Aut(Γ ) and ∆ = ∆1. We want to determine that in which cases of Theorem 1.2.3 do we have G = A.

In case (i), Aut(Kn,n) = Sn o Z2, so G = A only when n = 3. For the two graphs in case (ii), consider the standard double cover of the Petersen graph ﬁrst. By the same argument as in the proof of Corollary 1.2.2, we can see that either G = A or |A : G| = 4. Suppose that |A : G| = 4 and consider the set-wise

+ + + + ∼ + stabilizer A∆ = A . Now |A : G | = 4, where G = PGL(2, 5). It is clear that A acts primitively on ∆ and this action is faithful by the proof of the Theorem 4.3.3.

The core of G+ in A+ is PSL(2, 5) and A+ ≤ Aut(PSL(2, 5)) ∼= PGL(2, 5), by the same argument as in the proof of Corollary 1.2.2. This is a contradiction to |A+ : G+| = 4,

75 and therefore G = A. Similarly, we can show that G = A for the standard double cover of the Coxeter graph.

For the ﬁrst two graphs of case(iii), we can prove that G = A using the same argument as above. For the last graph Γ with valency 4, notice ﬁrst that if A 6= G then |A : G| = 2 by Lemma 4.3.4. In this case A = G × hσi, where σ ∈ A is an

∼ + involution with CG(K) = hσi, K = PSL(2, 27). We claim that σ∈ / A . To see

+ + + this, suppose that σ ∈ A . Then, as G ≤ A is primitive on ∆i for i = 1, 2 and σ

+ commutes with G , σ acts trivially on ∆i. Thus σ = 1, clearly a contadiction. So the only possibility is that σ exchanges ∆1 and ∆2. This implies that Γ is a standard double cover of a graph Σ with V (Σ) = ∆1. It is easy to see that Σ is primitive

+ (G , 3)-arc regular, but this is not possible by Theorem 1.2.2. Therefore G = A.

76 BIBLIOGRAPHY

[1] N. Biggs, Algebraic Graph Theory, Cambridge Uni. Press, New York (2th Ed. 1992).

[2] N. Blackburn and B. Huppert, Finite Groups III, Springer, Berlin, 1982.

[3] P. Cameron, Permutation Groups, London Mathematical Society Student Texts, 45. Cambridge University Press, Cambridge, 1999. x+220 pp.

[4] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of Finite Groups, Oxford Univ. Press, London/New York, 1985.

[5] J. D. Dixon and B.Mortimer, Permutation Groups, Graduate Texts in Mathe- matics, Springer-Verlag, New York, 1996.

[6] X. G. Fang, C. H. Li and J. Wang, Finite vertex primitive 2-arc regular graphs, J. Algebraic Combin. 25 (2007), no. 2, 125–140.

[7] X. G. Fang, C. E. Praeger, Finite two-arc transitive graphs admitting a Suzuki simple group, Comm. Algebra 27 (1999), no. 8, 3727–3754.

[8] M. Giudici, Maximal subgroups of almost simple groups with socle PSL(2, q). arXiv:math/0703685v1.

[9] D. Gorenstein, Finite Groups, Chelsea, New York. 1980.

[10] D. Gorenstein, The Classiﬁcation of Finite Simple Groups, Volume 1: Groups of Noncharacteristic 2 Type, Plenum Press, New York, 1983.

[11] D. Gorenstein, R. Lyons and R. Solomon, The classiﬁcation of the ﬁnite simple groups. Number 3. Part I. Chapter A, Almost Simple K-Groups. Mathematical Surveys and Monographs, 40.6. American Mathematical Society, Providence, RI, 2005.

77 [12] D. Gorenstein, R. Lyons and R. Solomon, The classiﬁcation of the ﬁnite simple groups. Number 6. Part IV. The special odd case. Mathematical Surveys and Monographs, 40.6. American Mathematical Society, Providence, RI, 2005.

[13] M. Hall, The Theory of Groups, AMS Chelsea Publishing, Rhode Island, 1999.

[14] C. H. Li, The ﬁnite vertex-primitive and vertex-biprimitive s-transitive graphs for s ≥ 4, Trans. Amer. Math. Soc. 353 (2001), 3511-3529.

[15] C. H. Li and A.´ Seress, Symmetrical path-cycle covers of a graph and polygonal graphs, J. Combin. Theory Ser. A 114 (2007), no. 1, 35–51.

[16] C. H. Li and J. Sir´aˇn,Mbiusˇ regular maps, J. Combin. Theory Ser. B 97 (2007), no. 1, 57–73.

[17] M. W. Liebeck, The aﬃne permutation groups of rank three, Proc. London Math. Soc. (3) 54 (1987), no. 3, 477–516.

[18] D. Maruˇsiˇc,Recent developments in half-transitive graphs, Discrete Math. 182 (1998), pp. 219231.

[19] C. E. Praeger, An O’Nan-Scott theorem for ﬁnite quasiprimitive permutation groups and an application to 2-arc transitive graphs, J. London Math. Soc. (2) 47 (1993), no. 2, 227–239.

[20] C. E. Praeger, On a reduction theorem for ﬁnite, bipartite 2-arc-transitive graphs, Australas. J. Combin. 7 (1993), 21–36.

[21] C. E. Praeger, Primitive permutation groups with a doubly transitive subcon- stituent, J. Austral. Math. Soc.Ser.A 45 (1988),66-77.

[22] R. Weiss, The nonexistence of 8-transitive graphs, Combinatorica 1 (1981), 309- 311.

[23] H. J. Zassenhaus, The Theory of Groups, Chelsea, New York, 1958.