THE COVERING NUMBERS of SOME FINITE SIMPLE GROUPS by Michael Epstein a Dissertation Submitted to the Faculty of the Charles E. S

THE COVERING NUMBERS OF SOME FINITE SIMPLE GROUPS by Michael Epstein

A Dissertation Submitted to the Faculty of The Charles E. Schmidt College of Science in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy

ii THE COVERING NUMBERS OF SOME FINITE SIMPLE GROUPS by Michael Epstein

This dissertation was prepared under the direction of the candidate's dissertation advisor, Dr. Spyros S. Magliveras, Department of Mathematical Sciences, and has been approved by the members of his supervisory committee. It was submitted to the faculty of the Charles E. Schmidt College of Science and was accepted in partial fulfillment of the requirements for the degree of Doctor of Philosophy.

SUPERVISORY COMMITTEE:

Dissertation Advisor

Hari Kalva, Ph.D. D&__~ c2A Lee Kling!~ Rainer Steinwandt, Ph.D. Chair, Department of Mathematical Sci ences Fred Richman, Ph.D.

ence

Khaled Sobhan, Ph.D. Interim Dean, Graduate College iii ACKNOWLEDGEMENTS

I am deeply indebted to my advisor Dr. Spyros Magliveras for his guidance, assistance, and endless patience. This dissertation certainly would not exist without his help. I am very fortunate to be able to count myself among his students. I would also like to thank the members of my supervisory committee: Dr. Hari Kalva, Dr. Lee Klingler, and Dr. Fred Richman. In addition I would like to recognize a number of other faculty members from the Department of Mathematical Sciences at FAU that have each contributed to my education in a meaningful way: Dr. Timothy Ford, Dr. William Kalies, Dr. Stephen Locke, Dr. Erik Lundberg, Dr. Yoram Sagher, Dr. Markus Schmidmeier, Dr. Tomas Schonbeck, Dr. Rainer Steinwandt, and Dr. Yuan Wang. I wish to thank my parents, David and Susan Epstein for their encouragement and for always believing in me. Finally, I want to thank my friends and classmates, particularly Yarema Boryshchak, Angela Robinson, Mayra Quiroga, Jessica Khera, Ashley Valentijn, Shaun Miller, Maxime Murray, and most of all Alexandra Milbrand.

iv ABSTRACT

Author: Michael Epstein Title: The Covering Numbers of some Finite Simple Groups Institution: Florida Atlantic University Dissertation Advisor: Dr. Spyros S. Magliveras Degree: Doctor of Philosophy Year: 2019

A finite cover C of a group G is a finite collection of proper subgroups of G such that G is equal to the union of all of the members of C. Such a cover is called minimal if it has the smallest cardinality among all finite covers of G. The covering number of G, denoted by σ(G), is the number of subgroups in a minimal cover of G. Here we determine the covering numbers of the projective special unitary groups

U3(q) for q ≤ 5, and give upper and lower bounds for the covering number of U3(q) when q > 5. We also determine the covering number of the McLaughlin sporadic simple group, and verify previously known results on the covering numbers of the Higman-Sims and Held groups.

v To Alex, my love. THE COVERING NUMBERS OF SOME FINITE SIMPLE GROUPS

List of Tables ...... ix

1 Introduction ...... 1

2 Preliminaries ...... 4 2.1 Basic Group Theory...... 4 2.2 Covering Numbers of Finite Groups...... 6 2.3 Fields...... 10 2.4 Trace and Semilinear Maps...... 11 2.5 Conjugate Symmetric Sesquilinear Forms...... 13 2.5.1 Deﬁnition and General Properties...... 13 2.5.2 Sesquilinear Forms over Finite Fields...... 15 2.6 A Bit of Projective Geometry...... 19 2.6.1 Fundamentals of Projective Geometry...... 19 2.6.2 Elations and Transvections...... 22 2.6.3 Unitary Polarities...... 25 2.7 Unitary Groups...... 27 2.7.1 Deﬁnition and General Properties...... 27

N 2.7.2 Embedding U3(q) into U3(q ), N odd...... 29

3 Covering Numbers of Small Projective Special Unitary Groups . 31

3.1 U3(2)...... 31

3.2 U3(3)...... 31

3.3 U3(4)...... 34

vii 3.4 U3(5)...... 38

4 The Covering Number of U3(q) with q > 5 ...... 42 4.1 Maximal Subgroups...... 42 4.2 A Rough Classiﬁcation of the Elements of G ...... 44 4.2.1 Type 1: Elements Fixing an Absolute Point...... 44 4.2.2 Sylow p-Subgroups...... 46 4.2.3 Type 2: Elements Fixing a Nonabsolute Point but no Absolute Points...... 49 4.2.4 Type 3: Elements Fixing no Points of PG(V )...... 53 4.3 Bounds on the Covering Number...... 60

5 McLaughlin Group ...... 66

6 Veriﬁcation of the covering numbers of the Higman-Sims and Held Groups ...... 72 6.1 Higman-Sims Group...... 72 6.2 Held Group...... 79

7 Conclusion and Open Problems ...... 91

Bibliography ...... 93

viii LIST OF TABLES

3.1 Conjugacy classes of elements of U3(3)...... 32

3.2 Conjugacy classes of maximal subgroups of U3(3)...... 32

3.3 Incidence matrix for U3(3)...... 33

3.4 Conjugacy classes of elements of U3(4)...... 35

3.5 Conjugacy classes of maximal subgroups of U3(4)...... 35

3.6 Incidence matrix for U3(4)...... 35

3.7 Conjugacy classes of elements of U3(5)...... 39

3.8 Conjugacy classes of maximal subgroups of U3(5)...... 39

3.9 Incidence matrix for U3(5)...... 39

5.1 Conjugacy classes of elements of the McLaughlin group...... 67 5.2 Conjugacy classes of maximal subgroups of the McLaughlin group.. 68 5.3 Incidence matrix for the McLaughlin Group...... 68

6.1 Conjugacy classes of elements of the Higman-Sims group...... 73 6.2 Conjugacy classes of maximal subgroups of the Higman-Sims group. 74 6.3 Incidence matrix for the Higman-Sims Group...... 74 6.4 Conjugacy Classes of Elements of the Held Group...... 80 6.5 Conjugacy Classes of Maximal Subgroups of the Held Group..... 81 6.6 Incidence matrix for the Held Group...... 81

ix CHAPTER 1 INTRODUCTION

A cover of a group G is defined to be a collection C of proper subgroups of G such that G = S C. Questions about such covers of groups go back at least as far as the 1950’s. It follows from a result of B. H. Neumann (specifically statement (4.4) from [32]) that a group has a finite cover if and only if it has a finite noncyclic homomorphic image. It is natural then to consider the number of subgroups used in a cover of such a group.

It is a standard exercise to prove that no group is the union of two proper subgroups. In 1959 Seymour Haber and Azriel Rosenfeld investigated groups which are the union of three proper subgroups in [18]. They proved that a group is a union of three proper subgroups if and only if it has the Klein 4-group as a homomorphic image. This fact was later reproven in [9] in 1970.

This might have been the end of this line of inquiry had J. H. E. Cohn not taken it back up in 1994. In [11] Cohn defines the covering number, denoted σ(G), of a group G which admits a finite cover to be the minimum of the set of cardinalities of the finite covers of G. Cohn then proves that a finite noncyclic supersolvable group G has covering number p + 1 where p is the least prime divisor of G such that G has more than one subgroup of index p. He then determines the covering numbers of A5 and S5, and characterizes the groups with covering number equal to four, five, or six. Cohn also makes two conjectures: first, that a finite noncyclic solvable group G has covering number of the form pa + 1, where p is prime and pa is the order of a chief factor of G, and second, that no group, solvable or otherwise, has covering 1 number equal to seven. Both conjectures were proven in [33] by M. J. Tomkinson, who suggested that covering numbers of simple groups should be studied.

Since then there have been a number of results regarding the covering numbers of simple and almost simple groups:

In [29], Attila Mar´otiproves several theorems on the covering numbers of alternating and symmetric groups. In particular, for the symmetric groups it is shown

n−1 that σ(Sn) = 2 if n is odd and n 6= 9 (it was later shown in [25] that this also holds

n−2 for n = 9), and σ(Sn) ≤ 2 if n is even. Regarding the alternating groups, Mar´oti

n−2 proves that for n 6= 7 or 9, σ(An) ≥ 2 with equality if and only if n ≡ 2 (mod 4) and gives upper bounds in the cases when equality does not hold. Further results on covering numbers of small alternating and symmetric groups can be found in [24, 25, 16].

R. A. Bryce, V. Fedri, and L. Serena investigate the covering numbers of the linear

groups GL2(q), SL2(q), P GL2(q), and L2(q) in [10]. Their main result is that if G is any of these groups, then, with a few small exceptions which are handled separately,

1 1 σ(G) = 2 q(q + 1) if q is even and σ(G) = 2 q(q + 1) + 1 if q is odd. J. R. Britnell, A. Evseev, R. M. Guralnick, P. E. Holmes, and A. Mar´oticonsidered the groups

GLn(q), SLn(q), P GLn(q), and PSLn(q) for larger n in [2,3], and in particular give the covering numbers of these groups when n ≥ 12.

P. E. Holmes determined the covering numbers of several of the sporadic simple groups in [21]. In particular, she proved that σ(M11) = 23, σ(M22) = 771, σ(M23) = 41079, σ(Ly) = 112845655268156, σ(O0N) = 36450855, 24541 ≤ σ(McL) ≤ 24553, and 5165 ≤ σ(J1) ≤ 5415. In [25], Luise-Charlotte Kappe, Daniela Nikolova-Popova,

2 and Erik Swartz showed that σ(M12) = 208 and 5281 ≤ σ(J1) ≤ 5414. In [22] Holmes and A. Marótigave upper and lower bounds for all of the remaining sporadic simple groups. They also determined the exact covering numbers for some of them; specifically they proved that σ(M24) = 3336, σ(F i22) = 221521, σ(He) = 464373, σ(Ru) = 12992175, σ(HS) = 1376, and σ(HN) = 229758831. The result on the covering number of M24 was also proved independently in [15] by the author and Spyros Magliveras (who were not aware of the aforementioned paper by Holmes and Maróti at the time).

3 CHAPTER 2 PRELIMINARIES

The reader is assumed to be familiar with abstract and linear algebra at the level of a solid graduate course. We refer the reader to [13] as a reference. In this chapter we recall a number of results that we will use in the later chapters.

2.1 BASIC GROUP THEORY

We will use standard group theoretic terminology and notation as in [13]. Vertical bars, e.g. |G|, |x|, etc., denote the cardinality of a set, the order of a group, or the order of an element of a group as appropriate. If G is a group and H is a subgroup of G, we write H ≤ G, and denote the index of H in G by [G : H]. We shall write

H E G if H is a normal subgroup of G. For g ∈ G, we denote by hgi the cyclic subgroup generated by g. If a group G acts on a set X, then for x ∈ X and g ∈ G, we denote the image of x under the action of g by g · x and the stabilizer of x in G is denoted Gx. As we make heavy use of the Atlas of ﬁnite groups [12] in some sections of this dissertation, it is convenient to adopt some of the notation and conventions used therein. In particular, we adopt the notation of the Atlas for conjugacy classes and for indi- cating algebraic conjugacy. The conjugacy classes of elements of order n in a group G will be denoted by n followed by a capital letter of the English alphabet, starting with ‘A’(e.g. 2A, 2B, 3A, etc.). Multiple letters following a number n indicate a union of conjugacy classes; for example ‘2AB’would denote the union of conjugacy classes 2A and 2B. If the elements from two conjugacy classes nX and nY generate

4 a single conjugacy class of cyclic subgroups, we say that nX and nY are algebraically conjugate and this relationship is indicated through the use of the algebraic conjugacy operators ∗, ∗∗, and ∗k as follows: we write nY = nX∗k to indicate that nY contains the kth powers of the elements of nX (where k is relatively prime to n), and nX∗∗ is short for nX∗(−1). In the case that the elements of nX are conjugate to their inverses and nY is the unique class other than nX which is algebraically conjugate to nX, we write nY = nX∗. We will also follow the Atlas’ conventions for group structures and extensions, except that we will denote a cyclic group of order n by Zn and an

k k elementary abelian group of order n by Zn. In particular, a general extension of N by H is denoted N.H, whereas split and non-split extensions are denoted N : H and N ·H respectively. Expressions involving three or more groups are left associated; for example A : B.C should be interpreted as (A : B).C.

The following results are well known and will be used throughout this dissertation.

Theorem 2.1.1 (Cauchy’s Theorem). If G is a ﬁnite group and p is a prime divisor of |G|, then G has an element of order p.

Theorem 2.1.2 (Lagrange’s Theorem). Let G be a ﬁnite group, and H be a subgroup of G. Then |G| = |H|[G : H].

Corollary 2.1.3. Let G be a ﬁnite group. Then for every x ∈ G, |x| divides |G|.

Theorem 2.1.4 (The Orbit-Stabilizer Theorem). Let G be a group which acts on a set X. For x ∈ X let Ox be the orbit of x under G. Then the map

{gGx : g ∈ G} → Ox

gGx 7→ g · x

is a well-deﬁned bijection. Consequently, |Ox| = [G : Gx].

5 We will make extensive use of Sylow subgroups in later chapters. We recall the deﬁnition and Sylow’s Theorem here.

Deﬁnition 2.1.5. Let G be a ﬁnite group of order pan where p is a prime not dividing n.A Sylow p-subgroup of G is a subgroup of order pa.

Theorem 2.1.6 (Sylow’s Theorem). Let G be a ﬁnite group of order pan, where p is a prime not dividing n. Then,

1. G has at least one Sylow p-subgroup.

2. Let P be a Sylow p-subgroup of G. If H is a subgroup of G, and |H| is a power of p, then H is contained in a conjugate of P . Consequently, any pair of Sylow p-subgroups of G are conjugate.

3. The number of Sylow p-subgroups of G divides n and is congruent to 1 modulo p.

2.2 COVERING NUMBERS OF FINITE GROUPS

The following basic results on covering numbers of groups are from [11]. We adopt the convention that if a group G does not admit a ﬁnite cover, then σ(G) = ∞.

Lemma 2.2.1. If G is a group and N E G, then σ(G) ≤ σ(G/N).

Lemma 2.2.2. Let H and K be groups. Then σ(H × K) ≤ min{σ(H), σ(K)}. Moreover, if H and K are ﬁnite and their orders are relatively prime, then σ(H×K) = min{σ(H), σ(K)}.

Theorem 2.2.3. If G is a ﬁnite, noncyclic p-group, then σ(G) = p + 1.

Theorem 2.2.4. If G is a ﬁnite, noncyclic, nilpotent group, then σ(G) = p + 1, where p is the least prime such that the Sylow p-subgroup of G is noncyclic.

6 Theorem 2.2.5. Let G be a ﬁnite, noncyclic, supersolvable group. Then σ(G) = p+1, where p is the least prime such that G has more than one subgroup of index p.

M. J. Tomkinson proved the following theorem on the covering numbers of ﬁnite solvable groups in [33].

Theorem 2.2.6. If G is a ﬁnite solvable group, then σ(G) = pa + 1, where pa is the order of the smallest chief factor of G with more than one complement in G.

We now give a deﬁnition and three simple observations that have been used in previous papers to determine the covering numbers of ﬁnite groups (see for example [16]). As the proofs are straightforward, we omit them.

Deﬁnition 2.2.7. Let x be an element of a group G. If hxi is maximal among the cyclic subgroups of G, i.e. whenever y ∈ G and hxi ≤ hyi we have hxi = hyi, then we say that x is a principal element of G and that hxi is a principal subgroup of G.

Lemma 2.2.8. A collection C of proper subgroups of a group G is a cover of G if and only if every principal element of G is contained in some member of C.

Lemma 2.2.9. If G is a ﬁnite noncyclic group, then there exists a minimal cover of G which contains only maximal subgroups of G.

Our method for determining the covering number of a given group G proceeds in two stages:

1. Find a small cover C of G.

2. Prove that no smaller cover exists and deduce that σ(G) = |C|.

The observations which form the content of Lemmas 2.2.8 and 2.2.9 greatly sim- plify this process: Lemma 2.2.8 is not only useful when proving that a collection of subgroups forms a cover, but (as will be seen below) underlies our strategy for proving

7 minimality. By Lemma 2.2.9 we need only ever consider covers consisting of maximal subgroups. In light of this it is critical to understand the incidence relation between the principal elements and the maximal subgroups of G.

Lemma 2.2.10. Let K be a conjugacy class of elements of a group G, and let M be a conjugacy class of subgroups of G. Then the numbers aM,K and bM,K deﬁned by

aM,K = |{I ∈ M : x ∈ I}|

bM,K = |{y ∈ K : y ∈ H}| where x is any ﬁxed element of K and H is any ﬁxed member of M, are independent of the choice of representatives x and H, and satisfy aM,K |K| = bM,K |M|.

We can construct two incidence matrices A and B with rows indexed by the conjugacy classes of principal elements of G and columns indexed by the conjugacy classes of maximal subgroups whose entries in row K and column M are the numbers a and b from Lemma 2.2.10 respectively. Throughout, we give only the matrix A for each group considered, as the entries of B can easily be computed from those of A, which are typically much smaller. The author has attempted, wherever possible, to avoid relying on the use of a computer to justify the results contained in this dissertation, and in this regard has found the following lemma useful for determining the entries of the incidence matrices in the following chapters.

Lemma 2.2.11. Let H be a subgroup of G, x ∈ G, K the conjugacy class of x, and θ be the permutation character of the action of G on the right cosets of H. Then,

|K| |K ∩ H| = θ(x) . [G : H]

Moreover, if H is self-normalizing, then the number of conjugates of H containing x is θ(x).

8 Proof. Recall that the permutation character of G on the right cosets of H is the

induced character obtained from the principal character 1H of H. Applying the formula for induced characters yields:

1 X θ(x) = 1 (g−1xg) |H| H g∈G |K ∩ H| = · |C (x)| |H| G |K ∩ H| |G| = · |H| |K| [G : H] = |K ∩ H| · |K|

from which the desired equation easily follows. The second part follows from the ﬁrst by Lemma 2.2.10 in the case that H is self-normalizing. Alternatively one can show directly that if H is self-normalizing then the actions of G on the right cosets of H by right multiplication and on the set of conjugates of H by conjugation are equivalent and have the same character.

We note that maximal subgroups of nonabelian simple groups are always self- normalizing, so for a (noncyclic) ﬁnite simple group G the entries of A are exactly the values of its primitive permutation characters on the principal elements of the group.

Once we have determined the incidence matrix A (or at least the relevant parts thereof), we can begin to look for small covers consisting of maximal subgroups. It is easy to see that a union of (whole) conjugacy classes of maximal subgroups of G forms a cover if and only if the sum of the corresponding columns of the matrix A has strictly positive entries. It will often, but not always, be the case that the minimal cover we construct will be of this form. In any case, once we have obtained a cover which we believe to be minimal we proceed to the second step in our strategy for determining covering numbers: proving minimality. 9 To do this we construct a system of linear inequalities in the variables xj =

|C ∩Mj|, where Mj denotes the jth conjugacy class of maximal subgroups of G, that must be satisﬁed for any cover C of G consisting of maximal subgroups. Typically, we begin with inequalities of the form

bi 1x1 + bi 2x2 + · · · ≥ |Ki|

where the bi j are the entries of the ith row of the incidence matrix B described above. However, one can often strengthen these inequalities by considering the actions of the group G on certain combinatorial objects such as the points of a ﬁnite geometry or the vertices of a strongly regular graph. Ultimately, this yields an integer linear P programming problem where the objective is to minimize the quantity |C| = j xj subject to the constraints given by these linear inequalities. The solution to this integer programming problem is always a lower bound for the covering number of G, but in many cases turns out to be equal to the covering number. In particular, if the lower bound obtained by considering these linear inequalities is equal to the cardinality of a cover C, then C is necessarily minimal.

2.3 FIELDS

We will make some use of the theory of fields, particularly finite fields, when dealing with unitary groups. We again refer the reader to [13], whose notation and terminology we will follow. If K is a field, K∗ denotes the group of nonzero elements of

K under multiplication. If q is a prime power, Fq denotes the field of order q. An extension F of a field K is any field which contains K as a subfield. We will write F/K to indicate that F is an extension of K.

10 2.4 TRACE AND SEMILINEAR MAPS

Definition 2.4.1. Let F/K be a Galois extension of fields. The trace, T r : F → K, is defined by X T r(α) = σ(α) . σ∈Gal(F/K) Note that this is not the most general definition of the trace (which can be defined for extensions which are not Galois extensions) but will be sufficient for our purposes here. Some basic facts about the trace:

• The trace is K-linear.

• The trace is onto, and has nonzero kernel (unless F = K).

• For α ∈ K, T r(α) = [F : K]α, where [F : K] is the degree of the ﬁeld extension.

Definition 2.4.2. Let F be a field, V,W be F -vector spaces, and σ be an automorphism of F . A function T : V → W is called σ-semilinear if for all x, y ∈ V and all α, β ∈ F , T (αx + βy) = σ(α)T (x) + σ(β)T (y). Such a map is called semilinear if it is σ-semilinear for some field automorphism σ.

Some facts about semilinear maps:

• ker(T ) and im(T ) are subspaces of V and W respectively.

• If T is σ-semilinear and τ-semilinear and T 6= 0, then σ = τ.

• Compositions of semilinear maps are semilinear: If T is σ-semilinear and R is τ-semilinear, then T ◦ R is σ ◦ τ-semilinear.

• Invertible semilinear maps send linearly independent sets to linearly independent sets.

11 Theorem 2.4.3 (Rank-Nullity Theorem for Semilinear Maps). Let V be a ﬁnite dimensional vector space and T : V → W be semilinear. Then dim(ker(T )) + dim(im(T )) = dim(V ).

Proof. We may as well assume that T is not linear, else the usual Rank-Nullity Theorem applies. So there is a unique (nonidentity) automorphism σ for which T is σ-semilinear. Let U be a complimentary subspace for ker(T ), and set k = dim(ker(T )), r = dim(im(T )), and n = dim(V ).

We ﬁrst show that r ≥ n − k: Let {u1, . . . , un−k} be a basis for U. We claim that P {T (u1),...,T (un−k)} is a linearly independent subset of im(T ). If αiT (ui) = 0,

P −1 P −1 P −1 then T ( σ (αi)ui) = 0 so that σ (αi)ui ∈ ker(T ). But also σ (αi)ui ∈ U,

−1 and U ∩ ker(T ) = 0. The linear independence of the ui’s forces σ (αi) = 0 for all

i and therefore αi = 0 for all i. Consequently, {T (u1),...,T (un−k)} is a linearly independent subset of im(T ), hence r ≥ n − k.

We now prove that r ≤ n − k. Let {w1, . . . , wr} be a basis for im(T ). We may pick a

preimage vi for each wi from U. An argument similar to that given above shows that

{v1, . . . , vr} is a linearly independent subset of U and hence r ≤ n − k.

Consequently, r = n − k and the theorem is proved.

Lemma 2.4.4. Let T : V → V be σ-semilinear. If there is a nonzero subspace U of V and nonzero scalar λ such that T (u) = λu for all u ∈ U, then σ is the identity automorphism and T is linear.

Proof. Let u ∈ U \{0}. For any α ∈ F ∗, σ(α)λu = σ(α)T (u) = T (αu) = λαu. Therefore, λ(σ(α) − α) = 0, and since λ is nonzero, it follows that σ(α) = α proving that σ is the identity map and therefore T is linear.

12 2.5 CONJUGATE SYMMETRIC SESQUILINEAR FORMS

2.5.1 Deﬁnition and General Properties

We begin with some deﬁnitions.

Deﬁnition 2.5.1. Let K be a ﬁeld, F a separable quadratic extension of K, and α 7→ α be the involution in the Galois group Gal(F/K) . Let V be a vector space over F .A conjugate symmetric sesquilinear form on V is a function f : V × V → F satisfying:

1. For all α, β ∈ F , and all x, y, z ∈ V , f(αx + βy, z) = αf(x, z) + βf(y, z), and

2. For all x, y ∈ V , f(y, x) = f(x, y).

Note that if f is such a form then for all α, β ∈ F , and all x, y, z ∈ V , f(x, αy + βz) = αf(x, y) + βf(x, z).

Deﬁnition 2.5.2. A conjugate symmetric sesquilinear form f is called degenerate if there exists y ∈ V \{0} such that for all x ∈ V , f(x, y) = 0. Otherwise the form is called nondegenerate.

From now on, we assume that K is a ﬁeld, F is a quadratic extension of K, V is a ﬁnite dimensional vector space with dimension n ≥ 1 over F and that f is a nondegenerate conjugate symmetric sesquilinear form on V . We will now establish some basic properties of f.

Lemma 2.5.3. For all x ∈ V , f(x, x) ∈ K.

Proof. Since f(x, x) = f(x, x), f(x, x) lies in the ﬁxed ﬁeld of the involution α 7→ α, which is K.

13 Lemma 2.5.4. For each y ∈ V , the map Ty : V → F given by Ty(x) = f(x, y) is a

surjective linear functional. Similarly, the map Ry : V → F given by Ry(x) = f(y, x) is a surjective semilinear map.

Note that Ty has rank 1 and dim(ker Ty) = n − 1.

Lemma 2.5.5. There exists x ∈ V with f(x, x) 6= 0.

Proof. Assume by way of contradiction that f(x, x) = 0 for all x ∈ V . Then for any y, z ∈ V ,

0 = f(y + z, y + z) = f(y, y) + f(y, z) + f(z, y) + f(z, z) = f(y, z) + f(z, y)

so that f(y, z) = −f(z, y). Since f is nondegenerate, there are y, z ∈ V such that f(y, z) 6= 0. Now for any α ∈ F ,

αf(y, z) = f(αy, z) = −f(z, αy) = −αf(z, y) = αf(y, z) .

since f(y, z) 6= 0, it follows that α = α for all α ∈ F , which contradicts our hypothesis that α 7→ α has order 2.

For a subset S of V we deﬁne S⊥ = {v ∈ V : f(v, s) = 0 for all s ∈ S}. This is always a subspace of V .

Lemma 2.5.6. Let W be a subspace of V . Then dim(W ) + dim(W ⊥) = dim(V ). Moreover, if W ∩ W ⊥ = 0, then V = W ⊕ W ⊥.

Proof. For a vector space U, let U ∗ denote its dual space. Consider the following maps: ψ : V → V ∗ given by ψ(x) = f(·, x), and ρ : V ∗ → W ∗ where ρ(g) is the restriction of g to W . Clearly ρ is a surjective linear map and ψ is a semilinear map

with respect to the automorphism α 7→ α. Now, since V is ﬁnite dimensional, V ∼= V ∗. By nondegeneracy, ker(ψ) = 0, so by the Rank-Nullity Theorem for Semilinear Maps, ψ must be surjvective. Hence φ = ρ ◦ ψ is a surjective semilinear map from V to W ∗. 14 Moreover, ker(φ) = W ⊥. Applying the Rank-Nullity Theorem for Semilinear Maps

again, and noting that W ∼= W ∗, dim(W ⊥) + dim(W ) = dim(V ). The other claim is then straightforward.

Given a subspace W of V , we may consider the restriction f|W ×W of f to W ×W .

It follows from the previous lemma that if f|W ×W is nondegenerate, so is f|W ⊥×W ⊥ .

Lemma 2.5.7. For a subspace W of V , W ⊥⊥ = W .

Proof. It is easy to see that W ⊆ W ⊥⊥. By the previous lemma, dim(W ⊥) + dim(W ⊥⊥) = dim(V ) = dim(W ) + dim(W ⊥). Thus dim(W ) = dim(W ⊥⊥), and therefore W = W ⊥⊥.

2.5.2 Sesquilinear Forms over Finite Fields

From this point forward we assume that K = Fq, F = Fq2 . The following lemma will be useful.

Lemma 2.5.8. The map φ : F ∗ → K∗ given by α 7→ αα is a homomorphism of groups with image K∗.

Proof. It is clear that φ is a homomorphism of groups. Now, if |K| = q then F ∗ and K∗ are cyclic groups of order q2 − 1 and q − 1 respectively, and α = αq for all α ∈ F . Then ker φ = {α ∈ F ∗ : αq+1 = 1} has order q + 1, and consequently im φ has order q − 1 by the First Isomorphism Theorem. But K∗ is the unique subgroup of F ∗ of order q − 1 and so im φ = K∗.

By analogy with the corresponding deﬁnition in an inner product space, we deﬁne

an orthonormal basis for V to be a basis {e1, e2, . . . , en} for V such that f(ei, ej) = 1

if i = j and f(ei, ej) = 0 otherwise. The following lemma proves that such a basis will always exist (when K and F are ﬁnite).

Lemma 2.5.9. V has an orthonormal basis with respect to f. 15 Proof. By Lemmas 2.5.5 and 2.5.8, there is x ∈ V with f(x, x) = 1. Let W = span(x).

⊥ Then, f|W ×W is nonsingular so f|W ⊥×W ⊥ is also nondegenerate and V = W ⊕ W . Inductively we can assume that W ⊥ has an orthonormal basis. Adding x to it gives an orthonormal basis for V .

Deﬁnition 2.5.10. We deﬁne a vector x ∈ V to be isotropic if f(x, x) = 0.

Over an inﬁnite ﬁeld it is possible that V has no nonzero isotropic vectors, e.g.

V = Cn with the usual complex inner product. Since we are assuming that F and K are ﬁnite ﬁelds, if n ≥ 2, then V always has nonzero isotropic vectors as will be seen below.

Lemma 2.5.11. If n ≥ 2 then V has nonzero isotropic vectors.

Proof. Let x ∈ V \{0}. If f(x, x) = 0 then x is a nonzero isotropic vector. If not, then with U = span (x) by Lemma 2.5.6, we get V = U ⊕ U ⊥, and since dim V ≥ 2, there is y ∈ U ⊥ such that y 6= 0. If f(y, y) = 0 then y is a nonzero isotropic vector. If not, by Lemma 2.5.8 there are α, β ∈ F ∗ such that αα = f(x, x)−1 and ββ = −f(y, y)−1. Then,

f(αx + βy, αx + βy) = ααf(x, x) + αβf(x, y) + βαf(y, x) + ββf(y, y) .

Since f(x, y) = f(y, x) = 0, we have f(αx + βy, αx + βy) = ααf(x, x) + ββf(y, y) = 1 − 1 = 0. Moreover, Since x and y are linearly independent, αx + βy is a nonzero isotropic vector.

Moreover, we can count the number of nonzero isotropic vectors in V , which is the content of the following lemma.

Lemma 2.5.12. The number of nonzero isotropic vectors in V is

δ(n, q) := (qn − (−1)n)(qn−1 − (−1)n−1) .

16 Proof. We proceed by induction on n. If n = 1, then V has no nonzero isotropic vectors and the lemma holds. Now, assume that n > 1 and choose x ∈ V with f(x, x) = 1. Let W = {x}⊥. The nonzero isotropic vectors of V are of two types: those in W , and those of the form αx+w, where α ∈ F ∗ and w ∈ W with f(w, w) 6= 0.

Now, f|W ×W is nondegenerate, so inductively there are δ(n − 1, q) nonzero isotropic vectors in W , and q2n−2 − 1 − δ(n − 1, q) nonisotropic vectors in W . For each of the latter, there are q + 1 choices for α ∈ F ∗ such that αx + w is an isotropic vector. Consequently, the number of nonzero isotropic vectors in V is

(q + 1)(q2n−2 − 1 − δ(n − 1, q)) + δ(n − 1, q), which one can easily show is equal to δ(n, q).

The following notion will prove useful.

Deﬁnition 2.5.13. A hyperbolic pair is a pair (c, d) ∈ V × V such that c and d are isotropic and f(c, d) = 1.

Lemma 2.5.14. Let c ∈ V be a nonzero isotropic vector. Then there are exactly q2n−3 vectors d ∈ V such that (c, d) is a hyperbolic pair.

Proof. Let X = {(x, λ) ∈ V × F : f(x, c) = 1, T r(λ) = f(x, x)}, and D = {d ∈ V : (c, d) is a hyperbolic pair}. Note that |X| = q2n−1. Deﬁne φ : X → D by (x, λ) = x − λc. A quick calculation shows that φ is well-deﬁned, i.e. that for (x, λ) ∈ X, (c, x − λc) is in fact a hyperbolic pair. Since d = φ(d, 0) for any d ∈ D, φ is onto. Moreover, for any d ∈ D the complete preimage of d under φ is {(d + λc, λ): λ ∈ F }. Consequently, φ is a q2 to 1 mapping, and the desired result follows.

We are mainly concerned in this work with the case that V is 3-dimensional, so we will assume for the remainder of this section that the dimension of V as an F -vector space is 3.

17 Lemma 2.5.15. Let c, d ∈ V be linearly independent isotropic vectors. Then f(c, d) 6= 0.

Proof. Suppose by way of contradiction that f(c, d) = 0. Let U = span({c, d}). Then {c}⊥ = {d}⊥ = U. By Lemma 2.5.14 there is e ∈ V such that (c, e) is a hyperbolic pair. Since e∈ / U, f(e, d) 6= 0. Let W = {e}⊥. Then V = W ⊕span(c) = W ⊕span(d). Then there are w ∈ W \{0}, and µ ∈ F ∗ such that c = w + µd. Now, w = c − µd ∈ W ∩ U, so w is orthogonal to each of c, d, and e. But c, d, and e span V , hence w is orthogonal to V , contradicting the nondegeneracy of the form f.

We now show that a hyperbolic pair can be extended to a particular type of basis for V that is convenient to work with when dealing with isotropic vectors.

Lemma 2.5.16. Let (c, d) be a hyperbolic pair. Then there exists x ∈ V such that f(x, c) = f(x, d) = 0, f(x, x) = 1, and {c, x, d} is linearly independent.

Proof. Let e be a nonisotropic vector in {c}⊥. By Lemma 2.5.8, we may rescale e, if necessary, so that f(e, e) = 1. Let λ = f(e, d) and x = e − λc. Then,

f(x, d) = f(e − λc, d) = f(e, d) − λf(c, d) = λ − λ = 0

f(x, c) = f(e − λc, c) = f(e, c) − λf(c, c) = 0 − 0 = 0 and

f(x, x) = f(e − λc, e − λc) = f(e, e) − λf(c, e) − λf(e, c) + λλf(c, c) = f(e, e) = 1.

Now, since c and e are linearly independent, so are c and x. And since d∈ / {c}⊥ = span({c, x}), it follows that {c, x, d} is a linearly independent set.

We will call a basis as in the previous lemma a unitary basis.

Lemma 2.5.17. Let x ∈ V with f(x, x) = 1. Then there exists a hyperbolic pair (c, d) such that {c, x, d} is a unitary basis for V . 18 ⊥ Proof. Let U = {x} . Now f|U×U is nondegenerate and U has dimension 2, so by

Lemma 2.5.11 there is an isotropic vector c ∈ U. Note that U * span{c, x} = {c}⊥, and it follows that there is d0 ∈ U such that f(d0, c) = 1. Choose λ such that T r(λ) = f(d0, d0) and set d = d0 − λc. Then, d ∈ U so that f(d, x) = 0,

f(d, c) = f(d0 − λc, c) = f(d0, c) − λf(c, c) = 1 , and

f(d, d) =f(d0 − λc, d0 − λc)

=f(d0, d0) − λf(c, d0) − λf(d0, c) + λλf(c, c)

=f(d0, d0) − T r(λ) = 0 .

Therefore, (c, d) is a hyperbolic pair with f(c, x) = f(d, x) = 0. Consequently {c, x, d} is a unitary basis for V .

Note that in the proof of this lemma we have shown that the orthogonal complement U = {x}⊥ of a nonisotropic vector x is spanned by a hyperbolic pair (c, d).

2.6 A BIT OF PROJECTIVE GEOMETRY

We will require some familiarity with projective geometry when investigating the covering numbers of unitary groups in chapter4. We include here only a brief introduction and refer the reader to [20] for further information.

2.6.1 Fundamentals of Projective Geometry

Let F be a field, and V an F -vector space of dimension n + 1 ≥ 2. Let ∼ be the equivalence relation on V \{0} defined by x ∼ y if and only if there is α ∈ F ∗ such that x = αy. Denote the equivalence class of a nonzero vector x by [x]. The projective geometry PG(V ) is the set of these equivalence classes, which are called the 19 points of the geometry. The dimension of PG(V ) is n, one less than the dimension of V as a vector space. In the case that F is a finite field of order q, we may write PG(n − 1, q) for PG(V ). If S ⊆ V we will denote by [S] the set {[x]: x ∈ S \{0}}. If W is a subspace of V of dimension k, then we say that [W ] is a subspace of PG(V ) of dimension k − 1. The points of the geometry are the 0-dimensional subspaces. Subspaces of PG(V ) of dimension 1 are called lines, those of dimension 2 are called planes, and those of dimension n − 1 are called hyperplanes.

Deﬁnition 2.6.1. Let V be an (n + 1)-dimensional vector space over the ﬁeld F . A frame, F, in PG(V ) is a set of n + 2 points of PG(V ) with the property that no hyperplane contains n + 1 points of F.

Deﬁnition 2.6.2. Let V and W be vector spaces of dimension n + 1 ≥ 2 over F . If n ≥ 2, a collineation from PG(V ) to PG(W ) is a bijection φ : PG(V ) → PG(W ) such that

1. the image of each subspace of PG(V ) is a subspace of PG(W ), and

2. for subspaces A and B of PG(V ), A ⊆ B if and only if φ(A) ⊆ φ(B).

In the case that n = 1, one deﬁnes a collineation from the projective line PG(V ) to the projective line PG(W ) by embedding both lines into planes and restricting the collineations of the planes which map PG(V ) onto PG(W ).

We note that any semilinear map T : V → W induces a collineation τ : PG(V ) → PG(W ) deﬁned by τ([x]) = [T (x)]. If T is in fact linear, then τ is called a projectivity. The following two theorems are well-known. The ﬁrst of these asserts that all collineations between two projective geometries are induced by semilinear maps.

Theorem 2.6.3 (The Fundamental Theorem of Projective Geometry). Let V and W be two n-dimensional vector spaces over a ﬁeld F , with n ≥ 2.

20 1. If τ : PG(V ) → PG(W ) is a collineation, then τ is induced by some semilinear map T : V → W .

0 2. If F = {P1,...,Pn+2} and F = {Q1,...,Qn+2} are frames of in PG(V ) and PG(W ) respectively, then there is a unique projectivity τ : PG(V ) → PG(W )

such that τ(Pi) = Qi for 1 ≤ i ≤ n + 2.

The last theorem in this subsection is a consequence of a more general result about automorphisms of symmetric designs (see, for example, Theorem 3.1 and Corollary 3.2 of [27] and Theorem 1.5.10 of [34]).

Theorem 2.6.4. A collineation from a ﬁnite projective geometry onto itself ﬁxes equally many points and hyperplanes. Moreover, such a collineation has the same cycle type when represented as a permutation of the hyperplanes of the geometry as when it is represented as a permutation of the points.

We end this subsection with the deﬁnitions of triangles and conics:

Deﬁnition 2.6.5. A triangle in a projective space is a set of three points which are not collinear. These points are called the vertices of the triangle. The sides of the triangle are the three lines determined by the three pairs of vertices of the triangle.

Deﬁnition 2.6.6. Let V be an n-dimensional vector space over a ﬁeld F , n ≥ 2 and let B = {b1, . . . , bn} be a basis for V . Let g ∈ F [x1, . . . , xn] be a homogeneous Pn polynomial, and let V (g) = {[ i=1 αibi] ∈ PG(V ): g(α1, . . . , αn) = 0}. Let I be the ideal of F [x1, . . . , xn] generated by g. The pair (V (g),I) is called a variety in PG(V ). The degree of the variety is the degree of g. The variety is called singular if there is Pn ∂g a nonzero vector a = i=1 αibi such that (α1, . . . , αn) = 0 for all 1 ≤ i ≤ n, and ∂xi is called nonsingular otherwise. A nonsingular variety of degree two is called a conic.

Conics are interesting objects in their own right, but for our present purposes we are more interested in their stabilizers in certain projective special unitary groups than in the conics themselves. 21 2.6.2 Elations and Transvections

Definition 2.6.7. A collineation ε of a projective space is called an elation if there is a hyperplane H fixed pointwise by ε and a point c on H such that ε fixes every line through c setwise. In this case H is called an axis and c is called a center for ε.

It is well known that an elation which is not the identity has a unique center and a unique axis.

Deﬁnition 2.6.8. Let V be a vector space of dimension at least 2. A linear map T : V → V such that rank(T − I) = 1 and (T − I)2 = 0 is called a transvection. The axis of the transvection is ker(T − I). The center of the transvection is im(T − I).

Note that the axis of the trasvection is ﬁxed pointwise by T , and any subspace of V which contains the center is ﬁxed setwise by T . Also note that for a transvection to exist, the dimension of the vector space must be at least 2.

Let T be a transvection. If e1 spans im(T − I) and e1, . . . , en−1 span the kernel, pick en such that (T −I)(en) = e1. Then the matrix for T −I in the basis {e1, . . . , en} has a 1 in position (1, n) and 0’s elsewhere. If we call this matrix U, then in the same basis T has matrix I + U. From this it is clear that det(T ) = 1, and therefore that T ∈ SL(V ). Moreover, if F is a ﬁnite ﬁeld of characteristic p, we can see that such a transvection has order p in SL(V ).

Lemma 2.6.9. Let φ : V → F be a nonzero linear functional and a ∈ ker(φ) \{0}. Then the map T : V → V given by x 7→ x + φ(x)a is a transvection. Conversely, every transvection is of this form.

Proof. Clearly T is linear, and T − I has rank 1. Let x ∈ V . Then (t − I)2(x) = (T − I)(φ(x)a) = φ(φ(x)a)a = φ(x)φ(a)a = 0 since a ∈ ker(φ). Hence, T is a transvection. Conversely, suppose that T is a transvection. Let a span the center of T , and note then that a ∈ ker(T − I). Deﬁne π : im(T − I) → F by π(αa) = α. Then

22 π ◦ (T − I) is a linear function on V , with [π ◦ (T − I)](a) = 0, and for any x ∈ V , (T − I)(x) = (π ◦ (T − I))(x)a, whence T (x) = x + (π ◦ (T − I))(x)a.

Lemma 2.6.10. The set of transvections in GL(V ) is a normal subset of GL(V ).

Proof. Let T (x) = x + φ(x)c be a transvection, S ∈ GL(V ). then S−1TS(x) = S−1(S(x) + φ(S(x))c) = x + S−1(φ(S(x))c) = x + φ(S(x))S−1(c). Now, φ ◦ S is a nonzero linear functional which is zero on S−1(c), hence by Lemma 2.6.9, S−1TS is a transvection.

Lemma 2.6.11. Let V be a vector space of dimension at least 3. The collineation of PG(V ) induced by a transvection is an elation, and conversely every nonidentity elation of PG(V ) is induced by a transvection of V .

Proof. It is straightforward to see that the collineation induced by a transvection is an elation. We will prove the converse. Let dim(V ) = n+1 ≥ 3 and let ε be an elation of PG(V ). Let [A] be the axis and [c] be the center of ε. By the Fundamental Theorem of Projective Geometry, there is a semilinear map τ such that the collineation induced P by τ is ε. Let {a1, . . . , an} be a basis for A and let a = ai. Since ε ﬁxes A pointwise,

∗ there are λ, λi ∈ F such that τ(a) = λa and τ(ai) = λiai for 1 ≤ i ≤ n. Then

X X X X λai = λ ai = τ(a) = τ(ai) = λiai ,

from which it follows by linear independence that λi = λ for each i. Let b ∈ A. Then there is some i for which ai and b are linearly independent. Without loss of generality (relabeling if necessary) we may assume that a1 and b are linearly independent.

∗ Extend {a1, b} to a basis {v1 = a1, v2 = b, . . . , vn} for A. As above, there if µ ∈ F such that τ(vi) = µvi for1 ≤ i ≤ n. Now, λa1 = τ(a1) = τ(v1) = µv1 = µa1, from which it follows that λ = µ. Thus τ(b) = τ(v2) = λv2 = λb, proving that τ(b) = λb for all b ∈ A. Note that by Lemma 2.4.4, τ is linear.

23 Now, let x ∈ V \ A. Since ε ﬁxes lines through [c], τ(x) ∈ span({x, c}). Write τ(x) = αx + βc. Now, dim(A) = n ≥ 2, so there is a ∈ A such that a and c are linearly independent. Consider τ(x+a). First, τ(x+a) = τ(x)+τ(a) = αx+βc+λa. On the other hand, since x + a∈ / A, τ(x + a) ∈ span({x + a, c}), whence we can write τ(x + a) = γ(x + a) + δc. Then, αx + βc + λa = γ(x + a) + δc. Now, since {x, a, c} is a linearly independent set, we have α = γ = λ. Consequently, τ(x) = λx + βc. Thus for all x ∈ V we can write τ(x) = λx + βxc, where βx varies linearly with x and such

−1 that βx = 0 for x ∈ A. It is not diﬃcult to see that by Lemma 2.6.9, λ I ◦ τ is a transvection which induces the same collineation as τ, namely ε.

Suppose now that there is a nondegenerate conjugate symmetric sesquilinear form f deﬁned on V . We will call a transvection T which is an isometry for f a unitary transvection. The following theorem characterizes the unitary transvections.

Theorem 2.6.12. A transvection T is a unitary transvection if and only if it has the form T (x) = x + αf(x, c)c for some nonzero isotropic vector c and nonzero scalar α with trace 0.

Proof. It is straightforward to check that any transvection of the given form is a unitary transvection. We prove the converse. Suppose that T (x) = x + φ(x)c is a

unitary transvection. Necessarily, c 6= 0. Note that since φ 6= 0 there is x0 ∈ V such

that φ(x0) 6= 0 and since f is nondegenerate, there is y0 ∈ V such that f(y0, c) = 1. Then for all x, y ∈ V ,

f(x, y) = f(T (x),T (y)) =f(x + φ(x)c, y + φ(y)c)

=f(x, y) + φ(y)f(x, c) + φ(x)f(y, c) + φ(x)φ(y)f(c, c) .

Consequently, for all x, y ∈ V ,

φ(y)f(x, c) + φ(x)f(y, c) + φ(x)φ(y)f(c, c) = 0 . (2.1) 24 Taking y = c and x = x0 in equation 2.1 and recalling that φ(c) = 0, we must

have 0 = φ(x0)f(c, c) = φ(x0)f(c, c). Since φ(x0) 6= 0, we must have f(c, c) = 0. This shows that the center of a unitary transvection must be spanned by isotropic vectors.

Now in light of the above, equation 2.1 becomes:

φ(y)f(x, c) + φ(x)f(y, c) = 0 . (2.2)

Setting y = y0 in equation 2.2, we have φ(y0)f(x, c) + φ(x) = 0 for all x ∈ V . Then φ

has the form φ(x) = αf(x, c) for all x ∈ V , where α = −φ(y0). Therefore condition 2.2 becomes αf(y, c)f(x, c) + αf(x, c)f(y, c) = 0 ,

which simpliﬁes to (α + α)f(x, c)f(y, c) = 0. Taking x = y = y0 in this last equation gives α + α = 0.

2.6.3 Unitary Polarities

Deﬁnition 2.6.13. A polarity is an inclusion-reversing permutation of the set of subspaces of a projective geometry which has order 2.

If ⊥ is a polarity on PG(V ), P is a point of PG(V ), and H is a hyperplane, then P ⊥ is a hyperplane called the polar of P and H⊥ is a point called the pole of H.

Observe that if V is equipped with a conjugate symmetric sesquilinear form f, it induces a polarity S 7→ S⊥ of PG(V ) where, if S = [W ] is the subspace of PG(V ) corresponding to the vector subspace W of V , S⊥ = [W ⊥], where W ⊥ is as deﬁned in subsection 2.5.1. Polarities of this type are called unitary polarities.

Deﬁnition 2.6.14. Let ⊥ be a polarity of PG(V ). A point P ∈ PG(V ) is called an absolute point of the polarity if P ∈ P ⊥.

25 Deﬁnition 2.6.15. A triangle in a PG(V ) is said to be self-polar (with respect to a polarity ⊥) if each side of the triangle is the polar line of the opposite vertex.

Note that the vertices of a self-polar triangle are necessarily nonabsolute.

Lemma 2.6.16. Let ⊥ be a unitary polarity on PG(V ), and P = [x] be a point of PG(V ). Then P is an absolute point if and only if x is an isotropic vector.

Proof. P is an absolute point if and only if [x] ∈ [x]⊥ = [{x}⊥], which holds if and only if x is an isotropic vector.

Lemma 2.6.17. Let ⊥ be a unitary polarity on PG(V ), where V has dimension n

∼ n n n−1 n−1 2 over F = Fq2 . Then PG(V ) has (q −(−1) )(q −(−1) )/(q −1) absolute points of the polarity ⊥.

Proof. This follows immediately from the previous lemma and Lemma 2.5.12.

In particular, we note that if n = 3, PG(V ) has q3 +1 absolute points with respect to a unitary polarity.

Theorem 2.6.18. Suppose PG(V ) ∼= PG(2, q2). A line ` of PG(V ) contains exactly 1 or q + 1 absolute points of a unitary polarity ⊥. More speciﬁcally, if ` = P ⊥ for some absolute point P , then P is the unique absolute point of `, and ` has q + 1 isotropic points otherwise.

Proof. Lemma 2.5.11 ensures that every 2-dimensional subspace of V contains a nonzero isotropic vector, and therefore that every line of PG(V ) contains an absolute point. The line ` corresponds to the subspace W = {x}⊥ for some x ∈ V \{0}. First, suppose that x is isotropic. By Lemma 2.5.15 it follows that the only isotropic vectors in W are the scalar multiples of x. Consequently the only absolute point on ` is [x] = P . If x is nonisotropic, then as noted above, W is spanned by a hyperbolic pair (c, d). If P is an absolute point on `, then either P = [d] or P has a unique vector

26 representative of the form c+αd for some α ∈ F with T r(α) = 0. There are q choices for such a scalar α since the kernel of the trace map from F to K is a one-dimensional K-vector space. Therefore, in this case ` contains exactly q + 1 absolute points.

2.7 UNITARY GROUPS

Here we will introduce the unitary groups over ﬁnite ﬁelds and describe some of their basic properties.

2.7.1 Deﬁnition and General Properties

Let K be a ﬁeld, F a quadratic extension of K, V a ﬁnite dimensional F -vector space, and f be a conjugate symmetric sesquilinear form on V . The general unitary group, GU(V ), is the group of all invertible linear maps T from V to V such that f(T x, T y) = f(x, y) for all x, y ∈ V . The special unitary group, SU(V ) is the subgroup of GU(V ) of consisting of those linear maps with determinant 1. The projective general unitary group, P GU(V ), and the projective special linear group, PSU(V ), are the groups of collineations of PG(V ) induced by GU(V ) and SU(V ), and are isomorphic to the quotients of GU(V ) and SU(V ) by their centers respectively. If K

n is a ﬁnite ﬁeld of order q and V = F we will write GUn(q), SUn(q), P GUn(q) and

(following the Atlas) Un(q) for GU(V ), SU(V ), P GU(V ) and PSU(V ) respectively.

The orders of these groups are well known:

Lemma 2.7.1. Let n ≥ 2 and q be a prime power. Then

n n(n−1) Y i i •| GUn(q)| = q 2 (q − (−1) ), i=1 |GU (q)| •| SU (q)| = |P GU (q)| = n , and n n q + 1 |SU (q)| •| U (q)| = n . n gcd(n, q + 1) 27 Lemma 2.7.2. Let T ∈ GU(V ) and λ = det T . Then λλ = 1.

Proof. Let {e1, . . . , en} be an orthonormal basis and let M be the matrix for T in this basis. Note that

  β1   X X X  .  f( αiei, βiei) = αiβi = [α1, . . . , αn]  .  .     βn P P It is not too hard to see then that for x = xiei and y = yiei

  y1   h i t  .  f(T x, T y) = x , x , ··· , x M · M  .  , 1 2 n     yn

where M t denotes the transpose of M, and M is the matrix obtained from M by replacing each entry m with its conjugate m. Since T is an isometry, we must have

    y1 y1     h i t  .  h i  .  x , x , ··· , x M · M  .  = x , x , ··· , x  .  . 1 2 n   1 2 n       yn yn

for all choices of the xi and yi. For any 1 ≤ i, j ≤ n choose xi = yj = 1 and all of the other components to be zero. Then the left hand side of the above equation gives the entry in row i and column j of M tM, and the right hand side is 1 if i = j and 0 otherwise. Hence we must have M tM = I. Since λ = det T = det M = det M t and det M = λ, it follows that λλ = 1.

We note that Un(q) is simple for all prime powers q and all n ≥ 2 except for

the cases U2(2), U2(3) and U3(2). Also, U2(q) is well known to be isomorphic to the projective special linear group L2(q), so the covering numbers of these are given in

[10]. Consequently, our focus here will be on the unitary groups U3(q). We note that

28 in general PSU(V ) acts transitively on both the absolute points and nonabsolute points of PG(V ), but if V is 3-dimensional, the action of PSU(V ) on the absolute points is doubly transitive.

Theorem 2.7.3. If V is 3-dimensional, then PSU(V ) is doubly transitive on the absolute points of PG(V ).

Proof. Let P, Q, P 0, and Q0 be absolute points with P 6= Q and P 0 6= Q0. Let c, d, c0, and d0 be vector representatives of P, Q, P 0, and Q0 respectively. Then c and d are linearly independent, as are c0 and d0. By Lemma 2.5.15 we may assume that these representatives are chosen so that (c, d) and (c0, d0) are hyperbolic pairs. By Lemma 2.5.16 there are x and x0 such that {c, x, d} and {c0, x0, d0} are unitary bases. Let T : V → V be the linear transformation deﬁned by T (c) = c0, T (x) = x0, and T (d) = d0. It is easy to see that T is an automorphism of V and preserves the form f. Hence T ∈ GU(V ). Let λ = det T . By Lemma 2.7.2, λλ = 1. Then also λ−1λ−1 = 1. It is not hard then to see that the linear transformation S : V → V given by S(c) = c0, S(x) = λ−1x0, and S(d) = d0 is an element of SU(V ). The collineation induced by S sends P to P 0 and Q to Q0.

N 2.7.2 Embedding U3(q) into U3(q ), N odd

∼ Let V be a 3-dimensional vector space over F = Fq2 . We note here that PSU(V ) can

N be embedded into a larger unitary group isomorphic to U3(q ), for any odd positive integer N, though later we will be mainly interested in the case N = 3.

Let N be an odd positive integer, E be an extension of the ﬁeld F of degree N, ˆ and B = {b0, b1, b2} be a basis for V . Let V be the E-vector space of all E-linear ˆ combinations of b0, b1, b2. Then V ⊆ V , and we can extend the form f to a conjugate symmetric sesquilinear form fˆ : Vˆ × Vˆ → E, by

29 2 2 ˆ X X X qN f( αibi, βjbj) = αiβj f(bi, bj) . i=0 j=0 i,j Furthermore, an isometry T ∈ SU(V ) extends to an isometry Tˆ ∈ SU(Vˆ ), where ˆ P2 P2 ˆ T ( i=0 αibi) = i=0 αiT (bi). Note that T has the same characteristic polynomial as T . Moreover, the map T 7→ Tˆ from SU(V ) into SU(Vˆ ) is a monomorphism, which in turn induces a monomorphism from PSU(V ) into PSU(Vˆ ), i.e. an embedding of

N U3(q) into U3(q ).

In particular, we note that PSU(V ) acts on PG(Vˆ ) by τ([x]) = [Tˆ(x)], where T ∈ SU(V ) is any isometry whose induced collineation is τ. We also note that this construction fails if N is even, for in that case the restriction of the form fˆ described above to V × V will not be a conjugate symmetric sesquilinear form, let alone be equal to f.

30 CHAPTER 3 COVERING NUMBERS OF SMALL PROJECTIVE SPECIAL UNITARY GROUPS

Here we consider projective special unitary groups U3(q) for small q.

3.1 U3(2)

The nonsimple case U3(2) is easily dealt with.

Theorem 3.1.1. The covering number of U3(2) is 3.

∼ 2 Proof. U3(2) = Z3 : Q8 has a homomorphic image isomorphic to the Klein 4-group, so σ(U3(2)) = 3.

3.2 U3(3)

∼ Let G = U3(3). We list the conjugacy classes of elements of G in table 3.1, and the conjugacy classes of maximal subgroups of G in table 3.2. We note that the subgroups from class M1 are normalizers of Sylow 3-subgroups and those from M3 are the centralizers of involutions.

31 Table 3.1: Conjugacy classes of elements of U3(3) Class Centralizer Order Class Size Cycle Type on 28 Points Principal 1A 6048 1 128 No 2A 96 63 14212 No 3A 108 56 1139 No 3B 9 672 1139 Yes 4A 96 63 1446 No 4B = 4A∗∗ 96 63 1446 No 4C 16 378 2246 Yes 6A 12 504 113164 No 7A 7 864 74 Yes 7B = 7A∗∗ 7 864 74 Yes 8A 8 756 122283 Yes 8B = 8A∗∗ 8 756 122183 Yes 12A 12 504 1131122 Yes 12B = 12A∗∗ 12 504 1131122 Yes

Table 3.2: Conjugacy classes of maximal subgroups of U3(3) Class Order Index Structure

1+2 M1 216 28 3+ : Z8

M2 168 36 L2(7)

· M3 96 63 Z4 S4

2 M4 96 63 Z4 : S3

32 Table 3.3: Incidence matrix for U3(3)

M1 M2 M3 M4 3B 1 3 0 3 4C 0 4 3 3 7AB 0 1 0 0 8AB 2 0 1 1 12AB 1 0 1 0

For 1 ≤ i ≤ 4, let θi be the permutation character of G acting on the right cosets of Hi ∈ Mi. The irreducible constituents of each of these are given in [12]. Explicitly we have:

θ1 = 1a + 27a

θ2 = 1a + 7bc + 21a

θ3 = 1a + 14a + 21a + 27a

θ4 = 1a + 7bc + 21a + 27a.

From these we can compute the entries in the incidence matrix given in table 3.3 by Lemma 2.2.11.

Proposition 3.2.1. C = M1 ∪ M2 is a cover of G by 64 subgroups.

Proof. This is clear from table 3.3.

We now show that that this cover is minimal.

Proposition 3.2.2. Let C be a cover of G consisting of maximal subgroups. Then

M2 ⊆ C.

33 Proof. From table 3.3 we see that the only maximal subgroups containing elements of

order 7 are those of class M2 and that each element of order 7 is contained in exactly

one maximal subgroup from class M2. Consequently any cover of G by maximal subgroups must contain all 36 subgroups from class M2.

Proposition 3.2.3. Let C be a cover of G. Then |C| ≥ 64.

Proof. By Lemma 2.2.9 we may assume without loss of generality that C consists only

of maximal subgroups of G. Let xi = |C ∩ Mi| for 1 ≤ i ≤ 4. By Proposition 3.2.2,

we have M2 ⊆ C, i.e. x2 = 36. Now, from table 3.3 only the maximal subgroups of G

from classes M1 and M3 contain elements of order 12, and moreover each element of

order 12 in G is contained in a unique member of class M1 and a unique member of

class M3. It follows from Lemma 2.2.10 that each subgroup from class M1 contains

36 elements of order 12, and each subgroup from M3 contains 16 elements of order

12. We deduce that 36x1 + 16x3 ≥ 1008, from which it follows that x1 + x3 ≥ 28. We conclude that |C| ≥ x1 + x2 + x3 ≥ 64.

By Propositions 3.2.1 and 3.2.3 we have:

Theorem 3.2.4. The covering number of U3(3) is 64.

3.3 U3(4)

∼ Let G = U3(4). The conjugacy classes of elements of of G are given in table 3.4 and the conjugacy classes of maximal subgroups are listed in table 3.5. We note that the

subgroups from class M2 are normalizers of cyclic subgroups generated by elements

from 5ABCD, and that those from M3 and M4 are the normalizers of the Sylow 5-subgroups and Sylow 13-subgroups of G respectively.

34 Table 3.4: Conjugacy classes of elements of U3(4) Class Centralizer Order Class Size Principal 1A 62400 1 No 2A 320 195 No 3A 15 4160 No 4A 16 3900 No 5A 300 208 No 5B = 5A∗∗ 300 208 Yes 5C = 5A∗2 300 208 Yes 5D = 5A∗3 300 208 Yes 5E 25 2496 Yes 5F = 5E∗ 25 2496 Yes 10A 20 3120 Yes 10B = 10A∗∗ 20 3120 Yes 10C = 10A∗7 20 3120 Yes 10D = 10A∗3 20 3120 Yes 13A 13 4800 Yes 13B = 13A∗∗ 13 4800 Yes 13C = 13A∗5 13 4800 Yes 13D = 13A∗8 13 4800 Yes 15A 15 4160 Yes 15B = 15A∗∗ 15 4160 Yes 15C = 15A∗2 15 4160 Yes 15D = 15A∗8 15 4160 Yes

Table 3.5: Conjugacy classes of maximal subgroups of U3(4) Class Order Index Structure 2+4 M1 960 65 2 : Z15 M2 300 208 Z5 : A5 2 M3 150 416 Z5 : S3 M4 39 1600 Z13 : Z3

Table 3.6: Incidence matrix for U3(4) M1 M2 M3 M4 4A 1 0 0 0 5EF 0 3 1 0 10ABCD 1 1 2 0 13ABCD 0 0 0 1 15ABCD 2 1 0 0

35 For 1 ≤ i ≤ 4, let θi be the permutation character of G acting on the right cosets of Hi ∈ Mi. The irreducible constituents of the ﬁrst three of these are given in [12].

The character θ4 is computed in Lemma 3.3.1, and its decomposition into irreducibles can then be found by computing inner products with the irreducible characters of G. Explicitly we have:

θ1 = 1a + 64a

θ2 = 1a + 39ab + 64a + 65a

θ3 = 1a + 39ab + 52abcd + 64a + 65a

θ4 = 1a + 13abcd + 39ab + 2 × 52abcd + 2 × 64a + 65abcde + 2 × 75abcd.

By Lemma 2.2.11 the values of these characters on the principal elements of G give the entries of the incidence matrix, which is given in table 3.6.

Lemma 3.3.1. The character θ4 takes the values 1600, 10, 1,and 0 on 1A, 3A, 13ABCD, and 2A ∪ 4A ∪ 5ABCDEF ∪ 10ABCD ∪ 15ABCD respectively.

Proof. Recall that by Lemma 2.2.11, for all g ∈ G, θ4(g) is the number of subgroups

from M4 which contain g. Of course the identity is contained in each of the 1600 members of M4, so θ4(1G) = 1600. Each element x of order 13 is contained in a unique Sylow 13-subgroup of G, namely hxi, and hence in a unique member of class

M4, namely NG(hxi). Hence θ4(x) = 1 for x ∈ 13ABCD. For any H ∈ M4, the 26 elements not contained in the Sylow 13-subgroup of H are from 3A. It follows from Lemma 2.2.11 that θ4(x) = 10 for x ∈ 3A and θ4(x) = 0 for x ∈ 2A ∪ 4A ∪ 5ABCDEF ∪ 10ABCD ∪ 15ABCD.

Proposition 3.3.2. Let C be a cover of G consisting of maximal subgroups. Then

M1 ∪ M4 ⊆ C. Moreover, M1 ∪ M4 is suﬃcient to cover all of the elements of G except those from 5EF.

36 Proof. Each element of G of order 4 or 13 is contained in a unique maximal subgroup

of G, which is from M1 for an element of order 4 and from M4 for an element of order

13. Since C covers these elements, it must contain each subgroup from M1 ∪ M4. The second part of the proposition is easily seen from table 3.6.

In light of the previous proposition the problem is reduced to ﬁnding a minimal

cover of the elements from 5EF. It suﬃces to use only subgroups from M2 for this purpose as follows:

Proposition 3.3.3. There exists a minimal cover G consisting of maximal subgroups which contains no subgroups from class M3.

Proof. Let C be any cover of G consisting of maximal subgroups. Note that each subgroup from M2 has exactly six Sylow 5-subgroups, which are also Sylow 5-subgroups of G. It follows that each Sylow 5-subgroup of G is contained in exactly three sub-

0 groups from class M2. We can obtain a new cover C by replacing each H ∈ C ∩ M3

0 0 by H , where H is any of the three members of M2 containing the unique Sylow 5-subgroup of H. Clearly C0 is a cover of G consisting of maximal subgroups from

0 M1 ∪ M2 ∪ M4 and |C | ≤ |C|.

Proposition 3.3.4. There is a set S ⊆ M2 with |S| = 80 whose members cover all

of the elements from 5EF, and no smaller subset of M2 will suﬃce for this purpose.

Proof. We construct a 1248 by 208 incidence matrix A with columns indexed by the subgroups from M2 and rows indexed by the cyclic subgroups generated by the elements from 5EF, with a 1 in the position corresponding to cyclic subgroup hgi and maximal subgroup H if hgi ≤ H, and a 0 otherwise. There is a natural one-to-one correspondence between the power set of M2 and the set of {0, 1}-vectors indexed by M2 given by S 7→ xS, where xS is the vector whose entry in position H is 1 if

H ∈ S and is 0 otherwise. It is easy to see that the subsets S of M2 which cover all of the elements of 5EF correspond to the vectors x such that each component 37 of the vector Ax is greater than or equal to 1. Finding a smallest set S which covers the elements from 5EF then amounts to solving an integer linear programming problem with binary variables. This problems was solved using the Gurobi [17] linear programming software, yielding an optimal solution of 80 subgroups.

Proposition 3.3.5. Let S be as in Proposition 3.3.4. Then C = M1 ∪ S ∪ M4 is a minimal cover of G consisting of 1745 subgroups.

Proof. This follows from Propositions 3.3.2 and 3.3.4.

Consequently,

Theorem 3.3.6. The covering number of U3(4) is 1745.

3.4 U3(5)

∼ Let G = U3(5). We list the conjugacy classes of elements of G in table 3.7, and the conjugacy classes of maximal subgroups of G in table 3.8. We note that in the action of G as a group of automorphisms of the Hoﬀman-Singleton graph, the subgroups of class M1 are the stabilizers of the vertices, and those from class M5 are the stabilizers of the edges. Also, the subgroups from class M4 are the normalizers of the Sylow

5-subgroups of G, and those from M8 are the centralizers of involutions.

38 Table 3.7: Conjugacy classes of elements of U3(5) Class Centralizer Order Principal 1A 126000 No 2A 240 No 3A 36 No 4A 8 No 5A 250 No 5B 25 Yes 5C 25 Yes 5D 25 Yes 6A 12 Yes 7A 7 Yes 7B = 7A∗∗ 7 Yes 8A 8 Yes 8B = 8A∗∗ 8 Yes 10A 10 Yes

Table 3.8: Conjugacy classes of maximal subgroups of U3(5) Class Order Index Structure M1 2520 50 A7 M2 2520 50 A7 M3 2520 50 A7 1+2 M4 1000 126 5+ : Z8 M5 720 175 M10 M6 720 175 M10 M7 720 175 M10 M8 240 525 2S5

Table 3.9: Incidence matrix for U3(5) M1 M2 M3 M4 M5 M6 M7 M8 5B 5 0 0 1 5 0 0 0 5C 0 5 0 1 0 5 0 0 5D 0 0 5 1 0 0 5 0 6A 1 1 1 0 0 0 0 3 7AB 1 1 1 0 0 0 0 0 8AB 0 0 0 2 1 1 1 1 10A 0 0 0 1 0 0 0 1

For 1 ≤ i ≤ 8, let θi be the permutation character of G acting on the right cosets 39 of Hi ∈ Mi. The irreducible constituents of the ﬁrst seven of these are given in [12]. Explicitly we have:

θ1 = 1a + 21a + 28a

θ2 = 1a + 21a + 28b

θ3 = 1a + 21a + 28c

θ4 = 1a + 125a

θ5 = 1a + 21a + 28a + 125a

θ6 = 1a + 21a + 28b + 125a

θ7 = 1a + 21a + 28c + 125a

From these we can compute the entries in the ﬁrst seven columns of table 3.9 by Lemma 2.2.11. We will need one further entry from table 3.9 which we justify with the following proposition:

Proposition 3.4.1. Each element of class 10A is contained in a unique subgroup from class M8.

5 Proof. For x ∈ 10A, CG(x) = hxi which contains a unique involution, namely x . The subgroups from class M8 are the centralizers of involutions in G, hence x is contained

5 in a unique member of M8, namely CG(x ).

The following cover of G was suggested to the author by Nicola Pace (personal communication, June 22, 2017):

Proposition 3.4.2. C = M1 ∪ M4 is a cover of G with |C| = 176.

Proof. This is clear in light of table 3.9.

40 We will prove that this cover is minimal.

Proposition 3.4.3. For any cover C of G, |C| ≥ 176.

Proof. By Lemma 2.2.9 we may assume without loss of generality that C is a cover of

G by maximal subgroups, and denote by xi the number of subgroups in C from class

Mi. As seen in table 3.9, only the 150 maximal subgroups isomorphic to A7 contain elements of order 7. Each of these contains 720 elements of order 7, and there are a total of 36000 such elements in G, so

720(x1 + x2 + x3) ≥ 36000.

The elements from class 10A must be covered by some collection of subgroups from classes M4 and M8. As seen in table 3.9, each element of order 10 appears in exactly one subgroup from class M4 and exactly one from M8. Thus each subgroup from class M4 must contain exactly 100 elements of order 10 and each subgroup from M8 contains exactly 24 elements of order 10. Thus we must have

100x4 + 24x8 ≥ 12600,

from which it follows that x4 +x8 ≥ 12600/100 = 126. Consequently, |C| ≥ 50+126 = 176.

By Propositions 3.4.2 and 3.4.3 we have the following theorem:

Theorem 3.4.4. The covering number of U3(5) is 176.

41 CHAPTER 4

THE COVERING NUMBER OF U3(q) WITH q > 5

a In this chapter we assume that p is a prime, q = p , K = Fq, F = Fq2 , V is a 3-dimensional F -vector space, f is a conjugate symmetric sesquilinear form on V , ⊥ ∼ is the associated polarity deﬁned on PG(V ), and G = PSU(V ) = U3(q).

4.1 MAXIMAL SUBGROUPS

The subgroup structure of the unitary groups U3(q) was investigated for odd q by H. H. Mitchell in [31], and for even q by R. W. Hartley in [19]. Some additional details about the subgroup structure of these groups can be found in [26]. For q odd, the subgroups of U3(q) are described by the following theorem:

Theorem 4.1.1 (Mitchell). Let H be a subgroup of U3(q), q odd. Then, H stabilizes a point, line, triangle, or imaginary triangle, or H is one of the following subgroups:

q3(q + 1)(q − 1) 1. the stabilizer of an absolute point, with order . gcd(3, q + 1) q(q + 1)2(q − 1) 2. the stabilizer of a nonabsolute point, with order . gcd(3, q + 1) 6(q + 1)2 3. the stabilizer of a triangle (in PG(2, q2)), with order . gcd(3, q + 1)

4. the stabilizer of an imaginary triangle (i.e. a triangle in PG(2, q6) but not 3(q2 − q + 1) PG(2, q2)), with order . gcd(3, q + 1)

5. the stabilizer of a conic, with order q(q + 1)(q − 1).

k 6. U3(q0), if q = q0 with k odd.

42 k 7. P GU3(q0), if q = q0 , k is odd, and 3 divides both k and q0 + 1.

8. the Hessian groups of order 216 (if 9 divides q + 1), 72 and 36 (if 3 divides q + 1).

9. L3(2), if −7 is not a square in Fq.

10. A6, if 5 is a square in Fq and Fq does not contain a primitive third root of unity.

k 11. A6.2, if q = 5 with k odd.

k 12. A7 if q = 5 with k odd.

Moreover, Mitchell proves that the subgroups of each type from the list above form a single conjugacy class if q + 1 is not divisible by 3, and form one or three conjugacy classes if q + 1 is divisible by 3.

If q is even, the maximal subgroups of U3(q) are as follows:

Theorem 4.1.2 (Hartley). Let q be even and H be a maximal subgroup of U3(q). Then H is one of the following subgroups:

k 5. U3(q0), if q = q0 , with k an odd prime.

3 3k 6. P GU3(q0), if q = q0 = 2 with k odd.

2 7. Z3 : Z4, if q = 2. 43 4.2 A ROUGH CLASSIFICATION OF THE ELEMENTS OF G

As Theorems 4.1.1 and 4.1.2 describe the maximal subgroups of G geometrically, it will be convenient for us to work with a rough characterisation of the elements of G in terms of the geometric objects they ﬁx. We consider three types of elements of G:

Type 1: Elements that ﬁx an absolute point of PG(V ).

Type 2: Elements that do not ﬁx an absolute point but do ﬁx a nonabsolute point of PG(V ).

Type 3: Elements that do not ﬁx any points of PG(V ).

Clearly any element of G is of exactly one of these types. We now consider each type of element in turn.

4.2.1 Type 1: Elements Fixing an Absolute Point

In this section we investigate the elements of G which fix an absolute point by first considering the elements of SU(V ) which fix an absolute point. We shall see, in particular, that every element of order p is of this type.

Lemma 4.2.1. Let P = [c] be an absolute point of PG(V ), T ∈ SL(V ) such that T ﬁxes the projective point P , and {c, x, d} be a unitary basis for V . Then T ∈ SU(V ) if and only if the matrix for T in this basis has the form   y α β     0 z γ  ,     0 0 w

where the following conditions hold:

1. yzw = 1

2. zz = 1, 44 3. T r(wβ) + γγ = 0,

4. yw = 1, and

5. αw + zγ = 0.

Proof. Suppose that T ∈ SU(V ). Note that since T ﬁxes P , c is an eigenvector for T . Since T must preserve {c}⊥ = span{c, x}, in the given basis T is represented by an upper triangular matrix, and hence has the form described above for some y,z,w,α,β,γ ∈ F . Now T ∈ SU(V ), and so 1 = det(T ) = yzw. The remaining properties follow easily from the fact that T is an isometry for the conjugate symmetric sesquilinear form f. Conversely, if the matrix for T in basis {c, x, d} is as above, it is easy to see that T is an isometry with respect to the form f and has determinant 1, so T ∈ SU(V ).

Note that from the conditions above, one can deduce the following equivalent conditions:

1. w ∈ F ∗,

2. y = w−1,

3. z = ww−1,

4. α = −w−1γ, and

5. γγ = −T r(wβ).

Lemma 4.2.2. For T as in Lemma 4.2.1, The matrix representing T k, k ≥ 2 in the basis {c, x, d} is   yk σ τ      0 zk π      0 0 wk

45 where,

X σ = αyizj i+j=k−1 i,j≥0 X X τ = βyiwj + αγyizjwm i+j=k−1 i+j+m=k−2 i,j≥0 i,j,m≥0 X π = γziwj . i+j=k−1 i,j≥0

The matrix for T −1 in this basis is   y−1 −(yz)−1α (yw)−1(αγz−1 − β)      0 z−1 −(zw)−1γ  .     0 0 w−1

Proof. The ﬁrst claim is proved by induction on k. The second is easily veriﬁed.

4.2.2 Sylow p-Subgroups

Fix a nonzero isotropic vector c, and choose vectors x, d ∈ V such that {c, x, d} is a unitary basis for V . Let S be the stabilizer of the projective point [c] in SU(V ), and let P be the set of all T ∈ S such that the matrix for T in this basis has the form   1 α β     0 1 γ .     0 0 1

Note that by Lemma 4.2.1 that T r(β) = −αα and γ = −α. It is easy to see that P is a nonabelian subgroup of S, and that any nonidentity element of P has order p if p is odd. If p = 2 then P has exponent 4. In particular, in this case a nonidentity element of P has order 2 if it is a transvection and has order 4 otherwise. The center Z(P ) of P consists of the identity and the unitary transvections with center spanned

46 by c, which is elementary abelian of order q. Since α may be chosen arbitrarily from F and for each α there are q choices for β satisfying T r(β) = −αα, it follows that P has order q3 and is a therefore a Sylow p-subgroup of SU(V ). On the other hand, suppose that T ∈ S has order pm. By Lemma 4.2.1, the matrix for T in the basis {c, x, d} has the form   y α β     0 z γ  ,     0 0 w

and by Lemma 4.2.2, it follows that ypm = zpm = wpm = 1. But y, z, w ∈ F ∗ and therefore have orders which are relatively prime to p. Thus y = z = w = 1 and T ∈ P . This proves that P is the unique Sylow p-subgroup of S.

Let T be a nonidentity element of P . Then the matrix for T in the basis {c, x, d} has the form   1 α β     0 1 γ     0 0 1 where T r(β) = −αα and γ = −α. Suppose that α = 0. Then γ = T r(β) = 0 but β 6= 0. In this case T is a unitary transvection with center spanned by c. It fixes precisely q2 + 1 points of PG(V ), namely [c] and the q2 nonabsolute points of [c]⊥. If [y] ∈ [c]⊥, then [y] is in fact fixed by all q of the transvections in P . If α 6= 0, then also β, γ 6= 0. It is easy to see in this case that T has no eigenvectors except those in span{c} and consequently T fixes no points of PG(V ) aside from [c]. We note that in either case, [c] is the only absolute point of PG(V ) fixed by T , so that P intersects the stabilizer of any absolute point other than [c] trivially. In particular, P is not contained in the stabilizer of any point of PG(V ) other than [c]. Since SU(V ) acts transitively by conjugation on both the Sylow p-subgroups and the stabilizers of the absolute points, it follows that SU(V ) has exactly q3 + 1 Sylow p-subgroups. Since 47 each of these contains q3 − 1 elements of order p, if p is odd, or of orders 2 and 4 if p = 2, and any two of them have a trivial intersection, the total number of elements of order p if p is odd, or of orders 2 and 4 if p = 2, is q6 − 1. We summarize these results in the following theorem:

Theorem 4.2.3. 1. If [c] is an absolute point of PG(V ), its stabilizer in SU(V ) contains a unique Sylow p-subgroup.

2. If P is a Sylow p-subgroup of SU(V ), then

(a) Every nonidentity element of P has order p if p is odd. If p = 2, every nonidentity element has order 2 or 4.

(b) There is a unique absolute point [c] whose stabilizer contains P ,

(c) P intersects each of the stabilizers of the q2 nonabsolute points on the line [c]⊥ in a subgroup of order q generated by the transvections in P , and

(d) P intersects the stabilizer of a other point other than those mentioned above trivially.

3. SU(V ) has exactly q3 + 1 Sylow p-subgroups, and has q6 − 1 elements of order p if p is odd. If p = 2, SU(V ) has (q3 + 1)(q − 1) involutions and (q3 + 1)(q3 − q) elements of order 4.

4. A transvection ﬁxes one absolute point and q2 nonabsolute points. An element of order p in SU(V ) which is not a transvection if p is odd, or of order 4 if p = 2, ﬁxes a unique point of PG(V ), and moreover this point is an absolute point.

From this theorem we obtain the following:

Corollary 4.2.4. 1. If [c] is an absolute point of PG(V ), its stabilizer G[c] in G contains a unique Sylow p-subgroup. 48 2. If P is a Sylow p-subgroup of G, then

(a) Every nonidentity element of P has order p if p is odd. If p = 2, every nonidentity element has order 2 or 4.

(b) There is a unique absolute point [c] whose stabilizer contains P ,

(c) P intersects each of the stabilizers of the q2 nonabsolute points on the line [c]⊥ in a subgroup of order q consisting of the elations in P , and

(d) P intersects the stabilizer of a point other than those mentioned above trivially.

3. G has exactly q3 + 1 Sylow p-subgroups, and G has q6 − 1 elements of order p if p is odd. If p = 2, G has (q3 + 1)(q − 1) involutions and (q3 + 1)(q3 − q) elements of order 4.

4. A nonidentity elation ﬁxes one absolute point and q2 nonabsolute points. An element of order p in G which is not an elation if p is odd, or of order 4 if p = 2, ﬁxes a unique point of PG(V ), and moreover this point is an absolute point.

4.2.3 Type 2: Elements Fixing a Nonabsolute Point but no Absolute Points

In this section we consider elements G which do not ﬁx any absolute points of PG(V ) but do ﬁx a nonabsolute point. First, we show that such elements exist.

Lemma 4.2.5. There exists τ ∈ G which ﬁxes a nonabsolute point of PG(V ) but does not ﬁx an absolute point.

∗ q+1 Proof. Let {b0, b1, b2} be an orthonormal basis for V . Let λ, µ ∈ F with λ = µq+1 = 1 and such that λ, µ and (λµ)−1 are all distinct. Note that such a choice of λ and µ always exists, for example we may take λ = 1 and µ to be any element of order 49 q +1 in F . Then the collineation τ induced by the transformation T ∈ SU(V ) deﬁned

−1 by T (b0) = λb0, T (b1) = µb1, and T (b2) = (λµ) b2 has the desired properties.

We observe that the collineation τ constructed in the proof of Lemma 4.2.5 actually ﬁxes three nonabsolute points which form a self-polar triangle. We will eventually prove that elements such as this are the only elements of G which ﬁx a nonabsolute point of PG(V ) but not any absolute points.

The following notion can be found in [35, 14].

n n−1 Deﬁnition 4.2.6. For a monic polynomial g(x) = x + αn−1x + ··· + α1x + α0 ∈

F [x] with α0 6= 0, we deﬁne

−1 n n−1 g˜(x) = α0 (α0x + α1x + ... αn−1x + 1) .

Lemma 4.2.7. If g, h ∈ F [x] are monic polynomials with g(0) 6= 0 and h(0) 6= 0,

then ghf(x) =g ˜(x)h˜(x).

Pn i Pm i Proof. Write g(x) = i=0 αix and h(x) = i=0 βix where αn = βm = 1, α0 6= 0, and β0 6= 0. Then, letting αj = 0 or βj = 0 if j > n or j > m respectively, we have

n+m i ˜ −1 −1 X X n+m−i g˜(x)h(x) = α0 β0 αjβi−jx i=0 j=0 n+m i ! −1 X X n+m−i = α0β0 αjβi−j x i=0 j=0

= ghf(x).

Lemma 4.2.8. If T ∈ GU(V ) and g(x) ∈ F [x] is the minimal polynomial for T , then g˜(x) = g(x).

50 Pn i Proof. Write g(x) = i=0 αix . Now, g is monic so αn = 1 and since T is invertible,

α0 6= 0. Since g(T ) is the zero linear transformation, it follows that for all v, w ∈ V ,

−1 −n f(α0 T g(T )v, w) = 0. Now,

n −1 −n X −1 i−n 0 = f(α0 T g(T )v, w) = α0 αif(T v, w) i=0 n X −1 n−i = α0 αif(v, T w) i=0 n X −1 n−i = f(v, α0 αiT w) i=0 = f(v, g˜(T )w).

Consequently, f(v, g˜(T )w)) = 0 for all v, w ∈ V , and by the nondegeneracy of the form f, we must haveg ˜(T ) = 0. Then g(x) dividesg ˜(x), and since the latter is monic of degree n, it follows thatg ˜(x) = g(x).

The following is Lemma 2 of [14].

Lemma 4.2.9. Suppose g(x) ∈ F [x] is monic of degree n with g(0) 6= 0. If g˜(x) = g(x), then n is odd and every root ξ of g(x) satisﬁes ξqn+1 = 1.

Theorem 4.2.10. If τ ∈ G fixes a nonabsolute point of PG(V ) but does not fix any absolute points, then τ fixes exactly three nonabsolute points of PG(V ) which form a self-polar triangle.

Proof. Let T ∈ SU(V ) be a transformation whose induced collineation is τ. By hypothesis, τ (and therefore also T ) ﬁxes a nonabsolute point P = [e0] ∈ PG(V ).

Without loss of generality we can assume that f(e0, e0) = 1. Choose an orthonormal

⊥ ⊥ basis {e1, e2} for {e0} . Since e0 is nonisotropic, e0 ∈/ {e0} = span{e1, e2}, and

{e0, e1, e2} is an orthonormal basis for V . In this basis the matrix for T has the form

51   λ 0 0     0 α β     0 γ δ where λq+1 = λ(αδ − βγ) = 1. The characteristic polynomial for T is c(x) = (x − λ)g(x), where g(x) = x2 − (α + δ)x + αδ − βγ is the characteristic polynomial

⊥ for T |{e0} . Note that as T ﬁxes no absolute points of PG(V ), T cannot have an eigenspace of dimension greater than one.

We claim that g(x) factors in F [x]. Suppose by way of contradiction that g(x) is irreducible over F . Then c(x) is in fact the minimal polynomial for T . By Lemma 4.2.8 we havec ˜(x) = c(x). It is easy to see that x − λ is also invariant under the ∼ operation. Then, in light of Lemma 4.2.7, it is easy to see thatg ˜(x) = g(x). Since g(x) is irreducible, by Lemma 4.2.9 the degree of g(x) must be odd. This is a contradiction since g(x) has degree 2. Consequently we conclude that g(x) factors in F [x].

Therefore we may write g(x) = (x − µ)(x − ν), µ, ν ∈ F . Neither µ nor ν can be equal to λ or else T would have a two dimensional eigenspace. We also claim that

⊥ µ 6= ν. Since µ is an eigenvalue for T |{e0} , there is a nonzero vector v ∈ span{e1, e2} such that T (v) = µv. Since T ﬁxes no absolute points, v is nonisotropic. Since

⊥ ⊥ ⊥ ⊥ T preserves both {e0} and {v} , T also preserves {e0} ∩ {v} , a 1-dimensional subspace of V . Let w be a generator for this subspace. Then w is an eigenvector

⊥ for T |{e0} . It is not possible for T (w) = µw, as then T would have a 2-dimensional eigenspace. So T (w) = νw and ν 6= µ. Again, since T fixes no absolute points, w is nonisotropic. Now e0, v, and w are eigenvectors of T corresponding to distinct eigenvalues, {e0, v, w} is a basis for V , so the points [e0,[v], and [w] form a triangle in PG(V ). Clearly T , and therefore also τ, fixes each of these points, and it follows from e0 ⊥ v, e0 ⊥ w, and v ⊥ w that the triangle {[e0], [v], [w]} is self-polar. It is 52 easy to see that τ fixes no other points of PG(V ).

4.2.4 Type 3: Elements Fixing no Points of PG(V )

Here we investigate the elements of G which do not fix any points of PG(V ). We will see that each such element fixes a unique imaginary triangle in a plane isomorphic to PG(2, q6). Our first goal is to prove that G contains elements which fix no points of PG(V ) when q > 2.

Lemma 4.2.11. Let r be the largest prime divisor of q2 − q + 1. If q > 2, then r ≥ 5 and gcd(r, q3(q − 1)(q + 1)2) = 1.

Proof. Note that q2 −q+1 = q(q−1)+1 = (q−2)(q+1)+3, from which it follows that gcd(q2 −q+1, q) = gcd(q2 −q+1, q−1) = 1 and gcd(q2 −q+1, q+1) = gcd(3, q+1) = 1 or 3. In the case that 3 divides q + 1, 3 also divides q − 2, so that q2 − q + 1 ≡ 3 (mod 9). Consequently, q2 − q + 1 is a power of 3 if and only if q = 2. Also note that q2 − q + 1 is odd for any q, so unless q = 2, in which case q2 − q + 1 = 3, there is a prime r ≥ 5 which divides q2 − q + 1. Since gcd(q2 − q + 1, q3(q − 1)(q + 1)2) = 1 or 3 and r > 3, it follows that gcd(r, q3(q − 1)(q + 1)2) = 1.

We note that the condition gcd(r, q3(q − 1)(q + 1)2) = 1 actually holds for any prime divisor r of q2 − q + 1 which is greater than 3.

Theorem 4.2.12. If q > 2 then G contains elements not ﬁxing any points of PG(V ).

Proof. Let r be the largest prime divisor of q2 − q + 1. By Cauchy’s Theorem,

q3(q2−1) G contains an element ρ of order r. Since r is relatively prime to gcd(3,q+1) and q(q−1)(q+1)2 gcd(3,q+1) , the orders of the stabilizers of the absolute and nonabsolute points of PG(V ) respectively, it follows that ρ is not contained in the stabilizer of any point of PG(V ).

We will accumulate and maintain some notation throughout this section. To begin with, fix an arbitrary element τ ∈ G which does not fix any point of PG(V ), and 53 let T ∈ SU(V ) be a transformation whose induced collineation is τ. Consider for a moment the characteristic polynomial c(x) of T . since τ and therefore T fixes no points of PG(V ), T has no eigenvalues in F . That is, c(x) has no roots in F , and therefore, being of degree 3, is irreducible over F . Let E be an extension of F of degree three. Then c(x) splits over E and has three distinct roots, necessarily of the

q2 q4 q4+q2+1 form α, α , α for some α ∈ E \ F satisfying α = 1. Let B = {b0, b1, b2} be a basis for V in which the matrix for T is in rational canonical form, i.e.

  0 0 1    2 4 2 4  1 0 −αq +1 − αq +q − αq +1 .     0 1 α + αq2 + αq4

Moreover, let V,ˆ f,ˆ and Tˆ as in subsection 2.7.2, with N = 3.

We now show that τ ﬁxes a unique imaginary triangle, i.e. a triangle with vertices in PG(Vˆ ) but not in PG(V ).

Theorem 4.2.13. τ ﬁxes exactly 3 points of PG(Vˆ ) which form a triangle.

ˆ Proof. Deﬁne vectors ei ∈ V , i = 0, 1, 2, as follows:

−1 2q4+q2 q4+2q2 2q2 2q4 q4 q2 e0 = D [(α − α )b0 + (α − α )b1 + (α − α )b2]

−1 q4+2 2q4+1 2q4 2 q4 e1 = D [(α − α )b0 + (α − α )b1 + (α − α )b2]

−1 2q2+1 q2+2 2 2q2 q2 e2 = D [(α − α )b0 + (α − α )b1 + (α − α)b2], where D = (αq4 − αq2 )(αq4 − α)(αq2 − α).

ˆ A straightforward calculation shows that ei is an eigenvector for T corresponding to eigenvalue αq2i . Since these are eigenvectors corresponding to distinct eigen- ˆ values, the set {e0, e1, e2} is linearly independent. Consequently, in PG(V ), the

points [e0], [e2], and [e3] are not collinear and therefore form a triangle. Moreover, 54 ˆ q2i τ([ei]) = [T (ei)] = [α ei] = [ei], and it is easy to see that τ ﬁxes no other points of PG(Vˆ ).

For the remainder of this section let ∆τ = {P0,P1,P2} be the imaginary triangle

ﬁxed by τ, where Pi = [ei] for i = 0, 1, 2 and the ei are as in the previous proof.

Furthermore, let `i be the side of ∆τ opposite vertex Pi for each i = 0, 1, 2. That is, ˆ `i is the line of PG(V ) passing through the points of {P0,P1,P2}\{Pi}. We note

that the change of basis matrix from the basis {b0, b1, b2} to the basis {e0, e1, e2} is the Vandermonde matrix

  1 α α2    2 2  A = 1 αq α2q  . (4.1)     1 αq4 α2q4 The inverse of A is given by

  α2q4+q2 − αq4+2q2 αq4+2 − α2q4+1 α2q2+1 − αq2+2   −1 −1  2 4 4 2  A = D  α2q − α2q α2q − α2 α2 − α2q  . (4.2)     αq4 − αq2 α − αq4 αq2 − α

Lemma 4.2.14. An odd number of the points Pi, i = 0, 1, 2, ﬁxed by τ are nonabsolute.

Proof. If all three of the Pi are nonabsolute, there is nothing to prove. So suppose that P0 is an absolute point. Clearly, each line ì is fixed by τ, and since τ fixes exactly three points, it follows from Theorem 2.6.4 that these are the only lines fixed by τ. On

⊥ ⊥ ⊥ ⊥ the other hand Pi must be ﬁxed by τ for i = 0, 1, 2, so {`0, `1, `2} = {P0 ,P1 ,P2 }.

⊥ ⊥ Since P0 is absolute, P0 ∈ P0 . Thus, P0 is either `1 or `2. Without loss of generality

55 ⊥ we may suppose that P0 = `1. By Theorem 2.6.18, P0 is the unique absolute point

⊥ ⊥ on P0 , and hence P2 is nonabsolute. Now, P2 is nonabsolute, so P2 = `2. But then

⊥ P1 = `0 which contains P1, proving that P1 is absolute.

Theorem 4.2.15. The imaginary triangle ∆τ ﬁxed by τ is self-polar.

Proof. By the previous lemma at least one of the Pi, say P0, is nonabsolute. Let T ∈ SU(V ) such that the collineation of PG(V ) induced by T is τ. Now,

ˆ ˆ ˆ ˆ ˆ q3+1 ˆ f(e0, e0) = f(T (e0), T (e0)) = F (αe0, αe0) = α f(e0, e0)

q3+1 ˆ from which it follows that (α − 1)f(e0, e0) = 0. Since e0 is nonisotropic, it follows that αq3+1 = 1. In fact, since αq4+q2+1 = 1 and gcd(q3 + 1, q4 + q2 + 1) = q2 − q + 1, we must have αq2−q+1 = 1. We also must have

ˆ ˆ ˆ ˆ ˆ q2 q5+1 ˆ f(e0, e1) = f(T (e0), T (e1)) = f(αe0, α e1) = α f(e0, e1)

q5+1 ˆ q5+1 from which we can deduce that (α − 1)f(e0, e1) = 0. We claim that α 6= 1. Suppose that αq5+1 does equal 1. Then since αq2−q+1 = 1 and gcd(q5 +1, q2 −q +1) = gcd(3, q + 1), it follows that the order of α in E∗ divides gcd(3, q + 1). But the unique subgroup of E∗ of order gcd(3, q + 1) is contained in F ∗, which contradicts

q5+1 ˆ the fact that α∈ / F . So α 6= 1, and f(e0, e1) = 0. A similar argument shows that ˆ ˆ f(e0, e2) = f(e1, e2) = 0. Since the ei are pairwise orthogonal, it follows from Lemma

2.5.15 that at most one of them is isotropic. Consequently at least two of the Pi are nonabsolute, but by Lemma 4.2.14, this forces all three to be nonabsolute. Therefore,

⊥ ⊥ ⊥ ⊥ ⊥ Pi ∈/ Pi , and since {`0, `1, `2} = {P0 ,P1 ,P2 }, we must have Pi = `i for i = 0, 1, 2,

proving that the triangle {P0,P1,P2} is self-polar.

Theorem 4.2.16. There is σ ∈ G such that

1. |σ| = (q2 − q + 1)/ gcd(3, q + 1),

56 2. τ ∈ hσi, and

3. σ ﬁxes ∆τ pointwise.

Proof. Let ζ ∈ E∗ be an element of order q2 − q + 1 and deﬁne the linear map

ˆ ˆ ˆ ˆ q2i ˆ Z : V → V by Z(ei) = ζ ei for i = 0, 1, 2. It is easily checked that Z is an isometry with respect to the form fˆ and is of order q2 −q +1. Since the order of α in E∗ divides q2 − q + 1, and ζ generates the unique subgroup of E∗ of order q2 − q + 1 it follows

ˆ ˆ q2i ˆ ˆ that α ∈ hζi. Since T is given by T (ei) = α ei, it follows that T ∈ hZi. We claim that the restriction Z of Zˆ to V is an element of SU(V ). Since Zˆ is an isometry for the extension fˆ of f, this will hold as long as im(Z) ⊆ V . Now, the matrix for Z in the basis {b0, b1, b2} is

  ζ 0 0   −1  2  A 0 ζq 0  A,     0 0 ζq4

where A and A−1 are as in equations 4.1 and 4.2, which is easily seen to have entries from F . Consequently, im(Z) ⊆ V and Z ∈ SU(V ). Since Tˆ ∈ hZˆi, we also have T ∈ hZi. If σ ∈ G is the collineation of PG(V ) induced by Z, then τ ∈ hσi and

ˆ q2i q2−q+1 σ(Pi) = [Z(ei)] = [ζ ei] = Pi for i = 0, 1, 2. It remains to prove that |σ| = gcd(3,q+1) . However this follows from the facts that |Z| = |Zˆ| = q2 − q + 1 and that if 3 divides q + 1 then ζ(q2−q+1)/3 = ζq2(q2−q+1)/3 = ζq4(q2−q+1)/3.

Lemma 4.2.17. The stabilizer of ∆τ in G is the normalizer of a Sylow r-subgroup of G, where r is the largest prime divisor of q2 − q + 1.

Proof. Let H be the full (setwise) stabilizer of ∆τ in G, and let Z and σ be as in the previous proof. Consider the linear transformation M ∈ SU(Vˆ deﬁned by

M(e0) = e1, M(e1) = e2, and M(e2) = e0. Now, the matrix for M in the basis

57 {b0, b1, b2} is   0 0 1   −1   A 1 0 0 A,     0 1 0 where A and A−1 are as in equations 4.1 and 4.2. A straightforward calculation shows that the entries of this matrix are in F . Consequently, M maps V to V and so the restriction of M to V is an element of SU(V ). Consequently, the collineation

µ induced by M|V is an element of G. Now, µ(P0) = M([e0]) = [M(e0)] = [e1] = P1.

Similarly, we have µ(P1) = P2, and µ(P2) = P0. Consequently, µ ∈ H. Note that H has order 3(q2 − q + 1)/ gcd(3, q + 1) by Theorem 4.1.1 if q is odd or by Theorem 4.1.2 is q is even. Therefore, H = hσ, µi. Now, M −1ZMˆ = Zˆq2 , so µ−1σµ = σq2 and hence hσi is normal in H. Note that hσi contains a unique Sylow r-subgroup R of G. Then, since R is a characteristic subgroup of hσi and hσi is normal in H, R is also normal in H. On the other hand, suppose that ν ∈ G normalizes R. Let ρ be a

k generator for R. Then for some integer k, ρν(Pi) = νρ (Pi) = ν(Pi), for i = 0, 1, 2. ˆ So ρ ﬁxes each of the points ν(Pi), but since the only points of PG(V ) ﬁxed by ρ are

P0, P1, and P2, it follows that ν permutes these and so ν ∈ H.

Since the stabilizers in G of the imaginary triangles ﬁxed by the elements which do not ﬁx any points of PG(V ) are the normalizers of the Sylow r-subgroups of G they are all conjugate. We will now show that these stabilizers are the unique maximal subgroups of G which contain the elements of order (q2 − q + 1)/ gcd(3, q + 1).

k 2 Lemma 4.2.18. Suppose q = q0 , where q0 ≥ 2, k ≥ 3, and k is odd. Then q − q + 1

3 3 2 does not divide 3q0(q0 + 1)(q0 − 1).

Proof. We will consider several cases.

Case 1: k ≥ 5

2 3 3 2 2k k 8 5 3 Consider the quantity q −q+1−3q0(q0 +1)(q0 −1) = q0 −q0 −3q0 +3q0(q0−1)+3q0 +1. 58 2k max(k,8) 2k k 8 5 3 Since k ≥ 5, q0 ≥ 4q0 and so q0 −q0 −3q0 ≥ 0. Since 3q0(q0 −1)+3q0 +1 > 0,

2 3 3 2 it follows that q − q + 1 > 3q0(q0 + 1)(q0 − 1) and the lemma holds in this case.

Case 2: k = 3 and q0 > 5

3 3 2 In this case the desired result follows from the fact that 3q0(q0 + 1)(q0 − 1) =

2 6 3 3 2 3 2 6 3 3(q0 − 1)(q0 − q + 1) + 3(2q0 − 1)(q0 − 1) and 0 < 3(2q0 − 1)(q0 − 1) < q0 − q0 + 1.

The lemma is easily checked to hold in the remaining cases, i.e. when k = 3 and

2 ≤ q0 ≤ 5.

Theorem 4.2.19. For q ≥ 7, each element of G of order (q2 − q + 1)/ gcd(3, q + 1) is contained in a unique maximal subgroup.

Proof. Let σ ∈ G be an element of order (q2 − q + 1)/ gcd(3, q + 1). The stabilizers of the absolute and nonabsolute points of PG(V ) have orders q3(q2 − 1)/ gcd(3, q + 1) and q(q + 1)2(q − 1)/ gcd(3, q + 1). Since (q2 − q + 1)/ gcd(3, q + 1) does not divide either of these σ is not contained in a stabilizer of either type. Consequently, σ is contained in the stabilizer of a unique imaginary triangle. We will prove that this is the only maximal subgroup containing σ. Note that (q2 −q+1)/ gcd(3, q+1) does not divide 6(q + 1)2/ gcd(3, q + 1) or q(q + 1)(q − 1), so σ is not contained in the stabilizer of a triangle (in PG(V )) or a conic. By Lemma 4.2.18,(q2 − q + 1)/ gcd(3, q + 1) does

3 3 2 k not divide 3q0(q0 + 1)(q0 − 1)/ gcd(3, q + 1) where q = q0 with k odd and k ≥ 3, so

σ is not contained in a maximal subgroup isomorphic to U3(q0) or P GU(3, q0). If q is even, this completes the proof. If q is odd, we must show that σ is not contained in any of the subgroups from Mitchell’s list of order 36, 72, 168, 216, 360, 720, or 2520.

Observe that if q ≥ 88, then (q2 − q + 1)/3 > 2520. Consequently, σ cannot be contained in a subgroup with order in {36, 72, 168, 216, 360, 720, 2520} in this case. So we turn our attention to the case 7 ≤ q < 88. Then q is not of the form 5k with 59 k odd, so there will be no subgroups of order 720 or 2520. Now, if q ≥ 34, then (q2 − q + 1)/3 > 360, so the theorem also holds when 34 ≤ q < 88. It is easy to see that (q2 − q + 1)/ gcd(3, q + 1) does not divide any of 36, 72, 168, 216, or 360 in the remaining cases, i.e. when q ∈ {7, 9, 11, 13, 17, 19, 23, 27, 29, 31}, and this completes the proof.

4.3 BOUNDS ON THE COVERING NUMBER

We are now ready to prove upper and lower bounds for the covering number of U3(q) for q ≥ 7.

Theorem 4.3.1. If q ≥ 7 then σ(G) ≤ q4 + 1 + q3(q + 1)2(q − 1)/3.

Proof. Recall that we have partitioned the elements of G into three types:

1. Type 1: Elements which ﬁx an absolute point of PG(V ).

2. Type 2: Elements which do not fix an absolute point of PG(V ) but do fix a nonabsolute point. Elements of this type were show to fix exactly 3 nonabsolute points which form a self-polar triangle.

3. Type 3: Elements which do not ﬁx any points of PG(V ). Elements of this type

were shown to ﬁx a unique imaginary triangle in PG(Vˆ ) ∼= PG(2, q6).

Let P be an absolute point of PG(V ), and let S be the set consisting of the stabilizers in U3(q) of all of the nonabsolute points of PG(V ) which are not on the line P ⊥. Note that |S| = q4 − q3 and every element of type 2 is contained in at least one of the members of S. Let S0 be the set consisting of the stabilizers of the q3 + 1 isotropic points of PG(2, q2), as well as the stabilizers of the q3(q + 1)2(q − 1)/3 imaginary triangles which are ﬁxed by the elements of type 3. Let C = S ∪ S0. Then |C| = q4 + 1 + q3(q + 1)2(q − 1)/3 and, in light of the classiﬁcation of the elements of G described above, it is clear that C is a cover of G. 60 We shall see that we can improve this upper bound a bit, but we cannot do very much better.

Lemma 4.3.2. If q ≥ 7 and C is any cover of G consisting of maximal subgroups, then C contains the q3(q + 1)2(q − 1)/3 stabilizers of the imaginary triangles ﬁxed by the elements of type 3 in G.

Proof. By Theorem 4.2.19, each element of order (q2 −q+1)/ gcd(3, q+1) is contained in the stabilizer of a unique imaginary triangle and in no other maximal subgroups of G. Consequently, since C covers all of these elements, C must contain the stabilizers of all of the q3(q + 1)2(q − 1)/3 imaginary triangles ﬁxed by these elements.

Theorem 4.3.3. If q ≥ 7 then 1 + q3(q + 1)2(q − 1)/3 ≤ σ(G). Moreover, if q is not a power of 3, then q3 + 1 + q3(q + 1)2(q − 1)/3 ≤ σ(G)

Proof. Let C be a minimal cover for G. Without loss of generality we can assume that C consists of maximal subgroups of G. By Lemma 4.3.2, C must contain the stabilizers of all q3(q + 1)2(q − 1)/3 imaginary triangles ﬁxed by the elements of G that ﬁx no point of PG(V ). Since these subgroups do not contain, for example, any of the involutions in G, we must have |C| > q3(q + 1)2(q − 1)/3.

For the remainder of the proof suppose that p 6= 3. If p = 2, let Ω be the set of elements of order 2 and 4 in G, and if p ≥ 5, let Ω be the set of all elements of order p in G. Also, let X1 be the set of stabilizers in G of the absolute points of PG(V ),

X2 be the set of stabilizers of the nonabsolute points, X3 be the set consisting of the stabilizers of the triangles in PG(V ), X4 be the set of the stabilizers of the imaginary triangles ﬁxed by the elements of type 3 in G, and X5 be the set of subgroups of G

k isomorphic to U3(q0) or P GU3(q0) where q = q0 with k odd and k ≥ 3. If p ≥ 5 let

X6 be the set of stabilizers of conics in G and X7 be the set of maximal subgroups of 61 G which are isomorphic to any of L3(2), A6, A6.2, A7, or the Hessian groups of order

216, 72, or 36. If p = 2, set X6 = X7 = ∅. Finally, for 1 ≤ i ≤ 7 let xi = |Xi ∩ C| and let ai = max|H ∩ Ω| if Xi 6= ∅ and ai = 0 otherwise. According to Theorems 4.1.1 H∈Xi 7 [ and 4.1.2 Xi contains all of the maximal subgroups of G. i=1

3 Now, every subgroup in X1 has exactly q − 1 elements of Ω, and every subgroup

2 from X2 has exactly q − 1 elements of Ω all of which are elations. Consequently

3 2 a1 = q − 1 and a2 = q − 1. The stabilizer of a triangle contains no elements of

2 Ω if p ≥ 5, in which case a3 = 0. If p = 2, 0 ≤ a3 < 6(q + 1) / gcd(3, q + 1) and

3 q ≥ 8. In both cases a3 < q − 1. Since we are assuming that p 6= 3, the stabilizer of an imaginary triangle contains no elements of Ω and a4 = 0. A subgroup

6 k of G isomorphic to U3(q0) of P GU3(q0) has q0 − 1 elements of Ω, but since q = q0

6 3 3 with k ≥ 3, q0 − 1 < q − 1, and we conclude that a5 < q − 1. If p = 2 then

2 a6 = a7 = 0. So suppose that p ≥ 5. The stabilizer of a conic has order q(q − 1),

2 3 3 so a6 ≤ (q − 1)(q − 1) < q − 1. Now since q ≥ 7, q − 1 ≥ 342, and the only subgroup in X7 with more than 342 elements of order p is A7 in the case that p = 5 which contains 504 elements of order 5. But if p = 5 and q ≥ 7, then in fact q ≥ 25. So 504 < 15624 ≤ q3 − 1 in this case. Consequently, we conclude that in all cases,

3 ai < q − 1 = a1 for 2 ≤ i ≤ 7.

7 [ 6 Since C is contained in Xi, C covers the q − 1 elements of Ω, and a4 = 0, we i=1 must have 7 X 6 aixi ≥ q − 1 . i=1 i6=4 3 Furthermore, since ai < q − 1 = a1 for 2 ≤ i ≤ 7, it follows that

7 7 3 X X 6 (q − 1) xi ≥ aixi ≥ q − 1 . i=1 i=1 i6=4 i6=4

62 7 X 3 3 2 Therefore xi ≥ q + 1, and since by Lemma 4.3.2 x4 = q (q + 1) (q − 1)/3, we i=1 i6=4 conclude that

7 X 3 3 2 σ(G) = |C| = xi ≥ q + 1 + q (q + 1) (q − 1)/3 . i=1

In light of Theorems 4.3.1 and 4.3.3 it follows immediately that

q6 Corollary 4.3.4. σ(U3(q)) ∼ 3 as q → ∞.

We can improve the upper bound given above by making use of some known results on polarity graphs.

Deﬁnition 4.3.5. Let Π be a projective plane and ⊥ be a polarity on Π. The polarity graph associated to the pair (Π, ⊥), is a graph G whose vertices are the points of Π, where there is an edge from P to Q if and only if P ∈ Q⊥.

This graph is not simple, since every absolute point has a loop, however we have no interest in these points in the present context and will ignore them. Note that triangles in G correspond to the self-polar triangles in the plane. The following theorem is a combination of Theorems 1.3, 4.7, and Remark 4.6 of [30].

Theorem 4.3.6. Let Π ∼= PG(2, q2), ⊥ be a unitary polarity on Π, and let G be the corresponding polarity graph. Then there exists a set S of m(q) nonabsolute points of Π such that the subgraph of G induced by S is triangle-free, where   4 q /2, if q is a power of 2   m(q) = q3 + 2q2 − 2q − 1, if q is a power of 3    4 kq /p, if q is a power of a prime p such that p = 3k ± 1 ≥ 5.

63 Theorem 4.3.7. If q ≥ 7, then σ(G) ≤ q4 + q2 + 1 − m(q) + q3(q + 1)2(q − 1)/3, where m(q) is as in the previous theorem. In particular,

σ(G) ≤ q4 − q3 − q2 + 2q + 2 + q3(q + 1)2(q − 1)/3.

Proof. Let S be a set as in the previous theorem, and let S0 be the set of all nonabsolute points of PG(V ) which are not in S. Note that |S0| = q4 − q3 + q2 − m(q). Since the subgraph of G induced by S is triangle-free, every self-polar triangle in PG(V ) has at least one vertex in S0. By Lemma 4.2.10, every element of G which ﬁxes a nonabsolute point of PG(V ) but not an absolute point is contained in the stabilizer in G of at least one point of S0. Consequently the set C consisting of

1. the q3 + 1 stabilizers of the absolute points of PG(V ),

2. the q4 − q3 + q2 − m(q) stabilizers of the nonabsolute points from S0, and

3. the stabilizers of the q3(q+1)2(q−1)/3 imaginary triangles ﬁxed by the elements of type 3 in G is a cover of G with |C| = q4 +q2 +1−m(q)+q3(q +1)2(q −1)/3. Note that for q ≥ 7, m(q) ≥ q3 + 2q2 − 2q − 1, so |C| ≤ q4 − q3 − q2 + 2q + 2 + q3(q + 1)2(q − 1)/3.

We end this chapter with the observation that the covering number of SUn(q)

equal to that of of Un(q), and in particular this is true for n = 3, so the upper and lower bounds given above hold for σ(SU3(q)) as well.

Theorem 4.3.8. σ(SUn(q)) = σ(Un(q)).

∼ Proof. If gcd(n, q + 1) = 1, then SUn(q) = Un(q) and there is nothing to prove. In particular this holds when n = 2 and q = 2. Therefore we may assume that gcd(n, q + 1) > 1. For the moment, assume also that (n, q) ∈ {(2, 3), (3, 2)}. Let

C be a minimal cover of SUn(q) consisting of maximal subgroups. We claim that

every maximal subgroup of SUn(q), and consequently every member of C, contains 64 the center Z of SU(V ). Suppose by way of contradiction that there is a maximal subgroup M of SUn(q) which does not contain Z. Then SUn(q) = MZ and M is ∼ normal in SUn(q). Moreover, the quotient SUn(q)/M = Z/M ∩ Z is abelian, and

therefore M contains the commutator subgroup of SUn(q). On the other hand, it

is known that SUn(q) is perfect (i.e. is equal to its own commutator subgroup) unless (n, q) ∈ {(2, 2), (2, 3), (3, 2)}. This yields a contradiction under the stated

assumptions, and the claim is proven. Now, let η : SUn(q) → Un(q) be the canonical

2 map which sends each linear transformation in SUn(q) to the collineation of PG(2, q ) it induces, and let C0 = {η(H): H ∈ C}. Since each H ∈ C is a proper subgroup of

SUn(q) containing Z = ker(η), η(H) is a proper subgroup of Un(q), and since C is

0 0 a cover of SUn(q) and η is a surjection, C is a cover of Un(q). Moreover, |C| = |C |. Consequently,

0 σ(SUn(q)) = |C| = |C | ≥ σ(Un(q)) .

The reverse inequality, σ(SUn(q)) ≤ σ(Un(q)), follows from Lemma 2.2.1. It remains to consider the cases (n, q) = (2, 3) and (n, q) = (3, 2). It follows from the results

in [10] that σ(SU2(3)) = σ(U2(3)) = 5, and SU3(2) has the Klein 4-group as a

homomorphic image, so σ(SU3(2)) = 3 = σ(U3(2)).

65 CHAPTER 5 MCLAUGHLIN GROUP

The McLaughlin group is a subgroup of index two in the full automorphism group of the McLaughlin graph, the unique (up to isomorphism) strongly regular graph with parameters (v, k, λ, µ) = (275, 112, 30, 56).

We note here for later use that the independence number of the McLaughlin graph is 22; this is stated without proof in [7], but can be easily deduced from the relevant parts of [8] as follows: First, it is clear from the construction of the graph given there that the McLaughlin graph has an independent set of size 22. To see that the independence number cannot be greater, we look at the eigenvalues of the graph. These are 112, 2, and −28 with multiplicities 1, 252, and 22 respectively. Since the graph has exactly 22 nonpositive eigenvalues (counted with multiplicity), it follows from the Cvetkovi´cbound (Theorem 3 of [8]) that the independence number can be no larger than 22.

In [21] P. E. Holmes considers the covering number of the McLaughlin group. There she gives a lower bound of 24541 and an upper bound of 24553 for the covering number using computational methods. Here we prove that the covering number of the McLaughlin group is in fact 24553.

Let G be the McLaughlin group. The conjugacy classes of elements of G are given in table 5.1, and the conjugacy classes of maximal subgroups of G are given in table 5.2. Table 5.3 gives the incidence between principal elements and the maximal 66 subgroups of G. We note that the subgroups from classes M1, M6, and M7 are the stabilizers of vertices, edges, and nonedges in the McLaughlin graph respectively.

Those from M2 and M3 are the stabilizers of independent sets of size 22. Also, the subgroups from class M8 are the centralizers of involutions from class 2A, and those from M12 are the normalizers of cyclic subgroups generated by elements from class 5A.

Table 5.1: Conjugacy classes of elements of the McLaughlin group Class Centralizer Order Class Size Principal 1A 898128000 1 No 2A 40320 22275 No 3A 29160 30800 No 3B 972 924000 No 4A 96 9355500 No 5A 750 1197504 No 5B 25 35925120 Yes 6A 360 2494800 No 6B 36 24948000 Yes 7A 14 64125000 No 7B = 7A∗∗ 14 64152000 No 8A 8 112266000 Yes 9A 27 33264000 Yes 9B = 9A∗∗ 27 33264000 Yes 10A 30 29937600 No 11A 11 81648000 Yes 11B = 11A∗∗ 11 81648000 Yes 12A 12 74844000 Yes 14A 14 64152000 Yes 14B = 14A∗∗ 14 64152000 Yes 15A 30 29937600 No 15B = 15A∗∗ 30 29937600 No 30A 30 29937600 Yes 30B = 30A∗∗ 30 29937600 Yes

67 Table 5.2: Conjugacy classes of maximal subgroups of the McLaughlin group

Class Order Index Structure M1 3265920 275 U4(3) M2 443520 2025 M22 M3 443520 2025 M22 M4 126000 7128 U3(5) 1+4 M5 58320 15400 3 : 2.S5 4 M6 58320 15400 Z3 .M10 M7 40320 22275 L3(4) : Z2 M8 40320 22275 Z2.A8 4 M9 40320 22275 Z2 : A7 4 M10 40320 22275 Z2 : A7 M11 7920 113400 M11 1+2 M12 3000 299376 5 : Z3 : Z8

Table 5.3: Incidence matrix for the McLaughlin Group M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12 5B 555305505551 6B 233321676660 8A 111222111124 9AB 200011000000 11AB 011000000000 12A 100032010002 14AB 000000111100 30AB 000010010001

For 1 ≤ i ≤ 12, let θi be the permutation character of G acting on the set of right cosets of Hi ∈ Mi. The ﬁrst four of these are given in [12], and the ﬁrst 11 of them can be found in [1]. Explicitly these are:

68 θ1 = 1a + 22a + 252a

θ2 = 1a + 22a + 252a + 1750a

θ3 = 1a + 22a + 252a + 1750a

θ4 = 1a + 22a + 252a + 1750a + 5103a

θ5 = 1a + 252a + 4500a + 5103a + 5544a

θ6 = 1a + 22a + 2 × 252a + 1750a + 3520a + 4500a + 5103a

θ7 = 1a + 22a + 2 × 252a + 2 × 1750a + 3520a + 5103a + 9625a

θ8 = 1a + 252a + 1750a + 5103a + 5544a + 9625a

θ9 = 1a + 22a + 2 × 252a + 2 × 1750a + 3520a + 5103a + 9625a

θ10 = 1a + 22a + 2 × 252a + 2 × 1750a + 3520a + 5103a + 9625a

θ11 = 1a + 22a + 2 × 252a + 3 × 1750a + 2 × 3520a + 4500a + 2 × 5103a

+ 8250ab + 3 × 9625a + 9856ab + 10395ab.

By Lemma 2.2.11 these characters give the entries in the ﬁrst 11 columns of the inicidence matrix given in table 5.3. We will require a few entries from the last column, which are justiﬁed by the following proposition. The remaining entries have been determined by computation for the sake of completeness but will not be needed here.

Proposition 5.0.9. Elements from 9AB ∪ 11AB ∪ 14AB are not contained in any

subgroups from M12.

Proof. This follows immediately from Corollary 2.1.3.

Proposition 5.0.10. There exists a cover C ⊆ M1 ∪ M2 ∪ M8 with |C| = 24533.

Proof. From table 5.3 we see that that M2 ∪ M8 is suﬃcient to cover all of the elements except those of order nine. These can be covered using subgroups from 69 class M1, so M1 ∪ M2 ∪ M8 is a cover of G consisting of 24575 maximal subgroups.

Recall that subgroups from classes M1, M6, and M7 are the stabilizers in G of the vertices, edges, and nonedges respectively of the McLaughlin graph. From the values of the corresponding permutation characters on the elements of order 9, we see that each ﬁxes exactly two vertices, one edge an no nonedges of the McLaughlin graph. Consequently, the stabilizers of any pair of distinct vertices will have elements of order nine in common if and only if the vertices are adjacent. Since the independence number of the McLauglin graph is 22, there is an independent set I consisting of 22

0 vertices. Letting M1 be the set of the stabilizers in G of those vertices which are not contained in I, it is clear that all of the elements of order nine in G are covered

0 0 by the subgroups of M1, and therefore that C = M1 ∪ M2 ∪ M8 is a cover of G by 24553 maximal subgroups and σ(G) ≤ 24553.

Note that this is the same cover as Holmes uses in [21] to establish her upper bound. We claim that this cover is minimal.

Proposition 5.0.11. For any cover C of G, |C| ≥ 24553.

Proof. Suppose that C is any minimal cover of G. We can assume without loss of

generality that C consists of maximal subgroups. For 1 ≤ i ≤ 12 let xi = |C ∩ Mi|. First consider the elements of order 11 in G. These elements only appear in the

subgroups from M2, M3, and M11. Each subgroup from M2 ∪ M3 contains 80640

elements of order 11, whereas each subgroup from M11 contains only 1440 of them.

Therefore 80640(x2 + x3) + 1440x11 ≥ 163296000, and consequently

x2 + x3 + x11 ≥ 2025 . (5.1)

The elements of order 14 are contained only in the subgroups from classes M7, M8,

M9, and M10, and each subgroup from any of these classes contains exactly 5760

70 elements of order 14. Thus, 5760(x7 + x8 + x9 + x10) ≥ 128304000 and

x7 + x8 + x9 + x10 ≥ 22275 . (5.2)

A similar argument applied to the elements of order nine yields the inequality x1 + x5 + x6 ≥ 138, but this can be improved by considering the way these elements act on the McLauglin graph. Note that each element of order nine in G ﬁxes exactly two vertices, one edge, and no nonedges of the McLaughlin graph, so the two vertices ﬁxed by any element of order nine must be adjacent. Let W be the set of vertices of the

McLaughlin graph whose stabilizers are not in C, and note that x1 + |W | = 275, the number of vertices in the McLaughlin graph. Now, in light of the previous discussion about the elements of order nine, an element x of order nine is not covered by the subgroups in M1 ∩ C if and only if both ends of the unique edge e ﬁxed by x are in W , i.e. if e is an edge of the subgraph induced by W . The only other maximal subgroups that cover elements of order nine are those from M5 and M6, and any subgroup from either of these classes contains exactly 4320 elements of order nine.

Since the subgroups from class M6 are the stabilizers of the edges of the McLaughlin graph, the total number of elements of order 9 left uncovered by M1 ∩C is 4032 times the number n of edges in the subgraph induced by W . Thus we must have at least n subgroups from M5 ∪ M6 to complete the cover. As noted above, the independence number of the McLaughlin graph is 22, which means that the subgraph induced by W can have no independent set of more than 22 vertices, and thus n ≥ |W | − 22.

Then x1 + x5 + x6 ≥ x1 + n ≥ x1 + |W | − 22 = 275 − 22 = 253, and so

x1 + x5 + x6 ≥ 253 . (5.3)

Consequently, from (5.1), (5.2), and (5.3) we must have |C| ≥ 253 + 2025 + 22275 = 24553.

Propositions 5.0.10 and 5.0.11 prove:

Theorem 5.0.12. The covering number of the McLaughlin group is 24553.

71 CHAPTER 6 VERIFICATION OF THE COVERING NUMBERS OF THE HIGMAN-SIMS AND HELD GROUPS

In this chapter we verify the results of [22] on the covering numbers of the Higman- Sims and Held sporadic simple groups.

6.1 HIGMAN-SIMS GROUP

The Higman-Sims group is named for Donald Higman and Charles Sims who discov- ered the group in 1968 as a group of automorphisms of the Higman-Sims graph, the unique strongly regular graph with parameters (v, k.λ, µ) = (100, 22, 0, 6). In 1969 Graham Higman independently constructed the Higman-Sims group as a doubly transitive group on 176 points.

Let G be the Higman-Sims group. We list the conjugacy classes of elements of G in table 6.1, and the conjugacy classes of maximal subgroups of G in table 6.2.

We note that the subgroups from classes M1 and M4 are the stabilizers of vertices and edges of the Higman-Sims graph respectively, and that those from M6 are the stabilizers of nonedges. Also, the subgroups from classes M10 and M11 are the centralizers of involutions from classes 2A and 2B respectively, and those from M12 are the normalizers in G of cyclic subgroups generated by elements from class 5B.

72 Table 6.1: Conjugacy classes of elements of the Higman-Sims group Class Centralizer Order Class Size Principal 1A 44352000 1 No 2A 7680 5775 No 2B 2880 15400 No 3A 360 123200 No 4A 3840 11550 No 4B 256 173250 No 4C 64 693000 No 5A 500 88704 No 5B 300 147840 No 5C 25 1774080 Yes 6A 36 1232000 Yes 6B 24 1848000 No 7A 7 6336000 Yes 8A 16 2772000 Yes 8B 16 2772000 Yes 8C 16 2772000 Yes 10A 20 2217600 No 10B 20 2217600 Yes 11A 11 4032000 Yes 11B = 11A∗∗ 11 4032000 Yes 12A 12 3696000 Yes 15A 15 2956800 Yes 20A 20 2217600 Yes 20B = 20A∗∗ 20 2217600 Yes

73 Table 6.2: Conjugacy classes of maximal subgroups of the Higman- Sims group

Class Order Index Structure M1 443520 100 M22 M2 252000 176 U3(5) : Z2 M3 252000 176 U3(5) : Z2 M4 40320 1100 L3(4) : Z2 M5 40320 1100 S8 4 M6 11520 3850 Z2 .S6 3 M7 10752 4125 Z4 : L3(2) M8 7920 5600 M11 M9 7920 5600 M11 · 4 · M10 7680 5775 Z4 Z2 S5 · 2 M11 2880 15400 Z2 × A6 Z2 M12 1200 36960 Z5 : Z4 × A5

Table 6.3: Incidence matrix for the Higman-Sims Group M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12 5C 511505055000 6A 033253000340 7A 211110200000 8A 200222100300 8B 022200140120 8C 022200104120 10B 022020000031 11AB 100000011000 12A 011011200102 15A 000010000001 20AB 011000000102

For 1 ≤ i ≤ 12, let θi be the permutation character of G acting on the set of right cosets of Hi ∈ Mi. Note that θ2 = θ3. The irreducible constituents of the ﬁrst ﬁve of these are given in [12]. The values of the characters θ1, θ4, θ5, θ7, θ8 θ9 and θ10 on the conjugacy classes of G can be found in [28]. Given in terms of their irreducible constituents these are:

74 θ1 = 1a + 22a + 77a

θ2 = 1a + 175a

θ3 = 1a + 175a

θ4 = 1a + 22a + 77a + 175a + 825a

θ5 = 1a + 77a + 154a + 175a + 693a

θ6 = 1a + 22a + 2 × 77a + 154a + 175a + 693a + 770a + 825a + 1056a

θ7 = 1a + 22a + 77a + 154a + 175a + 693a + 770a + 825a + 1408a

θ8 = 1a + 22a + 77a + 154b + 175a + 770a + 825a + 1056a + 2520a

θ9 = 1a + 22a + 77a + 154c + 175a + 770a + 825a + 1056a + 2520a

θ10 = 1a + 77a + 154a + 175a + 693a + 825a + 1056a + 1386a + 1408a

Using Lemma 2.2.11, these justify the ﬁrst 10 columns of table 6.3. The entries in the last two columns corresponding to elements of order 11, 15, or 20 can be determined from the next two propositions. The remaining entries have been included for the sake of completeness, but we will not need them here.

Proposition 6.1.1. Subgroups from M11 contain no elements from 11AB ∪ 15A ∪ 20AB.

Proof. Note that the subgroups from M11 are the centralizers of the involutions from class 2B. Elements of order 11 or 15 in G do not commute with any involutions so

cannot be contained in subgroups from M11. An element x ∈ 20AB commutes with a unique involution, x10, but this is in class 2A. Consequently, the elements of order

20 are not contained in the members of M10 either.

Proposition 6.1.2. Each subgroup from M12 contains no elements from 11AB, 80 elements from 15A, and 240 elements from 20AB.

75 Proof. (i) The subgroups from M12 have order 1200 which is not divisible by 11.

Consequently subgroups from M12 contain no elements from 11AB.

(ii) Note that the subgroups from M12 are the normalizers of cyclic subgroups generated by the elements from 5B. We claim that for x ∈ 5B and y ∈ 15A,

y ∈ NG(hxi) if and only if hxi ≤ hyi. One direction is obvious. For the other,

suppose that y ∈ NG(hxi). Now, CG(x) E NG(hxi) and has index 4, so reducing

modulo CG(x), y has order dividing 15 and 4, so y = 1 and so y ∈ CG(x). But

then x and y commute, so x ∈ CG(y) = hyi and consequently hxi ≤ hyi.

3 Therefore, for y ∈ 15A, the unique subgroup of M12 containing y is NG(hy i).

From this it follows from Lemma 2.2.10 that each subgroup from M12 contains 80 elements of order 15.

(iii) Let H ∈ M12. Then H is a direct product of normal subgroups N and K, where ∼ ∼ N = Z5 : Z4 and K = A5. Note that an element of order 5 which commutes with an element of order 4 must be the fourth power of an element of order 20, so in G only elements of class 5A commute with elements of order 4. Thus, all 24 elements of order 5 in K are from class 5A, whereas the four elements of order 5 from N are in class 5B.

We claim that no element of H of the form yz, where y ∈ N and z ∈ K both have order 5, commutes with an element of order 4 in H. Suppose that w ∈ H has order 4 and commutes with yz. We may write w = nk, where n ∈ N and k ∈ K. Now, since n commutes with z and k commutes with n−1yn, w−1yzw = yz implies that

y−1n−1yn = zk−1z−1k ∈ N ∩ K = {1}.

So k ∈ CK (z) = hzi. Since w has order 4, we must then have k = 1. Hence w = n, which commutes with z but not y, as y is in class 5B. Therefore w does 76 not commute with yz. This is a conradiction, and the claim is proven. Thus the only elements of order 5 in H which commute with an element of order 4 in H are the 24 elements of order 5 in K.

Now N has 10 elements of order 4, and K has none. Consequently H has 160 elements of order 4, each of the form nk, where n ∈ N has order 4 and k ∈ K has order 1 or 2. However, only those elements of order four from N commute with elements of order five in H, and each of these commutes with all 24 elements of order five in K. Consequently there are 240 pairs (n, k), where n, k ∈ H, such that n has order four, k has order 5, and nk = kn. The map from the set of these pair into the set of elements of order 20 in H defined by (n, k) 7→ nk is easily seen to be a bijection, so H contains exactly 240 elements of order 20.

Proposition 6.1.3. C = M1 ∪M2 ∪M5 is a cover of G by 1376 maximal subgroups.

Proof. This is clear in light of table 6.3.

We will prove that this cover is minimal.

Proposition 6.1.4. For any cover C of G, |C| ≥ 1376.

Proof. By Lemma 2.2.9 we may assume that C contains only maximal subgroups of

G. For 1 ≤ i ≤ 12 let xi = |C ∩ Mi|.

First, The only maximal subgroups containing elements of order 11 are those isomor-

phic to M11 or M22. Those isomorphic to M11 contain 1440 elements of order 11,

and those isomorphic to M22 contain 80640 of them. Consequently we must have

77 80640x1 + 1440(x8 + x9) ≥ 8064000, which implies that

x1 + x8 + x9 ≥ 100 . (6.1)

Consider the elements of order 15 in G. From table 6.3 we see that these elements appear only in the maximal subgroups from M5 and M12. By Proposition 6.1.2 each subgroup from M12 contains 80 elements of order 15, whereas each subgroup from 8 class M5, being isomorphic to S8, has 3 · 4! · 2! = 2688 elements of order 15. Thus we must have 2688x5 + 80x12 ≥ 2956800 and therefore:

168x5 + 5x12 ≥ 184800 . (6.2)

Finally, we turn our attention to the elements of order 20. From table 6.3, we know that of the only classes of maximal subgroups containing elements of order 20 are M2,

M3, M10, and M12. Since each element of order 20 is contained in a unique member of each of M2, M3, and M11, by Lemma 2.2.10 it follows that each subgroup from

M2 ∪ M3 contains exactly 25200 elements of order 20, while those from M10 each contain 768 elements of order 20. By Proposition 6.1.2, each subgroup from M12 contains exactly 240 elements of order 20. Thus, Since C covers all of the elements of order 20 in G we have 25200(x2 + x3) + 768x10 + 240x12 ≥ 4435200 and hence:

525(x2 + x3) + 16x10 + 5x12 ≥ 92400 . (6.3)

Now, multiplying (6.2) by 25 and (6.3) by 8 and adding the results yields 4200(x2 + x3 + x5) + 128x10 + 165x12 ≥ 5359200, from which it follows that

x2 + x3 + x5 + x10 + x12 ≥ 1276 . (6.4)

12 X Combining (6.1) and (6.4) yields xi ≥ 1376. i=1 Consequently we have:

Theorem 6.1.5. The covering number of the Higman-Sims group is 1376. 78 6.2 HELD GROUP

The Held group is named for Dieter Held who in 1969 predicted the existence of a new simple group that has a centralizer of an involution which is isomorphic to the

centralizer of an involution in M24 and in L5(2). Held was able to deduce a number of properties about this group such as its order and conjugacy classes. The Held group was then constructed by Graham Higman and John McKay.

Let G be the Held group. We list the conjugacy classes of maximal subgroups of G in table 6.5, and the conjugacy classes of elements of G in table 6.4. In the latter, we include the cycle types of the representatives of each conjugacy class acting on

2058 points as given in [1]. We note that the subgroups from class M5 are centralizers of involutions from class 2B, those from M7 are normalizers of cyclic subgroups generated by elements from class 3A, and the subgroups in classes M8 and M10 are the normalizers of cyclic subgroups of order 7 generated by elements from 7C and 7AB respectively.

79 Table 6.4: Conjugacy Classes of Elements of the Held Group Class Centralizer Order Class Size Principal Cycle Type 1A 4030387200 1 No 12058 2A 161280 24990 No 11542952 2B 21504 187425 No 14221008 3A 7560 533120 No 1423672 3B 504 7996800 No 3686 4A 672 5997600 No 1142704476 4B 384 10495800 No 1182124504 4C 128 31487400 No 122204504 5A 300 13434624 No 185410 6A 72 55977600 No 1102163486312 6B 24 167932800 No 3146336 7A 1176 3427200 No 7294 7B = 7A∗∗ 1176 3427200 No 7294 7C 1029 3916800 No 7294 7D 98 41126400 No 7294 7E = 7D∗∗ 98 41126400 No 7294 8A 16 251899200 Yes 124108252 10A 20 201519360 Yes 142553010190 12A 12 335865600 Yes 1224344862212156 12B 12 335865600 Yes 366412168 14A 56 71971200 No 72214136 14B = 14A∗∗ 56 71971200 No 72214136 14C 14 287884800 Yes 7614144 14D = 14C∗∗ 14 287884800 Yes 7614144 15A 15 268692480 Yes 12325815134 17A 17 237081600 Yes 1117121 17B = 17A∗ 17 237081600 Yes 1117121 21A 21 191923200 Yes 762196 21B = 21A∗ 21 191923200 Yes 762196 21C 21 191923200 Yes 2198 21D = 21C∗∗ 21 191923200 Yes 2198 28A 28 143942400 Yes 7214102868 28B = 28A∗∗ 28 143942400 Yes 7214102868

80 Table 6.5: Conjugacy Classes of Maximal Subgroups of the Held Group

Class Order Index Structure M1 1958400 2058 S4(4) : Z2 2 · M2 483840 8330 Z2 L3(4).S3 6 · M3 138240 29155 Z2 : 3 S6 6 · M4 138240 29155 Z2 : 3 S6 1+6 M5 21504 187425 2+ .L3(2) 2 M6 16464 244800 Z7 : 2L2(7) · M7 15120 266560 Z3 S7 1+2 M8 6174 652800 7+ :(S3 × Z3) M9 4032 999600 S4 × L3(2) M10 3528 1142400 Z7 : Z3 × L3(2) 2 M11 1200 3358656 Z5 : 4A4

Table 6.6: Incidence matrix for the Held Group M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 8A 22111400000 10A 41110020002 12A 21220020120 12B 00111000104 14CD 00001101000 15A 22220010000 17AB 10000000000 21AB 00000011020 21CD 01000002110 28AB 01000000110

For 1 ≤ i ≤ 11, let θi be the permutation character of G acting on the right cosets of Hi ∈ Mi. Note that θ4 = θ3. The irreducible constituents of θ1 are given in [12], however those given there for θ2 are known to be incorrect; θ2 is as given below, not 1a + 51ab + 680a + 1275a + 6272a (see for example [1], [4], or [6]). The irreducible constituents of θ3 = θ4 are given in [1], and those for θ5 and θ6 are given

81 in [4]. Explicitly we have:

θ1 = 1a + 51ab + 680a + 1275a

θ2 = 1a + 51ab + 680a + 1275a + 1920a + 4352a

θ3 = 1a + 51ab + 680a + 2 × 1275a + 1920a + 4352a + 7650a + 11900a

θ4 = 1a + 51ab + 680a + 2 × 1275a + 1920a + 4352a + 7650a + 11900a

θ5 = 1a + 680a + 2 × 1275a + 1920a + 4080a + 4352a + 6528a + 2 × 7650a

+ 10880a + 3 × 11900a + 2 × 13720a + 2 × 17439a + 21504ab

θ6 = 1a + 153ab + 1920a + 4080a + 4352a + 2 × 6272a + 6528a + 2 × 7650a

+ 2 × 10880a + 11475ab + 2 × 13720a + 2 × 14400a + 2 × 17493a + 20825a

+ 21504ab

From these we obtain the entries in the ﬁrst six columns of table 6.6. Many of the remaining entries can be deduced from the next few propositions by Lemma 2.2.10. Those few which are not handled by these propositions will not be needed for our present purposes and have been included only for the sake of completeness.

Proposition 6.2.1. Each subgroup from class M7 contains exactly 1512 elements from 10A, 1008 elements from 15A, 720 elements from each of 21A and 21B, and no elements from 8A ∪ 12B ∪ 14CD ∪ 17AB ∪ 21CD ∪ 28AB.

∼ · Proof. Let H ∈ M7. Then H = Z3 S7 and H = NG(hxi) for some x ∈ 3A. Note that CG(x) is a normal subgroup of H of index 2, so any element of H of odd order is in CG(x) and commutes with x.

(i) By Corollary 2.1.3, H cannot contain any elements of order 8, 14, 17, or 28, so H ∩ (8A ∪ 14CD ∪ 17AB ∪ 28AB) = ∅.

(ii) On the other hand, for p ∈ {5, 7} p divides |H|, so by Cauchy’s Theorem H has an element y of order p. As noted above, since y has odd order, y commutes 82 with x. Thus, xy ∈ H ∩ 15A if p = 5. If p = 7, then since (xy)7 = x ∈ 3A, so xy ∈ 21AB and y = (xy)15 ∈ 7C. Consequently we conclude that H ∩ 15A and H ∩ 21AB are nonempty, and H ∩ 7ABDE = H ∩ 21CD = ∅.

(iii) Now let z ∈ H ∩ (15A ∪ 21AB). Then z has odd order and commutes with

x, so hxi ≤ CG(z) = hzi. Consequently, the natural projection homomorphism H → H/hxi induces a one-to-one correspondence hzi 7→ hzi/hxi between the cyclic subgroups of order 3p in H and the Sylow p-subgroups of H/hxi, where

3 ∼ p = |z | ∈ {5, 7}. Now H/hxi = S7 which has 126 Sylow 5-subgroups and 120 Sylow 7-subgroups. Thus H has 126 cyclic subgroups of order 15, each with 8 generators from 15A, and 120 cyclic subgroups of order 21, each with 6 generators from 21A and 6 from 21B. Then |H ∩ 15A| = 1008, and |H ∩ 21A| = |H ∩ 21B| = 720.

(iv) Consider the natural projection homomorphism η : H → H/hxi. Since H/hxi ∼=

S7, H/hxi has 504 elements of order 10. Let a ∈ H/hxi be an element of order 10. Since η is a surjection, there is b ∈ H with η(b) = a. Then 10 divides |b|, and since G has no elements of order divisible by but greater than 10, we must have |b| = 10. Conversely, if b ∈ H has order 10, then hbi ∩ hxi = 1 so that |η(b)| = 10. Now, η is a 3-1 mapping, so H has three times as many elements

of order 10 as S7 does. Thus |H ∩ 10A| = 1512.

(v) We note that elements from 3A are fourth powers of those from class 12A, so

4 there is v ∈ 12A with v = x. Then v ∈ CG(x) ≤ H, so H ∩ 12A 6= ∅. Suppose that t ∈ H has order 12. Then hti acts as a group of automorphisms of hxi with

2 kernel Chti(x). Now, Aut(hxi) has order two, so t commutes with x. Suppose

2 2 by way of contradiction that t ∈ 12B so that t ∈ 6B. Then, |CG(t )| = 24 and

2 4 [CG(t ): hti] = 2. Now, ht i is the unique Sylow 3-subgroup of hti, which is a

2 4 normal subgroup of CG(t ). It follows that ht i is the unique Sylow 3-subgroup

83 2 4 8 2 of CG(t ), and both t , t ∈ 3B. But x ∈ 3A commutes with t , so hxi is also a

2 Sylow 3-subgroup of Cg(t ), a contradiction. Thus H ∩ 12B = ∅.

Proposition 6.2.2. Each subgroup from class M8 contains exactly 441 elements from each of 14C and 14D, 294 elements from each of 21A and 21B, 588 elements from each of 21C and 21D, and no elements from 8A ∪ 10A ∪ 12AB ∪ 15A ∪ 17AB ∪ 28AB.

∼ 1+2 Proof. Let H ∈ M8. Then there is P ∈ Syl7(G) such that H = NG(P ) = 7+ :

(S3 × Z3). Note that Z(P ) = hxi, where x ∈ 7C and H = NG(hxi).

(i) Corollary 2.1.3 implies that H contains no elements of order 8, 10, 12, 15, 17, or 28.

(ii) From [5] we know that each of 7A, 7B, 7D, and 7E intersects H in a single H-conjugacy class, while H ∩ 7C is a union of two H-conjugacy classes, and

that for y ∈ H ∩ 7DE we have CH (y) = CG(y). From this we may deduce that |H ∩ 7D| = |H ∩ 7E| = 63 and that for z ∈ 14CD, z ∈ H if and only if z2 ∈ H ∩ 7DE. Now for y ∈ 7DE, there are exactly seven z ∈ 14CD such that z2 = y, so we conclude that |H ∩ 14C| = |H ∩ 14D| = 441.

(iii) As noted above, H ∩7C is the union of two H-conjugacy classes, one of which is hxi\1. Let w be a representative of the other one. Suppose that v ∈ H ∩21AB.

3 3 Then v is conjugate to x or w in H. From [5], |CH (w)| = 49 whereas |CH (v )| is divisible by 3. Thus v3 is conjugate to x in H, i.e. v3 ∈ hxi. Conversely,

3 for any v ∈ 21AB such that v ∈ hxi, v ∈ CG(x) ≤ H. Consequently, the only elements from 21AB in H are the 588 such elements for which v3 ∈ hxi. Necessarily, half of these are from 21A and the other half from 21B.

(iv) Consider u ∈ H ∩ 7AB, and let C = CH (u). From [5] we have |C| = 147, so by Cauchy’s Theorem C contains an element t of order 3. Then tu ∈ C ∩ 21CD. 84 Now, C has a unique Sylow 7-subgroup N = hu, xi which is normal in C. Since t ∈ H, t normalizes hxi, but t cannot commute with x, for then we would ∼ have x ∈ CG(tu) = htui. Consequently, C = huihx, ti = Z7 × (Z7 : Z3). From this it is clear that C has exactly seven Sylow 3-subgroups, each cyclic of order 3, and therefore that there are exactly 14 elements of order 3 in H which commute with u. Moreover, by [5] we know that the G-conjugacy class of u intersects H in a single H-conjugacy class. We can therefore deduce that |H ∩ 7A| = |H ∩ 7B| = 42. Thus there are 1176 pairs (u, t) such that u, t ∈ H, u ∈ 7AB, |t| = 3, and ut = tu. Let S be the set of all such pairs. Then the map S → H ∩ 21CD given by (u, t) 7→ ut is a bijection, which can be seen as follows: First, for r ∈ H ∩ 21CD, (r15, r7) 7→ r, proving that the map is surjective.

Now, if u1t1 = u2t2, where (ui, ti) ∈ S for i = 1, 2 then t1 ∈ CG(u2t2) = hu2t2i

−1 which has exactly two elements of order three. Consequently, t1 ∈ {t2, t2 }. If

−1 2 2 t1 = t2 then u1 = t2u2, which is impossible since |u1| = 7 but |t2u2| = 21.

Therefore t1 = t2, and it follows immediately that u1 = u2. We conclude that |H ∩ 21C| = |H ∩ 21D| = 588.

Proposition 6.2.3. Each subgroup from class M9 contains exactly 336 elements from each of 12A and 12B, 192 elements from each of 21C and 21D, 144 elements from each of 28A and 28B, and no elements from 8A ∪ 10A ∪ 14CD ∪ 15A ∪ 17AB ∪ 21AB.

∼ Proof. Let H ∈ M9. Then H = S4 × L3(2), so there are normal subgroups K and L ∼ ∼ of H such that K = S4, L = L3(2), H = KL, and K ∩ L = 1.

(i) By order considerations H has no elements of order 10, 15, or 17.

(ii) Since L E H and has order divisible by seven, L contains all of the Sylow 7- subgroups of H, and consequently all of the elements of order seven in H. Let x ∈ L have order seven and y ∈ K have order four. Then x and y commute, 85 so xy ∈ 28AB, x ∈ 7AB, and y ∈ 4A. Moreover, H ∩ 7CDE = ∅ from which it follows that H also has no elements from 14CD or 21AB.

(iii) As seen above H ∩ 28AB 6= ∅, so let z ∈ H ∩ 28AB. Then z can be uniquely ∼ written as z = kl, where k ∈ K and l ∈ L. Now k ∈ K = S4 has order ∼ dividing 28, so |y| ∈ {1, 2, 4}. Similarly, l ∈ L = L3(2) has order dividing 28, so |l| ∈ {1, 2, 4, 7}. In order for kl to have order 28, the only possibility is |k| = 4

and |l| = 7. Now S4 has six elements of order four, and L3(2) has 48 elements of order 7, so H has 288 elements of order 28, half from 28A and the other half from 28B.

(iv) Now, K has elements of order three, which commute with the elements of order seven in L. Thus H has elements of order 21, which are necessarily from 21CD as noted previously. A similar line or reasoning to that given above for the elements of order 28 yields |H ∩ 21CD| = 384, from which it follows that |H ∩ 21C| = |H ∩ 21D| = 192.

(v) We claim that H has no elements of order eight. Suppose by way of contra-

diction that w ∈ H has order eight. Then we can uniquely write w = w1w2,

where w1 ∈ K and w2 ∈ L. Both of these must have order dividing eight, but

as neither S4 nor L3(2) has elements of order eight, |wi| ∈ {1, 2, 4} for i = 1, 2.

But then 8 = |w| divides the least common multiple of |w1| and |w2|, which is at most four. This contradiction proves the claim.

(vi) There are elements of order 12 from both classes 12A and 12B in H, which can

be seen as follows: Let u1 and u2 ∈ K and v1 and v2 ∈ L with |u1| = |v2| = 3

and |u2| = |v1| = 4. Then both u1v1 and u2v2 have order 12, but u2 commutes

with the elements of order seven in L and so |CG(u2)| is divisible by seven.

From this we deduce that u2 ∈ 4A, and so u2v2 ∈ 12A. Now, from [12] we know ∼ that L contains involutions of class 2B. Since L = L3(2) has a a unique class of 86 2 involutions, it follows that v1 ∈ 2B, so v1 ∈ 4BC. However, since v1 commutes

with an element u1 of order 3, we must have v1 ∈ 4B and u1v1 ∈ 12B. Now, K has eight elements of order three and six elements of order four, and L has 56 elements of order three and 42 elements of order four, from which we can deduce that H contains 336 elements from each of 12A and 12B.

Proposition 6.2.4. Each subgroup from class M10 contains exactly 588 elements from 12A, 336 elements from each of 21A and 21B, 168 elements from each of 21C and 21D, 126 elements from each of 28A and 28B, and no elements from 8A ∪ 10A ∪ 12B ∪ 14CD ∪ 15A ∪ 17AB.

∼ Proof. Let H ∈ M10. Then there is x ∈ 7A such that H = NG(hxi) and H = Z7 :

Z3 × L3(2). We may write H = NL, where N is a normal subgroup of H isomorphic to Z7 : Z3, L is a normal subgroup isomorphic to L3(2), and N ∩ L = 1. Note that

CG(x) = hxiL.

(i) By Corollary 2.1.3 H has no elements of order 10, 15, or 17.

(ii) It is easy to see that H has no elements of order eight as follows: Since L E H and [H : L] is relatively prime to two, L contains all of the Sylow 2-subgroups ∼ of H. Since L = L3(2) has no elements of order eight, neither does H.

(iii) Each element from H of order 4 is contained in L and commutes with x, which implies that each of these elements is from class 4A. Now the third powers of elements from 12B are in class 4B, so since H contains no elements from 4B, it follows that H contains no elements from 12B either.

(iv) From [5] we know there are three conjugacy classes of elementary abelian sub-

groups of G of order 49, with representatives E1, E2, and E3. Moreover, of

these only E1 contains elements from 7AB, and |E1 ∩ 7AB| = 42, |E1 ∩ 7C| = 6, 87 and E1 ∩ 7DE = ∅. Let P be a Sylow 7-subgroup of H. Then P is elementary

abelian of order 49, and x ∈ P ∩7A, so P is conjugate to E1 in G. Consequently, H ∩ 7CD = ∅, from which it follows that H ∩ 14CD = ∅.

(v) Let yi ∈ N and zi ∈ L for i = 1, 2 with |y1| = |z2| = 7 and |y2| = |z1| = 3. Let

wi = yizi for i = 1, 2. Then |w1| = |w2| = 21, and it is easy to see that every

element of order 21 in H is of one of these two types. Let us ﬁrst consider w1.

Since hxi is the unique Sylow 7-subgroup of N and y1 has order seven, y1 ∈ hxi

and so must be from 7AB, and consequently w1 ∈ 21CD. We will show that

w2 ∈ 21AB. First, note that L has two conjugacy classes of elements of order seven, but only one conjugacy class of cyclic subgroups of order seven, so either all of the elements of order seven in L are from 7C, or half are from 7A and the ∼ other half are from 7B. Since L ∩ 7DE = ∅ and NG(L) = H = Z7 : Z3 × L2(7) it follows that L is from the conjugacy class of subgroups from part (ii) of Theorem 23 in [5]. Consequently, all of the elements of order seven in L are

from 7C. In particular, z2 ∈ 7C, so w2 ∈ 21AB. Now, N has six elements of order seven, and 14 elements of order three, while L has 56 elements of order three and 48 elements of order seven. Thus H has 336 elements from 21CD, and 672 elements from 21AB.

(vi) H has elements of order 28 since L contains elements of order 4 which commute with x. Suppose that t ∈ H has order 28, and write t = uv where u ∈ N and v ∈ L. By considering the orders of the elements in N and L one can easily see that we must have |u| = 7 and |v| = 4. On the other hand, for any u ∈ N of order 7 and v ∈ L of order 4, uv has order 28, and it is not hard to see that the map (u, v) 7→ uv is a one-to-one correspondence between pairs (u, v) such that u ∈ N has order 7 and v ∈ L has order 4 and the elements of order 28 in H. Since N has 6 elements of order 7 and L has 42 elements of order 4, we conclude that H has 252 elements of order 28, from which it follows that 88 |H ∩ 28A| = |H ∩ 28B| = 126.

Proposition 6.2.5. The subgroups from class M11 contain no elements from 14CD∪ 17AB ∪ 21ABCD ∪ 28AB.

Proof. This follows immediately from Corollary 2.1.3 as the subgroups from M11 have order 1200.

We are now ready to construct a minimal cover of G.

Proposition 6.2.6. C = M1 ∪ M2 ∪ M5 ∪ M7 is a cover of G by 464373 subgroups.

Proof. This is clear from table 6.6.

Proposition 6.2.7. For any cover C of G, |C| ≥ 464373.

Proof. We may assume without loss of generality that C is a cover of G consisting of only maximal subgroups. For 1 ≤ i ≤ 11 let xi = |C ∩ Mi|. Then, since each element of order 17 is contained in a unique maximal subgroup of G which is from class M1, we must have M1 ⊆ C, i.e.

x1 = 2058. (6.5)

Consider the elements from 14C (or 14D). Subgroups from classes M5, M6, and M8 contain 1536, 1176, and 441 elements from 14C respectively, and none of the maximal subgroups from the other classes contain any elements from 14C. Since C covers all of the elements from 14C, we have

1536x5 + 1176x6 + 441x8 ≥ 287884800, or equivalently,

512x5 + 392x6 + 147x8 ≥ 95961600. (6.6)

89 Similar reasoning applied for classes 21A (or 21B) and 28A (or 28B) yield the following inequalities:

120x7 + 49x8 + 56x10 ≥ 31987200 (6.7)

960x2 + 8x9 + 7x10 ≥ 7996800. (6.8)

Now, multiplying both sides of the inequalities (6.6), (6.7), and (6.8) by 15, 64, and 8 respectively and adding the results gives

7680(x2 + x5 + x7) + 5880x6 + 5314x8 + 64x9 + 3640x10 ≥ 3550579200, from which we deduce

x2 + x5 + x6 + x7 + x8 + x9 + x10 ≥ 462315. (6.9)

Finally, from equation (6.5) and inequality (6.9) we conclude that

|C| ≥ x1 + x2 + x5 + x6 + x7 + x8 + x9 + x10 ≥ 464373.

Consequently we have:

Theorem 6.2.8. The covering number of the Held group is 464373.

90 CHAPTER 7 CONCLUSION AND OPEN PROBLEMS

In conclusion we summarize the results of the preceding chapters with the following theorems:

Theorem. σ(SUn(q)) = σ(Un(q)).

Theorem. σ(U3(2)) = 3, σ(U3(3)) = 64, σ(U3(4)) = 1745, and σ(U3(5)) = 176.

Theorem. For q ≥ 7 we have

3 2 4 2 3 2 k(q) + q (q + 1) (q − 1)/3 ≤ σ(U3(q)) ≤ q + q + 1 − m(q) + q (q + 1) (q − 1)/3 where   1, if q is a power of 3 k(q) =  3 1 + q , otherwise, and   4 q /2, if q is a power of 2   m(q) = q3 + 2q2 − 2q − 1, if q is a power of 3    4 kq /p, if q is a power of a prime p such that p = 3k ± 1 ≥ 5. In particular,

3 2 4 3 2 3 2 1 + q (q + 1) (q − 1)/3 ≤ σ(U3(q)) ≤ q − q − q + 2q + 2 + q (q + 1) (q − 1)/3 .

Theorem. σ(McL) = 24553.

In addition we have verified that σ(HS) = 1376 and σ(He) = 464373, first proved by Holmes and Marótiin [22].

91 There are still many open problems in this area. One could, for example, attempt to improve on the results given here and attempt to determine the exact covering

number of U3(q) for q ≥ 7. Alternatively one could consider the covering number of

Un(q) for n ≥ 4, or the covering numbers of other classical groups. Moreover, there are still a number of sporadic simple groups whose covering numbers are not known.

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