Maximal Cocliques in PSL2 (q)

1

INTRODUCTION

Maximal Cocliques in PSL2(q)
Jack Saunders1
School of Mathematics, University of Birmingham, B15 2TT
JPS675@bham.ac.uk

arXiv:1802.09449v2 [] 20 Mar 2018

Abstract
The generating graph of a finite group is a structure which can be used to encode
certain information about the group. It was introduced by Liebeck and Shalev and has been
further investigated by Lucchini, Maróti, Roney-Dougal and others. We investigate maximal
cocliques (totally disconnected induced subgraphs of the generating graph) in PSL2 (q) for q a
prime power and provide a classification for when q is prime. We then provide an interesting
geometric example which contradicts this result when q is not prime and illustrate why the
methods used for the prime case do not immediately extend to the prime-power case with
the same result.

1

Introduction

It is a well-known result of Steinberg (and later others) that every finite simple group may be
generated by just 2 elements, and from [8, 4] we know that if we pick two elements of the group
at random then the chance that they will generate the group is high. This then motivated the
definition of the generating graph by [7], a graph whose vertex set consists of the non-identity
elements of the group G and we draw an edge between two elements precisely when they generate
G.
It is then interesting to look at what information the generating graph alone can tell us about
the group. For example, in [11], Lucchini, Maróti and Roney-Dougal showed that the generating
graph determines the group up to isomorphism for sufficiently large simple groups and for symmetric groups. The structure of the graph itself has also been investigated, such as the clique
number, the presence of hamiltonian cycles and many other things (some examples: [9, 2, 10]).
In this case, we will be looking at what we may be able to identify in the group simply by
looking at maximal cocliques (totally disconnected induced subgraphs) of the generating graph for
PSL2 (q) for q a prime power, and we show that when q is prime the largest maximal cocliques are
either entirely made up of involutions or are maximal subgroups. We also indicate an interesting
geometry which crops up when q is no longer prime.
We currently believe that, in the prime power case, the example we have found and its conjugates
are the only sufficiently large exceptions and that otherwise a similar result to the prime case
should hold. The prime result tells us that, in particular, given the generating graph for PSL2 (p)
we can identify and distinguish the conjugacy class of all involutions and the Borel subgroups by
size alone. In the prime power case, it should be possible to identify and distinguish the Borel
subgroups, PSL2 (q).2 ≤ PSL2 (q 2 ) and the geometric example.
1 This

work was done as part of my PhD, thus I would like to thank my supervisor Corneliu Hoffman for
everything he has done for me thus far.

Jack Saunders

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Maximal Cocliques in PSL2 (q)

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INTRODUCTION

We proceed by examining the maximal subgroups of PSL2 (p) and their intersections in order to
provide a linear bound (in p) on those maximal cocliques which contain at least one element of
order greater than 2 and are not maximal subgroups. After this, we provide an example in the
case of PSL2 (q 2 ) which does not fit in with the result for PSL2 (p) to show that the result does
not generalise as-is and provide some evidence to suggest why using the same approach as in the
prime case is likely to be either very complicated or impossible.

The prime case
Notation
In what follows I shall use Oεn (q) to denote the corresponding general orthogonal group, D2n to
denote the dihedral group of order 2n, Cn to denote the cyclic group of order n and An and Sn to
denote the alternating and symmetric groups on n elements, respectively. We begin by defining
what we will be working with and proving some basic results which will be used throughout.
Definition 1.1
Let G be a group. Then A ⊆ G is a coclique if for all g, h ∈ A we have that hg, hi =
6 G. We say
that a coclique A is maximal if whenever B is a coclique we have that A ⊆ B =⇒ A = B.
Lemma 1.2
If G is a group and A ⊆ G is a maximal coclique then hAi =
6 G =⇒ hAi = A and A is a
maximal subgroup of G.
Proof: Suppose hAi =
6 G. If g ∈ hAi \ A then hA ∪ {g}i ⊆ hAi =
6 G. But then this contradicts
the maximality of A and so hAi = A. Also, if A ⊆ H G and we assume A 6= H then we may
pick g ∈ H \ A and again note that hA ∪ {g}i ⊆ hHi = H 6= G. In particular, hg, hi =
6 G for all
h ∈ A but this again contradicts the maximality of A and so A = H is a maximal subgroup. 
Lemma 1.3
If g ∈ G lies in a unique maximal subgroup M of G then any maximal coclique A containing g
must be equal to M .
Proof: Let h ∈
/ M and consider hg, hi. Since every proper subgroup is contained in a maximal
subgroup, we have that hg, hi lies in a maximal subgroup containing g. But M is the only such
subgroup. Thus we must have that hg, hi ⊆ M or hg, hi = G. But h ∈
/ M and so hg, hi = G.
Thus A ⊆ M and the maximality of A gives us that A = M .

For what we intend to do next, we will need to know the maximal subgroups of PSL2 (p) and
when they appear. We provide these below, one may find them in [3, §260 (p286)] or for a more
modern treatment look at the maximal subgroups for SL2 (p) in [1] and then simply quotient out
by its centre to obtain these subgroups.
Theorem 1.4 (Maximal subgroups of PSL2 (p))
The conjugacy classes of maximal subgroups of PSL2 (p) are as follows:
i) The Borel subgroups Cp ⋊ C 12 (p−1) appear for all p.
ii) Dp−1 appears for all p > 11.
iii) Dp+1 appears for all p > 7.
Jack Saunders

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Maximal Cocliques in PSL2 (q)

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INTRODUCTION

iv) A4 appears when p ≡ ±3, ±13 mod 40.
v) S4 appears when p ≡ ±1 mod 8 and has two conjugacy classes.
vi) A5 appears when p ≡ ±1 mod 10 and has two conjugacy classes.
We shall also make use of Aschbacher’s classification of maximal subgroups of classical groups
for the prime power case, so we shall include this here as well, though we ignore the classes which
do not appear. This is also found in [1].
Theorem 1.5 (Aschbacher’s Theorem for SL2 (q) and Ω−
4 (q))
The maximal subgroups of SL2 (q) and Ω−
(q)
(acting
naturally
on V = Fnq ) lie in one of the
4
following classes:
Class
C1
C2
C3
C5
C6
S

Description
Stabilisers of totally singular or non-singular subspaces.
L
Stabilisers of direct sum decompositions V = ti=1 Vi where the Vi all have the
same dimension.
Stabilisers of extension fields of Fq of index 2.
Stabilisers of subfields of Fq of prime index.
Normalisers of symplectic-type or extraspecial groups in absolutely irreducible
representations.
See [1, Definition 2.1.3], the specifics of this will not be used here.

Theorem 1.6
Let G = PSL2 (p) for some prime p > 2 and let A be a maximal coclique in G. Then A is either
a maximal subgroup, the conjugacy class of all involutions or the order of A is linear in p.
Proof: First we note that if hAi =
6 G then A is a maximal subgroup by Lemma 1.2. Otherwise,
if hAi = G then we have several cases to consider. In the first case, A may consist entirely of
involutions since the group generated by two involutions will always be dihedral and thus is never
equal to G. Otherwise, either A contains some element x of large order such that A \ hxi consists
entirely of involutions or it does not.
Case 1: Unique cyclic group
We first assume that A contains a cyclic group, hxi (where |x| ≥ 3), such that A \ hxi is made up
entirely of involutions. The size of such a set will depend on which maximal subgroups lie within
G and so we shall have to consider certain different choices of p separately. It is also important
to note that the order of x will restrict which maximal subgroups it may lie in.
Case 1a: |x| p + 1
We start by considering the case where |x| p + 1 and thus x may lie in Dp+1 , A4 , S4 or
A5 . If |x| > 5 then x may only lie in Dp+1 and thus lies in a unique maximal subgroup since
the intersection of conjugate dihedral groups in PSL2 (p) consists only of involutions. Then by
Lemma 1.3 we have that A must be a maximal subgroup. Otherwise, if |x| ≤ 5 then this splits
up dependent on which of the smaller maximal subgroups appear in G. We have 2 possibilites
to consider.
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INTRODUCTION

Subcase 1: G contains A4 as a maximal subgroup
If G contains A4 as a maximal subgroup then we must have that p ≡ ±3, ±13 mod 40. If
|x| > 3 then we are in the case discussed above, otherwise |x| = 3 and we must count all of the
involutions present in all subgroups in which x appears. We start off by noting that A4 contains
3 involutions and a dihedral group Dp+1 contains at most p+3
2 involutions (it contains one fewer
if p ≡ 3 mod 4). In order to figure out how many copies of A4 can contain x, we note that Dp+1
acts on the set of copies of A4 which contain x with stabiliser containing hxi. This action is
g
transitive since if x ∈ H1 ∩ H2 where H1 ∼
= A4 and H2 = H1 then there exists h ∈ H2 such
= H2 ∼
g
h
gh
h
that if hxi = hyi then hyi = hxi and so we have that H1gh = H2h = H2 and hxi = hyi = hxi.
gh
So we have that gh ∈ NG (x) and H1 = H2 .
Thus the orbit of any copy of A4 containing x under this action will have length at most p+1
3 .
Therefore from the entire conjugacy class of A4 in G we shall get a contribution of at most 3 p+1
3
involutions. Adding this up, we see that
p+1
Inv A4
3
where Inv H denotes the set of involutions in H, and we abuse notation so that k Inv A4 :=
S
k
∼
i=1 Inv Gi where Gi = A4 for all i. This gives
A ⊆ hxi ∪ Inv Dp+1 ∪

|A| ≤ 3 +

3(p + 1)
p+3
+p+1=
+ 4.
2
2

Subcase 2: G contains A5 or S4 as maximal subgroups.
Since we are only dealing with upper bounds, we shall combine the cases where either A5 , S4
or both appear as maximal subgroups of G as this encompasses the arguments used for each
individual case. Our worst case bound occurs when |x| = 3 and both A5 and S4 are maximal in
G, so we shall only consider this situation. There will be two conjugacy classes of both A5 and
S4 , so we must double the contribution we get from each type of subgroup. We act by Dp+1 as
before to see that
2(p + 1)
2(p + 1)
A ⊆ hxi ∪ Inv Dp+1 ∪
Inv S4 ∪
Inv A5
3
3
and so
p+3
33(p + 1)
|A| ≤ 3 +
+ 6(p + 1) + 10(p + 1) =
+ 4.
2
2
Case 1b: |x| p
When |x| p we clearly have that |x| = p since |x| > 2. But then x must lie in a unique Borel
subgroup and thus by Lemma 1.3 we have that A must be a maximal subgroup.
Case 1c: |x| p − 1
In this final case, we have the most subgroups to consider. If |x| p − 1 then x may lie in a Borel
subgroup, Dp−1 , A4 , S4 or A5 . We first need the following:
Sublemma 1.7
For an odd prime p, if p ≡ 1 mod 4 then the Borel subgroup B ∼
= Cp ⋊ C 21 (p−1) contains p
involutions, otherwise it has none.
Jack Saunders

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Maximal Cocliques in PSL2 (q)

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INTRODUCTION

Proof: We begin by noting that B may be regarded as the subgroup of upper or lower triangular
matrices with respect to some basis of F2p . Without loss of generality we shall assume that these
matrices are upper triangular and work in SL2 (p) where the Borel subgroup is instead isomorphic
to Cp ⋊ Cp−1 ∼
= (Fp , +) ⋊ F∗p and we shall simply ignore those matrices which differ by a sign.
Now, our goal is to find a matrix M such that M 2 = I2 or −I2 . We start by considering
2

M :=



a
0

b
a−1

2

=



a2
0

ab + ba−1
a−2



=



1 0
0 1



.

We see that for this we require a ∈ F∗p such that a2 = a−2 = 1, but the solution to this is unique
and given by a = a−1 = −1. We now consider the other possible solution
 2
 

a ab + ba−1
−1 0
2
M =
=
.
(⋆)
0
a−2
0 −1
When p ≡ 3 mod 4 we see that this has no solutions since we require a 4th root of unity in F∗p
which exists only when p ≡ 1 mod 4. We assume that p ≡ 1 mod 4 and let i be such a root of
unity. Then we see that i2 = i−2 = −1 and ib + bi−1 = ib − bi = 0 and so we see that (⋆) has
the following set of p solutions:



i b
b ∈ Fp .
0 −i
In SL2 (p) there are another p solutions given by the negations of these matrices, but upon
factoring out by the centre these sets of solutions are identified with one another.

We again start by considering the case where |x| > 5. Such an x will lie in one copy of Dp−1 and
two distinct Borel subgroups which will consist of the upper and lower triangular matrices with
respect to a basis in which x is diagonal. We need only count the involutions in these subgroups.
In this case we get
A ⊆ hxi ∪ Inv B1 ∪ Inv B2 ∪ Inv Dp−1
where B1 and B2 are the Borel subgroups in which x lies. This then gives us
|A| ≤ |x| + 2p +

p+1
5(p + 1)
=
+ |x| − 2.
2
2

When |x| ≤ 5 we must split this into several subcases as before.
Subcase 1: G contains A4 as a maximal subgroup
When G contains A4 as a maximal subgroup we need only concern ourselves with the case where
|x| = 3 and we count involutions as before. We may find out how many copies of A4 contain
x by using the action of Dp−1 on the set of A4 s containing x in a very similar way to previous
cases.
p−1
Inv A4
A ⊆ hxi ∪ Inv B1 ∪ Inv B2 ∪ Inv Dp−1 ∪
3
which then gives
p+1
7(p − 1)
|A| ≤ 3 + 2p +
+p−1=
+ 6.
2
2
Subcase 2: G contains A5 or S4 as maximal subgroups.

Jack Saunders

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Maximal Cocliques in PSL2 (q)

1

INTRODUCTION

As with the last time, we shall merge all of the cases where A5 , S4 or both appear as maximal
subgroups of G. Acting with Dp−1 we see that
A ⊆ hxi ∪ Inv B1 ∪ Inv B2 ∪ Inv Dp−1 ∪
and so
|A| ≤ 3 + 2p +

2(p − 1)
2(p − 1)
Inv A5 ∪
Inv S4
3
3

p+1
37(p − 1)
+ 10(p − 1) + 6(p − 1) =
+ 6.
2
2
Case 2: Multiple cyclic groups

We now consider the case where A contains multiple elements of order greater than 2 which do
not lie in some common cyclic group contained in A. We must first note that any element of A
must share some maximal subgroup with each other member of A. In particular, if A contains
any element of order greater than 5 then we need only consider contributions from the larger
maximal subgroups since clearly this element cannot lie in A4 , S4 or A5 . Thus A is mostly
determined by which subgroups this large element may lie in. If x ∈ A is an element of largest
order with |x| > 5 and |x| p + 1 then we see that x may only lie in Dp+1 . But then x lies in a
unique maximal subgroup and by Lemma 1.3 we are done. Similarly, if |x| p then |x| = p and
x lies in a unique Borel subgroup as before. Again, by Lemma 1.3 we are done.
Finally, if |x| p − 1 then x may lie in Dp−1 or two Borel subgroups, B1 and B2 . If A is not
one of these subgroups then A either consists of a few larger order elements along with all of the
involutions from the subgroups in which they lie or A is simply made up of the intersections of
these groups. In either case,
A ⊆ (Dp−1 ∩ B1 ) ∪ (Dp−1 ∩ B2 ) ∪ (B1 ∩ B2 ) ∪ Inv Dp−1 ∪ Inv B1 ∪ Inv B2 .
But then
|A| ≤ |Dp−1 ∩ B1 | + | Inv Dp−1 | + 2| Inv B1 | ≤ 3p
since Dp−1 has at most 12 (p + 1) involutions, and the intersection of the Borel subgroups and
Dp−1 is the cyclic group of order 12 (p − 1) consisting of those matrices which are diagonal with
respect to the basis given by the eigenvectors of the two Borel subgroups.
Otherwise, A contains several non-involutions of orders 3, 4 or 5 and we must consider the
possibility of contributions from A4 , S4 or A5 . The size of A will still, however, depend on the
large maximal subgroups that its elements may lie in. If the large order elements do not all have
orders simultaneously dividing p − 1, p or p + 1 then our only possibilities are an element x of
order 3 dividing p − 1 and an element y of order 5 dividing p + 1 both lying in a copy of A5 or
vice versa. In this case, we have that any element sharing maximal subgroups with x and y must
lie in one of the copies of S4 or A5 containing x or be an involution in Dp+1 . Since this will be
encompassed in the way the following cases are dealt with, it is therefore sufficient to consider a
single element x ∈ A with |x| > 2 and check whether |x| divides p + 1, p or p − 1.
Case 2a: |x| p + 1
We first consider the case where |x| p + 1. Here x may lie in Dp−1 , A4 , S4 or A5 . As before,
we have several subcases.
Jack Saunders

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Maximal Cocliques in PSL2 (q)

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INTRODUCTION

Subcase 1: G contains A4 as a maximal subgroup
Here we require |x| = 3 and note that any other element of A either lies in one of the copies of
A4 containing x or is an involution in Dp+1 , thus we have that
p+1
9(p + 1)
p+1
|A4 | + | Inv Dp+1 | ≤ 4(p + 1) +
+1=
+ 1.
3
2
2
Subcase 2: G contains A5 or S4 as maximal subgroups.
|A| ≤

Again, we merge the cases where any of these maximal subgroups appear. Our worst case of
course is when both A5 and S4 appear as maximal subgroups of G, so this is the one we shall
deal with. Here we fix up to three elements, x, y and z of respective orders 3, 4 and 5, lying in a
single copy of Dp+1 . Then any element of A will lie in one of the copies of S4 or A5 containing
any of x, y or z or be an involution in Dp+1 since Dp+1 has a unique cyclic subgroup of any
given order greater than 2. Thus,


!
[
[
[
[
A ⊆ hxi ∪ hyi ∪ hzi ∪ Inv Dp+1 ∪ 2 
S4 ∪
S4  ∪ 2
A5 ∪
A5 .
x∈S4

y∈S4

x∈A5

z∈A5

Using our normal approach to determine how many copies of S4 or A5 may contain a given
element, we obtain




185(p + 1)
p+1 p+1
p+1 p+1
p+1
+ 48
= 13 +
+ 120
+
+
.
|A| ≤ 3 + 4 + 5 +
2
3
5
3
4
2
Case 2b: |x| p

Here we have that |x| = p and |x| ≤ 5, so |x| = 3 = p. But then x lies in a unique maximal
subgroup (generated by itself) and so we are done by Lemma 1.3.
Case 2c: |x| p − 1
Here we need to consider contributions from two Borel subgroups, a copy of Dp−1 and any of
A4 , S4 or A5 which may arise.
Subcase 1: G contains A4 as a maximal subgroup
As usual, we need only concern ourselves with |x| = 3. We note again that Dp−1 contains a
unique cyclic subgroup of order 3 and also that the Borel subgroups each contain p distinct
cyclic subgroups of order 3. As usual, NG (x) acts on the set of A4 s containing x by conjugation
1
with orbit length at most |x|
(p − 1) =: kx .
Now, we note that any involutions which lie outside of the Borel subgroups must lie in some
shared maximal subgroup with every element of A of order 3. Thus we have that any such
involutions must lie in
kx
kx
[
\ [
A4 .
A4 ⊆
i=1

|x|=3 i=1
x∈A

Thus we get that
A ⊆ 2p hxi ∪ Inv B1 ∪ Inv B2 ∪ Inv Dp−1 ∪
Jack Saunders

7

p−1
Inv A4
|x|

Maximal Cocliques in PSL2 (q)

2

THE PRIME-POWER CASE

and so
|A| ≤ 6p + 2p +

p−1
p−1
p−1
p−1
p−1
+3
≤ 6p + 2p +
+3
= 19
+8
2
|x|
2
3
2

where we have taken 21 (p − 1) in place of | Inv Dp−1 | since the additional involution will be
contained in the Borel subgroups if it exists. Alternatively, we have
A ⊆ (B1 ∩ B2 ∩ Dp−1 ) ∪ 2p hxi ∪

kx
\ [

A4

|x|=3 i=1
x∈A

and from the above we may tidy up the final term to get
|A| ≤

p−1
p−1
p−1
p−1
p−1
+ 6p + 12
≤
+ 6p + 12
= 21
+ 6.
2
|x|
2
3
2

Subcase 2: G contains S4 or A5 as maximal subgroups.
As with the above cases, our worst bound comes from when both A5 and S4 are present so we
consider only this case and proceed in exactly the same way as before to obtain
A ⊆ Inv(B1 ∪ B2 ∪ Dp−1 ) ∪ 2p hxi ∪ 2p hyi ∪ 2p hzi ∪ 2

\

kx
[

|x|∈{3,4} i=1
x∈A

S4 ∪

\

kx
[

A5 .

|x|∈{3,5} i=1
x∈A

Bounding this as usual, we have
|A| ≤

p− 1
p−1
p−1
p−1
p−1
p−1
+ 2p + 6p + 8p + 10p + 48
+ 120
≤
+ 24p + 168
= 117
+ 26.
2
|x|
|x|
2
3
2

Combining all of the above, we obtain the desired result.

2



The prime-power case

We now consider the case of G := PSL2 (q0 ) for q0 = pn a prime power. We have little hope
of achieving the same result as before due to the existence of copies of PSL2 (r) for q0 = rs , s
prime, as a maximal subgroup of G. Indeed, simply trying to use the same methods as before
3
would give us bounds to the order of q02 for all even n in the case where the coclique was mostly
involutions with a single large order element. We also have an interesting geometry which crops
up for even n due to the fact that PSL2 (q 2 ) ∼
= PΩ−
4 (q) which we will investigate in this section.
At various points throughout this section, the maximal subgroups of G may be required, so I
shall leave them here for reference. From [1], as before, we know that the maximal subgroups
are the following:
i) The Borel subgroups, Eq2 ⋊ C 12 (q2 −1) .
ii) PSL2 (q).2 for q0 = q 2 , which splits into two conjugacy classes.
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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

iii) Dq0 −1 .
iv) Dq0 +1 .
v) PSL2 (q) where q0 = q r for an odd prime r.
vi) Two conjugacy classes of A5 for q0 = p2 , p ≡ ±3 mod 10 prime.
In terms of their classes in Aschbacher’s Theorem, we have that a Borel subgroup is the stabiliser
of an isotropic 1-space; the PSL2 (q).2 is the stabiliser of a non-isotropic 1-space; Dq0 −1 is the
stabiliser of a non-degenerate 2-space and the copies of PSL2 (q) are stabilisers of subfields of
prime index. We will not need to look into the classes of the other types of maximal subgroup.

2.1

Other cocliques of large order
3

We will now calculate explicitly the bound of order q02 given above, which is evidence that the
previous method may not work as well for this situation. We suppose that q is large enough that
all of the necessary maximal subgroups exist.
As before, we assume that A is a maximal coclique which contains an element x with |x| > 2 such
that A \ hxi consists entirely of involutions. If we suppose that x lies in PSL2 (q) and |x| q0 − 1
for q0 = q r , r an odd prime, then at the very least we may include all of the involutions from
this subgroup. We then also note that NG (x) acts on the set of copies of PSL2 (q) containing x
with orbit length
|NG (x)|
q0 − 1
=
.
|NH (x)|
q±1
Including all of the involutions from these other subgroups too, we count
(q0 − 1)| PSL2 (q)|
q0 − 1
q0 − 1
| Inv PSL2 (q)| =
=
q(q ∓ 1)
q±1
(q ± 1)(q ± 1)
q±1
involutions, but there is some multiple-counting which we must account for.
We now suppose that |x| = 21 (q + 1), so x ∈ Dq0 −1 . We then note that x lies in the intersection of
all copies of PSL2 (q) above and since x must lie in some maximal subgroup of PSL2 (q), we know
it must lie in a unique copy of Dq+1 . Then we have that the intersection of any two copies of
PSL2 (q) containing x must be either hxi or Dq+1 , but since Dq+1 normalises x we must also have
that Dq+1 ⊆ Dq0 −1 = NG (x), and in fact any such Dq+1 must lie in NDq0 −1 (NH (x)) ∼
= D2(q+1) .
If we assume the worst possible overcounting, then each one of the q + 1 involutions in this
1
(q0 − 1) times. Thus we know that our coclique must have at least
subgroup is counted |x|
2

q0 − 1
q0 − 1
q0 − 1 2
q(q + 1) − 2
(q + 1) = 2
(q − 1)
q+1
q+1
q+1

involutions. We don’t know if this coclique is necessarily maximal, but we at least have a lower
bound on its order.
1+ 1

So we obtain a bound of order q0 r where q0 = q r and r is the least odd prime such that this
happens and the least bound is obtained when we suppose that |x| = 21 (q ± 1). It’s possible that
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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

one could improve this with a good understanding of how these subfield stabilisers intersect, but
the smallest case in which this occurs is PSL2 (36 ) and is not easy to compute such things in.
3

Doing the same with PSL2 (q).2 for q0 = q 2 gives us the bound of order q02 mentioned above.
We suspect that a similar result will hold to in the prime case and that the geometric anomaly
described below is the only exception to what we had before, but the existence of this example
makes trying to use the previous
methods very complicated. We think that if a maximal coclique
p
A is suchpthat |A| > O( |G|) then it either consists of involutions or is a Borel subgroup, if
|A| = O( |G|) then it should be one of the subfield stabiliser maximal subgroups, and if it is
smaller we can’t really say much.

2.2

Construction

We first let q0 = q 2 and consider the action of G on F4q as PΩ−
4 (q). Fix some non-isotropic
vector v; to ease notation, we let v denote the Fq -span of v as well as the vector itself since the
distinction is not overly important here. Then F4q = v ⊕ v ⊥ . We wish to consider the elements of
G which have 2-dimensional eigenspaces lying in v ⊥ , so we must start by determining how many
such subspaces one may have.
We wish to collect all of these elements in order to obtain a large coclique with order cubic in q
since if one takes any two elements g, h with 2 dimensional eigenspaces Vg , Vh ⊆ v ⊥ then since
dim v ⊥ = 3 the intersection of any two 2-dimensional subspaces must be nontrivial. Thus hg, hi
must be contained in the stabiliser of Vg ∩ Vh 6= {0} and so hg, hi =
6 G.
3

−1
= q 2 + q + 1 subspaces of dimension 2, and as seen
A subspace of dimension 3 over Fq has qq−1
in [12] we have that if such a subspace (with symmetric bilinear form having gram matrix g) has
an orthonormal basis (det g a square) then it is of − type for q ≡ 3 mod 4 and if such a basis
does not exist (det g is a non-square) then it is of + type. If a space is of neither + nor − type
then it is degenerate (det g = 0). For q ≡ 1 mod 4, the 2-spaces with orthonormal bases are of
+ type and to obtain a − type space we must instead consider a space with basis {e, f } where
(e, e) = 1 and (f, f ) = α for some non-square α ∈ F∗q .

With this in mind, we wish to count the number of 2-dimensional subspaces of each given type
lying in v ⊥ . In order to do this, we must make use of Witt’s Theorem, an equivalent form of
Witt’s Cancellation Theorem [6, p12]:
Theorem 2.3 (Witt’s Cancellation Theorem)
If q, q1 and q2 are quadratic forms, then if the orthogonal direct sums q ⊥ q1 and q ⊥ q2 are
isometric, q1 and q2 are isometric.
Theorem 2.4 (Witt’s Theorem)
Any isometry between two subspaces of a nondegenerate quadratic space may be extended to an
isometry of the whole space.
Proof: Let U ∼
= W be two isometric subspaces of a quadratic space V . Then V = U ⊕ U ′ =
′
W ⊕ W for some (not necessarily orthogonal) complements U ′ , W ′ of U and W , respectively.
Clearly V is isometric to itself, so U ⊕ U ′ ∼
= W ⊕ W ′ where U and W are isometric. Using the
′ ∼
cancellation theorem above, we obtain U = W ′ . Let f : U → W and g : U ′ → W ′ be isometries,
then f ⊕ g : V → V is an extension of f to an isometry of V .

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THE PRIME-POWER CASE

Corollary 2.5
The orthogonal group Oεn (q) acts transitively on the set of subspaces of a given dimension and
type.
We now use this to count the number of 2-dimensional subspaces of v ⊥ by determining the
stabilisers of spaces of each type. We start with the stabiliser of a 2-space of + type. A matrix g
stabilising such a space also stabilises its orthogonal complement, which is simply a non-isotropic
1-space u. Since it acts as an orthogonal transformation on this 1-space, we have that g must
act on the space as ±1. Similarly, we have that g acts as an orthogonal transformation on the
original space of + type, thus we see that g must lie in O+
2 (q).2. Similarly, we obtain that the
stabiliser of a 2-space of − type in a 3-space is of the form O−
2 (q).2. Then the number of spaces
of + type is
| O3 (q)|
q+1
2q(q 2 − 1)
=q
=
+
4(q − 1)
2
| O2 (q).2|
and similarly there are q q−1
2 2-spaces of − type. Finally, we note that the remaining spaces must
q−1
all be degenerate, so there are q 2 + q + 1 − q q+1
2 − q 2 = q + 1 degenerate 2-spaces.
Knowing this, we may construct a coclique by taking the elements of G which have one of these
2-spaces as an eigenspace. Since any such element h must act as an orthogonal transformation
on these eigenspaces, we have that h may only act as ±1 on said spaces. Further, if h stabilises
a subspace U ⊆ v ⊥ then h must also stabilise U ⊥ and so if U is non-degenerate then our only
options are h acting as an element of SO(U ⊥ ) on U ⊥ and either I2 or −I2 on U .
First, we will determine the pointwise stabiliser of a degenerate subspace. For this, we take h to
be of the form


1 0
0 1 B 

A := 
0 0

C
0 0
and we take



0
0

k := 
1
0

0
1
0
0


1 0
0 0

0 0
0 α

as the matrix of our bilinear form, where α is such that −α is a non-square. A direct computation
shows that AkAT = k has solutions of the form


1 0 − 21 b2 b
0 1
0
0

A=
(⋆)
0 0
1
0
0 0
−b
1

and the group of matrices of this form is isomorphic to the additive group of the field, with each
matrix corresponding to b. We now need to check that the elements we are choosing actually lie
in G, and for this we need to use the spinor norm. From immediately after Proposition 2.5.7
of [5] we have that, for an element A = rv1 . . . rvk ∈ SOεn (q) written as a product of reflections,
where for (v, v) 6= 0 we define
(x, v)
v,
rv (x) := x − 2
(v, v)
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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

the spinor norm θ : SOεn (q) → F∗q /(F∗q )2 is defined by
θ(A) :=

k
Y

(vi , vi ) mod (F∗q )2 .

i=1

Then Ωεn (q) := ker θ. Thus we need to check that the elements of order q ± 1 with −1-eigenspaces
lying in v ⊥ lie in ker θ and also any of the elements of order p with degenerate eigenspaces.
We shall deal with the case of involutions first since this will help us with the stabilisers of
non-degenerate spaces.
Since G has a single conjugacy class of involutions we can immediately say that any involution
g must lie in a copy of Dq2 −1 and also be diagonalisable, thus we may write F4q = V+ ⊕ V− as
a decomposition into 1- and −1-eigenspaces of g. Since det g = 1 we must have that dim V+ is
either 0, 2 or 4, and thus it must be 2 since I4 is not an involution and −I4 ∈
/ G as G is centreless.
Further, this decomposition is orthogonal since if x ∈ V+ and y ∈ V− then (x, y) = (gx, gy) =
−(x, y) and so (x, y) = 0. Also, one of these eigenspaces must be of + type and the other of −
type.
To check whether g ∈ G we must check whether θ(g) is a square in Fq . To do this we suppose that
V+ = spanFq {f1 , f2 } and V− = spanFq {e1 , e2 }, where e1 and e2 are orthogonal and non-isotropic,
and note that g may be written as re1 re2 . Then θ(g) = (e1 , e1 )(e2 , e2 ) is a square in Fq if and
only if V− is of − type and q ≡ 3 mod 4 or V− is of + type and q ≡ 1 mod 4.
We can therefore see that the only involutions with 2-dimensional eigenspaces lying in v ⊥ are
those for which V+ ⊆ v ⊥ is of + type or V− ⊆ v ⊥ is of − type for q ≡ 3 mod 4, or vice versa
for q ≡ 1 mod 4.
For q ≡ 3 mod 4, we may take an element g of order q − 1 with −1-eigenspace V ⊆ v ⊥ to act
on V ⊥ as h2 , where SO(V ⊥ ) = hhi (we cannot take g generating the full SO(V ⊥ ) since then
1
/ G). As V is of − type, we may assume that it has an orthonormal basis {u, v}
g 2 (q−1) = −I4 ∈
and that we have a basis {e, f } of V ⊥ where (e, e) = (f, f ) = 0 and (e, f ) = 1. Then we may
express g as re−µ2 f ru re−f rv (where F∗q = hµi) and note that θ(g) = 4µ2 ∈ (F∗q )2 and so this
element lies in G. We then note that if g ′ is the element with the same action on V ⊥ but with
1
V as a 1-eigenspace then g ′ = g 2 (q+1) . Also, if we take g ′ instead to be the element acting as h
1
⊥
on V with 1-eigenspace V then the involution g ′ 2 (q−1) has a 1-eigenspace of − type and thus
does not lie in G.
In order to avoid a large mess of unnecessary matrices, we will use A ⊕ B to denote the matrix
with the 2 × 2 matrix A in the top left, the 2 × 2 matrix B in the bottom right and zeros
everywhere else.
Still taking q ≡ 3 mod 4, let V = V− ⊕ V+ where V− is 2-dimensional of − type and V+ is
2-dimensional of + type. Then let SO(V− ) = hhi and let g = h ⊕ −I2 and g ′ = h ⊕ I2 . Now note
that θ(g) = αθ(g ′ ) for some non-square α (coming from the reflection in the basis vector of nonsquare norm). Thus we have that g ∈ G ⇐⇒ g ′ ∈
/ G. Now let A := h ⊕ I2 and B := I2 ⊕ −I2 .
Then g = AB and g ′ = A. To confirm that g ∈ G it suffices to check whether θ(A) = 1 since
θ(B) = 1.
Now, θ(A) = θ− (h), where θ− is the restriction of θ to V− . Note that O(V− ) =< ru , rv > for some
u, v and that det ru = det rv = −1. Also, ru rv has order q + 1, so we see that SO(V− ) = hru rv i.
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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

We know that the index [SO(V− ) : Ω(V− )] = 2 and so we must have that θ(ru rv ) = θ(h) = 1.
Then θ(g ′ ) = 1 and θ(g) = 0. Thus g ∈ G.
If g ∈
/ G then we must have that some other such element, l := h2k ⊕ −I2 ∈ G, say. Then we
must also have that g ′ ∈ G, and so lg ′ = hh2k ⊕ −I2 = h2k+1 ⊕ −I2 ∈ G. This contradicts
h2k ⊕ −I2 ∈ G and so we must have that h2k+1 ⊕ −I2 ∈ G for 0 ≤ k ≤ 21 (q + 1). In particular
g ∈ G.
We now consider the case where q ≡ 1 mod 4 with g generating SO(V ) for some 2-space V of
− type where V ⊥ is a −1-eigenspace. We let g ′ be the corresponding element with V ⊥ as a
1-eigenspace. Then one can see that θ(g) = θ(g ′ ) and so g ∈ G ⇐⇒ g ′ ∈ G. To see that g ′ ∈
/G
1
we can then just note that the involution g ′ 2 (q+1) does not lie in G since it has a 1-eigenspace
of + type. We thus need to take g acting as h2 on V , where SO(V ) = hhi and as −1 on V ⊥ .
We note that there are other elements of order q + 1, all of which are powers of gg ′ = h2 ⊕ −I2 ,
which does lie in G since θ(gg ′ ) = θ(g)θ(g ′ ) ∈ (F∗q )2 since θ(g), θ(g ′ ) ∈
/ (F∗q )2 .
Next, we let q ≡ 1 mod 4 with g generating SO(V ) for some 2-space V of + type where V ⊥
is a −1-eigenspace. We let g ′ be the corresponding element with V ⊥ as a 1-eigenspace. Then
θ(g) = αθ(g ′ ) for some non-square α ∈ F∗q and so g ∈ G ⇐⇒ g ′ ∈
/ G. We can now confirm that
g ∈ G in a manner identical to the case where q ≡ 3 mod 4 above, writing g = re−µf ru re−f rv
and seeing that θ(g) = 4µα ∈ (F∗q )2 for a suitable basis of F4q .
We now consider the case where the eigenspace of g is degenerate with eigenvalue 1. We denote
this eigenspace by V and let V ⊥ := spanFq {e, f } where rad V = Fq (e + f ). Then the element
x := re rf is not the identity and has V as a 1-eigenspace, thus it lies in the pointwise stabiliser
of V and so has order p. Then θ(x) = (e, e)(f, f ) ∈ (F∗q )2 since we may choose a basis {e, f } for
V such that (e, e) = (f, f ) = 1. A suitable choice of e and f allows us to do the same for every
element of the pointwise stabiliser of V .
We wish to collect these elements to make a coclique, so we first check their intersections so
that we may determine the size of their union. Suppose that some element g has 2 distinct
2-dimensional eigenspaces inside of v ⊥ . Then clearly g must have a 3-dimensional 1-eigenspace
(namely v ⊥ ) and so g must also stabilise v and since we require det g = 1 we must have that
gv = v and g = I4 . Thus the intersection of any pair of groups generated by the elements of
the coclique of maximal order is trivial. Collecting all of these groups together, we see that we
obtain a set of size
(q − 2)

q(q − 1) q 2 (q + 1)
+
+ (q − 1)(q + 1) + 1 = q 3 + q
2
2

and so, as claimed, we obtain a coclique of order cubic in q.
Proof of maximality
We now show that the coclique obtained above is indeed maximal by considering the maximal
subgroups of PΩ−
4 (q). In what follows, when we refer to a ’point,’ what we shall actually mean
is a 1-space, which may be thought of as a projective point since we are working with PΩ.
This time around, we may ignore Dq2 +1 since for q > 3 we have q ± 1 > 2 and r|q − 1 and r|q 2 + 1
means that r|(q − 1)(q + 1) = q 2 − 1, thus r|q 2 + 1 − (q 2 − 1) = 2. A similar argument holds for
q + 1, and clearly the only natural number dividing both p and q 2 + 1 is 1. Then we may ignore
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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

A5 and the subfield stabilisers PSL2 (q) by simply choosing the stabilisers of non-degenerate 2spaces in the argument that follows since these are generated by elements of order q ± 1, which
are not found in A5 or PSL2 (q) for q > 3 and so if we take a generator h of the stabiliser of some
non-degenerate subspace and some other g ∈ G then we have that hg, hi cannot be contained in
either A5 or PSL2 (q) as these groups do not have elements of sufficiently large orders.
We note that the remaining three maximal subgroups are all of class C1 and so represent the
stabilisers of singular or non-singular subspaces. We thus look at how the stabilisers of various
subspaces of F4q interact with the coclique, which we shall call L for simplicity. We first note that
the dihedral subgroups Dq2 −1 represent the stabilisers of non-degenerate 2-dimensional subspaces
(as the stabiliser of a degenerate one lies in the point stabiliser of its radical) and recall that
the Borel subgroups and PSL2 (q).2 are the stabilisers of isotropic and non-isotropic 1-spaces,
respectively.
We also note that an element h ∈ L of maximal order stabilising a non-degenerate 2-space
U ⊆ v ⊥ will stabilise some point if and only if this point lies inside U . We first consider the
case where g ∈ G stabilises some non-degenerate 2-space. If g stabilises V ⊆ v ⊥ either V is
an eigenspace of g or there is some 1-space Fq u ⊆ V not fixed by g. If there is one such point
then there are many others, since if g fixes more than 2 points in the same 2-space then this
space would have to be an eigenspace for g, so we may choose some u ∈ V not fixed by g and
a corresponding x ∈ V ⊥ also not fixed by g such that spanFq {u, x} is non-degenerate. Then the
element of L of maximal order with eigenspace (spanFq {u, x})⊥ will be such that hg, hi stabilises
no proper nontrivial subspace of F4q and so must be equal to G.
Next, we consider the case where g stabilises some degenerate 2-space V ⊆ v ⊥ . Either V is an
eigenspace for g or there is a (non-isotropic) 1-space Fq u ⊆ V not fixed by g. Then, as in the
non-degenerate case, we may pick some non-isotropic vector x ∈ V ⊥ such that spanFq {u, x} is
non-degenerate and an element h ∈ L of maximal order with eigenspace (spanFq {u, x})⊥ such
that hg, hi = G.
Otherwise, if g has a 2-dimensional degenerate eigenspace inside of v ⊥ then either the corresponding eigenvalue is 1 and so g ∈ L or it is −1 and a direct computation gives solutions of the
form (⋆) with −1 on the diagonal and − 21 b2 replaced with 12 b2 on the top row. Squaring such a
matrix gives us an element of L and raising it to the power p (recall q = pk for some k) gives us
−I4 ∈
/ G, thus no such element exists in G.
We are now left with a number of simpler cases corresponding to the possible eigenspaces of a
general element g ∈ G. We use Vλ to denote the λ-eigenspace of g ∈ G in F4q =: V .
dim V λ ≥ 3.
We have that Vλ ∩ v ⊥ is an eigenspace of g of dimension at least 2 and so g ∈ L.
V = Vλ ⊕ Vµ where dim V λ = dim V µ = 2, λ 6= µ.
We ignore the case where either Vλ or Vµ lie in v ⊥ since this was dealt with earlier. Thus we
have that Vλ ∩ v ⊥ =: Fq x and Vµ ∩ v ⊥ =: Fq y are 1-spaces. Since an element h ∈ L of maximal
order fixes some point if and only if that point lies in its eigenspace U (or |h| = q − 1 and this
point is one of two isotropic eigenvectors), we may choose any h ∈ L of maximal order such that
x, y ∈
/ U and U is non-degenerate. We may do this since even in the worst case where v ⊥ has
Jack Saunders

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Maximal Cocliques in PSL2 (q)

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THE PRIME-POWER CASE

a basis of eigenvectors for distinct eigenvalues, there are at most 3(q + 1) 2-spaces containing
any of these three points, yet q 2 non-degenerate 2-spaces in v ⊥ and since we chose q > 3 earlier,
q 2 > 3(q + 1). Then hg, hi cannot stabilise any proper, nontrivial subspaces of V and so must be
equal to G.
V = Vλ ⊕ Vµ ⊕ Vν where dim V λ = 2, λ, µ, ν pairwise distinct.
We ignore the case where Vλ ⊆ v ⊥ since this has already been dealt with. As above, we simply
choose h ∈ L of maximal order stabilising a non-degenerate subspace U ⊆ v ⊥ which does not
intersect with any of these eigenspaces. The cases where Vµ or Vν are trivial and so we do not
have a full decomposition of V are also dealt with by the same choice of h.
V = Vλ ⊕ Vµ ⊕ Vν ⊕ Vκ , λ, µ, ν, κ pairwise distinct.
As with the above cases, we simply choose an element h ∈ L of maximal order stabilising a
non-degenerate subspace U ⊆ v ⊥ which does not intersect with any of these eigenspaces. The
cases where some of the eigenspaces are trivial and we do not have a full decomposition of V are
also dealt with by the same choice of h.
In all of the above cases, either we have that g ∈ L or hg, hi = G, thus L is a maximal coclique
and we are done.


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Maximal Cocliques in PSL2 (q)

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