ISSN 2590-9770
The Art of Discrete and Applied Mathematics 6 (2023) #P1.05
https://doi.org/10.26493/2590-9770.1494.1e4
(Also available at http://adam-journal.eu)
Some Erdős-Ko-Rado results for linear and
affine groups of degree two*
Karen Meagher† , Andriaherimanana Sarobidy Razafimahatratra
Department of Mathematics and Statistics, University of Regina, Regina,
Saskatchewan S4S 0A2, Canada
Received 16 October 2021, accepted 2 February 2022, published online 4 October 2022
Abstract
In this paper, we show that both the general linear group GL(2, q) and the special linear
group SL(2, q) have both the EKR property and the EKR-module property. This is done
using an algebraic method; a weighted adjacency matrix for the derangement graph for the
group is found and Hoffman’s ratio bound is applied to this matrix. We also consider the
group AGL(2, q) and the 2-intersecting sets in PGL(2, q).
Keywords: derangement graph, independent sets, Erdős-Ko-Rado Theorem, Symmetric group, Gen-
eral linear group, Affine linear group, Projective linear group
Math. Subj. Class.: Primary 05C35; Secondary 05C69, 20B05
1 Introduction
In this paper we will examine some Erdős-Ko-Rado properties for the general linear group
GL(2, q), the special linear group SL(2, q) and the affine general linear group AGL(2, q).
The Erdős-Ko-Rado (EKR) theorem for permutation groups has to do with finding max-
imum sets of permutations within a group so that any two permutations in the set inter-
sect. Let G ≤ Sym(n) be a transitive group of degree n and acting on the set [n] :=
{1, 2, . . . , n}. Permutations g and h ofG intersect if there exists some i ∈ {1, . . . , n} such
that g(i) = h(i), or equivalently, h−1g(i) = i. If g and h do not intersect, then h−1g is a
derangement.
For any positive integers n and k such that 2k ≤ n, the original EKR theorem [14]
gives the size of the largest collection of k-uniform intersecting sets of [n]. The maximum
*We are grateful to the anonymous referee for their feedback.
†Corresponding author. Research supported in part by an NSERC Discovery Research Grant, Application No.:
RGPIN-2018-03952.
E-mail addresses: meagherk@uregina.ca (Karen Meagher), sarobidy@phystech.edu (Andriaherimanana
Sarobidy Razafimahatratra)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Art Discrete Appl. Math. 6 (2023) #P1.05
size is achieved by taking all subsets that contain a fixed point. These collections are
called the canonical intersecting set systems. Analogously, the canonical intersecting sets
of permutations from a transitive group G ≤ Sym(n) are the sets of all permutations of G
that map some i to some j, where i, j ∈ [n]. The stabilizer in a group G of a point i will be
denoted by Gi. The canonical intersecting sets in G are the sets
Si,j := {g ∈ G : g(i) = j};
these are exactly the cosets of the point-stabilizers Gi. These are clearly intersecting sets
of permutations and it is easy to see that if G is a transitive group with degree n, then
|Si,j | = |G|n .
If S is an intersecting set in G, then for any element g ∈ G the set gS is also an
intersecting set; in particular, if S is a canonical intersecting set, then gS is also a canonical
intersecting set. The characteristic vector of a set S of permutations in a group G is a
length-|G| vector with entries indexed by the elements of the group and the g-entry is 1 if
g ∈ S and 0 otherwise. We will denote the characteristic vector of Si,j by vi,j (assuming
that the group G is clear from context). The group G acts on these vectors by its action
on the indices. Under this action, the vector space spanned by the characteristic vectors
of the canonical intersecting sets in a group G is a C[G]-module. We call this module the
EKR-module.
In this paper we consider the following three properties of a finite transitive group.
(1) EKR-property: A group has this property if the size of a maximum intersecting set
of permutations is the size of a canonical intersecting set.
(2) EKR-module property: A group has this property if the characteristic vector of any
maximum intersecting set is in the EKR-module.
(3) Strict-EKR property: A group has the strict-EKR property if the canonical inter-
secting sets are the only maximum intersecting sets.
The first result on the EKR-property for permutations dates back to the 1977 paper of
Deza and Frankl [16]. The EKR-property and the strict-EKR property have been an active
research area over the past decade [4, 12, 13, 17, 27, 30]. The EKR-module property is a
fairly new concept introduced by Ahmadi and Meagher in [6]. We state an observation on
the EKR-module property.
Observation 1.1. Let G be a transitive group. If G has the EKR-module property, then
the characteristic vector of any maximum intersecting set is a linear combination of the
characteristic vectors of the canonical intersecting set.
Note that if the transitive group G has the strict-EKR property, then it has the EKR-
module property. The notion of the EKR-module property is however more general; there
exist transitive groups that do not have the strict-EKR property, but have the EKR-module
property (the 2-transitive group PGL(n, q) for n ≥ 3 is an example of such groups). Fur-
ther, there are groups that do not have the EKR property, but admit the EKR-module prop-
erty (see [25] for examples of such groups).
It was shown in [28] that any 2-transitive group has the EKR-property, and in [26] it was
further shown that any 2-transitive group has the EKR-module property. There are several
papers showing different 2-transitive groups have the strict-EKR property [17, 27, 30]. In
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 3
many of these papers, the result follows using an algebraic method that relies on the 2-
transitivity of the group. In this paper we will show that this approach can be used for
GL(2, q) and SL(2, q) (which are transitive but not 2-transitive), where q is a prime power.
Theorem 1.2. For any prime power q, the groups GL(2, q) and SL(2, q) with their natu-
ral action on F2q \ {0} have the EKR-property and the EKR-module property. Moreover,
GL(2, q) does not have the strict-EKR property.
Let n ≥ 2 and q be a prime power. The EKR-property for GL(n, q) acting naturally
on the non-zero vectors of Fnq was first proved in [20]. Later, Ahanjideh and Ahanjideh [3]
proved a similar result and obtained various results about the EKR and strict-EKR prop-
erty of linear groups of degree 2. For instance, they proved that the permutation group
SL(2, q) has the EKR and strict-EKR property (therefore implying the EKR-module prop-
erty). However, this paper claimed that GL(2, q) also has the strict-EKR property, un-
fortunately the proof of this contained an error, as there is a maximum intersecting set of
GL(2, q) which is a subgroup not equal to a point-stabilizer (a corrected version has ap-
peared in [2]). The proof in [3] relies on the existence of Singer subgroups of GL(2, q),
i.e., regular subgroups of GL(2, q), to prove the EKR-property. Our proof also uses the
Singer subgroups, along with Hoffman’s ratio bound on a weighted adjacency matrix. This
approach puts into perspective the deep algebraic properties behind the EKR property. A
similar method is applied to obtain another proof that SL(2, q) has the EKR-module prop-
erty.
We also consider the affine group AGL(2, q), where q is a prime power. The action of
AGL(2, q) on the points of the affine plane is 2-transitive. Therefore, by [26] and [28], it
has both the EKR-property and the EKR-module property. In this paper, we prove however
that the action of AGL(2, q) on the lines of the affine plane does not have the EKR property.
Theorem 1.3. For any prime power q, the group AGL(2, q) acting on the lines of the affine
plane does not have the EKR property. However, if F ⊂ AGL(2, q) is intersecting, then
|F| ≤ q3(q − 1)2.
2 Module method
Throughout this section, we let G be a transitive permutation group of degree n. The
derangement graph of G, denoted by ΓG, has the elements of G as its vertices and two
vertices are adjacent if and only if they are not intersecting. The graph ΓG is the Cayley
graph of G, with connection set equal to the set of all derangements of G; we denote this
set by der(G). The derangement graph ΓG is defined so that a set S of G forms a coclique
in ΓG if and only if S is an intersecting set of G. Therefore, the largest intersecting sets of
G have size equal to the independence number α(ΓG) of the derangement graph ΓG.
The adjacency matrix A(X) of a graph X is a matrix in which rows and columns are
indexed by the vertices of X and the (i, j)-entry is 1 if i ∼ j, and 0 otherwise. The
eigenvalues of the graph are the eigenvalues of its adjacency matrix.
For any group G ≤ Sym(n), the connection set of the Cayley graph ΓG is the union
of all the conjugacy classes of derangements of G, so ΓG is a normal Cayley graph. The
eigenvalues of this graph can be calculated using the irreducible representations of G. This
follows from Babai’s formula for the eigenvalues of normal Cayley graphs [7] (this is also
in Diaconis and Shahshahani [11]).
4 Art Discrete Appl. Math. 6 (2023) #P1.05
Theorem 2.1. Let G be a permutation group. The eigenvalues of the Cayley graph ΓG are
given by
ηχ =
1
χ(id)
∑
x∈der(G)
χ(x),
where χ ranges over all irreducible characters of G. Moreover, the multiplicity of the
eigenvalue λ is
∑
χ χ(id)
2, where the sum is taken over all irreducible representations χ
with ηχ = λ.
The largest eigenvalue of ΓG is |der(G)|. This is the valency of the derangement graph
and it is afforded by the trivial character. Once we have the eigenvalues of a Cayley graph,
we can apply the ratio bound to find an upper bound on the size of a maximum coclique,
this bound is also known as Hoffman’s bound and Delsarte’s bound (a history of this bound
can be found in the lovely paper by Haemers [22]). We state the version of this bound for
a weighted graph, since this is the version that we will be using.
A weighted adjacency matrix AW (X) for a graph X is a symmetric matrix, with zeros
on the main diagonal, in which rows and columns are indexed by the vertices and the
(i, j)-entry is 0 if i ̸∼ j. A weighted adjacency matrix for a graph can be thought of as a
weighting on the edges of the graph, we note that the weight of an edge can be equal to 0.
For a d-regular graph, or for any weighted adjacency matrix with constant row sum equal
to d, the all-ones vector 1 is an eigenvector with eigenvalue d.
Theorem 2.2 (Delsarte-Hoffman bound or ratio bound [18, 22]). Let A be a weighted
adjacency matrix for a graphX on vertex set V (X). IfA has constant row sum d and least
eigenvalue τ , then the independence number α(X) of the graph X satisfies
α(X) ≤ |V (X)|
1− dτ
.
If equality holds for some coclique S with characteristic vector νS , then
νS −
|S|
|V (X)|
1
is an eigenvector with eigenvalue τ .
For any group G, the derangement graph ΓG is the union of graphs in the conjugacy
class association scheme; details can be found in [19]. The vertex set for this association
scheme is G, hence it has |G| vertices, and has rank equal to the number of conjugacy
classes of G. If C1, C2, . . . , Ck are the non-trivial conjugacy classes of G (so Ci ̸= {id},
for any i ∈ {1, 2, . . . , k}), then we let Ai be the adjacency matrix of the Cayley digraph
Cay(G,Ci), for any i ∈ {1, 2, . . . , k}. Let Ci1 , Ci2 , . . . , Ciℓ be the conjugacy classes of
derangements of G. We conclude that the adjacency matrix of ΓG is
A(ΓG) =
ℓ∑
j=1
Aij .
Since the conjugacy class association scheme of G is a commutative association scheme,
the matrices {Ai | i = 0, . . . , d} are simultaneously diagonalizable. Each of the common
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 5
eigenspaces is a union of irreducible C[G]-modules; so the eigenvalues can be found using
the irreducible representations of G. If χ is an irreducible representation of G, then the
eigenvalue of Ai belonging to χ is
|Ci|
χ(id)χ(xi) where xi is any element in Ci.
If a weighted adjacency matrix of a Cayley graph on G has the form A =
∑
i aiAi (so
the weights are constant on the conjugacy classes), then the eigenvalues of A are
1
χ(id)
∑
i
ai|Ci|χ(xi).
If the weights on all the conjugacy classes of derangements are 1, and 0 for all other con-
jugacy classes, then this equation is the formula in Theorem 2.1.
The permutation character of G is the character given by
fix(g) = |{i ∈ [n] : ig = i}|,
for g ∈ G. This is equal to the representation induced by the trivial representation on the
stabilizer of a point
ind(1Gi)
G(g) = fix(g).
The representation corresponding to the permutation character is called the permutation
representation. We call the C[G]-module of the permutation representation the permutation
module. A group is 2-transitive if and only if the permutation representation is the sum of
two irreducible representations, namely the trivial representation and another irreducible
representation. Thus, for a 2-transitive group the character χ(g) = fix(g)−1 is irreducible.
In this case the eigenvalue of the derangement graph afforded by χ is
ηχ =
−|der(G)|
n− 1
,
(see [5] for details). If G is 2-transitive then it is a straightforward calculation to show that
the dimension of the permutation module is 1+(n−1)2 and is spanned by {vi,j : i, j ∈ [n]}
(again, see [5] for details). So, in this case, the permutation module is isomorphic to the
EKR-module. Further, every 2-transitive group has the EKR-module property [26]. This
means that for any 2-transitive group the characteristic vector of the largest intersecting set
is a linear combination of the characteristic vectors of the canonical intersecting sets.
For any simply transitive group (i.e., transitive groups that are not 2-transitive), the
permutation module is still a C[G]-module, but it is the sum of more than two irreducible
C[G]-modules. In many cases the eigenvalues for the non-trivial irreducible modules in the
permutation module are not equal, and the ratio bound does not hold with equality.
The goal of this paper is to demonstrate how the algebraic approach can be applied to
some simply transitive groups, namely general linear group and the special linear group.
Our plan is to weight the adjacency matrix of the graph ΓG, forG = GL(2, q) and SL(2, q),
so that the eigenvalues for all non-trivial irreducible C[G]-modules in the permutation mod-
ule are equal and the ratio bound holds with equality. Then, we can conclude the group has
the EKR property, and we can further show that the characteristic vector for any maximum
coclique is in the EKR-module. This approach has been used for other simply transi-
tive groups, such as the transitive action of Sym(n) on both ordered and unordered tuples
in [12, 13], for n sufficiently large. This approach is also effectively used for the action of
Sym(n) on pairs [24] and 3-sets [9] of [n] for all n ≥ 5.
6 Art Discrete Appl. Math. 6 (2023) #P1.05
We will assign weights to the conjugacy classes of derangements in the group; the goal
is to find a weighting to get the best bound from the ratio bound. We will form a linear
program, in which the eigenvalues from all but the trivial representation are greater than or
equal to -1 and then we maximize the eigenvalue corresponding to the trivial representation
(this will be the largest eigenvalue). We consider the following setup.
Linear program. Let G ≤ Sym(n) be a transitive permutation group with conjugacy
classes of derangements D1, D2, . . . , Dk and trivial character χ0. Let g1, g2, . . . , gk be
representatives of the conjugacy classes D1, D2, . . . , Dk, respectively. We consider the
following optimization problem.
Maximize λχ0 =
k∑
i=1
ωi|Di|,
Subject to
λχ =
k∑
i=1
ωi|Di|χ(gi) ≥ −1, ∀χ ∈ Irr(G) \ {χ0},
ωi ∈ R, ∀i ∈ {1, 2, . . . , k}.
(2.1)
If the solution of the linear programming (LP) problem (2.1) is equal to n−1, then applying
the ratio bound, we have α(ΓG) ≤ |G|1−n−1−1
= |G|n . Hence, G would have the EKR property.
Lemma 2.3. LetG ≤ Sym(n) be a transitive group. If there is a weighing of the conjugacy
classes of derangements of G so that the non-trivial representations in the permutation
character give the least eigenvalue in LP (2.1), then the maximum LP (2.1) can give is
n− 1.
Proof. Assume that
fix(g) =
ℓ∑
j=0
mjχj(g)
where χ0 is the trivial representation and χ1, χ2, . . . , χℓ are the other constituents of the
permutation character of G and mj is the multiplicity of χj . Note
n = fix(id) =
ℓ∑
j=0
mjχj(id).
Assume that the conjugacy classes Di, for i = 1, . . . , k, are the conjugacy classes of
derangements of G. Let g1, g2, . . . , gk be representatives of D1, D2, . . . , Dk, respectively.
Let (ωi)i=1,...,k be a weighting on the conjugacy classes such that
−1 ≤ λχj =
1
χ(id)
k∑
i=1
ωi|Di|χj(gi),
for every irreducible character χj of G. In particular, for any j ∈ {1, 2, . . . , ℓ}, we write
−(χj(id)) ≤
k∑
i=1
ωi|Di|χj(gi).
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 7
Summing over all the χj ̸= χ0 (with their multiplicities) in the decomposition of fix(g),
we get
−(n− 1) = −
ℓ∑
j=1
mj(χj(id)) ≤
ℓ∑
j=1
k∑
i=1
ωi|Di|mjχj(gi) =
k∑
i=1
ωi|Di|(−1).
Therefore
∑k
i=1 ωi|Di| ≤ n − 1. Consequently, the maximum value of λχ0 for any such
weighting is n− 1.
3 The general linear group GL(2, q)
In this section q is assumed to be a prime power. The general linear group GL(2, q) acts
naturally on the non-zero vectors of F2q by left multiplication. The size of the maximum
intersecting sets in GL(2, q) have been determined in [3], and there are two intersecting
families of maximum size. The first family is the collection of canonical intersecting sets
(i.e., cosets of point-stablizers in GL(2, q)). The other family is the collection of all sub-
groups that are the stabilizer of a line and their cosets; these are the subgroups
Hℓ = {M ∈ GL(2, q) | ∀v ∈ F2q, Mv − v ∈ ℓ}
for some line ℓ of F2q , and their cosets. These two families are believed to be the only
maximum intersecting sets [2].
We will be using the notation from Adams [1]. The action of GL(2, q) on the q2 − 1
non-zero vectors in F2q is not 2-transitive, so this permutation group is simply transitive.
This action has q orbitals, which are described below.
(1) The diagonal
{
(v, v) | v ∈ F2q \ {0}
}
is clearly an orbital of size q2 − 1.
(2) There are q − 2 orbitals each of size (q2 − 1). The representatives of these orbitals
are (v, cv) where v ∈ F2q \ {0} and c ∈ Fq\{0, 1}.
(3) There is one final orbital of size (q2 − 1)(q2 − q), a representative of this orbital is
(u, v) where u and v are not co-linear elements of F2q \ {0}.
3.1 Conjugacy classes of GL(2, q)
Still following Adams’ [1] notation, the conjugacy classes of GL(2, q) can be divided into
four categories denoted by c1(x), c2(x), c3(x, y) and c4(z). The structure of the matrices
in these categories can be used to count the number of derangements in GL(2, q).
The first category is the matrices that are similar to the matrix of the form
c1(x) =
[
x 0
0 x
]
,
for some x ∈ F∗q . These matrices have one eigenvalue and are diagonalizable over Fq . Each
conjugacy class in this category has size 1. The conjugacy class that contains the identity
is c1(1). If x ̸= 1, then c1(x) is a conjugacy class of derangements. Thus, there are q − 2
conjugacy classes of derangements in this category, each of size 1.
8 Art Discrete Appl. Math. 6 (2023) #P1.05
The next category is the set of matrices similar to a matrix of the form
c2(x) =
[
x 1
0 x
]
,
for x ∈ F∗q . These matrices have only one eigenvalue and are not diagonalizable. Each
conjugacy class in this category has size q2 − 1. If x ̸= 1, then c2(x) is a conjugacy class
of derangements. Thus, there are q− 2 conjugacy classes of derangements in this category,
each of size q2 − 1.
The third category is the set of matrices similar to a matrix of the form
c3(x, y) = c3(y, x) =
[
x 0
0 y
]
for x, y ∈ F∗q and x ̸= y. These matrices have two distinct eigenvalues in Fq . Each
conjugacy class in this category has size q(q + 1). If x, y ̸= 1, then c3(x, y) is a conjugacy
class of derangements. Thus, there are
(
q−2
2
)
conjugacy classes of derangements in this
category, each of size q(q + 1).
The last category corresponds to matrices that do not have eigenvalues in Fq . Use Eq to
represent the unique quadratic extension of Fq , so these matrices have eigenvalues in Eq .
If A ∈ GL(2, q) is such matrix, then its characteristic polynomial f(t) = t2 + at + b is
irreducible over Fq . Hence, for q odd, ∆ = a2 − 4b is not a square in Fq . If δ2 = ∆, for
some δ ∈ Eq \Fq , then we can identify Eq to be Fq(δ). The map z = x+ δy 7→
[
x δy
y x
]
,
for x, y ∈ Fq with (x, y) ̸= (0, 0), is an embedding of E∗q into GL(2, q) (this matrix is
called the companion matrix of the element in E∗q). Thus, the final category of conjugacy
classes are the matrices similar to a matrix of the form
c4(z) =
[
x δy
y x
]
,
with δ ∈ Eq \Fq where z = x + δy ∈ Eq \Fq . Each conjugacy class in this category has
size q(q − 1). All of these conjugacy classes are classes of derangements. Thus, there are(
q
2
)
conjugacy classes of derangements in this category, each of size q(q − 1).
The total number of derangements in GL(2, n) is
(q−2)(1)+(q−2)(q2−1)+
(
q − 2
2
)
(q(q+1))+
(
q
2
)
(q(q−1)) = q(q3−2q2−q+3).
3.2 Irreducible representations of GL(2, q)
Again, we use the notation for the irreducible representations of GL(2, q) given in [1]
where more details can be found. The irreducible representations for GL(2, q) are, like the
conjugacy classes, split into four categories.
The first category is the set of degree-1 representations, denoted by ρ′(α), where α is
an irreducible representation of F∗q . If α is the trivial representation of F∗q , then ρ′(α) is
the trivial representation of GL(2, q). The next category are the degree-q representations
denoted by ρ(α), again α is an irreducible representation of F∗q . The third category are the
degree-(q− 1) representations denoted by π(χ). Here χ is an irreducible representation of
Eq with χ ̸= χ. The final category are the degree-(q+1) representations denoted by ρ(µ),
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 9
Number q − 1 q − 1
(q−1
2
) (q
2
)
Size 1 q2 − 1 q(q + 1) q(q − 1)
c1(x) c2(x) c3(x, y) c4(z)
Rep : Dim Number
ρ′(α) : 1
α = 1 1 1 1 1 1
α ̸= 1 q − 2 α(x2) α(x2) α(xy) α(Nz)
ρ(α) : q
α = 1 1 q 0 1 −1
α ̸= 1 q − 2 qα(x2) 0 α(xy) −α(Nz)
π(χ): q − 1
χ
(q
2
)
(q − 1)χ(x) −χ(x) 0 −χ(z)− χ(z)
ρ(µ) : q + 1
µ = (1, β) q − 2 (q + 1)β(x) β(x) µ(g) + µw(g) 0
µ = (α, β), α ̸= 1
(q−2
2
)
(q + 1)α(x)β(x) α(x)β(x) µ(g) + µw(g) 0
Table 1: Character Table for GL(2, q)
where µ is an irreducible representation of F∗q ×F
∗
q . This character is expressed in terms of
the norm map, N : E∗q → Fq with N(z) = zq+1.
The values of these characters on the four categories of conjugacy classes is given in
Table 1.
From Table 1 we can find the permutation character. Define a character χ by
χ := 1+ρ(1) +
∑
β
ρ(1, β). (3.1)
This character is the sum of the trivial representation, one representation of dimension
q, and q − 2 of the representations with dimension q + 1. These degree-1 and degree-
q representations both have α = 1. The q − 2 degree-(q + 1) representations are the
representations with α = 1 and β ̸= 1. The next result proves that the character χ is the
permutation character of GL(2, q).
Lemma 3.1. The character χ is the permutation character for the action of GL(2, q) on
the non-zero vectors of F2q .
Proof. Recall that the permutation character of the action of GL(2, q) on F2q \ {0} is given
by fix(A) = |
{
x ∈ F2q \ {0} | Ax = x
}
|, for any A ∈ GL(2, q). We will prove that χ is
equal to the permutation character on each conjugacy class of GL(2, q).
The value of χ on the conjugacy classes of type c1(x) is
1 + q(1) +
∑
β∈F̂q\{1}
(q + 1)(1)β(x) (3.2)
10 Art Discrete Appl. Math. 6 (2023) #P1.05
If x = 1, then this equals 1+q+(q−2)(q+1) = q2−1. For x = 1, c1(x) is the conjugacy
class containing the identity, so χ gives the number of fixed points for this conjugacy class.
If x ̸= 1, then χ is equal to 1 + q + (−1)(q + 1) = 0. For x ̸= 1, the class c1(x) is a
conjugacy class of derangements, so this number is correct.
The value of χ on c2(x) is
1 + 0 +
∑
β∈Irr(F∗q)\{1}
β(x).
If x = 1, this equals 1 + 0 + (q − 2) = q − 1; for x ̸= 1, this is equal to 1 + 0− 1 = 0. In
both cases, this is the number of fixed points of the elements in the conjugacy classes.
Since α = 1 in all the irreducible representations in χ, the value of χ on an element in
any conjugacy class in category c3(x, y) can be calculated to be
1 + 1(xy) +
∑
β∈Irr(F∗q)\{1}
(β(y) + β(x)) .
If x = 1, then
∑
β∈Irr(F∗q)\{1}
β(x) = q − 2; if x ̸= 1, the sum is −1. So if either x = 1
or y = 1, the value of χ on an element in c3(x, y) is equal to 1+1+(q−2)+−1 = q−1.
Similarly, if x and y are both not equal to 1, then this sum is 1 + 1 + −1 + −1 = 0. In
either case, the value of χ on an element of c3(x, y) is equal to the number of fixed points
of the element.
Finally for the conjugacy class c4(z) the value of this character is
1 + (−α(Nz)) + 0 = 1 + (−1) + 0 = 0
(again, with α = 1) for all z, which is correct since this is a conjugacy class of derange-
ments.
With these values, it is easy to see that the permutation representation is multiplicity-
free and to calculate the dimension of the permutation module of GL(2, q).
Corollary 3.2. The permutation representation is multiplicity-free and the permutation
module has dimension q3 + q2 − 3q − 1.
Proof. The dimension of the permutation module is the sum of the squares of the dimen-
sions of the irreducible representations in the decomposition. By Lemma 3.1, the dimen-
sion of the permutation module is
1 + q2 + (q − 2)(q + 1)2 = q3 + q2 − 3q − 1.
3.3 Cliques in ΓGL(2,q)
We can easily show that GL(2, q) has the EKR property using the well-known clique-
coclique bound. We will use a stronger version of this result from [18, Section 3.7] to
prove that the group also has the EKR-module property.
Theorem 3.3. Let {A0, A1, . . . , Ad} be an association scheme on v vertices and let X be
the union of some graphs in the association scheme. Let Ej , where j = 0, 1, . . . , d, be the
idempotents of the association scheme, with E0 = 1vJ .
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 11
If C is a clique and S is a coclique in X , then
|C| |S| ≤ v.
If equality holds, and x and y are the characteristic vectors of C and S respectively, then
xTEjx y
TEjy = 0 (i = 1, . . . , d).
The following lemma is well known and its proof is omitted.
Lemma 3.4. There is a subgroupH ofGL(2, q) isomorphic toCq2−1. Further,H contains
all the elements of conjugacy classes c1(x) and exactly two elements from each of the
conjugacy classes of type c4(z).
The following lemma was proved in [3, 20]. We include its proof here for completeness.
Lemma 3.5. The graph ΓGL(2,q) has a clique of size q2 − 1 and the group GL(2, q) has
the EKR-property.
Proof. From Lemma 3.4 the GL(2, q) has a subgroup H of size q2 − 1 in which all ele-
ments, except the identity, are derangements. For any distinct x, y ∈ H , it must be that
x−1y ∈ H and is a derangement, so if x ̸= y, then x and y are not intersecting. This im-
plies that the elements in H form a clique in ΓGL(2,q). Thus by the Theorem 3.3 a coclique
in ΓGL(2,q) can be no larger than
|GL(2,q)|
q2−1 = q(q−1). Since this is the size of the stabilizer
of a point, a canonical coclique is an intersecting set of maximum size.
Equality in the clique-coclique bound, implies a stronger result. If ψ is an irreducible
representation of GL(2, q), then the ψ-projection of a set S ∈ GL(2, q) is the projection of
the characteristic vector of S to the ψ-module. This projection is given by the matrix
Eψ(g, h) =
ψ(1)
|GL(2, q)|
ψ(hg−1),
and the projection of S to the ψ-module is EψvS . In particular, if ψ(S) =
∑
s∈S ψ(s) is
not zero, then the projection of S to the ψ module is not zero.
Lemma 3.6. The projection of any maximum coclique in ΓGL(2,q) to the C[GL(2, q)]-
module corresponding the character ρ(µ), where µ = (β, β) with β ̸= 1, or corresponding
to ρ(α), where α2 = 1 with α ̸= 1, equals 0.
Proof. The clique-coclique bound in Theorem 3.3 holds with equality, so for any irre-
ducible representation ψ of GL(2, q)
vTCEψvC v
T
SEψvS = 0
where C is a maximum clique and S is a maximum coclique. For any irreducible rep-
resentation ψ with EψvC ̸= 0, it must be that EψvS = 0. It suffices to show that
ψ(C) =
∑
c∈C ψ(c) is not zero for ψ = ρ(µ), with µ = (β, β) and β ̸= 1, and for
ψ = ρ(α), where α2 = 1, with α ̸= 1,
Let ρ(µ) be a character with µ = (β, β) and β ̸= 1. For the group H in Lemma 3.4
ρ(µ)(H) = (q + 1)(1)(q − 1) + 2(0)
(
q
2
)
= q2 − 1.
12 Art Discrete Appl. Math. 6 (2023) #P1.05
Similarly, let ρ(α) be the character of dimension q with α2 = 1. Then
ρ(α)(H) = q(1)(q − 1) + 2
(
(1)
q − 1
2
+ (1)
1
2
(
q − 1
2
)
+ (−1)1
2
(
q − 1
2
))
= q2 − 1.
Since the value of these characters over a maximum clique is non-zero, the projection of
any maximum coclique to these modules must be zero.
3.4 Eigenvalues for GL(2, q)
In this section we give a second proof that GL(2, q) has the EKR-property by calculating the
eigenvalues of the different classes in the conjugacy class association scheme on GL(2, q).
From Subsection 3.1 we know that the derangements of GL(2, q) belong to four fami-
lies of conjugacy classes: c1 = c1(x), with x ̸= 1; c2 = c2(x), with x ̸= 1; c3 = c3(x, y),
with x, y both not equal to one; and c4 = c4(z). Define Xi for i = 1, 2, 3, 4 to be the
graph with vertices indexed by elements of GL(2, q) and two vertices g, h are adjacent if
and only if gh−1 belongs to a conjugacy class in the family ci. Then ΓGL(2,q) =
∑4
i=1Xi.
The graphs Xi all belong to the conjugacy class association scheme for GL(2, q), so the
eigenvalues can be found using Table 1. These eigenvalues are given in Table 2, the rows
give the different types of representations, and the columns are the categories of conjugacy
classes of derangements. For each category of conjugacy class, we record the sum of the
value of the character over the different conjugacy classes of derangements in the category.
From this, we can easily calculate the eigenvalue of the derangement graph of GL(2, q);
these are given in the final column.
The spectrum of the derangement graph is (the raised number is the multiplicity)
q(q3−2q2−q+3)(1), q(q
4−2q3−2q2+4q+1), −q2+2q((q+1)
2(q−2)), −q2+q+1(q
2).
Note that the ratio bound gives a bound of
(q + 1)q(q − 1)2
1− q(q
3−2q2−q+3)
−q2+q+1
=
q(q2 − q − 1)
(q − 1)
,
which does not hold with equality. We next show that there is a weighted adjacency matrix
for which the ratio bound holds with equality. To get these weights, we set the eigenvalues
arising from non-trivial representations in the permutation representation to be equal to −1.
We weight the conjugacy classes of GL(2, q) with the weights in Table 3, and the
eigenvalues of the weighted adjacency matrix are given in Table 4.
The ratio bound on this weighted adjacency matrix gives
α(ΓGL(2,q)) ≤
|GL(2, q)|
1− q2−2−1
= q(q − 1).
This shows again that GL(2, q) has the EKR property.
Recall that the characters that sum up to the permutation character are the q − 2 repre-
sentations of dimension q + 1 with α = 1, the character of dimension q with α = 1 and
the trivial representation. All the non-trivial characters in the permutation character have
eigenvalue equal to −1 under this weighting. There are two other representation that also
give the eigenvalue −1; ρ′(α) and ρ(α) both with α2 = 1. To show that GL(2, q) has the
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 13
Category c1(x) c2(x) c3(x, y) c4(z)
x ̸= 1 x ̸= 1 x, y ̸= 1
Number q − 2 q − 2
(q−2
2
) (q
2
)
size 1 q2 − 1 q(q + 1) q(q − 1)
Rep : Dim number Eigenvalue
ρ′(α) : 1
α = 1 1 q − 2 q − 2
(q−2
2
) (q
2
)
q4−2q3−q2+ 3q
α2 = 1 1 q − 2 q − 2 − q−3
2
− q−1
2
q
else q − 3 −1 −1 1 0 q
ρ(α) : q
α = 1 1 q(q − 2) 0
(q−2
2
)
−
(q
2
)
−q2 + q + 1
α2 = 1 1 q(q − 2) 0 − q−3
2
q−1
2
q
else q − 3 −q 0 1 0 q
π(χ) : q − 1
χ = 1 a (q−1)(q−2) −(q − 2) 0 q − 1 q
χ ̸= 1 b −(q − 1) 1 0 0 q
ρ(µ) : q + 1
α = β c (q+1)(q−2) q − 2 −(q − 3) 0 q
α = 1 q − 2 −(q + 1) −1 −(q − 3) 0 −q2 + 2q
else d −(q + 1) −1 2 0 q
Table 2: The eigenvalues for the conjugacy classes of GL(2, q). Here a is either q−12 or
q
2 ;
b is either (q−1)
2
2 or
q(q−2)
2 ; c is
q−3
2 or
q−2
2 ; and d is either
(q−3)2
2 or
(q−2)(q−4)
2 , each
value depending on the parity of q.
EKR-module property, we will need to show that the projection of any maximum coclique
to these two modules is equal to 0. Lemma 3.6 implies this result for the representation
ρ(α), with α2 = 1. The representation ρ′(α) with α2 = 1, is a constituent of the repre-
sentation induced from the trivial representation on SL(2, q), so we need to consider the
subgroup SL(2, q).
3.5 The group SL(2, q)
The subgroup SL(2, q) of GL(2, q) also acts transitively on the non-zero vectors of F2q .
The conjugacy classes of derangements are essentially the same, but only the classes where
the determinants of the matrices are equal to 1 are included in SL(2, q). Many of the
irreducible characters of SL(2, q) are similar to the characters of GL(2, q). The character
table of SL(2, q) is given in [1]. Using this table it is possible to calculate the sum of the
value of all the irreducible character over the different conjugacy classes of derangements
in the different categories. The tables are slightly different for different values of q. We
report the values for q ≡ 1 (mod 4) and q ≡ 3 (mod 4) first, and then we discuss when q
is even.
14 Art Discrete Appl. Math. 6 (2023) #P1.05
Type c1(x), x ̸= 1 c2(x), x ̸= 1 c3(x, y), x, y ̸= 1 c4(z)
Weight − q−1q(q−2)
1
q(q−2)
1
q(q−3)
1
q(q−1)
Table 3: A weighting for the conjugacy classes of derangements in GL(2, q)
Rep:Dim number weighted eigenvalue
ρ′(α): 1
α = 1 1 q2 − 2
α2 = 1 (if q is odd) 1 -1
else q − 3 q−1q−2 +
q+1
q−3
π(χ): q − 1
χ = 1 q−12 or
q
2 q − 3
χ ̸= 1 (q−1)
2
2 or
q(q−2)
2
2
q−2
ρ(α): q
α = 1 1 -1
α2 = 1 1 -1
else q − 3 1q
(
q−1
q−2 +
q+1
q−3
)
ρ(µ): q + 1
α = β q−32 or
q−2
2 -1
α = 1 q − 2 -1
else (q−3)
2
2 or
(q−2)(q−4)
2
2
q−3
Table 4: Eigenvalues of the weighted adjacency graph for GL(2, q).
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 15
c1(x) c2(x) c3(x, y) c4(z)
size 1 (q2 − 1)/2 q(q + 1) q(q − 1)
Eigenvalue
Rep : Dim number of weighted matrix
ρ′(α) : 1
α = 1 1 1 2 q−3
2
q−1
2
q2 − 2
ρ(α) : q
α = 1 1 q 0 q−3
2
− q−1
2
-1
ρ(α) : q + 1
α(−1) = −1 q−1
4
−(q + 1) −2 0 0 -1
else q−5
4
(q + 1) 2 −2 0 -1
π(χ) : q − 1
χ(−1) = −1 q−1
4
−(q − 1) 2 0 0 q+1
q−1
χ(−1) = 1 q−1
4
(q − 1) −2 0 2 2 q
2−5
(q−1)2
π(χ) : q+1
2
w±e 2
q+1
2
1 -1 0 -1
π(χ) : q−1
2
w±0 2 −
q−1
2
1 0 0 q+1
q−1
Table 6: The eigenvalues for the conjugacy classes of SL(2, q) for q ≡ 1 (mod 4).
For q odd, the eigenvalues for the different conjugacy classes are recorded in Tables 6
and 7. Like the group GL(2, q), the ratio bound does not hold with equality for the group
SL(2, q), so a weighted adjacency matrix must be used. The weightings are given in Ta-
ble 5.
Type c1(x), x ̸= 1 c2(x) c3(x, y), x, y ̸= 1 c4(z)
Weight 0 1q−1
1
q
q2−3
q(q−1)2
Table 5: A weighting for the conjugacy classes of derangements in SL(2, q) with q odd.
The eigenvalues of the resulting weighted adjacency matrices are given in the final
columns of the Tables 6 and 7.
For q even, all the conjugacy classes of derangements are in either category c3(x, y) or
c4(z). Again we use the table of the irreducible characters is given in [1]. Table 9 records
the eigenvalues of the different conjugacy classes of derangements. The ratio bound does
not hold for the adjacency matrix, so a weighted adjacency matrix is used; these weights are
recorded in Table 8. The final column of Table 9 contains the eigenvalues of the weighted
adjacency matrix.
16 Art Discrete Appl. Math. 6 (2023) #P1.05
size 1 q
2−1
2
q(q + 1) q(q − 1)
c1(x) c2(x) c3(x, y) c4(z)
Eigenvalue
Rep : Dim number of weighted matrix
ρ′(α) : 1
α = 1 1 1 2 q−3
2
q−1
2
q2 − 2
ρ(α) : q
α = 1 1 q 0 q−3
2
− q−1
2
-1
ρ(α) : q + 1
α(−1) = −1 q−3
4
−(q + 1) −2 0 0 -1
else q−3
4
(q + 1) 2 −2 0 -1
π(χ) : q − 1
q+1
4
−(q − 1) 2 0 0 q+1
q−1
q−3
4
(q − 1) −2 0 2 2 q
2−5
(q−1)2
π(χ) : q+1
2
w±e 2 − q+12 -1 0 0 -1
π(χ) : q−1
2
w±0 2
q−1
2
-1 0 1 q
2−5
4
Table 7: The eigenvalues for the conjugacy classes of SL(2, q) for q ≡ 3 (mod 4).
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 17
size q(q + 1) q(q − 1)
c3(x, y) c4(z)
Eigenvalues
Rep : Dim number of weighted matrix
ρ′(α) : 1
α = 1 1 q−22
q
2 q
2 − 2
π(χ) : q − 1
χ q2 0 1
q+2
q
ρ(α) : q
α = 1 1 q−22 −
q
2 -1
ρ(α) : q + 1
α q−22 −1 0 -1
Table 9: The eigenvalues for the conjugacy classes of SL(2, q) for q even.
Type c3(x, y), x, y ̸= 1 c4(z)
Weight 1q
q+2
q2
Table 8: A weighting for the conjugacy classes of derangements in SL(2, q) with q even.
The decomposition of the permutation representation of SL(2, q) is similar to the per-
mutation representation of GL(2, q)—we omit the proof as it is very similar to the proof
for GL(2, q). For q odd it is the following
χ = 1+ρ(1) + 2
∑
α
ρ(α) + π(w±e ),
and for q even it is
χ = 1+ρ(1) + 2
∑
α
ρ(α).
Lemma 3.7. For all q the group SL(2, q) has the EKR property. Further, if S is a maximum
coclique in ΓSL(2,q), then the characteristic vector of S is in the permutation module.
Proof. For any value of q, the ratio between the largest eigenvalue and the least is −(q2−2)
in the weighted adjacency matrix. So for all of these weighted adjacency matrices, the ratio
bound gives
α(Γ) ≤ |SL(2, q)|
q2 − 1
= q,
which is exactly the order of the stabilizer of a point. Thus, the ratio bound holds with
equality for SL(2, q) for all q, so we conclude that SL(2, q) has the EKR property.
18 Art Discrete Appl. Math. 6 (2023) #P1.05
The ratio bound further implies if vS is the characteristic vector of S, then vS − 1q2−1 1
is a −1-eigenvector. For all values of q, the only representations that afford an eigenvalue
of −1 are representations in the permutation representation. This implies that vS is in the
permutation module.
3.6 GL(2, q) has the EKR-module property
In this next section we will prove that GL(2, q) has the EKR-module property.
Theorem 3.8. Let S be a maximum coclique in ΓGL(2,q). Then the characteristic vector of
S is in the permutation module.
Proof. The modules with eigenvalue −1 in the weighted adjacency matrix correspond to
the representations:
(1) ρ(µ) with µ = (β, β), or µ = (1, β)
(2) ρ(α) with α = 1, or α2 = 1
(3) ρ′(α) with α2 = 1.
By the ratio bound, the characteristic vector of any maximum coclique lies in the span of
these modules. The modules in the permutation representation are 1 = ρ′(1), ρ(1) and all
ρ(µ) with µ = (1, β). To prove this theorem is it necessary to show that the projection
of a maximum coclique to any of the modules with eigenvalue −1, that are not in the
decomposition of the permutation representation, is 0. By Lemma 3.6, the characteristic
vector of a maximum coclique cannot be in any ρ(µ) module with µ = (β, β) where β ̸= 1,
or in any the ρ(α) modules with α ̸= 1.
The last case to be considered is the degree 1 representation ρ′(α) with α2 = 1, and
α ̸= 1. The sum of all the degree 1 representations of GL(2, q) is the representation
induced from the trivial representation on SL(2, q). If T is a transversal for the cosets of
SL(2, q) in GL(2, q), then for α ̸= 1,
∑
x∈T ρ
′(α)(x) = 0.
Let S be a maximum coclique in GL(2, q), by Lemma 3.5, |S| = q(q − 1). Then
S ∩ SL(2, q) is a coclique of SL(2, q), and by Lemma 3.7, it cannot be larger than q.
Further, for any coset x SL(2, q) it must be that x−1(S ∩ xSL(2, q)) is also a coclique
in SL(2, q), and so |S ∩ xSL(2, q)| ≤ q. Since the sets S ∩ xSL(2, q) partition S and
|S| = q(q − 1), each |S ∩ xSL(2, q)| has size exactly q.
For any ρ′(α),
ρ′(α)(S) =
∑
x∈T
ρ′(α)(S ∩ xSL(2, q)) = q
∑
x∈T
ρ′(α)(x)
which equals 0, unless α = 1.
To prove that GL(2, q) has the EKR-module property, we will prove that the charac-
teristic vectors of the canonical intersecting sets form a spanning set for the permutation
module. By the Lemma 3.8, we have that the characteristic vector of any canonical inter-
secting set is in the permutation module. So we only need to show that the span of the these
vectors has the same dimension at the permutation module.
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 19
For x, y ∈ F2q , define v(x,y) be the length-|GL(2, q)| vector indexed by the elements in
GL(2, q). The g-entry of v(x,y) is 1 if g(x) = y, and 0 otherwise—these are the charac-
teristic vectors of the canonical cocliques. Next pick a set of pairwise non-colinear vectors
{xi : i = 1, 2, . . . , q + 1} from F2q \ {0}. For each xi with i = 1, 2, . . . , q + 1, define
Si = {v(xi,y) | y ∈ F
2
q \ {0}}.
This means that each Si is a set of q2 − 1 vectors, and there are q + 1 such sets.
Lemma 3.9. The set S1 ∪S2 ∪ · · · ∪Sq+1 is a spanning set for the permutation module of
GL(2, q).
Proof. Each canonical coclique is a maximum coclique and by Theorem 3.8 the vectors
v(x,y) are in permutation module. It only remains to show that the span of these vectors is
the entire module. From Corollary 3.2, it is sufficient to show that span of these vectors has
dimension q3 + q2 − 3q − 1.
Define a matrix N with columns the characteristic vectors in the sets Si, for i ∈
{1, . . . , q+1}. Order these vectors so that the vectors within a single set Si are consecutive,
and, within Si, the vectors v(i,y1) and v(i,y2) with y1 and y2 co-linear are consecutive.
The dot product of any v(i,j) and v(i,k) is q(q − 1) if j = k, and 0 otherwise. The dot
product of any two vectors v(i,j) and v(a,b) with i ̸= a, and j not co-linear with b is equal
to 1. Then NTN is
q(q − 1)I(q+1)(q2−1) +
(
(Jq+1 − Iq+1)⊗ ((Jq+1 − Iq+1)⊗ Jq−1)
)
This is a square matrix with (q + 1)(q2 − 1) rows and columns. The spectrum is
{q(q2 − 1)(1), (q2 − 1)(q
2), q(q − 1)((q−2)(q+1)
2), 0(2q)}
(the numbers in parentheses above the numbers is the multiplicity of the eigenvalue). Thus
the rank ofNNT , and henceN , is (q+1)(q2−1)−2q = q3+q2−3q−1, as required.
Since the characteristic vector of any maximum intersecting set in GL(2, q) is in the
permutation module and can be expressed as a linear combination of the canonical co-
cliques. So we conclude that GL(2, q) has the EKR-module property. We will prove the
same result of SL(2, q), using a slightly different approach.
Lemma 3.10. The group SL(2, q) has the EKR-module property.
Proof. Define the matrix N so that the rows correspond to the elements in SL(2, q) and the
columns pairs of elements from F2q . The (g, (i, j)) entry ofN is 1 if ig = j and 0 otherwise.
The columns of this matrix are the characteristic vectors of the canonical cocliques, by
Lemma 3.7 these are in the span of the q2 − 2- and −1-eigenspace. So it remains to prove
that the rank of this matrix is one more than the dimension of the -1-eigenspace of the
weighted adjacency matrix.
Consider the matrix NNT . The (g, h) entry of this matrix is the number elements on
which g and h agree. If g = h the entry is q2 − 1. The non-deragements in SL(2, q)
belong to the two conjugacy classes: c2(1, 1) and c2(1, γ). If gh−1 is in the conjugacy
class c2(1, 1) or c2(1, γ), then the (g, h)-entry is equal to q − 1. All other entries of NNT
are equal to 0.
20 Art Discrete Appl. Math. 6 (2023) #P1.05
Rep. ρ(α) ρ(1) ρ′(1) π(χ) ω±e ω
±
0 ω
±
Eigenvalue q − 1 0 q2 − 1 −(q + 1) q − 1 −(q + 1) 0
Multiplicity (q + 1)2 q−3
2
q2 1
(q−1)3
2
2( q+1
2
)2 2( q−1
2
)2 2q2
Table 10: Eigenvalues of A1 +A2.
This means that NNT is equal to
(q2 − 1)Iq(q2−1) + (q − 1)(A1 +A2),
where A1 and A2 are the adjacency matrices in the conjugacy class association scheme
corresponding to the conjugacy classes c2(1, 1) and c2(1, γ).
The eigenvalue of A1 + A2 can be calculated using the character table of SL(2, q) and
are given in Table 10.
From this it can be seen that eigenvalues of NNT are
(
((q2 − 1) + (q − 1)2)
(
(q−3)(q+1)2
2
)
, (q2 − 1)(2q
2), q(q2 − 1)(1), 0
(
(q−1)3
2
+2(
(q−1)
2
)2
))
.
The rank of N is (q − 1)q(q + 1)− q(q−1)
2
2 =
q(q−1)(q+3)
2 , which is one less than the
dimension of the -1-eigenspace of the weighted adjacency matrix for SL(2, q).
For q even there is only one conjugacy class of non-derangements. So NNT is equal
to
(q2 − 1)Iq(q2−1) + (q − 1)(A1),
where A1 is the adjacency matrix in the conjugacy class scheme that corresponds to the
single class of non-derangements. The eigenvalues of A1 are(
q2 − 1(1), (q − 1)
(
(q+1)2(q−2)
2
)
, 0(q
2),−(q + 1)
(
q(q−1)2
2
))
.
We deduce that eigenvalues of NNT , for q even, are(
q(q2 − 1)(1), ((q2 − 1) + (q − 1)2)
(
(q+1)2(q−2)
2
)
, (q2 − 1)(2q
2), 0
(
q(q−1)2
2
))
.
The rank of N is (q − 1)q(q + 1) − q(q−1)
2
2 =
q(q−1)(q+3)
2 , which is one less than the
dimension of the -1-eigenspace of the weighted adjacency matrix for SL(2, q).
Theorem 3.11. The group GL(2, q) does not have the strict-EKR property.
Proof. For a line ℓ, let Sℓ be the set of all M ∈ GL(2, q) with Mv − v ∈ ℓ for all v ∈ F2q .
This forms a group of size q(q − 1), this can be seen by counting the number of matrices
in Sℓ. Assume without loss of generality that ℓ is the line containing (0, 1) ∈ F2q . Then
any matrix in Sℓ has the (1, 1)-position equal to 1 and the (1, 2)-position equal to 0, then
there are (q − 1) choices for the (2, 2)-entry, since it cannot be 0, and q choices for the
(2, 1)-entry. Finally, from the structure of these matrices, it can be seen that each matrix in
Sℓ has a fixed point. Since Sℓ is a subgroup, for any M1,M2 ∈ Sℓ the matrix M1M−12 is
also in Sℓ so it has a fixed point. This shows that Sℓ is an intersecting set.
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 21
4 AGL(2, q) on lines
In this section we will examine two related imprimitive groups that do not have the EKR
property and for which the method used in Section 3 does not seem to produce good bounds.
The first group is the affine general linear group, AGL(2, q), with the action on lines, rather
than points. This action is related to PGL(2, q) acting on pairs of projective points; this is
the second group that we consider.
Recall that the affine plane AG(2, q), for any prime power q, is the incidence structure
(Vq,Lq,∼), where the set of points is Vq = F2q , the set of lines is
Lq =
{
Lu,v | u, v ∈ F2q, v ̸= 0
}
with Lu,v = {u+ tv | t ∈ Fq}, and for any x ∈ V , ℓ ∈ L, x ∼ ℓ if and only if x ∈ ℓ. The
permutation group AGL(2, q) consists of all affine transformations (M, z) : v 7→Mv+ z,
for any M ∈ GL(2, q) and z ∈ F2q . Hence, AGL(2, q) acts naturally on the vector space
F2q , which coincides with the points of AG(2, q). This action is 2-transitive and so, under
this action, AGL(2, q) has both the EKR-property and the EKR-module property.
The affine group AGL(2, q) also acts on the set of lines of AG(2, q) as follows: for any
(M, z) ∈ AGL(2, q), Lu,v ∈ Lq , we have (M, z)(Lu,v) = {(M, z)(u + tv) | t ∈ Fq}.
We will refer to this action as the action on the lines and this is the action we consider
in the section. This action is not 2-transitive, it is a rank 3 imprimitive action. There are
q + 1 blocks each of size q; each block is a set of parallel lines. This means each block
has exactly one line through 0 and all the other lines are shifts of this line. Since this group
is imprimitive of rank 3, the system of imprimitivity described above is the only system of
imprimitivity with this action.
4.1 Derangements in AGL(2, q)
In this section we will find the conjugacy classes of derangements in AGL(2, q).
Lemma 4.1. If M has no eigenvalues in Fq , then (M, z) is a derangement for any z ∈ F2q .
Proof. Let (M, z) be an element of AGL(2, q) that fixes the line ℓ = ℓ0+w where ℓ0 is the
line through zero given by ⟨v⟩. For any i ∈ Fq , the point iv+w is on ℓ, so (M, z)(iv+w)
is also on ℓ. Thus
(M, z)(iv + w) =M(iv + w) + z =M(iv) +M(w) + z =M(iv) + (M, z)(w).
Since (M, z)(w) ∈ ℓ, the vector M(iv) is the difference of two points both on the line ℓ.
This implies M(iv) is on the line ⟨v⟩ and v is an eigenvector for M .
Thus, if an element (M, z) is not a derangement, thenM has an eigenvector; the contra-
positive of this statement is that ifM has no eigenvalues, then (M, z) is a derangement.
Lemma 4.2. Assume M is not diagonalizable and has exactly one eigenvalue with corre-
sponding eigenvector s. Then (M, z) is a derangement if and only if the only eigenvalue of
M is equal to 1 and z ̸∈ ⟨s⟩.
Proof. First, if z ∈ ⟨s⟩ then (M, z) fixes the line through zero given by ⟨s⟩. So clearly in
this case (M, z) is not a derangement.
Assume that M has only one eigenvector s and the corresponding eigenvalue is µ ̸= 1.
Then the vector
w = (M − I)−1(s− z)
22 Art Discrete Appl. Math. 6 (2023) #P1.05
is defined and (M, z) is not a derangement since it fixes the line ⟨s⟩ + w. To see this
consider for any k,
(M, z)(ks+w) = µks+ (M − I)w+w+ z = µks+ s− z +w+ z = (µk + 1)s+w.
Assume Ms = s, z ̸∈ ⟨s⟩ and that (M, z) fixes the line ℓ = ℓ0 +w where ℓ0 is the line
through zero given by ⟨v⟩. Then (M, z)(w) = kv+w for some k, and that Mw−w+ z is
in the line ℓ0. Similarly, (M, z)(v +w) = k′v +w for some k′, so Mv +Mw −w + z is
also the line ℓ0. This implies that Mv is on ℓ0, so v is an eigenvector. As M has only one
eigenvector, ℓ0 = ⟨s⟩. Further Mw−w = (M − I)w, must be in ⟨s⟩, since the eigenvalue
corresponding to s is 1. But then the fact that Mw − w + z is on the line ℓ0 implies that z
is a multiple of s.
Lemma 4.3. If M has two distinct eigenvalues, then (M, z) is not a derangement, for any
z ∈ F2q .
Proof. Assume that v1 and v2 are eigenvectors of M with corresponding, distinct, eigen-
values µ1 and µ2. Since the eigenvalues are distinct, we can assume that µ2 ̸= 1.
Set ℓ0 = ⟨v1⟩ and express z = a1v1 + a2v2. We claim that (M, z) fixes the line
ℓ0 +
−1
µ2 − 1
z = ℓ0 +
−a2
µ2 − 1
v2.
To see this, consider:
(M, z)(ℓ0 +
−a2
µ2 − 1
v2) =M(ℓ0 +
−a2
µ2 − 1
v2) + (a1v1 + a2v2)
= ℓ0 +
−a2
µ2 − 1
µ2v2 + (a1v1 + a2v2)
= ℓ0 +
−a2
µ2 − 1
µ2v2 + a2v2
= ℓ0 +
−a2
µ2 − 1
v2.
Finally, we consider the case whereM is diagonalizable and both eigenvalues are equal,
so M is a scalar multiple of that identity matrix. In this case it is clear that (M, z) fixes the
line ⟨z⟩.
Lemma 4.4. If M is a scalar multiple of the identity matrix, then (M, z) is not a derange-
ment for any z.
In summary, in AGL(2, q) there is one conjugacy class of derangements of the form
(M, z) where M has 1 as its only eigenvalue and z is not an eigenvector for M . This class
has size (q2−1)(q2−q) and we denote it withC0. There is a family of
(
q
2
)
conjugacy classes
of derangements each of the form (M, z) where M has no eigenvalues. Each conjugacy
class in this family has size q3(q − 1), we will label these conjugacy classes by Ci with
i = 1, . . . ,
(
q
2
)
. Further, the permutations in conjugacy classes Ci with i = 1, . . . ,
(
q
2
)
fix
none of the blocks of imprimitivity of AGL(2, q).
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 23
4.2 Permutation representation of AGL(2, q)
Many of the irreducible representations of the group AGL(2, q) arise from a representation
on GL(2, q), of the representations that do not arise from an irreducible representation of
GL(2, q), there are q − 1 with dimension q2 − 1, and one with dimension (q − 1)(q2 − 1).
Since AGL(2, q) is a rank 3 imprimitive group, it is straightforward to find the permutation
representation of it.
Lemma 4.5. Let G be an imprimitive group with rank 3. Then, the permutation represen-
tation of G is the sum of three irreducible representations: the trivial representation, χ1
and χ2, where χ1 is the permutation representation from the action of G on the blocks.
Proof. Since the group has rank, 3 it is clear that the permutation representation is the sum
of 3 distinct irreducible representations, one of which must be the trivial character.
Let χ be the permutation representation of G, and χ1 the permutation representation of
G for the action of G on the blocks. Let G1 denote the stabilizer of a point in G and GB
the stabilizer of a block. Then
⟨χ, χ1⟩G = ⟨ind(1G1)G, ind(1GB )G⟩G = ⟨1G1 , res
(
(ind(1GB )
G)
)
G1
⟩G1
This equals the number of orbits G1 has on the blocks, which is 2. Both representations
include the trivial representation, so χ includes χ1 with multiplicity 1.
In particular, χ1 is the permutation representation from the action of AGL(2, q) on
the blocks, minus the trivial representation, so χ1(g) = fixblocks(g) − 1. This is the q-
dimensional representation arising from the representation ρ(1) of GL(2, q). Further, χ2 =
fix(g)−fixblocks(g) is an irreducible degree q2−1 permutation representation of AGL(2, q)
and χ2 restricted to GL(2, q) is the permutation representation on GL(2, q).
Since the permutations in the conjugacy classes Ci with i ∈ {1, . . . ,
(
q
2
)
} do not fix any
of the blocks, χ1(x) = −1 for each x ∈ Ci. Further, χ1(x) = 0 for any x ∈ C0, since
these permutations fix exactly one block.
To apply the method used for GL(2, q) and SL(2, q), a weighting must be found for the
conjugacy classes so that λχ1 ≥ −1 and λχ2 ≥ −1 with λ1 is maximized. It is possible to
give a formula for this eigenvalue where the weighting on Ci is denoted by ai:
λχ1 =
1
q
a0|C0|0 + (
q
2)∑
i=1
ai|Ci|(−1)
 = −q2(q − 1) (
q
2)∑
i=1
ai
and
λχ2 =
1
q2 − 1
a0|C0|(−1) + (
q
2)∑
i=1
ai|Ci|(0)
 = −a0(q2 − q).
It is straight-forward to see that an appropriate weighting will have both
a0 ≤
1
q2 − q
,
(q2)∑
i=1
ai ≤
1
q2(q − 1)
.
24 Art Discrete Appl. Math. 6 (2023) #P1.05
As predicted by Lemma 2.3, the value of the trivial character is
λχ1 =
a0|C0|1 + (
q
2)∑
i=1
ai|Ci|(1)
 ≤ (q2 − 1)(q2 − q)
q2 − q
+
q3(q − 1)
q2(q − 1)
= q2 − 1 + q.
The equation in the ratio bound gives
α ≤ (q − 1)q
3(q + 1)
1− q2+q−1−1
=
(q − 1)2q3(q + 1)
q2 + q
= (q − 1)2q2.
But this is not a bound on the size of a coclique, since we will see in the next section that
there is a larger coclique. The reason that this does not give a bound is that there will be
other irreducible characters with eigenvalue smaller than −1.
4.3 Intersecting sets in AGL(2, q)
In this section we prove Theorem 1.3. First we will give a weak upper bound on the size of
an intersecting set in AGL(2, q). Second, we will show that AGL(2, q) does not have the
EKR property by constructing cocliques in ΓAGL(2,q) that are larger than the stabilizer of
a point. First we note that there is a subgroup in AGL(2, q) in which every element except
the identity is a derangement.
Lemma 4.6. There is a subgroup in AGL(2, q) of size q + 1 in which all non-identity
elements are derangements.
Proof. Such a group is the cycle subgroup generated from any permutation that, when
restricted to the blocks, is a (q + 1)-cycle.
Just as in Lemma 3.4, this implies that the derangement graph ΓAGL(2,q) has a clique
of size q + 1. Theorem 1.3 follows from this lemma by the clique-coclique bound, Theo-
rem 3.3, since if F ⊂ AGL(2, q) is intersecting, then |F| ≤ |AGL(2,q)|q+1 = q
3(q − 1)2.
Next we construct a set of intersecting permutations from AGL(2, q) that is larger than
the canonical intersecting set. To do this we first need some facts.
Proposition 4.7. If (M, z) fixes the block B and ℓ = ⟨v⟩ is the line of B through 0, then v
is an eigenvector of M .
Proof. Assume that M fixes B and ℓ = ⟨v⟩ is the line of B that includes the zero vector.
Since (M, z) maps the 0-vector to z, we know that (M, z) maps the line ℓ to ℓ + z, in
particular, (M, z)(v) = Mv + z is on the line ℓ+ z. This means that Mv is on the line ℓ,
so v is an eigenvector of M .
Lemma 4.8. Let (M,w) be an element of AGL(2, q). If (M,w) fixes two of the blocks of
imprimitivity, then (M,w) fixes a line.
Proof. Assume that (M,w) fixes the blocks B1 and B2. Let ℓ1 and ℓ2 be the lines through
the zero vector in B1 and B2 (respectively). By Proposition 4.7, M is diagonalizable and
ℓ1 and ℓ2 are eigenspaces of M . Let µ1 and µ2 be the eigenvalues of M corresponding to
B1 and B2.
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 25
Case 1. Assume one of µ1 and µ2 is not equal to 1, so without generality we can assume
µ2 ̸= 1. Following the proof of Lemma 4.3, this implies (M,w) fixes the line ℓ1 + −1µ2−1w
in B1.
Case 2. If µ1 = µ2 = 1 then M is the identity matrix and (M,w) is a shift. Let ℓ be
the line that contains the zero vector and w. Then (M,w) fixes ℓ as well as every other line
in the block that contains ℓ. In fact, as long as w is not the zero vector, any such a (M,w)
fixes q lines.
Lemma 4.9. The stabilizer of the blocks in AGL(2, q) is an intersecting set of size
q2(q − 1).
Proof. The stabilizer of the blocks consists of all the elements (M, z) of AGL(2, q) where
M is a scalar multiple of the identity. The number of such elements is q2(q − 1). By
the previous result, every element has a fixed point. Since the stabilizer of the blocks is a
group, this implies that it is an intersecting set.
Let AGL(2, q)B denote the stabilizer of the blocks in AGL(2, q). Then, AGL(2, q)B,
and each of its cosets, is an intersecting set. Next will show that the union of a subset
of these cosets forms a larger intersecting set of permutations. Note that the quotient of
AGL(2, q) with the stabilizer of the blocks is isomorphic to the group PGL(2, q). A pair
of permutations (g, h) are 2-intersecting if there are two distinct points i and j so that
h−1g(i) = i and h−1g(j) = j; a set of permutations is 2-intersecting if any two elements
from the set are 2-intersecting.
Lemma 4.10. If S is a 2-intersecting set of permutations in PGL(2, q) (with the action on
the q + 1 blocks), then ⋃
x∈S
xAGL(2, q)B
is an intersecting set in AGL(2, q).
Proof. The action of PGL(2, q) is the action on the blocks, since S is 2-intersecting for any
two elements x, y ∈ S the permutation y−1x fixes two blocks. For any two permutations
xσ ∈ xAGL(2, q)B and yπ ∈ yAGL(2, q)B, the permutation π−1y−1xσ also fixes two
blocks and xσ and yπ are intersecting.
This motivates finding maximum 2-intersecting sets in PGL(2, q).
Theorem 4.11. If q is odd, then there is a set of 2-intersecting permutations in PGL(2, q)
with size (3q−5)/2. If q is even, there is a set of 2-intersecting permutations in PGL(2, q)
with size (3q − 4)/2.
Proof. We first consider the case when q is odd, the case for q even is almost identical.
We will construct a 2-intersecting set of permutations with size (3q − 5)/2. This set will
contain the identity, as well as (3q−7)/2 permutations with two fixed points. In PGL(2, q)
any non-identity element has at most 2 fixed points, so no two permutations can agree on
more than 2 elements. When q is odd there are (q − 3)/2 conjugacy classes, each of size
q(q + 1), of permutations each with exactly two fixed points in which the elements are not
involutions; call these the conjugacy classes of Type 1. There is another conjugacy class of
size q(q + 1)/2 with permutations that also have exactly two fixed points and in which the
elements are involutions; we will call this the conjugacy class of Type 2.
26 Art Discrete Appl. Math. 6 (2023) #P1.05
For any i ∈ {1, . . . , q + 1} and each conjugacy classes of Type 1 there are exactly 2q
permutations in the class that fix i. Further, for any j, k ∈ {1, . . . , q + 1}\[i], there are
exactly 2 permutations in each conjugacy classes of Type 1 that fix i and map j to k.
Let σ be a permutation in a conjugacy class of Type 1 that fixes i. Define the q pairs
P = {(j, k) : jσ = k, j, k ∈ {1, . . . , q + 1}\i}.
Consider any conjugacy class of Type 1 that does not contain σ. There are no permutations
π in this conjugacy class that fix i and have two distinct pairs (j1, k1), (j2, k2) ∈ P such
that jπ1 = k1 and j
π
2 = k2. Since there are exactly 2q permutations in the class that fix i,
for any pair (j, k) ∈ P , there are exactly two permutations in the conjugacy class that fix
i and map j to k. Since there are q pairs in P , counting shows that each permutation in
this conjugacy class that fixes i, must also map j to k for exactly one pair (j, k) ∈ P . In
general, this means that for any two permutations from different conjugacy classes of Type
1, if they both fix a common element, then they are actually 2-intersecting. We will use this
fact to build a 2-intersecting set of permutations.
For each conjugacy class of Type 1, let x be one of the two permutations that fix both 1
and 2. We can assume without loss of generality that x is the diagonal matrix with entries
1 and a. Fix h to be the unique permutation that maps 1 to 2, 2 to 3 and 3 to 1; h is the
product of 3-cycles and has order 3. Select the set of permutations {x, hxh−1, h−1xh}.
These three permutations all belong to the same conjugacy class and it can easily be seen
(by multiplying the matrix representative of the permutations) that they are pairwise 2-
intersecting. Further, x fixes the points 1 and 2, hxh−1 fixes the points 2 and 3 and h−1xh
fixes the points 1 and 3.
If we do this for each conjugacy class of Type 1, we get a set of 3 q−32 permutations. Any
two permutations from this set that belong to the same conjugacy class are 2-intersecting.
Any two permutations in this set that are from two different conjugacy classes agree on a
fixed point (every permutation in this set will fix at least two of the points 1, 2 and 3) so
they will also be 2-intersecting.
Finally, the permutations in the conjugacy class of Type 2 are involutions. There is
a set of three elements that fix at least two of 1, 2 or 3; this set forms a triangle in the
derangement graph. Using the same argument as for the conjugacy classes of Type 1, if
any permutation from a conjugacy class of Type 1 and a permutation from the conjugacy
class of Type 2 have a common fixed point, then they are 2-intersecting. So any one of
these 3 elements from the conjugacy class of Type 2 can be added to the set. Finally the
identity can be added to the set as well to produce a 2-intersecting set of size
3
q − 3
2
+ 1 + 1 =
3q − 5
2
.
The same argument work when q is even, but in this case the number of conjugacy
classes of Type 1 is q−22 and there is no conjugacy class of Type 2.
From these intersecting sets in PGL(2, q) it is possible to build intersecting sets in
AGL(2, q).
Theorem 4.12. For q is odd, there is an intersecting set in AGL(2, q) with size q
2(q−1)(3q−5)
2
and for q even, there is an intersecting set in AGL(2, q) with size q
2(q−1)(3q−4)
2 .
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 27
Since these sets are larger than the stabilizer of the point in AGL(2, q), this group does
not have the EKR property.
Corollary 4.13. The group AGL(2, q) acting on the lines does not have the EKR property.
From Theorem 4.11 and Theorem 1.3 we know that the size of the maximum intersect-
ing sets in AGL(2, q) is between q
2(q−1)(3q−5)
2 and q
3(q − 1)2, (for q odd). These two
bounds are quite far apart, and we believe that the actual size of the maximum intersecting
sets is much smaller than the bound in Theorem 1.3. We conjecture the following.
Conjecture 4.14. If F ⊂ AGL(2, q) is intersecting, then there exist a ∈ N and b ∈ Z such
that |F| ≤ q2(q − 1)(aq + b).
The group AGL(2, 3) and AGL(2, 4) are sufficiently small that it is possible to find the
size of the largest intersecting set using Grape [29]. For q = 3 the bound can be achieved
by taking 5 cosets of the stabilizer of the blocks. For q = 4 the bound can be achieved by
the union of four cosets of AGL(2, q)B.
Lemma 4.15. The size of the largest intersecting set in AGL(2, 3) is 45 and in AGL(2, 4)
it is 192.
It is straight-forward to solve the linear programming (LP) problem (2.1) for AGL(2, q)
with small values of q. Using Gurobi [21], the linear programming bound for AGL(2, 3)
give a maximum ratio of 5. This solution in the ratio bound proves that intersecting set in
AGL(2, 3) is no larger than 72. Similarly, for q = 4, solving the LP gives a maximum ratio
of 9, so the best result in the ratio bound is that an intersecting set is no larger than 288. For
AGL(2, 5) the best ratio is 9, giving a bound for an intersecting set of 1200. This group
was too large for an exhaustive search, but the largest set we were able to find was of size
500. For q = 7 the best ratio is 13, giving a bound of 7056; we were only able to find a set
of size 2352. It seems that the solution to the (LP) problem (2.1) for q odd gives a ratio of
2q − 1, and that this does not give an effective result in the ratio bound for q = 3, 4, 5, 7.
5 Further work
In this paper for GL(2, q) and SL(2, q) only the action on the vectors of F2q was considered.
Questions about the largest intersecting sets can be asked for any group action. Since any
group action, is equivalent to the action of the group on a set of cosets, this is a relatively
tractable problem. Most research on EKR-type results for groups focuses on well-known
group actions, there is a growing body of work considering all the actions of a group [8, 23].
Since the character table GL(2, q) is completely understood, it should be straightforward
to try this method for different actions of the general linear group.
Another interesting direction is to generalize what has been proved in this paper to
subgroups of the general linear group GL(n, q), where n ≥ 3. As mentioned in the in-
troduction, GL(n, q) has the EKR property due to the existence of Singer subgroups. It is
clear that this group does not have the strict-EKR property since the stabilizer of a vector in
Fnq \ {0} and the stabilizer of a hyperplane in Fnq are distinct intersecting sets of maximum
size. When n = 2, these two intersecting sets are the stabilizer of a point and the stabilizer
of a line. It would be interesting to know whether GL(n, q) has the EKR-module property
and to classify its maximum intersecting sets.
28 Art Discrete Appl. Math. 6 (2023) #P1.05
q 3 4 5 7 8 9 11 13
Size Max. Coclique 2 4 5 8 10 12 17 ≥ 19
Theorem 4.11 bound 2 4 5 8 10 11 14 17
Table 11: Size of maximum 2-intersecting set in PGL(2, q)
q 3 4 5 7 8 9 11 13
Size Max. Coclique 1 4 4 4 10 8 12 12
Table 12: Size of maximum 2-intersecting set in PSL(2, q)
It is known that if a group has a regular subgroup, then there is a weighting of the
conjugacy classes so that the ratio bound holds with equality [15, Theorem 3.5]. It would
be interesting to know of more cases where it can be shown that such a weighting exists.
Question 5.1. If a permutation group G has the EKR property, under what conditions is
there a weighting on the conjugacy classes so that ratio bound holds with equality?
The group AGL(2, q) does not have the EKR property, and we were able to construct
some larger intersecting set. But, we do not know if these sets are the largest possible, in
fact we do not even have good bounds on the size of these sets.
Question 5.2. What are the largest cocliques in the derangement graphs for AGL(2, q)?
We conjecture that the maximum intersecting set in AGL(2, q) will be formed by
unions of cosets of the stabilizers of the blocks. If this is true, then finding the maxi-
mum intersecting sets in AGL(2, q) reduces to finding the maximum 2-intersecting sets in
PGL(2, q), which leads to the next open question.
Question 5.3. For q > 3 what is the largest 2-intersecting set in PGL(2, q)?
Using Grape [29] the maximum 2-intersecting sets in PGL(2, q) and PSL(2, q) can be
determined for small values of q, these are recorded in Tables 11 and 12. Table 11 indicates
that the construction in Theorem 4.11 is not in general optimal.
Pablo Spiga noted that the set-wise stabilizer of subsets of size two of the projective line
in PSL(2, q) when q ≡ 1 (mod 4) forms a 2-intersecting set of size q − 1 [10, Problem
43]. Our results indicate that this is indeed the largest such intersecting set.
Conjecture 5.4. Let q ≡ 1 (mod 4), then the size of the maximum 2-intersecting set in
PSL(2, q) is (q − 1).
ORCID iDs
Karen Meagher https://orcid.org/0000-0002-7948-9149
Andriaherimanana Sarobidy Razafimahatratra https://orcid.org/0000-0002-3542-9160
K. Meagher et al.: Some Erdős-Ko-Rado results for linear and affine groups of degree two 29
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