Automorphism groups of the PSL2(𝑞) commuting involution graphs

James Bryden; Peter Rowley

doi:10.1515/jgth-2025-0143

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Automorphism groups of the PSL₂(𝑞) commuting involution graphs

Published/Copyright: February 12, 2026

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From the journal Journal of Group Theory

Abstract

Given a finite group 𝐺 and a conjugacy class of involutions 𝑋 of 𝐺, we define the commuting involution graph C ⁢ ( G , X ) to be the graph with vertex set 𝑋 and x , y ∈ X adjacent if and only if x ≠ y and x ⁢ y = y ⁢ x . In this paper, the automorphism group of the graph C ⁢ ( G , X ) is determined when G = PSL 2 ⁡ ( q ) .

1 Introduction

Given a finite group 𝐺 and a conjugacy class of involutions 𝑋 of 𝐺, the commuting involution graph on 𝑋, denoted C ⁢ ( G , X ) , is the graph with vertex set 𝑋 in which distinct x , y ∈ X are adjacent if and only if 𝑥 and 𝑦 commute. In [10], Fischer constructed his three sporadic simple groups using the automorphism groups of commuting involution graphs. Since then, the commuting involution graphs for other finite simple groups have been investigated. One can consult [3, 11] for symmetric and alternating groups, [2, 13, 14] for the sporadic groups, and [1, 4, 6, 7, 8, 9] for groups of Lie-type. The focus of much of this research is to describe the connectivity, diameters and disc structures of these graphs. Besides the work of Fischer and a result from [11], almost nothing is known about the automorphism groups of commuting involution graphs of simple groups. In this paper, we take a small step towards filling that void.

Theorem 1

Suppose that 𝑞 is a prime power and L = PSL 2 ⁡ ( q ) . Let 𝑋 denote the conjugacy class of involutions in 𝐿 and 𝐴 the automorphism group of the graph C ⁢ ( L , X ) . Then the following hold.

If 𝑞 is even, then A ≅ Sym ⁡ ( q − 1 ) ≀ Sym ⁡ ( q + 1 ) .
If 𝑞 is odd and q ≥ 7 , then A ≅ P ⁢ Γ ⁢ L 2 ⁢ ( q ) .

In the case when 𝑞 is even, C ⁢ ( L , X ) has q + 1 connected components each of which is a complete graph on q − 1 vertices. This observation immediately gives part (i) of Theorem 1, and so this paper is devoted to proving part (ii). We begin in Section 2 assembling properties of C ⁢ ( L , X ) which we need for our proof. There are two results which we single out on account of their playing an important role, namely Lemma 2 and Theorem 6. Lemma 2 indicates that vertex stabilizers in 𝐴 hold the key to determining 𝐴, while Theorem 6 features prominently in Lemmas 9 and 13. The threads proving Theorem 1 (ii) are drawn together in Section 3, where we have to subdivide into the two cases q ≡ 1 mod 4 and q ≡ 3 mod 4 . This is necessary as, depending on the congruence of 𝑞, C ⁢ ( L , X ) displays a number of different graph theoretic properties.

2 Preliminary results

Throughout the rest of this paper, q ≥ 7 denotes an odd prime power, L = PSL 2 ⁡ ( q ) , 𝑋 is the conjugacy class of involutions in 𝐿, and we set G = PGL 2 ⁡ ( q ) . Also set A = Aut ⁢ ( C ⁢ ( L , X ) ) . We need to work explicitly with the elements of 𝑋, and so we use representatives in SL 2 ⁡ ( q ) . Due to this choice, we have the peculiarity that, when x , y ∈ X are distinct, they commute in 𝐿 if and only if x ⁢ y = − y ⁢ x in SL 2 ⁡ ( q ) .

For x , y ∈ X , we use the notation d ⁢ ( x , y ) to be the graph distance metric on C ⁢ ( L , X ) . That is, d ⁢ ( x , y ) is the length of a shortest path from 𝑥 to 𝑦 in C ⁢ ( L , X ) and ∞ if no such path exists. By the diameter of C ⁢ ( L , X ) , we mean the quantity max ⁡ { d ⁢ ( x , y ) ∣ x , y ∈ X } . Additionally, for a vertex x ∈ X , we define the 𝑖th disc (about 𝑥) to be the set Δ i ⁢ ( x ) = { y ∈ X ∣ d ⁢ ( x , y ) = i } . Finally, A x denotes the stabilizer in 𝐴 of 𝑥.

2.1 The structure of C ⁢ ( L , X )

As is to be expected, many structural results for C ⁢ ( L , X ) can be obtained from basic properties of the involution centralizers in 𝐿. Whilst some of these observations are agnostic of which field we are working over, much of the structure of 𝐿, and so C ⁢ ( L , X ) , is very sensitive to changes in 𝑞. For our purposes, we divide our graphs into the two cases q ≡ 1 mod 4 and q ≡ 3 mod 4 . However, it is worth noting that, even within these two classes, the graphs exhibit quite different structures.

Before examining more graph-theoretic aspects of C ⁢ ( L , X ) , we discuss the visible symmetries C ⁢ ( L , X ) possesses. Notice that, as 𝐿 acts transitively on 𝑋, we have that C ⁢ ( L , X ) is a vertex transitive graph. Moreover, as L = ⟨ X ⟩ and 𝑋 is the unique involution class in 𝐿, we have that Aut ⁡ ( L ) ≅ P ⁢ Γ ⁢ L 2 ⁢ ( q ) acts faithfully on C ⁢ ( L , X ) . It is then the role of Theorem 1 (ii) to show that 𝐴 is no bigger than Aut ⁡ ( L ) . A useful observation to make here is that we get a nice factorization of 𝐴.

Lemma 2

For any vertex x ∈ X , we have A = A x ⁢ L .

Proof

This follows from 𝐿 being transitive on 𝑋. ∎

For a vertex x ∈ X , its stabilizer in 𝐿 is clearly C L ⁢ ( x ) which is a dihedral group. The neighbours of 𝑥 in C ⁢ ( L , X ) form the first disc, Δ 1 ⁢ ( x ) ⊆ C L ⁢ ( x ) , and L x = A x ∩ L partitions Δ 1 ⁢ ( x ) into two orbits. However, G x = A x ∩ G (also a dihedral group) acts transitively on Δ 1 ⁢ ( x ) . It is for this reason that we sometimes prefer to use G x instead of L x in our arguments as a known subgroup of A x .

The involution structure in 𝐿 forces vertex neighbourhoods in C ⁢ ( L , X ) to have very small intersections. As we tackle small values of 𝑞 computationally, we assume q > 11 in the next two lemmas, and hence L x is a maximal subgroup of 𝐿. However, we note that Lemmas 3 and 4 are true for all odd 𝑞.

Lemma 3

Assume q > 11 . Then, for all distinct x , y ∈ X , we have

| Δ 1 ⁢ ( x ) ∩ Δ 1 ⁢ ( y ) | ≤ 1 .

In particular, the graph C ⁢ ( L , X ) is 4-cycle free.

Proof

Let x , y ∈ X be distinct involutions. Then C L ⁢ ( x ) and C L ⁢ ( y ) are maximal subgroups of 𝐿 which are dihedral groups. Suppose that g ∈ C L ⁢ ( x ) ∩ C L ⁢ ( y ) is such that 𝑔 has order at least 3. Then we have N L ⁢ ( ⟨ g ⟩ ) ≥ C L ⁢ ( x ) , C L ⁢ ( y ) and thus N L ⁢ ( ⟨ g ⟩ ) = L . This contradicts the fact that 𝐿 is simple. Therefore, it follows that C L ⁢ ( x ) ∩ C L ⁢ ( x ) is either trivial or an elementary abelian 2-group of order 2 or 4. In the case that C L ⁢ ( x ) ∩ C L ⁢ ( y ) is a fours-group, we must have C L ⁢ ( x ) ∩ C L ⁢ ( y ) = { 1 , x , y , x ⁢ y } . In any case, | Δ 1 ⁢ ( x ) ∩ Δ 1 ⁢ ( y ) | ≤ 1 . In particular, C ⁢ ( L , X ) contains no subgraphs which are 4-cycles. ∎

The next lemma allows us to test whether two vertices are close together in C ⁢ ( L , X ) .

Lemma 4

Assume q > 11 and let x , y ∈ X be distinct involutions.

If q ≡ 1 mod 4 , then d ⁢ ( x , y ) ≤ 2 if and only if x ⁢ y has two distinct eigenvalues.
If q ≡ 3 mod 4 , then d ⁢ ( x , y ) ≤ 2 if and only if x ⁢ y does not have two distinct eigenvalues.

Proof

The group 𝐺 has two conjugacy classes of involutions and they are characterized by how many points of the projective line their elements fix. That is, one conjugacy class contains involutions fixing exactly two points of the projective line, and the other contains involutions which fix no points of the projective line. Suppose that x , y ∈ G are involutions such that x ⁢ y has two distinct eigenvalues. Then x ⁢ y fixes two points of the projective line, 𝑃 and 𝑄. Moreover, G { P , Q } = C G ⁢ ( z ) for some involution 𝑧 fixing 𝑃 and 𝑄. As C G ⁢ ( z ) is a dihedral group, and is maximal in 𝐺, x ⁢ y ∈ C G ⁢ ( z ) implies that x , y ∈ C G ⁢ ( z ) . Thus x ⁢ y has two distinct eigenvalues if and only if x , y ∈ C G ⁢ ( z ) for some involution 𝑧 fixing two projective line points.

In the case that q ≡ 1 mod 4 , the unique involution class of L ≤ G contains involutions fixing two points of the projective line. So, when x , y ∈ L are involutions, it follows that x ⁢ y has two distinct eigenvalues if and only if x , y ∈ C L ⁢ ( z ) for some involution z ∈ L , if and only if d ⁢ ( x , y ) ≤ 2 . In the case that q ≡ 3 mod 4 , the unique involution class of 𝐿 contains elements fixing no points of the projective line. As such, involutions x , y ∈ L satisfy x ⁢ y has two distinct eigenvalues if and only if d ⁢ ( x , y ) > 2 . ∎

In [4], it is shown that, when 𝑞 is not too small, the graph C ⁢ ( L , X ) has diameter 3. In addition, when q ≡ 1 mod 4 , the discs are determined explicitly. The next lemma collates this information and extends the explicit description of the discs to the case q ≡ 3 mod 4 . However, as C ⁢ ( L , X ) is vertex transitive, we have the freedom to choose any vertex to centre our discussion on. When q ≡ 1 mod 4 , we set

t = [ ι 0 0 − ι ] ∈ X ,

where ι ∈ GF ⁡ ( q ) is such that ι 2 = − 1 , and when q ≡ 3 mod 4 ,

t = [ 0 1 − 1 0 ] ∈ X .

Lemma 5

When q ≡ 1 mod 4 and q ≥ 17 or q ≡ 3 mod 4 and q ≥ 7 , the graph C ⁢ ( L , X ) has diameter 3. Moreover, if q ≡ 1 mod 4 , the discs of 𝑡 are as follows:

Δ 1 ⁢ ( t ) = { [ 0 ω − 1 / ω 0 ] | ω ≠ 0 } ,

Δ 2 ⁢ ( t ) = { [ σ μ 2 ⁢ τ τ − σ ] | σ , μ , τ ≠ 0 , σ 2 + μ 2 ⁢ τ 2 = − 1 } ,

Δ 3 ⁢ ( t ) = { [ ι α 0 − ι ] , [ ι 0 α − ι ] | α ≠ 0 } ∪ { [ σ μ ⁢ τ τ − σ ] | σ , μ , τ ≠ 0 , μ ⁢ is non-square , σ 2 + μ ⁢ τ 2 = − 1 } .

If q ≡ 3 mod 4 , the discs of 𝑡 are as follows:

Δ 1 ⁢ ( t ) = { [ σ τ τ − σ ] | σ 2 + τ 2 = − 1 } ,

Δ 2 ⁢ ( t ) = { [ − ( β + γ ) ⁢ τ 2 ⁢ σ β γ ( β + γ ) ⁢ τ 2 ⁢ σ ] | σ 2 + τ 2 = − 1 , ( β + γ ) 2 ⁢ τ 2 4 ⁢ σ 2 + β γ = − 1 , β ≠ γ } ,

Δ 3 ⁢ ( t ) = { 1 1 + α 2 ⁢ [ α ⁢ ( ω − 1 − ω ) α 2 ⁢ ω − 1 + ω − α 2 ⁢ ω − ω − 1 − α ⁢ ( ω − 1 − ω ) ] | α , ω ≠ 0 , ± 1 } .

Proof

In the case that q ≡ 1 mod 4 and q ≥ 17 , the diameter and discs are described by [4, Propositions 2.4, 2.5 and Theorem 2.7].

Suppose then that q ≡ 3 mod 4 and q ≥ 7 . Let x ∈ X be given by

[ α β γ − α ] .

If 𝑥 and 𝑡 commute in 𝐿, then in SL 2 ⁡ ( q ) , we have x ⁢ t = − t ⁢ x . This yields the relation β = γ . Thus

Δ 1 ⁢ ( t ) = { [ σ τ τ − σ ] | σ 2 + τ 2 = − 1 } .

Fix

s = [ σ τ τ − σ ] ∈ Δ 1 ⁢ ( t )

and suppose that x ∈ Δ 1 ⁢ ( s ) . Then α = − ( β + γ ) ⁢ τ 2 ⁢ σ . If β = γ , then x ∈ Δ 1 ⁢ ( t ) . Thus

Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) = { [ − ( β + γ ) ⁢ τ 2 ⁢ σ β γ ( β + γ ) ⁢ τ 2 ⁢ σ ] | ( β + γ ) 2 ⁢ τ 2 4 ⁢ σ 2 + β ⁢ γ = − 1 , β ≠ γ } ,

and so

Δ 2 ( t ) = { [ − ( β + γ ) ⁢ τ 2 ⁢ σ β γ ( β + γ ) ⁢ τ 2 ⁢ σ ] | σ 2 + τ 2 = − 1 , ( β + γ ) 2 ⁢ τ 2 4 ⁢ σ 2 + β γ = − 1 , β ≠ γ } .

Now, by [4, Theorem 1.1], every other vertex unaccounted for must lie in Δ 3 ⁢ ( t ) . We determine Δ 3 ⁢ ( t ) by determining G t = C G ⁢ ( t ) and its orbits on Δ 3 ⁢ ( t ) .

Suppose that

g = [ α β γ δ ]

is such that g ∈ C G ⁢ ( t ) . Then g ⁢ t = ϵ ⁢ t ⁢ g for some ϵ ∈ GF ( q ) ∗ . But (in GL 2 ⁡ ( q ) ), we have t − 1 = − t . Thus, as g t 2 = g , we must have ϵ 2 = 1 and so ϵ = ± 1 .

If g ⁢ t = t ⁢ g , then δ = α and γ = − β . If g ⁢ t = − t ⁢ g , then δ = − α and γ = β . Hence 𝑔 takes one of the two forms

[ α β − β α ] or [ α β β − α ] .

However, we can ignore scalars in 𝐺 and so 𝑔 can be represented by one of

g α : = [ 1 α − α 1 ] or h α : = [ α 1 1 − α ] .

Observe that

g α − 1 = 1 1 + α 2 ⁢ g − α and h α − 1 = 1 1 + α 2 ⁢ h α .

Letting

t ω = [ 0 ω − 1 / ω 0 ] ,

we have t = t 1 . Moreover, t ω ∈ Δ 3 ⁢ ( t ) for all ω ≠ 0 , ± 1 . Lastly, t ω and t λ are in the same G t -orbit if and only if λ ∈ { ± ω , ± 1 / ω } . Hence the t ω represent the G t -orbits of Δ 3 ⁢ ( t ) .

We have

t ω g α = 1 1 + α 2 ⁢ g − α ⁢ t ω ⁢ g α = 1 1 + α 2 ⁢ [ α ⁢ ( ω − 1 − ω ) α 2 ⁢ ω − 1 + ω − α 2 ⁢ ω − ω − 1 − α ⁢ ( ω − 1 − ω ) ] , t ω h α = 1 1 + α 2 ⁢ h α ⁢ t ω ⁢ h α = 1 1 + α 2 ⁢ [ α ⁢ ( ω − ω − 1 ) − α 2 ⁢ ω − ω − 1 α 2 ⁢ ω − 1 + ω − α ⁢ ( ω − ω − 1 ) ] .

Observe that if α ≠ 0 , then t ω g α = − t ω h 1 / α , whereby

t ω G t = { [ 0 − ω − 1 ω 0 ] , 1 1 + α 2 ⁢ [ α ⁢ ( ω − 1 − ω ) α 2 ⁢ ω − 1 + ω − α 2 ⁢ ω − ω − 1 − α ⁢ ( ω − 1 − ω ) ] | α } . ∎

2.2 Solutions of certain polynomials

Recalling that a polynomial over GF ⁡ ( q ) is said to be absolutely irreducible if it is irreducible as a polynomial over the algebraic closure of GF ⁡ ( q ) , we have the following appearing as [12, Theorem 6.57] which gives bounds on sizes of certain solution sets.

Theorem 6

Let m ∈ N and suppose that f ∈ GF ⁡ ( q ) ⁢ [ x ] with deg ⁡ f > 1 is such that y k − f ⁢ ( x ) is absolutely irreducible, where k = gcd ⁡ ( m , q − 1 ) . Then the number 𝑁 of solutions of the equation y m = f ⁢ ( x ) in GF ( q ) 2 satisfies

| N − q | ≤ ( k − 1 ) ⁢ ( d − 1 ) ⁢ q ,

where 𝑑 is the number of distinct roots of 𝑓 in its splitting field over GF ⁡ ( q ) .

3 The automorphism group of C ⁢ ( L , X )

In this section, we prove Theorem 1 (ii). As mentioned in Section 1, the proof breaks into two cases, q ≡ 1 mod 4 and q ≡ 3 mod 4 . However, in either case, the overall strategy is similar. For 𝑡 defined in Section 2, let 𝐾 be the normal subgroup of A t fixing Δ 1 ⁢ ( t ) vertex-wise. Then we investigate the orbits of 𝐾 on Δ 2 ⁢ ( t ) when q ≡ 1 mod 4 , and the orbits of 𝐾 on Δ 3 ⁢ ( t ) when q ≡ 3 mod 4 . With this information, we are then able to show t ∈ Z ⁢ ( A t ) , provided q ≥ 73 (for q ≡ 1 mod 4 ) and q ≥ 67 (for q ≡ 3 mod 4 ). Knowing that t ∈ Z ⁢ ( A t ) swiftly leads to pinning down 𝐴, and the small values of 𝑞 may be quickly checked using Magma [5].

3.1 The case q ≡ 1 mod 4

Since q ≡ 1 mod 4 , we may choose ι ∈ GF ⁡ ( q ) such that ι 2 = − 1 . Along with our involution 𝑡, we fix s ∈ Δ 1 ⁢ ( t ) , so as

t = [ ι 0 0 − ι ] and s = [ 0 1 − 1 0 ] .

Observe that, by Lemma 3, we have that every element of Δ 2 ⁢ ( t ) is adjacent to a unique vertex in Δ 1 ⁢ ( t ) . Moreover, A t ≥ G t is transitive on Δ 1 ⁢ ( t ) . Hence, to show that the 𝐾-orbits of Δ 2 ⁢ ( t ) take the form { v , v t } , it is sufficient to show that the 𝐾-orbits of Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) take this form.

We first describe Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) explicitly.

Lemma 7

We have

Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) = { [ σ τ τ − σ ] | σ , τ ≠ 0 , σ 2 + τ 2 = − 1 } .

Proof

Let

x = [ α β γ − α ] ∈ X .

Then x ∈ Δ 1 ⁢ ( s ) if and only if x ⁢ s = − s ⁢ x and x ≠ ± s . The equation x ⁢ s + s ⁢ x = 0 yields β = γ . So

x = [ σ τ τ − σ ] with ⁢ σ 2 + τ 2 = − 1 .

However, if 𝜎 or τ = 0 , clearly, we have x ∈ Δ 1 ⁢ ( t ) or x = ± t respectively. Therefore,

Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) = { [ σ τ τ − σ ] | σ , τ ≠ 0 , σ 2 + τ 2 = − 1 } ,

as desired. ∎

Define

s σ , τ = [ σ τ τ − σ ] ∈ Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t )

and further set E σ , τ = Δ 2 ⁢ ( s σ , τ ) ∩ Δ 1 ⁢ ( t ) . Moreover, notice that s σ , τ = s − σ , − τ and s σ , τ t = s σ , − τ .

The next two lemmas form the heart of our argument.

Lemma 8

Suppose that q > 11 and

t ω = [ 0 ω − 1 / ω 0 ] ∈ Δ 1 ⁢ ( t ) .

Then t ω ∈ E σ , τ if and only if ω 4 − ( 2 + 4 / τ 2 ) ⁢ ω 2 + 1 is a non-zero square in GF ⁡ ( q ) .

Proof

By Lemma 4, we have t ω ∈ E σ , τ if and only if t ω ⁢ s σ , τ has two distinct eigenvalues. Observe that the characteristic polynomial of t ω ⁢ s σ , τ is given by

χ ⁢ ( x ) = 1 + ( ω − 1 − ω ) ⁢ τ ⁢ x + x 2

which has discriminant

ϕ ⁢ ( ω ) = τ 2 ⁢ ( ω − 1 − ω ) 2 − 4 .

Now ϕ ⁢ ( ω ) is a non-zero square in GF ⁡ ( q ) if and only if

ω 2 τ 2 ⁢ ϕ ⁢ ( ω ) = ω 4 − ( 2 + 4 / τ 2 ) ⁢ ω 2 + 1

is a non-zero square in GF ⁡ ( q ) . This proves the lemma.∎

Lemma 9

Assume that q ≥ 73 . Then E ρ , μ = E σ , τ if and only if

( ρ , μ ) = ( ± σ , ± τ ) .

Proof

Fix s σ , τ and s ρ , μ in Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) . Notice that, as s σ , τ t = s σ , − τ , we have E σ , τ = E ± σ , ± τ .

Assume, for a contradiction, that ( ρ , μ ) ≠ ( ± σ , ± τ ) and E ρ , μ = E σ , τ . Define γ = − 2 − 4 / τ 2 , δ = − 2 − 4 / μ 2 ,

f γ ⁢ ( ω ) = ω 4 + γ ⁢ ω 2 + 1 , and f δ ⁢ ( ω ) = ω 4 + δ ⁢ ω 2 + 1 .

Then, by Lemma 8, f γ ⁢ ( ω ) is a non-zero square if and only if f δ ⁢ ( ω ) is a non-zero square. Therefore, f ( ω ) : = f γ ( ω ) f δ ( ω ) is a (possibly zero) square for all ω ∈ GF ⁡ ( q ) .

We now verify that 𝑓 is not the square of a polynomial in GF ⁡ ( q ) ̄ ⁢ [ ω ] and thus show that the polynomial y 2 − f ⁢ ( x ) ∈ GF ⁡ ( q ) ⁢ [ x , y ] is absolutely irreducible. Observe that any common factor of f γ , f δ is a factor of ( f γ − f δ ) ⁢ ( ω ) = ( γ − δ ) ⁢ ω 2 , and hence f γ and f δ are coprime. Moreover, as γ , δ ≠ ± 2 , we have that f γ and f δ are separable. Thus 𝑓 has no repeated linear factors and so cannot be a square in GF ⁡ ( q ) ̄ ⁢ [ ω ] . Hence y 2 − f ⁢ ( x ) ∈ GF ⁡ ( q ) ⁢ [ x , y ] is absolutely irreducible.

Define N = | { ( x , y ) ∈ GF ( q ) 2 ∣ y 2 = f ( x ) } | and observe that, as deg ⁡ f = 8 , we have N ≥ 2 ⁢ q − 8 . Applying Theorem 6 yields | N − q | ≤ 7 ⁢ q and hence

q 2 − 65 ⁢ q + 64 = ( q − 1 ) ⁢ ( q − 64 ) ≤ 0 ,

whereby, q ≤ 64 . However, by assumption, q ≥ 73 , which gives a contradiction. ∎

We require one more lemma before we can prove the q ≡ 1 mod 4 part of Theorem 1.

Lemma 10

If q ≥ 17 , then A t acts faithfully on Δ 2 ⁢ ( t ) .

Proof

Suppose that g ∈ A t fixes Δ 2 ⁢ ( t ) vertex-wise. By [4, Proposition 2.4], each element of Δ 2 ⁢ ( t ) ∪ Δ 3 ⁢ ( t ) is adjacent to a unique element of the form

[ ι α 0 − ι ] for ⁢ α ≠ 0 .

It follows that if

u = [ ι α 0 − ι ] ∈ Δ 3 ⁢ ( t ) ,

then

| Δ 1 ⁢ ( u ) ∩ Δ 2 ⁢ ( t ) | = q − 5 4 .

Since q ≥ 17 , we have | Δ 1 ⁢ ( u ) ∩ Δ 2 ⁢ ( t ) | ≥ 2 . Then, by Lemma 3, we must have u g = u . Suppose now that v ∈ Δ 1 ⁢ ( u ) ∩ Δ 3 ⁢ ( t ) . Then, as v ∈ Δ 3 ⁢ ( t ) , we have

v ′ ∈ Δ 2 ⁢ ( t ) ∩ Δ 1 ⁢ ( v ) .

So 𝑣 and v g both have 𝑢 and v ′ as neighbours. Thus Lemma 3 yields v = v g . It follows that g = 1 and A t acts faithfully on Δ 2 ⁢ ( t ) . ∎

We can now prove Theorem 1 (ii) when 𝑞 is large enough.

Lemma 11

If q ≥ 73 , then A = P ⁢ Γ ⁢ L 2 ⁢ ( q ) holds.

Proof

Suppose that q ≥ 73 and that K ⁢ ⊴ ⁢ A t is the vertex-wise stabilizer of Δ 1 ⁢ ( t ) in A t . By Lemma 9, we have that the 𝐾-orbits of Δ 1 ⁢ ( s ) ∩ Δ 2 ⁢ ( t ) all take the form { v , v t } . As A t acts transitively on Δ 1 ⁢ ( t ) , this implies that 𝐾 partitions Δ 2 ⁢ ( t ) into orbits of the form { v , v t } . Moreover, observe that t ∈ K fixes no vertices of Δ 2 ⁢ ( t ) . Suppose that g ∈ K also enjoys this property so that g ⁢ t fixes Δ 2 ⁢ ( t ) vertex-wise. Hence, by Lemma 10, we have g = t . That is, 𝑡 is the unique element of 𝐾 fixing no vertices of Δ 2 ⁢ ( t ) and hence ⟨ t ⟩ ⁢ ⊴ ⁢ A t , whereby, t A = t L by Lemma 2. As t L generates 𝐿, it follows that L ⁢ ⊴ ⁢ A . Moreover, clearly, C A ⁢ ( L ) = 1 as 𝐴 acts faithfully on the graph C ⁢ ( L , X ) . It then follows by the fact that Aut ⁡ ( L ) ≅ P ⁢ Γ ⁢ L 2 ⁢ ( q ) ≤ A that A = P ⁢ Γ ⁢ L 2 ⁢ ( q ) . This proves the result. ∎

3.2 The case q ≡ 3 mod 4

Suppose that q ≡ 3 mod 4 and let

t = [ 0 1 − 1 0 ] ∈ X

be our fixed involution.

Again, let K ⁢ ⊴ ⁢ A t be the subgroup of A t fixing Δ 1 ⁢ ( t ) vertex-wise. We argue to show that the 𝐾-orbits of Δ 3 ⁢ ( t ) all take the form { v , v t } .

Define s α , ω ∈ Δ 3 ⁢ ( t ) to be the involution with parameters 𝛼 and 𝜔 and define E α , ω = Δ 2 ⁢ ( s α , ω ) ∩ Δ 1 ⁢ ( t ) . Recall from the proof of Lemma 5 that every s α , ω is contained in a G t -orbit represented by s 0 , ω . As 𝐾 is normal in A t , it is therefore sufficient to show that the 𝐾-orbit containing s 0 , ω takes the form { s 0 , ω , s 0 , ω t } .

Lemma 12

Assume that q > 11 . The vertex

[ σ τ τ − σ ] ∈ Δ 1 ⁢ ( t )

belongs to E α , ω if and only if

4 ⁢ α ⁢ ( α 2 − 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ σ ⁢ τ + ( α 4 − 6 ⁢ α 2 + 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ τ 2 − 4 ⁢ ( α 4 + α 2 ⁢ ω 2 + α 2 ⁢ ω − 2 + 1 )

is not a non-zero square in GF ⁡ ( q ) .

Proof

By Lemma 4, we have

z = [ σ τ τ − σ ] ∈ Δ 1 ⁢ ( t )

belongs to E a , ω if and only if z ⁢ s a , ω does not have two distinct eigenvalues. Moreover, multiplying by a non-zero scalar does not change the number of distinct eigenvalues a matrix has. Whereby, z ⁢ s α , ω has two distinct eigenvalues if and only if ( 1 + α 2 ) ⁢ z ⁢ s α , ω has two distinct eigenvalues. The discriminant of the characteristic polynomial of ( 1 + α 2 ) ⁢ z ⁢ s α , ω is given by the stated polynomial. This proves the lemma. ∎

Lemma 13

If q ≥ 67 , then E α , ω = E α ′ , ω ′ if and only if

s α ′ , ω ′ = s α , ω or s α ′ , ω ′ = s α , ω t .

Proof

As we have noted before, as every s α , ω is in the same G t -orbit as s 0 , ω , it is sufficient to show that E α , ω = E 0 , λ if and only if α = 0 and ω = ± λ , ± λ − 1 .

One direction of the implication is trivial. Thus, towards a contradiction, assume that E α , ω = E 0 , λ but ( α , ω ) ≠ ( 0 , ± λ ± 1 ) . Then, by Lemma 12, it holds that, for all ( σ , τ ) with σ 2 + τ 2 = − 1 ,

(3.1) 4 ⁢ α ⁢ ( α 2 − 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ σ ⁢ τ + ( α 4 − 6 ⁢ α 2 + 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ τ 2 − 4 ⁢ ( α 4 + α 2 ⁢ ω 2 + α 2 ⁢ ω − 2 + 1 )

is a non-zero square in GF ⁡ ( q ) if and only if

(3.2) ( λ 2 + λ − 2 − 2 ) ⁢ τ 2 − 4

is a non-zero square in GF ⁡ ( q ) . It is possible to parameterize all but at most one point on the curve σ 2 + τ 2 = − 1 in terms of a single variable 𝜁 over GF ⁡ ( q ) . Fix a point ( σ 0 , τ 0 ) ∈ GF ( q ) 2 such that σ 0 2 + τ 0 2 = − 1 . Then, given any ζ ∈ GF ⁡ ( q ) , we recover a unique point ( σ , τ ) on the curve by taking

τ = − τ 0 − 2 ⁢ σ 0 ⁢ ζ + τ 0 ⁢ ζ 2 ζ 2 + 1 and σ = σ 0 − 2 ⁢ τ 0 ⁢ ζ − σ 0 ⁢ ζ 2 ζ 2 + 1 .

Substituting for 𝜎 and 𝜏 in (3.1) and (3.2) and multiplying by ( ζ 2 + 1 ) 2 , we obtain the two polynomials

θ α , ω ⁢ ( ζ ) = − 4 ⁢ α ⁢ ( α 2 − 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ ( σ 0 − 2 ⁢ τ 0 ⁢ ζ − σ 0 ⁢ ζ 2 ) ⁢ ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) + ( α 4 − 6 ⁢ α 2 + 1 ) ⁢ ( ω 2 + ω − 2 − 2 ) ⁢ ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) 2 − 4 ⁢ ( α 4 + α 2 ⁢ ω 2 + α 2 ⁢ ω − 2 + 1 ) ⁢ ( ζ 2 + 1 ) 2

and

θ 0 , λ ⁢ ( ζ ) = ( λ 2 + λ − 2 − 2 ) ⁢ ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) 2 − 4 ⁢ ( ζ 2 + 1 ) 2

respectively. Moreover, for all ζ ∈ GF ⁡ ( q ) , it holds that θ α , ω ⁢ ( ζ ) is a non-zero square in GF ⁡ ( q ) if and only if θ 0 , λ ⁢ ( ζ ) is a non-zero square in GF ⁡ ( q ) . Thus

f ( ζ ) : = θ α , ω ( ζ ) θ 0 , λ ( ζ )

is a square in GF ⁡ ( q ) for all ζ ∈ GF ⁡ ( q ) .

We now show that 𝑓 is not the square of a polynomial in GF ⁡ ( q ) ̄ ⁢ [ ζ ] and hence show that y 2 − f ⁢ ( x ) ∈ GF ⁡ ( q ) ⁢ [ x , y ] is absolutely irreducible. Observe that θ 0 , λ is the difference of two squares and hence has a factorization

θ 0 , λ ⁢ ( ζ ) = ( ( ( λ − λ − 1 ) ⁢ τ 0 − 2 ) + 2 ⁢ ( λ − λ − 1 ) ⁢ σ 0 ⁢ ζ − ( ( λ − λ − 1 ) ⁢ τ 0 + 2 ) ⁢ ζ 2 ) ( ( ( λ − λ − 1 ) ⁢ τ 0 + 2 ) + 2 ⁢ ( λ − λ − 1 ) ⁢ σ 0 ⁢ ζ − ( ( λ − λ − 1 ) ⁢ τ 0 − 2 ) ⁢ ζ 2 ) .

Moreover, observe that if θ 0 , λ has a repeated root ζ 0 , then ζ 0 2 = − 1 . However, one can easily check that no such root can exist, whereby, θ 0 , λ is separable.

If 𝑓 is the square of some polynomial in GF ⁡ ( q ) ̄ ⁢ [ ζ ] , then all of its linear factors must appear with even multiplicity. However, deg ⁡ θ 0 , λ = deg ⁡ θ α , ω and θ 0 , λ is separable. Hence 𝑓 is the square of a polynomial if and only if θ α , ω = κ ⁢ θ 0 , λ for some κ ∈ GF ⁡ ( q ) . Notice that ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) and 1 + ζ 2 are both irreducible in GF ⁡ ( q ) ⁢ [ ζ ] , and as such, there do not exist μ , η ∈ GF ⁡ ( q ) with

μ ⁢ ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) 2 + η ⁢ ( 1 + ζ 2 ) 2 = ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) ⁢ ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) .

So, as θ α , ω = k ⁢ θ 0 , λ , we must have 4 ⁢ α ⁢ ( α 2 − 1 ) ⁢ ( ω − ω − 1 ) 2 = 0 , and as ω ≠ ± 1 , we must have either α 2 = 1 or α = 0 . We consider each case in turn. If α 2 = 1 , then by examining the ( 1 + ζ 2 ) 2 coefficients of θ 0 , λ and θ α , ω , we see that we must have κ = ( ω + ω − 1 ) 2 . Then, from the ( τ 0 + 2 ⁢ σ 0 ⁢ ζ − τ 0 ⁢ ζ 2 ) 2 coefficients, we obtain

− 4 ⁢ ( ω 2 + ω − 2 − 2 ) = ( ω 2 + ω − 2 + 2 ) ⁢ ( λ 2 + λ − 2 − 2 ) .

Now, by re-arranging, we see that 𝜔 must be a root of the polynomial

(3.3) ( λ 2 + λ − 2 + 2 ) ⁢ ω 4 + 2 ⁢ ( λ 2 + λ − 2 − 6 ) ⁢ ω 2 + ( λ 2 + λ − 2 + 2 ) = 0 .

Clearly, if ω 0 is a root, then ω 0 2 is a root of (3.3) viewed as a quadratic in ω 2 . However, the discriminant of (3.3) as a quadratic equation in ω 2 is given by − 2 6 ⁢ ( λ − λ − 1 ) 2 which is a non-zero non-square in GF ⁡ ( q ) . Thus (3.3) has no solutions and we can conclude that α 2 = 1 is not a possible case.

This leaves only the case where α = 0 . By examining constant terms of θ 0 , λ and θ α , ω , it follows that κ = 1 . Hence ω − ω − 1 = λ − λ − 1 . There are then four possibilities for 𝜔. They are ± λ or ± λ − 1 . Thus ( α , ω ) = ( 0 , ± λ ) or ( α , ω ) = ( 0 , ± λ − 1 ) . However, we know this is not the case. We can then conclude that 𝑓 is not the square of a polynomial GF ⁡ ( q ) ̄ ⁢ [ ζ ] . Therefore, we have shown that

y 2 − f ⁢ ( x ) ∈ GF ⁡ ( q ) ⁢ [ x , y ]

is absolutely irreducible.

We can now finish the proof. Define N = | { ( x , y ) ∈ GF ( q ) 2 ∣ y 2 = f ( x ) } | and observe that, as deg ⁡ f = 8 , we get N ≥ 2 ⁢ q − 8 . We can then apply Theorem 6, which yields | N − q | ≤ 7 ⁢ q , whereby, ( q − 1 ) ⁢ ( q − 64 ) ≤ 0 and hence q ≤ 64 . Thus, as we have assumed that q > 64 , we get a contradiction. This proves the lemma. ∎

We require one more lemma before we can finish our argument.

Lemma 14

We have that 𝐾 acts faithfully on Δ 3 ⁢ ( t ) .

Proof

Let g ∈ K fix Δ 3 ⁢ ( t ) vertex-wise and suppose that there exists z ∈ Δ 2 ⁢ ( t ) such that z g ≠ z . If Δ 1 ⁢ ( z ) ∩ Δ 3 ⁢ ( t ) ≠ ∅ , then as Δ 1 ⁢ ( z ) ∩ Δ 1 ⁢ ( t ) ≠ ∅ , the graph C ⁢ ( L , X ) contains a 4-cycle. This contradicts Lemma 3 and thus we may suppose that Δ 1 ⁢ ( z ) ∩ Δ 3 ⁢ ( t ) = ∅ .

Choose any v ∈ Δ 3 ⁢ ( t ) and observe that because C ⁢ ( L , X ) has diameter 3,

2 ≤ d ⁢ ( z , v ) ≤ 3 .

If d ⁢ ( z , v ) = 2 , then there exists u ∈ Δ 1 ⁢ ( z ) ∩ Δ 1 ⁢ ( v ) , and as v ∈ Δ 1 ⁢ ( u ) ∩ Δ 3 ⁢ ( t ) , we must have u g = u . Hence C ⁢ ( L , X ) contains a 4-cycle on some vertex of Δ 1 ⁢ ( z ) ∩ Δ 1 ⁢ ( t ) , 𝑧, z g and 𝑢. This is a contradiction, whereby, d ⁢ ( z , v ) = 3 .

Let u 1 , u 2 be such that z , u 1 , u 2 , v is a path of length 3 from 𝑧 to 𝑣 in C ⁢ ( L , X ) . Note that, as Δ 1 ⁢ ( u 2 ) ∩ Δ 3 ⁢ ( t ) ≠ ∅ , we have u 2 g = u 2 . If u 1 = u 1 g , then we create a 4-cycle in C ⁢ ( L , X ) and thus we must have u 1 ≠ u 1 g . However, by Lemma 3, we have { u 2 } = Δ 1 ⁢ ( u 1 ) ∩ Δ 1 ⁢ ( u 1 g ) . But as u 1 , u 1 g ∈ Δ 2 ⁢ ( t ) and g ∈ K , we must have Δ 1 ⁢ ( u 1 ) ∩ Δ 1 ⁢ ( u 1 g ) ⊆ Δ 1 ⁢ ( t ) . Hence u 2 ∈ Δ 1 ⁢ ( t ) , which contradicts the fact that v ∈ Δ 3 ⁢ ( t ) . Therefore, for all z ∈ Δ 2 ⁢ ( t ) , we have Δ 1 ⁢ ( z ) ∩ Δ 3 ⁢ ( t ) ≠ ∅ and thus 𝐾 acts faithfully on Δ 3 ⁢ ( t ) . ∎

We can now prove the q ≡ 3 mod 4 part of Theorem 1 when 𝑞 is not too small.

Lemma 15

If q ≥ 67 holds, then A = P ⁢ Γ ⁢ L 2 ⁢ ( q ) .

Proof

Suppose that q ≥ 67 and that K ⁢ ⊴ ⁢ A t is the vertex-wise stabilizer of Δ 1 ⁢ ( t ) in A t . By Lemma 13, we have that the 𝐾-orbits of Δ 3 ⁢ ( t ) all take the form { v , v t } . Now we may proceed as in Lemma 11, using Lemma 14 in place of Lemma 10, to deduce that t ∈ Z ⁢ ( A t ) . Then it follows, as in Lemma 11, that A = P ⁢ Γ ⁢ L 2 ⁢ ( q ) . ∎

Lemmas 11 and 15 establish Theorem 1 (ii) provided q ≥ 73 . The cases when 5 < q < 73 can be settled using the following Magma code. This program takes roughly 30 seconds to run, and so the online version suffices!Q := [k : k in [ 7 .. 71] | (k mod 2 eq 1) and #Factorisation(k) eq 1]; for q in Q do L := PSL(2,q); X := {@ x : x in L | Order(x) eq 2 @}; Edges := {{i,j} : i,j in [1 .. #X] | i lt j and Order(X[i]*X[j]) eq 2}; Gam := Graph<#X | Edges>; A := AutomorphismGroup(Gam); print q, 1 eq #A/#PGammaL(2,q); end for;Therefore, the proof of Theorem 1 is complete.

Funding statement: The first author’s work was supported by the Heilbronn Institute for Mathematical Research.

Acknowledgements

We would like to thank Philip Dittmann for his help and advice relating to Theorem 6.

Communicated by: Andrea Lucchini

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Received: 2025-10-02

Revised: 2025-12-16

Published Online: 2026-02-12

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