# Fixed-point subgroups of GL3(𝑞)

John Cullinan
From the journal Journal of Group Theory

## Abstract

Let V be a vector space over a field k. We call a subgroup GGL(V) a fixed-point subgroup if det(1-g)=0 for all gG. Let q be a power of a prime. In this paper, we classify the fixed-point subgroups of GL3(q).

## 1 Introduction

### 1.1 Motivation

Let X/𝐐 be a smooth, projective algebraic variety and a rational prime. Then there are -adic representations

ρ:Gal(𝐐¯/𝐐)Aut(Héti(X𝐐¯,𝐐))

on the étale cohomology groups of X, and the image of such a representation can have interesting consequences for the arithmetic of the variety X. For example, when X is an elliptic curve and ρ is the -adic representation on the Tate module, then the image of ρ gives information on the -power torsion structure of X(𝐐). A concrete instance of this stems from a question of Lang, answered by Katz [10]. Namely, if X is an elliptic curve over 𝐐 such that, for all but finitely many primes p, the numbers #Xp(𝐅p) are divisible by n (where Xp denotes the reduction of X modulo a good primes p), then it is true that at least one of the curves X in the isogeny class of X has #X(𝐐) divisible by n. By translating to Galois representations, this result amounts to a classification of subgroups G of GL2(𝐙) such that det(1-g)0(modn) for all gG. One can ask for a similar classification of subgroups of symplectic similitude groups with a view towards higher-dimensional abelian varieties with divisibilities on their number of points modulo p; we provided such classifications in dimensions 4 and 6 in [3, 5, 4, 6] for the groups GSp4(𝐅) and GSp6(𝐅).

This raises a natural question: If k is a finite field, can one classify the irreducible subgroups of GLn(k) such that every element has a fixed point? (By “irreducible subgroup” we mean a subgroup GGLn(k) that acts irreducibly on the underlying vector space kn.) Let us call a subgroup G of GLn(k) a fixed-point subgroup if every element fixes a point in its natural representation.

By an exercise of Serre [14, Exercise 1], there are no irreducible fixed-point subgroups of GL2(k). One of the main results of [10] is that there are no irreducible fixed-point subgroups of GSp4(𝐅), where 𝐅 is the field of elements. In [3, 5, 4], we classified the fixed-point subgroups of GSp6(𝐅) and showed that none are irreducible. However, we recall an example of [4], originally communicated to us by Serre [15]. If L3(2) is the simple group of order 168, then the Steinberg representation

𝖲𝗍:L3(2)Sp8(𝐅2)

is absolutely irreducible, and 𝖲𝗍(L3(2)) is a fixed-point subgroup of Sp8(𝐅2). As an application of this observation, if A is an abelian fourfold defined over a number field K such that the image of the modulo-2 representation

ρ2¯:Gal(K¯/K)Aut(A[2])

coincides with 𝖲𝗍(L3(2)), then A has the property that, for all but finitely many primes 𝔭, the number of points #A¯𝔭(𝐅𝔭) on the reduction modulo 𝔭 of A is even, while no member of the isogeny class of A has an even number of K-rational torsion points. An interesting related question is whether such a fourfold can be realized over 𝐐.

Leaving the case of abelian varieties and symplectic groups, we focus on three-dimensional representations, which arise naturally in an arithmetic context as well. Using [7] as motivation, one can consider modular forms for congruence subgroups Γ0(N) of SL3(𝐙). Given a cuspidal eigenform fH3(Γ0(N),𝐂), let 𝐐f denote the number field generated by the Hecke eigenvalues of f. Let λ𝐐f be a prime dividing . Then we have attached to f the λ-adic Galois representation

ρλ:Gal(𝐐¯/𝐐)GL3(𝐐f,λ).

(See [7, § 3] for an explicit example of how such compatible families of representations arise.) The residual representation ρλ¯ then provides a natural setting for studying subgroups of GL3(k), where k is a finite field. In the aforementioned example, if imρλ¯ is a fixed-point subgroup of GL3(k), then we get additional information on congruence properties of the number of points on the variety modulo p, for all but finitely many p, by the Chebatorev density theorem. For this reason, and the ones mentioned above with respect to abelian varieties, the fixed-point subgroups of linear groups have special arithmetic interest.

In this paper, we continue our classification of fixed-point linear groups and determine all fixed-point subgroups of GL3(k), where k is a finite field. Unlike the classifications in [3, 5, 4] for the groups GSp4(𝐅) and GSp6(𝐅), the main theorem of this paper allows for k to be an arbitrary finite field of any characteristic.

### 1.2 The main theorem

We postpone a review of notation until the next section, except to remark that the maximal subgroups of a finite linear group fall into eight geometric classes 𝒞1,,𝒞8, together with a class 𝒮 of exceptional subgroups; we refer the reader to [1] for the details of the classification.

There are certain subgroups of GL3(q) that are easily identifiable as fixed-point subgroups, for example, those conjugate to

(1**0**0**)GL3(q),

and we do not wish to include them in our classification. We therefore declare a subgroup G of GL3(q) to be a trivial fixed-point group if the semisimplification of the underlying three-dimensional representation contains the trivial representation. Henceforth, we work exclusively with semisimple groups in this paper as these will have the same fixed-point properties as the parabolic groups they lie in. Our main theorem is as follows; see the following sections for all notational definitions.

### Theorem 1.1.

The maximal, non-trivial and semisimple fixed-point subgroups of GL3(q) are as follows:

 Dimensions of simple factors Isomorphism type Conditions (1,1,1) C2×C2 q odd (2,1) Dq-1 q odd Dq+1 q odd Irreducible Sym4 q odd SO3⁢(q) q odd Alt5 5∈𝐅q×2, p≠2,3

### Remark 1.2.

As a corollary of our theorem, we obtain that there are no irreducible fixed-point subgroups of GL3(q) in characteristic 2. In particular, the groups SO3(q) and Sym4, each of which is naturally a subgroup of GL3(q), fix a line in characteristic 2 (for the former, see [1, Theorem 1.5.41] while the latter is deduced from the Brauer table of Sym4 in characteristic 2).

### 1.3 Notation and setup

Let k be a finite field of characteristic p (we choose p instead of for consistency in the group theory literature), and write k=𝐅q, where q=pn. We follow the classification and notational scheme of [1], which is based on Aschbacher’s original classification of the subgroups of the finite classical groups.

Since GL3(q) is not a fixed-point group itself, any fixed-point subgroup must lie in a maximal subgroup of GL3(q) and hence in one of the eight geometric classes 𝒞1,,𝒞8, or the exceptional class 𝒮. We use the standard notation from finite group theory, basing much of our notational scheme on that of [1]. In particular, we set

1. Altn: the alternating group on n letters,

2. Symn: the symmetric group on n letters,

3. Cn: the cyclic group of order n,

4. Eq: the elementary abelian group of exponent p and rank n,

5. Am+n: if A is elementary abelian, then Am+n has elementary abelian normal subgroup Am and quotient An,

6. p+1+2n: the extra-special p group of order p1+2n and exponent p,

7. d: the center of SL3(q),

8. Z(q): the center (scalar matrices) of GL3(q),

9. Ln(q): the projective special linear group PSLn(q),

10. AB: the wreath product of A and B, where BPerm(A××A),

11. N.Q denotes a non-split extension of Q by N,

12. N:Q denotes a split extension of Q by N,

13. N.Q denotes an arbitrary extension of Q by N.

Our strategy for proving Theorem 1.1 is roughly as follows. Given a subgroup G of GL3(q), we intersect with SL3(q) and use the classification of maximal subgroups of SL3(q) outlined in [1, Chapter 2] to determine the fixed-point subgroups of SL3(q). We then lift back to GL3(q) to find the maximal fixed-point subgroups. The issue is that we may encounter novel subgroups – maximal subgroups M of GL3(q) such that MSL3(q) is not maximal in SL3(q). We will address any novelties as they arise. Toward that end, we record the maximal subgroups of SL3(q) in Table 1 below; see [1, Table 8.3] for complete details on the subgroup structure of SL3(q).

### Remark.

There is a typographical error in [1, Table 8.3]: in class 𝒞1, the group labelled Eq3:GL2(q) should be Eq2:GL2(q). We have corrected this in Table 1.

We also note that the papers [9, 13] provide a classification of the ternary linear groups over finite fields, from which one could recover [1, Table 8.3]; however, we prefer to begin with the classification scheme of [1] due to the modern notation and language.

Table 1

Maximal subgroups of SL3(q)

 Class Isomorphism type 𝒞1 Eq2⁢:⁢GL2⁢(q), Eq1+2⁢:⁢(q-1)2, GL2⁢(q) 𝒞2 (q-1)2⁢:⁢Sym3, q≥5 𝒞3 (q2+q+1)⁢.3, q≠4 𝒞5 SL3⁢(q0).(q-1q0-1,3) if q=q0r, r prime 𝒞6 3+1+2.Q8.(q-1,9)3 if p=q≡1(mod3) 𝒞8 d×SO3⁢(q) if q is odd (q0-1,3)×SU3⁢(q) if q=q02 𝒮 d×L2⁢(7) if q=p≡1,2,4(mod7), q≠2 3⁢.⁢A6 if q=p≡1,4(mod15) or q=p2, p=2,3(mod5), p≠3

The groups in class 𝒞1 are the parabolic subgroups of GL3(q) and we treat them separately in the next section. We then focus the rest of the paper on the irreducible fixed-point subgroups of GL3(q).

## 2 Parabolic fixed-point subgroups of GL3⁢(q)

Let G be a semisimple subgroup of GL3(q). We break the proof of Theorem 1.1 into two cases, depending on whether the action of G on k3 is reducible or irreducible. In case of a reducible representation, G lies in a parabolic subgroup (type 𝒞1) of GL3(q), and the irreducible factors are either all one-dimensional, or consist of a two-dimensional and one-dimensional factor. (In both cases, we replace the representations with their semisimplifications.) Moreover, we require the classification of subgroups of a direct product, given by Goursat’s lemma [2, p. 864].

## Theorem (Goursat’s lemma).

Let A and B be finite groups. The subgroups G of A×B are in one-to-one correspondence with the tuples (G1,G2,G3,ψ), where G1A, G2B, G3G2, and ψ:G1G2/G3 is a surjective homomorphism.

Beginning with the case where G is a subgroup of the diagonal subgroup Cq-13 of GL3(q), we write G(Cq-1×Cq-1)×Cq-1. We can describe G via two “Goursat tuples”

(H1,H2,H3,ψ),whereH1Cq-1×Cq-1,H2Cq-1,
(D1,D2,D3,ϕ),whereD1Cq-1,D2Cq-1,

and (D1,D2,D3,ϕ) is the Goursat tuple corresponding to H1Cq-1×Cq-1.

## Lemma 2.1.

Suppose G, acting diagonally on Fq3, is a fixed-point subgroup that does not fix a line. Then q is odd, and GC2×C2.

## Proof.

With all notation as above, we assume G(Cq-1×Cq-1)×Cq-1 is given by the Goursat tuple (H1,H2,H3,ψ), where H1(Cq-1×Cq-1), and H1 is given by the Goursat tuple (D1,D2,D3,ϕ). Let S be the subset of H1 consisting of pairs (x,y) such that neither x nor y is 1. We will show that, unless G is the group specified in the statement of the lemma, the size of S forces G to fix a line. We make several elementary observations:

1. S lies in kerψ (G is a fixed-point group).

2. H3 is trivial (G contains the elements of the form (kerψ,H3), Skerψ, and G is a fixed-point group).

3. Therefore, ψ:H1H2 is a surjective homomorphism.

We will give several estimates of #S below, and so we set the following notation:

hi=#Hi,di=#Di,k=#kerψ,l=#kerϕ.

Combining the observations, we immediately see that

(2.2)k=h1/h2#S+1,

where the “+1” is due to the identity of H1.

Since H1 is given by the Goursat tuple (D1,D2,D3,ϕ), we have h1=d1d3. We can estimate the size of S by writing #S=h1 minus the number of elements (x,y) of H1 with at least one x or y trivial; that is,

#S=d1d3-l-d3+1.

Comparing this to (2.2), we get our first estimate

(2.3)d1d3/h2d1d3-l-d3+2.

But since ld1, we can refine (2.3) to get our second estimate

(2.4)d1d3/h2d1d3-d1-d3+2=(d1-1)(d3-1)+1.

It is easy to check that the only integer triples

(d1,d3,h2)withd31andd1,h22

(recall that G is a non-trivial fixed-point group) satisfying (2.4) are of the form (d1,2,2) or (2,d3,2).

If q is even, then there is no such subgroup of G since q-1 is odd, so we suppose q is odd. We will work through the details of the case (d1,2,2) and omit those of the case (2,d3,2) since they are nearly identical. Therefore, we consider the group

H1={(g,ϕ(g))gD1andϕ:D1D2/{±1}}.

In general, there are 2d1/d2+1 pairs in H1 with a 1 in one of the components. Therefore, there are

2d1-(2d1/d2+1)

with both components non-trivial. We impose this condition on the estimate of k:

k2d1-(2d1/d2+1)+1=2d1-2d1/d2.

Notice that if d2>2, then G would be a trivial fixed-point group since we would have kerψ=H1, and so H2 would coincide with H3, which is trivial. Therefore, we may assume d2=2.

Since d2=d3=2, this means D2=D3={±1}, and so H1 is a direct product H1=D1×{±1}. Including the identity, there are at least d1 elements of H1 that must lie in kerψ:

(1,1),(g,-1),,(gd1-1,-1),

where g is a generator of D1. Since kerψ is a subgroup of H1, it follows that (g2,1)kerψ as well. Unless g2=1, this forces kerψ=H1 and G to be a trivial fixed-point group. We conclude that H1={±1}×{±1}. Together with

[H2:H3]=2andH3=1,

we get exactly the group C2×C2 as claimed in the lemma, which is given explicitly in terms of matrices as

(ϵ1ϵ2ϵ1ϵ2),

where ϵi{±1}. ∎

Next suppose that GGL2(q)×GL1(q) is semisimple with irreducible projection onto GL2(q). Since G is a subgroup of a direct product, it is given by a Goursat tuple (H1,H2,H3,ψ), with H1GL2(q). As above, we only classify those G which are not direct products, that is, H2H3.

## Observation.

If GGL2(q)×GL1(q) is a fixed-point subgroup that does not fix a line and is given by the Goursat tuple (H1,H2,H3,ψ), then H3 is trivial. This follows because any gH1 without a fixed point is paired via ψ with H3.

## Lemma 2.5.

With all notation as above, if G is a fixed-point subgroup of

GL2(q)×GL1(q)

that does not fix a line, then H1 is a proper subgroup of GL2(q).

## Proof.

It is an elementary counting problem to show that more than half the elements of GL2(q) do not have a fixed point once q>2 – use the fact that there are

1. q2(q-1)2/2 elements with eigenvalues in a quadratic extension,

2. q-2 non-trivial central elements,

3. (q-2)(q-1)(q+1) non-diagonalizable elements without a fixed point.

Dividing by the size of GL(2,q), we get

q2(q-1)2/2+(q-2)+(q-2)(q-1)(q+1)q4-q3-q2+q=q3-3q2q3-2q2-2q+2>12.

Thus if H1=GL2(q), then kerψ=GL2(q), and so H2=H3. But since H3 is trivial by the observation above, we have that H2 is trivial. If q=2, then H2 is trivial. ∎

By Lemma 2.5, H1 must lie in a proper subgroup of GL2(q) and hence lies in a maximal subgroup of GL2(q). By [12, Theorem 2.3], the subgroups H of GL2(q) not containing SL2(q) are described as follows (we use PH to denote the image of H in PGL2(q)):

1. If H contains an element of order q, then either G lies in a Borel subgroup or SL2(q)H;

2. PH is cyclic, and H is contained in a Cartan group;

3. PH is dihedral, and H is contained in the normalizer of a Cartan group but not in the Cartan subgroup itself;

4. PH is isomorphic to Alt4, Sym4 or Alt5.

Returning to our setup, if H1 lies in a Borel or a Cartan, then H1 is not irreducible. We therefore focus only on cases (3) and (4) of the subgroup classification of GL2(q). We recall from [16, § 3] the explicit description of the normalizers of the Cartan subgroups of GL2(q) and adopt that notation in what follows.

Let Cs(q) and Cns(q) denote the maximal split and non-split Cartan subgroups, respectively. Then Cs(q)𝐅q××𝐅q× and Cns(q)𝐅q2×, and each Cartan group has index 2 in its normalizer, which we denote by Cs+(q) and Cns+(q), respectively, borrowing the notation of [16]. Each normalizer has a distinguished dihedral subgroup, 𝖣s(q) and 𝖣ns(q), respectively, where

𝖣s(q)Cs(q)=Cs(q)SL2(q)and𝖣ns(q)Cns(q)=Cns(q)SL2(q).

That is, the “rotation” group of 𝖣s(q) (resp. 𝖣ns(q)) consists of the elements of Cs(q) (resp. Cns(q)) of determinant (norm) 1. It follows that

𝖣s(q)Dq-1and𝖣ns(q)Dq+1.

Each dihedral group admits a surjective homomorphism to C2 and, when q is odd, we can realize that homomorphism in the Goursat tuples

Dq-1(𝖣s(q),{±1},1,ψ)andDq+1(𝖣ns(q),{±1},1,ψ).

It is easy to check that both dihedral groups are fixed-point subgroups of

GL2(q)×GL1(q)

with irreducible projection to GL2(q) that do not fix a line in 𝐅q3. We will show in Proposition 2.6 below that these are the only such groups. In preparation for the proof we make some observations.

## Observation.

Let GGL2(q)×GL1(q) have Goursat tuple (H1,H2,H3,ψ), and suppose H1 is an irreducible subgroup of GL2(q) that normalizes a Cartan subgroup. Let G be a fixed-point group.

1. The normalizer of the split Cartan group has exactly 3q-4 elements with a fixed point; by fixing a basis, we can write these elements explicitly as

{(x00y):xory=1}{(0zz-10):z0}.
2. The normalizer of the non-split Cartan group has exactly q non-trivial elements with a fixed point, all of which belong to the non-trivial coset of the Cartan subgroup.

3. If G does not fix a line, then H1 must contain at least

#H1#H2-1#H2

elements with a fixed point.

## Proposition 2.6.

Let GGL2(q)×GL1(q) be a fixed point group with Goursat tuple (H1,H2,H3,ψ), and suppose H1 normalizes a Cartan subgroup.

If q is even, then H2 is trivial, and so G fixes a line in Fq3. If q is odd, then either H2 is trivial (and so G fixes a line), or G is dihedral with Goursat data (Ds(q),{±1},1,ψ) or (Dns(q),{±1},1,ψ).

## Proof.

We only sketch the proof since it comes down to an exercise in matrix manipulation. Suppose H2 is non-trivial. Because H1 normalizes a split Cartan subgroup, its maximal order is 2(3q-4) in the split case and 2(q+1) in the non-split case, by combining observations (2) and (3) above. When q is odd, in order to create a subgroup H1 (and not merely a subset) satisfying the hypotheses of the proposition, matrix manipulation shows that H1 must be a subgroup of 𝖣s(q) in the split case and 𝖣ns(q) in the non-split case and H2={±1}. When q is even, #H2 is odd, and so at least 23 of the elements of H1 must have a fixed point, and H1 must admit a cyclic odd-order quotient with all non-kernel elements having a fixed point. No such subgroup exists. ∎

We conclude this section by analyzing the subgroups of GL2(q) with projective image Alt4, Sym4 and Alt5. Let PH{Alt4,Sym4,Alt5}. The central extensions of PH are classified by the Schur multiplier. Neither Alt4 nor Alt5 has an ordinary two-dimensional irreducible representation; hence any central extension HGL2(q) of PH must be non-trivial for these groups. When PH=Sym4, the trivial central extensions of Sym4 do occur as subgroups of GL2(q).

In all cases, the Schur multiplier of PH has exponent 2; hence any central extension has the form 2.PH times a group of scalar matrices, the order of which can be deduced from [16, Lemma 3.21]. The isomorphism types of 2.PH that occur as subgroups of GL2(q) are

2.Alt4SL2(3),
2.Alt5SL2(5),
2.Sym4{21.Sym4Alt4C4,22.Sym4SL2(3).C2(non-split),23.Sym4GL2(3),24.Sym4C2×Sym4.

The complexity of the groups 2.Sym4 is due to the fact that the Schur multiplier H2(Sym4,C2)C2×C2. We now investigate the groups H for their fixed-point properties.

### 2.1 Projective image Alt4

Let q be coprime to 6. Let H be a subgroup of GL2(q) such that PHAlt4. There are three inequivalent absolutely irreducible ordinary representations σ1, σ2 and σ3 of SL2(3), with character values as follows (ω denotes a fixed primitive third root of unity):

 Class 1 2 3⁢A 3⁢B 4 6⁢A 6⁢B χ1 2 -2 -1 -1 0 1 1 χ2 2 -2 1+ω -ω 0 ω -1-ω χ3 2 -2 -ω 1+ω 0 -1-ω ω

The representation σ1 is defined over 𝐙, and σ1(SL2(3))SL2(q); σ2(SL2(3)) and σ3(SL2(3)) define subgroups of GL2(q) when q1(mod3). In any of the three representations, the only class with a fixed point is the identity.

### Lemma 2.7.

Let H be a maximal preimage of Alt4 in GL2(q). Let

GGL2(q)×GL1(q)

be a fixed point subgroup of GL3(q) with Goursat tuple (H1,H2,H3,ψ). Suppose H1 is an irreducible subgroup of H. Then H2 is trivial.

### Proof.

If H is a maximal preimage of Alt4, then it is a product of scalar matrices and the non-trivial extension 2.Alt4 of Alt4. Since all elements of H without a fixed point must belong to kerψ, it follows that 2.Alt4 is a subgroup of kerψ as well as the group of scalar matrices. Thus ψ is the trivial homomorphism; whence H2 is trivial. ∎

Now we consider the special cases of modular characteristic. If q is even, then any group H such that PH=Alt4 is not irreducible in GL2(q) [11, Lemma 6.1]. If q is a power of 3, then the isomorphism 2.Alt4SL2(3) shows that 2.Alt4 occurs naturally as a subfield subgroup of GL2(q). The same counting argument of Lemma 2.5 shows that more than half the elements of H do not have a fixed point, and hence H cannot give rise to a non-trivial fixed-point subgroup of GL3(q).

### 2.2 Projective image Alt5

Let q be coprime to 30. Then there are two inequivalent ordinary absolutely irreducible representations σ1 and σ2 of SL2(5) with the following character data:

 Class 1 2 3 4 5⁢A 5⁢B 6 10⁢A 10⁢B χ1 2 -2 -1 0 -1+52 -1-52 1 1+52 1-52 χ2 2 -2 -1 0 -1-52 -1+52 1 1-52 1+52

In both representations, the only element with a fixed point is the identity.

### Lemma 2.8.

Let H be a maximal preimage of Alt5 in GL2(q). Let

GGL2(q)×GL1(q)

be a fixed point subgroup of GL3(q) with Goursat tuple (H1,H2,H3,ψ). Suppose H1 is an irreducible subgroup of H. Then H2 is trivial.

### Proof.

The proof is identical to that of Lemma 2.7. ∎

In modular characteristic, if q is even, then the isomorphism

Alt5SL2(4)=PSL2(4)

shows that Alt5 occurs as a subfield subgroup of SL2(q) (once q>4). The same counting argument of Lemma 2.5 shows that more than half the elements of H do not have a fixed point, and hence H cannot give rise to a non-trivial fixed-point subgroup of GL3(q). The same argument applies when q is a power of 5 via the isomorphism 2.Alt5SL2(5).

If q is a power of 3, then 2.Alt5 only occurs as a subgroup of GL2(q) when q is an even power of 3 since it is required that 5(𝐅q×)2. And if q is an even power of 3, then 𝐅q contains 𝐅9, so it suffices to work in GL2(9). In GL2(9), the group 2.Alt5 has 15 elements without a fixed point; hence kerψ=2.Alt5, and so H2 is trivial.

### 2.3 Projective image Sym4

Let q be coprime to 6. We consider the four groups 2i.Sym4 for i=1,2,3,4 separately. The group 21.Sym4 has no faithful irreducible degree-2 ordinary representations, and we do not consider unfaithful representations in this analysis for fixed-point subgroups.

The group 22.Sym4 has two faithful irreducible ordinary degree-2 representations σ1, σ2 with the following character data:

 Class 1 2 3 4⁢A 4⁢B 6 8⁢A 8⁢B χ1 2 -2 -1 0 0 1 2 -2 χ2 2 -2 -1 0 0 1 -2 2

In the representations σ1 and σ2, the group 22.Sym4 has no non-trivial elements with a fixed point.

### Lemma 2.9.

Let H be a maximal preimage of Sym4 in GL2(q) that contains 22.Sym4. Let GGL2(q)×GL1(q) be a fixed point subgroup of GL3(q) with Goursat tuple (H1,H2,H3,ψ). Suppose H1 is an irreducible subgroup of H. Then H2 is trivial.

### Proof.

The proof is identical to that of Lemma 2.7. ∎

The group 23.Sym4 has two faithful irreducible ordinary degree-2 representations σ1, σ2 with the following character data:

 Class 1 2⁢A 2⁢B 3 4 6 8⁢A 8⁢B χ1 2 -2 0 -1 0 1 --2 -2 χ2 2 -2 0 -1 0 1 -2 --2

In both representations, there are exactly 35 elements without a fixed point.

### Lemma 2.10.

Let H be a maximal preimage of Sym4 in GL2(q) that contains 23.Sym4. Let GGL2(q)×GL1(q) be a fixed point subgroup of GL3(q) with Goursat tuple (H1,H2,H3,ψ). Suppose H1 is an irreducible subgroup of H. Then H2 is trivial.

### Proof.

Any element of 23.Sym4 without a fixed point belongs to kerψ; whence kerψ contains 23.Sym4 and the scalar matrices. Thus kerψ=H1, and so H2 is trivial. ∎

The group 24.Sym4 has two unfaithful irreducible degree-2 representations, and we do not consider unfaithful representations in this analysis.

We finish this section by considering the groups 22.Sym4 and 23.Sym4 in modular characteristic. If q is even, then neither 22.Sym4 nor 23.Sym4 is irreducible [11, Lemma 6.1]. If q is a power of 3, then the isomorphism 23.Sym4GL2(3) shows that H1 occurs as a subfield subgroup of GL2(q). The same counting argument of Lemma 2.5 shows that more than half the elements of H do not have a fixed point, and hence H cannot give rise to a non-trivial fixed-point subgroup of GL3(q). Finally, 22.Sym4 contains SL2(3) as index-2 subgroup, and the full group 22.Sym4 is contained in GL2(9). Again, the same counting argument of Lemma 2.5 shows that there are no non-trivial fixed-point subgroups in this case.

## 3 The irreducible fixed-point subgroups of GL3⁢(q)

In this section, we complete the proof of Theorem 1.1 in a case-by-case analysis based on the maximal subgroup classes.

### 3.1 Subgroups of type 𝒞2

The maximal subgroup of GL3(q) of type 𝒞2 is isomorphic to GL1(q)Sym3 as long as q5. If G is a subgroup of GL1(q)Sym3, then G fits into a split short exact sequence

1G0GP1,

where G0 is a subgroup of GL1(q)3 and P is subgroup of Sym3. If G is a fixed-point subgroup of GL1(q)Sym3, then so is G0. By Lemma 2.1, either G0 fixes a line or G0C2×C2.

### Lemma 3.1.

Suppose G0 fixes a line. Then any lift G of G0 to GL1(q)Sym3 fixes a line as well. Therefore, there are no irreducible fixed-point subgroups of type C2 when q is even, or when G0 fixes a line.

### Proof.

If G0 fixes a line, then consider the permutation group P. If P is trivial or has order 2, then G is reducible and fixes a line. So we assume P contains a 3-cycle. Choosing a basis with respect to which G0 fixes the first coordinate, we see that G contains matrices of the form

M(α,β)=def(1000α000β)ands=def(010001100).

In order for a matrix of the form M(α,β)s to have a fixed point, we must take αβ=1. Continuing, the product M(α,α-1)s2M(α,α-1)2s has a fixed point if and only if α{±1}. Finally,

sM(-1,-1)s2=(-1000-10001),

which shows that G0 can only contain the identity matrix M(1,1). But the full permutation group Sym3 fixes a line in this representation. Thus there are no irreducible fixed-point subgroups G such that G0 fixes a line. ∎

### Lemma 3.2.

Let q be odd. Suppose G is an irreducible fixed-point subgroup of GL1(q)Sym3. Then G is isomorphic to Alt4 or Sym4.

### Proof.

By Lemmas 2.1 and 3.1, we can assume G0C2×C2, given explicitly by

{(ϵ1ϵ2ϵ1ϵ2):ϵi{±1}}.

An easy calculation shows that the full wreath product (C2×C2)Sym3Sym4 is an irreducible fixed-point subgroup of GL1(q)S3, as well as its subgroup Alt4. ∎

### Remark.

It remains to discuss what happens for q<5. When q=2, the group of type 𝒞2 is not maximal in SL3(q), but belongs to the reducible maximal subgroup class of type 𝒞1. When q=3, the classes 𝒞2 and 𝒞8 coincide, in light of the isomorphism SO3(3)Sym4, so this group can be considered as an irreducible fixed-point subgroup of type 𝒞8 as well. When q=4, the group GL1(4)Sym3 is not a maximal subgroup of GL3(4) [1, Proposition 2.3.6].

### 3.2 Subgroups of type 𝒞3

There are no irreducible fixed-point subgroups of GL3(q) in this class, as we now show. The maximal subgroup of GL3(q) in this class is isomorphic to GL1(q3).3, with outer automorphisms given by the Galois group Gal(𝐅q3/𝐅q).

### Remark.

When q=4, the restriction of GL1(4).3 to SL3(4) is not maximal (see Table 1) in SL3(4), but GL1(4).3is maximal in GL3(4).

Let GGL1(q3).3 be a fixed-point subgroup. Then G fits into the short exact sequence

1NGQ1,

where N is a cyclic group of order dividing q3-1 and Q is either trivial or isomorphic to C3.

Let g be a generator for the group GL1(q3) and σ a generator of Gal(𝐅q3/𝐅q); in this representation, the eigenvalues of g have the form γ,γσ,γσ2. Because GL1(q3) is cyclic and because the eigenvalues of any power of g are powers of γ, γσ and γσ2, it follows that the only element of GL1(q3) with a fixed point is the identity. The trivial group lifts to a cyclic group of order 3 inside GL1(q3).3, and every element of such a C3 has a fixed point, but the group is not irreducible.

### 3.3 Subgroups of type 𝒞5

These are the field-restriction subgroups of GL3(q); that is, if we can write q=q0r, then GL3(q0) is naturally a subgroup of GL3(q). When r is prime, the group generated by GL3(q0) and the center Z(q) of GL3(q) is the maximal subgroup of GL3(q) of type 𝒞5.

Suppose r is prime, and let 𝒢=GL3(q0),Z(q). Let G be an irreducible fixed-point subgroup of 𝒢. Because no non-trivial element of Z(q) has a fixed point, it follows that G is an irreducible fixed-point subgroup of GL3(q0). Since we seek to classify the subgroups of type 𝒞5, we may assume (by descent) that G is an irreducible fixed-point subgroup of GL3(p), hence lies in a subgroup class other than 𝒞5. Therefore, the class 𝒞5 contains no irreducible fixed-point subgroups of GL3(q) that are not already contained in another class.

### 3.4 Subgroups of type 𝒞6

There are no irreducible fixed-point subgroups of GL3(q) in this class, as we now show. We first classify the fixed-point subgroups of SL3(q) in this class and then lift them to GL3(q). Recall from Table 1 that q=p1(mod3).

### Lemma 3.3.

Let G be a non-trivial fixed-point subgroup of

3+1+2.Q8.(q-1,9)3SL3(q).

Then GQ8 or GC3.

### Proof.

This is a finite computation, easily performed in Magma, and we omit the details. The result is that there are, up to isomorphism, two fixed-point subgroups of 3+1+2.Q8.(q-1,9)3: a cyclic group of order 3, and Q8. ∎

### Lemma 3.4.

There is no irreducible fixed-point subgroup of GL3(q) of type C6 that restricts to Q8.

### Proof.

The group Q8 is normal in any subgroup of GL3(q) that restricts to

Q8SL3(q).

The three-dimensional representation of Q8 decomposes into a two-dimensional factor and a one-dimensional factor. By Clifford’s theorem, any lift of Q8 to GL3(q) retains this decomposition; whence there are no irreducible subgroups of GL3(q) restricting to Q8. ∎

### Lemma 3.5.

There is no irreducible fixed-point subgroup of GL3(q) of type C6 that restricts to the fixed-point C33+1+2.Q8.(q-1,9)3SL3(q).

### Proof.

Because q1(mod3), the representation of the fixed-point C3 is completely reducible and decomposes into three one-dimensional representations, one of which is trivial. By Clifford’s theorem, the representation of any subgroup of GL3(q) restricting to C3 is either a sum of three one-dimensional representations, or is irreducible. If it were irreducible, the three one-dimensional representations of C3 (upon restriction) would be conjugate. Since only one of the three is trivial and a non-trivial representation cannot be conjugate to a trivial, it follows that the representation of any overgroup C3.m of C3 is not irreducible. This proves the lemma. ∎

### 3.5 Subgroups of type 𝒞8

There are two isomorphism types of maximal subgroups of SL3(q) of type 𝒞8, namely, d×SO3(q) and (q0-1,3)×SU3(q0) if q=q02. Moreover, this class contains no novel subgroups. We first consider the case of d×SO3(q).

The group SO3(q) is a fixed-point group [8, Proposition 6.10] and is irreducible in odd characteristic. If q1(mod3), then SO3(q) is maximal in SL3(q), while if q1(mod3), then d×SO3(q) is maximal, with d a scalar group of order 3. Because d is scalar, the maximal fixed-point subgroup of d×SO3(q) is SO3(q). Thus, for all q, the maximal fixed-point subgroup of SL3(q) of type 𝒞8 is SO3(q).

It remains to determine whether there exist fixed-point groups H that fit into the sequence SO3(q)HGL3(q) of proper containments. The groups of type 𝒞8 are scalar-normalizing [1, Definition 4.4.4] in the sense that any such group H has the presentation SO3(q)Z, where Z is a subgroup of the scalars of GL3(q). Thus any overgroup H properly containing SO3(q) is necessarily not a fixed-point group (some non-trivial element of Z multiplies the identity of SO3(q)). We therefore have the following result.

### Lemma 3.6.

Let q be a power of an odd prime. The maximal irreducible fixed-point subgroup of GL3(q) containing SO3(q) is SO3(q).

Next we consider the case of the subgroup (q0-1,3)×SU3(q0) of SL3(q) and, more generally, the subgroup GU3(q0) of GL3(q). We will show that there are no additional irreducible fixed-point subgroups arising in this class that have not already been classified. First the group GU3(q0) is not itself a fixed-point group; hence any irreducible fixed-point subgroup must lie in one of its maximal subgroups. We list the non-parabolic maximal subgroups of GU3(q0) and SU3(q0).

1. Type 𝒞2: GU1(q0)Sym3 is the maximal subgroup of GU3(q0) of type 𝒞2. The same argument as in Lemma 3.2 applies and shows the maximal irreducible fixed-point subgroup is isomorphic to Sym4.

2. Type 𝒞3: GU1(q03).3 is the maximal subgroup of GU3(q0) of type 𝒞3. The same argument as in Section 3.2 applies and shows there are no irreducible fixed point subgroups in this class.

3. Type 𝒞5: there are two subgroups of SU3(q0) in this class,

SU3(q1).(q+1q1+1,3)(ifq0=q1rfor primer)andSO3(q).
4. Type 𝒞6: there is one maximal subgroup of SU3(q0) in this class,

3+1+2.Q8.(q+1,9)3.
5. Type 𝒮: there are four isomorphism classes of subgroups of type 𝒮 of SU3(q):

1. d×L2(7) (d conjugates, q=p3,5,6(mod7), q5), where

d=gcd(q+1,3),
2. 3.Alt6 (3 conjugates, q=p11,14(mod15)),

3. 3.Alt6.23 (3 conjugates, q=5),

4. 3.Alt7 (3 conjugates, q=5).

Many of the same arguments as in the previous sections apply here as well. In particular,

1. in class 𝒞5, the same descent argument as in Section 3.3 shows that it is enough to classify the maximal subgroups of SU3(p);

2. in class 𝒞6, the same argument as in Section 3.4 applies as well: the only fixed-point subgroups of 3+1+2.Q8.(q+1,9)3 are C3 and Q8 and, by the same Clifford’s theorem argument, any lift to GU3(q) is reducible.

It remains to analyze the subgroups of type 𝒮. We delay our treatment of d×L2(7) and 3.Alt6 until the next section so that we can give a unified treatment of these two groups; they occur as maximal subgroups of SL3(q) for certain q and SU3(q) for others. We now consider the two subgroups 3.Alt6.23 and 3.Alt7 of SU3(5).

In both cases, we search in the subgroup lattices of 3.Alt6.23 and 3.Alt7 for fixed-point subgroups. One can check that the conjugacy classes of elements of order 1, 2, 5 have fixed points, while some of the classes of order 3, 4 and 6 do as well. The result of the search is that the following are the isomorphism types of fixed-point subgroups of 3.Alt6.23 and 3.Alt7:

{Cj}j=1,,5,F20.

Setting aside the group C5, each of the fixed-point groups listed above is reducible, and the semisimplification of each representation consists of three one-dimensional representations, one of which is trivial. The identical Clifford’s theorem argument of Section 3.4 shows that none of these groups lifts to an irreducible fixed-point subgroup of GL3(25). For the group C5, the semisimplification consists of three trivial representations, so the Clifford’s theorem argument does not immediately rule out an irreducible fixed-point subgroup of GL3(25). However, a search for all subgroups of GL3(25) of the form C5.m that are irreducible fixed-point subgroups reveals none. All computations for this section were performed in Magma.

### 3.6 Subgroups of type 𝒮

We complete the classification of irreducible fixed-point subgroups of GL3(q) with the groups of type 𝒮, and we incorporate two of the type 𝒮 subgroups of GU3(q) into this section as well. We recall the conditions under which each of these groups occur.

 Subgroup of Conditions SL3⁢(q) (q-1,3)×L2⁢(7) q=p≡1,2,4(mod7), q≠2 3⁢.⁢Alt6 q=p≡1,4(mod15) q=p2, p≡2,3(mod5), p≠3 SU3⁢(q) (q+1,3)×L2⁢(7) q=p≡3,5,6(mod7), q≠5 3⁢.⁢Alt6 q=p≡11,14(mod15)

The simple group L2(7) of order 168 has an absolutely irreducible three-dimensional representation over 𝐅q when -7(𝐅q×)2, and the group 3.Alt6 has an absolutely irreducible representation over 𝐅q when -3,5(𝐅q×)2. The conditions on q reflect these requirements. We start with d×L2(7).

### Lemma 3.7.

Let G be a maximal, irreducible fixed-point subgroup of

(q-1,3)×L2(7)SL3(q)𝑜𝑟(q+1,3)×L2(7)SU3(q),

subject to the conditions on q in the tables above. Then GSym4.

### Proof.

In either case, a fixed-point subgroup intersects the center trivially; hence GL2(7). The maximal subgroup Sym4 of L2(7) is an absolutely irreducible fixed-point subgroup, so it remains to show that no element of order 7 has a fixed point for any allowable q.

Fix a primitive seventh root of unity ω𝐅¯q. Then the characteristic polynomial on either class of order 7, evaluated at 1 is given by

43ω(ω-1)(ω4+2ω3+ω2+2ω+1)0.

The inequality follows from the observation that if ω4+2ω3+ω2+2ω+1=0, then the resultant

Res(ω4+2ω3+ω2+2ω+1,ω6+ω5+ω4+ω3+ω2+ω+1)=72=0,

which is impossible since q is coprime to 7. ∎

### Lemma 3.8.

Let G be a maximal, irreducible fixed-point subgroup of

3.Alt6SL3(q).

Then GAlt4 or Alt5.

### Proof.

Since 3.Alt6 contains the center of SL3(q), any fixed-point subgroup must be proper. There are five maximal subgroups of 3.Alt6:

3+1+2.4,d×Alt4(two copies),d×Alt5(two copies).

The group 3+1+2.4 was analyzed previously in Section 3.4 and does not possess any irreducible fixed-point subgroups. On the other hand, one can check that the (centerless) groups Alt4 and Alt5 of the remaining cases are each irreducible fixed-point subgroups. ∎

The groups of type 𝒮 are scalar normalizing [1, Proposition 4.5.2], and so there are no irreducible fixed-point overgroups HGL3(q) that properly contain the Alt4, Sym4 or Alt5 of the lemmas above. This completes the classification of non-trivial fixed-point subgroups of GL3(q) stated in Theorem 1.1.

Communicated by Nigel Boston

## Acknowledgements

We would like to thank the referee for pointing out several errors in an initial draft and for additional suggestions which improved the clarity and exposition of the paper.

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