Subgroups of GL(2,q)

Question

I have been wondering about something for a while now, and the simplest incarnation of it is the following question:

Find a finite group that is not a subgroup of any $GL_2(q)$.

Here, $GL_2(q)$ is the group of nonsingular $2 \times 2$ matrices over $\mathbb{F}_q$. Maybe I am fooled by the context in which this arose, but it seems quite unlikely that "many" finite groups are subgroups of $GL_2(q)$. Still, I can't seem to exclude a single group, but I am likely being stupid.

Motivation: This arose when I tried to do some explicit calculations related to Serre's modularity conjecture. I wanted to get my hands on some concrete Galois representations, and play around with the newforms associated to it. Call it recreational if you like (it certainly is!). In Serre's original paper [1], there is a wonderful treatment of some explicit examples. He uses the observation that the Galois group over $\mathbb{Q}$ of

$$x^7 -7x +3$$

is isomorphic to $PSL_2(7)$. This gives him a surjection $G_{\mathbb{Q}} := \mbox{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow PSL_2(7)$, which he combines with the character associated to $\mathbb{Q}(\sqrt{-3})$ to obtain a homomorphism $G_{\mathbb{Q}} \rightarrow PSL_2(7) \times C_2$. This homomorphism he can then lift to $GL_2(49)$ using a very clever calculation in the Brauer group.

The way I see it (without claiming this is justified) is that it is unclear or even provably impossible to embed $PSL_2(7)$ directly into some $GL_2(q)$, and for that reason we need a nice lifting.

This raises the natural question of when this little trick with the Brauer group is "necessary". What can be said in general about subgroups of $GL_2(q)$? Can we exclude any particular finite group or families of finite groups? Can we perhaps even classify the possible isomorphism types of such subgroups? Of course this is all becomes even more natural to ask in the light of the inverse Galois problem. How many Galois groups can we "see" in the two-dimensional representations?

Remark. Of course, the question could be approached the other way. If one has the (perhaps not unrealistic?) hope of realising every $GL_2(q)$ as a Galois group over $\mathbb{Q}$, how many groups do you automatically realise over $\mathbb{Q}$ by taking quotients of $GL_2(q)$?

Which groups $G$ admit a surjection $GL_2(q) \rightarrow G \rightarrow 1$ for some $q$?

Reference:

[1] Serre, Jean-Pierre, "Sur les représentations modulaires de degré 2 de Gal(Q/Q)", Duke Mathematical Journal 54.1, (1987): 179–230

Answer 1

Well, ${\rm GL}(2,q)$ has Abelian Sylow $p$-subgroups for every odd prime $p.$ The symmetric group $S_{n}$ has non-Abelian Sylow $p$-subgroups for each prime $p$ such that $p^2 \leq n.$ Hence thesymmetric group $S_{25}$ is not a subgroup of any ${\rm GL}(2,q)$ since it has non-Abelian Sylow $3$-subgroups and non-Abelian Sylow $5$-subgroups ( even the alternating group $A_{25}$ will do). To answer directly the question about ${\rm PSL}(2,7),$ if $q$ is a power of $2,$ then all odd order subgroups of ${\rm GL}(2,q)$ are Abelian, so ${\rm PSL}(2,7)$ can't be a subgroup of such a ${\rm GL}((2,q)$ since ${\rm PSL}(2,7)$ contains a non-Abelian subgroup of order $21.$ On the other hand, if $q$ is odd, then the alternating group $A_4$ is not a subgroup of ${\rm GL}(2,q)$ ( for example, any Klein 4-subgroup of such a ${\rm GL}(2,q)$ contains a cenral element of order $2,$ whereas $A_4$ has no central element of order $2$), but ${\rm PSL}(2,7)$ has a subgroup isomorphic to $A_4.$

Answer 2

We can generalize Ralph's answer to find the abelian groups that are $p$-groups for odd $p$ that cannot be subgroups of $GL_2$.

if $p|(q^2-1)(q^2-q)$ then exactly one of $p|(q+1)$, $p|(q-1)$ and $p|q$. If $p|q$ then $q=p^n$ and upper-triangular unipotent matrices provide enough, so they're a Sylow subgroup, $(\mathbb Z/p)^n$. If $p|(q-1)$ then the group of diagonal matrices contains a Sylow subgroup, which looks like $\mathbb Z/p^k \times \mathbb Z/p^k$. If $p|(q+1)$ then $\mathbb F_{p^2}^\times$ contains a Sylow subgroup, which looks like $\mathbb Z/p^l$.

Therefore $\mathbb Z/p^2 \times \mathbb Z/p \times \mathbb Z/p$ cannot be contained in any $GL_2(\mathbb F_q)$, while every abelian $p$-group not containing it can be.

Combined with Geoff's answer this gives a complete characterization of the solution to this problem for odd $p$-groups.

Answer 3

Claim 1: No group $$A_{a,b,,c} := \mathbb{Z}/2^a \times \mathbb{Z}/2^b \times \mathbb{Z}/2^c$$ with $a,b,c > 1$ is a subgroup of some $GL(2,q)$.

Claim 2: If $q \neq 2,3$, then the quotients of $GL(2,q)$ are exactly the subgroups of the cyclic group $\mathbb{F}^\times_q$ and the groups $GL(2,q)/D$ where $D \le \mathbb{F}^\times_q \cdot I$.

In case $GL(2,2) = S_3$, the only quotient is $\mathbb{Z}_2$.

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Proof: 1) Let $q$ be odd. We have the extension $$1 \to SL(2,q) \to GL(2,q) \to \mathbb{F}^\times_q \to 1$$ with $\mathbb{F}^\times_q$ cyclic (as multiplicative group of finite field). Let $P$ be a Sylow 2-subgroup $GL(2,q)$. Then we have the extension $$1 \to SL(2,q) \cap P \to P \to C \to 1$$ where $C$ is a cyclic 2-group. It's known that the Sylow 2-subgroups of of $SL(2,q)$ are cyclic or generalized quaternion (see this link). Hence $GL(2,q)$ has no subgroups of the form $A_{a,b,,c}$ with $a,b,c > 0$.

If $q = 2^n$ then $|GL(2,q)|= (q^2-1)(q^2-q) = q(q^2-1)(q-1)$. Hence a Sylow 2-subgroup $P$ is given by upper triangular matrices with unit diagonal. Thus $P \cong \mathbb{F}_q$ is an elementary abelian 2-subgroup of rank n.

2) Write $GL := GL(2,q)$, $SL := SL(2,q)$ and let $C = \langle \pm 1 \rangle$ be the center of $SL$.

Let $N$ be a normal subgroup of $GL$ and let be $H = N \cap SL$. Then $HC$ is a normal subgroup of $SL$ and $HC/C$ is a normal subgroup of the simple group $PSL = SL/C$. Hence $HC = C$ or $HC = SL$.

Case 1: $HC = C$. Hence $H \le C$. Fix $a \in N$. We want to show that $a$ is central in $GL$.

If $x \in GL$ then $xax^{-1}a^{-1} \in N \cap SL \le C$ (note: $SL$ is the commutator subgroup of $GL$). If $\operatorname{char}(\mathbb{F}_q) = 2$, then $C=1$ and we are done. So assume $\operatorname{char}(\mathbb{F}_q) > 2$. Define a map $f: GL \to C, x \mapsto [x,a]$. Since $C$ is central we have $$f(xy) = xyay^{-1}x^{-1}a^{-1}= xf(y)ax^{-1}a^{-1}=f(x)f(y)$$ i.e. $f$ is a group hom. into an abelian group. Hence $SL = [GL,GL]$ lies in the kernel of $f$, i.e. $SL$ commutes with $a$. Let $\alpha \in \mathbb{F}^\times_q$ be a generator and let $x_0 = \operatorname{diag}(\alpha,1)$. A direct calculation shows that $ax_0=-x_0a$ is not possible. Therefore $a,x_0$ commute and since $GL$ is generated by $SL$ and $x_0$, we conclude that $a$ is central. Hence $N$ is a central subgroup of $GL$.

Case 2: $HC=SL$. Suppose $C \nsubseteq H$. Since $C$ is a central subgroup of $SL$ of prime order, $H \cap C = 1$ and $SL=H \times C$ follows, contradicting the indecomposability of $SL$. Thus $H=SL$ and $SL \le N$. Therefore $\mathbb{F}^\times_q = GL/SL$ maps onto $GL/N$, i.e. $GL/N$ is a quotient of $\mathbb{F}^\times_q$ and as such isomorphic to a subgroup of $\mathbb{F}^\times_q$.