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 $q$?


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

  • 2
    $\begingroup$ To help answer your second question, if $H$ is a Sylow subgroup of $G$ then a Sylow subgroup of the inverse image of $H$ maps surjectively onto $H$ and is a subgroup of a Sylow subgroup of $H$. So $H$ is a quotient of a subgroup of a Sylow subgroup of $GL_2(q)$, so it's "smaller" than a Sylow subgroup of $GL_2(q)$, so you can use Geoff's, Ralph's, and my calculations to show that a group containing a large or non-abelian Sylow subgroup does not admit such a surjection. $\endgroup$ – Will Sawin Mar 20 '12 at 2:40

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 $A_4.$

  • $\begingroup$ This was overkill, actually! Only one non-Abelian Sylow subgroup for an odd prime would do. $\endgroup$ – Geoff Robinson Mar 19 '12 at 20:22
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    $\begingroup$ So any non-abelian $p$-group for odd $p$ works? $\endgroup$ – Will Sawin Mar 19 '12 at 20:43
  • $\begingroup$ Yes. In characteristic $2,$ finite $p$-subgroups are completely reducible when $p$ is odd, so passing to an extension field they are diagonalizable, hence Abelian. When $q$ is a power of $p,$ and $p$ is odd, the Sylow $p$-subgroups of ${\rm GL}(2,q)$ are elementary Abelian. $\endgroup$ – Geoff Robinson Mar 19 '12 at 21:16
  • $\begingroup$ In the above, I am talking about $p$-subgroups of ${\rm GL}(2,q).$ $\endgroup$ – Geoff Robinson Mar 19 '12 at 21:18
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    $\begingroup$ Accepted for the simplest criterion. I never realised this nice fact about odd Sylow subgroups of $GL_2(q)$! As an example, it follows from your answer that the only subgroups $A_n$ have $n\leq 5$, where we note that $A_5 \times C_3 \cong GL_2(4)$. Your arguments exclude the bigger ones, since $q$ would have to be even, resulting in abelian Sylow $2$-subgroups. This seems to suggest a general phenomenon, where only "small" groups will arise. In general the Sylows will be unpredictable and probably non-abelian, which gives a notion of "almost never" as in the question. Thanks! $\endgroup$ – Bertie Wooster Mar 20 '12 at 10:26

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.

  • $\begingroup$ I can't vote for this, or I would! $\endgroup$ – Geoff Robinson Mar 19 '12 at 23:07
  • $\begingroup$ @Geoff: if you register, then you will be able to vote. It will also safeguard you against losing your account if the cookie gets deleted for some reason or if you sit at a different computer. $\endgroup$ – Alex B. Mar 19 '12 at 23:29
  • $\begingroup$ It's more difficult to register on here than on math stack exchange. $\endgroup$ – Geoff Robinson Mar 20 '12 at 7:34
  • $\begingroup$ Since "most" groups with bounded order are $p$-groups, this deals with quite many of them. :-) Can anything be said about other families? For example, I think one has all the dihedral groups. Is there a positive answer for other families of non-abelian groups? $\endgroup$ – Bertie Wooster Mar 20 '12 at 10:47

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 $\mathbb{Z}_2$.

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 $\mathbb{F}^\times_q$.

  • $\begingroup$ Very nice answer, the point you make on the quotients is quite final. Thanks! $\endgroup$ – Bertie Wooster Mar 20 '12 at 10:35

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