$\newcommand{\Z}{\mathbf{Z}}$ Given a nice infinite collection of groups, for example the symmetric groups, one can ask whether any finite group is a subgroup of one of them. Of course any finite group acts on itself, so any finite group is a subgroup of a symmetric group. Similarly any finite group acts linearly on its group ring over a finite field, so given a field $k$, any finite group embeds into $GL(n,k)$ for some sufficiently large $n$ (as permutation matrices).

**Question 1:** What if we do what I ask in the title, and consider the groups $SL(2,R)$ as $R$ ranges over all commutative rings. Given an arbitrary finite group, can I find a commutative ring $R$ (with a 1) such that this group is a subgroup of $SL(2,R)$?

Of course this is inspired by this pesky question which, at the time of typing, seems to remain unsolved.

Here is a more specific question:

**Question 2:** Is there a commutative ring $R$ (with a 1) such that the symmetric group $S_4$ injects into $SL(2,R)$?

I haven't thought much about question 1 at all. I'll tell you what I know about question 2. Let's consider first the case where $R$ is an algebraically closed field. If the characteristic is zero, or greater than 3, then by character theory any map from $S_4$ into $SL(2,R)$ must contain $A_4$ in its kernel (the map must give a semisimple representation and the irreducible 2-dimensional one has non-trivial determinant).

If the characteristic is 3 then considering the restriction of a map $S_4\to SL(2,R)$ to a Sylow 2-subgroup we see again by character theory that the kernel must contain the central element. But the kernel is a normal subgroup of $S_4$ so it must contain $V_4$ and hence factors through a map $S_3\to SL(2,R)$. Now the image of an element of order 2 must be central and it's not hard to deduce that the 3-cycles must again be in the kernel.

In characteristic 2 there are more possibilities. If I got it right, the kernel of a map $S_4\to SL(2,R)$ ($R$ alg closed char 2) is either $S_4$, $A_4$ or $V_4$ and of course the representation can be non-semisimple this time.

We conclude from this case-by-case analysis that if $R$ is any ring and $S_4\to SL(2,R)$ is any map then the image of $V_4$ is in $1+M_2(J)$, where $J$ is the intersection of all the prime ideals, that is, the nilpotent elements of $R$.

I now wanted to consider the case $J^2=0$ and check that $V_4$ must be killed mod $J^2$ and then go by induction, but I couldn't bash it out and wonder whether it's true.

It's clear that one could brute-force the argument if one could do a Groebner basis calculation over the integers. I have tried one of these in my life---when trying to solve the open problem of whether every finite flat group scheme of order 4 was killed by 4. That latter question seems to be beyond current computers, but perhaps the one I'm raising here might not be. The problem would be that one has to work over $\Z$ and this slows things down greatly.

I then looked for counterexamples, but convinced myself that $S_4$ was not a subgroup of either $SL(2,\Z/4\Z)$ or $SL(2,\Z/2\Z[\epsilon])$ with $\epsilon^2=0$ [**edit**: I am wrong; $SL(2,\Z/2\Z[\epsilon])$ does work, as pointed out by Tim Dokchitser]. I don't know how to get a computer algebra system to check $SL(2,\Z/2\Z[\epsilon,\delta])$ so I gave up and asked here.

I suspect I am missing some standard fact :-/

commutativegroup scheme of order $n$ is killed by $n$: the "old-fashioned" proof that any commutative group of order $n$ is killed by $n$ (multiply all the elements together, call the result $x$, and note that $gx=x$) generalises very nicely. But there are non-commutative group schemes of order 4. The issue of course is a base with 2 locally nilpotent. If no conceptual proof is known one can try writing down the universal group scheme of order 4 and checking it on this, but the bottom line is that this "non-conceptual" approach involves writing down about 30 generators for an... – Kevin Buzzard Nov 30 '10 at 7:10