$\newcommand{\Z}{\mathbf{Z}}$
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 :-/
Best Answer
For Question 2, The central extension $\tilde{S}_4$ is certainly a subgroup of $\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]) \subset \mathrm{GL}_2(\mathbf{C})$. The image of the determinant is $\pm 1$. The image of $\tilde{S}_4$ in $$\mathrm{GL}_2(\mathbf{Z}[\sqrt{-2}]/2) = \mathrm{GL}_2(\mathbf{F}_2[x]/x^2)$$ is $S_4$, and all the elements have determinant one. It's easy to see that the central element $$\left( \begin{matrix} -1 & 0 \\\ 0 & -1 \end{matrix} \right)$$ lies in the kernel, so it suffices to note that nothing else does. Yet it's obvious that the map surjects onto $\mathrm{GL}_2(\mathbf{F}_2) = S_3$, and (from the character table) the image is larger than $S_3$, so the image is $S_4$.
For Question 1, if $G$ injects into $\mathrm{SL}_2(R)$ for some $R$ then it injects into such a ring where $R$ is Artinian. Here is the proof.
EDIT: Step 0. (This was in my head, but I forgot to mention it, as Kevin reminds me in the comments). One may replace $R$ by the subring generated by the images of the entries of $g-1$ for all $g \in G$, and hence assume that $R$ is finitely generated over $\mathbf{Z}$ and hence Noetherian. (The Krull intersection thm requires a Noetherian hypothesis.)
Step 1. If $x$ is a non-zero element of $R$, then there exists a maximal ideal $\mathfrak{m}$ of $R$ such that $x$ is non-zero in $R/\mathfrak{m}^k$ for some $k$. Proof: Let $\mathfrak{m}$ be some maximal ideal containing the annihilator of $x$. Then $x$ is non-zero in the localization $R_{\mathfrak{m}}$, and thus $x$ is non-zero in $R/{\mathfrak{m}^k}$ by the Krull intersection theorem.
Step 2. If $x_1, \ldots, x_n$ are non-zero elements of $R$, there exists an ideal $I$ such that each $x_i$ is non-zero in $R/I$ and $R/I$ is Artinian. Proof: Apply Step 1 to each $x_i$, and let $I = \bigcap \mathfrak{m}^{k_i}_i$.
Step 3. Suppose that $G$ has $n$ non-trivial elements. Let $x_1, \ldots, x_n$ denote a non-zero entry in the matrix $g - 1$ for each element of $g$. Apply Step 2 to deduce that $g$ is not the identity in $R/I$ for some Artinian quotient for all non-zero $g \in G$.
Remark: If $G$ is simple, then $G$ is actually a subgroup of $\mathrm{SL}(k)$ for some field $k$. Proof: Artinian rings are semi-local, so $G$ is a subgroup of $\bigoplus_{i=1}^{n} \mathrm{SL}(A_i)$ for Artinian rings $A_i$. Since $G$ is simple, it must be a subgroup of $\mathrm{SL}(A)$ for one such $A$. This latter group is filtered by the groups $\mathrm{SL}(k)$ and copies of $M_0(k)$ (trace zero matrices). Since the latter is abelian and $G$ is simple, we are done.
It's easy to find examples of groups which are not subgroups of $\mathrm{SL}_n(k)$ for all fields $k$ and some fixed integer $n$.