2 reduction to Noetherian case added

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 $n$.

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.

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 $n$.