Note that any finite abelian group $\Gamma$ can be embedded into $\operatorname{GL}(1,A)$ for some commutative ring $A$. Indeed, let $A=\mathbf Z[\Gamma]$ and $\Gamma\hookrightarrow \operatorname{GL}(1,\mathbf Z[\Gamma])$ be the obvious map. In particular, the symmetric group $S_2$ on two letters can be embedded into $\operatorname{GL}_1$ of a commutative ring. It is an easy fact (see for example this question) that the symmetric group $S_{n+1}$ embeds into $\operatorname{GL}(n,\mathbf Z)$. Moreover, the symmetric group $S_3$ does not embed into any $\operatorname{GL}(1,A)$ for stupid reasons ($S_3$ is not commutative, but $\operatorname{GL}_1$ of a commutative ring is commutative).
- Does the symmetric $S_{n+2}$ embed into $\operatorname{GL}_n$ of a commutative ring?
- In particular, does $S_4$ embed into any $\operatorname{GL}(2,A)$ for $A$ some commutative (unital) ring?
I've convinced myself that $S_4$ does not embed into $\operatorname{GL}(2,\mathbf C)$, so since any domain of finite type over $\mathbf Z$ embeds into $\mathbf C$, the ring $A$ in question 2 would have to be pretty "bad. "
In general, if the prime $p$ divides $n$, then the permutation representation of $S_n$ over the field ${\mathbb F}_p$ has two irreducible constituents of degree $1$ and one of degree $n-2$ and, for $n \ge 5$, this results in an embedding $S_n \to {\rm GL}(n-2,p)$.
Also, $S_4$ embeds into $GL(2,{\mathbb Z}_4)$: you can take the images of $(1,2)$ and $(1,2,3,4)$ to be $\left(\begin{array}{cc}3&1\\0&1\end{array}\right)$ and $\left(\begin{array}{cc}2&1\\1&0\end{array}\right)$.
It is known that there are no nontrivial representations of $S_n$ of degree less than $n-2$ over any field, and hence none over any domain.
I think it unlikely that there is such an embedding for any commutative ring $A$. Since $S_n$ does not embed into ${\rm GL}(n-3,K)$ for any domain $K$, such an embedding would have to have its image in the kernel of the map ${\rm GL}(n-3,A) \to {\rm GL}(n-3,A/I)$ for every prime ideal $I$ of $A$, which seems unlikely.