Homomorphic images of linear groups If $G$ is a linear group over a ring $R$, is every homomorphic image of $G$ again a linear group?
 A: No. For example, the free groups $F_{\infty}$ on countably many elements is a subgroup of $F_2$, which is linear (over $\mathbb{C}$ for example), so $F_{\infty}$ is linear. Its quotients are all countably-generated groups. It is straightforward to write down countably-generated groups which are not linear over a particular commutative ring, but to get one not linear over any commutative ring seems to require slightly more work.
Lemma: Let $A_{\infty}$ denote the group of alternating permutations (fixing finitely many elements) of $\mathbb{N}$. Then $A_{\infty}$ is simple.
Proof. Let $N$ be a normal subgroup and let $\rho : A_{\infty} \to A_{\infty}/N$ be the corresponding quotient. $A_{\infty}$ is the union of a chain of distinguished subgroups $A_n \subset A_{\infty}$ which are simple for $n \ge 5$; consequently their image in $A_{\infty}/N$ is either trivial or all of $A_n$ for $n \ge 5$, and if $A_n$ has nontrivial image then so does $A_m$ for all $m \ge \text{max}(n, 5)$ (by simplicity) and for all $m \le \text{max}(n, 5)$ (since these are contained in a sufficiently large $A_m$), hence for all $A_m$. Hence the image of all of $A_{\infty}$ is either trivial or all of $A_{\infty}$, so $N$ is either all of $A_{\infty}$ or trivial. $\Box$
Claim: $A_{\infty}$ is not linear over any commutative ring $R$.
Proof. Let $\rho : A_{\infty} \to \text{GL}_n(R)$ be a homomorphism. For any prime ideal $P$ of $R$, the image of $A_{\infty}$ in $\text{GL}_n(R/P)$ is either trivial or all of $A_{\infty}$ by the lemma. If the image is all of $A_{\infty}$ for some prime ideal $P$, then by passing to the fraction field of $R/P$ we may assume WLOG that $R$ is a field. Suppose $R$ has characteristic $p$. Since $A_{\infty}$ contains $S_n$ as a subgroup for every positive integer $n$, it contains every finite group as a subgroup, so in particular it contains $(\mathbb{Z}/q\mathbb{Z})^n$ as a subgroup for some prime $q \neq p$ and every positive integer $n$. By passing to the algebraic closure we may simultaneously diagonalize the action of such a group, and then taking $n$ sufficiently large gives a contradiction.
If no prime ideal $P$ has the above property, then $A_{\infty}$ is contained in the intersection of the kernels of the maps $\text{GL}_n(R) \to \text{GL}_n(R/P)$; consequently it is contained in the group of matrices congruent to $I \bmod N(R)$ where $N(R)$ denotes the nilradical. But this group has a filtration by normal subgroups consisting of the matrices congruent to $I \bmod N(R)^k$ for positive integers $k$, and all of the corresponding quotients are solvable (by inspecting where the commutators land). $A_{\infty}$ necessarily has nontrivial image in one of these quotients, which is a contradiction. $\Box$
A: In a somewhat different category, a meaningful (counter-) example is the quotient of (one form of) the Heisenberg group $N$ of upper-triangular nilpotent 3-by-3 real matrices by the subgroup of those with integer entries in the upper right corner and 0's immediately superdiagonal. (The Lie algebra did not change, etc.)
