TITLE: Finitely presented sub-groups of GL(n,C)
QUESTION [19 upvotes]: Here are two questions about finitely generated and finitely presented groups (FP):
1) Is there an example of an FP group that does not admit a homomorphism to $GL(n,C)$ with trivial kernel for any n?
The second question is modified according to the sujestion of Greg below.
2) For which $n$ given two subgroups of $GL(n,C)$ generated by explicit lists of matrices, together with finite lists of relations and the promise that they are sufficient, is there an algorithm to determine if they are isomorphic as groups?" 
In both cases we don't impose any conidtion on the group (apart from been FP), in particular it need not be discrete in $GL(n,C)$.

REPLY [10 votes]: I just wanted to make a comment on Mal'cev's theorem (if I could leave this as a comment, I would).
Mal'cev's paper is a great exposition of the theorem, as well as a lot of other related material, all written in a basic yet enlightening style.
Also, if you know a little commutative algebra (as in the Nullstellensatz, the one given in Eisenbud pg. 132), there is quick and easy proof of Mal'cev's theorem. I could sketch it if necessary, but I am right now in the process of LaTeX-ing it, so I'll probably just come back and post a link.
Steve
EDIT - a sketch of the argument:
Mal'cev's theorem says a finitely generated linear group is residually finite. So let $X\subset GL(n,F)$ be a finite subset of the general linear group over some field $F$, and $G=\langle X \rangle$.  First, make $X$ symmetric, so that if $x\in X$ then also $x^{-1}\in X$.  Each $x\in X$ is an $n\times n$ matrix, so we can assemble all entries from all elements of $X$, getting a finite subset of $F$.  Let $R$ denote the subring of $F$ generated by this subset (along with $1$).  Then $R$ is a Jacobson ring, and since it's a subring of $F$, it's Jacobson radical is $0$.  Now $G$ is  a subgroup of $GL(n,R)$;  let $g\in G$ be a non-identity element, so that $g-I_n\neq 0$, where $I_n$ is the identity matrix.  Thus $g-I_n$ has a non-zero element, and thus there is some maximal ideal $m\subset R$ not containing this non-zero element.  The matrix ring homomorphism $M_n(R)\rightarrow M_n(R/m)$ (reducing everything mod $m$) induces a group homomorphism $G\rightarrow GL(n,R/m)$, where $g$ is not in the kernel. But $R/m$ is finite (by the Nullstellensatz), so $GL(n,R/m)$ is a finite group.