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.