If every finite quotient group of a finitely generated linear group G is solvable, then G is solvable

finitely-generatedgroup-theoryinfinite-groupslinear-groupssolvable-groups

For this question, I was able to show that each finite quotient is polycyclic: Suppose $N$ is a normal subgroup of finite index. Then, all subgroups of $G/N$ are finite, so $G/N$ is Noetherian. A solvable group is polycyclic iff it is Noetherian. At this point, I don't know how to continue or if I'm even going in the right direction. I am familiar with Theorems by Mal'cev, Milnor, Wolf etc.

I would appreciate any help!

Best Answer

Let $G \leq \mathrm{GL}_n(K)$ be your group. Let $f(n)$ be the maximal derived length of a soluble subgroup of $\mathrm{GL}_n(F)$ for any field $F$ (Zassenhaus's theorem). Suppose $g \in G^{(f(n))}$ is nontrivial. By Mal'cev's theorem, there is a finite field $F$ and a homomorphism $\pi : G \to \mathrm{GL}_n(F)$ such that $\pi(g) \neq 1$. But $\pi(g) \in \pi(G)^{(f(n))}$, and $\pi(G)$ is finite and therefore soluble by hypothesis, so this is a contradiction. Hence $G^{(f(n))} = 1$.