If $G$ is a group let $(G^{(m)})_{m \geq 0}$ be the derived series. If there is some $m$ such that $G^{(m+1)} = G^{(m)}$, call the smallest such $m$ the derived length of $G$. I am interested in conditions which control the derived length of finitely generated $G \leq \mathrm{GL}_n(K)$, where $K$ is an arbitrary field. Here are a couple.
- If $G$ is actually soluble, then the derived length is bounded in terms of $n$. This was proved by Zassenhaus and sharpened by Newman and others. The sharp bound has the form $O(\log n)$.
- Suppose $G$ is finite. Let $P$ be the last term in the derived series. Then any minimal subgroup $S \leq G$ such that $G = PS$ must be soluble, and $G^{(m)} \leq P S^{(m)}$ for each $m$. By 1 this implies that $P = G^{(m)}$ for some $m = O(\log n)$.
On the other hand, any nonabelian free group has infinite derived length, and there are nonabelian free subgroups of $\mathrm{GL}_2(\mathbf{Q})$.
Here are a couple specific questions, to make my question concrete. Let $G \leq \mathrm{GL}_n(K)$ be finitely generated.
- If $G$ has finite derived length (aka perfect-by-soluble), must its derived length be bounded in terms of $n$?
- If $G$ is soluble-by-perfect, same question.
