The center of a linear group with finite conjugacy classes has finite index Given a linear group $G\subset GL_n(k)$ with the property that every conjugacy class is finite (or equivalently, each centraliser has finite index), show that it follows that the center has finite index.
The converse is true for any group because the order of any conjugacy class is bounded by the index of the center by orbit-stabilisor.
So far I could only prove the result when $G$ is finitely generated, for then $Z(G)=\cap_{i=1}^nC_G{(g_i)}$ for some $g_i\in G$, and thus $|G:Z(G)|\le \prod_i|G:C_G{(g_i)}|<\infty$.
It was thinking maybe to somehow apply Schur's result on the Burnside problem (a periodic linear group is locally finite), but I am not sure this helps.
 A: Each element of $G$ has a finite index centralizer, and hence every finitely generated subgroup of $G$ has finite index centralizer.
Let $F$ be a finite subset of $G$ whose centralizer $C$ in $M_n(k)$ has minimal dimension (observe that this centralizer is a subalgebra, hence a linear subspace). I claim that $C$ is the centralizer of $G$. Indeed, if $g\in G$, the centralizer of $F\cup\{g\}$ is contained in the centralizer $C$ of $F$, and hence equals $C$ by minimality of the dimension. Thus $C$ centralizes $g$. Since this holds for all $g$ we get the result.
Hence the center $C\cap G$ of $G$ coincides with the centralizer of $F$, and thus has finite index.
More generally, if $G$ is any FC-group (group all of whose conjugacy classes are finite) and satisfying the descending chain condition on centralizers (every descending chain of centralizers stabilizes), then $G$ has center of finite index. The descending chain condition on centralizers is satisfied by linear groups (in a strong sense, since there's a uniform bound on lengths of chains).
