Assume that $G$ is a group that fits into a short exact sequence
$$1 \longrightarrow F \longrightarrow G \longrightarrow \mathbb{Z}^n \longrightarrow 1$$
with $F$ a finite group. For $f \in F$ not the identity, can we find some finite-index subgroup $A\_f \subset G$ such that $f \notin A\_f$?


Actually, the answer to this is yes : By assumption we have that $G$ is quasi-isometric to $\mathbb{Z}^n$, and it can easily be derived from Gromov's theorem on groups with polynomial growth that $G$ is virtually abelian. But this is attacking an ant with a sledgehammer. Does anyone know a more elementary proof?



Separating elements in abelian by finite groups
It follows easily from the result that (for finitely generated $G$) $G'$ finite implies $|G:Z(G)|$ finite.


If that is still too sledgehammer-like, then by replacing $G$ by $C\_G(F)$ we can assume $F$ is central and hence $G$ is class 2 nilpotent. Let $|F|=k$, and let $x\_1,\ldots,x\_n$ generate $G$ modulo $F$. Since the commutator map is bilinear in class 2 nilpotent groups, we have $[x\_i^k,x\_j] = [x\_i,x\_j]^k = 1$ for all $i,j$, so the subgroup $\langle x\_1^k,\ldots,x\_n^k \rangle$ is central and has finite index in $G$.


