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?