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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?

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  • $\begingroup$ @Todd Trimble : Yes, of course. I corrected the question. $\endgroup$
    – Curtis
    Commented Jan 17, 2013 at 20:59
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    $\begingroup$ I'd call these groups 'finite-by-abelian'. The distinction is quite important in this case, since the abelian-by-finite case is trivial. $\endgroup$
    – HJRW
    Commented Jan 18, 2013 at 10:42

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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 $G$.

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  • $\begingroup$ Doesn't this solution assume that $F$ is abelian? I'm not assuming that. $\endgroup$
    – Curtis
    Commented Jan 17, 2013 at 21:40
  • $\begingroup$ @ Curtis: there is an action of $G$ on $F$ by conjugation, giving a homomorphism $G \to Aut(F)$. The kernel of this action is the centralizer $C_G(F)$ of $F$, and is clearly finite index. $\endgroup$
    – Ian Agol
    Commented Jan 17, 2013 at 22:35
  • $\begingroup$ Assuming $F$ is central so $G$ is 2-step nilpotent, there are homomorphisms $G \to F$ given by $a \mapsto [a,x_i]$, $i=1,\ldots,n$, since $[ab,x]=[a,x][b,x]$. The intersection of the kernels of these maps for all $i$ gives $Z(G)$, and therefore is finite index since $F^n$ is finite. $\endgroup$
    – Ian Agol
    Commented Jan 17, 2013 at 23:03
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    $\begingroup$ What is sledgehammerish about the first statement? If $G'$ is finite, then any conjugacy class is finite (the conjugates of $x$ are contained in $xG'$). Thus for every generator $g_i$, its conjugacy class is finite, so its centralizer $C_G(g_i)$ is finite index. The intersection of these is the center $Z(G)$, so as long as $G$ is finitely generated, $Z(G)$ is finite index. When $G$ is not finitely generated, instead you can show that the second center is finite index. $\endgroup$
    – Steve D
    Commented Jan 17, 2013 at 23:11

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