Let $G$ be a finitely generated, infinite, virtually abelian group; that is,
- $G$ is infinite, but
- there is a finite set $X\subseteq G$ generating $G$, and
- there exists abelian $H\leq G$ with $[G:H]<\infty$.
Does there exist a surjective homomorphism $f\in G\to\mathbb{Z}$?
I ask because a key claim in Tao's proof of Gromov's theorem on groups of polynomial growth is that such a homomorphism exists (for $G$ a subset of a compact Lie group — I don't know if this matters). (Having reread the post, Tao does not make that claim; I am still interested in the question, though). And yet it is not obvious to me, nor can I think of a counterexample.
I can prove the claim when $G$ is virtually central (i.e., $[G:Z(G)]<\infty$). For: since $G$ is infinite but $[G:Z(G)]$ is not, so must $Z(G)$ be. In any group, $Z(G)$ injects into the abelianization $G^\text{ab}=G/[G,G]$. Of course, there are infinite abelian groups with no direct factor of infinite order (e.g. $\mathbb{Q}/\mathbb{Z}$), but none are finitely generated. And $G^\text{ab}$ is finitely generated by $X^\text{ab}$.
Now, $$H/[G,H]\subseteq Z(G/[G,H])$$ so that $G/[G,H]$ is virtually central, but it isn't clear to me that $Z(G/[G,H])$ should be an infinite group anymore.
Best Answer
As Moishe Kohan points out in the comments, the infinite dihedral group $$D_{\infty}=\langle\rho,\sigma:\sigma^2=1,\sigma\rho\sigma=\rho^{-1}\rangle$$ is a counterexample.
The group $D_{\infty}$ is infinite and finitely generated, with index-$2$ abelian subgroup $\langle\rho\rangle$. But the commutator subgroup of $D_{\infty}$ is $\langle\rho^2\rangle$, so that $D_{\infty}$ has finite abelianization.
More generally, let $V$ be any infinite abelian group generated by finite $S$ and let finite $G\leq\mathrm{Aut}(V)$ act on $V$ such that, for any $s\in S$, there exists $T\in G$ and $v\in V$ with $Tv=v+s$ (in fancy language: as a $\mathbb{Z}[G]$-module, $V$ is trivial modulo the augmentation ideal). Then consider the semidirect product $H=V\rtimes G$. $H$ is then infinite, virtually abelian, and finitely generated. For any $v\in V$ and $T\in G$, we have $$TvT^{-1}v^{-1}=Tv-v$$ Thus then $V\subseteq[H,H]$, so that $H^\text{ab}$ is finite.