Groups quasi-isometric to $\mathbb{Z}^n$ I am interested by the following result: A group quasi-isometric to $\mathbb{Z}^n$ is virtually $\mathbb{Z}^n$.
I know the article Harmonic analysis, cohomology, and the large-scale geometry of amenable groups by Yehuda Shalom, but currently I don't have any access to it.
Do you know another document on this subject?
 A: This can be proved using the fact that the Hirsch length of a polycyclic group is a quasi-isometric invariant (this is a result of Bridson). Let's denote the Hirsch length of a group by $h(G)$.
Now Bass proved that a virtually nilpotent group $G$ has polynomial growth, and the degree of that growth is given by
$$ d(G) = \displaystyle\sum rk(G_i/G_{i+1})i.$$
Here $G_i$ is the $i$-th term of the lower central series.
The converse is Gromov's polynomial growth theorem, that groups with polynomial growth are virtually nilpotent.
So now let's put all this together.  Let $G$ be a finitely generated group quasi-isometric to $\mathbb{Z}^n$. Then by Gromov's theorem there is a subgroup of finite index $H\le G$, which is nilpotent (and so polycyclic).
Now it is easy to check that $h(\mathbb{Z}^n)=d(\mathbb{Z}^n)=n$. So $h(H)=d(H)=n$. But this implies (using Bass's formula) that $H'$ is finite. It is easy to then see that $[H:Z(H)]$ is finite (see below).
Now $Z(H)$ is an abelian group, also quasi-isometric to $\mathbb{Z}^n$.  Thus, again by Bass's formula, $Z(H)\cong \mathbb{Z}^n\times T$, for some finite abelian group $T$.
But then $[G:\mathbb{Z}^n] = |T|\cdot[H:Z(H)]\cdot[G:H]$, which is finite.
Proof that $|H'|$ finite implies $[H:Z(H)]$ finite.
Let $H$ be generated by $h_1,h_2,\ldots,h_m$. Now the conjugates of $h_i$ are contained in the set $h_iH'$, which by assumption is finite. Thus $h_i$ has finitely many conjugates, and so $C_H(h_i)$ has finite index in $H$.  But then $Z(H)=\cap C_H(h_i)$ is also finite index.
