Torsion-free virtually $\mathbb{Z}^n$ is $\mathbb{Z}^n$?

Question

Let $G$ be a torsion-free group with a subgroup $H$ of finite index isomorphic to $\Bbb Z^n$. Is $G$ isomorphic to $\Bbb Z^n$? In this question we know the anwser is true for $n=1$.

Answer 1

Here is a nonabelian example with $|G:H|=2$.

The group $G$ is generated by the subgroup $H$ and $t$, where $H = \langle a,b,c \rangle$ is free abelian of rank $3$, $t^2=a$, $t^{-1}bt=c$, and $t^{-1}ct=b$.

Added later: There is a slightly simpler example also with $|G:H|=2$ and $G = \langle H,t \rangle$, but with $H = \langle a,b \rangle$ free abelian of rank $2$, in which $t^2=a$ and $t^{-1}bt=b^{-1}$. This group is metacyclic and has the simple presentation $\langle b,t \mid t^{-1}bt=b^{-1} \rangle$.

Answer 2

Bieberbach showed that a compact $n$-dimensional flat manifold $M$ is finitely covered by a (flat) torus. It follows that $\pi_1(M)$ fits into an exact sequence $0 \to \mathbb{Z}^n \to \pi_1(M) \to F \to 0$ where $F$ is a finite group; that is, $\pi_1(M)$ is virtually $\mathbb{Z}^n$. As the universal cover of the torus is $\mathbb{R}^n$, the same is true of $M$ so it is an aspherical manifold (a manifold with contractible universal cover).

An aspherical manifold is an example of an Eilenberg-MacLane space and is in particular determined up to homotopy by its fundamental group.

Claim: An aspherical manifold has torsion-free fundamental group.

If $\Gamma$ is a subgroup of $\pi_1(M)$ isomorphic to $\mathbb{Z}_k$, then $M' := \widetilde{M}/\Gamma$ is an $n$-dimensional aspherical manifold with fundamental group $\mathbb{Z}_k$. But the infinite lens space $S^{\infty}/\mathbb{Z}_k$ also has fundamental group $\mathbb{Z}_k$ and contractible universal cover, namely $S^{\infty}$, so it is homotopy equivalent to $M'$. But this is impossible as $M'$ is an $n$-dimensional manifold while $S^{\infty}/\mathbb{Z}_k$ has unbounded cohomology, as is discussed here.

Therefore the fundamental group of every compact flat manifold is an example of a group which is torsion-free and virtually $\mathbb{Z}^n$. The groups that arise this way are precisely the torsion-free crystallographic groups, also known as Bieberbach groups. An example of such a group is the fundamental group of the Klein bottle as suggested by Moishe Cohen in the comments above.

Bieberbach showed that there are finitely many such manifolds in any given dimension; see A058104 on OEIS. Furthermore, each one has a different fundamental group and hence gives rise to another example; in particular, every compact flat $n$-dimensional manifold which isn't the torus gives a non-trivial example.