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 $\langle b,t \mid t^{-1}bt=b^{-1} \rangle$.