Is there a finite index **torsion-free** subgroup $G$ of $GL\_n(\mathbb Z)$, where $n\ge 3$,
such that the coinvariants group $\mathbb Z^n\_G$ is finite? 


Here $G$ acts on $\mathbb Z^n$ in the standard way, and $\mathbb Z^n\_G$ by definition is the quotient of $\mathbb Z^n$ by the subgroup generated by the set $\{gz-z: g\in G, z\in\mathbb Z^n\}$. 


If $G=GL\_n(\mathbb Z)$, then $\mathbb Z^n\_G$ is finite because G contains $-I\_n$, but then $GL\_n(\mathbb Z)$ isn't torsion-free.



Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants
For any finite index subgroup $G$ there is a nonzero integer $m$ so that $G$ contains the elementary matrices $e\_{ij}(m)$ that have ones on the diagonal, $m$ at the $(i,j)$ entry and zeroes elsewhere. So the span of $\{gz-z: g\in G, z\in\mathbb Z^n\}$ contains all multiples of $m$
and ${\mathbb Z}^n\_G$ is finite. Now just take a torsion-free finite index subgroup $G$.


**Edit** (by Igor Belegradek): I add some detail for my records. 
If $e\_r$ is a vector in the standard basis, then 
it is easy to write $me\_r$ in the form $v-e\_{ij}(m)v$ for
some $i\neq j$ and $v\in\mathbb Z^n$. It remains to show that
for any finite index subgroup $G$ of $GL\_n(\mathbb Z)$
all such $e\_{ij}(m)$ lie in $G$ for some $m$. By making $G$ smaller
we can assume it is normal and of index $k$. Since 
$e\_{ij}(m)=e\_{ij}(1)^m$, it suffices to assume that $k$ divides $m$.


