Finite generation of homology of a group with cocompact group action Let $G$ be a group that acts on a space $X$, and the quotient $X/G$ is compact. Then any torsion-free subgroup of finite index of $G$ is the fundamental group of an acyclic space with only finitely many cells in each dimension. The homology of $G$ is then finitely generated in all dimensions. Why is it so?

Edit: apparently the general statement is false. Here's the context of the question:
Let $\text{Out}(F_n) = \text{Aut}(F_n)/\text{Inn}(F_n)$ be the outer automorphism group of the free group on $n$ generators. The spine of the outer space $K_n$ (defined by Vogtmann and Culler in their 1986 paper to study automorphism groups of free groups) is a contractible space that acts on $\text{Out}(F_n)$ freely. The quotient $K_n/\text{Out}(F_n)$ is finite and thus compact. 
The statement in the paper is as follows: the fact that $K_n/\text{Out}(F_n)$ is compact implies immediately that any torsion-free subgroup of finite index of $\text{Out}(F_n)$ is the fundamental group of an acyclic space with only finitely many cells in each dimension. The homology of $\text{Out}(F_n)$ is then finitely generated in all dimensions. 
It seems that the statement about the fundamental group follows easily from the fact that the quotient is compact, but I don't see why. 
Thank you!
 A: I think I can show that any torsion-free subgroup $\Gamma\subset Out(F_n)$ with finite index is the fundamental group of an aspherical finite CW complex (i.e. having a potentially non-trivial fundamental group but no higher homotopy groups).
$Out(F_n)$ acts on $K_n$ with finite stabilizers, so if $\Gamma$ is a torsion-free subgroup then $\Gamma$ acts freely on $K_n$. Since $K_n$ is contractible and $Out(F_n)$ (and hence $\Gamma$) is discrete, it follows that $K_n/\Gamma \simeq K(\Gamma,1)$, i.e. it is an aspherical space with fundamental group isomorphic to $\Gamma$. If $\Gamma$ has finite index in $Out(F_n)$ then the map $K_n/\Gamma \to K_n/Out(F_n)$ is a finite covering space, so if $K_n/Out(F_n)$ is a finite cell complex then $K_n/\Gamma$ is as well.
I think the statement "with finitely many cells in each degree" might be superfluous, because it seems like $K_n/\Gamma$ is actually a finite complex. In fact in Vogtmann's notes on page 12 I found the following:

Since the quotient of the spine $K_n$ by $Out(F_n)$ is finite, so is
  the quotient by any finite index subgroup $H$

so this should apply in particular to our torsion-free $\Gamma$ of finite index.
