# 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!

• Welcome to math.stackexchange! Don't forget to check out the guide for asking questions – William Apr 15 at 14:03
• Can you provide more context for your question, maybe the original source? As stated it is not true. For the first sentence you could take $X=pt$ and $G$ any group (or even $X=G$ if you need a free action). For the second sentence, we could then take $G = \mathbb{Z}$ with subgroup $H=2\mathbb{Z}$ so that $BH = S^1$ which is not acyclic; in fact only perfect groups can be the fundamental group of an acyclic space since $H_1$ is the abelianization of $\pi_1$. The third sentence then cannot be true in this generality since there are (finitely presented) groups with infinitely generated $H_3$. – William Apr 15 at 14:23
• Thanks for the edit, that's much more specific. I don't have access to their paper, but I did find expository notes by Vogtmann, which on page 13 says "Since Outer space and its spine $K_n$ are contractible, and since $Out(F_n)$ acts with finite stabilizers, the quotient of $K_n$ by any torsion-free subgroup $Γ$ of finite index is an aspherical space". Do you mean "aspherical" instead of "acyclic"? In that case I think I can answer. – William Apr 15 at 21:51
• The original statement can be found on page 4 of pi.math.cornell.edu/~vogtmann/papers/ICM/kvogtmann.pdf under 2.3 Finite generation of homology. I think it does mean acyclic instead of aspherical... the question you're talking about is quite different. – yshen Apr 15 at 22:28

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.