Suppose I have a group ${ G }$ acting freely on the sphere ${ S^n }$ for ${ n \geq 2 }$. Does this somehow get me a free resolution of ${ \mathbb{Z} }$ as a ${ \mathbb{Z}[G] }$-module?
Forgive me if the question is obvious. I was reading Brown's book on Cohomology of Groups, and I saw there that the previous statement holds true if we replace ${ S^n }$ with a contractible CW space ${ X }$ (Or anything which is acyclic).
I saw someone mention this statement as a comment on this MSE question Proving that there doesn't exist a free $\mathbb{Z}_2\times \mathbb{Z}_2$ action on $S^n$. which makes me think that a proof of it would be trivial, but I just can't figure it out!
Suppose you have a $G$-CW complex structure on $S^n$ for your free action. Then cellular chains are a complex of free $G$-modules. This chain complex fails to be a resolution of $H_0(S^n)=\mathbb{Z}$ because $H_n(S^n)=\mathbb{Z}$. But we can just glue together another copy of the chains to kill the kernel $C_n(S^n)\rightarrow C_{n-1}(S^n)$, ad infinitum. Hence, we get a periodic resolution of $\mathbb{Z}$ as a $G$-module.