# Free action of Group $G$ on $S^n$ gives free resolution of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module.

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.
• Please may you add a few more details?, how does procedure use the free $G$-action? Oct 16 at 8:49