Cohomology result (reference request)

Question

I want to have reference for this result.

Let $G$ be a group of order $p^k$ where $k\geq0$ and $A$ be a $G$-module. If for all positive $r$ and for all subgroup $H$ of $G$, $$H^r(H,A)=0$$ then for all $i\in \mathbb{Z}$, $H^i(G,A)=0$.

I read this in a paper which says this is a well known result but I have very limited knowledge in cohomology so I don't even know where to look. Could you let me know (as specific as possible would be nice but just the text would be nice). I have tried looking in Hatchers but I couldn't find the result.

Answer 1

This isn't true for ordinary cohomology of groups. If $A$ is an injective $\mathbb{Z}[G]$-module, then $H^r(H,A)=0$ for all $H$ and all $r>0$. But the category of $\mathbb{Z}[G]$-modules has enough injectives, so there is an injective module $A$ with a nontrivial map $\mathbb{Z}\to A$ from the fixed module $\mathbb{Z}$. For such $A$, $H^0(G,A)$ will be nontrivial.

However, this is true if $H^i(G,A)$ refers to Tate cohomology. In fact, you only need to assume that there exists an $r$ such that $H^r(G,A)=H^{r+1}(G,A)=0$ (so you don't need all subgroups, or all positive $r$). A nice proof can be found in section 1.11 of these notes (specifically, a stronger result is stated as Theorem 1.11.11, but note that to get your result, you only need the material in that section up to Proposition 1.11.7 together with the proof of (ii)$\Rightarrow$(iii) in Theorem 1.11.11).

Answer 2

If G is a finite group and A is a G-module, A is called cohomologically trivial iff for any subgroup H of G and any integer r in Z, the Tate cohomology groups $H^r$(H, A) (usually written with a "hat") are null. In your case where G is a p-group,the most complete characterization of cohomological triviality can be found in Serre's "Local Fields", chapter 9, §5, thm. 8 : A is G-cohomologically trivial iff there exists an integer r such that $H^r$(G, A) = $H^{r+1}$(G, A) = $0$ , iff A has finite projective dimension over the ring Z[G].