Cohomology of finite p-groups

Question

Given a finite abelian $p$-group $A$ acted on by a finite $p$-group $G$.

Under the assumption $\operatorname{H}^1(G,A_1)=0$, where $A_1$ is the set of elements of $A$ having order at most $p$, what can one say about $\operatorname{H}^n(G,A)$, $n$ arbitrary?

From a result of Gaschutz, we can deduce that $\operatorname{H}^n(G,A_1)=0$, for all $n>0$ (and even for all integer $n$ in dealing with Tate cohomology). Together with the exact sequence $0 \rightarrow A_1 \rightarrow A \rightarrow pA \rightarrow 0$, yield $\operatorname{H}^n(G,A) \cong \operatorname{H}^n(G,pA)$, for all $n>0$.

What can one say more than this?

Can we hope that $\operatorname{H}^n(G,A)=0$?

Thanks in Advance.

Answer 1

How about $p=2$, $A= \langle x,y \mid 2x=4y=0 \rangle$, $G=\langle g | g^2=1 \rangle$, where $gx=x+2y$, $gy=y$.

Then $A_1 = \langle x,2y \rangle$, and $H^1(G,A_1)=0$, but $H^1(G,A) \ne 0$.