TITLE: Group cohomology of product with swapping (twisting) factors
QUESTION [5 upvotes]: Let $M$ be a $G$-module, where $G = \Bbb Z / 2 \Bbb Z$. Define a $G$-module structure on $A = M \oplus M$ by $g \cdot (a,b) = (g \cdot b, g \cdot a)$.
What is the group cohomology $H^*(G, A)$ is terms of $H^*(G,M)$ ? 
Thoughts:
I know that $H^n(G, -)$ commutes with products, but here the action on $A$ is not the diagonal action, so I'm not sure what to do here. Can we at least compute $H^1(G, A)$, provided that $H^1(G, M)=0$ ?

REPLY [2 votes]: For even $n>0$, $H^n$ is the kernel of $1-g$ factored by the image of $1+g$.
The kernel of $1-g$ is the set of solutions of $(a,b)=(g\cdot b,g\cdot a)$
which is equivalent to $b=g\cdot a$. Thus the typical element of the 
kernel is $(a,g\cdot a)=(a,0)+g\cdot(a,0)$ lying in the image of $1+g$.
Therefore $H^n(G,M)=0$ for even $n>0$.
A similar calculation gives  $H^n(G,M)=0$ for odd $n$.