Group cohomology of product with swapping (twisting) factors 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 $H^1(G, M)=0$ ?
 A: 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 $n$.
A: What you have written down is the $\Bbb Z/2$-module $M \otimes \Bbb Z[\Bbb Z/2]$. 
The $G$-module $M \otimes \Bbb Z[G]$ is actually independent of the choice of action on $M$, up to isomorphism: the map $m \otimes [g] \mapsto g^-1m \otimes [g]$ gives a $G$-equivariant isomorphism to $M \otimes \Bbb Z[G]$ with the trivial action on $M$. In particular we may as well assume the action is trivial. 
If you choose a resolution $0\to A_1 \to A_2 \to M \to 0$ of $M$ by free abelian groups, this gives a short exact sequence $0 \to A_1[\Bbb Z/2] \to A_2[\Bbb Z/2] \to M[\Bbb Z/2] \to 0$, so there is an associated long exact sequence in cohomology.
If I can convince you that the cohomology of $\Bbb Z^\infty[\Bbb Z/2]$ is zero in degrees larger than $0$, and that $H^0(\Bbb Z/2; A[\Bbb Z/2]) \cong A$ naturally in free abelian groups $A$, then the long exact sequence tells us that $H^*(\Bbb Z/2; M[\Bbb Z/2]) = 0$ in degrees larger than 0, and gives $H^0 \cong M$. 
But observe that $$H^*(\Bbb Z/2; \Bbb Z^\infty[\Bbb Z/2]) \cong H^*(\Bbb Z/2; \Bbb Z[\Bbb Z/2])^\infty$$ and this may easily be seen to be zero at the level of the definition (in degrees larger than 0). $H^0$ is just the fixed point space, which should make the statements about it clear. 
