I have a problem with computation of $H^2(G, \mathbb Z_2G)$, where $G=\mathbb Z\times \mathbb Z$.
I know that I have to find a nice projective resolution for $\mathbb Z$ and then by taking the functor Hom(,), the second cohomology group is obtained.
It is too much calculation.
Is there a shorter argument for that?
Check Shapiro's lemma in Kenneth Brown's book Cohomology of Groups, page 73. Take $H = \mathbb{Z}$. I hope it helps.
If $X$ is a contractible, free $G$-CW-complex with compact quotient, then there is an isomorphism
$$ H^k(G;\mathbb{Z}_2G) \cong H^k_c(X;\mathbb{Z}_2) $$
where the group on the right denotes cohomology with compact support. This is essentially Proposition 7.5 in Brown's "Cohomology of groups" book (the argument also works for $\mathbb{Z}_2$).
In your case with $G=\mathbb{Z} \times \mathbb{Z}$ we can use $X=\mathbb{R} \times \mathbb{R}$ with the action given by coordinate-wise translation. This is a free action, $X$ is a $G$-CW-complex and $X/G \cong S^1 \times S^1$, which is compact. Therefore
$$ H^2(G;\mathbb{Z}_2 G) \cong H^2_c(\mathbb{R} \times \mathbb{R};\mathbb{Z}_2) \cong H^2(S^2;\mathbb{Z}_2) \cong \mathbb{Z}_2 $$
where the second isomorphism comes from $S^2$ being the one-point compactification of $\mathbb{R} \times \mathbb{R}$.
