# Kunneth formula for group cohomology over a field

This is Exercise 6.1.10 from Weibel's Introduction to homological Algebra.

Let $G$ and $H$ be two groups, if $k$ is a field, show that $$H^n(G\times H;k) \cong\bigoplus H^p(G;k)\otimes H^{n-p}(H; k)$$ where $H^i(G;k)$ means the $\text{Ext}^i_{kG}(k)$.

Let $P_*$ be a $kG$ projective resolution of $k$, $Q_*$ be a $kH$ projective resolution of $k$, then $P_*\otimes_kQ_*$ is a $k(G\times H)$ projective resolution of $k$.

Then $$H^n(G\times H;k)=H^n\text{Hom}_{G\times H}(P_*\otimes Q_* , k)$$ and $$\bigoplus H^p(G;k)\otimes H^{n-p}(H;k)=\bigoplus H^p\text{Hom}_G(P_* , k)\otimes H^{n-p}\text{Hom}_H(Q_* , k)\cong H^n\bigg (\text{Hom}_G(P_* , k)\otimes \text{Hom}_H(Q_* , k)\bigg)$$ where the last isomorphism follows from Kunneth formula for complexes.

What remains is to show that $H^n\text{Hom}_{G\times H}(P_*\otimes Q_* , k)\cong H^n\bigg (\text{Hom}_G(P_* , k)\otimes \text{Hom}_H(Q_* , k)\bigg)$.

If $P_*$ consists of finitely generated $kG$ modules, then I can show this isomorphism, but I think we may not be able to choose $P_*$ to be finitely generated, at least when $G$ is not finitely generated.

Any help or hints for this problem would be appreciated, thank you in advance.