For a trivial action on the coefficient, we have the following Kuenneth formula
for group cohomology:


$$
H^n(G\_1 \times G\_2; M) \cong 
[\oplus\_{i= 0}^n H^i(G\_1;M) \otimes\_M H^{n-i}(G\_2;M)] 
\oplus [\oplus\_{p =0}^{n+1} \text{Tor}^M(H^p(G\_1;M),H^{n+1-p}(G\_2;M))]
$$
where $G\_1$ and $G\_2$ are finite groups and/or compact Lie groups --Edit-- and $M$ is a PID such as $Z$?


If the group action on the coefficient $M$ is non-trivial, does the above Kuenneth formula remain valid?



Kuenneth-formula for group cohomology with nontrivial action on the coefficient
I learned this through Ken Brown's textbook "Cohomology of Groups" (and studying under him): I basically use the beginning of his Chapter 5 and solve the two exercises in that section.


Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the map $(F\otimes\_GM)\otimes(F'\otimes\_{G'}M')\rightarrow (F\otimes F')\otimes\_{G\times G'}(M\otimes M')$ given by $(x\otimes m)\otimes(x'\otimes m')\mapsto(x\otimes x')\otimes(m\otimes m')$. 


Note that $(F\otimes\_GM)$ $\otimes(F'\otimes\_{G'}M')$ = $(F\otimes M)\_G \otimes(F'\otimes M')\_G$ ,


which is the quotient of $F\otimes M\otimes F'\otimes M'$ by the subgroup generated by elements of the form $gx\otimes gm\otimes g'x'\otimes m'$. The isomorphism $F\otimes M\stackrel{\cong}{\rightarrow}M\otimes F$ of chain complexes ($M$ in dimension $0$) is given by $x\otimes m\mapsto (-1)^{deg(m)\cdot deg(x)}m\otimes x=m\otimes x$, and so the aforementioned quotient is isomorphic to $F\otimes F'\otimes M\otimes M'$ modulo the subgroup generated by elements of the form $(gx\otimes g'x'\otimes gm\otimes g'm')=(g,g')\cdot(x\otimes x'\otimes m\otimes m')$ where this latter action is the diagonal $(G\times G')$-action. Now this is precisely


$$(F\otimes F'\otimes M\otimes M')\_{G\times G'} = (F\otimes F')\otimes\_{G\times G'}(M\otimes M')$$ and hence the considered map is an isomorphism.


Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus\_{p=0}^nH\_p(G,M)\otimes H\_{n-p}(G',M')\rightarrow H\_n(G\times G',M\otimes M')$


$\rightarrow\bigoplus\_{p=0}^{n-1}Tor\_1^\mathbb{Z}(H\_p(G,M),H\_{n-p-1}(G',M'))$ by Proposition I.0.8[Brown]. Note that in order to apply the proposition we needed one of the chain complexes (say, $F\otimes\_G M$) to be dimension-wise $\mathbb{Z}$-free (and so with a free resolution $F$ this means we needed $M$ to be $\mathbb{Z}$-free).


Actually, the general Künneth theorem has a more relaxed condition and it suffices to choose $M$ (or $M'$) as a $\mathbb{Z}$-torsion-free module}.




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Cohomology Künneth Formula (no proofs, just statements/notes)


Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the cochain cross-product $Hom\_G(F,M)\otimes Hom\_{G'}(F',M')\rightarrow Hom\_{G\times G'}(F\otimes F',M\otimes M')$ which maps the cochains $u$ and $u'$ to $u\times u'$ given by $\langle u\times u',x\otimes x'\rangle=(-1)^{deg(u')\cdot deg(x)}\langle u,x\rangle\otimes\langle u',x'\rangle$. This map is an isomorphism under the hypothesis that either $H\_i(G,M)$ or $H\_i(G',M')$ is of finite type, that is, the $i$th-homology group is finitely generated for all $i$ (alternatively, we could simply require the projective resolution $F$ or $F'$ to be finitely generated). 


For example, if $M=\mathbb{Z}$ then $Hom\_G(-,\mathbb{Z})$ commutes with finite direct sums, so we need only consider the case $F=\mathbb{Z}G$. An inverse to the above map is given by $t\mapsto \varepsilon\otimes \phi$, where $\varepsilon$ is the augmentation map and $\phi:F'\rightarrow M'$ is given by $\phi(f')=t(1\otimes f')$. Note that this does not hold for infinitely generated $P=\bigoplus^\infty \mathbb{Z}G$ because $Hom\_G(P,\mathbb{Z})\cong\prod^\infty \mathbb{Z}$ is not $\mathbb{Z}$-projective (i.e. free abelian).


Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus\_{p=0}^nH^p(G,M)\otimes H^{n-p}(G',M')\rightarrow H^n(G\times G',M\otimes M')\rightarrow$


$\bigoplus\_{p=0}^{n+1}Tor\_1^\mathbb{Z}(H^p(G,M),H^{n-p+1}(G',M'))$ by Proposition I.0.8[Brown].


