I've been looking for any kind of universal coefficient theorem for group homology and cohomology, including dual universal coefficient theorems. However, the only things I can find are ones where the group action on the coefficients is trivial. As such, my question is:


Let $G$ be a group, $M$ be a $G$-module. Is there any universal coefficient theorem for $H^\*(G, M)$ or $H\_\*(G, M)$?


My specific interest, as in [Cohomology of lattice with coefficients in field of rational functions](https://mathoverflow.net/q/199120/44191), is where $G = \mathbb{Z}^n$, but I'd love if there were a more general answer.



Universal coefficient theorem for group homology and cohomology
You're only finding UCT in the literature for trivial group actions, *because there is no general UCT for nontrivial group actions*:


The general Kunneth formula does not hold for arbitrary groups and actions. But it does hold a good amount of times, and I elaborated on this here: [Kuenneth-formula for group cohomology with nontrivial action on the coefficient](https://mathoverflow.net/questions/75472/kuenneth-formula-for-group-cohomology-with-nontrivial-action-on-the-coefficient/75485#75485).


Now the UCT, which relates $H\_\*(G,M)$ to $H\_\*(G,\mathbb{Z})$, only follows from the Kunneth formula *for trivial group actions*. The Kunneth theorem considers the tensor product $C\_\*\otimes D\_\*$ of two chain complexes, and the special case for UCT is $D\_\*=M$ for some $R$-module. To guarantee that the images of the boundary maps are $R$-projective, some extra assumptions are needed (like $R$ is a PID). For group homology, we work with $F\_\*\otimes\_{\mathbb{Z}G}M$ where $F\_\*$ is a free resolution of $\mathbb{Z}$ as a $\mathbb{Z}G$-module. We cannot take $R=\mathbb{Z}G$ (otherwise the assumption about the boundaries would imply that all homology groups are trivial), so we take $R=\mathbb{Z}$. But then $M$ must be trivial as a $\mathbb{Z}G$-module in order to express $F\_\*\otimes\_{\mathbb{Z}G}M$ as $C\_\*\otimes\_\mathbb{Z}M$. In this case, $F\_\*\otimes\_{\mathbb{Z}G}M=(F\_\*\otimes\_{\mathbb{Z}G}\mathbb{Z})\otimes\_\mathbb{Z}M$ and we can apply the UCT.


Morally, in lieu of Qiaochu's remark you must ask: Given any data involving the $G$-action, how would the operators $\oplus,\otimes,\text{Tor},\text{Ext}$ encode such information? And as shown above, you can't use that information to pass from $M$ to $\mathbb{Z}$. For example, let $\mathbb{Z}\_2$ act on $M=\mathbb{Z}\_2\oplus\mathbb{Z}\_2$ by swapping the generators of the summands. Where would this maneuver exist on the coefficient $\mathbb{Z}$ or on any homological object? We can't simply "forget" the action, because $H^1(\mathbb{Z}\_2,M\_\text{nontriv})=0$ while $H^1(\mathbb{Z}\_2,M\_\text{triv})=M$.


