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Assume that $G$ is a finite group and that $A$ is an arbitrary $G$-module. Then we know that can define the cohomology groups of $G$ with coefficients in $A$ in the usual way and we denote the latter by $H^{∗}(G;A)$. However, I never read that for an arbitrary $G$-module we can define a "product" on cohomology groups that gives the structure of a graded ring, whilst we are talking about the associative graded-commutative ring (given by the so-called cup product) whether $A=Z,F$, with the latter being an arbitrary field and both they've been treated in that case as trivial GG-modules. Why we don't have ring structure in the case where the coefficients are given by non-trivial $G$-modules?

P.S. Excuse me if you find the question simple, but I am looking desperately for an answer. Thank you!

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    $\begingroup$ You always get an exterior product $H^*(G;A)\otimes H^*(G;B)\to H^*(G;A\otimes B)$. When $A$ is a ring in $G$-modules (that is a ring which is a G-module such that $g(x)g(y)=g(xy)$) you can postcompose that to get a ring map $H^*(G;A)\otimes H^*(G;A)\to H^*(G;A\otimes A)\to H^*(G;A)$. $\endgroup$
    – Denis Nardin
    Jun 25, 2016 at 20:31
  • $\begingroup$ @DenisNardin thank you for your response! whilst, when you don't have a ring the above is not possible? what if $A$ is a ring and $G$ isn't acting trivial? $\endgroup$
    – mayer_vietoris
    Jun 25, 2016 at 20:33
  • $\begingroup$ All that you need is that $G$ acts with ring maps (so the action does not need to be trivial). If $A$ is not a ring you always have the exterior product $H^*(G;A)\otimes H^*(G;A)\to H^*(G;A\otimes A)$. Just think about $H^0$, you are asking something that generalizes a ring structure on $A^G$, in general that has no right to exist. $\endgroup$
    – Denis Nardin
    Jun 25, 2016 at 20:39
  • $\begingroup$ @DenisNardin I think that I get this.. however, the problem is that I read somewhere that cohomology of a group has a ring structure whether $G$ acts trivially i.e. the $\mathbf{untwisted}$ $\mathbf{case}$, and only in that case. It was wrong this? or I am missing something subtle here? $\endgroup$
    – mayer_vietoris
    Jun 25, 2016 at 20:44
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    $\begingroup$ That's doubly wrong, @mayer_vietoris! If $G$ acts trivially on some abelian group $A$, $H^*(G;A)$ does not automatically get a ring structure: you still need $A$ to be a ring! And if $G$ acts on a ring $A$ via ring homomorphisms, $H^*(G;A)$ does get a ring structure, as Denis explained, even if that action is not trivial. In short: ring structure in cohomology comes, unsurprisingly, from rings not from triviality of actions. $\endgroup$
    – Omar Antolín-Camarena
    Jun 25, 2016 at 20:49

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