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The group cohomology of cyclic groups can be computed easily due to the periodity. Now how can one compute the group cohomology $H^r(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},M)$? As least in some special case, e.g. $M$ has trivial action, or even $M=\mathbb{Z}$? I would like to know the case that $M$ is not $\mathbb{Z}$.

Furthermore, can one compute $H^r(\mathbb{Z}/n\mathbb{Z}\times\cdots\times \mathbb{Z}/n\mathbb{Z},M)$?

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  • $\begingroup$ @PaulBroussous Thanks for the ideal! I can try this. $\endgroup$
    – Heer
    Commented Apr 5, 2018 at 14:20
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    $\begingroup$ I am surprised that this question was closed. $\endgroup$
    – Derek Holt
    Commented Apr 5, 2018 at 15:09
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    $\begingroup$ @DerekHolt: I voted to close because the question as asked is fully answered by the Kunneth formula. If the asker is more interested in the case when $M$ has torsion, or a non-trivial action, then they should say so in the question. $\endgroup$
    – Mark Grant
    Commented Apr 6, 2018 at 6:20
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    $\begingroup$ @MarkGrant But the question was phrased more generally and mentioned the case of trivial action as a special case. I would have been interested myself to answers to more general cases. It seems to me that questions that are more elementary than this one (such as one that I answered myself recently) do not get closed, but I guess that is a result of everyone being familiar with different topics. $\endgroup$
    – Derek Holt
    Commented Apr 6, 2018 at 7:36

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If $M$ is torsion-free (as a $\mathbb{Z}-$module), then the Kunneth formula still holds. If $M = \mathbb{Z}$ with trivial group action, then it holds by the corresponding result for the CW-complex category and abstract nonsense; more generally, it holds for chain complexes under some fairly mild restrictions (including having finite homology in each dimension, but that's clear here.)

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  • $\begingroup$ @YCor: $M = \mathbb{Z}$ with trivial $G$-action. Edited. $\endgroup$
    – anomaly
    Commented Apr 4, 2018 at 21:41
  • $\begingroup$ Thanks for you reply. What if $M$ is not $\mathbb{Z}$? $M=\mathbb{Z}$ is not enough for my case! $\endgroup$
    – Heer
    Commented Apr 5, 2018 at 10:20
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    $\begingroup$ It's fine as long as $M$ is torsion-free and the homology is finite in each dimension. Brown's book on group cohomology has a proof of the Kunneth formula in the category of general chain complexes. $\endgroup$
    – anomaly
    Commented Apr 5, 2018 at 21:42

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