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Let $G=<\sigma>$ be a cyclic group of order $n$. For any $\mathbb{Z}[G]$ module $M$ it is known that the group cohomology

$$ H^i(G, M) = \begin{cases} M^G &\text{ if } i = 0 \\ M^G/NM &\text{ if } i \text{ odd} \\ ker(N)/(\sigma -1)M &\text{ if $i$ even} \end{cases}$$

where $N$ is the norm map appearing in the free resolution of $\mathbb{Z}$. I am wondering if there is a similar description for the group $G \times G$ and any $\mathbb{Z}[G\times G]$ module $M$.

In particular, if $G$ is a cyclic group of order $2$, $H^i(G,M) = M^G/NM$ for all $i>0$. I am wondering if in the case of $H^*(G\times G, M)$ the cohomology is also a quotient of $H^*(G; k) \otimes_k M^{G\times G}$ where $M$ is now a $k[G\times G]$ module and $k$ is a field with two elements.

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    $\begingroup$ Is $M$ a $G$-module or a $G\times G$-module ? In the first case what does $H^i(G\times G, M)$ mean; and in the second one, what does $M^G$ mean ? $\endgroup$ – Max Feb 18 at 9:09
  • $\begingroup$ I will clarify it. Thanks. $\endgroup$ – C. Zhihao Feb 18 at 14:59
  • $\begingroup$ This is not true. Try the trivial module and apply the Kunneth theorem. $\endgroup$ – user98602 Feb 18 at 15:26
  • $\begingroup$ does not the Kunneth theorem apply for cohomologies of the form $H^*(G \times H; M \otimes N)$? $\endgroup$ – C. Zhihao Feb 18 at 16:07
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    $\begingroup$ @MikeMiller I now get what you meant, Thanks for pointing that out. I will formulate an output that will make more sense. $\endgroup$ – C. Zhihao Feb 19 at 20:14

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