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I know if $k$ is a field such that $char(k)$ divides $|G|$ (a finite group), then finding the ring structure on $H^\ast(G,k)$ can be very, very hard.

But what about when $k=\mathbb{Z}$? Is the computation easier? I know of several "modern" techniques to compute the cohomology groups, but have never encountered a detailed computation of the ring structure.

Thanks!

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1 Answer 1

up vote 2 down vote accepted

This answers a previous version of the question...

If $G$ is a finite group and $k$ is a field such that $\operatorname{char}k\nmid|G|$, then $H^p(G,k)=0$ for all $p>0$, so that the only non-zero cohomology group us $H^0(G,k)\cong k$, and in fact this isomorphism is an isomorphism of $k$-algebras.

In other words: nothing interesting happens!

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ah, sorry, I messed that up. I meant to be talking about over $\mathbb{Z}$. –  user641 Apr 24 '12 at 5:17
    
There is no need to any «modern techniques» for this: this was known aproximately before group cohomology was defined :D –  Mariano Suárez-Alvarez Apr 24 '12 at 5:18
    
Sorry about that! Perhaps it is too late for me to be posting questions... :) –  user641 Apr 24 '12 at 5:20

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