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I wonder if the group cohomology of a finite group $G$ with coefficients in $\mathbb{Z}$ is finite. This statement may be too strong. I am interested in, for instance, dihedral group. $$ G=D_{2n}=\langle a,b | \ a^n=b^2=abab=e \rangle $$ Assume that $a$ acts trivially and $b$ acts as $-id$ on $\mathbb{Z}$.

First cohomology is $\mathbb{Z}^G=0$. The second cohomology already seems quite involved to me.

I read several post about group cohomology on StackExchange and MathOverflow, but I still have trouble computing explicit example and getting intuition behind the concept.

Thank you for your assistance.

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    do you mean each cohomology group $H^p(G;\mathbb{Z})$ is a finite group or do you mean the whole cohomology ring $H^*(G;\mathbb{Z})$? The latter statement is of course false, seen e.g. in the finite cyclic groups.2012-08-10
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    Sorry for the confusion. I menat that each $H^{p}(G,M)$ is finite.2012-08-10
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    There are projective resolutions for the dihedral group (due to Wall) that can be used to compute the cohomology for every coefficient module. In particular it shouldn't be to hard to figure out $H^2(D_{2n};-)$.2012-08-10
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    I will check it up. Thanks, Ralph.2012-08-11

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The answer is yes, since it's torsion (killed by the order of $G$) and finitely generated (since you can pick a resolution by finitely generated abelian groups).

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    (I mean, more precisely, that the underlying abelian groups of the free $ZG$-modules appearing in the standard resolution of $Z$ are finitely generated.)2012-08-10
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    Each term of the standard resolution is a direct sum of finite $\mathbb{Z}[G]$s but I don't quite see why it's torsion.2012-08-10
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    in general if $G$ is a finite group, the order of $G$ kills any cohomology group $H^i(G, M)$. To see this, use the fact that there are corestriction and restriction maps for any normal subgroup $H$ of $G$ such that $H^i(G, M) \to H^i(H, M) \to H^i (G, M)$ is multiplication by $[G:H]$. Now apply this in the special case where $H$ is trivial to get that multiplication by the order of $G$ is the zero map.2012-08-10
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    I didn't know the sequence. Thanks ^_^2012-08-10
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    (i didn't word that well - it's not an exact sequence, just "when you restrict, then you corestrict, you multiply by the index" combined with $H^i({1}, M) = 0$.)2012-08-10
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    An alternative argument to see that $H^i(G;M)$ is annulated by $|G|$ for $i > 0$ is to consider $\hat{H}^\ast(G;M)$ as (unitary) module over $\hat{H}^\ast(G;\mathbb{Z})$. Then it's clear because $1 \in \hat{H}^0(G;\mathbb{Z})=\mathbb{Z}/|G|$.2012-08-10
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    What is the "hat" $\hat{H}^{*}(G;M)$?2012-08-11
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    It's Tate cohomology (see the book Brown: Cohomology of Groups, VI.4). It can be seen as a collection of homology and cohomology as $\hat{H}^i(G;M) = H^i(G;M)$ if $i > 0$ and $\hat{H}^{-i}(G;M) = H_{i-1}(G;M)$ if $i > 1$.2012-08-11
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    OK. I will take a look at the reference. Thanks, Ralph.2012-08-12