Group cohomology of finite groups

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.

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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. – mland Aug 10 '12 at 14:32
Sorry for the confusion. I menat that each $H^{p}(G,M)$ is finite. – Michel Aug 10 '12 at 17:20
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 $H^2(D_{2n};-)$. – Ralph Aug 10 '12 at 23:05
I will check it up. Thanks, Ralph. – Michel Aug 11 '12 at 0:30

1 Answer

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.) – countinghaus Aug 10 '12 at 13:16
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. – Michel Aug 10 '12 at 17:34
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. – countinghaus Aug 10 '12 at 18:49
I didn't know the sequence. Thanks ^_^ – Michel Aug 10 '12 at 19:13

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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 $1 \in \hat{H}^0(G;\mathbb{Z})=\mathbb{Z}/|G|$. – Ralph Aug 10 '12 at 23:02
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