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The group of cohomology $H^3(G,\mathbb{Z})$ is finite when G is finite.

I am not sure how this is finite. We use the definite as follows:

$H^n(G,K) = Ext_\mathbb{Z}^n$$_G (\mathbb{Z}, K)$ and we use the $G$-free resolution of $\mathbb{Z}$.

Any help would be appreciated!

Kenta S
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scsnm
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    With other definitions of group cohomology you get that $H^3$ is a sub-quotient of $\mathbb{Z}[G^3]$ (I think, perhaps it’s $G^4$) which is a finitely generated abelian group; moreover, it’s known that multiplication by $|G|$ vanishes $H^*(G,—)$. So your $H^3$ is torsion and a finitely generated abelian group, thus finite. – Aphelli Aug 07 '20 at 11:27

1 Answers1

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When $G$ is a finite group of order $n$, then $H^k(G,A)$ is $n$-torsion for all $k\ge1$.

From the standard resolution, $H^k(G,A)$ is finitely generated whenever $A$ is finitely generated as an Abelian group. As $H^k(G,A)$ is both finitely generated and torsion, it is a finite Abelian group.

Angina Seng
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  • Hello Angina! thank you very much! Would you please expand a bit on the statement "From the standard resolution, ...Abelian group." I'd really appreciate it! – scsnm Aug 07 '20 at 11:32
  • I seem to have it. not sure though. So the homomorphisms from ZG to A are determined by how the elements of ZG are mapped to the generators of A. Is it correct? Thank you – scsnm Aug 07 '20 at 11:40