12
$\begingroup$

Is there an example of a finite group $G$ and an action on $M=\mathbb{Z}^n$ such that $H^2(G,M)$ has exponent greater than the exponent of $G$?

(Especially, can we have $G=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and some free $\mathbb{Z}$ module $M$ with $G$ action, such that $H^2(G,M)$ has elements of order $4$?)

$\endgroup$

1 Answer 1

21
$\begingroup$

For each finite group $G$ there is a $G$-module $M$ that is a free abelian group of finite rank such that $H^2(G,M)=\mathbb{Z}/|G|$.

Proof: Let $I$ be the augmentation ideal of $\mathbb{Z}G$. Then $H^1(G,I)=\mathbb{Z}/|G|$. $\,\,I$ is a finitely generated $\mathbb{Z}G$-module (it's f.g. even as free abelian group). Hence there is a short exact sequence of $\mathbb{Z}G$-modules $$0 \to M \to F \to I \to 0$$ where $F$ is a free $\mathbb{Z}G$-module of finite rank. In particular, $F$ is a free abelian group of finite rank and so is the subgroup $M$. The long exact cohomology sequence now yields $H^2(G,M) \cong H^1(G,I)$. q.e.d.


Added: If $\Omega^n(\mathbb{Z})$ denotes the kernel of $P_{n-1} \to P_{n-2}$ of a projective resolution $P \to \mathbb{Z}$ then the same argument shows $H^n(G,\Omega^n(\mathbb{Z}))=\mathbb{Z}/|G|$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy