Homology of abelian groups and their finite-index subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:00:54Z http://mathoverflow.net/feeds/question/89834 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89834/homology-of-abelian-groups-and-their-finite-index-subgroups Homology of abelian groups and their finite-index subgroups Ron 2012-02-29T06:04:22Z 2012-02-29T09:01:54Z <p>Fix some $1 \leq k \leq n$. I'm looking for finite-dimensional vector spaces $M_{n,k}$ over $\mathbb{Q}$ on which $\mathbb{Z}^n$ acts such that the natural map $H_k(\ell \mathbb{Z}^n,M_{n,k}) \rightarrow H_k(\mathbb{Z}^k,M_{n,k})$ is not an isomorphism for some $\ell \geq 2$. Here $\ell \mathbb{Z}^n$ is the subgroup of $\mathbb{Z}^n$ consisting of vectors each of whose entries is divisible by $\ell$ and the map on group homology is induced by the inclusion $\ell \mathbb{Z}^n \hookrightarrow \mathbb{Z}^n$.</p> http://mathoverflow.net/questions/89834/homology-of-abelian-groups-and-their-finite-index-subgroups/89847#89847 Answer by Johannes Ebert for Homology of abelian groups and their finite-index subgroups Johannes Ebert 2012-02-29T09:01:54Z 2012-02-29T09:01:54Z <p>$k=n=1$, $M_{n,k}=M=\mathbb{Q}$. Let $\mathbb{Z}$ act on $\mathbb{Q}$ by $n \cdot q = (-1)^n q$. </p> <p>The homology is $H_i(\mathbb{Z};M)=\mathbb{Q}$ for $i=1$ and $0$ otherwise. The subgroup $2 \mathbb{Z}$ acts trivially on $M$ and so $H_i (2 \mathbb{Z};M)=\mathbb{Q}$ for $i=0$ and $0$ otherwise.</p>