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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 $\ell \mathbb{Z}^n \hookrightarrow \mathbb{Z}^n$.

flag	asked Feb 29 2012 at 6:04 Ron 13●2

	In your target homology group, I think $\mathbb{Z}^k$ should be $\mathbb{Z}^n$. – Mark Grant Feb 29 2012 at 13:50
	Thanks Mark!!!! – Ron Mar 2 2012 at 6:21

Johannes Ebert · Accepted Answer · 2012-02-29 09:01:54Z UTC

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$k=n=1$, $M_{n,k}=M=\mathbb{Q}$. Let $\mathbb{Z}$ act on $\mathbb{Q}$ by $n \cdot q = (-1)^n q$.

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.

answered Feb 29 2012 at 9:01

Johannes Ebert
11.5k●33●55

I think you can generalize your example to all $n,l$ by letting $\mathbb{Z}$ act on $\mathbb{C}$ by $n \cdot z = \zeta_l^nz$. Then extend the action to an action of $\mathbb{Z}^n$ on $\mathbb{C} \otimes \cdots \otimes \mathbb{C}$. The case for a fixed $k$ should then be manageable by dimension-shifting. – Ralph Feb 29 2012 at 10:29

Homology of abelian groups and their finite-index subgroups

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