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I would like to ask the following: if for a group $G$ the homology $H_n(G,\mathbb{Z})$ is $\mathbb{Z}$-torsion for every $n\geq n_0$, then what can be said concerning $\mathbb{Z}$-torsion for $H_k(G,M)$ where M is a $\mathbb{Z}G$-module? For example I know that if M is a trivial $G$-module, then $H_n(G,M)\simeq H_n(G,\mathbb{Z})\otimes_{\mathbb{Z}}M\oplus Tor_1^{\mathbb{Z}}(H_{n-1}(G,\mathbb{Z}),M)$ [Weibel, Th. 6.1.12] and hence $H_n(G,M)$ is $\mathbb{Z}$-torsion if $H_n(G,\mathbb{Z})$ is. What happens if M is not a trivial $G$-module?

Weibel, "An introduction to homological Algebra"

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There exists an acyclic group $G$ which has the property that there exists a finite-index normal subgroup $H\lhd G$ such that $\mathbb{Z} \leq H_1(H;\mathbb{Z})$. In particular, then $H_1(G;\mathbb{Z}[G/H]) \cong H_1(H;\mathbb{Z})$ is not torsion (by Shapiro's lemma).

The example is the fundamental group of the complement of a wild arc. Examples in the linked paper, such as "Fox's stitch", are infinite cyclic covers of a hyperbolic 2-component link complement. There are such examples which are arithmetic, and therefore have a finite-index subgroup which is "RFRS", by a theorem of mine. Thus, for any element in the group, there is a finite-index subgroup for which it is homologically non-trivial. This property passes to subgroups, in which case the fundamental group of the complement of Fox's stitch has the property that there is a finite-index subgroup with infinite abelianization (in fact, a non-trivial homomorphism to $\mathbb{Z}$).

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  • $\begingroup$ I guess the OP was primarily interested in $H_k(G;M)$ for $k >> n_0$ (as his universal coefficient example suggests). $\endgroup$
    – Ralph
    Oct 13 '11 at 8:22
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This Kunneth formula still holds, I proved it here:

Kuenneth-formula for group cohomology with nontrivial action on the coefficient

which holds in our situation for all nontrivial $G$-modules.

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  • $\begingroup$ Actually, you cannot use this to get your Universal Coefficients Formula unless the actions are trivial! So nevermind. $\endgroup$ Nov 14 '11 at 20:03

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