Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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"

share|improve this question
add comment

2 Answers

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 $\mathbb{Z}$).

share|improve this answer
I guess the OP was primarily interested in $H_k(G;M)$ for $k >> n_0$ (as his universal coefficient example suggests). –  Ralph Oct 13 '11 at 8:22
add comment

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.

share|improve this answer
Actually, you cannot use this to get your Universal Coefficients Formula unless the actions are trivial! So nevermind. –  Chris Gerig Nov 14 '11 at 20:03
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.