Let $G$ be a graph group and let $S$ be a finitely generated subgroup of $S$. What torsion can $H_1(S)$ have?

Let me put this in context: Let $Y$ be a graph, then the corresponding graph group has a generator for each vertex and for each edge we add the relation that the corresponding generators commute. Such groups are also known as right angled Artin groups. If $Y$ has no edges the graph group is the free group, if $Y$ is a complete graph, then the graph group is a free abelian group.

It is known by work of Crisp and Wiest that non-orientable surface groups embed in graph groups. Such surface groups can have 2-torsion in their abelianization. Is that the only type of torsion which can appear in the abelianization of the subgroup of a graph group?

My interest in this question comes from the recent announcement by Dani Wise that `most' hyperbolic 3-manifold groups embed in a graph group.


1 Answer 1


This isn't a proper answer, but here are some remarks.

  • Wise's work actually purports to tell you that most (all?) hyperbolic 3-manifold groups virtually embed in graph groups - ie, a finite-index subgroup embeds. So you might lose a lot of torsion when you pass to this finite-index subgroup.
  • Haglund and Wise have a very nice criterion for injecting fundamental groups of cube complexes into graph groups. It would be completely reasonable to try to use this to construct interesting examples of torsion in the homology of subgroups of graph groups.
  • Because the embeddings in Haglund--Wise are quasiconvex, it follows from work of Haglund that every such subgroup H is a virtual retract - that is, there is a finite-index subgroup K that contains H and the inclusion map has a left inverse K->H. In particular, any torsion you see in the homology of H will also show up in the homology of K. It would be easy to use a computer algebra package like GAP to look for torsion in finite-index subgroups of graph groups.


In a comment below, Stefan points out a very nice way of constructing torsion in finite-index subgroups of graph groups, due to Bridson and Miller. As I'm familiar with the argument, I'll explain it. The graph group in question will always be a direct product of two free groups.

Let Q be any group you like and let q:F->Q be a surjection from a free group. Now take the fibre product K of q with itself; in other words, consider the subgroup

$K= \{(g,h)\in F\times F\mid q(g)=q(h)\}~.$

There is a standard five-term exact sequence in homology that derives from the map $q:F\to Q$, which reduces to

$0\to H_2(Q)\to H_0(Q,H_1(\ker f))\to H_1(F)\to H_1(Q)$

where, by definition, $H_0(Q,H_1(\ker f))$ is the quotient of the abelianisation of $\ker f$ by the natural action of $Q$ by conjugation. On the other hand, projecting onto a factor decomposes $K$ as $\ker f\rtimes F$ (where the action is, again, by conjugation), from which it follows that

$H_1(K) = H_0(Q,H_1(\ker f))\oplus H_1(F)~.$

Putting these together, we see that $H_2(Q)$ embeds into $H_1(K)$.

The fibre product $K$ can also be characterised as the preimage of the diagonal subgroup of $Q\times Q$ in $F\times F$, so if $Q$ is finite then $K$ is of finite index in $F\times F$. This proves the following.

Proposition. For any finite group $Q$ there is a finite-index subgroup $K$ of $F\times F$ with $H_2(Q)\subseteq H_1(K)$.

This shows that all sorts of torsion can arise in $H_1$ of subgroups of graph groups.

Remark. We took $Q$ to be finite because otherwise $K$ isn't quasiconvex. But the rest of the argument goes through.

  • $\begingroup$ you are right of course, I was sloppy in my question, "most" hyperbolic embed virtually into graph groups. Otherwise the question would be stupid. Thanks for the comments. Since the only graph groups are understand are free groups and free abelian groups my intuition for graph groups is not well developed. $\endgroup$ Feb 12, 2010 at 20:20
  • 1
    $\begingroup$ Minasyan in his paper `Hereditary conjugacy separability of right angled Artin groups and its applications' in Example 1.2 gives another construction which might well be generalized to creating all kinds of torsion. $\endgroup$ Feb 13, 2010 at 16:11

Your Answer

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

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