MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've heard it conjectured that a finitely presentable group $G$ is hyperbolic if it satisfies the following two conditions.

  1. $G$ contains no subgroup isomorphic to a Baumslag-Solitar group $BS(n,m)$ (including $BS(1,1) \cong \mathbb{Z}^2$).
  2. $G$ is rationally of finite type in the sense that all the groups $H_k(G;\mathbb{Q})$ are finite-dimensional and $H_k(G;\mathbb{Q})=0$ for $k \gg 0$.

Question : Can someone tell me an example of a finitely presentable group that satisfies $1$ but but not $2$? All the examples of finitely presentable groups I know of that don't satisfy $2$ actually have plenty of copies of $\mathbb{Z}^2$ in them.

share|cite|improve this question
You probably know this, but if you relax f.p. to f.g. then I believe the first Grigorchuk group is an example (right?). – Khalid Bou-Rabee Feb 8 '13 at 2:58
@Khalid Bou-Rabee : That's right; it has infinite $\mathbb{Q}$-cohomological dimension (and it might have infinite rank $H_2$, though I don't know off the top of my head). – Steven Feb 8 '13 at 3:05
Steven: Your finiteness conditions are not quite right, in the conjecture you should assume instead that your finitely-presentable group $G$ has type $FP$ over ${\mathbb Q}$, i.e., ${\mathbb Q}$ admits a finite resolution by finitely-generated projective $G[{\mathbb Q}]$-modules, see Brown's book "Cohomology of groups". This is a much stronger assumption than your assumption on homology groups with trivial coefficients. If $G$ is torsion-free, you can simply say that $G$ admits a finite $K(G,1)$. Even then, this conjecture is widely expected to fail. – Misha Feb 8 '13 at 7:21
If you're interested in this problem, you may also want to take a look at the idea to find a counterexample in this paper: Benson Farb and Lee Mosher, Convex cocompact subgroups of mapping class groups, Geom. Topol. 6 (2002), 91–152 (electronic). MR MR1914566. – Richard Kent Feb 8 '13 at 14:40
up vote 7 down vote accepted

Noel Brady's finitely presented non-hyperbolic group embeds in a hyperbolic group (and hence satisfies (1)), but has infinitely generated third integral homology.

I'm guessing, but don't remember with certainty, that the rational third homology is infinitely generated.

Noel Brady, Branched coverings of cubical complexes and subgroups of hyperbolic groups, J. London Math. Soc. (2) 60 (1999), no. 2, 461–480. MR MR1724853

share|cite|improve this answer
I'm fairly certain that this is the only known example of a group with this property. I don't know a reference for this fact (though asking Noel would be a good start) - I thought this was stated explicitly in Bridson's problem list on the AIM 'Open Problems in Geometric Group Theory' wiki, but the wiki seems to have vanished! – HJRW Feb 8 '13 at 10:24
Great, thanks!! – Steven Feb 8 '13 at 15:24

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.