# Nonhyperbolic groups that contain no free abelian groups or Baumslag-Solitar groups

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.

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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