Recently I've been reading about cohomological finiteness conditions for groups, my main source being Brown's book "Cohomology of Groups".

One of the first things one learns is that a group with finite cohomological dimension necessarily is torsionfree. It is easy to come up with an example of a torsionfree group with infinite cohomological dimension: non-finitely generated free abelian group immediately springs to mind. A little bit of googling revealed an example of an infinite-dimensional torsionfree $FP_{\infty}$ group.$^1$ (A group is said to be of type $FP_{\infty}$ if there exists a projective $ZG$-resolution {$P_i$} of $Z$ such that each $P_i$ is finitely generated.) This example is Thompson's group $F$. However, it is well-known that $F$ contains a non-finitely generated free abelian group.

So, the question is: what is an example of a torsionfree group with infinite cohomological dimension and no inifinitely generated free abelian subgroup (that is, if there exists one)?

$^1$ In case anyone is interested: K. S. Brown, R. Geoghegan, Invent. Math. 77, 367--381.


2 Answers 2


You can take a union of torsion-free word-hyperbolic groups with cohomological dimension approaching infinity to get such an example. Word-hyperbolic groups are rank one, so one can show that any abelian subgroup must intersect each subgroup in a cyclic group, and therefore the subgroup cannot be an infinitely generated free abelian subgroup.

Here's an explicit sequence. Take a sequence of cocompact torsion-free lattices in $\Gamma_n< Isom(\mathbb{H}^n)$, such that $\Gamma_n < \Gamma_{n+1}$ from the natural embedding $Isom(\mathbb{H}^n) < Isom(\mathbb{H}^{n+1})$. The limit $\Gamma_{\infty}=\underset{n\to\infty}{\lim} \Gamma_n$ is a group of cohomological dimension $\infty$ since $\Gamma_n$ is of cohomological dimension $n$.

For example, consider the quadratic form over $\mathbb{Q}(\sqrt{5})$ which is $q_n(x_0,x_1,\ldots,x_n)=((1-\sqrt{5})/2 )x_0^2+x_1^2+\cdots+x_n^2$. This gives a quadratic form over $\mathbb{R}$ of signature $(n,1)$. Consider the group $O(q_n,\mathbb{Z}[\phi])< GL(n+1,\mathbb{Q}(\sqrt{5}))$, where $\phi=(1+\sqrt{5})/2$, the group of integral matrices preserving this quadratic form. Then there exists a prime ideal $\mathcal{P}<\mathbb{Z}[\phi]$ such that $\Lambda_n=Ker\{ GL(n+1,\mathbb{Z}[\phi])\to GL(n+1,\mathbb{Z}[\phi]/\mathcal{P})\}$ is torsion-free for all $n$. Let $\Gamma_n=O(q_n,\mathbb{Z}[\phi])\cap \Lambda_n$. Then $\Gamma_n$ is a torsion-free hyperbolic group acting cocompactly on $\mathbb{H}^n$ (by considering the Lorentzian model of hyperbolic space), and clearly $\Gamma_n<\Gamma_{n+1}$.

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    $\begingroup$ Here is another way to get an increasing sequence of hyperbolic groups $G_n$ of cohomological dimension $n$: using strict relative hyperbolization one can show that any closed aspherical $n$-manifold with hyperbolic fundamental group is a retract of a closed aspherical $(n+1)$-manifold with hyperbolic fundamental group, see arxiv.org/abs/math/0509504 Now as in your argument the union of $G_n$'s is the desired example. I wonder if there is an $FP_\infty$ example. $\endgroup$ Apr 4, 2011 at 17:47
  • $\begingroup$ @ Igor: I imagine the Higman embedding theorem should allow one to embed these examples into a finitely presented group that has the same property (no infinite rank abelian subgroup). However, I'm not sure this will be $FP_{\infty}$. $\endgroup$
    – Ian Agol
    Apr 4, 2011 at 18:47

Ian Agol's answer is complete and achieves more than is required. It is perhaps worth pointing out that if you just want an example that meets the Title Criteria you can take $G$ to be the free product of the groups in the sequence $(A_n)_{n\in \mathbb N}$ where the $n$th group $A_n$ is free abelian of rank $n$. This might be useful if you wanted a quick example for a lecture course or some related purpose.


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