I seek an example of an amenable group of finite cohomological dimension that is not virtually polycyclic and has finite abelianization.


**Remarks.**


1. Elementary amenable groups of finite cohomological dimension are virtually solvable.
2. There are lots of virtually abelian groups of finite cohomological dimension that have finite abelianization. The simplest one is the fundamental group of a certain flat 3-manifold, the Hantsche-Wendt manifold.
3. Virtually polycyclic groups are finitely presented, so perhaps an example can be found among virtually solvable groups that are not finitely presentable, which do exist.
4. Elementary amenable groups of cohomological dimension two are classified [here](http://arxiv.org/abs/math/0401312), see Theorem 3, and they have infinite abelianization (except for the trivial group).
5. Another idea would be to search for an amenable group that is not elementary amenable.
There are very few known examples, and I do not know if any of the examples have finite cohomological dimension and finite abelianization. See [here](https://mathoverflow.net/questions/53803/amenable-groups-with-finite-classifying-space/53812#53812) for a list of torsion-free examples (of course, finite cohomological dimension implies torsion-free.



Amenable groups of finite cohomological dimension
Igor, what about doing the following. 


Let $A$ be a virtually polycyclic group of finite cohomological dimension whose abelianization is finite, but has has ${\mathbb Z}\_n$, $n$ even, among its cyclic factors. Say, the fundamental group of a Hantsche-Wendt manifold would do the job. Next, take $B$, the semidirect product of $\mathbb Q$ and of ${\mathbb Z}$, where ${\mathbb Z}$ acts (through ${\mathbb Z}\_2$) by $q\to -q$. Now, make $A$ act on $B$ where the action on $\mathbb Q$ is trivial and the action on ${\mathbb Z}$ is by $n\to -n$ (this is indeed an action as $\mathbb Z$ is abelian). Lastly, let $G$ the semidirect product of $B$ and $A$ using the above action. By construction $G$ has finite abelianization. Also $G$ has finite cohomological dimension because its semidirect factors have this property ($\mathbb Q$ is locally cyclic). Finally, $G$ is not virtually polycyclic as it contains $\mathbb Q$, and subgroups of virtually polycyclic groups are finitely generated. (This paragraph is edited by Igor Belegradek).


You can also imitate the construction of H-W groups, where ${\mathbb Z}\_2^n$ would act 
on ${\mathbb Q}^n$ rather than ${\mathbb Z}^n$, but I did not think about the details. 


