Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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, 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 for a list of torsion-free examples (of course, finite cohomological dimension implies torsion-free.

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

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.

share|improve this answer
    
Thank you, this is a great idea, but I do not see why the group $\mathbb Q^\times$ has finite cohomological dimension (which is surely needed in order for your example to have finite cohomological dimension). The cohomological dimension of $\mathbb Q^\times$ is $>2$. Is it finite? –  Igor Belegradek Mar 8 '12 at 14:03
    
Igor, your first check that ${\mathbb Q}$ has homological dimension $1$ (every locally free group does). Then cohomological dimension of any group is at most homological dimension $+1$. (See Bieri's notes on cohomological dimension.) –  Misha Mar 8 '12 at 14:23
    
Oops sorry, I confused additive and multiplicative rational numbers. Instead of ${\mathbb Q}^\times$, use the multiplicative group of the numbers of the form $2^n$, $n$ is integer. This group is locally free (of rank $1$). –  Misha Mar 8 '12 at 14:28
    
Thanks, either the rationals or dyadic rationals work instead of $\mathbb Q^\times$. The "multiplicative group of the numbers of the form $2^n$, $n$ is integer" is infinite cyclic, so it must be a typo. I am accepting the answer (which I shall edit by adding some details, if you don't mind). I think using the same idea it should be possible to find a finitely generated example. –  Igor Belegradek Mar 8 '12 at 15:51
    
Yes, of course. –  Misha Mar 8 '12 at 16:37
add comment

Your Answer

 
discard

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.