Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question was inspired by the recent post: The centralizer of an element x in free group is cyclic.

Is it true that non-identity elements have abelian centralizer in free metabelian groups?

This is probably very simple, but I couldn't find myself a proof.

I don't know if it helps but I was able to reduce the problem to the following assertion: $$\forall a\in F, \forall b\in F, [a\in (F-F'), b\in (F'-F'') \Longrightarrow [a,b]\notin F'']$$ (here $F$ is a free group (not metabelian)).

$\hskip250pt$ Thanks for any ideas!

share|improve this question
2  
$F'/F''$ can be regarded as a ${\mathbb Z}Q$-module, where $Q=F/F'$ is free abelian. It is the so-called relation module of $Q$. In general the relation module is isomorphic to a submodule of a free module. So $F'/F''$ is a submodule of a free ${\mathbb Z}Q$- module. I think your assertion above follows from that. –  Derek Holt Jul 3 '12 at 8:24
    
@Derek: Thanks a lot! But I still need more time to understand what you wrote. –  Fred Jul 4 '12 at 2:36
    
The result that $F'/F''$ is a submodule of a free ${\mathbb Z}Q$-module in proved in Chapter 11 of "Presentations of Groups", D.L. Johnson, London Math Soc Student Texts 15. –  Derek Holt Jul 4 '12 at 13:42

1 Answer 1

Free metabelian groups have faithful embeddings into the affine group $K\rtimes K^*$ of some field $K$ (Magnus embedding), or equivalently matrices $\begin{pmatrix}a & b\\ 0 & 1\end{pmatrix}$ with $a\in K^*$ and $b\in K$, and the latter group has the property (straightforward exercise) that the centralizer of any nontrivial element is abelian. This passes to subgroups and answers your question.

share|improve this answer

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.