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!

  • 2
    $\begingroup$ $F'/F''$ 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. $\endgroup$ – Derek Holt Jul 3 '12 at 8:24
  • $\begingroup$ @Derek: Thanks a lot! But I still need more time to understand what you wrote. $\endgroup$ – Fred Jul 4 '12 at 2:36
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Jul 4 '12 at 13:42

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.


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