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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!

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    $F'/F''$ 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.2012-07-03
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    @Derek: Thanks a lot! But I still need more time to understand what you wrote.2012-07-04
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    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 15.2012-07-04

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