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We say a group $ G $ is a Wang group if it fits into a SES $$ 0 \to N \to G \to \mathbb{Z}^k \to 0 $$ where $ N $ is nilpotent finitely generated and torsion free. Such a group $ G $ is always solvable finitely generated and torsion free.

Is it true that the commutator subgroup of a Wang group $ G $ is finite if and only if $ G $ is abelian?

when $ N $ is abelian the answer seems to be yes see the comment from markvs here

Semidirect product of free abelian groups

I am interested in the more general case where $ N $ is nilpotent but perhaps not abelian.

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    $\begingroup$ "is torsion": you mean torsion-free? And for a torsion-free group, clearly "finite commutator group" means "abelian". $\endgroup$
    – YCor
    Commented Feb 14, 2022 at 23:40

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As Ycor points out this question is easy: A torsion free group (Wang or otherwise) is abelian if and only if its commutator subgroup is finite.

For the first direction, abelian implies trivial commutator subgroup so certainly the commutator subgroup is finite.

Reverse impliciation: since $ G $ is torsion free then any finite subgroup is trivial. So if the commutator subgroup is finite then actually the commutator subgroup is trivial and $ G $ is abelian.

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