Product of commutators in a group with abelian commutator subgroup

Question

My question is inspired by this: Questions about finitely generated nilpotent groups. Let $G$ be a group, generated by $n$ elements, such that $[G,G]$ is abelian. Can we use the results stated in the quoted question to show that any element $h \in [G,G]$ can be expressed as a product of no more than $2n$ commutators?

Answer 1

I don't know, but apparently there is a result by Stroud from a 1966 PhD thesis that every finitely generated virtually abelian-by-nilpotent group is verbally elliptic. This class certainly includes finitely generated metabelian groups, and verbally elliptic implies that in any such group there is a bound on the number of commutators needed to express any element of $G'$. So, if $f(n)$ is the bound for the free metabelian group of rank $n$, then in any $n$-generator metabelian group every element of $G'$ can be expressed as a product of $f(n)$ commutators.

You are asking whether $f(n) \le 2n$, and I have no idea!

This stuff seems hard to me, but if you want to learn about it, then the best place to start is probably the book "Words Notes on Verbal Width in groups" by Dan Segal, LMS Lecture Note Series 361. (I found a downloadable version online, but that's probably illegal.)