Nilpotent and torsion-free not implies torsion-free abelianization

Statement

It is possible to have a torsion-free nilpotent group $G$ such that the abelianization of $G$ has $p$-torsion for every prime $p$.

Related facts

Converse

• Nilpotent with torsion-free abelianization not implies torsion-free

Proof

Let $G$ be the central product of unitriangular matrix group:UT(3,Z) with the group of rational numbers, where the center of the former is identified with a copy of $\mathbb{Z}$ in the latter. Then,

• $G$ is torsion-free.
• $G^{\prime }$ is isomorphic to $\mathbb{Z}$, and $G/G^{\prime }$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Q}/\mathbb{Z}$. This has $p$-torsion for all primes $p$.