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 $$p$$.

In particular, this shows that the derived subgroup in a nilpotent group is not necessarily a quotient-torsion-freeness-closed subgroup.

Dual fact
Somewhat surprisingly, the dual fact to this is not true. The dual fact, if true, would state that the center of a divisible nilpotent group need not be divisible (and in particular, that the center need not be divisibility-closed in a nilpotent group). This is false. In fact, upper central series members are completely divisibility-closed in nilpotent group.

Converse

 * Nilpotent group 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.
 * $$G'$$ is isomorphic to $$\mathbb{Z}$$, and $$G/G'$$ is isomorphic to $$\mathbb{Z} \times \mathbb{Z} \times \mathbb{Q}/\mathbb{Z}$$. This has $$p$$-torsion for all primes $$p$$.