Nilpotent and torsion-free not implies torsion-free abelianization

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) does not always satisfy a particular subgroup property (i.e., quotient-torsion-freeness-closed subgroup)
View subgroup property satisfactions for subgroup-defining functions $|$ View subgroup property dissatisfactions for subgroup-defining functions

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.

Related facts

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

Further information: central product of UT(3,Z) and Q

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 $p$.