Nilpotent with rationally powered center not implies rationally powered

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., center) does not always satisfy a particular subgroup property (i.e., powering-faithful subgroup)
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Statement

It is possible to have a nilpotent group $G$ such that the center $Z(G)$ is rationally powered, but such that $G$ is not powered over any prime.

In particular, it is possible to have a nilpotent group $G$ such that the center $Z(G)$ is not a powering-faithful subgroup of $G$.

Related facts

• Center is powering-faithful in finite nilpotent group

Related facts

Dual fact

The dual fact to this is nilpotent group with rationally powered abelianization need not be rationally powered.

Similar facts

• Nilpotent and torsion-free not implies torsion-free abelianization
• Nilpotent group with rationally powered abelianization need not be rationally powered

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, $Z(G)$ is isomorphic to the group of rational numbers, hence is powered over every prime. However, $G$ is not powered over any prime (we can see this from the fact that $G/Z(G)\cong \mathbb {Z} \times \mathbb {Z}$ is not $p$-divisible for any prime $p$).