Nilpotent with rationally powered center not implies rationally powered

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

Dual fact
The dual fact to this is dual::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 $$p$$).