Nilpotent with rationally powered center not implies rationally powered

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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

Related facts

Dual fact

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

Similar facts

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).