Derived subgroup not is local divisibility-closed in nilpotent group

Statement
It is possible to have a nilpotent group $$G$$ (in fact, in our example, $$G$$ is a group of nilpotency class two) such that the [derived subgroup]] $$G'$$ is not a local divisibility-closed subgroup, i.e., there exists an element of $$G'$$ that has $$n^{th}$$ roots in $$G$$ but no $$n^{th}$$ roots in $$G'$$.

Opposite facts

 * Derived subgroup is divisibility-closed in nilpotent group, hence it is also a powering-invariant subgroup and quotient-powering-invariant subgroup (using that normal subgroup contained in the hypercenter satisfies the subgroup-to-quotient powering-invariance implication).

Proof
The simplest examples are:


 * $$G$$ is particular example::dihedral group:D8, so $$G'$$ is the particular example::center of dihedral group:D8. The non-identity element of $$G'$$ has two square roots in $$G$$ but none in $$G'$$.
 * $$G$$ is particular example::quaternion group, so $$G'$$ is the particular example::center of quaternion group. The non-identity element $$-1$$ of $$G'$$ has six square roots in $$G$$ but none in $$G'$$.
 * For any prime, $$G$$ is the semidirect product of cyclic group of prime-square order and cyclic group of prime order. $$G'$$ is a group of prime order, and it lives inside a cyclic subgroup of order $$p^2$$, hence its non-identity elements have $$p^{th}$$ roots, none of which are in $$G$$.