Local powering-invariant and normal not implies quotient-local powering-invariant

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ of $$G$$ that is a local powering-invariant normal subgroup (it is both a local powering-invariant subgroup and a normal subgroup) but is not a quotient-local powering-invariant subgroup.

Facts used

 * 1) uses::Center not is quotient-local powering-invariant

Example using the center
See Fact (1). The example used appears to have derived length three.

Examples of the infinite dihedral group
Let $$G$$ be the infinite dihedral group:

$$G := \langle a,x \mid x = x^{-1}, xax^{-1} = a^{-1} \rangle$$

Let $$H$$ be the subgroup $$\langle a \rangle$$. Then:


 * $$H$$ is a local powering-invariant in $$G$$: Note that every non-identity element has at most one $$p^{th}$$ root for every $$p$$, and that root must be in $$H$$. The identity element has a unique $$p^{th}$$ root (namely itself) for each odd $$p$$, and infinitely many square roots. Hence, the local powering-invariance condition applies.
 * $$H$$ is not quotient-local powering-invariant in $$G$$: The non-identity element $$a^2$$ has a unique square root in $$G$$, namely $$a$$. But its image in $$G/H$$ is the identity element of cyclic group:Z2, hence has two square roots.

Note that this example is simpler than the example used for the center. $$G$$ here is a metacyclic group and $$H$$ is a characteristic subgroup of $$G$$ on account of being the centralizer of derived subgroup.