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. Center not is quotient-local powering-invariant

Proof

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.