Complemented normal not implies local powering-invariant

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., complemented normal subgroup) need not satisfy the second subgroup property (i.e., local powering-invariant subgroup)
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Statement

It is possible to have a group $G$ and a complemented normal subgroup $H$ of $G$ such that $H$ is not a local powering-invariant subgroup of $G$ , i.e., there exists an element $h \in H$ and a natural number $n$ such that there is a unique solution $u \in G$ to $u^{n} = h$ , but $u \notin H$ .

Related facts

Opposite facts

Complemented normal implies quotient-powering-invariant and quotient-powering-invariant implies powering-invariant

Proof

Let $G$ be the infinite dihedral group:

$G : = ⟨ a, x ∣ x^{2} = e, x a x = a^{- 1} ⟩$ .

Here, $e$ denotes the identity element.

Let $H$ be the subgroup $⟨ a^{2}, x ⟩$ .

Then the following are true:

$H$ is a complemented normal subgroup of $G$ . It is normal because it is a subgroup of index two and it has a permutable complement $⟨ a x ⟩$ of order two.
$H$ is not local powering-invariant in $G$ . For the element $h = a^{2}$ and $n = 2$ , the only solution to $u^{2} = h$ for $u \in G$ is $u = a$ , and $a \notin H$ .