Complemented normal not implies local powering-invariant

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 $$u \notin H$$.

Opposite facts

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

Proof
Let $$G$$ be the particular example::infinite dihedral group:

$$G := \langle a,x \mid x^2 = e, xax = a^{-1} \rangle$$.

Here, $$e$$ denotes the identity element.

Let $$H$$ be the subgroup $$\langle a^2,x \rangle$$.

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 $$\langle ax \rangle$$ 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 $$a \notin H$$.