Finite index not implies local powering-invariant

Statement
It is possible to have a group $$G$$ and a subgroup of finite index $$H$$ 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 $$u \notin H$$.

In fact, we can choose $$H$$ to satisfy any of these additional constraints:


 * We can choose $$H$$ to be a characteristic subgroup of finite index; in fact, we can choose $$H$$ to be the derived subgroup.
 * We can choose $$H$$ to be a subgroup of index two.

Example with subgroup of index two
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 subgroup of index two in $$G$$.
 * $$H$$ 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 $$a \notin H$$.

Example with characteristic subgroup of finite index that uses derived subgroup
Consider the infinite dihedral group, given by the presentation:

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

where $$e$$ denotes the identity of $$G$$. We find that:

$$[G,G] = \langle a^2 \rangle$$

is an infinite cyclic group.

Now consider the element $$h = a^2$$. Let $$n = 2$$. We note that all elements outside $$\langle a \rangle$$ have order two, hence any element $$u$$ with $$u^2 = h$$ must be inside $$\langle a \rangle$$. The only possibility is thus $$u = a$$, which is outside $$H$$. Thus, the element $$h = a^2$$ has a unique square root in $$G$$, but this is not in $$H$$, completing the proof.