Powering-invariant not implies local powering-invariant

Statement
It is possible to have a group $$G$$ and a powering-invariant subgroup $$H$$ of $$G$$ that is not local powering-invariant. In other words, the following are true:


 * 1) $$H$$ is powering-invariant in $$G$$: For any prime number $$p$$, if $$G$$ is powered over $$p$$, $$H$$ is also powered over $$p$$.
 * 2) $$H$$ is not local powering-invariant in $$G$$: There exists a natural number $$n$$ and an element $$g \in H$$ such that there is a unique $$x \in G$$ satisfying $$x^n = g$$, but $$x \notin H$$.

Proof
Set $$G = \mathbb{Z}$$ and $$H$$ as the subgroup $$2\mathbb{Z}$$.


 * $$H$$ is powering-invariant in $$G$$: $$G$$ is not powered over any prime, so $$H$$ is is powering-invariant in $$G$$ for vacuous reasons.
 * $$H$$ is not local powering-invariant in $$G$$: The element $$2 \in H$$ has a unique square root (corresponding to $$n = 2$$) in $$G$$, but this square root is not in $$H$$.