Powering-invariant 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., powering-invariant subgroup) need not satisfy the second subgroup property (i.e., local powering-invariant)
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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 $H$.