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

It is possible to have an abelian group $G$ and a verbal subgroup $H$ of $G$ that is not a local powering-invariant subgroup of $G$: in other words, it is possible to have $h\in H$ and $n\in \mathbb{N}$ are such that there is a unique $x\in G$ satisfying $x^n=h$, and despite the uniqueness, we have $x\notin H$.

Related facts

Similar facts

• Characteristic subgroup of abelian group not implies local powering-invariant

Opposite facts

• Verbal subgroup of abelian group implies divisibility-closed
• Characteristic subgroup of abelian group implies powering-invariant

Proof

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

• $H$ is verbal in $G$: It is the set of elements expressible as squares, hence is verbal.
• $H$ 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 $H$.