Verbal subgroup of abelian group not implies local powering-invariant

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 $$x \notin H$$.

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 $$H$$.