Local powering-invariant not implies local divisibility-closed

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ satisfying the following:


 * 1) $$H$$ is a local powering-invariant subgroup of $$G$$: If an element of $$H$$ has a unique $$n^{th}$$ root in $$G$$, that root is in $$H$$.
 * 2) $$H$$ is not a local divisibility-closed subgroup of $$G$$: There exists a natural number $$n$$ and an element of $$H$$ that has a (non-unique) $$n^{th}$$ root in $$G$$ but no $$n^{th}$$ root in $$H$$.

Proof
We can take any example of a subgroup of finite group where the subgroup is not local divisibility-closed. Some simple examples are below:


 * Z2 in Z4: Let $$G$$ be cyclic group:Z4 and $$H$$ be Z2 in Z4, the unique cyclic subgroup of order two. The non-identity element of $$H$$ has square roots in $$G$$ but not in $$H$$, so $$H$$ is not local divisibility-closed. However, being a subgroup of a finite group, it is divisibility-closed.
 * Center of dihedral group:D8
 * Center of quaternion group