Divisibility-closed 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 divisibility-closed subgroup of $$G$$: If $$n$$ is a natural number such that every element of $$G$$ has a $$n^{th}$$ root in $$G$$, then every element of $$H$$ has a $$n^{th}$$ root in $$H$$.
 * 2) $$H$$ is not a local divisibility-closed subgroup of $$G$$: There exists a natural number $$n$$ and an element $$g \in H$$ such that $$x^n = g$$ has solutions for $$x \in G$$ but no solution for $$x \in H$$.

Related facts

 * Derived subgroup not is local divisibility-closed in nilpotent group whereas derived subgroup is divisibility-closed in nilpotent group

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