Divisibility-closed implies powering-invariant

Statement
Suppose $$G$$ is a subgroup and $$H$$ is a divisibility-closed subgroup of $$G$$, i.e., if $$n$$ is a natural number such that every element of $$G$$ has an $$n^{th}$$ root in $$G$$, then every element of $$H$$ has a $$n^{th}$$ root in $$H$$. Then, $$H$$ is also a powering-invariant subgroup of $$G$$, i.e., if $$p$$ is a prime number such that every element of $$G$$ has a unique $$p^{th}$$ root, then every element of $$H$$ has a unique $$p^{th}$$ root in $$H$$.

Converse

 * Powering-invariant not implies divisibility-closed

Proof
This statement is pretty direct. Note that existence follows from being divisibility-closed, and uniqueness in the whole group implies uniqueness in the subgroup as well.