Powering-invariant not implies divisibility-closed

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ such that:


 * 1) $$H$$ is a powering-invariant subgroup of $$G$$: If $$n$$ is a natural number such that every element of $$G$$ has a unique $$n^{th}$$ root, then every element of $$H$$ has a unique $$n^{th}$$ root within $$H$$.
 * 2) $$H$$ is not a divisibility-closed subgroup of $$G$$: There exists a natural number $$n$$ such that every element of $$G$$ has a $$n^{th}$$ root (not necessarily unique) but not every element of $$H$$ has a $$n^{th}$$ root within $$H$$.

Related facts

 * Center not is divisibility-closed

Proof idea
The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-closed. We will construct an example where the subgroup is finite.

Proof details
For any prime number $$p$$:


 * Let $$G$$ be the $$p$$-quasicyclic group.
 * Let $$H$$ be the subgroup comprising the elements of order 1 or $$p$$.

Clearly:


 * $$H$$, being finite, is powering-invariant (in fact, both $$G$$ and $$H$$ are powered over precisely the set of primes other than $$p$$).
 * However, $$H$$ is not divisibility-closed: $$G$$ is $$p$$-divisible, but $$H$$ is not.