Finite not implies divisibility-closed in abelian group

Statement
It is possible to have a group $$G$$ (in fact, we can choose $$G$$ to be an abelian group) and a finite subgroup $$H$$ of $$G$$ such that $$H$$ is not a divisibility-closed subgroup of $$G$$. In other words, there exists a prime number $$p$$ such that $$G$$ is a $$p$$-divisible group but $$H$$ is not a $$p$$-divisible group.

In fact, we can construct an example of this sort for each prime number $$p$$.

Proof
For any prime number $$p$$:


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

Clearly:


 * $$H$$ is a finite subgroup. In fact, it has order $$p$$.
 * However, $$H$$ is not divisibility-closed: $$G$$ is $$p$$-divisible, but $$H$$ is not.