Fully invariant subgroup of abelian group not implies divisibility-closed

Statement
It is possible to have an abelian group $$G$$ and a fully invariant 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 $$p$$-divisible but $$H$$ is not.

Opposite facts

 * Characteristic subgroup of abelian group implies powering-invariant

Proof
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:


 * $$H$$ is a fully invariant subgroup of $$G$$.
 * However, $$H$$ is not divisibility-closed: $$G$$ is $$p$$-divisible, but $$H$$ is not.