Fully invariant subgroup of abelian group not implies divisibility-closed

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup of abelian group) need not satisfy the second subgroup property (i.e., divisibility-closed subgroup)
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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a abelian group. That is, it states that in a abelian group, every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup) need not satisfy the second subgroup property (i.e., divisibility-closed subgroup)
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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.

Related facts

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.