Finite not implies divisibility-closed in abelian group

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., finite subgroup) need not satisfy the second subgroup property (i.e., divisibility-closed subgroup)
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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 $p$-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.