Divisibility-closedness is not finite-intersection-closed

Statement
It is possible to have a group $$G$$ and subgroups $$H,K$$ of $$G$$ such that both $$H$$ and $$K$$ are divisibility-closed subgroups of $$G$$ but the intersection of subgroups $$H \cap K$$ is not.

In fact, we can choose our example so that $$G$$ is a divisible abelian group, and hence, so are $$H$$ and $$K$$.

Related facts

 * Powering-invariance is strongly intersection-closed
 * Divisibility-closedness is not finite-join-closed
 * Powering-invariance is not finite-join-closed

Proof
Let $$G$$ be the external direct product of the (additive) group of rational numbers $$\mathbb{Q}$$ and the group of rational numbers modulo integers $$\mathbb{Q}/\mathbb{Z}$$. In other words, we have:

$$G = \mathbb{Q} \times \mathbb{Q}/\mathbb{Z}$$

Consider the subgroups $$H$$ and $$K$$ defined as follows:


 * $$H = \{ (a,0) \mid a \in \mathbb{Q} \}$$ is the first direct factor.
 * Let $$\pi:\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$$ be the natural quotient map. Define:

$$K = \{ (a,\pi(a)) \mid a \in \mathbb{Q} \}$$.

Now, the following are true:


 * $$G$$ is a divisible abelian group (i.e., $$n$$-divisible for all natural numbers $$n$$), because both its direct factors are.
 * $$H$$ and $$K$$ are both isomorphic to $$\mathbb{Q}$$, hence are both divisible abelian groups (i.e., $$n$$-divisible for all $$n$$), and hence, are divisibility-closed in $$G$$.
 * The intersection $$H \cap K$$ is the subgroup:

$$\{ (a,0) \mid a \in \mathbb{Z} \}$$

This is isomorphic to the group of integers, which is not divisible by any $$n > 1$$. Hence, it is not divisibility-closed in $$G$$.