Divisibility-closedness is not finite-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., divisibility-closed subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about divisibility-closed subgroup|Get more facts about finite-intersection-closed subgroup property|

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

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 $G$ .