# Divisibility-closedness is not finite-intersection-closed

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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).
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## 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 $K$.

## 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 $G$.