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.

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.