Subgroup of abelian group not implies abelian-extensible automorphism-invariant

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., subgroup of abelian group) need not satisfy the second subgroup property (i.e., abelian-extensible automorphism-invariant subgroup)
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Statement

It is possible to have an abelian group $G$ and a subgroup $H$ of $G$ such that $H$ is not an abelian-extensible automorphism-invariant subgroup of $G$, i.e., there exists an abelian-extensible automorphism $\sigma$ of $G$ such that $\sigma (H)$ is not equal to $H$.

Facts used

1. Divisible abelian group implies every automorphism is abelian-extensible

Proof

Integers in rationals

Let $\!G$ be the additive group of rational numbers, i.e., the abelian group $(\mathbb {Q} ,+)$, $\!H$ be the subgroup $\mathbb {Z}$ of integers, and $\!\sigma$ be the automorphism of $G$ given by $x\mapsto x/2$.

Then:

• By fact (1), $\sigma$ is abelian-extensible, because $G$ is a divisible abelian group.
• $\sigma$ does not send $H$ to itself.

Rationals in two copies

Let $\!G$ be the direct sum of two copies of the rational numbers, i.e., the abelian group $\mathbb {Q} \oplus \mathbb {Q}$, let $H$ be the first direct factor, and $\!\sigma$ be the automorphism of $G$ that interchanges the two coordinates. Then:

• $G$ is a divisible abelian group, so by fact (1), $\sigma$ is abelian-extensible.
• $\sigma$ sends $H$ to the other direct factor, so $\sigma$ does not preserve $H$.