AEP does not satisfy intermediate subgroup condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., AEP-subgroup) not satisfying a subgroup metaproperty (i.e., intermediate subgroup condition).
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Statement

Property-theoretic statement

The subgroup property of being an AEP-subgroup does not satisfy the subgroup metaproperty of the intermediate subgroup condition.

Statement with symbols

It is possible to have groups $H\leq K\leq G$ such that $H$ is an AEP-subgroup of $G$ but $H$ is not an AEP-subgroup of $K$.

Proof

Example of an Abelian group

Let $A$ and $B$ be isomorphic copies of $\mathbb {Z} /4\mathbb {Z}$. Let $C$ and $D$ be subgroups of order two in $A$ and $B$ respectively. Then, define:

$G=A\times B,H=C\times D,K=C\times B$.

We claim that:

• $H\leq K\leq G$: This is clear from the definition.
• $H$ is an AEP-subgroup of $G$
• $H$ is not an AEP-subgroup of $K$: Consider the automorphism of $H$ that exchanges the generators of $C$ and $D$. This cannot extend to an automorphism of $K$, because in $K$, the generator of $D$ is the double of an element, while the generator of $C$ is not the double of anything.