AEP does not satisfy intermediate subgroup condition

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 $$H \le K \le G$$ such that $$H$$ is an AEP-subgroup of $$G$$ but $$H$$ is not an AEP-subgroup of $$K$$.

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:


 * $$H \le K \le G$$: 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.