AEP does not satisfy intermediate subgroup condition

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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 \le K \le 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 \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.