Cofactorial automorphism-invariance is not transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., cofactorial automorphism-invariant subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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Statement

It is possible to have groups $H\leq K\leq G$ such that $H$ is a cofactorial automorphism-invariant subgroup of $K$ and $K$ is a cofactorial automorphism-invariant subgroup of $G$, but $H$ is not a cofactorial automorphism-invariant subgroup of $G$.

The key observation here is that the set of prime divisors of the order of $K$ may be a proper subset of the set of prime divisors of the order of $G$. Shrinkage in the set of prime divisors is necessary to construct a counterexample.

Facts used

1. p-automorphism-invariant not implies characteristic

Proof

By Fact (1), we can find a prime number $p$ and $p$-groups $H\leq K$ such that $H$ is $p$-automorphism-invariant in $K$ but not characteristic in $K$. In particular, we can find an automorphism of $K$ of order $p'$ relatively prime to $p$ such that $H$ is not invariant under that automorphism. Define $G$ as the semidirect product of $K$ by the cyclic group generated by that automorphism. Then:

• $H$ is cofactorial automorphism-invariant in $K$: $p$ is the only prime dividing the order of $K$, and $H$ is invariant under $p$-automorphisms.
• $K$ is cofactorial automorphism-invariant in $G$: In fact, $K$ is a normal Sylow subgroup of $G$ and hence a characteristic subgroup of $G$. So it is invariant under all automorphisms of $G$.
• $H$ is not cofactorial automorphism-invariant in $G$: It is not invariant under the conjugation operation by the generator of the cyclic $p'$-group.

Concretely, using the minimal example provided in the proof of Fact (1), we obtain that for any odd prime, we can construct $H,K,G$ to have orders $p^{3},p^{6},2p^{6}$ respectively.