Left-transitively permutable implies characteristic

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., left-transitively permutable subgroup) must also satisfy the second subgroup property (i.e., characteristic subgroup)
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Statement

Suppose $H$ is a subgroup of a group $K$ such that whenever $K$ is a Permutable subgroup (?) of a group $G$ , $H$ is also a permutable subgroup of $G$ . Then, $H$ is a characteristic subgroup of $K$ .

Facts used

Every group is normal fully normalized in its holomorph

Proof

Given: A subgroup $H$ of a group $K$ such that whenever $K$ is a permutable subgroup of a group $G$ , then $H$ is permutable in $G$ .

To prove: $H$ is characteristic in $K$ : for any automorphism $σ$ of $K$ , and any $g \in H$ , $σ (g) \in H$ .

Proof: Let $G$ be the holomorph of $K$ ; in other words, we have:

$G = K ⋊ A u t (K)$ .

Let $φ : G \to A u t (K)$ be the natural retraction with kernel $K$ .

Now, consider $σ$ as an element of $G$ (via its membership in $A u t (K)$ , and $g$ also as an element of $G$ . Let $A$ be the cyclic subgroup of $G$ generated by $σ$ . Since by assumption $H$ is permutable in $G$ , we have:

$A H = H A$

In particular, consider the product $σ g$ . This is in $A H$ , hence it is also in $H A$ , so there exists $h \in H, α \in A$ such that $σ g = h α$ . Applying $φ$ to both sides, we get:

$φ (σ g) = φ (h α)$

This yields:

$σ = α$ .

In particular, we obtain that:

$σ g = h σ$ .

Thus, we obtain that:

$σ g σ^{- 1} = h \in H$

Since the action of $A u t (K)$ is by conjugation, this yields:

$σ (g) = σ g σ^{- 1} \in H$ .