Automorph-conjugacy is transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Statement

Suppose $H \le K \le G$ are groups such that $H$ is an automorph-conjugate subgroup of $K$, and $K$ is an automorph-conjugate subgroup of $G$. Then, $H$ is an automorph-conjugate subgroup of $G$.

Proof

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Given: Groups $H \le K \le G$ such that $H$ is an automorph-conjugate subgroup of $K$ and $K$ is an automorph-conjugate subgroup of $G$. An automorphism $\sigma$ of $G$.

To prove: There exists $x \in G$ such that $\sigma(H) = xHx^{-1}$

Proof:

Step no. Assertion/construction Given data used Previous steps used Explanation
1 $\sigma(H) \le \sigma(K)$ $H \le K$ given-direct
2 There exists $g \in G$ such that $g K g^{-1} = \sigma(K)$ $K$ is an automorph-conjugate subgroup of $G$ $\sigma$ is an automorphism of $G$
given-direct
3 Denote by $c_g$ the map $u \mapsto gug^{-1}$. Then $c_g^{-1} \circ sigma$ is an automorphism of $G$ that restricts to an automorphism of $K$. Step (2) direct from the step
4 There exists $k \in K$ such that $(c_g^{-1} \circ \sigma)(H) = kHk^{-1}$. $H$ is an automorph-conjugate subgroup of $K$ Step (3) Step-given combination direct
5 Setting $x = gk$, we get that $\sigma(H) = xHx^{-1}$ Step (4) Simple algebraic manipulation gives $\sigma(H) = c_g(kHk^{-1})$ $= gkHk^{-1}g^{-1} = (gk)H(gk)^{-1}$