Automorph-conjugacy is transitive

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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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Get more facts about automorph-conjugate subgroup |Get facts that use property satisfaction of automorph-conjugate subgroup | Get facts that use property satisfaction of automorph-conjugate subgroup|Get more facts about transitive subgroup property


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}