Automorph-conjugacy is normalizer-closed

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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., normalizer-closed subgroup property)
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Suppose G is a group and H is an automorph-conjugate subgroup of G, then N_G(H) (the normalizer of H in G) is also an automorph-conjugate subgroup of G.

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Proof idea

The key idea behind the proof is that taking the normalizer commutes with automorphisms. In other words, for any automorphism \sigma, we have \sigma(N_G(H)) = N_G(\sigma(H)).

Proof details

Given: A group G, an automorph-conjugate subgroup H.

To prove: N_G(H) is an automorph-conjugate subgroup: given any automorphism \sigma of G, N_G(H) and \sigma(N_G(H)) are conjugate.

Proof: \sigma(N_G(H)) = N_G(\sigma(H)). By assumption, H is automorph-conjugate, so there exists a g \in G such that \sigma(H) = gHg^{-1}. Thus, \sigma(N_G(H)) = N_G(gHg^{-1}) = gN_G(H)g^{-1}, and thus, N_G(H) and \sigma(N_G(H)) are conjugate by g.