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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Statement

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.

Related facts

Generalizations

Proof

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.