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This article gives a proof/explanation of the equivalence of multiple definitions for the term automorph-conjugate subgroup
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Contents 1 Statement 1.1 For automorph-conjugate subgroups 1.2 For Sylow subgroups 1.3 For characteristic subgroups of Sylow subgroups 2 Facts used 3 Proof 3.1 Proof for automorph-conjugate subgroups 3.2 Proof for Sylow subgroups 3.3 Proof for characteristic subgroups

Statement

For automorph-conjugate subgroups

Let H be a normal subgroup of G and P an automorph-conjugate subgroup of H. Then:

HN_G(P) = G

where N_G(P) denotes the normalizer of P in G.

For Sylow subgroups

Let H be a normal subgroup of G and P a Sylow subgroup of H. Then:

HN_G(P) = G

where N_G(P) denotes the normalizer of P in G.

For characteristic subgroups of Sylow subgroups

Let H be a normal subgroup of G, P be a Sylow subgroup of H, and K be a characteristic subgroup of P. In other words, K is a characteristic subgroup of Sylow subgroup of H. Then:

HN_G(K) = G.

Facts used

1. Sylow implies automorph-conjugate
2. Characteristic implies automorph-conjugate
3. Automorph-conjugacy is transitive

Proof

Proof for automorph-conjugate subgroups

(This proof uses the left action convention)

Given: H a normal subgroup of G. P an automorph-conjugate subgroup of H.

To prove: HN_G(P) = G.

Proof: Let . Consider gPg ^{− 1}. Since H is normal, the map is an automorphism restricted to H. Since P is automorph-conjugate in H, there exists such that hPh ^{− 1} = gPg ^{− 1}.

Then, , and hence g = hx ^{− 1}, with . thus, every element of G can be expressed as the product of an element of H and an element of N_G(P), and we are done.

Proof for Sylow subgroups

This follows from the statement for automorph-conjugate subgroups and fact (1).

Proof for characteristic subgroups

By facts (1), (2) and (4), every characteristic subgroup of a Sylow subgroup is automorph-conjugate, so this statement again follows from the statement for automorph-conjugate subgroups.