Frattini's argument

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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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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 g \in G. Consider gPg^{-1}. Since H is normal, the map x \mapsto gxg^{-1} is an automorphism restricted to H. Since P is automorph-conjugate in H, there exists h \in H such that hPh^{-1} = gPg^{-1}.

Then, g^{-1}h  = x\in N_G(P), and hence g = hx^{-1}, with h \in H, x^{-1} \in N_G(P). 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.