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

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.