# Frattini's argument

This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
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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: $HN_G(K) = G$.

## 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.