Frattini's Argument

Lemma

Suppose $G$ is a transitive on $\Omega$. Then $H<G$ is transitive on $\Omega$ iff $G=H{G}_{\omega }$ for $\omega \in \Omega$.

\begin{proof} Suppose that $H$ is transitive on $\Omega$. For any $\omega \in \Omega$ and any $g\in G$, define ${\omega }^{\prime }:={\omega }^g$. Since $H$ is transitive, there is $h\in H$ such that ${\omega }^h={\omega }^{\prime }={\omega }^g$. So we have $h^{-1}g\in G_{\omega }$ and $g=h(h^{-1}g)\in H{G}_{\omega }$. Therefore, $G=H{G}_{\omega }$.

Conversely, assume that $G=H{G}_{\omega }$. For any $\omega ,{\omega }^{\prime }\in \Omega$, there is $g\in G$ such that ${\omega }^g={\omega }^{\prime }$. As $g$ can be written as $g=xh$ with $x\in G_{\omega }$ and $h\in H$, there is ${\omega }^g={\omega }^{xh}={\omega }^h={\omega }^{\prime }$, i.e., $H$ acts on $\Omega$ transitively. \end{proof}

Here is another version of Frattini’s argument from wiki. In fact, it is a special case of ^9wyb80. Let $\Omega$ be the set of Sylow subgroups of $H⊲G$. Then both $G$ and $H$ act on $\Omega$ by conjugation transitively, and the point stabilizer $G_P=N_G(P)$ for a $P\in \Omega$.

Corollary

If $H⊲G$ and $P$ is a Sylow subgroup of $H$, then $G=H{N}_G(P)$.

\begin{proof} Let $g\in G$. Since $H⊲G$ and $P⩽H$, the group $gP{g}^{-1}⩽H$ is also a Sylow $p$-subgroup of $H$. By ^ueqhoc, there is a $h\in H$ such that $gP{g}^{-1}=hP{h}^{-1}$. Then $h^{-1}g\in N_G(P)$ and $g$ can be written as $g=h(h^{-1}g)\in H{N}_G(P)$. Hence $G=H{N}_G(P)$. \end{proof}