On the normal closure of a subgroup Let $G=P \ltimes O_{p^{\prime}}(G)$ be a finite group, where $P \in {\rm Syl}_p(G)$. Let $q\in\pi(O_{p^{\prime}}(G))$. I know Frattini argument.
How we conclude from Frattini argument that there exists a Sylow $q$-subgroup 
$Q$ of $O_{p^{\prime}}(G)$ with $Q=Q^P$?
 A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$An alternative proof goes like this.
Let $\Omega$ be the set of Sylow $q$-subgroups of $H = O_{p^{\prime}}(G)$ (hence of $G$). Let $S \in \Omega$.
We have $\Size{\Omega} = \Size{H : N_{H}(S)} \mid \Size{H}$, hence $p \nmid  \Size{\Omega}$.
Now $P$ acts by conjugation on $\Omega$, and the orbits have length a divisor of $P$, hence a power of $p$. If all the orbits have length greater than $1$, then $p \mid \Size{\Omega}$, a contradiction.
Hence $P$ has an orbit $\Set{Q}$ of length $1$, that is $Q^{P} = Q$.
A: Let $G=p^em$ with $p\nmid m$.
$O_{p'}(G)$ is normal in $G$.
Let $Q$ be a Sylow $q$-subgroup of $O_{p'}(G)$.
By the Frattini argument, $G=N_G(Q)O_{p'}(G)$.
Since $p$ does not divide $|O_{p'}(G)|$, $p^e$ divides $|N_G(Q)|$. Let $R$ be a Sylow $p$-subgroup of $N_G(Q)$. As $|R|=p^e$, $R$ is also a Sylow $p$-subgroup of $G$.
By Sylow's Theorem, there is some $g\in G$ with $R^g=P$. Therefore $P\subseteq N_G(Q^g)$.
As $O_{p'}(G)$ is normal in $G$, $Q^g$ is a Sylow $q$-subgroup of $O_{p'}(G)$ and we are done.
