Use Sylow's theorem to show that $G = HN_G(P)$

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Let $G$ be a group of finite order. Let $H$ be a normal subgroup of $G.$ Let $P$ be a Sylow $p$-subgroup of $H.$ Using Sylow's theorem, show that $G = H N_G(P).$

2 Answers

Hint:

1- $|P|=|P^g|$ where $g\in G$ and $P^g=gPg^{-1}$.

2- So $P$ and $P^g$ are $p-$sylows of $H$. Note that $P^g<H$.

3- Use the second Sylow theorem for $H$.

There is a general fact that goes as follows

Theorem Let $G$ be a group, $X$ a $G$-set (i.e. a set on which $G$ acts), $H$ a subgroup and suppose that $H$ acts (so we can restrict the action to one of $H$) transitively on $X$. Then for any $x\in X$, $G={\rm stab}_G\; x\;H$.

Proof Pick $g\in G$. We seek to write $g=kh$ where $kx=x$. Now $H$ acts transitively on $X$, so taking the element $g^{-1}x$ there exists $h$ such that $hg^{-1}x=x$. But then $hg^{-1}$ stabilizes $x$; then so does $gh^{-1}$, and $g=gh^{-1}h=kh$, as we wanted.

Corollary (The Frattini Argument) Let $G$ be a finite group, $H$ a normal subgroup and $P$ a Sylow $p$-subgroup of $H$. Then $G=H\; N_G(P)$.

Proof Consider the set $X$ of Sylow subgroups of $H$. Then $G$ acts on $X$ by conjugation (since $H$ is normal) and $H$ acts transitively on $X$ by virtue of (one of) Sylow's theorem(s). By the argument above, we may choose any Sylow subgroup of $H$, call it $P$. Then ${\rm stab}_GP=N_G(P)$ and hence $G=N_G(P)\, H=H\, N_G(P)$.

A nice result is obtained from this.

Corollary (Frattini) The Frattini subgroup of a finite group is nilpotent.

Proof We show that every Sylow subgroup of $\Phi(G)$ is normal in $\Phi(G)$. We already know $\Phi(G)\lhd G$. Pick a Sylow subgroup $P$. By the Frattini Argument, $G=\Phi(G)N_G(P)$. But $\Phi(G)$ is the set of nongenerators, so $G=N_G(P)$. Hence $P\lhd G$; in particular $P\lhd \Phi(G)$. Thus $\Phi(G)$ is the direct product of its Sylow subgroups, which are $p$-groups. Since finite $p$-groups are nilpotent and direct products of nilpotent groups are nilpotent, we get the result.