1
$\begingroup$

I am stumped on this question. Does anyone have some helpful hints or a solution to this question? Thanks!

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 $G = H N_G(P).$

$\endgroup$
4
$\begingroup$

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 $H$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.