fourth Sylow theorem

We are going to reconstruct the salient of a part of the classical theorem:

Each $H$ p-S of $G$ is contained into a $Q$ p-SS of $G$

Here the reference frame:

1. $\#$p-SS divides $|G|$:

The set p-SS must be considered as an orbit of the action $G\times$ p-SS $\to$ p-SS via $g\cdot Q=gQg^{-1}$.  Since from sylow II we know that each two are conjugated the there is only one orbit

p-SS $={\rm Orb}_G(Q)\leftrightarrow G/{\rm St}_G(Q)$

giving us a trivial  orbital partition. Then $\#$p-SS=$|{\rm Orb}_G(Q)|$, i.e.

$\#$p-SS=$=[G:{\rm St}_G(Q)]=[G:N_G(Q)]$,

because for the isotropy group is ${\rm St}_G(Q)=\{\!x\!\in\!G: xQx^{-1}\!=\!Q\}=N_G(Q)\!\}$.

Then $|G|=\#{\rm p\!-\!SS}|N(Q)|$.

2. Observe also that $p\not|$#p-SS:

since $|G|=p^mr$ and $|G|=|N_G(Q)|[G:N_G(Q)]$,  as  far as $|N_G(Q)|=|Q|[N_G(Q):Q]$

so

$p^mr=|G|$

$=|Q|[G:N_G(Q)][N_G(Q):Q]$

$=p^m[N_G(Q):Q]\cdot\#$p-SS

which implies that $r=[N_G(Q):Q]\cdot\#$p-SS, then $p\!\!\not|\#$p-SS.

Proof of Theorem:

Considering the conjugation sub-action

$H\times$ p-SS$\to$ p-SS,

we get a orbit decomposition p-SS = ${\rm Orb}_H(Q_1)\sqcup\cdots\sqcup{\rm Orb}_H(Q_t)$ with the corresponding class equation

#p-SS = $[H:N_H(Q_1)]+\cdots+[H:N_H(Q_t)]$,

but asumming that $|H|=p^{\alpha}$ then

#p-SS = $\frac{p^{\alpha}}{|N_H(Q_1)|}+\cdots+\frac{p^{\alpha}}{|N_H(Q_t)|}$.

Now since $p\!\!\not|$#p-SS then there is some $Q_i$ for which $|N_H(Q_i)|=|H|$, so

$H=N_H(Q_i)$.

With that, it is easy these: $HQ_i and $Q_i$ is normal in $HQ_i$.

Also $\pi:H\to HQ_i/Q_i$ given by $x\mapsto\pi(x)=xQ_i$  is an epimorphism, so by the fundamental theorem of group-morphisms we have

$\frac{HQ_i}{Q_i}\cong\frac{H}{H\cap Q_i}$,

Observing that $H,\ Q_i and $|HQ_i|=[H:H\cap Q_i]|Q_i|$ then

$|HQ_i|=\frac{|H||Q_i|}{|H\cap Q_i|}=p^{m+\alpha-\beta}$,

where $|H\cap Q_i|=p^{\beta}$,

but $m$ is maximal, then $\alpha-\beta=0$. Hence $|H|=|H\cap Q_i|$ and

$H=H\cap Q_i.

$\Box$

Confer Milne Chapter 5

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