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<G 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<HQ_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<Q_i.

\Box

Confer Milne Chapter 5

数学

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