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**slevvio** hello, thanks for your response, however I dont think i was very clear. This proof is meant to be the first part in proving the conjugacy theorem.

So far I have proved :

1) if the order of G is $\displaystyle p^ms$, where p does not divide s, then there exists a Sylow-p subgroup of G

2) there exists a subgroup of G of any p-power order, as long as that power of p divides $\displaystyle p^m$.

Now I am trying to show that 3) Any p-subgroup of G is contained in a Sylow p-subgroup of G (I think this is sayign that that subgroup is maximal)

So of course I haven't proved the conjugacy theorem yet. I will show you the exact way the proof is written out in the notes, the way they say that p does not divide r seems as if it is almost offhand

(iii) Let

P be a Sylow p-subgroup. Let Q be any p-subgroup. Let $\displaystyle \mathcal{P}$ be the set of all

conjugates of P. Then G acts on $\displaystyle \mathcal{P}$ by conjugation. By the orbit–stabilizer theorem, the number

r of conjugates of P is $\displaystyle [ G : N_G(P) ] $. Note that $\displaystyle p \not| r$. Then Q acts on $\displaystyle \mathcal{P}$ by conjugation. Each

Q-orbit has size a power of P. There must be an orbit whose order is not divisible by p. However the only power of p which is not divisible by p is

$\displaystyle p^0$ = 1. So there is an orbit of size 1. Let $\displaystyle gPg^{-1}$ element in an orbit of size 1. Then Q normalizes $\displaystyle gPg^{-1}$. Therefore $\displaystyle gPQg^{-1}$ is a p-group. Hence$\displaystyle gPQg^{-1} = gPg^{-1}$ and so $\displaystyle Q \subseteq gPg^{-1}$.

(iv) Assume now that

Q is a Sylow p-subgroup. Then it follows that $\displaystyle Q = gPg^{-1}$. Thus every Sylow p-subgroup is conjugate to P.

I don't realy understand any of it yet but I am stuck on the p does not divide r part