# Thread: Part of Sylow Theorems

1. ## Part of Sylow Theorems

I am trying to prove, or understand the proof of, the following part of the sylow theorems (according to my notes anyway):

Let G be a finite group and let p be a prime which divides the order of G. Then

Every p-subgroup of G is contained in a Sylow p-subgroup of G.

Proof: Let P be a Sylow p-subgroup. Let Q be any p-subgroup. Let $\mathcal{P}$ be the set of all conjugates of P. Then G acts on $\mathcal{P}$ by conjugation. By the orbit stabiliser theorem, the number of conjugates of P is r = | $\mathcal{P}$| = [G: $N_G(P)$ ]. Note that p does not divide r.

but I don't understand why p does not divide r! could anyone help me with out with this? thanks very much

2. Originally Posted by slevvio
I am trying to prove, or understand the proof of, the following part of the sylow theorems (according to my notes anyway):

Let G be a finite group and let p be a prime which divides the order of G. Then

Every p-subgroup of G is contained in a Sylow p-subgroup of G.

Proof: Let P be a Sylow p-subgroup. Let Q be any p-subgroup. Let $\mathcal{P}$ be the set of all conjugates of P. Then G (P?) acts on $\mathcal{P}$ by conjugation. By the orbit stabiliser theorem, the number of conjugates of P is r = | $\mathcal{P}$| = [G: $N_G(P)$ ]. Note that p does not divide r.

but I don't understand why p does not divide r! could anyone help me with out with this? thanks very much
Lemma . If a group H of order $p^n$ ( p prime) acts on a finite set $\mathcal{P}$ and if $S_0 = \{x \in \mathcal{P} | hx = x \text{ for all } h \in H\},$ then $|\mathcal{P}| = |S_0| (\text{mod p})$ (Hungerford, p93).

Let $\mathcal{P}$ be the set of all conjugates of P and let P acts on $\mathcal{P}$ by conjugation. An element T (sylow P-subgroup) in $\mathcal{P}$ satisfies $T \in S_0$ in the above lemma iff $xTx^{-1} = T$ for all $x \in P$. We see that $P < N_G(T)$. We also know that T is normal in $N_G(T)$. By the second Sylow theorem, P and T should be conjuate in $N_G(T)$. This forces T= P and $|S_0| = 1 \text{ mod (p) }$. Since $|\mathcal{P}| = [G: N_G(P)]$ and p does not divide $|S_0|$, we conclude that $p \nmid \mathcal{P}$.

3. 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 $p^ms$, 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 $p^m$.

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 $\mathcal{P}$ be the set of all
conjugates of P. Then G acts on $\mathcal{P}$ by conjugation. By the orbit–stabilizer theorem, the number
r of conjugates of P is $[ G : N_G(P) ]$. Note that $p \not| r$. Then Q acts on $\mathcal{P}$ 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
$p^0$ = 1. So there is an orbit of size 1. Let $gPg^{-1}$ element in an orbit of size 1. Then Q normalizes $gPg^{-1}$. Therefore $gPQg^{-1}$ is a p-group. Hence $gPQg^{-1} = gPg^{-1}$ and so $Q \subseteq gPg^{-1}$.
(iv) Assume now that
Q is a Sylow p-subgroup. Then it follows that $Q = gPg^{-1}$. 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

4. Originally Posted by 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 $p^ms$, 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 $p^m$.

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 $\mathcal{P}$ be the set of all
conjugates of P. Then G acts on $\mathcal{P}$ by conjugation. By the orbit–stabilizer theorem, the number
r of conjugates of P is $[ G : N_G(P) ]$. Note that $p \not| r$. Then Q acts on $\mathcal{P}$ 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
$p^0$ = 1. So there is an orbit of size 1. Let $gPg^{-1}$ element in an orbit of size 1. Then Q normalizes $gPg^{-1}$. Therefore $gPQg^{-1}$ is a p-group. Hence $gPQg^{-1} = gPg^{-1}$ and so $Q \subseteq gPg^{-1}$.
(iv) Assume now that
Q is a Sylow p-subgroup. Then it follows that $Q = gPg^{-1}$. 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
In case G acts on $\mathcal{P}$ by conjugation, there is only one orbit in $\mathcal{P}$ under G such that $r | | G |$, where $r = [ G : N_G(P) ] = |\mathcal{P}|$ and $|G|=p^ms$ , p does not divide s. Here, r cannot be equal to p and $r | p^ms$, so $p \nmid r$.