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Math Help - Part of Sylow Theorems

  1. #1
    Senior Member slevvio's Avatar
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    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
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  2. #2
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    Quote Originally Posted by slevvio View Post
    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} .
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  3. #3
    Senior Member slevvio's Avatar
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    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
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  4. #4
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    Quote Originally Posted by slevvio View Post
    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.
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