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Sylow theorem part 2

  1. Apr 30, 2007 #1
    okay so I'm going through the proof of sylow part 2, i.e. the bit that says if
    [tex] N_p is the number of subgroups of G of order p^n then N_p \equiv1modp

    now I have got to the part where I have taken the subgroup P of order P^n that you get from sylow part 1 and I have to show that it is a unique fixed point W.r.t conjugation of P on S={Q: Q subgroup of G: |Q|=p^n}....anyway for those of you who know the proof, I am trying to find the order of the subgroup PQ, where Q is the supposed subgroup used in the contradiction argument......

    so anyway |PQ| is what I am trying to deduce, I have noethers 1st isomorphism theorem

    [tex] PQ/Q isomorphic to P/P \capQ Q [/tex]


    the next line asserts [tex] |P\cap Q| =p^e for som e e\leq n [/tex]

    how is this so is P intersection Q a subgroup of P??:confused:
  2. jcsd
  3. Apr 30, 2007 #2


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    You really should explain what you're trying to do more clearly since we haven't all seen the particular proof you're going from. I think I've worked out what you're doing, and even though you probably know all this, I'll go over it again in case I'm wrong, or if someone else wants to follow along.

    We have |S|=np, and if we let P act on S by conjugation, we get a partition of S into distinct orbits. One of these will contain P by itself, since conjugation of P by any element in P gives back P.

    For any other orbit O, say, let Q be a representative from O. Then we have [itex]|O|=[P:N_G(Q) \cap P ][/itex] (do you see why?), and since |P| is a power of p, |O| must be as well.

    If |O|=1, this would mean P is contained in [itex]N_G(Q)[/itex]. Then the first isomorphism theorem says:

    [tex] PQ/Q \cong P/P \cap Q [/tex]

    or, taking the orders, using the fact that |G/H|=|G|/|H| for finite groups, and rearranging, we get:

    [tex] |PQ| = \frac{|P||Q|}{|P \cap Q|} [/tex]

    (note this is also true from a more basic argument even when PQ isn't a subgroup).

    Since all groups on the RHS have order a power of p (since, yes, [itex]P \cap Q[/itex] is a subgroup of P, which I hope isn't your whole question, because it's the easiest part of this proof), so does PQ. But by assumption, P (as well as Q) is a maximal p-subgroup, so we must have PQ=P, ie, P=Q.

    To finish up, since np is the sum of the sizes of the different orbits, and since exactly one of these is 1, with all the others are divisible by p (since they are powers of p greater than 1), we see np=1 (mod p).
    Last edited: Apr 30, 2007
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