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Normalizers an p-Sylow subgroups

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Let G be a finite group, and H be a normal subgroup of G. Let B be a p-Sylow subgroup of H, for some p dividing |H|. Show that G=HN_G(B).

    (N_G(B) is the normalizer of B in G, that is the biggest subgroup of G which contains B and B is normal in it. Equivalently N_G(B)={g\inG|gBg^-1=B})

    2. Relevant equations

    The Sylow theorems.

    3. The attempt at a solution

    To tell the truth I'm pretty stumped with this question. I know that the index of the normalizer is the number of p-Sylow subgroups, but B is a p-Sylow subgroup of H, and the normalizer is in G. I also know that B must be the only p-Sylow subgroup of its normalizer. I can't see how this adds up to a solution. I also tried considering a left action of H on G, but things seem to abstract to get something from that.
  2. jcsd
  3. Dec 5, 2008 #2


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    Homework Helper

    This is a classical result, called Frattini's argument.

    The proof is pretty straightforward. So if you don't want to look it up, here's a hint: take a g in G and consider g-1Bg. There are two important things we can say about this - what are they?
  4. Dec 5, 2008 #3
    well, first, since H is normal, gBg^-1 is a subgroup of H.
    second, it's also a p-Sylow subgroup of H, so it's conjugate to B.

    Oh! so it's also of the form hBh^-1!
    great, I'll take it from here. thanks!
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