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The normalizer of the normalizer of a p-sylow supgroup

  1. Oct 29, 2009 #1
    Im trying to prove N(N(P)) = N(P)

    So N(P) = set oh h where h^-1Ph = p

    Then N(N(P)) = k where k^-1hk = h

    the fact that p is a p sylow subgroup gives me what information? I am unsure.

    Thanks in advance!
     
  2. jcsd
  3. Oct 29, 2009 #2

    MathematicalPhysicist

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    Gold Member

    Well I guess if you can prove that one is a subset of the other, and both of them have the same cardinality, then they are equal.

    Do you see how you can prove this?
     
  4. Oct 29, 2009 #3
    Not really sorry. Both are subgroups of G yes?

    If I let H = the normalizer of P
    Can I say P is normal in H and that as all sylow subgroups are conjugate H contains no other sylow subgroups then if N(N(P)) moves P somewhere else then there would be 2 P sylow subgroups in H?

    Note: Havent been told that P is normal in H


    Thanks very much for you reply
     
    Last edited: Oct 29, 2009
  5. Oct 29, 2009 #4
    If P is the only subgroup of H, then P is normal in H
     
  6. Nov 2, 2009 #5
    Thanks! So I can do it this way then with that result.
     
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