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Sylow p-subgroups

  1. Mar 4, 2008 #1
    [SOLVED] Sylow p-subgroups

    1. The problem statement, all variables and given/known data
    Let G be a finite group and let primes p and q \neq p divide |G|. Prove that if G has precisely one proper Sylow p-subgroup, it is a normal subgroup, so G is simple.

    EDIT: that should say "G is not simple"


    2. Relevant equations



    3. The attempt at a solution
    I don't see the point of q. If G has precisely one proper Sylow p-subgroup, then you can conjugate with all the elements of the group and you cannot get of the subgroup or else you would have another Sylow-p-subgroup, right? So, it must be normal, right?
     
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2
    anyone?
     
  4. Mar 4, 2008 #3

    NateTG

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    Yes. (Is it possible that you misread the question - if there is a normal subgroup, the group can't be simple.)
     
    Last edited: Mar 4, 2008
  5. Mar 4, 2008 #4
    See the EDIT. But am I right about the part about q being unnecessary? You can induce the existence of q by the fact there is a proper Sylow p-subgroup, right?
     
  6. Mar 4, 2008 #5

    morphism

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    If |G| doesn't have another prime divisor, then we can't say for sure that G isn't simple (e.g. if |G|=p^2).
     
  7. Mar 4, 2008 #6
    But it must have another prime divisor if it has a proper p-Sylow subgroup.
     
  8. Mar 4, 2008 #7

    morphism

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    Ah, you're right. Missed the word proper.
     
  9. Mar 5, 2008 #8

    NateTG

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    Yep.
     
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