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Math Help - proving the sylow p-subgroups are abelian

  1. #1
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    proving the sylow p-subgroups are abelian

    Assume that p is an odd prime and G is a finite simple group with exactly 2p+1 Sylow p-subgroups. Prove that the Sylow p-subgroups are abelian.

    I haven't made much progress: It's easy if p^2 doesn't divide the order of G, since then the Sylow p-subgroups have order p and are therefore cyclic.

    Thanks,
    Hollywood
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  2. #2
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    Re: proving the sylow p-subgroups are abelian

    i figured it out:

    let G act on the sylow p-subgroups by conjugation.

    this gives a homomorphism of G into S2p+1.

    since G is simple, and the action is transitive, the kernel of this action must be trivial.

    therefore |G| divides (2p+1)!.

    now we can ask, what is the highest possible power of p that divides (2p+1)! ?

    well, the only integers 2 ≤ n ≤ 2p+1 that p divides are p and 2p. this means that the highest possible power of p that divides (2p+1)! is p2.

    thus the sylow p-subgroups of G have either order p, or p2, and are abelian in either case.
    Thanks from hollywood
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  3. #3
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    Re: proving the sylow p-subgroups are abelian

    Brilliant!!

    Thanks,
    Hollywood
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  4. #4
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    Re: proving the sylow p-subgroups are abelian

    Deveno - well done!
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