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Math Help - prime index

  1. #1
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    prime index

    Prove that if G is a finite group and H\leq G whose index is the least prime dividing the order of G, then H\triangleleft G.

    Proof:

    This is all I could come up with-
    |G|=|G:H|\times |H|, |G:H|=p, and let |H|=m.

    Now, |G|=pm and p| \ |G|.

    Now what do I do?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: prime index

    Quote Originally Posted by dwsmith View Post
    Prove that if G is a finite group and H\leq G whose index is the least prime dividing the order of G, then H\triangleleft G.

    Proof:

    This is all I could come up with-
    |G|=|G:H|\times |H|, |G:H|=p, and let |H|=m.

    Now, |G|=pm and p| \ |G|.

    Now what do I do?
    Here is the classic way to do this problem. Let G act on G/H (the coset space, not the quotient group) in the usual way. This gives a homomorphism \phi:G\to S_{[G:H]}. It's trivial that \ker\phi\leqslant H. That said, note that since [G:\ker\phi] divides |G| it cannot be divisible by any smaller prime than [G:H] yet since [G:\ker\phi] divides |S_{[G:H]}|=[G:H]! (by FIT) it clearly then follows that [G:\ker\phi]=[G:H] and so |\ker\phi|=|H|. Since \ker\phi\leqslant H this implies that \ker\phi=H and so H\unlhd G.


    A more interesting proof, if you know double cosets, can be found here on my blog.
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