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Thread: prime index

  1. #1
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    prime index

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

    Proof:

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

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

    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 $\displaystyle H\leq G$ whose index is the least prime dividing the order of G, then $\displaystyle H\triangleleft G$.

    Proof:

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

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

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


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