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Thread: Orders of simple groups and groups conjugate to normal subgroups

  1. #1
    Newbie diehardwalnut's Avatar
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    Orders of simple groups and groups conjugate to normal subgroups

    Let $G$ be a simple group and let $H$ be a proper subgroup of $G$. If $p$ is the largest prime that divides $G$ and $k$ is the number of subgroups of $G$ conjugate to $H$, prove $p \leq k$. A nudge in the right direction would be appreciated as I don't know how to start.
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  2. #2
    MHF Contributor
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    Re: Orders of simple groups and groups conjugate to normal subgroups

    Prove the following and you're done:

    1. Let $N=N_G(H)=\{g\in G\,:\,gHg^{-1}=H\}$. Then the number k of conjugates of H in G is the index of N in G; i.e. the number of left cosets of N in G.

    2. The Cayley representation $\alpha$ of G on the left cosets of N is defined by: if $g\in G$ and $xN$ is a left coset, $\alpha(g)(xN)=gxN$. Then $\alpha$ is a homomorphism of G into the full symmetric group of k objects (cosets).

    3. If $k<p$, $p$ can't divide $k!$ and so $p$ must divide the order of Ker($\alpha$), contradiction.
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