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Thread: G cannot be the union of conjugates

  1. #1
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    G cannot be the union of conjugates

    If $\displaystyle G $ is a finite group and $\displaystyle H$ is a subgroup of $\displaystyle G$ then prove that $\displaystyle \displaystyle G \neq \bigcup\limits_{a \in G} aHa^{-1}$
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    If $\displaystyle G $ is a finite group and $\displaystyle H$ is a subgroup of $\displaystyle G$ then prove that $\displaystyle \displaystyle G \neq \bigcup\limits_{a \in G} aHa^{-1}$
    Of course, we assume H is a PROPER subgroup.

    Let G act by conjugation on the set X of all its proper subgroups; then we get that

    $\displaystyle s:=|Orb(H)|=[G:N_G(H)]\leq [G:H]=r$ , say, and since $\displaystyle |aHa^{-1}|=|H|\,\,\,\forall\,a\in G$ ,we get that:

    $\displaystyle |\bigcup aHa^{-1}|\leq 1+s(|H|-1)\leq 1+r(|H|-1)=1+r|H|-r=|G|-(r-1)<|G|$

    (Question: why $\displaystyle |\bigcup aHa^{-1}|\leq 1+s(|H|-1)$ ??)

    Tonio
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