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Math Help - One more normalizer q

  1. #1
    Senior Member sfspitfire23's Avatar
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    One more normalizer q

    If H is a subgroup of G, show that H is normal to the normalizer (N(H))
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    H is clearly a subgroup of N_G(H).

    To see that H \lhd N_G(H), take any x \in N_G(H). By definition, xHx^{-1}=H. Since this is true for all x \in N_G(H), the statement follows.
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