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Help with Normalizers

  1. Feb 15, 2008 #1
    For a subgroup H of G and a fixed element a ∈ G,
    let H^a = {x∈ G / axa^-1 ∈ H}, it's normalizer N(H) = {a∈G / H^a=H}

    Show that for any subgroup H of G, N(H) is a subgroup of G.

    I know that for the first one I need to show that closure holds, an identity exists, and inverses exist. But I don't even know where to start with closure!!!
  2. jcsd
  3. Feb 16, 2008 #2


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    If a and b are in N(H), write down what H^(ab) is. Now maybe you can try showing that [itex]H^{ab} \subset H[/itex] and [itex]H \subset H^{ab}[/itex].
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