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Math Help - Urgent - normal subgroup proof

  1. #1
    pkr
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    Urgent - normal subgroup proof

    Stuck on the following question;

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  2. #2
    Super Member Gamma's Avatar
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    i \Rightarrow ii
    Ng=gN \Rightarrow g^{-1}Ng=N (multiply on the left by g inverse)

    In particular this means
    g^{-1}Ng \subset N

    that is precisely the statement of ii

    ii \Rightarrow i
    Assume the hypothesis. For any g, it is clear that we have g^{-1}Ng \subset N \Rightarrow Ng \subset gN.

    But then we also have
    (g^{-1})^{-1}Ng^{-1}\subset N \Rightarrow gNg^{-1} \subset N \Rightarrow gN \subset Ng

    Thus gN=Ng as desired.
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