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Normalizer subgroup proof proving the inverse

  1. Sep 17, 2009 #1
    This specifically relates more towards the argument as to why an inverse exists.

    First the problem

    The normalizer is defined as follows, NG(H)={g-1Hg=H} for some g in NG(H). I get why identity exists and why the operation is closed. It is in arguing that an inverse exists that I have beef. Specifically this argument:

    eHe= (g-1)-1g-1Hgg-1=(g-1)-1Hg-1

    so g-1 is in NG(H)

    This above proof I found in the book Groups, rings, and fields by D.A.R. Wallace.

    I wondering why this is more valid than this:


    So g-1 is in NG(H)

    I'm sorry if this question is so dense someone has an aneurysm
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 18, 2009 #2
    I have another proof, but I'm not sure why this is any more valid. It follows the same route as the first one I presented up to this point:

    This is equivalent to

    Therefore g^{-1} is in N_G(H)
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