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Showing this is surjective

  1. Nov 21, 2011 #1
    I have a mapping A_g:G ---> G defined by
    A_g(x) = g^-1(x)g (for all x in G)

    and as part of showing it is an automorphism i need show it is surjective.

    I'm not entirely sure how to do this but have made an attempt and would appreciate and feedback or hints to what I actually need to show. I know the definition of surjectivity and also that a mapping is surjective iff Im(of mapping) = G

    My attempt:
    It is surjective since if x is in G, then gxg^-1 is in G and then A_g(gxg^-1) = (g^-1)gx(g^-1)g = idxid = x.

    Thanks in advance
     
  2. jcsd
  3. Nov 22, 2011 #2

    CompuChip

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    Yep, surjective just means that every element has a pre-image, and you have shown that by writing down the pre-image explicitly.
     
  4. Nov 22, 2011 #3
    thanks
     
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