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Homework Help: Show isomorphism for element g in group G

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.


    2. Relevant equations



    3. The attempt at a solution
    not even really understanding the question. can someone break it down for me, and explain what the question is asking? am i trying to find a function, or rather show that i(x) preserves the structure?
     
    Last edited: Feb 16, 2010
  2. jcsd
  3. Feb 16, 2010 #2
    Re: isomorphism

    Does it specify what is F? Or is g an element of G?

    Also, does it specify what exactly is g' in ig(x) = gxg'?
     
  4. Feb 16, 2010 #3
    Re: isomorphism

    i'm sorry, it's supposed to be:
    let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.
    g' is the inverse of g, i think.
     
  5. Feb 16, 2010 #4
    Re: isomorphism

    I am assuming that g is an element of G and that g' is the inverse of g.

    The problem is giving you a function and asking you to show that it is an isomorphism. To show that it is an isomorphism, you must show it is a bijection and that for all x, y in G, ig(xy) = ig(x)ig(y)
     
  6. Feb 16, 2010 #5
    Re: isomorphism

    so for the latter part:
    ig(xy)=gxyg'
    =gxg'gyg'
    =ig(x)*ig(y)
    is that right?
     
    Last edited: Feb 16, 2010
  7. Feb 16, 2010 #6
    Re: isomorphism

    and for the bijection:

    injection
    let ig(x) = ig(y) for x,y in G. then
    gxg'=gyg'
    g'gxg'=g'gy'g inverses
    xg'=yg'
    x=y by right cancellation

    surjection
    let x,y in G. then
    y=gxg'
    yg=gxg'g
    yg=gx
    g'yg=g'gx
    g'yg=x

    is this right? i kind of feel like i'm using normal multiplication to do this, ignoring commutativity.
     
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