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A group isomorphism

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex](G,\cdot)[/itex] be a group. Defining the new operation * such that [itex] a*b = b \cdot a [/itex] it is pretty easy to show that [itex](G,*)[/itex] is a group. Show that this new group is isomorphic to the old one.

    2. Relevant equations

    3. The attempt at a solution

    I have been experimenting with the possibility to define an isomorphism [itex] \phi (a*b) = b a [/itex] but can't really seem to get it right. Can anyone give me a hint on how to find the isomorphism?

  2. jcsd
  3. Jul 21, 2011 #2
    Will the conjugate action work?
  4. Jul 21, 2011 #3
    Hm, actually after doing some more exercises I figured out a function that works. I am not quite sure of what you mean by conjugate action but I post my solution here:

    Let [itex] \phi: (G,*) \rightarrow (G, \cdot) [/itex] such that [itex] \phi (a) = a^{-1} [/itex]. Then [itex] \phi(a) \phi(b) = a^{-1} b^{-1} [/itex]. Consider now [itex] \phi(a*b) = (ba)^{-1} = a^{-1} b^{-1} [/itex] and we see that [itex] \phi(a) \phi(b) = \phi(a*b) [/itex].
  5. Jul 21, 2011 #4
    Ooo nice. I was thinking of something like a --> ga(g-1). But that doesnt work.
  6. Jul 21, 2011 #5
    This group is called the opposite group. It is important for studying the relation between left- and right- group actions. Just thought I'd throw that fun fact in there.
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