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Proof of a linear operator acting on an inverse of a group element

  1. Oct 7, 2013 #1
    Hey guys!

    Basically, I was wondering how to prove the following statement. Ive seen it in the Hamermesh textbook without proof, so I wanted to know how you go about doing it.

    Let's say you have a group element [itex]g_{1}[/itex], which has a corresponding inverse [itex]g_{1}^{-1}[/itex]. Let's also define a linear transformation D for this group element.

    So what im trying to prove is that

    [itex]D(g_{1}^{-1}) = [D(g_{1})]^{-1} [/itex]

    Can u guys point me in the right direction?

    Thanks!
     
  2. jcsd
  3. Oct 7, 2013 #2

    Office_Shredder

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    By linear transformation do you mean group homomorphism? If so just consider
    [tex] D(g_1 g_1^{-1}) = D(g_1) D(g_1^{-1}) [/tex]
     
  4. Oct 7, 2013 #3
    yes it is a homomorphism. But what u wrote in the previous message, where do I go from there? cos [itex]D(g_{1}g_{1}^{-1})[/itex] is just [itex]D(E)[/itex] where E is the identity, right? o.o
     
  5. Oct 7, 2013 #4

    Office_Shredder

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    Yes, and what is D(E) equal to?
     
  6. Oct 7, 2013 #5
    OHH I get it. Basically, because [itex]D(E) = E[/itex], we can say that

    [itex]E = D(g_{1})D(g_{1}^{-1})[/itex]
    Then by multiplying both sides on the left by [itex][D(g_{1})]^{-1}[/itex] we get

    [itex][D(g_{1})]^{-1} = D(g_{1}^{-1})[/itex]....I think...
     
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