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Math Help - Automorphisms of a group G

  1. #1
    Super Member Bernhard's Avatar
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    Automorphisms of a group G

    Dummit and Foote Section 4.4 Automorphisms Exercise 1 reads as follows:

    Let \sigma \in Aut(G) and \phi_g is conjugation by g, prove that \sigma \phi_g \sigma^{-1} = \phi_{\sigma (g)}.

    ================================================== ====

    A start to the proof is as follows:

    \sigma \phi_g \sigma^{-1}

    = \sigma (\phi_g (\sigma^{-1} (x)))

    = \sigma (g. \sigma^{-1} (x). g^{-1} )

    = \sigma (g). x . \sigma (g^{-1} )

    Now we have completed the proof if \sigma (g^{-1}) = ( {\sigma (g) )}^{-1}

    But why is \sigma (g^{-1}) = ( {\sigma (g) )}^{-1} in this case??

    Can anyone please help clarify this situation?

    Peter
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  2. #2
    Super Member Bernhard's Avatar
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    Re: Automorphisms of a group G

    I have now thought about this and suspect that the answer to my own question is as follows:

    For an automorphism we have \phi (e) = e

    Thus \phi (e) = \phi (g g^{-1}) = \phi (g) \phi( g^{-1} ) = e

    and from this it follows that \phi (g^{-1}) = {[\phi (g)]}^{-1}

    Am I correct?

    Peter
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Re: Automorphisms of a group G

    Quote Originally Posted by Bernhard View Post
    I have now thought about this and suspect that the answer to my own question is as follows:

    For an automorphism we have \phi (e) = e

    Thus \phi (e) = \phi (g g^{-1}) = \phi (g) \phi( g^{-1} ) = e

    and from this it follows that \phi (g^{-1}) = {[\phi (g)]}^{-1}

    Am I correct?

    Peter
    Yes, that is correct.
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