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Is that mapping the homomorphism?

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex]G[/itex] be a group and let [itex]Aut(G)[/itex] be the group of automorphisms of [itex]G[/itex].

    (a) For any [itex]g \in G[/itex], define [itex]\phi_{g}(x) = g^{-1}xg[/itex]. Check that [itex]\phi_{g}(x)[/itex] is an automorphism.

    (b) Consider the map:

    [itex]\Phi:G \rightarrow Aut(G)[/itex]
    [itex]g \mapsto \phi_{g}[/itex]

    Check that [itex]\Phi[/itex] is a homomorphism.

    2. The attempt at a solution

    Evaluate [itex]\Phi[/itex] at gh with arbitrary x, so:

    [itex]\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}g^{-1}xgh = \phi_{h}(\phi_{g}(x))[/itex]
    [itex]= \phi_{h} \circ \phi_{g} (x) = \Phi(h) \circ \Phi(g)[/itex]

    But the operation is reversed for this situation. So this is considered to be an antihomomorphism.

    Any comments?
     
  2. jcsd
  3. Apr 27, 2013 #2

    Fredrik

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    In question (a), you probably meant "check that ##\phi_g## is an automorphism". Do you need help with (a)? You have only shown your work for (b).

    At the far left, you need to write ##\Phi(gh)(x)=\phi_{gh}(x)=##. The x appears out of nowhere in your calculation, and then it magically disappears at the end. So you need to do something similar at the end.

    I agree that we end up with ##\Phi(gh)=\Phi(h)\circ\Phi(g)##.
     
  4. Apr 27, 2013 #3

    Office_Shredder

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    The multiplication on the Automorphism group is sometimes written backwards to accomodate this (for example on the wikipedia page about this homomorphism) - where
    [tex] \left(\phi_{g} \phi_{h}\right)(x) = \phi_h \left( \phi_g(x) \right) [/tex]
    so you should check to make sure that multiplication is composition read from right to left

    If this isn't what it's supposed to be then probably there's a type and the homomorphism is supposed to be
    [tex] g \mapsto \phi_{g^{-1}} [/tex]
     
  5. Apr 27, 2013 #4
    Nope, don't need help with part (a). Sorry to indicate that beforehand.
     
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