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Exhibiting a homomorphism f: G-> Aut(G)

  1. Sep 18, 2011 #1
    I'm watching Harvard's abstract algebra course online and the professor says that the map f(g)(h)=ghg-1 is a homomorphism between the groups G and Aut(G). the thing that I don't understand is that whether ghg-1 is an element of Aut(G) or not. It must be an element of Aut(G) because It's in the image of f but I can't figure out how it is an element of Aut(G). Can someone clarify what f(g)(h) means at the first place? if h is the input, then what is f(g)? Is f(g)(h) the same thing as fog(h)? Shouldn't our input be from G and our output in Aut(G)? so if h is in G, Is ghg-1 an automorphism on the group G under its operation? I'm confused :confused:
  2. jcsd
  3. Sep 18, 2011 #2


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    Do you agree that [itex]f_g(h)= ghg^{-1}[/itex] (I prefer that notation so we don't get "g" and "h" confused) is "auto"- that is, that it maps amember of the group to a member of the group? That should be clear because the group is closed under the group operation.

    So the only thing left to prove is that it is a homomorphism: that is, that [itex]f_g(hk)= f_g(h)f_g(k)[/itex].

    Okay, [itex]f_g(h)f_g(k)= (ghg^{-1})(gkg^{-1})[/itex]. What does that reduce to?
    Last edited by a moderator: Sep 18, 2011
  4. Sep 18, 2011 #3
    well, I think the main confusion was about the notation he had used. so, ghg-1 is indeed an automorphism therefore it's in Aut(G). now It's clear that f is a group homomorphism. Thanks for the help HallsofIvy.
  5. Sep 18, 2011 #4
    No, it's not. [itex]ghg^{-1}[/itex] is an element of G, not of Aut(G).
    The map

    [tex]f_g:G\rightarrow G:h\rightarrow ghg^{-1}[/tex]

    is in Aut(G). And the map

    [tex]f:G\rightarrow Aut(G):g\rightarrow f_g[/tex]

    is the homomorphism.
  6. Sep 18, 2011 #5
    Yea, I meant the same thing. Thanks for pointing out the slight difference in the way I had phrased my sentence. I meant ghg-1 defines a set theoretic automorphism on G (It's one-to-one and onto) and fg (the function defined by that relation) satisfies fg(h.h')=fg(h) . fg(h').
    Now as you said, the map [tex]f:G\rightarrow Aut(G):g\rightarrow f_g[/tex] is a homomorphism because we have: fg.g'(h)=(g.g').h.(g.g')-1=(g.g').h.(g')-1g-1=g(g'h(g')-1)g-1=g(fg'(h))g-1=fg(fg'(h))=fgofg'(h).

    the thing is that Benedict Gross first misinterpreted the problem, so I too got confused what he was talking about and lost the path we were following.
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