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Proving that Aut(G) is a subgroup of Bij(G)

  1. Apr 16, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let G be a group. An Isomorphism Φ: G --> G is called an automorphism of G. Let Aut(G) denote the set of all automorphisms of G.

    Prove that Aut(G) is a subgroup of Bij(G).

    2. Relevant equations
    For it to be a subgroup we need to show:
    i) e ∈ Aut(G)
    ii) For all x,y ∈ Aut(G), xy∈ Bij(G)
    iii) For all x ∈Aut(G), x^-1 ∈ Aut(G)

    3. The attempt at a solution

    i) Clearly the identity, e: G--> G is an automorphism. This is trivial.

    ii) Φ(xy) = Φ(x) Φ(y) ??
    Let Φ = Φ2⋅Φ1

    Then Φ(xy) = Φ2( Φ1(xy)) = Φ2( Φ1(x) Φ1(y)) = Φ2( Φ1(x)) Φ2( Φ1(y)) = Φ(x) Φ(y)

    iii) Let u,v ∈ G, want to show that Φ^-1(uv) = Φ^1(u)⋅ Φ^-1(v)
    Let x = Φ^-1(u) and y = Φ^-1(v) so that Φ(x) = u and Φ(y) = v.

    Φ^-1(uv) = Φ^-1( Φ(x) Φ(y)) = Φ^-1( Φ(xy)) = xy = Φ^-1(u) Φ^-1(v)

    Thus, Φ^-1 is an automorphism.

    Therefore, by i, ii, iii, Aut(G) is a subgroup of Bij(G).


    Correct? :D
     
  2. jcsd
  3. Apr 16, 2016 #2

    LCKurtz

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    You abandon threads like https://www.physicsforums.com/threads/describe-the-kernel-and-image.866462 and expect us to continue helping you?
    I'm not really interested in this thread other than to note you haven't told us what Bij(G) is. Similar to what you did in this thread:
    https://www.physicsforums.com/threads/describe-the-kernel-and-image.866462
    which you then abandoned. You have been around long enough to know better.
     
  4. Apr 16, 2016 #3

    RJLiberator

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    What's wrong with 'abandoning' threads as before?

    Unfortunately, I still don't know the answer to the previous thread (we didn't go over that particular problem when we did our homework review).

    Bij(G) https://en.wikipedia.org/wiki/Bijection
     
  5. Apr 16, 2016 #4

    andrewkirk

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    Some comments.

    Here you use x,y for maps between group elements
    and later on you use x for a group element, while using ##\Phi_1,\Phi_2## for maps. It will avoid confusion (both for you and for the reader) if you keep consistent notation.

    Here the goal is too weak:
    it needs to be xy∈ Aut(G). That follows very easily from what you've written. But that step, however easy, should not be omitted.

    In your proofs of (ii) and (iii) you have shown that ##\Phi_1\circ\Phi_2## and ##\Phi^{-1}## are homomorphisms. You need to show that they are automorphisms. Given they are bijections, that follows easily, but you still need to show that step.

    Also, you need to show (or at least state) that ##\Phi^{-1}\circ\Phi=e##. This is very easy given what you've already done, but the step needs to be shown.
     
  6. Apr 16, 2016 #5

    jim mcnamara

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    What's wrong with abandoning a thread?
    Good answers are not simply one off every time. @andrewkirk put some effort into his answer. Disjointed or partial nonsense questions require even more: Sorting out the good from the bad and from the egregious.

    A lot of times I only answer part of what is asked to see if the OP takes the hint. Obviously you can't be bothered to read and acknowledge answers.

    You can behave as you choose; you can also find that the only people answering your questions are math newbies and do not know what they're doing. Get the point? Your choice. I think this was what @LCKurtz politely tried to mention. You probably lost someone who knows his material. And he called you on it.
     
  7. Apr 16, 2016 #6

    RJLiberator

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    bow.gif

    Thank you Andrew, that helps complete my understanding with this proof. I now see the connection between general subgroup definition and the definition needed here that I had wrote confusingly.

    Thanks for the advice, but you guys can over analyze as you'd like. Unfortunately, I don't have the time to at the moment. Stuck behind a boatload of homework! Good luck.
     
  8. Apr 17, 2016 #7

    micromass

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    It is extremely extremely rude.

    Very soon you may find nobody willing to answer your threads. I already know of some who deliberately avoid your threads.
     
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