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Math Help - group of Möbius transformations isomorphic to S3 group

  1. #1
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    group of Möbius transformations isomorphic to S3 group

    I have problem with understanding the following problem. Show that the group of Möbius transformations is isomorphic to S_{3}
    We are also supposed to identify involutions i.e elements of order 2.
    The hint to the problem is :show that the group permutes the elements 0,1,\infty What isomorphism can be constructed here?

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  2. #2
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    Re: group of Möbius transformations isomorphic to S3 group

    Which Mobius transformations, explicitly, are you talking about? The general set of Mobius functions is infinite, you must have a particular 6 in mind....I suspect you mean:

    f(x) = x
    g(x) = 1/x (this is an involution)
    h(x) = 1 - x (this is also clearly an involution)
    k(x) = x/(x - 1) (this is not so obviously an involution)
    m(x) = 1/(1 - x)
    n(x) = (x - 1)/x

    to construct an isomorphism with S3, follow the hint: do the involutions fix one of {0,1,∞}? Which functions correspond to 3-cycles?
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    Re: group of Möbius transformations isomorphic to S3 group

    Quote Originally Posted by Deveno View Post

    to construct an isomorphism with S3, follow the hint: do the involutions fix one of {0,1,∞}? Which functions correspond to 3-cycles?
    the id function f(x)=x fixes all these 3 elements 0,1,\infty

    g(x)=\frac{1}{x} fixes 1, \infty ?

    I still do not know how to construct a function that will be bijective and will give me my isomorphism..
    Last edited by rayman; September 8th 2013 at 07:25 AM.
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    Re: group of Möbius transformations isomorphic to S3 group

    No, g(x) = 1/x does NOT fix ∞, it "swaps" 0 and ∞. If we ORDER our 3-element set like so:

    {0,1,∞}

    then it stands to reason that g acts on this set the same way the transposition (1 3) acts on the (ordered) set {1,2,3}.

    Clearly the identity is f. So if we call our bijective mapping φ: we should have:

    φ(f) = id
    φ(g) = (1 3)

    Any thoughts on how to continue?
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