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Show something's a group

  1. Sep 28, 2006 #1
    Let G = {e^itheta);theta in R}

    Show that G is isomorphic to the group of rotations in the plane given by 2x2 matrices.

    Define phi:G->R(theta)

    1-1: Consider e^ix, e^iy in G

    Assume phi(expix)) = phi(expiy)) and we want to show exp(iy)=exp(iy)

    Group ofrotations R(theta) is the matrix:

    cosx -sinx = cosy -siny
    sinx cosx siny cosy

    But that implies cosx = cosy
    which is not necessarily true.

    What's wrong here?
     
  2. jcsd
  3. Sep 28, 2006 #2

    matt grime

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    Really you're better off saying

    G:= {exp(it) : t in [0,2pi) }

    since exp is periodic on R, thus the elements when theta equals x x+2pi, x+4pi, etc, are all the same in G, and that might be confusing you.

    I see no problem with the fact that those two matrices being equal implies cos(x)=cos(y) and sin(x)=sin(y).
     
  4. Sep 28, 2006 #3

    HallsofIvy

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    I don't think that's what you meant to say. The group of rotations consists of all matrices of the form
    [tex]\left[\begin{array}{cc}cos \theta & -sin \theta \\ sin \theta & cos \theta\end{array}\right][/tex]
    It is not required that two matrices be equal nor is there any x or y.
    You know that [itex]e^{i\pi\theta}e^{i\pi\phi}= e^{i\pi(\theta+\phi)}[/itex]. What is the product
    [tex]\\left[\begin{array}{cc}cos \theta & -sin \theta \\ sin \theta & cos \theta\end{array}\right]\left[\begin{array}{cc}cos \phi & -sin \phi \\ sin \phi & cos \phi\end{array}\right]?
    Now apply the sine and cosine sum formulas:
    [tex]sin(\theta+ \phi)= cos(\theta)sin(\phi)+ sin(\theta)cos(\phi)[/tex]
    [tex]cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)[/tex]
     
  5. Sep 28, 2006 #4
    To show that something is one-to-one, I'm not sure why you multiplied exp(i*pie*theta) with exp(i*pie*phi).

    How can not showing that phi(x) = Phi(b) => a = b, not be required?
     
    Last edited: Sep 28, 2006
  6. Sep 29, 2006 #5

    matt grime

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    It is required. However, it is not at all clear what your confusion is. I still have no idea what you mean by 'that is not necessarily true' in you post. If phi(a)=phi(b), then obvisouly cos(a)=cos(b) and sin(a)=sin(b) which is iff and onl if a=b (mod 2pi, like I said, your G is not very well defined).
     
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