Is there a mistake in the proof for G being isomorphic to R(theta)?

[Nusc]

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?

 

[matt grime]

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).

 

[HallsofIvy]

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?

I don't think that's what you meant to say. The group of rotations consists of all matrices of the form
$$\left[\begin{array}{cc}cos \theta & -sin \theta \\ sin \theta & cos \theta\end{array}\right]$$
It is not required that two matrices be equal nor is there any x or y.
You know that $e^{i\pi\theta}e^{i\pi\phi}= e^{i\pi(\theta+\phi)}$. 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]?<br /> Now apply the sine and cosine sum formulas:<br /> $$sin(\theta+ \phi)= cos(\theta)sin(\phi)+ sin(\theta)cos(\phi)$$<br /> $$cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)$$[/tex]

 

[Nusc]

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?

 

[matt grime]

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).