I need to show that (C-{0},*)/{x+yi|x^2+y^2=1} is isomorphic to (R+,*)

[catherinenanc]

1. I need to show that the quotient group G/ℝ+ is isomorphic to U using the Fundamental Theorem on group homomorphisms. I believe I can get there if someone can just help me find what ψ, the homomorphism function is supposed to be.
G=(ℂ-{0},*)
U={x+yi|x^2+y^2=1}

Homework Equations

Fundamental Theorem on group homomorphisms

The Attempt at a Solution

Note that ℝ⁺ is the subset of G where for any (x+yi)∈G, y=0.
First, we need to find a homomorphism ϕ:G→U such that ker(ϕ)=ℝ⁺.
Need to find: a,b such that (x+yi)⋅(a+bi)=(x+yi) given x²+y²=1 and a²+b²=1.
(x+yi)⋅(a+bi)=ax+bxi+ayi-by=(ax-by)+(bx+ay)i=(x+yi)
We also know (ax-by)²+(bx+ay)²=1.
Then (ax-by)=x and (bx+ay)=y.
So, a=1 and b=0, so e_{U}=1+0i.
Therefore, ϕ(x+yi)=1+yi, since only y=0 will give you 1+0i, which is e_{U}, which makes R⁺ be the ker(ϕ).
This ϕ is not onto.
ϕ(x+yi) is a homomorphism since for any a+bi,c+di∈G, ϕ((a+bi)⋅(c+di))=ϕ((ac-bd)+(ad+bc)i)=1+(ad+bc)i≠***(1-bd)+(b+d)i=(1+bi)⋅(1+di)=ϕ(a+bi)⋅ϕ(c+di). ***this is not equal
Then, by Thm 13.2 (Fundamental Theorem on group homomorphisms), ϕ(G)=G/ker(ϕ). So, (1+yi)≅G/R⁺.
****I went wrong somewhere since it's supposed to be U, not (1+yi).

 

[rasmhop]

Note that ℝ⁺ is the subset of G where for any (x+yi)∈G, y=0.

and x > 0 I assume. At least if you use standard notation.

First, we need to find a homomorphism ϕ:G→U such that ker(ϕ)=ℝ⁺.

This is the right idea. Then show it is onto (surjective).

Need to find: a,b such that (x+yi)⋅(a+bi)=(x+yi) given x²+y²=1 and a²+b²=1.
(x+yi)⋅(a+bi)=ax+bxi+ayi-by=(ax-by)+(bx+ay)i=(x+yi)
We also know (ax-by)²+(bx+ay)²=1.
Then (ax-by)=x and (bx+ay)=y.
So, a=1 and b=0, so e_{U}=1+0i.

I don't understand why you try to find such a,b. Also since you know \mathbb{C}\setminus\{0\} (or U) is a group it is immediate that the only such a,b is (1,0) since you may just cancel out (x+iy).

Therefore, ϕ(x+yi)=1+yi, since only y=0 will give you 1+0i, which is e_{U}, which makes R⁺ be the ker(ϕ).
This ϕ is not onto.

and it is not a group homomorphism, nor is it a function to U.Try to see if you can think of a simpler way to construct a group homomorphism. Remember that G is the punctured complex plane and U is the unit circle. Do you happen to know a standard way to associate a unit vector to any non-zero vector? Try that for your \varphi.

 

[catherinenanc]

"Try to see if you can think of a simpler way to construct a group homomorphism. Remember that G is the punctured complex plane and U is the unit circle. Do you happen to know a standard way to associate a unit vector to any non-zero vector? Try that for your φ."

Ok, so you're saying that my homomorphism could be ψ(x+yi)=(x+yi)/√(x^2+y^2), right?
I haven't checked that that's a homomorphism, but assuming it is, that would be onto!

 

[catherinenanc]

Sorry, it won't let me use the right symbol, so for now I'm using ψ.

 

[rasmhop]

That is a good idea. I would however suggest writing
|z| = \sqrt{x^2+y^2}
because then you can use identities like |zw|=|z||w| and your group homomorphism is just z \mapsto z/|z| (this is of course the exact same thing just written in more concise notation).

The reason why we might suspect that it is a homomorphism is that multiplication of complex numbers just adds their angles and multiply their magnitudes. The map you proposed would simply consist of forgetting magnitudes. Of course this is not a formal argument, but it suggests why it should be true.

 

[catherinenanc]

Thank you so much! I'm pretty sure I have it now. You are awesome!

 

[Deveno]

i am confused.

the title of the thread says that you need to show that

C*/U ≅ R⁺ (where multiplication is the group operation for both groups)

which is true, but the proper homomorphism has not been discussed here.

subsequent discussion has focused on trying to show that:

C*/R⁺ ≅ U which is not quite the same thing.

 

[catherinenanc]

You are correct, sorry. I must have been looking at part a of my homework problem when I wrote the title and part b when I wrote out the problem. I had done part a already, as it was the easier of the two, and thanks to this thread, I got part b as well!