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

  1. Feb 28, 2012 #1
    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}

    2. Relevant equations
    Fundamental Theorem on group homomorphisms

    3. 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).
     
  2. jcsd
  3. Feb 28, 2012 #2
    and x > 0 I assume. At least if you use standard notation.

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

    I don't understand why you try to find such a,b. Also since you know [itex]\mathbb{C}\setminus\{0\}[/itex] (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).

    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 [itex]\varphi[/itex].
     
  4. Feb 28, 2012 #3
    "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!
     
  5. Feb 28, 2012 #4
    Sorry, it won't let me use the right symbol, so for now I'm using ψ.
     
  6. Feb 28, 2012 #5
    That is a good idea. I would however suggest writing
    [tex]|z| = \sqrt{x^2+y^2}[/tex]
    because then you can use identities like |zw|=|z||w| and your group homomorphism is just [itex]z \mapsto z/|z|[/itex] (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.
     
  7. Feb 28, 2012 #6
    Thank you so much! I'm pretty sure I have it now. You are awesome!
     
  8. Feb 29, 2012 #7

    Deveno

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    Science Advisor

    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.
     
  9. Feb 29, 2012 #8
    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!
     
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