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Quotient Group is isomorphic to the Circle Group

  1. Dec 2, 2012 #1
    A portion of a homework problem was given me to solve for practice. I have solved some but not all of the homework problem and I hope you all can help.

    Here is the problem:

    1. For each x [tex]\in[/tex] R it is conventional to write cis(x) = cos(x) + i sin(x). Prove that cis(x+y) = cis(x) cis(y).

    Let x, y [tex]\in[/tex] R. We want to prove that cis(x+y) = cis(x) cis(y). Thus,
    cis(x+y) = cos(x+y) + i sin(x+y)
    = (cos(x) cos(y) - sin(x)sin(y)) + i(cos(x)sin(y) + sin(x)cos(y))
    = cos(x) cos(y) - sin(x) sin(y) + i sin(x) cos(y) + i sin(y) cos(x)
    = (cos(x)+ i sin(x))(cos(y) + i sin(y))
    = cis(x) cis(y)

    2. Let T designate the set {cis(x) : x [tex]\in[/tex] R}, that is, the set of all the complex numbers lying on the unit circle, with the operation of multiplication. Use part 1 to prove that T is a group.

    3. Use the FHT to conclude that T isomorphic R/<2[tex]\pi[/tex]>

    4. Prove that g(x) = cis(2[tex]\pi[/tex]x) is a homomorphism from R onto T, with kernel Z

    Let g: R -> T by g(x) = cis(2[tex]\pi[/tex]x).
    g is subjective since ever element of T is of the from cis(a) = cis(2[tex]\pi[/tex](a/2[tex]\pi[/tex])) = g(a/2[tex]\pi[/tex]) for some a [tex]\in[/tex] R. The kernel of g is the set of x [tex]\in[/tex] R such that cis(2[tex]\pi[/tex]x) = 1. This equation only holds true if and only if 2[tex]\pi[/tex]x = 2[tex]\pi[/tex]k for some k [tex]\in[/tex] Z. Divide by 2[tex]\pi[/tex] then you get ker(g) = Z. Hence, g is a homomorphism with a kernel of Z

    5. Conclude that T is isomorphic R/Z

    By the FHT g: R -> T is a homomorphism and R is subjective to T. Since Z is the kernel of g then H is isomorphic to R/Z


    I really have no clue how to do 2 and 3 so any help would be great on that. Also if you can verify that my other three are correct that would be great. Thanks for the help in advance
     
  2. jcsd
  3. Dec 2, 2012 #2
    for #2 you need to verify the group axioms.

    pick [tex] z,w,v \in T [/tex]

    you need to show:

    1. [tex] zw \in T [/tex]

    by the first problem you have proved... we have

    [tex] zw=cis(x)cis(y)=cis(x+y) \in T [/tex]

    cis(x+y) is in T by definition since x+y is in R

    2. [tex] (zw)v=z(wv) [/tex]

    not hard to show... (since T is a subset of the complex numbers)

    3. there exists a multiplicative identity in T

    ...its 1( =cis(0) ) !

    4. there exists an inverse for z in T

    [tex] z \in T \Rightarrow |z|=1 [/tex]

    use [tex] \frac{1}{z} = \frac{\bar z}{ |z|^2} [/tex]

    and |z|=1 to show that 1/z is in T.
     
  4. Dec 2, 2012 #3
    For #3 define:

    Note that R here is treated as an additive group!

    [tex] f: R \rightarrow T [/tex]

    by [tex] x \rightarrow cis(x) [/tex]

    problem 1 shows its an additive homomorphism

    you need to show that the kernel of f is [tex] < 2 \pi > [/tex]

    do you know how to do this?
     
  5. Dec 2, 2012 #4
    Not really!! I have taught myself this whole course
     
  6. Dec 2, 2012 #5
    What does this mean?
     
  7. Dec 2, 2012 #6
    [tex] 1=cis(0) \in T [/tex]

    the multiplicative identity in T is the number 1 which is e^0=cos0+isin0=cis(0)
    1 is in T since 1=cis(0) and 0 is a real number
     
  8. Dec 2, 2012 #7
    Take an arbitrary element of [tex] <2 \pi > [/tex] and show it is in the kernel.

    Next take an arbitrary element of the kernel and show it is in [tex] <2 \pi > [/tex]

    This is a technique called double containment and it is how we normally show that sets are equal.
     
  9. Dec 2, 2012 #8

    micromass

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