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Isomorphism of quotient group

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Let G be the group of real numbers under addition and let N be the subgroup of G consisting of all the integers. Prove that G/N is isomorphic to the group of all complex numbers of absolute value 1 under multiplication.

    Hint: consider the mapping f: R-->C given by f(x)=e^[2pi(ix)]


    3. The attempt at a solution
    So this says that a subgroup of Z is normal in R.
    G/N is the quotient group of left cosets of N in G.
    And I want to prove that G/N is isomorphic to (a+bi)(c+di) <---not sure if this is what I want to prove...but if it is then...it equals ac+adi+bci-bd= +/- 1
    Which implies
    ac+adi+bci-bd=ac-bd=+/- 1

    Am I thinking of this right so far?

    I'm not sure how to use the hint.
     
    Last edited: Oct 9, 2008
  2. jcsd
  3. Oct 8, 2008 #2

    CompuChip

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    The hint gives you a function from R to C. You want a function from R/N to some subset of C, however. Can you construct such a function from the given one? Can you prove it is an isomorphism?
     
  4. Oct 8, 2008 #3
    The given function is cos2(pi)x+isin2(pi))x...I'm not sure how to construct a mapping for this....I'm pretty lost.
     
  5. Oct 8, 2008 #4

    morphism

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    Do you notice anything special about f, such as say, if it's a homomorphism (between appropriate groups)..?
     
  6. Oct 9, 2008 #5
    Between appropriate groups? But how is G/N defined in this problem...I don't understand.
    I know e^(2pi)(ix) = cos2(pi)x+isin2(pi)x...
     
  7. Oct 9, 2008 #6

    morphism

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    f is a homomorphism between G and the group of all complex numbers of absolute value 1 under multiplication.

    What's its kernel?
     
  8. Oct 9, 2008 #7
    The kernel is all the elements of the image of the mapping?
     
  9. Oct 9, 2008 #8

    morphism

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    No. The kernel is the set of elements in G that get mapped to the identity.
     
  10. Nov 27, 2011 #9
    What it means for a complex number to have absolute value 1, is: given $a+bi \in \C$, we say that $a+bi$ has absolute value $1$ if $a^2+b^2 =1$.

    We can see that the function in your hint, $e^{2\pi (ix)} = \cos{2 \pi x} + i\sin{2 \pi x}$. This function is well-defined, as it maps elements of $G/N$ to elements of $\C^*$ ($\C^*$ is the group of complex numbers of absolute value 1) with an absolute value of $1$ because $\cos^2{x}+\sin^2{x} =1$ for all $x$

    Now:

    To show $\phi$ is a homomorphism, let $x,y \in G/N$, and
    \begin{align*}
    \phi(x+y) &= e^{2 \pi [i(x+y)]}\\
    &= e^{2 \pi(ix) + 2 \pi(iy)}\\
    &=e^{2 \pi(ix)}e^{2 \pi(iy)}\\
    &=\phi(x)\phi(y)
    \end{align*} and so, we see $\phi$ is a homomorphism.

    To show something that is injective or surjective is fairly straightforward so I'll leave that to you.
     
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