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Math Help - Find Isomorphisms

  1. #1
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    Find Isomorphisms

    Show that:

    1. The additive group  \mathbb {R} / \mathbb {Z} is isomorphic to the multiplicative group of all complex numbers of modulus 1.

    2. The additive group  \mathbb {Q} / \mathbb {Z} is isomorphic to the multiplicative group of all complex roots of unity.

    Okay, so I understand that I need find a mapping between these groups, prove they are well-defined, linear, and one to one and onto, but would anyone please explain to me how the complex numbers of mod 1 and roots of unity going to behave?

    Complex numbers of mod 1, wouldn't that be any combination of a(i)?

    Roots of unity, well, I know how it would work in integers, but complex, I'm a bit lost on that.

    Thanks for all your help!!!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Show that:

    1. The additive group  \mathbb {R} / \mathbb {Z} is isomorphic to the multiplicative group of all complex numbers of modulus 1.

    2. The additive group  \mathbb {Q} / \mathbb {Z} is isomorphic to the multiplicative group of all complex roots of unity.
    Define \theta: \mathbb{R}\to \mathbb{C}^{\times} by \theta ( a) = e^{\pi i a}.
    Now use the fundamental homomorphism theorem.

    Similarly with \mathbb{Q}.

    And the race is on ....! Who will post first? TPH or NonCommAlg. Egg on my face if both pull out. It's neck-and-neck at the moment.
    What is this?
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    Show that:

    1. The additive group  \mathbb {R} / \mathbb {Z} is isomorphic to the multiplicative group of all complex numbers of modulus 1.
    let \mathbb{C}_1 be the multiplicative group of all complex numbers of modulus 1. define f: \mathbb{R} \longrightarrow \mathbb{C}_1 by f(x)=e^{2x\pi i}. clearly f is a homomorphism. now if z \in \mathbb{C}, then z=\cos \theta +i \sin \theta = e^{i \theta}=f(\theta/2\pi).

    so f is onto. also x \in \ker f \Longleftrightarrow e^{2x \pi i}=1 \Longleftrightarrow 2x \pi i = 2k \pi i, \ \text{for some} \ k \in \mathbb{Z} \Longleftrightarrow x=k \in \mathbb{Z}. so \ker f = \mathbb{Z}. \ \ \ \ \square


    2. The additive group  \mathbb {Q} / \mathbb {Z} is isomorphic to the multiplicative group of all complex roots of unity.
    let \mathbb{C}_2 be the the multiplicative group of all complex roots of unity. define g: \mathbb{Q} \longrightarrow \mathbb{C}_2 by g(x)=e^{2x \pi i}. the same as above g is a homomorphism and \ker g = \mathbb{Z}. now choose z \in \mathbb{C}_2. then we have

    z=e^{2k \pi i/n} for some integers k,n. hence z=g(k/n). so g is onto. Q.E.D.
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