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Math Help - Surjective homomorphism on roots of unity

  1. #1
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    Surjective homomorphism on roots of unity

    Let G = \{ z \in \mathbb{C} |z^n = 1 \text{ for some n } \in \mathbb{Z} ^+ \}. Prove that for any fixed integer k > 1 the map from G to itself defined by z \mapsto z^k is a surjective homomorphism but is not injective.
    So let
    \phi _k : G \to G : g \mapsto g^k.

    Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that x^k = y^k is true when x is not equal to y. My other question about this is do we need to find an x, y for each k? For example, (-1)^k = (1)^k is only true for even k.

    I'm just not wrapping this one around in my head today.

    Thanks!
    -Dan
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  2. #2
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    Re: Surjective homomorphism on roots of unity

    Quote Originally Posted by topsquark View Post
    So let
    \phi _k : G \to G : g \mapsto g^k.

    Proving the homomorphism is trivial. It's the surjection part that's getting to me. (I haven't tackled the injection yet. Please leave that out of the replies for now.) What I am looking at is the following...I want to show that there is at least one pair of x, y in G such that x^k = y^k is true when x is not equal to y.
    No, you don't! That would be showing this is a NOT an injection. It would not be showing this is a surjection. To show this is a surjection you need to show that if z is a complex number such that z^n= 1 for some n then there exist a complex number w such that w^n= 1 such that w^k= z for this specific z.

    My other question about this is do we need to find an x, y for each k? For example, (-1)^k = (1)^k is only true for even k.
    For z= -1, take w= -1.

    I'm just not wrapping this one around in my head today.

    Thanks!
    -Dan
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  3. #3
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    Re: Surjective homomorphism on roots of unity

    1. Surjective: if z^n=1 then z=e^{2mi\pi/n} for some m with 0\leq m<n. Can't you find w with w^{kn}=1 and w^k=z?

    2. Injective: Think about the kernel of the homomorphism.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Surjective homomorphism on roots of unity

    Sorry I didn't get back sooner. I just finished the problem, both surjective and injective. I appreciate the help.

    -Dan
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