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Thread: Roots of Unity as a group

  1. #1
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    Roots of Unity as a group

    Let $\displaystyle G=\{z\in\mathbb{Z}: \ z^n=1, \ n\in\mathbb{N}\}$.
    Prove $\displaystyle (G,*)$ is an abelian group under multiplication.

    This makes sense since the units $\displaystyle \mathbb{Z}_p$ are group under multiplication.

    $\displaystyle 1+0\mathbf{i}$ is the identity.

    How do I find the inverse?
    Last edited by dwsmith; Aug 27th 2011 at 03:14 PM.
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  2. #2
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    Re: Roots of Unity as a group

    Quote Originally Posted by dwsmith View Post
    Let $\displaystyle G=\{z\in\mathbb{Z}: \ z^n=1, n\in\mathbb{N}\}$.
    Prove $\displaystyle (G,*)$ is an abelian group under multiplication.
    $\displaystyle 1+0\mathbf{i}$ is the identity.
    How do I find the inverse?
    Here is a hint:
    $\displaystyle \exp \left( {\frac{{\pi i}}{5}} \right)$ is a $\displaystyle 10^{th}$ root of unity.

    $\displaystyle \exp \left( {\frac{{9\pi i}}{5}} \right)$ is its inverse.
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    Re: Roots of Unity as a group

    Quote Originally Posted by Plato View Post
    Here is a hint:
    $\displaystyle \exp \left( {\frac{{\pi i}}{5}} \right)$ is a $\displaystyle 10^{th}$ root of unity.

    $\displaystyle \exp \left( {\frac{{9\pi i}}{5}} \right)$ is its inverse.
    So the inverse is
    $\displaystyle \exp\left(\frac{2\pi i(n-k)}{n}\right)$
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  4. #4
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    Re: Roots of Unity as a group

    For associative and commutative, that is trivial true since $\displaystyle \mathbb{C}$ is a field, correct?
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    Re: Roots of Unity as a group

    Quote Originally Posted by dwsmith View Post
    For associative and commutative, that is trivial true since $\displaystyle \mathbb{C}$ is a field, correct?
    Well it is a field after all!
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