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Math Help - finite subgroup of C*

  1. #1
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    finite subgroup of C*

    Let G be a finite subgroup of \mathbb{C}^*. For example, \{ 1,-1,i,-i \} is a subgroup of order 4.
    (a) Prove that G must be a subset of T=\{ z \in \mathbb{C}^* | \text{ } |z|=1  \}.
    (b) Prove that G is cyclic and is generated by an element of form e^{\frac{2 \pi i}{n}}.
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  2. #2
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    Quote Originally Posted by numbertheory12 View Post
    (a) Prove that G must be a subset of T=\{ z \in \mathbb{C}^* | \text{ } |z|=1  \}.
    (b) Prove that G is cyclic and is generated by an element of form e^{\frac{2 \pi i}{n}}.
    Finite subfields of the multiplicative group of a field must be cyclic subgroups.
    From here it should be clear why (a) and (b) follows.


    But you can also see (a) because if re^{i\theta} \in G then it means that r^k e^{ik\theta} \in G for all k>0. But if r\not = 1 then r^k \to \infty if r>1 or r^k \to 0 if r<1. Therefore r=1 and so the complex number in G must lie on the unit circle. To show (b) let G = \{ e^{i\theta_1}, ... , e^{i\theta_m} | \theta_j \in [0,2\pi)\}. By Lagrange's theorem we have (e^{i\theta_j})^m = 1 \implies \theta_j m = 2\pi r, r\in \mathbb{Z}^+. Therefore, \theta_j = \tfrac{2\pi r i}{m} for some r\in \mathbb{Z}^+. Define H = \left< e^{2\pi i/m} \right>, this subgroup is cyclic and contains G, but, |G| = |H|, from here it follows that G=H. Thus, G must be a cyclic subgroup.
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