finite subgroup of C*

**numbertheory12** · December 9th 2008, 03:08 PM

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 $e^{\frac{2 \pi i}{n}}$ .

**ThePerfectHacker** · December 9th 2008, 04:07 PM

Originally Posted by numbertheory12

(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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