> [!NOTE] Theorem > Let $n\in\mathbb{N}^{+}.$ Let $U_{n}$ denote the [[Roots of Unity|nth roots of unity]]. Then $(U_{n},\times)$ is a [[Cyclic Group|cyclic group]]. **Proof**: By [[Roots of Unity in Exponential Form]], $U_{n}=\{ \zeta^{k}\mid k\in \mathbb{\mathbb{Z}} \}$ where $\zeta=e^{2\pi i/k}/.$ Let $\zeta^{m}, \zeta^{n}\in U_{n}.$ Then $\zeta^{m}\zeta^{n}=\zeta^{m+n}\in U_{n}$ since $m+n\in \mathbb{Z}.$ Note that multiplication of complex numbers is associative. Also $\zeta^{0}=1 \in U_{n}.$ Furthermore if $\zeta^{m}\in U_{n}$ then $\zeta^{-m}\in U_{n}$ since $-m\in \mathbb{Z}.$ Therefore $U_{n}$ is a [[Groups|group]]. Now $U_{n}$ is [[Generated Subgroup|generated]] by $\zeta$ thus $U_{n}$ is cyclic.