> [!NOTE] Theorem > Let $n\in\mathbb{N}^{+}.$ Let $S_{n}$ denote the [[Symmetric Groups of Finite Degree|nth symmetric group]]. Let $\pi\in S_{n}$ be a [[Cyclic Permutation of n Letters|cyclic permutation on n letters]] of length $k.$ Then the [[Order of Group Element|order]] of $\pi$ is $k.$ **Proof**: Let $\pi=(a_{0},a_{1},\dots,a_{k-1}).$ For all $n\in \mathbb{N}$ and $0\leq j\leq k-1,$ $\pi^{n}(a_{j})=a_{j+n \pmod{k}}$ which is true by induction on $n$ since $\pi(a_{j+n \pmod{k}})=a_{j+n+1 \pmod{k}}.$ Thus $\pi^{k}(a_{j})=a_{j}$ and since $\pi$ fixes elements not in $\{ a_{0},a_{1},\dots a_{k-1} \},$ $\pi^{k}=\text{Id}_{S_{n}}.$ Furthermore, for all $1\leq l<k,$ $\pi^{l}\neq \text{Id}.$ # Applications See [[Order of Product of Disjoint Permutations]].