> [!NOTE] Theorem > Let $n\in\mathbb{N}^{+}.$ Let $S_{n}$ denote the [[Symmetric Groups of Finite Degree|nth symmetric group]]. Let $\rho$ be a [[Permutation of Finite Degree|permutation of n letters]] such that $\rho=\sigma_{1}\sigma_{2}\dots\sigma_{k}$ is a product of disjoint [[Cyclic Permutation of n Letters|cycles]] of lengths $m_{1},m_{2},\dots,m_{k}.$ Then the [[Order of Group Element|order]] of $\rho$ is $\text{lcm}(m_{1},\dots,m_{k})$ where $\text{lcm}$ denotes [[Lowest Common Multiple (LCM)|lowest common multiple]]. **Proof**: Suppose $\alpha$ and $\beta$ are disjoint permutations of order $m$ and $n.$ Let $k=\text{lcm}(m,n).$ Since [[Disjoint Permutations Commute]], $(\alpha \beta)^{k}=\alpha^{k}\beta^{k}=e.$Thus the order of $(\alpha \beta),$ denote $t$ must divide $k.$ But then $\alpha^{t}\beta^{t}=e$ so that $\alpha^{t}=\beta^{-t}.$ However if $\alpha$ and $\beta$ are disjoint then so are $\alpha^{t}$ and $\beta^{-t}.$ But if $\alpha^{t}$ and $\beta^{-t}$ are equal and have no common symbol, they must both be the identity. It follows that $m$ and $n$ must divide $t.$ This means that $k=t.$ Thus $k=t.$ By [[Order of k-Cycle is k]], $\sigma_{1},\sigma_{2},\dots,\sigma_{k}$ have orders $m_{1},m_{2},\dots m_{k}$ respectively thus the theorem is true.