The strategy is to show (1) $(\sigma\tau)^n = e$ where $n = \operatorname{lcm}(o(\sigma), o(\tau))$, and (2) no smaller positive integer works. The key ingredients are commutativity of disjoint cycles and the [Order Division Lemma](/theorems/771). **Step 1: $(\sigma\tau)^n = e$ for $n = \operatorname{lcm}(o(\sigma), o(\tau))$.** Let $a = o(\sigma)$ and $c = o(\tau)$, so $a \mid n$ and $c \mid n$. Write $n = ab = cd$ for appropriate positive integers $b, d$. Since $\sigma$ and $\tau$ are disjoint, they commute by the [Disjoint Cycles Commute](/theorems/774) theorem. Therefore: \begin{align*} (\sigma\tau)^n = \sigma^n \tau^n = (\sigma^a)^b (\tau^c)^d = e^b \cdot e^d = e. \end{align*} **Step 2: No smaller positive integer works.** [claim:Lcm Is Minimal] If $(\sigma\tau)^m = e$ for some $m \geq 1$, then $n \mid m$. [/claim] [proof] Since $\sigma$ and $\tau$ commute, $(\sigma\tau)^m = \sigma^m \tau^m = e$. Now $\sigma^m$ and $\tau^{-m}$ are equal: $\sigma^m = \tau^{-m}$. But $\sigma$ acts as the identity on the support of $\tau$, so $\sigma^m$ acts as the identity there too. Meanwhile $\tau^{-m}$ acts as the identity on elements outside the support of $\tau$. For $\sigma^m = \tau^{-m}$ to hold on all of $\{1, \ldots, n\}$, both must equal the identity. So $\sigma^m = e$ and $\tau^m = e$. By the [Order Division Lemma](/theorems/771), $a \mid m$ and $c \mid m$, hence $\operatorname{lcm}(a, c) = n \mid m$. [/proof] **Step 3: Conclusion.** By Step 1, $o(\sigma\tau) \mid n$. By Step 2, $n \mid o(\sigma\tau)$. Therefore $o(\sigma\tau) = n = \operatorname{lcm}(o(\sigma), o(\tau))$.