Order of a permutation is the lcm of its cycle lengths

Statement

Let $\sigma \in S_n$, and let $\sigma = c_1 c_2 \cdots c_r$ be its decomposition into disjoint cycles of lengths $k_1, \dots, k_r$. Then the order of $\sigma$ is $\operatorname{lcm}(k_1, \dots, k_r)$.

Remarks

A $k$-cycle has order $k$. The disjoint cycles commute, so $\sigma^m = c_1^m \cdots c_r^m$, and this is the identity exactly when each $c_i^m$ is the identity. For example, $(1\ 2)(3\ 4\ 5)$ has order $6$.

Explanations

L1 human

A permutation returns to the identity when each of its cycles is back at the start. A cycle of length $k$ is back at the start after each multiple of $k$ steps. So you need the first common multiple of all the cycle lengths.