Order of Product of Disjoint Permutations

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\pi$ be a product of disjoint cycles of orders $k_1, k_2, \ldots, k_r$.

Then:
 * $\left|{\pi}\right| = \operatorname{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$

Proof
The order of a $k$-cycle is clearly $k$.

Now suppose $\pi = \rho_1 \rho_2 \cdots \rho_r$ where each $\rho_s$ is of order $k_s$, and $\rho_1$ to $\rho_r$ are mutually disjoint.

Let $t = \operatorname{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$.

Then $\pi^t = \rho_1^t \rho_2^t \cdots \rho_r^t$.

As $\forall k: 1 \le k \le r: \exists m \in \Z: t = m k$, we have $\forall k: \rho_k^t = e$.

Thus $\pi^t = e$ and certainly $\left|{\pi}\right| \mathop \backslash t$.

But if $\pi^u = e$, then each $\rho_s = e$ so $k_s \mathop \backslash u$ and thus $u \mathop \backslash t$.

Thus:
 * $\left|{\pi}\right| = t = \operatorname{lcm} \left\{{k_1, k_2, \ldots, k_r}\right\}$