# About the Proof for “the Order of a Permutation $\sigma$ is the $lcm$ of the Orders of its Disjoint Cycles.”

The order of a permutation $$\sigma \in S_n$$ is the least common multiple of the orders of its disjoint cycles.

I didn't get the proof which consists of the following reasoning:

Let $$\sigma = \sigma_1 ... \sigma_r$$ with $$r \le n$$ and $$\{ \sigma_i \}_{1 \le i \le r}$$ disjoint cycles in question. Since every $$\sigma_i$$ commute in this case, we have: $$\sigma^m = \sigma_{1}^{m} ... \sigma_{r}^{m}$$, for all $$m \in \mathbb{Z}$$, and $$\sigma^m = (1)$$ iff $$\sigma_{i}^{m} = (1)$$ for all $$i$$. Therefore, $$\sigma^m = (1)$$ iff $$| \sigma_i |$$ divides $$m$$ for all $$i$$ (for $$\sigma$$ is cyclic). Since $$| \sigma |$$ is the least such $$m$$, the conclusion follows.

I have big trouble to understand why we needed to use this $$m$$-th power with, furthermore, considering the case about $$\sigma^m$$ being the identity permutation.

Thank you,

• I'm not sure where your confusion lies. The $m$-th power is needed because one is seeking to find the order of $\sigma$ in $S_n$. – user458276 Dec 27 '18 at 19:32
• What do you think that “order of $\sigma$” means? – José Carlos Santos Dec 27 '18 at 19:43
• I'm totally ok with your comments. I hadn't considered enough the question before posting it. But, I wonder if we wouldn't have here a slice of reasoning missing still? If not, I get it. – freehumorist Dec 27 '18 at 20:14

As for all groups, the order of a permutation $$\sigma\in S_n$$ is, by definition, the least positive integer $$m$$ such that $$\sigma^n=\operatorname{id}$$.
In other words, it is the positive generator of the group homomorphism \begin{align} (\bf Z,+)&\longrightarrow (S_n,\circ)\\ k &\longmapsto \sigma^k=\underbrace{\sigma\circ\sigma\circ\dots\circ\sigma}_{k \:\text{ factors}}\end{align}