About the lemma below:

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,

  • 4
    $\begingroup$ 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 $S_n$. $\endgroup$ – user458276 Dec 27 '18 at 19:32
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    $\begingroup$ What do you think that “order of $\sigma$” means? $\endgroup$ – José Carlos Santos Dec 27 '18 at 19:43
  • $\begingroup$ 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. $\endgroup$ – 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}


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