Instead of trying multiplication again and again until I get $(1)(2)(3)(4)(5)(6)(7),$ is there an efficient, logical method to compute order of $(157)(134)(12)$ of $S_{10}$?

Is there some relation to determine it?

Also, I am aware there is some relation with order of group, but don't remember exactly. I would be grateful if you could help.

Thank you.


We can easily compute the order of a permutation which is written as the product of disjoint cycles.

So first we need to "multiply out" (compose) the permutation expressed as the product of non-disjoint cycles you posted. Let's call your permutation $\alpha$:

$$\alpha = (\color{blue}{\bf 1}57)(\color{blue}{\bf 1}34)(\color{blue}{\bf 1}2) \in S_{10}$$

I've highlighted in blue why $\alpha$ fails to be the product of disjoint cycles: $1$ appears in each of the three factor-cycles, and in each, is permuted to different values: $1\to 5,\;1\to 3, \;1\to 2$

So, when we compose these cycles that are each a factor of $\alpha$, we obtain the one-cycle that caveman provides: $$\alpha = (157)(134)(12) = (6)(1\,5\,7\,3\,4\,2)(8)(9)(10) = (1\,5\,7\,3\,4\,2).$$

Now we can address the order of a permutation:

We define the length of a cycle to be the number of elements in the cycle.

We define the $|$ order $|$ of a permutation written as the product of disjoint cycles to be the least common multiple $(\operatorname{lcm})$ of the lengths of those cycles. So for $\rho \in S_n$, written as the product of $k$ disjoint cycles:

$$\operatorname{order}(\rho) = |\rho| = |\sigma_1\sigma_2\cdots\sigma_k| = \operatorname{lcm}(|\sigma_1|, |\sigma_2|, \ldots , |\sigma_k|)$$

Since $\alpha =(1\,5\,7\,3\,4\,2)$ is a "one-cycle", its order is equal to its length, which is $6$.

If we had the permutation, e.g.: $$\beta = (1\,2\,3)(4\,5)(6\,7\,8),$$ then the order of $\beta$ would be $\operatorname{lcm}(3, 2, 3) = 3\cdot 2 = 6$.


$$(157)(134)(12) = (157342)$$ so it has order $6$.

  • 1
    $\begingroup$ woops, i was looking at the wrong thing. $\endgroup$ – rotten Mar 16 '13 at 18:10
  • $\begingroup$ @MaisamHedyelloo, I checked with computer seems like I am right. $\endgroup$ – user58512 Mar 16 '13 at 18:40
  • $\begingroup$ Just to add more information, for any permutation $\tau\in\mathcal S_n$, we may write $\tau=\sigma_1\sigma_2\cdots\sigma_k$, where $\sigma_k$ are disjoint cycles in $\mathcal S_n$. Then $|\tau|=$lcm$\{|\sigma_i|\}$ where $i$ ranges over $1,\ldots,k$. $\endgroup$ – Ian Coley Mar 16 '13 at 19:25

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