5
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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 $\operatorname{lcm}(3, 2, 3) = 3\cdot 2 = 6$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.