Exercise 1.3.3 (Orders of some permutations)

Proof. Consider the first example $\sigma =(135)(24)$. Since the cycles are disjoints, for all $k\in \mathbb{Z}$,

so $|\sigma |=6$. More generally, the order of a permutation is the l.c.m. of the lengths of the cycles in its cycle decomposition(see Ex.15). These gives the following results.

Ex.1

$$\begin{array}{ccc}\hfill \hfill & \hfill \text{cycles}\hfill & \hfill \text{order}\hfill \\ & & \\ \hfill \sigma \hfill & \hfill (135)(24)\hfill & \hfill 6\hfill \\ \hfill \tau \hfill & \hfill (15)(23)\hfill & \hfill 2\hfill \\ \hfill {\sigma }^2\hfill & \hfill (153)\hfill & \hfill 3\hfill \\ \hfill \mathit{\sigma \tau }\hfill & \hfill (2534)\hfill & \hfill 4\hfill \\ \hfill \mathit{\tau \sigma }\hfill & \hfill (1243)\hfill & \hfill 4\hfill \\ \hfill {\tau }^2\sigma \hfill & \hfill (135)(24)\hfill & \hfill 6\hfill \end{array}$$

Ex.2

$$\begin{array}{ccc}\hfill \hfill & \hfill \text{cycles}\hfill & \hfill \text{order}\hfill \\ & & \\ \hfill \sigma \hfill & \hfill (113510)(3158)(414117129)\hfill & \hfill 12\hfill \\ \hfill \tau \hfill & \hfill (114)(2915134)(310)(5127)(811)\hfill & \hfill 30\hfill \\ \hfill {\sigma }^2\hfill & \hfill (15)(1310)(3815)(41112)(7914)\hfill & \hfill 6\hfill \\ \hfill \mathit{\sigma \tau }\hfill & \hfill (1113)(24)(59871015)(1314))\hfill & \hfill 6\hfill \\ \hfill \mathit{\tau \sigma }\hfill & \hfill (14)(29)(3131215115)(81014)\hfill & \hfill 6\hfill \\ \hfill {\tau }^2\sigma \hfill & \hfill (1215834141112137510)\hfill & \hfill 13\hfill \end{array}$$