0 votes 0 votes Let $\sigma$ be the permutation: $$\begin{array} {}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 5 & 6 & 2 & 4 & 9 & 8 & 7 & 1, \end{array}$$ $I$ be the identity permutation and $m$ be the order of $\sigma$ i.e. $m=\text{min}\{\text{positive integers }n: \sigma ^n=I \}$. Then $m$ is $8$ $12$ $360$ $2520$ Combinatory isi2015-mma combinatory + – Arjun asked Sep 23, 2019 recategorized Nov 17, 2019 by Lakshman Bhaiya Arjun 734 views answer comment Share Follow See 1 comment See all 1 1 comment reply `JEET commented Nov 18, 2019 reply Follow Share https://math.stackexchange.com/questions/231143/finding-the-order-of-permutations-in-s-8 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Disjoint cycles are (1,3,6,9),(2,5,4),(7,8) length of them are respectively : { 4,3,2} so m is lcm(4,3,2) = 12 Amartya answered May 22, 2020 Amartya comment Share Follow See all 0 reply Please log in or register to add a comment.