Let $\sigma$ be the permutation: [ \begin{array}{lllllllll} 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. $m=\min \left\{\text { positive integers } n: \sigma^{n}=I\right\} .$ Then $m$ is
Disjoint cycles are (1,3,6,9),(2,5,4),(7,8) length of them are respectively : { 4,3,2} Can m be lcm(4,3,2) = 12? How is this formula generated?
On
Say the cycle decomposition is $\sigma = \sigma_1 \sigma_2 \cdots \sigma_k$.
If you raise it to the $n$'th power you get
$$ \sigma^n = \sigma_1^n \sigma_2^n \cdots \sigma_k^n $$
because all the disjoint cycles commute so you can gather them together next to each other.
Now if this is to be the identity permutation, each of the $\sigma_i^n$ had better be the identity. That is because they are on disjoint parts, so there is no way for them to cancel with each other.
So from the first condition, you know that $n$ has to be a multiple of the length of $\sigma_1$.
From the second condition, you know that $n$ has to be a multiple of the length of $\sigma_2$.
Continue this way and you see that $n$ has to be a common multiple of all the lengths and that is why you get $lcm$.
your answer is correct. If is a permutation of the elements in then the order of denoted is the smallest positive integer such that where is the identity permutation.