TITLE: Express the permutation $\sigma = \left({}^1_5\,{}^2_8 \,{}^3_3\,{}^4_6\,{}^5_7\,{}^6_4\,{}^7_1\,{}^8_2\right)$ as a product of transpositions
QUESTION [5 upvotes]: $\newcommand{\lcm}{\operatorname{lcm}}$
I was hoping to get some feedback.

Consider the permutation
  $$\sigma = \begin{pmatrix}
1&2&3&4&5&6&7&8\\
5&8&3&6&7&4&1&2
\end{pmatrix}.$$
  (a) Describe the orbits of $\sigma$.
  (b) Express $\sigma$ as a product of disjoint cycles, and then as a product of transpositions.
  (c) What is the order of $\sigma$? Explain.


(a) The orbits of $\sigma$ are $\{1,5,7\}, \{2,8\},\{3\},\{4,6\}$.
(b) As a product of disjoint cycles: $\sigma = (1\;5\;7)(2\;8)(3)(4\;6)$
As a product of transpositions: $\sigma = (1\;7)(1\;5)(2\;8)\boxed{(3)\,}(4\,6)$
I am not sure if I should include $(3)$ since transpositions are cycles of length $2$.
(c) I don't know what the order of a permutation is. My first guess was that it is the number of orbits of $\sigma$. So I would say $4$. But then I saw a post that used the $\lcm$. I infer that it is the $\lcm$ of the lengths of the orbits of $\sigma$; hence $\lcm(2,3) = 6$. Is this correct? If so, why is it the $\lcm$?

REPLY [5 votes]: (a) -- Correct.
(b) -- No, you don't need to include $(3)$.
(c) -- The order of a permutation $\sigma$ is the smallest positive integer $\ell$ so that $\sigma^\ell=1$. Since the disjoints cycles of $\sigma$ are mutually commutative, $\ell$ is divisible by the order of each cycle. Therefore, $\ell$ is the l.c.m of those orders, which are exactly the lengths of the cycles.