How to write permutations as product of disjoint cycles and transpositions

$$\sigma=\begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 4&2&9&10&6&5&11&7&8&1&3 \end{pmatrix}$$

(1) I am asked to write this permutation in $S_{11}$ as a product of disjoint cycles and also as a product of transpositions.

(2) Also find the order of the element. Is this permutation even or odd?

i think these are the disjoint cycles

$E_{1}=(1,4,10)$, $\;\operatorname{order}E_1 =3$

$E_{2}= (3,9,8,7,11),\;$ $\operatorname{order}E_{2} =5$

$E_{3}=(5,6),\;$ $\operatorname{order}E_{3} =2$

$S_{11}$= $E_{1} \cdot E_{2}\cdot E_{3}$

$\operatorname{order}E_{3}$ is even so the order of the permutation is even. Why are they asking this? and what is the significance of it being even or odd?

transpositions i have read this a couple times but the one example in my textbook is rather unclear, i am not sure what this means?

i think the transposition for $E_{1}$ is $(1,4)(1,10)$ but im not sure what this means.

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I’ll use a longer cycle to help describe two techniques for writing disjoint cycles as the product of transpositions:

Let’s say $\tau = (1, 3, 4, 6, 7, 9) \in S_9$

Then, note the patterns:

Method 1: $\tau = (1, 3, 4, 6, 7, 9) = (1, 9)(1, 7)(1, 6)(1, 4)(1, 3)$

Method 2: $\tau = (1, 3, 4, 6, 7, 9) = (1, 3)(3, 4)(4, 6)(6, 7)(7, 9)$

Both products of transpositions, method $1$ or method $2$, represent the same permutation, $\tau$. Note that the order of the disjoint cycle $\tau$ is $6$, but in both expressions of $\tau$ as the product of transpositions, $\tau$ has $5$ (odd number of) transpositions. Hence $\tau$ is an odd permutation.


Now, don’t forget to multiply the transpositions you obtain for each disjoint cycle so you obtain an expression of the permutation $S_{11}$ as the product of the product of transpositions, and determine whether it is odd or even:

$\sigma = (1, 4, 10)(3, 9, 8, 7, 11)(5, 6)$.

  • The order of $\sigma = \operatorname{lcm}(3, 5, 2) = 30$.
  • Expressing $\sigma $ as the product of transpositions:

    • $\sigma =(1, 4)(4, 10)(3, 9)(9, 8)(8, 7)(7, 11)(5, 6):\quad 7$ transpositions in all, so $\sigma $ is an odd permutation (which happens to be of even order).