How to write permutations as product of disjoint cycles and transpositions

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).

If you missed the class on cycles, you can use the following mechanized approach to write $\sigma$ as a product of transpositions:

$(1 \; 4) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 4&2&9&10&6&5&11&7&8&1&3 \end{pmatrix} =$ $\quad \quad \quad \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&9&10&6&5&11&7&8&4&3 \end{pmatrix}$

$(3 \; 9) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&9&10&6&5&11&7&8&4&3 \end{pmatrix} =$ $\quad \quad \quad \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&10&6&5&11&7&8&4&9 \end{pmatrix}$

$(4 \; 10) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&10&6&5&11&7&8&4&9 \end{pmatrix} =$ $\quad \quad \quad\; \, \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&6&5&11&7&8&10&9 \end{pmatrix}$

$(5 \; 6) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&6&5&11&7&8&10&9 \end{pmatrix} = $ $\quad \quad \quad \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&11&7&8&10&9 \end{pmatrix}$

$(7 \; 11) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&11&7&8&10&9 \end{pmatrix} = $ $\quad \quad \quad\; \, \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&7&11&8&10&9 \end{pmatrix}$

$(8 \; 11) \circ \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&7&11&8&10&9 \end{pmatrix} = $ $\quad \quad \quad \; \, \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&7&8&11&10&9 \end{pmatrix}$

$(9 \; 11) \circ\begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&7&8&11&10&9 \end{pmatrix} = $ $\quad \quad \quad \; \, \begin{pmatrix} 1 & 2 &3 & 4& 5& 6&7 &8 &9 &10 & 11 \\ 1 & 2&3&4&5&6&7&8&9&10&11 \end{pmatrix}$

$ $

$\tag{ANS} \sigma = (1 \; 4)\,(3 \; 9)\,(4 \; 10)\,(5 \; 6)\,(7 \; 11)\,(8 \; 11)\,(9 \; 11)$