Finding $\sigma^3$ and $\sigma^{-1}$ of non disjoint cycles

Question

Let $\sigma=(257)(423)(57)(3416)\in S_8$ find $\sigma^3$ and $\sigma^{-1}$

So it is easier to calculate a product of disjoint cycles so we will first write the permutation as a product of disjoint cycle as any permeation can be written in this form

$$(3416)=\begin{pmatrix} 3 & 4 & 1 & 6\\ 4 & 1 & 6 & 3 \end{pmatrix} $$

$$(57)(3416)=\begin{pmatrix} 3 & 4 & 1 & 6 & 5 & 7 & 8\\ 4 & 1 & 6 & 3 & 7 & 5 & 8 \end{pmatrix}$$

$$(423)(57)(3416)=\begin{pmatrix} 3 & 4 & 1 & 6 & 5 & 7 & 2 & 8\\ 4 & 1 & 6 & 3 & 7 & 5 & 3& 8\\ 2 & \# & \# & 4 & \# & \# & \# & \# \end{pmatrix}$$

$$(257)(423)(57)(3416)=\begin{pmatrix} 3 & 4 & 1 & 6 & 5 & 7 & 2 & 8\\ 4 & 1 & 6 & 3 & 7 & 5 & 3& 8\\ 2 & \# & \# & 4 & \# & \# & \#& \#\\ 5 & \# & \# & \# & 2 & 7 & \# & \# \end{pmatrix}$$

So overall $$(257)(423)(57)(3416)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ 6 & 3 & 5 & 1 & 2 & 4 & 7 & 8\\ \end{pmatrix}=(164)(235)$$

Now it is easier to find the inverse:

$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ 4 & 5 & 2 & 6 & 3 & 1 & 7 & 8\\ \end{pmatrix}$$

Which is: $$(146)(253)$$

Is there an easy way to find $\sigma^{3}$?

Answer 1

The advantage in cycle notation (in contrast to the two line arrangements you write) is that it facilitates calculation.

You can simplify the original product of nondisjoint cycles by multiplying it out, reading from left to right:

1 to 6
6 to 3
3 to 4 to 1

so the product begins $$ (163)\ldots $$ It ends just the way you found out with the long calculation.

Since it's the product of two disjoint three cycles its cube is the identity and its inverse is its square, which you get by writing it backwards.

You could find the inverse (in cycle notation) directly just by writing the cycles backwards in reverse order.

$$ [(257)(423)(57)(3416)]^{-1} = (6143)(75)(324)(752) . $$ To find the cube, just read the changes in the direction you have established by convention. For left to right in your example you would have (for the cube)

1 to 6 then 6 to 3 then 3 to 4 to 1
2 to 5 to 7 then 7 to 2 to 3 to 4 then 4 to 2
3 to 4 to 1 then 1 to 6 then 6 to 3

so the cube begins $$ (1)(2)(3)\ldots $$ In this case we already know the cube turns out to be the inverse.