\(\displaystyle \tau= \begin{pmatrix}123456789 \\ 327651498\end{pmatrix}\) which means 1 goes to 3, 2 stays 2, 3 goes to 7, 4 goes to 6, 5 stays 5, six goes to 1, 7 goes to 4, 8 goes to 9 and 9 goes to 8. \(\displaystyle \tau^{-1}\) is the opposite: 1 goes to 6, 2 stays 2, 3 goes to 1, 4 goes to 7, 5 stays 5, 6 goes to 4, 7 goes to 3, 8 goes to 9, and 9 goes to 8. That can be written as \(\displaystyle \tau^{-1}= \begin{pmatrix}\frac{123456789}{621754398}\end{pmatrix}\).
(To write \(\displaystyle \tau\) in cycle notation, 1 goes to 3 goes to 7 goes to 4 goes to 6 goes to 1: (13746). stays 2. 5 stays 5. 8 goes to 9 goes 2 8: (89). ]tex]\tau= (13746)(89)[/tex]. \(\displaystyle \tau^{-1}\) takes 1 to 6 to 4 to 7 to 3 to 1: (16473). (We don't have to start at 1. If we started at 6 we would get (64731), just the reverse of (16573). \(\displaystyle \tau^{-1}= (16473)(89)\).)
\(\displaystyle \sigma= \begin{pmatrix}123456789 // 584316729\end{pmatrix}\). 1 goes to 5, 5 goes to 1. 2 goes to 8, 8 goes to 2. 3 goes to 4, 4 goes to 3. 6 stays 6, 7 stays 7 and 9 stays 9! That can be written cycle notation as (15)(28)(34). That's good because it makes it very easy to find the 5 power. If we do each of those 5 times, and odd number of times, its just as if we did just 1: \(\displaystyle \sigma^5= \begin{pmatrix}123456789 // 584316729\end{pmatrix}\)!
Now, what \(\displaystyle \sigma^5\tau^{-1}\) actually is depends, as Jomo said, on whether your convention is "right to left" and "left to right". Going "left to right" we would first do \(\displaystyle \sigma^5\) which, as I said before, just \(\displaystyle \sigma\) again. And, as before, in cycle notation (which I find easier to use), that is (15)(28)(34). \(\displaystyle \tau^{-1}\), in cycle notation, was (16473)(89). \(\displaystyle \sigma^5\tau^{-1}= [(15)(28)(34)][(16473)(89)\). Combining them, 1 goes to 5 which stays 5, 2 goes to 8 which goes to 9. 3 goes to 4 that goes to 7. 4 goes to 3 that goes to 1. 5 goes to 1 which goes to 6. 6 stays 6 which goes to 4. 7 stays 7 which goes to 3. 8 goes to 2 which stays 2, 9 stays 9 which goes to 8: 1 goes to 5, 2 goes to 9, 3 goes go 7, 4 goes to 1, 5 goes to 6, 6 goes to 4, 7 goes to 3, 8 goes to 2, and 9 goes to 8: \(\displaystyle \begin{pmatrix}123456789 \\ 597164328\end{pmatrix}\). In cycle notation that is (1564)(298)(37).
"Right to left" would be "1 to 6 stays 6", "2 stays 2 goes to 8", "3 goes to 1 goes to 5", "4 goes to 7 stays 7", "5 stays 5 goes to 1", "6 goes to 4 goes to 3", "7 goes to 3 goes to 4", "8 goes to 9 stays 9", 9 goes to 8 goes to 2". That is 1 goes to 6, 2 goes to 8, 3 goes to 5, 4 goes to 7, 5 goes to 1, 6 goes to 3, 7 goes to 4, 8 goes to 9, 9 goes to 2: \(\displaystyle \begin{pmatrix}123456789 \\ 685713492\end{pmatrix}\). In cycle notation that is (1635)(289)(47).
