Definition 1.4.1: A permutation of a set is a bijection (one-to-one and onto) : → .
Solve $\sigma x = \tau$ which means I need to solve it the following way:
$\sigma^1 \sigma x = \sigma^{-1} \tau \to I x = \sigma^{-1} \tau \to x = \sigma^{-1} \tau$
let $\sigma = (1,3,2)(4,5)$
which translates to:
$\sigma =$ $$ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 5 & 4 \\ \end{matrix} $$
let $\tau =(2,1,3)$
then
$\tau =$ $$ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 2 & 4 & 5 \\ \end{matrix} $$
I am not looking to solve the equation in cycle notation until I get more comfortable with the material.
There fore $\sigma^{-1} =$
$$ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 5 & 4\\ \end{matrix} $$
so $\sigma^{-1} \tau=$
$$ \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 5 & 4 \\ \end{matrix} $$
Are there any mistakes? Is it fine to do $\sigma^{-1} \tau$ rather than $\tau \sigma^{-1}$
when I multiplied back my value for $x$ with sigma I got $\tau$ back