Hello,
I am stuck on part three of this question.
i) $\displaystyle \sigma $ = (1 6) (1 4) (1 2) (3 7) (3 5) (8 9) = even permutation
ii) $\displaystyle \sigma^{2} $ = 1 2 3 4 5 6 7 8 9
4 6 7 1 3 2 5 8 9
III) how do I find T??
You do it digit by digit.
You want: $\displaystyle \tau \sigma(1) = 2$. You have $\displaystyle \tau \sigma(1) = \tau(\sigma(1)) = \tau(2)$, so $\displaystyle \tau(2) = 2$.
You want: $\displaystyle \tau \sigma(2) = 1$. You have $\displaystyle \tau \sigma(2) = \tau(\sigma(2)) = \tau(4)$, so $\displaystyle \tau(4) = 1$.
Etc.
Turning two-line notation into three-line notation:
$\displaystyle \tau \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}$
Essentially, the first two lines show where $\displaystyle \sigma$ sends numbers while the second two lines show where $\displaystyle \tau$ sends numbers. So, the first and last line show where the composition $\displaystyle \tau \sigma$ sends numbers. To write $\displaystyle \tau$, just write the last two lines, but order the columns so that the top row is in order:
$\displaystyle \tau = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 2 & 7 & 1 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$
In case it is not obvious to the OP, $\displaystyle \sigma^{-1}$ is extremely easy to calculate. You are given that $\displaystyle \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}$, so its inverse is just flipping the two rows:
$\displaystyle \sigma^{-1} = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$
So, now, the product (composition) of the two permutations: $\displaystyle \tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}$ is easy to calculate (and gives the same answer as above). Generally, as you continue in mathematics, Plato's solution is going to be easier to apply.
$\displaystyle \tau\sigma = (1 2)$, so multiplying both sides by $\displaystyle \sigma^{-1}$ on the right, you get $\displaystyle \tau \sigma \sigma^{-1} = (1 2) \sigma^{-1}$. On the LHS, $\displaystyle \sigma \sigma^{-1}$ is the identity permutation, so the LHS simplifies to $\displaystyle \tau$. So, it is not that $\displaystyle \sigma \sigma^{-1} = \tau$, but that you are multiplying $\displaystyle \tau$ by the identity permutation, so you are not changing it. If you send 1 to 2, then send 2 back to 1, it is the same as if you never sent it anywhere at all. That is how multiplying a permutation by its inverse gives the identity permutation.