Composition of Permutations Suppose that

$$\sigma = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 \\ 2 &amp; 7 &amp; 4 &amp; 1 &amp; 6 &amp; 5 &amp; 3   \end{pmatrix}$$

$$\tau = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 \\ 6 &amp; 2 &amp; 5 &amp; 4 &amp; 3 &amp; 7 &amp; 1 \end{pmatrix} $$

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

Compute $\tau\mu\sigma^{-1}$

Workings:

$\sigma = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 \\ 2 &amp; 7 &amp; 4 &amp; 1 &amp; 6 &amp; 5 &amp; 3   \end{pmatrix}$

$\sigma^{-1} = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 \\ 4 &amp; 1 &amp; 7 &amp; 3 &amp; 6 &amp; 5 &amp; 3  \end{pmatrix}$

Now I only know how to do the composition of permutations if they are of the same size. Not different sizes.

So I don't know how $\mu\sigma^{-1}$ is supposed to be computed.

If I were to guess

$\mu\sigma^{-1} = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp;5 &amp; 6 &amp; 7 \\ 2 &amp; 4 &amp; 5 &amp; 3 &amp; 8 &amp; 6 &amp; 3\end{pmatrix}$

Any help will be appreciated
Before multiplying, convert the smaller permutations to the size of the larger:
$$\sigma = \pmatrix{1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;\color{blue}8\\2&amp;7&amp;4&amp;1&amp;6&amp;5&amp;3&amp;\color{blue}8}$$
$$\sigma^{-1} = \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 \\ 4 &amp; 1 &amp; 7 &amp; 3 &amp; 6 &amp; 5 &amp; \color{red}2 &amp; 8\end{pmatrix}$$

Then proceed as usual.
$$\mu\sigma^{-1} = \pmatrix{1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;\color{red}8\\2&amp;4&amp;5&amp;3&amp;8&amp;6&amp;\color{red}1&amp;\color{red}7}$$
(red here is an error in your post).
Writing $\sigma$, $\tau$, $\mu$ as products of disjoint cycles, you have:


$\sigma=(1,2,7,3,4)(5,6)$
$\tau=(1,6,7)(3,5)$
$\mu=(1,4,2)(5,6,8,7)$


Now proceed as usual for compositions of permutations....