Help with writing a permutation in a symmetric group as a product of disjoint cycles

Question

I don't follow how the step from the product of permutations to the two-line notation is done. Any help with this would be greatly appreciated!

Answer 1

The product is taken right to left. If $f=(2467)$, $g=(12365)$, and $h=(1375)$, then each element gets fed into the product like so:

\begin{align} 1&\stackrel{f}{\mapsto}1\stackrel{g}{\mapsto}2\stackrel{h}{\mapsto}2 \\ 2&\stackrel{f}{\mapsto}4\stackrel{g}{\mapsto}4\stackrel{h}{\mapsto}4 \\ 3&\stackrel{f}{\mapsto}3\stackrel{g}{\mapsto}6\stackrel{h}{\mapsto}6 \\ &\vdots \end{align}

Can you continue from here?

Answer 2

Apply the permutation $(1375)(12365)(2467)$ on the LHS, say, to the digit $1$. Then $(2467)$ fixes it, and then $(12365)$ sends it to $2$, where it stays under $(1375)$. So $1$ is mapped to $2$ under $(1375)(12365)(2467)$. And so on with the other digits. The result is $(124)(365)$.