Simplify $(1\, 2)(3\, 4)(1\, 3)(2\, 4)$ into composite of two transpositions

Question

What I did was writing $(1\, 2)(3\, 4)(1\, 3)(2\, 4)$ as $\bigl(\begin{smallmatrix} 1& 2 &3 & 4\\ 2& 1 &4 & 3 \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix} 1& 2 &3 &4 \\ 3& 4& 1 & 2 \end{smallmatrix}\bigr)$ which is equal to $(1 4)(2 3)$

Is what I'm doing right ? And isn't there any other method for it ?

Answer 1

What you have done is correct.

Alternatively, you could make use of the fact that disjoint cycles commute and, instead, compute $(34)(12)(24)(13)$. This makes little if any difference at all.

Answer 2

Is what I'm doing right ?

Yes.

And isn't there any other method for it ?

Yes, in particular you can skip the matrix notation. Assume your permutation acts from left, so you can write it as $\sigma=\sigma_4\sigma_3\sigma_2\sigma_1$, where the indexes denote what factor acts first. Now, let's follow the story of $1$: what's the first factor doing something on ("moving") $1$? $\sigma_2$, which sends $1$ to $3$; then, $\sigma_3$ sends $3$ to $4$, and finally $\sigma_4$ does nothing on ("fixes") $4$. So, for the time being, $\sigma=(14)\dots$. Now, follow likewise yourself the story of $2$.