Express the following permutation as a product of disjoint $2$ cycles :$(4,2,5,3)(1,5,2,4)$. 
Express the following permutation as a product of disjoint $2$ cycles:
$$(4,2,5,3)(1,5,2,4).$$

I can find out the permutation as $(1,5,3)(2)(4)$. We can write $$(1,5,3)(2)(4)=(1,5)(1,3)(2)(4).$$
But how to express them as disjoint $2$ cycles? I am not quite getting it.
Is there a general way to do these problems? There may be posts in this site concerning the same topic probably, but I can't seem to find it . . .
 A: Given are two permutations $\pi,\sigma$ with
\begin{align*}
\color{blue}{\pi}&\color{blue}{=(4,2,5,3)}\\
\color{blue}{\sigma}&\color{blue}{=(1,5,2,4)}\\
\end{align*}
Composition of $\pi\sigma$ (order right to left) gives using two notations
\begin{align*}
\color{blue}{\pi\sigma}&=(4,2,5,3)(1,5,2,4)=(1,3,4)(2)(5)\color{blue}{=(1,3,4)}\tag{*1*}\\
\pi\sigma
&=\begin{pmatrix}
1&2&3&4&5\\
1&5&4&2&3\\
\end{pmatrix}
\begin{pmatrix}
1&2&3&4&5\\
5&4&3&1&2\\
\end{pmatrix}
=
\begin{pmatrix}
1&2&3&4&5\\
3&2&4&1&5\\
\end{pmatrix}
\end{align*}
The right-most permutation in (*$1$*) is written without using cycles of length $1$. They can be omitted since they provide no additional information. A cycle of length $n>1$ can always be written as product of $n-1$ transpositions, i.e. cycles of length $2$. We obtain from (*$1$*)
\begin{align*}
\color{blue}{\pi\sigma}&=(1,3,4)\,\,\color{blue}{=(1,3)(3,4)}\tag{*2*}\\
\end{align*}
We do not need in (*$2$*) the usage of cycles of length $1$. We can write $\pi\sigma$ as product of two transpositions.
