How to decompose permutations? In algebra, we have seen theorems such as 

Every permutation is the product of disjoint cycles of length $\geq 2$.

I don't really know how to apply this, so I looked at its proof hoping it would be helpful.

Proof: Decompose $\{1, \dots, n\}$ disjointly in orbits of $\langle \sigma \rangle$.

This didn't help me at all. I don't really know what is meant by this. Can anyone explain to me the proof and also how to actually do it? For example, how can I find:
$$\begin{pmatrix} 1 \end{pmatrix} \begin{pmatrix} 2 & 4 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$$
Thanks in advance for any help.

The key to decomposing cycles is to trace the "orbit" of each element under the permutation.
So, for example, let's decompose 
$$
\sigma= \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}
$$
We begin by finding $\sigma(1)$.  Applying the permutations from right to left, we find $1\to2$ under the right-most cycle, $2$ in turn stays the same under the middle cycle, and $2\to1$ under the leftmost cycle.  So, $\sigma(1)=1$.
Since we ended where we began, our first cycle is $\pmatrix{1}$.  We move on to the next element
Now, find $\sigma(2)$.  We find $2\to3$, $3\to4$, $4\to4$.  Thus, $\sigma(2)=4$.
Now, find $\sigma(4)$.  We find $4\to4\to3\to3$.  Thus, $\sigma(4)=3$.
Now, find $\sigma(3)$. We find $3\to1\to1\to2$.  Thus, $\sigma(3)=2$.
Since we ended where we began, our second cycle is $\pmatrix{2&4&3}$.  Since there are no more elements to permute, we are done.
Thus, we find $\sigma = \pmatrix{1}\pmatrix{2&4&3}$.
