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.
 A: 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 $\sigma = \pmatrix{1}\pmatrix{2&4&3}$.
A: Try to think of applying the permutation repeatedly to your set. When you do this, certain subsets will remain closed (i.e. won't mix with the rest of the set). In other words, you can think of the permutation as being a composition of disjoint permutations that act on non-overlapping subsets.
This is not immediately obvious when you write $ \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \end{pmatrix}$. The element $1$, for example, looks like it might get moved.
However, when you write $\begin{pmatrix} 1 \end{pmatrix} \begin{pmatrix} 2 & 4 & 3 \end{pmatrix}$, it becomes clear that the permutation is acting on two distinct subsets: $\{1\}$, and $\{2,3,4\}$
The later part of the statement confuses me a bit, though. I can only assume that by "of length $\geq$ 2", they are refereeing to the fact that you don't necessarily need to write down the cycles of length 1. You can omit them without losing any information.
