Permutations disjoint cycle Write the following permutation into a product of disjoint cycles?
(145)(256)(35)
I'm unsure as to how do this?
 A: From right to left follow the "cycle" of each element, and when it gets closed take the next unused number:
$$1\to4\;,\;\;4\to 5\;,\;\;5\to3\;,\;\;3\to 5\to 6\;,\;\;6\to2\;,\;\;2\to5\to1...\text{closed}$$
and we already have the cycle $\;(1\;4\;5\;3\;6\;2)\;$...and since no digit between $\;1\;$ to $\;6\;$ is unused we're done.
If we had for example $\;(1347)(2537) = (1\;3)(2\;5\;4\;7)\;$ . Why? Because
$$1\to3\;,\;\;3\to7\to1...\text{closed, so we take}\;\;2\to5\;,\;\;5\to3\to 4\;,\;\;4\to 7\;,\;\;7\to2...\text{closed}$$ 
and we get both cycles above.
A: Just see where each element goes. Follow the cycles from right to left (I hope that this is the convention of your textbook), and inside a  cycle follow the elements from left to right.
$1\mapsto 4$
$2\mapsto5\mapsto 1$
$3\mapsto 5\mapsto 6$
$4\mapsto5$
$5\mapsto3 $
$6\mapsto2$
Hence in two-row notation we have
$$\begin{pmatrix}
1&2&3&4&5&6\\
4&1&6&5&3&2
\end{pmatrix}$$
As a product of disjoint cycles we have
$$(145362)$$
so just one cycle!
