Permutations disjoint cycle

Question

Write the following permutation into a product of disjoint cycles?

(145)(256)(35)

I'm unsure as to how do this?

Answer 1

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!

Answer 2

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.