Permutations in products of disjointed cycles

Question

How do I calculate the following permutation in the symmetric group $S_6$ giving the answers as products of disjoint cycles: $$(2,3,5,6)(1,6,2,4)$$

I have tried following this question but I don't understand the process of reading right -> left as it seems to change direction and I don't truely understand why.

Answer 1

The product (composition) of these two permutations looks like this.

$$\begin{array}{ccccccr} 1&2&3&4&5&6& \\ \downarrow&\downarrow&\downarrow&\downarrow&\downarrow&\downarrow& (1,6,2,4)\\ 6&4&3&1&5&2&\\ \downarrow&\downarrow&\downarrow&\downarrow&\downarrow&\downarrow&(2,3,5,6)\\ 2&4&5&1&6&3 \end{array}$$

So in the product you have $(1,2,4)(3,5,6)$.

With a bit of practice, you don't need to write out the complete functions.

Answer 2

Begin with one: "one goes to six (on the right, of course), which then goes to 2 (on the left), so one goes to two" : $\;1\to 2\;$ .

Now with two: "two goes to four which then doesn't move" , so $\;2\to 4\;$ .

Continue as above until a cycle is closed and then begin with the smallest number left, if any, and etc.:

$$(2\;3\;5\;6)(1\;6\;2\;4)=(1\;2\;4)(3\;5\;6)$$

Answer 3

Cycles are functions. The product of two cycles is just a composition of two functions, thus we read right to left. First look at $(4 1 3)$, then $(1 2 3 5)$. But inside the cycle you read left to right, that is $4 \rightarrow 1 \rightarrow 3 \rightarrow 4$.