Product of disjoint cycle I've found the question to find product cycle of let be $\phi$ = (2, 4, 9, 7,) (6, 4, 2, 5, 9) (1, 6) (3, 8, 6) in S9. I know how to express the permutation of S9 as product of disjoint cycle, like the identity of S9 can be expressed by (1)(2), But I've 2 point that I don't really understand of this question.


*

*Why we must express $\phi$ as a product of disjoint cycle again, while (2, 4, 9, 7,) (6, 4, 2, 5, 9) (1, 6) (3, 8, 6) is "disjoint cycle" itself.

*In the first cycle 2 ---> 4, but in the second 2 ---> 5, I don't understand.


Any help would be appreciated.
 A: The permutation $\phi \in S_9$ is the product of those four cycles, which are not disjoint.  Thinking of permutations as bijective functions from a set of nine elements back to itself, the product is a composition.
Consider the number $1$.  Reading the cycles from right to left (as functions) and noting that a number is fixed if it's not mentioned in a cycle, we get
$$
1 \mapsto 1 \mapsto 6 \mapsto 4 \mapsto 9
$$
so $\phi(1) = 9$.  Now, what is $\phi(9)$?  Continue like that until you get back to $1$.  This will be one of the disjoint cycles in $\phi$. 
A: To use my above comment as an answer, they want to to take the cycles
$$(2 4 9 7)(6 4 2 5 9)(1 6)(3 8 6)$$
and write them as disjoint cycles.
So, going from left to right $$1\rightarrow 6 \rightarrow3$$ $$2\rightarrow4\rightarrow2$$
$$3 \rightarrow8$$ $$4 \rightarrow 9 \rightarrow 6 \rightarrow1$$
$$5 \rightarrow9$$ $$6 \rightarrow4$$
$$7 \rightarrow 2 \rightarrow5$$
$$8\rightarrow6$$
$$9 \rightarrow7$$
So, all told we have $$(1 3 8 6 4)(2)(5 9 7)$$ in disjoint cycle notation. 
A: 1 > 1 > 6 > 4 > 9
9 > 9 > 9 > 6 > 6
6 > 3 > 3 > 3 > 3
3 > 8 > 8 > 8 > 8
8 > 6 > 1 > 1 > 1
thus we have (19638)
2 > 2 > 2 > 5 > 5
5 > 5 > 5 > 9 > 7
7 > 7 > 7 > 7 > 2
thus we have (257)
and together we have (19638)(257)
which is a product of disjoint cycles
