Disjoint cycles vs product of cycles I just want to address the idea of disjoint cycles and product of cycles because I believe I am confusing the idea.
My professor said:
Write $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ as a product of disjoint cycles in $S_7$ 
where the answer is:
$$(2\:5)(6\:4\:7)(2\:4\:5\:3)=(2\:7\:6\:4)(3\:5)$$
but the idea here is he moved from right to left.
Now I'm working out problems and I encountered this problem:
Consider the permutation $a=(1\:4\:6\:3\:7)(2\:8)$ and $b=(2\:5\:3)(4\:7\:8\:1)$ in $S_8$
now calculate: 
$$ab=(1\:4\:6\:3\:7)(2\:8)(2\:5\:3)(4\:7\:8\:1)=(1\:7\:4\:6\:2)(3\:8\:5)$$
but in order to calculate the product, the work was done from left to right?
So my question is why for disjoint cycles, you move from right to left but for products left to right?
 A: For the first one, writing $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ as a product of disjoint cycles means that you must write it in a way no number is repeated in any other cycle. In the cycles $(2\:5)$ and $(2\:4\:5\:3)$ we clearly see that $2$ and $5$ are repeated. So what can we do? Firstly write this product as a cycle. Take any number, for example $2$. We have $(2\:5)(6\:4\:7)(2\:4\:5\:3)$. Thus the image of 2 according to $(2\:4\:5\:3)$ is 4 and the image of 4 according to $(6\:4\:7)$ is 7 and the image of 7 according to $(2\:5)$ is itself,i.e, 7 since it doesn't appear in the cycle. So the image of 2 according to $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ is 7. Start writing $(2\:7$. Now what's the image of 7? Do the same think, you'll get 6. $(2\:7\:6$. The image of 6 is 4. $(2\:7\:6\:4$. The image of 4 is 2. $(2\:7\:6\:4\:2$. WAIT A MINUTE! 2 again! So we stop here: $(2\:7\:6\:4)$. But in $(2\:5)(6\:4\:7)(2\:4\:5\:3)$ there are numbers not in $(2\:7\:6\:4)$ which are 3 and 5. According to $(2\:5)(6\:4\:7)(2\:4\:5\:3)$, the image of 3 is 5 and the image of 5 is 3, $(3\:5)$. We have all the numbers here: $(2\:7\:6\:4)$ and $(3\:5)$. Thus:$$(2\:5)(6\:4\:7)(2\:4\:5\:3)=(3\:5)(2\:7\:6\:4)=(2\:7\:6\:4)(3\:5)$$
We have commutivity because the cycles are disjoint.
Edit: For your question left or right, it's comes from commutativity of disjoint cycles:$$a=(1\:4\:6\:3\:7)(2\:8)=(2\:8)(1\:4\:6\:3\:7)$$
