Product of disjoint cycles in Abstract Algebra I've got some permutations:
$\left( \begin{smallmatrix}
  1 & 2 & 3 & 4 & 5 & 6 \\
 3 & 5 & 4 & 1 & 6 & 2 \end{smallmatrix} \right) $ and $\left( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 5 & 4 & 6 \end{smallmatrix} \right)$
I've found the cycles to be: $(1,3,4)(2,5,6)(1,3)(2)(4,5)(6)$ but now I want to decompose this into product of disjoint cycles.
I have no idea how to do that. After I do that, I want to write them as a product of transposition. I've searched high and low for a good explanation on how to do this but none seems to help.
 A: This is a decomposition into product of disjoint cycles and into product of transpositions
$$\left( \begin{smallmatrix}
  1 & 2 & 3 & 4 & 5 & 6 \\
 3 & 5 & 4 & 1 & 6 & 2 \end{smallmatrix} \right)=(1\ 3\ 4)(2\ 5\ 6)=(1\ 4)(1\ 3)(2\ 6)(2\ 5) $$
Can you do the same for the second permutation?
A: You have two separate permutations. $$\left( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 5 & 4 & 6 \end{smallmatrix} \right),\quad \text{ and }\quad\left( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 1 & 5 & 4 & 6 \end{smallmatrix} \right)$$
The first permutation, written as the product of disjoint cycles, is given by $$(1,3,4)(2,5,6)$$
The second, by $$(1,3)(2)(4,5)(6) = (1, 3)(4, 5)$$
To write a product of disjoint cycles as a product of transpositions, there are any number of ways to do so. One algorithm, for example, is used here: $$(a_1, a_2, a_3)(a_4, a_5, a_6, a_7) = (a_1, a_3)(a_1, a_2)(a_4, a_7)(a_4, a_6)(a_4, a_5)$$
Using that approach on your first permutation, we decompose $$(1,3,4)(2,5,6) = (1, 4)(1, 3)(2, 6)(2, 5)$$
Note that your second permutation, $(1,3)(2)(4,5)(6) = (1, 3)(4, 5)$, is already the product of transpositions.
One-easy-to-remember algorithm, which I'll apply to your first permutation, is as follows, $$(1, 3, 4)(2, 5, 6) = (1, 3)(3, 4)(2, 5)(5, 6)$$

Note: Do not be troubled that there are two correct decompositions here. You just need to know that composing any correct-but-non-unique decompositions will yield the same permutation you have prior to decomposing it into the product of transpositions.
