Understanding how to get composition of disjoint cycles from composition of nondisjoint cycles in a proof I am looking at the proof of the theorem given on page 19 of this document as:
"Writing $\sigma \in S_n$ as a product of transpositions in different ways, $\sigma$ is either always composed of an even number of transpositions, or always an odd number of transpositions."
One confusing part of this proof to me is the line:
$$(c \enspace a_2 \enspace \cdots \enspace a_{k-1} \enspace d \enspace a_{k+1} \enspace \cdots \enspace a_{k+l})(c \enspace d) = (d \enspace a_2 \enspace \cdots \enspace a_{k-1})( c \enspace a_{k+1} \enspace \cdots \enspace   a_{k+l}) $$
How do I understand this composition of cycles on the left-hand side as the composition of cycles on the right-hand side? Is there a step-by-step way to show how to get from one composition to the other?
Thanks.
 A: I explain how I do it here for specific permutations. For generic ones like this one the process is similar. It's important to know if you compose permutations left-to-right or right-to-left, though... Your equation suggests you compose them right-to-left. Meaning that if you have two cycles/permutations, $\sigma_1$ and $\sigma_2$, and you write $\sigma_1\sigma_2$, then you mean that $\sigma_2$ is applied first, and then $\sigma_2$.
Consider your left hand side (I will use commas instead of forcing space):
$$(c, a_2, \ldots, a_{k-1}, d,a_{k+1}, \ldots,a_{k+l})(c, d)$$
What happens to $c$? First we apply the transposition $(c,d)$, which sends $c$ to $d$. Then we apply the cycle, which sends $d$ to $a_{k+1}$. Thus, $c$ is sent to $a_{k+1}$. So we start writing the result as:
$$(c,a_{k+1},$$
Next, $a_{k+1}$ is left alone by $(c,d)$, and then sent to $a_{k+2}$; then $a_{k+2}$ is left alone by $(c,d)$, and then sent to $a_{k+3}$, and so on until we get to $a_{k+l}$. So now we have:
$$(c,a_{k+1},a_{k+2},\ldots,a_{k+l},$$
Next, $a_{k+l}$ is left fixed by $(c,d)$, and then sent to $c$ by the long cycle, so this "closes" this cycle and we get
$$(c,a_{k+1},\ldots,a_{k+l})\cdots$$
Next, what happens to $d$? First it is sent to $c$, then $c$ is sent to $a_2$; so $d$ is sent to $a_2$:
$$(c,a_{k+1},\ldots,a_{k+l})(d,a_2,$$
then $a_2$ is fixed by $(c,d)$ and sent then sent to $a_3$, and so on until we get to $a_{k-1}$. Then $a_{k-1}$ is fixed by $(c,d)$, and sent to $d$ by the cycle, so we get
$$(c,a_{k+1},\ldots,a_{k+l})(d,a_2,a_3,\ldots,a_{k-1}).$$
And this covers all the elements that appear in either of the two permutation, so we get the final result:
$$(c,a_2,\ldots,a_{k-1},d,a_{k+1},\ldots,a_{k+l})(c,d) = (c,a_{k+1},\ldots,a_{k+l})(d,a_2,a_3,\ldots,a_{k-1}).$$
Which is what you had.
If after completing the second cycle there were still terms left over, you would start "running them through" to see what happens to them, leading to further cycles on the right hand side.
