Are there any nontrivial ways to factor n-cycles into a product of cycles? I was reading a proof here about the simplicity of $A_n (n \ge 5)$. It states (and proves) a lemma about 3-cycles:
A 3-cycle $(a, b, c)$ may be written as $(a, b, c) = (1, 2, a)^{-1}(1, 2, c)(1, 2, b)^{-1}(1,2, a)$ (here, multiplication is right to left). 
Later on, the author shows that a 3-cycles $(1, 2, 3)^{-1}$ and $(1, 2, k)$ are conjugate in $A_n$, decomposing the former into a product of the latter and some transpositions: $$(1, 2, k) = ((1, 2)(3, k))(1, 2, 3)^{-1}((1, 2)(3, k))^{-1}.$$
This certainly makes sense now that I see it, but how is this arrived at a priori? Is there a way to "factorize" cycles into other cycles? 
Thank you so much.

I believe you need another way of looking on conjugation:
If the group $S_n$ is realized as acting on $n$ points, then conjugating by some permutation is just renumerating points. (i.e. if you have a cycle $\sigma=(123)$ and a renumeration $\tau=(12)$ then the new cycle will be $\sigma=(213)$ and that is exactly the conjugation by $\tau$). Now it is very easy to construct the element in $S_n$ that changes two $3$-cycles and all you need is to check, that this element is even. Well, you have a concrete formula for the element so that's easy.
