Are there any nontrivial ways to factor n-cycles into a product of cycles?

Question

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.

Answer 1

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.

Answer 2

If the permutation $x$ maps $i$ to $j$, then conjugation by $g$ (i.e. $gxg^{-1}$) maps $g(i)$ to $g(j)$ since $gxg^{-1}(g(i)) = gx(i) = g(j)$. Thus, the permutation $gxg^{-1}$ has the same cycle structure as $x$, but with the points $i$ replaced by their image $g(i)$.

For example if the 3-cycle $x=(132)$ is already in the normal subgroup $H$ of $A_n$, and we want to obtain another 3-cycle, say $(2k1)$, then we look for a map $g$ that takes $1$ to $2$, $3$ to $k$ and $2$ to $1$; for example, consider $g=(12)(3k)$. Observe then that $gxg^{-1}$ is the 3-cycle $(12k)$, and by normality $gxg^{-1}$ is required to be in the subgroup $H$. Thus, if $H$ contains the 3-cycle $(123)$ then $H$ also contains the 3-cycle $(12k)$, and so on. We try to get all the 3-cycles in this manner by considering action by conjugation, which expresses the new 3-cycles as a product of three factors that have the form $gxg^{-1}$.