Working with the knowledge that the set of 3-cycles generates $A_{n}$, the basic idea is to express any 3-cycle as a word in $(123)$ and $(12…n)$ when $n$ is odd.
Not knowing how to progress, I decided to work with the concrete group $A_{5}$. I've listed all 3-cycles of $A_{5}$ below with each element followed by its inverse: $$(123),(132),(124),(142),(125),(152),(134),(143),(135),(153),(145),(154),(234),(243),(235),(253),(245),(254),(345),(354)$$ Now, I am able to generate the following elements $(234),(345),(451),(512)$ (and, correspondingly, their inverses) using the following rule:$$(12345)^{i}(123)(12345)^{-i}$$ for $i=1,2,3,4$ (for $i=5$ we get back $(123)$).
Promising as this looks, I am not able to extend the idea to cover the other 3-cycles in $A_{5}$.
Suggestions and hints are appreciated. Would prefer to answer myself.
Best Answer
Note that
\begin{align*} (1\ 2\ \dots\ n)(1\ 2\ 3)(1\ 2\ \dots\ n)^{-1} &= (2\ 3\ 4)\\ (1\ 2\ \dots\ n)(2\ 3\ 4)(1\ 2\ \dots\ n)^{-1} &= (3\ 4\ 5)\\ &\ \ \vdots \end{align*}
More generally, given a permutation $\sigma$ and a cycle $(a_1\ a_2\ \dots\ a_k)$, we have
$$\sigma(a_1\ a_2\ \dots\ a_k)\sigma^{-1} = (\sigma(a_1)\ \sigma(a_2)\ \dots\ \sigma(a_k)).$$
Finally, use the fact that $(a\ b\ c)^{-1} = (a\ c\ b)$.