Showing that $A_n$ is generated by the 3-cycles in $S_n$

Question

I am trying to show that $A_n$ is generated by the 3-cycles in $S_n$. It seems that every 3-cycle of the form $(a_1,a_2,a_3)$ can just be written as $(a_1,a_3)(a_1,a_2)$ so every 3-cycle turns into an even number of transpositions (2-cycles). Is this sufficient?

Answer 1

An element of $A_n$ is a product of an even number of transpositions. For any pair of transpositions, find one or two 3-cycles whose product is equal to their product. After doing this you can generate all products of an even number of transpositions.

Answer 2

Any element of the alternating group $\alpha \in A_n$ can be written as a product of an even number of transpositions with the decomposition $\alpha = (i_1 i_2 \dots i_r) = (i_1 i_r)(i_1 i_{r-1}) \dots (i_1 i_2)$ where $r$ is odd. Then we have that any pair of transpositions with a common element can be written as a 3-cycle $(i_1 i_k)(i_1 i_{k-1}) = (i_1 i_{k-1} i_k)$. By iterating on pairs of tranposition in the aforementioned decomposition, we can rewrite $\alpha$ as a product of 3-cycles.