The Alternating group $A_n$ is generated by its $3$-cycles. That is, every element of $A_n$ can be made by multiplying together $3$-cycles only.

Proof

The product of two transpositions is a product of $3$-cycles:

• $(ij)(kl) = (ijk)(jkl)$
• $(ij)(jk) = (ijk)$
• $(ij)(ij) = e$.

Therefore any permutation which is a product of evenly-many transpositions (that is, all of $A_n$) is a product of $3$-cycles, because we can group up successive pairs of transpositions.

Conversely, every $3$-cycle is in $A_n$ because $(ijk) = (ij)(jk)$.