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:

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

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 $\left(ijk\right)=\left(ij\right)\left(jk\right)$.