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)$~$.