The alternating group 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:

$(i j) (k l) = (i j k) (j k l)$

$(i j) (j k) = (i j k)$

$(i j) (i j) = e$ .

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 $(i j k) = (i j) (j k)$ .