Prove the alternating group on \(n\) letters is generated by the 3-cycles for \(n \geq 3\).
Solution
Every element of \(A_n\) is a product of an even number of transpositions, so it suffices to express a product of two transpositions as a product of \(3\)-cycles.
If the transpositions share one letter, then \[ (ab)(bc)=(abc). \] If they are disjoint, then \[ (ab)(cd)=(abc)(bcd). \] If they are equal, their product is the identity. Hence every even permutation is a product of \(3\)-cycles.
Conversely, every \(3\)-cycle is even because \[ (abc)=(ac)(ab). \] Therefore the subgroup generated by all \(3\)-cycles is exactly \(A_n\) for \(n\ge3\).