Show that \[ [S_n,S_n]=A_n. \]
Solution
The sign homomorphism \[ \operatorname{sgn}:S_n\to\{\pm1\} \] has abelian target, so every commutator lies in its kernel. Hence \[ [S_n,S_n]\subseteq A_n. \]
For the reverse inclusion, recall that \(A_n\) is generated by \(3\)-cycles. Every \(3\)-cycle is a commutator of transpositions. For example, \[ [(12),(23)] =(12)(23)(12)(23) =(132). \] Conjugating gives every \(3\)-cycle, so all generators of \(A_n\) lie in \([S_n,S_n]\). Therefore \[ A_n\subseteq[S_n,S_n]. \]
Thus \[ [S_n,S_n]=A_n. \] For \(n=1,2\) both sides are trivial, so the statement remains valid.