Problem 5.4.5 (Commutator subgroup of $S_n$)

Proof. Let $S_n^{\prime }$ be the commutator subgroup of $S_n$.

As in Exercise 4, since $\mathrm{sgn}(a^{-1}{b}^{-1}\mathit{𝑎𝑏})=1$ for all $a,b\in S_n$, every commutator is in $A_n$, thus the subgroup generated by the commutators is contained in $A_n$:

(Alternatively, since $S_n∕A_n\simeq Z_2$ is abelian, then $S_n^{\prime }\le A_n$ by Proposition 7 (4).)

Conversely, for all $a,b,c$,

therefore every $3$-cycle is a commutator. Since $A_n$ is generated by the $3$-cycles, $A_n\le S_n^{\prime }$, so