Show the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$ Show that for $n \geq 5$, the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$. 

I'm not sure on how to handle this problem. I know that the commutator subgroup $G'$ will consist of products of commutators $[\sigma_{i}, \tau_{i}]$ for permutations $\sigma_{i}, \tau_{i} \in S_{n}$. However, I don't know where to start. What would be the best way to approach this?

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*$A_n$ is generated by elements of the type $(abc)$ (i.e. 3-cycles)

*Each $3$-cycle is a commutator: $(abc)=(ab)(ac)(ab)(ac)$.

*$[S_n:A_n]=2$.
Can you complete the proof now?
