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?

The fact that $A_n$ is simple for $n\geq 5$ can be used to answer this question in quite a different way from the other answers. Let $N=[S_n, S_n]$ denote the commutator subgroup. Then:


*

*(General fact about groups.) Note that $G/H$ is abelian if and only if $[G, G]\leq H$.

*$N\leq A_n$ as $S_n/A_n$ is abelian. Note that as $N$ is normal in $S_n$ it is also normal in $A_n$.

*$N$ is non-trivial as $S_n$ is non-abelian.

*Therefore, $N$ is a non-trivial, normal subgroup of $A_n$. As $A_n$ is simple we have that $N=A_n$, as required.
