# Commutator subgroup of Sn

It depends what $n$ is. If $n\geq 5$ then $(S_n)' = A_n$, to see a proof go here. If $n=4$ then $(S_4)'$ must be a normal subgroup. This means $(S_4) ' = S_4, A_4,V$ where $V=\{(1),(12)(34),(13)(24),(14)(23)\}$. It remains for you to eliminate which of these are not possible. We begin by noting that $(S_4)'\not = S_4$ because $S_4/A_4$ is abelian which would imply that $S_4\subseteq A_4$ a contradiction. Now for $S_4/V$ to be abelian it would imply it is a cyclic group of order six. But it is impossible for $\sigma V\in S_4/V$ to have order $6$ since $\sigma$ is a cycle. Thus, $S_4/V\simeq S_3$ is not abelian which implies $A_4$ must be the commutator subgroup. Try doing it for $(S_3)'$.