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