Commutator Group of $S_4$ and $A_4$

Question

This is directly out of Dummit and Foote S5.5 Q4. In class on Friday my professor gave this question to us as an exercise, with the answer being that all 3-cycles$\in S_4$ was the commutator group for both.

So far I have gotten to saying that because $A_4$ is normal and $S_4 / A_4 \cong \mathbb{Z} _ 2$, which is abelian so $S'_4 \leq A_4$.

I am confused on where to go next and the process needed to get to the final answer. Any help is appreciated, thanks.

Answer 1

@mesel's answer is correct, but here's a (maybe?) more easy proof.

You already proved $S_{4}' \leq A_{4}$, so let's try to prove $A_{4} \leq S_{4}'$. As mentionned in the comment, $A_{4}$ is generated by the $3$- cycles of $S_{4}$, thus you only have to show that each $3$-cycle is a commutator of $S_{4}$.

But $(abc) = (acb)(acb) = (ac)(bc)(ac)(bc) = [(ac), (bc)] \in S_{4}'$

Answer 2

A different way,

$S_3$ is a subgroup of $S_4$ and $A_3=S_3'\leq S_4'$. Thus, $|S_4'|$ is divisible by $3$. As it is contained by $A_4$ ,$|S_4'|\in \{3,6,12\}$. it is not $3$ as the subgroups of order $3$ is not normal and $A_4$ does not have a subgroup of order $6$. Hence, $S_4'=A_4$.