Calculate the commutator subgroup of $S_4$

Question

So I have been tasked with calculating the commutator subgroup of $S_4$. As a warmup, I was able to calculate the commutator subgroup of $S_3$ through brute force calculations as there were only $6^2$ possibilities. I found that ${S_3}^{'}=\{e,(1\,\,2\,\,3),\,(1\,\,3\,\,2)\}$.

For $S_4$, I clearly do not want to attempt all $24^2$ computations, so what kind of strategy could I employ to get this done in a reasonable amount of time?

Answer 1

First note that all commutators will be even permutations.

Then note that $[(a,c),(b,c)] = (a, b, c)$, if $a, b, c$ are distinct.

So in $S_{4}'$ you find all the $3$-cycles.

Answer 2

First, observe that each basic commutator $\;[a,b]\;,\;\;a,b\in S_4\;$ is in fact in $\;A_4\;$ . This can also be seen as follows :

$\;S_4/A_4\cong C_2\;$ is abelian and thus $\;[S_4,S_4]\le A_4\;$.

Now just count up some elements and deduce $\;S_4'=A_4\;$ , for example taking into account that $\;A_4\;$ has no subgroup of order six.