Commutator series of S4 I saw Don Antonio's answer Calculate the commutator subgroup of $S_4$ and got my hint. Can you see if I am correct? I use 1 theorem: G Abelian iff G' = 1.
S4 has 3 proper normal subgroups: A4, K4, 1.
Since S4 is nonAbelian, S4' =/= 1.
Since [(12),(13)] = (123), not in K4, S4' =/= K4.
Therefore, S4'=A4.
We WTF A4'.
A4 has only 2 proper normal subgroups, K4, 1.
But since A4 nonAbelian, A4' =/= 1, and so A4' = K4.
Then since K4 is Abelian, K4' = 1.
Answer: S4 > A4 > K4 > 1.
Is this OK?
 A: You are quite correct. It is first of all true that for any set $A$ the derived subgroup of the finitary symmetric group on $A$ (the group of all permutations of finite support) has the alternate group on $A$ as its derived subgroup.
In the particular case of $\mathrm{A}_4$, since you know that the Klein-type subgroup $K$ is normal and that any element of $\mathrm{A}_4$ is either a bi-transposition (term coined up ad-hoc by me by which I mean a product of $2$ disjoint transpositions) in $K$ or a $3$-cycle, you only need to assess the commutator between two $3$-cycles (a commutator which has at least one term in $K$ will itself belong to $K$ by normality); two $3$-cycles of equal support are either equal or one is the square of the other, so at any rate they commute and their commutator is therefore trivial; if not, then by the inclusion-exclusion principle their supports intersect necessarily in a two element set $\{a, b\}$, such that one of the cycles is of support $\{a, b, c\}$ and the other one of support $\{a, b, d\}$; without any loss of generality then the commutator studied will be either of the type:
$$[(abc)(abd)]=(ab)(cd)$$
or
$$[(abc)(bad)]=(ac)(bd)$$
and at any rate will belong to $K$. Hence, on the one hand $\mathrm{D}(\mathrm{A}_4) \leqslant K$ and the reverse inclusion can be established either by your line of reasoning or by noticing that the above commutation relations express all elements of $K$ as commutators between some pairs of $3$-cycles.
