= Solution
Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a <soluble group> if this <derived series> reaches the identity after finitely many steps. Its <derived length> is the least $d$ with $G^{(d)}=1$.
Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\leq A_4$. This <Klein four-group> is a <normal subgroup> and Abelian; $A_4/V$ has order three, so $A_4'\subseteq V$. The group $A_4$ is nonabelian, so its derived <subgroup> is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived <subgroup> is a <normal subgroup>, containing any one forces it to contain all three. Thus $A_4'=V$, and $V'=1$. Therefore \b[$A_4$ has <derived length> two].
If $H\leq G$, every <commutator> in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a <subgroup> of a <solvable group> is solvable, with no larger <derived length>.
The sign map gives $S_4'\subseteq A_4$. The <commutator> $[(12),(23)]$ is a three-cycle. Normality of $S_4'$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4'=A_4$, so the complete <derived series> is
$$
\boxed{S_4\supset A_4\supset V\supset1.}
$$
Every inclusion is strict, proving solvability and <derived length> three.
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