= Solution
{c}
A finite <soluble group> is one whose <derived series> $G^{(0)}=G$, $G^{(j+1)}=[G^{(j)},G^{(j)}]$ reaches the trivial group. Equivalently it has a finite normal series with abelian factors. If $H\leq G$, induction using <commutator> containment gives $H^{(j)}\leq G^{(j)}$ for every $j$. Thus termination for $G$ implies termination for $H$, proving \b[every <subgroup> of a <soluble group> is soluble].
For $S_4$, the chain
$$
\{1\}\triangleleft V_4\triangleleft A_4\triangleleft S_4
$$
has abelian factors $V_4$, $C_3$ and $C_2$. Here $V_4$ consists of the identity and the three double transpositions. It is normal because conjugation permutes those double transpositions. Hence \b[$S_4$ is soluble].
An example neither soluble nor simple is \b[$S_5$]. Its <subgroup> $A_5$ is proper, nontrivial and normal, so $S_5$ is not simple. The group $A_5$ is nontrivial and perfect: its generating three-cycles are commutators of double transpositions, using two additional symbols, so $[A_5,A_5]=A_5$. For example the <commutator> of $(12)(45)$ and $(23)(45)$ is a three-cycle, and conjugation and inversion give all three-cycles. Thus $A_5$ is not soluble, and <subgroup> inheritance then proves $S_5$ is not soluble either.
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