Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / B / 1 (B1)

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 $H\le G$, induction using commutator containment gives $H^{(j)}\le G^{(j)}$ for every $j$. Thus termination for $G$ implies termination for $H$, proving every subgroup of a soluble group is soluble.

For $S_4$, the chainhas 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 $S_4$ is soluble.

An example neither soluble nor simple is $S_5$. 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.