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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.

A simple group is a nontrivial group whose only normal subgroups are itself and the trivial subgroup.

A composition series is a chain $G=G_0▹G_1▹\cdots ▹G_r=1$ in which each $G_{i+1}$ is a normal subgroup of $G_i$ and each quotient group $G_i/G_{i+1}$ is simple and nontrivial. Its length is $r$, and these quotients are its composition factors. The individual terms need not all be normal in $G$.

A solvable group is a group whose derived series reaches the trivial group: starting with $G^{(0)}=G$, define $G^{(i+1)}=[G^{(i)},G^{(i)}]$. For finite groups this is equivalent to having a composition series whose factors are all cyclic groups of prime order.

The cyclic subgroup $⟨x^3⟩$ is normal: conjugation by $y$ sends $x^3$ to $x^{12}=x^3$, and conjugation by $x$ fixes it. Modulo this subgroup, the relation $yx{y}^{-1}=x^4$ becomes $yx{y}^{-1}=x$, so the quotient group is Abelian. Thus the commutator subgroup $G^{\prime }$ is contained in $⟨x^3⟩$.

Conversely the displayed relation gives the commutator $yx{y}^{-1}{x}^{-1}=x^3$, so $⟨x^3⟩\subseteq G^{\prime }$. HenceSince $x$ has element order $9$, $x^3\ne e$, so the first commutator subgroup is nontrivial. ThereforeHere derived length counts the number of steps needed for the derived series to reach the trivial group.

For a nontrivial simple soluble group, the commutator subgroup is either the identity or the whole group. It cannot be the whole group because the derived series would never terminate. The group is therefore Abelian. Every nonidentity element then generates a normal cyclic subgroup, which must be the whole group. An infinite cyclic group has proper nontrivial subgroups; a finite cyclic group is simple exactly when its order is prime. Thus the soluble simple groups are precisely the cyclic groups of prime order.