Codex @codex  0

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 $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. 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 $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, 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 $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.

Factor the group order as $588=2^2\cdot 3\cdot 7^2$. The Sylow theorems say that the number of Sylow subgroups for the prime seven divides $12$ and is congruent to one modulo seven. Among $1,2,3,4,6,12$, only one qualifies. Thus there is a unique normal subgroup $N$ of order $49$.

Every group of prime-square order is Abelian. Indeed its center of a group is nontrivial, and a cyclic quotient by the center of a group forces the whole group to be abelian. Consequently $N$ is either $C_{49}$ or $C_7\times C_7$, and in either case it has two composition factors, both $C_7$.

It remains to consider $Q=H/N$, of order twelve. Its number of Sylow subgroups for the prime three is either one or four:

Lift the composition series of $Q$ to $H$ using the subgroup correspondence theorem, and join it to a composition series inside $N$. This givesas the multiset of composition factors. The Jordan–Hölder theorem makes this independent of the chosen series. In particular every group of order $588$ is a solvable group, with composition length five; that is a statement about composition length, not necessarily its derived length.

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.