= Solution
A <subnormal series> of a <finite group> is a chain $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_r=G$. A <normal series of a group> requires every $G_i$ normal in the whole group. This distinction matters: a <composition series> is a <subnormal series> with nontrivial simple factors, whereas a <chief series> is a strict, unrefinable <normal series of a group>, whose factors are <minimal normal subgroups> of the appropriate ambient quotients. Both exist in a <finite group>, by successively choosing maximal proper <normal subgroups> or a maximal chain of ambient-normal subgroups. The direct-product result proved above shows that <chief factors> are powers of isomorphic <simple groups>, rather than necessarily simple themselves.
The <Jordan–Hölder theorem> says that any two <composition series> have the same multiset of simple factors, up to isomorphism and order. Here is an induction proof. The assertion is trivial for the identity group. Let $A,B$ be the penultimate terms of two series for $G$; they are maximal proper <normal subgroups>. If $A=B$, apply induction there. Otherwise $AB=G$, since its image is a nontrivial <normal subgroup> in each simple quotient. Put $C=A\cap B$. The maps to the quotients give
$$
A/C\cong G/B,\qquad B/C\cong G/A.
$$
For example, $a\mapsto aB$ has kernel $C$ and is onto because $AB=G$. Take a <composition series> of $C$ and append $A$, obtaining a <composition series> of $A$ because $A/C$ is simple; do the same with $B$. Induction inside these smaller groups shows that the factors in the two original series are, in both cases, the factors of $C$ together with $G/A$ and $G/B$. This proves the theorem and justifies the phrase <Jordan–Hölder factors> independently of a chosen series.
The <derived series> is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are <characteristic subgroups>, hence normal. A group is soluble if this series reaches the identity. Equivalently it admits a <subnormal series> with abelian factors. One direction uses the <derived series> itself. For the other, note the extension principle: if $N\triangleleft G$ has derived length $a$ and $G/N$ has derived length $b$, then $G^{(b)}\le N$, so $G^{(a+b)}=1$. Applying this principle along an abelian-factor series proves solubility.
Subgroups and quotients of a <soluble group> are soluble: $H^{(i)}\le G^{(i)}$ for a subgroup, and the <commutator subgroup> of a quotient is the image of the <commutator subgroup>. A finite soluble <simple group> is abelian, since its <commutator subgroup> is normal and cannot equal the whole group if the <derived series> terminates. An abelian simple <finite group> is cyclic of prime order. Therefore \b[a <finite group> is soluble exactly when all its <Jordan–Hölder factors> are cyclic of prime order]: necessity follows from the subgroup/quotient property, and sufficiency follows from the extension principle. Its <chief factors> are elementary abelian, by the minimal-normal-subgroup result.
For nilpotence, the <lower central series> is $\gamma_1(G)=G$, $\gamma_{i+1}(G)=[\gamma_i(G),G]$. A group is nilpotent if $\gamma_{c+1}=1$ for some $c$. An ascending <central series> $1=N_0\le\cdots\le N_c=G$ satisfies $[N_{i+1},G]\le N_i$. Reversing a terminating <lower central series> gives such a <central series>. Conversely, induction gives $\gamma_i(G)\le N_{c-i+1}$, so any such <central series> forces $\gamma_{c+1}=1$. The <upper central series> $Z_0=1$, $Z_{i+1}/Z_i=Z(G/Z_i)$ contains every ascending <central series> term by term, and thus reaches $G$ exactly in the nilpotent case. Because $[\gamma_i,\gamma_i]\le[\gamma_i,G]=\gamma_{i+1}$, induction gives $G^{(i)}\le\gamma_{i+1}$: \b[<nilpotent groups> are soluble].
For <finite groups> there is the stronger structural characterization: \b[a <finite group> is nilpotent exactly when it is the <direct product of groups> formed from its <Sylow subgroups>]. To prove the forward direction, the <upper central series> gives the <normalizer> condition. If $H<G$, choose its first term $Z_i$ not contained in $H$, with $Z_{i-1}\le H$. An element of $Z_i\setminus H$ has all <group commutators> with $H$ in $Z_{i-1}$, so normalizes $H$. Hence $H<N_G(H)$.
For a <Sylow subgroup> $P$, its <normalizer> is self-normalizing. Indeed, if $x$ normalizes $N_G(P)$, the two <Sylow subgroups> $P^x,P$ of that <normalizer> are conjugate there; adjusting $x$ by an element of $N_G(P)$ puts it in $N_G(P)$ already. Applying the strict <normalizer> condition to $N_G(P)$ shows that it cannot be proper, so $P$ is normal. Normal <Sylow subgroups> of distinct primes commute, since their <group commutators> lie in their trivial intersection. Their product has <cardinality> $|G|$ and is direct.
Conversely, a nontrivial finite $p$-group has nontrivial centre: in the class equation every noncentral <conjugacy class> has size divisible by $p$, so $p$ divides the centre's order. Induct on its order through the quotient by its centre to obtain an <upper central series>. Thus every finite $p$-group is nilpotent, and a finite <direct product of groups> of them is nilpotent because its <upper central series> is obtained componentwise.
A finite <supersolvable group> has a <normal series of a group> with cyclic prime-order factors. It is soluble, but the ambient-normal condition is stronger than the condition on a <composition series>. Finite <nilpotent groups> are supersolvable: a $p$-group has a central subgroup of order $p$, and induction in its quotient builds an ambient-normal prime-factor series; <direct products of groups> of the Sylow factors combine these series. The converses fail. The group $S_3$ is supersolvable via $1<C_3<S_3$, but is not nilpotent because its Sylow $2$-subgroups are not normal. The group $A_4$ is soluble via $1<V_4<A_4$, but is not supersolvable: no subgroup of order two is normal, since the three involutions are conjugate, and no subgroup of order three is normal, since there are four such subgroups. In this example $1<C_2<V_4<A_4$ is a <composition series> whose first $C_2$ is not normal in $A_4$, while $1<V_4<A_4$ is a <chief series> with a nonsimple <chief factor> $V_4$.
Finally, a <perfect group> satisfies $G'=G$. In any <finite group>, the <derived series> eventually stabilizes at a perfect <characteristic subgroup>, because a descending sequence of finite subgroup orders eventually stops decreasing. It is trivial exactly when the group is soluble. A nonabelian <simple group> is perfect: its nontrivial <commutator subgroup> is normal and must be the whole group. Thus series distinguish successively stronger notions, summarize the building blocks through <Jordan–Hölder factors>, and separate soluble behaviour from a persistent perfect part.
Back to article page
