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In a finite group, choose a maximal chain of normal subgroups from $1$ to $G$. Finiteness ensures such a chain exists, and maximality leaves no normal subgroup of $G$ strictly between successive terms. This is a chief series.

The infinite cyclic group has no chief series. In any finite chain starting at zero, the first nonzero subgroup is $n\mathbb{Z}$. But $2n\mathbb{Z}$ is a nonzero proper subgroup of it and is normal in $\mathbb{Z}$. Thus zero is not maximal among the normal subgroups in that first term, contradicting the definition.

For any chief series, $G_{n-1}$ is a maximal proper normal subgroup of $G$. The subgroup correspondence theorem identifies normal subgroups of $G/G_{n-1}$ with normal subgroups of $G$ containing $G_{n-1}$. Consequently the top chief factor is simple.

Other chief factors need only have no proper nonzero subgroup invariant under the action of the whole group. In $A_4$,is a chief series, where $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Conjugation in $A_4$ permutes the three involutions transitively, so none of its three order-two subgroups is normal in $A_4$. Thus $V_4$ is a minimal nontrivial normal subgroup; the top quotient is $C_3$. But the bottom chief factor $V_4/1\cong C_2\times C_2$ is not simple, since it has proper order-two normal subgroups as an abstract group.

A composition series is a finite chainwhose inclusions are strict and whose quotients $G_i/G_{i-1}$ are simple. These quotients are the composition factors. The normal subgroup condition is relative to the next member, rather than necessarily to all of $G$.

Induct on the order of a finite group. The trivial group has the empty composition series. Otherwise choose a maximal normal subgroup $N⪇G$, which exists because there are finitely many subgroups. The subgroup correspondence theorem identifies normal subgroups of $G/N$ with normal subgroups of $G$ containing $N$, so maximality makes $G/N$ a simple group. By induction $N$ has a composition series; appending $G$ gives one for $G$.

For example, the nonabelian symmetric group $S_3$ haswith composition factors $C_3$ and $C_2$, both abelian.

Write $C_n=⟨g⟩$ and set $N=⟨g^p⟩$. The order of $g^p$ is $n/\gcd (n,p)=m$, so $N\cong C_m$. Every subgroup of an abelian group is a normal subgroup. The quotient group $C_n/N$ is generated by $gN$ and has order $n/m=p$, hence is $C_p$ and is a simple group. By the subgroup correspondence theorem, there is no proper normal subgroup strictly between $N$ and $C_n$. Therefore $N$ is a maximal normal subgroup with the required quotient group.

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.