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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.

In the Klein four-group $V=\{1,a,b,ab\}$, the chainsare distinct composition series. All their composition factors are $C_2$.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same composition factors up to permutation and group isomorphism. Thus the subgroup chains may differ, while the multiset of simple quotients is intrinsic.

Supposeare two factorizations into primes. Repeatedly apply the cyclic-subgroup construction of part (c), first removing $p_r$, then $p_{r-1}$, and so on. This constructs a composition series for $C_n$ whose composition factors are $C_{p_1},\dots ,C_{p_r}$. The second factorization constructs another composition series with composition factors $C_{q_1},\dots ,C_{q_s}$.

By the Jordan–Hölder theorem, $r=s$ and these composition factors agree up to group isomorphism and permutation. Since isomorphic finite groups have the same order, the corresponding primes agree. This proves uniqueness of prime factorization up to permutation, using existence of prime factorizations but not their uniqueness.

A composition series is a finite chain $1=G_0⊲G_1⊲\cdots ⊲G_r=G$ in which each inclusion is proper, each subgroup is normal in the next, and every quotient $G_i/G_{i-1}$ is a nontrivial simple group. These quotients are the composition factors.

For a finite group, use induction on its order. The trivial group has the empty series. Otherwise choose a maximal proper normal subgroup $N$; such a subgroup exists because there are finitely many subgroups and $1$ is proper. The correspondence theorem says normal subgroups of $G/N$ correspond to normal subgroups of $G$ containing $N$, so maximality makes $G/N$ simple. Inductively construct a composition series of the smaller group $N$ and append $G$. This proves existence.

A nonabelian example with abelian factors isBoth factors are simple because their orders are prime.

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.

Let $1◃N◃G$ be the given length-two composition series. Thus $N$ and $G/N$ are simple groups. Let $M$ be any proper nontrivial normal subgroup of $G$, such as the penultimate term of another composition series.

If $M=N$, its two factors are already the given ones. Suppose $M\ne N$. Since $M\cap N$ is normal in the simple group $N$, it is either $1$ or $N$. The latter would mean $N\subseteq M$, but then $M/N$ would be a proper nontrivial normal subgroup of $G/N$, impossible. Thus $M\cap N=1$.

Also $M$ cannot be contained in $N$, since it would then be a nontrivial normal subgroup of $N$ distinct from $N$. The product $MN$ is a normal subgroup of $G$, strictly containing $N$. Simplicity of $G/N$ therefore forces $MN=G$. The second isomorphism theorem for groups givesIn particular $M$ is simple. Every other composition series has such an $M$ as its last proper term, and no nontrivial refinement inside a simple $M$ is possible. Hence every composition series has length two.

The general Jordan–Hölder theorem states that any two composition series of a finite group have the same length and the same composition factors up to group isomorphism and permutation. The argument above proves both assertions in this length-two case: another series either uses $N$ itself or has factors isomorphic to the original pair in the opposite order. No general uniqueness theorem was used to prove this special case.

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.

Suppose the $S_i$ are pairwise nonisomorphic simple modules. Any submodule $W$ of their finite direct sum that projects onto every summand is the entire sum. Induct on $k$. Projection onto the first $k-1$ factors is onto by induction. Its kernel is a submodule of $S_k$, hence zero or all of $S_k$. If zero, $W$ is isomorphic to the first $k-1$ factors but also has quotient $S_k$, contradicting their distinct composition factors and the Jordan–Hölder theorem. Therefore the kernel is $S_k$, proving the assertion. Consequently, if $S_i=A{x}_i$, the map $a\mapsto (a{x}_i{)}_i$ is onto and gives $A/({\bigcap }_i{\mathrm{Ann}}_A(x_i))\cong {⨁}_i{S}_i$ as left modules. The annihilators of these vectors need only be left ideals.

For a normal subgroup $N$ of $G$, projection to the quotient group gives a bijection between subgroups of $G$ containing $N$ and subgroups of $G/N$. The maps are $H\mapsto H/N$ and inverse image under projection. Their two composites are identities because each relevant subgroup is a union of cosets of $N$. They preserve inclusions and normality; consequently a composition series in a quotient lifts to a subnormal chain above $N$ with the same composition factors.