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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.

Any two finite subnormal series have refinements with the same successive factors up to ordering and group isomorphism. For descending chains $A_i$ and $B_j$, insert $A_{i+1}(A_i\cap B_j)$ between $A_i$ and $A_{i+1}$ and insert $B_{j+1}(B_j\cap A_i)$ in the other chain. The Zassenhaus butterfly lemma identifies the factor indexed by $(i,j)$ in the first refinement with the factor indexed by $(j,i)$ in the second. If both chains are composition series, their simple factors admit no nontrivial refinement, giving the Jordan–Hölder theorem.

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.