 Solution

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.