We prove the theorem by induction on the length $n$ of a composition series.

Base case ($n = 0$): $G = \{e\}$, trivially true.

Base case ($n = 1$): $G$ is simple. Any composition series is $\{e\} \trianglelefteq G$ with one factor $G$. Unique.

Inductive step: Let $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G$ and $\{e\} = H_0 \trianglelefteq H_1 \trianglelefteq \cdots \trianglelefteq H_m = G$ be two composition series.

Case 1: $G_{n-1} = H_{m-1}$. Then the two series share the same maximal normal subgroup, and by induction on $G_{n-1}$, the portions below $G_{n-1}$ have the same factors. The top factors are both $G/G_{n-1}$, so the full series are equivalent.

Case 2: $G_{n-1} \neq H_{m-1}$. Since both are maximal normal subgroups of $G$ (as the quotients are simple), and $G_{n-1} \neq H_{m-1}$, the product $G_{n-1} H_{m-1}$ is a normal subgroup of $G$ strictly containing both. By maximality: $G_{n-1} H_{m-1} = G$.

Let $K = G_{n-1} \cap H_{m-1}$. By the second isomorphism theorem:

$G/G_{n-1} = G_{n-1} H_{m-1}/G_{n-1} \cong H_{m-1}/K,$ $G/H_{m-1} = G_{n-1} H_{m-1}/H_{m-1} \cong G_{n-1}/K.$

Now $K$ has composition series (being a subgroup of $G_{n-1}$). Let $\{e\} = K_0 \trianglelefteq \cdots \trianglelefteq K_\ell = K$ be one.

Then:

• $\{e\} = K_0 \trianglelefteq \cdots \trianglelefteq K_\ell = K \trianglelefteq G_{n-1} \trianglelefteq G$ is a composition series (factors of $K$, then $G_{n-1}/K \cong G/H_{m-1}$, then $G/G_{n-1}$).
• $\{e\} = K_0 \trianglelefteq \cdots \trianglelefteq K_\ell = K \trianglelefteq H_{m-1} \trianglelefteq G$ is a composition series (factors of $K$, then $H_{m-1}/K \cong G/G_{n-1}$, then $G/H_{m-1}$).

By induction on $G_{n-1}$: the original series through $G_{n-1}$ has the same factors as the series through $K$ then $G_{n-1}$. By induction on $H_{m-1}$: similarly for the series through $H_{m-1}$.

Comparing: both original series have the same multiset of composition factors as the "bridge" series through $K$ (the factors of $K$, plus $G/G_{n-1}$ and $G/H_{m-1}$, which simply swap positions). $\blacksquare$