Theorem (Jordan-Holder) Any two composition series are equivalent.
proof: Induction of the length of the series. Suppose we have $$1 \lhd \cdots \lhd L \lhd G$$ and $$1 \lhd \cdots \lhd K \lhd G$$ then $L \lhd KL \lhd G$ implies that $KL/L$ is a normal subgroup of the composition factor $G/L$ which being a simple group implies that $$KL = L\text{ or }G$$ and similarly $$KL = K\text{ or }G.$$
Suppose $KL \not = G$, then $L = K$ and we are done by induction. Suppose $KL = G$ then by noether2 we have $$\frac{G}{L} \simeq \frac{KL}{L} \simeq \frac{K}{K \cap L}$$ and $$\frac{G}{K} \simeq \frac{KL}{K} \simeq \frac{L}{K \cap L}$$ therefore the composition series $$
\begin{array}{c}
1 &\lhd& \cdots &\lhd& L \lhd G \\
1 &\lhd& \cdots \lhd L \cap K &\lhd& L \lhd G \\
1 &\lhd& \cdots \lhd L \cap K &\lhd& K \lhd G \\
1 &\lhd& \cdots &\lhd& K \lhd G \\
\end{array}$$ are equivalent.
