Proof of the Jordan Holder theorem from Serge Lang 
Why is there precisely one index $j$ such that $G_i/ G_{i+1} = G_{ij}/G_{i,j+1}$? How does the conclusion follow?

This is my understanding of it.
To say $G_i/G_{i + 1}$ is simple means that there it has no normal subgroups other than $\{e\}$ and $G_i/G_{i + 1}$. This implies that there are no normal subgroups $G_i \unrhd H \unrhd G_{i + 1}$ other than $G_i$ and $G_{i+1}$ because $H/G_i \unlhd G_{i+1}/G_i$.
Therefore, if we take a normal tower $$G = G_0 \rhd G_1 \rhd G_2 \rhd \cdots \rhd G_m = \{e\}, $$
where $G_i/G_{i+1}$ is simple, then any refinement must be obtained by adding copies of $G_0$ or $G_1$ between $G_0$ and $G_1$ and adding copies of $G_1$ or $G_2$ between $G_1$ or $G_2$ and so on. But there has to be some unique place where in the refined tower $G_{ij} = G_i$ and $G_{i,j+1} = G_{i+1}$.
