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?

To make it clear, I'm trying to answer this question by finishing a complete version of this proof, using the symbol and terminologies he used in his Algebra, Revised Third Edition.

Let $G$ be a group, and let
$$
    G=G_1\supset G_2 \supset \cdots \supset G_r = \{e\}
$$
be the normal tower mentioned in this theorem. Then let
$$
    G=H_1 \supset H_2 \supset \cdots \supset H_s = \{e\}
$$
be another normal tower. Define
$$
    G_{ij}=G_{i+1}(H_j \cap G_i)  .
$$
Then we have a refinement of the tower $\{G_i\}$
$$
    G=G_{11} \supset \cdots \supset G_{1,s-1} \supset G_{21} \supset \cdots \supset G_{r-1,s-1}\supset \{e\}
$$
Now consider a tower indexed by some $i=1,2,\cdots,r-1$
$$
    G_{i} \supset G_{i1}\supset G_{i2} \supset \cdots \supset G_{i,s-1} \supset G_{is} \supset G_{i+1}
$$
Since $G_i/G_{i+1}$ is simple, there must be a positive integer $j$ (this is the $j$ we wanted!) such that $G_{it}=G_i$ for $t=1,2,\cdots,j$ and $G_{it}=G_{i+1}$ for $t=j+1,\cdots,s$. Therefore we have $G_i/G_{i+1}=G_{ij}/G_{i,j+1}$. After removing duplicates from $\{G_{ij}\}$, we obtained the identical tower with the same property.
(To clarify the existence of $j$, take a look at the assumption of this theorem. If there exists e a tower $G_{i} \supset H \supset G_{i+1}$ where $G_{i}\neq H$ and $G_{i+1} \neq H$, then $G_{i}/G_{i+1}$ is not simple, since the nontrivial subgroup $H/G_{i+1}$ is normal in $G_{i}/G_{i+1}$. Hence the existence and uniqueness of $j$ follows from the fact that $G_{i}\neq G_{i+1}$ and subgroup relationship inside the tower.)
