TITLE: The Jordan-Hölder Theorem
QUESTION [5 upvotes]: I want to solve the following exercise from Dummit & Foote's Abstract Algebra (exercise 10 in page 106):

Prove part (2) of the Jordan-Hölder Theorem by induction on $\min\{r,s\}$. [Apply the inductive hypothesis to $H=N_{r-1} \cap M_{s-1}$ and use the preceding exercises.]

The Jordan-Hölder Theorem as stated in the book is:

Let $G$ be a finite group with $G \neq 1$. Then 

$G$ has a composition series and
The composition factors in a composition series are unique, namely, if $1=N_0 \leq N_1 \leq \dots \leq N_r=G$ and $1=M_0 \leq M_1 \leq \dots \leq M_s=G$ are two composition series for $G$, then $r=s$ and there is some permutation $\pi$ of $\{1,2,\dots,r\}$ such that $$M_{\pi(i)} / M_{\pi(i)-1} \cong N_i/N_{i-1}, \; \; \; \; \; \; \; \; \; \; \; 1 \leq i \leq r.  $$



My attempt:
We induct on $n:=\min\{r,s\}$. The case $n=2$ was proven in Exercise 9. We assume the induction hypothesis holds for all groups with two composition series the shorter length of which is $<n$. Let $G$ be a group with two composition series as stated in the theorem such that $\min\{r,s\}=n$. The hint suggests to consider $H=N_{r-1} \cap M_{s-1}$. However, I don't see how: I don't have two composition series for $H$ in mind, and even if I had, I (seemingly) couldn't deduce anything about the composition factors of $G$.
Please help me complete this proof, Thanks!

REPLY [3 votes]: Hint: Observe that when $N_{r-1} \neq M_{s-1}$, we have $N_{r-1}M_{s-1} = G$. Then draw the chains side by side and investigate $H = N_{r-1} \cap M_{s-1}$, which the two chains share (Can you say more about this, what does its factor say? Can you apply 2. isomorphism theorem?).