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!
If $N_{r-1}=M_{s-1}$ then you can use induction, so assume not.
Choose a composition series of $H$. Now we have four composition series of $G$, your two original ones, and two that go through $H$ (it's best to draw a picture). Two of the series go through $N_{r-1}$ and the other two go through $M_{s-1}$, so you can apply induction to both of those.
To finish off the proof, you need the isomorphisms $G/N_{r-1} \cong M_{s-1}/H$ and $G/M_{s-1} \cong N_{r-1}/H$, which come from the Second Isomorphism Theorem.