Exercise 3.4.9 (Special case of Jordan-Hölder Theorem)

Proof.

By hypothesis $\{1\}=M_0⊴M_1⊴M_2=G$ is a composition series, so $M_1$ and $G∕M_1$ are simple.

First note that $r>1$, otherwise $1=N_0⊴N_1=G$ is a composition series, so $G$ is simple, in contradiction with $\{1\}\ne M_0⊴G$.

Since $N_{r-1}⊴N_r=G$ and $M_1⊴G$, then $N_{r-1}\cap M_1⊴M_1$ (see Exercise 3.1.24).

But $M_1$ is simple, therefore $N_{r-1}\cap M_1=\{1\}$ or $N_{r-1}\cap M_1=M_1$.

$\bullet$ If $N_{r-1}\cap M_1=M_1$, then $M_1\le N_{r-1}$, where $M_1$ is a normal subgroup of $G$, thus

By the Lattice Isomorphism Theorem (part (5)),

But $G∕M_1$ is simple, thus $N_{r-1}∕M_1=\{\bar{1}\}$ or $N_{r-1}∕M_1=G∕M_1$, so $N_{r-1}=M_1$ or $N_{r-1}=G=N_r$. Since ${(N_i)}_{0\le i\le r}$ is a composition series, $N_{r-1}=N_r$ is impossible, so

If $r>2$, then $\{1\}\ne N_{r-2}⊴N_{r-1}=M_1$. This is impossible, since $M_1$ is simple, so $r=2$, and the composition series ${(N_i)}_{0\le i\le r}$ is $\{1\}⊴N_1=M_1⊴G$, so the two composition series are identical, and the composition factors are the same.

$\bullet$ We suppose now that $N_{r-1}\cap M_1=\{1\}$.

Since $N_{r-1}⊴G$ and $M_1⊴G$, then $N_{r-1}{M}_1$ is a subgroup of $G$, and

(If $g\in G$, $n\in N_{r-1}$ and $m\in M_1$, then $g\mathit{𝑛𝑚}{g}^{-1}=\mathit{𝑔𝑛}{g}^{-1}\mathit{𝑔𝑚}{g}^{-1}\in N_{r-1}{M}_1$.)

By the part 5 of the Lattice Isomorphism Theorem, $(N_{r-1}{M}_1)∕M_1⊴G∕M_1$, where $G∕M_1$ is simple, thus $N_{r-1}{M}_1=M_1$ or $N_{r-1}{M}_1=G$. In the first case, $N_{r-1}\le M_1$ and $\{1\}=N_{r-1}\cap M_1=N_{r-1}$. This is impossible, because $r>1$, so $N_{r-1}\ne \{1\}$. This proves

By the Second Isomorphism Theorem, $G∕M_1\simeq N_{r-1}$, so $N_{r-1}$ is simple. But the only simple group among the $N_i$ is $N_1$, thus $r-1=1$ and $r=2$. So the composition series is $\{1\}=N_0⊴N_1⊴N_2=G$, and the composition factors are $N_1\simeq G∕M_1$, and by the Second isomorphism Theorem $G∕N_1\simeq M_1$. So the composition factors are the same, in reverse order. □