Exercise 3.4.10 (Proof of Jordan-Hölder Theorem)

Proof. The part (2) of the Jordan-Hölder Theorem is true if $n=\min <!--\; FUNCTION\; APPLICATION\; -->(r,s)=1$. In this case $G$ is simple, so $G$ has no other composition series than $\{1\}⊴G$ (any other refinement will produce a proper non trivial normal subgroup of $G$: this is impossible (the case $n=2$, proved is Exercise 9, is not essential here).

Let $n>1$ be an integer. Assume that the part (2) of the Jordan-Hölder Theorem is true for any composition series of length $r,s$ such that $\min <!--\; FUNCTION\; APPLICATION\; -->(r,s)<n$. Consider now two composition series ${(N_i)}_{0\le i\le r}$ and ${(M_j)}_{0\le j\le s}$ such that $\min <!--\; FUNCTION\; APPLICATION\; -->(r,s)=n$ (then $r>1$ and $s>1$). Set $H=N_{r-1}\cap M_{s-1}$. Then by Exercise 6, $H$ has a composition series ${(H_k)}_{0\le k\le t-2}$, where $H_0=\{1\}$ and $H_{t-2}=H=N_{r-1}\cap M_{s-1}$.

Since $s>1$, $M_{s-1}\ne \{1\}$. We note that $M_{s-1}$ cannot be a proper subgroup of $N_{r-1}$, otherwise by the Lattice Isomorphism Theorem $\{\bar{1}\}\ne N_{r-1}∕M_{s-1}⊴G∕M_{s-1}$, where $G∕M_{s-1}$ is simple, so $N_{r-1}=G$: this is impossible because ${(N_i)}_{0\le i\le r}$ is a composition series of $G$. Similarly $N_{r-1}$ is not a proper subgroup of $M_{s-1}$. So there are only two possibilities : $N_{r-1}=M_{s-1}$, or $N_{r-1}\nleqq M_{s-1}$ and $N_{r-1}\nleqq M_{s-1}$ (see these two cases in the schema below).

In the first case, since $\min <!--\; FUNCTION\; APPLICATION\; -->(r-1,s-1)<n$, there are two composition series ${(N_i)}_{0\le i\le r-1}$ and ${(M_j)}_{0\le j\le s-1}$of $N_{r-1}=M_{s-1}$. The induction hypothesis shows that $r-1=s-1$, so $r=s$ and the factors are the same, up to ordering. If we add the last factor $G∕N_{r-1}$, we obtain the same factors for the two composition series ${(N_i)}_{0\le i\le r}$ and ${(M_j)}_{0\le j\le s}$ of $G$.

Consider now the case $N_{r-1}\ne M_{s-1}$, where $N_{r-1}\nleqq M_{s-1}$ and $N_{r-1}\nleqq M_{s-1}$. We show that $G=N_{r-1}{M}_{s-1}$. Since $N_{r-1}⊴G$ and $M_{s-1}⊴G$, then $N_{r-1}{M}_{s-1}$ is a normal subgroup of $G$ (same argument as in Exercise 9). If $G\ne N_{r-1}{M}_{s-1}$, then we obtain a refinement of the series $(N_r)$ given by $N_r⊴N_{r-1}{M}_{s-1}⊴G$, which gives $(N_{r-1}{M}_{s-1})∕N_{s-1}⊴G∕N_{r-1}$, where $G∕N_{r-1}$ is simple and $N_{r-1}{M}_{s-1}\ne G$, thus $N_{r-1}{M}_{s-1}=N_{s-1}$, which implies $M_{s-1}\le N_{s-1}$. This is impossible, because $M_{s-1}$ is not a subgroup of $N_{s-1}$. This proves that

Since $N_{r-1}⊴G$ and $M_{s-1}⊴G$, by the second isomorphism Theorem $N_{r-1}\cap M_{s-1}⊴N_{r-1}$, $N_{r-1}\cap M_{s-1}⊴M_{s-1}$, and

which are simple groups.

So $G$ has four composition series, given by

The composition series (1) and (2) pass through $N$. By the induction hypothesis, the two extract compositions series for $N_{r-1}$ have same length $r-1=t-1$, so $r=t$. Similarly, the two given compositions series of $M_{s-1}$ have the same length $t-1=s-1$, so $s=t$. This gives $r=s$.

Moreover, by the induction hypothesis, the factors of the two composition series

are the same (up to ordering), and the factors of the two composition series

are also the same.

Thus the composition series (2) and (3) of $G$ have the same factors, and also the compositions series (4) and (5).

Since

we obtain that the composition series (3) and (4) have the same factors. Therefore the four composition series have the some factors, up to ordering. The induction is done, which proves part (2) of the Jordan-Hölder Theorem. □