# 合成列

## 群的情形

${\displaystyle G}$ 为群，${\displaystyle G}$ 的合成列是对应于一族子群

${\displaystyle \{e\}=H_{0}\subset H_{1}\subset \cdots \subset H_{n}=G}$

## 模的情形

${\displaystyle \{0\}=J_{0}\subset \cdots \subset J_{n}=M}$

## 例子

${\displaystyle C_{1}\triangleleft C_{2}\triangleleft C_{6}\triangleleft C_{12))$,
${\displaystyle C_{1}\triangleleft C_{2}\triangleleft C_{4}\triangleleft C_{12))$,
${\displaystyle C_{1}\triangleleft C_{3}\triangleleft C_{6}\triangleleft C_{12))$

${\displaystyle C_{2},C_{3},C_{2))$
${\displaystyle C_{2},C_{2},C_{3))$
${\displaystyle C_{3},C_{2},C_{2))$

## 若尔当-赫尔德定理

${\displaystyle \{0\}=M_{0}\subset \cdots \subset M_{r}=M}$
${\displaystyle \{0\}=M'_{0}\subset \cdots \subset M'_{s}=M}$

${\displaystyle \mathrm {min} (r,s)}$数学归纳法。若 ${\displaystyle \mathrm {min} (r,s)=0}$${\displaystyle M=0}$，若 ${\displaystyle \mathrm {min} (r,s)=1}$${\displaystyle M}$单模。以下假定 ${\displaystyle r,s\geq 2}$

${\displaystyle M_{r-1}=M_{s-1))$，据归纳法假设，${\displaystyle r-1=s-1}$${\displaystyle M_{i+1}/M_{i))$${\displaystyle M'_{i+1}/M'_{i))$${\displaystyle 0\leq i\leq r-2}$）之间仅差排列。此外 ${\displaystyle M/M_{r-1}=M_{/}M'_{s-1))$，故定理成立。

${\displaystyle M_{r-1}\neq M'_{s-1))$。此时必有 ${\displaystyle M_{r-1}+M'_{s-1}=M}$。置 ${\displaystyle N:=M_{r-1}\cap M'_{s-1))$，于是

${\displaystyle M/M_{r-1}=(M_{r-1}+M'_{s-1})/M_{r-1}\simeq M'_{s-1}/N}$
${\displaystyle M/M'_{s-1}=(M_{r-1}+M'_{s-1})/M'_{s-1}\simeq M_{r-1}/N}$

${\displaystyle N}$ 的合成列 ${\displaystyle \{0\}=K_{0}\subset \cdots \subset K_{t}=N}$，依上式知

${\displaystyle \{0\}=K_{0}\subset \cdots \subset K_{t}=N\subset M_{r-1}\subset M\quad \ldots (*)}$
${\displaystyle \{0\}=K_{0}\subset \cdots \subset K_{t}=N\subset M'_{s-1}\subset M\quad \ldots (**)}$