# Abstract Algebra/Composition series

Definitions 2.7.1:

Let ${\displaystyle G}$ be a group. A normal series of ${\displaystyle G}$ are finitely many subgroups ${\displaystyle N_{1},\ldots ,N_{n}}$ of ${\displaystyle G}$ such that

${\displaystyle \{e\}=N_{n}\triangleleft N_{n-1}\triangleleft \cdots \triangleleft N_{1}=G}$

Two normal series ${\displaystyle N_{1},\ldots ,N_{n}}$ and ${\displaystyle M_{1},\ldots ,M_{k}}$ of ${\displaystyle G}$ are equivalent if and only if ${\displaystyle n=k}$ and there exists a bijective function ${\displaystyle \sigma :\{1,\ldots ,n\}\to \{1,\ldots ,n\}}$ such that for all ${\displaystyle j\in \{1,\ldots ,n-1\}}$:

${\displaystyle N_{j}/N_{j+1}\cong M_{\sigma (j)}/M_{\sigma (j)+1}}$

A normal series ${\displaystyle N_{1},\ldots ,N_{n}}$ of ${\displaystyle G}$ is a composition series of ${\displaystyle G}$ if and only if for each ${\displaystyle j\in \{1,\ldots ,n-1\}}$ the group

${\displaystyle N_{j}/N_{j+1}}$

is simple.

Theorem 2.7.2:

Let ${\displaystyle G}$ be a finite group. Then there exists a composition series of ${\displaystyle G}$.

Proof:

We prove the theorem by induction over ${\displaystyle |G|}$.

1. ${\displaystyle |G|=1}$. In this case, ${\displaystyle G}$ is the trivial group, and ${\displaystyle M_{1}}$ with ${\displaystyle M_{1}=G}$ is a composition series of ${\displaystyle G}$.

2. Assume the theorem is true for all ${\displaystyle n\in \mathbb {N} }$, ${\displaystyle n<|G|}$.

Since the trivial subgroup ${\displaystyle \{e\}\subset G}$ is a normal subgroup of ${\displaystyle G}$, the set of proper normal subgroups of ${\displaystyle G}$ is not empty. Therefore, we may choose a proper normal subgroup ${\displaystyle N}$ of maximum cardinality. This must also be a maximal proper normal subgroup, since any group in which it is contained must have at least equal cardinality, and thus, if ${\displaystyle M}$ is normal such that

${\displaystyle N\subsetneq M\subsetneq G}$

, then

${\displaystyle |M|>|N|}$

, which is why ${\displaystyle N}$ is not a proper normal subgroup of maximal cardinality.

Due to theorem 2.6.?, ${\displaystyle G/N}$ is simple. Further, since ${\displaystyle |N|<|G|}$, the induction hypothesis implies that there exists a composition series of ${\displaystyle N}$, which we shall denote by ${\displaystyle N_{2},\ldots ,N_{n}}$, where

${\displaystyle \{e\}=N_{n}\triangleleft N_{n-1}\triangleleft \cdots \triangleleft N_{2}=N}$

. But then we have

${\displaystyle \{e\}=N_{n}\triangleleft \cdots \triangleleft N_{2}=N\triangleleft N_{1}:=G}$

, and further for each ${\displaystyle m\in \{1,\ldots ,n-1\}}$:

${\displaystyle N_{m}/N_{m+1}}$ is simple.

Thus, ${\displaystyle N_{1},\ldots ,N_{n}}$ is a composition sequence of ${\displaystyle G}$.${\displaystyle \Box }$

Our next goal is to prove that given two normal sequences of a group, we can find two 'refinements' of these normal sequences which are equivalent. Let us first define what we mean by a refinement of a normal sequence.

Definition 2.7.3:

Let ${\displaystyle G}$ be a group and let ${\displaystyle N_{1},\ldots ,N_{n}}$ be a normal sequence of ${\displaystyle G}$. A refinement of ${\displaystyle N_{1},\ldots ,N_{n}}$ is a normal sequence ${\displaystyle N_{1}',\ldots ,N_{k}'}$ such that

${\displaystyle \{N_{1},\ldots ,N_{n}\}\subseteq \{N_{1}',\ldots ,N_{k}'\}}$

Theorem 2.7.4 (Schreier):

Let ${\displaystyle G}$ be a group and let ${\displaystyle N_{1},\ldots ,N_{n}}$, ${\displaystyle M_{1},\ldots ,M_{k}}$ be two normal series of ${\displaystyle G}$. Then there exist refinements ${\displaystyle N_{1}',\ldots ,N_{m}'}$ of ${\displaystyle N_{1},\ldots ,N_{n}}$ and ${\displaystyle M_{1}',\ldots ,M_{l}'}$ of ${\displaystyle M_{1},\ldots ,M_{k}}$ such that ${\displaystyle N_{1}',\ldots ,N_{m}'}$ and ${\displaystyle M_{1}',\ldots ,M_{l}'}$ are equivalent.

Proof:

Theorem 2.7.5 (Jordan-Hölder):

Let ${\displaystyle G}$ be a group and let ${\displaystyle N_{1},\ldots ,N_{n}}$ and ${\displaystyle M_{1},\ldots ,M_{k}}$ be two composition series of ${\displaystyle G}$. Then ${\displaystyle N_{1},\ldots ,N_{n}}$ and ${\displaystyle M_{1},\ldots ,M_{k}}$ are equivalent.

Proof:

Due to theorem 2.6.?, all the elements of ${\displaystyle \{N_{1},\ldots ,N_{n}\}}$ must be pairwise different, and the same holds for the elements of ${\displaystyle \{M_{1},\ldots ,M_{k}\}}$.

Due to theorem 2.7.4, there exist refinements ${\displaystyle N_{1}',\ldots ,N_{m}'}$ of ${\displaystyle N_{1},\ldots ,N_{n}}$ and ${\displaystyle M_{1}',\ldots ,M_{l}'}$ of ${\displaystyle M_{1},\ldots ,M_{k}}$ such that ${\displaystyle N_{1}',\ldots ,N_{m}'}$ and ${\displaystyle M_{1}',\ldots ,M_{l}'}$ are equivalent.

But these refinements satisfy

${\displaystyle \{N_{1}',\ldots ,N_{m}'\}=\{N_{1},\ldots ,N_{n}\}}$

and

${\displaystyle \{M_{1}',\ldots ,M_{l}'\}=\{M_{1},\ldots ,M_{k}\}}$

, since if this were not the case, we would obtain a contradiction to theorem 2.6.?.

We now choose a bijection ${\displaystyle \sigma :\{1,\ldots ,m\}\to \{1,\ldots ,m\}}$ such that for all ${\displaystyle j\in \{1,\ldots ,m-1\}}$:

${\displaystyle N_{j}/N_{j+1}\cong M_{\sigma (j)}/M_{\sigma (j)+1}}$