# Abstract Algebra/Composition series

Definitions 2.7.1:
Let $G$ be a group. A normal series of $G$ are finitely many subgroups $N_{1},\ldots ,N_{n}$ of $G$ such that
    $\{e\}=N_{n}\triangleleft N_{n-1}\triangleleft \cdots \triangleleft N_{1}=G$
Two normal series $N_{1},\ldots ,N_{n}$ and $M_{1},\ldots ,M_{k}$ of $G$ are equivalent if and only if $n=k$ and there exists a bijective function $\sigma :\{1,\ldots ,n\}\to \{1,\ldots ,n\}$ such that for all $j\in \{1,\ldots ,n-1\}$:
    $N_{j}/N_{j+1}\cong M_{\sigma (j)}/M_{\sigma (j)+1}$
A normal series $N_{1},\ldots ,N_{n}$ of $G$ is a composition series of $G$ if and only if for each $j\in \{1,\ldots ,n-1\}$ the group
    $N_{j}/N_{j+1}$
is simple.
Theorem 2.7.2:
Let $G$ be a finite group. Then there exists a composition series of $G$.
Proof:
We prove the theorem by induction over $|G|$.
1. $|G|=1$. In this case, $G$ is the trivial group, and $M_{1}$ with $M_{1}=G$ is a composition series of $G$.
2. Assume the theorem is true for all $n\in \mathbb {N}$, $n<|G|$.
Since the trivial subgroup $\{e\}\subset G$ is a normal subgroup of $G$, the set of proper normal subgroups of $G$ is not empty. Therefore, we may choose a proper normal subgroup $N$ 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 $M$ is normal such that
    $N\subsetneq M\subsetneq G$
, then
    $|M|>|N|$
, which is why $N$ is not a proper normal subgroup of maximal cardinality.
Due to theorem 2.6.?, $G/N$ is simple. Further, since $|N|<|G|$, the induction hypothesis implies that there exists a composition series of $N$, which we shall denote by $N_{2},\ldots ,N_{n}$, where
    $\{e\}=N_{n}\triangleleft N_{n-1}\triangleleft \cdots \triangleleft N_{2}=N$
. But then we have
    $\{e\}=N_{n}\triangleleft \cdots \triangleleft N_{2}=N\triangleleft N_{1}:=G$
, and further for each $m\in \{1,\ldots ,n-1\}$:
    $N_{m}/N_{m+1}$ is simple.
Thus, $N_{1},\ldots ,N_{n}$ is a composition sequence of $G$.$\Box$
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 $G$ be a group and let $N_{1},\ldots ,N_{n}$ be a normal sequence of $G$. A refinement of $N_{1},\ldots ,N_{n}$ is a normal sequence $N_{1}',\ldots ,N_{k}'$ such that
    $\{N_{1},\ldots ,N_{n}\}\subseteq \{N_{1}',\ldots ,N_{k}'\}$
Theorem 2.7.4 (Schreier):
Let $G$ be a group and let $N_{1},\ldots ,N_{n}$, $M_{1},\ldots ,M_{k}$ be two normal series of $G$. Then there exist refinements $N_{1}',\ldots ,N_{m}'$ of $N_{1},\ldots ,N_{n}$ and $M_{1}',\ldots ,M_{l}'$ of $M_{1},\ldots ,M_{k}$ such that $N_{1}',\ldots ,N_{m}'$ and $M_{1}',\ldots ,M_{l}'$ are equivalent.
Proof:
Theorem 2.7.5 (Jordan-Hölder):
Let $G$ be a group and let $N_{1},\ldots ,N_{n}$ and $M_{1},\ldots ,M_{k}$ be two composition series of $G$. Then $N_{1},\ldots ,N_{n}$ and $M_{1},\ldots ,M_{k}$ are equivalent.
Proof:
Due to theorem 2.6.?, all the elements of $\{N_{1},\ldots ,N_{n}\}$ must be pairwise different, and the same holds for the elements of $\{M_{1},\ldots ,M_{k}\}$.
Due to theorem 2.7.4, there exist refinements $N_{1}',\ldots ,N_{m}'$ of $N_{1},\ldots ,N_{n}$ and $M_{1}',\ldots ,M_{l}'$ of $M_{1},\ldots ,M_{k}$ such that $N_{1}',\ldots ,N_{m}'$ and $M_{1}',\ldots ,M_{l}'$ are equivalent.
But these refinements satisfy
    $\{N_{1}',\ldots ,N_{m}'\}=\{N_{1},\ldots ,N_{n}\}$
and
    $\{M_{1}',\ldots ,M_{l}'\}=\{M_{1},\ldots ,M_{k}\}$
, since if this were not the case, we would obtain a contradiction to theorem 2.6.?.
We now choose a bijection $\sigma :\{1,\ldots ,m\}\to \{1,\ldots ,m\}$ such that for all $j\in \{1,\ldots ,m-1\}$:
    $N_{j}/N_{j+1}\cong M_{\sigma (j)}/M_{\sigma (j)+1}$