Abstract Algebra/Composition series

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

$N⊊M⊊G$

, 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 $N_2,\dots ,N_n$, where

$\{e\}=N_n◃N_{n-1}◃\cdots ◃N_2=N$

. But then we have

$\{e\}=N_n◃\cdots ◃N_2=N◃N_1:=G$

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

$N_m/N_{m+1}$ is simple.

Thus, $N_1,\dots ,N_n$ is a composition sequence of $G$.$◻$

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.

Proof:

Proof:

Due to theorem 2.6.?, all the elements of $\{N_1,\dots ,N_n\}$ must be pairwise different, and the same holds for the elements of $\{M_1,\dots ,M_k\}$.

Due to theorem 2.7.4, there exist refinements ${N_1}^{\prime },\dots ,{N_m}^{\prime }$ of $N_1,\dots ,N_n$ and ${M_1}^{\prime },\dots ,{M_l}^{\prime }$ of $M_1,\dots ,M_k$ such that ${N_1}^{\prime },\dots ,{N_m}^{\prime }$ and ${M_1}^{\prime },\dots ,{M_l}^{\prime }$ are equivalent.

But these refinements satisfy

$\{{N_1}^{\prime },\dots ,{N_m}^{\prime }\}=\{N_1,\dots ,N_n\}$

and

$\{{M_1}^{\prime },\dots ,{M_l}^{\prime }\}=\{M_1,\dots ,M_k\}$

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

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

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