Finite Group has Composition Series

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Theorem

Let $G$ be a finite group.


Then $G$ has a composition series.


Proof 1

Let $G$ be a finite group whose identity is $e$.

Either $G$ has a proper non-trivial normal subgroup or it does not.

If not, then:

$\set e \lhd G$

is the composition series for $G$.

Otherwise, $G$ has one or more proper non-trivial normal subgroup.

Of these, one or more will have a maximum order.

Select one of these and call it $G_1$.

Again, either $G_1$ has a proper non-trivial normal subgroup or it does not.

If not, then:

$\set e \lhd G_1 \lhd G$

is a composition series for $G$.

By the Jordan-Hölder Theorem, there can be no other composition series which is longer. As $G_1$ is a proper subgroup of $G$:

$\order {G_1} < \order G$

where $\order G$ denotes the order of $G$.

Again, if $G_1$ has one or more proper non-trivial normal subgroup, one or more will have a maximum order.

Select one of these and call it $G_2$.

Thus we form a normal series:

$\set e \lhd G_2 \lhd G_1 \lhd G$


The process can be repeated, and at each stage a normal subgroup is added to the series of a smaller order than the previous one.

This process cannot continue infinitely.

Eventually a $G_n$ will be encountered which has no proper non-trivial normal subgroup.


This needs considerable tedious hard slog to complete it.
In particular: Proof or explanation is required for why the repeated process cannot continue indefinitely. In particular, explain why eventually there will be no non-trivial normal subgroup
Seriously? You need it explained why reducing the size of a finite number cannot go on for ever>?
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Thus a composition series:

$\set e \lhd G_n \cdots \lhd G_2 \lhd G_1 \lhd G$

will be the result.

$\blacksquare$


Proof 2


Work In Progress
In particular: Bring up to site standards
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it.
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Let $G$ be a finite group whose identity is $e_G$. We shall use induction on $|G|$. If $G$ is trivial ($|G|=1$), then its composition series is

$G=\{e_G\}$.

Suppose $G$ has a composition series if $|G|<n$, then it suffices to construct a composition series for $G$ with order $n$. If $G$ is simple, then its composition series is

$G\supsetneq \{e_G\}$.

Otherwise, $G$ has one or more proper non-trivial normal subgroup. Let $S$ denote the set of all non-trivial normal proper subgroup of $G$. Notice that $S$ is non-empty, ordered by inclusion, and $G$ is an upper bound of every chain in $S$, then by Zorn's lemma, $S$ has at least one maximal element, denoted $H$. Since $H\in S$, $H$ is a non-trivial normal proper subgroup of $G$, in particular, $|H|<|G|$, so by the induction hypothesis, $H$ has a composition series. Now we have

$G\supsetneq H\supsetneq H_1\supsetneq \cdots \supsetneq H_n\supsetneq \{e_G\}$.

In order to show that it is indeed a composition series of $G$, it suffices to check that $G/H$ is simple, which is a direct consequence of the maximal property of $H$, so we are done. $\blacksquare$


Sources

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