Composition Series

Relevant Courses: Math 493

Definition: Composition Series

Let $G$ be an arbitrary group. A composition series is a normal series $(e)=G_0◃G_1◃\cdots ◃G_k=G$ with the added condition that $G_{i+1}/G_i$ is simple.

We can then also show the following:

Lemma

All finite groups admit a composition series.

Proof

Let us prove via induction, on the order of the group $G$.

Base Case: $|G|=1$. This means that the group $G$ must be trivial, so it has only one normal subgroup (itself, or $e$). Thus, $G_k/G_0=G_k$, which has order 1 so has no proper normal subgroups so is simple. Thus its composition series is the trivial composition series $(e)=G_0=G.$

Inductive Assumption: For all groups $K$ such that $|K|<n$, we have that $K$ admits a composition series.

We can split this up into two cases: $G$ is simple or $G$ is not simple.

Suppose $G$ is simple. Then, it has no normal subgroups except $(e)$ and itself. The only proper subgroup then is $(e)=G_0$, and—since $G$ is simple—$G/G_0\simeq G$ must also be simple, giving us that the composition series exists and is simple $(e)=G_0◃G_1=G.$ Contrarily, suppose that $G$ is not simple. Then, $G$ must have at least one nontrivial proper normal subgroup, which we will denote $N$ (meaning $N◃G$). Since $N$ is a nontrivial proper subgroup, it has order strictly less than $G$ and thus has a composition series $(e)=N_0◃\cdots ◃N_k=N.$ Additionally, $G/N$ has order strictly less than $G$, so must also have a composition series $(e)=N/N={\overline{G}}_0◃{\overline{G}}_1◃\cdots ◃{\overline{G}}_l=G/N.$ By The Third Isomorphism Theorem, there is a bijection between subgroups of $G$ containing $N$, and subgroups of $G/N$ that would map $G_i\to {\overline{G}}_i$. Moreover, ${\overline{G}}_{i+1}/{\overline{G}}_i\simeq G_{i+1}/G_i$ More importantly, ${\overline{G}}_l/{\overline{G}}_{l-1}$ is simple, which means that $G_l/G_{l-1}$ is also simple.

Therefore, we can see that $N◃G_1◃\cdots ◃G_{l-1}◃G_l=G.$ And so, we get the composition series $(e)=N_0◃\cdots N_k=N◃G_1◃\cdots ◃G_{l-1}◃G_l=G.$

Thus, we have constructively shown that $G$ must have a composition series.

This proof is complete by induction.

One of the most important things about the composition series is its relation shipw ith the Jordan-Hölder Theorem.