Composition Series in a Group

# Composition Series in a Group

 Definition: Let $G$ be a group. A Composition Series for $G$ is a (finite) chain of successive subgroups of $G$, denoted by $\{ e \} = G_0 \leq G_1 \leq ... \leq G_n = G$ with the following properties: 1) $G_i$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq n-1$. 2) $G_{i+1}/G_i$ is a simple group for all $0 \leq i \leq n-1$. The Length of the composition series is the number $n$, and the Composition Factors of the composition series are the quotient groups $G_{i+1}/G_i$.

Recall that a group $G$ is said to be a simple group if it has no nontrivial proper normal subgroups. Thus condition (2) requires each of the composition factors $G_{i+1}/G_i$ to have no nontrivial proper normal subgroups.

For example, consider $G = \mathbb{Z}_{12}$, the cyclic group of order $12$, and write $\mathbb{Z}_{12} = \langle a \rangle$ so that $a^{12} = 1$. Recall that every subgroup of a cyclic group is cyclic. Thus, all of the subgroups of $G = \mathbb{Z}_{12}$ are:

(1)
\begin{align} \quad \mathbb{Z}_1 &= \{ 1 \} = \langle 1 \rangle \\ \quad \mathbb{Z}_2 &\cong \{ 1, a^6 \} = \langle a^6 \rangle \\ \quad \mathbb{Z}_3 &\cong \{ 1, a^4, a^8 \} = \langle a^4 \rangle \\ \quad \mathbb{Z}_4 &\cong \{ 1, a^3, a^6, a^9 \} = \langle a^3 \rangle \\ \quad \mathbb{Z}_6 &\cong \{ 1, a^2, a^4, a^6, a^8, a^{10} \} = \langle a^2 \rangle \end{align}

Then $\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4$ and $\mathbb{Z}_6$ are subgroups of $\mathbb{Z}_{12}$. Since every cyclic group is abelian, we have these each of these groups are normal subgroups. Consider the following chain of subgroups:

(2)
\begin{align} \quad \{ e \} &= \mathbb{Z}_1 \leq \mathbb{Z}_2 \leq \mathbb{Z}_4 \leq \mathbb{Z}_{12} = G \\ \quad \{ e \} &= \mathbb{Z}_1 \leq \leq \mathbb{Z}_2 \leq \mathbb{Z}_6 \leq \mathbb{Z}_{12} = G \\ \quad \{ e \} &= \mathbb{Z}_1 \leq \mathbb{Z}_3 \leq \mathbb{Z}_6 \leq \mathbb{Z}_{12} = G\\ \end{align}

The three series above are all composition series for $G = \mathbb{Z}_{12}$. Indeed, each subgroup is normal in the next since all of the groups are abelian. Moreover, the composition factors in all three composition series are isomorphic to either $\mathbb{Z}_2$ or $\mathbb{Z}_3$ - both of which are simple groups as they have no nontrivial proper subgroups - and hence, no nontrivial proper normal subgroups.