Composition Series in a Group

Composition Series in a Group

Definition: Let $G$ be a group and let $1$ denote the identity in $G$. A subnormal series $\{ 1 \} = G_0 \triangleleft G_1 \triangleleft ... \triangleleft G_k = G$ is a Composition Series if all of the factors $G_{i+1}/G_i$ for $i \in \{ 0, 1, ..., k-1 \}$ are simple groups.

Recall that a group is said to be simple if it has no nontrivial proper normal subgroups.

For example, consider the cyclic group of order $12$, denote it $C_{12} = (a)$ where $a^{12} = 1$. Let:

(1)
\begin{align} \quad C_1 &= \{ 1 \} = (1) \\ \quad C_2 &= \{ 1, a^6 \} = (a^6) \\ \quad C_3 &= \{ 1, a^4, a^8 \} = (a^4) \\ \quad C_4 &= \{ 1, a^3, a^6, a^9 \} = (a^3) \\ \quad C_6 &= \{ 1, a^2, a^4, a^6, a^8, a^{10} \} = (a^2) \end{align}

Then $C_1, C_2, C_3, C_4$ and $C_6$ are subgroups of $C_{12}$ and each of these groups are isomorphic to the cyclic group of order $1$, $2$, $3$, $4$, and $6$ respectively. Since every cyclic group is abelian, we have these each of these groups are normal subgroups. Consider the following subnormal series:

(2)
\begin{align} \quad C_1 \triangleleft C_2 \triangleleft C_4 \triangleleft C_{12} \quad (*) \end{align}
(3)
\begin{align} \quad C_1 \triangleleft C_2 \triangleleft C_6 \triangleleft C_{12} \quad (**) \end{align}
(4)
\begin{align} \quad C_1 \triangleleft C_3 \triangleleft C_6 \triangleleft C_{12} \quad (***) \end{align}

It is not hard to see that the factors of all three subnormal series are isomorphic to the cyclic group of order $2$ and the cyclic group of order $3$. But by Lagrange's Theorem, the cyclic groups of order $2$ and $3$ have no nontrivial proper subgroups. So the cyclic groups of order $2$ and $3$ are normal. So indeed, all of the subnormal series above are composition series.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License