# 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)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)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.