A subnormal series of a group $G$ is a finite sequence of subgroups \begin{equation*}G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \},\end{equation*} where $H_i$ is a normal subgroup of $H_{i+1}$. If each subgroup $H_i$ is normal in $G$, then the series is called a normal series. The length of a subnormal or normal series is the number of proper inclusions.

##### Example13.11

Any series of subgroups of an abelian group is a normal series. Consider the following series of groups: \begin{gather*} {\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\},\\ {\mathbb Z}_{24} \supset \langle 2 \rangle \supset \langle 6 \rangle \supset \langle 12 \rangle \supset \{0\}. \end{gather*}

##### Example13.12

A subnormal series need not be a normal series. Consider the following subnormal series of the group $D_4$: \begin{equation*}D_4 \supset \{ (1), (12)(34), (13)(24), (14)(23) \} \supset \{ (1), (12)(34) \} \supset \{ (1) \}.\end{equation*} The subgroup $\{ (1), (12)(34) \}$ is not normal in $D_4$; consequently, this series is not a normal series.

A subnormal (normal) series $\{ K_j \}$ is a refinement of a subnormal (normal) series $\{ H_i \}$ if $\{ H_i \} \subset \{ K_j \}$. That is, each $H_i$ is one of the $K_j$.

##### Example13.13

The series \begin{equation*}{\mathbb Z} \supset 3{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 90{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}\end{equation*} is a refinement of the series \begin{equation*}{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}.\end{equation*}

The best way to study a subnormal or normal series of subgroups, $\{ H_i \}$ of $G$, is actually to study the factor groups $H_{i+1}/H_i$. We say that two subnormal (normal) series $\{H_i \}$ and $\{ K_j \}$ of a group $G$ are isomorphic if there is a one-to-one correspondence between the collections of factor groups $\{H_{i+1}/H_i \}$ and $\{ K_{j+1}/ K_j \}$.

##### Example13.14

The two normal series \begin{gather*} {\mathbb Z}_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \{ 0 \}\\ {\mathbb Z}_{60} \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{ 0 \} \end{gather*} of the group ${\mathbb Z}_{60}$ are isomorphic since \begin{gather*} {\mathbb Z}_{60} / \langle 3 \rangle \cong \langle 20 \rangle / \{ 0 \} \cong {\mathbb Z}_{3}\\ \langle 3 \rangle / \langle 15 \rangle \cong \langle 4 \rangle / \langle 20 \rangle \cong {\mathbb Z}_{5}\\ \langle 15 \rangle / \{ 0 \} \cong {\mathbb Z}_{60} / \langle 4 \rangle \cong {\mathbb Z}_4. \end{gather*}

A subnormal series $\{ H_i \}$ of a group $G$ is a composition series if all the factor groups are simple; that is, if none of the factor groups of the series contains a normal subgroup. A normal series $\{ H_i \}$ of $G$ is a principal series if all the factor groups are simple.

##### Example13.15

The group ${\mathbb Z}_{60}$ has a composition series \begin{equation*}{\mathbb Z}_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \langle 30 \rangle \supset \{ 0 \}\end{equation*} with factor groups \begin{align*} {\mathbb Z}_{60} / \langle 3 \rangle & \cong {\mathbb Z}_{3}\\ \langle 3 \rangle / \langle 15 \rangle & \cong {\mathbb Z}_{5}\\ \langle 15 \rangle / \langle 30 \rangle & \cong {\mathbb Z}_{2}\\ \langle 30 \rangle / \{ 0 \} & \cong {\mathbb Z}_2. \end{align*} Since ${\mathbb Z}_{60}$ is an abelian group, this series is automatically a principal series. Notice that a composition series need not be unique. The series \begin{equation*}{\mathbb Z}_{60} \supset \langle 2 \rangle \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{ 0 \}\end{equation*} is also a composition series.

##### Example13.16

For $n \geq 5$, the series \begin{equation*}S_n \supset A_n \supset \{ (1) \}\end{equation*} is a composition series for $S_n$ since $S_n / A_n \cong {\mathbb Z}_2$ and $A_n$ is simple.

##### Example13.17

Not every group has a composition series or a principal series. Suppose that \begin{equation*}\{ 0 \} = H_0 \subset H_1 \subset \cdots \subset H_{n-1} \subset H_n = {\mathbb Z}\end{equation*} is a subnormal series for the integers under addition. Then $H_1$ must be of the form $k {\mathbb Z}$ for some $k \in {\mathbb N}$. In this case $H_1 / H_0 \cong k {\mathbb Z}$ is an infinite cyclic group with many nontrivial proper normal subgroups.

Although composition series need not be unique as in the case of ${\mathbb Z}_{60}$, it turns out that any two composition series are related. The factor groups of the two composition series for ${\mathbb Z}_{60}$ are ${\mathbb Z}_2$, ${\mathbb Z}_2$, ${\mathbb Z}_3$, and ${\mathbb Z}_5$; that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

##### Proof

A group $G$ is solvable if it has a subnormal series $\{ H_i \}$ such that all of the factor groups $H_{i+1} / H_i$ are abelian. Solvable groups will play a fundamental role when we study Galois theory and the solution of polynomial equations.

##### Example13.19

The group $S_4$ is solvable since \begin{equation*}S_4 \supset A_4 \supset \{ (1), (12)(34), (13)(24), (14)(23) \} \supset \{ (1) \}\end{equation*} has abelian factor groups; however, for $n \geq 5$ the series \begin{equation*}S_n \supset A_n \supset \{ (1) \}\end{equation*} is a composition series for $S_n$ with a nonabelian factor group. Therefore, $S_n$ is not a solvable group for $n \geq 5$.