## Section13.2Solvable Groups

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 \}\text{,} \end{equation*}

where $$H_i$$ is a normal subgroup of $$H_{i+1}\text{.}$$ If each subgroup $$H_i$$ is normal in $$G\text{,}$$ 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\}\text{.} \end{gather*}

### Example13.12.

A subnormal series need not be a normal series. Consider the following subnormal series of the group $$D_4\text{:}$$

\begin{equation*} D_4 \supset \{ (1), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \supset \{ (1), (1 \, 2)(3 \, 4) \} \supset \{ (1) \}\text{.} \end{equation*}

The subgroup $$\{ (1), (1 \, 2)(3 \, 4) \}$$ is not normal in $$D_4\text{;}$$ 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 $$\{ H_i \} \subset \{ K_j \}\text{.}$$ That is, each $$H_i$$ is one of the $$K_j\text{.}$$

### 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\}\text{.} \end{equation*}

The best way to study a subnormal or normal series of subgroups, $$\{ H_i \}$$ of $$G\text{,}$$ is actually to study the factor groups $$H_{i+1}/H_i\text{.}$$ 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 $$\{ K_{j+1}/ K_j \}\text{.}$$

### 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\text{.} \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\text{.} \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 $$n \geq 5\text{,}$$ 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 $$k \in {\mathbb N}\text{.}$$ 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 $${\mathbb Z}_{60}\text{,}$$ it turns out that any two composition series are related. The factor groups of the two composition series for $${\mathbb Z}_{60}$$ are $${\mathbb Z}_2\text{,}$$ $${\mathbb Z}_2\text{,}$$ $${\mathbb Z}_3\text{,}$$ and $${\mathbb Z}_5\text{;}$$ that is, the two composition series are isomorphic. The Jordan-Hölder Theorem says that this is always the case.

We shall employ mathematical induction on the length of the composition series. If the length of a composition series is 1, then $$G$$ must be a simple group. In this case any two composition series are isomorphic.

Suppose now that the theorem is true for all groups having a composition series of length $$k\text{,}$$ where $$1 \leq k \lt n\text{.}$$ Let

\begin{gather*} G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\\ G = K_m \supset K_{m-1} \supset \cdots \supset K_1 \supset K_0 = \{ e \} \end{gather*}

be two composition series for $$G\text{.}$$ We can form two new subnormal series for $$G$$ since $$H_i \cap K_{m-1}$$ is normal in $$H_{i+1} \cap K_{m-1}$$ and $$K_j \cap H_{n-1}$$ is normal in $$K_{j+1} \cap H_{n-1}\text{:}$$

\begin{gather*} G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\\ G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \}\text{.} \end{gather*}

Since $$H_i \cap K_{m-1}$$ is normal in $$H_{i+1} \cap K_{m-1}\text{,}$$ the Second Isomorphism Theorem (Theorem 11.12) implies that

\begin{align*} (H_{i+1} \cap K_{m-1}) / (H_i \cap K_{m-1}) & = (H_{i+1} \cap K_{m-1}) / (H_i \cap ( H_{i+1} \cap K_{m-1} ))\\ & \cong H_i (H_{i+1} \cap K_{m-1})/ H_i\text{,} \end{align*}

where $$H_i$$ is normal in $$H_i (H_{i+1} \cap K_{m-1})\text{.}$$ Since $$\{ H_i \}$$ is a composition series, $$H_{i+1} / H_i$$ must be simple; consequently, $$H_i (H_{i+1} \cap K_{m-1})/ H_i$$ is either $$H_{i+1}/H_i$$ or $$H_i/H_i\text{.}$$ That is, $$H_i (H_{i+1} \cap K_{m-1})$$ must be either $$H_i$$ or $$H_{i+1}\text{.}$$ Removing any nonproper inclusions from the series

\begin{equation*} H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\text{,} \end{equation*}

we have a composition series for $$H_{n-1}\text{.}$$ Our induction hypothesis says that this series must be equivalent to the composition series

\begin{equation*} H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\text{.} \end{equation*}

Hence, the composition series

\begin{equation*} G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \} \end{equation*}

and

\begin{equation*} G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \} \end{equation*}

are equivalent. If $$H_{n-1} = K_{m-1}\text{,}$$ then the composition series $$\{H_i \}$$ and $$\{ K_j \}$$ are equivalent and we are done; otherwise, $$H_{n-1} K_{m-1}$$ is a normal subgroup of $$G$$ properly containing $$H_{n-1}\text{.}$$ In this case $$H_{n-1} K_{m-1} = G$$ and we can apply the Second Isomorphism Theorem once again; that is,

\begin{equation*} K_{m-1} / (K_{m-1} \cap H_{n-1}) \cong (H_{n-1} K_{m-1}) / H_{n-1} = G/H_{n-1}\text{.} \end{equation*}

Therefore,

\begin{equation*} G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \} \end{equation*}

and

\begin{equation*} G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \} \end{equation*}

are equivalent and the proof of the theorem is complete.

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), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \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 $$n \geq 5\text{.}$$