## Section10.3Solvable Groups

### Worksheet10.3.1Activity: (Sub)Normal Series

Given a group, we can look at subgroups. We say that a sequence of subgroups

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

is a subnormal series provided each $H_i$ is normal in $H_{i+1}\text{,}$ and a normal series if each $H_i$ is normal in $G\text{.}$

A non-trivial group $G$ is called simple provided it has no non-trivial normal subgroups. We say that a subnormal series is a composition series and that a normal series is a principle series if every quotient group $H_{i+1}/H_i$ is simple.

###### 1.

Find a subnormal series for $D_4\text{.}$ Is it a normal series?

###### 2.

Find two different normal series for $\Z_{60}$ of length 3 (length is the number of proper inclusions).

###### 3.

Find the quotient groups $H_{i+1}/H_i$ for both series above. How are these related? Are the series composition series?

###### 4.

Find a composition series for $\Z_{60}\text{.}$ Can you take it to be a refinement of the normal series you found above?

### Subsection10.3.2Series of Subgroups

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.

###### Example10.21.

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*}
###### Example10.22.

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), (12)(34), (13)(24), (14)(23) \} \supset \{ (1), (12)(34) \} \supset \{ (1) \}\text{.} \end{equation*}

The subgroup $\{ (1), (12)(34) \}$ 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 $K_j\text{.}$

###### Example10.23.

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 $\{ K_{j+1}/ K_j \}\text{.}$

###### Example10.24.

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 group is called simple provided it contains no non-trivial normal subgroups. For series, we care whether the factor groups are simple or not.

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.

###### Example10.25.

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.

###### Example10.26.

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.

###### Example10.27.

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.

###### Proof.

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 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.

###### Example10.29.

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 $n \geq 5\text{.}$

### Exercises10.3.3Collected Homework

###### 1.

Consider the normal series below for $\Z_{24}\text{:}$

\begin{equation*} \Z_{24} \supset \langle 12 \rangle \supset \{0\}\text{.} \end{equation*}
1. Find the two quotient groups for the series. Find the “standard” abelian groups each is isomorphic to.

2. For the quotient group that is not simple found above, find a non-trivial normal subgroup (of the quotient group). Then realize the subgroup as a quotient group $G'/\langle 12\rangle$ for some $G'$

3. Demonstrate/explain how this shows us how to build a longer normal series for $\Z_{24}\text{.}$

4. Find two different composition series for $\Z_{24}$ (one can be an extension of what you were working on above). Then use quotient groups to demonstrate that the two series are “isomorphic” (and explain what this means).