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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 \},\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 \(n \geq 5\).