13.2: Solvable Groups
A subnormal series of a group \(G\) is a finite sequence of subgroups
\[ G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\text{,} \nonumber \]
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.
Example \(13.11\)
Any series of subgroups of an abelian group is a normal series.
Solution
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*}
Example \(13.12\)
A subnormal series need not be a normal series. Consider the following subnormal series of the group \(D_4\text{:}\)
\[ D_4 \supset \{ (1), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \supset \{ (1), (1 \, 2)(3 \, 4) \} \supset \{ (1) \}\text{.} \nonumber \]
Solution
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 \(K_j\text{.}\)
Example \(13.13\)
The series
\[ {\mathbb Z} \supset 3{\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 90{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\} \nonumber \]
Solution
is a refinement of the series
\[ {\mathbb Z} \supset 9{\mathbb Z} \supset 45{\mathbb Z} \supset 180{\mathbb Z} \supset \{0\}\text{.} \nonumber \]
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 \(\{ K_{j+1}/ K_j \}\text{.}\)
Example \(13.14\)
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
Solution
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.
Example \(13.15\)
The group \({\mathbb Z}_{60}\) has a composition series
\[ {\mathbb Z}_{60} \supset \langle 3 \rangle \supset \langle 15 \rangle \supset \langle 30 \rangle \supset \{ 0 \} \nonumber \]
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*}
Solution
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
\[ {\mathbb Z}_{60} \supset \langle 2 \rangle \supset \langle 4 \rangle \supset \langle 20 \rangle \supset \{ 0 \} \nonumber \]
is also a composition series.
Example \(13.16\)
For \(n \geq 5\text{,}\) the series
\[ S_n \supset A_n \supset \{ (1) \} \nonumber \]
Solution
is a composition series for \(S_n\) since \(S_n / A_n \cong {\mathbb Z}_2\) and \(A_n\) is simple.
Example \(13.17\)
Not every group has a composition series or a principal series. Suppose that
\[ \{ 0 \} = H_0 \subset H_1 \subset \cdots \subset H_{n-1} \subset H_n = {\mathbb Z} \nonumber \]
is a subnormal series for the integers under addition.
Solution
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.
Theorem \(13.18\). Jordan-Hölder
Any two composition series of \(G\) are isomorphic.
- 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 (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
\[ H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \}\text{,} \nonumber \]
we have a composition series for \(H_{n-1}\text{.}\) Our induction hypothesis says that this series must be equivalent to the composition series
\[ H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \}\text{.} \nonumber \]
Hence, the composition series
\[ G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \} \nonumber \]
and
\[ G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \} \nonumber \]
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,
\[ K_{m-1} / (K_{m-1} \cap H_{n-1}) \cong (H_{n-1} K_{m-1}) / H_{n-1} = G/H_{n-1}\text{.} \nonumber \]
Therefore,
\[ G = H_n \supset H_{n-1} \supset H_{n-1} \cap K_{m-1} \supset \cdots \supset H_0 \cap K_{m-1} = \{ e \} \nonumber \]
and
\[ G = K_m \supset K_{m-1} \supset K_{m-1} \cap H_{n-1} \supset \cdots \supset K_0 \cap H_{n-1} = \{ e \} \nonumber \]
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.
Example \(13.19\)
The group \(S_4\) is solvable since
\[ S_4 \supset A_4 \supset \{ (1), (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \} \supset \{ (1) \} \nonumber \]
has abelian factor groups;
Solution
however, for \(n \geq 5\) the series
\[ S_n \supset A_n \supset \{ (1) \} \nonumber \]
is a composition series for \(S_n\) with a nonabelian factor group. Therefore, \(S_n\) is not a solvable group for \(n \geq 5\text{.}\)