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Abstract Algebra: Theory and Applications
Thomas W. Judson
IndexPrevUpNext
PrevUpNext
1 Preliminaries
A Short Note on Proofs
Sets and Equivalence Relations
Reading Questions
References and Suggested Readings
Sage Exercises
2 The Integers
The Division Algorithm
Programming Exercises
Integer Equivalence Classes and Symmetries
Definitions and Examples
Additional Exercises: Detecting Errors
4 Cyclic Groups
Cyclic Subgroups
Multiplicative Group of Complex Numbers
The Method of Repeated Squares
5 Permutation Groups
Definitions and Notation
Dihedral Groups
6 Cosets and Lagrange's Theorem
Cosets
Lagrange's Theorem
Fermat's and Euler's Theorems
7 Introduction to Cryptography
Private Key Cryptography
Public Key Cryptography
Additional Exercises: Primality and Factoring
8 Algebraic Coding Theory
Error-Detecting and Correcting Codes
Linear Codes
Parity-Check and Generator Matrices
Efficient Decoding
9 Isomorphisms
Definition and Examples
Direct Products
10 Normal Subgroups and Factor Groups
Factor Groups and Normal Subgroups
The Simplicity of the Alternating Group
11 Homomorphisms
Group Homomorphisms
The Isomorphism Theorems
Additional Exercises: Automorphisms
12 Matrix Groups and Symmetry
Matrix Groups
13 The Structure of Groups
Finite Abelian Groups
Solvable Groups
14 Group Actions
Groups Acting on Sets
The Class Equation
Burnside's Counting Theorem
Programming Exercise
References and Suggested Reading
15 The Sylow Theorems
The Sylow Theorems
Examples and Applications
16 Rings
Integral Domains and Fields
Ring Homomorphisms and Ideals
Maximal and Prime Ideals
An Application to Software Design
17 Polynomials
Polynomial Rings
Irreducible Polynomials
Additional Exercises: Solving the Cubic and Quartic Equations
18 Integral Domains
Fields of Fractions
Factorization in Integral Domains
19 Lattices and Boolean Algebras
Lattices
Boolean Algebras
The Algebra of Electrical Circuits
20 Vector Spaces
Subspaces
Linear Independence
21 Fields
Extension Fields
Splitting Fields
Geometric Constructions
22 Finite Fields
Structure of a Finite Field
Polynomial Codes
Additional Exercises: Error Correction for BCH Codes
23 Galois Theory
Field Automorphisms
The Fundamental Theorem
A GNU Free Documentation License
B Hints and Answers to Selected Exercises
C Notation
Authored in PreTeXt
Section 13.2 Solvable 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.
Example 13.11.
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*}
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 \(K_j\text{.}\)
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 \(\{ K_{j+1}/ K_j \}\text{.}\)
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.
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.
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.
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.
Theorem 13.18. Jordan-Hölder.
Any two composition series of \(G\) are isomorphic.
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*}
\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*}
\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.
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
is a composition series for \(S_n\) with a nonabelian factor group. Therefore, \(S_n\) is not a solvable group for \(n \geq 5\text{.}\)