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Abstract Algebra: The Basic Graduate Year

Robert B. Ash

PREFACE

This is a text for the basic graduate sequence in abstract algebra, oļ¬ered by most
universities. We study fundamental algebraic structures, namely groups, rings, ļ¬elds and
modules, and maps between these structures. The techniques are used in many areas of
mathematics, and there are applications to physics, engineering and computer science as
well. In addition, I have attempted to communicate the intrinsic beauty of the subject.
Ideally, the reasoning underlying each step of a proof should be completely clear, but the
overall argument should be as brief as possible, allowing a sharp overview of the result.
These two requirements are in opposition, and it is my job as expositor to try to resolve
the conļ¬‚ict.
My primary goal is to help the reader learn the subject, and there are times when
informal or intuitive reasoning leads to greater understanding than a formal proof. In the
text, there are three types of informal arguments:

1. The concrete or numerical example with all features of the general case. Here, the
example indicates how the proof should go, and the formalization amounts to substi-
tuting Greek letters for numbers. There is no essential loss of rigor in the informal
version.
2. Brief informal surveys of large areas. There are two of these, p-adic numbers and
group representation theory. References are given to books accessible to the beginning
3. Intuitive arguments that replace lengthy formal proofs which do not reveal why a
result is true. In this case, explicit references to a precise formalization are given. I am
not saying that the formal proof should be avoided, just that the basic graduate year,
where there are many pressing matters to cope with, may not be the appropriate place,
especially when the result rather than the proof technique is used in applications.

I would estimate that about 90 percent of the text is written in conventional style,
and I hope that the book will be used as a classroom text as well as a supplementary
reference.
Solutions to all problems are included in the text; in my experience, most students
ļ¬nd this to be a valuable feature. The writing style for the solutions is similar to that
of the main text, and this allows for wider coverage as well as reinforcement of the basic
ideas.
Chapters 1ā“4 cover basic properties of groups, rings, ļ¬elds and modules. The typi-
cal student will have seen some but not all of this material in an undergraduate algebra
course. [It should be possible to base an undergraduate course on Chapters 1ā“4, traversed
at a suitable pace with detailed coverage of the exercises.] In Chapter 4, the fundamental
structure theorems for ļ¬nitely generated modules over a principal ideal domain are de-
veloped concretely with the aid of the Smith normal form. Students will undoubtedly be
comfortable with elementary row and column operations, and this will signiļ¬cantly aid
the learning process.
In Chapter 5, the theme of groups acting on sets leads to a nice application to com-
binatorics as well as the fundamental Sylow theorems and some results on simple groups.
Analysis of normal and subnormal series leads to the Jordan-HĀØlder theorem and to solv-
o
able and nilpotent groups. The ļ¬nal section, on deļ¬ning a group by generators and
relations, concentrates on practical cases where the structure of a group can be deduced
from its presentation. Simplicity of the alternating groups and semidirect products are
covered in the exercises.
Chapter 6 goes quickly to the fundamental theorem of Galois theory; this is possible
because the necessary background has been covered in Chapter 3. After some examples
of direct calculation of a Galois group, we proceed to ļ¬nite ļ¬elds, which are of great
importance in applications, and cyclotomic ļ¬elds, which are fundamental in algebraic
number theory. The Galois group of a cubic is treated in detail, and the quartic is
covered in an appendix. Sections on cyclic and Kummer extensions are followed by Galoisā™
fundamental theorem on solvability by radicals. The last section of the chapter deals with
transcendental extensions and transcendence bases.
In the remaining chapters, we begin to apply the results and methods of abstract
algebra to related areas. The title of each chapter begins with āIntroducing . . . ā, and the
areas to be introduced are algebraic number theory, algebraic geometry, noncommutative
algebra and homological algebra (including categories and functors).
Algebraic number theory and algebraic geometry are the two major areas that use the
tools of commutative algebra (the theory of commutative rings). In Chapter 7, after an
example showing how algebra can be applied in number theory, we assemble some algebraic
equipment: integral extensions, norms, traces, discriminants, Noetherian and Artinian
modules and rings. We then prove the fundamental theorem on unique factorization of
ideals in a Dedekind domain. The chapter concludes with an informal introduction to
p-adic numbers and some ideas from valuation theory.
Chapter 8 begins geometrically with varieties in aļ¬ne space. This provides moti-
vation for Hilbertā™s fundamental theorems, the basis theorem and the Nullstellensatz.
Several equivalent versions of the Nullstellensatz are given, as well as some corollaries
with geometric signiļ¬cance. Further geometric considerations lead to the useful algebraic
techniques of localization and primary decomposition. The remainder of the chapter is
concerned with the tensor product and its basic properties.
Chapter 9 begins the study of noncommutative rings and their modules. The basic
theory of simple and semisimple rings and modules, along with Schurā™s lemma and Ja-
cobsonā™s theorem, combine to yield Wedderburnā™s theorem on the structure of semisimple
rings. We indicate the precise connection between the two popular deļ¬nitions of simple
ring in the literature. After an informal introduction to group representations, Maschkeā™s
theorem on semisimplicity of modules over the group algebra is proved. The introduction
of the Jacobson radical gives more insight into the structure of rings and modules. The
chapter ends with the Hopkins-Levitzki theorem that an Artinian ring is Noetherian, and
the useful lemma of Nakayama.
In Chapter 10, we introduce some of the tools of homological algebra. Waiting until
the last chapter for this is a deliberate decision. Students need as much exposure as
possible to speciļ¬c algebraic systems before they can appreciate the broad viewpoint of
category theory. Even experienced students may have diļ¬culty absorbing the abstract
deļ¬nitions of kernel, cokernel, product, coproduct, direct and inverse limit. To aid the
reader, functors are introduced via the familiar examples of hom and tensor. No attempt
is made to work with general abelian categories. Instead, we stay within the category of
modules and study projective, injective and ļ¬‚at modules.
In a supplement, we go much farther into homological algebra than is usual in the basic
algebra sequence. We do this to help students cope with the massive formal machinery
that makes it so diļ¬cult to gain a working knowledge of this area. We concentrate on
the results that are most useful in applications: the long exact homology sequence and
the properties of the derived functors Tor and Ext. There is a complete proof of the
snake lemma, a rarity in the literature. In this case, going through a long formal proof is
entirely appropriate, because doing so will help improve algebraic skills. The point is not
to avoid diļ¬culties, but to make most eļ¬cient use of the ļ¬nite amount of time available.

Robert B. Ash
October 2000

Further Remarks

Many mathematicians believe that formalism aids understanding, but I believe that
when one is learning a subject, formalism often prevents understanding. The most im-
portant skill is the ability to think intuitively. This is true even in a highly abstract ļ¬eld
such as homological algebra. My writing style reļ¬‚ects this view.
Classroom lectures are inherently ineļ¬cient. If the pace is slow enough to allow
comprehension as the lecture is delivered, then very little can be covered. If the pace
is fast enough to allow decent coverage, there will unavoidably be large gaps. Thus
the student must depend on the textbook, and the current trend in algebra is to produce
massive encyclopedias, which are likely to be quite discouraging to the beginning graduate
student. Instead, I have attempted to write a text of manageable size, which can be read
by students, including those working independently.
Another goal is to help the student reach an advanced level as quickly and eļ¬ciently as
possible. When I omit a lengthy formal argument, it is because I judge that the increase
in algebraic skills is insuļ¬cient to justify the time and eļ¬ort involved in going through
the formal proof. In all cases, I give explicit references where the details can be found.
One can argue that learning to write formal proofs is an essential part of the studentā™s
mathematical training. I agree, but the ability to think intuitively is fundamental and
must come ļ¬rst. I would add that the way things are today, there is absolutely no danger
that the student will be insuļ¬ciently exposed to formalism and abstraction. In fact there
is quite a bit of it in this book, although not 100 percent.
I oļ¬er this text in the hope that it will make the studentā™s trip through algebra more
enjoyable. I have done my best to avoid gaps in the reasoning. I never use the phrase
āit is easy to seeā under any circumstances. I welcome comments and suggestions for
improvement.
Copyright c 2000, by Robert B. Ash
Paper or electronic copies for noncommercial use may be made freely without explicit
permission of the author. All other rights are reserved.
ABSTRACT ALGEBRA: THE BASIC GRADUATE YEAR

CHAPTER 0 PREREQUISITES
0.1 Elementary Number Theory
0.2 Set Theory
0.3 Linear Algebra

CHAPTER 1 GROUP FUNDAMENTALS
1.1 Groups and Subgroups
1.2 Permutation Groups
1.3 Cosets, Normal Subgroups and Homomorphisms
1.4 The Isomorphism Theorems
1.5 Direct Products

CHAPTER 2 RING FUNDAMENTALS
2.1 Basic Deļ¬nitions and Properties
2.2 Ideals, Homomorphisms and Quotient Rings
2.3 The Isomorphism Theorems For Rings
2.4 Maximal and Prime Ideals
2.5 Polynomial Rings
2.6 Unique Factorization
2.7 Principal Ideal Domains and Euclidean Domains
2.8 Rings of Fractions
2.9 Irreducible Polynomials

CHAPTER 3 FIELD FUNDAMENTALS
3.1 Field Extensions
3.2 Splitting Fields
3.3 Algebraic Closures
3.4 Separability
3.5 Normal Extensions

CHAPTER 4 MODULE FUNDAMENTALS
4.1 Modules and Algebras
4.2 The Isomorphism Theorems For Modules
4.3 Direct Sums and Free Modules
4.4 Homomorphisms and Matrices
4.5 Smith Normal Form
4.6 Fundamental Structure Theorems
4.7 Exact Sequences and Diagram Chasing

CHAPTER 5 SOME BASIC TECHNIQUES OF GROUP THEORY
5.1 Groups Acting on Sets
5.2 The Orbit-Stabilizer Theorem
5.3 Applications to Combinatorics
5.4 The Sylow Theorems
5.5 Applications of the Sylow Theorems
5.6 Composition Series
5.7 Solvable and Nilpotent Groups
5.8 Generators and Relations

CHAPTER 6 GALOIS THEORY
6.1 Fixed Fields and Galois Groups
6.2 The Fundamental Theorem
6.3 Computing a Galois Group Directly
6.4 Finite Fields
6.5 Cyclotomic Fields
6.6 The Galois Group of a Cubic
6.7 Cyclic and Kummer Extensions
6.9 Transcendental Extensions
Appendix to Chapter 6

CHAPTER 7 INTRODUCING ALGEBRAIC NUMBER THEORY
7.1 Integral Extensions
7.2 Quadratic Extensions of the Rationals
7.3 Norms and Traces
7.4 The Discriminant
7.5 Noetherian and Artinian Modules and Rings
7.6 Fractional Ideals
7.7 Unique Factorization of Ideals in a Dedekind Domain
7.8 Some Arithmetic in Dedekind Domains

CHAPTER 8 INTRODUCING ALGEBRAIC GEOMETRY
8.1 Varieties
8.2 The Hilbert Basis Theorem
8.3 The Nullstellensatz: Preliminaries
8.4 The Nullstellensatz: Equivalent Versions and Proof
8.5 Localization
8.6 Primary Decomposition
8.7 Tensor Product of Modules Over a Commutative Ring
8.8 General Tensor Products

CHAPTER 9 INTRODUCING NONCOMMUTATIVE ALGEBRA
9.1 Semisimple Modules
9.2 Two Key Theorems
9.3 Simple and Semisimple Rings
9.4 Further Properties of Simple Rings, Matrix Rings, and Endomorphisms
9.5 The Structure of Semisimple Rings
9.6 Maschkeā™s Theorem
9.8 Theorems of Hopkins-Levitzki and Nakayama

CHAPTER 10 INTRODUCING HOMOLOGICAL ALGEBRA
10.1 Categories
10.2 Products and Coproducts
10.3 Functors
10.4 Exact Functors
10.5 Projective Modules
10.6 Injective Modules
10.7 Embedding into an Injective Module
10.8 Flat Modules
10.9 Direct and Inverse Limits
Appendix to Chapter 10

SUPPLEMENT
S1 Chain Complexes
S2 The Snake Lemma
S3 The Long Exact Homology Sequence
S4 Projective and Injective Resolutions
S5 Derived Functors
S6 Some Properties of Ext and Tor
S7 Base Change in the Tensor Product

SOLUTIONS TO PROBLEMS
Chapter 0

Prerequisites

All topics listed in this chapter are covered in A Primer of Abstract Mathematics by
Robert B. Ash, MAA 1998.

0.1 Elementary Number Theory
The greatest common divisor of two integers can be found by the Euclidean algorithm,
which is reviewed in the exercises in Section 2.5. Among the important consequences of
the algorithm are the following three results.

0.1.1
If d is the greatest common divisor of a and b, then there are integers s and t such that
sa + tb = d. In particular, if a and b are relatively prime, there are integers s and t such
that sa + tb = 1.

0.1.2
If a prime p divides a product a1 Ā· Ā· Ā· an of integers, then p divides at least one ai

0.1.3 Unique Factorization Theorem
If a is an integer, not 0 or Ā±1, then

(1) a can be written as a product p1 Ā· Ā· Ā· pn of primes.
(2) If a = p1 Ā· Ā· Ā· pn = q1 Ā· Ā· Ā· qm , where the pi and qj are prime, then n = m and, after
renumbering, pi = Ā±qi for all i.

[We allow negative primes, so that, for example, ā’17 is prime. This is consistent with the
general deļ¬nition of prime element in an integral domain; see Section 2.6.]

1
2 CHAPTER 0. PREREQUISITES

0.1.4 The Integers Modulo m
If a and b are integers and m is a positive integer ā„ 2, we write a ā” b mod m, and say
that a is congruent to b modulo m, if a ā’ b is divisible by m. Congruence modulo m
is an equivalence relation, and the resulting equivalence classes are called residue classes
mod m. Residue classes can be added, subtracted and multiplied consistently by choosing
a representative from each class, performing the appropriate operation, and calculating
the residue class of the result. The collection Zm of residue classes mod m forms a
commutative ring under addition and multiplication. Zm is a ļ¬eld if and only if m is
prime. (The general deļ¬nitions of ring, integral domain and ļ¬eld are given in Section 2.1.)

0.1.5
(1) The integer a is relatively prime to m if and only if a is a unit mod m, that is, a has
a multiplicative inverse mod m.
(2) If c divides ab and a and c are relatively prime, then c divides b.
(3) If a and b are relatively prime to m, then ab is relatively prime to m.
(4) If ax ā” ay mod m and a is relatively prime to m, then x ā” y mod m.
(5) If d = gcd(a, b), the greatest common divisor of a and b, then a/d and b/d are relatively
prime.
(6) If ax ā” ay mod m and d = gcd(a, m), then x ā” y mod m/d.
(7) If ai divides b for i = 1, . . . , r, and ai and aj are relatively prime whenever i = j, then
the product a1 Ā· Ā· Ā· ar divides b.
(8) The product of two integers is their greatest common divisor times their least common
multiple.

0.1.6 Chinese Remainder Theorem
If m1 , . . . , mr are relatively prime in pairs, then the system of simultaneous equations
x ā” bj mod mj , j = 1, . . . , r, has a solution for arbitrary integers bj . The set of solutions
forms a single residue class mod m=m1 Ā· Ā· Ā· mr , so that there is a unique solution mod m.
This result can be derived from the abstract form of the Chinese remainder theorem;
see Section 2.3.

0.1.7 Eulerā™s Theorem
The Euler phi function is deļ¬ned by Ļ•(n) = the number of integers in {1, . . . , n} that
are relatively prime to n. For an explicit formula for Ļ•(n), see Section 1.1, Problem 13.
Eulerā™s theorem states that if n ā„ 2 and a is relatively prime to n, then aĻ•(n) ā” 1 mod n.

0.1.8 Fermatā™s Little Theorem
If a is any integer and p is a prime not dividing a, then apā’1 ā” 1 mod p. Thus for any
integer a and prime p, whether or not p divides a, we have ap ā” a mod p.
For proofs of (0.1.7) and (0.1.8), see (1.3.4).
0.2. SET THEORY 3

0.2 Set Theory
0.2.1
A partial ordering on a set S is a relation on S that is reļ¬‚exive (x ā¤ x for all x ā S),
antisymmetric (x ā¤ y and y ā¤ x implies x = y), and transitive (x ā¤ y and y ā¤ z implies
x ā¤ z). If for all x, y ā S, either x ā¤ y or y ā¤ x, the ordering is total.

0.2.2
A well-ordering on S is a partial ordering such that every nonempty subset A of S has a
smallest element a. (Thus a ā¤ b for every b ā A).

0.2.3 Well-Ordering Principle
Every set can be well-ordered.

0.2.4 Maximum Principle
If T is any chain (totally ordered subset) of a partially ordered set S, then T is contained
in a maximal chain M . (Maximal means that M is not properly contained in a larger
chain.)

0.2.5 Zornā™s Lemma
If S is a nonempty partially ordered set such that every chain of S has an upper bound
in S, then S has a maximal element.
(The element x is an upper bound of the set A if a ā¤ x for every a ā A. Note that
x need not belong to A, but in the statement of Zornā™s lemma, we require that if A is a
chain of S, then A has an upper bound that actually belongs to S.)

0.2.6 Axiom of Choice
Given any family of nonempty sets Si , i ā I, we can choose an element of each Si .
Formally, there is a function f whose domain is I such that f (i) ā Si for all i ā I.
The well-ordering principle, the maximum principle, Zornā™s lemma, and the axiom of
choice are equivalent in the sense that if any one of these statements is added to the basic
axioms of set theory, all the others can be proved. The statements themselves cannot be
proved from the basic axioms. Constructivist mathematics rejects the axiom of choice
and its equivalents. In this philosophy, an assertion that we can choose an element from
each Si must be accompanied by an explicit algorithm. The idea is appealing, but its
acceptance results in large areas of interesting and useful mathematics being tossed onto
the scrap heap. So at present, the mathematical mainstream embraces the axiom of
choice, Zornā™s lemma et al.
4 CHAPTER 0. PREREQUISITES

0.2.7 Proof by Transļ¬nite Induction
To prove that statement Pi holds for all i in the well-ordered set I, we do the following:
1. Prove the basis step P0 , where 0 is the smallest element of I.
2. If i > 0 and we assume that Pj holds for all j < i (the transļ¬nite induction
hypothesis), prove Pi .
It follows that Pi is true for all i.

0.2.8
We say that the size of the set A is less than or equal to the size of B (notation A ā¤s B) if
there is an injective map from A to B. We say that A and B have the same size (A =s B)
if there is a bijection between A and B.

0.2.9 SchrĀØder-Bernstein Theorem
o
If A ā¤s B and B ā¤s A, then A =s B. (This can be proved without the axiom of choice.)

0.2.10
Using (0.2.9), one can show that if sets of the same size are called equivalent, then ā¤s
on equivalence classes is a partial ordering. It follows with the aid of Zornā™s lemma that
the ordering is total. The equivalence class of a set A, written |A|, is called the cardinal
number or cardinality of A. In practice, we usually identify |A| with any convenient
member of the equivalence class, such as A itself.

0.2.11
For any set A, we can always produce a set of greater cardinality, namely the power set 2A ,
that is, the collection of all subsets of A.

0.2.12
Deļ¬ne addition and multiplication of cardinal numbers by |A|+|B| = |AāŖB| and |A||B| =
|A Ć— B|. In deļ¬ning addition, we assume that A and B are disjoint. (They can always be
disjointized by replacing a ā A by (a, 0) and b ā B by (b, 1).)

0.2.13
If ā„µ0 is the cardinal number of a countably inļ¬nite set, then ā„µ0 + ā„µ0 = ā„µ0 ā„µ0 = ā„µ0 . More
generally,
(a) If Ī± and Ī² are cardinals, with Ī± ā¤ Ī² and Ī² inļ¬nite, then Ī± + Ī² = Ī².
(b) If Ī± = 0 (i.e., Ī± is nonempty), Ī± ā¤ Ī² and Ī² is inļ¬nite, then Ī±Ī² = Ī².

0.2.14
If A is an inļ¬nite set, then A and the set of all ļ¬nite subsets of A have the same cardinality.
0.3. LINEAR ALGEBRA 5

0.3 Linear Algebra
It is not feasible to list all results presented in an undergraduate course in linear algebra.
Instead, here is a list of topics that are covered in a typical course.

1. Sums, products, transposes, inverses of matrices; symmetric matrices.
2. Elementary row and column operations; reduction to echelon form.
3. Determinants: evaluation by Laplace expansion and Cramerā™s rule.
4. Vector spaces over a ļ¬eld; subspaces, linear independence and bases.
5. Rank of a matrix; homogeneous and nonhomogeneous linear equations.
6. Null space and range of a matrix; the dimension theorem.
7. Linear transformations and their representation by matrices.
8. Coordinates and matrices under change of basis.
9. Inner product spaces and the projection theorem.
10. Eigenvalues and eigenvectors; diagonalization of matrices with distinct eigenvalues,
symmetric and Hermitian matrices.

A more advanced course might cover the following topics:

12. Generalized eigenvectors and the Jordan canonical form.
13. The minimal and characteristic polynomials of a matrix; Cayley-Hamilton theorem.
14. The adjoint of a linear operator.
15. Projection operators.
16. Normal operators and the spectral theorem.
Chapter 1

Group Fundamentals

1.1 Groups and Subgroups
1.1.1 Deļ¬nition
A group is a nonempty set G on which there is deļ¬ned a binary operation (a, b) ā’ ab
satisfying the following properties.
Closure: If a and b belong to G, then ab is also in G;
Associativity : a(bc) = (ab)c for all a, b, c ā G;
Identity : There is an element 1 ā G such that a1 = 1a = a for all a in G;
Inverse: If a is in G, then there is an element aā’1 in G such that aaā’1 = aā’1 a = 1.
A group G is abelian if the binary operation is commutative, i.e., ab = ba for all a, b
in G. In this case the binary operation is often written additively ((a, b) ā’ a + b), with
the identity written as 0 rather than 1.
There are some very familiar examples of abelian groups under addition, namely the
integers Z, the rationals Q, the real numbers R, the complex numers C, and the integers
Zm modulo m. Nonabelian groups will begin to appear in the next section.
The associative law generalizes to products of any ļ¬nite number of elements, for exam-
ple, (ab)(cde) = a(bcd)e. A formal proof can be given by induction. If two people A and
B form a1 Ā· Ā· Ā· an in diļ¬erent ways, the last multiplication performed by A might look like
(a1 Ā· Ā· Ā· ai )(ai+1 Ā· Ā· Ā· an ), and the last multiplication by B might be (a1 Ā· Ā· Ā· aj )(aj+1 Ā· Ā· Ā· an ).
But if (without loss of generality) i < j, then (induction hypothesis)
(a1 Ā· Ā· Ā· aj ) = (a1 Ā· Ā· Ā· ai )(ai+1 Ā· Ā· Ā· aj )
and
(ai+1 Ā· Ā· Ā· an ) = (ai+1 Ā· Ā· Ā· aj )(aj+1 Ā· Ā· Ā· an ).
By the n = 3 case, i.e., the associative law as stated in the deļ¬nition of a group, the
products computed by A and B are the same.

1
2 CHAPTER 1. GROUP FUNDAMENTALS

The identity is unique (1 = 1 1 = 1), as is the inverse of any given element (if b and b
are inverses of a, then b = 1b = (b a)b = b (ab) = b 1 = b ). Exactly the same argument
shows that if b is a right inverse, and a a left inverse, of a, then b = b .

A subgroup H of a group G is a nonempty subset of G that forms a group under the
binary operation of G. Equivalently, H is a nonempty subset of G such that if a and b
belong to H, so does abā’1 . (Note that 1 = aaā’1 ā H; also, ab = a((bā’1 )ā’1 ) ā H.)
If A is any subset of a group G, the subgroup generated by A is the smallest subgroup
containing A, often denoted by A . Formally, A is the intersection of all subgroups
containing A. More explicitly, A consists of all ļ¬nite products a1 Ā· Ā· Ā· an , n = 1, 2, . . . ,
where for each i, either ai or aā’1 belongs to A. To see this, note that all such products
i
belong to any subgroup containing A, and the collection of all such products forms a
subgroup. In checking that the inverse of an element of A also belongs to A , we use
the fact that

(a1 Ā· Ā· Ā· an )ā’1 = aā’1 Ā· Ā· Ā· aā’1
n 1

which is veriļ¬ed directly: (a1 Ā· Ā· Ā· an )(aā’1 Ā· Ā· Ā· aā’1 ) = 1.
n 1

The groups G1 and G2 are said to be isomorphic if there is a bijection f : G1 ā’ G2 that
preserves the group operation, in other words, f (ab) = f (a)f (b). Isomorphic groups are
essentially the same; they diļ¬er only notationally. Here is a simple example. A group G
is cyclic if G is generated by a single element: G = a . A ļ¬nite cyclic group generated
by a is necessarily abelian, and can be written as {1, a, a2 , . . . , anā’1 } where an = 1, or in
additive notation, {0, a, 2a, . . . , (n ā’ 1)a}, with na = 0. Thus a ļ¬nite cyclic group with
n elements is isomorphic to the additive group Zn of integers modulo n. Similarly, if G
is an inļ¬nite cyclic group generated by a, then G must be abelian and can be written as
{1, aĀ±1 , aĀ±2 , . . . }, or in additive notation as {0, Ā±a, Ā±2a, . . . }. In this case, G is isomorphic
to the additive group Z of all integers.
The order of an element a in a group G (denoted |a|) is the least positive integer n such
that an = 1; if no such integer exists, the order of a is inļ¬nite. Thus if |a| = n, then the
cyclic subgroup a generated by a has exactly n elements, and ak = 1 iļ¬ k is a multiple
of n. (Concrete examples are more illuminating than formal proofs here. Start with 0 in
the integers modulo 4, and continually add 1; the result is 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, . . . .)
The order of the group G, denoted by |G|, is simply the number of elements in G.

1.1.4 Proposition
If G is a ļ¬nite cyclic group of order n, then G has exactly one (necessarily cyclic) subgroup
of order n/d for each positive divisor d of n, and G has no other subgroups. If G is an inļ¬-
nite cyclic group, the (necessarily cyclic) subgroups of G are of the form {1, bĀ±1 , bĀ±2 , . . . },
where b is an arbitrary element of G, or, in additive notation, {0, Ā±b, Ā±2b, . . . }.
1.1. GROUPS AND SUBGROUPS 3

Proof. Again, an informal argument is helpful. Suppose that H is a subgroup of Z20 (the
integers with addition modulo 20). If the smallest positive integer in H is 6 (a non-divisor
of 20) then H contains 6, 12, 18, 4 (oops, a contradiction, 6 is supposed to be the smallest
positive integer). On the other hand, if the smallest positive integer in H is 4, then H =
{4,8,12,16,0}. Similarly, if the smallest positive integer in a subgroup H of the additive
group of integers Z is 5, then H = {0, Ā±5, Ā±10, Ā±15, Ā±20, . . . }. ā™£

If G = {1, a, . . . , anā’1 } is a cyclic group of order n, when will an element ar also have
order n? To discover the answer, letā™s work in Z12 . Does 8 have order 12? We compute
8, 16, 24 (= 0), so the order of 8 is 3. But if we try 7, we get 7, 14, 21, . . . , 77, 84 = 7 Ć— 12,
so 7 does have order 12. The point is that the least common multiple of 7 and 12 is
simply the product, while the lcm of 8 and 12 is smaller than the product. Equivalently,
the greatest common divisor of 7 and 12 is 1, while the gcd of 8 and 12 is 4 > 1. We have
the following result.

1.1.5 Proposition
If G is a cyclic group of order n generated by a, the following conditions are equivalent:

(a) |ar | = n.
(b) r and n are relatively prime.
(c) r is a unit mod n, in other words, r has an inverse mod n (an integer s such that
rs ā” 1 mod n).

Furthermore, the set Un of units mod n forms a group under multiplication. The order
of this group is Ļ•(n) = the number of positive integers less than or equal to n that are
relatively prime to n; Ļ• is the familiar Euler Ļ• function.

Proof. The equivalence of (a) and (b) follows from the discussion before the statement
of the proposition, and the equivalence of (b) and (c) is handled by a similar argument.
For example, since there are 12 distinct multiples of 7 mod 12, one of them must be 1;
speciļ¬cally, 7 Ć— 7 ā” 1 mod 12. But since 8 Ć— 3 is 0 mod 12, no multiple of 8 can be
1 mod 12. (If 8x ā” 1, multiply by 3 to reach a contradiction.) Finally, Un is a group
under multiplication because the product of two integers relatively prime to n is also
relatively prime to n. ā™£

Problems For Section 1.1
1. A semigroup is a nonempty set with a binary operation satisfying closure and asso-
ciativity (we drop the identity and inverse properties from the deļ¬nition of a group).
A monoid is a semigroup with identity (so that only the inverse property is dropped).
Give an example of a monoid that is not a group, and an example of a semigroup
that is not a monoid.
2. In Z6 , the group of integers modulo 6, ļ¬nd the order of each element.
3. List all subgroups of Z6 .
4 CHAPTER 1. GROUP FUNDAMENTALS

4. Let S be the set of all n by n matrices with real entries. Does S form a group under
5. Let S ā— be the set of all nonzero n by n matrices with real entries. Does S ā— form a
group under matrix multiplication?
6. If H is a subgroup of the integers Z and H = {0}, what does H look like?
7. Give an example of an inļ¬nite group that has a nontrivial ļ¬nite subgroup (trivial
means consisting of the identity alone).
8. Let a and b belong to the group G. If ab = ba and |a| = m, |b| = n, where m and n
are relatively prime, show that |ab| = mn and that a ā© b = {1}.
9. If G is a ļ¬nite abelian group, show that G has an element g such that |g| is the least
common multiple of {|a| : a ā G}.
10. Show that a group G cannot be the union of two proper subgroups, in other words, if
G = H āŖ K where H and K are subgroups of G, then H = G or K = G. Equivalently,
if H and K are subgroups of a group G, then H āŖ K cannot be a subgroup unless
H ā K or K ā H.
11. In an arbitrary group, let a have ļ¬nite order n, and let k be a positive integer. If
(n, k) is the greatest common divisor of n and k, and [n, k] the least common multiple,
show that the order of ak is n/(n, k) = [n, k]/k.
12. Suppose that the prime factorization of the positive integer n is

n = p e1 p e 2 Ā· Ā· Ā· p er
r
12

and let Ai be the set of all positive integers m ā {1, 2, . . . , n} such that pi divides m.
Show that if |S| is the number of elements in the set S, then

n
|Ai | = ,
pi
n
|Ai ā© Aj | = for i = j,
pi pj
n
|Ai ā© Aj ā© Ak | = for i, j, k distinct,
pi p j pk

and so on.
13. Continuing Problem 12, show that the number of positive integers less than or equal
to n that are relatively prime to n is

1 1 1
Ļ•(n) = n(1 ā’ )(1 ā’ ) Ā· Ā· Ā· (1 ā’ ).
p1 p2 pr

14. Give an example of a ļ¬nite group G (of order at least 3) with the property that the
only subgroups of G are {1} and G itself.
15. Does an inļ¬nite group with the property of Problem 14 exist?
1.2. PERMUTATION GROUPS 5

1.2 Permutation Groups
1.2.1 Deļ¬nition
A permutation of a set S is a bijection on S, that is, a function Ļ : S ā’ S that is one-
to-one and onto. (If S is ļ¬nite, then Ļ is one-to-one if and only if it is onto.) If S is not
too large, it is feasible to describe a permutation by listing the elements x ā S and the
corresponding values Ļ(x). For example, if S = {1, 2, 3, 4, 5}, then

1 2 3 4 5
3 5 4 1 2

is the permutation such that Ļ(1) = 3, Ļ(2) = 5, Ļ(3) = 4, Ļ(4) = 1, Ļ(5) = 2. If we
start with any element x ā S and apply Ļ repeatedly to obtain Ļ(x), Ļ(Ļ(x)), Ļ(Ļ(Ļ(x))),
and so on, eventually we must return to x, and there are no repetitions along the way
because Ļ is one-to-one. For the above example, we obtain 1 ā’ 3 ā’ 4 ā’ 1, 2 ā’ 5 ā’ 2.
We express this result by writing

Ļ = (1, 3, 4)(2, 5)

where the cycle (1, 3, 4) is the permutation of S that maps 1 to 3, 3 to 4 and 4 to 1,
leaving the remaining elements 2 and 5 ļ¬xed. Similarly, (2, 5) maps 2 to 5, 5 to 2, 1 to 1,
3 to 3 and 4 to 4. The product of (1, 3, 4) and (2, 5) is interpreted as a composition, with
the right factor (2, 5) applied ļ¬rst, as with composition of functions. In this case, the
cycles are disjoint, so it makes no diļ¬erence which mapping is applied ļ¬rst.
The above analysis illustrates the fact that any permutation can be expressed as a
product of disjoint cycles, and the cycle decomposition is unique.

A permutation Ļ is said to be even if its cycle decomposition contains an even number
of even cycles (that is, cycles of even length); otherwise Ļ is odd. A cycle can be de-
composed further into a product of (not necessarily disjoint) two-element cycles, called
transpositions. For example,

(1, 2, 3, 4, 5) = (1, 5)(1, 4)(1, 3)(1, 2)

where the order of application of the mappings is from right to left.
Multiplication by a transposition changes the parity of a permutation (from even to
odd, or vice versa). For example,

(2, 4)(1, 2, 3, 4, 5) = (2, 3)(1, 4, 5)
(2, 6)(1, 2, 3, 4, 5) = (1, 6, 2, 3, 4, 5);

(1, 2, 3, 4, 5) has no cycles of even length, so is even; (2, 3)(1, 4, 5) and (1, 6, 2, 3, 4, 5) each
have one cycle of even length, so are odd.
Since a cycle of even length can be expressed as the product of an odd number of
transpositions, we can build an even permutation using an even number of transpositions,
6 CHAPTER 1. GROUP FUNDAMENTALS

and an odd permutation requires an odd number of transpositions. A decomposition into
transpositions is not unique; for example, (1, 2, 3, 4, 5) = (1, 4)(1, 5)(1, 4)(1, 3)(1, 2)(3, 5),
but as mentioned above, the cycle decomposition is unique. Since multiplication by a
transposition changes the parity, it follows that if a permutation is expressed in two
diļ¬erent ways as a product of transpositions, the number of transpositions will agree in
parity (both even or both odd).
Consequently, the product of two even permutations is even; the product of two odd
permutations is even; and the product of an even and an odd permutation is odd. To
summarize very compactly, deļ¬ne the sign of the permutation Ļ as

+1 if Ļ is even
sgn(Ļ) =
ā’1 if Ļ is odd
Then for arbitrary permutations Ļ1 and Ļ2 we have
sgn(Ļ1 Ļ2 ) = sgn(Ļ1 ) sgn(Ļ2 ).

There are several permutation groups that are of major interest. The set Sn of all per-
mutations of {1, 2, . . . , n} is called the symmetric group on n letters, and its subgroup An
of all even permutations of {1, 2, . . . , n} is called the alternating group on n letters. (The
group operation is composition of functions.) Since there are as many even permutations
as odd ones (any transposition, when applied to the members of Sn , produces a one-to-one
correspondence between even and odd permutations), it follows that An is half the size
of Sn . Denoting the size of the set S by |S|, we have
|Sn | = n!, |An | = 1 n!
2

We now deļ¬ne and discuss informally D2n , the dihedral group of order 2n. Consider
a regular polygon with center O and vertices V1 , V2 , . . . , Vn , arranged so that as we move
counterclockwise around the ļ¬gure, we encounter V1 , V2 , . . . in turn. To eliminate some of
the abstraction, letā™s work with a regular pentagon with vertices A, B, C, D, E, as shown
in Figure 1.2.1.

n C yyyy
nnn yyy
n
nnn yyy
nnn yyy
nnn y
Dd O B
dd ˜˜
dd ˜˜
dd ˜˜
˜
E A

Figure 1.2.1

The group D10 consists of the symmetries of the pentagon, i.e., those permutations
that can be realized via a rigid motion (a combination of rotations and reļ¬‚ections). All
symmetries can be generated by two basic operations R and F :
1.2. PERMUTATION GROUPS 7

R is counterclockwise rotation by 360 = 360 = 72 degrees,
5
n
F (āļ¬‚ipā) is reļ¬‚ection about the line joining the center O to the ļ¬rst vertex (A in this
case).
The group D2n contains 2n elements, namely, I (the identity), R, R2 , . . . , Rnā’1 , F ,
RF , R2 F , . . . , Rnā’1 F (RF means F followed by R). For example, in the case of the
pentagon, F = (B, E)(C, D) and R = (A, B, C, D, E), so RF = (A, B)(C, E), which is
the reļ¬‚ection about the line joining O to D; note that RF can also be expressed as F Rā’1 .
In visualizing the eļ¬ect of a permutation such as F , interpret F ā™s taking B to E as vertex
B moving to where vertex E was previously.
D2n will contain exactly n rotations I, R, . . . , Rnā’1 and n reļ¬‚ections F, RF, . . . , Rnā’1 F .
If n is odd, each reļ¬‚ection is determined by a line joining the center to a vertex (and pass-
ing through the midpoint of the opposite side). If n is even, half the reļ¬‚ections are
determined by a line passing through two vertices (as well as the center), and the other
half by a line passing through the midpoints of two opposite sides (as well as the center).

1.2.4 An Abstract Characterization of the Dihedral Group
Consider the free group with generators R and F , in other words all ļ¬nite sequences whose
components are R, Rā’1 , F and F ā’1 . The group operation is concatenation, subject to
the constraint that if a symbol and its inverse occur consecutively, they may be can-
celled. For example, RF F F F ā’1 RF Rā’1 RF F is identiļ¬ed with RF F RF F F , also written
as RF 2 RF 3 . If we add further restrictions (so the group is no longer āfreeā), we can
obtain D2n . Speciļ¬cally, D2n is the group deļ¬ned by generators R and F , subject to the
relations
RF = F Rā’1 .
F 2 = I,
Rn = I,
The relations guarantee that there are only 2n distinct group elements I, R, . . . , Rnā’1 and
F, RF, . . . , Rnā’1 F . For example, with n = 5 we have
F 2 R2 F = F F RRF = F F RF Rā’1 = F F F Rā’1 Rā’1 = F Rā’2 = F R3 ;
also, R cannot be the same as R2 F , since this would imply that I = RF , or F = Rā’1 =
R4 , and there is no way to get this using the relations. Since the product of two group
elements is completely determined by the deļ¬ning relations, it follows that there cannot
be more than one group with the given generators and relations. (This statement is true
āup to isomorphismā; it is always possible to create lots of isomorphic copies of any given
group.) The symmetries of the regular n-gon provide a concrete realization.
Later we will look at more systematic methods of analyzing groups deļ¬ned by gener-
ators and relations.

Problems For Section 1.2
1. Find the cycle decomposition of the permutation
1 2 3 4 5 6
4 6 3 1 2 5
and determine whether the permutation is even or odd.
8 CHAPTER 1. GROUP FUNDAMENTALS

2. Consider the dihedral group D8 as a group of permutations of the square. Assume that
as we move counterclockwise around the square, we encounter the vertices A, B, C, D
in turn. List all the elements of D8 .
3. In S5 , how many 5-cycles are there; that is, how many permutations are there with
the same cycle structure as (1, 2, 3, 4, 5)?
4. In S5 , how many permutations are products of two disjoint transpositions, such as
(1, 2)(3, 4)?
5. Show that if n ā„ 3, then Sn is not abelian.
6. Show that the products of two disjoint transpositions in S4 , together with the identity,
form an abelian subgroup V of S4 . Describe the multiplication table of V (known as
the four group).
7. Show that the cycle structure of the inverse of a permutation Ļ coincides with that
of Ļ. In particular, the inverse of an even permutation is even (and the inverse of an
odd permutation is odd), so that An is actually a group.
8. Find the number of 3-cycles, i.e., permutations consisting of exactly one cycle of
length 3, in S4 .
9. Suppose H is a subgroup of A4 with the property that for every permutation Ļ in A4 ,
Ļ 2 belongs to H. Show that H contains all 3-cycles in A4 . (Since 3-cycles are even,
H in fact contains all 3-cycles in S4 .)
10. Consider the permutation

1 2 3 4 5
Ļ= .
2 4 5 1 3

Count the number of inversions of Ļ, that is, the number of pairs of integers that are
out of their natural order in the second row of Ļ. For example, 2 and 5 are in natural
order, but 4 and 3 are not. Compare your result with the parity of Ļ.
11. Show that the parity of any permutation Ļ is the same as the parity of the number
of inversions of Ļ.

1.3 Cosets, Normal Subgroups, and Homomorphisms
Let H be a subgroup of the group G. If g ā G, the right coset of H generated by g is

Hg = {hg : h ā H};

similarly, the left coset of H generated by g is

gH = {gh : h ā H}.

It follows (Problem 1) that if a, b ā G, then

if and only if abā’1 ā H
Ha = Hb
1.3. COSETS, NORMAL SUBGROUPS, AND HOMOMORPHISMS 9

and
if and only if aā’1 b ā H.
aH = bH
Thus if we deļ¬ne a and b to be equivalent iļ¬ abā’1 ā H, we have an equivalence relation
(Problem 2), and (Problem 3) the equivalence class of a is
{b : abā’1 ā H} = Ha.
Therefore the right cosets partition G (similarly for the left cosets). Since h ā’ ha,
h ā H, is a one-to-one correspondence, each coset has |H| elements. There are as many
right cosets as left cosets, since the map aH ā’ Haā’1 is a one-to-one correspondence
(Problem 4). If [G : H], the index of H in G, denotes the number of right (or left) cosets,
we have the following basic result.

1.3.2 Lagrangeā™s Theorem
If H is a subgroup of G, then |G| = |H|[G : H]. In particular, if G is ļ¬nite then |H|
divides |G|, and
|G|
= [G : H].
|H|
Proof. There are [G : H] cosets, each with |H| members. ā™£

1.3.3 Corollary
Let G be a ļ¬nite group.
(i) If a ā G then |a| divides |G|; in particular, a|G| = 1. Thus |G| is a multiple of the
order of each of its elements, so if we deļ¬ne the exponent of G to be the least common
multiple of {|a| : a ā G}, then |G| is a multiple of the exponent.
(ii) If G has prime order, then G is cyclic.
Proof. If the element a ā G has order n, then H = {1, a, a2 , . . . , anā’1 } is a cyclic subgroup
of G with |H| = n. By Lagrangeā™s theorem, n divides |G|, proving (i). If |G| is prime
then we may take a = 1, and consequently n = |G|. Thus H is a subgroup with as many
elements as G, so in fact H and G coincide, proving (ii). ā™£
Here is another corollary.

1.3.4 Eulerā™s Theorem
If a and n are relatively prime positive integers, with n ā„ 2, then aĻ•(n) ā” 1 mod n. A
special case is Fermatā™s Little Theorem: If p is a prime and a is a positive integer not
divisible by p, then apā’1 ā” 1 mod p.
Proof. The group of units mod n has order Ļ•(n), and the result follows from (1.3.3). ā™£
We will often use the notation H ā¤ G to indicate that H is a subgroup of G. If H is
a proper subgroup, i.e., H ā¤ G but H = G, we write H < G.
10 CHAPTER 1. GROUP FUNDAMENTALS

1.3.5 The Index is Multiplicative
If K ā¤ H ā¤ G, then [G : K] = [G : H][H : K].

Proof. Choose representatives ai from each left coset of H in G, and representatives bj
from each left coset of K in H. If cK is any left coset of K in G, then c ā ai H for some
unique i, and if c = ai h, h ā H, then h ā bj K for some unique j, so that c belongs to
ai bj K. The map (ai , bj ) ā’ ai bj K is therefore onto, and it is one-to-one by the uniqueness
of i and j. We therefore have a bijection between a set of size [G : H][H : K] and a set
of size [G : K], as asserted. ā™£

Now suppose that H and K are subgroups of G, and deļ¬ne HK to be the set of all
products hk, h ā H, k ā K. Note that HK need not be a group, since h1 k1 h2 k2 is not
necessarily equal to h1 h2 k1 k2 . If G is abelian, then HK will be a group, and we have the
following useful generalization of this observation.

1.3.6 Proposition
If H ā¤ G and K ā¤ G, then HK ā¤ G if and only if HK = KH. In this case, HK is the
subgroup generated by H āŖ K.

Proof. If HK is a subgroup, then (HK)ā’1 , the collection of all inverses of elements of HK,
must coincide with HK. But (HK)ā’1 = K ā’1 H ā’1 = KH. Conversely, if HK = KH,
then the inverse of an element in HK also belongs to HK, because (HK)ā’1 = K ā’1 H ā’1 =
KH = HK. The product of two elements in HK belongs to HK, because (HK)(HK) =
HKHK = HHKK = HK. The last statement follows from the observation that any
subgroup containing H and K must contain HK. ā™£

The set product HK deļ¬ned above suggests a multiplication operation on cosets. If
H is a subgroup of G, we can multiply aH and bH, and it is natural to hope that we get
abH. This does not always happen, but here is one possible criterion.

1.3.7 Lemma
If H ā¤ G, then (aH)(bH) = abH for all a, b ā G iļ¬ cHcā’1 = H for all c ā G. (Equiva-
lently, cH = Hc for all c ā G.

Proof. If the second condition is satisļ¬ed, then (aH)(bH) = a(Hb)H = abHH = abH.
Conversely, if the ļ¬rst condition holds, then cHcā’1 ā cHcā’1 H since 1 ā H, and
(cH)(cā’1 H) = ccā’1 H(= H) by hypothesis. Thus cHcā’1 ā H, which implies that
H ā cā’1 Hc. Since this holds for all c ā G, we have H ā cHcā’1 , and the result fol-
lows. ā™£

Notice that we have proved that if cHcā’1 ā H for all c ā G, then in fact cHcā’1 = H
for all c ā G.
1.3. COSETS, NORMAL SUBGROUPS, AND HOMOMORPHISMS 11

1.3.8 Deļ¬nition
Let H be a subgroup of G. If any of the following equivalent conditions holds, we say
that H is a normal subgroup of G, or that H is normal in G:

(1) cHcā’1 ā H for all c ā G (equivalently, cā’1 Hc ā H for all c ā G).
(2) cHcā’1 = H for all c ā G (equivalently, cā’1 Hc = H for all c ā G).
(3) cH = Hc for all c ā G.
(4) Every left coset of H in G is also a right coset.
(5) Every right coset of H in G is also a left coset.

We have established the equivalence of (1), (2) and (3) above, and (3) immediately
implies (4). To show that (4) implies (3), suppose that cH = Hd. Then since c belongs
to both cH and Hc, i.e., to both Hd and Hc, we must have Hd = Hc because right
cosets partition G, so that any two right cosets must be either disjoint or identical. The
equivalence of (5) is proved by a symmetrical argument.
Notation: H G indicates that H is a normal subgroup of G; if H is a proper
normal subgroup, we write H G.

1.3.9 Deļ¬nition of the Quotient Group
If H is normal in G, we may deļ¬ne a group multiplication on cosets, as follows. If aH
and bH are (left) cosets, let

(aH)(bH) = abH;

by (1.3.7), (aH)(bH) is simply the set product. If a1 is another member of aH and b1
another member of bH, then a1 H = aH and b1 H = bH (Problem 5). Therefore the set
product of a1 H and b1 H is also abH. The point is that the product of two cosets does
not depend on which representatives we select.
To verify that cosets form a group under the above multiplication, we consider the
four deļ¬ning requirements.

Closure: The product of two cosets is a coset.

Associativity : This follows because multiplication in G is associative.

Identity : The coset 1H = H serves as the identity.

Inverse: The inverse of aH is aā’1 H.

The group of cosets of a normal subgroup N of G is called the quotient group of G
by N ; it is denoted by G/N .
Since the identity in G/N is 1N = N , we have, intuitively, āset everything in N equal
to 1ā.
12 CHAPTER 1. GROUP FUNDAMENTALS

1.3.10 Example
Let GL(n, R be the set of all nonsingular n by n matrices with real coeļ¬cients, and let
SL(n, R) be the subgroup formed by matrices whose determinant is 1 (GL stands for
āgeneral linearā and SL for āspecial linearā). Then SL(n, R) GL(n, R), because if A is
a nonsingular n by n matrix and B is n by n with determinant 1, then det(ABAā’1 ) =
det A det B det Aā’1 = det B = 1.

1.3.11 Deļ¬nition
If f : G ā’ H, where G and H are groups, then f is said to be a homomorphism if for all
a, b in G, we have

f (ab) = f (a)f (b).

This idea will look familiar if G and H are abelian, in which case, using additive notation,
we write

f (a + b) = f (a) + f (b);

thus a linear transformation on a vector space is, in particular, a homomorphism on the
underlying abelian group. If f is a homomorphism from G to H, it must map the identity
of G to the identity of H, since f (a) = f (a1G ) = f (a)f (1G ); multiply by f (a)ā’1 to get
1H = f (1G ). Furthermore, the inverse of f (a) is f (aā’1 ), because

1 = f (aaā’1 ) = f (a)f (aā’1 ),

so that [f (a)]ā’1 = f (aā’1 ).

1.3.12 The Connection Between Homomorphisms and Normal
Subgroups
If f : G ā’ H is a homomorphism, deļ¬ne the kernel of f as

ker f = {a ā G : f (a) = 1};

then ker f is a normal subgroup of G. For if a ā G and b ā ker f , we must show that
abaā’1 belongs to ker f . But f (abaā’1 ) = f (a)f (b)f (aā’1 ) = f (a)(1)f (a)ā’1 = 1.
Conversely, every normal subgroup is the kernel of a homomorphism. To see this,
suppose that N G, and let H be the quotient group G/N . Deļ¬ne the map Ļ : G ā’ G/N
by Ļ(a) = aN ; Ļ is called the natural or canonical map. Since

Ļ(ab) = abN = (aN )(bN ) = Ļ(a)Ļ(b),

Ļ is a homomorphism. The kernel of Ļ is the set of all a ā G such that aN = N (= 1N ),
or equivalently, a ā N . Thus ker Ļ = N .
1.3. COSETS, NORMAL SUBGROUPS, AND HOMOMORPHISMS 13

1.3.13 Proposition
A homomorphism f is injective if and only if its kernel K is trivial, that is, consists only
of the identity.

Proof. If f is injective and a ā K, then f (a) = 1 = f (1), hence a = 1. Conversely, if K is
trivial and f (a) = f (b), then f (abā’1 ) = f (a)f (bā’1 ) = f (a)[f (b)]ā’1 = f (a)[f (a)]ā’1 = 1,
so abā’1 ā K. Thus abā’1 = 1, i.e., a = b, proving f injective. ā™£

1.3.14 Some Standard Terminology
A monomorphism is an injective homomorphism
An epimorphism is a surjective homomorphism
An isomorphism is a bijective homomorphism
An endomorphism is a homomorphism of a group to itself
An automorphism is an isomorphism of a group with itself
We close the section with a result that is applied frequently.

1.3.15 Proposition
Let f : G ā’ H be a homomorphism.
(i) If K is a subgroup of G, then f (K) is a subgroup of H. If f is an epimorphism
and K is normal, then f (K) is also normal.
(ii) If K is a subgroup of H, then f ā’1 (K) is a subgroup of G. If K is normal, so is
f ā’1 (K).

Proof. (i) If f (a) and f (b) belong to f (K), so does f (a)f (b)ā’1 , since this element coin-
cides with f (abā’1 ). If K is normal and c ā G, we have f (c)f (K)f (c)ā’1 = f (cKcā’1 ) =
f (K), so if f is surjective, then f (K) is normal.
(ii) If a and b belong to f ā’1 (K), so does abā’1 , because f (abā’1 ) = f (a)f (b)ā’1 , which
belongs to K. If c ā G and a ā f ā’1 (K) then f (cacā’1 ) = f (c)f (a)f (c)ā’1 , so if K is
normal, we have cacā’1 ā f ā’1 (K), proving f ā’1 (K) normal. ā™£

Problems For Section 1.3
In Problems 1ā“6, H is a subgroup of the group G, and a and b are elements of G.

1. Show that Ha = Hb iļ¬ abā’1 ā H.
2. Show that āa ā¼ b iļ¬ abā’1 ā Hā deļ¬nes an equivalence relation.
3. If we deļ¬ne a and b to be equivalent iļ¬ abā’1 ā H, show that the equivalence class
of a is Ha.
4. Show that aH ā’ Haā’1 is a one-to-one correspondence between left and right cosets
of H.
14 CHAPTER 1. GROUP FUNDAMENTALS

5. If aH is a left coset of H in G and a1 ā aH, show that the left coset of H generated
by a1 (i.e., a1 H), is also aH.
6. If [G : H] = 2, show that H is a normal subgroup of G.
7. Let S3 be the group of all permutations of {1, 2, 3}, and take a to be permutation
(1, 2, 3), b the permutation (1, 2), and e the identity permutation. Show that the
elements of S3 are, explicitly, e, a, a2 , b, ab and a2 b.
8. Let H be the subgroup of S3 consisting of the identity e and the permutation b = (1, 2).
Compute the left cosets and the right cosets of H in S3 .
9. Continuing Problem 8, show that H is not a normal subgroup of S3 .
10. Let f be an endomorphism of the integers Z. Show that f is completely determined
by its action on 1. If f (1) = r, then f is multiplication by r; in other words, f (n) = rn
for every integer n.
11. If f is an automorphism of Z, and I is the identity function on Z, show that f is
either I or ā’I.
12. Since the composition of two automorphisms is an automorphism, and the inverse of
an automorphism is an automorphism, it follows that the set of automorphisms of a
group is a group under composition. In view of Problem 11, give a simple description
of the group of automorphisms of Z.
13. Let H and K be subgroups of the group G. If x, y ā G, deļ¬ne x ā¼ y iļ¬ x can be
written as hyk for some h ā H and k ā K. Show that ā¼ is an equivalence relation.
14. The equivalence class of x ā G is HxK = {hxk : h ā H, k ā K}, called a double coset
associated with the subgroups H and K. Thus the double cosets partition G. Show
that any double coset can be written as a union of right cosets of H, or equally well
as a union of left cosets of K.

1.4 The Isomorphism Theorems
Suppose that N is a normal subgroup of G, f is a homomorphism from G to H, and Ļ is
the natural map from G to G/N , as pictured in Figure 1.4.1.

GH
f
G
za
z
z
Ļ
z f
G/N

Figure 1.4.1

We would like to ļ¬nd a homomorphism f : G/N ā’ H that makes the diagram com-
mutative. Commutativity means that we get the same result by traveling directly from G
to H via f as we do by taking the roundabout route via Ļ followed by f . This requirement
translates to f (aN ) = f (a). Here is the key result for ļ¬nding such an f .
1.4. THE ISOMORPHISM THEOREMS 15

1.4.1 Factor Theorem
Any homomorphism f whose kernel K contains N can be factored through G/N . In other
words, in Figure 1.4.1 there is a unique homomorphism f : G/N ā’ H such that f ā—¦ Ļ = f .
Furthermore:

(i) f is an epimorphism if and only if f is an epimorphism;
(ii) f is a monomorphism if and only if K = N ;
(iii) f is an isomorphism if and only if f is an epimorphism and K = N .

Proof. If the diagram is to commute, then f (aN ) must be f (a), and it follows that f , if
it exists, is unique. The deļ¬nition of f that we have just given makes sense, because if
aN = bN , then aā’1 b ā N ā K, so f (aā’1 b) = 1, and therefore f (a) = f (b). Since

f (aN bN ) = f (abN ) = f (ab) = f (a)f (b) = f (aN )f (bN ),

f is a homomorphism. By construction, f has the same image as f , proving (i). Now the
kernel of f is

{aN : f (a) = 1} = {aN : a ā K} = K/N.

By (1.3.13), a homomorphism is injective, i.e., a monomorphism, if and only if its kernel
is trivial. Thus f is a monomorphism if and only if K/N consists only of the identity
element N . This means that if a is any element of K, then the coset aN coincides with
N , which forces a to belong to N . Thus f is a monomorphism if and only if K = N ,
proving (ii). Finally, (iii) follows immediately from (i) and (ii). ā™£

The factor theorem yields a fundamental result.

1.4.2 First Isomorphism Theorem
If f : G ā’ H is a homomorphism with kernel K, then the image of f is isomorphic
to G/K.

Proof. Apply the factor theorem with N = K, and note that f must be an epimorphism
of G onto its image. ā™£

If we are studying a subgroup K of a group G, or perhaps the quotient group G/K,
we might try to construct a homomorphism f whose kernel is K and whose image H has
desirable properties. The ļ¬rst isomorphism theorem then gives G/K ā¼ H (where ā¼ is
= =
our symbol for isomorphism). If we know something about H, we may get some insight
into K and G/K.
We will prove several other isomorphism theorems after the following preliminary
result.
16 CHAPTER 1. GROUP FUNDAMENTALS

1.4.3 Lemma
Let H and N be subgroups of G, with N normal in G. Then:

(i) HN = N H, and therefore by (1.3.6), HN is a subgroup of G.
(ii) N is a normal subgroup of HN .
(iii) H ā© N is a normal subgroup of H.

Proof. (i) We have hN = N h for every h ā G, in particular for every h ā H.
(ii) Since N is normal in G, it must be normal in the subgroup HN .
(iii) H ā© N is the kernel of the canonical map Ļ : G ā’ G/N , restricted to H. ā™£

The subgroups we are discussing are related by a āparallelogramā or ādiamondā, as
Figure 1.4.2 suggests.

HN q
qq
ww qq
ww qq
ww qq
ww
w
H qq N
w
qq ww
qq ww
qq ww
q ww

Figure 1.4.2

1.4.4 Second Isomorphism Theorem
If H and N are subgroups of G, with N normal in G, then

H/(H ā© N ) ā¼ HN/N.
=

Note that we write HN/N rather than H/N , since N need not be a subgroup of H.

Proof. Let Ļ be the canonical epimorphism from G to G/N , and let Ļ0 be the restriction
of Ļ to H. Then the kernel of Ļ0 is H ā©N , so by the ļ¬rst isomorphism theorem, H/(H ā©N )
is isomorphic to the image of Ļ0 , which is {hN : h ā H} = HN/N . (To justify the last
equality, note that for any n ā N we have hnN = hN ). ā™£

1.4.5 Third Isomorphism Theorem
If N and H are normal subgroups of G, with N contained in H, then

G/H ā¼ (G/N )/(H/N ),
=

a ācancellation lawā.
1.4. THE ISOMORPHISM THEOREMS 17

Proof. This will follow directly from the ļ¬rst isomorphism theorem if we can ļ¬nd an
epimorphism of G/N onto G/H with kernel H/N , and there is a natural candidate:
f (aN ) = aH. To check that f is well-deļ¬ned, note that if aN = bN then aā’1 b ā N ā H,
so aH = bH. Since a is an arbitrary element of G, f is surjective, and by deļ¬nition of
coset multiplication, f is a homomorphism. But the kernel of f is

{aN : aH = H} = {aN : a ā H} = H/N. ā™£

Now suppose that N is a normal subgroup of G. If H is a subgroup of G containing
N , there is a natural analog of H in the quotient group G/N , namely, the subgroup H/N .
In fact we can make this correspondence very precise. Let

Ļ(H) = H/N

be a map from the set of subgroups of G containing N to the set of subgroups of G/N .
We claim that Ļ is a bijection. For if H1 /N = H2 /N then for any h1 ā H1 , we have
h1 N = h2 N for some h2 ā H2 , so that hā’1 h1 ā N , which is contained in H2 . Thus
2
H1 ā H2 , and by symmetry the reverse inclusion holds, so that H1 = H2 and Ļ is
injective. Now if Q is a subgroup of G/N and Ļ : G ā’ G/N is canonical, then

Ļ ā’1 (Q) = {a ā G : aN ā Q},

a subgroup of G containing N , and

Ļ(Ļ ā’1 (Q)) = {aN : aN ā Q} = Q,

proving Ļ surjective.
The map Ļ has a number of other interesting properties, summarized in the following
result, sometimes referred to as the fourth isomorphism theorem.

1.4.6 Correspondence Theorem
If N is a normal subgroup of G, then the map Ļ : H ā’ H/N sets up a one-to-one
correspondence between subgroups of G containing N and subgroups of G/N . The inverse
of Ļ is the map Ļ„ : Q ā’ Ļ ā’1 (Q), where Ļ is the canonical epimorphism of G onto G/N .
Furthermore:

(i) H1 ā¤ H2 if and only if H1 /N ā¤ H2 /N , and, in this case,

[H2 : H1 ] = [H2 /N : H1 /N ].

(ii) H is a normal subgroup of G if and only if H/N is a normal subgroup of G/N .

More generally,

(iii) H1 is a normal subgroup of H2 if and only if H1 /N is a normal subgroup of H2 /N ,
and in this case, H2 /H1 ā¼ (H2 /N )/H1 /N ).
=
18 CHAPTER 1. GROUP FUNDAMENTALS

Proof. We have established that Ļ is a bijection with inverse Ļ„ . If H1 ā¤ H2 , we have
H1 /N ā¤ H2 /N immediately, and the converse follows from the above proof that Ļ is
injective. To prove the last statement of (i), let Ī· map the left coset aH1 , a ā H2 , to the
left coset (aN )(H1 /N ). Then Ī· is a well-deļ¬ned injective map because

aā’1 b ā H1
aH1 = bH1 iļ¬
iļ¬ (aN )ā’1 (bN ) = aā’1 bN ā H1 /N
iļ¬ (aN )(H1 /N ) = (bN )(H1 /N );

Ī· is surjective because a ranges over all of H2 .
To prove (ii), assume that H G; then for any a ā G we have

(aN )(H/N )(aN )ā’1 = (aHaā’1 )/N = H/N

so that H/N G/N . Conversely, suppose that H/N is normal in G/N . Consider the
homomorphism a ā’ (aN )(H/N ), the composition of the canonical map of G onto G/N
and the canonical map of G/N onto (G/N )/(H/N ). The element a will belong to the
kernel of this map if and only if (aN )(H/N ) = H/N , which happens if and only if
aN ā H/N , that is, aN = hN for some h ā H. But since N is contained in H, this
statement is equivalent to a ā H. Thus H is the kernel of a homomorphism, and is
therefore a normal subgroup of G.
Finally, the proof of (ii) also establishes the ļ¬rst part of (iii); just replace H by H1
and G by H2 . The second part of (iii) follows from the third isomorphism theorem (with
the same replacement). ā™£
We conclude the section with a useful technical result.

1.4.7 Proposition
If H is a subgroup of G and N is a normal subgroup of G, we know by (1.4.3) that HN ,
the subgroup generated by H āŖ N , is a subgroup of G. If H is also a normal subgroup
of G, then HN is normal in G as well. More generally, if for each i in the index set I, we
have Hi G, then Hi , i ā I , the subgroup generated by the Hi (technically, by the set
āŖiāI Hi ) is a normal subgroup of G.
Proof. A typical element in the subgroup generated by the Hi is a = a1 a2 Ā· Ā· Ā· an where ak
belongs to Hik . If g ā G then

g(a1 a2 Ā· Ā· Ā· an )g ā’1 = (ga1 g ā’1 )(ga2 g ā’1 ) Ā· Ā· Ā· (gan g ā’1 )

and gak g ā’1 ā Hik because Hik G. Thus gag ā’1 belongs to Hi , i ā I . ā™£

Problems For Section 1.4
1. Let Z be the integers, and nZ the set of integer multiples of n. Show that Z/nZ
is isomorphic to Zn , the additive group of integers modulo n. (This is not quite a
tautology if we view Zn concretely as the set {0, 1, . . . , nā’1}, with sums and diļ¬erences
reduced modulo n.)
1.5. DIRECT PRODUCTS 19

2. If m divides n then Zm ā¤ Zn ; for example, we can identify Z4 with the subgroup
{0, 3, 6, 9} of Z12 . Show that Zn /Zm ā¼ Zn/m .
=
3. Let a be an element of the group G, and let fa : G ā’ G be āconjugation by aā, that
is, fa (x) = axaā’1 , x ā G. Show that fa is an automorphism of G.
4. An inner automorphism of G is an automorphism of the form fa (deļ¬ned in Prob-
lem 3) for some a ā G. Show that the inner automorphisms of G form a group under
composition of functions (a subgroup of the group of all automorphisms of G).
5. Let Z(G) be the center of G, that is, the set of all x in G such that xy = yx for all y
in G. Thus Z(G) is the set of elements that commute with everything in G. Show that
Z(G) is a normal subgroup of G, and that the group of inner automorphisms of G is
isomorphic to G/Z(G).
6. If f is an automorphism of Zn , show that f is multiplication by m for some m relatively
prime to n. Conclude that the group of automorphisms of Zn can be identiļ¬ed with
the group of units mod n.
7. The diamond diagram associated with the second isomorphism theorem (1.4.4) illus-
trates least upper bounds and greatest lower bounds in a lattice. Verify that HN is the
smallest subgroup of G containing both H and N , and H ā© N is the largest subgroup
of G contained in both H and N .
8. Let g be an automorphism of the group G, and fa an inner automorphism, as deļ¬ned
in Problems 3 and 4. Show that g ā—¦ fa ā—¦ g ā’1 is an inner automorphism. Thus the group
of inner automorphisms of G is a normal subgroup of the group of all automorphisms.
9. Identify a large class of groups for which the only inner automorphism is the identity
mapping.

1.5 Direct Products
1.5.1 External and Internal Direct Products
In this section we examine a popular construction. Starting with a given collection of
groups, we build a new group with the aid of the cartesian product. Letā™s start with
two given groups H and K, and let G = H Ć— K, the set of all ordered pairs (h, k),
h ā H, k ā K. We deļ¬ne multiplication on G componentwise:

(h1 , k1 )(h2 , k2 ) = (h1 h2 , k1 k2 ).

Since (h1 h2 , k1 k2 ) belongs to G, it follows that G is closed under multiplication. The
multiplication operation is associative because the individual products on H and K are
associative. The identity element in G is (1H , 1K ), and the inverse of (h, k) is (hā’1 , k ā’1 ).
Thus G is a group, called the external direct product of H and K.
We may regard H and K as subgroups of G. More precisely, G contains isomorphic
copies of H and K, namely

H = {(h, 1K ) : h ā H} and K = {(1H , k) : k ā K}.
20 CHAPTER 1. GROUP FUNDAMENTALS

Furthermore, H and K are normal subgroups of G. (Note that (h, k)(h1 , 1K )(hā’1 , k ā’1 ) =
(hh1 hā’1 , 1K ), with hh1 hā’1 ā H.) Also, from the deļ¬nitions of H and K, we have
G = H K and H ā© K = {1}, where 1 = (1H , 1K ).
If a group G contains normal subgroups H and K such that G = HK and H ā©K = {1},
we say that G is the internal direct product of H and K.
Notice the key diļ¬erence between external and internal direct products. We construct
the external direct product from the component groups H and K. On the other hand,
starting with a given group we discover subgroups H and K such that G is the inter-
nal direct product of H and K. Having said this, we must admit that in practice the
distinction tends to be blurred, because of the following result.

1.5.2 Proposition
If G is the internal direct product of H and K, then G is isomorphic to the external direct
product H Ć— K.
Proof. Deļ¬ne f : H Ć— K ā’ G by f (h, k) = hk; we will show that f is an isomorphism.
First note that if h ā H and k ā K then hk = kh. (Consider hkhā’1 k ā’1 , which belongs
to K since hkhā’1 ā K, and also belongs to H since khā’1 k ā’1 ā H; thus hkhā’1 k ā’1 = 1,
so hk = kh.)
(a) f is a homomorphism, since
f ((h1 , k1 )(h2 , k2 )) = f (h1 h2 , k1 k2 ) = h1 h2 k1 k2 = (h1 k1 )(h2 k2 ) = f (h1 , k1 )f (h2 , k2 ).
(b) f is surjective, since by deļ¬nition of internal direct product, G = HK.
(c) f is injective, for if f (h, k) = 1 then hk = 1, so that h = k ā’1 .Thus h belongs
to both H and K, so by deļ¬nition of internal direct product, h is the identity, and
consequently so is k. The kernel of f is therefore trivial. ā™£
External and internal direct products may be deļ¬ned for any number of factors. We
will restrict ourselves to a ļ¬nite number of component groups, but the generalization to
arbitrary cartesian products with componentwise multiplication is straightforward.

If H1 , H2 , . . . Hn are arbitrary groups, the external direct product of the Hi is the cartesian
product G = H1 Ć— H2 Ć— Ā· Ā· Ā· Ć— Hn , with componentwise multiplication:
(h1 , h2 , . . . , hn )(h1 , h2 , . . . hn ) = (h1 h1 , h2 h2 , . . . hn hn );
G contains an isomorphic copy of each Hi , namely
H i = {(1H1 , . . . , 1Hiā’1 , hi , 1Hi+1 , . . . , 1Hn ) : hi ā Hi }.

As in the case of two factors, G = H 1 H 2 Ā· Ā· Ā· H n , and H i G for all i; furthermore, if
g ā G then g has a unique representation of the form
g = h1 h2 Ā· Ā· Ā· hn where hi ā H i .
1.5. DIRECT PRODUCTS 21

Speciļ¬cally, g = (h1 , . . . , hn ) = (h1 , 1, . . . , 1) . . . (1, . . . , 1, hn ). The representation is
unique because the only way to produce the i-th component hi of g is for hi to be the ith
component of the factor from H i . If a group G contains normal subgroups H1 , . . . , Hn
such that G = H1 Ā· Ā· Ā· Hn , and each g ā G can be uniquely represented as h1 Ā· Ā· Ā· hn with
hi ā Hi , i = 1, 2, . . . , n, we say that G is the internal direct product of the Hi . As in the
case of two factors, if G is the internal direct product of the Hi , then G is isomorphic
to the external direct product H1 Ć— Ā· Ā· Ā· Ć— Hn ; the isomorphism f : H1 Ć— Ā· Ā· Ā· Ć— Hn ā’ G
is given by f (h1 , . . . , hn ) = h1 Ā· Ā· Ā· hn . The next result frequently allows us to recognize
when a group is an internal direct product.

1.5.4 Proposition
Suppose that G = H1 Ā· Ā· Ā· Hn , where each Hi is a normal subgroup of G. The following
conditions are equivalent:

(1) G is the internal direct product of the Hi .
j=i Hj = {1} for i = 1, . . . , n; thus it does not matter in which order the Hi
(2) Hi
are listed.
iā’1
Hj = {1} for i = 1, . . . , n.
(3) Hi j=1

Proof. (1) implies (2): If g belongs to the product of the Hj , j = i, then g can be written
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