Algebra 1 Book.pdf

Instructor's Preface Vll
Student's Preface xi
Dependence Chart Xlll
o Sets and Relations 1
GROUPS AND SUBGROUPS
1 Introduction and Examples 1 1
2 Binary Operations 20
3 Isomorphic Binary Structures 28
4 Groups 36
5 Subgroups 49
' 6 Cyclic Groups 59
7 Generating Sets and Cayley Digraphs 68
PERMUTATIONS, CoSETS, AND DIRECT PRODUCTS
8 Groups of Permutations 75
9 Orbits, Cycles, and the Alternating Groups 87
10 Cosets and the Theorem of Lagrange 96
11 Direct Products and Finitely Generated Abelian Groups 104
t12 Plane Isometries 1 14
7
11
75
iii
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iv Contents
HOMOMORPHISMS AND FACTOR GROUPS
13 Homomorphisms 125
14 Factor Groups 135
15 Factor-Group Computations and Simple Groups 144
+16 Group Action on a Set 154
t17 Applications of G-Sets to Counting 161
RINGS AND FIELDS
18 Rings and Fields 167
19 Integral Domains 177
20 Fermat's and Euler's Theorems 184
21 The Field of Quotients of an Integral Domain 190
22 Rings of Polynomials 198
23 Factorization of Polynomials over a Field 209
t24 Noncommutative Examples 220
t25 Ordered Rings and Fields 227
IDEALS AND FACTOR RINGS
26 Homomorphisms and Factor Rings 237
27 Prime and Maximal Ideals 245
t28 Grabner Bases for Ideals 254
EXTENSION FIELDS
29 Introduction to Extension Fields 265
30 Vector Spaces 274
31 Algebraic Extensions 283
t32 Geometric Constructions 293
33 Finite Fields 300
ADVANCED GROUP THEORY
34 Isomorphism Theorems 307
35 Series of Groups 3 1 1
36 Sylow Theorems 321
37 Applications of the Sylow Theory 327
125
167
237
265
307

Contents v
38 Free Abelian Groups 333
39 Free Groups 341
40 Group Presentations 346
GROUPS IN TOPOLOGY
41 Simplicial Complexes and Homology Groups 355
42 Computations of Homology Groups 363
43 More Homology Computations and Applications 371
44 Homological Algebra 379
FACTORIZATION
45 Unique Factorization Domains 389
46 Euclidean Domains 401
47 Gaussian Integers and Multiplicative Norms 407
AUTOMORPHISMS AND GALOIS THEORY
48 Automorphisms of Fields 415
49 The Isomorphism Extension Theorem 424
50 Splitting Fields 431
51 Separable Extensions 436
t52 Totally Inseparable Extensions 444
53 Galois Theory 448
54 Illustrations of Galois Theory 457
55 Cyclotomic Extensions 464
56 Insolvability of the Quintic 470
Appendix: Matrix Algebra 477
Bibliography 483
Notations 487
Answers to Odd-Numbered Exercises Not Asking for Definitions or Proofs 491
Index 513
j Not required for the remainder of the text.
I This section is a prerequisite for Sections 17 and 36 only.
355
389
415

This is an introduction to abstract algebra. It is anticipated that the students have studied
calculus and probably linear algebra. However, these are primarily mathematical ma
turity prerequisites; subject matter from calculus and linear algebra appears mostly in
illustrative examples and exercises.
As in previous editions of the text, my aim remains to teach students as much about
groups, rings, and fields as I can in a first course. For many students, abstract algebra is
theirfirstextendedexposure to an axiomatic treatment ofmathematics. Recognizing this,
I have included extensive explanations concerning what we are trying to accomplish,
how we are trying to do it, and why we choose these methods. Mastery of this text
constitutes a finn foundation for more specialized work in algebra, and also provides
valuable experience for any further axiomatic study of mathematics.
Changes from the Sixth Edition
The amount of preliminary material had increased from one lesson in the first edition
to four lessons in the sixth edition. My personal preference is to spend less time before
getting to algebra; therefore, I spend little time on preliminaries. Much of it is review for
many students, and spending four lessons on it may result in their not allowing sufficient
time in their schedules to handle the course when new material arises. Accordingly, in
this edition, I have reverted to just one preliminary lesson on sets and relations, leaving
other topics to be reviewed when needed. A summary of matrices now appears in the
Appendix.
The first two editions consisted of short, consecutively numbered sections, many of
which could be covered in a single lesson. I have reverted to that design to avoid the
cumbersome and intimidating triple numbering of definitions, theorems examples, etc.
In response to suggestions by reviewers, the order of presentation has been changed so
vii

viii Instructor's Preface
that the basic material on groups, rings, and fields that would normally be covered in a
one-semester course appears first, before the more-advanced group theory. Section 1 is
a new introduction, attempting to provide some feeling for the nature of the study.
In response to several requests, I have included the material on homology groups
in topology that appeared in the first two editions. Computation of homology groups
strengthens students' understanding of factor groups. The material is easily accessible;
after Sections 0 through 15, one need only read about free abelian groups, in Section 38
through Theorem 38.5, as preparation. To make room for the homology groups, I have
omitted the discussion of automata, binary linear codes, and additional algebraic struc
tures that appeared in the sixth edition.
I have also included a few exercises asking students to give a one- or two-sentence
synopsis of a proof in the text. Before the first such exercise, I give an example to show
what I expect.
Some Features Retained
I continue to break down most exercise sets into parts consisting of computations, con
cepts, and theory. Answers to odd-numbered exercises not requesting a proof again
appear at the back of the text. However, in response to suggestions, I am supplying the
answers to parts a), c), e), g), and i) only of my lO-part true-false exercises.
The excellent historical notes by Victor Katz are, ofcourse, retained. Also, a manual
containing complete solutions for all the exercises, including solutions asking forproofs,
is available for the instructor from the publisher.
A dependence chart with section numbers appears in the front matter as an aid in
making a syllabus.
Acknowledgments
I am very grateful to those who have reviewed the text or who have sent me suggestions
and corrections. I am especially indebted to George M. Bergman, who used the sixth
edition and made note of typographical and other errors, which he sent to me along
with a great many other valuable suggestions for improvement. I really appreciate this
voluntary review, which must have involved a large expenditure of time on his part.
I also wish to express my appreciation to William Hoffman, Julie LaChance, and
Cindy Cody of Addison-Wesley for their help with this project. Finally, I was most
fortunate to have John Probst and the staff at TechBooks handling the production of the
text from my manuscript. They produced the most error-free pages I have experienced,
and courteously helped me with a technical problem I hadwhile preparing the solutions
manual.
Suggestions for New Instructors of Algebra
Those who have taught algebra several times have discovered the difficulties and devel
oped their own solutions. The comments I make here are not for them.
This course is an abrupt change from the typical undergraduate calculus for the
students. A graduate-style lecture presentation, writing out definitions and proofs on the
board for most of the class time, will not work with most students. I have found it best

Instructor's Preface ix
to spend at least the first half of each class period answering questions on homework,
trying to get a volunteer to give a proofrequested in an exercise, and generally checking
to see if they seem to understand the material assigned for that class. Typically, I spent
only about the last 20 minutes ofmy 50-minute time talking about new ideas for the next
class, and giving at least one proof. From a practical point of view, it is a waste of time
to try to write on the board all the definitions and proofs. They are in the text.
I suggest that at least half of the assigned exercises consist of the computational
ones. Students are used to doing computations in calculus. Although there are many
exercises asking for proofs that we would love to assign, I recommend that you assign
at most two or three such exercises, and try to get someone to explain how each proof is
performed in the next class. I do think students should be asked to do at least one proof
in each assignment.
Students face a barrage of definitions and theorems, something they have never
encountered before. They are not used to mastering this type ofmaterial. Grades on tests
that seem reasonable to us, requesting a few definitions and proofs, are apt to be low and
depressing for most students. My recommendation for handling this problem appears in
my article, Happy Abstract Algebra Classes, in the November 2001 issue of the MAA
FOCUS.
At URI, we have only a single semester undergraduate course in abstract algebra.
Our semesters are quite short, consisting of about 42 50-minute classes. When I taught
the course, I gave three 50-minute tests in class, leaving about 38 classes for which the
student was given an assignment. I always covered the material in Sections 0-1 1, 13-15,
18-23, 26, 27, and 29-32, which is a total of 27 sections. Of course, I spent more than
one class on several of the sections, but I usually had time to cover about two more;
sometimes I included Sections 16 and 17. (There is no point in doing Section 16 unless
you do Section 17, or will be doing Section 36 later.) I often covered Section 25, and
sometimes Section 12 (see the Dependence Chart). The job is to keep students from
becoming discouraged in the first few weeks of the course.

This course may well require a different approach than those you used in previous math
ematics courses. You may have become accustomed to working a homework problem by
turning back in the text to find a similar problem, and then just changing some numbers.
That may work with a few problems in this text, but it will not work for most of them.
This is a subject in which understanding becomes all important, and where problems
should not be tackled without first studying the text.
Let me make some suggestions on studying the text. Notice that the text bristles
with definitions, theorems, corollaries, and examples. The definitions are crucial. We
must agree on terminology to make any progress. Sometimes a definition is followed
by an example that illustrates the concept. Examples are probably the most important
aids in studying the text. Pay attention to the examples. I suggest you skip the proofs
of the theorems on your first reading of a section, unless you are really "gung-ho" on
proofs. You should read the statement of the theorem and try to understand just what it
means. Often, a theorem is followed by an example that illustrates it, a great aid in really
understanding what the theorem says.
In summary, on your first reading of a section, I suggest you concentrate on what
information the section gives, and on gaining a real understanding of it. If you do not
understand what the statement of a theorem means, it will probably be meaningless for
you to read the proof.
Proofs are very basic to mathematics. After you feel you understand the information
given in a section, you should read and try to understand at least some of the proofs.
Proofs of corollaries are usually the easiest ones, forthey often follow very directly from
the theorem. Quite a lot of the exercises under the "Theory" heading ask for a proof. Try
not to be discouraged at the outset. It takes a bit of practice and experience. Proofs in
algebra can be more difficult than proofs in geometry and calculus, for there are usually
no suggestive pictures that you can draw. Often, a proof falls out easily if you happen to
xi

xii Student's Preface
look at just the right expression. Of course, it is hopeless to devise a proof if you do not
really understand what it is that you are trying to prove. For example, if an exercise asks
you to show that given thing is a member of a certain set, you must know the defining
criterion to be a member of that set, and then show that your given thing satisfies that
criterion.
There are several aids for your study at the back of the text. Of course, you will
discoverthe answers to odd-numbered problems not requesting a proof. If you run into a
notation such as Zn that you do not understand, look in the list of notations that appears
after the bibliography. If you run into terminology like inner automorphism that you do
not understand, look in the Index for the first page where the term occurs.
In summary, although an understanding of the subject is important in every mathe
matics course, it is really crucial to your performance in this course. May you find it a
rewarding experience.
Narragansett, RI I.B.P.

Dependence Chart
0-11
1:2
13-15
16 34
� I
I
17 36 35
I
37
38 39
I I
41-44 40
18-23
2�j
26
he
27
l;-h-42
29-33
I
48-51
j;,
53-54
55 56

SETS AND RELATIONS
On Definitions, and the Notion of a Set
Many students do not realize the great importance of definitions to mathematics. This
importance stems from the need for mathematicians to communicate with each other.
If two people are trying to communicate about some subject, they must have the same
understanding ofits technical terms. However, there is an important structural weakness.
It is impossible to define every concept.
Suppose, for example, we define the term set as "A set is a well-defined collection of
objects." One naturally asks what is meant by a collection. We could define it as "A
collection is an aggregate of things." What, then, is an aggregate? Now our language
is finite, so after some time we will run out of new words to use and have to repeat
some words already examined. The definition is then circular and obviously worthless.
Mathematicians realize that there must be some undefined or primitive concept with
which to start. At the moment, they have agreed that set shall be such a primitive concept.
We shall not define set, but shall just hope that when such expressions as "the set of all
real numbers" or "the set of all members ofthe United States Senate" are used, people's
various ideas of what is meant are sufficiently similar to make communication feasible.
We summarize briefly some of the things we shall simply assume about sets.
1. A set S is made up of elements, and if a is one of these elements, we shall
denote this fact by a E S.
2. There is exactly one set with no elements. It is the empty set and is denoted
by 0.
3. We may describe a set either by giving a characterizing property of the
elements, such as "the set of all members of the United States Senate," or by
listing the elements. The standard way to describe a setby listing elements is
to enclose the designations of the elements, separated by commas, in braces,
for example, { I, 2, IS}. If a set is described by a characterizing property P(x)
of its elements x, the brace notation {x I P(x)} is also often used, and is read
"the set of all x such that the statement P(x) about x is true." Thus
{2, 4, 6, 8} = {x I x is an even whole positive number :s 8}
= {2x I x = 1 , 2, 3, 4}.
The notation {x I P(x)} is often called "set-builder notation."
4. A set is well defined, meaning that if S is a set and a is some object, then
either a is definitely in S, denoted by a E S, or a is definitely not in S, denoted
by a f/:. S. Thus, we should never say, "Consider the set S of some positive
numbers," for it is not definite whether 2 E S or 2 f/:. S. On the other hand, we
1

2 Section 0 Sets and Relations
can consider the set T of all prime positive integers. Every positive integer is
definitely either prime or not prime. Thus 5 E T and 14 rf:. T. It may be hard to
actually determine whether an object is in a set. For example, as this book
goes to press it is probably unknown whether 2(265) + 1 is in T. However,
2(265) + 1 is certainly either prime or not prime.
It is not feasible for this text to push the definition of everything we use all the way
back to the concept of a set. For example, we will never define the number Tr in terms of
a set.
Every definition is an if and only if type of statement.
With this understanding, definitions are often stated with the only if suppressed, but it
is always to be understood as part of the definition. Thus we may define an isosceles
triangle as follows: "A triangle is isosceles if it has two sides of equal length," when we
really mean that a triangle is isosceles if and only if it has two sides of equal length.
In our text, we have to define many terms. We use specifically labeled and numbered
definitions for the main algebraic concepts with which we are concerned. To avoid an
overwhelming quantity of such labels and numberings, we define many terms within the
body of the text and exercises using boldface type.
Boldface Convention
A term printed in boldface in a sentence is being defined by that sentence.
Do not feel that you have to memorize a definition word for word. The important
thing is to understand the concept, so that you can define precisely the same concept
in your own words. Thus the definition "An isosceles triangle is one having two equal
sides" is perfectly correct. Of course, we had to delay stating our boldface convention
until we had finished using boldface in the preceding discussion of sets, because we do
not define a set!
In this section, we do define some familiar concepts as sets, both for illustration and
for review of the concepts. First we give a few definitions and some notation.
0.1 Definition A set B is a subset of a set A, denoted by B <;; A or A :2 B, if every element of B is in
A. The notations B C A or A ::) B will be used for B <;; A but B i= A. •
. Note that according to this definition, for any set A, A itself and 0 are both subsets of A.
0.2 Definition If A is any set, then A is the improper subset of A. Any other subset of A is a proper
subset of A. •

Sets and Relations 3
0.3 Example Let S = {I, 2, 3}. This set S has a total of eight subsets, namely 0, {I}, {2}, {3},
{I, 2}, {I, 3}, {2, 3}, and {I, 2, 3}. ..
0.4 Definition Let A and B be sets. The set A x B = {(a, b) I a E A and b E B} is the Cartesian
product of A and B. •
0.5 Example If A = {I, 2, 3} and B = {3, 4}, then we have
A x B = {(1 , 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}.
Throughout this text, much work will be done involving familiar sets of numbers.
Let us take care of notation for these sets once and for all.
Z is the set of all integers (that is, whole numbers: positive, negative, and zero).
Ql is the set ofallrationalnumbers (that is, numbers thatcanbe expressed as quotients
min of integers, where n -::J. 0).
JR is the set of all real numbers.
22:+, Ql+, and JR+ are the sets of positive members of 22:, Ql, and Rrespectively.
C is the set of all complex numbers.
22:*, Ql*, JR*, and C* are the sets ofnonzero members of 22:, Ql, Rand C, respectively.
0.6 Example The set JR x JR is the familiar Euclidean plane that we use in first-semester calculus to
draw graphs of functions. ..
Relations Between Sets
We introduce the notion of an element a of set A being related to an element b of set B,
which we might denote by a .jIB b. The notation a .jIB b exhibits the elements a and b in
left-to-right order, just as the notation (a, b) for an element in A x B. This leads us to
the following definition of a relation .jIB as a set.
0.7 Definition A relation between sets A and B is a subset � of A x B. We read (a, b) E .� as "a is
related to boo and write a ,j'B b. •
0.8 Example (Equality Relation) There is one familiar relation between a set and itself that we
consider every set S mentioned in this text to possess: namely, the equality relation =
defined on a set S by
= is the subset {(x, x) I X E S} of S x S.
Thus for any X E S, we have x = x, but if x and yare different elements of S, then
(x, y
) (j. = and we write x -::J. y. ..
We willrefer to any relation between a set S and itself, as in the preceding example,
as a relation on S.
0.9 Example Thegraph ofthe function f where f(x) = x3 forall x E JR, is the slbset {(x, x3) I x E JR}
of JR x R Thus it is a relation on R The function is completely determined by its graph.
..

The preceding example suggests that rather than define a "function" y = f(x) to
be a "rule" that assigns to each x E JR exactly one y E JR, we can easily describe it as a
certain type of subset of JR x JR, that is, as a type of relation. We free ourselves from JR
and deal with any sets X and Y.
0.10 Definition A function cp mapping X into Y is a relation between X and Y with the property that
each x E X appears as the first member of exactly one ordered pair (x, y) in cpo Such a
function is also called a map or mapping ofX into Y. We write cp : X --+ Y and express
(x, y) E cp by cp(x) = y. The domain of cp is the set X and the set Y is the codomain of
cpo The range of cp is cp[X] = {cp(x) I x E X}. •
0.11 Example We can view the addition of real numbers as a function + : (JR x JR) --+ lFt, that is, as a
mapping of JR x JR into R For example, the action of + on (2, 3) E JR x JR is given in
function notation by +«2, 3)) = 5. In set notation we write «2, 3), 5) E +. Of course
our familiar notation is 2 + 3 = 5. £.
Cardinality
The number of elements in a set X is the cardinality of X and is often denoted by IXI.
For example, we have I{2, 5, 7} I = 3.1t willbe important for us to know whether two sets
havethe same cardinality. Ifboth sets are finite there is no problem; we can simply count
the elements in each set. But do Z, Q, and JR have the same cardinality? To convince
ourselves that two sets X and Y have the same cardinality, we try to exhibit a pairing of
each x in X with only one y in Y in such a way that each element of Y is alsoused only
once in this pairing. For the sets X = {2, 5, 7} and Y = {?, !, #}, the pairing
2++?, 5++#, 7++!
shows they have the same cardinality. Notice that we could also exhibit this pairing as
{(2, ?), (5, #), (7, !)} which, as a subset of X x Y, is a relation between X and Y. The
pairing
1
t
o
2
t
-1
3
t
1
4
t
-2
5
t
2
6
t
-3
7
t
3
8
t
-4
9
t
4
10
t
-5
shows that the sets Z and Z+ have the same cardinality. Such a pairing, showing that
sets X and Y have the same cardinality, is a special type of relation++between X and
Y called a one-to-one correspondence. Since each element x of X appears precisely
once in this relation, we can regard this one-to-one correspondence as afunction with
domain X. The range of the function is Y because each y in Y also appears in some
pairing x++y. We formalize this discussion in a definition.
0.12 Definition *A function cp : X --+ Y is one to one if CP(XI) = CP(X2) only when Xl = X2 (see Exer-
cise 37). The function cp is onto Y if the range of cp is Y. •
* We should mention another terminology, used by !be disciples of N. Bourbaki, in case you encounter it
elsewhere. In Bourbaki's terminology, a one-to-one map is an injection, an onto map is a surjection, and a
map that is both one to one and onto is a bijection,

If a subset ofX x Y is a one-fo-one function if;mapping Xonto Y, then each x E X
appears as the first member ofexactly one ordered pair in if;and also each y E Y appears
as the second member of exactly one ordered pair in if;. Thus if we interchange the first
and second members ofallorderedpairs (x , y) in if;to obtain a set oforderedpairs (y , x),
we get a subset of Y x X, which gives a one-to-one function mapping Y onto X. This
function is called the inverse function of if;, and is denoted by if;-l. Summarizing, if
if;maps X one to one onto Y and if;(x) = y, then if;-l maps Y one to one onto X, and
if;-l(y) = X.
0.13 Definition Two sets X and Y have the same cardinality ifthere exists a one-to-one function mapping
X onto Y, that is, if there exists a one-to-one correspondence between X and Y. •
0.14 Example The function f : JR --+ JR where f(x) = x2 is not one to one because f(2) = f(-2) = 4
but 2 #- -2. Also, it is not onto JRbecause therange is the proper subset ofallnonnegative
numbers in R However, g : JR --+ JR defined by g(x) = x3 is both one to one and onto
JR. �
We showed that Z and Z+ havethe same cardinality. We denote this cardinalnumber
by �o, so that IZI = IZ+ I = �o. It is fascinating that a proper subset of an infinite set
may have the same number of elements as the whole set; an infinite set can be defined
as a set having this property.
We naturally wonder whether all infinite sets have the same cardinality as the set Z.
A set has cardinality �o if and only ifall of its elements could be listed in an infinite row,
so that we could "number them" using Z+. Figure 0. 15 indicates that this is possible
for the set Q. The square array of fractions extends infinitely to the right and infinitely
downward, and contains all members of Q. We have shown a string winding its way
through this array. Imagine the fractions to be glued to this string. Taking the beginning
of the string and pulling to the left in the direction of the arrow, the string straightens
out and all elements of Q appear on it in an infinite row as 0, !, - !, 1 , - 1 , �, . . . . Thus
IQI = �o also.
0.15 Figure

Ifthe set S = {x E JR. I 0 < x < I} has cardinality �o, all its elements could be listed
as unending decimals in a column extending infinitely downward, perhaps as
0.3659663426 · . .
0.7103958453 · . .
0.0358493553 · . .
0.9968452214 · . .
We now argue that any such array must omit some number in S. Surely S contains a
number r having as its nth digit after the decimal point anumber different from 0, from 9,
andfromthe nth digit ofthe nth number in this list. For example, r might start .5637· . . .
The 5 rather than 3 after the decimal point shows r cannot be the first number in S
listed in the array shown. The 6 rather than 1 in the second digit shows r cannot be the
second number listed, and so on. Because we could make this argument with any list,
we see that S has too many elements to be paired with those in Z+. Exercise 15 indicates
that JR. has the same number of elements as S. We just denote the cardinality of JR. by
IJR.I. Exercise 19 indicates that there are infinitely many different cardinal numbers even
greater than IJR.I.
Partitions and Equivalence Relations
Sets are disjoint if no two of them have any element in common. Later we will have
occasion to break up a set having an algebraic structure (e.g., a notion of addition) into
disjoint subsets that become elements in a related algebraic structure. We conclude this
section with a study of such breakups, or partitions of sets.
0.16 Definition A partition of a set S is a collection of nonempty subsets of S such that every element
of S is in exactly one of the subsets. The subsets are the cells ofthe partition. •
When discussing apartition ofa set S, we denote byx the cell containingthe element
x of S.
0.17 Example Splitting Z+ into the subset ofeven positive integers (those divisible by 2) and the subset
of odd positive integers (those leaving a remainder of 1 when divided by 2), we obtain
a partition of Z+ into two cells. For example, we can write
14 = {2, 4, 6, 8, 10, 12, 14, 16, 18, . . .}.
We could also partition Z+ into three cells, one consisting of the positive integers
divisible by 3, another containing all positive integers leaving a remainder of 1 when di
vided by 3, and the last containing positive integers leaving a remainder of 2 when
divided by 3.
Generalizing, for each positive integer 17 , we can partition Z+ into 17 cells according
to whether the remainder is 0, 1, 2, . . . , 17 - 1 when a positive integer is divided by n.
These cells are the residue classes modulo 17 in Z+. Exercise 35 asks us to display these
partitions for the cases n = 2, 3, and 5. ...

Each partition of a set S yields a relation .A3 on S in a natural way: namely, for
x, Y E S, let x .n Y if and only if x and y are in the same cell of the partition. In set
notation, we would write x .n y as (x, y) E ,n (see Definition 0.7). A bit of thought
shows that this relation .n on S satisfies the three properties of an equivalence relation
in the following definition.
0.18 Definition An equivalence relation .A3 on a set S is one that satisfies these three properties for all
x, y, z E S.
1. (Reflexive) x .A3 x.
2. (Symmetric) Ifx .A3 y, then y .A3 x.
3. (Transitive) If x .n y and y .JoB z then x .A3 z. •
To illustrate why the relation .Y13 corresponding to a partition of S satisfies the
symmetric condition in the definition, we need only observe that if y is in the same cell
as x (that is, if x .A3 y), then x is in the same cell as y (that is, y .A3 x). We leave the
similar observations to verify the reflexive and transitive properties to Exercise 28.
0.19 Example For any nonempty set S, the equality relation = defined by the subset {(x , x) I X E S} of
S x S is an equivalence relation. £.
0.20 Example (Congruence Modulo n) Let n E Z+. The equivalence relation on Z+ corresponding
to the partition of Z+ into residue classes modulo n, discussed in Example 0.17, is
congruence modulo n. It is sometimes denoted by =/1' Rather than write a -nb, we
usually write a == b (mod n), read, "a is congruent to b modulo n." For example, we
have 15 == 27 (mod 4) because both 15 and 27 haveremainder 3when divided by 4. £.
0.21 Example Let a relation ,A3 on the set Z be defined by n .:72 m if and only if nm ::: 0, and let us
determine whether .JoB is an equivalence relation.
Reflexive a .:72 a, because a2 ::: 0 for all a E Z.
Symmetric If a .A3 b, then ab ::: 0, so ba ::: 0 and b ,Y13 a.
Transitive If a .Y13 b and b .A3 c, then ab ::: 0 and bc ::: O. Thus ab2c = acb2 ::: O.
If we knew b2 > 0, we could deduce ac ::: 0 whence a ,713 c. We have to examine the
case b = 0 separately. A moment of thought shows that -3.:72 0 and 0 ,A3 5, but we do
nothave -3 .JoB 5. Thus the relation .JoB is not transitive, and hence is not an equivalence
��. £.
We observed above that a partition yields a natural equivalence relation. We now
show that anequivalencerelation ona set yields anaturalpartition ofthe set. The theorem
that follows states both results for reference.
0.22 Theorem (Equivalence Relations and Partitions) Let S be a nonempty set and let � be an
equivalence relation on S. Then � yields a partition of S, where
a= {x E S l x �a}.

Also, each partition of S gives rise to an equivalence relation � on S where a � b if and
only if a and b are in the same cell ofthe partition.
Proof We must show that the different cells a = {x E Six � a} for a E S do give a partition
of S, so that every element of S is in some cell and so that if a Eb, then a = b. Let
a ES. Then a Ea by the reflexive condition 0), so a is in at least one cell.
Suppose now that a were in a cell b also. We need to show that a = Jj as sets; this
will show that a cannot be in more than one cell. There is a standard way to show that
two sets are the same:
Show that each set is a subset of the other.
We show thata S b. Let x Ea. Then x � a. But a E b, so a � b. Then, by the transitive
condition (3), x � b, so x E b. Thus as b. Now we show that b S a. Let Y Eb. Then
Y � b. But a Eb, so a � b and, by symmetry (2), b � a. Then by transitivity (3), Y � a,
so yEa. Hence b S a also, so b = a and our proof is complete. •
Each cell in the partition arising from an equiValence relation is an equivalence
class.
EXERCISES 0
In Exercises 1 through 4, describe the set by listing its elements.
1. {x E lR 1 x2 = 3} 2. {m E Z 1m2 = 3}
3. {m E Z 1mn = 60 for some n E Z}
In Exercises 5 through 10, decide whether the object described is indeed a set (is well defined). Give an alternate
description of each set.
5. {n E Z+ 1n is a large number}
6. {n E Z 1n2 < O}
7. {n EZ 139 < n3 < 57}
8. {x E Q 1x is almost an integer}
9. {x EQ 1x may be written with denominator greater than 100}
10. {x E Q 1x may be written with positive denominator less than 4}
11. List the elements in {a, b, c} x {I, 2, c}.
12. Let A = {I, 2, 3} and B = {2, 4, 6}. For each relation between A and B given as a subset of A x B, dccide
whether it is a function mapping A into B. If it is a function, decide whether it is one to one and whether it is
onto B.
a. (Cl, 4), (2, 4), (3, 6)}
c. {(1 , 6), (1, 2), (1, 4)}
e. (Cl, 6), (2, 6), (3, 6)}
b. (Cl, 4), (2, 6), (3, 4)}
d. {(2, 2), (1, 6), (3, 4)}
t {(I, 2), (2, 6), (2, 4)}
13. Illustrate geometrically that two line segments AB and CD ofdifferent length have the same number of points
by indicating in Fig. 0.23 what point y of CD might be paired with point x of AB.

Exercises 9
0.23 Figure
14. Recall that for a, b E lR and a < b, the closed interval [a, b] in lR is defined by [a, b] = {x E lR I a::: x::: b}.
Show that the givenintervals have the same cardinality by giving aformulaforaone-to-onefunction f mapping
the first interval onto the second.
a. [0,I ] and [0, 2] b. [I,3] and [5, 25] c. [a,b] and [c,d]
15. Show that S = {x E lR I ° < x < I } has the same cardinality as R [Hint: Find an elementary function of
calculus that maps an interval one to one onto lR, and then translate andscale appropriately to make thc domain
the set S.]
For any set A, we denote by .7' (A) the collection of all subsets of A. For example, if A = {a, b, c, d}, then
{a, b, d} E .�(A). The set .7'(A) is the power set of A. Exercises 16 through 19 deal with the notion of the power
set of a set A.
16. List the elements of the power set of the given set and give the cardinality of the power set.
a. 0 b. {a} c. {a, b} d. {a, b, c}
17. Let A be a finite set, and let IAI = s. Based on the preceding exercise, make a conjecture about the value of
1.7'(A)I. Then try to prove your conjecture.
18. For any set A, finite or infinite, let BA be the set of all functions mapping A into the set B = {O, I }. Show that
the cardinality of BA is the same as the cardinality of the set .3"(A). [Hint: Each element of BA determines a
subset of A in a natural way.]
19. Show that the power set of a set A,finite or infinite, has too many elements to be able to be put in a one-to-one
correspondence with A. Explain why this intuitively means that there are an infinite number ofinfinite cardinal
numbers. [Hint: Imagine a one-to-one function <p mapping A into .7'(A) to be given. Show that <p cannot be
onto .:7'(A) by considering, for each x E A, whether x E <p(x) and using this idea to define a subset S of A that
is not in the range of <p.] Is the set o
feverything a logically acceptable concept? Why or why not?
20. Let A = {I, 2} and let B = {3, 4, 5}.
a. Illustrate, using A and B, why we consider that 2 + 3 = 5. Use similar reasoning with sets of your own
choice to decide what you would consider to be the value of
i. 3 + �o, ii. �o + �o·
b. Illustrate why we consider that 2 . 3 = 6 by plotting the points of A x B in the plane lR x R Use similar
reasoning with a figure in the text to decide what you would consider to be the value of �o . �o.
21. How many numbers in the interval °::: x::: I can be expressed in the form .##, where each # is a digit
0, 1,2, 3, .
. . , 9? How many are there of the form .#####? Following this idea, and Exercise IS, decide what
you would consider to be the value of lO�o. How about 12�() and 2�O?
22. Continuing the idea in the preceding exercise and using Exercises 18 and 19,use exponential notation to fill in
the three blanks to give a list of five cardinal numbers, each of which is greater than the preceding one.
�o, IlRl. -
, - , _.

In Exercises 23 through 27, find the number of different partitions of a set having the given number of elements.
23. 1 element 24. 2 elements 25. 3 elements
26. 4 elements 27. 5 elements
28. Consider apartition of a set S. The paragraph following Definition 0.1 8 explained why the relation
x .Jl3 y if and only if x and y are in the same cell
satisfies the symmetric condition for an equivalence relation. Write similar explanations of why the reflexive
and transitive properties are also satisifed.
In Exercises 29 through 34, determine whether the given relation is an equivalencerelation on the set. Describe the
partition arising from each equivalence relation.
29. n.Y(, m in Z if nm > 0 30. x.JB)' in lR? if x :::: y
31. x.Jl3 y in lR? if Ixl = Iyl 32. x.jl(, y in lR? if Ix - yl ::S 3
33. n � m in Z+ if n and m have the same number of digits in the usual base ten notation
34. n.3i3 m in Z+ if n and m have the same final digit in the usual base ten notation
35. Using set notation ofthe form {#, #. #.. . .} foraninfinite set, write the residue classes modulo n in Z+ discussed
in Example 0.17 for the indicated value of n.
L n=2 �n=3 � n=5
36. Let n E Z+ and let � be defined on Z by r � s if and only if r - s is divisible by n, that is, if and only if
r - s = nq for some q E Z.
a. Show that � is an equivalence relation on Z. (It is called "congruence modulo n" just as it was for Z+. See
part b.)
b. Show that, when restricted to the subset Z+ of Z. this � is the equivalence relation, congruence modulo n,
of Example 0.20.
c. The cells of this partition of Z are residue classes modulo n in Z. Repeat Exercise 35 for the residue classes
modulo in Z rather than in Z+ using the notation {.. . , #, #, #, . . .} for these infinite sets.
37. Students often ITlisunderstand the concept of a one-to-one function (mapping). I think I know the reason. You
see, a mapping ¢ : A � B has a direction associated with it, from A to B. It seems reasonable to expect a
one-to-one mapping simply to be a mapping that carries one point of A into one point of B, in the direction
indicated by the arrow. But of course, evel}' mapping of A into B does this, and Definition 0.1 2 did not say
that at all. With this unfortunate situation in mind, make as good a pedagogical case as you can for calling the
functions described in Definition 0.12 two-to-twofunctions instead. (Unfortunately, it is almost impossible to
get widely used terminology changed.)

Groups and Subgroups
Section 1 Introduction and Examples
Section 2 Binary Operations
Section 3 Isomorphic Binary Structures
Section 4 G roups
Section 5 Subgroups
Section 6 Cyclic Groups
Section 7 Generating Sets and Cayley Digraphs
"SEd'ION1 INTRODUCTION AND EXAMPLES
In this section, we attempt to give you a little idea of the nature of abstract algebra.
We are all familiar with addition and multiplication of real numbers. Both addition
and multiplication combine two numbers to obtain one number. For example, addition
combines 2 and 3 to obtain 5. We consider addition and multiplication to be binary
operations. In this text, we abstract this notion, and examine sets in which we have one
or more binary operations. We think of a binary operation on a set as giving an algebra
on the set, and we are interested in the structuralproperties of that algebra. To illustrate
what we mean by a structural property with our familiar set lR. of real numbers, note
that the equation x + x = a has a solution x in lR. for each a E lR., namely, x = a12.
However, the corresponding multiplicative equation x . x = a does not have a solution
in lR. if a < O. Thus, lR. with addition has a different algebraic structure than lR. with
multiplication.
Sometimes two different sets with what we naturally regard as very different binary
operations tum out to have the same algebraic structure. For example, we will see in
Section 3 that the set lR. with addition has the same algebraic structure as the set lR.+ of
positive real numbers with multiplication!
This section is designed to get you thinking about such things informally. We will
make everything precise in Sections 2 and 3. We now tum to some examples. Multipli
cation of complex numbers of magnitude 1 provides us with several examples that will
be useful and illuminating in our work. We start with a review of complex numbers and
their multiplication.
11

12 Part I Groups and Subgroups
Complex Numbers
yi
1.1 Figure
A real number can be visualized geometrically as a point on a line that we often regard
as an x-axis. A complex number can be regarded as a point in the Euclidean plane, as
shown in Fig. 1. 1. Note that we label the vertical axis as the yi -axis rather than just the
y-axis,
Cartesian coordinates (a, b) is labeled a + bi in Fig. 1.1. The set C ofcomplex numbers
is defined by
C = {a + bi I a. b E lR}.
We consider lR to be a subset of the complex numbers by identifying a real number r
with the complex number r + Oi. For example, we write 3 + Oi as 3 and - Jr + Oi as -Jr
and 0 + Oi as O. Similarly, we write 0 + Ii as i and 0 + si as si .
Complex numbers were developed after the development of real numbers. The
complex number i was inventedto provide a solution to the quadratic equation x2 = -1,
so we require that
(1)
Unfortunately, i has been called an imaginary number, and this terminology has led
generations of students to view the complex numbers with more skepticism than the real
numbers. Actually, allnumbers, such as 1, 3, Jr, --13, and i are inventions of our minds.
There is no physical entity that is the number 1. If there were, it would surely be in a
place ofhonor in some great scientific museum, and past it would file a steady stream of
mathematicians, gazing at 1 in wonder and awe. A basic goal ofthis text is to show how
we can invent solutions of polynomial equations when the coefficients ofthe polynomial
may not even be real numbers!
Multiplication of Complex Numbers
The product (a + bi)(c + di) is defined in the way it must be if we are to enjoy the
familiar properties of real arithmetic and require that i2 = -1, in accord with Eq. (1).

Section 1 Introduction and Examples 13
Namely, we see that we want to have
(a + bi)(e + di) = ae + adi + bei + bdi2
= ae + adi + bei + bd(-I)
= (ae - bd) + (ad + be)i.
Consequently, we define multiplication of Zl= a + be and Z2 = e + di as
Zl Z2 = (a + bi)(e + di) = (ae - bd) + (ad + be)i,
.
(2)
which is of the form r + si with r = ae - bd and s = ad + be. It is routine to check
that the usual properties ZlZ2 = Z2 ZI, Zl ( Z2 Z3) = ( ZI Z2) Z3 and Zl( Z2 + Z3) = Zl Z2 + Zl Z3
all hold for all Zl , Z2, Z3 E C.
1.2 Example Compute (2 - 5i)(8 + 3i).
Solution We don't memorize Eq. (2), but rather we compute the product as we did to motivate
that equation. We have
(2 - 5i)(8 + 3i) = 16 + 6i - 40i + 15 = 3 1 - 34i.
To establish the geometric meaning of complex multiplication, we first define the abso
lute value la + bi I of a + bi by
(3)
This absolute value is a nonnegative real number and is the distance from a + bi to the
origin in Fig. 1.1. We can now describe a complex number Z in the polar-coordinate form
Z= Izl(cos e + i sin e), (4)
where e is the angle measured counterclockwise from the x-axis to the vector from 0 to
z, as shown in Fig. l.3. A famous formula due to Leonard Euler states that
eiB = cos e + i sin e .
Euler's Formula
We ask you to derive Euler's formula formally from the power series expansions for
ee, cos e and sin e in Exercise 4l. Using this formula, we can express Zin Eq. (4) as
yi
z = Izl(cos e + i sine)
i Izl sine
-¥'----'---'---------e---� x
o Izl cos e
1.3 Figure

Z = Izlei8. Let us set
and
and compute their product in this form, assuming that the usual laws of exponentiation
hold with complex number exponents. We obtain
Z1Z2 = IZlleiellz2leie, = IZ11IZ2Iei(81+82)
(5)
Note that Eq. 5 concludes in the polar form of Eq. 4 where IZlz21 = IZ111z21 and the
polar angle 8 for Z1Z2 is the sum 8 = 81 + 82, Thus, geometrically, we multiply complex
numbers by multiplying their absolute values and adding their polar angles, as shown
in Fig. 104. Exercise 39 indicates how this can be derived via trigonometric identities
without recourse to Euler's formula and assumptions about complex exponentiation.
-2 -1 0
yi
1.4 Figure
2
yi
----'-t---'-----'---� x
1.5 Figure
Note that i has polar angle rr/2 and absolute value 1, as shown in Fig. 1.5. Thus i2
has polar angle 2(rr/2) = rr and 11 . 11 = 1, so that i2 = -1.
1.6 Example Find all solutions in C of the equation Z2 = i.
Solution Writing the equation Z2 = i in polar form and using Eq. (5), we obtain
Id(cos 28 + i sin 28) = 1(0 + i).
Thus IzI2 = 1, so Izi = 1. The angle 8 for z must satisfy cos 28 = 0 and sin 28 = l.
Consequently, 28 = (rr/2) + n(2rr), so 8 = (rr/4) + nrr for an integer n. The values of
n yielding values 8 where 0 :::: 8 < 2rr are 0 and 1, yielding 8 = rr/4 or 8 = 5rr/4. Our
solutions are
Z1 = 1 (cos : + i sin :) and ( 5rr 5rr
)
Z2 = 1 cos 4 + i sin 4
or
and
-1
Z2 = -J20 + i).

1.7 Example Find all solutions of Z4 = -16.
Solution As in Example 1.6 we write the equation in polar form, obtaining
IzI4(COS 48 + i sin48) = 16(-1 + Oi).
Consequently, Izl4 = 16, so Izl = 2 while cos 48 = -1 and sin48 = O. We find that
48 = n + n(2n), so 8 = (n/ 4) + n(n/2) for integers n. The different values of 8 ob
tained where 0 ::::: 8 < 2n are n/4, 3n/ 4, 5n/4, and 7n/4. Thus one solution of Z4 =
-16 is
In a similar way, we find three more solutions,
Y'2(-1 + i), Y'2(-1 - i), and Y'20- i).
Thelasttwo examples illustrate thatwe canfindsolutionsofan equationzn = a + bi
by writing the equation in polar form. There will always be n solutions, provided that
a + bi =I O. Exercises 16 through 21 ask you to solve equations of this type.
We will not use addition or division of complex numbers, but we probably should
mention that addition is given by
(a + bi) + (e + di) = (a + e) + (b + d)i.
and division ofa + bi by nonzero e + di can be performed using the device
a + bi
e + di
Algebra on Circles
a + bi e - di
e + di e - di
(ae + bd) + (be - ad)i
ae + bd be - ad
o ? + ? 0 i.
c- + d- c- + d�
(6)
(7)
Let U = {z Eel Izl = I}, so that U is the circle in the Euclidean plane with center at
the origin and radius 1, as shown in Fig. 1.8. The relation IZIZ21 = IZ111z21 shows that
the product of two numbers in U is again a number in U; we say that U is closed under
multiplication. Thus, we can view multiplication in U as providing algebra on the circle
in Fig. 1.8.
As illustrated in Fig. 1.8, we associate with each z = cos 8 + i sin8 in U a real
number 8 E oc that lies in the half-open interval where 0 ::::: 8 < 2n . This half-open
interval is usually denoted by [0, 2n), but we prefer to denote it by OC2rr for reasons
that will be apparent later. Recall that the angle associated with the product Z1Z2 of two
complex numbers is the sum 81 + 81 ofthe associated angles. Of course if 81 + 82 :::: 2n

yi
---+------��---+------.---�x
2
1.8 Figure
then the angle inlRzrr associated with ZIZZ is 81 + 8z - 2rr. This gives us an addition
modulo 27l' onlR2rr. We denote this addition here by +Zrr.
1 9 E I I TTll h 3rr
+ 5rr 11rr 2 3rr
. xamp e n l&Zrr, we ave T 2rr ""4 = 4 - rr = ""4'
There was nothing special about the number 2rr that enabledus to define addition on
the half-open intervallR2rr. We can use any half-open intervallRc = {x ElR I 0 ::; x < c}.
1.10 Example InlR23, wehave 16 +23 19 = 35 - 23 = 12.InlRs.5, wehave 6 +S.5 8 = 14 - 8.5 = 5.5.
JJ...
Now complex number multiplication on the circle U where Izi = I and addition
modulo 2rr onlRl:T have the same algebraicproperties. We have the natural one-to-one
correspondence Z B- 8 between Z E U and 8 ElR2rr indicated in Fig. 1.8. Moreover, we
deliberately defined +2rr so that
if ZI B- 81 and Z2 B- 82, then ZI' Zz B- (81 +2rr 82). (8)
isomorphism
The relation (8) shows that if we rename each ZE U by its corresponding angle 8
shown in Fig. 1 .8, then the product of two elements in U is renamed by the sum of the
angles for those two elements. Thus U with complex number multiplication andlR2rr
with additionmodulo2rr musthave the same algebraicproperties. They differonly in the
names ofthe elements and thenames ofthe operations. Suchaone-to-one correspondence
satisfying the relation (8) is called an isomOlphism. Names of elements and names of
binary operations are not important in abstract algebra; we are interested in algebraic

properties. We illustrate what we mean by saying that the algebraic properties of U and
of �2J[ are the same.
1.11 Example In U there is exactly one element e such that e . Z = z for all z E U, namely, e = l .
The element 0 in �2J[ that corresponds to 1 E U is the only element e in �2J[ such that
e +2JT x = x for all x E �2JT. ...
1.12 Example The equation z . z . z . z = 1 in U has exactly four solutions, namely, 1, i, - 1, and -i.
Now 1 E U and 0E�2J[ correspond, and the equationx +2rr X +2rr X +2,,- X = 0in�2"
has exactly four solutions, namely, 0, rr/2, rr, and 3rr/2, which, of course, correspond
to 1, i, -1, and -i, respectively. ...
Becauseour circle U has radius 1, ithas circumference 2rr andthe radian measure of
an angle e is equal to the length ofthe arc the angle subtends. Ifwe pick up ourhalf-open
interval �2rr, put the 0 in the interval down on the 1 on the x -axis and wind it around the
circle U counterclockwise, it will reach all the way back to 1. Moreover, each number
in the interval will fall on the point of the circle having that number as the value of the
central angle e shown in Fig. 1.S. This shows that we could also think of addition on
�2rr as being computed by adding lengths of subtended arcs counterclockwise, starting
at z = 1, and subtracting 2rr ifthe sum of the lengths is 2rr or greater.
If we think of addition on a circle in terms of adding lengths of arcs from a starting
point P on the circle and proceeding counterclockwise, we can use a circle of radius
2, which has circumference 4rr, just as well as a circle of radius 1. We can take our
half-open interval �4"- and wrap it around counterclockwise, starting at P; it will just
coverthe whole circle. Addition of arcs lengths gives us a notion of algebra forpoints on
this circle ofradius 2, which is surely isomorphic to �,,- with addition +4,,-. However,
if we take as the circle Iz 1 = 2 in Fig. I .S, multiplication of complex numbers does not
give us an algebra on this circle. The relation IZlz2 1 = IZ1 1 1z2 1 shows that the product of
two such complex numbers has absolute value 4 rather than 2. Thus complex number
multiplication is not closed on this circle.
'
The preceding paragraphs indicate that a little geometry can sometimes be of help
in abstract algebra. We can use geometry to convince ourselves that �2JT and �4rr are
isomorphic. Simply stretch out the interval �2rr uniformly to cover the interval �4J[, or,
ifyou prefer, use a magnifier ofpower 2. Thus we set up the one-to-one correspondence
a *+ 2a between a E �27T and 2a E �4rr. The relation (S) for isomorphism becomes
if a *+ 2a and b *+ 2b then (a +2JT b) *+ (2a +4,,- 2b). (9)
isomorphism
This is obvious if a + b :s 2rr. If a + b = 2rr + c, then 2a + 2b = 4rr + 2c, and the
final pairing in the displayed relation becomes c *+ 2c, which is true.
1.13 Example x +4,,- x +4n X +4)"[ x = 0 in �rr has exactly four solutions, namely, 0, rr, 2rr, and 3rr,
which are two times the solutions found for the analogous equation in �2"- in Exam
ple 1.12. ...

There is nothing special about the numbers 27f and 47f in the previous argument.
Surely, IRe with +e is isomorphic to IRd with +d for all c, d E IR+. We need only pair
x E IRe with (d/c)x E IRd.
Roots of Unity
The elements ofthe set U" = {z E e l Z
"
= I} are called the nth roots ofunity. Using the
technique of Examples 1.6 and 1.7, we see that the elements of this set are the numbers
for m = 0, 1 , 2, · · · , n - 1 .
They all have absolute value 1, so U" c U. If we let i; = cos 2; + i sin 2;, then these
nth roots ofunity can be written as
(10)
Because i;" = 1, these n powers of i; are closed under multiplication. For example, with
n = 10, we have
Thus we see that we can compute i;
i
i;
j by computing i +11j, viewing i and j as elements
of IRI1.
Let :£11 = {O, L 2, 3, . . . , n - I}. We see that :£11 c IRI1 and clearly addition modulo
n is closed on :£11'
1.14 Example The solution of the equation x + 5 = 3 in :£8 is x = 6, because 5 +8 6 = 1 1 - 8 = 3.
...
Ifwe rename each ofthe nth roots ofunity in (10) by its exponent, we use fornames
all the elements of :£" . This gives a one-to-one correspondence between Un and :£n.
Clearly,
if i;
i
++i and i;j++j, then (i;i . i;
j )++(i +n j). (11)
isomorphism
Thus U" with complex number multiplication and :£11 with addition +n have the same
algebraic properties.
1.15 Example It can be shown that there is an isomorphism of Us with :£8 in which i; = ei2rrj8++5.
Under this isomorphism, we must then have i;2 = i; . i;++5 +s 5 = 2. ...
Exercise 35 asks you to continue the computation in Example 1.15, finding the
elements of :£s to which each ofthe remaining six elements of Us correspond.

Section 1 Exercises 19
EXERCISES 1
In Exercises 1 through 9 compute the given arithmetic expression and give the answer in the form a + bi for
a, b E R
3. i23
4. (_i)35 6. (8 + 2i)(3 - i)
7. (2 - 3i)(4 + i) + (6 - 5i)
10. Find I3 - 4i l .
5. (4 - i)(5 + 3i)
8. (1 + i)3 9. (1 - i)5 (Use the binomial theorem.)
11. Find 16 + 4i I.
In Exercises 12 through 15 write the given complex number Z in the polar form Izl(p + qi) where Ip + qi 1 = 1 .
12. 3 - 4i 13. - 1 + i 14. 12 + 5i 15. -3 + 5i
In Exercises 16 through 21, find all solutions in <C of the given equation.
16. z4 = 1 17. z4 = - 1 18. z3 = -8
20. Z
6
= 1 21. Z
6
= -64
19. Z3 = -27i
In Exercises 22 through 27, compute the given expression using the indicated modular addition.
22 10 + 16 23. 8 +10 6 24. 20.5 +25 19.3
• 17
25. � +1 � 26. 3: +2" 3;
28. Explain why the expression 5 +6 8 in lR6 makes no sense.
In Exercises 29 through 34, find all solutions x of the given equation.
30 + 3" 3;r · l11l
• X 2IT 2' = 4" m m,.2,,-
32. x +7 x +7 X = 5 in LZ7
27. 2v'2 +J32 3v'2
29. X +15 7 = 3 in LZ15
31. x +7 x = 3 in LZ7
33. x +12 x = 2 in LZ!2 34. X +4 x +4 X +4 X = ° in LZ4
35. Example 1.15 asserts that there is an isomorphism of Us with LZg in which I; = ei(Jr/4) B- 5 and 1;2 B- 2. Find
the element of LZg that corresponds to each of the remaining six elements 1;'" in Us for m = 0, 3, 4, 5, 6,
and 7.
36. There is an isomorphism of U7 with LZ7 in which I; = ei(hj7) B- 4. Find the element in LZ7 to which I;m must
correspond for m = 0, 2, 3, 4, 5, and 6.
37. Why can there be no isomorphism of U6 with LZ6 in which I; = ei(:T/3) corresponds to 4?
38. Derive the formulas
sin(a + b) = sin a cos b + cos a sin b
and
cos(a + b) = cos a cos b - sina sin b
by using Euler's formula and computing eiCieib.
39. Let Zl = IZI I (cos81 + i sin81) and Z2 = IZ2 1(cos 82 + i sin 82). Use the trigonometric identities in Exercise 38
to derive ZIZ2 = IZl l lz2 1 [cos(81 + 82) + i sin(81 + 82)] .
40. a. Derive a formula for cos 38 in terms of sin 8 and cos 8 using Euler's formula.
b. Derive the formula cos 38 = 4 cos3 8 - 3 cos 8 frompart (a) andthe identity sin
2
8 + cos
2
8 = 1 . (We will
have use for this identity in Section 32.)

41. Recall the power series expansions
x2 X3 X4 x·
eX = I + x + - + - + - + . . . + - + . . .
2! 3! 4! n!
x3 xS x7 X21l-1
sinx = x - - + - - - + · · · + (- lr-1 + . . .
3 ! 5 ! 7 ! (2n - I)! '
x2 X4 x6 x2•
cosX = I - - + - - - + . . . + (- It -- + . . .
2! 4! 61 (2n)1
and
from calculus. Derive Euler's formula e
i
e = cos e + i sin e formally from these three series expansions.
BINARY OPERATIONS
Suppose that we are visitors to a strange civilization in a strange worldand are observing
one of the creatures of this world drilling a class of fellow creatures in addition of
numbers. Suppose also that we have not been told that the class is learning to add, but
werejust placed as observers inthe room where this was going on. Weare asked to give a
reporton exactlywhat happens. The teachermakes noises that sound tous approximately
like gloop, poyt. The class responds with bimt. The teacher then gives ompt, gajt, and the
class responds with poyt. What are they doing? We cannot report that they are adding
numbers, for we do not even know that the sounds are representing numbers. Of course,
we do realize that there is communication going on. All we can say with any certainty is
that these creatures know some rule, so that when certain pairs ofthings are designated
in their language, one after another, like gloop, poyt, they are able to agree on a response,
bimt. This same procedure goes on in addition drill in our first grade classes where a
teacher may sayfour, seven, and the class responds with eleven.
In our attempt to analyze addition and multiplication ofnumbers, we are thus led to
the ideathat addition isbasicallyjust arule thatpeoplelearn, enabling themto associate,
with two numbers in a given order, some number as the answer. Multiplication is also
such a rule, but a different rule. Note finally that in playing this game with students,
teachers have to be a little careful of what two things they give to the class. If a first
grade teacher suddenly inserts ten, sky, the class will be very confused. The rule is only
defined for pairs of things from some specified set.
Definitions and Examples
As mathematicians, let us attempt to collect the core of these basic ideas in a useful
definition, generalizing the notions of addition and multiplication of numbers. As we
remarked in Section 0, we do not attempt to define a set. However, we can attempt to
be somewhat mathematically precise, and we describe our generalizations asfunctions
(see Definition 0.10 and Example 0. 1 1) ratherthan as rules. Recall from Definition 0.4
thatforany set S, the set S x S consists of all ordered pairs (a, b) forelements a and b
of S.
2.1 Definition A binary operation * on a set S is a function mapping S x S into S. For each (a, b) E
S x S, we will denote the element *((a, b)) of S by a * b. •

Section 2 Binary Operations 21
Intuitively, we may regard a binary operation * on S as assigning, to each ordered
pair (a, b) of elements of S, an element a * b of S. We proceed with examples.
2.2 Example Our usual addition + is a binary operation on the set R Our usual multiplication · is a
different binary operation on R In this example, we could replace JR by any of the sets
C, Z, JR+, or Z+. ...
Note that we require a binary operation on a set S to be defined for every ordered
pair (a, b) of elements from S.
2.3 Example Let M(JR) be the set of all matricest with real entries. The usual matrix addition + is not
a binary operation on this set since A + B is not defined for an ordered pair (A, B) of
matrices having different numbers of rows or of columns. ...
Sometimes a binary operation on S provides a binary operation on a subset H of S
also. We make a formal definition.
2.4 Definition Let * be a binary operation on S and let H be a subset of S. The subset H is closed
nnder * iffor all a, b E H we also havea * b E H. In this case, the binary operation on
H given by restricting * to H is the induced operation of * on H. •
By our very definition of a binary operation * on S, the set S is closed under *, but
a subset may not be, as the following example shows.
2.5 Example Our usual addition + on the set JR of real numbers does not induce a binary operation
on the set JR* of nonzero real numbers because 2 E JR* and -2 E JR*, but 2 + (-2) = 0
and 0 rt JR*. Thus JR* is not closed under *. ...
In our text, we will often have occasion to decide whether a subset H of S is closed
under a binary operation * on S. To arrive at a correct conclusion, we have to know what
it meansfor an element to be in H, and to use this fact. Students have trouble here. Be
sure you understand the next example.
2.6 Example Let + and . be the usual binary operations of addition and multiplication on the set
Z, and let H = {n2ln E Z+}. Determine whether H is closed under (a) addition and
(b) multiplication.
Forpart (a), weneed onlyobservethat 12 = 1 and22 = 4 arein H, butthat I + 4 = 5
and 5 rt H. Thus H is not closed under addition.
For part (b), suppose that r E H and s E H. Using what it means for r and s to be
in H, we see that there must be integers n and m in Z+ such that r = n2 and s = m2•
Consequently, rs = n2m2 = (nm)2. By the characterization of elements in H and the
fact that nm E Z+, this means that rs E H, so H is closed under multiplication. ...
j Most students of abstract algebra have studied lin
ear algebra and are familiar with matrices and matrix
operations. For the benefit of those students, examples in
volving matrices are often given. The reader who is
not familiar with matrices caneither skip all references to them or tum to the Appendix at the back of the text,
where there is a short summary.

2.7 Example Let F be the set ofallreal-valuedfunctions fhaving as domainthe setIR ofreal numbers.
We are familiar from calculus with the binary operations +, -, " and 0 on F. Namely,
for each ordered pair (f, g) of functions in F, we define for each x EIR
and
f + g by (f + g)(x) = f(x) + g(x)
f - g by (f - g)(x) = f(x) - g(x)
f . g by (f . g)(x) = f(x)g(x)
f o g by (f 0 g)(x) = f(g(x))
addition,
subtraction,
multiplication,
composition.
All four of these functions are again real valued with domain IR, so F is closed under all
four operations +, -, " and o. ...
The binary operations described in the examples above are very familiar to you.
In this text, we want to abstract basic structural concepts from our familiar algebra.
To emphasize this concept of abstraction from the familiar, we should illustrate these
structural concepts with unfamiliar examples. We presented the binary operations of
complex number multiplication on U and Un, addition +n on Zn, and addition +e on IRe
in Section 1.
The most important method ofdescribing a particular binary operation * on a given
set is to characterize the element a * b assigned to each pair (a, b) by some property
defined in terms of a and b.
2.8 Example On Z+, we define a binary operation * by a * b equals the smaller of a and b, or the
common value if a = b. Thus 2 * 1 1 = 2; 15 * 10 = 10; and 3 * 3 = 3. ...
2.9 Example On Z+, we define a binary operation *' by a *' b = a. Thus 2 *' 3 = 2, 25 *' 10 = 25,
and 5 *' 5 = 5. ...
2.10 Example On Z+, we define a binary operation *" by a *" b = (a * b) + 2, where * is defined in
Example 2.8. Thus 4 *" 7 = 6; 25 *" 9 = 1 1 ; and 6 *" 6 = 8. ...
It may seem that these examples are of no importance, but consider for a moment.
Suppose we go into a store to buy a large, delicious chocolate bar. Suppose we see two
identical bars side by side, the wrapper of one stamped $1.67 and the wrapper of the
other stamped $1.79. Of course we pick up the one stamped $1.67. Our knowledge of
which one we want depends on the fact that at some time we learned the binary operation
* of Example 2.8. It is a very important operation. Likewise, the binary operation *' of
Example 2.9 is defined using our ability to distinguish order. Think what a problem we
would have if we tried to put on our shoes first, and then our socks! Thus we should
not be hasty about dismissing some binary operation as being of little significance. Of
course, our usual operations of addition and multiplication ofnumbers have a practical
importance well known to us.
Examples 2.8 and 2.9 were chosen to demonstrate that a binary operation may or
may not depend on the order of the given pair. Thus in Example 2.8, a * b = b * a for
all a, bEZ+', and in Example 2.9 this is not the case, for 5 *' 7 = 5 but 7 *' 5 = 7.

Section 2 Binary Operations 23
2.11 Definition A binary operation * on a set S is commutative if (and only if) a * b = b * a for all
a, b E S. •
As was pointed out in Section 0, it is customary in mathematics to omit the words
and only if from a definition. Definitions are always understood to be if and only if
statements. Theorems are not always ifand only ifstatements, and no such convention
is ever usedfor theorems.
Now suppose we wish to consider an expression of the form a * b * c. A binary
operation * enables usto combine onlytwo elements, andherewe havethree. Theobvious
attempts to combine the three elements are to form either (a * b) * c or a * (b * c). With
* defined as in Example 2.8, (2 * 5) * 9 is computed by 2 * 5 = 2 and then 2 * 9 = 2.
Likewise, 2 * (5 * 9) is computed by 5 * 9 = 5 and then 2 * 5 = 2. Hence (2 * 5) * 9 =
2 * (5 * 9), and it is not hard to see that for this *,
(a * b) * c = a * (b * c),
so there is no ambiguity in writing a * b * c. But for *" ofExample 2.10,
(2 *" 5) *" 9 = 4 *" 9 = 6,
while
2 *" (5 *" 9) = 2 *" 7 = 4.
Thus (a *" b) *" c need not equal a *" (b *" c), and an expression a *" b *" c may be
ambiguous.
2.12 Definition A binary operation on a set S is associative if(a * b) * c = a * (b * c) for all a, b, c E S.
•
It can be shown that if * is associative, then longer expressions such as a * b *
c * d are not ambiguous. Parentheses may be inserted in any fashion for purposes of
computation; the final results of two such computations will be the same.
Composition of functions mapping lR into lR was reviewed in Example 2.7. For any
set S and any functions f and g mapping S into S, we similarly define the composition
f o g ofg followedby f asthefunctionmapping S into S suchthat (f 0 g)(x) = f(g(x))
for all X E S. Some of the most important binary operations we consider are defined
using composition of functions. It is important to know that this composition is always
associative whenever it is defined.
2.13 Theorem (Associativity ofComposition) Let S be a set and let f, g, and h be functions mapping
S into S. Then f 0 (g 0 h) = (f 0 g) 0 h.
Proof To show these two functions are equal, we must show thatthey give the same assignment
to each X E S. Computing we find that
(f 0 (g 0 h))(x) = f((g 0 h)(x)) = f(g(h(x)))
and
((f 0 g) 0 h)(x) = (f 0 g)(h(x)) = f(g(h(x))),
so the same element f(g(h(x))) of S is indeed obtained. •

2.14 Example
As an example of using Theorem 2.13 to save work, recall that it is a fairly painful
exercise in summation notation to show that multiplication of n x n matrices is an
associative binary operation. If, in a linear algebra course, we first show that there
is a one-to-one correspondence between matrices and linear transformations and that
multiplication of matrices corresponds to the composition of the linear transformations
(functions), we obtain this associativity at once from Theorem 2.13.
Tables
For a finite set, a binary operation on the set can be defined by means of a table in which
the elements of the set are listed across the top as heads of columns and at the left side
as heads ofrows. We always require that the elements ofthe set be listed as heads across
the top in the same order as heads down the left side. The next example illustrates the
use of a table to define a binary operation.
Table 2.15 defines the binary operation * on S = {a, b, c} by the following rule:
2.15 Table
(ith entry on the left) * (jth entry on the top)
= (entry in the ith row andjth column ofthe table body).
*
a
b
c
a b
b c
a c
c b
c
b
b
a
Thus a * b = c and b * a = a, so * is not commutative.
We can easily see that a binary operation defined by a table is commutative ifand
only ifthe entries in the table are symmetric with respect to the diagonal that starts at
the upper left comer ofthe table and terminates at the lower right comer.
2.16 Example Complete Table 2.17 so that * is a commutative binary operation on the set S =
{a, b, c, d}.
Solution From Table 2.17, we see that b * a = d. For * to be commutative, we must have a * b =
2.17 Table d also. Thus we place d in the appropriate square defining a * b, which is located
* a b
a b
b d a
c a c
d a b
2.18 Table
c d
d
b c
symmetrically across the diagonal in Table 2. 18 from the square defining b * a. We
obtain the rest of Table 2. 18 in this fashion to give our solution. ..
Some Words of Warning
Classroom experience shows the chaos that may result if a student is given a set and
asked to define some binary operation on it. Remember that in an attempt to define a
binary operation * on a set S we must be sure that
1. exactly one element is assigned to each possible orderedpair ofelements of S,
2. for each orderedpair ofelements of S, the element assignedto it is again in S.
Regarding Condition 1 , a student will often make an attempt that assigns an element
of S to "most" ordered pairs, but for a few pairs, determines no element. In this event,
* is not everywhere defined on S. It may also happen that for some pairs, the at
tempt could assign any of several elements of S, that is, there is ambiguity. In any case

of ambiguity, * is not well defined. If Condition 2 is violated, then S is not closed
under *.
Following are several illustrations of attempts to define binary operations on sets.
Some of them are worthless. The symbol * is used for the attempted operation in all
these examples.
2.19 Example On Ql, let a * b = a/b. Here * is not everywhere defined on Ql, for no rational number is
assigned by this rule to the pair (2, 0). ...
2.20 Example On Ql+, let a * b = a/b. Here both Conditions 1 and 2 are satisfied, and * is a binary
operation on Ql+. ...
2.21 Example On Z+, let a * b = a/b. Here Condition 2 fails, for 1 * 3 is not in Z+. Thus * is not a
binary operation on Z+, since Z+ is not closed under *. ...
2.22 Example Let F be the set of all real-valued functions with domainlR as in Example 2.7. Suppose
we "define" * to give the usual quotient of f by g, that is, f * g = h, where hex) =
f(x)/g(x). Here Condition 2 is violated, for the functions in F were to be defined for
all real numbers, and for some g E F, g(x) will be zero for some values of x inlR and
hex) would not be defined at those numbers in R For example, if f(x) = cos x and
g(x) = x2, then h(O) is undefined, so h tJ- F. ...
2.23 Example Let F be as in Example 2.22 and let f * g = h, where h is the function greater than
both f and g . This "definition" is completely worthless. In the first place, we have not
defined what it means for one function to be greater than another. Even if we had, any
sensible definition would result in there being many functions greater than both f and
g, and * would still be not well defined. ...
2.24 Example Let S be a set consisting of 20 people, no two of whom are of the same height. Define
* by a * b = c, where c is the tallest person among the 20 in S. This is aperfectly good
binary operation on the set, although not a particularly interesting one. ...
2.25 Example Let S be as in Example 2.24 and let a * b = c, where c is the shortest person in S who
is taller than both a and b. This * is not everywhere defined, since if either a or b is the
tallest person in the set, a * b is not determined. ...
II EXERelSE S 2
Computations
Exercises 1 through 4 concern the binary operation * defined on S = {a, b, c, d, e} by means of Table 2.26.
1. Compute b * d, c * c, and [(a * c) * el * a.
2. Compute (a * b) * c and a * (b * c). Can you say on the basis of this computations whether * is associative?
3. Compute (b * d) * c and b * Cd * c). Can you say on the basis ofthis computation whether * is associative?

2.26 Table
* a b c d
a a b c b
b b c a e
c c a b b
d b e b e
e d b a d
4. Is * commutative? Why?
e
d
c
a
d
c
2.27 Table 2.28 Table
* a b c d * a b
a a b c a a b
b b d c b b a
c c a d b c c d
d d a d
5. Complete Table 2.27 so as to define a commutative binary operation * on S = {a, b,c, d}.
c d
c d
c d
c d
6. Table 2.28 can be completed to define an associative binary operation * on S = {a, b, c, d}. Assume this is
possible and compute the missing entries.
In Exercises 7 through 1 1, determine whether the binary operation * defined is commutative and whether * is
associative.
7. * defined on Z by letting a * b = a - b
8. * defined on Q by letting a * b = ab + 1
9. * defined on Q by letting a * b = ab/2
10. * defined on Z+ by letting a * b = 2ab
11. * defined on Z+ by letting a * b = ab
12. Let S be a sethaving exactly one element. How many differentbinary operations can be defined on S? Answer
the question if S has exactly 2 elements; exactly 3 elements; exactly n elements.
13. How many different commutative binary operations can be defined on a set of 2 elements? on a set of 3
elements? on a set of n elements?
Concepts
In Exercises 14 through 16, correct the definition of the italicized term without reference to the text, if correction
is needed, so that it is in a form acceptable for publication.
14. A binary operation * is commutative if and only if a * b = b * a.
15. A binary operation * on a set S is associative if and only if, for all a, b, c E S, we have
(b * c) * a = b * (c * a).
16. A subset H of a set S is closed under a binary operation * on S if and only if (a * b) E H for all a, b E S.
In Exercises 17 through 22, determine whether the definition of * does give a binary operation on the set. In the
event that * is not a binary operation, state whether Condition 1,Condition 2, or both ofthese conditions on page 24
are violated.
17. On Z+, define * by letting a * b = a - b.
18. On Z+, define * by letting a * b = ab.
19. On lR, define * by letting a * b = a - b.
20. On Z+, define * by letting a * b = c,where c is the smallest integer greater than both a and b.

21. On Z+, define * by letting a * b = e, where e is at least 5 more than a + b.
22. On Z+, define * by letting a * b = e, where e is the largest integer less than the product of a and b.
23. Let H be the subset of M2(JR) consisting of all matrices ofthe form [� -!Jfor a , b E JR. Is H closed under
a matrix addition? b matrix multiplication?
24. Mark each of the following true or false.
___ a. If * is any binary operation on any set S, then a * a = a for all a E S.
___ b. If * is any commutative binary operation on any set S, then a * (b * c) = (b * c) * a for all a , b,
e E S.
c. Ih is any associativebinary operation on any set S, then a * (b * c) = (b * c) * a for all a, b, e E S.
d. The only binary operations of any importance are those defined on sets of numbers.
e. A binary operation * on a set S is commutative if there exist a, b E S such that a * b = b * a .
f. Every binary operation defined on a set having exactly one element is both commutative and
associative.
___ g. A binary operation on a set S assigns at least one element of S to each ordered pair of elements
ofS.
___ h. A binary operation on a set S assigns at most one element of S to each ordered pair of elements of
S.
___ i. A binary operation on a set S assigns exactly one element of S to each ordered pair of elements
of S.
___ j. A binary operation on a set S may assign more than one element of S to some ordered pair of
elements of S.
25. Give a set different from any of those described in the examples of the text and not a set of numbers. Define
two different binary operations * and *' on this set. Be sure that your set is well defined.
Theory
26. Prove that if * is an associative and commutative binary operation on a set S, then
for all a , b, e, d E S. Assume the associative law only for triples as in the definition, that is, assume only
(x * y) * z = x * (y * z)
for all x, y, Z E S.
In Exercises 27 and 28, either prove the statement or give a counterexample.
27. Every binary operation on a set consisting of a single element in both commutative and associative.
28. Every commutative binary operation on a set havingjust two elements is associative.
Let F be the set of all real-valued functions having as domain the set JR of all real numbers. Example 2.7 defined
the binary operations +, - , " and 0 on F. In Exercises 29 through 35, either prove the given statement or give a
counterexample.
29. Function addition + on F is associative.
30. Function subtraction - on F is commutative

31. Function subtraction - on F is associative.
32. Function multiplication · on F is commutative.
33. Function multiplication . on F is associative.
34. Function composition 0 on F is commutative.
35. If * and *' are any two binary operations on a set S, then
a * (b *' c) = (a * b) *
'
(a * c) for all a, b, c E S.
36. Suppose that * is an associative binary operation on a set S. Let H = {a E S I a * x = x * a for all X E S}.
Show that H is closed under *. (We think of H as consisting of all elements of S that commute with every
element in S.)
37. Suppose that * is an associative and commutativebinary operationonaset S. Show that H = {a E S I a * a = a}
is closed under *. (The elements of H are idempotents ofthe binary operation *.)
ISOMORPIDC BINARY STRUCTURES
Compare Table 3.1 for the binary operation * on the set S = {a, b, c} with Table 3.2 for
the binary operation *' on the set T = {#, $, &}.
Notice that if, in Table 3.1, we replace all occurrences of a by #, every b by $, and
every c by & using the one-to-one correspondence
we obtain precisely Table 3.2. The two tables differ only in the symbols (or names)
denotingthe elements andthe symbols * and *
'
forthe operations. Ifwe rewrite Table 3.3
with elements in the order y, x, z, we obtain Table 3.4. (Here we did not setup any one
one-correpondence; we just listed the same elements in different order outside the heavy
bars of the table.) Replacing, in Table 3.1, all occurrences of a by y, every b by x, and
every c by z using the one-to-one correspondence
a *+ y b *+ x c *+ z
we obtain Table 3.4. We think ofTables 3.1, 3.2, 3.3, and 3.4 as being structurally alike.
These four tables differ only in the names (or symbols) for their elements and in the
order that those elements are listed as heads in the tables. However, Table 3.5 for binary
operation * and Table 3.6 for binary operation *' on the set S = {a, b, c} are structurally
different from each other and from Table 3.1. In Table 3.1, each element appears three
times in the body ofthe table, while the body of Table 3.5 contains the single element b.
In Table 3.6, for all s E S we get the same value c for s *' s along the upper-left to lower
right diagonal, while we get three different values in Table 3.1. Thus Tables 3.1 through
3.6 give just three structurally different binary operations on a set of three elements,
provided we disregard the names of the elements and the order in which they appear as
heads in the tables.
The situation we have just discussed is somewhat akin to children in France and in
Germany learning the operation of addition on the set Z+. The children have different

3.1 Table
* a b c
a c a b
b a b c
c b c a
3.4 Table
*" y x z
y z y x
x y x Z
Z x Z y
Section 3 Isomorphic Binary Structures 29
3.2 Table 3.3 Table
*' # $ & *" x y Z
# & # $ x x y z
$ # $ & y y Z x
& $ & # Z Z x y
3.5 Table 3.6 Table
'* a b c * a b c
a b b b a c a b
b b b b b b c a
c b b b c a b c
names (un, deux, trois, . . . versus ein, zwei, drei . . .) forthe numbers, buttheyarelearning
the same binary structure. (In this case, they are also using the same symbols for the
numbers, so their addition tables would appear the same if they list the numbers in the
same order.)
We are interested in studying the different types ofstructuresthatbinary operations
can provide on sets having the same number of elements, as typified by Tables 3.4, 3.5,
and 3.6. Let us consider a binary algebraic structuret (5, *) tobe a set 5 together with
a binary operation * on S. In order for two such binary structures (S, *) and (S', *') to
be structurally alike in the sense we have described, we would have to have a one-to-one
correspondence between the elements x of 5 and the elements x' of 5' such that
if X B- X' and Y B- Y', then x * Y B- X' *' y'. (1)
A one-to-one correspondence exists if the sets 5 and 5' have the same number of
elements. It is customary to describe a one-to-one correspondence by giving a ane
ta-one function ¢ mapping 5 onto 5' (see Definition 0. 12). For such a function ¢, we
regard the equation ¢(x) = x' as reading the one-to-one pairing x B- x in left-to-right
order. In terms of ¢, the final B- correspondence in (1), which asserts the algebraic
structure in S' is the same as in 5, can be expressed as
¢(x * y) = ¢(x) *' ¢(y).
Such, a function showing that two algebraic systems are structurally alike is known as
an isomorphism. We give a formal definition.
3.7 Definition Let (S, *) and (5', *') be binary algebraic structures. An isomorphism of S with S' is a
one-to-one function ¢ mapping S onto S' such that
¢(x * y) = ¢(y) *' ¢(y) for all x, Y E S.
homomorphismproperty
(2)
t Remember that boldface type indicates that a term is being defined.

If such a map ¢ exists, then S and S' are isomorphic binary structures, which we
denote by S :::::: S', omitting the * and *' from the notation. •
You may wonder why we labeled the displayed condition in Definition 3.7 the ho
momorphismproperty rather than the isomorphismproperty. The notion ofisomorphism
includes the idea of one-to-one correspondence, which appeared in the definition via the
words one-to-one and onto before the display. In Chapter 13, we will discuss the rela
tion between S and S' when ¢ : S � S' satisfies the displayed homomorphism property,
but ¢ is not necessarily one to one; ¢ is then called a homomorphism rather than an
isomorphism.
It is apparent that in Section 1, we showed that the binary structures (U, .) and
(JRc, +c) are isomorphic for all c E JR+. Also, (Un, .) and ('1',n , +n) are isomorphic for
each n E '1',+.
Exercise 27 asks us to show that for a collection ofbinary algebraic structures, the
relation :::::: in Definition 3.7 is an equivalence relation on the collection. Our discussion
leading to the preceding definition shows thatthebinary structures defined by Tables 3.1
through 3.4 are in the same equivalence class, while those given by Tables 3.5 and 3.6 are
in different equivalence classes. We proceed to discuss how to try to determine whether
binary structures are isomorphic.
How to Show That Binary Structures Are Isomorphic
We now give an outline showing how to proceed from Definition 3.7 to show that two
binary structures (S, *) and (S', *') are isomorphic.
Step 1 Define thefunction ¢ that gives the isomorphism ofS with S'. Now this
means that we have to describe, in some fashion, what ¢(s) is to be for every s E S.
Step 2 Show that ¢ is a one-to-onefunction. That is, suppose that ¢(x) = ¢(y)
in S' and deduce from this that x = y in S.
Step 3 Show that ¢ is onto S'. That is, suppose that s' E S' is given and show that
there does exist s E S such that ¢(s) = s'.
Step 4 Show that ¢(x * y) = ¢(x) *' ¢(y)for all x, Y E S. This is just a question
of computation. Compute both sides of the equation and see whether they are the
same.
3.8 Example Let us show that the binary structure (R +) with operation the usual addition is isomor
phic to the structure (JR+ , . ) where · is the usual multiplication.
Step 1 We have to somehow convert an operation of addition to mUltiplication.
Recall from ah+c = (ah)(ac) that addition of exponents corresponds to
multiplication oftwo quantities. Thus we try defining ¢ : JR � JR+ by
¢(x) = eX for x E R Note that eX > 0 for all x E JR, so indeed,
¢(x) E JR+ .
Step 2 If ¢(x) = ¢(y), then eX = eY• Taking the natural logarithm, we see that
x = y, so ¢ is indeed one to one.

Step 3 If r E JR+, then In(r) E JR and ¢(In r) = e1n r = r. Thus ¢ is onto JR+.
Step 4 For x, Y E JR, we have ¢(x + y) = eX+Y = eX . eY = ¢(x) · ¢(y). Thus we
see that ¢ is indeed an isomorphism. ..
3.9 Example Let 2Z = {2n I n E Z}, so that 2Z is the set of all even integers, positive, negative, and
zero. We claim that (Z, +) is isomorphic to (2Z, +) , where + is the usual addition. This
will give an example ofa binary structure (Z, +) that is actually isomorphic to a structure
consisting of a proper subset under the induced operation, in contrast to Example 3.8,
where the operations were totally different.
Step 1 The obvious function ¢ : Z ---+ 2Z to try is given by ¢(n) = 2n for n E Z.
Step 2 If ¢(m) = ¢(n), then 2m = 2n so m = n. Thus ¢ is one to one.
Step 3 If n E 2Z, then n is even so n = 2m for m = nl2 E Z. Hence
¢(m) = 2(nI2) = n so ¢ is onto 2Z.
Step 4 Let m, n E Z. The equation
¢(m + n) = 2(m + n) = 2m + 2n = ¢(m) + ¢(n)
then shows that ¢ is an isomorphism.
How to Show That Binary Structures Are Not Isomorphic
We now tum to thereverse question, namely:
How do we demonstrate that two binary structures (S, *) and (S', *') are not
isomorphic, ifthis is the case?
This wouldmeanthat there is no one-to-one function ¢ from S onto S' withthe property
¢(x * y) = ¢(x) *' ¢(y) for all x, y E S. In general, it is clearly not feasible to try every
possible one-to-one function mapping S onto S' and test whether it has this property,
except in the case where there are no such functions. This is the case precisely when S
and S' do not have the same cardinality. (See Definition 0.13.)
3.10 Example Thebinary structures (Q, +) and (JR, +) are not isomorphic because Q has cardinality �o
while IJRI =f. �o. (See the discussion following Example 0. 13.) Note that it is not enough
to say that Q is a proper subset of R Example 3.9 shows that a proper subset with the
induced operation can indeed be isomorphic to the entire binary structure. ..
A structural property of a binary structure is one that must be shared by any
isomorphic structure. It is not concerned with names or some other nonstructural char
acteristics of the elements. For example, the binary structures defined by Tables 3.1 and
3.2 are isomorphic, although the elements are totally different. Also, a structural prop
erty is not concerned with what we consider to be the "name" of the binary operation.
Example 3.8 showed that a binary structure whose operation is ourusual addition can be
isomorphic to one whose operation is our usual multiplication. The number of elements
in the set S is a structural property of (8, *).

In the event that there are one-to-one mappings of S onto S', we usually show that
(S, *) is not isomorphic to (8', *') (if this is the case) by showing that one has
some structural property that the other does not possess.
3.11 Example The sets Z and Z+ both have cardinality �o, and there are lots of one-to-one functions
mapping Z onto Z+ . However, the binary structures (Z, .) and (Z+, .), where · is the
usual multiplication, are not isomorphic. In (Z, .) there are two elements x such that
x . x = x, namely, 0 and 1 . However, in (Z+, .), there is only the single element 1. .A.
We list afew examples of possible structural properties andnonstructuralproperties
of a binary strUcture (8, *) to get you thinking along the right line.
Possible Structural Properties
1. The set has 4 elements.
2. The operation is commutative.
3. x * x = x for all X E S.
4. The equation a * x = b has a
solution x in S for all a , b E S.
Possible Nonstructural Properties
a. The number 4 is an element.
b. The operation is called "addition."
c. The elements of S are matrices.
d. S is a subset of C.
We introduced the algebraic notions ofcommutativity and associativity in Section 2.
One other structural notion that will be of interest to us is illustrated by Table 3.3,
where for the binary operation *" on the set {x, y, z}, we have x *" u = u *" x = u
for all choices possible choices, x, y, and z for u . Thus x plays the same role as 0 in
(R+) where 0 + U = U + 0 = U for all U E �, and the same role as 1 in (�, .) where
1 . U = U • 1 = U for all U E R Because Tables 3.1 and 3.2 give structures isomorphic
to the one in Table 3.3, they must exhibit an element with a similar property. We see that
b * U = U * b = u for all elements u appearing in Table 3. 1 and that $ *' u = u *' $ = u
for all elements u in Table 3.2. We give a formal definition of this structural notion and
prove a little theorem.
3.12 Definition Let (S, *) be a binary structure. An element e of S is an identity element for * if
e * s = s * e = s for all s E S. •
3.13 Theorem (Uniqueness of Identity Element) A binary structure (S, *) has at most one identity
element. That is, if there is an identity element, it is unique.
Proof Proceeding in the standard way to show uniqueness, suppose that both e and e are ele
ments of S serving as identity elements. We let them compete with each other. Regarding
e as an identity element, we must have e * e = e. However, regarding e as an identity
element, we must have e * e = e. Wethus obtain e = e, showing that an identity element
must be unique. •

If you now have a good grasp of the notion of isomorphic binary structures, it
should be evident that having an identity element for * is indeed a structural property of
a structure (S, *). However, we know from experience that manyreaders will be unable
to see the forest because of all the trees that have appeared. For them, we now supply a
careful proof, skipping along to touch those trees that are involved.
3.14 Theorem Suppose (S, *) has an identity element e for *. If¢ : S -+ S' is an isomorphism of (S, *)
with (S', *'), then ¢(e) is an identity element for the binary operation *' on S'.
Proof Let s' E Sf. We must show that¢(e) *' s' = s' *' ¢(e) = Sf. Because ¢is an isomorphism,
it is a one-to-one map of S onto S'. In particular, there exists s E S such that ¢(s) = Sf.
Now e is an identity element for * so that we know that e * s = s * e = s. Because ¢ is
a function, we then obtain
¢(e * s) = ¢(s * e) = ¢(s).
Using Definition 3.7 of an isomorphism, we can rewrite this as
¢(e) *
'
¢(s) = ¢(s) *
'
¢(e) = ¢(s).
Remembering that we chose s E S such that ¢(s) = Sf, we obtain the desired relation
¢(e) *' s' = s' *' ¢(e) = Sf. •
We conclude with three more examples showing via structural properties that cer
tain binary structures are not isomorphic. In the exercises we ask you to show, as in
Theorem 3.14, that the properties we use to distinguish the structures in these examples
are indeed structural. That is, they must be shared by any isomorphic structure.
3.15 Example We show that the binary structures (1Qi, +) and (Z, +) under the usual addition are not
isomorphic. (Both IQi and Z have cardinality �o, so there are lots of one-to-one functions
mapping IQi onto Z.) The equation x + x = c has a solution x for all c E 1Qi, but this is
not the case in Z. For example, the equation x + x = 3 has no solution in Z. We have
exhibited a structural property that distinguishes these two structures. ..
3.16 Example The binary structures (C, .) and (jE., .) under the usual multiplication are not isomorphic.
(It can be shown that C and � have the same cardinality.) The equation x . x = c has a
solution x for all C E C, but x . x = - 1 has no solution in R ..
3.17 Example Thebinary structure (M2(�), .) of2 x 2realmatrices with theusualmatrixmultiplication
is notisomorphic to (jE., .) withtheusualnumbermultiplication. (It canbe shownthatboth
sets have cardinality I�I.) Multiplication ofnumbers is commutative, but multiplication
of matrices is not. ..

EXERCISES 3
In all the exercises, + is the usual addition on the set where it is specified, and · is the usual multiplication.
Computations
1. What three things must we check to determine whether a function ¢: S -+ S' is an isomorphism of a binary
structure (S, *) with (S', *')?
In Exercises 2 through 10, determine whether the given map ¢ is an isomorphism of the first binary structure with
the second. (See Exercise 1 .) If it is not an isomorphism, why not?
2. (Z, +) with (Z, +) where ¢(n) = -n for n E Z
3. (Z, +) with (Z, +) where ¢(n) = 2n for n E Z
4. (Z, +) with (Z, +) where ¢(n) = n + 1 for n E Z
5. (([Jl, +) with (([Jl, +) where ¢(x) = x/2 for x E ([Jl
6. (([Jl, .) with (([Jl, .) where ¢(x) = x2 for x E ([Jl
7. (lR, .) with (lR, .) where ¢(x) = x3 for x E JR
8. (M2(JR), ·) with (lR, .) where ¢(A) is the determinant of matrix A
9. (MJ (JR), .) with (JR, .) where ¢(A) is the determinant of matrix A
10. (lR, +) with (JR+ , .) where ¢(r) = OS for r E JR
In Exercises 1 1 through 15, let F be the set of all functions f mapping JR into JR that have derivatives of all orders.
Follow the instructions for Exercises 2 through 10.
11. (F, +) with (F, +) where ¢(f) = 1', the derivative of f
12. (F, +) with (lR, +) where ¢(f) = 1'(0)
13. (F, +) with (F, +) where ¢(f)(x) = fat f(t)dt
14. (F, +) with (F, +) where ¢(f)(x) = �lt Uot f(t)dt]
15. (F, ·) with (F, .) where ¢(f)(x) = x . f(x)
16. The map ¢ : Z -+ Z defined by ¢(n) = n + 1 for n E Z is one to one and onto Z. Give the definition of a
binary operation * on Z such that ¢ is an isomorphism mapping
a. (Z, +) onto (Z, *), b. (Z, *) onto (Z, +).
In each case, give the identity element for * on Z.
17. The map ¢ : Z -+ Z defined by ¢(n) = n + 1 for n E Z is one to one and onto Z. Give the definition of a
binary operation * on Z such that ¢ is an isomorphism mapping
a. (Z, ·) onto (Z, *), b. (Z, *) onto (Z, ·).
In each case, give the identity element for * on Z.
18. The map ¢ : ([Jl -+ ([Jl defined by ¢(x) = 3x - 1 for x E ([Jl is one to one and onto ([Jl. Give the definition of a
binary operation * on ([Jl such that ¢ is an isomorphism mapping
a. (([Jl, +) onto (([Jl, *), b. (([Jl, *) onto (([Jl, +).
In each case, give the identity element for * on ([Jl.

19. The map ¢ : QJ -+ QJ defined by ¢(x) = 3x - 1 for x E QJ is one to one and onto QJ. Give the definition of a
binary operation * on QJ such that ¢ is an isomorphism mapping
a. (QJ. .) onto (QJ, *) , b. (QJ, *) onto (QJ, .) .
In each case, give the identity element for * on QJ.
Concepts
20. The displayed homomorphism condition for an isomorphism ¢ in Definition 3.7 is sometimes summarized
by saying, "¢ must commute with the binary operation(s)." Explain how that condition can be viewed in this
manner.
In Exercises 21 and 22, correct the definition of the italicized term without reference to the text, if correction is
needed, so that it is in a form acceptable for publication.
21. A function ¢ : S -+ S' is an isomorphism if and only if ¢(a * b) = ¢(a) *' ¢(b).
22. Let * be a binary operation on a set S. An element e of S with the property s * e = s = e * s is an identity
elementfor * for all s E S.
�
ProofSynopsis
A good test of your understanding of a proof is your ability to give a one or two sentence synopsis of it, explaining
the idea of the proof without all the details and computations. Note that we said "sentence" and not "equation."
From now on, some of our exercise sets may contain one or two problems asking for a synopsis of a proof in the
text. It should rarely exceed three sentences. We should illustrate for you what we mean by a synopsis. Here is our
one-sentence synopsis of Theorem 3.14. Read the statement ofthe theorem now, and then our synopsis.
Representing an element of Sf as ¢(s) for some s E S, use the homomorphism property
of ¢ to carry the computation of ¢(e) *' ¢(s) back to a computation in S.
That is the kind of explanation that one mathematician might give another if asked, "How does the proof go?"
We did not make the computation or explain why we could represent an element of Sf as ¢(s). To supply every
detail would result in a completely written proof. We just gave the guts of the argument in our synopsis.
23. Give a proof synopsis ofTheorem 3. l3.
Theory
24. An identity element for a binary operation * as described by Definition 3.12 is sometimes referred to as "a
two-sided identity element." Using complete sentences, give analogous definitions for
a. a le
ft identity element eL for *, and b. a right identity element eR for *.
Theorem 3.13 shows that if atwo-sided identity element for * exists, itis unique. Is the same true for a one-sided
identity element youjust defined? If so, prove it. If not, give a counterexample (S, *) for a finite set S and find
the first place where the proof of Theorem 3.l3 breaks down.
25. Continuing the ideas of Exercise 24 can a binary structure have a left identity element eL and a right identity
element eR where eL =I- eR? If so, give an example, using an operation on a finite set S. If not, prove that it is
impossible.

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