University of Toronto Mississauga
Department of Mathematical and Computational Sciences
Groups and Symmetries
Third Edition (Revised)
Ali Mousavidehshikh
c 2018/2020 Ali Mousavidehshikh, All rights reserved.
Dedicated to my parents Zahra and Seyedalizaman,
my brother and sister Mahmoud and Maryam,
and our cats Blitz, Shireen, and Susan.
Contents
Preface
3
1 Preliminaries
6
1.1
Well Ordering Principle and Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.2
Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.3
Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
1.4
Maps and Well Defined Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
1.5
Well Defined Operations and Modular Arithmetic . . . . . . . . . . . . . . . . . . . .
20
1.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
1.7
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
2 Groups
28
2.1
Definition and Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.2
Order of a Group, Order of an Element . . . . . . . . . . . . . . . . . . . . . . . . . .
34
2.3
Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.4
Cayley Table for Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.5
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
2.6
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
3 Generators, Relations, and Cyclic Groups
48
3.1
Definition of Generators and Relations . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.2
Cyclic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
3.3
Subgroup Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
3.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
3.5
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
4 Dihedral, Symmetric, and Alternating Groups
62
4.1
Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
4.2
Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
4.3
Alternating Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
4.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
4.5
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
5 Group Homomorphisms and Isomorphisms
1
78
5.1
Homomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
5.2
Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
5.3
Automorphisms, Inner Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . .
84
5.4
Cayley’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
5.5
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
5.6
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
6 Cosets and Lagrange’s Theorem
92
6.1
Definition and Properties of Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
6.2
Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
6.3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6.4
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
7 Normal Subgroups and Factor Groups
104
7.1
Definition and Properties of Normal Subgroups . . . . . . . . . . . . . . . . . . . . . 104
7.2
Factor (Quotient) Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
7.3
Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
7.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
7.5
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
8 Direct Products and Sums
119
8.1
External Direct Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
8.2
Internal Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
8.3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
8.4
Challenging Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
2
Preface
This book was written with the intention of introducing students to the wonderful world of Algebra.
More specifically, the book introduces students to the notion of groups, and how these abstract sets
can be used to describe the movements of certain geometric objects. This book is my attempt to
make groups and their applications (mathematically speaking) lovable.
Every chapter ends with two Exercise sections. The first section contains simple to mediocre level
exercises, and the second section contains challenging questions (some open). I believe doing the
exercises is the only way to learn mathematics. Serge Lang set the following exercise on page 105 of
his book, Algebra (with a little paraphrasing):
Take any book in Algebra and prove all the theorems without looking at the proofs given in the
book.
Taken literally, the statement of the exercise is absurd. But its spirit is absolutely accurate; the
subject only appears difficult, but once broken down and understood, it is actually quite beautiful.
I would like to thank my colleagues Peter Crooks, Payman Eskandari, and Tyler Holden for reading through this book, spotting several errors and misprints, and for making useful suggestions on
improving the book. Also, I would like to thank former students for making useful suggestions on
improving the book and spotting several errors.
Please keep in mind that this book is still under development. Students, instructors, and mathematicians are encouraged to contact me with comments, suggestions, and any errors found in the
book.
Ali Mousavidehshikh
ali.mousavidehshikh@utoronto.ca
University of Toronto, Mississauga Campus
3
One of my former students (in MAT301 and also a reading course) wrote the following wonderful
piece that I think is important to share with students. The way she explains her thought process
about drawing and its beauty to her is how I feel about proofs in mathematics.
Convince you of the beauty of Drawing
I want you to a think about the feeling when you are proud of yourself for completing something,
whether it’s a math problem you recently solved or a goal you finally accomplished. Hold on to
that f eeling of accomplishment.
Now, I want you to imagine the most beautiful thing you can. It could be some sort of code with
a complex algorithm or a flower you picked off a bouquet or maybe a scenic trail you once walked
through. You know it’s beautiful but you don’t quite understand the extent of the beauty it holds
at first, but it intrigues you. It intrigues you so much that you examine every single detail. You take
your time to really look at it and look through it, to the point where you know it inside out. You
have examined it so much that there is nothing left to scrutinize, add, or question. You’ve studied
it so much that you have now found imperfections you never knew were there at first. All that’s left
to do is admire the perf ections within the imperf ections.
Now imagine, you created that beauty. You took the time to add every layer of complexity. You
have accepted the imperfections and notice the allurement of every detail. The final details are what
make it beautiful, where your admiration stems f rom.
Now, I’ll walk you through how I feel when I draw something.
Your starting point is a blank page with a final product in mind. You don’t know where to start, so
you start with the bare lines. This raw state is dull, but you know you can make it into something
beautiful, even though you erase more than you leave on the page. You cannot finish it all in one
sitting because you’ll end up rushing the piece out of frustration. This frustration leads to a distorted
final product, which is not what you envisioned. So, you take a breather.
After the first breather you notice the piece holds a different kind of beauty, this beauty is in it’s
ugliest layer. It isn’t quite where you want it to be, but it has more character. It’s hard for you to
look at the piece as a whole because it lacks the beauty of your envisioned final product. You just
want to get started on adding the final details so the bare lines lose your interest. But in order to
add the final details you need the initial structure, something to build off - the bare lines.
You’ve pushed through and now you have the structure you need to make the final product come to
life. This is when you start on the details you become lost, addicted, in your own world, and fixated
on a tiny part of the page. You have adopted a new way to look at the details, you can blur a small
area to get rid of the noise from other parts of the page. Doing this allows you to see a version of the
overall piece and not fixate on the imperfections but rather find the beauty in the imperfections, it
allows sanity in the process. You learn patience and precision in the most natural way. Adding the
details takes you, what feels like about an hour, in reality 3 hours have actually passed. You took
your time and made every part of the picture as beautiful as you possibly could.
Now, the details are finished, and you can finally look at the picture as a whole and not be embarrassed
by the bare lines. It’s at that point where you finally appreciate the process, and it’s at that point,
that you impress yourself with something you took the time to create. The best part is, you do it for
yourself, no one else, and it’s your own creation. F or me, the process in drawing is it0 s own beauty.
Nadia Battistella
4
5
1
Preliminaries
1.1
Well Ordering Principle and Divisibility
In this section, we state and prove some of the properties of the integers. We will be using the
following standard notations:
N = {1, 2, 3, . . . , }
Z = {0, ±1, ±2, . . . , }
na
o
Q =
: a ∈ Z, b ∈ N
b
R = all the real numbers
Natural numbers
Integers
Rational numbers
Real numbers
We start with an important and useful property of the natural numbers.
Well Ordering Principle
Axiom 1.1. Every non-empty set of positive integers contains a smallest member.
The Well Ordering Principle can be taken as an axiom or as a theorem. This is because the Well
Ordering Principle cannot be proved from the usual properties of arithmetic. However, those with a
strong background in set theory or analysis can argue (and rightly so) that it should be a theorem.
In set theory one can use the notion of inductive sets, and in analysis one can use the completeness
property of the real numbers to prove the Well Ordering Principle. This is one of very few occasions
in this book where we are less concerned with proving a mathematical result than with applying it.
Divisibility
Given integers m and n, we say m divides n, and write m|n, if there exist an integer u such
that n = mu. If m does not divide n, we write m - n.
If m|n, we say m is a divisor of n. We now establish a fundamental property of the integers that
will be used frequently.
Division Algorithm
Theorem 1.2. Let a, b ∈ Z with b > 0. Then there exist unique integers q, r such that
a = bq + r, where 0 ≤ r < b.
We call r the remainder and q the quotient.
Proof. Fix a, b ∈ Z with b > 0. First we prove there exist q and r such that a = bq+r, where 0 ≤ r < b.
Then we prove that q and r are unique. Let S(a, b) = {c : c ≥ 0 and c = a − bk for some k ∈ Z}.
We claim that S(a, b) 6= ∅. To see this we consider three cases for a:
Case 1. If a = 0, take k = −b,
Case 2. If a > 0, take k = 0,
Case 3. If a < 0, take k = a.
6
If 0 ∈ S(a, b), there exist k ∈ Z such that
a − bk = 0 ⇒ a = bk ⇒ b|a,
and we can take k = (a/b) and r = 0.
If 0 ∈
/ S(a, b), then a 6= 0 and S(a, b) ⊆ N. By the Well Ordering Principle, there exist a smallest
element in S(a, b), call it r. Then, r = a − bk for some k ∈ Z. It follows that a = bk + r and r > 0
(since every element of S(a, b) is positive). If r ≥ b, then
a − b(k + 1) = a − bk − b = r − b ≥ 0 ⇒ a − b(k + 1) ∈ S(a, b).
However,
a − b(k + 1) = a − bk − b |{z}
< a − bk = r,
since b>0
contradicting the minimality of r. Hence, 0 < r < b and letting q = k establishes existence.
For uniqueness, suppose bq + r = a = bq 0 + r0 with 0 ≤ r < b, 0 ≤ r0 < b. Note that r0 ≤ r or r ≤ r0 .
Without loss of generality, we assume r ≤ r0 . Then b(q − q 0 ) = r0 − r, which implies that b|(r0 − r).
Since 0 ≤ r0 − r ≤ r0 < b, we have r0 − r = 0. Hence, r0 = r, from which we get q 0 = q, completing
the proof.
For example, if a = 57 and b = 14, then 57 = 14(4) + 1. So, q = 4 and r = 1 (notice that 0 ≤ r < 14).
Greatest Common Divisor
Given a, b ∈ Z with at least one of a, b non-zero, the greatest common divisor of a and b is
the largest integer that divides both a and b, and is denoted by gcd(a, b). More precisely,
gcd(a, b) = d if and only if the following two conditions are satisfied;
(1) d|a and d|b, and
(2) If d0 |a and d0 |b, then d0 ≤ d.
We now state some of the facts about the gcd of two integers a and b (at least one of them non-zero):
(1) gcd(a, b) = gcd(b, a).
(2) gcd(a, b) ≥ 1.
(3) gcd(0, b) = |b| for any b ∈ Z − {0}.
We say two integers a and b are relatively prime if gcd(a, b) = 1. For example, note that 4 and 7
are relatively prime, i.e., gcd(4, 7) = 1. Similarly, 15 and 22 are relatively prime. However, 10 and
45 are not relatively prime (as gcd(10, 45) = 5).
The following proposition tells us the relationship between the greatest common divisor of two numbers and the two numbers themselves.
Proposition 1.3. Given a, b ∈ Z with at least one of a or b non-zero, there exist s, t ∈ Z such
that gcd(a, b) = as + bt. Moreover, gcd(a, b) is the smallest positive integer of the form as + bt.
7
Proof. Fix a, b ∈ Z with at least one of a or b non-zero. Let S(a, b) = {am+bn > 0 : m, n ∈ Z}. Note
that am + bn is non-zero for some choice of m and n. If am + bn is positive for this choice of m and
n, then S(a, b) is non-empty. If am + bn is negative for this choice of m and n, then a(−m) + b(−n)
is positive and S(a, b) is again non-empty. It follows that S(a, b) is non-empty in all cases.
By the Well Ordering Principle, S(a, b) has a smallest element, say d = as+bt. We claim d = gcd(a, b).
By the Division Algorithm, there exist unique integers q and r such that a = dq + r, where 0 ≤ r < d.
If r > 0, then
r = a − dq = a − (as + bt)q = a − asq − btq = a(1 − sq) + b(−tq),
from which it follows that r ∈ S(a, b), contradicting the minimality of d. Hence, r = 0, and d|a. A
similar proof shows that d|b.
Now suppose e|a and e|b. Then, a = el and b = ek for some l, k ∈ Z. In particular,
d = as + bt = els + ekt = e(ls + kt) ⇒ e|d ⇒ e ≤ d,
completing the proof.
For example, if a = 24 and b = 132, then gcd(24, 132) = 12. Moreover, 24(−5) + 132(1) = 12. In the
proof of the previous proposition, we have proven the following two useful corollaries.
Corollary 1.4. Suppose gcd(a, b) = d. If e|a and e|b, then e|d.
Corollary 1.5. gcd(a, b) = 1 if and only if there exist integers s and t such that as + bt = 1.
Proposition 1.3 tells us that gcd(a, b) = as + bt for some integers s and t. On the other hand,
Corollary 1.5 tells us that if we can find s and t with as + bt = 1, then gcd(a, b) = 1, in essence
giving a sort of converse of Proposition 1.3. One should be careful here. It is not true in general
that if as + bt = n for n > 1, then n = gcd(a, b). Take a = 2 and b = 4, then 2(3) + 4(5) = 26, but
gcd(2, 4) 6= 26. So why is this the case when n = 1? Answer: In this case, note that 1 is the smallest
positive integer of the form as + bt, and it then follows by Proposition 1.3 that gcd(a, b) = 1.
Recall that a prime number p is a natural number greater than 1 whose only divisors are ±1 and ±p.
We are now in position to prove one of the most important properties of prime numbers.
Corollary 1.6. Given integers a and b, if p is a prime number and p|(ab), then p|a or p|b.
Proof. Suppose p divides ab, i.e., ab = pn for some n ∈ Z. If p|a, there is nothing to prove. If p - a,
then gcd(p, a) = 1 (since p is prime). By Corollary 1.5, there exist integers s and t such that
1 = as + pt ⇒ b = asb + ptb = pns + ptb = p(ns + tb) ⇒ p | b,
completing the proof.
The fact that p is prime in the previous corollary is very important. For example 6|(9 × 4), however,
6 does not divide 9 nor does it divide 4. The previous corollary says this cannot happen for a prime
number.
8
Least Common Multiple
The least common multiple of two non-zero integers a and b is the smallest positive integer
that is a multiple of both a and b, denoted by lcm(a, b).
For example, lcm(6, 15) = 30 and lcm(12, 30) = 60.
1.2
Induction
In this section, we state two forms of proof that are equivalent to the Well Ordering Principle. In
fact, the following are equivalent;
(a) Well Ordering Principle,
(b) Mathematical Induction (ordinary induction),
(c) Mathematical Induction (strong induction),
(d) Axiom of Choice,
(e) Zorn’s Lemma.
The Axiom of Choice and Zorn’s Lemma are some other fundamental axioms that are often used in
mathematics but are outside the scope of this course.
A mathematical statement is a sentence that can be either true or false (not both). For example, “if
x is a real number, then x2 ≥ 0” is a mathematical statement. However, “2+3” is not a mathematical
statement.
Principle of Mathematical Induction (Ordinary Induction)
Theorem 1.7. Let a ∈ Z and P (a), P (a + 1), P (a + 2), . . . be an infinite sequence of mathematical statements. Suppose this sequence satisfies the following two properties:
(1) P (a) is true, and
(2) for each integer k ≥ a, P (k) ⇒ P (k + 1).
Then, P (n) is true for all integers n ≥ a.
The fact that this is equivalent to the Well Ordering Principle is left as an exercise (see Exercise
1.29). In Step (1) of Mathematical Induction is called the base case. In step (2), P (k) is called the
induction hypothesis. We now give examples of proofs which are done via ordinary induction.
Example 1.8. Prove that
n
X
i=1
i=
n(n + 1)
for all natural numbers n.
2
9
Solution. For each natural number n, we can think of P (n) as being the statement
n
X
i=
i=1
n(n + 1)
.
2
We use mathematical induction to prove P (n) is true for all natural numbers n. Since the natural
numbers start at 1, we first show that P (1) is true. This follows from the fact that
1
X
i=1=
i=1
1(1 + 1)
.
2
Next is our induction hypothesis: let k be a natural number and suppose P (k) is true. Then,
k+1
X
i=
i=1
k
X
i + (k + 1) =
i=1
k(k + 1)
k(k + 1) + 2(k + 1)
(k + 1)(k + 2)
+ (k + 1) =
=
.
2
2
2
In particular, P (k + 1) is true. By Mathematical Induction, P (n) is true for all natural numbers n.♣
When P (n) is clear from content, we will usually omit it.
Example 1.9. Prove that
n
X
i=0
i2 =
n(n + 1)(2n + 1)
for all non-negative integers.
6
Solution. The base case follows from the fact that
0
X
i=0
i2 = 0 =
0(0 + 0)(2(0) + 1)
.
6
Assume the result is true for n = k, where k ∈ N. Then,
k+1
X
i=0
2
i =
k
X
i2 + (k + 1)2 =
i=0
=
=
=
=
=
k(k + 1)(2k + 1)
+ (k + 1)2
6
by the induction hypothesis
k(k + 1)(2k + 1) + 6(k + 1)2
6
(k + 1)(k(2k + 1) + 6(k + 1))
6
2
(k + 1)(2k + 7k + 6)
6
(k + 1)(k + 2)(2k + 3)
6
(k + 1)((k + 1) + 1)(2(k + 1) + 1)
.
6
So the result is true for n = k + 1, and by Mathematical Induction,
n
X
i=1
non-negative integers.
10
i2 =
n(n + 1)(2n + 1)
for all
6
♣
Notice that (n + 1)2 ≥ 2n for 1 ≤ n ≤ 5. However, (n + 1)2 = n2 + 2n + 1 is a polynomial and 2n is
an exponential function, and exponential functions eventually dominate polynomials (calculus).
Example 1.10. Prove that
(n + 1)2 < 2n for all natural numbers n ≥ 6.
Solution. The base case, which is n = 6, follows from the fact that 49 < 64. Assume the result is
true for n = k, where k is natural number greater than or equal to 6. Then,
2k+1 = 2(2k ) > 2(k + 1)2 = 2k 2 + 4k + 2 = k 2 + 4k + 4 + k 2 − 2 > k 2 + 4k + 4 = (k + 2)2 .
♣
The result follows by Mathematical Induction.
We now give another version of the Principle of Mathematical Induction, most commonly called
Strong Induction.
Principle of Mathematical Induction (Strong Induction)
Theorem 1.11. Let a ∈ Z and P (a), P (a + 1), P (a + 2), . . . be an infinite sequence of mathematical statements. Suppose this sequence satisfies the following two properties:
(1) P (a) is true, and
(2) for each k ≥ a, (P (a) ∧ P (a + 1) ∧ . . . ∧ P (k)) ⇒ P (k + 1).
Then, P (n) is true for all integers n ≥ a.
Again, step (1) is called the base case. The difference between induction and strong induction is
in the induction hypothesis. The induction hypothesis in strong induction is as follows: assume
P (a), P (a + 1), . . . , P (k) are true. This is where strong induction gets its name from. The induction
hypothesis in strong induction is much stronger than induction. After all, in induction we only
assume P (k) is true. In strong induction we assume that P (a), P (a + 1), . . . P (k) are all true.
Example 1.12. Prove that every natural number n can be written in the form 2a b where a is
a non-negative integer and b is a positive odd integer.
Solution. We prove this by strong induction. The base case follows from the fact that 1 = 20 (1)
(that is, take a = 0 and b = 1). Assume the result is true for n = 1, 2, . . . , k. If k + 1 is odd, then
take a = 0 and b = k + 1 (for this part we do not need induction nor strong induction). If k + 1 is
even, then k + 1 = 2m for some integer 1 ≤ m ≤ k. By the induction hypothesis, m = 2c d for some
non-negative integer c and positive odd integer d. Then,
k + 1 = 2m = 2(2c d) = 2c+1 d.
So we can take a = c + 1 and b = d. That is, the result holds for n = k + 1, and by Strong Induction
the result holds for all natural numbers.
♣
To see the necessity for strong induction (rather than induction) in the previous example, one should
notice that m is some integer between 1 and k (inclusive). But, we have no idea where m is. More
11
specifically, if we only used induction, the hypothesis would only tell us the result holds for k.
However, m is almost never equal to k. In fact, m = k gives k + 1 = 2k, which is equivalent to k = 1.
So our induction hypothesis would be useless for every k 6= 1.
We now prove one of the most important theorems in arithmetic.
Fundamental Theorem of Arithmetic (Part 1)
Theorem 1.13. Every natural n ≥ 2 is a prime or product of primes.
Proof. The proof is by strong induction on n. Our base case is n = 2, which is prime. So the base
case holds. Assume the result holds for n = 2, 3, . . . , k. If k + 1 is prime, the result holds. If k + 1
is not prime, then k + 1 = ab for some natural numbers a and b with 2 ≤ a ≤ k and 2 ≤ b ≤ k. By
induction, a and b can both be written as product of primes, say a = p1 p2 . . . pm and b = q1 q2 . . . qr .
So k + 1 = ab = p1 p2 . . . pm q1 q2 . . . qr , which is a product of primes. By strong induction, the result
holds for all natural numbers greater than or equal to 2.
Notice that the Fundamental Theorem of Arithmetic tells us that if n is a natural number greater
than or equal to 2, then n has a prime factor. Of course if n is prime, it has only one prime factor,
namely, itself.
One of the consequences of the Fundamental Theorem of Arithmetic is that there are infinitely many
primes, a very famous theorem proved by Euclid.
Infinitely Many Primes
Theorem 1.14. There are infinitely many prime numbers.
Proof. Suppose there are finitely many primes. Then we can enumerate them in ascending
order.
!
n
Y
Let p1 < p2 < . . . < pn be the list of all primes in ascending order. Set a =
pi + 1. Notice
i=1
!
n
Y
that a > pn . Moreover, 1 = a −
pi . If pi | a for some i, then pi | 1 (see Exercise 1.2), a
i=1
contradiction. In particular, a is not divisible by any prime number, contradicting the Fundamental
Theorem of Arithmetic.
By the Fundamental Theorem of Arithmetic, every natural number n ≥ 2 can be factored into primes.
The natural question is: Is the factorization unique? For example,
105 = 3 × 5 × 7 = 3 × 7 × 5 = 7 × 3 × 5 = 7 × 5 × 3 = 5 × 3 × 7 = 5 × 7 × 3.
If we don’t care about the order in which the primes are written, 105 has one prime factorization.
That is, the prime factorization is unique up to reordering the primes. The following theorem tells
us up to reordering the prime factors, there is only one prime factorization for all natural numbers
n ≥ 2. That is, once we have factored a number into primes, the primes used never changes in any
factorization of that number, and the number of times each prime is used never changes. The only
thing that can change is the order in which we write the primes.
12
Fundamental Theorem of Arithmetic (Part 2)
Theorem 1.15. The product in Theorem 1.13 is unique up to reordering.
Proof. The proof is left as an exercise (see Exercise 1.9).
Corollary 1.16. Given non-zero integers a, b, and c, prove that gcd(a, bc) = 1 if and only if
gcd(a, b) = 1 = gcd(a, c).
Proof. Let d = gcd(a, b), e = gcd(a, c), f = gcd(a, bc). Suppose f = 1. Since d | b, we have d|(bc).
In particular, d|f (by Corollary 1.4). Thus d = 1. A similar proof shows that e = 1.
Conversely, suppose d = e = 1 and f > 1. By Theorem 1.13, f = p1 p2 ...pk (pi -prime). Since f | a
and f | bc we have p1 | a and p1 | bc. Thus p1 | b or p1 | c (by Corollary 1.6). If p1 | b, then p1 | d (by
Corollary 1.4), consequently, p1 |1, a contradiction. If p1 | c, a similar argument gives a contradiction.
Hence, f = 1.
There is a very nice relationship between the least common multiple of two numbers and their greatest
common divisor.
Corollary 1.17. Given two non-zero integers a and b, prove that
lcm(a, b) =
ab
.
gcd(a, b)
Proof. The proof is left as an exercise (see Exercise 1.13).
1.3
Equivalence Relations
In high school we are taught about similar triangles. More specifically, two triangle are similar if
their corresponding angles are congruent and the corresponding sides are in proportion. In other
words, similar triangles are the same shape, but not necessarily the same size. The notion of similar
triangles extends (or generalizes) the notion of equality on triangles. Of course, equal triangles are
similar, but not vice versa.
In general, the notion of equality in mathematics is extremely strict. When we say two things are
equal, we mean they are the same in every possible way (mathematically speaking). In this chapter
we introduce a notion, called an equivalence relation, which aims to generalize the notion of equality.
Or if you like, loosen the definition of equality. For example, we can think of similar triangles as
being equivalent.
We begin with the formal definition of an equivalence relation.
13
Equivalence Relation
An equivalence relation on a set S is a set R ⊆ S × S satisfying the following three properties:
1. (a, a) ∈ R for all a ∈ S (reflexive).
2. If (a, b) ∈ R, then (b, a) ∈ R (symmetric).
3. If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R (transitive).
If R is an equivalence relation on S, the following all mean the same thing: (a, b) ∈ R, a ∼ b,
a ≡ b. Moreover, when R is an equivalence relation and (a, b) ∈ R, we say a and b are
equivalent.
Notice that an equivalence relation R on a set S is a subset of S × S. In practice, it is helpful to
think of (a, b) ∈ R as meaning that “a is related to b”. One can think of the reflexive property as
saying: every element in our set is related to itself. The symmetric property says that if a is related
to b, then b is related to a. The transitive property is telling us that if a is related to b and b is
related to c, then a is related to c.
Example 1.18. Which one of the following relations is an equivalence relation?
(a) Define the following relation on the integers: a ∼ b ⇔ a < b.
(b) Let S be the set of all polynomials with real coefficients. Define the following relation on
S: f ∼ g if and only if f 0 = g 0 (the first derivative).
(c) Let S = Z and n ∈ N with n ≥ 2. Define a ≡ b (mod n) if and only if n|(b − a). Another
notation frequently used for a ≡ b (mod n) is a mod n = b mod n.
Solution. (a) This is not an equivalence relation (reflexivity and symmetry both fail, but transitivity
holds).
(b) This is an equivalence relation. The reflexive and symmetric property are straightforward. If
f ∼ g and g ∼ h, then f 0 = g 0 = h0 . Hence, the relation is also transitive.
(c) This is an equivalence relation. Fix an integer n ≥ 2. For any integer a, n|(a − a), so a ≡ a
(mod n).
If a ≡ b (mod n), then n|(a − b), which in turn implies n|(b − a). Thus b ≡ a (mod n).
If a ≡ b (mod n) and b ≡ c (mod n), then n|(a − b) and n|(b − c). In particular, there exist integers
m and k such that a − b = nm and b − c = nk. So a − c = (a − b) + (b − c) = n(m + k). That is,
n|(a − c) and a ≡ c (mod n).
♣
Equivalence Classes
When ∼ is an equivalence relation on a set S, for any a ∈ S, [a] = {x ∈ S : x ∼ a} is called
the equivalence class of a in S, and a is called a representative for [a].
Notice that the equivalence class of a consists of all elements which are equivalent to a under the
relation ∼.
14
Example 1.19. In Example 1.18 we showed that parts (b) and (c) defined an equivalence
relation. For part (b), find the equivalence class of each polynomial. For part (c), if n = 5,
find the equivalence classes of [1] and [2].
Solution. For part (b), the equivalence class of a polynomial f is [f ] = {f + c : c a constant} (as two
functions Rhave the same derivative if and only if they differ by a constant). For those familiar with
calculus, f 0 = [f ]. For part (c), if we let n = 5, then
[1] = {m ∈ Z : m ≡ 1(mod 5)} =
=
=
=
{m ∈ Z : 5|(m − 1)}
{m ∈ Z : m − 1 = 5k, k ∈ Z}
{m ∈ Z : m = 5k + 1, k ∈ Z}
{5k + 1 : k ∈ Z}
Similarly, [2] = {5k + 2 : k ∈ Z}.
♣
Partition
A partition of a set S is a collection of non-empty disjoint subsets of S whose union is S.
The set {[0, ∞), (−∞, 0)} is a partition for R. However, the set {[0, ∞), (−∞, 0]} is not a partition
for R (both sets contain zero, thus they are not disjoint).
Many students are familiar with the union (or intersection) of two sets. We now extend these notions
to arbitrary unions (or intersections) of sets.
Arbitrary Union and Intersection
Let I be a set (finite or infinite), and suppose Ai is a set (finite or infinite) for each i ∈ I. We
define the arbitrary union and intersection as follows:
[
Ai = {x : x ∈ Ai for some i ∈ I} (union),
i∈I
\
Ai = {x : x ∈ Ai for all i ∈ I} (intersection).
i∈I
The set I is called an index set.
When the set I in the previous definition is the set of the natural numbers, we use the following
notation:
[
i∈I
Ai =
∞
[
i=1
Ai and
\
i∈I
Ai =
∞
\
Ai .
i=1
Proposition 1.20. The equivalence classes of an equivalence relation on a set S constitute
a partition of S. Conversely, for any partition P of S, there is an equivalence relation on S
whose equivalence classes are the elements of P .
15
Proof. Let ∼ be an equivalence relation on S. For any a ∈ S, a ∈ [a] (by reflexivity). In particular,
[
S=
[a].
a∈S
We claim that for all a, b ∈ S, either [a] = [b] or [a] ∩ [b] = ∅. Suppose [a] ∩ [b] 6= ∅. We show that
[a] = [b]. Since [a] ∩ [b] is not empty, there exist a c ∈ [a] ∩ [b]. In particular, a ∼ c and c ∼ b. By
transitivity, a ∼ b. If x ∈ [a], then x ∼ a and a ∼ b, which gives x ∼ b, thus x ∈ [b]. Hence, [a] ⊆ [b].
Similarly, if x ∈ [b], then x ∼ b and b ∼ a, which gives x ∼ a, thus x ∈ [a]. Hence, [b] ⊆ [a].
Now we show that any partition P of a set S corresponds to an equivalence relation on S whose
equivalence classes are the elements of P . Suppose P = {Pi : i ∈ I}, where I is an index set. Define
a ∼ b if and only if there exist a j ∈ I such that a, b ∈ Pj (notice that each element is exactly in one
Pi ). A routine check shows that this defines an equivalence relation on S and by construction the
equivalence classes are precisely the elements of P .
It should not come as any surprise that finite sets are much easier to work with then infinite sets.
We now give an example that shows that equivalence relations can be used to go from an infinite set
to a finite set.
Example 1.21. Fix an integer n ≥ 2. Define Zn = {[a]| a ∈ Z}, where [a] is the equivalence
class of a under the relation in Example 1.18(c). Prove that Zn = {[0], [1], ..., [n − 1]}.
Solution. Notice that {[0], [1], ..., [n − 1]} ⊆ Zn . If [a] ∈ Zn , by the Division Algorithm there exist
unique q and r such that a = nq + r with 0 ≤ r < n. Therefore, n|(a − r) ⇒ a ≡ r (mod n) ⇒
[a] = [r] (by Proposition 1.20). We have established the inclusion Zn ⊆ {[0], [1], . . . , [n]}. Hence,
Zn = {[0], [1], ..., [n − 1]}. The proof of the fact that these equivalence classes are distinct is left to
the reader.
♣
1.4
Maps and Well Defined Functions
A map f from a set√A to a set B is a rule that assigns to each element a ∈ A element(s) in B. For
example, f (x) = ± x is a map from the non-negative real numbers to the real numbers;
√
f (0) = 0, f (1) = ±1, f (2) = ± 2, f (4) = ±2, f (25) = ±5.
Well Defined Functions
A well defined function from a set A to a set B is a map that assigns to each element a ∈ A a
unique element f (a) ∈ B, called the image of a under f . When a map is a well defined function
f
from A to B, we write f : A → B or A → B, and simply say f is a function.
One should notice that the only difference between a map and a function between two sets is that a
map can have multiple outputs for a given input. However, a function
has only one output for each
√
input. So all functions√are maps but not vice versa; f (x) = ± x is a map but not a function, and
we say that f (x) = ± x is not well-defined.
All of the following are functions from R to R:
16
(a) f (x) = 2,
(b) f (x) = x + 4,
(c) f (x) = x2 + 4x + 3.
There is another useful notation used to illustrate the rule of a function. If f : A → B is a function
and f (a) = b, we write a 7→ b. This sometimes allows us to describe the rule of the function. For
example, if f : R → R is the function f (x) = x2 , we describe the rule of this function as follows: let
f : R → R be defined via x 7→ x2 .
Domain/Co-domain/Image of A Function
If f : A → B is a function, the set A is called the domain of f and B is called the co-domain of
f . The image of a function, denoted by f (A) or Im(f ), is the set of all points that are actually
realized as outputs of f . That is,
f (A) = {f (a) : a ∈ A}.
If C ⊆ A, we define f (C) = {f (x) : x ∈ C}.
Example 1.22. Find the image of the following functions.
(a) f : R → R defined via x 7→ x2 .
(b) g : N → N defined via n 7→ n + 1.
Solution. (a) Since the square of every real number is non-negative, f (R) ⊆ [0, ∞). Conversely,
√
√
√
given y ∈ [0, ∞), y ∈ R and y = ( y)2 = f ( y) ∈ f (R). That is, [0, ∞) ⊆ f (R). Hence, the image
of f is [0, ∞).
(b) A similar argument to the one given in part (a) shows that f (N) = {2, 3, 4, . . .}.
Example 1.23. Let f : N×N → N be defined by f (a, b) =
♣.
a(a + 1)b
. Find f ({1, 2, 4}×{3, 5}).
2
Solution. Notice that {1, 2, 4} × {3, 5} = {(1, 3), (1, 5), (2, 3), (2, 5), (4, 3), (4, 5)}. Moreover,
f (1, 3) = 3, f (1, 5) = 5, f (2, 3) = 9, f (2, 5) = 15, f (4, 3) = 30, f (4, 5) = 50.
Hence, f ({1, 2, 4} × {3, 5}) = {3, 5, 9, 15, 30, 50}.
♣
One should notice that a function comes with three pieces of data; its domain, co-domain, and rule.
So for two functions to be equal, we need all these things to be the same. We make this into a formal
definition.
Equality of Functions
Let f : A → B and g : C → D be functions. We say that f and g are equal if and only if
A = C, B = D, and f (a) = g(a) for all a ∈ A. When two functions f and g are equal, we
write f = g.
17
In our definition, A = C tells us that the domain of the two functions are equal. Similarly, B = D
tells us that the co-domains are equal. The fact that f (a) = g(a) for all a ∈ A tells us that the rule
of the two functions are the same.
Consider the functions f : R → R defined via x 7→ x2 and g : R → [0, ∞) defined via x 7→ x2 . It is
very tempting to say that these two functions are equal. While it is true that these two functions
have the same domain and rule, they do not have the same co-domain. Hence, they are not equal.
Composition of Functions
Let f : A → B and g : B → C. The composition g ◦ f is the function from A to C defined
by (g ◦ f )(a) = g(f (a)). When there is no confusion with function multiplication, we write gf
instead of g ◦ f .
For example, if f, g : R → R via f (x) = x2 and g(x) = x + 1, then both f g and gf are defined.
Moreover,
(f g)(x) = f (g(x)) = f (x + 1) = (x + 1)2 = x2 + 2x + 1 and (gf )(x) = g(f (x)) = g(x2 ) = x2 + 1.
This shows that f g does not have to equal gf (as (f g)(1) = 4, but (gf )(1) = 2).
One should be very careful with composition. The composition gf is defined if and only
co-domain of f is equal to the domain of g, and the composition f g is defined if and only
co-domain of g is equal to the domain of f . Given two functions f and g, it is possible for one
compositions to be defined but not the other. For example, if f : R → Z is defined via x 7→
g : R → R is defined via x 7→ 1, then f g is defined, but gf is not defined.
if the
if the
of the
1 and
Notice that function composition is associative; that is, if f : A → B, g : B → C, and h : C → D,
then
h(gf ) = (hg)f.
The proof of this fact is left to the reader.
Injection/Surjection/Bijection
Let f : A → B be a function.
(a) f is called one-to-one (or injective) if and only if,
for any x, y ∈ A, f (x) = f (y) ⇒ x = y.
(b) f is called surjective (or onto) if and only if, for all b ∈ B, there exist a ∈ A such that
f (a) = b.
(c) f is called a bijection if and only if it is one-to-one and surjective. In this case, define
g : B → A via
f (a) = b ⇔ g(b) = a.
The function g is called the inverse of f , and is denoted by f −1 .
18
The fact that the map g in part (c) of the previous definition is a function is a direct consequence of
f being a bijection. In fact, a function is a bijection if and only if g = f −1 is a function (the proof is
left to the reader).
Proposition 1.24. Let f : A → B and g : B → C.
(a) If f and g are injective, prove that their composition gf is injective.
(b) If f and g are surjective, prove that their composition gf is surjective.
(c) If f and g are bijections, prove that their composition gf is a bijection, and also that
(gf )−1 = f −1 g −1 .
Proof. Notice that gf : A → C.
(a) If (gf )(a) = (gf )(b) where a, b ∈ A, then g(f (a)) = g(f (b)). Since g is injective, f (a) = f (b),
and injectivity of f implies that a = b. Hence, gf is injective.
(b) Given c ∈ C, there exist b ∈ B such that g(b) = c (as g is surjective). Since f is surjective,
there exist an a ∈ A such that f (a) = b. In particular, (gf )(a) = g(f (a)) = g(b) = c. Hence, gf is
surjective.
(c) If f and g are bijections, then they are both injective and surjective. By parts (a) and (b), gf is
both injective and surjective, and thus a bijection.
Now we prove the second part of the statement. First notice that (gf )−1 : C → A and f −1 g −1 :
C → A. So the domains and co-domains of the two functions are equal. Given c ∈ C, we show that
(gf )−1 (c) = (f −1 g −1 )(c). Let (gf )−1 (c) = a. It follows that c = (gf )(a) = g(f (a)). Let f (a) = b.
Then g(b) = c and
(f −1 g −1 )(c) =
=
=
=
f −1 (g −1 (c))
f −1 (b)
a
(gf )−1 (c).
since g(b) = c
since f (a) = b
Hence, f −1 g −1 = (gf )−1 .
The previous Theorem says that the composition of two injective functions is injective, the composition of two surjective functions is surjective, and the composition of two bijections is a bijection.
Example 1.25. Define f : N → N via n 7→ n + 1. Is f injective? Is f surjective? Is f a
bijection?
Solution. Given natural numbers n and m, if f (n) = f (m), then
n + 1 = m + 1 ⇒ n = m.
Thus f is injective. Furthermore, Im(f ) = {2, 3, 4, . . .} 6= N. Hence, f is not surjective. It follows
that f is not a bijection.
♣
19
1.5
Well Defined Operations and Modular Arithmetic
Operation on a Set
Given a set S, we say · is an operation on S if there exist a set A such that a · b ∈ A for any
a, b ∈ S. Equivalently, · : S × S → A is a map that sends (a, b) to a · b.
For example, addition is an operation on the integers (in fact, on all the real numbers). Using the
notation in the preceding definition, S = Z = A. Given any two integers a and b, a + b ∈ Z. Another
example is the operation of division on the natural numbers. In this case, S = N, and we can take
A = Q (one can also take A to be the set of the positive rational numbers).
For most students, it is taken for granted that an operation on a given set is well defined. For
example, if somebody asks you to compute 2 + 3 (where + is understood to be normal addition),
most people (if not all) would answer 5, and they would be correct. The whole idea here is that there
is no ambiguity about the numbers 2 and 3, nor about the operation of addition on the integers. So
the answer is obvious to most people. We say such an operation is well defined (shortly we will define
this more precisely).
Now one might ask; “when is this not the case?” Let S = {A, B}, where A = {1, 3, 5} and B =
{2, 4, 6}. Define the operation · on S as follows;
A · B = a + b where a ∈ A, b ∈ B, and a + b is normal addition.
As it stands, we are allowed to pick any element in a ∈ A and any element in b ∈ B and A · B = a + b.
But if this is the case, then A · B has multiple answers. For example, we can take a = 1 and b = 2,
which gives A · B = 3. However, we can also take a = 1 and b = 4, which gives A · B = 5. This
shows that the way the operation is defined, A · B has multiple answers. When this occurs, we say
the operation · is not well defined.
One might argue and say we should put more restrictions on how we pick the element a ∈ A and the
element b ∈ B. For example, suppose we alter the definition of · as follows;
A · B = a + b where a is the first element in A and b is the first element in B.
While this is now more restrictive than before, the problem that arises now is that the same set can
be written in many different ways (as long as it has more than one element). In our example,
A · B = {1, 3, 5} · {2, 4, 6} = 1 + 2 = 3,
A · B = {3, 1, 5} · {2, 4, 6} = 3 + 2 = 5.
While the second definition makes more precise how we pick the elements from each set, what it does
not take into account is the fact that an element in S can be written in more than one way.
From this example we see that for an operation to be well defined on a set, the operation cannot be
vague in the way it is defined, and must take into account the possible distinct representations of the
elements in the set that it is being defined on. This leads us to the following definition.
Definition 1: Well Defined Operations
An operation · on a set S is said to be well defined if it satisfies the following property;
If x, w, y, z ∈ S with x = y and w = z, then x · w = y · z.
20
One can also think of an operation being well defined in terms of functions. For example, the
operation of addition on the integers being well defined is equivalent to the map f : Z × Z → Z given
by (a, b) 7→ a + b being a function (that is, each input gives a unique output).
Definition 2: Well Defined Operations
Suppose · is an operation on a set S such that there exist a set A with the property that
a · b ∈ A for any a, b ∈ S. We say that · is well defined if and only if the map f : S × S → A
given by (a, b) 7→ a · b is a function.
We now turn our attention to modular arithmetic. Given an integer n ≥ 2, let ≡ be the relation on
the integers defined in Example 1.18(c). As shown in that example, this is an equivalence relation.
The equivalence class of an element a ∈ Z is
[a] = {b ∈ Z : a ≡ b
(mod n)}
We say [a] is the equivalence class of a mod n. Since equivalence classes partition a set, [a] = [b] if
and only if a ≡ b (mod n), and this in turn is equivalent to a − b = nk for some integer k.
We first define operations on Zn (n ≥ 2), and show that they are well defined.
Proposition 1.26. For a fixed integer n ≥ 2, the two operations
[a] + [b] = [a + b]
[a] · [b] = [ab]
addition,
multiplication,
on Zn are well defined.
Proof. Suppose [a] = [c] and [b] = [d]. Since [a] = [c] and [b] = [d], there exist integers u and v such
that a − c = nu and b − d = nv. Then,
(a + b) − (c + d) = (a − c) + (b − d) = nu + nv = n(u + v), and
(ab) − (cd) = ab − bc + bc − cd = b(a − c) + c(b − d) = bnu + cnv = n(bu + cv).
This shows that [a] + [b] = [a + b] = [c + d] = [c] + [d], [a] · [b] = [ab] = [cd] = [c] · [d], and the result
follows.
The previous proposition generalizes from the case of two element [a] and [b] to the case of finitely
many [a1 ], [a2 ], . . . , [ak ] (induction).
We usually write [a][b] instead of [a] · [b]. These operations are called addition and multiplication
mod n, respectively. Notice that the proof of Proposition 1.26 shows that if a ≡ c (mod n) and b ≡ d
(mod n), then a + b ≡ c + d (mod n) and ab ≡ cd (mod n). This is a very powerful property of
modular arithmetic.
21
Example 1.27. Compute the following in Z8 .
(a) [4] + [7].
(b) [4] · [7].
(c) [2]1000 .
(d) [3]501 .
(e) 25([5]) = [5] + [5] + . . . + [5].
|
{z
}
25 terms
Solution. By Example 1.21, Z8 = {[0], [1], [2], [3], [4], [5], [6], [7]}.
(a) [4] + [7] = [11] = [3] (since 11 ≡ 3 (mod 8)).
(b) [4] · [7] = [28] = [4] (since 28 ≡ 4 (mod 8)).
(c) Observe that 23 ≡ 8 mod 8 = 0 mod 8. So [2]3 = [0], and
[2]1000 = [2]999 [2] = ([2]3 )333 [2] = [0][2] = [0].
(d) Observe that [3]2 = [9] = [1]. Hence,
[3]501 = [3]500 [3] = ([3]2 )250 [3] = [1][3] = [3].
(e)
25([5]) = 24([5]) + [5] = 12([5] + [5]) + [5] = 12([2]) + [5] =
=
=
=
3([2] + [2] + [2] + [2]) + [5]
3([0]) + [5]
[0] + [5]
[5].
♣
Example 1.28. Define f : Z2 → Z2 via [a] 7→ [a]2 + [a] and g : Z2 7→ Z2 via [a] 7→ [0]. Prove
that f = g.
Solution. It suffice to show that f ([a]) = g([a]) for all [a] ∈ Z2 (since the domains and co-domains
of the two functions are equal). Observe that
f ([0]) = [0]2 + [0] = [0] = g([0]), and f ([1]) = [1]2 + [1] = [2] = [0] = g([1]).
♣
Hence, f = g.
22
1.6
Exercises
Exercise
1.1. Let a, b ∈ Z (with at least one of a, b non-zero) with d = gcd(a, b). Prove that
a b
gcd
,
= 1.
d d
Exercise 1.2. Suppose a|b and a|c. Prove that a|(mb + nc) for any integers m and n.
Exercise 1.3. Prove or disprove. Let a, b, c ∈ Z. If gcd(a, b) = 1 and c|(ab), then c|a or c|b.
Exercise 1.4. (a) Let a, b, c ∈ Z. If a|c, b|c, and gcd(a, b) = 1, prove that (ab)|c.
(b) Show, by example, that part (a) is false if one removes the assumption gcd(a, b) = 1.
Exercise 1.5. Suppose a, b ∈ Z with at least one of them non-zero. Prove that gcd(a − kb, b) =
gcd(a, b) for any integer k.
Exercise 1.6. (Euclidean Algorithm) Let a, b ∈ Z such that 0 < b ≤ a. By the Division Algorithm
there exist unique integers q and r1 such that a = bq + r1 with 0 ≤ r1 < b. Now apply the Division
Algorithm to b and r1 to obtain integers q2 and r2 such that b = r1 q2 +r2 with 0 ≤ r2 < r1 . Continuing
in this fashion we obtain the following:
0 ≤ r1 < b
0 ≤ r2 < r1
0 ≤ r3 < r2
a = bq + r1
b = r1 q2 + r2
r1 = r2 q3 + r3
..
.
This gives us the following sequence of non-negative integers:
b > r1 > r2 > r3 > . . . .
(1.1)
(a) Prove that the sequence in (1.1) must stop at zero. That is, there exist a natural number k such
that rk = 0.
(b) If k = 1, prove that gcd(a, b) = b. If k > 1, prove that gcd(a, b) = rk−1 . Hint. Use Exercise 1.5.
Exercise 1.7. (a) Let a = 228 and b = 58. Use the Euclidean Algorithm to compute gcd(a, b).
Moreover, find integers u and v such that au + bv = gcd(a, b). Hint. Look closely at the numbers
you get in the Euclidean Algorithm.
(b) Let a = 124256 and b = 1210. Use the Euclidean Algorithm to compute gcd(a, b). Moreover, find
integers u and v such that au + bv = gcd(a, b).
(c) Let a = 560 and b = 10240. Use the Euclidean Algorithm to compute gcd(a, b). Moreover, find
integers u and v such that au + bv = gcd(a, b).
Exercise 1.8. Use induction to prove the following identities for all natural numbers n.
(a) Prove that
n
X
i=1
(b) Prove that
n
X
i=1
(2i − 1)2 =
4n3 − n
.
3
1
n
=
.
i(i + 1)
n+1
23
(c) Prove that
n
X
i ≤ n2 .
i=1
(d) Prove that
√
n
X
√
1
√ ≤ 2 n.
n≤
i
i=1
(e) Let 0 < a < 1. Prove that
√
n
X
√
1
√ ≤ 2 n.
n≤
i
i=1
(f) Prove that n3 + 2n is divisible by 3.
(g) Prove that 42n − 1 is divisible by 5.
Exercise 1.9. Prove Theorem 1.15. Hint. Use strong induction.
Exercise 1.10. Let a ∈ Z and P (a), P (a + 1), P (a + 2), . . . be an infinite sequence of mathematical
statements. Use Mathematical induction to prove the following (these are variations of mathematical
induction).
(a) Suppose m is an even integer greater than or equal to a,
(1) P (m) is true, and
(2) for each k ≥ m, P (k) ⇒ P (k + 2).
Prove that P (n) is true for every even integer n ≥ m.
(b) Suppose m is an odd integer greater than or equal to a,
(1) P (m) is true, and
(2) for each k ≥ m, P (k) ⇒ P (k + 2).
Prove that P (n) is true for every odd integer n ≥ m.
(c) Suppose m is an odd integer greater than or equal to a,
(1) P (m) and P (m + 1) are true, and
(2) for each k ≥ m, P (k) ⇒ P (k + 2).
Prove that P (n) is true for every integer n ≥ m.
Exercise 1.11. (a) Prove that 23n−1 + 5 · 3n is divisible by 11 for all even natural numbers.
(b) Is part (a) true if we replace even by odd? Explain.
Exercise 1.12. For each integer n ≥ 2, prove that there are n consecutive numbers none of which
are prime (that is, they are all composite).
Exercise 1.13. Prove Corollary 1.17. Hint. Use the Fundamental Theorem of Arithmetic.
Exercise 1.14. Define the following relation on the integers:
a ∼ b ⇔ a − b is even.
Prove that this is an equivalence relation on the integers. List all the distinct equivalence classes.
24
Exercise 1.15. Define the following relation on the real numbers:
a ∼ b ⇔ |a| = |b|.
Prove that this is an equivalence relation on the real numbers. Find the equivalence classes of [0], [1],
[π], [−1], [−π].
Exercise 1.16. Define the following relation on Z − {0}:
a ∼ b ⇔ gcd(a, b) = 1.
Determine whether this relation is reflexive? symmetric? transitive? Is this an equivalence relation?
Justify your answer.
Exercise 1.17. Let S be a non-empty set, A be the set of all functions from S to S, and B be the
set of all bijective functions from S to S. Define the following relation on A:
f ∼ g ⇔ ∃ h ∈ B such thatf h = g.
Is this an equivalence relation on A? Justify your answer.
Exercise 1.18. Suppose A and B are finite sets, and A ⊆ B. Let |A| and |B| be the number of
elements in A and B, respectively. Prove that if |A| = |B|, then A = B.
Exercise 1.19. (a) Let A and B be two finite sets with |A| = |B|, and f : A → B a function. Prove
that f injective if and only if surjective.
(b) Give an example showing that part (a) is false if we remove the hypothesis that A and B are
finite sets.
Exercise 1.20. Let A be the set of all finite non-empty subsets of the natural numbers (this set is
countable). Determine which of the following maps from A to Z are functions.
(a) f (A) is the sum of all the elements in A, where A ∈ A. For example, f ({1, 2, 4}) = 1 + 2 + 4 = 7.
(b) f (A) is the product of all the elements in A, where A ∈ A.
(c) Given A ∈ A,
(
1
if |A| = 1, 2
f (A) =
the sum of any three elements in A if |A| ≥ 3.
Exercise 1.21. Find the image of the following functions. Justify your answers.
(a) f : N × N → N defined via (a, b) 7→ a + b.
(b) f : N × N → R defined via (a, b) 7→ a − b.
(c) f : Z × Z → R defined via (a, b) 7→
a(a + 1)b
.
2
(d) f : N × N → R defined via (a, b) 7→
a(a + 1)b
.
2
25
(e) f : Z × Z → R defined via (a, b) 7→
(f) f : R → R defined via x 7→
(g) f : R → R defined via x 7→
a+b
.
2
x
.
|x| + 1
x2
x
.
+1
(h) f : (0, ∞) → R defined via x 7→
4x
+ 1.
x+3
Exercise 1.22. For each part of Exercise 1.21, determine if the function is injective/surjective/bijective.
Exercise 1.23. Define f : R × R → R via (a, b) 7→ a − b.
(a) Find f ([−1, 1] × (0, 4)).
(b) Find f ((0, 3) × {4}).
(c) Find f ((1, 5) × [1, 5)).
Exercise 1.24. Fix an integer n ≥ 2. Define the following operation on Zn :
[a] + [b] = a + b, where + on the right of the equality is usual addition.
Is this operation well defined? Justify your answer.
Exercise 1.25. (a) What digit does the number 3237 end in?
(b) What digit does the number 21020 end in?
Exercise 1.26. (a) Prove that two natural numbers a and a + 1 are prime if and only if a = 2.
(b) Suppose p is a natural number with p, p + 2, and p + 4 all prime. Prove that p = 3.
Exercise 1.27. (a) Given an integer a, prove that a2 ≡ 0 or 1 (mod 4).
(b) Given three integers a, b, and c, prove that a2 + b2 + c2 is not a perfect square. Note: An integer
n is called a perfect square if there exist an integer m such that n = m2 .
(c) Prove that the equation a2 + b2 = 3c2 has no solutions in non-zero integers a, b, and c.
Exercise 1.28. Prove that an integer n is divisible by 3 if and only if the sum of its digits is divisible
by 3.
1.7
Challenging Exercises
Exercise 1.29. Prove that the Well Ordering Axiom and Principle of Mathematical Induction (ordinary or strong) are equivalent.
Exercise 1.30. Let F1 = 1 = F2 , and Fn = Fn−1 + Fn−2 for all integers n ≥ 3. Prove that
Fn =
Φn − φn
√
,
5
√
√
1+ 5
1− 5
where Φ =
and φ =
. Hint. First show that 1 + Φ = Φ2 and 1 + φ = φ2 .
2
2
26
Exercise 1.31. Fix an integer n ≥ 1. Given non-negative real numbers x1 , x2 , . . . , xn , prove that
(x1 x2 · · · xn )1/n ≤
x1 + x2 + . . . + xn
.
n
Exercise 1.32. Given seven distinct integers a1 , a2 , a3 , a4 , a5 , a6 , a7 , prove that 10|(ai −aj ) or 10|(ai +
aj ) for some i 6= j, i, j ∈ {1, 2, 3, 4, 5, 6, 7}.
Exercise 1.33. (Open question) Given a natural number p, p and p + 2 are called twin primes if
they are both prime. How many twin primes are there? It is conjectured that there are infinitely
many.
27
2
Groups
2.1
Definition and Properties
In this section, we introduce the basic algebraic structure to be studied in this course, called a group.
We then prove some of the properties enjoyed by groups.
Binary Operation
Given a non-empty set G, a binary operation on G is a function from G × G → G.
Usual addition (or multiplication) on the integers (or real numbers) is a binary operation. Division
is not a binary operation on the natural numbers, nor is it a binary operation on the integers.
Group
Let G be a set together with a binary operation ·, that assigns to (a, b) ∈ G × G an element
a · b in G (usually written as ab). We say (G, ·) is a group if it satisfies the following three
properties:
1. Associativity; (ab)c = a(bc) for all a, b, c ∈ G.
2. Identity; there exist an element e ∈ G such that ae = a = ea for all a ∈ G. The element
e is called the identity element in G.
3. Inverses; for all a ∈ G, there exist b ∈ G such ab = e = ba. We denote b by a−1 and call
it the inverse of a.
Associativity allows us to drop the brackets. In particular, instead of writing (ab)c we simply write
abc, and this can be extended to any finite number of elements in G (induction). A group G is called
trivial if and only if |G| = 1, that is, G = {e}. We will shortly prove that the identity element and
inverses in a group are unique.
Example 2.1. Which one of the following is a group?
(a) (Z, +), where + is usual addition.
(b) (Z, ×), where × is usual multiplication.
(c) (N, +), where + is usual addition.
Solution. (a) Since the sum of two integers is an integer, the operation of addition as binary.
Moreover, addition on the integers is associative. The identity element in this set is the number zero.
For any a ∈ Z, a + (−a) = 0, so a−1 = −a ∈ Z. Hence, the integers are a group under addition.
(b) This is not a group. To see this, notice that 1 is the identity element in Z under the operation of
multiplication. However, there does not exist an integer b such that 2b = 1, i.e., 2−1 does not exist.
(c) This is not a group. This set has no identity element (notice that 0 ∈
/ N). Also, inverses do not
exist.
♣
28
Unless otherwise stated when we say Z is a group we assume the operation is usual addition.
Abelian Group
A group G is called Abelian (or commutative) if ab = ba for all a, b ∈ G.
Example 2.2. Which one of the groups in example 2.1 are Abelian?
Solution. (Z, +) is an Abelian group. The other two are not groups.
♣
We now state some of the properties of a group.
Properties of A Group
Theorem 2.3. Let G be a group, e the identity element in G, and a, b ∈ G. Then,
(1) e is unique, i.e., e is the unique element of G satisfying ea = a = ae for all a ∈ G.
(2) If ab = ac, then b = c. Similarly, if ba = ca, then b = c.
(3) a−1 is unique, i.e., a−1 is the unique element of G satisfying aa−1 = e = a−1 a.
(4) (ab)−1 = b−1 a−1 .
(5) (a−1 )−1 = a.
Proof. (1) Suppose e and f are both identity elements of G. Then, ae = a = ea and af = a = f a
for all a ∈ G. In particular, e = ef = f .
(2) If ab = ac, then
a−1 (ab) = a−1 (ac) ⇒ (a−1 a)b = (a−1 a)c ⇒ eb = ec ⇒ b = c.
A similar argument shows that if ba = ca, then b = c.
(3) Suppose b and c are both inverses of a. Then, ba = e = ca, and by part (2), b = c.
(4) Notice that (ab)(b−1 a−1 ) = a(bb−1 )a−1 = aea−1 = aa−1 = e. Similarly, (b−1 a−1 )(ab) = e. Hence,
(ab)−1 = b−1 a−1 .
(5) This follows from the fact that aa−1 = e = a−1 a.
One of the consequences of part (2) of the previous proposition is that if a ∈ G with ab = e or ba = e,
then b = a−1 , i.e., we do not have to show both equations hold. Also, if for some b ∈ G, ab = a (or
ba = a), then b must be the identity, i.e., we do not have to check bx = xb = x for all x ∈ G.
The rest of this section consists of examples of groups.
The Abelian Group Zn
Example 2.4. Given an integer n ≥ 2, Zn is an Abelian group under the operation [a] + [b] =
[a + b].
29
Solution. Given [a], [b] ∈ Zn , [a]+[b] = [a+b] ∈ Zn , so the operation is binary. For any [a], [b], [c] ∈ Zn ,
[a] + ([b] + [c]) = [a] + [b + c] = [a + (b + c)] = [(a + b) + c] = [a + b] + [c] = ([a] + [b]) + [c].
We have just shown associativity. Since
[a] + [0] = [a] = [0] + [a] for all [a] ∈ Zn ,
the identity element in Zn is [0]. Moreover,
[a] + [−a] = [a − a] = [0] = [−a + a] = [−a] + [a].
That is, [a]−1 = [−a] ∈ Zn . Hence, Zn is a group. Furthermore,
[a] + [b] = [a + b] = [b + a] = [b] + [a] for any [a], [b] ∈ Zn .
♣
The Abelian Group U (n)
Example 2.5. Given an integer n ≥ 2, define
U (n) = {[a] ∈ Zn : [ax] = [1] for some integer x}.
Show that U (n) is an Abelian group under the operation [a][b] = [ab].
Solution. Given [a], [b] ∈ U (n), there exist integers x and y such that [ax] = [1] = [by]. This is
equivalent to ax + nt = 1 and by + ns = 1, for some integers t and s. By Corollary 1.5, gcd(a, n) =
1 = gcd(b, n), and by Corollary 1.16, gcd(ab, n) = 1. Using Corollary 1.5 we get (ab)u + nv = 1 for
some integers u and v, and this in turn is equivalent to [(ab)u] = [1]. Hence, [a][b] = [ab] ∈ U (n),
which implies that the operation is binary. For any [a], [b], [c] ∈ U (n),
[a]([b][c]) = [a][bc] = [a(bc)] = [(ab)c] = [ab][c] = ([a][b])[c].
We have just shown associativity. The identity element in this group is [1] (as [a][1] = [a] = [1][a] for
all [a] ∈ U (n)). Moreover, given [a] ∈ U (n), there exist an integer b such that
[a][b] = [ab] = [1] = [ba] = [b][a].
Notice that [b] ∈ U (n) (as [ba] = [ab] = [1]). So [a]−1 = [b] ∈ U (n). Hence, U (n) is a group and
[a][b] = [ab] = [ba] = [b][a] for all [a], [b] ∈ U (n).
♣
The students should check that the set U (n) can also be written as follows:
U (n) = {[a] ∈ Zn : gcd(a, n) = 1}.
Unless otherwise stated the operations on Zn and U (n) are the operations defined in the preceding
two examples, respectively. Moreover, we abuse notation and write a instead of [a] in these groups.
It should be understood that the elements in these groups are not integers (they are equivalence
classes mod n). For example, Z4 = {0, 1, 2, 3} and U (10) = {1, 3, 7, 9}. The operations in Zn and
U (n) are called addition mod n and multiplication mod n, respectively.
30
Example 2.6. Let
G=
a a
a a
: a ∈ R, a 6= 0 .
Prove that G is an Abelian group under the usual matrix multiplication.
Solution. From linear algebra we know matrix multiplication is associative. Moreover,
a a
b b
2ab 2ab
=
∈G
a a
b b
2ab 2ab
Notice that we have used that fact that a, b 6= 0 implies that 2ab 6= 0. We have just proven the
operation is binary. Since
1 1 1 1 a a
a a
a a
2
2
2
2
=
= 1 1
,
1
1
a a
a a
a a
2
2
2
2
1 1
the identity element in G is 21 12 . Furthermore,
2
2
−1 1
a a
= 4a
1
a a
4a
1
4a
1
4a
∈G
Therefore, G is a group under matrix multiplication. Furthermore,
a a
b b
2ab 2ab
b b
a a
=
=
.
a a
b b
2ab 2ab
b b
a a
♣
Hence, G is Abelian.
The previous example seems to contradict what many of us learned in linear algebra: an n × n matrix
A has an inverse if and only if det A 6= 0. All the matrices in the previous example have determinant
zero. So what is happening here? How is it possible that these matrices have inverses? The answer
while simple is very subtle. The theorems in linear algebra are based on the fact that the entries in
our matrices is the set of all real numbers with no relation between the entries themselves. However,
in the previous example, the restriction on the matrices is much more severe: every entry must be
the same. This condition changes the identity element in this set, which in turn changes the notion
of inverses. So theorems that we knew from linear algebra do not apply in the previous example.
ab
(where ab is the usual product of the
2
rational numbers). Show that (G, ∗) is an Abelian group.
Example 2.7. Let G = Q − {0}, and define a ∗ b =
p
c
pc
, b =
where p, q, c, d ∈ Z − {0}. Then, a ∗ b =
∈ G.
q
d
2qd
Moreover, for any a, b, c ∈ G, we have
Solution. Given a, b ∈ G, let a =
a ∗ (b ∗ c) = a ∗
a(bc)
(ab)c
(a ∗ b)c
bc
=
=
=
= (a ∗ b) ∗ c.
2
4
4
2
31
For any a ∈ G, we have
a∗2=
a(2)
2a
=a=
=2∗a
2
2
4
4
4
Hence, the identity element in G is 2. Furthermore, ∈ G (since a ∈ G) and a ∗ = 2 = ∗ a.
a
a
a
4
−1
That is, a = . Hence, (G, ∗) is a group. Also,
a
a∗b=
ab
ba
=
= b ∗ a for any a, b ∈ G,
2
2
♣
showing that G is Abelian.
Example 2.8. Let G = R − {−1}, and define x ∗ y = x + y + xy (the operations on the right
of the equality are normal addition and multiplication). Prove (G, ∗) is an Abelian group.
Solution. Given x, y ∈ G, x ∗ y ∈ R (since R is closed under addition and multiplication). Moreover,
if x ∗ y = −1, then
x + y + xy = −1 ⇒
⇒
⇒
⇒
x + y + xy + 1 = 0
x(1 + y) + 1 + y = 0
(x + 1)(y + 1) = 0
x = −1 or y = −1,
a contradiction. Therefore, ∗ is a binary operation on G. For any x, y, z ∈ G,
(x ∗ y) ∗ z = (x + y + xy) ∗ z = x + y + xy + z + z(x + y + xy)
= x + y + xy + z + zx + zy + zxy,
and
x ∗ (y ∗ z) = x ∗ (y + z + yz) = x + y + z + yz + x(y + z + yz)
= x + y + z + yz + xy + xz + xyz.
In particular, (x ∗ y) ∗ z = x ∗ (y ∗ z). Hence, ∗ is an associative operation on G. For any x ∈ G,
we have x ∗ 0 = x = 0 ∗ x. Thus the identity element in G is 0. Moreover, by solving the equation
x ∗ y = 0 for y we get
x−1 = y =
−x
1+x
(notice that x 6= −1).
We need to show y ∈ G, i.e., y 6= −1. If y = −1, then −(1 + x) = −x, which in turn gives −1 = 0, a
contradiction. Hence, (G, ∗) is a group. Moreover,
x ∗ y = x + y + xy = y + x + yx = y ∗ x,
♣
showing that G is Abelian.
32
Example 2.9. Let
a b
G=
: a + d = b + c, a, b, c, d ∈ R
c d
Show that G is an Abelian group under (the usual) matrix addition.
Solution. From linear algebra we know matrix addition is associative. Given A, B ∈ G, let
x y
x1 y1
A=
, B=
,
z w
z1 w1
where x, y, z, w, x1 , y1 , z1 , w1 ∈ R, and
x + w = y + z, x1 + w1 = y1 + z1 ⇒ (x + x1 ) + (w + w1 ) = (y + y1 ) + (z + z1 )
In particular,
A+B =
x y
x1 y1
x + x1 y + y 1
+
=
∈G
z w
z1 w1
z + z1 w + w1
This proves the operation is binary. Let
E=
0 0
0 0
∈ G.
Notice that A + E = A = E + A for all A ∈ G. Hence, E is the identity element of G. Moreover, if
x y
A=
∈ G, where x, y, z, w ∈ R, x + w = y + z,
z w
then −x − w = −y − z. Thus −A ∈ G, and A−1 = −A (as A + (−A) = E = (−A) + A). Hence, G
is a group under matrix addition. This group is Abelian:
x y
x1 + x y 1 + y
x1 y1
x + x1 y + y 1
x y
x1 y 1
+
,
=
=
+
=
z + z1 w + w1
z1 + z w1 + w
z1 w1
z w
z1 w1
z w
♣
showing that G is Abelian.
The next example gives a necessary and sufficient condition for a group to be Abelian.
Example 2.10. Let G be a group. Then, G is Abelian if and only if (ab)−1 = a−1 b−1 for all
a, b ∈ G.
Solution. If G is Abelian, then all the elements in G commute with each other. Combining this with
Theorem 2.3(4), for any a, b ∈ G we have (ab)−1 = b−1 a−1 = a−1 b−1 .
Conversely, suppose (ab)−1 = a−1 b−1 for all a, b ∈ G. Given x, y ∈ G,
xy =
=
=
=
(x−1 )−1 (y −1 )−1
(x−1 y −1 )−1
(y −1 )−1 (x−1 )−1
yx
by Theorem 2.3(5)
by our assumption
by Theorem 2.3(4)
by Theorem 2.3(5),
33
♣
showing that G is Abelian
If g ∈ G, we define g 0 = e, and for any natural number n, we define
g n = g · g · · · · · g and g −n = g −1 · g −1 · · · · · g −1 .
|
{z
}
|
{z
}
n−times
n−times
It should be noted that in a group we do not permit non-integer exponents. Notice that when the
operation in your group is addition, the preceding notation translates to g m = mg for any g ∈ G and
m ∈ Z. Moreover, if a, b ∈ G and the operation is addition, then ab−1 = a − b.
Example 2.11. Let G be a group. Suppose that x2 = e for all x ∈ G. Prove that G is Abelian.
Solution. Given a, b ∈ G, we have a = a−1 , b = b−1 , and ab = (ab)−1 . Hence,
ab = (ab)−1
= b−1 a−1 by Theorem 2.3(4)
= ba,
♣
showing that G is an Abelian group.
2.2
Order of a Group, Order of an Element
In this section, we introduce the notion of order for a group and its elements.
Order of A Group
If G is a group, the number of elements in G is called the order of G. The order of a group G
is denoted by |G|. If G has infinitely many elements, countable or not, we write |G| = ∞.
For example, the integers under addition are an infinite group (i.e., |Z| = ∞). Whereas the group
U (6) = {1, 5} is a group of order 2 (i.e., |U (6)| = 2).
Order of An Element
Given an element g in a group G, the order of g is the smallest positive integer n such that
g n = e. If no such n exist, we say that g has infinite order. The order of an element g ∈ G is
denoted by |g|.
A few quick facts that follow from the definition of the order of an element g in a group G are:
(1) |g| = 1 if and only if g = e.
(2) |g| = ∞ is equivalent to the following statement: g n = e if and only if n = 0.
(3) |g| = n is equivalent to: (i) g n = e, and (ii) if g m = e for some natural number m, then n ≤ m.
Example 2.12. List the elements in U (10). Find the order of U (10) and every element in it.
34
Solution. The group U (10) = {1, 3, 7, 9}. So |U (10)| = 4. The identity element in U (10) is 1. Notice
that
71
72
73
74
7 6= 1,
49 = 9 6= 1,
7 · 72 = 7 · 9 = 63 = 3 6= 1,
72 · 72 = 9 · 9 = 81 = 1.
=
=
=
=
Hence, |7| = 4. A similar computation shows that |1| = 1, |3| = 4, |9| = 2.
♣
Example 2.13. List the elements in Z8 and find the order of every element.
Solution. The group Z8 = {0, 1, 2, 3, 4, 5, 6, 7}. The identity element in Z8 is 0. Notice that
21
22
23
24
=
=
=
=
2,
2 + 2 = 4,
2 + 2 + 2 = 6,
2 + 2 + 2 + 2 = 8 = 0.
Hence, |2| = 4. Similar computations show that
|0|
|1|
|4|
|6|
=
=
=
=
1
8 = |3| = |5| = |7|
2
4
♣
Example 2.14. Find the order of every element in Z.
Solution. As we mentioned at the beginning of this section, the integers are an infinite group under
addition. Since zero is the identity element, |0| = 1. If n 6= 0, we show that |n| = ∞. Assume to
the contrary, say |n| = m. Then mn = nm = 0. This implies that n = 0 or m = 0, a contradiction.
Hence, every non-identity element in Z has infinite order.
♣
Example 2.15. Let a, b ∈ G with |a| = 4, |b| = 2, and a3 b = ba. Find |ab|.
Solution. We do this by process of elimination. If |ab| = 1, then ab = e, equivalently a = b−1 . By
Exercise 2.18, |a| = |b|, a contradiction. Thus |ab| ≥ 2. Since
(ab)2 = (ab)(ab) = a(ba)b = a(a3 b)b = a4 b2 = ee = e,
we get |ab| = 2.
♣
Example 2.16. Let a, b ∈ G with |a| = 2, b 6= e, |ab| = 2, and aba = b2 . Determine |b|.
35
Solution. Since b 6= e, |b| ≥ 2. Notice that |a| = 2 implies that a = a−1 . If |b| = 2, then
e = b2 = aba ⇒ b = a−1 a−1 = a2 = e, a contradiction.
Thus |b| ≥ 3. Since
b3 = b2 b = (aba)b = (ab)2 = e,
we get |b| = 3.
♣
The following corollary gives us a necessary and sufficient condition for ak = e whenever a has finite
order.
Corollary 2.17. Given a ∈ G with |a| = n < ∞, ak = e if and only if n|k. In other words,
ak = e if and only if k is an integer multiple of n.
Proof. If n|k, then k = nm for some integer m. In particular, ak = anm = (an )m = em = e.
Conversely, if ak = e, by the Division Algorithm, there exist unique integers q and r such that
k = nq + r with 0 ≤ r < n. It follows that
ar = ak−nq = ak (an )−q = ee−q = e.
Minimality of n implies that r = 0, and the result follows.
Example 2.18. Suppose x is an element of a group G and xn = e = xm for some integers n, m
(not both zero). Prove that |x| |d, where d = gcd(n, m).
Solution. Notice that |x| is finite, as xn = e (can also use xm = e). By Proposition 1.3, there exist
integers u and v such that d = mu + nv. In particular,
xd = xmu+nv = (xm )u (xn )v = eu ev = e.
♣
The result now follows by Corollary 2.17.
2.3
Subgroups
The set of even integers E is a group under addition (proof of this is left to the reader). This set is
also a subset of a bigger set which is also a group under addition, namely, the integers. Of course,
the integers are also a subset of the rationals, which is also a group under addition. One can extend
this to the real numbers; that is, we have the following chain of groups ordered by inclusion:
E ⊆ Z ⊆ Q ⊆ R.
This situation arises so often that we introduce a special term to describe it.
Subgroup
A subset H of a group G is called a subgroup of G, if H itself is a group under the binary
operation of G. If H is a subgroup of G, we write H G.
36
If H G, then it must have an identity element, and thus it must be non-empty. Moreover, the
identity element of G is also the identity element of H. To see this, since H is not empty, there exist
an a ∈ H. Since H is a subgroup, a−1 ∈ H. Since our operation is binary, e = aa−1 ∈ H, and the
uniqueness of the identity element proves the result.
It is very easy to prove that {e} G for any group G. This is called the trivial subgroup of G.
In particular, for any group G, the sets {e} and G are subgroups of G. If H is a subgroup of G
but is not equal to G itself, we write H ≺ G. Such a subgroup is called a proper subgroup of G.
For example, the set of even integers is a proper subgroup of the integers, the integers are a proper
subgroup of the rationals, and the rationals are a proper subgroup of the reals.
Example 2.19. Is the set of odd integers a subgroup of the integers under addition?
Solution. No. The sum of two odd integers is even. Thus the operation is not binary. One can also
use the fact that there is no identity element in the set of odd integers.
♣
Example 2.20. Is the set H = (R − Q) ∪ {1} a subgroup of G = R − {0} under multiplication?
Solution. This is not a subgroup. To see this, notice that
is, the operation is not binary.
√
√ √
2 ∈ H, however, 2 2 = 2 ∈
/ H. That
♣
Given a group G, the relation on subgroups of G is a reflexive and transitive relation. However, it
is almost never a symmetric relation. The only time it is a symmetric relation is when G = {e}.
Proposition 2.21. Let G be a group and ∅ =
6 H ⊆ G. If H satisfies the following property:
a, b ∈ H ⇒ ab−1 ∈ H,
then H G.
Proof.
Then,
a = e,
get xy
Associativity follows from the fact that H ⊆ G. Since H is non-empty, there exist an x ∈ H.
e = xx−1 ∈ H (by taking a = x = b). If x ∈ H, then x−1 = ex−1 ∈ H (by taking
b = x). Suppose x, y ∈ H. By our preceding argument y −1 ∈ H. Taking a = x, b = y −1 we
= x(y −1 )−1 ∈ H. Hence, H is a group, showing that H G.
Example 2.22. Let G be an Abelian group and define H = {x ∈ G| x2 = e}. Prove that
H G.
Solution. By definition H ⊆ G. Moreover, e = e2 ∈ H, showing that H is not empty. Given
x, y ∈ H, we have x = x−1 and y = y −1 . This gives
(xy −1 )2 = xy −1 xy −1 =
=
=
=
x2 (y −1 )2 as G is Abelian
x2 y 2
ee
e.
Hence, xy −1 ∈ H, and the result follows by Proposition 2.21.
37
♣
The following example shows that the intersection of two subgroups is again a subgroup.
Example 2.23. Suppose H, K G. Prove that H ∩ K G.
Solution. Notice that H ∩ K ⊆ H ⊆ G. Moreover, e ∈ H ∩ K (as e ∈ H and e ∈ K). Thus
H ∩ K is not empty. Given x, y ∈ H ∩ K, the elements x and y are in H and in K. Since H and
K are subgroups of G, we have xy −1 is in H and K. That is, xy −1 ∈ H ∩ K. By Proposition 2.21,
H ∩ K G.
♣
It is not true that the union of two subgroups has to be a subgroup (see Exercise 2.25).
Proposition 2.24. Let G be a group and let H be a non-empty subset of G. If H satisfies the
following two properties:
1. a, b ∈ H ⇒ ab ∈ H (H is closed under the operation of the group), and
2. a ∈ H ⇒ a−1 ∈ H (H is closed under inverses),
then H G.
Proof. Suppose a, b ∈ H. Since H is closed under inverses, b−1 ∈ H. Since H is closed under
multiplication we have ab−1 ∈ H, and the result follows by Proposition 2.21.
Example 2.25. Suppose G is an Abelian group and H, K G. Prove that
HK = {hk : h ∈ H, k ∈ K} G.
Solution. Notice that e = ee ∈ HK and HK ⊆ G (since the operation in G is binary). Given
a, b ∈ HK, then a = xy and b = zw for some x, z ∈ H and y, w ∈ K. Since G is Abelian and H and
K are subgroup of G,
ab = xyzw = (xz)(yw) ∈ HK, and
a
= (xy)−1 = y −1 x−1 = x−1 y −1 ∈ HK.
−1
By Proposition 2.24, HK G.
♣
Example 2.26. Let H be a subgroup of a group G, and for each g ∈ G, define
gHg −1 = {ghg −1 : h ∈ H}.
Prove that gHg −1 G for all g ∈ G.
Solution. Given g ∈ G, e = geg −1 ∈ gHg −1 (thus gHg −1 is not empty). Since the operation in G is
binary, gHg −1 ⊆ G. Given a, b ∈ gHg −1 , a = ghg −1 and b = gh1 g −1 for some h, h1 ∈ H. It follows
that
ab = (ghg −1 )(gh1 g −1 ) = g(hh1 )g −1 ∈ gHg −1 , and
a−1 = (ghg −1 )−1 = (g −1 )−1 h−1 g −1 = gh−1 g −1 ∈ gHg −1 .
38
By Proposition 2.24, gHg −1 G.
♣
The following sets (which we will show are subgroups) play a crucial role in group theory.
Center/Centralizer/Normalizer
Let G be a group.
• The center of G, denoted by Z(G), is the set of all elements in G that commute with
every element of G;
Z(G) = {g ∈ G : gx = xg ∀x ∈ G}.
• Given an element a ∈ G, the centralizer of a ∈ G, denoted by CG (a), is the set of elements
in G that commute with a;
CG (a) = {g ∈ G : ga = ag}.
• If H G, the centralizer of H in G, denoted by CG (H), is the set of elements in G that
commute with every element in H;
CG (H) = {g ∈ G : gh = hg ∀h ∈ H}.
• If H G, the normalizer of H in G is the set
NG (H) = {g ∈ G : gHg −1 = H}.
It follows by definition that G is Abelian if and only if Z(G) = G. Moreover, CG (G) = Z(G).
Theorem 2.27. Let G be a group. Then,
1. Z(G) CG (a) G for all a ∈ G.
2. Z(G) CG (H) NG (H) G for all H G.
3. G is Abelian if and only if Z(G) = CG (a) for all a ∈ G.
\
4. Z(G) =
CG (a).
a∈G
Proof. The proofs of (2), (3), and (4) are left as an exercise (see Exercise 2.17). The elements of the
center of a group commute with every element in the group. In particular, given a ∈ G, g ∈ Z(G),
we have ga = ag. This is equivalent to g ∈ CG (a). Since a ∈ G and g ∈ Z(G) were chosen arbitrarily,
we have Z(G) ⊆ CG (a) for any a ∈ G. It now suffices to show that Z(G) G and CG (a) G.
Since ea = ae for all a ∈ G, i.e., e ∈ Z(G). Given x, y ∈ Z(G), for any a ∈ G,
(xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy).
39
Thus xy ∈ Z(G). Moreover,
xa = ax for all a ∈ G ⇔ x−1 (xa)x−1 = x−1 (ax)x−1 for all a ∈ G ⇔ ax−1 = x−1 a for all a ∈ G.
That is, x−1 ∈ Z(G), and the result follows by Proposition 2.24, Z(G) G.
We now show CG (a) G for all a ∈ G. Given a ∈ G, the identity element e is in CG (a) (as ea = ae).
So CG (a) 6= ∅. Moreover, given x, y ∈ CG (a),
(xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy) ⇒ xy ∈ CG (a), and
xa = ax ⇔ x−1 (xa)x−1 = x−1 (ax)x−1 ⇔ ax−1 = x−1 a ⇔ x−1 ∈ CG (a).
By Proposition 2.24, CG (a) G.
Example 2.28. Let
GL(2, R) =
a b
: a, b, c, d ∈ R, ad − bc 6= 0 .
c d
This
usual matrix
set is
a group under
multiplication (see Exercise 2.1(b)). Find the centralizer
1 1
1 1
of
, i.e., CGL(2,R)
.
1 0
1 0
Solution.
CGL(2,R)
1 1
1 0
=
=
=
=
a b
a b
1 1
1 1
a b
∈ GL(2, R) :
=
c d
c d
1 0
1 0
c d
a b
a+b a
a+c b+d
∈ GL(2, R) :
=
c d
c+d c
a
b
a b
∈ GL(2, R) : a = b + d, b = c
c d
b+d b
2
: b, d ∈ R, (b + d)d 6= b .
b
d
♣
Example 2.29. Let
SL(2, R) =
a b
: a, b, c, d ∈ R, ad − bc = 1 .
c d
This set is a group under usual matrix multiplication (see Exercise 2.30). Find the center of
SL(2, R), i.e., Z(SL(2, R)).
Solution.
We need to find all matrices that commute with every element in SL(2, R). Suppose
a b
∈ Z(SL(2, R)). Then,
c d
a b
0 1
0 1
a b
a b
a b
=
⇒ a = d and c = −b ⇒
=
.
c d
−1 0
−1 0
c d
c d
−b a
40
Furthermore,
1 1
1 1
a
a b
=
−b a
−1 0
−1 0
−b
a b
a 0
Hence,
=
with a2 = 1 (as our matrix must
c d
0 a
a 0
Z(SL(2, R)) ⊆
: a ∈ R,
0 a
b
a
⇒ b = 0.
be an element of SL(2, R)). This proves
a =1 .
2
The reverse inclusion is obtained by noticing
a 0
x y
ax ay
x y
a 0
=
=
.
0 a
z w
az aw
z w
0 a
Hence,
Z(SL(2, R)) =
a 0
1 0
−1 0
2
: a ∈ R, a = 1 =
,
.
0 a
0 1
0 −1
♣
Example 2.30. Suppose x, y ∈ G with xy ∈ Z(G). Prove that xy = yx.
Solution. Since xy ∈ Z(G) it commutes with every element in G. Hence,
(xy)x−1 = x−1 (xy) = (x−1 x)y = y ⇒ xy = yx.
♣
Proposition 2.31. Suppose H is a non-empty subset of a group G and |H| < ∞. If H is
closed under the operation of G (i.e., the operation on G defines a binary operation on H) then
H G.
Proof. We only need to show that H is closed under inverses. Let a ∈ H. If a = e, then
a−1 = e−1 = e = a ∈ H.
If a 6= e, consider the sequence a, a2 , a3 , .... Since H is closed under the operation, these are all in
H. Since H is finite, not all of these elements are distinct. In particular, ai = aj for some i 6= j.
Without loss of generality we can assume i > j. Then ai−j = e. Since a 6= e, we have i − j > 1.
Thus, a(ai−j−1 ) = e ⇒ a−1 = ai−j−1 ∈ H. Hence, H G. Remark. The condition that H is finite
is a necessary one (see Exercise 2.13).
2.4
Cayley Table for Groups
If G is a finite group under the operation ·, a very nice way to arrange a · b for all a, b ∈ G is via
a table, called a Cayley table. The entry a · b appears in the box corresponding to the row a and
column b and the entry b · a appears in the box corresponding to the row b and column a. Note that
a · b may not equal to b · a. For example, if G = {e, a, b, c} is a group under the operation ·, then its
Cayley table is
41
·
e
e
e
a a·e
b b·e
c c·e
a
b
e·a e·b
a·a a·b
b·a b·b
c·a c·b
c
e·c
a·c
b·c
c·c
Example 2.32. Give the Cayley table for Z4 .
Solution.
+
0
1
2
3
0
0
1
2
3
1
1
2
3
0
2
2
3
0
1
3
3
0
1
2
♣
Example 2.33. Give the Cayley table for U (10).
Solution.
·
1
3
7
9
1
1
3
7
9
3
3
9
1
7
7
7
1
9
3
9
9
7
3
1
♣
Example 2.34. Give the Cayley table for U (7).
Solution.
·
1
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
1
3
5
3
3
6
2
5
1
4
4
4
1
5
2
6
3
5
5
3
1
6
4
2
6
6
5
4
3
2
1
♣
42
2.5
Exercises
Unless otherwise stated G is a group.
Exercise 2.1. Determine which of the following are groups under the defined operation.
(a) The set of 2 × 2 matrices with entries in R under matrix addition.
a b
(b) The set GL(2, R) =
: det A = ad − bc 6= 0, a, b, c, d ∈ R under usual matrix mulc d
tiplication. This group is called the General Linear Group.
(c) The set R× = R − {0} under usual multiplication.
(d) The set R>0 = {x ∈ R : x > 0} under usual multiplication.
(e) The set Q>0 = {x ∈ Q : x > 0} under usual multiplication.
(f) The set Z × Z under the operation (a, b)(c, d) = (ad + bc, bd).
(g) The set Q under the operation a · b =
addition.
a+b
, where addition on the right of the equality is usual
7
Exercise 2.2. Determine which groups in Exercise 2.1 are Abelian.
Exercise 2.3. Is the set of the integers under the operation a · b = a + b − 2 a group? Justify your
answer. If yes, is it Abelian? Justify your answer.
Exercise 2.4. Is the set of the integers under the operation a · b = a − b + 1 a group? Justify your
answer. If yes, is it Abelian? Justify your answer.
a b
Exercise 2.5. Let G =
: a, b, c, d ∈ R, bc = 0 . Is G a group under matrix addition?
c d
Justify your answer.
Exercise 2.6. Let
a b
G=
: a, b, c, d ∈ R, ad = 0 .
c d
Is G a group under usual matrix addition? Justify your answer.
Exercise 2.7. Find the inverse and order of every element in U (7)? Do the same for U (10).
Exercise 2.8. Is Z6 a group under the operation [a][b] = [ab]? Justify your answer.
Exercise 2.9. Let M2×2 (R) be the set of all 2 × 2 matrices with entries from R. The trace of a 2 × 2
matrix A, denoted tr(A), is defined to be the sum of its main diagonal. That is,
a b
tr(A) = a + d where A =
.
c d
Let G = {A ∈ M2×2 (R) : tr(A) = 0}. Is G a group under usual matrix addition (justify your answer)? If yes, is it an Abelian Group (justify your answer)?
43
Exercise 2.10. Is the set G in Exercise 2.9 a group under matrix multiplication? Justify your
answer.
Exercise 2.11. Let
G=
a b
: a, b, c ∈ R .
c a
Is G a group under usual matrix addition (justify your answer)? If yes, is it an Abelian Group (justify
your answer)?
Exercise 2.12. Let G = {5, 15, 25, 35}. Prove that G is a group under multiplication modulo 40.
Find the order of each element in this group.
Exercise 2.13. Give an example of a non-empty subset H of a group G that is closed under the
operation of G but is not a subgroup of G.
Exercise 2.14. (a) Find the order of A, B ∈ GL(2, R), where
0
1
0 −1
A=
, and B =
.
−1 −1
1 0
(b) Find the order of |AB|. Is |AB| = |A| |B|?
Exercise 2.15. Find the order of all the elements in Z10 and U (20).
Exercise 2.16. (a) Suppose a ∈ G with |a| = 9. Prove that there exist x ∈ G such that x5 = a2 .
(b) Suppose a ∈ G with |a| = 11. Prove that there exist x ∈ G such that x6 = a.
Exercise 2.17. Prove the rest of Theorem 2.27.
Exercise 2.18. Show that |a| = |a−1 | for all a ∈ G.
Exercise 2.19. Let g ∈ G such that |g| = n < ∞.
(a) Prove that if n = 2k + 1 for some integer k ≥ 0, then g i 6= g −i for all i = 1, 2, . . . , n − 1.
(b) Prove that if n = 2k for some integer k ≥ 1 and 1 ≤ i < n, then g i = g −i if and only if i = k.
Exercise 2.20. Suppose a, b ∈ G such that |a| = 2, |b| ≥ 2, and b = ab2 a. Determine |b|.
Exercise 2.21. In this exercise we prove that if G is a finite group, then ever element in G has finite
order. N ote. Exercise 2.46 gives an example of an infinite group in which every element has finite
order.
(a) Suppose g ∈ G with |g| = ∞. Prove that g n = g m if and only if n = m.
(b) Conclude that if a group has an element of infinite order, then the group itself has infinite order.
Exercise 2.22. (a) Prove that CG (a) = CG (a−1 ) for all a ∈ G.
(b) Suppose a ∈ G with |a| = n, and k is an integer with gcd(n, k) = 1. Prove that CG (a) = CG (ak ).
Exercise 2.23. Suppose H and K are subgroups of G. Prove that HK is a subgroup of G if and
only if HK = KH.
44
Exercise 2.24. Suppose H is a subgroup of G. Prove that HZ(G) is a subgroup of G.
Exercise 2.25. (a) Give an example of a group G with subgroups H and K such that H ∪ K is not
a subgroup of G.
(b) Suppose H1 , H2 , H3 , . . . is an infinite collection of subgroups of a group G such that Hi ⊆ Hi+1
for all i ∈ N. Prove that
∞
[
Hi
i=1
is a subgroup of G.
Exercise 2.26. Suppose H1 , H2 , H3 , . . . is an infinite collection of subgroups of a group G. Prove
that
∞
\
Hi
i=1
is a subgroup of G.
Exercise 2.27. Let H G and define a relation ∼ on G by
a ∼ b ⇔ b−1 a ∈ H.
Prove that ∼ is an equivalence relation on G.
Exercise 2.28. Define the following operation on R2 = {(x, y) : x, y ∈ R}:
(a, b) + (c, d) = (a + c, b + d).
(a) Prove that (R2 , +) is a group.
(b) Let H = {(x, y) ∈ R2 : xy ≥ 0}. Is H a subgroup of R2 ?
(c) Let K = {(x, y) ∈ R2 : x2 + y 2 = 1}. Is K a subgroup of R2 ? Justify your answer.
Exercise 2.29. Suppose G is an Abelian group. Prove that H = {g 2 : g ∈ G} is a subgroup of G
(see Exercise 4.25).
Exercise 2.30. Let
SL(2, R) =
a b
: a, b, c, d ∈ R, ad − bc = 1 .
c d
This is called the special linear group. Show that SL(2, R) is a group under matrix multiplication
and it is a subset of GL(2, R). Conclude that SL(2, R) GL(2, R).
Exercise 2.31. Show that the set Q× = Q − {0} is a subgroup of R×
Exercise 2.32. Show that the set Q>0 = {x ∈ Q : x > 0} is a subgroup R>0 .
45
Exercise 2.33. Let G be an Abelian group and n a fixed positive integer.
(a) Prove that
H = {g ∈ G : g n = e} G.
(b) Prove that
K = {g n : g ∈ G} G.
Exercise 2.34. Suppose G is a group of order n ≥ 3. Prove that G does not have a subgroup of
order n − 1.
Exercise 2.35. Let H and K be subgroups of G. Prove that H ∪ K is a subgroup if and only if
either H ⊆ K or K ⊆ H.
Exercise 2.36. Suppose x, y ∈ G with |x| = |y| = |xy| = 2. Prove that x and y commute with each
other, i.e., xy = yx.
Exercise 2.37. Prove or disprove. If |a3 | = |b3 |, then |a| = |b|.
Exercise 2.38. Prove that G is Abelian if and only if (xy)2 = x2 y 2 for all x, y ∈ G.
Exercise 2.39. Suppose G is a finite group with even order. Prove that G must have an element of
order 2.
Exercise 2.40. Prove that H CG (H) if and only if H is an Abelian subgroup of G.
Exercise 2.41. (a) Find Z(GL(2, R)).
0 −1
(b) Find CGL(2,R)
.
−1 1
1 −1
(c) Find CGL(2,R)
.
−1 0
Exercise 2.42. (a) Suppose that the table shown below is a Cayley table for a group G. Fill in the
blank entries.
· e
e e
a
b
c
d
a
b
c
d
d
c
e
a
e
a
(b) Use the table to compute the following.
(1) CG (b).
(2) Z(G).
(3) CG (d).
46
(4) a(bc).
(5) b2 ac2 d.
(6) (ab)2 .
(7) The order of every element in G.
(8) |abcd|.
2.6
Challenging Exercises
×
Exercise 2.43. Fix an integer n ≥ 2. Let Z×
n = Zn − {0}, and define the following operation on Zn :
×
[a][b] = [ab]. Prove that Zn is a group under this operation if and only if n is prime.
Exercise 2.44. Fix an integer n ≥ 2. Let
Gn = {f : Z → Z : f is a bijection and f (i + n) = f (i) + n ∀i ∈ Z}, and
Hn = {f ∈ Gn : f (i) ≡ i(mod n) for all i ∈ Z}.
(a) Prove that Gn is a group under function composition.
(b) Prove that Hn Gn .
Exercise 2.45. Suppose a, b ∈ G such that |a| = 2k + 1 for some integer k ≥ 0, and ab = b−1 a.
Show that b = b−1 .
Exercise 2.46. In this exercise we give an example of an infinite Abelian group with every element
having finite order. Let G = {x ∈ Q : 0 ≤ x < 1}.
(a) Prove that G is an infinite Abelian group under the operation
(
a+b
if 0 ≤ a + b < 1
a·b=
a + b − 1 if 1 ≤ a + b < 2.
(b) Prove that every element in G has finite order.
Exercise 2.47. (a) Prove that the set of rational numbers Q under addition is a group.
(b) Suppose H is a subgroup of Q (under addition) with the following property:
1
∈ H for every non-zero x ∈ H.
x
Prove that H = {0} or H = Q.
Exercise 2.48. Let G be an Abelian group and H = {x ∈ G : |x| = 2k + 1 for some k ≥ 0}. Prove
that H G.
Exercise 2.49. Fix an integer n ≥ 1. Suppose G is a finite group such that an bn = bn an for all
a, b ∈ G. Let
H = {x ∈ G : gcd(|x|, n) = 1}.
Prove that H is an Abelian subgroup. Hint. First show that H is Abelian, then show its a subgroup.
Note: this does not say G is Abelian when n ≥ 2.
47
3
Generators, Relations, and Cyclic Groups
3.1
Definition of Generators and Relations
Recall that the integers under addition are a group, and every integer is a power of 1 under this
operation, i.e., n = 1n . We say that the integers are generated by 1 under addition, and say 1 is
a generator for the integers. Notice that the integers are also generated by −1 (as n = (−1)−n ).
This shows that it is possible for two different elements to generate the same group. Moreover, most
groups are generated by more than one element (see Dihedral Groups). In this sense, the integers
are very special. We begin with the formal definition of generators.
Generators
Let A be a non-empty set. Define
hAi = {an1 1 an2 2 · · · anmm : m ∈ N, ai ∈ A, ni ∈ Z}.
We call the elements of A generators, and say hAi is the set generated by A.
In the previous definition, the ai ’s in an1 1 an2 2 · · · anmm need not be distinct. Combining exercises 2.26
and 3.5 proves that if A is a non-empty subset of a group G, then hAi is a subgroup of G. When
A is a finite set {a1 , a2 , . . . , an }, we write ha1 , a2 , . . . , an i for the subgroup generated by A instead of
h{a1 , a2 , . . . , an }i. In this course we are mostly interested in the case when A is a finite non-empty
subset of a group G. Expressions of the form an1 1 an2 2 · · · anmm are called words in the elements of A.
The set A can be finite or infinite (countable or uncountable). For example, if A = {a, b, c}, then
some of the elements of this group are:
acb, a2 b2 ca3 , ac3 b2 , abcacbcab, a2 cbcaba3 .
If A is a set consisting of a single element a, then
hAi = hai = {an : n ∈ Z}.
Example 3.1. Find the subgroup generated by A = {12, 18} in Z. Find the subgroup h6i. Is
there any relation between h12, 18i and h6i?
Solution. Since Z is an Abelian group,
h12, 18i = {12n + 18m : n, m ∈ Z} and h6i = {6n : n ∈ Z}.
We now prove that these two sets are in fact equal. To see this, notice that 6 = 12(−1) + 18(1).
So 6 ∈ h12, 18i. Since h12, 18i is a subgroup of Z (in particular, the operation is binary), we have
h6i ⊆ h12, 18i.
Conversely, 12, 18 ∈ h6i (as 12 = 6 × 2 and 18 = 6 × 3). Since h6i is a subgroup of Z, any integer
linear combination of 12 and 18 will be in h6i. That is, h12, 18i ⊆ h6i, completing the proof.
♣
Now we turn our attention to the notion of relations. It is possible that in a group elements interact
with each other in a certain way. For example, it is possible that ab = ba for two elements a and b
48
in a group G. Also, it is possible that ab2 = b2 a−1 in a group. The relations between the elements of
a non-empty subset A of a group G play an important role in the subgroup they generate.
If there is no relation between the elements in A, we write the subgroup generated by A as hAi (as
we did above). However, if there are relations between the elements in A, we write the set generated
by A as follows;
hA : Ri
where R is the set of relations between the ai ’s in A. For example, suppose a is an element of a group
G such that |a| = 4. We write the subgroup generated by a in G as
ha : |a| = 4i.
When there is no confusion about the order of a, we simply write hai.
If A contains more than one element and there are relations, then we always write the subgroup
generated by A as hA : Ri. For example, suppose A = {a, b, c}, where |a| = 2 = |b|, |c| = 3, ab = ba,
and bc = cb. We write the group generated by A as follows:
ha, b, c : |a| = 2 = |b|, |c| = 3, ab = ba, bc = cbi
It is then understood that the elements in this subgroup are strings (or words) consisting only of the
letters a, b, and c (and any power of them) obeying the relations stated. Some of the elements of this
group are:
(abc2 )(cba) = abc3 ba = ab2 a = a2 = e
(abac)(bac) = a2 bcbac = bcbac = b2 cac = cac
(c2 ab)(cba) = c2 acb2 a = c2 aca.
As mentioned in the beginning of this section, if A is a non-empty subset of a group G, combining
Exercises 2.26 and 3.5 proves that hAi is a subgroup of G. We now give an alternate proof of this
fact when A = {a}.
hai is a subgroup
Proposition 3.2. For all a ∈ G we have hai G.
Proof. Since a ∈ hai, hai is not empty. Since the operation in G is binary, hai ⊆ G. Given x, y ∈ hai,
then x = am and y = an for some integers m, n ∈ Z. Then,
xy −1 = am (an )−1 = am−n ∈ hai.
By Proposition 2.21, hai G.
We now show that the order of an element in a group is the same as the order of the subgroup that
element generates in the group.
Theorem 3.3. Let a ∈ G.
1. If |a| = ∞ then, ai = aj if and only if i = j.
2. If |a| = n then, ai = aj if and only if n|(i − j). Moreover, hai = {e, a, a2 , a3 , . . . , an−1 }.
49
Proof. (1) Notice that |a| = ∞ is equivalent to am = e if and only if m = 0. It follows that
ai = aj ⇔ ai−j = e ⇔ i − j = 0 ⇔ i = j.
(2) We first prove ai = aj if and only if n|(i − j). If n|(i − j), then i − j = nk for some integer k, and
ai = ank+j = ank aj = (an )k aj = eaj = aj .
Conversely, if ai = aj , then ai−j = e. By Corollary 2.17, n|(i − j).
From the definition we have {e, a, a2 , a3 , . . . , an−1 } ⊆ hai. Conversely, given ak ∈ hai, by the Division
Algorithm there exist unique integers q and r such that k = nq + r with 0 ≤ r < n. In particular,
ak = anq+r = (an )q ar = ar ∈ {e, a, a2 , a3 , . . . , an−1 },
showing that hai ⊆ {e, a, a2 , a3 , . . . , an−1 }, completing the proof.
Corollary 3.4. |a| = |hai|.
Proof. This follows from Theorem 3.3.
Combining Corollary 3.4 and Proposition 3.2 gives |a| ≤ |G|. The subgroup hai is called the cyclic
subgroup of G generated by a. For example, in Z, h2i is the set of even integers, h3i is the set of all
multiples of 3, and Z = h1i = h−1i. In Z8 , h4i = {0, 4}, h2i = {0, 2, 4, 6}, and
h1i = h3i = h5i = h7i = Z8 .
3.2
Cyclic Groups
In this section, we turn our attention to groups which can be generated by a single element. If
G = hai for some a ∈ G, we say that G is a cyclic group and a is a generator for G. To see where
the name comes from, assume G = hai such that |a| = 5. By Corollary 3.4, |G| = 5. By Theorem
3.3(2), G = {e, a1 , a2 , a3 , a4 }. We can draw this group as follows;
a5
a4
a1
a3
a2
50
Where a5 = e. In general, when |a| = n and G = hai, we can think of G geometrically as a cycle
starting at a and going clockwise (rotating by 360/n each time), increasing the power of a one at
time. The cycle completes with an = e connecting back to a. Below we draw the cycle for G = hai
where |a| = 7:
a6
a7
a5
a1
a4
a2
a3
If the order of a is infinite, the cycle never gets completed. The next example illustrates that a cyclic
group may have many different generators.
Example 3.5. Show that U (10) is cyclic and find all the elements that generate U (10).
Solution. Notice that U (10) = {1, 3, 7, 9}. Moreover U (10) = h3i = h7i. Hence, U (10) is a cyclic
group and 3 and 7 are generators for it. Notice that h9i = {1, 9} =
6 U (10), so 9 is not a generator for
U (10).
♣
A group G is cyclic if and only if there exist an element a ∈ G such that G = hai. In particular, to
show that a group is not cyclic, it suffices to show that G 6= hai for any a ∈ G.
Example 3.6. U (8) is not cyclic.
Solution. Notice that U (8) = {1, 3, 5, 7}, and
|1| = 1, |3| = |5| = |7| = 2.
Since |U (8)| = 4, we have U (8) 6= hai for any a ∈ U (8). Hence, U (8) is not cyclic.
♣
Smallest Subgroup
Let H be a subgroup of a group G and A a subset of G. We say that H is the smallest subgroup
of G containing A if it has the following property:
(K G and A ⊆ K ⊆ H) ⇒ K = H.
If H is a subgroup of G and a ∈ H, every power of a is an element of H (since the operation is
binary). In particular, hai ⊆ H. This shows that hai is the smallest subgroup of G containing a (or
51
more precisely, {a}). A similar proof shows that hAi is the smallest subgroup of G containing A (see
Exercise 3.5).
The next few results focus on the order of elements in subgroups generated by a single element.
Proposition 3.7. Suppose a ∈ G with |a| = n < ∞. Given a natural number k,
hak i = hagcd(k,n) i and |ak | =
n
.
gcd(k, n)
Proof. Set d = gcd(k, n). Then k = dm for some integer m. By Proposition 1.3, there exist integers
u and v such that d = ku + nv. If b ∈ hak i, there exist an integer s such that b = (ak )s = aks =
adms = (ad )ms ∈ had i. Conversely, suppose b ∈ had i. Then b = (ad )z for some integer z. Therefore,
b = adz = akuz+nvz = (ak )uz (an )vz = (ak )uz ∈ hak i. Hence, hak i = had i. This completes the proof of
the first part of the proposition.
Notice that if l|n, then
(al )n/l = an = e ⇒ |al | ≤ (n/l).
In particular, if i < (n/l), then (al )i 6= e (by the minimality of n, since li < n). Hence, |al | = (n/l).
n
Now we prove the second part of the proposition: |ak | = |hak i| = |had i| = |ad | = .
d
Corollary 3.8. Suppose G is a finite cyclic group of order n. Then |a| |n for all a ∈ G.
Proof. Let G = hgi with n = |G| = |g|. Given a ∈ G, we have a = g k for some integer k, and by
n
n
n
= gcd(k, n).
Proposition 3.7,
= k =
n
|a|
|g |
gcd(k,n)
The following corollary characterizes when two cyclic groups are equal.
Corollary 3.9. Let |a| = n < ∞. Then,
(1) hai i = haj i if and only if gcd(i, n) = gcd(j, n).
(2) |ai | = |aj | if and only if gcd(i, n) = gcd(j, n).
(3) hai = haj i if and only if gcd(j, n) = 1.
(4) |a| = |aj | if and only if gcd(j, n) = 1.
Proof. We prove (1) and (2). The proofs of (3) and (4) follow from (1) and (2), respectively (by
letting i = 1).
(1) Suppose gcd(i, n) = gcd(j, n). By Proposition 3.7,
hai i = hagcd(i,n) i = hagcd(j,n) i = haj i.
52
Conversely, if hai i = haj i, then Corollary 3.4 and Proposition 3.7 imply that
n
n
= |ai | = |hai i| = |haj i| = |aj | =
,
gcd(i, n)
gcd(j, n)
and the result follows.
(2) By Proposition 3.7, gcd(i, n) = gcd(j, n) ⇔
n
n
=
⇔ |ai | = |aj |.
gcd(i, n)
gcd(j, n)
Given n ≥ 2, Zn is a cyclic group with Zn = h1i. Since |1| = n, the preceding corollary tells us that
every number relatively prime to n generates Zn . For examples, Z8 is generated by 1, 3, 5, and 7.
Fundamental Theorem of Cyclic Groups
Theorem 3.10.
1. Every subgroup of a cyclic group is cyclic.
2. If |hai| = n and H hai, then |H| | n.
3. Suppose |a| = n. For each k ∈ N with k|n, the group hai has exactly one subgroup of
order k, namely han/k i.
Notice that this theorem tells us how many subgroups a finite cyclic group has and how to find them
(they correspond to the divisors of the order of the group).
Proof. (1) Let G = hai for some a ∈ G and suppose H G. We show that H is cyclic. If H is the
trivial subgroup, then H = {e} = hei. Suppose H 6= {e}. Then there exist a non-identity element
x ∈ H. Since a is a generator for G, x = ab for some non-zero integer b. Notice that x−b ∈ H, and
either b or −b > 0. Let m be the smallest positive integer such that am ∈ H (Well Ordering Principle
guarantees the existence of such an m). Since H is a subgroup of G and am ∈ H, ham i ⊆ H. Now
we show the reverse inclusion, which will complete the proof. If h ∈ H ⊆ G = hai, then h = aα for
some α ∈ Z. By the Division Algorithm, there exist unique integers q and r such that α = mq + r
with 0 ≤ r < m. Thus
aα = amq+r = amq ar ⇒ ar = a−mq aα = (am )−q h ∈ H.
Minimality of m implies that r = 0. Hence,
α = mq ⇒ h = aα = (am )q ∈ ham i.
(2) If H is a subgroup of hai, then H is cyclic and H = ham i, where m is the smallest positive integer
such that am ∈ H (by part (1)). Hence,
|H| = |ham i| = |am | =
n
⇒ |H| |n.
gcd(m, n)
(3) If k|n, then han/k i hai, and
|han/k i| =
n
n
=
= k.
gcd(n, n/k)
n/k
53
We will show that this is the only subgroup of order k. Suppose H hai with |H| = k. By part (1)
H = ham i, where m is the smallest positive integer such that am ∈ H. By the Division Algorithm,
there exist unique integers q and r such that n = mq + r with 0 ≤ r < m. Thus
ar = a−mq an = a−mq ∈ H,
and minimality of m implies that r = 0. That is, m|n. Hence,
m|n ⇒ m = gcd(m, n) ⇒ k = |H| = |ham i| = |agcd(n,m) | =
⇒ m=
n
n
=
gcd(n, m)
m
n
⇒ H = ham i = han/k i.
k
Corollary 3.11. For each positive divisor k of n, the set hn/ki is the unique subgroup of Zn
of order k. Moreover, these are the only subgroups of Zn .
Proof. The group Zn is cyclic, and the result follows by Theorem 3.10.
For example, consider the group Z15 . The subgroups of Z15 are as follows:
h15/15i
h15/5i
h15/3i
h15/1i
=
=
=
=
Z15 ,
h3i,
h5i,
h0i,
order 15,
order 5,
order 3,
order 1,
and these are all the subgroups of Z15 .
Example 3.12. Suppose G is a finite cyclic group and has three subgroups: G itself, {e} and
a subgroup of order 5. What is |G|?
Solution. Let |G| = n. Since G has three distinct subgroups, |G| ≥ 3. By the Fundamental Theorem
of Cyclic Groups, n = 5k for some natural number 5k with 1 < k < n (for if k = 1, the only
subgroups of G would be the trivial subgroup and itself). It follows that G must have a subgroup of
order k. If k 6= 5, then G would have at least four distinct subgroups (one of each of the following
sizes; 1, 5, k, and n), a contradiction. Hence, k = 5 and |G| = 25.
♣
We now introduce an important number-theoretic function called the Euler phi f unction.
Euler Phi Function
Define φ : N → N as follows:
φ(1) = 1, and φ(n) = |U (n)| for all natural numbers n ≥ 2.
In particular, for n ≥ 2, the Euler phi function denotes the number of positive integers less
than n and relatively prime to n.
54
By the fundamental Theorem of Cyclic groups, the order of the elements in the group must divide
the order of the group, provided the group is finite. The next theorem tells us how many elements
of a given order there are in a finite cyclic group.
Theorem 3.13. If d ∈ N and d|n, then the number of elements of order d in a cyclic group of
order n is φ(d).
Proof. Let G = hai be a cyclic group of order n and d|n. By the Fundamental Theorem of Cyclic
Groups G has exactly one subgroup H of order d, and H is cyclic. Let H = hbi. Suppose x ∈ G with
|x| = d, and let H 0 = hxi. Then H 0 is a subgroup of G of order d, thus H 0 = H. In particular, any
element of order d is in H, and every such element generates H (by Corollary 3.4). But, hbk i = hbi
if and only if gcd(k, d) = 1, and there are φ(d) such natural numbers.
Example 3.14. How many elements in Z6 generate Z6 . Find all of them.
Solution. Notice that |Z6 | = 6 and Z6 = h1i. So we need to find all the elements in Z6 that have
order 6. By Theorem 3.13, there are φ(6) = 2 of them. By Corollary 3.9, these are the numbers that
are relatively prime to 6, namely, 1 and 5.
♣
Corollary 3.15. In a finite group the number of elements of order d is a multiple of φ(d) (we
no longer assume our group is cyclic).
Proof. If a finite group has no elements of order d, then 0 = 0 × φ(d), and we are done. Now suppose
there exist a ∈ G such that |a| = d. By our preceding theorem we know that hai has φ(d) elements
of order d (since hai is a cyclic group). If all such elements of G are in hai, then we get φ(d) elements
of orders d in G and we are done. Suppose that there exist b ∈ G such that |b| = d and b ∈
/ hai.
Then, hbi also has φ(d) elements of order d. If x ∈ hai ∩ hbi with |x| = d, then hai = hxi = hbi,
a contradiction. In particular, hai and hbi share no element of order d. Thus, we have found 2φ(d)
elements of order d. If all elements of order d in G are in hai or hbi, we are done. If not, continue
this process. Since G is finite the process stops, and the result follows.
Theorem 3.16. For p-prime, n ∈ N, φ(pn ) = pn − pn−1 . Furthermore, If gcd(a, b) = 1, then
φ(ab) = φ(a)φ(b).
The proof of the above result is number theoretic in nature, and is omitted. You can find the proof
in most elementary number theory books. We can use Theorem 3.16 to find φ(n) for any natural
number n. For example,
φ(25) = φ(52 ) = 52 − 5 = 20
φ(100) = φ(22 )φ(52 ) = (22 − 2)(52 − 5) = 40
φ(105) = φ(3)φ(5)φ(7) = 2 × 4 × 6 = 48.
55
3.3
Subgroup Lattice
The relationship between the various subgroups of a group can be illustrated with a subgroup lattice
of the group. This is a diagram that includes all the subgroups of the group and connects a subgroup
H at one level to a subgroup K at a higher level with a sequence of line segments if and only if H is
a proper subgroup of K. It should be noted that these diagrams can become very big if the group
has many subgroups. The following examples give the subgroup lattices for Z10 , U (8), and Z30 .
Example 3.17. Draw the Subgroup lattice of Z10 .
Solution. By the Fundamental Theorem of Cyclic Groups Z10 has only 4 subgroups: the trivial
subgroup, h2i, h5i, and Z10 .
Z10 = h1i
h2i
h5i
h0i
♣
Example 3.18. Draw the Subgroup lattice of U (8).
Solution. The subgroup lattice of U (8) is
U (8)
h3i
h5i
h7i
h1i
♣
56
Example 3.19. Draw the Subgroup lattice of Z30 .
Solution. By the Fundamental Theorem of Cyclic Groups, all the subgroups of Z30 correspond to
the divisors of 30. So the subgroup lattice looks like
Z30 = h1i
h2i
h3i
h5i
h6i
h10i
h15i
h0i
♣
Example 3.20. The Klein 4-group , denoted by V4 , is the group of order 4 with the following
Cayley table:
· 1
1 1
a a
b b
c c
a b c
a b c
1 c b
c 1 a
b a 1
Draw the subgroup lattice of V4 .
Solution.
V4
hai
hbi
hci
h1i
♣
57
3.4
Exercises
Unless otherwise stated G is a group.
Exercise 3.1. In the set Z under addition find the following subgroups.
(a) h2, 4i.
(b) h21, 38i.
(c) h6, 15, 21i.
(d) h9, 14, 28i.
Exercise 3.2. Suppose G = hai and |a| = 20, find the following subgroups of G.
(a) ha2 i.
(b) ha3 i.
(c) ha4 i.
(d) ha5 i.
(e) ha6 i.
(f) ha9 i.
Exercise 3.3. U (15) is a group under multiplication modulo 15 and 4, 7 ∈ U (15).
(a) Is h4i = U (15)? Justify your answer.
(b) Is h7i = U (15)? Justify your answer.
Exercise 3.4. Suppose G = ha, b : |a| = |b| = 3 and |ab2 | = 1i. Find a5 b2 .
Exercise 3.5. Let A be a non-empty subset of a group G, and define
A = {an1 1 an2 2 · · · anmm : m ∈ N, ai ∈ A, ni = ±1}.
(a) Prove that hAi = A.
(b) Prove that
hAi =
\
H.
A⊆H
HG
Conclude that hAi = A if and only if A G.
(c) Prove that if H is a subgroup of G containing A, then hAi ⊆ H. This shows that hAi is the
smallest subgroup of G containing A.
Exercise 3.6. Prove that hai = ha−1 i for all a ∈ G.
58
Exercise 3.7. Suppose a ∈ G such that |a| = ∞. If m is a non-zero integer, prove that han i ⊆ ham i
if and only if m|n.
Exercise 3.8. (a) Prove that a group of order 3 must be cyclic.
(b) Prove that a group of order 5 must be cyclic.
(c) Give an example of a group of order 4 that is not cyclic.
Exercise 3.9. Suppose that G is a group that has exactly one non-trivial proper subgroup. Prove
that G is cyclic.
Exercise 3.10. Let
G=
1 0
:n∈Z .
n 1
Prove that G is a cyclic group under matrix multiplication.
Exercise 3.11. Suppose that a finite cyclic group G has exactly three distinct subgroups: G itself,
the trivial subgroup, and a subgroup of order p, where p is prime. Find |G|.
Exercise 3.12. Let G be a cyclic group of order 15. Suppose x ∈ G and exactly two of x3 , x5 , and
x9 are equal. Find |x12 | and |x14 |.
Exercise 3.13. (a) Suppose a, b ∈ G such that |a| = 8 and |b| = 19. Prove that hai ∩ hbi = {e}.
(b) Suppose a, b ∈ G such that |a| = n, |b| = m, and gcd(n, m) = 1. Prove that hai ∩ hbi = {e}.
Exercise 3.14. Suppose G is a cyclic group and it has at least one element of infinite order. Prove
that every non-identity element in G has infinite order.
Exercise 3.15. Prove that no group can have exactly two elements of order 2.
Exercise 3.16. Suppose G is a cyclic group with order 99. Prove that G must have an element of
order 3.
Exercise 3.17. Draw the subgroup lattice for the following groups.
(a) U (25).
(b) Z15
(c) Z20 .
Exercise 3.18. Draw the subgroup lattice for U (40). Which of its subgroups are cyclic?
Exercise 3.19. Prove that no group is the union of two proper subgroups.
Exercise 3.20. (a) Suppose H, K Q (under addition), H 6= {0}, and K 6= {0}. Prove that
H ∩ K 6= {0}.
(b) By example show the statement in part (a) is false if one replaces Q with R in part (a).
Exercise 3.21. Suppose G is a group with p elements of order p, where p is prime. Prove that G is
not cyclic.
59
3.5
Challenging Exercises
Exercise 3.22. Let a, b ∈ G. Prove that
ha, b : |a| = |b| = |ab| = 2i = {e, a, b, ab}.
Exercise 3.23. Let
G = ha, b : ab2 = b3 a, ba3 = a2 bi.
What are the possible orders of G (justify your answer)?
Exercise 3.24. Let
G = ha, b : a3 = b2 = (ab)3 = ei.
What are the possible orders of G (justify your answer)?
Exercise 3.25. Let
G = hx, y : x4 = y 3 = e, xy = y 2 x2 i.
(a) Prove that y 2 = y −1 and yx3 = x3 y. Hint. For the second part show that y 2 x3 y = x3 .
(b) Show that xy = yx.
(c) Show that xy = e.
(d) Show that x = y = e, and conclude that G is the trivial group.
Exercise 3.26. In Exercise 2.46 we gave an example of an infinite group whose elements all had
finite order. This showed that it is possible for all elements in a group to have finite order but the
group itself having infinite order. In this question we show that if the group has a finite number of
subgroups, then the group must be finite.
(a) Suppose a ∈ G with |a| = ∞. Prove that hai i = haj i if and only if i = ±j.
(b) Suppose G has a finite number of subgroups. Prove that every element in G has finite order.
Hint. Use part (a).
(c) Suppose G has a finite number of subgroups. Prove that |G| < ∞.
Exercise 3.27. Prove that a group G is the union of proper subgroups if and only if G is not cyclic.
Exercise 3.28. Prove that Q under addition is not cyclic.
Exercise 3.29. A group G is called divisible if it satisfies the following two properties:
1. G is an Abelian group, and
2. Given x ∈ G and a non-zero integer n, there exist a y ∈ G such that y n = x.
Notice that if the operation in your group is addition, y n = x means ny = x.
(a) Prove that Q and R are divisible groups under addition.
60
(b) Prove that Z under addition is not a divisible group.
(c) Prove that the set of the positive rationals under multiplication is not a divisible group.
Exercise 3.30. A groups G is said to be f initely generated if G = hAi, where A is a finite subset
of G. If a group is divisible, prove that it is not finitely generated. Conclude that Q and R are not
finitely generated under addition.
Exercise 3.31. (a) Prove that every finite group is finitely generated.
(b) Prove that every finitely generated subgroup of the additive group Q is cyclic.
Exercise 3.32.
Prove that the multiplicative
group of positive rational numbers (Q>0 ) is generated
1
by the set
: p is a prime number .
p
Exercise 3.33. Let G be a finite Abelian group and suppose that there exist an a ∈ G such that
|g| ≤ |a| for all g ∈ G. Prove that |g| divides |a| for all g ∈ G.
61
4
4.1
Dihedral, Symmetric, and Alternating Groups
Dihedral Groups
An important family of examples of groups is the class of groups whose elements are symmetries of
geometric objects. Suppose we wanted to move an equilateral triangle in such a way that it occupies
the exact same place in space. One can think of removing an equilateral triangle region from a
plane (a piece of paper), move it in some way, then put the triangle back into the space it originally
occupied. In this section we describe all possible ways in which this can be done for a triangle, and
more generally, for an n-gon.
First we look at the equilateral triangle that we mentioned above. What are the possible movements
that we can do to it? There are 6 possible movements, and they are as follows:
(1) Rotation by zero degrees counter clockwise, denoted by R0 :
A
B
A
C
B
C
(2) Rotation by 120 degrees counter clockwise, denoted by R120 :
A
B
C
C
A
B
(3) Rotation by 240 degrees counter clockwise, denoted by R240 :
A
B
B
C
C
A
There are also three reflections (a reflection is a rotation by 180 degrees about a give axis);
(4) We denote the first one by V1 :
62
A
B
A
C
C
B
(5) We denote the second one by V2 ;
A
B
C
C
B
A
(6) We denote the third one by V3 :
A
B
B
C
A
C
Remark. The axis of reflection in (4), (5), and (6) do not move with the point that they go through.
For example, V1 goes through point A in the picture in (4). However, it is understood that V1 always
represents that axis of reflection. To illustrate this, we can apply V1 to the triangle
B
C
A
to get
B
A
C
63
We define D3 = {R0 , R120 , R240 , V1 , V2 , V3 }. We can compose elements in this set. For example,
R120 V1 is (composition is from right to left):
A
A
B
V1
−→
B
C
R120
−→
C
B
A
C
In particular, R120 V1 = V3 . Similarly, we can compute V1 R120 :
A
C
C
R120
−→
B
C
V1
−→
A
B
B
A
Thus V1 R120 = V2 . This shows that the elements in D3 do not necessarily commute. In particular,
D3 is a non-Abelian group. Below we give the Cayley table for D3 :
·
R0
R120
R240
V1
V2
V3
R0
R0
R120
R240
V1
V2
V3
R120
R120
R240
R0
V2
V3
V1
R240
R240
R0
R120
V3
V1
V2
V1
V1
V3
V2
R0
R240
R120
V2
V2
V1
V3
R120
R0
R240
V3
V3
V2
V1
R240
R120
R0
Notice that R0 is the identity element in D3 . Moreover, |R0 | = 1, |R120 | = |R240 | = 3, and |V1 | =
|V2 | = |V3 | = 2. From the Cayley table we get the following:
R120 V1
R240 V1
R120 V2
R240 V2
R120 V3
R240 V3
=
=
=
=
=
=
V1 R240
V1 R120
V2 R240
V2 R120
V3 R240
V3 R120
−1
= V1 R120
−1
= V1 R240
−1
= V2 R120
−1
= V2 R240
−1
= V3 R120
−1
= V3 R240
.
Let r = R120 (or R240 ) and s = V1 (or V2 or V3 ). Then
e = r0 , r = R120 , r2 = R240 ,
s = V1 , sr = V2 , sr2 = V3 .
So the Cayley table above becomes;
64
·
e
r
r2
s
sr
sr2
e
e
r
r2
s
sr
sr2
r
r
r2
e
sr
sr2
s
r2
r2
e
r
sr2
s
sr
s
s
sr2
sr
e
r2
r
sr
sr
s
sr2
r
e
r2
sr2
sr2
sr
s
r2
r
e
In particular,
D3 = hr, s : |r| = 3, |s| = 2, rs = sr−1 i.
Given an integer n ≥ 3, it is customary to let r = R360/n (though any rotation with order n works),
and let s be any of the reflection axes of the n-gon that leaves it occupying the same space. It can
be shown (see Exercise 4.26) that
Dn = hr, s : |r| = n, |s| = 2, rs = sr−1 i.
This group is called the Dihedral group of order 2n (the operation is composition).
Example 4.1. Draw the subgroup lattice of D4 = hr, s : |r| = 4, |s| = 2, rs = sr−1 i.
Solution.
D4
hsi
hs, r2 i
hri
hrs, r2 i
hr2 si
hr2 i
hrsi
hr3 si
hei
♣
4.2
Symmetric Groups
A permutation of a set A is a function from A to A that is both one to one and onto (i.e., a bijection).
A permutation group of a set A is a set of permutations of A that forms a group under function
composition. Although groups of permutations of any non-empty set exist, we will focus on the case
where our set A is finite. It is customary, as well as convenient, to take A = {1, 2, ..., n}, where
n ∈ N.
65
Symmetric Group
If A = {1, 2, ..., n}, the set of all permutations of A is called the symmetric group of degree n,
denoted by Sn .
Sn Is a Group
Proposition 4.2. The symmetric group of degree n is a group under composition.
Proof. The identity map on A, denoted by ε, is a bijection (thus a permutation). That is, ε ∈ Sn .
This map is the identity element in Sn .
• Associativity follows from the fact that function composition is associative.
• The operation is binary, since the composition of two bijections is a bijection.
• A function is a bijection if and only if it has an inverse (which is also a function). Moreover,
the inverse is also a bijection. This show that Sn satisfies the inverse property for a group.
Hence, Sn is a group under composition.
An integer a ∈ {1, 2, . . . , n} is said to be f ixed by a permutation α ∈ Sn if α(a) = a.
m-Cycle
An m-cycle in Sn is a string of integers of length m, written (a1 a2 . . . am ), where ai ∈
{1, 2, . . . , n}, that corresponds to the permutation in Sn that sends ai to ai+1 for 1 ≤ i ≤ m−1,
am to a1 , and fixes all the other integers in {1, 2, . . . , n} that do not appear in the cycle. We
say m is the length of the cycle. One can think of an m-cycle as follows;
a1
a2
···
a3
am−1
am ,
with the understanding that any integer not in the cycle is fixed (that is, it is sent back to
itself).
It should be noted that when working with cycles it is very important to know which Sn we are
working in. For example, the element (1 2 3) can be thought of as an element of Sn for any n ≥ 3.
But the permutation it corresponds to depends on the value of n. The table below gives a description
of (1 2 3) in Sn for n = 3, 4, 5, 6:
S3
S4
S5
S6
1 → 2, 2 → 3, 3 → 1
1 → 2, 2 → 3, 3 → 1, and fixes 4.
1 → 2, 2 → 3, 3 → 1, and fixes 4 and 5.
1 → 2, 2 → 3, 3 → 1, and fixes 4, 5, and 6.
66
In general, for each α ∈ Sn the numbers 1 to n will be rearranged and grouped into k cycles of the
form
(a1 a2 . . . am1 ) ◦ (am1 +1 am1 +2 . . . am2 ) ◦ . . . ◦ (amk−1 +1 amk−1 +2 . . . amk ).
For example, the element (1 2)(3 4) ∈ S5 sends 1 to 2, 2 to 1, 3 to 4, 4 to 3, and fixes 5. Similarly,
the element (1 3 4)(2 5) ∈ S7 sends 1 to 3, 3 to 4, 4 to 1, 2 to 5, 5 to 2, and fixes 6 and 7. Below we
draw a picture of these maps, respectively;
(1 2)(3 4) ∈ S5 :
1
(1 3 4)(2 5) ∈ S7 :
2
1
3
3
4
4
2
5
Recall that function composition is the operation in Sn . So how do we compose cycles? Exactly the
same way we compose functions (after all, permutations are functions that happen to be bijections).
For example, suppose α = (1 4 5)(6 2) and β = (6 5 1)(3 4 2) in S6 . We will compute α ◦ β in two
ways. First we use the method we have used since high school. Notice that
α(1) = 4, α(2) = 6, α(3) = 3, α(4) = 5, α(5) = 1, α(6) = 2, and
β(1) = 6, β(2) = 3, β(3) = 4, β(4) = 2, β(5) = 1, β(6) = 5.
So
(α ◦ β)(1)
(α ◦ β)(2)
(α ◦ β)(3)
(α ◦ β)(4)
(α ◦ β)(5)
(α ◦ β)(6)
=
=
=
=
=
=
α(β(1)) = α(6) = 2,
α(β(2)) = α(3) = 3,
α(β(3)) = α(4) = 5,
α(β(4)) = α(2) = 6,
α(β(5)) = α(1) = 4,
α(β(6)) = α(5) = 1.
In particular,
α◦β :
1
2
3
5
4
6
Thus, α ◦ β = (1 2 3 5 4 6). As you can see, this method is very tedious, specially if we start working
in Sn for large n. Notice that function composition goes from right to left. The same idea goes when
we compose permutations written in cycle notation. So for us to compute (α ◦ β)(a), we look at our
cycle for β and see where it sends a, and then we look at our cycle for α and see where it sends β(a),
and this gives us where α ◦ β sends a. Below we draw a picture of this process for a = 1, 2 and 3:
(1
4
5)(6
2)
◦
(6
5
1)(3
67
4
2)
=⇒ (α ◦ β)(1) = 2
(1
4
5)(6
2)
◦
(6
5
1)(3
4
2)
=⇒ (α ◦ β)(2) = 3
(1
4
5)(6
2)
◦
(6
5
1)(3
4
2)
=⇒ (α ◦ β)(3) = 5
Notice that there is no dashed arrow in the second diagram. The reason for this is 3 does not appear
in the cycles of α, so we know α fixes 3. A similar computation for the other entries shows that
α ◦ β = (1 2 3 5 4 6) (of course, same answer as above). Notice that β ◦ α = (1 2 5 6 3 4). In
particular, α ◦ β 6= β ◦ α. This should not come as a surprise, since function composition is not a
commutative operation.
Much like we did for function composition (in Chapter 1), we write αβ instead of α ◦ β.
Example 4.3. Construct the Cayley table for S3 .
Solution. Notice that S3 = {ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}. The Cayley table for S3 is:
◦
ε
(1 2)
(1 3)
(2 3) (1 2 3) (1 3 2)
ε
ε
(1 2)
(1 3)
(2 3) (1 2 3) (1 3 2)
(1 2)
(1 2)
ε
(1 3 2) (1 2 3) (2 3)
(1 3)
(1 3)
(1 3)
(123)
ε
(1 3 2) (1 2)
(2 3)
(2 3)
(2 3) (1 3 2) (1 2 3)
ε
(1 3)
(1 2)
(1 2 3) (1 2 3) (1 3)
(2 3)
(1 2) (1 3 2)
ε
(1 3 2) (1 3 2) (2 3)
(1 2)
(1 3)
ε
(1 2 3)
♣
Example 4.4. List the elements in S4 . Let α = (1 2)(3 4) and β = (1 2 4 3). Find α2 , αβ,
and β 2 .
Solution.
S4 = {ε, (1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3), (1 3 2), (1 2 4),
(1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2)(3 4), (1 3)(2 4),
(1 4)(2 3), (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)}.
Moreover,
α2 = ε, αβ = (2 3), β 2 = (1 4)(2 3).
We now focus on the properties of Sn and its elements.
Proposition 4.5. The group Sn has n! elements.
68
♣
Proof. The permutations of {1, 2, . . . , n} are precisely the injective functions on this set to itself
because it is finite (see Exercise 1.19) and we can count the number of injective functions. An
injective function α can send the number 1 to any of the n element in {1, 2, . . . , n}; α(2) can then
be any of the elements of this set except α(1), so there are n − 1 choices for α(2); similarly, for α(3)
we have n − 2 choices, and so on. Thus there are precisely n × (n − 1) × (n − 2) × . . . × 2 × 1 = n!
possible injective functions from {1, 2, . . . , n} to itself. Since each of these injections corresponds to
a bijection, |Sn | = n!.
Sn Is A Non-Abelian Group For All n ≥ 3
Proposition 4.6. The group Sn is non-Abelian for all n ≥ 3.
Proof. Notice that (1 2) and (1 3) are both in Sn for n ≥ 3. Moreover,
(1 2)(1 3) = (1 3 2) and (1 3)(1 2) = (1 2 3).
That is, (1 2)(1 3) 6= (1 3)(1 2), and the result follows.
Proposition 4.7. If α = (a1 a2 ... am ), then α−1 = (am am−1 ... a1 ).
Proof. The proof follows from the fact that (a1 a2 ... am )(am am−1 ... a1 ) = ε.
For example, the inverse of (1 2 3) (in any Sn with n ≥ 3) is (3 2 1) and the inverse of (1 3 4 7 5)
(in any Sn with n ≥ 7) is (5 7 4 3 1). Notice that if α = (1 3)(2 4 3) ∈ S4 , then α−1 6= (3 1)(3 4 2).
To compute the inverse we use that fact that (ab)−1 = b−1 a−1 . So α−1 = (2 4 3)−1 (1 3)−1 , and
now the preceding proposition gives us α−1 = (3 4 2)(3 1). In general, if α = α1 α2 · · · αk , then
α−1 = αk−1 · · · α2−1 α1−1 .
Disjoint Cycles
Two cycles are said to be disjoint if they have no numbers in common.
For example, (1 2) and (3 4 5) in S5 are disjoint cycles. However, (1 2 3) and (3 4 5) in S5 are not
disjoint cycles. As we saw before, two elements in Sn do not necessarily commute. However, under
certain special circumstances they do commute.
Proposition 4.8. Disjoint cycles commute.
Proof. Let α = (a1 a2 ... am ) and β = (b1 b2 ... bn ) with no common entries. If c does not appear in
either of the two cycles, then they both fix c, and αβ(c) = c = βα(c). For each ai ,
αβ(ai ) = α(ai ) = ai+1 = α(ai ) = β(α(ai )) for i = 1, 2, . . . , m − 1, and
αβ(am ) = α(am ) = a1 = α(am ) = β(α(am )).
Similarly, αβ(bi ) = βα(bi ) for each bi , completing the proof.
69
Remark. The converse of the previous proposition is false. For example, no cycle is ever disjoint
from itself, however, every cycle commutes with itself. Exercise 4.29 gives a necessary and sufficient
condition for two cycles to commute with each other.
Theorem 4.9. Every permutation of a finite set can be written as a cycle or as a product of
disjoint cycles.
Proof. Let α be a permutation on A = {1, 2, . . . , n}. To write α in disjoint cycle form, we start by
choosing any member of A, say a1 , and let a2 = α(a1 ), a3 = α(a2 ) = α(α(a1 )) = α2 (a1 ), and so
on, until we arrive at a1 = αm (a1 ) for some natural number m (we know such an m exist since A
is a finite group and a1 , α(a1 ), α2 (a1 ), . . . are all in A). Hence, we get an m-cycle (a1 a2 . . . am ) with
ai+1 = αi (a1 ) for 1 ≤ i ≤ m − 1 and a1 = αm (a1 ). If all the elements in A appear in this cycle, we
get
α = (a1 a2 . . . am ),
and the proof is done. If not, pick b1 not appearing in the cycle and repeat the process with b1 to
obtain an m1 -cycle (b1 b2 . . . bm1 ). These cycles are disjoint (since α is a bijection). If all the elements
in A appear in one of these two cycles, we get
α = (a1 a2 . . . am )(b1 b2 . . . bm1 ),
and the proof is done. If not, repeating this process and using the fact that A is a finite set (implying
that this process must stop after a finite number of iterations) we get a disjoint set of cycles whose
composition equals α.
The cycle decomposition of a permutation is its decomposition into disjoint cycles. By the previous
theorem, every element in Sn has a cycle decomposition. Since disjoint cycles commute, it does not
matter in what order we write the cycles that appear in the cycle decomposition of a given element in
Sn . Moreover, this cycle decomposition is unique up to reordering (proof of this is left to the reader).
Theorem 4.10. (1) If α ∈ Sn is an m-cycle, then |α| = m.
(2) The order of a permutation of a finite set written in disjoint cycles (i.e., its cycle decomposition) is the least common multiple of the length of the cycles.
Proof. The proof is left as an exercise.
Example 4.11. Let α = (1 2 4)(1 3 5)(2 4 6) ∈ S6 . Find |α| and α−1 .
Solution. Notice that α = (1 2 4)(1 3 5)(2 4 6) = (1 3 5 2)(4 6). In particular, |α| = lcm(2, 4) = 4.
Moreover, α−1 = (4 6)−1 (1 3 5 2)−1 = (4 6)(2 5 3 1) = (2 5 3 1)(4 6).
♣
Notice that (1 2 3) = (3 1 2) = (2 3 1). In general, there are m ways to write an m-cycle;
(a1 a2 a3 . . . am−1 am ) = (am a1 a2 a3 . . . am−1 )
= (am−1 am a1 a2 . . . am−2 )
..
.
= (a2 a3 . . . am−1 am a1 ).
70
These are called the cyclic permutations of the m-cycle (a1 a2 a3 . . . am−1 am ). These cyclic
permutations play a big role in proving a very famous theorem in finite group theory; Cauchy’s
Theorem (see Exercise 7.34).
4.3
Alternating Groups
In this section, we define even permutations and show that they form a subgroup in Sn . We start
with a definition.
Transposition
A 2-cycle is called a transposition.
For example, in S3 we have the following transpositions: (1 2), (1 3), (2 3). The next proposition
shows that the set of transpositions in Sn generate Sn .
Proposition 4.12. Every permutation in Sn (n > 1) is a product of transpositions (not necessarily disjoint).
Proof. Since every permutation is a product of disjoint cycles (by Theorem 4.12), it suffices to
show that every cycle is a product of transpositions. Given an m-cycle (a1 a2 ... am ) we have
(a1 a2 ... am ) = (a1 am )(a1 am−1 )...(a1 a2 ), completing the proof.
Every element in Sn (n ≥ 2) can be written in many different ways as a non-disjoint product of
transpositions. For example, in S3 ,
(1 2 3) = (1 2)(2 3) = (1 3)(1 2) = (1 2)(1 3)(1 2)(1 3).
Theorem 4.13. If ε = β1 β2 ...βr , where βi are transpositions, then r is even.
Proof. The proof is left as an exercise.
Even and Odd Permutations
A permutation that can be expressed as a product of an even number of transpositions is called
an even permutation. A permutation that can be expressed as a product of an odd number of
transpositions is called an odd permutation.
A natural questions is: “given an element α ∈ Sn , is it possible for one of its decompositions into
transpositions to use an even number of transpositions and another decomposition into transpositions
use an odd number of them?” The next propositions states that this cannot occur.
Proposition 4.14. If a permutation α can be expressed as a product of an even (odd) number
of transpositions, then every decomposition of α into a product of transpositions must have an
even (odd) number of transpositions.
71
Proof. Notice that if γ is a transposition, then γ −1 = γ. Suppose
β1 β2 . . . βr = α = γ1 γ2 . . . γs ,
then
−1
(β1 β2 . . . βr )(γ1 γ2 . . . γs )−1 = ε ⇒ β1 β2 . . . βr γs−1 γs−1
. . . γ1−1 = ε
⇒ β1 β2 . . . βr γs γs−1 . . . γ1 = ε.
This is a decomposition of ε as a product of transpositions. By Theorem 4.13, r + s is even. Hence,
either both r and s are even or both are odd.
One should notice that the set of odd permutations does not form a subgroup in Sn provided n ≥ 2.
For example, the operation is not binary on the odd permutations (the composition of two odd
permutations gives an even permutation). Also, the identity is not an odd permutation. However,
the set of even permutation do form a subgroup.
Alternating Group
The set of the even permutations in Sn , denoted by An , is called the alternating group of degree
n.
Even Permutations Form A Subgroup
Proposition 4.15. For n ≥ 1, the set of even permutations in Sn is a subgroup of Sn .
Proof. By Theorem 4.13, ε is an even permutation. By definition, An ⊆ Sn . Moreover, if α and β
are even permutations, then
α = α1 α2 . . . αm and β = β1 β2 . . . βk ,
|
{z
}
| {z }
m even
k even
where αi and βi are transpositions. It follows that
αβ −1 = (α1 α2 . . . αm )(β1 β2 . . . βk )−1 = α1 α2 . . . αm βk−1 . . . β2−1 β1−1 = α1 α2 . . . αm βk . . . β2 β1 ..
|
{z
}
m+k even
Hence, αβ −1 is an even permutation, and the result follows by Proposition 2.21.
We now prove that the number of even permutations and odd permutations in Sn is the same. In
particular, they each constitute half of Sn .
Order of Alternating Group
Proposition 4.16. For n ≥ 2, |An | =
|Sn |
n!
= .
2
2
72
Proof. Let On be the set of odd permutations in Sn , and define f : An → On via α → (1 2)α. Then
f is a bijection (proof of this is left to the reader). In particular, |An | = |On |. Since {An , On } is a
partition for Sn , we get
|Sn | = |An | + |On | = 2|An |.
Hence, |An | =
|Sn |
n!
= .
2
2
A very useful way to tell if an m-cycle is even or odd is by looking at m.
Proposition 4.17. An m-cycle in Sn is an even permutation if and only if m is odd.
Proof. Let α = (a1 a2 . . . am ). Notice that
α = (a1 am )(a1 am−1 ) . . . (a1 a2 ) .
{z
}
|
m−1 terms
The result follows from the observation that m is odd if and only if m − 1 is even.
For example, α = (1 2 3) is an even permutation in Sn for n ≥ 3. Similarly, β = (1 2 4 3) is an odd
permutation in Sn for n ≥ 4.
Example 4.18. Construct the Cayley table for A3 .
Solution. Notice that S3 = {ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}, (1 2 3) = (1 3)(1 2), and (1 3 2) =
(1 2)(1 3). That is, (1 2 3) and (1 3 2) are even permutations. In particular, A3 = {ε, (1 2 3), (1 3 2)}.
The Cayley table for A3 is:
◦
ε
(1 2 3) (1
(1 3 2) (1
ε
(1 2 3) (1
ε
(1 2 3) (1
2 3) (1 3 2)
3 2)
ε
(1
3 2)
3 2)
ε
2 3)
♣
Example 4.19. (1) List the elements in A4 .
(2) Find the order of each element in A4 .
(3) Draw the subgroup lattice for A4 .
Solution. (1)
A4 = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2),
(1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)}
(2)
|ε| = 1
|(1 2)(3 4)| = |(1 3)(2 4)| = |(1 4)(2 3)| = 2
|(1 2 4)| = |(1 4 2)| = |(1 3 4)| = |(1 4 3)| = |(2 3 4)| = |(2 4 3)| = 3.
73
(3)
A4
h(1 2)(3 4), (13)(24)i
h(1 2 3)i
h(1 2)(3 4)i
h(1 3)(2 4)i
h(1 2 4)i
h(1 3 4)i
h(2 3 4)i
h(1 4)(2 3)i
hεi
♣
74
4.4
Exercises
Unless otherwise stated G is a group.
Exercise 4.1. Find all the symmetries of the square (draw pictures), that is, the elements of D4 .
Find the Cayley table for D4 . Also, prove that D4 = hr, s : |r| = 4, |s| = 2, rs = sr−1 i, where r
is a rotation by 90 degrees counter clockwise, and s is the reflection of the square along the main
diagonal of the square.
Exercise 4.2.
(a) Compute the order of each element in D4 and D5 .
(b) Find Z(D3 ), Z(D4 ), Z(D5 ).
Exercise 4.3. D3 = hr, s : |r| = 3, |s| = 2, rs = sr−1 i.
(a) Find CD3 (r2 ).
(b) Find CD3 (s).
(c) Let H = hri. Find rHr−1 and sHs−1 .
Exercise 4.4. D4 = hr, s : |r| = 4, |s| = 2, rs = sr−1 i.
(a) Find CD4 (r2 ).
(b) Find CD4 (s).
(c) Let H = hr2 i. Find CD4 (H) and ND4 (H).
Exercise 4.5. The set Hn = {g 2 : g ∈ Dn } (where n is an integer greater than or equal to 3) is a
subset of Dn = hr, s : |r| = n, |s| = 2, rs = sr−1 i. Prove that Hn = hr2 i.
Exercise 4.6. Construct the Cayley table for S4 and A4 . Find the center of each group.
Exercise 4.7. The set H = {ε, (1 2)} is a subgroup of S4 .
(a) Find CS4 (H).
(b) Find NS4 (H).
Exercise 4.8. Prove Theorem 4.10.
Exercise 4.9. Prove Theorem 4.13.
Exercise 4.10. (a) Draw the subgroup lattice of S3 and A3 .
(b) Draw the subgroup lattice of S4 (at least try).
(c) Draw the subgroup lattice of S5 and A5 (at least try).
Exercise 4.11. Find the following.
(a) CS3 ((1 3 2)).
75
(b) CS4 ((1 3 2)).
(c) CS4 ((1 2)(3 4)).
(d) CS4 ((1 4 3)).
(e) CA4 ((1 3 2)).
(f) CA3 ((1 3 2)).
Exercise 4.12. In our example for the triangle given at the start of this chapter, if we think of A as
the number 1, B as the number 2, and C as the number 3, show that every element in D3 corresponds
to an element in S3 .
Exercise 4.13. If α ∈ S7 , what are possible values for |α|? Do the same thing in S8 and S10 .
Exercise 4.14. (a) Find all numbers n such that S4 has an element of order n.
(b) Find all numbers n such that S5 has an element of order n.
(c) Find all numbers n such that S6 has an element of order n.
Exercise 4.15. Repeat all parts of the previous exercise for A4 , A5 , and A6 .
Exercise 4.16. Let H = {α2 : α ∈ S4 }. Prove that H = A4 .
Exercise 4.17. Let H = {α2 : α ∈ S5 }. Prove that H = A5 .
Exercise 4.18. Let H = {α2 : α ∈ S6 }. Prove that H 6= A6 .
Exercise 4.19. Prove that Z(Sn ) = {ε} for n ≥ 3.
Exercise 4.20. Prove that Z(An ) = {ε} for n ≥ 4.
Exercise 4.21. Show that every element in An for n ≥ 3 can be expressed as a 3-cycle or a product
of 3-cycles.
Exercise 4.22. Given that β and γ are in S4 with βγ = (1 3 4 2), γβ = (1 4 2 3), and γ(2) = 3,
determine β and γ.
Exercise 4.23. Suppose that H Sn , where n ≥ 2. Prove that either H An or half of the
elements of H are even permutations.
Exercise 4.24. Suppose that H Sn , where n ≥ 2. If |H| = 2k + 1 for some integer k ≥ 0, prove
that H An . Hint. Use Exercise 4.23.
Exercise 4.25. Give an example of a non-Abelian group G such that H = {g 2 : g ∈ G} is not a
subgroup of G (see Exercise 2.29).
76
4.5
Challenging Exercises
Exercise 4.26. Given an integer n ≥ 3, let r = R360/n be a rotation of the n-gon by 360/n degrees
counter clockwise (this can also be clockwise), and let s be any of the reflection axes of the n-agon
that leave it occupying the same space. Prove that
Dn = hr, s : |r| = n, |s| = 2, rs = sr−1 i.
Hint. Show that if x is any element of Dn which is not a power of r, then rx = xr−1 . Also show
that every such x must have order 2. Deduce that Dn is generated by the two elements s and sr,
and use this to prove the result.
Exercise 4.27. Given β ∈ Sn , the cycle type of β is (n1 , n2 , . . . , nk ), where β = β1 β2 . . . βk is the cycle
decomposition of β with cycles of length 1 included, ni is the length of βi , and n1 ≤ n2 ≤ . . . ≤ nk .
For example, the cycle type of α = (3)(1 2)(4 6 7)(8 5 9) ∈ S9 is (1, 2, 3, 3). Given σ, α ∈ Sn , prove
that σασ −1 and α have the same cycle type.
Exercise 4.28. If n = 2k for some integer k ≥ 2, show that Z(Dn ) = {R0 , R 360k }. If n ≥ 3 is odd,
n
prove that Z(Dn ) = {R0 }. In the even case, R 360k = R180 .
n
Exercise 4.29. Let α, β ∈ Sn (where n ∈ N). Prove that α and β commute if and only if α and β
are disjoint or one is a power of the other.
Exercise 4.30. Given an integer n ≥ 2, prove that Sn = h(1 2), (1 2 3 . . . n)i.
77
5
Group Homomorphisms and Isomorphisms
5.1
Homomorphism
In this section, we introduce one of the most fundamental concepts in algebra - a homomorphism.
Group Homomorphism
Let (G, ·), (G0 , ?) be two groups. A function φ : G → G0 is called a group homomorphism if
φ(x · y) = φ(x) ? φ(y) for all x, y ∈ G. We say that φ preserves the group operation in G.
The term homomorphism comes from the Greek words homo, meaning like, and morphe, meaning
form. If φ : (G, ·) → (G0 , ?) is a homomorphism, we write
φ(xy) = φ(x)φ(y) instead of φ(x · y) = φ(x) ? φ(y).
It is understood that the multiplication xy is taking place in G, while the multiplication φ(x)φ(y) is
taking place in G0 .
Example 5.1. Show that φ : Z → Z defined via x 7→ 2x is a group homomorphism.
Solution. Given x, y ∈ Z, φ(x + y) = 2(x + y) = 2x + 2y = φ(x) + φ(y).
♣
Example 5.2. Show that φ : R× → R× defined via φ(x) = 2x is not a group homomorphism.
The set R× = R − {0} is a group under multiplication.
Solution. Notice that φ(3 · 4) = φ(12) = 24. However, φ(3)φ(4) = 6 · 8 = 48. This can also be seen
by the fact that φ(xy) = 2xy but φ(x)φ(y) = (2x)(2y) = 4xy.
♣
Example 5.3. Show that φ : Z12 → Z10 defined via x 7→ 3x is not a group homomorphism.
Solution. Notice that φ(5 + 7) = φ(12) = φ(0) = 0. However, φ(5) + φ(7) = 15 + 21 = 36 = 6. In
fact, this function is not well defined. To see this, 0 = φ(0) = φ(12) = 36 = 6.
♣
Kernel and Image
If φ : G → G0 is a homomorphism, the kernel of φ is the set of elements in G that get mapped
to the identity element in G0 . The kernel of φ is denoted by ker φ. In set notation,
ker φ = {g ∈ G : φ(g) = eG0 },
where eG0 is the identity element of G0 . The image of φ is the set of all possible outputs of φ.
The image of φ is denoted by Im(φ). In set notation,
Im(φ) = {φ(g) : g ∈ G}.
The following proposition states some of the properties enjoyed by group homomorphisms.
78
Proposition 5.4. Suppose (G, ·) and (G0 , ?) are groups, and φ : G → G0 a group homomorphism. Then,
(1) φ(eG ) = eG0 , where eG and eG0 are the identity elements of G and G0 , respectively.
(2) For any g ∈ G and n ∈ Z, φ(g n ) = (φ(g))n .
(3) For any g ∈ G with |g| < ∞ we have |φ(g)| | |g|. That is, the order of φ(g) divides the
order of g.
(4) ker φ G and Im(φ) G0 .
(5) If H G, then φ(H) G0 , where φ(H) = {φ(h) : h ∈ H}.
(6) If H is a cyclic subgroup of G, then φ(H) is cyclic subgroup of G0 .
(7) If H is an Abelian subgroup of G, then φ(H) is an Abelian subgroup of G0 .
(8) If K G0 , then φ−1 (K) G, where φ−1 (K) = {a ∈ G : φ(a) ∈ K}.
(9) φ is injective if and only if ker φ = {eG }.
Proof. The proofs of parts (2), (4)-(7), and (9) are left as exercises.
(1) Notice that φ(eG ) = φ(eG · eG ) = φ(eG ) ? φ(eG ). Hence,
eG0 = (φ(eG )) ? (φ(eG ))−1 = (φ(eG ) ? φ(eG )) ? (φ(eG ))−1 = φ(eG ) ? (φ(eG ) ? (φ(eG ))−1 )
= φ(eG ) ? eG0
= φ(eG )
(3) Suppose |g| = k. Then, (φ(g))k = φ(g k ) = φ(eG ) = eG0 . Hence, |φ(g)| < ∞, and the result follows
by Corollary 2.17.
(8) Since φ(eG ) = eG0 ∈ K, we have eG ∈ φ−1 (K). Given a, b ∈ φ−1 (K), we get φ(a), φ(b) ∈ K. Part
(2) implies that φ(b−1 ) = (φ(b))−1 ∈ K, and this gives
φ(ab−1 ) = φ(a)φ(b−1 ) = φ(a)(φ(b))−1 ∈ K ⇒ ab−1 ∈ φ−1 (K).
By Proposition 2.21, φ−1 (K) G.
Example 5.5. Suppose that φ : Z25 → Z25 (operation is addition) is a homomorphism and
φ(6) = 18. Determine a formula for φ(x) (as a function of x).
Solution. Notice that
φ(x) = φ(1| + 1 +
{z. . . + 1}) = |φ(1) + φ(1){z+ . . . + φ(1)} = xφ(1)
x terms
x terms
for any x ∈ Z25 . So it suffices to find φ(1);
φ(1) = −φ(−1) = −φ(24) = −4φ(6) = −4(18) = −72 = 3 (operations in mod 25).
79
Hence, φ(x) = 3x for all x ∈ Z25 .
♣
Example 5.6. Define φ : Z12 → Z12 by x 7→ 3x. Show that φ is a homomorphism. Find
ker φ, |2| and |φ(2)|.
Solution. For any x, y ∈ Z12 , φ(x + y) = 3(x + y) = 3x + 3y = φ(x) + φ(y), and
ker φ = {x ∈ Z12 : φ(x) = 0} = {x ∈ Z12 : 3x = 0} = {0, 4, 8}.
Notice that |2| = 6 and |φ(2)| = |6| = 2. In particular, |2| =
6 |φ(2)|, however, |φ(2)| | |2|.
♣
Example 5.7. Show that φ : R× → R× defined via φ(x) = x2 is a group homomorphism.
Moreover, show that this map is not injective.
Solution. Given x, y ∈ R× , φ(xy) = (xy)2 = x2 y 2 = φ(x)φ(y). This proves that φ is a group
homomorphism. We show φ is not injective in two ways. Notice that φ(1) = 1 = φ(−1) but 1 6= −1.
Alternatively, ker φ = {±1}, so by Proposition 5.4(9), φ is not injective.
♣
Example 5.8. Show that if φ : Q → Q (both under addition) is a group homomorphism, then
φ(x) = xφ(1) for all x ∈ Q.
Solution. If x = 0, then 0·φ(1) = 0 = φ(0) (by Proposition 5.4(1)). First we show that φ(nr) = nφ(r)
for any integer n and any rational number r (this is already true for n = 0). Given a rational number
r, if n ∈ N, then
φ(nr) = φ (r + r + . . . r) = φ(r) + φ(r) + . . . + φ(r) = nφ(r).
{z
} |
{z
}
|
n terms
n terms
If n is a negative integer, then by the preceding argument,
φ(nr) = φ(−(−n)r)) = −φ((−n)r) = −(−n)φ(r) = nφ(r).
Now we turn our attention back to the question at hand. Suppose x is a non-zero rational number,
a
say x = , where a and b are non-zero integers. The above argument gives
b
a
1
1
φ(x) = φ
=φ a
= aφ
.
b
b
b
Moreover,
b
1
1
1
1
φ(1) = φ
=φ b
= bφ
⇒φ
= φ(1).
b
b
b
b
b
a
1
a
Hence, φ(x) = φ
= aφ
= φ(1) = xφ(1), completing the proof.
b
b
b
♣
For those familiar with linear algebra, if T : V → W is a linear transformation and {v1 , v2 , . . . , vn } is a
basis for V (even a set that spans V ), then T is completely determined by its values on v1 , v2 , . . . , vn .
80
This says that once we know T (v1 ), T (v2 ), . . . , T (vn ), we know T (v) for all v ∈ V . There is an
analogous result in group theory.
Proposition 5.9. Suppose G = ha1 , a2 , ..., ak i. If φ : G → G0 is a group homomorphism, then
φ is completely determined by its values on a1 , a2 , ..., ak .
Proof. Given x ∈ G, x = cn1 1 cn2 2 · · · cnmm , where ni ∈ Z and ci ∈ {a1 , a2 , . . . , ak }. Then,
φ(x) = φ(cn1 1 cn2 2 · · · cnmm ) = (φ(c1 ))n1 (φ(c2 ))n2 · · · (φ(cm ))nm .
Since φ(ci ) ∈ {φ(a1 ), φ(a2 ), . . . , φ(ak )}, if we know the value of φ at a1 , a2 , . . . , ak , then we know the
value of φ(x).
To see the real power of the preceding proposition consider a homomorphism φ : Z → G where G is
some group. Then the map φ is completely determined by φ(1), equivalently φ(−1) (this is because
Z = h1i = h−1i). That is, once we know φ(1), we know φ everywhere.
5.2
Isomorphism
Is it possible for two things to look very different but yet mean the same thing? The answer is yes.
For example, consider the phrases in different languages which have the same meaning. There are
many languages across the world. All designed for us to be able to communicate. Now suppose you
are in a room with two other people, one speaks only Arabic, the other speaks only Spanish, and you
speak only English. If you are only interested in the meaning of the things they are saying, what
do you do? You probably get a dictionary, a computer, or an app to translate each language for
you. The same thing often happens in group theory (in fact, in many parts of mathematics). More
specifically, in mathematics we are interested in the characteristics of the space we are in, not how
the elements of the space are represented. In this section, we give a formal method for determining
when two groups that look different are actually the same.
Isomorphism
Let (G, ·) and (G0 , ?) be groups. A map φ : G → G0 is called an isomorphism if φ is a
homomorphism and a bijection. If there exists an isomorphism φ : G → G0 , we say G and G0
are isomorphic and write (G, ·) ∼
= (G0 , ?). If there is no ambiguity about the operations, we
0
∼
write G = G .
Since every isomorphism is a homomorphism, all the properties of homomorphisms carry over to
isomorphisms (but not vice versa). In particular, all the properties stated in Proposition 5.4 also
hold for isomorphisms.
If φ : G → G0 is an isomorphism, one can think of it as the thing that transforms objects in G into
objects in G0 while preserving all algebraic structure in both groups. In essence, φ plays the role
between two groups that a dictionary plays between two languages with the additional property that
it respects the grammar in each language while forming sentences (a little more to it, but you get the
idea). In most parts of mathematics we only care about the meaning of things rather than how they
are presented, so the idea of isomorphisms plays a crucial role in almost all parts of mathematics (the
81
definitions vary depending on the field, but the idea is the same). For this reason, mathematicians
often think of isomorphic objects as the same thing, and when doing so they use the phrase “up to
isomorphism”.
Example 5.10. Show that (R, +) ∼
= (R>0 , ·), where R>0 is the set of positive real numbers
and · is usual multiplication.
Solution. It suffices to find an isomorphism between the two groups. Define f : R → R>0 via
f (x) = ex . If f (x) = f (y), then ex = ey and taking the natural logarithm (base e) of both sides gives
x = y. This shows that our function is injective. Given x ∈ R>0 , ln x ∈ R and f (ln x) = eln x = x
(showing surjectivity). Moreover, If x, y ∈ R, then
f (x + y) = ex+y = ex ey = f (x)f (y).
Thus, f is a homomorphism. Hence, f is an isomorphism, completing the proof.
♣
Notice that for two groups G and G0 to be isomorphic there must exist an isomorphism between
G and G0 . The direction of the isomorphism does not matter, since the inverse of an isomorphism
is also an isomorphism (see Proposition 5.14). This does not say that every map between the two
groups has to be an isomorphism. For example, every group is isomorphic to itself (via the identity
map). However, some maps from a group back to itself need not be isomorphisms (see Example 5.7
or the next example).
Example 5.11. Show that the map φ : S4 → S4 defined by α 7→ (123)α is a bijection but not
an isomorphism.
Solution. If φ(x) = φ(y), then (123)x = (123)y. Multiplying both sides on the left by (123)−1 = (321)
gives x = y. Hence, φ is injective. Given y ∈ S4 , (321)y ∈ S4 and φ((321)y) = (123)(321)y = y.
Hence, φ is surjective. This proves that φ is a bijection.
If φ were an isomorphism, it would have to map the identity in S4 to the identity in S4 (Proposition
5.4(1)). However, φ(ε) = (123) 6= ε. Hence, φ is not an isomorphism.
Remark. You can also show that there exist a, b ∈ S4 with φ(ab) 6= φ(a)φ(b). For example take
a = (12) and b = (23).
♣
Example 5.12. Show that the function φ : Z → Z given by x 7→ 2x is not an isomorphism.
Solution. We have already seen this map is a homomorphism (Example 5.1). Notice that this map
is injective. However, it is not surjective (the image of this function is the set of even integers). ♣
Example 5.13. Show that the following groups are not isomorphic.
(a) U (10) and U (12).
(b) (Q, +) and (Q× , ·), where Q× is the set of the non-zero rationals and · is usual multiplication.
Solution. (a) Notice that x2 = 1 for all x ∈ U (12) (you can check this). We now suppose that there
exists an isomorphism f : U (10) → U (12). Then f (9) = f (3 · 3) = (f (3))2 = 1 = f (1), contradicting
the fact that f is injective.
82
(b) Suppose they were isomorphic. Let φ : Q → Q× be an isomorphism. Then there exist a ∈ Q
such that
2
1
1
1
−1 = φ(a) = φ
a+ a = φ
a
,
2
2
2
♣
a contradiction.
We now state and prove some of the properties of isomorphisms.
Properties of Isomorphisms
Proposition 5.14. Suppose φ : G → G0 is an isomorphism. Then,
(1) φ−1 : G0 → G is an isomorphism.
(2) gh = hg if and only if φ(g)φ(h) = φ(h)φ(g). Consequently, G is Abelian if and only if
G0 is Abelian.
(3) G = hgi if and only if G0 = hφ(g)i. In particular, G is cyclic if and only if G0 is cyclic.
(4) If K G, then φ(K) ∼
= K, where φ(K) = {φ(x) : x ∈ K}.
(5) |g| = |φ(g)| for all g ∈ G.
(6) φ(Z(G)) = Z(G0 ).
(7) Fix k ∈ Z and g ∈ G. The equation xk = g has the same number of solutions in G as
the equation xk = φ(g) in G0 .
(8) Given n ∈ N, the number of elements in G of order n equals the number of elements in
G0 of order n.
(9) |G| = |G0 |.
Proof. We prove (1), (2), (3), and (9) and leave the rest as an exercise.
(1) Since φ is a bijection, so is φ−1 . It suffices to show that φ−1 is a homomorphism. Given x, y ∈ G0 ,
there exist a, b ∈ G such that φ(a) = x and φ(b) = y. In particular, xy = φ(a)φ(b) = φ(ab), so
φ−1 (xy) = ab = φ−1 (x)φ−1 (y).
(2) Given g, h ∈ G, if gh = hg, then φ(g)φ(h) = φ(gh) = φ(hg) = φ(h)φ(g). Notice that the
preceding argument is valid for any homomorphism.
Conversely, suppose φ(g)φ(h) = φ(h)φ(g). Then φ(gh) = φ(hg), and since φ is injective we get
gh = hg.
(3) Suppose G = hgi. Since φ(g) ∈ G0 , hφ(g)i ⊆ G0 . For the reverse inclusion, if x ∈ G0 , then there
exist a y ∈ G such that φ(y) = x. Since G = hgi, there exist an integer k such that y = g k . Hence,
x = φ(y) = φ(g k ) = (φ(g))k ∈ hφ(g)i.
If G0 is cyclic, a similar argument to the one in the preceding paragraph (using φ−1 in place of φ)
gives the desired result.
(9) This follows from the fact that φ is a bijection.
83
Theorem 5.15. Suppose G is a cyclic group.
(1) If G is infinite, then G ∼
= Z.
(2) If G is finite of order n, then G ∼
= Zn .
Proof. We prove (1) and leave (2) as an exercise. Let G = hgi with |G| = ∞. Define φ : Z → G via
n 7→ g n . We show that φ is an isomorphism. Given a, b ∈ Z,
φ(a + b) = g a+b = g a g b = φ(a)φ(b).
If φ(a) = φ(b), then g a = g b . In particular, g a−b = e and since |g| = |G| = ∞, we have a = b. We
have just proven that f is a homomorphism and injective. To see that φ is surjective, if y ∈ G then
y = g a for some integer a. So φ(a) = g a = y.
The preceding theorem says that up to isomorphism there is only one finite cyclic group of any given
order n (namely, Zn ), and there is only one infinite cyclic group (namely, Z).
Example 5.16. Show that U (10) ∼
= Z4 .
Solution. Notice that |U (10)| = 4 and U (10) = h[3]i. By Theorem 5.15, U (10) ∼
= Z4 .
5.3
♣
Automorphisms, Inner Automorphisms
Isomorphisms whose domain and co-domain are equal are called automorphisms. We make this
notion into a formal definition.
Automorphisms
An isomorphism from a group G to itself is called an automorphism.
It should be noted that the identity map from a group G back to itself is always an automorphism.
Example 5.17. Define ψ : R2 → R2 (under componentwise addition) via (a, b) 7→ (b, a). Then
ψ is an automorphism.
Solution. It suffices to show that ψ is an isomorphism. Given (a, b), (c, d) ∈ R2 ,
ψ((a, b) + (c, d)) = ψ(a + c, b + d) = (b + d, a + c) = (b, a) + (d, c) = ψ(a, b) + ψ(c, d).
This shows that ψ is a homomorphism. Moreover,
(a, b) ∈ ker ψ ⇔ ψ(a, b) = (0, 0) ⇔ (b, a) = (0, 0) ⇔ a = 0 = b ⇔ (a, b) = (0, 0).
That is, ker ψ = {(0, 0)}. By Proposition 5.4(9), ψ is injective. Given (a, b) ∈ R2 , (b, a) ∈ R2 and
ψ(b, a) = (a, b), proving surjectivity. Hence, ψ is a bijection, completing the proof.
♣
The maps that we are about to define play a very important role in group theory (in fact, abstract
algebra).
84
Inner Automorphism Induced By An Element
Let G be a group and g ∈ G. Define φg : G → G via x 7→ gxg −1 . This map is called the inner
automorphism induced by g.
The next proposition shows why the word automorphism is used in the previous definition.
Proposition 5.18. For each g ∈ G, φg is an automorphism.
Proof. Given g ∈ G, it suffices to show that φg is an isomorphism. If x, y ∈ G, then
φg (xy) = g(xy)g −1 = (gxg −1 )(gyg −1 ) = φg (x)φg (y).
This shows that φg is a homomorphism. Moreover,
x ∈ ker φg ⇔ φg (x) = eG ⇔ gxg −1 = eG ⇔ x = eG .
That is, ker φg = {eG }. By Proposition 5.4(9), φg is injective. Given y ∈ G, g −1 yg ∈ G and
φg (g −1 yg) = g(g −1 yg)g −1 = y,
proving surjectivity. Hence, φg is a bijection, completing the proof.
Example 5.19. Let G = S3 . Find φ(1
2)
: S3 → S3 .
Solution. Notice that φ(1 2) (α) = (1 2)α(1 2)−1 = (1 2)α(1 2). Hence,
φ(1 2) (ε)
φ(1 2) ((1 2))
φ(1 2) ((1 3))
φ(1 2) ((2 3))
φ(1 2) ((1 2 3))
φ(1 2) ((1 3 2))
=
=
=
=
=
=
(1
(1
(1
(1
(1
(1
2)ε(1 2) = ε
2)(1 2)(1 2) = (1 2)
2)(1 3)(1 2) = (2 3)
2)(2 3)(1 2) = (1 3)
2)(1 2 3)(1 2) = (1 3 2)
2)(1 3 2)(1 2) = (1 2 3).
The next proposition proves a very important property about the maps φg .
Proposition 5.20. For each g, h ∈ G, φgh = φg φh , where φg φh means φg ◦ φh .
Proof. Given g, h, x ∈ G, φgh (x) = (gh)x(gh)−1 = g(hxh−1 )g −1 = φg (hxh−1 ) = (φg φh )(x).
85
♣
Automorphism and Inner Automorphism Group
Given a group G, define
Aut(G) = {φ : G → G : φ an isomorphism},
Inn(G) = {φg : g ∈ G}.
Aut(G) is called the automorphism group of G and Inn(G) is called the inner automorphism
group of G.
Exercise 5.19 asks you to prove the following;
1. Aut(G) is a group under composition.
2. Inn(G) is a subgroup of Aut(G).
Notice that Proposition 5.20 tells us that the map G → Aut(G) given by g 7→ φg is a group
homomorphism.
Theorem 5.21. For all natural numbers n > 1 we have Aut(Zn ) ∼
= U (n).
Proof. Since Zn = h1i, by Proposition 5.9, for any ψ ∈ Aut(Zn ), ψ is completely determined by
its value on 1. In particular, ψ = φ if and only if ψ(1) = φ(1). Given ψ ∈ Aut(Zn ), we have
Zn = h1i = hψ(1)i = h1ψ(1) i (isomorphisms send generators to generators). Since the order of 1 in
Zn is n, by Corollary 3.9, gcd(n, ψ(1)) = 1. That is, ψ(1) ∈ U (n).
Define, Φ : Aut(Zn ) → U (n) via ψ 7→ ψ(1). Our first paragraph implies that Φ is well defined and
one to one. Given r ∈ U (n), define α : Zn → Zn via s 7→ sr (mod n). The fact that α ∈ Aut(Zn ) is
left as an exercise. Moreover, Φ(α) = α(1) = r. Hence, Φ is onto.
Given α, β ∈ Aut(Zn ),
β(1) (mod n) terms
}|
{
z
Φ(αβ) = (αβ)(1) = α(β(1)) = α (1 + 1 + ... + 1) =
=
=
β(1) (mod n) terms
}|
{
z
α(1) + α(1) + ... + α(1)
α(1)β(1) (mod n)
Φ(α)Φ(β).
Hence, Φ is an isomorphism.
Example 5.22. (a) Prove that if (G, ∗) ∼
= (H, ?), then Aut(G) ∼
= Aut(H).
(b) Show that the converse of part(a) is not true in general.
Solution. (a) Since H ∼
= G, there exist an isomorphism f : H → G. Given ψ ∈ Aut(H), f ψf −1 :
G → G. Moreover, f ψf −1 is a bijection (since composition of bijections is a bijection), and it is a
homomorphism (since composition of homomorphisms is a homomorphism, see Exercise 5.3). Hence,
f ψf −1 ∈ Aut(G). Define Φ : Aut(H) → Aut(G) via ψ 7→ f ψf −1 . We now show that this function is
an isomorphism.
86
Suppose ψ, φ ∈ Aut(H), then
Φ(ψφ) = f (ψφ)f −1 = (f ψf −1 )(f φf −1 ) = Φ(ψ)Φ(φ).
It should be noted that the operations in automorphism groups is composition. The previous argument shows that Φ is a homomorphism.
If Φ(ψ) = Φ(φ), then f ψf −1 = f φf −1 . Since f is a bijection, ψ = φ. That is, Φ is one-toone. Moreover, if ϕ ∈ Aut(G), then f −1 ϕf ∈ Aut(H) (similar to the argument given in the first
paragraph), and
Φ(f −1 ϕf ) = f (f −1 ϕf )f −1 = ϕ
That is, Φ is onto. Hence, Φ : Aut(H) → Aut(G) is an isomorphism, and the result follows.
(b) By Theorem 5.21 and Example 5.16, Aut(Z10 ) ∼
= U (10) ∼
= Z4 . Since U (5) is a cyclic group of
∼
order 4, Theorems 5.15 and 5.21 imply that Aut(Z5 ) ∼
U
(5)
=
= Z4 . However, Z10 Z5 (they don’t
even have the same order).
♣
5.4
Cayley’s Theorem
Cayley’s Thoerem
Theorem 5.23. Let G be a group and let S(G) denote the group of permutations of G. Then
G is isomorphic to a subgroup of S(G).
Proof. Given a group G and and element g ∈ G, define Tg : G → G via x 7→ gx (called left
multiplication by g). The fact that Tg is a bijection (i.e., a permutation of G) is left as an exercise.
Let G0 = {Tg : g ∈ G}. Notice that G0 ⊆ S(G). Given g, h, x ∈ G,
Tg Th (x) = Tg (Th (x)) = Tg (hx) = g(hx) = (gh)(x) = Tgh (x).
In particular, Tg Th = Tgh for all g, h ∈ G. This will enable us to show that G0 is a group under
composition, and hence a subgroup of S(G). To see this, given Tg , Th , Tk ∈ G0 ,
1. Tg Th = Tgh ∈ G0 , proving the operation is binary.
2. The identity element in G0 is Te (where e is the identity element in G). This follows from the
fact that Tg Te = Tge = Tg .
3. Since Tg Tg−1 = Tgg−1 = Te , (Tg )−1 = Tg−1 ∈ G0 .
4. Tg (Th Tk ) = Tg Thk = Tg(hk) = T(gh)k = Tgh Tk = (Tg Th )Tk , proving associativity.
Define Φ : G → G0 via g 7→ Tg . We show that Φ is an isomorphism.
(1) Φ(g) = Φ(h) ⇒ Tg = Th ⇒ Tg (e) = Th (e) ⇒ g = ge = he = h. Hence, Φ is injective.
(2) By construction Φ is onto.
(3) Suppose g, h ∈ G, then Φ(gh) = Tgh = Tg Th = Φ(g)Φ(h). Hence, Φ is a homomorphism.
Therefore, G ∼
= G0 . The group G0 constructed is called the left regular representation of G.
87
Example 5.24. Consider U (10) = {1, 3, 7, 9}. Find the maps T1 , T3 , T7 , T9 which are constructed in Cayley’s Theorem.
Solution. Consider T3 : U (10) → U (10). Then
T3 (1) = 3, T3 (3) = 9, T3 (7) = 1, T3 (9) = 7 ⇒ T3 = (1 3 9 7).
In general, T1 = identity, T3 = (1 3 9 7), T7 = (1 7 9 3), T9 = (1 9)(3 7).
88
♣
5.5
Exercises
Unless otherwise stated G and G0 are groups.
Exercise 5.1. Let eG0 be the identity element in G0 . Prove that φ : G → G0 given by g 7→ eG0 is a
homomorphism. This is called the trivial homomorphism.
Exercise 5.2. (a) Prove that the map from G to itself defined by g 7→ g 2 is a homomorphism if and
only if G is Abelian.
(b) Prove that the map from G to itself defined by g 7→ g −1 is a homomorphism if and only if G is
Abelian. This is called the inverse map.
Exercise 5.3. (a) Suppose φ : G → H and γ : H → K are homomorphisms. Prove that γφ : G → K
is a homomorphism. This shows that the composition of homomorphisms is a homomorphism.
(b) If φ and γ in part (a) are isomorphisms, prove that γφ is an isomorphism. This shows that the
composition of isomorphisms is an isomorphism.
Exercise 5.4. Fix a natural number n. Let G be a group with the following property: |g| |n for all
g ∈ G.
(a) If gcd(n, m) = 1, prove that the function Φ : G → G given by g 7→ g m is injective.
(b) Prove that the function in (a) is a homomorphism if G is Abelian.
Exercise 5.5. Prove the remaining parts of Proposition 5.4.
Exercise 5.6. Suppose that φ : Z25 → Z20 is a group homomorphism with φ(3) = 4.
(a) Determine φ(x).
(b) Determine the image and kernel of φ.
Exercise 5.7. Prove or disprove. Suppose that φ : G → G0 is a group homomorphism
(a) If G0 is Abelian, then G is Abelian.
(b) If G0 is cyclic, then G is cyclic.
Exercise 5.8. Prove the remaining parts of Proposition 5.14
Exercise 5.9. Prove Theorem 5.15(2).
Exercise 5.10. Define the following relation on groups: G ∼ H if and only if G ∼
= H. Show that
this relation is an equivalence relation.
Exercise 5.11. Suppose φ, ψ ∈ Aut(G). Prove that H = {g ∈ G : φ(g) = ψ(g)} is a subgroup of G.
Exercise 5.12. Let n and m be non-zero integers. Prove that the subgroups hni and hmi of Z (under
addition) are isomorphic.
Exercise 5.13. (a) Is S3 ∼
= D3 ? Justify your answer.
(b) Is S4 ∼
= D4 ? Justify your answer.
89
Exercise 5.14. Let (G, ·) be a group. Let G0 = G (as sets) and define the following operation on
G0 : a ? b = b · a.
(a) Prove that (G0 , ?) is a group.
(b) Prove that (G, ·) ∼
= (G0 , ?).
Exercise 5.15. Let G = D3 . Find the map φr : D3 → D3 .
Exercise 5.16. Let G = S3 .
(a) Find φ(1
2 3) .
(b) Find φ(1
3 2) .
(c) Find φ(2 3) .
Exercise 5.17. Let G = S4 .
(a) Find φ(1
2 3) .
(b) Find φ(1
3 2) .
(c) Find φ(2 3) .
(d) Find φ(1 2) .
(e) Find φ(1
3 2 4) .
(f) Find φ(1
3)(2 4) .
Exercise 5.18. Find the order of the following elements in Inn(S5 ).
(a) φ(1
2 4 3) .
(b) φ(1
2)(4 3) .
(c) φ(1
3 2)(2 3 4) .
(d) φ(1
3)(1 2 4 5) .
Exercise 5.19. Prove that Aut(G) is a group under function composition. Moreover, prove that
Inn(G) is a subgroup of Aut(G).
Exercise 5.20. (a) Prove that a group G is Abelian if and only if Inn(G) = {φe }.
(b) Give an example of a group with Inn(G) 6= Aut(G).
Exercise 5.21. Let H = {α ∈ S7 : α(3) = 3} and K = {α ∈ S7 : α(5) = 5}. Prove that H ∼
= K.
Exercise 5.22. Show that the map α defined in Theorem 5.21 is in Aut(Zn ).
Exercise 5.23. Let g ∈ G and h ∈ Z(G). Prove that φg = φhg .
Exercise 5.24. Show that the map Tg defined in Theorem 5.23 is a bijection for each g ∈ G.
Exercise 5.25. Let G = U (8).

 1
Exercise 5.26. Let G = 0

0
Find T1 , T3 , T5 , and T7 .


a b


1 c : a, b, c ∈ Q .

0 1
(a) Prove that G is a group under matrix multiplication.
(b) Find Z(G).
(c) Prove that Z(G) ∼
= Q, where Q is under addition.
90
5.6
Challenging Exercises
Exercise 5.27. Suppose φ : Q → Z is a homomorphism, where both Q and Z are under addition.
Prove that φ is the zero map (i.e. φ(x) = 0 for all x ∈ Q).
Exercise 5.28. Prove that Q>0 (the set of positive rational numbers) under multiplication is isomorphic to a proper subgroup of itself. Hint. Use Exercise 2.29.
Exercise 5.29. Prove that Q under addition is not isomorphic to a proper subgroup of itself.
Exercise 5.30. Recall that R× = R − {0} is a group under multiplication. Suppose φ ∈ Aut(R× ).
Prove that φ((0, ∞)) = (0, ∞) and φ((−∞, 0)) = (−∞, 0).
91
6
Cosets and Lagrange’s Theorem
6.1
Definition and Properties of Cosets
In this section, we introduce a new and powerful tool for analysing a group, and state some of its
properties.
Cosets
Let G be a group and H a non-empty subset of G. For any a ∈ G, we define
aH = {ah : h ∈ H},
Ha = {ha : h ∈ H},
aHa−1 = {aha−1 : h ∈ H}.
When H G, aH is called a left coset of H in G. Similarly, Ha is called a right coset of H in
G. The element a is called a coset representative of aH (or Ha).
It should be understood that aH (or Ha) is written in multiplication notation (since we always write
the operation in an arbitrary group using multiplication). For example, when the operation in our
group is addition, we get aH = a + H, Ha = H + a, and aHa−1 = a + H − a.
Example 6.1. Let H = {0, 4} Z8 . Find all the left and right cosets of H in Z8 .
Solution. Since our operation is addition modulo 8, aH = a + H = H + a = Ha. The left (and
right) cosets of H in Z are as follows;
0+H
1+H
2+H
3+H
=
=
=
=
4+H
5+H
6+H
7+H
= H,
= {1, 5},
= {2, 6},
= {3, 7}.
♣
Hence, H has 4 left cosets in Z8 .
Example 6.2. Let G = S3 and H = {ε, (1 2)} S3 . Find all the left and right cosets of H in
G.
Solution. The left cosets of H in G are as follows
εH = (1 2)H = H,
(1 3)H = (1 2 3)H = {(1 3), (1 2 3)},
(2 3)H = (1 3 2)H = {(2 3), (1 3 2)}.
Hence, H has 3 left cosets in S3 . The right cosets of H in G are as follows
Hε = H(1 2) = H,
H(1 3) = H(1 3 2) = {(1 3), (1 3 2)},
H(2 3) = H(1 2 3) = {(2 3), (1 2 3)}.
92
Hence, H has 3 right cosets. Notice that (1 3)H 6= H(1 3).
♣
The previous two examples illustrate the following:
1. It is possible aH = bH with a 6= b.
2. Cosets need not be subgroups.
3. In general, the left coset aH may not equal the right coset Ha.
A fourth observation one might want to make is that the number of left (right) cosets of a subgroup
H of a group G is less than the order of the group G. However, this turns out to be true only under
certain circumstances. In general, if G is a finite group and H is a non-trivial subgroup of G, then
the number of left (right) cosets is less than or equal to the order of G (this is a consequence of
Lagrange’s Theorem which is proved in the next section). However, if G is an infinite group and H
is a subgroup of G, then the number of left (right) cosets may be finite or infinite (depending on H
and G). We will shortly give an example of each scenario. First, we state and prove some of the
properties enjoyed by cosets of a subgroup in a group.
Properties of Cosets
Proposition 6.3. Let H be a subgroup of a group G and a, b ∈ G. Then;
1. a ∈ aH and a ∈ Ha.
2. aH = H ⇔ a ∈ H ⇔ Ha = H.
3. (ab)H = a(bH) and H(ab) = (Ha)b.
4. aH = bH ⇔ a ∈ bH. Similar result for right cosets. That is, Ha = Hb ⇔ a ∈ Hb.
5. aH = bH or aH ∩ bH = ∅. Similar result for right cosets. That is, Ha = Hb or
Ha ∩ Hb = ∅.
6. aH = bH ⇔ a−1 b ∈ H ⇔ b−1 a ∈ H. Similarly, Ha = Hb ⇔ ab−1 ∈ H ⇔ ba−1 ∈ H.
7. |aH| = |bH|. Similar result for right cosets. That is, |Ha| = |Hb|.
8. aH = Ha ⇔ H = aHa−1 .
9. aH G ⇔ a ∈ H ⇔ Ha G.
Proof. We prove all the results for left cosets. A similar proof works for the right cosets.
(1) Since e ∈ H we have a = ae ∈ aH.
(2) Notice that
aH = H ⇒ a = ae ∈ aH = H ⇒ a ∈ H.
Conversely, suppose a ∈ H. Since the operation in H is binary, aH ⊆ H. Moreover, given h ∈ H,
a−1 h ∈ H ⇒ h = a(a−1 h) ∈ aH.
93
Hence, H ⊆ aH. In particular, aH = H.
(3) (ab)H = {(ab)h : h ∈ H} = {a(bh) : h ∈ H} = a({bh : h ∈ H}) = a(bH).
(4) Suppose aH = bH. Then, a ∈ aH = bH, i.e., a ∈ bH. Conversely, suppose a ∈ bH. Then, a = bh
for some h ∈ H. Given g ∈ H, ag = (bh)g = b(hg) ∈ bH. That is, aH ⊆ bH. Moreover, for any
c ∈ bH, we have c = bx for some x ∈ H. In particular,
c = bx = (ah−1 )x = a(h−1 x) ∈ aH.
That is, bH ⊆ aH. Hence, aH = bH.
(5) If aH ∩ bH 6= ∅, there exist a c ∈ aH ∩ bH. This implies that c ∈ aH and c ∈ bH, and by (4),
aH = cH = bH.
(6) This follows from the following equivalences:
aH = bH ⇔ b ∈ aH ⇔
⇔
⇔
⇔
b = ah for some h ∈ H
a−1 b = h for some h ∈ H
a−1 b ∈ H
b−1 a ∈ H
(since H G).
A similar proof works for the right cosets.
(7) Define f : aH → bH via ah 7→ bh. We claim that f is a bijection. Notice that every element in
aH has a unique representation ah for some h ∈ H. In particular, if ah = ag for some h, g ∈ H, then
h = g (multiply both sides on the left by a−1 ). So the function f is well defined. If f (ah) = f (ah1 ),
then
bh = bh1 ⇒ h = h1 ⇒ ah = ah1 .
Given bh ∈ bH, ah ∈ aH and f (ah) = bh. We have shown that f is a bijection, so |aH| = |bH|.
(8) This follows from the following equivalences:
aH = Ha ⇔ (aH)a−1 = (Ha)a−1 ⇔ aHa−1 = H.
(9) If a ∈ H, then
aH ∩ H 6= ∅ ⇒ aH = H G.
Conversely,
aH G ⇒ e ∈ aH ⇒ aH ∩ H 6= ∅ ⇒ aH = H ⇒ a ∈ H,
completing the proof.
Combining properties (1) and (5) in the previous proposition shows that the left cosets {aH : a ∈ G}
form a partition of G. Likewise, the right cosets {Ha : a ∈ G} form a partition of G.
Remark. If a group G is Abelian, then aH = Ha for all a ∈ G and H G.
Example 6.4. Let G = Z and H = {2k : k ∈ Z} Z. Find the cosets of H in G.
94
Solution. By Proposition 6.3(6), a + H = b + H if and only if a − b ∈ H, i.e., a − b is even. Hence,
the cosets of H in G are H and 1 + H. In particular, H has 2 left cosets in Z.
♣
Example 6.5. Let G = Q (under addition) and H = Z Q . How many cosets does H have
in G?
Solution. By Proposition 6.3(6), a + H = b + H if and only if a − b ∈ H, i.e., a − b is an integer. In
particular,
1
1
+ H 6=
+ H for all distinct natural numbers n and m.
n
m
That is, Z has the same number of cosets as the order of Q.
♣
In general, it seems like knowing the left cosets of a subgroup H in a group G does not tell us much
about its right cosets. As we have already illustrated, aH is not necessarily equal to Ha. However,
it turns out that the number of left cosets is always equal to the number of right cosets.
Number of Left Cosets Equals The Number of Right Cosets
Theorem 6.6. Suppose H is a subgroup of a group G (not necessarily finite), and let
G/H = {gH : g ∈ G}, and H \ G = {Hg : g ∈ G}.
Then, |G/H| = |H \ G|.
Proof. Define Φ : G/H → H \ G via gH 7→ Hg −1 . We show that this map is a well defined function
which is also a bijection. Notice that
gH = g1 H ⇔ g1−1 g ∈ H ⇔ g1−1 (g −1 )−1 ∈ H ⇔ Hg −1 = Hg1−1 ⇔ Φ(gH) = Φ(g1 H).
This proves that Φ is a well defined function, and that it is injective. Moreover, given Hg ∈ H \ G,
g −1 H ∈ G/H, and Φ(g −1 H) = H(g −1 )−1 = Hg. Hence, Φ is surjective, completing the proof.
Index of H in G
Let H be a subgroup of a group G (not necessarily finite). The index of H in G, denoted by
[G : H], is the common value of |G/H| and |H \ G| (this value can be infinite).
Suppose G = S3 and H = {ε, (1 2)}. In example 6.2 we showed that
S3 /H = {H, (1 3)H, (2 3)H}, and H \ S3 = {H, H(1 3), H(2 3)}.
Notice that H 6= H(1 3), (1 3)H 6= H(1 3), and (2 3)H 6= H(1 3). In particular, S3 /H 6= H \ S3 .
However, |S3 /H| = |H \ S3 | = 3. Hence, [S3 : H] = 3.
Remark. Given sets A and B, sometimes in mathematics A \ B = {x ∈ A : x ∈
/ B} = A − B. It
should be clear from the context when we write A \ B if we are referring to A − B or the set of right
cosets of A in B.
95
6.2
Lagrange’s Theorem
In this section, we state and prove the most important theorem in finite group theory. We then state
and prove some of its consequences.
Lagrange’s Theorem
Theorem 6.7. Let H be a subgroup of a finite group G. Then |H| | |G| (i.e. the order of H
must divide the order of G), and
[G : H] =
|G|
.
|H|
Proof. We prove the result for left cosets. A similar proof works for the right cosets. Let
a1 H, a2 H, . . . , ar H
denote the distinct left cosets of H in G (notice that there are a finite number of them as G is a finite
group). By Proposition 6.3 parts (1) and (5),
G=
r
[
ai H, where ai H ∩ aj H = ∅ for i 6= j.
i=1
That is, |G| =
r
X
|ai H|. By Proposition 6.3(7), we have |ai H| = |H| for all i = 1, 2, ..., r. Hence,
i=1
|G| =
r
X
i=1
|ai H| =
r
X
|H| = r|H| ⇒ |H| | |G|, and r =
i=1
|G|
.
|H|
Since r is the number of distinct left cosets our proof is complete.
Since H is always one of the cosets of H, we get the following useful result: [G : H] = 1 if and only
if G = H.
It should be noted that Lagrange’s Theorem gives us a list of candidates for the order of the subgroups
of a group. For example, if the order of our group is 10, then the group may have subgroups of order
1, 2, 5, or 10, but no others.
As we will shortly see, the converse of Lagrange’s Theorem is false. That is, it is possible for a number
to divide the order of a finite group yet the group have no subgroup of that order (see Example 6.15).
The following three corollaries are a direct consequence of Lagrange’s Theorem, one of which classifies
all groups of prime order.
In a finite group the order of an element divides the order of the group
Corollary 6.8. If |G| < ∞, then |a| | |G| for all a ∈ G.
Proof. Lagrange’s Theorem implies that |hai| | |G| (as hai G), and the result follows by Corollary
3.4.
96
Groups of prime order are cyclic
Corollary 6.9. Groups of prime order are cyclic.
Proof. Let G be a group with order p, where p is a prime number. Pick a non-identity element a ∈ G.
By Corollary 6.8, |a| = p (as a 6= e). Hence, G = hai.
As a consequence of the previous corollary and Theorem 5.15, groups of order p (where p is a prime
number) are isomorphic to Zp .
Corollary 6.10. Suppose |G| = n < ∞. Then an = e for all a ∈ G.
Proof. Given a ∈ G, by Corollary 6.8, n = |a|k for some integer k. Hence, an = (a|a| )k = ek = e.
Example 6.11. Suppose G is a group with order pq, where p and q are distinct primes, and
a, b ∈ G with |a| = p and |b| = q. If H is a subgroup of G with a, b ∈ H, prove that H = G.
Solution. Let H G and a, b ∈ H. Since |a| = p > 1 and |b| = q > 1, we have a 6= e, and b 6= e. By
Lagrange’s Theorem, |H| = 1, p, q, or pq. If |H| = 1, then a = b = e, a contradiction. If |H| = p,
then |a| = p = |b|, implying that p = q, a contradiction. If |H| = q, then |a| = q = |b|, implying that
p = q, a contradiction. Hence, |H| = pq. Since H ⊆ G and |H| = |G| < ∞, we have H = G.
♣
We now use group theory to prove a number theoretic result, known as Fermat’s Little Theorem.
Fermat’s Little Theorem
Corollary 6.12. For any a ∈ Z, and p prime, we have ap ≡ a(mod p).
Proof. Let a ∈ Z, p a prime number. If p|a, then p|(a(ap−1 − 1)). In particular, p|(ap − a), and thus
ap ≡ a (mod p). If p - a, then gcd(a, p) = 1 (as p is prime). It follows that a ∈ U (p), and by Corollary
6.10, ap−1 = a|U (p)| ≡ 1 (mod p). Hence, ap ≡ a (mod p), completing the proof.
Corollary 6.8 shows that the order of every element in a finite group must divide the order of the
group. The next two examples show how this result can be used to impose conditions on the elements
of a finite group.
Example 6.13. Prove that a group of order 8 must have an element of order 2.
Solution. By Corollary 6.8, every non-identity element in G has order 2, 4, or 8. If G has an element
of order 8, then G is cyclic and by the Fundamental Theorem of Cyclic groups G has an element of
order 2. If G has an element of order 4, say g, then
|g 2 | =
4
= 2 by Proposition 3.7.
gcd(2, 4)
Of course, if G has an element of order 2 there is nothing to prove.
97
♣
Example 6.14. Suppose G is a finite group of order n and gcd(m, n) = 1. If g ∈ G with
g m = e, then g = e.
Solution. By Corollary 1.5, there exist s and t such that ms + nt = 1. By Corollary 6.10, g n = e.
Hence,
g = g ms+nt = (g m )s (g n )t = es et = e,
♣
completing the proof.
We now show that the converse of Lagrange’s Theorem is false.
The Converse of Lagrange’s Theorem is False
Example 6.15. Show that A4 has no subgroup of order 6.
Solution. The group A4 has order 12. Notice that A4 has eight elements of order 3, namely,
(1 2 3), (2 4 3), (1 4 2), (1 3 4), (1 3 2), (1 4 3), (2 3 4), (1 2 4).
Suppose H A4 and |H| = 6. First we show that every element of order 3 in A4 is in H.
Assume to the contrary. That is, assume a ∈ A4 , |a| = 3, and a ∈
/ H. Then H ∩ aH = ∅. Moreover,
|H| = |aH| = 6. In particular, A4 = H ∪ aH (as A4 has only 12 elements). Since A4 is a group, we
have a2 ∈ A4 = H ∪ aH;
• If a2 ∈ H, then a = a4 = (a2 )2 ∈ H, a contradiction.
• If a2 ∈ aH, then a2 = ah for some h ∈ H, i.e., a = h ∈ H, a contradiction.
We have just shown that every element of order 3 in A4 must be in H. But |H| = 6 and we have 8
such elements, a contradiction. Therefore, no such subgroup of A4 exists.
♣
We now state a theorem that relates the order of the product of two subgroups with the order of the
subgroups themselves. Much like Lagrange’s Theorem, this theorem imposes powerful restrictions on
the existence of subgroups of certain order (see Example 6.17).
|HK|
Proposition 6.16. Suppose G is a group and H, K G. Recall that
HK = {hk : h ∈ H, k ∈ K}.
(a) If H or K is an infinite group, then HK is an infinite set.
(b) If |H| < ∞ and |K| < ∞, then
|HK| =
|H| |K|
.
|H ∩ K|
Proof. The proof is left as an exercise (see Exercise 6.11).
98
Example 6.17. A group G of order 75 can have at most one subgroup of order 25.
Solution. Let H and K be two subgroups of G with order 25. Recall that H ∩ K is a group (since
H and K are groups). Moreover, H ∩ K H and H ∩ K K. By Lagrange’s theorem, |H ∩ K| = 1
or 5 or 25. By Proposition 6.16,
|HK| =
625
|H| |K|
=
.
|H ∩ K|
|H ∩ K|
If |H ∩ K| = 1 or 5, it would imply that |HK| = 625 or 125, respectively, a contradiction (since
|HK| ≤ |G| = 75). If |H ∩ K| = 25 then,
|H ∩ K| = |H| ⇒ H ∩ K = H ⇒ H ⊆ K.
Similarly, H ∩ K = K implying that K ⊆ H. Thus H = K, completing the proof.
♣
We now classify all groups with order 2p, where p is an odd prime. We will take care of the case
p = 2 in Theorem 8.17.
Classification of groups of order 2p (where p is an odd prime number)
Theorem 6.18. Let G be a group with |G| = 2p, where p is an odd prime. Then G ∼
= Z2p or
∼
G = Dp .
Proof. Suppose G Z2p . By Lagrange’s Theorem, the order of every non-identity element is either
2 or p.
For the moment, assume that the order of every non-identity element is 2. Then G is Abelian (see
Example 2.11). Pick two non-identity element a, b ∈ G and note that H = {e, a, b, ab} G. That is,
G has a subgroup of order 4, this implies that 4|(2p), which in turn gives 2|p, contradicting the fact
that p is an odd prime. Therefore, there exists x ∈ G with |x| = p.
Let b ∈ G with b ∈
/ hxi (there exist such an element as order of G is 2p and order of x is p).
By Lagrange’s Theorem |b| = 2 or p. Because hxi ∩ hbi hxi, Lagrange’s Theorem implies that
|hxi ∩ hbi| | p. In particular, |hxi ∩ hbi| = 1 or p. If |hxi ∩ hbi| = p = |x|, then hxi ∩ hbi = hxi (since
both are finite sets). In particular, hxi ⊆ hbi and |b| ≥ p. But this means |b| = p = |x|, which yields
hxi = hbi, a contradiction. So, |hxi ∩ hbi| = 1. By Proposition 6.16,
2p = |G| ≥ |hxihbi| =
|hxi| |hbi|
= |x| |b| = p |b| ⇒ 2 ≥ |b| ⇒ |b| = 2
|hxi ∩ hbi|
We have in fact proven that every element not in hxi has order 2.
If xb ∈ hxi, then xb = xk for some integer k. Then b = x−1 = xk−1 ∈ hxi, a contradiction. Thus
xb ∈
/ hxi and by the preceding paragraph |xb| = 2. In particular,
(xb)2 = e ⇒ xb = (xb)−1 = b−1 x−1 = bx−1
(6.1)
This relation completely determines the multiplication table (see Exercise 6.23). This argument
shows that any group of order 2p must have two generators, one of order 2, one of order p, and
99
satisfy the relation in (6.1). That is, all such groups are isomorphic. One such group is the dihedral
group of order 2p. Hence, if |G| = 2p, then
G∼
= hx, b : |x| = p, |b| = 2 , xb = bx−1 i = Dp ,
completing the proof.
Example 6.19. (a) Prove that any non-Abelian group G of order 10 is isomorphic to D5 .
(b) Prove that any non-Abelian group G of order 2p, where p is an odd prime, is isomorphic
to Dp .
Solution. (a) By Theorem 6.18, G ∼
= Z10 or D5 . Since G is not cyclic (as it is non-Abelian), G ∼
= D5 .
(b) Proof is very similar to part (a). By Theorem 6.18, G ∼
= Z2p or Dp . Since G is not cyclic (as it
∼
is non-Abelian), G = Dp .
♣
100
6.3
Exercises
Unless otherwise stated G is a group.
Exercise 6.1. Let H = {4k : k ∈ Z}. Find all the left (right) cosets of H in Z.
Exercise 6.2. Find all the left (right) cosets of {1, 10} in U (11).
Exercise 6.3. Find all the left and right cosets of H = h(1 3 2)i in S3 .
Exercise 6.4. Find all the left and right cosets of H = h(1 4 2 3)i in S4 .
Exercise 6.5. (a) Let |a| = 40. How many left cosets of ha6 i in hai are there? List them.
(b) Let |a| = n. How many left cosets of hak i in hai are there? Justify your answer.
Exercise 6.6. For each of the following give an example satisfying the conditions described.
(a) Find an element a in a group G and a subgroup H of G such that aH 6= Ha.
(b) Find elements a, b in a group G and a subgroup H such that aH ∩ Hb 6= ∅ and aH 6= Hb. Why
does this not contradict Proposition 6.3(5)?
Exercise 6.7. If H and K are subgroups of G and g ∈ G, show that g(H ∩ K) = gH ∩ gK.
Exercise 6.8. Give an example of a group G with subgroups H and K such that HK is not a
subgroup of G.
Exercise 6.9. Let a, b ∈ G, and H, K G. If aH = bK, prove that H = K.
Exercise 6.10. Prove that φ(n) is even for all n ≥ 3.
Exercise 6.11. Prove Proposition 6.16.
Exercise 6.12. Let |G| = 100, H G with |H| = 25. If g ∈ G with |g| = 5k for some non-negative
integer k, prove that g ∈ H.
Exercise 6.13. (a) If |G| = 21, prove that every proper subgroup of G is cyclic.
(b) Is part (a) true if we assume |G| = pq, where p and q are prime (not necessarily distinct).
Exercise 6.14. Suppose that |G| > 1 and G has no proper non-trivial subgroups. Prove that |G| = p,
where p is prime. Hint. First show that G must be finite.
Exercise 6.15. Let |G| = pq, where p and q are distinct primes. If G has only one subgroup of order
p and only one of order q, prove that G is cyclic.
Exercise 6.16. Let G be a group of order 25. Prove that either G is cyclic or |g| ≤ 5 for all g ∈ G.
Exercise 6.17. Let G be finite group with subgroups H and K such that H ⊆ K.
(a) Prove that
[G : H] = [G : K][K : H].
(b) If G is a finite group, prove that [G : Z(G)] cannot be prime.
101
Exercise 6.18. If |G| < 100 and G has subgroups of order 10 and 25, prove that |G| = 50.
Exercise 6.19. Let |G| = pqr, where p, q, and r are distinct primes. If H and K are subgroups of
G with |H| = pr and |K| = qr, prove that |H ∩ K| = r.
Exercise 6.20. Suppose H and K are subgroups of a group G with |H| = 105 and |K| = 14. Prove
that H ∩ K is cyclic.
Exercise 6.21. Suppose H and K are subgroups of a group G with |H| = 33 and |K| = 21. Prove
that H ∩ K is cyclic.
Exercise 6.22. Suppose φ : G → G0 is a homomorphism with |G| = p, where p is prime. Prove that
φ is either one-to-one or φ(g) = eG0 for all g ∈ G, where eG0 is the identity element of G0 .
Exercise 6.23. Suppose G is a group with order 2p, where p is an odd prime number. Use the
relation (6.1) stated in the proof of Theorem 6.18, |x| = p, and |b| = 2 to write the following elements
in the form xn bm , where n ∈ {0, 1, . . . , p − 1} and m ∈ {0, 1}.
(a) x2 (b3 x4 ).
(b) x3 (bx2 ).
(c) x4 (b2 x).
0 −1
0
1
Exercise 6.24. Let A =
,B =
∈ GL(2, R). In exercise 2.14 you showed that
1 0
−1 −1
|A| = 4, |B| = 3, and |AB| = ∞.
(a) Find |hAihBi|.
(b) Prove that hAihBi is not a subgroup of GL(2, R).
Exercise 6.25. Let H Dn . If |H| = 2k + 1 for some non-negative integer k, prove that H is cyclic.
6.4
Challenging Exercises
Exercise 6.26. Let |G| < ∞, and H G. If |G| = n|H|, prove that g n! ∈ H for all g ∈ G.
Exercise 6.27. Let H and K be subgroups of G such that [G : K] < ∞
(a) Prove that [H : H ∩ K] < ∞.
(b) Prove that [H : H ∩ K] ≤ [G : K].
(c) Prove that equality holds in (b) if and only if G = HK.
Exercise 6.28. Let C be a non-empty set and G S(C) (the group of all permutations of C). For
each i ∈ C, define the following two sets;
stabG (i) = {φ ∈ G : φ(i) = i}, orbG (i) = {φ(i) : φ ∈ G}.
The set stabG (i) is called the stabilizer of i in G. The set orbG (i) is called the orbit of i under G.
102
(a) Prove that stabG (i) G for any i ∈ C.
(b) If G is a finite group of permutations of a non-empty set S, prove that |G| = |orbG (i)| |stabG (i)|
for any i ∈ C. Hint. Let A be the set of left cosets of stabG (i). Build a bijection between A
and orbG (i).
Exercise 6.29. Let C = {1, 2, 3, 4, 5} and G = {ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} S5 .
(a) Compute stabG (2) and orbG (2). Confirm that |G| = |orbG (2)| |stabG (2)|.
(b) Compute stabG (3) and orbG (3). Confirm that |G| = |orbG (3)| |stabG (3)|.
103
7
Normal Subgroups and Factor Groups
7.1
Definition and Properties of Normal Subgroups
Suppose H is a subgroup of a group G. A natural question to ask is when does the set of left (right)
cosets of H in G form a group? And if they do, what is the operation? It turns out that we can
make the set of left cosets of H into a group provided aH = Ha for all a ∈ G. A word of caution:
the reader might think that this condition should be true for any subgroup of a group. But this is
far from the truth. For example, if G = S3 , H = {ε, (1 2)}, and a = (2 3), then
(2 3)H = {(2 3), (1 3 2)} =
6 {(2 3), (1 2 3)} = H(2 3).
Normal Subgroup
A non-empty subset H of a group G is called a normal subgroup if it satisfies the following
two conditions:
1. H G, and
2. aH = Ha for all a ∈ G
In this case we write H E G.
Notice that the second condition is equivalent to aHa−1 = H for all a ∈ G. Caution: aH = Ha does
not mean ah = ha for all h ∈ H. It means that for each h ∈ H, there exist h1 , h2 ∈ H such that
ah = h1 a and ah2 = ha. In particular, the elements h1 and h2 need not necessarily equal to h.
Proposition 7.1. H E G if and only if gHg −1 ⊆ H for all g ∈ G.
Proof. If H is a normal subgroup of G, then gHg −1 = H for all g ∈ G. Conversely, suppose
gHg −1 ⊆ H for all g ∈ G. Given a ∈ G, aHa−1 ⊆ H. Moreover, a−1 H(a−1 )−1 ⊆ H (taking g = a−1 ).
In particular, H ⊆ aHa−1 . Hence, aHa−1 = H, showing that H is normal in G.
Example 7.2. The set H = {ε, (1 2)} is a subgroup of S4 . Show that H is not normal in S4 .
Solution. Notice that (1 3)H = {(1 3), (1 2 3)}, meanwhile H(1 3) = {(1 3), (1 3 2)}. In particular,
(1 3)H 6= H(1 3), showing that H is not normal in S4 .
♣
Example 7.3.
(a) Every subgroup of an Abelian group is normal.
(b) If H Z(G), then H E G. Consequently, Z(G) E G.
(c) Let H E G and K G. Then HK G.
Solution. (a) If G is an Abelian group and H is a subgroup of G, then xh = hx for all x ∈ G and
h ∈ H. Hence, xH = {xh : h ∈ H} = {hx : h ∈ H} = Hx for all x ∈ G.
104
(b) Suppose H Z(G). Given x ∈ G and h ∈ H, xh = hx (every element in the center commutes
with every element in G). In particular, xH = Hx for all x ∈ G (similar to part (a)). Hence, H E G.
The second part follows from the fact that Z(G) Z(G).
(c) Notice that e = ee ∈ HK, showing that HK is not empty. Since the operation in G is binary,
we have HK ⊆ G. Given x, y ∈ HK, x = ab and y = cd for some a, c ∈ H and b, d ∈ K. Since H is
normal in G, bc ∈ bH = Hb. In particular, there exist c1 ∈ H such that bc = c1 b. Using this and the
fact that H and K are subgroups gives
xy = (ab)(cd) = a(bc)d = a(c1 b)d = ac1 |{z}
bd ∈ HK.
|{z}
∈H
∈K
Normality of H gives b−1 H = Hb−1 . That is, b−1 a−1 = hb−1 for some h ∈ H. Since K is a subgroup
and b ∈ K, we have b−1 ∈ K. Thus
x−1 = (ab)−1 = b−1 a−1 = hb−1 ∈ HK.
♣
By Proposition 2.24, HK is a subgroup of G.
Notice that there are subgroups H and K of a group G for which HK is not a subgroup (Exercise
6.8). Much like the center of a group, the kernel of a homomorphism is always a normal subgroup.
In fact, the converse of this statement is also true. That is, N is a normal subgroup of G if and only
if it is the kernel of some homomorphism φ : G → H (see Proposition 7.15).
Kernels are Normal Supgroups
Proposition 7.4. Let φ : G → H be a group homomorphism. Then ker φ E G.
Proof. By Proposition 5.4(4), ker φ G. Given a ∈ G and x ∈ ker φ,
φ(axa−1 ) = φ(a)φ(x)(φ(a−1 )) = φ(a)(φ(a))−1 = eH .
That is, axa−1 ∈ ker φ. Since a ∈ G and x ∈ ker φ were chosen arbitrarily, a(ker φ)a−1 ⊆ ker φ for all
a ∈ G, completing the proof.
Index 2
Proposition 7.5. If H G with [G : H] = 2, then H E G.
Proof. It suffices to prove aH = Ha for all a ∈ G. Given a ∈ G, we consider two cases;
Case 1. If a ∈ H, then aH = H = Ha (Proposition 6.3(2)).
Case 2. If a ∈
/ H, then aH 6= H and H 6= Ha (Proposition 6.3(2)). Since [G : H] = 2, the number of
left (right) cosets is 2. That is, {H, aH} is the set of left cosets of H in G and {H, Ha} is the set of
right cosets of H in G. Since the set of left (right) cosets of H partition G (Proposition 6.3(1) and
(5)),
x ∈ aH ⇔ x ∈
/ H ⇔ x ∈ Ha.
Hence, aH = Ha, completing the proof.
105
An E Sn
Corollary 7.6. An E Sn for all integers n ≥ 2.
Proof. Notice that An Sn and [Sn : An ] = 2. By Proposition 7.5, An E Sn .
Example 7.7. Show that the converse of Proposition 7.5 is false.
Solution. Let G = Z8 and H = h4i. Notice that H E G (as G is Abelian). However, [G : H] = 4. ♣
Theorem 7.8. Suppose G and G0 are groups with φ : G → G0 a group homomorphism. Then,
(1) φ(a) = φ(b) ⇔ a ker φ = b ker φ.
(2) If φ(g) = h, then φ−1 (h) = {x ∈ G : φ(x) = h} = g ker φ.
(3) If K E G, then φ(K) E φ(G) (notice that φ(K) may not be normal in G0 ).
(4) If | ker φ| = n < ∞, then φ is an n to 1 mapping from G onto φ(G).
(5) If K is a subgroup of G with |K| = n, then |φ(K)| | n.
(6) If K 0 E G0 , then φ−1 (K 0 ) E G.
Proof. (1) This follows from the following equivalences:
φ(a) = φ(b) ⇔ φ(b−1 a) = eG0 ⇔ b−1 a ∈ ker φ ⇔ a ker φ = b ker φ.
(2) Suppose φ(g) = h (this implies that g ∈ φ−1 (h)). Given x ∈ φ−1 (h),
φ(x) = h = φ(g) ⇒ g ker φ = x ker φ ⇒ x ∈ g ker φ.
Conversely, if a ∈ g ker φ then, a = gy for some y ∈ ker φ and
φ(a) = φ(gy) = φ(g)φ(y) = φ(g) = h ⇒ a ∈ φ−1 (h).
Hence, g ker φ = φ−1 (h).
(3) Suppose K E G. Given x ∈ φ(G), there exist some g ∈ G with x = φ(g). This yields
xφ(K)x−1 = {xφ(k)x−1 : k ∈ K} =
=
=
=
{φ(g)φ(k)(φ(g))−1 : k ∈ K}
{φ(gkg −1 ) : k ∈ K}
{φ(k) : k ∈ K}
φ(K).
since K E G
(4) By part (1), two elements under φ have the same image if and only if they belong to the same
coset of ker φ. Since all the cosets of φ have the same number of elements (namely, | ker φ| = n), the
result follows. Notice that this says n|φ(G)| = |G|.
106
(5) Suppose K G and let |K| = n. Define φK : K → φ(K) via h 7→ φ(h). The map φK is called
the restriction map of φ to K. Since K is a finite group so is ker φk , say | ker φk | = t. By construction
φK is onto, and by part (4), φK is a t to 1 mapping and t|φ(K)| = |K| = n.
(6) Given x ∈ G and a ∈ φ−1 (K 0 ), we have
φ(xax−1 ) = φ(x)φ(a)(φ(x))−1 ∈ K 0 (since K 0 E G0 ) ⇒ xax−1 ∈ φ−1 (K 0 ).
That is, xφ−1 (K 0 )x−1 ⊆ φ−1 (K 0 ) (as a ∈ φ−1 (K 0 ) was chosen arbitrarily). Furthermore, x ∈ G was
chosen arbitrarily, thus xφ−1 (K 0 )x−1 ⊆ φ−1 (K 0 ) for any x ∈ G. Hence, φ−1 (K 0 ) E G.
We saw in chapter 2 that the subgroup relation is a transitive relation. This is not the case for
normality.
Example 7.9. Give an example of groups K, H and G, such that K E H, H E G, but K is
not normal in G.
Solution. Let K = {ε, (1 2)(3 4)}, H = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} and G = A4 . Notice
that K E H (since [H : K] = 2), and H E G (this can be easily checked by brute force). However,
(1 2 3)((1 2)(3 4))(1 2 3)−1 = (1 3)(2 4) ∈
/K
♣
In particular, K is not normal in G.
We now show that normality is transitive if one assumes the middle group (H in the previous example)
is cyclic.
Example 7.10. Suppose H is a cyclic normal subgroup of a group G, prove that any subgroup
of H is also a normal subgroup of G.
Solution. Let H = hai for some a ∈ G, and N H. Notice that N is automatically normal in H
(as H is Abelian). By the Fundamental Theorem of Cyclic Groups N is cyclic, say N = hgi for some
g ∈ H. Since H is generated by a and N is a subset of H, we get g = ar for some r ∈ N. That is,
N = har i.
We now show that xN x−1 ⊆ N for all x ∈ G. Given x ∈ G, n ∈ N , we have n = (ar )m for some
integer m. In particular, xnx−1 = x(ar )m x−1 = x(am )r x−1 = (xam x−1 )r . Since H E G, xam x−1 ∈ H.
So xam x−1 = ak for some integer k. Thus
xnx−1 = (ak )r = (ar )k ∈ N.
♣
Hence, N E G.
7.2
Factor (Quotient) Groups
In this section, we focus on defining an operation on the left (right) cosets of a subgroup H of a
group G and show that under this operation the set of left (right) cosets of H in G become a group.
Theorem 7.11. Let G be a group and H E G. The set G/H = {gH : g ∈ G} is a group under
the operation of (aH)(bH) = (ab)H.
107
Proof. First we show that the operation defined is well defined. Suppose aH = cH and bH = dH.
Then, a ∈ cH and b ∈ dH, i.e., a = ch and b = dh1 for some h, h1 ∈ H. Thus
(aH)(bH) = (ab)(H) = ((ch)(dh1 ))H =
=
=
=
=
=
(chd)h1 H
(chd)H
(h1 H = H)
(cdh2 )H
(H E G)
(cd)h2 H
(cd)H
(h2 H = H)
(cH)(dH).
Given aH, bH, cH in G/H,
1. (aH)(bH) = (ab)H ∈ G/H, proving the operation is binary.
2. (aH)((bH)(cH)) = (aH)((bc)H) = (a(bc))H = ((ab)c)H = ((ab)H)(cH) = ((aH)(bH))(cH),
proving associativity.
3. (aH)(H) = aH = (H)(aH), and H = eH ∈ G/H. That is, H is the identity element in G/H.
4. (aH)(a−1 H) = (aa−1 )H = eH = H, and a−1 H ∈ G/H. Hence, (aH)−1 = a−1 H.
Hence, G/H is a group under the operation (aH)(bH) = (ab)H.
When the operation in the group is addition, it is understood that aH = a + H. Notice that if H is
a normal subgroup of G, then G/H = H \ G.
Factor (Quotient) Groups
If H is a normal subgroup of G, the group G/H is called the f actor (or quotient) group of G
by H.
Why are factor groups important? When |G| < ∞ and H 6= {e}, G/H is smaller than G and its
structure is usually less complicated than that of G (since it has fewer elements). Since H is the
identity element in G/H, one can think of the entire set H as being zero (under addition) or 1 (under
multiplication) in the group G/H.
A word of caution about notation. If H is a normal subgroup of G, when we ask for |aH| we are
asking for the order of aH in G/H.
Example 7.12. Let G = Z18 and H = h6i. List the elements in G/H, and find |3 + H|.
Solution. Observe that H = {0, 6, 12}. Hence,
G/H = {H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H}.
Moreover,
3 + H 6= H, and (3 + H)2 = (3 + H) + (3 + H) = 6 + H = H.
108
Thus |3 + H| = 2.
♣
Example 7.13. Let G be a group.
(1) If G is Abelian, then G/K is Abelian for any K G.
(2) If G is cyclic, then G/K is cyclic for any K G.
Solution. (1) Notice that since G is Abelian, every subgroup of G is normal. Given K G, and
aK, bK ∈ G/K,
(aK)(bK) = (ab)K = (ba)K = (bK)(aK).
(2) Suppose G = hai and K G (notice that K is normal in G as G is Abelian). Since G/K
is a group and aK ∈ G/K, we have haKi ⊆ G/K. We now prove the reverse inclusion. Given
gK ∈ G/K, g = an for some integers n. Hence,
gK = an K = (aK)n ∈ haKi.
That is, G/K ⊆ haKi, completing the proof.
♣
Example 7.14. Let G = U (32) and K = {1, 15}. Then K E G (as G is Abelian). Prove that
G/K ∼
= Z8 .
Solution. Notice that |G/K| =
or 8. Furthermore,
|G|
|K|
=
16
2
= 8 and 3K ∈ G/K. By Lagrange’s Theorem, |3K| = 1, 2, 4
3K 6= K, (3K)2 = 9K 6= K, (3K)4 = 81K = 17K 6= K.
Hence, |3K| = 8 which implies that G/K is cyclic. Hence, G/K ∼
= Z8 .
♣
Proposition 7.15. If N E G, then there exist a group homomorphism φ : G → H (H some
group) such that ker φ = N .
Proof. Since N is normal, G/N is a group. Let H = G/N , and define φ : G → H via g 7→ gN . Given
a, b ∈ G,
φ(ab) = (ab)N = (aN )(bN ) = φ(a)φ(b).
By Proposition 6.3(2),
ker φ = {a ∈ G : φ(a) = N } = {a ∈ G : aN = N } = {a ∈ G : a ∈ N } = N,
completing the proof.
Theorem 7.16. If G/H is cyclic, where H Z(G), then G is Abelian.
109
Proof. Suppose G/H = haHi, where H Z(G). Given x, y ∈ G, we have xH, yH ∈ G/H. So there
exist integers m and n such that xH = (aH)n = an H and yH = (aH)m = am H. By Proposition
6.3(4), x ∈ an H and y ∈ am H. This implies that x = an h and y = am h1 for some h, h1 ∈ H. Notice
that h, h1 ∈ H ⊆ Z(G). That is, h and h1 commute with every element in G. Hence,
xy = an ham h1 = an am hh1 = am an h1 h = am h1 an h = yx,
completing the proof.
The contrapositive of the above theorem is that if G is a non-Abelian group, then G/H is not cyclic
for any H Z(G).
Example 7.17. Suppose that G is a non-Abelian group, and |G| = pq, where p and q are
prime. Then Z(G) = {e}.
Solution. By Lagrange’s Theorem, |Z(G)| = 1, p, q, or pq. Since G is not Abelian, |Z(G)| 6= pq.
If |Z(G)| = p, then |G/Z(G)| = q. By Corollary 6.9, G/Z(G) is cyclic. Theorem 7.16 implies that
G is Abelian, a contradiction. Hence, |Z(G)| =
6 p. A similar proof shows that |Z(G)| =
6 q. Hence,
|Z(G)| = 1, completing the proof.
♣
One must take caution with factor groups. It is very tempting to say that if H and K are subgroups
of a group G and H ∼
= K, then G/H ∼
= G/K. However, this assertion is wrong (see Example 8.8).
We now state and prove one of the more famous results in finite group theory, known as Cauchy’s
Theorem.
Cauchy’s Theorem for Abelian Groups
Theorem 7.18. Suppose G is an Abelian Group with 2 ≤ |G| < ∞, p a prime number with
p| |G|. Then there exist g ∈ G such that |g| = p.
Proof. Let |G| = n. We apply strong induction on n. If n = 2, then G is cyclic. In particular,
G = hai for some non-identity element a ∈ G, and |a| = n = 2. Now suppose the result holds for all
groups whose order is less than n and greater than 1 (with n > 2). We prove the result holds for a
group of order n. Pick a prime number p|n. First we show that G has an element of prime order.
Let x ∈ G be a non-identity element. Since the order of the group is finite, so is the order of x. Let
|x| = m = q1 q2 ...qr , where qi are prime. If r = 1, then x itself has prime order. If r > 1, then
|xq2 q3 ...qr | =
|x|
q1 q2 ...qr
=
= q1 .
gcd(m, q2 ...qr )
q2 q3 ...qr
In particular, xq2 q3 ...qr has prime order. Using the above notation, let y = xq2 q3 ...qr . If p = q1 , then y
has order p and we are done.
Suppose p 6= q1 . Since G is Abelian, hyi is a normal subgroup of G (in fact, every subgroup of G is
normal). Set G1 = G/hyi. Then
|G1 | =
|G|
|G|
=
⇒ p| |G1 | (since p 6= q1 and p| |G|).
|y|
q1
110
Since |G1 | < |G|, there exist an element in G1 with order p (induction hypothesis), say bhyi, where
b ∈ G. Notice that bhyi =
6 hyi, since the order of bhyi is bigger than 1 and the order of hyi is 1. In
particular, b 6= e. Moreover,
hyi = (bhyi)p = bp hyi ⇒ bp ∈ hyi ⇒ |bp | = 1 or q1 (as |y| = q1 ).
If |bp | = 1, then bp = e, which implies that |b| = 1 or p. Because b 6= e, we have |b| = p and we are
done. If |bp | = q1 , then
(bq1 )p = bpq1 = (bp )q1 = e ⇒ |bq1 | | p ⇒ |bq1 | = 1 or p.
We claim that |bq1 | = p, which will complete the proof. If bq1 = e, then
(bhyi)q1 = bq1 hyi = hyi ⇒ p|q1 ,
a contradiction. Hence, |bq1 | = p.
It should be noted that Cauchy’s Theorem is in fact true for any finite group; Abelian or not (see
Exercise 7.34). We now use normal subgroups to give an alternate proof of the fact that A4 has no
subgroup of order 6.
A4 has no subgroup of order 6
Example 7.19. The group A4 has no subgroups of order 6.
Solution. Suppose H A4 with |H| = 6. Then [A4 : H] = 2 and by Proposition 7.5, H E A4 . Since
[A4 : H] = 2, we have (aH)2 = H for all a ∈ A4 . This is equivalent to a2 ∈ H for all a ∈ A4 . A quick
computation in A4 shows that A4 has 9 elements of the form a2 , and by above they must all be in
H. A contradiction (since order of H is 6). Hence, no such H exist.
♣
Theorem 7.20. G/Z(G) ∼
= Inn(G).
Proof. Define Φ : G/Z(G) → Inn(G) via gZ(G) 7→ φg (recall that φg (x) = gxg −1 for all x ∈ G). If
gZ(G) = hZ(G), then h−1 g ∈ Z(G). In particular, (h−1 g)x = x(h−1 g) for all x ∈ G, and this is
equivalent to hxh−1 = gxg −1 for all x ∈ G, that is, φg = φh , which gives Φ(gZ(G)) = Φ(hZ(G)).
This shows that Φ is well defined. Suppose Φ(gZ(G)) = Φ(hZ(G)), then,
φg = φh ⇒ gxg −1 = hxh−1 ∀x ∈ G ⇒ (h−1 g)x = x(h−1 g) ∀x ∈ G
⇒ h−1 g ∈ Z(G)
⇒ gZ(G) = hZ(G).
Therefore, Φ is injective. Given φg ∈ Inn(G), we have gZ(G) ∈ G/Z(G), and this gives Φ(gZ(G)) =
φg . Thus, Φ is surjective. Furthermore,
Φ((gZ(G))(hZ(G))) = Φ((gh)(Z(G)) = φgh = φg φh = Φ(gZ(G))Φ(hZ(G)),
showing that Φ is a homomorphism, and completing the proof.
111
7.3
Isomorphism Theorems
In general, it is much easier to construct homomorphisms than isomorphisms. The next theorem
allows us to build an isomorphism from a homomorphism.
First Isomorphism Theorem
Theorem 7.21. Let φ : G → K be a group homomorphism. Then G/ ker φ ∼
= φ(G).
Proof. Define Φ : G/ ker φ → φ(G) via g ker φ 7→ φ(g).
(1) Φ is well defined and injective:
g ker φ = h ker φ ⇔
⇔
⇔
⇔
⇔
h−1 g ∈ ker φ
φ(h−1 g) = eK
(φ(h))−1 φ(g) = eK
φ(h) = φ(g)
Φ(g ker φ) = Φ(h ker φ).
(2) Φ is onto: given φ(g) ∈ φ(G) we have g ker φ ∈ G/ ker φ and Φ(g ker φ) = φ(g).
(3) Φ is a homomorphism:
Φ((x ker φ)(y ker φ)) = Φ((xy) ker φ) = φ(xy) = φ(x)φ(y) = Φ(x ker φ)Φ(y ker φ).
Hence, Φ is an isomorphism, showing that G/ ker φ ∼
= φ(G).
Corollary 7.22. If φ : G → K is a group homomorphism with G and K finite groups. Then
|φ(G)| | |G| and |φ(G)| | |K|.
Proof. Since φ(G) = Im(φ) K, by Lagrange’s theorem |φ(G)| | |K|. Moreover,
φ(G) ∼
= G/ ker φ ⇒ |φ(G)| = |G/ ker φ| =
|G|
|G|
⇒ | ker φ| =
⇒ |φ(G)| | |G|,
| ker φ|
|φ(G)|
completing the proof.
By Theorem 2.27, CG (H) is a subgroup of NG (H). The following corollary tells us that CG (H) is the
kernel of a homomorphism from NG (H) to Aut(H), and thus CG (H) E NG (H). Moreover, it shows
that NG (H)/CG (H) is actually a subgroup of the automorphism group of H (up to isomorphism).
Corollary 7.23. If H is a subgroup of a group G, then NG (H)/CG (H) is isomorphic to a
subgroup of Aut(H).
112
Proof. Define Φ : NG (H) → Aut(H) via g 7→ φg where φg (x) = gxg −1 for all x ∈ H (notice that
φg ∈ Aut(H) as g ∈ NG (H) implies that gHg −1 = H). Since Φ(xy) = φxy = φx φy = Φ(x)Φ(y), the
map Φ is a homomorphism. Furthermore,
ker Φ =
=
=
=
{g ∈ NG (H) : Φ(g) = φeH } where φeH is the identity map on H
{g ∈ NG (H) : gxg −1 = x for all x ∈ H}
{g ∈ NG (H) : gx = xg for all x ∈ H}
CG (H)
By the First Isomorphism Theorem, NG (H)/CG (H) ∼
= Φ(NG (H)) Aut(H).
Second Isomorphism Theorem
Theorem 7.24. If H is a subgroup of a group G and N is a normal subgroup of G, then H ∩N
is normal in H, N is normal in HN , and H/(H ∩ N ) ∼
= HN/N .
Proof. We first show that H ∩ N is normal in H. Given x ∈ H and n ∈ H ∩ N , n ∈ H and n ∈ N .
Since H is a subgroup of G, xnx−1 ∈ H. Normality of N implies that xnx−1 ∈ N . So xnx−1 ∈ H ∩N .
Hence, x(H ∩ N )x−1 ⊆ H ∩ N for all x ∈ H. This shows that H ∩ N is normal in H.
Normality of N in HN follows from the fact that N is normal in G and HN ⊆ G. Define
Φ : HN → H/(H ∩ N ) via hn 7→ h(H ∩ N ).
Notice that h(H ∩ N ) = hH ∩ hN = H ∩ hN (Proposition 6.3(2) and Exercise 6.7). We show that
this map is a well defined homomorphism with kernel N and image H/(H ∩ N ). After doing so the
result follows by the First Isomorphism Theorem. If hn = h1 n1 , then h = h1 n1 n−1 and
Φ(hn) = h(H ∩ N ) =
=
=
=
=
=
(h1 n1 n−1 )(H ∩ N )
(h1 n1 n−1 )H ∩ (h1 n1 n−1 )N
hH ∩ h1 N
H ∩ h1 N
h1 (H ∩ N )
Φ(h1 n1 )
(Exercise 6.7)
(Proposition 6.3(2))
(Proposition 6.3(2))
(Exercise 6.7)
We have just proven that Φ is a well defined map. Given hn, h1 n1 ∈ HN , normality of N implies
that there exist n2 ∈ N such that nh1 = h1 n2 . This yields
Φ((hn)(h1 n1 )) = Φ(h(nh1 )n1 ) =
=
=
=
Φ((hh1 )(n2 n1 ))
(hh1 )(H ∩ N )
(h(H ∩ N ))(h1 (H ∩ N ))
Φ(h)Φ(h1 ).
Given h(H ∩ N ) ∈ H/(H ∩ N ), he ∈ HN (where e is the identity element in G and hence, in N )
and Φ(he) = h(H ∩ N ). Thus Im(Φ) = H/(H ∩ N ).
Given hn ∈ ker Φ, we have
Φ(hn) = H ∩ N ⇒ h(H ∩ N ) = H ∩ N ⇒ h ∈ H ∩ N ⇒ h ∈ N ⇒ hn ∈ N.
113
It follows that ker Φ ⊆ N . Conversely, given n ∈ N , we have n = en ∈ HN . So,
Φ(n) = Φ(en) = e(H ∩ N ) = H ∩ N.
Hence, n ∈ ker Φ, completing the proof.
Third Isomorphism Theorem
Theorem 7.25. If N and H are normal subgroups of a group G such that N H, then N EH,
H/N is normal in G/N and
(G/N )/(H/N ) ∼
= G/H.
Proof. The proof is left as an exercise.
114
7.4
Exercises
Unless otherwise stated G is a group.
Exercise 7.1. Let H = {ε, (1 2 3), (1 3 2)}.
(a) Prove that H is a subgroup of Sn for n ≥ 3.
(b) Is H a normal subgroup of S3 ?
(c) Is H a normal subgroup of S4 ?
(d) Is H a normal subgroup of S5 ?
Exercise 7.2.
(a) Prove that A5 has a subgroup of order 12.
(b) Prove that A5 does not have a subgroup of order 15 to 20, nor a subgroup of order 30.
Exercise 7.3. Let H be a subgroup of a group G.
(a) Prove that |gHg −1 | = |H|.
(b) If |H| = n < ∞, and H is the only subgroup of G with order n, prove that H E G.
Exercise 7.4. Prove that SL(2, R) E GL(2, R).
a b
Exercise 7.5. (a) Let H =
: a, b, d ∈ R, ad 6= 0 . Is H a normal subgroup of GL(2, R)?
0 d
a b
(b) Let K =
: a, b, c ∈ R, bc 6= 0 . Is K a normal subgroup of GL(2, R)?
c 0
Exercise 7.6. If N is an Abelian group with N E G and H is any subgroup of G, prove that
N ∩ H E N H.
Exercise 7.7. Given an integer n ≥ 2, prove that Z/hni ∼
= Zn .
Exercise 7.8. If H E G, prove that CG (H) E G.
Exercise 7.9. Suppose that φ : Z20 → Z20 is a homomorphism with ker φ = {0, 5, 10, 15}. If
φ(7) = 3, determine all elements that map to 3 under φ.
Exercise 7.10. Suppose that φ : U (24) → U (24) is a homomorphism with ker φ = {1, 11, 19}. If
φ(5) = 19, determine all elements that map to 19 under φ.
Exercise 7.11. Suppose that H and K are subgroups G.
(a) Prove that HK = H if and only if K ⊆ H.
(b) Suppose that N E G, |G| < ∞, and H G. If |G/N | is prime, prove that H ⊆ N or N H = G.
Exercise 7.12. Show that if G is non-Abelain, then Aut(G) is not cyclic.
115
Exercise 7.13. Suppose H G such that φ(H) = H for all φ ∈ Aut(G). Prove that H is normal
in G.
Exercise 7.14. Let G be a finite group, H G, and N E G. If gcd(|H|, |G/N |) = 1, prove that
H N.
Exercise 7.15. Let G be a finite group, g ∈ G, and N E G. Prove that the order of gN in G/N
must divide the order of g in G.
Exercise 7.16. Prove that if N is a normal subgroup of the finite group G and gcd(|N |, |G/N |) = 1,
then N is the unique subgroup of G with order |N |.
Exercise 7.17. (a) Prove that every element in Q/Z has finite order. Notice that Q/Z itself is an
infinite group.
(b) Prove that the only element in R/Q with finite order is the identity element.
Exercise 7.18. Suppose H G and |G| = 2p, where p is an odd prime number. Prove that either
|H| = 2 or H E G.
Exercise 7.19. Let G be a finite group, H G, |H| = 2k + 1 for some non-negative integers k, and
[G : H] = 2. Show that the product of all the elements in G (taken in any order) is not an element
of H.
Exercise 7.20. A subgroup H of G is said to be maximal if it satisfies the following two properties:
(1) H 6= G, and
(2) If K G with H ( K ⊆ G, then K = G.
Prove that the center of a group is never a maximal subgroup of a group.
Exercise 7.21. (a) Prove that h2i is a maximal subgroup in Z (under addition).
(b) Prove that h3i is a maximal subgroup in Z.
(c) Show that h4i is not a maximal subgroup in Z.
(d) Prove that hni is a maximal subgroup of Z if and only if n is prime.
Exercise 7.22. Let G = Z/h10i and H = h5i/h10i. List the elements of H and G/H. Show that
G/H ∼
= Z5 .
Exercise 7.23. Let G be an Abelian group and let n ∈ N. Let Gn = {g ∈ G : g n = e} and
Gn = {g n : g ∈ G}. Prove that G/Gn ∼
= Gn .
Exercise 7.24. (a) Suppose φ : Z36 → G is a surjective homomorphism, where |G| = 12. Find ker φ.
(b) (a) Suppose φ : Z25 → G is a surjective homomorphism, where |G| = 5. Find ker φ.
Exercise 7.25. A subgroup H of a group K is called a characteristic subgroup of K if φ(H) = H
for all φ ∈ Aut(K). Prove that if H is a characteristic subgroup of K, and K is a normal subgroup
of G, then H is a normal subgroup of G.
116
Exercise 7.26. Find the order of the following elements in their respective groups.
(a) Order of 5h13i in U (14)/h13i.
(b) Order of 4h20i in U (21)/h20i.
(c) Order of 4h9i in U (17)/h9i.
(d) Order of 4h2i in U (9)/h2i.
(e) Order of 4h2i in U (7)/h2i.
(f) Order of 9h5i in U (14)/h5i.
(g) Order of 3h9i in U (16)/h9i.
7.5
Challenging Exercises
Exercise 7.27. Let G be a finite group and let x and y be distinct elements of order 2 in G that
generate G. Prove that G ∼
= Dn , where n = |xy|.
Exercise 7.28. Find a group G with the following property: for all integers n ≥ 2, G has subgroup
isomorphic to Zn .
Exercise 7.29. Let G be a group.
(a) Given ψ ∈ Aut(G) and g ∈ G, prove that ψ ◦ φg ◦ ψ −1 = φψ(g) .
(b) Prove that Inn(G) E Aut(G).
Exercise 7.30. Suppose G is a finite group, H, K G, H 6= K with [G : H] = 2 = [G : K]. Prove
the following:
(a) (H ∩ K) E G.
(b) [G : H ∩ K] = 4.
(c) G/(H ∩ K) is not cyclic.
Exercise 7.31. Suppose |G| = 35. We show that G is cyclic. Assume to the contrary.
(a) Prove that G must have an element of order 5 and an element of order 7.
(b) Prove that G has only one subgroup of order 7, call it H. Hint. Assume to the contrary and
use Proposition 6.16.
(c) Use part (b) and Exercise 7.3 to show that NG (H) = G (i.e. H E G).
(d) Show that H CG (H). Hint. Use the fact that H has prime order.
(e) Use part (d) and Lagrange’s Theorem to conclude |CG (H)| = 7 or 35.
(f) Show that if |CG (H)| = 35, then G is cyclic, contradicting our assumption. Conclude that
CG (H) = H. Hint. If |CG (H)| = 35, prove that H Z(G), and use this to prove that G is
cyclic.
117
(g) Use part (f) and Corollary 7.23 to get a contradiction.
Exercise 7.32. Prove the Third Isomorphism Theorem (Theorem 7.25).
Exercise 7.33. Let A = {xyx−1 y −1 : x, y ∈ G}.
(a) Prove that hAi is normal in G.
(b) Prove that G/hAi is an Abelian Group.
(c) If N is a normal subgroup of G and G/N is Abelian, prove that hAi N .
(d) Prove that if hAi H G, then H E G.
The subgroup hAi is called the commutator of G.
Exercise 7.34. In this question we prove Cauchy’s Theorem for any finite group (see Theorem 7.18).
This proof is due to James McKay (Another proof of Cauchy’s group theorem, Amer. Math, Monthly,
66(1959), p.119).
Cauchy’s Theorem: Let G be a finite group and let p| |G|, where p is prime. Then there exist g ∈ G
with |g| = p.
(1) Suppose G is a finite group and p| |G|, where p is a prime number. Let A = {(g1 , g2 , . . . , gp ) :
g1 g2 · · · gp = e}. Show that p| |A|. Hint. Prove that |A| = |G|p−1 .
(2) Show that a cyclic permutation of an element in A is again in A.
(3) Define the following relation on A:
α ∼ β ⇔ α is a cyclic permutation of β.
Prove that ∼ is an equivalence relation on A.
(4) Prove that an equivalence class contains a single element if and only if it is of the form
{(g, g, . . . , g)} with g p = e.
(5) Prove that every equivalence class has order 1 or p (use the fact that p is prime).
(6) From (5) deduce that |G|p−1 = a + pb, where a is the number of equivalence classes of size 1
and b is the number of equivalence classes of size p.
(7) Use (6) to show that p|a. Conclude that a > 1.
(8) Since {(e, e, . . . , e)} is an equivalence class of size 1, conclude from (7) that there must be a
non-identity element g in G with equivalence class {(g, g, . . . , g)} and g p = e, completing the
proof (by Lagrange’s Theorem, |g| = p).
118
8
Direct Products and Sums
8.1
External Direct Sums
In this section, we show how to piece together groups to make larger groups.
External Direct Sums
Let G1 , G2 , ..., Gn be a finite collection of groups. The external direct sum of G1 , G2 , ..., Gn ,
written
G1 ⊕ G2 ⊕ ... ⊕ Gn =
n
M
Gi ,
i=1
is the set of all n-tuples with the i-th component an element of Gi , and the operation done
componentwise. More specifically,
G1 ⊕ G2 ⊕ ... ⊕ Gn =
n
M
Gi = {(g1 , g2 , ..., gn ) : gi ∈ Gi },
i=1
and
(g1 , g2 , ..., gn )(h1 , h2 , ..., hn ) = (g1 h1 , g2 h2 , ..., gn hn ),
with the understanding that gi hi is performed under the operation in Gi .
This construction is not new. For example, the plane R2 is simply R ⊕ R, where the operation is
componentwise addition (we ignore scalar multiplication for the time being).
Theorem 8.1. Let G1 , G2 , . . . , Gn be a finite collection of groups. Then;
1.
n
M
Gi is a group.
i=1
2.
n
M
i=1
Gi =
n
Y
|Gi |.
i=1
Proof. (1) Given (g1 , g2 , . . . , gn ), (h1 , h2 , . . . , hn ), (k1 , k2 , . . . , kn ) ∈
n
M
Gi ,
i=1
(g1 , g2 , . . . , gn )(h1 , h2 , . . . , hn ) = (g1 h1 , g2 h2 , . . . , gn hn ) ∈
n
M
i=1
119
G,
proving the operation is binary. Moreover,
((g1 , g2 , . . . , gn )(h1 , h2 , . . . , hn ))(k1 , k2 , . . . , kn ) =
=
=
=
=
(g1 h1 , g2 h2 , . . . , gn hn )(k1 , k2 , . . . , kn )
((g1 h1 )k1 , (g2 h2 )k2 , . . . , (gn hn )k1 )
(g1 (h1 k1 ), g2 (h2 k2 ), . . . , gn (hn k1 ))
(g1 , g2 , . . . , gn )(h1 k1 , h2 k2 , . . . , hn kn )
(g1 , g2 , . . . , gn )((h1 , h2 , . . . , hn ))(k1 , k2 , . . . , kn )),
proving associativity. Notice that (g1 , g2 , . . . , gn )(e1 , e2 , . . . , en ) = (g1 , g2 , . . . , gn ), where ei is the
n
M
identity element in Gi . Hence, (e1 , e2 , . . . , en ) is the identity element of
Gi . Moreover,
i=1
(g1 , g2 , . . . , gn )(g1−1 , g2−1 , . . . , gn−1 ) = (e1 , e2 , . . . , en ).
−1
In particular, (g1 , g2 , . . . , gn )
=
(g1−1 , g2−1 , . . . , gn−1 )
∈
n
M
Gi .
i=1
(2) Notice that |Gi | ≥ 1. If |Gj | = ∞ for some j = 1, 2, . . . , n, then
n
Y
|Gi | = ∞. Moreover,
i=1
(eG1 , eG2 , . . . , g , . . . , eGn ) ∈
|{z}
∈Gj
In particular,
n
M
Gi = ∞ =
i=1
n
Y
n
M
Gi .
i=1
|Gi |. If |Gi | = mi < ∞ for each i ∈ {1, 2, . . . n}, then for the first
i=1
coordinate we have m1 choices, the second coordinate we have m2 choices, and so on. Hence,
n
M
Gi = m1 × m2 × . . . × mn =
i=1
n
Y
|Gi |,
i=1
completing the proof.
Example 8.2. List the elements in U (4) ⊕ U (10). Find the order of (3, 3).
Solution. The set U (4) ⊕ U (10) = {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3), (3, 7), (3, 9)}. Notice that
(1, 1) is the identity element in U (4) ⊕ U (10). Moreover,
(3, 3)1
(3, 3)2
(3, 3)3
(3, 3)4
=
=
=
=
(3, 3) 6= (1, 1),
(9, 9) = (1, 9) 6= (1, 1),
(27, 27) = (3, 7) 6= (1, 1),
(81, 81) = (1, 1).
Hence, |(3, 3)| = 4.
♣
Example 8.3. List the elements of Z2 ⊕ Z3 . What group is Z2 ⊕ Z3 isomorphic to?
120
Solution. The set Z2 ⊕ Z3 = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}. This is an Abelian group of
order 6. Moreover, |(1, 1)| = 6. Hence, Z2 ⊕ Z3 = h(1, 1)i, and Z2 ⊕ Z3 ∼
♣
= Z6 .
Theorem 8.4. Suppose G1 , G2 , . . . , Gm are groups with |Gi | = ni < ∞. Then
|(g1 , g2 , ..., gm )| = lcm(|g1 |, |g2 |, ..., |gm |).
It should be noted that |(g1 , g2 , ..., gm )| is the order of (g1 , g2 , ..., gm ) in
m
M
Gi , meanwhile |gi | is the
i=1
order of gi in Gi .
Proof. Let (g1 , g2 , ..., gn ) ∈
m
M
Gi . Notice that
i=1
n
M
i=1
Gi =
m
Y
ni < ∞ ⇒ |(g1 , g2 , ..., gm )| < ∞.
i=1
Moreover, |gi | ≤ |Gi | < ∞ for i = 1, 2, . . . , n. Let s = lcm(|g1 |, |g2 |, ..., |gm |) and t = |(g1 , g2 , ..., gm )|.
For each i = 1, 2, . . . , m, s = |gi |ki for some integer ki . In particular,
|g |
|g |
s
|gm | km
(g1 , g2 , ..., gm )s = (g1s , g2s , ..., gm
) = ((g1 1 )ki , (g2 2 )k2 , . . . , (gm
) ) = (e1 , e2 , ..., em ).
Thus t ≤ s. Furthermore,
t
(g1t , g2t , ..., gm
) = (g1 , g2 , ..., gm )t = (e1 , e2 , ..., em ) ⇒ git = ei for each i = 1, 2, . . . , m
⇒ |gi | |t for each i = 1, 2, ..., m.
This in turn implies that t is common multiple of |g1 |, |g2 |, ..., |gm |. That is, s ≤ t. Hence, s = t.
Example 8.5. Find the order of (2, 2, (1 3 2)) ∈ Z4 ⊕ U (7) ⊕ S4 .
Solution. The order of 2 ∈ Z4 is 2, the order of 2 ∈ U (7) is 3, and the order of (1 3 2) ∈ S4 is 3. By
Theorem 8.4, |(2, 2, (1 3 2)| = lcm(2, 3, 3) = 6.
♣
Example 8.6. How many elements of order 5 are there in Z25 ⊕ Z10 .
Solution. By Theorem 8.4, (a, b) has order 5 if and only if 5 = lcm(|a|, |b|). In particular, |a| = 5
and |b| = 1 or 5, or |a| = 1 and |b| = 5 (remember |a| |25 and |b| |10).
Case 1. |a| = 5 and |b| = 1 or 5. We have φ(5) = 4 choice for a. We have one choice for b if |b| = 1
and 4 choices for b if |b| = 5, giving a total of 5 choices for b. Therefore, there are 4 × 5 = 20 such
elements.
Case 2. |a| = 1 and |b| = 5. Then we have one choice for a (namely, a = 0) and 4 choices for b.
Therefore, we have 1 × 4 = 4 such elements.
Hence, Z25 ⊕ Z10 has 20 + 4 = 24 elements of order 5.
121
♣
Example 8.7. Determine the number of cyclic subgroups of order 10 in Z100 ⊕ Z25 .
Solution. To do this, we first need to find the number of elements of order 10 in Z100 ⊕ Z25 . Then,
find out how many of them give us distinct cyclic subgroups. A similar argument to the one given
in the previous example shows that Z100 ⊕ Z25 has 24 elements of order 10. By Theorem 3.13, each
= 6 cyclic
cyclic group of order 10 has φ(10) = 4 elements which generate it. Therefore, there are 24
4
subgroups of order 10.
♣
We now use external direct sums to show that it is possible for two subgroups H and K to be
isomorphic, but G/H and G/K not be isomorphic.
Example 8.8. Let G = Z4 ⊕ U (4), H = h(2, 3)i, and K = h(2, 1)i. Prove that H ∼
= K but
G/H G/K.
Solution. Notice that H and K are both cyclic groups of order 2. Hence, H ∼
= Z2 ∼
= K. Moreover,
G/K = {K, (1, 1)K, (1, 3)K, (2, 3)K}, G/H = {H, (1, 1)H, (2, 1)H, (3, 1)H}
The order of all non-identity elements in G/K is 2. However, |(1, 1)H| = 4, that is, G/H = h(1, 1)Hi.
Hence, G/K is not isomorphic to G/H.
♣
Proposition 8.9. Let G and H be groups. Define Φ1 : G ⊕ H → G, Φ2 : G ⊕ H → H
via Φ1 (g, h) = g and Φ2 (g, h) = h. These maps are called projection maps onto G and H,
respectively. Prove that Φ1 and Φ2 are group homomorphisms which are surjective with
ker Φ1 = {eG } ⊕ H, and ker Φ2 = G ⊕ {eH }.
Proof. Given (a, b), (c, d) ∈ G ⊕ H,
Φ1 ((a, b)(c, d)) = Φ1 (ac, bd) = ac = Φ1 (a, b)Φ1 (c, d).
Given g ∈ G, the element (g, eH ) is in G ⊕ H and Φ1 (g, eH ) = g. We have just proven that Φ1 is a
surjective group homomorphism. Moreover,
ker Φ1 =
=
=
=
{(g, h) ∈ G ⊕ H : Φ1 (g, h) = eG }
{(g, h) ∈ G ⊕ H : g = eG }
{(eG , h) : h ∈ H}
{eG } ⊕ H.
A similar proof gives the desired results for Φ2 .
Proposition 8.10. Prove that G1 ⊕ G2 ∼
= G2 ⊕ G1 , where G1 and G2 are groups.
Proof. Define Φ : G1 ⊕ G2 → G2 ⊕ G1 via Φ(g1 , g2 ) = (g2 , g1 ). Given (a, b), (x, y) ∈ G1 ⊕ G2 ,
Φ((a, b)(x, y)) = Φ(ax, by) = (by, ax) = (b, a)(y, x) = Φ(a, b)Φ(x, y).
122
If Φ(a, b) = Φ(x, y), then (b, a) = (y, x), which in turn gives b = y and a = x. That is, (a, b) = (x, y).
Given (x, y) ∈ G2 ⊕ G1 , the element (y, x) is in G1 ⊕ G2 and Φ(y, x) = (x, y).
The previous proposition can be extended to any finite number of groups (induction). For example,
Z5 ⊕ U (10)
R⊕Z
Z5 ⊕ U (5) ⊕ R
∼
= U (10) ⊕ Z5
∼
= Z⊕R
∼
= U (5) ⊕ R ⊕ Z5
Proposition 8.11. Suppose H1 and H2 are normal subgroups of groups G1 and G2 , respectively.
Then H1 ⊕ H2 E G1 ⊕ G2 . Moreover, (G1 ⊕ G2 )/(H1 ⊕ H2 ) ∼
= (G1 /H1 ) ⊕ (G2 /H2 ).
Proof. The fact that H1 ⊕ H2 G1 ⊕ G2 is left as an exercise (see Exercise 8.3). Given (g1 , g2 ) ∈
G1 ⊕G2 , suppose a ∈ (g1 , g2 )H1 ⊕H2 , then a = (g1 , g2 )(h1 , h2 ) for some (h1 , h2 ) ∈ H1 ⊕H2 . Normality
of H1 and H2 implies that
a = (g1 h1 , g2 h2 ) = (h01 g1 , h02 g2 ) = (h01 , h02 )(g1 , g2 ) for some h01 ∈ H1 , h02 ∈ H2 .
In particular, a ∈ H1 ⊕ H2 (g1 , g2 ). Hence, (g1 , g2 )H1 ⊕ H2 ⊆ H1 ⊕ H2 (g1 , g2 ). A similar proof shows
the reverse containment. That is, (g1 , g2 )H1 ⊕ H2 = H1 ⊕ H2 (g1 , g2 ). Since (g1 , g2 ) ∈ G1 ⊕ G2 was
chosen arbitrarily, we have H1 ⊕ H2 E G1 ⊕ G2 . Define Φ : G1 ⊕ G2 → (G1 /H1 ) ⊕ (G2 /H2 ) via
Φ(g1 , g2 ) = (g1 H1 , g2 H2 ). Then, Φ is a surjective homomorphism, and ker Φ = H1 ⊕ H2 (see Exercise
8.4). The result follows by the First Isomorphism Theorem.
Theorem 8.12. Suppose G and H are finite cyclic groups. Then, G ⊕ H is cyclic if and only
if gcd(|G|, |H|) = 1.
Proof. Let |G| = m < ∞ and |H| = n < ∞. Notice that |G ⊕ H| = mn. Let d = gcd(m, n). Suppose
G ⊕ H is cyclic. Then there exist (g, h) ∈ G ⊕ H with G ⊕ H = h(g, h)i. It follows that |(g, h)| = mn.
Moreover,
(g, h)mn/d = (g mn/d , hmn/d ) = ((g m )n/d , (hn )m/d ) = (eG , eH ).
It follows that mn = |(g, h)| ≤
mn
.
d
Thus d = 1.
Conversely, if d = 1, then
|(g, h)| = lcm(m, n) =
mn
= mn = |G ⊕ H|,
gcd(m, n)
where G = hgi and H = hhi. Hence, G ⊕ H = h(g, h)i, completing the proof.
Notice that the previous theorem tells us that if G = hgi and H = hhi are finite cyclic groups with
gcd(|G|, |H|) = 1, then (g, h) generates G ⊕ H, that is, G ⊕ H = h(g, h)i.
123
Corollary 8.13. Suppose |Gi | < ∞, for i = 1, 2, ..., m with Gi cyclic. Then,
m
M
Gi is cyclic if and only if gcd(|Gi |, |Gj |) = 1 for i 6= j.
i=1
Proof. Use induction and the previous theorem.
Q
Lk
Corollary 8.14. Let m = ki=1 ni , where ni ∈ N and ni ≥ 2. Then, Zm ∼
= i=1 Zni if and
only if gcd(ni , nj ) = 1 for i 6= j.
Proof. Let Gi = Zni in Corollary 8.13.
For example,
Z30 ∼
= Z2 ⊕ Z15 ∼
= Z2 ⊕ Z3 ⊕ Z5 ,
Z60 Z10 ⊕ Z6 .
We now combine external direct products with the isomorphism theorems.
Example 8.15. If H and K are normal subgroups of a group G and H ∩ K = {e}, prove that
G is isomorphic to a subgroup of G/H ⊕ G/K.
Solution. Define Φ : G → G/H ⊕ G/K via g 7→ (gH, gK). For any g, g1 ∈ G,
Φ(gg1 ) = (gg1 H, gg1 K) = ((gH)(g1 H), (gK)(g1 K)) = (gH, gK)(g1 H, g1 K) = Φ(g)Φ(g1 ).
Moreover,
ker Φ =
=
=
=
=
=
{g ∈ G : Φ(g) = (H, K)}
{g ∈ G : (gH, gK) = (H, K)}
{g ∈ G : gH = H and gK = K}
{g ∈ G : g ∈ H and g ∈ K}
H ∩K
{e}.
by Proposition 6.3(2)
Since G ∼
= G/{e} (proof left to the reader),
G∼
= G/{e} = G/ ker Φ ∼
= Φ(G)
G/H ⊕ G/K
First Isomorphism Theorem
by Proposition 5.4(4),
♣
completing the proof.
124
8.2
Internal Direct Products
Given a finite number of groups (the groups themselves can be infinite), the external direct sum
enables us to construct larger groups. In this section, we aim to reverse this process. More precisely,
given a group G, we aim to break it down into a product of subgroups in such a way that we could
extract many of the properties of the group from the properties of the component pieces. Of course,
this is not always possible, but under certain nice circumstances it is.
Internal Direct Product Of Two Subgroups
Suppose H and K are subgroups of a group G. We say that G is the internal direct product
of H and K and write G = H × K if the following three properties hold:
(1) H, K E G.
(2) G = HK.
(3) H ∩ K = {e}.
For example, Z6 = H × K, where H = {0, 2, 4} and K = {0, 3}.
Internal Direct Product Of n Subgroups
G = H1 × H2 × ... × Hn is defined as follows:
(1) Hi E G for all i = 1, 2, . . . , n.
(2) G = H1 H2 · · · Hn .
(3) (H1 H2 · · · Hi ) ∩ Hi+1 = {e} for i = 1, 2, ..., n − 1.
Notice that condition (3) implies that Hi ∩ Hj = {e} provided i 6= j. To see this, we can assume
without any loss of generality that i < j. If g ∈ (Hi ∩ Hj ), then g ∈ Hi , which implies that
g = e...e
g e . . . e ∈ H1 H2 . . . Hi−1 Hi Hi+1 . . . Hj−1 .
|{z}
i−th
position
In particular,
g ∈ (H1 H2 . . . Hi−1 Hi Hi+1 . . . Hj−1 ) ∩ Hj ,
and by condition (3), the set on the right is trivial, and thus g = e.
We now show that the external and internal direct sums are isomorphic.
Theorem 8.16. If G is the internal direct product of a finite number of subgroups
H1 , H2 , . . . , Hn , then G = H1 × H2 × ... × Hn ∼
= H1 ⊕ H2 ⊕ ... ⊕ Hn .
Proof. Let G = H1 × H2 × ... × Hn . We first show that the normality of the Hi ’s together with the
third condition of the definition implies that the elements of Hi commute with the elements of Hj for
125
i 6= j. To do so, suppose hi ∈ Hi and hj ∈ Hj with i 6= j. Then by normality of Hi and properties
of cosets we have
−1
−1
−1 −1
−1 −1
(hi hj h−1
i )hj ∈ Hj hj = Hj and (hi hj hi )hj = hi (hj hi hj ) ∈ hi Hi = Hi
In particular,
−1
−1 −1
hi hj h−1
i hj ∈ Hi ∩ Hj = {e} ⇒ hi hj hi hj = e ⇒ hi hj = hj hi ,
as desired.
Now we show that each member of G can be expressed uniquely in the form h1 h2 · · · hn , where
hi ∈ Hi . Condition 2 of the definition guarantees that there is at least one such expression. Suppose
h1 h2 · · · hn = g = k1 k2 · · · kn , where hi , ki ∈ Hi for i = 1, 2, ..., n. By the first paragraph we have
−1
−1
−1
−1 −1
−1
−1
kn h−1
n = (k1 k2 · · · kn−1 ) h1 h2 · · · hn−1 = kn−1 · · · k2 k1 h1 h2 · · · hn−1 = k1 h1 k2 h2 · · · kn−1 hn−1 .
In particular,
−1
kn h−1
n ∈ H1 H2 · · · Hn−1 ∩ Hn = {e} ⇒ kn hn = e ⇒ hn = kn
That is, h1 h2 ...hn−1 = k1 k2 ...kn−1 . Repeating the process, shows that hi = ki for i = 1, 2, ..., n.
Hence, the expression is unique.
Define
Φ : G → H1 ⊕ H2 ⊕ ... ⊕ Hn via h1 h2 ...hn 7→ (h1 , h2 , ..., hn ).
The preceding paragraph shows that this map is well defined. The proof of the fact that Φ is an
isomorphism is left as an exercise.
One might ask why we construct two things that are isomorphic? After all, algebraically they are the
same. What we have shown is kind of misleading. The isomorphism is valid provided our collection
of groups is finite. If we have an infinite number of groups (or subgroups) Gi , then the isomorphism
no longer holds, and the two groups are no longer isomorphic.
Groups of order p2
Theorem 8.17. Suppose |G| = p2 , where p is a prime number. Then, G ∼
= Zp2 or Zp ⊕ Zp .
Proof. Suppose G is not cyclic. We show that G ∼
= Zp ⊕ Zp . Since G is not cyclic there does not
2
exist g ∈ G with |g| = p . By Lagrange’s Theorem, all non-identity elements in G have order p. Let
x ∈ G with x 6= e. Then,
|x| = p ⇒ |hxi| = p < p2 .
Pick y ∈
/ hxi. Then, hyi is a subgroup of G with order p. Since p is a prime number, hxi ∩ hyi = {e}.
Moreover,
|hxihyi| =
|x| |y|
= p2 ⇒ G = hxihyi.
1
By Exercise 8.20, hxi and hyi are normal subgroups of G. Hence, G = hxi × hyi, and the result
follows by Theorems 5.15 and 8.16.
126
The previous theorem classifies all groups whose size is the square of a prime number.
Corollary 8.18. If |G| = p2 , where p is a prime number, then G is Abelian.
Proof. By Theorem 8.17, G ∼
= Zp2 or G ∼
= Zp ⊕ Zp . In the first case it is cyclic, and hence, Abelian.
In the second case, it is the external direct sum of two Abelian groups, and hence, Abelian (see
Exercise 8.1).
127
8.3
Exercises
Unless otherwise stated G is a group.
Exercise 8.1. Let G1 , G2 , . . . , Gn be a finite collection of groups. Prove that
n
M
Gi is Abelian if
i=1
and only if each Gi is Abelian.
Exercise 8.2.
(a) Suppose Gi = hgi i for i = 1, 2, . . . , m, |Gi | < ∞. Prove that
gcd(|Gi |, |Gj |) = 1 for i 6= j if and only if
m
M
Gi = h(g1 , g2 , . . . , gm )i.
i=1
(b) Suppose that
m
M
Gi is cyclic. Prove that each Gi is cyclic.
i=1
(c) Prove that Z ⊕ Z is not cyclic. This shows that the converse of part (b) is false.
Exercise 8.3. (a) If H1 and H2 are subgroups of groups G1 and G2 , respectively, prove that H1 ⊕H2 G1 ⊕ G2 .
n
n
M
M
(b) If Hi Gi for i = 1, 2, . . . , n, prove that
Hi Gi .
i=1
i=1
Exercise 8.4. Prove that the map Φ in Proposition 8.11 is a surjective homomorphism with kernel
H1 ⊕ H2 .
Exercise 8.5. (a) Is Z4 ⊕ Z6 isomorphic to Z24 ?
(b) Is Z3 ⊕ Z7 isomorphic to Z21 ?
(c) Is Z10 ⊕ Z20 isomorphic to Z200 ?
Exercise 8.6. Let G be a group.
(a) Prove that G ⊕ {e} ∼
= G.
(b) Prove that G/{e} ∼
= G.
(c) Prove that G × {e} = G.
Exercise 8.7. (a) Determine the number of cyclic subgroups of order 10 in Z30 ⊕ Z120 . Find all the
possible generators.
(b) Determine the number of cyclic subgroups of order 5 in Z5 ⊕ Z25 ⊕ Z125 . Find all the possible
generators.
Exercise 8.8. Let p be a prime. Prove that Zp ⊕ Zp has exactly p + 1 subgroups of order p.
Exercise 8.9. Let (g1 , g2 , . . . , gm ) ∈
m
M
Gi . Prove that |(g1 , g2 , . . . , gm )| = ∞ if and only if |gi | = ∞
i=1
for some i = 1, 2, . . . , m.
Exercise 8.10. Prove that the map Φ in Theorem 8.16 is an isomorphism.
128
Exercise 8.11. Suppose |G/Z(G)| = p2 , where p is prime. Prove that G/Z(G) ∼
= Zp ⊕ Zp .
Exercise 8.12. Let φ : Z ⊕ Z → Z via (a, b) 7→ a + b.
(1) Prove that φ is a homomorphism.
(2) Find ker φ.
(3) Find φ−1 (2), φ−1 (5), and φ−1 (10).
Exercise 8.13. (a) Is S4 ∼
= Z4 ⊕ U (7)? Justify your answer.
(b) Is S4 ∼
= Z6 ⊕ U (5)? Justify your answer.
Exercise 8.14. Suppose φ : Z ⊕ Z → Q is a homomorphism such that φ(2, 5) =
Determine φ(0, 7).
2
3
and φ(3, 4) = 12 .
Exercise 8.15. Suppose φ : Z4 ⊕ Z7 → Z28 is an isomorphism with φ(1, 3) = 5. Find (a, b) ∈ Z4 ⊕ Z7
such that φ(a, b) = 7.
Exercise 8.16. (a) Determine Aut(Z2 ⊕ Z2 ).
(b) Determine Aut(Z3 ⊕ Z3 ).
Exercise 8.17. (a) How many (distinct) isomorphisms are there from Z6 to Z2 ⊕ Z3 ?
(b) How many (distinct) isomorphisms are there from Z20 to Z4 ⊕ Z5 ?
Exercise 8.18. Let H = h(12)i and K = h(123)i in S3 . Is S3 = H × K?
Exercise 8.19. Let G = Z under addition, H = h8i and K = h3i.
(a) Prove that G = HK. Notice that hk = h + k, since the operation is addition.
(b) Is G = H × K? Justify your answer.
Exercise 8.20. Suppose G is a non-cyclic group of order p2 , where p is a prime number. Prove that
every subgroup of G is normal. Do not assume G is Abelian.
Exercise 8.21. Suppose H and K are subgroups of a group G with |H| = 28 and |K| = 32. Prove
that H ∩ K is Abelian.
Exercise 8.22. Let G = Z ⊕ Z5 and let H = {g ∈ G : |g| = ∞ or |g| = 1}. Is H a subgroup of G?
If yes, prove it. If no, give a counter example showing one of the properties of a subgroup fails.
Exercise 8.23. Let G be a group. Suppose that G = H × K and that N is a normal subgroup of
H. Prove that N is a normal subgroup of G.
Exercise 8.24. Let G be a group and let H = {(g, g) ∈ G ⊕ G : g ∈ G}.
(a) Prove that H is a subgroup of G ⊕ G.
(b) Prove that H is a normal subgroup of G ⊕ G if and only if G is Abelian.
129
8.4
Challenging Exercises
Exercise 8.25. Prove that if p is a prime number, then U (p) is cyclic. Give an example showing the
converse of this statement is false.
Exercise 8.26. Let G = Z ⊕ Z.
(a) Prove that (1, 0) + h(2, 2)i has infinite order in G/h(2, 2)i.
(b) Prove that (1, 1) + h(2, 2)i =
6 h(2, 2)i.
(c) Prove that the order of (1, 1) + h(2, 2)i in G/h(2, 2)i is 2.
(d) Use the preceding parts to prove that G/h(2, 2)i is not cyclic.
Exercise 8.27. Determine the order of (Z ⊕ Z)/h(3, 2)i. Is the group cyclic?
Exercise 8.28. (a) Describe the set h(1, 0)i in Z ⊕ Z. Determine if the group Z ⊕ Z/h(1, 0)i is cyclic.
(b) Describe the set h(2, 4)i in Z ⊕ Z. Determine if the group Z ⊕ Z/h(2, 4)i is cyclic.
(c) Describe the set h(3, 3)i in Z4 ⊕ Z6 . Determine if the group Z4 ⊕ Z6 /h(3, 3)i is cyclic.
Exercise 8.29. Suppose G and H are groups. Prove that
(G ⊕ H)/({eG } ⊕ H) ∼
= G and (G ⊕ H)/(G ⊕ {eH )} ∼
=H
Exercise 8.30. Prove that every group of order 77 is cyclic.
Exercise 8.31. Suppose G is a finite Abelian group that does not contain a subgroup isomorphic to
Zp ⊕ Zp for any prime number p. Prove that G is cyclic.
Exercise 8.32. Let N be a normal subgroup of G and let H be a subgroup of G. If N is a subgroup
of H, prove that H/N is a normal subgroup of G/N if and only if H is a normal subgroup of G.
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