Without beauty, we are lost.

MAS276: Rings and Groups
Lecturer’s version
Dr E. Cheng
J24 Hicks Building
Semester 2, 2011–12
http://cheng.staff.shef.ac.uk/mas276/
Without beauty, we are lost.
Contents
Introduction 4
I Rings 8
1 Introduction to rings 8
1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.2 Subrings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Division 19
2.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2 Zero-divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Integral domains . . . . . . . . . . . . . . . . . . . . . . . . . 31
3 Factorisation 37
3.1 Unique factorisation . . . . . . . . . . . . . . . . . . . . . . . 38
3.2 Euclidean domains . . . . . . . . . . . . . . . . . . . . . . . . 47

2 CONTENTS
II Groups 55
4 Revision 55
4.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . 55
4.2 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.3 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 62
5 Quotient groups 63
5.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
5.2 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
5.3 Quotient groups . . . . . . . . . . . . . . . . . . . . . . . . . 70
6 Conjugacy 72
6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
6.2 The class equation . . . . . . . . . . . . . . . . . . . . . . . . 79
7 Homomorphisms 88
7.1 Kernels and images . . . . . . . . . . . . . . . . . . . . . . . . 88
7.2 Quotient groups revisited . . . . . . . . . . . . . . . . . . . . 96
7.3 First isomorphism theorem . . . . . . . . . . . . . . . . . . . 102
III Homework and Tutorial Questions 109
8 Homework questions 110
9 Tutorial questions 120

CONTENTS 3
Information
Please see the course webpage for answers to some Frequently Asked Ques-
tions on matters such as what to do if you miss a lecture. If you email me
with a question that is already answered in the FAQ, I will not reply.
• Homework is due every week whether or not you have a tutorial. There
is also tutorial work for every week whether or not you have a tutorial.
These exercises and instructions are at the back of the booklet.
• There will be an online test each week. These will count for
The deadline for these is:
...........................................
...........................................
• There will be “office hours” each week when you can come and ask for
extra help with any lecture material or exercises. These are at
...........................................
• In week #n should should do tutorial sheet #n, which will help you
with
1. Online Test Week #n, due on Thursday of week #n, and
2. Homework Sheet #n, due on Monday of week #(n + 1).
• For this course you have two hours of lectures per week and one hour
of tutorial per fortnight. We expect you to do at least three hours of
private study per course per week as well. If you do not do this
amount, you are likely to find that there is too much material
to learn all at once before the exams.

4 Introduction
Introduction
You have already met groups.
Groups are sets of things where those things interact with each other in
nice ways.
Some key examples are:
• symmetry groups of various objects (rotations and reflections)
• the integers under addition
• similarly Q, R, C
• n × n matrices under addition
• polynomials under addition
But in the cases apart from symmetries, saying something is a group simply
doesn’t capture everything that’s going on. It’s a bit like saying
A frying pan is a lump of metal.
Beer is a liquid.
In Numbers and Proofs you studied various properties of integers:
• addition and subtraction
• multiplication
• divisibility—when can we divide?
• division with remainder
• highest common factor, Euclid’s algorithm
• prime numbers
• unique factorisation into primes

Introduction 5
• modular arithmetic
What we’re going to do now is find out what basic properties about Z
enabled us to do all that.
Question: Why do we want to know this?
It’s like saying: Why might we want to know how a car/computer works?
• sheer curiosity
• so that if it goes wrong we can fix it
• it might help us drive better, especially in bad weather
• it helps us be able to use a different one
• we can show off to our friends
Numbers are great. Numbers are everywhere. It would be nice to understand
them better.
But also, there are other number-like situations where it would be nice to be
able to use all the techniques we know about integers—so we have to know
whether that will work or not. Will everything go horribly wrong??
Example: Fermat’s Last Theorem
Example: Will there be some strange loopholes e.g. doctors vs nurses
football?

can always divide
e.g. Q, R, C
6 Introduction
fields are vacuously all of those
Groups
Rings
Fields Integral domains
Key examples
Unique factorisation domains
Euclidean domains
these are about what
happens when you can’t
divide by some non-zero
things
• Z Q R C
divide can do
analysis
can solve
poly
equs
• R[i] throw some extra thing in and stir it up
Z[i] = the “Gaussian integers”
Z[ √ d ] where d is a square-free integer
+, −,0
also × and a distributive law
no zero-divisors
unique factorisations
can do Euclid’s algorithm
• Zn gives us different behaviour depending on whether n is prime or
not
• R[x] = polynomials with coefficients in R

Introduction 7
• R[x1,... xn] = polynomials in n variables
• Matn(R) = n × n matrices with coefficients in R

8 Section 1. Introduction to rings
Part I
Rings
1 Introduction to rings
1.1 Definitions
Think about Z and what we can do with it:
• add
• subtract
• multiply
• we can’t necessarily divide
• we have special numbers 0 and 1 — what is special about them?
Definition 1.1.1. A ring is a set R equipped with two operations: addition
+ and multiplication × (or just .) i.e.
for all a,b ∈ R we have an element a + b ∈ R and a × b ∈ R
satisfying the following axioms:
1. associativity of + ∀a,b,c ∈ R (a + b) + c = a + (b + c)
2. additive identity ∃0 ∈ R s.t. ∀a ∈ R a + 0 = a = 0 + a
3. additive inverse ∀a ∈ R ∃(−a) ∈ R s.t. a + (−a) = 0
4. commutativity of + ∀a,b ∈ R a + b = b + a
these four say R is an Abelian group under +
5. associativity of × ∀a,b,c ∈ R (ab)c = a(bc)
6. multiplicative identity ∃1 ∈ R s.t. ∀a ∈ R a.1 = a = 1.a
these two say R is a monoid under ×
7. distributive law ∀a,b,c ∈ R a(b + c) = ab + ac
∀a,b,c ∈ R (b + c)a = ba + ca

1.1 Definitions 9
Examples 1.1.2. Some of these examples we’ll do in more detail later.
1. Our most basic intuitive examples are
Z ⊂ Q ⊂ R ⊂ C
—we’ll later see that these are subrings as shown.
2. Note that the even numbers do not form a ring—why?
there’s no unit
They’re something else useful called an “ideal” which you’ll meet later
if you do Rings and Modules.
3. 2 × 2 real matrices form a ring.
What is 1? What is 0?
4. Polynomials with real coefficients form a ring.
What is +? What is ×? What is 1? What is 0?
5. Zn is a ring for each natural number n.
Definition 1.1.3. A commutative ring is a ring R in which × is commu-
tative, i.e.
∀a,b ∈ R ab = ba
Note that in this case we can drop some parts of the axioms—which?
half of multiplicative identity, half of distributivity
Remarks 1.1.4. Here are some remarks about the definition which are
boring, but it would be somehow wrong not to mention them.
1. Some people don’t demand a 1 in their definition of ring.
2. Some people don’t allow 0 = 1 in their definition of ring.
3. Some people always demand commutativity of × in their definition of
ring.
Remarks 1.1.5. Here are some more interesting remarks.

10 Section 1. Introduction to rings
1. We don’t demand multiplicative inverses. If we have them in a ring
R, that ring is called a field.
2. Many of the most obvious things we think are true are not actually
axioms—we have to prove them from scratch.
3. Identities and inverses are characterised by the axiom they obey, as
we will now see.
Lemma 1.1.6. “Identities are unique”
Let R be a ring. Suppose we have u,u ′ ∈ R such that
1. ∀x ∈ R u + x = x i.e. u is an additive identity, and
2. ∀x ∈ R u ′ + x = x i.e. u ′ is an additive identity.
Then u = u ′ .
Proof.
Putting x = u ′ in (1) gives u + u ′ = u ′ .
Putting x = u in (2) gives u ′ + u = u.
So we have
u ′ = u + u ′
= u ′ + u by commutativity of +
= u
as required.
Since identities are unique we can “name this baby”. We name it 0, and
then any element satisfying this property must be 0.
Next we can do something very similar for additive inverses.
Lemma 1.1.7. “Inverses are unique”
Let R be a ring, and let x ∈ R.
Suppose we have a,b ∈ R such that
1. x + a = 0 i.e. a is inverse to x
2. x + b = 0 i.e. b is inverse to x

1.1 Definitions 11
Then a = b.
Proof.
But also
x + b = 0 =⇒ a + (x + b) = a + 0
a + (x + b) = (a + x) + b by
= (x + a) + b by
= 0 + b by
= b by
= a by definition of 0
associativity of +
.................................................
commutativity of +
.................................................
assumption 1
.................................................
definition of 0
.................................................
Thus a = a + (x + b) = b as required.
So, like with the identities, we can “name this baby”. We call it −x, and
anything satisfying this property must be −x.
Note 1.1.8.
a − b is defined to be a + (−b).
Lemma 1.1.9. Let R be a ring and x ∈ R.
Then 0.x = 0.
Proof.
0x + 0x = (0 + 0)x by
= 0x by
Now, subtracting 0.x from both sides gives
(0x + 0x) − 0x = 0x − 0x
= 0 by
distributive law
.................................................
definition of 0
.................................................
definition of −
.................................................

12 Section 1. Introduction to rings
But
(0x + 0x) − 0x = 0x + (0x − 0x) by
Thus
= 0x + 0 by
= 0x by
0x = (0x + 0x) − 0x
= 0
associativity of +
.................................................
definition of −
.................................................
definition of 0
.................................................
as required.
Lemma 1.1.10. Let R be a ring and x ∈ R. Then
−(−x) = x
Proof. We need to show that x is the additive inverse of −x, i.e.
x + (−x) = 0.
But this is true by definition of −x, so x = −(−x) as required.
This is like Cinderella. Incidentally (and irrelevantly), did you know that
in the original, her slipper wasn’t made of glass at all?
Lemma 1.1.11. Let R be a ring and a,b ∈ R. Then
Proof. We need to show that
i.e. (−a)b + (ab) = 0
(−a)b = −(ab).
(−a)b is the additive inverse of (ab).
i.e. .................................................................................................................

1.1 Definitions 13
Now by the distributive law
(−a)b + (ab) ((−a) + a).b
LHS................................. = ................................
0.b
= .......................... by definition of additive inverse
0
= ................... by Lemma 1.1.9
Thus (−a)b = −(ab) as required.
Corollary 1.1.12. Let R be a ring and a ∈ R. Then
Proof.
Question: Can you tell the time?
(−1)a = −a.
(−1)a = −(1.a) by Lemma 1.1.11
Definition 1.1.13. Integers mod n
= −a by definition of 1.
Let n ∈ N. Then the ring Zn or Z/nZ is defined as follows:
• its set of elements is { 0, 1,... , n − 1 }
• + is addition mod n
• × is multiplication mod n
Definition 1.1.13 by equivalence classes.
Let x,y ∈ Z. We say x ≡ y (mod n) if n|x − y.
This is an equivalence relation.
Equivalence classes are represented by the elements {0,1,... n − 1}

14 Section 1. Introduction to rings
The equivalence classes form a ring: we define operations
and so on. This ring is Zn.
[x] + [y] = [x + y]
Definition 1.1.14. Polynomial rings
Let R be a ring.
Then we define R[x] to be the set of all polynomials with coefficients in R,
i.e.
for any n ∈ N0, where each ai ∈ R.
This is a ring.
Given
we define
So the first few terms of f.g are:
a0 + a1x + a2x 2 + · · · + anx n
f = a0 + a1x + a2x 2 + · · · + anx n
g = b0 + b1x + b2x 2 + · · · + bmx m
f + g =
(ak + bk)x k
k
f.g =
( aibj)x k
k
i+j=k
a0b0 + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x 2 + · · ·
...........................................................................................................................
We also need a 0 and a 1. These are:
Definition 1.1.15. Matrix rings
Let R be a ring, n ∈ N.
0 is 0, 1 is 1
.................
We write Matn(R) for the set of all n × n matrices with coefficients in R.
This is a ring with with the usual matrix addition and multiplication.
0 and 1 are:
⎛ ⎞ ⎛ ⎞
0 0 1 0
⎝ ⎠ and⎝
⎠
0 0 0 1

1.2 Subrings 15
Note that this is a non-commutative ring. For example:
⎛ ⎞ ⎛ ⎞
1 1 0 0
⎝ ⎠ and⎝
⎠
0 1 0 1
do not commute: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 0 0 0 1
⎝ ⎠ ⎝ ⎠ = ⎝ ⎠
0 1 0 1 0 1
but ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 0 1 1 0 0
⎝ ⎠ ⎝ ⎠ = ⎝ ⎠
0 1 0 1 0 1
1.2 Subrings
Remember subgroups? Subrings are very similar.
Definition 1.2.1. Subring
Let R be a ring. A subring of R is a subset S ⊂ R such that S is a ring
with the same +, ×,0,1 as R.
Explicitly, this means for all a,b ∈ S:
a + b ∈ S
−a ∈ S
0 ∈ S —automatic, but it’s good to remember it
those say that S is a subgroup of R
ab ∈ S
1 ∈ S
Examples 1.2.2. Some examples and non-examples of subrings:
1. Z ⊂ Q ⊂ R ⊂ C
2. Is Z2 a subring of Z?
No: eg 1 + 1 = 0 ∈ Z2 but 1 + 1 = 2 ∈ Z.
Compare with: Zn is not a subgroup of Z. It is a quotient group.

16 Section 1. Introduction to rings
3. Is Z2 a subring of Z4?
No: same counterexample.
4. Are the odd numbers a subring of Z?
No: eg they are not closed under addition.
5. Are the even numbers a subring of Z?
No: eg they do not contain 1.
The only subring of Z is Z itself.
6. Define R = { a
b
This is a subring of Q.
∈ Q | b is odd }.
Given a c
b , d ∈ R we can check:
• a c ad + bc
+ =
b d bd
which is in R because
• − a
b
• a
b .c
ac
=
d bd
is in R because
which is in R because
• 1 = 1
1
odd × odd = odd
........................................
b is odd
...........................................
which is in R because 1 is odd.
7. Is the trivial ring 0 a subring of R?
odd × odd = odd
........................................
No, unless R = 0 as well, because in 0, 0=1, which is only true if the
only element is 0.
8. Is Matn(R) a subring of R?
No, as it isn’t even a subset of R.
9. Is R[x] a subring of R? R is a subring of R[x]?
1. No, as again it isn’t even a subset of R. 2. Yes
10. Given a ring R and a natural number n, let S be the subset of R[x]
consisting of polynomicals of degree at most n. Is this a subring of R?
Only if n is 0, otherwise it won’t be closed under multiplication.

1.2 Subrings 17
Definition 1.2.3. Z[ √ d] BRING YOGHURT
Given d ∈ Z, the ring Z[ √ d] is defined to be all numbers of the form
a + b √ d
for any a,b ∈ Z. It is defined as a subring of C, so + and × are as in C.
Note we need to check that Z[ √ d] is a ring: is it closed under the + and ×
we have defined?
√ √ √
Given a1 + b1 d,a2 + b2 d ∈ Z[ d], we have
√ √
• a1 + b1 d + a2 + b2 d = (a1 + a2) + (b1 + b2) √ d
√ √
• (a1 + b1 d)(a2 + b2 d) = (a1a2 + b1b2d) + (b1a2 + a1b2) √ d
√ √
• −(a1 + b1 d) = −a − b d
• 1 = 1 + 0. √ d
So this really is a subring of C.
We are taking the integers and “throwing something in”.
So the interesting cases are when √ d is not an integer—the most interesting
cases are when √ d is irrational, which is why we use the following condition.
Definition 1.2.4. Square-free integers
An integer d is called square-free if it has no repeated prime factors.
Equivalently:
n ∈ Z and n 2 |d =⇒ n = ±1.
Note that 0 is not square-free.
So, examples of square-free integers are: 2, 10, 15, .........................
Examples of integers that are not square-free are: 8, 9, 12, 18, ...................
Actually what we usually end up needing to use is:
Lemma 1.2.5. Let d be square-free. Then for any a ∈ Z
d|a 2 ⇒ d|a.

18 Section 1. Introduction to rings
Lemma 1.2.6. Let d = 1 and square-free. Then √ d ∈ Q.
Proof. Both of these are on Tutorial sheet 3.
Proposition 1.2.7. Let d = 0,1 and square-free.
Then every element of Z[ √ d] has a unique expression as a + b √ d.
This should remind you of a basis for a vector space. It is very similar. It is
very different from the situation with polynomials—with polynomials, none
of the fruit gets stirred back in.
Proof. By contradiction.
√ √
Suppose a1 + b1 d = a2 + b2 d with b1 = b2.
Then
⇒
a1 − a2 = (b2 − b1) √ d
√ a1 − a2
d =
b2 − b1
since b1 = b2
⇒ √ d is rational # contradicts Lemma 1.2.6
So we must have b1 = b2. But this also gives a1 = a2.
Definition 1.2.8. Gaussian integers
When d = −1 we get Z[ √ d] = Z[i].
This is called the Gaussian integers and consists of all complex numbers
a + bi where a,b ∈ Z.
Finally here’s a definition that we won’t use, but it’s here for completeness.
Definition 1.2.9. Ring homomorphism Given rings R,S, a ring ho-
momorphism is a function f : R −→ S such that
1. ∀a,b ∈ R f(a + b) = f(a) + f(b) group homomorphism
2. ∀a,b ∈ R f(ab) = f(a).f(b)
3. f(1) = 1

Remarks 1.2.10.
1. Just like for group homomorphisms, we also have f(0) = 0, but we
don’t have to include this as an axiom because it follows from (1),
using additive inverses.
f(a) + f(0) = f(a + 0) = f(a), so can add −f(a) to both sides.
If we had multiplicative inverses, we wouldn’t have to demand f(1) =
1.
2. This definition ensures that if R is a subring of S, the inclusion R −→ S
is a ring homomorphism.
2 Division
Division is hard when you teach it to small children. We didn’t demand it
in a ring. Anyway, what is division?
Compare this with the question: What is subtraction?
adding the additive inverse
Answer: .................................................................
So division is defined to be multiplication by the multiplicative inverse
—if it exists!
For example given x ∈ Z, 1
x
—in fact it’s only in Z if x = ±1.
might not be in Z
But given x = 0 ∈ Q we definitely have 1
x ∈ Q
and likewise for R, C.
Now 1
x
then we can “divide by x” by multiplying by 1
x .
is “number such that when you multiply it by x you get 1” —and
The elements we can divide by are called units.
19

20 Section 2. Division
2.1 Units
Definition 2.1.1. Unit
Let R be a ring and x ∈ R. Then x is called a unit if it has a multiplicative
inverse, i.e.
∃y ∈ R s.t. xy = 1 = yx.
Note: We proved that if it a multiplicative inverse, then it is unique (this
is the same proof as the additive case, Lemma 1.1.7). So we can “name this
baby”, and we name it x −1 .
Examples 2.1.2.
1. In Z the only units are ±1. (They are very unit-like.)
—in fact ±1 are always units.
2. In Q every non-zero x is a unit: x −1 = 1
x .
Likewise in R and C.
3. In Matn(R) the units are the
invertible matrices.
......................................
4. On Tutorial sheet #1 you found the units in Zn for some n.
On Tutorial sheet #2 you saw that the units are in Zn are easy to
find.
Lemma 2.1.3. If x and y are units then xy is a unit.
Proof. We show that xy has an inverse given by: y −1 x −1
(y −1 x −1 )(xy) = 1
(xy)(y −1 x −1 ) = 1
Theorem 2.1.4. Units in Zn.
Let n > 1 be an integer and let a ∈ Zn. Then
a is a unit in Zn ⇔ hcf(a,n) = 1

2.1 Units 21
Proof. Recall from Numbers and Proofs:
hcf(a,n) = 1 ⇐⇒ ∃r,s ∈ Z s.t. ar + ns = 1
⇐⇒ ∃r,s ∈ Z s.t. ar − 1 = ns
⇐⇒ ∃r ∈ Z s.t. n|ar − 1
⇐⇒ ∃r ∈ Z s.t. ar ≡ 1 (mod n)
⇐⇒ ∃y ∈ Zn s.t. ay = 1 ∈ Zn
i.e. hcf(a,n) = 1 iff a is a unit in Zn.
This theorem gives us a way of not only finding if something has an inverse
in Zn, but of actually finding it.
Example 2.1.5. Is 47 a unit in Z157? If so, find its inverse.
1. We use Euclid’s algorithm to find hcf(157,47):
157 = 3.47 + 16
47 = 2.16 + 15
16 = 1.15 + 1
15 = 15.1 + 0
So hcf(157,47) = 1
......... and 47 is/is not* a unit in Z157.
2. We feed the numbers back in to find r,s ∈ Z s.t. ar + ns = 1:
Line 1: 16 = 157 − 3.47
Line 2: 15 = 47 − 2.16
= 47 − 2.(157 − 3.47)
= 7.47 − 2.157
Line 3: 1 = 16 − 1.15
= (157 − 3.47) − (7.47 − 2.157)
= 3.157 − 10.47
So −10.47 ≡ 1 (mod 157).
−10 ≡ 147
and .......................... is the inverse of 47 in Z157.

22 Section 2. Division
Proposition 2.1.6. The units of a ring R form a group U(R) under ×.
Proof. We need to show
1. x,y ∈ U(R) =⇒ x.y ∈ U(R)
2. x ∈ U(R) =⇒ x −1 ∈ U(R).
Then we automatically get 1 ∈ U(R). Now
1.
is Lemma 2.1.3
.....................................................................
2. We need to show that x−1 is a unit, i.e., that it has an inverse. We
x
show that its inverse is ............ :
x −1 .x = 1 = x.x −1
by definition of x −1 .
Examples 2.1.7.
1. In Z we have U(Z) = {1, −1} which is a group, isomorphic to C2, the
cyclic group of order 2.
2. In Q the non-zero elements form a multiplicative group.
3. In Z8 the units are {1,3,5,7}. Every element has order 2.
2.2 Zero-divisors
How do we usually solve quadratics? E.g.
x 2 + 3x = 2 = 0 as in HW # 1
(x + 1)(x + 2) = 0
⇒ x + 1 = 0 or x + 2 = 0
⇒ x = −1, −2

2.2 Zero-divisors 23
We crucially used the fact that
ab = 0 ⇒ a = 0 or b=0.
.........................................................................
“You can’t multiply non-zero things and get zero.”
Whereas in Z6
ab = 0 ⇒ 1. a = 0
or 2. b = 0
or 3. {a,b} = {2,3}
or 4. {a,b} = {3,4}
So if we try to solve that quadratic we get
(x + 1)(x + 2) = 0 ⇒ 1. x + 1 = 0 i.e. x = 5
or 2. x + 2 = 0 i.e. x = 4
or 3. {x + 1,x + 2} = {2,3} i.e. x = 1
or 4. {x + 1,x + 2} = {3,4} i.e. x = 2
So the solutions are x = 1,2,4,5 —there are four solutions!
What about in Z9? Z12? Z30? this is long
Z9 3,3,
Z12 2,2,3
Z30 2,3,5
x + 1,x + 2 x
x + 1 = 0 8
x + 2 = 0 7
x + 1,x + 2 x
x + 1 = 0 11
x + 2 = 0 10
3,4 2
8,9 7

24 Section 2. Division
x + 1,x + 2 x
x + 1 = 0 29
x + 2 = 0 28
5,6 4
9,10 8
14,15 13
15,16 14
20,12 19
24,25 23
The moral is that our methods for solving quadratics are rather more com-
plicated if we have more options for ab = 0. In this example the problem
was caused by the fact that 2.3 = 0 and also 3.4 = 0.
In fact, it’s not just quadratics that go wrong, e.g.
has two solutions:
or more simply
x ≡ 2,5
.........................
2x ≡ 4 (mod 6)
2x ≡ 0 (mod 6)
x ≡ 0,3
has two solutions: .........................
So even solving linear equations is a bit strange. We also know that
does not necessarily mean a ≡ b e.g.
3a ≡ 3b (mod 6)
3.1 ≡ 3.5 , or 0,2; 2,4; 0,4; 3,5
...................................................
All this is because in Z6 we have the possibility of multiplying non-zero a
and b getting zero.
These are called zero-divisors.

2.2 Zero-divisors 25
Definition 2.2.1. Let R be a ring and r ∈ R.
Then r is called a left zero-divisor if
r is called a right zero-divisor if
∃s = 0 ∈ R s.t. rs = 0.
∃s = 0 ∈ R s.t. sr = 0.
If R is both a left and a right zero-divisor we simply say it is a zero-divisor;
in particular if R is commutative then we don’t need to distinguish between
left and right.
If R = 0 then 0 is certainly a zero-divisor. This is a “boring” case, so if
r = 0 and is a zero-divisor we call it a proper zero-divisor.
In this course almost all the rings we deal with will be commutative. The
main exception is matrix rings.
Examples 2.2.2.
1. The zero-divisors in Z, Q, R, C are:
Only 0, so there are no proper ones.
2. The zero-divisors in Z4 are: 0,2
The zero-divisors in Z6 are: 0,2,3,4
The zero-divisors in Z8 are: 0,2,4,6
Can you guess what the pattern is?
⎛ ⎞
1 1
3. In Mat2(Z), ⎝ ⎠ is a left and right zero-divisor: e.g.
2 2
⎛ ⎞⎛
⎞
1 1 1
⎝ ⎠⎝
1
⎠ = 0 and
2 2 −1 −1
⎛ ⎞⎛
⎞
−2 1 1 1
⎝ ⎠⎝
⎠ = 0
−2 1 2 2
In fact any matrix with determinant 0 is a zero-divisor.
Theorem 2.2.3. Zero-divisors in Zn. (See also Tutorial sheet #2.)
Let n > 1 be an integer and a ∈ Zn. Then

26 Section 2. Division
1. a is a unit in Zn ⇐⇒ hcf(a,n) = 1, and
2. a is a zero-divisor in Zn ⇐⇒ hcf(a,n) = 1.
N.B. So every element of Zn is a unit or a zero-divisor and not both.
We will see that there are some rings with elements that are neither a unit
or a zero-divisor, but it is impossible to be both a unit and a zero-divisor.
Before proving this theorem let’s look at some examples.
Example 2.2.4. Consider Z12.
Can we multiply 2 by something non-zero and get 0? 6
Let’s try and do it a bit more systematically.
3 4
4 3
6 2
8 3
9 4
10 6
Example 2.2.5. n = 12,a = 8
Now hcf(8,12) = 4
.......... so we expect 8 to be a zero-divisor in Z12.
So can we find b such that 8b ≡ 0? We write
12 = 3.4
8 = 2.4
so 2.4.3 ≡ 0 (mod 12)
i.e. 8.3 ≡ 0 (mod 12)
That example was easy enough to do just by staring, but for bigger numbers
this technique is much more efficient than staring.

2.2 Zero-divisors 27
Example 2.2.6. n = 201, a = 63
First we find hcf(63,201) e.g. by finding their prime factorisations:
201 = 3.67
63 = 3.3.7
So hcf(63,201) = 3
................ so we expect 63 to be a zero-divisor in Z201.
So can we find b such that 63b ≡ 0? We have
3.3.7.67 ≡ 0 (mod 201)
i.e. 63.67 ≡ 0 (mod 201)
Before we prove the theorem, the following result is useful.
Lemma 2.2.7. “You can’t be both a zero-divisor and a unit” boring
lecture?
Let R be a ring and a ∈ R. Then
a is a unit =⇒ a is not a zero-divisor
Equivalently (using the contrapositive):
a is a zero-divisor =⇒ a is not a unit.
Proof. Suppose a is a unit and ar = 0. Then
r = a −1 .ar
= a −1 .0
= 0
so a is not a zero-divisor.
Proof of Theorem 2.2.3.
1. Done as Theorem 2.1.4.
2. Write hcf(a,n) = d > 1 and seek b = 0 such that ab ≡ 0 (mod n).
Now
n = n
a
.d, a =
d d .d
a n
d , d ∈ Z

28 Section 2. Division
Put b = n
. Then ab = a.n
d d
a
= .n ≡ 0 (mod n).
d
since a
d ∈ Z. Finally we must check b = 0. Now
0 < n
d
< n
since d > 1, so
n
≡ 0
d
(mod n).
For the converse we need to show that if a is a zero-divisor then
hcf(a,n) = 1.
Now, if a is a zero-divisor then by Lemma 2.2.7 it cannot be a unit,
so by Theorem 2.1.4 we have hcf(a,n) = 1 as required.
Example 2.2.8. In Z6[x], the element 1 + 3x is neither a unit nor a
zero-divisor.
For 1 + 3x to be a unit we need
(1 + 3x)(a0 + a1x + a2x 2 + · · · anx n ) = 1 ∈ Z6[x]
Comparing coefficients, this gives us
comparing constants a0 ≡ 1 (mod 6)
comparing coeffs of x i 3ai−1 + ai ≡ 0 (mod 6) ∀0 < i ≤ n
comparing coeffs of x n+1 3an ≡ 0 (mod 6)
Now 3a0 + a1 ≡ 0 =⇒ 3 + a1 ≡ 0 =⇒ a1 ≡ 3.
Similarly 3a1 + a2 ≡ 0 =⇒ 3 + a2 ≡ 0 =⇒ a2 ≡ 3.
Similarly ai ≡ 3 for all 0 < i ≤ n.
But also 3an ≡ 0 #.
Hence 1 + 3x is not a unit.
To show that 1 + 3x is not a zero-divisor, consider
(1 + 3x)(a0 + a1x + a2x 2 + · · · anx n ) = 0 ∈ Z6[x]

2.2 Zero-divisors 29
Comparing coefficients, this gives us
comparing constants a0 ≡ 0 (mod 6)
comparing coeffs of x i 3ai−1 + ai ≡ 0 (mod 6) ∀0 < i ≤ n
comparing coeffs of x n+1 3an ≡ 0 (mod 6)
Now 3a0 + a1 ≡ 0 =⇒ a1 ≡ 0.
Similarly 3a1 + a2 ≡ 0 =⇒ a2 ≡ 0.
Similarly ai ≡ 0 for all 0 < i ≤ n.
Hence 1 + 3x is not a zero-divisor.
Remark 2.2.9. Note that to prove r is not a zero-divisor we prove
rs = 0 =⇒ s = 0.
So in general we have the following picture.
In Zn it happens to be like this:
R
units zero-divisors
R
units zero-divisors

30 Section 2. Division
We’ll see that in a field it’s like this:
units
and in an integral domain it’s like this:
units
We can still do something like divide by r even if r is not a unit—as long as
it isn’t a zero-divisor.
Example 2.2.10. In Z6 suppose we are given
Can we deduce
No: e.g. 3.1 ≡ 3.5 but 1 ≡ 5.
However we can do 3a − 3b ≡ 0
so 3(a − b) ≡ 0
R
R
3a ≡ 3b (mod 6)
a ≡ b (mod 6) ?
so if 3 is not a zero-divisor we can deduce a − b ≡ 0 i.e. a ≡ b.
Lemma 2.2.11. Cancellation
Let R be a ring. Suppose r = 0 ∈ R is not a zero-divisor, and ra = rb.
Then a = b.
0
0

2.3 Integral domains 31
Proof. ra = rb ⇒ ra − rb = 0
⇒ r(a − b) = 0
⇒ a − b = 0 since r is not a zero-divisor
⇒ a = b
Effectively we have “cancelled” r.
Remember how in Numbers and Proofs you solved things like 2x ≡ 4 (mod 6) ?
—You had to start by taking the hcf of 2 and 6. Effectively, that was to see
if 2 was a zero-divisor or not (although they didn’t tell you at the time).
We have seen that zero-divisors are a bit annoying. In the next section,
we imagine we’re in a glorious world in which there are no zero-divisors.
Such a world is called an integral domain.
2.3 Integral domains
We have been studying various things that go a bit pear-shaped if you have
zero-divisors e.g. solving all kinds of equations. If we want to exclude that
possibility we go into “integral domains”.
Note that from this point on, all our rings are commutative.
Definition 2.3.1. An integral domain (ID) is a commutative ring with
no proper zero-divisors i.e. the only zero-divisor is 0.
Examples 2.3.2.
1. Z is an ID. So are Q, R, C.
2. Z[ √ d] ⊂ C and we will see that this means it must be an ID.
3. We will see: if R is an ID then R[x] is an ID (and the converse).
N.B. In an ID we can cancel by anything non-zero, since by Lemma 2.2.11
we can cancel by anything that isn’t a zero-divisor.
In the case of Zn it is nice and easy to tell whether or not we have an ID.

32 Section 2. Division
Theorem 2.3.3. Zn is an ID ⇐⇒ n is prime.
Proof. Follows from Theorem 2.2.3:
Zn is ID ⇔ ∀ 1 ≤ a ≤ n − 1, a is not a zero-divisor
⇔ ∀ 1 ≤ a ≤ n − 1,hcf(a,n) = 1 by Theorem 2.2.3
⇔ n is prime
Examples 2.3.4. So for example
• Z2, Z3, Z5, Z7, Z11,... are IDs
• Z4, Z6, Z8, Z9,... are not IDs.
Lemma 2.3.5. “A subring of ID is ID”
Let S be a subring of R.
Then R is ID =⇒ S is ID.
Proof.
Let a = 0,b = 0 ∈ S.
Then a = 0,b = 0 ∈ R so ab = 0 ∈ R
Then a = 0,b = 0 ∈ R so ab = 0 ∈ S
since S has the same multiplication as R.
Examples 2.3.6.
1. Z[ √ d] is a subring of C so is always an ID.
2. Any ring R is a subring of R[x].
So if R[x] is an ID then R must be an ID.
In practice we often use the contrapositive.
The converse of that second part is more profound, and harder to prove:
Theorem 2.3.7. Let R be a ring. Then R is ID ⇐⇒ R[x] is ID.
Proof.
“⇐=” is by Lemma 2.3.5 since R is a subring of R[x].

2.3 Integral domains 33
“=⇒”
Suppose R is ID.
Let f,g = 0 ∈ R[x]. We need to show fg = 0.
Let deg(f) = m, deg(g) = n.
Recall deg(f) = largest m s.t. coeff of x m is not 0.
So write f(x) = a0 + · · · + amx m
g(x) = b0 + · · · + bnx n
Now am, bn = 0 =⇒ ambn = 0 since R is ID.
But ambn is the coefficient of x m+n in fg.
So fg = 0.
Now we turn our attention to finding out what the units are in various IDs.
Theorem 2.3.8. Units in polynomial rings.
Let R be an ID.
Then the units in R[x] are just the units in R (so in particular they are all
constants).
Proof. —similar to Theorem 2.3.7.
Consider f,g ∈ R[x] s.t. fg = 1.
As before let deg(f) = m, deg(g) = n and write
so in particular am, bn = 0.
f(x) = a0 + · · · + amx m
g(x) = b0 + · · · + bnx n
Now we know that the coefficient of x m+n in fg is ambn
and this is non-zero as R is an ID.
But comparing coefficients in the equation fg = 1 we must have m + n = 0,
since that is the only non-zero coefficient on the RHS.
But m + n = 0 =⇒ m = 0 and n = 0.

34 Section 2. Division
So we know that f = a0 and g = b0, constants.
Moreover we know that a0b0 = 1 as constants, i.e. they are units in R.
Next we investigate units in Z[ √ d]. These are not very obvious at first sight.
To help with this we introduce something called a “norm”, which is a bit
like a way of measuring how “big” elements of Z[ √ d] are. Eventually we’ll
show that the units are precisely those elements of norm 1, but it will take
us a few more lemmas to get there.
Definition 2.3.9. Norm in Z[ √ d].
Let d = 1 be a square-free integer, and let r = a + b √ d ∈ Z[ √ d].
We define the norm of r as
N(r) = |a 2 − db 2 | = |(a + b √ d)(a − b √ d)|.
This is well-defined by uniqueness of a and b (Proposition 1.2.7).
Note that this means N(r) = ±(a + b √ d)(a − b √ d).
The following properties of the norm are extremely useful.
Lemma 2.3.10. Let d = 1 be a square-free integer, and r,s ∈ Z[ √ d]. Then
1. N(r) = 0 ⇐⇒ r = 0
2. N(rs) = N(r)N(s).
Proof.
1. “⇐=” is by the definition of norm.
Now for “=⇒”, put r = a + b √ d and suppose N(r) = 0
i.e. |a 2 − b 2 d| = 0
so a 2 = b 2 d. (1)
If b = 0 we have √ d = ± a
∈ Q #
b
since d is square-free (Lemma 1.2.6).
So b = 0, and by (1) we then get a = 0.
So r = 0 as required.

2.3 Integral domains 35
√ √
2. We simply calculate: Put r = a1 + b1 d, s = a2 + b2 d. Then
N(r)N(s) = |a 2 1 − b2 1 d|.|a2 2 − b2 2 d|
= |a 2 1 a2 2 + b2 1 b2 2 d2 − a 2 1 b2 2 d − a2 2 b2 1 d|
N(rs) = N (a1a2 + b1b2d) + (a1b2 + b1a2) √ d
= |(a1a2 + b1b2d) 2 − (a1b2 + b1a2) 2 d|
= |a 2 1 a2 2 + 2a1a2b1b2d + b 2 1 b2 2 d2 − a 2 1 b2 2 d − 2a1a2b1b2d − b 2 1 a2 2 d|
= |a 2 1 a2 2 + b2 1 b2 2 d2 − a 2 1 b2 2 d − b2 1 a2 2 d|
= N(r)N(s)
PS That was not fun to type. boring lecture?
We can now prove that the norm really does tell us which elements are units.
Theorem 2.3.11. Let d = 1 be a square-free integer and r ∈ Z[ √ d]. Then
Proof.
“⇐=”
Consider r = a + b √ d with N(r) = 1.
r is a unit in Z[ √ d] ⇐⇒ N(r) = 1.
Then N(r) = |(a + b √ d)(a − b √ d)| = 1
so r.(a − b √ d) = ±1
so either (a − b √ d) or −(a − b √ d) is an inverse for r, i.e. r is a unit.
“=⇒”
Conversely suppose r is a unit in R. Then
N(r)N(r −1 ) = N(rr −1 ) by Lemma 2.3.10
= N(1)
= 1
Now N(r) and N(r −1 ) are both non-negative integers so must both be 1
so in particular N(r) = 1 as required.

36 Section 2. Division
Examples 2.3.12. We can use this to “test” whether a given element is a
unit or not:
1. Consider Z[ √ 2], r = 7 + 5 √ 2. Then N(r) = |49 − 25.2| = 1
so 7 + 5 √ 2 isn’t a unit / is a unit* with inverse −(7 − 5√ 2)
....................
2. Consider Z[ √ 3], r = 4 + 7 √ 3. Then N(r) = |16 − 49.3| = 131 = 1
so 4 + 7 √ 3 isn’t a unit / is a unit* with inverse
3. Consider Z[ √ 3], r = 7 + 4 √ 3. Then N(r) = |49 − 16.3| = 1
so 7 + 4 √ 3 isn’t a unit / is a unit* with inverse
*delete as appropriate
not a unit
....................
7 − 4 √ 3
....................
Note that this does give us a way of testing if a given element is a unit, but
it doesn’t give us a way of actuallly finding all the units. In fact, finding all
the units is hard in general—the following theorem tells us when it’s easy
and when it’s hard.
Theorem 2.3.13. Units in Z[ √ d]. Let d = 1 be a square-free integer.
1. If d < −1 then U(Z[ √ d]) = {1, −1}.
2. If d = −1 then U(Z[ √ d]) = {1, −1,i, −i}.
3. If d > 1 then U(Z[ √ d]) is infinite.
Proof.
1. Put d = −m, with m > 1.
Then N(a + b √ d) = |a 2 − b 2 d| = a 2 + b 2 m.
But m > 1 so a 2 + b 2 m = 1 ⇒ b 2 = 0 and a 2 = 1
⇒ b = 0 and a = ±1
dummy text

2. (Gaussian integers)
N(a + bi) = |a 2 − b 2 .(−1)| = a 2 + b 2 .
Now a 2 + b 2 = 1 ⇒ {a 2 = 1 and b 2 = 0} or {a 2 = 0 and b 2 = 1}
37
⇒ {a = ±1 and b = 0} or {a = 0 and b = ±1}
⇒ r = ±1 or ± i
dummy text
3. We will only prove this for the case d = 3.
We know r = 7 + 4 √ 3 is a unit (see Examples 2.3.12).
We also know that the units are closed under multiplication
(Lemma 2.1.3), so r k is a unit for all k ∈ N.
Now r > 1 so
i.e. the numbers r k are distinct
1 < r < r 2 < r 3 < · · ·
so U(Z[ √ d]) is infinite.
Note that to make the proof of part (3) work for any other value of d, we
just have to find one unit in Z[ √ d] other than ±1.
Definition 2.3.14. A ring is called a field if every non-zero element is a
unit.
Example 2.3.15. Q, R, C are all fields.
Lemma 2.3.16. Zn is a field if and only if n is prime.
Proof.
An element a = 0 ∈ Zn is a unit if and only if hcf(a,n) = 1 (Theorem 2.1.4).
This is true for all a = 0 ∈ Zn if and only if hcf(a,n) = 1 for all 1 < a < n
i.e. if and only if n is prime.
3 Factorisation
Now that we know a bit about dividing, we can think about factorisation.

38 Section 3. Factorisation
3.1 Unique factorisation
One of the most crucial things about the integers is unique prime factorisa-
tion, that is:
“Every integer has a unique factorisation into a product of prime
numbers.”
Question: What does “unique” mean? What about these different factori-
sations of 30
30 = 2 × 3 × 5 = (−2) × (−3) × 5 = 5 × 3 × 2?
There could be some ±1 floating around, and we could change the order of
things, but that doesn’t count as genuinely different.
We’ll see that we have to be even more careful about this in rings with many
units. But first we have to make sure we know what “prime” means in an
arbitrary ring. The crucial property we want is that a prime number can’t
be factorised any more. This is actually called “irreducibility”.
Definition 3.1.1. “Irreducible”
Let R be a commutative ring and r = 0 ∈ R. Then r is irreducible in R if
1. r is not a unit, and
2. r cannot be written as a product of non-units, i.e.
r = st =⇒ s is a unit or t is a unit.
Example 3.1.2. We can check that −3 is irreducible in Z:
1. −3 is not a unit because the units in Z are just
1,-1
................

3.1 Unique factorisation 39
2. If −3 = st then the possibilities for s,t are:
1,-3; -3,1; -1, 3; 3, -1
....................................................................
and in each case s or t is a unit.
Example 3.1.3. Suppose we remove the number 2 from existence.
Then the following numbers become “irreducible”:
4,6,8,10,14,...
and we get some non-unique factorisations, e.g.
Remarks 3.1.4.
24 = 6.4 = 3.8
1. The irreducibles in Z are ±p where p is prime.
2. The condition saying r is not a unit is like the fact that 1 is not
considered to be a prime number.
3. Do the irreducibles form a group? Of course not—we can’t multiply
them!
4. Dire warning: in Z any prime p has the following property:
p|ab =⇒ p|a or p|b.
This property is called being prime. This is not the same as being
irreducible in general, but in Z the notions happen to coincide. This
is not true in all rings!
Theorem 3.1.5. Criterion for irreducibility in Z[ √ d].
Let d = 1 be a square-free integer, and r ∈ Z[ √ d]. Then
Proof.
N(r) is prime =⇒ r is irreducible in Z[ √ d].
Let r ∈ Z[ √ d] with clN(r) not prime.
So certainly N(r) = 1 so r is not a unit, by Theorem 2.3.11.

40 Section 3. Factorisation
Now suppose r = st. We aim to deduce that s or t must be a unit.
Then N(r) = N(s)N(t) by Lemma 2.3.10.
But N(r) is prime, so we must have N(s) = 1 or N(t) = 1
so by Theorem 2.3.11 s or t is a unit.
Hence r is irreducible as required.
Note that this condition is sufficient but not necessary, that is, the con-
verse of the theorem is not true. i.e.
There exists a square-free integer d and r ∈ Z[ √ d] such that r is irreducible
in Z[ √ d] but N(r) is not prime.
Examples 3.1.6.
1. In Z[ √ −7], N(2 + √ −7) =
prime
so 2 + √ −7 is irreducible in Z[ √ −7].
2. In Z[ √ −7] there is no element of norm 2:
|4 + 7| = 11
................................................. which is
|a 2 + 7b 2 | = 2 =⇒ b = 0 otherwise a 2 + 7b 2 ≥ 7
=⇒ a 2 = 2 # contradicts a ∈ Z
This means that any element of norm 4 or 8 must be irreducible:
if r has norm 4 it is certainly not a unit.
Now suppose r = st, so N(r) = N(s)N(t).
But there is no element of norm 2
so we must have
either N(s) = 1 and N(t) = 4, in which case s is a unit
or N(s) = 4 and N(t) = 1, in which case t is a unit.
A similar argument works for 8.
3. On Tutorial sheet 5 you’ll show that in Z[ √ −11] there are no elements
of norm 3 or 23, and hence an element is irreducible if it has norm
3 2 , 23 2 , 3.23, 3.3.23, 3.23.23, 3 3 , 23 3

3.1 Unique factorisation 41
For example N(5 + 2 √ −11) =
4. In Z[ √ −7] we can factorise 11 as
Now
|25 + 44| = 69 = 3.23
.....................................
11 = (2 + √ −7)(2 − √ −7).
N(2 + √ −7) = N(2 − √ −7) = 11,
so neither of these factors is a unit
so we have factorised 11 as a product of non-units
i.e. 11 is not irreducible in Z[ √ −7].
One moral of this is that a number can be irreducible in one ring but
not another, so it is important to say what ring we’re working in when
we’re talking about irreducibles.
5. In general if N(r) = k then k is not irreducible, as in the above exam-
ple.
Now we need to think about what “uniqueness” means for unique factori-
sations. We know that changing the order of the factors shouldn’t count
as genuinely different, but here’s something else that might confuse us es-
pecially if we’re in a ring with many units.
We might factorise an element r as
r = p.q
but then given a unit u we could also put
r = (pu).(u −1 q).
This is cheating! This should not count as a “different” factorisation. For
example in Z[i]
this is (3 + 2i)(−i) × (i)(3 − 2i)
13 = (3 + 2i)(3 − 2i) = (2 − 3i)(2 + 3i)
We say that (3+2i) is an “associate” of (2 − 3i) because they only differ by
a factor of a unit. That is, they’re not genuinely different. Likewise, (3-2i)
is an “associate” of (2 + 3i).

42 Section 3. Factorisation
However the following example has genuinely different factors: check irr
Definition 3.1.7. Associates
6 = 2 × 3 = (1 + √ −5)(1 − √ −5)
Elements s and t in a commutative ring R are said to be associates if there
is a unit u such that t = us.
(Note that this definition is symmetric, since t = us ⇐⇒ s = u −1 t.)
Examples 3.1.8. It follows that associates are easiest to spot in rings with
obvious units:
1. −r is always an associate of r, as −1 is always a unit.
2. In Z, the only units are ±1, so the only associates of r are ±r.
This is also the case in Z[ √ d] when d < −1.
3. In Z[i] the units are ±1, ±i, so the associates of a + bi are
−a − bi
−b + ai
b − ai
4. In Z[ √ d] with d > 1 there are infinitely many units, so every element
has infinitely many associates.
5. In Z[ √ d], if s and t are associates then they have the same norm,
because:
t = us gives N(t) = N(u)N(s) = 1.N(s) = N(s)
.....................................................................
Is the converse true?
No: it is possible to have the same norm but not be associates eg in
Z[i] obviously 2 + 3i and 2 − 3i have the same norm, but can’t be
associates as the only units are 1, −1,i, −i.

3.1 Unique factorisation 43
Lemma 3.1.9. Let R be a commutative ring.
If r is irreducible in R then all its associates are also irreducible in R.
Proof. Let p be an associate of r, so p = ur where u is a unit.
1. p cannot be a unit, since otherwise r = u −1 p would be a unit, but r is
irreducible so is not a unit (by definition).
2. We need to show: p = st ⇒ s or t is a unit.
Now p = st means ur = st
⇒ r = u −1 st = (u −1 s).t
⇒ u −1 s is a unit or t is a unit (since r is irreducible)
Now u −1 s is a unit ⇒ s = u.(u −1 s) is a unit.
So we have s is a unit or t is a unit.
Now, given any factorisation of an element s into irreducibles
s = p1p2 · · · pk
we could replace each pi by an associate of it, and that would be really
boring. This does not count as different, just like in our example
s = pq = (pu).(u −1 q).
Example 3.1.10. In this example we’ll work in Z[ √ 3] and find two factori-
sations of 20+7 √ 3 that look different, but are actually related by associates.
That is, we’ll find irreducibles p1,p2,q1,q2 with
p1p2 = q1q2 = 20 + 7 √ 3
and a unit u such that q1 = p1u, q2 = u −1 p2.
For the first factorisation, take p1p2 to be
(1 + 2 √ 3)(2 + 3 √ 3) = 20 + 7 √ 3.
We need to check that this is a factorisation into irreducibles:

44 Section 3. Factorisation
• N(1 + 2 √ 3) = 1 − 4.3
= 11
N(1 + 2 √ 3) = ................................. ........................
• N(2 + 3 √ 3) = 4 − 9.3
= 23
N(2 + 3 √ 3) = ................................. ........................
so 1 + 2 √ 3 and 2 + 3 √ 3 are both irreducible
11 and 23 are both prime (Theorem 3.1.5)
because ..........................................................................................
Now we pick a unit u = 7 + 4 √ 3. First we check it is a unit:
N(7 + 4 √ 3) = .............1
7 − 4 √ 3
so 7 + 4 √ 3 is a unit, with inverse u −1 = ........................................
Now consider
31 + 18 √ 3
q1 = p1u = (1 + 2 √ 3)(7 + 4 √ 3) = ...............................
7 − 4 √ 3 −22 + 13 √ 3
q2 = u −1 p2 = (...................)(2 + 3 √ 3) = ...............................
20 + 7 √ 3
Now check q1q2 = .............................................
So we have two factorisations into irreducibles:
20+7 √ 3 = (........................)(........................) = (........................)(........................)
But these are not “really different” factorisations because
they are related by associates
......................................................................................................................
We make this notion of “not really different factorisations” precise with the
following definition of “equivalence” of factorisations.

3.1 Unique factorisation 45
Definition 3.1.11. Equivalent factorisations
Let R be a ring, r ∈ R.
Let p1 · · · pn and q1 · · · qm be two factorisations of r into irreducibles.
These are called equivalent if n = m and the pi’s and qi’s can be reordered
so that for each i, pi and qi are associates.
Definition 3.1.12. Unique factorisation domain
An integral domain is called a unique factorisation domain (UFD) if
1. every non-zero non-unit has a factorisation into irreducibles, and
2. any two such factorisations are equivalent.
Examples 3.1.13.
1. Z is a UFD. This is the “fundamental theorem of arithmetic”, although
personally I want to throw up every time I hear the words “fundamen-
tal theorem of...”. Seriously, the fundamentality of a theorem should
speak for itself, shouldn’t it?
2. Q, R, C are UFD’s rather stupidly—there aren’t any non-zero non-
units, so there are no elements that have to satisfy the condition.
Therefore, they all do! We say the condition is “vacuously satisfied”.
3. Z[ √ −5] and Z[ √ −11] are not UFDs. We’ll work through one of these
examples a bit later.
4. Z[i] is a UFD.
5. Recall that if S ⊂ R and R is ID, then S is ID.
Is the same true for UFDs? Why?
No, because the irreducibles could be different—in a subring, it is likely
that more things will be irreducible cf Example 3.1.3
.....................................................................................................................
Question: Is it going to be easier to show that something is or isn’t a UFD?
To show something isn’t, we only have to exhibit one thing with non-unique
factorisations. To show something is, we have to work much harder.
Z[ √ −11] and Z[ √ −13] are on the homework and tutorial sheets.
Now we will do Z[ √ −5].

46 Section 3. Factorisation
Example 3.1.14. Z[ √ −5] is not a UFD since we have
6 = 2.3 = (1 + √ −5)(1 − √ −5).
We have to check that 2, 3, (1+ √ −5), (1− √ −5) are irreducible in Z[ √ −5],
and are not associates. So we take norms:
N(2) = 4
N(3) = 9
N(1 + √ −5) = N(1 − √ −5) = 1 + 5 = 6
• First we show that in Z[ √ −5] there are no elements of norm 2 or 3:
|a 2 + 5b 2 | = 2 =⇒ b = 0 otherwise a 2 + 5b 2 ≥ 5
and similarly for norm 3.
=⇒ a 2 = 2 # contradicts a ∈ Z
• Next we deduce that any element of norm 4, 9 or 6 must be irreducible:
If r has norm 4 it is certainly not a unit.
Now suppose r = st, so N(r) = N(s)N(t).
But there is no element of norm 2
so we must have
either N(s) = 1 and N(t) = 4, in which case s is a unit
or N(s) = 4 and N(t) = 1, in which case t is a unit.
A similar argument works for 9 and 6.
• Next we show that these irreducibles are not associates:
The units in Z[ √ −5] are just ±1, so 2 can’t be an associate of (1 ±
√ −5).
Also: associates have the same norm, since units have norm 1,
so N(pu) = N(p).

3.2 Euclidean domains 47
So we have found two non-equivalent factorisations of 6 into irreducibles in
Z[ √ −5, showing that this ring is not a UFD.
Example 3.1.15.
Does the above example exhibit C as a non-UFD?
No, because those elements are not irreducibles in C—they are units.
Z[ √ −5] is a subring of C which is (vacuously) a UFD.
...........................................................................
In order to show that something is a UFD, it would help to know how
to factorise it. In Z we have Euclid’s algorithm, and so we might ask
ourselves whether we can do Euclid’s algorithm in other rings. The answer
is: sometimes, but not always. When we can, the ring is called a Euclidean
domain.
3.2 Euclidean domains
We usually do Euclid’s algorithm in Z≥0. How does it work?
It depends on the crucial fact:
∀a,b ∈ Z≥0 ∃ q,r ∈ Z≥0 s.t. a = qb + r and 0 ≤ r < b.
This means that when we do the algorithm, the r’s will get smaller and
smaller and will eventually have to become 0.
In Numbers and Proofs you also did Euclid’s algorithm on polynomials,
where in that case it was the degree of the polynomial remainder that got
smaller and smaller.
In general we can do Euclid’s algorithm in a ring R if there’s some way of
measuring the “size” of elements, so that we can make remainders that get
smaller and smaller.
We’ve already seen one notion of “size” of elements of some rings:
The norm in Z[ √ d].
...................................................................................................

48 Section 3. Factorisation
We’ll see that this does enable us to do Euclid’s algorithm on the Gaussian
integers.
Definition 3.2.1. Euclidean domain
A Euclidean domain (ED) is an integral domain R such that there is a
function
such that for all a,b ∈ R \ {0}
δ : R \ {0} −→ Z≥0
1. ∃q,r ∈ R s.t. a = qb + r and r = 0 or δ(r) < δ(b)
2. if b|a in R then δ(b) ≤ δ(a).
Such a function is called a Euclidean function.
recall b|a means ∃k ∈ R s.t. a = kb
Examples 3.2.2.
1. Z is ED with δ(a) = |a|.
2. R[x] is ED with δ(f) = deg(f).
3. Z[i] is ED with δ(r) = N(r) = |r| 2
(remembering Z[i] ⊂ C).
We won’t prove these, but we’ll look at some examples. The point is that in
an ED we can use Euclid’s algorithm to find highest common factors, and
do that ar + bs = h thing which is kind of fiddly and annoying but useful.
Definition 3.2.3. Highest common factor
Let R be an integral domain, a,b ∈ R.
Then h is a highest common factor (hcf) of a and b if
1. h|a and h|b ←− “it is a common factor”
2. if d|a and d|b then d|h ←− “it is the highest one”
Remarks 3.2.4.
1. Hcf’s don’t necessarily exist if you’re not in a UFD.
eg in Z[ √ −3: 4 = 2.2 = (1 + √ −3)(1 − √ −3) and (1 + √ −3).2

3.2 Euclidean domains 49
2. Hcf’s aren’t unique: if h is one hcf and u is a unit, then uh is also an
hcf. However in an integral domain all hcf’s of a given pair of elements
must be associates.
Example 3.2.5. Hcf in R[x].
We use Euclid’s algorithm to find the hcf of
x 3 + 7x 2 + 14x + 8 and x 3 + 6x 2 + 11x + 6 ∈ R[x].
x 3 + 7x 2 + 14x + 8 = 1.(x 3 + 6x 2 + 11x + 6) + (x 2 + 3x + 2)
x 3 + 6x 2 + 11x + 6 = (x + 3)(x 2 + 3x + 2)
So the hcf is x 2 + 3x + 2.
Example 3.2.6. Hcf in Z[i].
(This is extremely tedious and you should feel very glad that I’m not actually
going to ask you to do it. This is probably the kind of thing it would be
better to ask Maple to do, but I don’t believe in getting computers to do
things for me—if I can’t do it myself I just do something more interesting
instead! It’s possible I will get into trouble for saying that.)
17+43i
11+3i
11+3i
4+4i
17+43i = 11+3i .11−3i
11−3i
11+3i = 4+4i .4−4i
4−4i
in C in Z[i] in Z[i]
= 316
130
422 + 130i −→ 17 + 43i = (2 + 3i)(11 + 3i) + (4 + 4i)
↑ ↑
∼ 2.4 ∼ 3.2
= 56
32
↑
∼ 1.75
So hcf is −1 − i or indeed 1 + i.
Check that 1 + i really goes into both of those:
and:
17+43i
1+i
11+3i
1+i
17+43i = 1+i . 1+i 60
1−i = 2
11+3i = 1+i .1−i
1−i = 7 + 4i.
32 − 32i −→ 11 + 3i = (20 − i)(4 + 4i) + (−1 − i)
26 + 2 i = 30 + 13i
−→ 4 + 4i = −4(−1 − i)

50 Section 3. Factorisation
Have you any idea how long it took to type that one example?
two and a half hours.
Answer: .................................
One of the main points is to show that if we have Euclid’s algorithm then
we have a UFD. In order to do this we need to know:
1. In an ED, any a and b have an hcf.
Moreover we have ar + bs = h for some r,s ∈ R.
(The proof of this is exactly the same as in Z —we feed the numbers
back into Euclid’s algorithm backwards.)
2. In an ED, if p is irreducible then hcf(a,p) is 1 or p
(or u or up where u is a unit).
3. In an ED, if p is irreducible then
i.e. “irreducible ⇒ prime”.
—this follows from (1) and (2)
p|ab ⇒ p|a or p|b
4. In an ED, if r = ab for non-units a and b then
δ(a) < δ(r)
δ(b) < δ(r)
strictly less—would only be equal if the other element is a unit
We can then show that everything is a product of irreducibles, by contra-
diction: suppose not, and take the “smallest” element that is not a product
of irreducibles. In particular it is not irreducible—so it can be expressed as
a product of “smaller” things. Contradiction. Boom. We then use (2) to
show that the factorisation is unique.
Theorem 3.2.7.
Let R be a Euclidean Domain, p irreducible in R, a ∈ R.
Then either 1 or p is an hcf of a and p.

3.2 Euclidean domains 51
Proof.
• Suppose p|a. Then p is an hcf of a and p
since clearly p|a and p|p
and {d|a and d|p} ⇒ d|p
• Suppose p ∤ a. Then claim: 1 is an hcf of a and p.
Clearly 1|a and 1|p.
Now suppose d|a and d|p
i.e. a = k1d and p = k2d for some k1,k2 ∈ R.
But p irreducible =⇒ k2 is a unit or d is a unit.
If k2 is a unit then d = k −1
2 p
so a = k1k −1
2 p
and thus p|a # so k2 is not a unit.
So d must be a unit, in which case d|1
hence 1 is an hcf as claimed.
Theorem 3.2.8. “In an ED, irreducible implies prime”
Let R be a Euclidean Domain, p irreducible in R, a,b ∈ R.
Then p|ab =⇒ p|a or p|b
Proof.
Suppose p|ab and p ∤ a. We need to show p|b.
Now, by Theorem 3.2.7 we know 1 must be an hcf of a and p.
So we have ar + ps = 1 for some r,s ∈ R.
Now multiplying by b gives bar + bps = b.
But p|ab so p|abr + bps
i.e. p|b.
Definition 3.2.9. Prime
Let R be a commutative ring. An element p ∈ R is called prime if
p|ab =⇒ p|a or p|b.

52 Section 3. Factorisation
Theorem 3.2.10.
Let R be a Euclidean domain.
If a = st for non-units s,t then
δ(s) < δ(a)
δ(t) < δ(a)
Note that the definition gives us ≤ immediately, but we want strictly

3.2 Euclidean domains 53
Now a cannot be irreducible, by hypothesis
so we must have a = rs where r,s are non-units.
But then by Theorem 3.2.10 we have
δ(r) < δ(a)
δ(s) < δ(a)
By (1) this means r and s have factorisation into irreducibles
r = r1r2 · · · rk
s = s1s2 · · · sl
But then a = r1r2 · · · rk.s1s2 · · · sl and is factorised. #.
So every non-unit has at least one factorisation into a product of irreducibles.
Next we show uniqueness of factorisations:
Suppose there are non-equivalent factorisations of a.
Choose the smallest n with
a = p1p2 · · · pn = q1q2 · · · qm
non-equivalent factorisations into irreducibles.
Now pn|q1q2 · · · qm
so by Theorem 3.2.8 pn must divide one of the qi.
Wlog pn|qm.
But qm is irreducible so we must have qm = upn for some unit u.
Now we have
p1p2 · · · pn−1pn = q1q2 · · · qm−1(u.pn)
so p1p2 · · · pn−1 = q1q2 · · · qm−1u
(We can cancel pn since R is a Euclidean domain, which means in particular
it is an integral domain.)
But these are both factorisations into irreducibles so must be equivalent,
since n is the smallest n with non-equivalent factorisations into irreducibles.

54 Section 3. Factorisation
But if these are equivalent then
p1p2 · · · pn−1pn = q1q2 · · · qm−1(u.pn)
must also be equivalent. #

Part II
Groups
4 Revision
All of this section is supposedly “revision” from Groups and Symmetries. If
you’re not sure about this material then you should revise it.
4.1 Definitions and examples
Definition 4.1.1. Group
A group G is a set equipped with a binary operation . satisfying
1. associativity ∀a,b,c ∈ G (ab)c = a(bc)
2. identity ∃e ∈ G s.t. ∀g ∈ G g.e = g = e.g
3. additive inverse ∀g ∈ G ∃g −1 ∈ G s.t. g.g −1 = g −1 .g = e
Definition 4.1.2. Abelian group
A group G is called Abelian if for all g,h ∈ G we have gh = hg.
Definition 4.1.3. Order
The order of a group G is: |G| = the number of elements of G
The order of an element g ∈ G is the smallest natural number k > 0 such
that g n = e.
Examples 4.1.4. Permutation groups
Sn is the group of permutations of n elements.
55

56 Section 4. Revision
|Sn| = n! HHH 1 : n, 2 : n!, 3 : n!
2 , 4 : 2n, 5 : infinite
This is also called the nth symmetric group.
Recall that there are two different types of notation:
• two-row notation ⎛
⎞
1 2 3 4 5 6
⎝ ⎠
2 4 5 1 3 6
• disjoint cycle notation:
(124)(35)
This element has cycle type: 3,2
Recall that every element can be written as a product of transpositions:
(12)(24)(35)
These are no longer necessarily disjoint.
(12)(24)(35)(16)(16)
Also, this isn’t unique e.g. ..............................................
but all expressions for the same element will have the same parity.
If it’s even, the element is called an even permutation.
If it’s odd, the element is called an odd permutation.
Do the even permutations form a subgroup of Sn? yes
Do the odd permutations form a subgroup of Sn? no
The alternating group An is the subgroup of Sn given by:
the even permutations
...........................................................................
|An| = n!
2 HHH
Is Sn Abelian? No for n > 2. HHH
(12)(23) = (123)
(23)(12) = (132)

4.1 Definitions and examples 57
Definition 4.1.5. Subgroup
Let G be a group.
A subgroup of G is a subset H ⊂ G such that H is a group with the same
operation.
Example 4.1.6. Cyclic groups
The cyclic group of order n consists of elements
with a n = 1.
1,a,a 2 ,...a n−1
Any group of order n is cyclic if it has an element of order n.
Examples 4.1.7. Symmetry groups
1. Dn is the group of symmetries of a regular n-gon.
(n = 1,2 are slightly peculiar though.)
|Dn| = 2n HHH
It consists of ........... rotations and ............ reflections n and n HHH
The group is generated by one rotation and one reflection.
The rotations form a subgroup of Dn isomorphic to Cn.
2. O2 is the group of symmetries of the circle.
|O2| = ∞ HHH
There are rotations rotα
and reflections refα
α
This group can be generated from one reflection and all the rotations.
α
2

58 Section 4. Revision
3. SO2 is the rotation group of the circle.
SO2 is a subgroup of O2.
4. Note that
D1 ∼ = C2 ∼ = S2
D3 ∼ = S3
5. The smallest non-abelian group is D3 = S3.
|D3| = 6 HHH
Examples 4.1.8. Groups of matrices
Do the n × n real matrices form a group under multiplication?
No—we might not have inverses. HHH
1. GLn(R) = the group of invertible real n × n matrices under multi-
plication
similarly we have GLn(F) for any field F.
This is called the general linear group
2. SLn(F) = matrices of determinant 1 with entries in a field F.
This is a subgroup of GLn(F) and is called the special linear group.
4.2 Basic theory
Definition 4.2.1. Orders of elements
Recall that the order of g ∈ G is the least n ∈ N such that g n = e if such
an n exists; otherwise g has infinite order.
Examples 4.2.2.
1. What is the order of rot2π
3
What is the order of rot2π
n
in O2? 3 HHH
in O2? n HHH
What is the order of refα in O2? 2 HHH
2. What is the order of a 3 in C12? 4 HHH
What is the order of a 4 in C6? 3 HHH

4.2 Basic theory 59
What is the order of a m in Cn?
n
hcf(m,n)
explain!! HHH
3. Every element g ∈ G generates a cyclic subgroup of G:
〈g〉 = {1,g 2 ,... ,g n−1 }
The order of this subgroup is the order of g.
Definition 4.2.3. Cosets maltesers!!
Let G be a group, a ∈ G and H a subgroup of G
The left coset aH is given by
aH = {ah | h ∈ H}
and is a subset of elements of G (not a subgroup).
The right coset Ha is given by
Ha = {ha | h ∈ H}.
Can we have aH = bH if a = b? yes HHH
aH = bH ⇐⇒ a ∈ bH or equivalently b ∈ aH
⇐⇒ ∃h ∈ H s.t. a = bh or equivalently b = ah
⇐⇒ b −1 a ∈ H compare b − a ∈ nZ for Zn
For example if H = 〈a 2 〉 ⊂ C6 with C6 = {1,a,a 2 ,...a 5 } then aH is the
same as
a 3 H,a 5 H
..........................................................................
Do we have aH = Ha? In general no HHH If it’s equal for all a then it’s
a normal subgroup.
Example 4.2.4.
Let G = S3, H = A3 ⊂ S3.
How many left cosets are there? 2 HHH

60 Section 4. Revision
Note:
• |aH| = |H| for all a ∈ G.
• Every element of G is in some coset since g ∈ gH.
• a ∈ bH ⇐⇒ aH = bH.
So if |G| is finite, the cosets form a partition of G into equal sized sets, which
gives us Lagrange’s Theorem:
Theorem 4.2.5. Lagrange
For any group G and subgroup H
Corollary 4.2.6.
|H| | |G|.
Given any g ∈ G, the order of g divides |G|
since the order of g is the order of 〈g〉 ⊂ G.
Corollary 4.2.7.
If |G| is prime then G must be cyclic.
Proof.
Let |G| = p prime and let g be any non-identity element.
Then by Corollary 4.2.6 we know that the order of g must divide p
so must be 1 or p since p is prime.
But g is not the identity, so the order of g cannot be 1
so the order of g is p, showing that G is cyclic.
Definition 4.2.8. index
The index of H in G is |G|
|H| .
Definition 4.2.9. Group actions
A group G acts on a (non-empty) set X if we have for each g ∈ G a function
X −→ X
x ↦→ g ∗ x

4.2 Basic theory 61
interacting properly with the group structure, i.e.
Example 4.2.10.
1. e ∗ x = x for all x ∈ X
2. g ∗ (h ∗ x) = gh ∗ x for all g,h ∈ G, x ∈ X
D4 acts on the vertices of the square.
D4 also acts on the edges of the square.
Definition 4.2.11. Orbit and stabiliser
Given an action of G on X as above, the orbit and stabiliser of x ∈ X are
given by:
• orb(x) = { g ∗ x | g ∈ G } ⊂ X
• stab(x) = { g ∈ G | g ∗ x = x } ⊂ G means subgroup
Note that
orb(x) = orb(y) ⇐⇒ x ∈ orb(y)
so we get an equivalence relation ∼ on X
x ∼ y ⇐⇒ they are in the same orbit.
Theorem 4.2.12. Orbit-stabiliser
Let G be a finite group acting on a set X. Then for any x ∈ X
|orb(x)|.|stab(x)| = |G|
Example 4.2.13. D4 acting on the square
Recall that D4 acts on the corners of the square:
4
Then orb(1) = {1,2,3,4} HHH
stab(1) = {e,ref13} HHH
So we can check the orbit-stabiliser theorem holds:
3
2
1
4.2 = 8 = |D4|
......................

62 Section 4. Revision
Note that there is a more delicate version of this theorem which also works
for infinite groups. It involves cosets. Cosets are traditionally unpopular
among students.
orb(x)
∼
−→ cosets of stab(x)
y ↦→ {g ∈ G | g ∗ x = y} “all ways of sending x to y”
4.3 Homomorphisms
Definition 4.3.1. Homomorphism
A group homomorphism is a function
such that for all x,y ∈ G
θ : G −→ H
θ(xy) = θ(x).θ(y)
Note: it follows that θ(e) = e and θ(x −1 ) = (θ(x)) −1 .
θ(x).θ(e) = θ(x) so θ(e) = e and θ(x).θ(x −1 ) = θ(e) = e
Definition 4.3.2. Group isomorphism
A group isomorphism θ : G −→ H is a homomorphism that is bijective.
Equivalently: there exists a homomorphism φ : H −→ G such that
Examples 4.3.3.
φ ◦ θ = idH
θ ◦ φ = idG.
1. If H is a subgroup of G, the inclusion H −→ G is a homomorphism.
2. Do we have a homomorphism yes HHH
Z −→ Zn
a ↦→ a (mod n) ?

3. Do we have a homomorphism no HHH
Zn −→ Z
a (mod n) ↦→ a ?
No—just like in the ring case, because addition isn’t the same
eg (n − 1) + 1 = 0
4. How many group homomorphisms are there HHH 1, 2, 3, 4, 5:many
Answer: one for each a ∈ Z
Z
θ
−→ Z ?
1 −→ a
n ↦→ na = a + a + · · · + a or θ(n) = an
There is always a group homomorphism G −→ 0 = {e}.
Also, given any groups G,H there is always a trivial homomorphism
5 Quotient groups
G −→ H
g ↦→ e
We’re going to see that we can sort of multiply and divide groups under the
right circumstances.
5.1 Subgroups
Definition 5.1.1. Subgroup
H is a subgroup of a group G if it is a subset of G and a group under the
same operation. Equivalently:
1. for all a,b ∈ H we have ab ∈ H,
2. e ∈ H, and
3. for all a ∈ H we have a −1 ∈ H.
We write H ⊆ G unless confusion is likely.
63

64 Section 5. Quotient groups
Examples 5.1.2.
1. Z ⊂ Q ⊂ R ⊂ C under addition
2. Is Zn a subgroup of Z? No, just like for rings.
3. An ⊂ Sn
Cn ⊆ Sn
Dn ⊆ Sn
Cn ⊆ Dn
Dn ⊆ O2
4. Is D3 ⊆ D4? No: 6 ∤ 8
Is S3 ⊆ S4? Yes: always for n < m, Sn ⊂ Sm
5. SLn(F) ⊆ GLn(F)
6. The trivial group 0 = {e} is a subgroup of every group.
5.2 Cosets
Definition 5.2.1. Cosets
Let H be a subgroup of G.
Then the left coset aH is the set: { ah | h ∈ H }.
The right coset Ha is the set: { ha | h ∈ H }.
For example if H = {e,h1,h2,h3} then
aH = {a,ah1,ah2,ah3}
bH = {b,bh1,bh2,bh3}
but now we actually multiply those elements out.

5.2 Cosets 65
Example 5.2.2.
Consider the cyclic group C12 with elements {e,a,a 2 ,a 3 ,... a 11 }
so a 12 = a 0 = e.
Let H = 〈a 4 〉
so H has elements ............................................ e,a 4 ,a 8
So we have the following cosets:
eH = {e,a 4 ,a 8 } a 6 H = {a 6 ,a 10 ,a 2 }
aH = {a,a 5 ,a 9 } a 7 H = {a 7 ,a 11 ,a 3 }
a 2 H = {a 2 ,a 6 ,a 10 } a 8 H = {a 8 ,e,a 4 }
a 3 H = {a 3 ,a 7 ,a 11 } a 9 H = {a 9 ,a,a 5 }
a 4 H = {a 4 ,a 8 ,e } a 10 H = {a 10 ,a 2 ,a 6 }
a 5 H = {a 5 ,a 9 ,a } a 11 H = {a 11 ,a 3 ,a 7 }
|C12| = ...................12 |H| = ....................3
So the number of distinct cosets = ................ |C12 |
|H|
The following cosets are the same:
eH = a 4 H = a 8 H
aH = a 5 H = a 9 H
a 2 H = a 6 H = a 10 H
a H = a 7 H = a 11 H
= 4

66 Section 5. Quotient groups
Example 5.2.3. Cosets in D3
Recall that D3 is the symmetry group of the equilateral triangle:
1
a is rot anticlockwise, bi is ref through line through i (draw in dotted)
We have elements: e,a,a 2 ,b1,b2,b3
Note b2 = ab1
b3 = a 2 b1.
We have a subgroup H = {e,b1}.
Now let’s find the left cosets:
2
eH = {e,b1}
aH = {a,ab1} = {a,b2}
a 2 H = {a 2 ,a 2 b1} = {a 2 ,b3}
b1H = {b1,e}
So the following cosets are equal:
b2H = {b2,b2b1} = {b2,a}
b3H = {b3,b3b1} = {b3,a 2 }
3
H = b1H
aH = b2H
a 2 H = b3H
The number of distinct cosets is .............................. 3

5.2 Cosets 67
Lemma 5.2.4.
Let H be a subgroup of G and a ∈ G.
Then the coset aH has the same number of elements as H.
Proof.
Certainly aH can’t have more elements that H.
It can only have fewer if ah = ah ′ for some h = h ′ ∈ H.
But ah = ah ′ =⇒ h = h ′ .
Lemma 5.2.5.
Let H be a subgroup of G.
If there is an element x ∈ G such that x ∈ aH and x ∈ bH
then aH = bH.
Proof.
x ∈ aH so x = ah1, say.
x ∈ bH so x = bh2, say.
Then a = bh2h −1
1 . (1)
We show aH ⊆ bH: y ∈ aH =⇒ y = ah for some h ∈ H
Similarly bH ⊆ aH
=⇒ y = bh2h −1
1 .h by (1)
=⇒ y ∈ bH
so aH = bH as required.

68 Section 5. Quotient groups
Corollary 5.2.6. ABSOLUTELY KEY, MUST REMEMBER
Let H be a subgroup of G and a,b ∈ G.
Then aH = bH if and only if any one of the following equivalent conditions
holds:
• b ∈ aH • a ∈ bH
• ∃ h ∈ H s.t. b = ah • ∃ h ∈ H s.t. a = bh
• a −1 b ∈ H • b −1 a ∈ H
We can then define an equivalence relation on the elements of G:
a ∼ b ⇐⇒ aH = bH.
Note that for right cosets it’s all the other way round:
• b ∈ Ha • a ∈ Hb
• ∃ h ∈ H s.t. b = ha • ∃ h ∈ H s.t. a = hb
• ba −1 ∈ H • ab −1 ∈ H
Example 5.2.7. Modular arithmetic bring sheet with nos
Let G = Z.
Put H = 4Z = { all multiples of 4 ∈ Z }
= { k ∈ Z s.t. 4|k }
= { 4n | n ∈ Z }
= { ... , −8, −4,0,4,8,12,... }
This is an additive group so the cosets are a + H.
e.g. 1 + H = {... , −7, −3,1,5,9,13,...} = { k | k ≡ 1 (mod 4) }
2 + H = {... , −6, −2,2,6,10,14,...} = { k | k ≡ 2 (mod 4) }
3 + H = {... , −5, −1,3,7,11,14,...} = { k | k ≡ 3 (mod 4) }
4 + H = H
5 + H = 1 + H

5.2 Cosets 69
So 1 + H = 5 + H = 9 + H = 13 + H = · · ·
So a ∼ b ⇐⇒ a + H = b + H
⇐⇒ b − a ∈ H
⇐⇒ 4 | b − a
⇐⇒ a ≡ b (mod 4)
So we can achieve modular arithmetic by faffing around with cosets. Of
course, that’s a bit over the top for modular arithmetic, but it shows us how
we can generalise the idea of modular arithmetic to other (less obvious)
groups. This is called quotient groups. In the above example we started
with Z and “quotiented” (divided) by 4Z. In general we can quotient by
any normal subgroup.
Definition 5.2.8. Normal subgroup
A subgroup H ⊆ G is a normal subgroup if
equivalently: be ready to explain this
∀ a ∈ G aH = Ha
∀ a ∈ G, h ∈ H aha −1 ∈ H.
Informally “H is stable under conjugation”.
We write H G.
Examples 5.2.9.
1. If G is abelian then all subgroups are normal since
aha −1 = aa −1 h = h ∈ H
.....................................................................
2. Any group G has normal subgroups 0 and G.
3. An Sn (See Tutorial #7, q.1)
4. Cn Dn but C2 is not. Recall from Example 5.2.3:
H = {e,b1} ∼ = C2

70 Section 5. Quotient groups
use b1a = b3,b1a 2 = b2
eH = {e,b1} He = {e,b1}
aH = {a,ab1} = {a,b2} Ha = {a,b1a} = {a,b3}
a 2 H = {a 2 ,a 2 b1} = {a 2 ,b3} Ha 2 = {a 2 ,b1a 2 } = {a 2 ,b2}
b1H = {b1,e} Hb1 = {b1,e}
b2H = {b2,b2b1} = {b2,a} Hb2 = {b2,b1b2} = {b2,a 2 }
b3H = {b3,b3b1} = {b3,a 2 } Hb3 = {b3,b1b3} = {b3,a}
So the left cosets do not equal the right cosets, so H is not a normal
subgroup.
5. In fact if H is a subgroup of index 2 then it must be a normal sub-
group. (See Tutorial #7, q.1)
Later we’ll see that homomorphisms give rise to normal subgroups in an
important way.
Question: Which makes you feel more like a rat up a drainpipe—cosets or
equivalence relations?
5.3 Quotient groups
Quotient groups are a really important piece of mathematics. I would say
they are the most important and beautiful piece of theory in this course
(though of course others might disagree).
The point is that if we have a normal subgroup H G then the cosets form
a group. Later we’ll see how to think of this as an equivalence relation.
Definition 5.3.1. Quotient group
Let H G. Then the cosets of H form a group called the quotient group
G/H.

5.3 Quotient groups 71
• The elements are the cosets aH.
Remember g1H = g2H ⇐⇒ g −1
2 g1 ∈ H.
How many elements are there?
• We define the group operation by
• The identity element is eH = H.
We have to check this makes sense i.e.
|G/H| = |G|
|H|
(aH).(bH) = (ab)H
if a1H = a2H and b1H = b2H (1)
then we must have (a1b1)H = (a2b2)H. (2)
Once we have checked this, the group axioms follow immediately.
Now, by Corollary 5.2.6 we know that (1) means a −1
2 a1 ∈ H and b −1
2 b1 ∈ H.
Also (2) means (a2b2) −1 (a1b1) ∈ H.
Now (a2b2) −1 (a1b1) = b −1
2 a−1
2 .a1b1
= b −1
2 (a −1
2 a1)
∈H
b2
∈H
. b −1
2 b1
∈H
trick
Note that this crucially depended on H being a normal subgroup.
Examples 5.3.2.
1. nZ = { nk | k ∈ Z }
This is a normal subgroup of Z since Z is abelian.
We formed the quotient group before: Z/nZ = Zn
2. Sn/An ∼ = C2
Dn/Cn ∼ = C2.
3. R 2 /R gives us lines in R 2 draw pic

72 Section 6. Conjugacy
4. R 3 /R gives us lines in R 3 draw pic
5. We know that 0 G for any group. So put H = 0.
{a}
Then for any a ∈ G the coset aH is ................................
So the quotient group G/0 is G
.......................................
6. G/G = 0
.............
Here’s something else we won’t really be using, but it’s included for com-
pleteness. It’ll be very important in the future.
Definition 5.3.3. Product Given groups A and B we can form a new
group A × B as follows.
• The elements are { (a,b) | a ∈ A,b ∈ B }
• Multiplication is “pointwise”: (a1,b1).(a2,b2) = (a1a2,b1b2).
If it’s an additive group we have
(a1,b1) + (a2,b2) = (a1 + a2,b1 + b2)
—does this remind you of anything? R × R = R 2
6 Conjugacy
Conjugacy is a relationship between elements that means they behave in
similar ways.
E.g. getting past someone into a seat.

6.1 Definitions 73
6.1 Definitions
Definition 6.1.1. Conjugate
Let G be a group and a,g ∈ G.
The conjugate of a by g is gag −1 .
Where have you seen conjugates before?
Linear: conjugate matrices, diagonalisation etc
..........................................................................
Examples 6.1.2.
1. In an Abelian group gag −1 = a
so everything is only conjugate to itself.
2. In a dihedral group and O2, conjugating a rotation through α by any
reflection gives rotation through −α.
actually this is the defining property of dihedral groups
3. In Sn, conjugates are all those elements of the same cycle type.
Lemma 6.1.3.
Given a,g ∈ G, the order of gag −1 equals the order of a.
Proof.
Suppose the order of a is n, so n is the smallest natural number such that
a n = e.
First we check that (gag −1 ) n = e.
Now
(gag −1 ) n = gag −1 .gag −1 .... .gag −1
n times
= ga n g −1
= geg −1
= e

74 Section 6. Conjugacy
Now we check that n is the smallest natural number such that
(gag −1 ) n = e.
Suppose there is a smaller one i.e. 0 < k < n such that (gag −1 ) k = e.
Then
(gag −1 ) k = ga k g −1
= e
so a k = g −1 g
So such a k cannot exist.
= e # contradicts the order of a being n
So the order of gag −1 is n as required.
Definition 6.1.4. Conjugacy class, centraliser
Let G be a group and a ∈ G.
The conjugacy class of a in G is
The centraliser of a in G is
conj G(a) = “all those elements conjugate to a”
= { gag −1 | g ∈ G }
centG(a) = “all those elements that commute with a
= { g ∈ G | gag −1 = a }
Does this remind you anything? group actions, orbit/stabiliser

6.1 Definitions 75
Theorem 6.1.5. “Conjugation is a group action”
Defining
g ∗ a = gag −1 ∀g ∈ G
gives a group action of G on its set of elements.
Then for all a ∈ G: orb(a) = conj G(a)
stab(a) = centG(a)
So if G is finite, by the orbit-stabiliser theorem we have
|conj G(a)| . |centG(a)| = |G|.
.........................................................................
In particular
|conj G(a)| | |G|
and we can partition G into conjugacy classes. not all equal sizes
Also, it follows that centG(a) is a subgroup of G.
Proof.
eae −1
1.∀a ∈ G : e ∗ a = ...............................
= a tick
g ∗ (hah −1 )
2.∀g,h,a ∈ G : g ∗ (h ∗ a) = ............................... substitute definition of h∗
g(hah −1 )g −1
= ............................... substitute definition of g∗
(gh).a.(gh) −1
= ............................... re-associate, and
use definition of (gh) −1
= (gh) ∗ a

76 Section 6. Conjugacy
Theorem 6.1.6. Conjugacy classes in Sn
Given an element a ∈ Sn, the conjugacy class of a consists of all elements
of the same cycle type as a.
We won’t quite prove it, but we will show how to “do” it.
Example 6.1.7.
In S6 consider a = (123)(45).
The conjugacy class is all elements of cycle type 3,2.
(213)(36), (314)(25),...
E.g. .....................................................................
How many such elements are there?
6.5.4
3
So we have |conjS6 (a)| = 120
. 3 = 120
6! 6!
|centS6 (a)| = = = 3.2 = 6
120 6.5.4
In the next example we’ll actually exhibit the conjugacy.

6.1 Definitions 77
Example 6.1.8. Conjugacy in S9
Consider α = ( 1 9 6 3 )( 2 4 8 )( 5 7 ) ∈ S9
β = ( 3 1 )( 9 2 4 7 )( 6 5 8 ) ∈ S9
We claim that these are conjugate.
To show this, we need to find θ ∈ S9 s.t. β = θαθ −1 .
1. Line up the two permutations by re-ordering the disjoint cycles:
α = 1 9 6 3 2 4 8 5 7
β =
(9247)(658)(31)
2. Re-order the vertical pairs of number to give a standard two-row no-
tation:
967534182
θ =
⎛
⎜
⎝
3. Check your answer:
i) Write θ in cycle notation:
ii) Write θ −1 in cycle notation:
1 2 3 4 5 6 7 8 9
⎞
⎟
⎠
(19264537)
.........................................
(73546291)
.........................................
iii) Calculate θαθ −1 (which should come out to be β):
(19264537) . (1963)(248)(57) . (73546291)
= (13)(2479)(586)
= β
So we have exhibited θ such that β = θαθ −1 ,
showing that α and β really are conjugate.

78 Section 6. Conjugacy
Example 6.1.9. Conjugacy classes in D4 bring plastic square
Recall that D4 is the group of symmetries of the square:
In D4 we have elements e,a,a 2 ,a 3 ,b1,b2,b3,b4
a is rot anticlockwise, bi is ref in the line i
○4
We know ref.rotα.ref −1 = rot−α
• We know that e is conjugate to .................... only e
○3
○2
○1
a, since rotations commute
• Conjugating a by a rotation gives ..................................
a −1 = a 3 by the above
Conjugating a by a reflection gives .................................
{a,a 3 }
So the conjugacy class of a is .......................................
a 2 , since rotations commute
• Conjugating a 2 by a rotation gives .................................
(a 2 ) −1 = a 2 by the above
Conjugating a 2 by a reflection gives ................................
{a 2 }
So the conjugacy class of a 2 is ......................................
• For b1 we calculate (noting that b −1
i = bi): just use plastic square
eb1e = b1 b1b1b1 = b1
ab1a −1 = b3 b2b1b2 = b3
a 2 b1a −2 = b1 b3b1b3 = b1
a 3 b1a −3 = b3 b4b1b4 = b3
{b1,b3}
So the conjugacy class of b1 is ......................................

6.2 The class equation 79
{b2,b4}
• It follows that the remaining conjugacy class is .....................
How do we know? What sort of elements are in conjugacy classes of
their own? —only those that commute with everything, and we know
that reflections don’t commute with rotations.
6.2 The class equation
We found that in D4 the conjugacy class sizes are:
1, 1, 2, 2, 2 total elements = 8
We can learn a surprising amount about a group just by looking at this
apparently stupid equation
Definition 6.2.1. Class equation
1 + 1 + 2 + 2 + 2 = 8
Let G be a finite group with conjugacy classes
Then the class equation for G is
C1,C2,... ,Ck.
|C1| + |C2| + · · · |Ck| = |G|.
Note that the class equation has the following features:
• 1 must appear at least once (for e)
• each number must divide |G|.
Example 6.2.2. Class equation for S3
We know that elements of S3 are conjugate precisely if they have the same
cycle type, so all we have to do to find the conjugacy class sizes is
1. write down every possible cycle type in S3, and
2. count how many permutations there are of each type.

80 Section 6. Conjugacy
So in S3 we have class equation
1 + 2 + 3 = 6
..........................................
cycle type typical element no. of elements
3 (123) 2
2,1 (12) 3
1,1,1 e 1
Example 6.2.3. Class equation for S5 and A5
See Tutorial #7 q.4
Cycle types in S5:
cycle type typical element # ODD EVEN
5 (12345) = (12)(23)(34)(45) 5!
5 24 X
4,1 (1234) = (12)(23)(34) 5!
5 30 X
3,2 (123)(45) = (12)(23)(45) 5.4.3.2
4 20 X
3,1,1 (123) = (12)(23) 5.4.3
3 .2.1
2 20 X
5.4
2,2,1 (12)(34) 2 .3.2
2 .1
2 15 X
5.4
2,1,1,1 (12) 2 10 X
1,1,1,1 e 1 X
• The class equation for S5 is
• The class equation for A5
1 + 10 + 15 + 20 + 20 + 30 + 24 = 120
60
5! = 120 Total even = ............
can’t be 1 + 15 + 20 + 24 = 60 because those numbers don’t all divide
60
.....................................................................

6.2 The class equation 81
Example 6.2.4.
Suppose |G| = 7.
Then the class equation must be:
1 + 1 + 1 + 1 + 1 + 1 + 1 = 7
all the numbers must divide 7.
since .....................................................................
This holds for groups of order any prime p.
Example 6.2.5.
Suppose |G| = 8.
Then the possibilities are:
1 1 1 1 1 1 1 1
1 1 1 1 1 1 2
1 1 1 1 2 2
1 1 1 1 4
1 1 2 2 2 ←− this one is D4
1 1 2 4
Question: What do those 1’s tell us? That is, what kind of element is in
a conjugacy class of its own?
Answer: An element a such that
“a commutes with everything”
∀g ∈ G gag −1 = a
i.e. ga = ag

82 Section 6. Conjugacy
Definition 6.2.6. Centre
Let G be a group and a ∈ G.
We say a is central in G if: ∀g ∈ G ga = ag.
The set of all central elements is called the centre of G:
Z(G) = { a ∈ G | ∀g ∈ G ga = ag }
Equivalent ways of saying a is central:
∀g ∈ G, a ∈ centG(g)
Examples 6.2.7.
= { a ∈ G | ∀g ∈ G a ∈ centG(g) }
=
centG(g)
g
∀g ∈ G, ga = ag ∀g ∈ G, gag −1 = a
centG(a) = G
1. Z(D4) = { e, rotπ } —see Example 6.1.9.
2. Z(D3) = { e } by studying multiplication table
3. If G is abelian then Z(G) = G.
4. Z(G) always contains e.
Proposition 6.2.8.
conj G(a) = {a}
For any group G, Z(G) is a normal subgroup of G. Z(G) G
Proof.
• Z(G) is a subgroup of G because it is the intersection of subgroups
centG(g). We can also check directly: given a,b ∈ Z(G) and g ∈ G,
g.ab.g −1 = gag −1 .gbg −1 trick
= ab since a,b ∈ Z(G)

6.2 The class equation 83
• Now we show it is a normal subgroup.
We need to show:
for all a ∈ Z(G) and g ∈ G, gag −1 ∈ Z(G).
.....................................................................
a ∈ Z(G), since a ∈ Z(G)
But gag −1 =........................................................
So we have a normal subgroup as required.
Note that looking at the class equation, we see Z(G) appearing in the form
of all the 1’s.
Examples 6.2.9. Centres and the class equation
1. The class equation for D4 is
1 + 1 + 2 + 2 + 2 = 8
1
+ 1
+2 + 2 + 2 = 8
centre
These 1’s are the classes of e and rotπ.
2. The class equation for S5 is:
1 + 10 + 15 + 20 + 20 + 30 + 24 = 120
trivial
so the centre is .....................................................
3. Let |G| = 8. We saw all the possible class equations in Example 6.2.5.
We deduce that the centre must be non-trivial.

84 Section 6. Conjugacy
Proposition 6.2.10.
Let G be a group of order p k where p is prime.
Then Z(G) must be non-trivial.
Proof.
The only possible numbers in the class equation are:
1 or powers of p, since the numbers must divide p k .
...........................................................................
We know we must have at least one 1, because
e is in a conjugacy class by itself
...........................................................................
Now if we have only one 1:
• the LHS of the equation will be congruent to .....1 (mod p) , whereas
• the RHS is congruent to .....0 (mod p) #.
So we must have more than one 1 in the class equation.
This shows that the centre is non-trivial because
each 1 corresponds to an element in the centre
...........................................................................
We can also detect other normal subgroups.
Theorem 6.2.11.
Let G be a group and H a subgroup.
Then H is normal if and only if it is a union of conjugacy classes.
We’ll prove this after looking at some examples.

6.2 The class equation 85
Example 6.2.12. Normal subgroups of D4
For D4 we have
1 + 1 + 2 + 2 + 2
e a 2 {a,a 3 } {b1,b3} {b2,b4}
We know that the possible orders of subgroups of D4 are:
1,2,4,8
...........................................................................
by ...................................................Theorem Lagrange’s
• We know we have normal subgroups 0 and G.
—and clearly these are unions of conjugacy classes
• For subgroups of order 2, there is only one possible union of conjugacy
classes:
{e} ∪ {a 2 }, since any subgroup must contain e
.....................................................................
This is indeed a subgroup of D4, so is a normal subgroup.
• For subgroups of order 4, we have the following possible unions of
conjugacy classes:
{e} ∪ {a 2 } ∪ {a,a 3 }
{e} ∪ {a 2 } ∪ {b1,b3}
{e} ∪ {a 2 } ∪ {b2,b4}
since e must be contained in any subgroup.
We can check that these are all subgroups, so must be normal sub-
groups.
• So the following subgroups are not normal:
{e,bi} for each i
.....................................................................

86 Section 6. Conjugacy
Example 6.2.13. Normal subgroups of S5
In S5 we know that conjugacy classes are given by
elements of the same cycle type
...........................................................................
• Is 〈 (123) 〉 a normal subgroup?
No, because for example it doesn’t contain the conjugate cycle (345).
Besides, its order is 3, whereas there are 5.4.3
3
cycle type, so some must be missing.
= 20 elements of this
.....................................................................
• Is 〈 (12345) 〉 a normal subgroup?
No, because its order is 5 whereas there are 5!
5
cycle type, so some must be missing.
= 24 elements of this
.....................................................................
Proof of Theorem 6.2.11.
• Suppose H is a subgroup of G and is a union of conjugacy classes.
We aim to show that H is a normal subgroup.
So, given h ∈ H, g ∈ G, we need to show ghg −1 ∈ H.
Now we know that conj G(h) ⊆ H
so ghg −1 ∈ conj G(h) ⊆ H.
So H is a normal subgroup as required.
• Conversely suppose H G.
Given h ∈ H we want to show conj G(h) ⊆ H.
So consider x ∈ conj G(h)
i.e. x = ghg −1 for some g ∈ G.
But ghg −1 ∈ H since H is normal, so conj G(h) ⊆ H as required.

6.2 The class equation 87
Example 6.2.14. Conjugacy in A4.
Consider the conjugacy class of α = (123) in S4.
β θ s.t. θαθ −1 = β θ
ODD EVEN
(123) e X
(132) (23) X
(124) (34) X
(142) (243) X
(134) (234) X
(143) (24) X
(234) (1234) X
(243) (124) X
This conjugacy class splits into two classes in A4:
• Elements conjugate to α in S4 via an even cycle:
{(123), (142), (134), (243)}
.....................................................................
• Elements conjugate to α in S4 via an odd cycle:
{(132), (124), (143), (234)}
.....................................................................
Theorem 6.2.15. Conjugacy in An
If there is an odd permutation θ such that θα = αθ then
conj An(α) = conj Sn (α).
Otherwise the conjugacy class splits in two in An.

88 Section 7. Homomorphisms
7 Homomorphisms
Recall that a group homomorphism is a function θ : G −→ H such that:
∀x,y ∈ G, θ(xy) = θ(x)θ(y).
...........................................................................
Here G and H are of course groups.
Here’s a “schematic diagram” of the group homomorphism axiom:
multiply in G
G H
do θ
x y θ(x) θ(xy) θ(y)
xy
Remember it follows that
• θ(e) = e
• θ(a −1 ) = (θ(a)) −1
do θ
θ(xy)
Homomorphisms produce some very interesting subgroups.
7.1 Kernels and images
multiply in H
Homomorphisms between groups are much more powerful than functions
between sets. They draw their power from the group structure of the groups
in question.
Definition 7.1.1. Kernel and image
Let G and H be groups, and θ : G −→ H be a homomorphism.

7.1 Kernels and images 89
The kernel of θ is: Kerθ = “everything that is sent to e” ⊆ G
= { g ∈ G | θ(g) = e }
The image of θ is: Imθ = “everything that is hit by θ” ⊆ H
Lemma 7.1.2.
= { θ(g) | g ∈ G }
Let θ : G −→ H be a homomorphism. Then
1. Kerθ is a subgroup of G.
2. Imθ is a subgroup of H.
= { h ∈ H | h = θ(g) for some g ∈ G }
also written θ(G)
Note that to show that an element of G is in Kerθ we
apply θ to it and check that the answer is e.
...........................................................................
To show that an element of H is in Imθ we
find an element of G that get sent to it by θ
...........................................................................
Proof.
1. • e ∈ Kerθ since .......................... θ(e) = e
• Given a and b ∈ Kerθ we have ab ∈ Kerθ since
θ(a)θ(b)
θ(ab) = ................................... by definition of group homomorphism
e.e
= ........................ since a,b ∈ Kerθ
e
= ........................ by definition of e.

90 Section 7. Homomorphisms
• Given a ∈ Kerθ we have a −1 ∈ Kerθ since
(θ(a)) −1
θ(a −1 ) = ................................... by standard result for group homomorphisms
e −1
= ........................ since a ∈ Kerθ
e
= ........................ by definition of e.
2. • e ∈ Imθ since ............................ θ(e) = e
• Given x and y ∈ Imθ we show xy ∈ Imθ:
Put x = θ(a), y = θ(b), say.
Then
θ(a)θ(b)
xy = ................................... (substitute)
θ(ab)
= ........................ by definition of group homomorphism
So xy ∈ Imθ as required.
• Given x ∈ Imθ we show x −1 ∈ Imθ:
Put x = θ(a), say.
Then
(θ(a)) −1
x −1 = ................................... (substitution)
θ(a −1 )
= ........................ by standard result for group homomorphisms
Example 7.1.3.
So x −1 ∈ Imθ as required.
Let G = GLn(R), the group of invertible real n × n matrices.
Let H = R \ {0} under multiplication.
We define a group homomorphism θ : G −→ H by
θ(A) = detA.

7.1 Kernels and images 91
This is a homomorphism since
det(AB) = detA.detB
...........................................................................
• Kerθ = { A ∈ GLn(R) | detA = 1 }
• Imθ = H
= SLn(R) by definition
We can always find A with detA = x eg ( x 0
0 1 )
Example 7.1.4.
D3 acts on the vertices of an equilateral triangle giving a homomorphism
D3 −→ S3.
• Kerθ = {e} which symmetries fix every vertex?
• Imθ = S3 which perms of the vertices can be achieved by symmetries?
We will see that this shows D3 ∼ = S3.
In general Dn acts on the vertices of the regular n-gon, giving a homomor-
phism
• Kerθ = {e}
Dn −→ Sn.
• Imθ = not the whole of Sn unles n = 3
eg can’t switch two adjacent vertices and nothing else, except on the
triangle
We will see that Imθ ∼ = Dn.
More generally, if G acts on a set of n things, we get a homomorphism
G −→ Sn.

92 Section 7. Homomorphisms
Example 7.1.5.
Consider the following polynomials in x1,x2,x3,x4:
Then as S4 permutes x1,x2,x3,x4
it also permutes p1,p2,p3
p1 = x1x2 + x3x4
p2 = x1x3 + x2x4
p3 = x1x4 + x2x3
E.g. (123) : x1x2 + x3x4 p1 also p2 p3
p3
x2x3 + x1x4 ......... p1 p2
(14)(23) : x1x2 + x3x4 p1 p2 p3
This gives rise to a homomorphism
x4x3 + x2x1 p1 p2 p3
............................ ......... ......... .........
θ : S4 −→ S3
and Kerθ is “all elements of S4 that fix p1, p2, and p3”
(14)(23), (12)(34), (13)(24), e
= .........................................................................
N.B. There are four other elements that fix p1:
(12),(34),(1324),(1423)
.........................................................................
but they don’t fix p2 and p3, so they aren’t in the kernel.
The kernel and image are intimately related to injectivity and surjectivity.
In fact, they basically are injectivity and surjectivity.

7.1 Kernels and images 93
Remember what “injective” and “surjective” mean?
A function f : A −→ B is
f(a1) = f(a2) =⇒ a1 = a2
• injective if ........................................................
∀b ∈ B ∃a ∈ A s.t. f(a) = b
• surjective if .......................................................
If you can’t remember this then write a 5-step plan for how you’re going to
remember it:
1. .....................................................................
2. .....................................................................
3. .....................................................................
4. .....................................................................
5. .....................................................................
Do you think I’m joking? Yes/No (delete as appropriate)
Lemma 7.1.6. Let θ : G −→ H be a homomorphism. Then
Proof.
“ =⇒ ”
Suppose θ is injective.
Let g ∈ Kerθ, so θ(g) = e.
We need to show g = e:
Now θ(g) = e = θ(e)
θ is injective ⇐⇒ Kerθ = 0 = {e}
but θ is injective so this implies g = e.

94 Section 7. Homomorphisms
“⇐=”
Conversely suppose Kerθ = 0.
Let g1,g2 ∈ G with θ(g1) = θ(g2). (1)
We need to show g1 = g2:
Now we have
θ(g1) −1 .θ(g2) = θ(g −1
1 ).θ(g2) standard property of homs
= θ(g −1
1 .g2) by definition of hom
But also θ(g1) −1 .θ(g2) = e by (1)
So θ(g −1
1 .g2) = e
i.e. g −1
1 .g2 ∈ Kerθ.
Thus g −1
1 .g2 = e since Kerθ = {e}.
So g1 = g2.
Lemma 7.1.7.
Let θ : G −→ H be a homomorphism. Then
θ is surjective ⇐⇒ Imθ = H
I’m not sure that even deserves to be called a Lemma, as it’s just the defi-
nition of Im. Oh well, it deserves emphasis.
We are going to have an amazing relationship between these things:
This tells us many things e.g.
G Kerθ ∼ = Imθ
|Imθ| = |G|
|Kerθ|
Moreover, it means that to understand “multiple hits” we only have to
understand what gets sent to e.
We’d better recall quotient groups, and also check that Kerθ is a normal
subgroup, so that we can quotient by it.

7.1 Kernels and images 95
Proposition 7.1.8.
Let θ : G −→ H be a homomorphism.
Then Kerθ is a normal subgroup of G.
Proof.
We already know Kerθ is a subgroup of G (Lemma 7.1.2).
To show it is normal, we need to show:
∀a ∈ Kerθ and g ∈ G, gag −1 ∈ Kerθ
...........................................................................
Now
θ(g).θ(a).θ(g −1 )
θ(gag −1 ) = .......................................... since θ is a homomorphism
θ(g).e.θ(g −1 )
= .......................................... since a ∈ Kerθ
θ(g).θ(g −1 )
= .......................................... by definition of e
θ(g.g −1 )
= ........................ since θ is a homomorphism
θ(e)
= ................... by definition of inverse
e
= ............... since θ is a homomorphism
So gag −1 ∈ Kerθ as required.

96 Section 7. Homomorphisms
7.2 Quotient groups revisited
We will treat quotient groups slightly differently this time.
Let H G. We define the quotient group G H
• The first important thing to remember about quotient groups is: what
are the elements?
The elements are the cosets aH
.....................................................................
• The next important thing to remember is: what is the group opera-
tion?
How can we multiply cosets?? If you had to multiply aH and bH,
what would the answer be in your wildest dreams?
(ab)H
..............................
In fact we can multiply any subsets of a group:
if A and B are subsets of G then
AB = { ab | a ∈ A, b ∈ B }
Note that some of these elements may be the same as each other, as in the
next example.
Example 7.2.1. Multiplication of cosets
• Consider the cyclic group C12 with elements
Let A = {a,a 2 ,a 3 }
B = {a 4 ,a 5 ,a 6 }
{e,a,a 2 ,...,a 11 }.
{a 5 ,a 6 ,a 7 ,a 8 ,a 9 }
Then AB =.........................................................

7.2 Quotient groups revisited 97
• Now let H be the subgroup {e,a 4 ,a 8 }, and put
a,a 5 ,a 9
A = aH = ......................
a 2 ,a 6 ,a 10
B = a 2 H = ......................
{a 3 ,a 7 ,a 11 }
Then AB =.........................................................
a 3 H
So (aH).(a 2 H) is the coset ...........................
This works because
H C12 since C12 is abelian.
.....................................................................
This makes your life easier so please remember
easy
it:
If H is a normal subgroup of G then it is to
multiply cosets. You do not actually have to go round multiplying every
element by every other because:
(aH)(bH) = (ab)H

98 Section 7. Homomorphisms
Example 7.2.2.
Consider the group D3 with multiplication table (with the usual notation):
e a a 2 b1 b2 b3
e e a a 2 b1 b2 b3
a a a 2 e b2 b3 b1
a 2 a 2 e a b3 b1 b2
b1 b3 b4 b2 e a 2 a
b2 b2 b1 b3 a e a 2
b3 b3 b2 b1 a 2 a e
1. Let H be the not normal subgroup {e,b1}.
i) The coset aH = {ae,ab1} = { a,b2 }.
ii) The set (eH).(aH) has elements
iii) Is this a coset of H?
2. Let V be the normal subgroup {e,a,a 2 }.
a
ea = ...........
b2
eb2 = ...........
b4
b1a = ...........
a 2
b1b2 = ...........
No—wrong number of elements
.............................................
i) The coset b1V = {b1e,b1a,b1a 2 } = { b1,b3,b2 }
ii) (eV ).(b1V ) has elements
eb1 = b1 ......... ab1 = b2 ......... a2b1 = b3 .........
eb2 = b2 ......... ab2 = b3 ......... a2b2 = b1 .........
eb3 = b3 ......... ab3 = b1 ......... a2b3 = b2 .........
iii) So (eV ).(b1V ) = { b1,b2,b3 } and is the coset ......................... b1V

7.2 Quotient groups revisited 99
Definition 7.2.3. Quotient group
Let H G.
Then the cosets of H form a group called the quotient group denoted
with
G H
• group operation given by multiplication of cosets, so
• identity given by H = eH since
(aH).(bH) = (ab)H
H.(aH) = (aH).H = aH
• inverses given by (aH) −1 = (a −1 )H since
(aH)(a −1 H) = eH = H
(a −1 H)(aH) = eH = H
Another way of thinking about quotient groups is by equivalence rela-
tions. We use the equivalence relation
x ∼ y ⇐⇒ y −1 x ∈ H.
Then the equivalence classes form a group! The operation is
[x].[y] = [xy]
which is well-defined because H is normal. same proof as for when we first
defined quotient groups
Actually these are just like cosets in disguise:
[x] = [y] ⇐⇒ xH = yH
cosets equivalence relation
multiplication (aH)(bH) = (ab)H [a][b] = [ab]
identity eH = H [e]
inverse of aH (a −1 )H [a −1 ]

100 Section 7. Homomorphisms
Here are a couple of applications of quotient groups.
Theorem 7.2.4.
Let G be a group with G Z(G) cyclic.
Then G is abelian.
Proof.
Write Z = Z(G)
and pick a generator gZ of G/Z. NB g is now fixed.
This means every element of G/Z is of the form (gZ) n = g n Z.
Now, given an element a ∈ G it is in some coset g n Z, say
i.e. a = g n z for some n ∈ N0,z ∈ Z.
Now we want to show ∀a,b ∈ G, ab = ba.
Put a = g m z1
b = g n z2
Then ab = g m z1.g n z2
= g m g n z2z1 since
= g n g m z2z1 since
z1 ∈ Z so commutes with everything
...............................................
g m g n = g m+n = g n g m
...............................................
= g n z2.g m z1 since moving z2 this time
...............................................
= ba

7.2 Quotient groups revisited 101
Theorem 7.2.5.
If p is prime then every group of order p 2 is abelian.
Note we already know that every group of order p is
Proof.
We aim to show Z(G) = G.
cyclic
.....................
We use Proposition 6.2.10 which says that in a group of order p k (with p
prime), the centre must be non-trivial. from class equation
So Z(G) is a non-trivial subgroup of G
so by Lagrange’s Theorem its order must be
Suppose |Z(G)| = p. aim for a contradiction
Then |G/Z(G)| = p so G/Z(G) is cyclic.
So by Theorem 7.2.4 G is abelian
i.e. |Z(G)| = p2
............................... #
p or p 2
.............................
So we must have |Z(G)| = p 2 i.e. G is abelian as required.

102 Section 7. Homomorphisms
7.3 First isomorphism theorem
This is the really big theorem of the course.
Theorem 7.3.1. The First Isomorphism Theorem for Groups
Let θ : G −→ H be a group homomorphism. Then
G Kerθ ∼ = Imθ.
Before we prove this, let’s think about why it’s great.
If we only had a function of sets, how could we ever tell anything about
anything??
stuff about even covering, e pulling things in
Remarks 7.3.2.
Note that we have some special cases:
1. If θ is injective we have Kerθ = 0
so the First Isomorphism Theorem says:
G ∼ = Imθ.
2. If θ is surjective we have Imθ = H
so the First Isomorphism Theorem says:
3. So if θ is an isomorphism
G Kerθ ∼ = H.
the First Isomorphism Theorem says:
as expected.
G ∼ = H
4. If we have N G then we have a homomorphism
G −→ G/N
whose kernel is N and image is G/N.

7.3 First isomorphism theorem 103
5. As a special case of the above, given a homomorphism θ : G −→ H we
know we have Kerθ G so we get a homomorphism
G −→ G Kerθ ∼ = Imθ
whose kernel is still Kerθ, but it is now surjective.
we threw away everything that wasn’t hit
Example 7.3.3.
Fix n ∈ N.
Dn acts on the vertices of a regular n-gon as in Example 7.1.4, giving a
homomorphism
We know that the kernel is 0............
θ : Dn −→ Sn.
so by the First Isomorphism Theorem
Dn 0 ∼ = Imθ = Dn
.........................................
Example 7.3.4.
Fix n ∈ N.
Define θ : O2 −→ O2 by
Then
rotα ↦→ rotnα
refα ↦→ refnα
• Imθ = O2 since we can hit any α from α
n
.............................................................
• Kerθ = 〈rot2π 〉
n
∼ = Cn
............................................................
So by the First Isomorphism Theorem: think harder about this
We can picture the cosets as
O2 Cn
∼= O2
.........................................
•
•
•
•
•
•
•
• • • • •

104 Section 7. Homomorphisms
Example 7.3.5.
Define θ : R −→ SO2 by
Then
α ↦→ rotα
• Imθ = SO2
.............................................................
• Kerθ = {0, ±2π, ±4π,...} ∼ = 〈2π〉 ∼ = Z
............................................................
So by the First Isomorphism Theorem:
Example 7.3.6.
Define θ : Dn −→ Z2 by
Then
α ↦→
R Z ∼ = SO2
..................................................
⎧
⎨
⎩
0 if α is a rotation
1 if α is a reflection
• Imθ = Z2
.............................................................
• Kerθ =
the rotations i.e. Cn
............................................................
So by the First Isomorphism Theorem:
Dn Cn
∼= Z2
..................................................

7.3 First isomorphism theorem 105
Example 7.3.7.
D6 acts on the diagonals of the regular hexagon.
So we get a homomorphism θ : D6 −→ S3.
Then
○3
○2
• Kerθ = {e,rotπ}
............................................................
• We can deduce what the image is from the First Isomorphism Theo-
rem, which says:
D6 Kerθ
order = 12
2
So Imθ has order
so must be
○1
∼= Imθ ⊆ S3
↑ ↑
= 6 order = 6
6
....... and is a subgroup of S3,
the whole thing.
.................................

106 Section 7. Homomorphisms
Finally here’s the proof of the First Isomorphism Theorem.
This is not examinable.
Proof of First Isomorphism Theorem.
We exhibit a group isomorphism
f : G Kerθ
∼
−→ Imθ.
For convenience we write K = Kerθ throughout.
1. We define f(aK) = θ(a) ∈ Imθ.
We check that this is well-defined
i.e. aK = bK =⇒ θ(a) = θ(b).
Now aK = bK ⇐⇒ a −1 b ∈ K
=⇒ θ(a −1 b) = e
=⇒ θ(a −1 ).θ(b) = e
=⇒ θ(a) −1 .θ(b) = e
=⇒ θ(a) = θ(b).
2. We show that this is a group homomorphism.
i.e. f((aK)(bK)) = f(aK).f(bK).
Now (aK)(bK) = (ab)K
so f((aK)(bK)) = f((ab)K)
= θ(ab)
= θ(a).θ(b) since θ is a homomorphism
= f(aK).f(bK)

7.3 First isomorphism theorem 107
3. We show that f is surjective.
i.e. for all y ∈ Imθ, ∃x ∈ G Kerθ s.t. f(x) = y.
Now y ∈ Imθ ⇐⇒ y = θ(a) for some a ∈ G.
Then putting x = aK ∈ G Kerθ we have
4. Finally we show that f is injective
f(x) = f(aK)
= θ(a)
= y.
i.e. Kerf = {e} (using Lemma 7.1.6).
Let x ∈ Kerf
so x = aK for some a ∈ G, and f(aK) = e.
Now f(aK) = θ(a) and
θ(a) = e =⇒ a ∈ Kerθ
which is the identity in G Kerθ .
=⇒ aK = K
So f is a group isomorphism as required.

108 Section 7. Homomorphisms

Part III
Homework and Tutorial Questions
109
• Homework #n is to be done in week n and handed in at the lecture
on Monday of week #(n + 1).
• Tutorial #n is to be done in week n regardless of whether or not you
have a tutorial that week. In each tutorial you can get help with the
previous two weeks’ tutorial work (and earlier if necessary).
• In non-tutorial weeks you can come to my office hours to get help
with that week’s tutorial work.
• Please make sure you read any hints given in the questions. It’s
extraordinary number of times tutors get asked questions which are in
fact answered in the hints...
• You should always justify all your answers. If a question says “Is 14
a zero-divisor?” don’t just answer yes or no – say how you decided
that was the answer. When you write out your working, include some
explanation about what you did. This is good practice for exams, but
is also useful for the marker so they can understand what you were
doing (and therefore be in a better position to help you if it went
wrong). It’s also useful for you, because when you’re revising you
should be going back over your homework to make sure you understand
it better than you did the first time round.

110 Section 8. Homework questions
8 Homework questions
Homework #1
This set looks long, but it isn’t actually long. There’s a lot stuff here that isn’t so much
question as some explanation about why these questions are useful things for you to
think about.
1. On the next page is a page of statements I have copied from some Semester One
exams I have marked. Imagine this is work that a student has handed in. Mark it in
the way that you would like to see your homework marked. (Pen or pencil? Marks
out of 10 or a letter grade? Encouragement or criticism? Helpful comments?)
This exercise is partly to help you think about how you should read over your own
work to see if it is correct, partly so I can see what you consider to be useful
feedback on your work, and partly to make sure you won’t ever make any of the
same mistakes listed here.
2. This question is to jog your memory about modular arithmetic. Let n ∈ N and
a ∈ Z.
• Recall that n|a means ∃k ∈ Z s.t. a = kn.
• Recall that “x ≡ y (mod n)” means n|(x − y).
Suppose x ≡ y (mod n). Show that:
i) a + x ≡ a + y (mod n), and
ii) ax ≡ ay (mod n).
3. This question is about attempting to work in Z6. Taking square roots and solving
quadratic equations don’t necessarily work the way we’re expecting...
i) In a ring R, x is said to be a square root of y when x 2 = y. Find the elements
of Z6 that have square roots i.e. find all the elements a ∈ Z6 such that there
exists x ∈ Z6 s.t. x 2 with a (mod 6)
You might find it useful to use the multiplication table from the tutorial sheet.
You should find that not all elements have square roots: there are only four
that do. Are you surprised?
ii) Find all solutions of x 2 + 3x + 2 = 0 in Z6. You will probably need to do it
by trial and error, because our usual methods for solving quadratic equations
won’t work here. Why? Note how many solutions there are. Did it surprise
you?

1.
1 1 1
+ =
a b a + b
2. (a + b) n = a n + b n
3. (a − b) 3 = a 3 − 3a 2 b − 3ab 2 − b 3
4. 3x = 2
3
=⇒ x = 2
−1 3
n + n
5. 4 = 2(n + n
2
−1 ) 3
1 1
1− − 6. e 2 = e 2
7. e 3
1 2 2
− + = e
3 9 27
3
1 2
−
3 9
8. (e x + e −x ) 3 = e 3x + e −3x
9. 3e x + e −3x = 0
10. tanx = 1 =⇒ x = 0.785 to 3 d.p.
111

Homework #2
112 Section 8. Homework questions
1. Match the following informal statements with their formal counterparts. Note that
these are all facts that could be true in a ring R, but aren’t necessarily true about
all rings.
i) The additive inverse of −1 is 1.
ii) The multiplicative inverse of −1 is −1.
iii) 0 has no multiplicative inverse in R.
iv) If x has a multiplicative inverse then so does −x.
v) If x and y are in R it isn’t necessarily true that xy is non-zero.
vi) Every non-zero element of R has a multiplicative inverse.
A) (−1)(−1) = 1
B) ∃x,y ∈ R s.t. xy = 0
C) ∀x = 0 ∈ R ∃y ∈ R s.t. xy = yx = 1
D) −(−1) = 1
E) ∀x ∈ R, 0x = 1
F) ∃y ∈ R s.t. xy = yx = 1 ⇒ ∃z ∈ R s.t. (−x)z = z(−x) = 1
2. Find all the units in Z10 and fill in a multiplication table for them.
3. Let R = Z[ √ 7]. Let r = 8 − 3 √ 7 and s = 8 + 3 √ 7. Compute rs and deduce that
r and s are units in R. Hence solve this equation in R:
(8 + 3 √ 7)x = 3
This should rather remind you of the process of “rationalising the denominator”.

Homework #3
113
1. Match the following statements with their negations. Note that I’m trying to catch
you out.
i) ∀x ∈ R x.0 = 0
ii) ∃x = y ∈ R s.t. xy = 0
iii) x ∈ R ⇒ 1.x = x
iv) ∀x = y ∈ R, xy = 0
v) ∃x ∈ R s.t. 1.x = x
vi) ∃x ∈ R s.t. x.0 = 0
A) ∀x ∈ R, x.0 = 0
B) ∃x ∈ R s.t. x.0 = 0
C) x ∈ R ⇒ 1.x = x
D) ∀x = y ∈ R, xy = 0
E) ∃x = y ∈ R s.t. xy = 0
F) ∃x ∈ R s.t. 1.x = x
2. For each of the following, determine whether or not a is a unit in Zn and, if it is
a unit, use Euclid’s algorithm to find its inverse.
i) n = 90, a = 25
ii) n = 91, a = 26
iii) n = 47, a = 25
Hint: if you think you’ve found the inverse for a, multiply it back with a and check
that you get something that’s congruent to 1 mod n. It’s so easy to check that
something really is an inverse in Zn, you should never ever get the wrong answer
– you might get stuck and not be able to find an answer (we all have bad days) but
you shouldn’t ever get the wrong answer.
3. For this question you can copy the method from tutorial sheet #3.
i) Show, by finding its inverse, that 2 + 3x is a unit in Z9[x].
ii) Show, by finding its inverse, that 3 + 4x is a unit in Z16[x].

Homework #4
114 Section 8. Homework questions
1. What does “hence” mean in a maths question? eg “i) Prove this. ii) Hence do
that.” By contrast, what does “hence or otherwise” mean?
You might think this is a silly question, but the reason I’m asking it is that loads
of people always get it wrong in exams.
2. For every element a in Z24 determine whether a is a unit or a zero-divisor. If a is
a unit find its inverse, and if a is a zero-divisor find b = 0 such that ab = 0. Write
your answers out in a table, so that it’s easier to read.
3. Is 523 is a unit or a zero-divisor in Z570? If it is a unit find its inverse, and if it is
a zero-divisor find b = 0 such that 523b = 0.
Homework #5
1. Find a unit r ∈ Z[ √ 11] such that r > 1. Hence show that the group of units of
Z[ √ 11] is infinite.
2. This question is about Z[ √ −13].
i) Show that Z[ √ −13] has no element of norm 2 or 11.
ii) Hence show that any element of Z[ √ −13] with norm 4, 22 or 121 is irreducible
in Z[ √ −13].
iii) Calculate the norm of (3 + √ −13).
iv) Hence express 22 as a product of two irreducible factors in Z[ √ −13] in two
different ways, and deduce that Z[ √ −13] is not a unique factorisation domain.
Hint: copy the procedure on Example 3.1.12. You must show that your two
factorisations are non-equivalent, and that all your factors are irreducible.
3. If R is a unique factorisation domain and S is a subring of R, does it follow that
S is a unique factorisation domain?

Homework #6
115
This whole sheet is revision from Groups and Symmetries, so you may need to look things
up from your old notes. I know it’s hard to remember things from previous modules, so
this homework is to help you jog your memory in preparation for the groups part of our
course.
1. Write down the whole multiplication table for D3, the group of symmetries of the
equilateral triangle. Then look it up on Wikipedia to check your answer.
2. Consider the square with lines of symmetry labelled 1, 2, 3 and 4 as below.
○4
○3
Recall that D4 is the group of symmetries of the square, and observe that it acts
on the lines 1,2,3,4.
i) For each element of D4, write down the corresponding permutation of 1,2,3,4
in a table like the one below. In this table, e is the identity of the group, a is
rotation through π
2 anti-clockwise, and bi is reflection in the line i.
e
a
a 2
a 3
ii) Find all the elements of the orbit of 1.
b1
b2
b3
b4
○2
○1
1 2 3 4
iii) Find all the elements of the stabiliser of 1.

116 Section 8. Homework questions
iv) Verify that this satisfies the orbit-stabiliser theorem.
v) Now consider the circle with similar looking lines on it as below:
○4
○3
Are these lines are acted on by O2 (the group of symmetries of the circle) as
for the square above?
○2
○1

Homework #7
117
1. Recall that D3 is the group of symmetries of the equilateral triangle. Consider the
equilateral triangle with corners labelled as below.
○2
1
3
○3
i) For each element of D3, write down the corresponding permutation of 1,2,3
in a table like the one below. In this table, e is the identity of the group, a is
rotation through 2π
3 anti-clockwise, and bi is reflection in the line i.
ii) This defines a function
e
a
a 2
b1
b2
b3
2
○1
1 2 3
θ : D3 −→ S3.
Show that θ is a group isomorphism. This means it is a group homomorphism
that is also a bijection.
2. Go through your notes from the Rings part of the course and write a list of every
definition you think you will need to learn for the exam.

Homework #8
118 Section 8. Homework questions
1. Write down all possible cycle types in S6. For each cycle type, write down a typical
permutation of that type, and calculate the number of permutations of that type.
Please try to lay out your answers nicely so that they’re easy to read.
2. Now consider the symmetric group S9 and let
Find |conjS9 (α)| and |centS9 (α)|.
⎛
⎞
1 2 3 4 5 6 7 8 9
α = ⎝ ⎠.
9 4 1 8 7 5 2 6 3

Homework #9
119
Note that Q.2 is just like the last question on Tutorial #9, so you can follow
that model answer through.
1. i) Write down all possible cycle types in S4, and the number of elements of S4
of each type.
ii) Using your answer to part (i), write down the class equation for S4.
iii) Identify the conjugacy classes consisting of even permutations. These numbers
cannot give the class equation for A4. Why?
iv) Recall that A4 is a normal subgroup of S4. Use the class equation for S4 to
show that if there is another non-trivial normal subgroup of S4 it must have
order 4.
Recall: to be a normal subgroup, its order must both divide 24 and be a sum
of conjugacy class sizes including one class of size 1 for the identity.
v) Write down the four elements that would have to be the four elements of this
group, according to the conjugacy class sizes you found in part (i). Verify that
these elements do in fact form a subgroup of S4.
2. Let G be a group of order 39 and suppose that the centre of G is {e}.
i) Determine the class equation for G, justifying your answer carefully.
ii) Find the number of elements of order 3 and the number of elements of order
13 in G.
iii) Let h ∈ G be an element of order 13. Let H = 〈h〉, the subgroup generated
by h. Use the class equation to show that H is a normal subgroup of G.
iv) Show that H is the only normal subgroup of G other than the trivial group
{e} and G itself.

120 Section 9. Tutorial questions
9 Tutorial questions
Tutorial questions #1
1. We write Zn for the ring of integers mod n. Fill in the following multiplication
tables for Z3, Z4, Z5 and Z6:
× 0 1 2
0
1
2
× 0 1 2 3
0
1
2
3
× 0 1 2 3 4
0
1
2
3
4
× 0 1 2 3 4 5
2. In a ring R, b is said to be a multiplicative inverse for a if ab = ba = 1. Use the
tables you filled in above to find the multiplicative inverses for the elements of Z3,
Z4, Z5 and Z6 if they exist.
3. A field is a ring in which every non-zero element has a multiplicative inverse.
Which of the above rings is a field? State precisely what it means for a ring not
to be a field.
4. An integral domain is a ring in which 1 = 0 and
a = 0 and b = 0 ⇒ ab = 0.
Which of the above rings is an integral domain? State precisely what it means for
a ring not to be an integral domain.
5. In the Countdown video we watched, the last stage of James Martin’s calculation
was to divide 23,800 by 25. Carol Vorderman vaguely says out loud “Well to do
that you multiply by 4...” which shows she’s using the trick that dividing by 25
is the same as multiplying by 4 and dividing by 100 (both of which are probably
easier to do in your head than dividing by 25 directly). Which ring axiom is she
using here?
6. We will define a binary operation ⊗ on Z by
a ⊗ b = ab + 1.
0
1
2
3
4
5

121
Show that this operation is not associative. Can you think of a binary operation
on Z that is not commutative?
7. In any ring R, we can prove using only the ring axioms that −(−1) = 1. Is the
following a valid proof?
∀x ∈ R, (−1)x = −x.
Therefore − (−1) = (−1)(−1)
but we know (−1)(−1) = 1.
Observe that the result is true – but this doesn’t mean the proof is valid. Moreover
observe that all the equalities valid – but this still doesn’t mean the proof is valid.
What is wrong with it? Can you write a valid proof?
If you found the earlier questions easy, see if you can prove that the following are true
in every ring R. You should prove them directly from the ring axioms, without using
any other “facts” that you know to be true, even if they seem very obvious!
1. ∀x ∈ R, 0.x = 0.
2. ∀x,y ∈ R, (−1)x = −x.
3. ∀a,b,c ∈ R, a + b = 0 and a + c = 0 ⇒ b = c.

122 Section 9. Tutorial questions
Tutorial questions #2
1. The following is a proof that (x − y)(x + y) = x 2 − y 2 in any commutative ring.
However, the justification for each line of argument has not been included. Please
fill them in. You may only use ring axioms and Lemmas from immediately after
them in the notes.
(x − y)(x + y) = x(x + y) + (−y)(x + y) by ...........................................(1)
= (x 2 + xy) + (−y).x + (−y).y) by ...........................................(2)
= x 2 + xy + ((−y).x + (−y).y
= x 2 + (xy + (−y).x) + (−y).y
= x 2 + (xy + (−(yx))) + (−y 2 )
= x 2 + (xy + (−(xy))) + (−y 2 )
by ...........................................(3)
by ...........................................(4)
by ...........................................(5)
by ...........................................(6)
= x 2 + (0 + (−y 2 )) by ...........................................(7)
= x 2 + (−y 2 ) by ...........................................(8)
= x 2 − y 2 by definition of subtraction
Note that proofs should always include a justification of how each line of the argu-
ment followed from the previous line.
2. i) Write out a multiplication table for Z8.
ii) Find the units in Z8. Recall that the units are those elements with a multi-
plicative inverse.
iii) Now write out a multiplication table just for the units in Z8. Use this to show
that the units of Z8 form a multiplicative group.
3. If x and y are units is x + y necessarily a unit?
The previous question about the units in Z8 may help you with this.
4. Let R be a commutative ring and x ∈ R. Prove that x has at most one multiplica-
tive inverse, that is, if xa = 1 and xb = 1 then a = b. Just copy the proof of the
result for additive inverses, but make it multiplicative instead.

123
5. This question is to give you some practice with the rings Z[ √ d], which consists of
all numbers of the form a + b √ d where a,b ∈ Z. We’ll look at the case d = 2.
These might look a bit like polynomial rings but actually they’re a bit different.
i) Let a,b ∈ Z. Calculate (a + bx)(a − bx) ∈ Z[x].
ii) Let a,b ∈ Z. Calculate (a + b √ 2)(a − b √ 2) ∈ Z[ √ 2].
iii) These two situations might not look very different so far. But now do it for
a = 5,b = 3. What happens?
iv) Can you think of some non-zero values of a and b that give (a+b √ 2)(a−b √ 2) =
1?
v) Can you think of some non-zero values of a and b that give (a+bx)(a−bx) = 1?
vi) What’s the point I’m trying to make here??
6. i) Let a,n,r,s be integers such that
ar + ns = 1.
Show that a and r are mutually inverse in Zn.
ii) Deduce that a is a unit in Zn if and only if the highest common factor of a
and n is 1.
Can you remember a result from Numbers and Proofs that links the second
part of this question to the first part, and thus enables you to deduce the result
you’re asked for...??
If you found the previous questions easy, try these:
1. Let R be a ring. Show that if 1 = 0 ∈ R then x = 0 for all x ∈ R.
2. Let R be a ring. Recall that R is called an integral domain if
so R is not an integral domain if
{ a = 0 and b = 0 } ⇒ ab = 0
∃a = 0 and b = 0 s.t. ab = 0.
Recall also that on the Tutorial Sheet 1 you found that Z3 and Z5 are integral
domains, but Z4 and Z6 are not integral domains.
Show that Zn is not an integral domain for n = 8,9,10. Can you guess what
property of n determines whether or not Zn is an integral domain? Can you prove
it?

124 Section 9. Tutorial questions
Tutorial questions #3
1. The following are some submitted “proofs” from homework #1. The question
asked you to prove: if x ≡ y (mod n) then ax ≡ ay (mod n). Which of the
following is a valid proof? What is wrong with the others?
• “Proof” A:
• “Proof” B
ax ≡ ay (mod n) =⇒ n|ax − ay
=⇒ n|a(x − y)
=⇒ n|x − y
=⇒ x ≡ y (mod n)
x ≡ y (mod n) so multiplying both sides by a gives
• “Proof” C
• “Proof” D
ax ≡ ay (mod n).
x ≡ y (mod n) = n|x − y
= n|a(x − y)
= n|ax − ay
x ≡ y (mod n) =⇒ n|x − y
= ax ≡ ay (mod n)
=⇒ x − y = kn for some k ∈ Z
=⇒ ax − ay = akn
=⇒ n|ax − ay
=⇒ ax ≡ ay (mod n)
2. i) Show that (x + 1)(x + 2)(x + 3) ≡ x 3 − x ∈ Z6[x].
ii) What are the roots of (x + 1)(x + 2)(x + 3) in R?
iii) What are the roots of x 3 − x in R?
iv) What are the roots of (x + 1)(x + 2)(x + 3) in Z6?

v) Did this surprise you? Can you see why it’s true?
125
To find roots you need to find values of x for which the given formula comes out
to 0.
3. Show (1 + 3x) is a unit in Z9[x].
Hint: you need to find integers a and b such that
(1 + 3x)(a + bx) ≡ 1 ∈ Z9[x].
4. Here is a method for solving the quadratic equation x 2 + 3x + 2 = 0 in the reals,
producing two solutions. In the homework you should have seen that there are
four solutions in Z6, which means that this method doesn’t work in Z6. Which
step goes wrong?
x 2 + 3x + 2 = (x + 1)(x + 2)
(x + 1)(x + 2) = 0 =⇒ x + 1 = 0 or x + 2 = 0
=⇒ x = −1 or x = −2
5. For any integer n > 0, define φ(n) to be the number of integers coprime to n in
the set
{1,2,3,... ,n}.
Fill in the following table. In the second column you should write all the integers
between 1 and n and coprime to n that you will be counting in the third column.
This function φ is called Euler’s Totient Function, and is quite a useful tool. In
the next few questions we’ll see that there are clever ways of calculating φ(n) other
than writing down all the relevant coprime integers and counting them.

126 Section 9. Tutorial questions
n coprime integers φ(n)
1 1 1
2 1 1
3 1,2 2
4 1,3 2
5 1,2,3,4 4
6
7
8
9
10
11
12
6. Show that if p is prime, φ(p) = p − 1.
7. Show that if p and q are distinct prime numbers
φ(pq) = φ(p)φ(q).
(Why do we need p and q to be distinct?)
8. Let d > 1 be a square-free integer, i.e. its prime factorisation has no repeated
factors, or equivalently for all n ∈ Z
n 2 |d ⇒ n 2 = 1.
i) Show that if d|a 2 then d|a. Hint: think about the prime factors of d.
ii) Hence prove that √ d is irrational.
Hint: Copy the proof overleaf that √ 2 is irrational, but be careful about where
you have to use part (i).

Proof that √ 2 is irrational: by contradiction
Suppose √ 2 is rational.
Put √ 2 = a
b
i.e. hcf(a,b) = 1.
Then 2 = a2
b 2
=⇒ 2b 2 = a 2
=⇒ 2|a 2
where a,b ∈ Z and the fraction is in its lowest terms
=⇒ 2|a since 2 is prime
=⇒ a = 2k for some k ∈ Z
=⇒ 2b 2 = (2k) 2 = 4k 2
=⇒ b 2 = 2k 2
=⇒ 2|b 2
=⇒ 2|b since 2 is prime
So 2|a and 2|b # contradicts hcf(a,b) = 1.
127
Hence √ 2 is irrational as required.

128 Section 9. Tutorial questions
Tutorial questions #4
1. This question is about the group of units in Zn again, but this time we’ll go a step
further and analyse what the group of units actually is. This is also to jog your
memory a bit about some group stuff from MAS175. “Recall” that a cyclic group
is one in which there is an element a such that every other element of the group is
a power of a. Such an a is called a generator.
i) On homework #2 you found the multiplication table for the unit group of Z10.
Find an element of the unit group that has order 4. Remember, the order of
an element a is the smallest integer k > 0 such that a k = 1.
ii) Deduce that this group is cyclic.
Oh OK I’ll do it for you: a group of order n is cyclic if and only if it has an
element of order n. In fact any element of order n is a generator.
iii) The above was the boringly efficient way of showing that the unit group in this
case was cyclic. Something a bit more fun is to rewrite the multiplication table.
Take your element of order 4 – let’s call it a. Then reorder your multiplication
table like this:
× 1 a a 2 a 3
1
a
a 2
a 3
except that you’ll put the actual numbers in, instead of a,a 2 ,a 3 . Now fill in
the table, and you should see the nice swirly pattern associated with a cyclic
group.
iv) Now, since I practically did that whole thing for you, do the whole thing for
Z9. Start by writing down the units and their multiplication table, then show
that the group is cyclic, and reorder the table to get the cyclic pattern.
2. For every element a in Z15 determine whether a is a unit or a zero-divisor. If a is
a unit find its inverse, and if a is a zero-divisor find b = 0 such that ab = 0.
Hints:

129
• You could of course do this by filling in a multiplication table, but I hope
you’ll agree that that would be a bit of a long and boring way to do it.
• We know we can find inverses using Euclid’s algorithm, but for numbers as
small as this you may find it easier to do it by trial and error/staring. Don’t
forget that if you can think of b such that ab ≡ −1 then you know that
a.(−b) ≡ 1. This can sometimes help.
• Note that you only really need to work all this out for half of the numbers,
because once you’ve done a you can immediately write down the answer for
−a in either case, using (−a)(−b) = ab.
3. For each of the following values of a, determine whether a is a unit or a zero-divisor
in Z570. Again, if a is a unit find its inverse, and if a is a zero-divisor find b = 0
such that ab = 0.
i) a = 46 ii) a = 299 iii) a = 105
4. Let d be a square-free integer, and consider Z[ √ d]. Given an element r = a+b √ d ∈
Z[ √ d] define
N(r) = |a 2 − db 2 |.
This is called the norm of an element, and we’ll be seeing more of this useful gadget
soon.
Calculate N(r) for the following elements.
i) 3 + 2 √ 2 ∈ Z[ √ 2]
ii) 8 + 3 √ 7 ∈ Z[ √ 7]
iii) 17 − 12 √ 2 ∈ Z[ √ 2]
5. i) Show that all the above elements are units in their respective rings.
ii) What point do you think I’m trying to make here?
6. This question is related to Homework #2, question 3.
i) Let R be a ring, and a,b ∈ R. What does a
mean ?
b
ii) On the homework, you solved the equation
(8 + 3 √ 7)x = 3.
Below is a solution of the apparently similar equation
(10 + 3 √ 7)x = 3.

130 Section 9. Tutorial questions
This method is the one that some people used in the homework. However
here it does not give a solution in Z[ √ 7], which means that something in the
solution method isn’t valid in Z[ √ 7]. What is it? Hint: part (i) of this question
is supposed to be relevant. The point of this question is to get you to think a
bit more about what division “is”.
(10 + 3 √ 7)x = 3
x =
=
3
10 + 3 √ 7
3
10 + 3 √ 7 . 10 − 3√7 10 − 3 √ 7
= 30 − 9√ 7
37
iii) Do you think question 5 sheds any light on this matter?
7. Do question 2 again but for Z21.

Tutorial questions #5
131
1. Recall that the norm of a + b √ d ∈ Z[ √ d] is |a 2 − b 2 d|. Find the norm of each the
following elements:
i) (1 + 2 √ 3), (1 − 2 √ 3), (−1 + 2 √ 3), (−1 − 2 √ 3) ∈ Z[ √ 3]
ii) 3 ∈ Z[ √ 2], 3 ∈ Z[ √ 3], 3 ∈ Z[ √ −7], 3 ∈ Z[ √ −11]
iii) (5 + 2 √ 6), (5 − 2 √ 6), (−5 + 2 √ 6), (−5 − 2 √ 6) ∈ Z[ √ 6]
iv) 3 + 2 √ 2 ∈ Z[ √ 2]
v) (5 + 2 √ −11), (2 + 5 √ −11) ∈ Z[ √ −11]
2. Write down three more elements with the same norm as
6174952171 + 6174955132 √ 410533
Those are the phone and fax numbers of the Harvard maths department, in case
you were wondering.
3. Which of the elements in Question 1 are units of their respective rings? Write
down 4 more units in Z[ √ 2].
Remember that r is a unit iff N(r) = 1.
4. Write a list of all possible norms less than 20 of elements in Z[ √ −7].
Norms have to be non-negative integers, but not all non-negative integers are valid
norms in all rings. This gives us important information when we’re analysing
those rings, sometimes, as we’ll see later.
5. An element r ∈ R is called irreducible if r is not a unit, and
r = st ⇒ s is a unit or t is a unit.
i) Show that if N(st) is prime then either N(s) = 1 or N(t) = 1.
Hint: use N(st) = N(s)N(t)
ii) What does this tell us about r if N(r) is prime?
Hint: what does N(s) = 1 tell us about s?
iii) Which of the elements in Question 1 can we now deduce is definitely irre-
ducible?
6. This question is about the converse of the above question. We now know that
N(r) prime ⇒ r irreducible.

132 Section 9. Tutorial questions
But it turns out that r can be irreducible even if N(r) is not prime, and this
depends on the fact that not all integers are valid values of N(r), as we saw in
question 4.
i) Show that there is no element in Z[ √ −11] whose norm is 3.
Hint: to get a norm of 3, we’d have to find a,b ∈ Z such that a 2 + 11b 2 = 3.
Is that possible?
ii) Consider s,t ∈ Z[ √ −11]. Show that if N(st) = 9 then N(s) = 1 or N(t) = 1.
iii) Deduce that 3 is irreducible in Z[ √ −11] even though N(3) is not prime.
iv) Is 3 irreducible in Z[ √ −3]?
7. This question pushes the previous question a bit further and finds some non-unique
factorisation in Z[ √ −11].
i) Show that there is no element of norm 23 in Z[ √ −11].
ii) Show that if N(r) = 69 then r must be irreducible.
Hint: 69 = 3 × 23.
iii) Show that (5+2 √ −11), (5−2 √ −11), 3, and 23 are all irreducible in Z[ √ −11].
iv) Write down two different factorisations of 69 into irreducibles in Z[ √ −11].
That is, find rs = r ′ s ′ = 69 where r,s,r ′ ,s ′ are all irreducible, and r ′ ,s ′
cannnot be obtained from r,s by multiplying by units. This shows that
Z[ √ −11] is not a UFD.
To do the last part you need to remember what the units in Z[ √ −11] are.
8. This question is about expressing integers as the sum of two squares.
Consider r = (1 + 2i)(3 + 4i) = −5 + 10i ∈ Z[i]. Now N(r) = 5 2 + 10 2 = 125 but
also we know that (1 + 2i)(3 − 4i) must have the same norm as r. (Why??) And
(1 + 2i)(3 − 4i) = 11 + 2i
so taking the norm of that, we get 11 2 + 2 2 = 125. Expressing things as the sum
of two squares is Interesting, and look! We’ve done it in two different ways.
Now you do it – consider
(2 + 3i)(4 − 5i) = 23 + 2i
and copy the above procedure to express 533 as a the sum of two squares in two
different ways.

Tutorial questions #6
Note that this whole sheet is revision about groups.
You should bring your Groups and Symmetries notes to the tutorial.
133
1. This question is to jog your memory about the symmetric groups Sn. We’ll do S9.
i) Write down the following permutation in disjoint cycle notation:
⎛
⎞
1 2 3 4 5 6 7 8 9
⎝ ⎠ .
9 4 1 8 7 5 2 6 3
ii) Write down the following permutation in two-row notation:
iii) Let
and
(1 5 3)(2 7).
⎛
⎞
1 2 3 4 5 6 7 8 9
α = ⎝ ⎠
9 4 1 8 7 5 2 6 3
⎛
⎞
1 2 3 4 5 6 7 8 9
β = ⎝ ⎠.
6 3 1 2 7 4 9 5 8
Calculate βα. Remember this means do α first and then β afterwards.
iv) Let α = (1 5 3) and β = (2 5 1). Calculate αβ and βα, and observe that they
are not equal. Permutations do not commute, so Sn is not in general abelian.
v) Write (1 5 4 7)(2 9) as a product of transpositions. Remember a transposition
is a permutation that just swaps two numbers eg (1 6).
2. Recall that a permutation is called even if, when it is written as a product of
transpositions, it turns out to have an even number of transpositions.
i) Show that the even permutations form a subgroup of Sn. This is called the
alternating group An.
ii) Do the odd permutations form a subgroup of Sn?
3. Recall that D3 is the group of symmetries of the equilateral triangle.
i) Write down all the elements of D3.

134 Section 9. Tutorial questions
ii) Pick one reflection and one non-trivial rotation in D3, and express all the
non-identity elements of D3 in terms of these two.
This means that these two elements “generate” the group.
iii) Find six subgroups of D3. (This is all the subgroups of D3).
4. This question is about cosets. Recall that given a subgroup H of a group G, for
each g ∈ G the left coset gH is the set
{gh : h ∈ H}.
Recall also that aH = bH ⇐⇒ a −1 b ∈ H.
Now consider the following subgroup of S4:
H = {e,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
i) Let a ∈ S4 be the permutation (1 2). Write down all the elements of the coset
aH.
ii) Show that (1 4) and (2 3) are in the same left coset of H.
iii) How many different left cosets of H are there?
5. A subgroup H of G is called a normal subgroup if
g ∈ G and h ∈ H ⇒ g −1 hg ∈ H.
i) Show that if G is abelian, all its subgroups are normal.
ii) Let H be the subgroup of D3 containing just the rotations. Convince yourself
that H is a normal subgroup of D3.
iii) Show that An is a normal subgroup of Sn. (See question 2 for definitions.)
We will later show that if H is a normal subgroup, we can make the cosets of H
into a group called the quotient group. This is one of the many reasons normal
subgroups are important. “Normal” doesn’t mean “ordinary” here!
6. Recall that the orthogonal group O2 is the group of symmetries of the circle. The
elements of this group can be expressed as
rotα = rotate anticlockwise through angle α
refα = reflect in line through the centre at an angle of α
to the horizontal
2
i) Here’s another way of expressing this group. Write α for rotα, and M = ref0
ie reflection in the “x-axis”. In this new notation, what is refα?

135
ii) Show that O2 has a subgroup isomorphic to D3. You can do this by algebra
or by geometry.
iii) Write down four other subgroups of O2 other than the trivial group and the
whole group.

136 Section 9. Tutorial questions
Tutorial questions #7
1. This question is about the symmetric group S3, but the analogous result is true
for Sn in general.
i) Write down all the elements of S3, in cycle notation. (Check you have the
right number of them.)
ii) Write down all the elements of S3 as products of transpositions. Hence, find
the even permutations i.e. the elements of A3.
iii) Find all the left cosets of A3 in S3.
iv) Check that, for all α ∈ S3, αA3 = A3α. This means that A3 is a normal
subgroup of S3.
v) When H is a normal subgroup of G then we can make the left cosets of H
into a group called the “quotient group” G/H. What is the quotient group
S3/A3? Hint: what is its order?
2. i) Write 4Z for the subset of Z given by
{k ∈ Z s.t. 4|k}.
Show that 4Z is a subgroup of Z. Remember: Z is a group under addition.
ii) What are all the left cosets of 4Z in Z? Note that since Z is a group under
addition, a left coset of a subgroup H will be written a + H instead of aH.
iii) We can define an operation ⊕ on the cosets of H = 4Z by
(a + H) ⊕ (b + H) = (a + b) + H.
This makes the cosets of H into a group. In what guise have you seen this
group before?
3. i) Consider D6, the symmetry group of the regular hexagon. What happens if
you conjugate a rotation by a reflection? That is, if a is a rotation and b is a
reflection, what is bab −1 ?
ii) What happens in O2, the symmetry group of the circle?
4. This question is about cycle types in S5. The “cycle type” of an element α of Sn
is found as follows: write α in disjoint cycle form, with the cycles in descending
order of length. Then list the lengths of the cycles (including the trivial ones of
length 1 at the end). For example, the element (1 3)(5 2 6) ∈ S6 can be re-written
(5 2 6)(1 3)(4)

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and has cycle type (3,2,1). The element (1 6)(2 4) has cycle type (2,2,1,1).
i) Find all the possible cycle types of elements in S5.
ii) Which ones are odd and which ones are even? Hint: to work this out, you
need to rewrite the cycles as products of tranpositions.
iii) How many elements of each cycle type are there?
5. Conjugate the element (1 2 3)(4 5) by the following elements. What do you notice
about the resulting cycle types?
i) (3 4) ii) (1 4 5) iii) (1 2 3 4)

138 Section 9. Tutorial questions
Tutorial questions #8
1. Calculate θαθ−1 ⎛
⎞
1 2 3 4 5 6 7 8 9
where θ = ⎝ ⎠ and α = (1 9 6 3)(2 4 8)(5 7).
4 1 9 6 5 2 3 7 8
2. For each of the following elements α of S9, find the number of elements of S9 of
the same cycle type.
i) (1 2)
ii) (1 2 3)
iii) (1 2 3 4)
iv) (1 2 3 4 5)
v) (1 2 3 4)(5 6 7)
vi) (1 9 6 3)(2 4 8)(5 7)
vii) (1 2)(3 4)
3. For each of the elements α in question 2, find |conjS9 (α)| and |centS9 (α)|. Re-
member:
• In Sn, the elements conjugate to an element α are those with the same cycle
type.
• In any finite group G, |conj G(α)|.|centG(α)| = |G|.
4. Let A,B be the following matrices with entries in Z7:
⎛ ⎞ ⎛ ⎞
1 2
A = ⎝ ⎠ ,
4 4
B = ⎝ ⎠ .
3 4 2 6
i) Find the determinant of each of these matrices, to check that A ∈ GL2(Z7)
and B ∈ GL2(Z7).
Recall that GL2(Z7) is the group of invertible 2 × 2 matrices with entries in
Z7.
ii) Show that
iii) Compute the conjugate ABA −1 .
A −1 ⎛ ⎞
5 1
= ⎝ ⎠.
5 3
iv) Find the order of ABA −1 and hence find the order of B.
5. In this question you should use the following facts:

139
• The order of each conjugacy class must divide |G|, the order of the group G.
• Every element is in precisely one conjugacy class, so orders of the conjugacy
classes must add up to |G|. So for a group of order 12, for example, we could
have 1,3,4,4 or 1,2,3,6 or 1,2,3,3,3 but not 1,3,3,3.
i) What is the order of the conjugacy class of the identity e in any group? Hint:
how many elements b,g can you find such that g = beb −1 ?
ii) In a group of order 6, is it possible to have three conjugacy classes of order 2?
Is it possible to have two conjugacy classes of order 3? What are the possible
combinations of conjugacy class orders?
iii) What are the possible combinations for a group of order 8?
iv) What are the possible combinations for a group of order 10?

140 Section 9. Tutorial questions
Tutorial questions #9
1. Find the following conjugates in O2:
i) rotφ rotθ(rotφ) −1
ii) refφ rotθ(refφ) −1
iii) rotφ refθ(rotφ) −1
iv) refφ refθ(refφ) −1
Hint: it may help you to remember refφ = rotφ ref0.
2. In each of the following either find θ ∈ S7 such that θαθ −1 = β or explain why no
such θ exists:
i) α = (1 3 5 2 4)(6 7), β = (1 2)(3 4 5 6 7)
ii) α = (1 2)(3 4)(5 6), β = (1 2)(6 7)
3. Calculate θαθ−1 ⎛
⎞
1 2 3 4 5 6 7 8 9
where θ = ⎝ ⎠ and α = (1 9 6 3 5)(2 4 8).
4 1 9 6 5 2 3 7 8
You should be able to write this straight down, without even working out what θ −1
is!
4. You are given that
H = {e,(1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}
is a normal subgroup of S4, where e is the identity element of S4. Show that the
following hold in S4/H:
i) the inverse of (1 2)H is (3 4)H
ii) (1 2 3)H(1 3 4)H = (1 3 4)H(1 2 3)H
Hint: it will help you to use the fact that the identity element in the quotient group
is eH which is the coset H.
5. Let G be a group of order 21 and suppose that the centre of G is {e}.
i) Determine the class equation for G.
ii) Find the number of elements of order 3 and the number of elements of order
7 in G.
Hint: let a be an element of a conjugacy class of size 3. Now a is also an
element of its own centraliser, which is a subgroup of G. We know what
the order of this subgroup is, therefore we can work out the order of a. Do
something similar for elements in conjugacy classes of size 7.

141
iii) Let h ∈ G be an element of order 7. Let H = 〈h〉, the subgroup generated by
h. Use the class equation to show that H is a normal subgroup of G.
iv) Show that H is the only normal subgroup of G other than the trivial group
{e} and G itself.
v) Can you think of some other n (other than 21) where you could use this sort
of argument on a group of order n?

142 Section 9. Tutorial questions
Tutorial questions #10
1. Consider the cyclic group C12 with elements
2. Let
i) Let
Are A and B subgroups?
ii) What is the set AB?
{e,a,a 2 ,a 3 ,a 4 ,a 5 ,a 6 ,a 7 ,a 8 ,a 9 ,a 10 ,a 11 }
A = {a 2 ,a 3 ,a 4 }
B = {a 3 ,a 4 ,a 5 }.
iii) Let H be the subgroup {e,a 4 ,a 8 }. Write down the elements of the cosets a 2 H
and a 3 H.
iv) What is the set (a 2 H)(a 3 H)? First work this out by multiplying every element
of a 2 H by every element of a 3 H. Then check that it is the coset you were
expecting.
v) What is the quotient group C12/H?
p1 = (x1 + x2)(x3 + x4)
p2 = (x1 + x3)(x2 + x4)
p3 = (x1 + x4)(x2 + x3).
The symmetric group S4 acts on these polynomials by permuting the variables,
that is, for α ∈ S4
α ∗ (xi + xj)(xk + xl) = (x α(i) + x α(j))(x α(k) + x α(l)).
This definition of how the symmetric group acts on the polynomials is very formal
and a bit impenetrable. So in this question we’re going to sit down and fiddle
around with it to see what it actually does
i) The element (2 3) acts on each polynomial by switching 2 and 3, so in effect
x2 becomes x3 and x3 becomes x2. Write down the result of doing this to the
polynomial
(x1 + x2)(x3 + x4).

143
ii) What you have just done is apply the element (2 3) to the polynomial p1.
Which of p1,p2,p3 was the result?
iii) Now apply (2 3) to p2 and to p3 and see which polynomial results in each case.
iv) This means that the element (2 3) has permuted the polynomials {p1,p2,p3}.
So we have produced a permutation in S3. Which one?
v) Now try it for the element (1 2)(3 4). What happens if you apply this to each
of p1,p2,p3?
vi) So, what permutation of S3 have you produced from (1 2)(3 4)?
vii) Find all the permutations in S4 that leave p1 fixed. This is the stabiliser of
p1.
viii) Find all the permutations that fix each of p1, p2, p3.
ix) The above method gives a homomorphism f : S4 −→ S3. What is its kernel?
x) What does the First Isomorphism Theorem tell us about this situation?