Lecture 1

Recall that if A is an arbitrary set and f : An → A then f is called an operation on A. For n = 1, f

is a unary operation, for n = 2, f is a binary operation, for n = 3, f is a ternary operation, etc. The

modern point of view on algebra is that algebra is a part of mathematics that studies operations.

The most important for us will be binary operations.

Examples:

1. N—the set of all natural numbers; +, ·, − binary operations on N (we assume that 0 ∈ N)

− is not a binary operation on N since it is not defined for all pairs 〈n,m〉 ∈ N2: 2 − 3 /∈ N.

2. Z—the set of all integers; +, ·,− are binary operations on Z , but division / is not a binary

operation on Z.

3. Q,R,C—the sets of rationals, reals and complex numbers, +, ·,− are operations on Q,R,C but

division is not an operation, on any of these sets—we cannot divide by zero. If K is one of

the sets Q,R,C, denote by K∗ = K\{0}. Then the division a/b is an operation on K∗.

Are + and · -operations on K∗?

4. The square root is not a unary operation on Q,R,C - on Q and R it is not everywhere defined

and on C it has two values.

The cubic root is not an operation on Q, but it is an operation on R. It is not an operation

on C by the same reasons as above.

5. Let Matm,n(K) (K-is an example 3) be the set of all m × n matrices with the entries in K.

Then the addition of matrices is a binary operation on Matm,n(K):

A =

a11 . . . a1n

−−−−−−am1 . . . amn

, B =

b11 . . . b1n

−−−−−−bm1 . . . bmn

,

A+B =

a11 + b11, . . . , a1n + b1n

−−−−−−−−−−am1 + bm1, . . . , amn + bmn

for any fixed λ ∈ K the multiplication by λ is a unary operation on Matm,n(K).

λA =

λan . . . λa1n

−−−−−−λam1 . . . λamn

The multiplication of matrices is a binary operation on Matm,n(K) only if m = n. We write

in this case Matn(K).

A =

a11 . . . a1n

−−−−−−an1 . . . ann

B =

b11 . . . b1n

−−−−−−bn1 . . . bnn

A · B =

c11 . . . c1n

−−−−−−cn1 . . . cnn

where

cij = ai1b1j + ai2b2j + . . .+ ainbnj

6. Let B be an arbitrary nonempty set then BB-is the set of all mapping f : B → B. The

composition of mappings is an operation on BB. Recall that if f ∈ BB, i.e., f : B → B, and

g ∈ BB, i.e., g : B → B then f ◦ g : B → B, which is defined by the formula:

f ◦ g(b) = f (g(b))

f : R → R f(x) = x2 f ◦ g(x) = f (g(x)) = (sin x)2 = sin2 x

g : R → R g(x) = sin x g ◦ f(x) = sin(x2)

7. Let S(B) be the set of all bijective mappings from BB. Since the composition of two bijective

mappings is bijective the composition ◦ is a binary operation on S(B). An important case is

B = {1, 2, . . . , n}. Then S(B) is denoted by Sn. The elements of Sn are called permutations

of 〈1, 2, . . . , n〉. Usually the permutations are denoted by

π =(

1 2 . . . ni1i2 . . . in

)(∗)

Now it is easy to determine the number of elements (the cardinality) of Sn. Each permutation

is uniquely determined by the lower row of the table (∗).

2

The element i1 can be chosen arbitrary, so we have n possibilities to choose i1. Whenever i1

is chosen we have only n−1 possibilities to choose i2 since our map π is injective. So we have

n · (n− 1) possibilities to choose the pair 〈i1, i2〉. Whenever this pair is chosen we have only

(n− 2) possibilities to choose i3. And thus we have n(n− 1)(n− 2) possibilities to choose the

triple 〈i1i2i3〉. Continuing this consideration we see that the lower row 〈i1, i2, i3, . . . , in〉 can

be chosen by n(n− 1)(n− 2) . . . 1 = n! ways. Thus we have proved the following

Theorem. |Sn| = n!

π1 =(

1 2 3 4 52 1 4 3 5

)π2 =

(1 2 3 4 53 2 1 5 4

)

π1 ◦ π2 =(

1 2 3 4 54 1 2 5 3

)

Exercises:

1. List all elements of S2, S3, S4.

2. Find π2 ◦ π1 as the example above.

3. Find the permutation of π such that

(1 2 3 4 5 66 3 4 5 2 1

)◦ π =

(1 2 3 4 5 62 3 1 6 4 5

)

4. Find the permutation π such that

π ◦(

1 2 3 4 5 66 3 4 5 2 1

)=(

1 2 3 4 5 62 3 1 6 4 5

).

3

Lecture 2

Main properties of binary operations.

If ◦ : A× A→ A we write a ◦ b instead of ◦(a, b).

Commutativity.

We say that ◦ is commutative if ∀a, b ∈ A a ◦ b = b ◦ a.

Examples:

1. Addition and multiplication on N, Z,Q,R,C, addition on Matm,n(K) are commutative opera-

tions.

2. Composition ◦ on BB is in general no commutative operation. Recall the example of Lecture 1:

f : R → R f(x) = sin x; g : R → R, g(x) = x2

f ◦ g(x) = sin(x2), g ◦ f(x) = sin2 x.

Permutations also may not commute:

(1 2 33 1 2

)(1 2 32 1 3

)=(

1 2 31 3 2

)6=(

1 2 32 1 3

)(1 2 33 1 2

)=(

1 2 33 2 1

)

3. Multiplication of Matn(K) is not commutative.

A =(

1 02 1

)B =

(1 20 1

)A · B =

(1 22 5

)B · A =

(5 212 5

)

Sometimes the so called symmetrized product of matrices is useful:

{A,B} =1

2(A ·B +B · A)

Obviously {A,B} = {B,A}. So this operation is commutative.

{(1 02 1

),(

1 20 1

)}=(

3 27 5

)

4. − and / are obviously non-commutative operations: a− b = −(b− a) a/b =1

(b/a).

4

Associativity:

We say that ◦ is associative of ∀a, b, c ∈ A (a ◦ b) ◦ c = a ◦ (b ◦ c). By induction, we see that

in the product of n terms it does not matter where the parenthesis are located so we can write

a1 ◦ a2 ◦ . . . ◦ a12 without parenthesis.

Examples:

1. Addition and multiplication on R,C,N,Q, Z,MatR(K) are associate operations.

2. Composition on BB is an associative operation (thus, composition on Sn is an associative

operation).

3. − is not an associative operation

a− (b− c) = a− b+ c (a− b) − c = a− b− c.

4. / is not an associative operation.

a

(b/c)=ac

b

(a/b)

c=

a

bc

5. {A,B} is not an associative operation.

Homework: Take for c =(

1 22 1

)and A,B the same as in example 3 and check the associativity

of { , }. Here associativity is written in the form of:

{{A,B}, C} = {A, {B,C}}

Neutral (unit) element.

We say that e ∈ A is a neutral (unit) element for the operation ◦ : A2 → A if ∀a ∈ A a ◦ e =

e ◦ a = a.

Proposition: A neutral element is unique.

Proof.

Let e1, e2 ∈ A be neutral elements. Then e1 ◦ e2 = e1 since e2 is a neutral element. On the other

hand, e1 ◦ e2 = e2 since e1 is a neutral element. Thus e1 = e2.

Examples:

5

1. 0 is the neutral element for addition on N, Z,Q,R,C.

2. 1 is the neutral element for multiplication on N, Z,Q,R,C.

3. The identity map idB : B → B, such that ∀b ∈ B idB(b) = b, is the neutral element for

composition on BB. For example(

1 2 . . . n1 2 . . . n

)is the unit element for composition of

permutations.

4. The unit matrix In ∈Matn(K) is the unit element for matrix multiplication. Recall that

In =

1 0 0 . . . 00 1 0 . . . 00 0 1 . . . 0−−−−−−−−0 0 0 . . . 1

We know from linear algebra that for any A ∈Matn(K) A ◦ In = In ◦ A = A.

5. Cancellation laws. We say that the operation ◦A2 → A satisfies the left cancellation law if

∀a, x, y ∈ A

a ◦ x = a ◦ y ⇒ x = y

◦A2 → A satisfies the right cancellation law of ∀a, x, y ∈ A

x ◦ a = y ◦ a⇒ x = y

For commutative operations left and right cancellation laws are the same and we speak simply

about the cancellation law.

Examples:

1. Addition and multiplication on N, Z,Q,R,C satisfy the cancellation law.

2. The composition on BB satisfies neither right nor left cancellation laws:

(a) f : R → R f(x) = x2

g : R → R g(x) = x g = idR

6

h : R → R h(x) = −x

f ◦ g(x) = x2 f ◦ h(x) = (−x)2 = x2. So f ◦ g = f ◦ h, but g 6= h.

(b) Let g(x) =

{ √1 − x2 |x| ≤ 1

0 |x| > 1h(x) =

{ √1 − x2 |x| ≤ 1

1 |x| > 1

Then g(x) 6= h(x). Let f(x) = sin x. Then g ◦ f = h ◦ f g ◦ f(x) = h ◦ f(x) = | cosx|.

3. The composition on S(B) satisfies both the left and and right cancellation laws. Indeed we

know that for any bijection f ∈ S(B) there exists the inverse bijection f−1 ∈ S(B) such that

f ◦ f−1 = f−1 ◦ f = idB. So if f ◦ g = f ◦ h then f−1 ◦ (f ◦ g) = f−1 ◦ (f ◦ h) by

associativity:

(f−1 ◦ f) ◦ g = (f−1 ◦ f) ◦ h

i.e.

id13 ◦ g = idB ◦ h

g′′ = h′′

Homework: Prove the right cancellation law. For π ∈ Sn the inverse can be determined in the

following way:

π =(

1 2 . . . ni1 i2 . . . in

)π−1 =

(i1 i2 . . . in1 2 . . . n

)

Now we have only to rearrange columns to obtain (1 2 . . . n) in the upper row.

π =(

1 2 3 4 53 1 2 5 4

)π−1 =

(3 1 2 5 41 2 3 4 5

)=(

1 2 3 4 52 3 1 5 4

)

7

Lecture 3

4. Returning to example 2 of the previous lecture, let f : B → B be injective. We know from

Math 2800 that this holds iff f has a inverse mapping f such that f ◦ f = idB. Now if

f ◦ g = f ◦ h, f ◦ (f ◦ g) = f ◦ (f ◦ h) ⇒ (f ◦ f) ◦ g = (f ◦ f) ⇒ idB ◦ g = idB ◦ h⇒ g = h.

So we proved that:

(a) If f is injective then it satisfies the left cancellation law: f ◦ g = f ◦ h⇒ g = h.

Homework: Prove the following statement.

(b) If f is subjective then it satisfies the right cancellation law: g ◦ f = h ◦ f ⇒ g = h.

Hint: Use the fact that subjective mappings have right inverse.

Indeed the statements inverse to (a) and (b) are true also and, thus, the following theorem

holds.

Indeed the statements inverse to (a) and (b) are true also and, thus, the following theorem

holds.

Theorem. A mapping f is injective iff the left cancellation law holds for f ; f is surjective

iff the right cancellation law holds for f .

Homework: Functions f in the examples 2(a) and (b)

(a) f(x) = x2, (b) f(x) = sin x are neither injective or surjective. Construct for f(x) = x2

functions g and h such that the right cancellation law fails and for f(x) = sin x – function g

and h such that the left cancellation law fails.

5. Let U be a set. Then ∩, ∪ are binary operation on P(U). These operations are commutative:

A ∩B = B ∩A; A ∪B = B ∪A. The set U is a neutral element for ∩: A ∩ U = A, while the

emptyset is the neutral element for ∪: A ∪ ∅ = A. The operations ∪,∩ do not satisfied the

cancellation law.

(A ∪B) ∩ A = A ∩B(A ∩B) ∩ B = A ∩B

8

Thus (A ∩ B) ∩ A = (A ∩ B) ∩B even when A 6= B.

Homework: Show that the cancellation law fails for the union.

6. The left and right cancellation law fail for matrix multiplication:

Let A =(

1 00 0

), B =

(0 00 1

). The A ·B =

(0 00 0

). It is easy to see that A ·

(0 00 0

)=

(0 00 0

)and

(0 00 0

)·B =

(0 00 0

). Let O =

(0 00 0

). So we have A ·B = A ·O but B 6= O

and O · B = A · B, but O 6= A.

Inverse elements.

Let ◦ : A × A → A have the unit element e. We say that an element b is an inverse to a if

a ◦ b = b ◦ a = e.

Proposition.

If ◦ is an associative operation then each element a has no more than one inverse element.

Proof.

Let b and b1 be inverse elements to a. Then a ◦ b = e. Multiply this equality by b1 from the left

b, ◦(a ◦ b) = b1 ◦ e = b1. But b1 ◦ (a ◦ b) = (b1 ◦ a) ◦ b = e ◦ b = b. So b1 = b. Here we used that b1 is

an inverse to a. �

Usually the inverse element to an element a is denoted by a−1. So a ◦ a−1 = a−1 ◦ a = e.

(a−1)−1 = a.

Now we come to our main definition.

Definition.

We say that a set G with a binary operation ◦ : G× G → G is a group if the following properties

hold:

1. ◦ is associative

2. ◦ has the neutral element e, which is called simply the unit.

3. each element g ∈ G has the inverse element g−1.

9

Examples of the groups and non-groups

1. (N,+) is not a group 1–2 hold, but 3 fail.

Remark: If (G, ◦) satisfies 1) then it is called a semigroup; if 1) and 2)—then a semigroup

with the unit.

2. (N, ·) – a semigroup with the unit.

3. (Z,+) = a group: ◦ is the unit, −a is the inverse to a.

Definition.

If in a group (G, ◦) the operation ◦ is commutative then (G, ◦) is called a commutative or

abelian group. (The honor of great Norwegian Mathematician Nils Keurik Abel (1802–1828).)

So, (Z,+) is an abelian group.

4. (Z, ·) is a semigroup with the unit.

5. (Q,+) is an abelian group.

6. (Q, ·) is not a group—the element 0 has no inverse.

Theorem. Let (S, ◦) be a semigroup with the unit. Denote by S∗ the set of all elements

s ∈ S that have the inverse (the invertible elements). Then S∗ is a group under the same

operation ∗.

Proof.

First of all we have to show that S∗ is closed under ◦. It means that if s1 ∈ S∗ and s2 ∈ S∗

then s1 ◦ s2 ∈ S∗. So given that s−11 and s−1

2 exist we have to show that (s1 ◦ s2)−1 exists.

Consider the element s−12 ◦ s−1

1 . Then

(s1 ◦ s2) ◦ (s−12 ◦ s−1

1 ) = s1 ◦((s2 ◦ s−1

2 ) ◦ s−11

)= s1 ◦ (e ◦ s−1

1 ) = s1 ◦ s−11 = e.

Homework: (s−12 ◦ s−1

1 ) ◦ (s1 ◦ s2) = e.

10

These equalities show that s−12 ◦ s−1

1 = (s1 ◦ s2)−1. So S∗ is closed under ◦. This operation is

associative in S∗ since it is associative in S, the unit e ∈ S∗ since, obviously e−1 = e. And

each element of S∗ has the inverse by definition. Thus (S∗, 0) is a group. �

7. Z∗ = {−1, 1} is a group under multiplication.

8. Q∗ = Q\{0} is a group under multiplication.

9. R∗ = R\{0} is a group under multiplication

10. C∗ = C\{0} is a group under multiplication

Let us recall how it works in the last example. Let Z = a+bi. Then Z = a−bi (complex conjugate)

Z · Z = (a + bi)(a − bi) = a2 − (bi)2 = a2 − (i2)b2 = a2 + b2 = |z|2. Thus Z · Z · 1

|Z|2 = 1 and

Z−1 =1

Z=

1

|Z|2 · Z.

For example, (3 + 4i)−1 =1

25(3 − 4i). Indeed: (3 + 4i) · 1

25(3 − 4i) =

1

25(9 + 16) = 1

In the proof of this theorem we obtained the following important formula

(a ◦ b)−1 = b−1 ◦ a−1

11

Lecture 4

1. Consider one more interesting example.

H = {a + bi + cj + dk|a, b, c, d ∈ R}, i, j, k are some formal symbols. Addition in H is

defined component twice. To define multiplication we have to define first multiplication of the

elements i, j, k. Put i · j = k, j · k = i, k · i = j, i2 = j2 = k2 = −1

i · j = −ji k · j = −kj, i · k = −ki.

Now (a + bi + cj + dk) · (a′ + b′i + c′j + d′k) = −(aa′ + bb′ + cc′ + dd′) + (ab′ + a′b + cd′ −

c′d)i+ (ac′ + a′c+ db′ − b′b)j + (ad′ + a′d+ bc′ − b′c)k.

If α = a + bi+ cj + dk then α = a− bi− cj − dk is a complex conjugate quaternion.

Homework: Show that

2. α · α = a2 + b2 + c2 + d2

3. α · β = β · α

4. (α · β) · γ = α · (β · γ)

By definition: |α|2 = α · α = a2 + b2 + c2 + d2

|α · β|2 = α · β · α · β = α · β · β · α = α · |β|2 · α = |β|2 · α · α = |β|2 · |α|2. Thus

|α · β| = |α| · |β|

We used here that if t ∈ R, α ∈ H then tα = αt. Now α · α = |α|2 so α · α

|α|2 = 1, i.e α−1 =1

|α|2α.

Thus H∗ = H\{0}. The group of quaternions. One more interesting property of quaternions.

If a = 0, then α = bi + cj + dk is called a purely imaginary quaternion. We may consider it as

a vector in R3, a ijk form the orthonormal basis in R3. let β = b′i+ c′j + d′k be another vector in

R3. Then

α · β = −(bb′ + cc′ + dd′) + (cd′ − c′d)i+ (db′ − d′b)j + (bc′ − b′c)k

(α, β) = bb′ + cc′ + dd′ – the dot-product of vectors α, β

12

α× β =

∣∣∣∣∣∣∣

i i kb c db′ c′ d′

∣∣∣∣∣∣∣= (cd′ − c′d)i+ (b′d− bd′)j + (bc′ − b′c)k.

α · β = −(α, β) + α× β

This formula will give a very interesting connection between the quaternion algebra and the vector

algebra.

H∗ is our first example of a non-abelian group.

2. S(B) – the family of all bijections on B is a group under composition. This group is also

non-abelian. In particular Sn gives us an example of finite non-abelian group.

3. Since the matrix multiplication is associative Matn(K) is a semigroup with the unit. Here

K = Q or R or C.

Matn(K)∗ is the group of invertible matrices. Recall that A ∈ Matn(K) has the inverse matrix

A−1. (A ·A−1 = A−1 ·A = In) iff det(A) 6= 0. Thus Matn(K)∗ = {A ∈Matn(K))| detA 6= 0}.

This group has the special notation GLn(K). Recall that

A−1 =1

detA(Aji,

where Aij = (−1)i+jMij and Mij is obtained from detA by renewal of i-th row and j-th

column. If A =(a bc d

)then A−1 =

1

(ad− bc)

(d −b−c a

)A =

−1 2 00 1 −11 0 −1

.

Homework: Find A−1 and check your answer. A lot of other examples of groups are based on

then option of a subgroup.

Definition.

Let (G, ◦) be a group and H ⊆ G. We say that H is a subgroup of G if H is a group under the

same operation ◦.

Proposition 1.

The following statements are equivalent.

13

1. H is a subgroup of G.

2. ∀a, b ∈ H(a ◦ b ∈ H ∧ a−1 ∈ H)

3. ∀a, b ∈ H a ◦ b−1 ∈ H

Proof.

1.⇒2. Let H be a subgroup of G. Then H is closed under ◦, i.e. a, b,∈ H ⇒ a ◦ b ∈ H . The

operation ◦ is associative on H since it is associative on G and H is a part of G. Since H is

a group under ◦ then there exists eh ∈ H such that h ◦ eh = h. Since there exists h−1 ∈ G

we have in G: h−1 ◦ (h ◦ eh) = (h−1 ◦ h) ◦ eh = e ⇒ eh = e, and thus e ∈ H . Again, since H

is a group and a ∈ H then there exist c ∈ H such that a ◦ c = c ◦ a = e. So c is the inverse

element to a in G, and thus c = a−1 (the inverse element is unique).

2.⇒3. If a, b ∈ H then b−1 ∈ H by 2). Then a ◦ b−1 ∈ H since H is closed under ◦.

3.⇒1. Let a ∈ H . Then a ◦ a−1 = e ∈ H (take b = a) by 3. Since e, a ∈ H , by 3. e ◦ a−1 = a−1 ∈ H .

Since a, b ∈ H then a, b−1 ∈ H and by 3. a ◦ (b−1)−1 = a ◦ b ∈ H . So H is closed under ◦,

contains e, an a−1 together with any a ∈ H . ◦ is associative in H . Since it is associative in

G. Thus (H, ◦) is a group.

Usually the inverse element to an element a is denoted by a−1. So a ◦ a−1 = a−1 ◦ a = e

(a−1)−1 = a. Now we come to our main definition.

Definition.

We say that a set G with a binary operation ◦ : G× G → G is a group if the following properties

hold:

1. ◦ is associative

2. ◦ has the neutral element e, which is called simply the unit.

3. each element g ∈ G has the inverse element g−1

14

Examples.

1. (N,+) is not a group 1–2 hold but 3 fail.

Remark. If (G, ◦) satisfies 1., then it is called a semigroup; if 1. and 2.–then a semigroup with

the unit.

2. (N, ·) – a semigroup with the unit

3. (Z,+) – a group: ◦ is the unit, −a is the inverse to a.

Examples of subgroups and non-subgroups.

1. Let (Z,+) be a subgroup of (Q,+), (Q,+) is a subgroup of (R,+), (R,+) is a subgroup of

(C,+), (C,+) is a subgroup of (H,+).

2. (Q∗, ·) is a subgroup of (R∗, ·), (R∗, ·) is a subgroup of (C, ·), (C∗, ·) is a subgroup of (H∗, ·).

3. Let A ⊆ Z, A is the set of all even numbers. Since the sum of two even numbers is an

even number and if n ∈ A thus −n ∈ A, we see that A is a subgroup of Z. More generally,

let n ∈ N and denote by (n) = {k ∈ Z|n|k}. If k ∈ (n), and ℓ =∈ (n), i.e., n|k and n|ℓ,

then by properties of divisibility n|k − ℓ. So k − ℓ ∈ (n) and (n) is a subgroup of (Z,+) by

Proposition 1(3), A = (2).

4. Recall that each rational number r has the canonically noncancellable representation r =a

b,

where ged(a, b) = 1 and b > 0. Fix a prime number p and consider the Q(p) such that r ∈ Q(p)

iff the canonical representation of r is of the forma

pk, k ∈ N.

Leta

pk,b

pℓ∈ Z. Consider

a

pk− b

pℓ.

(a) k = ℓ, thena

pk− b

pℓ=a− b

pkand after cancellation by p in an appropriate degree we

obtain thata− b

pk∈ Q(p).

(b) k > ℓa

pk− b

pℓ=

a

pk− b · pk−ℓ

pk=a− b · pk−ℓ

pk.

15

(c) Since p 6 \a, but p|b·pk−ℓ, we have p 6 \a−bpk−ℓ, i.e. (p, a−b·pk−ℓ) = 1 anda

pk− b

pℓ∈ Q(p).

So Q(p) is a subgroup of (Q,+).

16

Lecture 5

We continue to consider examples of subgroups and non-subgroups.

5. The set B ⊆ Z of all odd numbers is not a subgroup of (Z,+) since it is not closed under the

operation +: the sum of two odd numbers is an even number.

6. Let R∗

+ ={t ∈ R

∣∣∣t > 0}. Since if t > 0, and s > 0 then

t

s= t · s−1 > 0 we have t ∈ R∗

+ ∧ s ∈

R∗

+ → t · s−1 ∈ R∗

+ and thus R∗

+ is a subgroup of (R∗, ·).

x

y

a

b

M7. Consider the set S ={z ∈ C

∣∣∣|z| = 1}. if z1, z2 ∈ S1, then

|z1 · z2| ∈ |z1| · |z2| = | · | = 1, and thus z1 · z2 ∈ S1. If

z = a + bi ∈ S1, i.e. |z| = 1 then z−1 =1

2= 2 = a − bi

|z| =√a2 + b2 = |z| = 1, i.e., z−1 ∈ S1. S1 is a subgroup

of (C∗, ·). You know that a complex number z = a + bi is

identified with the point M(a, b) on R2 or with the vector−→

OM . The length of this vector |−→

OM | = |z|. Then the set S1 can be identified with the set of

all points M such that |OM | = 1, i.e., with the unit circle that is called also the 1-dimensional

sphere.

x

y

S1

1

The set of all points P (a, b, c) in R3 such that |OP | = 1 is a 2-

dimensional sphere that is denoted by S2.

S ={P (a, b, c)|a2 + b2 + c2 = 1

}. By analogy the points P (a, b, c, d) ∈

R4 such that a2 + b2 + c2 + d2 = 1 for a 3-dimensional sphere S3. In

general n-dimensional sphere Sn ={〈x0, x1, . . . , xn〉 ∈ Rn

∣∣∣x20 + x2

1 + . . .+ x2n = 1

}.

The points 〈abcd〉 ∈= R4 can be identified with quaternions a+ bi+ cj + dk. Thus S3 = {α ∈

H||α|+ = 1}.

17

x

y

S1

z

Homework: Show that S3 is a subgroup of (H∗, ·). We see that on

the points of the unit circle we can define an operation such that the

unit circle together with this operation is a group. This operation

is continuous. The same can be done on the 3-dimensional sphere.

Can we define a continuous operation on S2 such that S2 with the

operation is a group? The answer is negative. There is a very difficult theorem due to Adams

in modern mathematics that says that only on S1, S3 and S7 the continuous operations can be

defined that these sphere with these operations are groups.

8. Consider the set Aff(R) ={f : R → R

∣∣∣f(x) = ax+ b, a 6= 0}. If f ∈ Aff(R), f(x) = ax + b, then

f−1(x) =1

ax − b

a. We see that Aff(R) ⊆ S(R), since any element of Aff(R) has the inverse and,

thus, is a bijection. Moreover, we have shown that if f ∈ Aff(R) then f−1 ∈ Aff(R). If g ∈ Aff(R)

and g(x) = cx+ d, then f ◦ g(x) = f (g(x)) = a(cx+ d) + b = acx+ ad+ b. Since a 6= 0, c 6= 0 we

have ac 6= 0 and thus f ◦ g ∈ Aff(R), we proved that Aff(R) is a subgroup of (S(R), 0).

9. Homework. Consider the set G ={f : R → R

∣∣∣f ′(x) > 0, x > 0}. Show that G is a subgroup of

S(R).

10. Consider the setGL2(R)+ ={(

a, bc, c

)∈ GL(R)

∣∣∣a, b, c, d ≥ 0}

It is easy to see that if A, B ∈ GL2(R)+

then A ·B ∈ GL2(R)+ but GL2(R)+ is not a subgroup of GL2(R) since A−1 not always is in GL2(R)+

if A ∈ GL(R)+.

Example:(

1 20 1

)−1

=(

1 −20 1

).

11. Recall that ifA =

a11 a12 . . . a1n

a21 a22 . . . a2n

an1 an2 . . . ann

then the transposed matrixAT =

a11 a12 . . . an1

a121 a22 . . . an2

a1n a2n . . . ann

.

In the course of linear algebra it was proved that

(AB)T = BT · AT .

A matrix A is called symmetric if AT = A. It is easy to see that if A is symmetric then A−1 is also

symmetric:

18

A · A−1 = In = A−1 ·A

(A−1)T · AT = ITn = In = AT (A−1)T , so (A−1)T = (AT )−1

And if AT = A then (A−1)T = A−1. But the set Sn ={A ∈ GLn(K)

∣∣∣AT = A}

is not a subgroup of

GLn(K): the product of two symmetric matrices may not be symmetric:

(1 00 2

)(0 11 0

)=(

0 12 0

)

12. Recall that det(A·B) = detA·detB. Let SLn(K) ={A∣∣∣ detA = 1

}. If A,B ∈ SLn(K) det(A·B) =

detA·detB = 1 A·B ∈ SLn(K) det(A−1) =1

det(A), so A−1 ∈ SLn(K). Thus SLn(K) is a subgroup

of GLn(R).

13. Homework. Is the set{A ∈ GLn(R)

∣∣∣ detA > 0}

a subgroup of GLn(R).

14. Let O(n) ={A ∈ GLn(R)

∣∣∣AT = A−1}. Let us prove that O(n) is a subgroup of GLn(R). Let

A ∈ O(n), B ∈ O(n), i.e. AT = A−1, BT = B−1.

(AB−1)T = (B−1)T · AT = (BT )−1A−1 = (B−1)−1A−1 = BA−1

(AB−1)−1 = (B−1)−1A−1 = BA−1. So (AB−1)T = (AB−1)−1, i.e. AB−1 ∈ O(n). By proposition

1(3) O(n) is a subgroup of GLn(K)

15. SO(n) ={A ∈ O(n)

∣∣∣detA = 1}. So SO(n) = O(n) ∩ SLn(K).

SO(n) is a subgroup of GLn(K) by the following.

Proposition 2. Let{Hα

∣∣∣α ∈ I}

be a family of subgroups G. Then H =⋂

α∈I

Hα is a subgroup of G.

In particular if H1 and H2 are subgroups of G then H = H1 ∩H2 is a subgroup of G.

Proof. Let a, b ∈ H then for any α ∈ I, a, b ∈ Hα, since Hα is a subgroup of Gab−1 ∈ Hα. Since

this holds ∀α ∈ I we have ab−1⋂

α∈I

Hα = H .

19

Lecture 6

Let us describe the matrices from O(2) and SO(2). Let A =(a bc d

)∈ O(2). Then A · AT = I2,

i.e.(a bc d

)(a cb d

)=(

1 00 1

)

a2 + b2 = c2 + d2 = 1 ac + bd = 0

ac + bd = 0 ⇐⇒ d

a= −b

c= λ

Then d = λa, c = −λb, 1 = d2 + c2 = λ2a2 + λ2b2 = λ2(a2 + b2) = λ2. So λ2 = 1, λ = ±1. Thus we

have matrices of two types(a b−b a

)or(a bb −a

), where a2 + b2 = 1.

a2 + b2 = 1 ⇐⇒ a = cosϕ ∧ b = sinϕ. So A =(

cosϕ sinϕ− sinϕ cosϕ

)or(

cosϕ sinϕ− sinϕ cosϕ

)

A ∈ SO(2) detA =∣∣∣A =

(cosϕ sinϕ− sinϕ cosϕ

)

For the complex numbers it is more neutral to consider not transposed matrices, but conjugate

matrices.

Let A ∈ GLn(C), A = (Zij)ni,j=1, Zij ∈ C.

Then the conjugate matrix A∗ = (Zji)ni,j=1, i.e., to obtain A∗ we have to transpose A and to

replace all entries by conjugate. Example:

A =(

2 i1 + i −i

)A∗ =

(2 1 − i−i i

)

As above (A ·B)∗ = B∗ ·A∗. The similar considerations show that U(n) ={A ∈ GLn(C)

∣∣∣A∗ = A−1}

is a subgroup of GLn(C).

SU(n) ={A ∈ U(n)

∣∣∣ det(A) = 1}

is a subgroup of U(n).

It can be shown that A ∈ SU(2) iff A =(a+ bi c+ di−c + di a− bi

), where a2 + b2 + c2 + d2 = 1. This

group has very important applications in theoretical physics.

Further properties of groups:

Proposition. In any group G each of the equations

a× x = b (1)x× a = b (2)

20

has the unique solution for any a, b ∈ G.

Proof. (1) Let us show that x = a−1 · b is a solution of (1) a · x = a(a−1 · b) = (a · a−1) · b = e · b = b,

so we have proved that (1) has a solution. Let us show that this solution is is unique.

Let x0 ∈ G be an arbitrary solution of (1), i.e.

a−1|ax0 = b ⇒ a−1(ax0) = a−1b⇒ x0 = a−1ba−1(ax0) = (a−1a) · x0 = ex0 = x0

Homework 1: Prove the proposition for the equation (2).

Homework 2: Find the solution of the equation ax · b = c in an arbitrary group.

Homework 3:

(a) Find X in S5.

(1 2 3 4 53 1 4 2 5

)·X ·

(1 2 3 4 54 1 3 5 2

)=(

1 2 3 4 55 4 3 2 1

)

(b) Find X in H∗ (2 − 3i+ 4j + k) ·X · (3 − i− j − k) = (1 + i+ j + k).

(c) Find X in GL2(C)(i− 11 − i

)X(

1 + i 2 − 3i−2 − 3i 1 − i

)=(

1 + i 2 − i−2 + 1 −1 + i

)

Proposition 1 shows that a group G satisfies left and right cancellation laws:

ax = ay ⇒ x = yx · b = y · b ⇒ x = y

If a set G, on which the binary operation ◦ is defined, is finite then this operation can be described

by the following table, G = {g1g2 . . . gn}:

◦ g1 g2 . . . gn

g1 gi1 gi2 . . . gin

g2 gj1 gj2 . . . gjn

...gn gk1

gk2. . . gkn

This table is called the Cayley Table. Proposition 1 shows that if G is a group, then each row

〈gi, gi2 . . . gin〉, 〈gj1, gj2 . . . , gjn〉, . . ., 〈gk1

, gk2, . . . , gkn

〉 as well as each column

gi1

gj1...gk1

,

gj2

gj2...gk2

, . . . ,

gin

gjn

...gkn

21

is a permutation of elements g1, g2, . . . gn. An (n×n)-matrix in which each row and in each column

is a permutation of elements {1, 2, . . . , n} is called the Latin Square. So the Cayley Table of a finite

group is a Latin Square, but not any Latin Square is an element of a finite group.

◦ 1 2 31 1 2 32 2 3 13 3 1 2

This is a group.Check the law ofassociativity.

◦ 1 2 31 3 1 22 2 3 13 1 2 3

This is nota group—no unit, andassociativity law fails:(1 ◦ 2) ◦ 3 = 2, 10(2 ◦ 3) = 3

Homework:∗ Construct two Latin squares on {1, 2, 3, 4} such that they both are Cayley Tables of

binary operations with the unit. 1, but one of them is the group and the other is not.

Rings and Fields.

Definition. A set R with two binary operations + – the addition and ◦ – the multiplication is

called a ring if

1. (R,+) is an Abelian group.

2. Multiplication is associative a ◦ (b ◦ c) = (a ◦ b) ◦ c.

3. Multiplication has the unit 1: 1 · a = a · 1 = a

4. The distributive laws hold:

a ◦ (b+ c) = a ◦ b+ a ◦ c(a + b) ◦ c = a ◦ c + b ◦ c

If a ◦ b = b ◦ a, ∀a, b ∈ R, then R is called a commutative ring.

Proposition 2. ∀a ∈ R a ◦ o = o ◦ a = o

Proof.

a ◦ b = a ◦ (b+ o) = a ◦ b+ a ◦ o, i.e. //////a ◦ b = //////a ◦ b +a ◦ o⇒ a ◦ o = o.

Homework: Show that o ◦ a = o.

Examples: Z,Q,R,C,H,Matn(K), (n) ⊆ Z is a ring itself.

A subset that is a ring under the same operations is called a subring.

22

Rings of continuous functions C[a, b], differentiable functions C[a, b] etc. If R is a ring then

R[x] ={a0 + a1x+ . . .+ anX

n∣∣∣ai ∈ R, n ∈ N

}is called the ring of polynomials over R. We will

discuss it later.

Since (R, ·) is a semigroup, R∗ ={a∣∣∣∃a−1(a · a−1 = a−1 · a = 1

}is a group.

Definition. If R∗ = R\{0} i.e. ∀a ∈ R (a 6= 0 → ∃b ∈ R(a · b = b · a = 1)), then R∗ is called a field.

We will consider mainly commutative fields. Examples Q,R,C,H.

23

Lecture 7

Let A be a set, ◦ : A2 ⇒ A – a binary operation on A and ∼ – an equivalence relation or A.

Recollection: A binary relation ∼ on A is a subset ∼⊆ A2. Instead of writing 〈a, b〉 ∈∼ we write

a ∼ b. ∼ is an equivalence relation if:

1. a ∼ a (reflexive)

2. a ∼ b⇒ b ∼ a (symmetric)

3. a ∼ b ∧ b ∼ c⇒ a ∼ c (transitive)

For a ∈ A define a∼ ={b∣∣∣b ∼ a

}– the equivalence class of properties:

1. a ∈ a∼

2. a∼ = b∼ ⇔ a ∼ b

3. a 6= b⇒ a ∩ b = ∅

The family of all equivalence classes A/ ∼={a∣∣∣a ∈ A

}is called the quotient set of A by ∼ or the

factor-set of A by ∼. This is a partition of A: each element of A belongs to one and only one class

from a partition.

a

b

c

A

d

eDefinition 1. We say that an equivalence relation ∼ is a congruence

relation for a binary operation f if a ∼ a′ ∧ b ∼ b′ ⇒ a ◦ b ∼ a′ ◦ b′.

If ∼ is a congruence relation for ◦ then the operation ◦ is defined by

A/ρ by the following rule:

a◦b = a ◦ b.

We have to show that this definition is correct. For this is necessary and sufficient that if a = c,

b = d then a◦b = c◦d, i.e. (a ◦ b)∼ = (c ◦ d)∼. But a = c ∧ b = d⇒ a ∼ c ∧ b ∼ d⇒ a ◦ b ∼ c ◦ d⇒

a ◦ b = c ◦ d. Here we used that ∼ is a congruence relation for ◦.

Example 1. Consider That ∼⊆ R2 such that x ∼ y iff x− y ∈= Z. It is an equivalence relation:

24

1. x− x = 0 ∈ Z ⇒ x ∼ x

2. y − x− (x− y) ⇒ (x− y ∈ Z ⇒ y − x ∈ Z) ⇒ (x ∼ y ⇒ y ∼ x)

3. x ∼ y ∧ y ∼ x⇒ x− y ∈ Z ∧ y = z ∈ Z ⇒ (x− y) + (y − 2) = x− z ∈ Z ⇒ x ∼ 2.

If x ∼ x′ and y ∼ y′ then x− x′ ∈ Z ∧ y − y′ ∈ Z ⇒ (x− x′) + (y − y′) = (x+ y) − (x′ + y′) ∈ Z ⇒

x+ y ∼ x′ + y′. This is not a congruence relation for ·

Indeed: 2.7 ∼ 1.7, 0.3 ∼ 1.3 but 1.7 · 1.3 = 2.21 6∼ 2.7 · 0.3 = .81 Consider now a quotient set

R/ ∼={t∣∣∣ t ∈ R

}. This is not a good form of notation for this set, because t runs over R some

elements t occur not once: 2.7, 0.7,−1.3 is only one element. So we have to find such a subset

A ⊆ R that

1. ∀t ∈ R∃a ∈ A t ∼ a

2. ∀a, b ∈ A(a 6= b⇒ a 6∼ b)

In this case R/ ∼={a∣∣∣ a ∈ A

}, but each class occurs only once. This A is called a full system of

representatives of equivalence classes. Let us find one such full system of representatives. In our

case A = [0, 1). For all t ∈ R t ∼ ({t} – fractional part of t ({t} = t− [t])

If α ∈ [0, 1), β ∈ [0, 1), α 6= β then∣∣∣α− β

∣∣∣ < 1, i.e. α− β 6= Z, i.e. α 6∼ β.

In our case, 0.5 + 0.8 = 1.3 = 0.3.

If a, b ∈ A then we can find c ∈ A such that c ∼ a+ b and this c is unique:

a + b = c

In our case

c = {a + b}

So we may consider + as a binary operation on A−⊕: a + b = a⊕ b. In our case a⊕ b = {a+ b}

– addition module 1.

Example 2. Consider Z, fix a positive n and let x ≡ y(n) iff n∣∣∣x− y. This is an equivalence relation.

We proved it in Math 2800. Show that this is a congruence relation both for + and ·. For + the

25

proof is similar to the previous one (HW). Let us prove it for ◦.

x ≡ x′(modn) ⇒ n∣∣∣x− x′

y ≡ y′(modn) ⇒ n∣∣∣y − y′

xy−x′y′ = xy−xy′+xy′−x′y′ = x(y−y′)+y′(x−x′). n∣∣∣y−y′∧n

∣∣∣x−x′ ⇒ n∣∣∣x(y−y′)+y′(x−x′) =

xy − x′y′ ⇒ xy ≡ x′y′(n). By the division theorem

x = nq + r, 0 ≤, 0 ≤ r < n, i.e. r = 0, 1, . . . , n− 1

Zn = {0, 1, . . . , n − 1} is full system of representatives Z/(n). Indeed x ∼ rem(x, n) ∈ Zn, and if

a 6= b ∈ Zn then 0 <∣∣∣a − b| < n so n

∣∣∣(a − b). Now the quotient operations n+m, n·m can be

represented as the operations on ⊕ ⊙ on Zn

a⊕ b = rem(a + b, n)a⊙ b = rem(a · b, n)

Let us show that (Zn·,⊕,⊙) is a cummutative ring.

1. (Zn⊕) is an albian group.

(a) a⊕ b = rem(a + b, n) = rem(b+ a, n) = b⊕ c

(b) (a⊕ b) = oplusc = a⊕ (b⊕ c)

(a+b)+c = (a+ b)+c = ˜(a+ b) + c = ˜a+ (b+ c) = a+ ˜(b+ c) = a+(b+c)

Homework. The same proof for associativity of multiplication similar for distributivity.

(c) a⊕ 0 = rem(a + 0, n) = rem(an, ) − 0 − neutral

(d) ⊖a = n− a a⊕ n− a = rem(a+ n− a, n) = rem(n, n) = 0.

So we proved that Zn is a ring. Now we are going to describe the group Z∗

n of inveptible elements.

Theorem. a ∈ Z∗

n iff gcd(a1n) = 1.

Proof.

26

1. Let a ∈ Z∗

n ⇒ ∃b ∈ Zna⊙ b = 1

rem(a · b, n) = 1 ⇒ ab = qn + 1. Let gcd(a, n) = d > 1 then d|a ⇒ d|ab, d|n ⇒ d|qn ⇒

d|ab− qn = 1. Contradiction.

2. Let gcd(an, ) = 1 ⇒ ∃uv ∈ Z · au+ vn = 1 ⇒ au− 1 = (−v) · n⇒ n|au− 1 ⇒ a · u ≡ 1(n)

b = rem(u, n) ⇒ b ∈ Zn ⇒ b ≡ u(n) ⇒ a · b ≡ au(n) ⇒ a · b ≡ 1(n) ⇒ rem(a · b, n) = a⊙ b =

1 ⇒ b = a−1.

Examples: 13−1 in (25)

13 · 2 − 25 = 1 ⇒ 13.2 ≡ 1(25) ⇒ 13−1 = 2.

Corollary. If p is a prime number then Zp is a field.

Indeed 1, 2, . . . , p− 1 6 \p so ∀a ∈ Zp, a 6= 0 gcd(a, p) = 1 Z∗

p = Zp\{0}.

Examples:2

3in Z7

2

3= 2 ⊙ 3−1 what is 3−1 : (3 · 5 ≡ 1(7). So 3−1 = 5 ⇒ 2 ⊙ 5 = 3. So

2

3= 3 ⇒ 2 = 3 ⊙ 3 since

3 · 3 = g ≡ 2 (mod 7).

Homework. (a) 7−1 in Z11, (b)3

5in Z13 (c)

Find Z∗

15 and all inverse elements.

27

Lecture 8

The cardinality of Z∗

n is denoted by ϕ(n) and is called the Euler function, i.e.

ϕ(n) = |{k|0 < k < n ∧ gcd(k, n) = 1}|

Example 1 If p is a prime then ϕ(p) = p− 1. Since 1, 2, . . . , p− 1 are co-prime with p.

Example 2 If p is a prime number then ϕ(pα) = pα − pα−1 = pα

(1 − 1

p

).

Proof.

m–is not co-prime with pα iff p|m so 1 · p, 2 · p, 3 · p, . . . , pα−1 · p is the list of all numbers ≤ pα

and non-co-prime with pα. We see they are pα−1 such numbers. Thus there are pα − pα−1 numbers

< pα and co-prime with pα.

Theorem. If gcd(a, b) = 1 then ϕ(a · b) = ϕ(a) · ϕ(b). Proof in the course of Number Theory.

Now if n = pα1

1 pα2

2 . . . pαm

m where pi are distinct primes. Then ϕ(n) = ϕ(pα1

1 )ϕ(pα2

2 ) . . . ϕ(pαm

m ) =

pα1

1

(1 − 1

p1

)pα2

2

(1 − 1

p2

). . . pαm

m

(1 − 1

pm

)= pα1

1 . . . pαm

m

(1 − 1

p1

). . .

(1 − p

pm

).

ϕ(n) = n

(1 − 1

p1

)(1 − 1

p2

). . .

(1 − 1

pm

)

Example. ϕ(12) = ϕ(3 · 4) = ϕ(3) · ϕ(4) = ϕ(3) · ϕ(22) = (3 − 1) · (22 − 21) = 2 · 2 = 4

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Homework. ϕ(20), ϕ(100), ϕ(2004), ϕ(30), ϕ(48). LIST ALL CO-PRIMES.

Euler’s Theorem If gcd(a,m) = 1 then aϕ(m) ≡ 1(modm)

Before we prove this theorem let us prove the following statements:

1. If gcd(am) = 1, then ax ≡ ay(modm) ⇒ x ≡ y.

Proof. Let k = rem(a,m) i.e. a = qm + k. Then gcd(k,m) = 1, indeed if not then

1 < d|m ∧ d|k ⇒ d(a) contradiction. Thus k ∈ Z∗

m. Since a ≡ k(m), ax ≡ kx(m)ay ≡ ky(m)

, hence

kx ≡ ky(m), Let ℓ ∈ Z∗

m, ℓ = k−1, ℓ ◦ kx ≡ x, ℓ ◦ ky ≡ y, ℓ ◦ k = 1, i.e. ℓ ◦ k ≡ 1(m).

kx = ℓky(m) ⇒ x ≡ y(modm).

28

2. Let gcd(u,m) = 1, gcd(v,m) = 1. Then gcd(u · v,m) = 1

Proof. If gcd(u·v,m) ⇒ 1 then ∃ prime p such that p|u·v∧p|m, but since p is prime p|u·v ⇒ p|u∨p|v

If p|u ∧ p|m→ gcd(u,m) > 1

If p|v ∧ p|m→ gcd(v,m) > 1

}contradiction.

Homework. What basic methods are used in this proof?

Proof of the Euler’s Theorem.

Let {k1, k2, . . . kϕ(m)} = Z∗

m. Enough to consider this theorem for a < m. Indeed if a > m then

a = qm + k, 0 < k < m, a ≡ k(modm), aϕ(m) ≡ kϕ(m)(m). gcd(k,m) = 1, proved above. Thus

k ∈ Z∗

m, k ◦ k1, k ◦ k2 . . . k ◦ kϕ(m) is a permtation of elements of Z∗

m since Z∗

m since Z∗

m is group. We

have

k ◦ k1 = ki1(m)k ◦ k2 = ki2(m)

......

...k ◦ km = kiϕ(m)(m)

=⇒

k · k1 = ki1(m)k · k2 = ki2(m)

......

...k · km = kiϕ(m)(m)

=⇒ (multiply all congruences)

kϕ(m) (k1 · k2 . . . km)︸ ︷︷ ︸b

≡ ki1 . . . kiϕ(m)︸ ︷︷ ︸b

(m) and (b,m) = 1 by statement 2.

By statement 1, we may cancel and thus

kϕ(m) ≡ 1(m).

Corollary 1. If p is a prime and p 6 \a then ap−1 ≡ 1(mod p). This corollary is called the small

Fermat’s theorem.

Corollary 2. If a ∈ Z∗

m then k = rem(aϕ(m)−1, m

)= a−1.

Proof. by Euler’s Theorem

aϕ(m)−1 · a ≡ 1(modm)

Since k ≡ aϕ(m)−1(m), k · a ≡ 1(modm), thus k ◦ a = 1 in Z∗

m thus k = a−1.

Examples.

1. Find 17−1 in Z∗

37 ϕ(37) = 37 − 1 = 36, so we have to find rem(1735, 37).

29

17 ≡ 17(37) 17−1 = 24172 = 289 ≡ 30 ≡ (−7)(37) 24 · 17 ≡ 400 = 11 · 37 + 1174 ≡ 49 ≡ 12(37) 24 · 17 ≡ 1(37)178 ≡ 144 ≡ 33 ≡ −4(37) 24 ◦ 17 = 11716 ≡ 16(37)1732 ≡ 256 ≡ 34 ≡ −3(37)1734 ≡ 21(37)1735 ≡ 21 · 17 ≡ 357 ≡ 24(37)

Homework:

1. Determine whether a ∈ Z∗

m and, if yes, then find a−1:

(a) a = 15, m = 64; (b) a = 14, m = 189; (c) a = 19, m = 20

2. Solve the following system of equations in the field Z29:{

14x + 23y = 1826x − 7y = 15

Integral domains.

Let R be a ring, we say that elements a, b ∈ R are divisors of zero iff a, b 6= 1, but a · b = 0.

Example. 3, 4 ∈ Z12 are divisors of zero since 3, 4 6= 0 in Z12, but 3 ⊙ 4 = 0 (3 · 4 = 12 ≡ 0(12)).

More generally if n is a composite number, i.e., n = a · b, 1 < a, b < n then a 6= 0, b 6= 0 in Z12, but

a⊙ b = 0, since a · b = n ≡ 0(n).

Def. If a ring R does not contain any divisors of zero then R is called an integral domain. Obviously

any field F is an integral domain. Indeed, if a, b ∈ F and a 6= 0, b 6= 0 ⇒ a · b 6= 0.

Proof. Assume that a · b = 0. Since a 6= 0∃a−1

a−1|a · b = 0 = a−10 (a · b) = a−1 · 0 = 0

(a−1a)11b = 1 · b = b

Thus b = 0 and we obtain a contradiction. So, if R is a subring of a field F (e.g., Z is a subring of

Q,R, and C) then R is an integral domain. For commutative ring the inverse statement holds also.

Theorem. A commutative ring R is an integral domain iff it is a subring of a field.

Proof. Consider the set M ={〈a, b〉

∣∣∣b 6= O, a, b,∈ R}. It is convenient for me to write

a

binstead of

〈a, b〉. Define operations + and · on M by the formulas

a

b+c

d=ad+ bc

bd;

a

b· cd

=ac

bd.

30

Notice thata

b+c

d∈ M since b · d 6= 0 : b 6= 0, d 6= 0 and R–an integral domain.

a

b· cd∈ M by the

same reason. Consider the equivalence relation ∼ on M such thata

b∼ a′

b′iff ab′ = a′b.

Show that this is an equivalence relation:a

b∼ a

b⇔ ab = ab holds

a

b∼ a′

b′⇒ ab′ = a′b;

a′

b′∼ a

b⇔

a′b = ab′ holdsa

b∼ a′

b′,a′

b′∼ a′′

b′′⇒ a

b∼ a′′

b′′.

1. a′ 6= 0. ab′ = a′b a′b′′ = a′′b′∣∣∣a⇒ a′b′′a = a′′ab′ = a′′a′b⇒ a′(ab′′ − a′′b) = 0.

Since a′ 6= 0 ⇒ ab′′ − a′′b = 0 ⇒ a′′b = ab′′ ⇒ a

b∼ a′′

b′′

2. a′ = 0. ab′ = a′b⇒ ab′ = 0.

Since b′ 6= 0, we have a = 0. Similarly, from the second condition a′′ = 0. So ab′′ = 0 and

a′′b = 0 ⇒ ab′′ = a′′b. Proved

31

Lecture 9

For commutative ring the inverse statement holds also.

Theorem. A commutative ring R is an integral iff it is a surfing of a field.

Proof.

Consider the set M = {〈a, b, 〉|b 6= O, a, b ∈ R}. It is convenient for me to writea

binstead of 〈a, b〉.

Define operations + and · on M by the formulas:

a

b+c

d=ad+ bc

bd;

a

b· cd

=ac

bd.

Notice thata

b+c

d∈ M since b · d 6= 0: b 6= 0, d 6= 0 and R–an integral domain.

a

b· cd∈ M by the

same reason. Consider the equivalence relation ∼ on M such thata

b∼ a′

b′iff ab′ = a′b.

Show that this is an equivalence relation:

a

b∼ a

b⇔ ab = ab — holds

a

b∼ a′

b′⇒ ab′ = a′b;

a′

b′∼ a

b⇔ a′b = ab′ – holds

a

b∼ a′

b′,a′

b′∼ a′′

b′′⇒ a

b∼ a′′

b′′

1. a′ 6= 0, ab′ = a′b a′b′′ = a′′b′|a⇒ a′b′′a = a′′ab′ = a′′a′b ⇒ a′(ab′′ − a′′b) = 0

Since a′ 6= 0 ⇒ ab′′ − a′′b = 0 ⇒ a′′b = ab′′ ⇒ a

b∼ a′′

b′′

2. a′ = 0, ab′ = a′b⇒ ab′ = 0. Since b′ = 0, we have a = 0.

Similarly, from the second condition a′′ = 0. So ab′′ = 0 and a′′b = 0 ⇒ ab′′ = a′′b. Proved.

Let us show that ∼ is a congruence relation for +, ·a

b· a

′

b′;c

d∼ c′

d′. We have to prove

a

b+c

d∼ a′

b′+c′

d′and

a

b· cd∼ a′

b′· c

′

d′.

1. + Given: ab′ = a′b; cd′ = c′d

Prove (ad+ bc)b′d′ = (a′d′ + c′b′)bd.

(ab+bc)b′d′−(a′d′+c′b′)bd = adb′d′+bcb′d′−a′d′bd−c′b′bd = dd′(ab′ − a′b︸ ︷︷ ︸0

)+bb′(cd′ − c′d︸ ︷︷ ︸0

) = 0.

Proved.

32

2. Homework: · Let us show that(M

∼ , +, ·)

is a field.

Obviously + and · are commutative and · is associative. Thus +,· are commutative and · is

associative.

HW. Prove that + is associative in M .

To prove the distributivity law consider:

a

b·(c

d+e

f

)=a

b· cf + ed

df=acf + aed

bdf

and

a

b· cd

+ae

bf=acbf + aebd

b2df

We see that the top and the bottom have common factor that can be cancelled. Why?

acbf + aebd

b2df∼ acf + aed

bdf

Indeed:

(acbf + aebd) bdf = (acf + aed)b2df

Existence of O.

0

b∼ 0

b′, indeed: 0 ·b′ = 0 ·b. Let us show that

(0

b

)∼

= Oc

d+

0

b=cb

db∼ c

d, i.e.

(c

d

)∼

+(

0

b

)∼

=(c

d

)∼

. Proved.

Leta

b=

0

d⇒ ad = 0. Since d 6= 0 ⇒ a = 0. We proved that

(a

b

)∼

= 0 iff a = 0.

Existence of 1.

a

b· dd

=ad

bd∼ a

b⇒(a

b

)∼

·(d

d

)∼

=(a

b

)∼

Thus

(d

d

)∼

= 1

If(a

b

)∼

6= O then ∃(c

d

)(a

b

)∼

·(c

d

)∼

= 1(a

b

)∼

6= 0 ⇒ a 6= 0− ⇒ b

a∈M

a · bb · a =

a · ba · b ∼ 1

1proved.

Now we have to show that R can be considered as a part of M .

Let R ={(

a

1

)∼ ∣∣∣a ∈ R

}.

33

a

1∼ b

1⇔ a = b

a

1+b

1=a+ b

1=a

1· b

1=a · b1

So we see that a ↔(a

1

)∼

is one-to-one correspondence and(a

1

)∼

+

(b

1

)∼

=

(a+ b

1

)∼ (

a

1

)∼(b

1

)∼

=

(ab

1

)∼

so R can be identified with R. Theorem is

proved. The field Q is constructed from R in this way.

Trigonometric Representation of Complex Numbers

y

xϕ

a

rb

z = a + bi r = |z| ϕ = arg(z) a = r cosϕ b = r sinϕ

r =√a2 + b2 tanϕ =

b

a

z = r(cosϕ+ i sinϕ)

If ϕ − Ψ = 2πk, k ∈ Z then r(cosϕ + i sinϕ) = r(cosψ +

i sinψ)

Ex:

1. z = 1 + i =√

2(cos

π

4+ i sin

π

4

)

2. z = 1 − i =√

2(cos

(−π

4

)+ i sin

(−π

4

))=

√2(cos

π

4+ i sin

7π

4

)

3. z = −1 − i =√

2(cos

(5π

4

)+ i sin

5π

4

)

4. z = −1 − i =√

2(cos

3π

4+ i sin

3π

4

)

5. z = 1 = cos 0 + i sin 0

6. z = −1 = cos π + i sin π

7. z = i = cosπ

2+ i sin

π

2

Homework:

1. z = −i

2. z = ±1 ± i√

3

34

3. z = ±√

3 ± i

z ∈ S′ i.e., |z| = 1 iff z = cosϕ+ i sinϕ. Thus arbitrary z ∈ C can be represented in the form

z = r ∈ ǫ

where r ≥ 0, ǫ ∈ S′. This representation is unique for r > 0 when z = 0, r = 0 and ǫ can be taken

arbitrary.

Let ǫ1 = cosϕ+ i sinϕ, ǫ2 = cosψ + i sinψ

e1 · ǫ2 = (cosϕ+ i sinϕ)(cosψ + i sinψ) =

(cosϕ cosψ − sinϕ sinψ) + i(sinϕ cosψ + sinψ cosϕ)

(cosϕ+ i sinϕ)(cosψ + i sinψ) = cos(ϕ+ ψ) + i sin(ϕ+ ψ)

z1 = r1ǫ1, z2 = r2ǫ2 ⇒ z1 · z2 = r1 · r2ǫ1 · ǫ1ǫ1 = cosϕ− i sinϕ = cos(−ϕ) + i sin(−ϕ)

ǫ1ǫ2

= ǫ1 · ǫ2 = cos(ϕ− ψ) + i sin(ϕ− ψ)

z1z2

=r1r2ǫ1 · ǫ2

If z = a+ bi = r(cosϕ+ i sinϕ)ǫ = cosψ + i sinψ

z · ǫ = r [cos(ϕ+ ψ) + i sin(ϕ+ ψ)]y

xϕ

ψa+ bi

Geometrically: to multiply z = a + bi by ǫ = cosψ + i sinψ is

the same as to rotate the vector {a, b} by an angle ψ.

DeMoivre Formula:

[r(cosϕ+ i sinϕ)]n = rn(cosnϕ + i sinnϕ)

This formula can be used for finding the nth roots of a complex number.

Let z = r(cosϕ + i sinϕ). We have to find u ∈ C such that un = z. Let u = s(cosψ + i sinψ).

Then un = sn(cosnψ + i sin nψ) sn = r nψ = ϕ u0n√r(cos

ϕ

n+ i sin

ϕ

n

)if nψ = ϕ + 2π un =

r [cos(ϕ+ 2π) + i sin(ϕ+ 2π)]

35

u1 = n√r(cos

ϕ+ 2π

n+ i sin

ϕ+ 2π

n

)is also an nth root of 2. The same holds for

u2 = n√r(cos

ϕ+ 2π · 2n

+ i sinϕ+ 2π · 2

n

)

u3 = n√r(cos

ϕ+ 2π · 3n

+ i sinϕ+ 2π · 3

n

)

un−1 = n√r

(cos

ϕ+ 2π(n− 1)

n+ i sin

ϕ+ 2π(n− 1)

n

)

un = n√r(cos

ϕ+ 2πn

n+ i sin

ϕ+ 2πn

n

)= n

√r(cos

(ϕ

n+ 2π

)+ i sin

(ϕ

n+ 2π

))= u0.

Thus

n√

2 = n√r

[cos

ϕ+ 2πk

n+ i sin

ϕ+ 2πk

n

]

k = 0, 1, . . . , n− 1

36

Lecture 10

Example 1:

4√−1 = cos

π + 2πk

4+ i sin

π + 2πk

4k = 0, 1, 2, 3

y

x

u1

u2 u3

u0

A Square

u0 = cosπ

4+ i sin

π

4=

√2

2+ i

√2

2=

√2

2(1 + i)

u1 = cos3π

4+ i sin

3π

4= −

√2

2+ i

√2

2=

√2

2(−i+ 1)

u2 = cos5π

4+ i sin

5π

4= −

√2

2− i

√2

2=

√2

2(−1 − i)

u3 = cos7π

4+ i sin

7π

4=

√2

2− i

√2

2=

√2

2(1 − i)

Homework:3√

1 + i,3√i,

5

√√2 + i

Example 2:

6√

1 = cos2πk

6+ i sin

2πk

6, k = 0, 1, 2, 3, 4, 5

u0 = cos 0 + i sin 0 = 1

u1 = cosπ

3+ i sin

π

3=

1

2+

√3

2i

u2 = cos2π

3+ i sin

2π

3= −1

2+

√3

2i

u3 = cosπ + i sin π = −1

u4 = cos4π

3+ i sin

4π

3= −1

2−

√3

2i

u5 = cos5π

3+ i sin

5π

3=

1

2−

√3

2i

π

3 Regular Hexagon

Roots of unit.

We say that ǫ is a root of unit if ǫN = 1.

Theorem 1. Let G be the set of all roots of unit then G is a subgroup of S′.

Proof. ǫ1 ∈ G, ǫ2 ∈ G⇒ ∃N1ǫN1

1 = 1∃N2ǫN2

2 = 1.

(ǫ1ǫ2)N1·N2 =

(ǫN1

1 = 1)N2 ·

(ǫN2 = 1

)N= 1 ⇒ ǫ1ǫ2 ∈ G

(ǫ−11

)N1

= ǫN1

1 = ǫN1

1 = T = 1 ⇒ ǫ−11 ∈ G

37

Theorem 2. UN ={ǫ|ǫN − 1

}=

{cos

2πk

N+ i sin

2πk

N|k = 0, 1, . . .N

}is a subgroup of G.

(ǫ1 · ǫ2)N = ǫN1 · ǫN2 = 1 · 1 = 1(ǫ−11 )N = 1 (see above)

Regular n-gon.

Theorem 3. Let ǫk = cos2πk

N+ i sin

2πk

N. Then

N−1∑

k=0

ǫk = 0

Proof. Let S =N−1∑

k=0

ǫk ǫi · ǫj = ǫi+j(N).

ǫ · S = ǫ1(ǫ0 + ǫ1 + ǫ2 + . . .+ ǫN−1 =

ǫ1 + ǫ2 + ǫ2 + . . .+ ǫN−1 + ǫ0 = S ⇒ ǫ1S = S ⇒

S(1 − ǫ1) = 0 ⇒ S = 0 since ǫ1 6= 1

Second Proof. We have ǫk = ǫk1 k = 0, 1, 2, 3, . . . , N − 1.

S =N−1∑

k=0

ǫk = 1 + ǫ1 + ǫ21 + ǫ21 + . . .+ ǫN−11 =

(ǫ1)N − 1

ǫ1 − 1=

1 − 1

ǫ1 − 1= 0.

(1 + x+ x2 + . . .+ xn−1) =xn − 1

x− 1

Third Proof. Rotate the picture by the angle2n

n. Since each vector turns on this angle then the

unit seems to also turn on2π

n. But this sum does not change!

Return to our Example 2. We see that u3 = −1 is the 6th root of 1. But, indeed, u23 = 1. So it

is the square root of one.

If um = 1 then umd = 1. Thus m-th root is also m-th root for all d.

Definition 1. We say that ǫ is a primitive root of the unit of degree N if ǫN = 1, but for any

0 < m < N ǫm 6= 1. Let again ǫk = cos2πk

N+ i sin

2πk

N.

Theorem 4.

1. ǫk is a primitive N -th root of the unit iff gcd(k,N) = 1.

2. If gcd(k,N) = d, then ǫk isN

d-th root of the unit, and

N

dis the least m such that ǫmk = 1.

38

Proof. We have ǫk = ǫk1.

Statement 1 follows from statement 2 (put d = 1). So we have to prove only statment 2.

LetN

d= m,

k

d= ℓ. Then (ǫk)

N

d = (ǫk1)N

d = ǫkN

d

1 = ǫℓN1 = (ǫN1 )ℓ = 1ℓ = 1. Let (ǫk)s = 1 ⇒

ǫks1 = 1 ⇒ cos

2πks

N+ i sin

2πks

N= 1 ⇒ ks

N= t ∈ Z ⇒ tN = ks|d ⇒ tm = ℓs ⇒ m|ℓs. But

since gcd(N, k) = d ⇒ gcd

(N

d,k

d

)= gcd(m, ℓ) = 1. Indeed: uN + vk = d ⇒ u

N

d+ v

k

d= 1

m|ℓs ∧ gcd(m, ℓ) = 1 ⇒ m|s⇒ m ≤ s.

Algebraic computation of some roots of unit.

3√

1; ǫ0, ǫ1, ǫ2. ǫ0 = 1, ǫ1 = cos2π

3+ i sin

2π

3= −1

2+ i

√3

2, ǫ2 = cos

2π

3+ i sin

4π

3= −1

2− i

√3

2.

For N = 6 we obtained already.

For N = 44√

1 = ±√

2

2± i

√2

2. So we have to find roots of the equation.

Not so trivial is for N = 5. (x5 − 1) = (x− 1)(x4 + x3 + x2 + x+ 1), x4 + x3 + x2 + x+ 1 = 0.

Divide both parts by x2. We come to the equation x2 + x−1 + x1 + x−1 + 1 = 0.

Put z = x+ x−1 then z2 = x2 + 2 + x−2 Thus

x2 + x−2 + x1 + x−1 + 1 = z2 + z − 1

z2 + z − 1 = 0 z1 = −1

2+

√5

2z2 = −1

2−

√5

2.

x+ x−1 = z1 ⇒ x2 − z1x+ 1 = 0 ⇒ x =z1 ± i

√4 − z2

1

2.

x+ x−1 = z2 ⇒ x2 − z2x+ 1 = 0 ⇒ x =z2 ± i

√4 − z2

2

2

We used that z21 < 4 and 22

2 < 4.

x1,2 =

√5 − 1

4± i

√10 + 2

√5

4x3,4 = −

√5 − 1

4± i

√10 − 2

√5

4

ǫ1 = x1 (I quadrant) ǫ2 = x3, ǫ3 = x4 ǫ5 = x2

ǫk = cos2πk

5+ i sin

2πk

5= cos 72◦k + i sin 72◦ · k

cos 72◦ =

√5 − 1

4

Let us show how to obtain the similar expressions for N = 17. We’ll write ǫ for ǫ1, ǫk for ǫk.

ǫ−k = ǫ17−k ǫk + ǫ−k = 2 cos2πk

17

39

16∑

k=1

ǫk = −1 (by theorem 3).∞∑

k=1

(ǫk + ǫ−k) = −1.

40