1Groups and homomorphisms

IA Groups

1.1 Groups

Definition

(Binary operation)

.

A (binary) operation is a way of combining two

elements to get a new element. Formally, it is a map ∗ : A × A → A.

Definition

(Group)

.

A group is a set

G

with a binary operation

∗

satisfying

the following axioms:

1. There is some e ∈ G such that for all a, we have

a ∗ e = e ∗ a = a. (identity)

2. For all a ∈ G, there is some a

−1

∈ G such that

a ∗ a

−1

= a

−1

∗ a = e. (inverse)

3. For all a, b, c ∈ G, we have

(a ∗ b) ∗ c = a ∗ (b ∗ c). (associativity)

Definition

(Order of group)

.

The order of the group, denoted by

|G|

, is the

number of elements in G. A group is a finite group if the order is finite.

Note that technically, the inverse axiom makes no sense, since we have not

specified what

e

is. Even if we take it to be the

e

given by the identity axiom,

the identity axiom only states there is some

e

that satisfies that property, but

there could be many! We don’t know which one

a ∗ a

−1

is supposed to be equal

to! So we should technically take that to mean there is some

a

−1

such that

a ∗ a

−1

and

a

−1

∗ a

satisfy the identity axiom. Of course, we will soon show that

identities are indeed unique, and we will happily talk about “the” identity.

Some people put a zeroth axiom called “closure”:

0. For all a, b ∈ G, we have a ∗ b ∈ G. (closure)

Technically speaking, this axiom also makes no sense — when we say

∗

is a

binary operation, by definition,

a ∗ b

must be a member of

G

. However, in

practice, we often have to check that this axiom actually holds. For example, if

we let G be the set of all matrices of the form

1 x y

0 1 z

0 0 1

under matrix multiplication, we will have to check that the product of two such

matrices is indeed a matrix of this form. Officially, we are checking that the

binary operation is a well-defined operation on G.

It is important to know that it is generally not true that

a∗b

=

b∗a

. There is

no a priori reason why this should be true. For example, if we are considering the

symmetries of a triangle, rotating and then reflecting is different from reflecting

and then rotating.

However, for some groups, this happens to be true. We call such groups

abelian groups.

Definition (Abelian group). A group is abelian if it satisfies

4. (∀a, b ∈ G) a ∗ b = b ∗ a. (commutativity)

If it is clear from context, we are lazy and leave out the operation

∗

, and

write

a ∗ b

as

ab

. We also write

a

2

=

aa

,

a

n

=

aaa · · · a

| {z }

n copies

,

a

0

=

e

,

a

−n

= (

a

−1

)

n

etc.

Example. The following are abelian groups:

(i) Z with +

(ii) Q with +

(iii) Z

n

(integers mod n) with +

n

(iv) Q

∗

with ×

(v) {−1, 1} with ×

The following are non-abelian groups:

(vi)

Symmetries of an equilateral triangle (or any

n

-gon) with composition.

(D

2n

)

(vii) 2 × 2 invertible matrices with matrix multiplication (GL

2

(R))

(viii) Symmetry groups of 3D objects

Recall that the first group axiom requires that there exists an identity element,

which we shall call

e

. Then the second requires that for each

a

, there is an inverse

a

−1

such that

a

−1

a

=

e

. This only makes sense if there is only one identity

e

, or

else which identity should a

−1

a be equal to?

We shall now show that there can only be one identity. It turns out that the

inverses are also unique. So we will talk about the identity and the inverse.

Proposition. Let (G, ∗) be a group. Then

(i) The identity is unique.

(ii) Inverses are unique.

Proof.

(i)

Suppose

e

and

e

0

are identities. Then we have

ee

0

=

e

0

, treating

e

as an

inverse, and ee

0

= e, treating e

0

as an inverse. Thus e = e

0

.

(ii)

Suppose

a

−1

and

b

both satisfy the inverse axiom for some

a ∈ G

. Then

b = be = b(aa

−1

) = (ba)a

−1

= ea

−1

= a

−1

. Thus b = a

−1

.

Proposition. Let (G, ∗) be a group and a, b ∈ G. Then

(i) (a

−1

)

−1

= a

(ii) (ab)

−1

= b

−1

a

−1

Proof.

(i) Given a

−1

, both a and (a

−1

)

−1

satisfy

xa

−1

= a

−1

x = e.

By uniqueness of inverses, (a

−1

)

−1

= a.

(ii) We have

(ab)(b

−1

a

−1

) = a(bb

−1

)a

−1

= aea

−1

= aa

−1

= e

Similarly, (

b

−1

a

−1

)

ab

=

e

. So

b

−1

a

−1

is an inverse of

ab

. By the uniqueness

of inverses, (ab)

−1

= b

−1

a

−1

.

Sometimes if we have a group

G

, we might want to discard some of the

elements. For example if

G

is the group of all symmetries of a triangle, we might

one day decide that we hate reflections because they reverse orientation. So

we only pick the rotations in

G

and form a new, smaller group. We call this a

subgroup of G.

Definition

(Subgroup)

.

A

H

is a subgroup of

G

, written

H ≤ G

, if

H ⊆ G

and

H with the restricted operation ∗ from G is also a group.

Example.

– (Z, +) ≤ (Q, +) ≤ (R, +) ≤ (C, +)

– (e, ∗) ≤ (G, ∗) (trivial subgroup)

– G ≤ G

– ({±1}, ×) ≤ (Q

∗

, ×)

According to the definition, to prove that

H

is a subgroup of

G

, we need to

make sure

H

satisfies all group axioms. However, this is often tedious. Instead,

there are some simplified criteria to decide whether H is a subgroup.

Lemma (Subgroup criteria I). Let (G, ∗) be a group and H ⊆ G. H ≤ G iff

(i) e ∈ H

(ii) (∀a, b ∈ H) ab ∈ H

(iii) (∀a ∈ H) a

−1

∈ H

Proof. The group axioms are satisfied as follows:

0. Closure: (ii)

1.

Identity: (i). Note that

H

and

G

must have the same identity. Suppose that

e

H

and

e

G

are the identities of

H

and

G

respectively. Then

e

H

e

H

=

e

H

.

Now

e

H

has an inverse in

G

. Thus we have

e

H

e

H

e

−1

H

=

e

H

e

−1

H

. So

e

H

e

G

= e

G

. Thus e

H

= e

G

.

2. Inverse: (iii)

3. Associativity: inherited from G.

Humans are lazy, and the test above is still too complicated. We thus come

up with an even simpler test:

Lemma (Subgroup criteria II). A subset H ⊆ G is a subgroup of G iff:

(I) H is non-empty

(II) (∀a, b ∈ H) ab

−1

∈ H

Proof. (I) and (II) follow trivially from (i), (ii) and (iii).

To prove that (I) and (II) imply (i), (ii) and (iii), we have

(i) H must contain at least one element a. Then aa

−1

= e ∈ H.

(iii) ea

−1

= a

−1

∈ H.

(ii) a(b

−1

)

−1

= ab ∈ H.

Proposition.

The subgroups of (

Z,

+) are exactly

nZ

, for

n ∈ N

(

nZ

is the

integer multiples of n).

Proof.

Firstly, it is trivial to show that for any

n ∈ N

,

nZ

is a subgroup. Now

show that any subgroup must be in the form nZ.

Let

H ≤ Z

. We know 0

∈ H

. If there are no other elements in

H

, then

H = 0Z. Otherwise, pick the smallest positive integer n in H. Then H = nZ.

Otherwise, suppose (

∃a ∈ H

)

n - a

. Let

a

=

pn

+

q

, where 0

< q < n

. Since

a − pn ∈ H

,

q ∈ H

. Yet

q < n

but

n

is the smallest member of

H

. Contradiction.

So every

a ∈ H

is divisible by

n

. Also, by closure, all multiples of

n

must be in

H. So H = nZ.