1Groups and homomorphisms

IA Groups

1.2 Homomorphisms

It is often helpful to study functions between different groups. First, we need to

define what a function is. These definitions should be familiar from IA Numbers

and Sets.

Definition

(Function)

.

Given two sets

X

,

Y

, a function

f

:

X → Y

sends each

x ∈ X

to a particular

f

(

x

)

∈ Y

.

X

is called the domain and

Y

is the co-domain.

Example.

–

Identity function: for any set

X

, 1

X

:

X → X

with 1

X

(

x

) =

x

is a function.

This is also written as id

X

.

–

Inclusion map:

ι

:

Z → Q

:

ι

(

n

) =

n

. Note that this differs from the

identity function as the domain and codomain are different in the inclusion

map.

– f

1

: Z → Z: f

1

(x) = x + 1.

– f

2

: Z → Z: f

2

(x) = 2x.

– f

3

: Z → Z: f

3

(x) = x

2

.

– For g : {0, 1, 2, 3, 4} → {0, 1, 2, 3, 4}, we have:

◦ g

1

(x) = x + 1 if x < 4; g

1

(4) = 4.

◦ g

2

(x) = x + 1 if x < 4; g

1

(4) = 0.

Definition

(Composition of functions)

.

The composition of two functions is a

function you get by applying one after another. In particular, if

f

:

X → Y

and

G : Y → Z, then g ◦ f : X → Z with g ◦ f(x) = g(f (x)).

Example. f

2

◦ f

1

(

x

) = 2

x

+ 2.

f

1

◦ f

2

(

x

) = 2

x

+ 1. Note that function

composition is not commutative.

Definition

(Injective functions)

.

A function

f

is injective if it hits everything

at most once, i.e.

(∀x, y ∈ X) f(x) = f(y) ⇒ x = y.

Definition

(Surjective functions)

.

A function is surjective if it hits everything

at least once, i.e.

(∀y ∈ Y )(∃x ∈ X) f(x) = y.

Definition

(Bijective functions)

.

A function is bijective if it is both injective

and surjective. i.e. it hits everything exactly once. Note that a function has an

inverse iff it is bijective.

Example. ι

and

f

2

are injective but not subjective.

f

3

and

g

1

are neither. 1

X

,

f

1

and g

2

are bijective.

Lemma. The composition of two bijective functions is bijective

When considering sets, functions are allowed to do all sorts of crazy things,

and can send any element to any element without any restrictions. However, we

are currently studying groups, and groups have additional structure on top of

the set of elements. Hence we are not interested in arbitrary functions. Instead,

we are interested in functions that “respect” the group structure. We call these

homomorphisms.

Definition

(Group homomorphism)

.

Let (

G, ∗

) and (

H, ×

) be groups. A

function f : G → H is a group homomorphism iff

(∀g

1

, g

2

∈ G) f(g

1

) × f(g

2

) = f(g

1

∗ g

2

),

Definition

(Group isomorphism)

.

Isomorphisms are bijective homomorphisms.

Two groups are isomorphic if there exists an isomorphism between them. We

write G

∼

=

H.

We will consider two isomorphic groups to be “the same”. For example, when

we say that there is only one group of order 2, it means that any two groups of

order 2 must be isomorphic.

Example.

– f

:

G → H

defined by

f

(

g

) =

e

, where

e

is the identity of

H

, is a

homomorphism.

–

1

G

:

G → G

and

f

2

:

Z →

2

Z

are isomorphisms.

ι

:

Z → Q

and

f

2

:

Z → Z

are homomorphisms.

– exp : (R, +) → (R

+

, ×) with exp(x) = e

x

is an isomorphism.

–

Take (

Z

4

,

+) and

H

: (

{e

ikπ/2

:

k

= 0

,

1

,

2

,

3

}, ×

). Then

f

:

Z

4

→ H

by

f(a) = e

iπa/2

is an isomorphism.

– f

:

GL

2

(

R

)

→ R

∗

with

f

(

A

) =

det

(

A

) is a homomorphism, where

GL

2

(

R

)

is the set of 2 × 2 invertible matrices.

Proposition. Suppose that f : G → H is a homomorphism. Then

(i) Homomorphisms send the identity to the identity, i.e.

f(e

G

) = e

H

(ii) Homomorphisms send inverses to inverses, i.e.

f(a

−1

) = f(a)

−1

(iii) The composite of 2 group homomorphisms is a group homomorphism.

(iv) The inverse of an isomorphism is an isomorphism.

Proof.

(i)

f(e

G

) = f(e

2

G

) = f(e

G

)

2

f(e

G

)

−1

f(e

G

) = f(e

G

)

−1

f(e

G

)

2

f(e

G

) = e

H

(ii)

e

H

= f(e

G

)

= f(aa

−1

)

= f(a)f(a

−1

)

Since inverses are unique, f(a

−1

) = f(a)

−1

.

(iii)

Let

f

:

G

1

→ G

2

and

g

:

G

2

→ G

3

. Then

g

(

f

(

ab

)) =

g

(

f

(

a

)

f

(

b

)) =

g(f(a))g(f (b)).

(iv) Let f : G → H be an isomorphism. Then

f

−1

(ab) = f

−1

n

f

f

−1

(a)

f

f

−1

(b)

o

= f

−1

n

f

f

−1

(a)f

−1

(b)

o

= f

−1

(a)f

−1

(b)

So

f

−1

is a homomorphism. Since it is bijective,

f

−1

is an isomorphism.

Definition

(Image of homomorphism)

.

If

f

:

G → H

is a homomorphism, then

the image of f is

im f = f(G) = {f(g) : g ∈ G}.

Definition (Kernel of homomorphism). The kernel of f, written as

ker f = f

−1

({e

H

}) = {g ∈ G : f(g) = e

H

}.

Proposition.

Both the image and the kernel are subgroups of the respective

groups, i.e. im f ≤ H and ker f ≤ G.

Proof.

Since

e

H

∈ im f

and

e

G

∈ ker f

,

im f

and

ker f

are non-empty. Moreover,

suppose

b

1

, b

2

∈ im f

. Now

∃a

1

, a

2

∈ G

such that

f

(

a

i

) =

b

i

. Then

b

1

b

−1

2

=

f(a

1

)f(a

−1

2

) = f(a

1

a

−1

2

) ∈ im f.

Then consider

b

1

, b

2

∈ ker f

. We have

f

(

b

1

b

−1

2

) =

f

(

b

1

)

f

(

b

2

)

−1

=

e

2

=

e

. So

b

1

b

−1

2

∈ ker f.

Proposition.

Given any homomorphism

f

:

G → H

and any

a ∈ G

, for all

k ∈ ker f, aka

−1

∈ ker f.

This proposition seems rather pointless. However, it is not. All subgroups

that satisfy this property are known as normal subgroups, and normal subgroups

have very important properties. We will postpone the discussion of normal

subgroups to later lectures.

Proof. f(aka

−1

) = f(a)f(k)f (a)

−1

= f(a)ef(a)

−1

= e. So aka

−1

∈ ker f.

Example. Images and kernels for previously defined functions:

(i) For the function that sends everything to e, im f = {e} and ker f = G.

(ii) For the identity function, im 1

G

= G and ker 1

G

= {e}.

(iii) For the inclusion map ι : Z → Q, we have im ι = Z and ker ι = {0}

(iv) For f

2

: Z → Z and f

2

(x) = 2x, we have im f

2

= 2Z and ker f

2

= {0}.

(v)

For

det

:

GL

2

(

R

)

→ R

∗

, we have

im det

=

R

∗

and

ker det

=

{A

:

det A

=

1} = SL

2

(R)

Proposition. For all homomorphisms f : G → H, f is

(i) surjective iff im f = H

(ii) injective iff ker f = {e}

Proof.

(i) By definition.

(ii)

We know that

f

(

e

) =

e

. So if

f

is injective, then by definition

ker f

=

{e}

. If

ker f

=

{e}

, then given

a, b

such that

f

(

a

) =

f

(

b

),

f

(

ab

−1

) =

f(a)f(b)

−1

= e. Thus ab

−1

∈ ker f = {e}. Then ab

−1

= e and a = b.

So far, the definitions of images and kernels seem to be just convenient

terminology to refer to things. However, we will later prove an important

theorem, the first isomorphism theorem, that relates these two objects and

provides deep insights (hopefully).

Before we get to that, we will first study some interesting classes of groups

and develop some necessary theory.