4Quotient groups

IA Groups

4.3 The Isomorphism Theorem

Now we come to the Really Important Theorem

TM

.

Theorem

(The Isomorphism Theorem)

.

Let

f

:

G → H

be a group homomor-

phism with kernel K. Then K C G and G/K

∼

=

im f.

Proof.

We have proved that

K C G

before. We define a group homomorphism

θ : G/K → im f by θ(aK) = f(a).

First check that this is well-defined: If a

1

K = a

2

K, then a

−1

2

a

1

∈ K. So

f(a

2

)

−1

f(a

1

) = f(a

−1

2

a

1

) = e.

So f(a

1

) = f(a

2

) and θ(a

1

K) = θ(a

2

K).

Now we check that it is a group homomorphism:

θ(aKbK) = θ(abK) = f(ab) = f(a)f(b) = θ(aK)θ(bK).

To show that it is injective, suppose

θ

(

aK

) =

θ

(

bK

). Then

f

(

a

) =

f

(

b

). Hence

f(b)

−1

f(a) = e. Hence b

−1

a ∈ K. So aK = bK.

By definition,

θ

is surjective since

im θ

=

im f

. So

θ

gives an isomorphism

G/K

∼

=

im f ≤ H.

If

f

is injective, then the kernel is

{e}

, so

G/K

∼

=

G

and

G

is isomorphic to

a subgroup of

H

. We can think of

f

as an inclusion map. If

f

is surjective, then

im f = H. In this case, G/K

∼

=

H.

Example.

(i)

Take

f

:

GL

n

(

R

)

→ R

∗

with

A 7→ det A

,

ker f

=

SL

N

(

R

).

im f

=

R

∗

as for

all

λ ∈ R

∗

,

det

λ 0 · · · 0

0 1 · · · 0

.

.

.

.

.

.

.

.

.

.

.

.

0 0 0 1

=

λ

. So we know that

GL

n

(

R

)

/SL

n

(

R

)

∼

=

R

∗

.

(ii)

Define

θ

: (

R,

+)

→

(

C∗, ×

) with

r 7→ exp

(2

πir

). This is a group homomor-

phism since

θ

(

r

+

s

) =

exp

(2

πi

(

r

+

s

)) =

exp

(2

πir

)

exp

(2

πis

) =

θ

(

r

)

θ

(

s

).

We know that the kernel is

Z C R

. Clearly the image is the unit circle

(S

1

, ×). So R/Z

∼

=

(S

1

, ×).

(iii) G

= (

Z

∗

p

, ×

) for prime

p 6

= 2. We have

f

:

G → G

with

a 7→ a

2

. This

is a homomorphism since (

ab

)

2

=

a

2

b

2

(

Z

∗

p

is abelian). The kernel is

{±

1

}

=

{

1

, p −

1

}

. We know that

im f

∼

=

G/ ker f

with order

p−1

2

. These

are known as quadratic residues.

Lemma.

Any cyclic group is isomorphic to either

Z

or

Z/

(

nZ

) for some

n ∈ N

.

Proof.

Let

G

=

hci

. Define

f

:

Z → G

with

m 7→ c

m

. This is a group

homomorphism since

c

m

1

+m

2

=

c

m

1

c

m

2

.

f

is surjective since

G

is by definition

all c

m

for all m. We know that ker f C Z. We have three possibilities. Either

(i) ker f = {e}, so F is an isomorphism and G

∼

=

Z; or

(ii) ker f = Z, then G

∼

=

Z/Z = {e} = C

1

; or

(iii) ker f

=

nZ

(since these are the only proper subgroups of

Z

), then

G

∼

=

Z/(nZ).

Definition

(Simple group)

.

A group is simple if it has no non-trivial proper

normal subgroup, i.e. only {e} and G are normal subgroups.

Example. C

p

for prime

p

are simple groups since it has no proper subgroups

at all, let alone normal ones.

A

5

is simple, which we will prove after Chapter 6.

The finite simple groups are the building blocks of all finite groups. All finite

simple groups have been classified (The Atlas of Finite Groups). If we have

K C G

with

K 6

=

G

or

{e}

, then we can “quotient out”

G

into

G/K

. If

G/K

is not simple, repeat. Then we can write

G

as an “inverse quotient” of simple

groups.