IB Groups, Rings and Modules

1Groups

1.2

Normal subgroups, quotients, homomorphisms, iso-

morphisms

We all (hopefully) recall what the definition of a normal subgroup is. However,

instead of just stating the definition and proving things about it, we can try to

motivate the definition, and see how one could naturally come up with it.

Let

H ≤ G

be a subgroup. The objective is to try to make the collection of

cosets

G/H = {gH : g ∈ G}

into a group.

Before we do that, we quickly come up with a criterion for when two cosets

and

are equal. Notice that if

, then

g ∈ g

. So

·h

for

some

. In other words, (

)

−1

· g

h ∈ H

. So if two elements represent the

same coset, their difference is in

. The argument is also reversible. Hence two

elements g, g

represent the same H-coset if and only if (g

)

−1

g ∈ H.

Suppose we try to make the set

G/H

{gH

g ∈ G}

into a group, by the

obvious formula

H) · (g

H) = g

However, this doesn’t necessarily make sense. If we take a different representative

for the same coset, we want to make sure it gives the same answer.

If g

H = g

H, then we know g

= g

· h for some h ∈ H. So

H) · (g

H) = g

H = g

hH = g

H = (g

H) · (g

H).

So all is good.

What if we change g

? If g

H = g

H, then g

= g

· h for some h ∈ H. So

H) · (g

H) = g

H = g

Now we are stuck. We would really want the equality

H = g

to hold. This requires

)

−1

∈ H.

This is equivalent to

−1

∈ H.

So for

G/H

to actually be a group under this operation, we must have, for any

h ∈ H and g ∈ G, the property g

−1

hg ∈ H to hold.

This is not necessarily true for an arbitrary

. Those nice ones that satisfy

this property are known as normal subgroups.

Definition (Normal subgroup). A subgroup

H ≤ G

is normal if for any

h ∈ H

and g ∈ G, we have g

−1

hg ∈ H. We write H C G.

This allows us to make the following definition:

Definition (Quotient group). If

H C G

is a normal subgroup, then the set

G/H

of left H-cosets forms a group with multiplication

H) · (g

H) = g

with identity eH = H. This is known as the quotient group.

This is indeed a group. Normality was defined such that this is well-defined.

Multiplication is associative since multiplication in

is associative. The inverse

of gH is g

−1

H, and eH is easily seen to be the identity.

So far, we’ve just been looking at groups themselves. We would also like to

know how groups interact with each other. In other words, we want to study

functions between groups. However, we don’t allow arbitrary functions, since

groups have some structure, and we would like the functions to respect the group

structures. These nice functions are known as homomorphisms.

Definition (Homomorphism). If (

G, ·, e

) and (

H, ∗, e

) are groups, a function

φ : G → H is a homomorphism if φ(e

) = e

, and for g, g

∈ G, we have

φ(g · g

) = φ(g) ∗φ(g

If we think carefully,

(

) =

can be derived from the second condition,

but it doesn’t hurt to put it in as well.

Lemma. If φ : G → H is a homomorphism, then

φ(g

−1

) = φ(g)

−1

Proof. We compute φ(g ·g

−1

) in two ways. On the one hand, we have

φ(g · g

−1

) = φ(e) = e.

On the other hand, we have

φ(g · g

−1

) = φ(g) ∗φ(g

−1

By the uniqueness of inverse, we must have

φ(g

−1

) = φ(g)

−1

Given any homomorphism, we can build two groups out of it:

Definition (Kernel). The kernel of a homomorphism φ : G → H is

ker(φ) = {g ∈ G : φ(g) = e}.

Definition (Image). The image of a homomorphism φ : G → H is

im(φ) = {h ∈ H : h = φ(g) for some g ∈ G}.

Lemma. For a homomorphism

G → H

, the kernel

ker

(

) is a normal

subgroup, and the image im(φ) is a subgroup of H.

Proof. There is only one possible way we can prove this.

To see ker(φ) is a subgroup, let g, h ∈ ker φ. Then

φ(g · h

−1

) = φ(g) ∗φ(h)

−1

= e ∗ e

−1

= e.

So gh

−1

∈ ker φ. Also, φ(e) = e. So ker(φ) is non-empty. So it is a subgroup.

To show it is normal, let

g ∈ ker

(

). Let

x ∈ G

. We want to show

−1

gx ∈ ker(φ). We have

φ(x

−1

gx) = φ(x

−1

) ∗ φ(g) ∗ φ(x) = φ(x

−1

) ∗ φ(x) = φ(x

−1

x) = φ(e) = e.

So x

−1

gx ∈ ker(φ). So ker(φ) is normal.

Also, if φ(g), φ(h) ∈ im(φ), then

φ(g) ∗φ(h)

−1

= φ(gh

−1

) ∈ im(φ).

Also, e ∈ im(φ). So im(φ) is non-empty. So im(φ) is a subgroup.

Definition (Isomorphism). An isomorphism is a homomorphism that is also a

bijection.

Definition (Isomorphic group). Two groups

and

are isomorphic if there

is an isomorphism between them. We write G

∼

Usually, we identify two isomorphic groups as being “the same”, and do not

distinguish isomorphic groups.

It is an exercise to show the following:

Lemma. If φ is an isomorphism, then the inverse φ

−1

is also an isomorphism.

When studying groups, it is often helpful to break the group apart into smaller

groups, which are hopefully easier to study. We will have three isomorphism

theorems to do so. These isomorphism theorems tell us what happens when we

take quotients of different things. Then if a miracle happens, we can patch what

we know about the quotients together to get information about the big group.

Even if miracles do not happen, these are useful tools to have.

The first isomorphism relates the kernel to the image.

Theorem (First isomorphism theorem). Let

G → H

be a homomorphism.

Then ker(φ) C G and

ker(φ)

∼

im(φ).

Proof.

We have already proved that

ker

(

) is a normal subgroup. We now

have to construct a homomorphism

G/ ker

(

)

→ im

(

), and prove it is an

isomorphism.

Define our function as follows:

f :

ker(φ)

→ im(φ)

g ker(φ) 7→ φ(g).

We first tackle the obvious problem that this might not be well-defined, since we

are picking a representative for the coset. If

g ker

(

) =

ker

(

), then we know

−1

· g

∈ ker(φ). So φ(g

−1

· g

) = e. So we know

e = φ(g

−1

· g

) = φ(g)

−1

∗ φ(g

Multiplying the whole thing by

(

) gives

(

) =

(

). Hence this function is

well-defined.

Next we show it is a homomorphism. To see

is a homomorphism, we have

f(g ker(φ) · g

ker(φ)) = f(gg

ker(φ))

= φ(gg

)

= φ(g) ∗φ(g

)

= f(g ker(φ)) ∗ f(g

ker(φ)).

So f is a homomorphism. Finally, we show it is a bijection.

To show it is surjective, let

h ∈ im

(

). Then

(

) for some

. So

h = f(g ker(φ)) is in the image of f .

To show injectivity, suppose

(

g ker

(

)) =

(

ker

(

)). So

(

) =

(

). So

(

−1

· g

) =

. Hence

−1

· g

∈ ker

(

), and hence

g ker

(

) =

ker

(

). So

done.

Before we move on to further isomorphism theorems, we see how we can use

these to identify two groups which are not obviously the same.

Example. Consider a homomorphism

C → C \ {

}

given by

z 7→ e

. We

also know that

z+w

= e

This means φ is a homomorphism if we think of it as φ : (C, +) → (C \ {0}, ×).

What is the image of this homomorphism? The existence of

log

shows that

φ is surjective. So im φ = C \ {0}. What about the kernel? It is given by

ker(φ) = {z ∈ C : e

= 1} = 2πiZ,

i.e. the set of all integer multiples of 2πi. The conclusion is that

(C/(2πiZ), +)

∼

(C \ {0}, ×).

The second isomorphism theorem is a slightly more complicated theorem.

Theorem (Second isomorphism theorem). Let

H ≤ G

and

K C G

. Then

HK = {h ·k : h ∈ H, k ∈ K} is a subgroup of G, and H ∩ K C H. Moreover,

∼

H ∩ K

Proof. Let hk, h

∈ HK. Then

(hk)

−1

= h

−1

= (h

−1

)(hk

−1

The first term is in

, while the second term is

−1

∈ K

conjugated by

, which also has to be in

be normality. So this is something in

times

something in K, and hence in HK. HK also contains e, and is hence a group.

To show

H ∩K C H

, consider

x ∈ H ∩K

and

h ∈ H

. Consider

−1

. Since

x ∈ K

, the normality of

implies

−1

xh ∈ K

. Also, since

x, h ∈ H

, closure

implies h

−1

xh ∈ H. So h

−1

xh ∈ H ∩ K. So H ∩ K C H.

Now we can start to prove the second isomorphism theorem. To do so, we

apply the first isomorphism theorem to it. Define

φ : H → G/K

h 7→ hK

This is easily seen to be a homomorphism. We apply the first isomorphism

theorem to this homomorphism. The image is all

-cosets represented by

something in H, i.e.

im(φ) =

Then the kernel of φ is

ker(φ) = {h ∈ H : hK = eK} = {h ∈ H : h ∈ K} = H ∩ K.

So the first isomorphism theorem says

H ∩ K

∼

Notice we did more work than we really had to. We could have started by

writing down

and checked it is a homomorphism. Then since

H ∩ K

is its

kernel, it has to be a normal subgroup.

Before we move on to the third isomorphism theorem, we notice that if

K C G

, then there is a bijection between subgroups of

G/K

and subgroups of

containing K, given by

{subgroups of G/K} ←→ {subgroups of G which contain K}

X ≤

−→ {g ∈ G : gK ∈ X}

≤

←− K C L ≤ G.

This specializes to the bijection of normal subgroups:

{normal subgroups of G/K} ←→ {normal subgroups of G which contain K}

using the same bijection.

It is an elementary exercise to show that these are inverses of each other.

This correspondence will be useful in later times.

Theorem (Third isomorphism theorem). Let

K ≤ L ≤ G

be normal subgroups

of G. Then



∼

Proof. Define the homomorphism

φ : G/K → G/L

gK 7→ gL

As always, we have to check this is well-defined. If

, then

−1

∈

K ⊆ L. So gL = g

L. This is also a homomorphism since

φ(gK · g

K) = φ(gg

K) = gg

L = (gL) ·(g

L) = φ(gK) · φ(g

K).

This clearly is surjective, since any coset

is the image

(

). So the image

is G/L. The kernel is then

ker(φ) = {gK : gL = L} = {gK : g ∈ L} =

So the conclusion follows by the first isomorphism theorem.

The general idea of these theorems is to take a group, find a normal subgroup,

and then quotient it out. Then hopefully the normal subgroup and the quotient

group will be simpler. However, this doesn’t always work.

Definition (Simple group). A (non-trivial) group

is simple if it has no normal

subgroups except {e} and G.

In general, simple groups are complicated. However, if we only look at abelian

groups, then life is simpler. Note that by commutativity, the normality condition

is always trivially satisfied. So any subgroup is normal. Hence an abelian group

can be simple only if it has no non-trivial subgroups at all.

Lemma. An abelian group is simple if and only if it is isomorphic to the cyclic

group C

for some prime number p.

Proof.

By Lagrange’s theorem, any subgroup of

has order dividing

Hence if

is prime, then it has no such divisors, and any subgroup must have

order 1 or

, i.e. it is either

{e}

itself. Hence in particular any normal

subgroup must be {e} or C

. So it is simple.

Now suppose

is abelian and simple. Let

e 6

g ∈ G

be a non-trivial element,

and consider

{··· , g

−2

, g

−1

, e, g, g

, ···}

. Since

is abelian, conjugation

does nothing, and every subgroup is normal. So

is a normal subgroup. As

is simple,

{e}

. Since it contains

g 6

, it is non-trivial. So we

must have H = G. So G is cyclic.

is infinite cyclic, then it is isomorphic to

. But

is not simple, since

2Z C Z. So G is a finite cyclic group, i.e. G

∼

for some finite m.

n | m

, then

m/n

generates a subgroup of

of order

. So this is a normal

subgroup. Therefore

must be

or 1. Hence

cannot be simple unless

has

no divisors except 1 and m, i.e. m is a prime.

One reason why simple groups are important is the following:

Theorem. Let G be any finite group. Then there are subgroups

G = H

B H

B ··· B H

= {e}.

such that H

i+1

is simple.

Note that here we only claim that

i+1

is normal in

. This does not say

that, say, H

is a normal subgroup of H

Proof. If G is simple, let H

= {e}. Then we are done.

is not simple, let

be a maximal proper normal subgroup of

. We

now claim that G/H

is simple.

G/H

is not simple, it contains a proper non-trivial normal subgroup

L C G/H

such that

L 6

{e}, G/H

. However, there is a correspondence

between normal subgroups of

G/H

and normal subgroups of

containing

must be

K/H

for some

K C G

such that

K ≥ H

. Moreover, since

non-trivial and not

G/H

, we know

is not

. So

is a larger normal

subgroup. Contradiction.

So we have found an

C G

such that

G/H

is simple. Iterating this process

gives the desired result. Note that this process eventually stops, as

i+1

< H

, and hence |H

i+1

| < |H

|, and all these numbers are finite.