1

1

Normal Group & Factor Group

Chapter 11

2

Normal GroupDefinition: Let G be a group and H be its subgroup. For any a∈G, define set Ha ={ha: h∈H}. We say that Ha is a right coset of H.

Definition: If for every a∈G , left coset aH and right coset Haare same. Then H is called a normal subgroup of G.

Theorem 1: If group G is commutative, then every subgroup of G is a normal subgroup.

Definition Let G be a group and H be its subgroup. For any a∈G, define set aHa−1 ={aha−1 : h∈H}.

2

3

Theorem 2: Let H be a subgroup of G. Then for any a∈G, setaHa−1 is a subgroup of G.Proof: We only need to show that aHa−1 is closure under multiplication and inverse operations.Let x, y ∈ aHa−1 , then for some h1, h2 ∈H

x = ah1a−1 and y = ah2a−1

Then xy = (ah1a−1)(ah2a−1)

= ah1h2a−1 ∈ aHa−1

and x−1= (ah1a−1)−1 [use formula (xy)−1 = y−1x−1]

= (a−1)−1(h1)−1a−1

= ah1−1a−1 ∈ aHa−1

So aHa−1 is a subgroup of G.Note: aHa−1 is called a conjugate group of H by element a

4

Theorem 3: H is a normal subgroup of G if and only if for everya∈G we have H = aHa−1

Proof: Let H be a normal subgroup of G and a∈G . If x∈ aHa−1 then for some h1∈H, x=ah1a−1

Because aH=Ha , for some h2∈H, we have ah1 =h2a.Then x=ah1a−1 = h2aa−1 = h2 ∈ H. So aHa−1 ⊆ HIf h∈H, similarly for some h3∈H, we have ah3 =ha.then h = h(aa−1)= (ha)a−1= (ah3)a−1 ∈ aHa−1

So H ⊆ aHa−1 which implies H = aHa−1

Conversely, if for any a∈G H = aHa−1 then H = a−1HaIf x∈ aH then for some h1∈H ,

x= ah1 = ah1(a−1a)= (ah1a−1)a ∈ HaIf y∈ Ha then for some h2∈H ,

y= h2a = (aa−1)h2a = a(a−1h2a)∈ aHSo aH =Ha. H is a normal subgroup.

3

5

Theorem 4: Let H be a subgroup of G. If the size of G is double of the size of H, then H is a normal subgroup of G.Note: The ratio of the size of G over the size of H is called the index of G over H, written as [G :H].Proof: Pick any a∈G. If a∈H, then aH=H=HaIf a∉H , then

G = H ∪aH and H ∩ aH= empty setSimilarly

G = H ∪Haand H ∩ Ha= empty setTherefore we must have aH = Ha.So H is a normal subgroup of G.

6

Theorem 5: Let θ be a homomorphism from group G to group H Then the kernel K of θ is a normal subgroup of G.Proof: Pick any a∈G. Let x∈ aKa−1

then for some k∈ K x = aka−1

θ(x) = θ(aka−1)=θ(a)θ(k)θ(a−1)=θ(a)eHθ(a)−1

= θ(a)θ(a)− 1 = eH

Then x∈K So aKa−1 ⊆ KSimilarly

a−1Ka⊆ K � K ⊆ aKa−1

Therefore K = aKa−1

So K is a normal group of G.

4

7

Example

Now look at subgroup

( ) ( ) ( ) ( ) ( ) ( ){ }31 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 31 2 3 2 3 1 3 1 2 1 3 2 3 2 1 1 1 3

, , , ,S =

H1 is a normal subgroup because that index [S3 : H1 ] = 2

We know that permutation group

It has a subgroup

( ) ( ) ( ){ }11 2 3 1 2 3 1 2 31 2 3 2 3 1 3 1 2

, ,H =

( ) ( ){ }21 2 3 1 2 31 2 3 1 3 2

,H =

8

( ) ( ) ( ) ( ){ }( )( ) ( )( ){ }( ) ( ){ }

( ) ( ) ( ){ }( )( )( ) ( )( ){ }( ) ( ){ }

2

2

1 2 33 2 1

1 2 3 1 2 33 1 2 3 1 2

1 2 3 1 2 33 1 2 3 1 2

1 2 33 1 2

1

1 2 3 1 2 31 2 3 1 3 2

1 2 3 1 2 31 2 3 1 3 2

1 2 3 1 2 31 2 3 1 3 2

1 2

2 3 1 2 33 1 2 3 1 2

1 2 3 1 2 33 1

3 1 2 31 2 3 1 3 2

1 2 32

2

1 3

3 1 2

1 2 33 1 2

,

,

,

,

,

,

H

H

=

=

=

=

=

=

They are not same, so H2 is not a normal subgroup of S3

Now calculate the left coset and the right coset.

5

9

Coset OperationLet G = {1, 2, 3, 4, 5, 6}be the multiplicative group of Z7 andH = {1,6}be its subgroup. The all cosets of H are

{1, 6} {2, 5} {3, 4}Define set F = { {1, 6}, {2, 5}, {3, 4} }Now we define the operation * in F as the following:

{1, 6} * {2, 5}={x×y: x∈{1, 6} and y∈{2, 5}}={1×2, 1×5, 6×2, 6×5}= {2, 5, 12, 30}(mod 7) = {2, 5}

This is because 12(mod 7) = 7 and 30(mod 7) = 2Similarly we can get {1, 6}*{3, 4}={3, 4}

{2, 5} *{3, 4}={1, 6} {1, 6} * {1, 6}={1, 6}{2, 5} *{2, 5}={3, 4} {3, 4} *{3, 4}={2, 5}

Because the operation is commutative, no more calculations.

10

So we can draw an operation table

* {1, 6} {2, 5} {3, 4}{1, 6} {1, 6} {2, 5} {3, 4}{2, 5} {2, 5} {3, 4} {1, 6}{3, 4} {3, 4} {1, 6} {3, 4}

From table, we can see that{1, 6}is identity element, {2, 5} and {3, 4}are inverse each other.Therefore F is a group.

If we use notation e={1, 6}and g={2, 5}, then

{3, 4}={2, 5}*{2, 5}= g2

g3=g *g2= {2, 5}*{3, 4}={1, 6}= e

So that F={{1, 6}{2, 5}{3, 4}}={e, g, g2}is a cyclic group.

This group is called the Factor Group G over H denoted as/F G H=

6

11

Theorem 6: Let H be a normal subgroup of G. For any two elements a, b∈G define operation * between cosets aH and bHas (aH) * (bH) ={xy: x∈ aH, y∈ bH}

Then (aH) * (bH) = (ab)HProof: Because a∈aH, then

(aH)*(bH) ⊇ a *(bH)= (ab)HPick any g ∈ (aH)*(bH) then for some h1 , h2 ∈H,

g=ah1bh2= a(h1b)h2

Because bH =HbWe can find h3 ∈H such that h1b=bh3

So g= a(bh3)h2 = (ab)(h3h2 ) ∈ (ab)HTherefore (aH) * (bH) = (ab)HNote: If H is not a normal subgroup of G, then (aH) * (bH) might equal (ab)H

12

Factor GroupTheorem 7: Let H be a normal subgroup of G. Let F be set of all cosets of H. Define operation among cosets as

(aH) * (bH) = (ab)HThen {F , *} is a group. This group is called the Factor Groupof G over H denoted as F=G/H.Proof. The associativity is directly from group G associativityamong its elements.The identity element in {F , *} is HThis is because

(H) * (aH) = (eH) * (aH) = (ea)H= aH(aH) * (H) = (aH) * (eH) = (ae)H= aH

The inverse element of aH in {F , *} is a−1HThis is because (aH) * (a−1H)= (aa−1H)= (eH) =H