# Group Homomorphisms and Normal Subgroup

**Prerequisite :** Groups

**Homomorphism : **

A mapping (function) *f* from a group *(G,*) to a* group *(G’,+) *is called a group homomorphism (or group morphism) from *G *to *G’ *if, *f*(a*b) = *f *(a)* + f*(b) , for all a, b belong to set G.

**Example –**

- The function f(x)=a
^{x}from (R,+) to (R_{o},*), here R is the set of all real numbers and R_{o}is the set of all real numbers excluding 0. as for any n,m ∈ R f(n+m)= a^{n+m}= a^{n }* a^{m }=f(n)*f(m) , which is homomorphism.

- For cyclic group Z
_{3 }= {0, 1, 2} and the Z(set of integers) with addition,

The function f(x)=x mod3 from Z_{3}to (Z,+) is a group homomorphism.

**NOTE*** – *for a homomorphism

*f:G*→G’

*f*is*a monomorphism*if*f*is injective (one-one).*f*is*Epimorphism,*if*f*is surjective (onto).*f*is Endomorphism if*G = G’.*- G’ is called the homomorphic image of the group G.

**Theorems Related to Homomorphism:****Theorem 1 – **If f is a homomorphism from a group (G,*) to (G’,+) and if e and e’ are their respective identities, then

f(e) = e’. f(n^{-1}) = f(n)^{-1} ,n ∈ G .**Proof –**

1. Let n ∈ G, then n * e = a = e * a

=> f(n * e) = f(a) = f(e * n)

=> f(n) + f(e) = f(n) = f(e) + f(n) , because f is a homomorphism

=> f(e) is the identity in G’

=> f(e) = e’. So it is proved that the image of the identity of G under the group morphism f is the identity of G’.

2. Let n’ be the inverse of n ∈ G, then a * a’ = e = a’ * a

=> f(a * a’) = f(e) = f(a’ * a)

=> f(a) + f(a’) = e’ = f(a’) + f(a), because f is a homomorphism

=> f(a’) = f(a)’.

So it is proved the image of the inverse of any element of G under the group morphism f is the inverse of the image of a in G’.

**Theorem 2 – **If f is a homomorphism of a group (G,*) to a group (G’,+), then

H is a subgroup of G => f(H) is a subgroup of G’ .

H’ is a subgroup of G’ => f'(H’) ={x ∈ G / f(x) ∈ H’} is a subgroup of G.

Proof –

as f(H) is a subgroup of G’ and f(H) ≠ ∅ because e ∈ H => f(e) = e’ ∈ f(H) where e’ is identity in G’ .

If a’, b’ ∈ f(H), then a’ + b’ ∈ f(H)

=> there exists a, b in H such that f(a) = a’ and f(b) = b’

=> a’ + (b’)’ = f(a) + f(b)’ = f(a) + f(b’), because f(b)’ = f(b’)

= f(a * b’), because f is a homomorphism.

But a ∈ H , b’ ∈ H => a * b’ ∈ H

=> f(a * b’) ∈ f(H)

=> f(a * b’) = a’ + (b’)’ ∈ f(H). Therefore, f(H) is a subgroup of G’.

2. as f(H’)’ is a subgroup of G and f(H)’ ≠ ∅ because at least e ∈ f(H’)’

If a, b ∈ f(H’)’ , then a, b ∈ f(H’)’ , then f(a) ∈ H’ and f(b) ∈ H’

=> f(a) + f(b)’ ∈ H, therefore H is a subgroup of G.

=>f(a) + f(b’) ∈ H’ => f(a * b’) ∈ H’ , therefore f is homomorphism.

**Normal Subgroup :**

A subgroup (*N*,*) of a group (*G*,*) is called a normal subgroup of (*G*,*) if, for g ∈ *G* and n ∈ N, we have g*n*g^{-1} ∈ *N*.^{ }

We write it as H<IG.**Example – **

^{-1}= 1*1*(1^{-1}) = 1 ∈ N

Also, for g=-1 and n=1 here g*n*g^{-1}=1 ∈ N.- group N=({1,-1},*) is a normal subgroup of group G=({1,-1,i,-i},*).
- A group containing the identity element {e} of any group G, is normal to G.

**NOTE – **

- If N is a subgroup of G, we can say g*N=N*g ,∀g∈ G where g*N is the left coset and N*g is the right coset of H.
- If all the subgroups of a non-Abelian group are normal, it is called the Hamiltonian
**group**. - For any group G 1.) G itself and 2.) {e} where e is, identity element, are called improper normal subgroups and other than these two are called proper groups.
- A group having no proper normal subgroups is called a simple group.
- The intersection of two normal subgroups is also a normal subgroup.
- Every subgroup of a cyclic group is a normal subgroup.

**Theorems Related to normal subgroup :****Theorem 1 – **All the subgroups of an Abelian group are normal.**Proof –**

Let N be any subgroup of any abelian group G.

if g ∈ G (so we also have g^{-1} as the inverse of g ) and n ∈ N, then, g*n*g^{-1} = (n*g)*g^{-1} , because G is abelian so commutative

=n*(g*g^{-1}) , because G is abelian, so associative

=n*e = n ∈ N

Therefore, g ∈ G , n ∈ N => g*n*g^{-1} ∈ N , so H is a normal subgroup of G.

**Theorem 2 – **A subgroup (N,*) of a group (G,*) is normal if and only if g*N*g^{-1} = N for all g ∈ G.**Proof – **

Let N be a normal subgroup of G. Then , for every g ∈ G, n ∈ N => g*n*g^{-1} ∈ N

=> g*N*g^{-1} ⊆ N ( 1)

Also, for all g ∈ G, g*N*g^{-1} ⊆ N

=> g^{-1}*N*(g^{-1})^{-1} ⊆ N , as g^{-1} ∈ G.

=> g^{-1}*N*g ⊆ N

=> g*(g^{-1}*N*g)*g^{-1} ⊆ g*N*g^{-1}

=> (g*g^{-1})*N*(g*g^{-1}) ⊆ g*N*g^{-1}

=> N ⊆ g*N*g^{-1} (2)

from equation 1 and 2 ,

g*N*g^{-1} =N , ∀ g ∈ G

**Conversely – ** let g*N*g^{-1} =H , ∀ g ∈ G

then g*N*g^{-1} => g*N*g^{-1} ⊆ N

=> g*n*g^{-1} ∈ N , for n ∈ N.

Thus, it is proved that if and only if N is a subset of G, then g*N*g^{-1} =N , ∀ g ∈ G.