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)=ax from (R,+) to (Ro,*), here R is the set of all real numbers and Ro is the set of all real numbers excluding 0. as for any n,m ∈ R f(n+m)= an+m = an * am =f(n)*f(m) , which is homomorphism.
- For cyclic group Z3 = {0, 1, 2} and the Z(set of integers) with addition,
The function f(x)=x mod3 from Z3 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 –
-
group N=({1},*) is the normal subgroup of group G=({1,-1},*) because for g=1 and n= 1, g*n*g-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.