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Home / Modern Abstract Algebra / Homomorphism and Isomorphism

# Homomorphism and Isomorphism

Definition of Homomorphism: Let (G, *) and (Gʹ, 0) be any two groups. Then a mapping f : G → Gʹ is said to be a homomorphism if

f(a * b) = f(a) o f(b), ∀ a, b ∈ G.

Definition: A homomorphism f of a group G to a group Gʹ is called

## (ii) a monomorphism if f is one-one.

Definition of Isomorphism: Let (G, *) and (Gʹ, 0) be any two groups. Then a mapping f : G → Gʹ is said to be a Isomorphism if

f(a * b) = f(a) o f(b), ∀ a, b ∈ G and if f is one-one and onto i.e., bijective.

Definition of automorphism: An isomorphism of a group G with itself is called automorphism of G. In other words every one-one homomorphism of a group G onto itself is called automorphism of G.

Also any homomorphism of a group G to itself is called an endomorphism of G.

### Theorem: If f is a homomorphism of a group G to a group Gˊ, then f(G) the homomorphic image of G, is a subgroup of Gˊ.

Proof: Let e and eˊ be the identity elements of G and Gˊ respectively. Then f(e) = eˊ ∈ f(G) ⊂ Gˊ.

Let ?, ? ∈ f(G). Then ? = f(a) and ? = f(b) for some a, b ∈ G.

Now ??-1 = f(a)f(b)-1 = f(ab-1).

Since G is a group, a, b ∈ G ⟹ ab-1 ∈ G and so ??-1 = f(ab-1) ∈ f(G) .

Thus f(G) is a subgroup of Gˊ. (Proved)

Kernel of a homomorphism: If f is a homomorphism of a group G to a group Gˊ, then kernel of f, denoted by Kerf or Kf is defined by

Kerf = {x ∈ G| f(x) = eˊ, the identity of Gˊ}

= the set of all those elements of G which are mapped on the identity eˊ of G.

Note that Kerf is the pre image of {eˊ}

i.e. Kerf = f-1(eˊ).

#### Theorem: If f is a homomorphism of a group G to group Gˊ, then Kerf is a normal subgroup of G.

Proof: Let e and eˊ be the identity elements of G and Gˊ respectively. We have f(e) = eˊ ⟹ e ∈ Kerf

⟹ Kerf is a non empty subset of G.

Let x, y ∈ Kerf. Then f(xy-1) = f(x) f(y-1) = f(x) [f(y)]-1 = eˊ (eˊ)-1 = eˊ

⟹ xy-1 ∈ Kerf

Thus Kerf is a subgroup of G.

Again for any x ∈ Kerf and for any g ∈ G.

f(g-1xg) = f(g-1) f(x) f(g) = [f(g)]-1 eˊ f(g), [∵ f(x) = eˊ]

= [f(g)]-1f(g) = eˊ

Thus g-1 x g ∈ Kerf. This proves that Kerf is a normal subgroup of G.  (Proved)

### Theorem: A homomorphism f of a group G to a group Gˊ is one-one if and only if Kerf = {e}, e being the identity of G.

Proof: Let f be one-one. Let x ∈ Kerf. Then f(x) = eˊ, the identity of Gˊ. Also f(e) = eˊ, thus f(x) = eˊ = f(e). Since f is one-one, f(x) = f(e)

⟹ x = e. Hence Kerf = {e}.

Conversely let Kerf = {e}

For any x, y ∈ G, f(x) = f(y) ⟹ f(x)[f(y)]-1 = eˊ

⟹ f(x) f(y)-1 = eˊ

⟹ f(xy-1) = eˊ

⟹ xy-1 ∈ Kerf

⟹ xy-1 = e

⟹ x = y

Hence f is one-one.

The proof is complete. (Proved)

## Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the ...