**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

## (i) an epimorphism if f is onto i.e., if f(G) = Gʹ. In this case we say that Gʹ is an homomorphic image of G.

## (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 K_{f} 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^{-1}xg) = f(g^{-1}) f(x) f(g) = [f(g)]^{-1} eˊ f(g), [∵ f(x) = eˊ]

= [f(g)]^{-1}f(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)**