Question

Define , fundamental theorem of homomorphism of groups

Answer 1

The first group isomorphism theorem,also known as the fundamental homomorphism theorem,states that if is a group homomorphism,then and ,where indicates that is normal subgroup of,denotes the group kernel and indicates that and are isomorphic groups.
Hope it help u

Answer 2

Statement :

Every homomorphic image of a group G is isomomorphic to some quotient group of G .

Proof :

Let f be a group homomorphism from a group G to G'. Then G' is a homomorphic image of the group G .

Let K be the kernel of f , then we shall prove that

G/K ≅ G' .

Now ,

We define a mapping ∅ from G/K to G' such that

∅(Ka) = f(a) ∀ a ∈ G .

First we show that the mapping ∅ is well defined .

Let Ka = Kb , a , b ∈ G

→ ab⁻¹ ∈ K

→ f(ab⁻¹) = e' , where e' is the identity element in G'

→ f(a)f(b⁻¹) = e'

→ f(a)[f(b)]⁻¹ = e'

→ f(a)[f(b)]⁻¹f(b) = e'f(b)

(post multiplying with f(b) both sides)

→ f(a) = f(b)

→ ∅ is well defined .

Now , we show that the mapping ∅ is one-one .

Let Ka , Kb ∈ G/K such that ∅(Ka) = ∅(Kb)

→ f(a) = f(b)

→ f(a)[f(b)]⁻¹ = f(b)[f(b)]⁻¹

(post multiplying with [f(b)]⁻¹ both sides)

→ f(a)f(b⁻¹) = e'

→ f(ab⁻¹) = e'

→ ab⁻¹ ∈ K

→ Ka = Kb

→ ∅ is one-one .

Now , we show ∅ is onto .

Let y ∈ G' be any arbitrary element , then y = f(a) for some a ∈ G . Hence , Ka ∈ G/K .

So , we have ∅(Ka) = f(a) = y

→ ∅ is onto .

Now ,

We show that the mapping ∅ is homomorphism .

Let Ka , Kb ∈ G/K , then

→ ∅(KaKb) = ∅(Kab)

→ ∅(KaKb) = f(ab)

→ ∅(KaKb) = f(a)f(b)

→ ∅(KaKb) = ∅(Ka)∅(Kb)

→ ∅ is homomorphism .

Since ∅ is one-one onto homomorphism , hence it is an isomomorphism from G/K to G' .

Thus , G/K ≅ G' .

Hence proved .