Fundamental Theorem Of Homomorphism Of Group

Fundamental Theorem Of Homomorphism Of Group :

          Let Φ be a homomorphism of G onto G̅ with kernel K .Then G/K ≈ G̅ .

Proof:

              Let G̅ be the homomorphie image of a group G and Φ be the corresponding homomorphie . Then K is normal subgroup of G . 

     To prove that G/K ≈ G̅ .

If   a ∈ G , the Ka ∈ G /K and Φ(a) ∈G̅ 

Let ψ : G/K →G̅ such that ψ(Ka) = Φ(a) ∀ a∈G

 Where Ka is called right coset and Kb is called left coset.

To Show The Mapping ψ is well defined :

      i.e if a,b ∈G and Ka = Kb ,then 

ψ(Ka) = ψ(Kb) 

We have Ka = Kb => ab⁻¹ ∈ K 

                 => Φ(ab⁻¹) = e̅     (identity of G̅)

               =>Φ(a) Φ(b⁻¹) = e̅

             => Φ(a) [Φ(b)]⁻¹ = e̅

            => Φ(a) [Φ(b)]⁻¹Φ(b) = e̅ Φ(b) 

          => Φ(a) e̅ = Φ(b)

          => Φ(a) = Φ(b)

         => ψ(Ka) = ψ(Kb)

        => ψ is well defined

To Prove ψ is one _one :

           We have ψ(Ka) = ψ(Kb)

                         => Φ(a) = Φ(b)

                        => Φ(a) [Φ(b)]⁻¹ = Φ(b) [Φ(b)]⁻¹

                       => Φ(a) Φ(b⁻¹) = e̅

                      => Φ(ab⁻¹) = e̅

                      => ab⁻¹ ∈ K [∵ K is Kernel ]

                     => Ka =Kb

                     => ψ is one _one

To Prove ψ is Onto :

                 Let y be any element of G̅ , then we have y = Φ(a) for some a∈G as Φ is onto  G̅ . Now Ka ∈ G/K and we  have 

                ψ (Ka) = Φ (a) = y

              => ψ is onto 

To Prove ψ is a Homomorphism :

        ψ [ (Ka) (Kb)] = ψ (Kab)

                                = Φ (ab)

                                = Φ(a) Φ(b)

                               = ψ(Ka) ψ(Kb) 

=> ψ is an isomorphism of G / K onto G̅ .

Hence   G/ K ≈ G̅  (proved)