Three Theorems of Isomorphism

Three Theorems of Isomorphism:

First Theorem of Isomorphism:

             If f: G →G' be an onto homomorphism with kernel K = ker f , then G/K ≈ G' 

In other words , every homomorphic image of a group G is isomorphic to a quotient group of G. 

Proof : 

     Define a map Φ : G/K →G' such that ,
                                      Φ (Ka) = f(a) , a∈G

We Show Φ  is an isomorphism :

       That Φ is well defined follows by 

                          Ka = Kb 

                     ⇒ab⁻¹ ∈K = Ker f

                     ⇒f(ab⁻¹) = e' 

                    ⇒f(a) (f(b))⁻¹ = e'

                    ⇒f(a)  = f(b) 

                    ⇒Φ(Ka) = Φ(Kb) 

By retracing the steps backward , we will prove that Φ is one _one . Again as 

        Φ(KaKb) = Φ(Kab) = f(ab) = f(a)f(b) = Φ(Ka) Φ(Kb)  

       we find Φ is a homomorphism .

To Check That Φ Is Onto :

          Let g' ∈ G' be any element . Since 

      f : G → G' is onto , ∃ g∈ G , such that 

                            f(g) = g' 

         Now Φ(Kg) = f(g) = g' 

Showing thereby that Kg is the required pre image of g under Φ .

Hence Φ is an isomorphism. (proved) 

Remark: 

           The above theorem is also called Fundamental Theorem Of Group Homomorphism. It can also be stated as : 

If f: G →G ' is a homomorphism with K = Ker f , then G/ Ker f = f( G) .

Second Theorem Of Isomorphism :

       Let H and K be two subgroups of a group G , where H is normal in G , then 

               HK/H ≈ K/ H∩K 

Proof : 

    Clearly  H∩K is a normal subgroup of K and as H ⊂ HK ⊂ G , H will be normal in HK.

Let us define a map f: K →HK / H such that 

                                    f(K) = HK 

then as k₁ = k₂ ⇒ Hk₁ = Hk₂ ⇒f(k₁) = f(k₂)

we find f is well defined .

Again f(k₁k₂) = Hk₁k₂ = Hk₁Hk₂ = f(k₁) f(k₂)

It shows f is a homomorphism .

That f is onto is obvious  and thus using Fundamental Theorem of Group Homomorphism , we find 

                HK / H = K / Ker f 

Since k ∈ ker f ⇔ f(k) = H

                        ⇔ Hk = H 

                       ⇔ k ∈ H

                      ⇔ k ∈ H ∩ K  ( k∈ K as Ker f ⊂ K)

We find Ker f = H ∩ K and our theorem is proved . 

Lemma :

            If H and K are two normal subgroups of a group G such that H ⊂ K , then K / H is a normal subgroup of G / K  , and conversely .

Proof :

    K / H is  a non empty subset of G/K , by definition  , For any  Hk₁ , Hk₂ ∈ K / H 

    (Hk₁)(Hk₂)⁻¹ = (Hk₁)(Hk₂⁻¹) = Hk₁k₂⁻¹ ∈ K/H

i.e K/ H is a subgroup . 

Again , for any Hk ∈ K /H and Hg ∈ G/H , we notice  (Hg)⁻¹(Hk)(Hg) = Hg⁻¹HkHg

                                         = Hg⁻¹kg ∈ K/H 

as g∈G , k∈K , is normal in G gives g⁻¹Kg ∈K.

Third Theorem Of Isomorphism :

           If H and K are two normal subgroups of G such that H⊂ K , then 

                       G/K ≈ G/H / K/H

Proof :

           We know the above lemma ensures that K/H is a normal subgroup of G/H and therefore , we can talk of G/H / K/H , define a map f : G/H → G/ K such that

            f(Ha) = Ka , a∈G

f is well defined as  Ha = Hb

                            ⇒ ab⁻¹ ∈H ⊂ K

                            ⇒Ka = Kb

                            ⇒f(Ha) = f(Hb)

f is a homomorphism as

               f ( HaHb) = f(Hab) = Kab = KaKb =                                                                f(Ha) f(Hb)

ontoness of f is obvious ,

Using Fundamental Theorem of group homomorphism we can say

                 G/ K ≈ G/H / Ker f

We claim Ker f = K/H

A member of Ker f will be some member of G/H .   

Now,Ha ∈ Ker f ⇔f(Ha) = k (identity of G/K)

                             ⇔ Ka = K

                             ⇔ a∈ K

                             ⇔Ha ∈ K/H

Hence proves our result . It is also called

   Freshman's Theorem .

           

About Scientist : 

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