Answer to Question #192144 in Abstract Algebra for Arvind kumar

Question

Use the fundamental theorem ofhomomorphism for groups to prove

Answer 1

 zassehaus (buttergly lemma)

statement:- let H,K be subgroups of a group G. let H',K' be normal subgroup of H and K, respectively, then

(i) (H∩K')H' ⊲ (H∩K)H'

(ii) (H'∩K)K' ⊲ (H∩K)K'

(iii)(H∩K)H'/ (H∩K')H' ≅ (H∩K)K'/(H'∩K)K'

proof:- since H,K are subgroup of G, (H∩K) is also a subgroup of G. Moreover, (H∩K)⊂K, so (H∩K) is a subgroup of K also.

No K'is normal (H∩K) = (H∩K)∩K'is a normal subgroup d (H∩K). Similarly, (H'∩K) is normal in (H∩K). 

Consequently product of two normal subgroups,(H∩K') (H'∩K)= L, is a normal subgroup of (H∩K). 

Now, (H∩K)CH,H'<H implies that (H∩K) H' is a subgroup of H. Similarly, (H∩K)K' is a subgroup of K.

Define a map

ϕ: (H∩K)H'→ (H∩K)/L

by

ϕ(xy) = xL,

x∈H∩K,y∈H'

ϕ is well defined, since for x,x₁ ∈H∩K,y,y₁∈H'

if

xy =x₁y₁

x₁^-1x=y₁y^-1∈(H∩K)∩H'= (H'∩K) 

x₁^-1x∈(H'∩K)⊂L

x₁^-1x∈L

xL=x₁L

ϕ (xy) =ϕ (x₁y₁)

 ϕ is clearly onto.

Now for x,x₁ ∈H∩K,y,y'∈H'

 ϕ [(xy)(x₁y₁)]= ϕ [xx₁(x₁^-1yx₁)y₁]= ϕ [xx₁y₂y₁]

                          where y₂ =x₁^-1yx₁∈H'

=xx₁ L 

=xLx₁ L 

=ϕ (xy) ϕ (x₁y₁)

so ϕ is an epimorphism. 

Further, 

xy ∈ker ϕ ⟺ϕ (xy) = L 

         ⟺xL =L 

         ⟺x∈L 

         ⟺x=x₁y₁, with x₁ ∈H∩K', y₁ ∈H'∩K

thus xy ∈ker ϕ if only if 

xy=(x₁y₁)y

   =x₁(y₁y) ∈(H∩K')H'

Therefore, ker ϕ = (H∩K') H'. 

This implies that (H∩K')H'⊲ (H∩K)H'. 

Finally, by Fundamental Homomorphism Theorem, 

we have (H∩K)H'/(H ∩K')H'≅ (H∩K)/L. 

A symmetric argument shows that 

(H'∩K)K'⊲ (H∩K)K', and (H∩K)K'/(H'∩K)K'≅(H∩ K)/L. 

Consequently, (H∩ K) H'/(H∩K')H'≅ (H∩K) K'/(H'∩K) K'.

  

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