Question

Use the Fundamental Theorem of Homomorphism for Groups to prove the following 

theorem, which is called the Zassenhaus (Butterfly) Lemma: 

Let H and K be subgroups of a group G and H′ and K′ be normal subgroups of H 

and K, respectively. Then 

 i) H′(H ∩ K′) H′(H ∩ K)

H (H K)

′ ∩ ∩ ′

∩ − ′ ′∩

′ ∩ − ′ ∩ ′

′ ∩ (15)

The situation can be represented by the subgroup diagram below, which explains the

name ‘butterfly’

Answer 1

Let H and K be subgroups of a group G and H′ and K′ be normal subgroups of H

and K, respectively. Then

i) H′(H ∩ K′) H′(H ∩ K)

ii) K′(H′∩ K) K′(H ∩ K)

Fundamental Theorem of Homomorphism for Groups:

Given two groups G and H and a group homomorphism f : G→H, let K be a normal subgroup in G and φ the natural surjective homomorphism G→G/K (where G/K is a quotient group). If K is a subset of ker(f) then there exists a unique homomorphism h:G/K→H such that f = h φ.

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.

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'.