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Home / Modern Abstract Algebra / Correspondence theorem on homomorphism

# Correspondence theorem on homomorphism

Let f be a homomorphism of a group G onto a group Gʹ with Kernel K(=Kerf).

Let Hʹ be a subgroup of Gʹ and H = {x ∊G | f(x) ∊ Hˊ} = f -1(Hʹ). Then H is a subgroup of G containing K.

If Hʹ is a normal subgroup of Gʹ then H is normal in G. Further if L is any other subgroup of G containing K such that f(L) = Lʹ then L = H.

Proof: (i) Let e and eʹ be the identities in G and Gʹ respectively. Hʹ being a subgroup of Gʹ, eʹ ∊ Hˊ.

Also f(e) = eˊ and so e ∊ H. Thus H is a non empty subset of G. Let x, y ∊ H.

Then f(x), f(y) ∊ Hˊ. Since Hˊ is a subgroup of Gˊ, f(x) f(y)-1 ∊ Hˊ

⟹f(x)f(y)-1 ∊ Hˊ.

Thus f(xy -1) ∊ Hˊ ⟹ xy -1∊ H.

Since x, y ∊H ⟹ xy -1∊ H, H is a subgroup of G.

Now K = Kerf = {x ∊ G | f(x) = eˊ ∊ Hˊ} ⟹ K ⊂ H.

# Thus H is a subgroup of G containing K.

(ii) Let Hˊ be a normal subgroup of Gˊ.

Then we are to show that H is a normal subgroup of G.

Let h ∊ H and x ∊ G.

Then f(x -1hx) = f(x -1)f(h)f(x)

Now f(h) ∊ Hˊ, f(x) ∊ Gˊ and since Hˊ is a normal in Gˊ.

f(x)-1f(h)f(x) = f(x -1hx) ∊ Hˊ

⟹ x-1hx ∊ H,

## Thus H is normal in G.

(iii) Let L be any subgroup of G containing K such that f(L) = Hˊ.

We show that L = H.

Every ℓ ∊ L ⟹ f(ℓ )∊ f(L) = Hˊ ⟹ ℓ ∊ H.

Hence L⊂ H.

Conversely let h ∊ H.

Then f(h) ∊ Hˊ = f(L)

⟹ f(h) = f(ℓ) for some ℓ ∊ L.

⟹ f(h) f(ℓ)-1 = e′

⟹ f(h) f(ℓ-1) = e′

⟹f(h ℓ-1) = e′

⟹ h ℓ-1 ∊ K

⟹ h ℓ-1 ∊ L since K ⊂ L.

⟹ h ℓ-1 = ℓ′ for some ℓ′ ∊ L.

⟹ h = ℓ′ ℓ ∊ L, since L being a subgroup ℓ′, ℓ ∊ L.

⟹ ℓ′ ℓ ∊ L

Thus H ⊂ L.

### Hence H = L.

The proof is complete.

## Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the ...