**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 ^{-1}hx) = f(x^{ -1})f(h)f(x)

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

f(x)^{-1}f(h)f(x) = f(x ^{-1}hx) ∊ Hˊ

⟹ x^{-1}hx ∊ 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.**