Proof: Let H be a normal subgroup og G.

Then g^{-1}hg ∈ H, ∀ g ∈ G and ∀ h ∈ H.

Since g^{-1} Hg = {g^{-1} hg | g ∈ h and h ∈ H},

g^{-1} Hg ⊂ H ————————(i)

Again for every h ∈ H and for every g ∈ G

We have h = g^{-1}g hg^{-1}g = g^{-1}(ghg^{-1})g ∈ g ^{–1} Hg, since gh g ^{-1}= (g ^{-1})^{-1}hg^{ – 1}∈ H

∴ H ⊂ g^{ -1} Hg ———————–(ii)

*From (i) and (ii) we get*

g ^{-1} Hg = H.

#### Conversely let g^{ -1} Hg = H, ∀ g ∈ G

Then for every g ∈ G and for every h ∈ H,

we have g ^{-1}hg ∈ g ^{-1}Hg and so g^{-1}hg ∈ H.