i.e., Every left coset of H in G is a right coset of H in G.

**Proof:** Let H be a normal subgroup of G.

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

Let a ∈ G and h ∈ H. Then a^{-1} ha ∈ H.

Thus a^{-1} ha = h_{1} for some h_{1} ∈ H.

⟹ ha = ah_{1} ∈ aH

Thus ha ∈ Ha ⟹ ha ∈ aH, and so Ha ⊂ aH

Now let ah ⊂ aH with h ∈ H.

Then ah = ah a^{-1}a = (aha^{-1})a = ((a^{-1})^{-1}ha^{-1})a ∈ Ha,

since (a^{-1})^{-1}ha^{-1}∈ Ha

Thus aH ⊂ Ha.

Hence aH = Ha, ∀ a ∈ G.

**Conversely let aH =Ha , ∀ a ∈ G.**

Let g ∈ G and h ∈ H.

Then g^{-1}hg = g^{-1}(hg).

Now hg ∈ Hg = gH ⟹ hg = g h_{1} for some h_{1} ∈ H.

^{-1}hg = g

^{-1}(gh

_{1}) = (g

^{-1}g)h

_{1}= eh

_{1}= h

_{1}]

This implies that g^{-1}hg ∈ H.