Prove that if (ab)^-1 = a^-1b^-1, for all a, b ϵG then G is abelian

Prove that if (ab)^-1 = a^-1b^-1, for all a, b ϵG then G is abelian

Proof:

Given (ab)^-1 = a^-1.b^-1…………………..(1), for all a, b ϵG

Since G is a group, we have (ab)^-1 = b^-1.a^-1 ………….(2), for all a, b ϵ G

From (1) and (2), b^-1a^-1 = a^-1b^-1

Taking the inverse

(b^-1a^-1)^-1 = (a^-1 b^-1)^-1

(b^-1)^-1(a^-1)^-1 = (a^‑1)^-1(b^-1)^-1 [since (a^-1)^-1 = a, for all a ϵG]

⇒ba = ab

⸫ G is an abelian group.