**Prove that if (ab) ^{-1} = a^{-1}b^{-1}, for all a, b **ϵ

**G 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^{-1}a^{-1} = a^{-1}b^{-1}

Taking the inverse

(b^{-1}a^{-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.

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