If G is a group in which (ab)^{1} = a^{1} b^{1} for three consecutive integers I for all a, b ∈G, show that G is abelian.

**Solution:** We have

(ab)^{1} = a^{1} b^{1} ——————————-(i)

(ab)^{i+1} = a^{i+1} b^{i+1} ————————-(ii)

(ab)^{i+2} = a^{i+2} b^{i+2} ————————-(iii)

*From (ii), we have*

(ab)^{i+1} = a^{i+1} b^{i+1}

⟹(a^{i}b^{i}) (ab) = a^{i+1} b^{i+1} [by(i)]

⟹ a^{i}(b^{i}a)b = a^{i}(ab^{i})b

⟹ b^{i}a = ab^{i} ——————————-(iv) [by cancellation law]

Similarly from (iii) we have,

(ab)^{i+2} = a^{i+2} b^{i+2}

⟹ (a^{i+1}b^{i+1}) (ab) = a^{i+2} b^{i+2}

⟹ a^{i+1}(b^{i+1}a)b = a^{i+1}(ab^{i+1})b

⟹ b^{i+1}a = ab^{i+1}

⟹ b^{i+1}a = (ab^{i})b = (b^{i}a)b [by (iv)]

⟹ b^{i}(ba) = b^{i} (ab)

⟹ ba = ab [by cancellation law]

⟹ G is abelian. (Proved)