**If G is a group that (ab) ^{n} = a^{n}.b^{n} , for all a, b **ϵ

**G and for three consecutive integers n. Then prove that G is an abelian group.**

**Proof: **

Let a and b be any two elements of G.

Suppose n, n+1 , n + e are three consecutive integers such that

(ab)^{n} = a^{n}.b^{n} …………….(1)

(ab)^{n+1} = a^{n+1}.b^{n+1} ………..(2)

(ab)^{n+2} = a^{n+2}. b^{n+2}……………(3)

Equation (2) can also be written as,

(ab)(ab)^{n }= a.a^{n}.b.b^{n}

a.b.a^{n}.b^{n} = a.a^{n}.b.b^{n}

b.a^{n} = a^{n}.b [by cancellation law]

Equation (3) can also be written as,

(ab)(ab)^{n+1 }= a.a^{n+1}.b.b^{n+1}

a.b.a^{n+1}.b^{n+1} = a.a^{n+1}.b.b^{n+1}

b.a^{n+1} = a^{n+1}.b [by cancellation law]

b.a^{n}.a = a^{n}.a.b

a^{n}.b.a = a^{n}.a.b

⇒ ba = ab [by left cancellation law]

Therefore, G is an abelian group.

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