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Prove that if $(ab)^i = a^ib^i \forall a,b\in G$ for three consecutive integers $i$ then G is abelian

How can I prove $G$ is abelian if ($a \cdot b)^i = a^i \cdot b^i $ holds for three consecutive integers, i?

My attempt

(1) $(a \cdot b) ^i = a^i \cdot b^i $

(2) $(a \cdot b) ^{i+1} = a^{i+1} \cdot b^{i+1} $

(3) $(a \cdot b) ^{i+2} = a^{i+2} \cdot b^{i+2} $

To prove $a \cdot b = b\cdot a$

Multiplying eqn 1 and 3, we get $(a \cdot b) ^i \cdot (a \cdot b) ^{i+2} = (a \cdot b) ^{2i+2}$ $ =((a \cdot b) ^{i+1})^2$ $ = (a^{i+1} \cdot b^{i+1})^2$ $ =(a^{i+1} \cdot b^{i+1}) \cdot (a^{i+1} \cdot b^{i+1})$

I am stuck at this

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    Oldie but goodie. Asked and answered before [here.](http://math.stackexchange.com/questions/40996/prove-that-if-abi-aibi-forall-a-b-in-g-for-three-consecutive-integers)2012-07-27
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    @AndréNicolas Hi it might sound very basic but can you help me understand how did we arrive at $a^i \cdot b=b \cdot a^i$ from the step before? Would much appreciate it2012-07-27
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    $(ab)^{i+2} = (ab)^{i+1}ab = a^{i+1}b^{i+1}ab$ and $(ab)^{i+2} = a^{i+2}b^{i+2} = a^{i+1}ab^{i+1}b$. Here you can cancel to get $b^{i+1}a = ab^{i+1}$. Hopefully this was the unclear part.2012-07-27

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