$(G,\cdot)$ a group. If $\exists n\in \mathbb{Z} $ such that $(a\cdot b)^{n+i}=a^{n+i}\cdot b^{n+i}$ for $i=0,1,2.$ $\forall a,b \in G$. Prove that $(G,\cdot)$ is Abelian. I'm not sure how to prove this. Why $i$ must be 0 ,1 or 2?
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Given that $(a\cdot b)^{n+2}=a^{n+2}\cdot b^{n+2}$ we can multiply by $a^{-1}$ and $b^{-1}$ on the left and right to get $$(b\cdot a)^{n+1}=a^{n+1}\cdot b^{n+1}=(a\cdot b)^{n+1},$$ the latter equality holding by assumption. Repeating this trick we get $$(b\cdot a)^n=a^n\cdot b^n=(a\cdot b)^n,$$ and dividing these two left hand sides and right hand sides of the above by eachother yields $$b\cdot a = a\cdot b.$$
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