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

1 Answers 1

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Here $(ab)^{i+2}=(ab)^{i+1}.(ab)$ $\Rightarrow a^{i+2}b^{i+2}=a^{i+1}b^{i+1}ab$ $\Rightarrow aa^{i+1}.b^{i+1}.b=aa^{i}b^{i} b ab $ $\Rightarrow a^{i+1}b^{i+1}=a^{i}b^{i}ba\ (\because \text{By right and left cancellatin law})$ $\Rightarrow (ab)^{i+1}=(ab)^{i}(ba)$ $\Rightarrow (ab)^{i}(ab)=(ab)^{i}(ba)$ Thus $ab=ba\ (\because \text{By left cancellation law}) $ It follows that $G$ is abelian.

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    Hi Thanks a lot. I also explored along the same lines, before hitting a road block and explored in a different line. My confusion was: On page 28 of I.N.Herstein's Topics in Algebra first para, where he introduces the concept of exponents, he specifically writes $a^k = a \cdot a^{k-1}$ Doesnt this presume order in the way an exponent is decomposed in terms of lower exponents? As I understand in step 3 you have equivalently expanded such a power in both the ways i.e $a^k = aa^{k-1}= a^{k-1}a$ Is it okay?2012-07-27