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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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Soham
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    Oldie but goodie. Asked and answered before here. – André Nicolas Jul 27 '12 at 06:58
  • @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 it – Soham Jul 27 '12 at 07:52
  • $(ab)^{i+2} = (ab)^{i+1}ab = a^{i+1}b^{i+1}ab$ 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. – Karolis Juodelė Jul 27 '12 at 08:57

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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.

Kns
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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? – Soham Jul 27 '12 at 10:30