Question #350792

2.6. If G is a group in which (ab)i = aibi for three consecutive integers i for all a, b ∈ G, show that G is abelian.


1
Expert's answer
2022-06-21T06:42:33-0400

Observe that if there exist two consecutive integers nn , n+1n+1 such that (ab)n=anbn(ab)^n=a^nb^n and (ab)n+1=an+1bn+1(ab)^{n+1}=a^{n+1}b^{n+1} for all a,bGa,b\in G, then an+1bn+1=(ab)nab=anbnaba^{n+1}b^{n+1}=(ab)^{n}ab=a^nb^nab. Then we obtain an+1bn+1=anbnaba^{n+1}b^{n+1}=a^nb^nab. Now by multiplying this equation from left by ana^n and right by b1b^{-1} we obtain abn=bnaab^n=b^na.

In our case taking n=in=i and n=i+1n=i+1, we have abi=biaab^i=b^ia and by taking n=i+1n=i+1 and i+2i+2 we have abi+1=bi+1aab^{i+1}=b^{i+1}a.

This shows that abi+1=bi+1a=bbia=babiab^{i+1}=b^{i+1}a=bb^ia=bab^i and now multiplying from right by bib^i we obtain ab=baab=ba. Hence GG is abelian


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