Answer to Question #350792 in Abstract Algebra for Eliz

Question

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.

Answer 1

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

In our case taking $n=i$ and $n=i+1$, we have $ab^i=b^ia$ and by taking $n=i+1$ and $i+2$ we have $ab^{i+1}=b^{i+1}a$.

This shows that $ab^{i+1}=b^{i+1}a=bb^ia=bab^i$ and now multiplying from right by $b^i$ we obtain $ab=ba$. Hence $G$ is abelian