I’ve been working on this problem listed in Herstein’s

Topics in Algebra(Chapter 2.3, problem 4):If G is a group such that (ab)i=aibi for three consecutive integers i for all a,b∈G, show that G is abelian.

I managed to prove it, but I’m not very happy with my result (I think there’s a neater way to prove this). Anyway, I’m just looking to see if there’s a different approach to this.

My approach:Let j=i+1,k=i+2 for some i∈Z.

Then we have that (ab)i=aibi, (ab)j=ajbj and (ab)k=akbk.

If (ab)k=akbk, then ajbjab=ajabjb.

We cancel on the left and right and we have bja=abj, that is biba=abj.

Multiply both sides by ai on the left and we get aibiba=ajbj, so (ab)iba=(ab)j.

But that is (ab)iba=(ab)iab.Cancelling on the left yields ab=ba, which holds for all a,b∈G, and therefore, G is abelian.

Thanks!

**Answer**

A very similar solution, but maybe slightly shorter: again let i, i+1 and i+2 be the three consecutive integers that work for a and b. From ai+1bi+1=(ab)i+1=(ab)(ab)i=abaibi we get aib=bai. The same proof with i replaced by i+1 gives

ai+1b=bai+1. Now ab=abaia−i=aaiba−i=ai+1ba−i=bai+1a−i=ba.

**Attribution***Source : Link , Question Author : Fernando Martin , Answer Author : Robert Israel*