# Prove that if $(ab)^i = a^ib^i \forall a,b\in G$ for three consecutive integers $i$ then G is abelian

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 $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, 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 $i\in \mathbb{Z}$.

Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.

If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.

We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.
Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$.
But that is $(ab)^iba = (ab)^iab$.

Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.

Thanks!

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I think you actually have the most obvious way of doing this –  Jay Kopper May 24 '11 at 3:08
I think this is the only way. What motivates you that there could be an another proof of this. –  user9413 May 24 '11 at 3:09
I don't know if there's another way; I just wanted to check if there was a simpler way because the other exercises from this unit were relatively straightforward and this one took me some time, so maybe I was overlooking a simpler solution. –  Fernando Martin May 24 '11 at 3:15
That's also the way I did it. It would be less confusing, though, if you wrote $i+1$ and $i+2$ in full. –  Yuval Filmus May 24 '11 at 3:24

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