# Prove that if (ab)i=aibi∀a,b∈G(ab)^i = a^ib^i \forall a,b\in G for three consecutive integers ii 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!

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 $a^{i+1} b^{i+1} = (ab)^{i+1} = (ab)(ab)^i = aba^i b^i$ we get $a^i b = b a^i$. The same proof with $i$ replaced by $i+1$ gives
$a^{i+1} b = b a^{i+1}$. Now $a b = a b a^i a^{-i} = a a^i b a^{-i} = a^{i+1} b a^{-i} = b a^{i+1} a^{-i} = b a$.