An abelian group

Let $G$ be a group such that for (some) three consecutive integer numbers $i$ and for all $a, b \in G$ , it holds that:

$$(ab)^i=a^i b^i$$

Prove that $G$ is abelian.

Solution

Let $i, \; i+1, \; i+2$ be the three consecutive numbers. Then:

• $$\begin{align*}
  \left ( ab \right )^{i+2} &=a^{i+2}b^{i+2} \\
   &= aa^{i+1} b^{i+1}b\\
   &= a\left ( ab \right )^{i+1} b\\
   &= a\left ( ab \right )^i \left ( ab \right ) b\\
   &= aa^i b^i abb
  \end{align*}$$
•  \begin{align*}
  \left ( ab \right )^{i+2} &= \left ( ab \right )^{i+1}\left ( ab \right )\\
   &= a^{i+1} b^{i+1} ab
  \end{align*}

Combining these two results we have that:

\begin{align*}
a a^i b^i abb = a^{i+1}b^{i+1}ab &\Leftrightarrow \left ( a^{i+1} \right )^{-1} a a^i b^i  abb = \left ( a^{i+1} \right )^{-1} a^{i+1} b^{i+1} ab \\
 &\Leftrightarrow  b^i abb = b^{i+1}a b\\
 &\Leftrightarrow  b^i ab = b^{i+1}a\\
 &\Leftrightarrow ab =ba
\end{align*}

Hence $G$ is abelian.