Non-abelian group $G$ satisfying $(a \cdot b)^i=a^i \cdot b^i,$ for two consecutive integers.

Question

Given an example of a non-abelian group $G$ satisfying $(a \cdot b)^i=a^i \cdot b^i, \forall a, b \in G$ for two consecutive integers.

This is question 5 from Herstein Page 35. I have proved that conclusion holds for three positive integers.Thanks.

Answer 1

This is actually possible for any finite nonabelian group. To see this, suppose $G$ is a finite nonabelian group with $|G|=n$. Then the order of any element divides $n$, hence for any element $x\in G$ we have $x^n=e$. Thus for $a,b\in G$ we have $(ab)^n=a^nb^n=e$. Furthermore, $a^{n+1}=a$, $b^{n+1}=b$, $(ab)^{n+1}=ab=a^{n+1}b^{n+1}$.