I am reading I.N.Herstein-Topics in Algebra
I am able to solve the following exercise:
If $G$ is a group in which $(a.b)^i=a^i.b^i$ for three consective integers $i$ for all $a,b\in G$, $G$ is abelian.
For solving this problem, I arrived at an intermediate step/expression $(a^{i-1}.b=b.a^{i-1})$ of the proof using two consecutive integers for which the relation holds.
Question:
Show that the above conclusion does not hold if we assume the relation $(a.b)^i=a^i.b^i$ holds only for two consecutive integers.
Here, I think I need to find an example where the expression $a^{i-1}.b=b.a^{i-1}$ holds but $b.a\neq a.b$.
So, I think that a subgroup of $M_n(\mathbb{F})$ should be an example where $N^{i-1}=I\enspace \forall N\in M_n(\mathbb{F})$.
But I am unable to find the above-desired group.
Can anyone help me to find it?
Also, other examples would also be appreciated.
