G is a group in which (ab)^{i} = a^{i}b^{i}for three consecutive integers i
and for every a,b Є G. Show that G is abelian.
Give an example to show that this result does not holds forsemi groups.
Suppose
$$(1)\qquad (ab)^{k-1}=a^{k-1}b^{k-1}$$ $$(2)\qquad (ab)^{k}=a^{k}b^{k},$$ $$(3)\qquad (ab)^{k+1}=a^{k+1}b^{k+1}.$$
Using $(1)$ and $(2):$ $$\displaystyle\begin{aligned}
&(ab)^{k-1}=a^{k-1}b^{k-1}=a^{-1}a^kb^kb^{-1}=a^{-1}(ab)^kb^{-1}\\
&=a^{-1}(ab)(ab)\ldots (ab)b^{-1}=(ba)(ba)\ldots (ba)=(ba)^{k-1}.
\end{aligned}$$ Using $(2)$ and $(3):$ $(ab)^{k}=a^{k}b^{k}$ y $(ab)^{k+1}=a^{k+1}b^{k+1}:$
$$\displaystyle\begin{aligned}
&(ab)^{k}=a^{k}b^{k}=a^{-1}a^{k+1}b^{k+1}b^{-1}=a^{-1}(ab)^{k+1}b^{-1}\\
&=a^{-1}(ab)(ab)\ldots (ab)b^{-1}=(ba)(ba)\ldots (ba)=(ba)^{k}.
\end{aligned}$$ Using $(ab)^{k-1}=(ba)^{k-1}:$ $$\left \{ \begin{matrix} (ab)^k=(ab)^{k-1}(ab)\\ (ba)^k=(ba)^{k-1}(ba)=(ab)^{k-1}(ba)\end{matrix}\right.$$ and by the equality $(ab)^{k}=(ba)^{k},$ we have $(ab)^{k-1}(ab)=(ab)^{k-1}(ba).$
By the cancellation law: $ab=ba$ for all $a,b\in G.$
Please, show some work.Give an example to show that this result does not holds forsemi groups.