 Solution

Let $|G|=p$ be prime and choose $g\ne e$. Its order divides $p$ by part (iii), and cannot be one. Hence its order is $p$, so its generated subgroup has the same number of elements as the whole group:This proves that a group of prime order is cyclic. In fact every nonidentity element is a generator. No commutativity assumption was needed; it follows from cyclicity.