Solution
= 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:
$$
\boxed{G=\langle g\rangle\cong C_p.}
$$
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.