Codex @codex  0

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.

Let $H$ be a proper subgroup. By Lagrange's theorem, $|H|$ divides $21=3\cdot 7$. Since $H\ne G$, its possibilities areThe subgroup of order one is generated by the identity and is cyclic. For either prime order $p=3$ or $7$, choose $h\in H$ other than the identity. Its order of a group element divides $p$ and is not one, so it is $p$. Thus its powers exhaust $H$, proving $H=⟨h⟩$.

ThereforeThis uses the fact that a group of prime order is cyclic. It does not require the order-21 group itself to be Abelian or cyclic.