= Solution
Let $H$ be a proper <subgroup>. By <Lagrange's theorem>, $|H|$ divides $21=3\cdot7$. Since $H\ne G$, its possibilities are
$$
|H|\in\{1,3,7\}.
$$
The 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 $H=\langle h\rangle$.
Therefore
$$
\boxed{\text{Every proper subgroup of }G\text{ is cyclic, of order }1,3\text{ or }7.}
$$
This 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.
Back to article page
