Let $n$ be the order of a finite cyclic group $G$ . For every positive divisor $m$ of $n$ , there exists one and only one subgroup of order $m$ of $G$ . The group $G$ has no other subgroups.

Proof. If $g$ is a generator of $G$ and $n = mk$ , then $g^k$ generates the subgroup $\langle g^k\rangle$ , the order of which is equal to the order of $g^k$ , i.e. equal to $m$ . Any subgroup $H$ of $G$ is cyclic (see this entry). If $|H| = m$ , then $H$ must have a generator of order $m$ ; thus apparently $H = \langle g^{\pm k}\rangle = \langle g^k\rangle$ .