Proof. Let $g$ be a generator of $G$ . Then $|g|=|\langle g \rangle |=|G|$ . Let $z \in {\mathbb Z}$ such that $nz=|G|=|g|$ . Consider $\langle g^z \rangle$ . Since $g \in G$ , then $g^z \in G$ . Thus, $\langle g^z \rangle \le G$ . Since $\displaystyle |\langle g^z \rangle |=|g^z|=\frac {|g|}{\gcd(z,|g|)}=\frac {nz}{\gcd(z,nz)}=\frac {nz}{z}=n$ , it follows that $\langle g^z \rangle$ is a subgroup of $G$ of order $n$ .