Subgroups of a cyclic group

Subgroup existence. Let $d\mid n$. Then $g^{n/d}$ has order $d$ (since $(g^{n/d})^d=g^n=e$ and no smaller power works), so $\langle g^{n/d}\rangle$ is a subgroup of order $d$.

Uniqueness. Let $H$ be a subgroup of order $d$. $H$ is cyclic (subgroup of a cyclic group), say $H=\langle g^k\rangle$. Then the order of $g^k$ is $n/\gcd(k,n)=d$, so $\gcd(k,n)=n/d$. In particular, $n/d\mid k$, so $g^k\in\langle g^{n/d}\rangle$, hence $H\subseteq\langle g^{n/d}\rangle$. Since both have $d$ elements, $H=\langle g^{n/d}\rangle$.

Conclusion. There is a bijection between subgroups of $G$ and divisors of $n$.