Subgroups of cyclic groups are cyclic

Proposition (Subgroups of cyclic groups are cyclic). Let $G$ be a group that is cyclic, meaning $G=\langle g\rangle$ for some $g\in G$ (so $G$ is a cyclic group ). Let $H\le G$ be a subgroup .

1. If $G$ is infinite cyclic (isomorphic to $\mathbb Z$), then either $H=\{e\}$ or there exists a unique integer $m\ge 1$ such that $H=\langle g^m\rangle$.
2. If $|G|=n<\infty$, then there exists a divisor $d\mid n$ such that $|H|=d$, and moreover $H=\langle g^{n/d}\rangle$.

Context. This result gives a complete classification of subgroups of cyclic groups: they are completely controlled by divisibility (finite case) or by a single positive integer (infinite case). It is frequently paired with Lagrange's theorem in finite group computations.

Proof sketch. (1) Consider the set $S=\{k\in\mathbb Z_{\ge 0}: g^k\in H\}$. If $H\neq\{e\}$ then $S$ contains a positive integer; let $m$ be the least positive element. Show $H=\langle g^m\rangle$ by Euclidean division: write $k=qm+r$ with $0\le r<m$, and use $g^k\in H$ to force $g^r\in H$, hence $r=0$. (2) Reduce to (1) in the finite cyclic group of order $n$ by working with exponents modulo $n$, and use Lagrange to identify $|H|$ as a divisor of $n$.