Proof. Let $G$ be a cyclic group and $H \le G$ If $G$ is trivial, then $H=G$ and $H$ is cyclic. If $H$ is the trivial subgroup, then $H=\{e_G\}=\langle e_G \rangle$ and $H$ is cyclic. Thus, for the remainder of the proof, it will be assumed that both $G$ and $H$ are nontrivial.

Let $g$ be a generator of $G$ Let $n$ be the smallest positive integer such that $g^n \in H$

Claim: $H=\langle g^n \rangle$ Let $a \in \langle g^n \rangle$ Then there exists $z \in {\mathbb Z}$ with $a=(g^n)^z$ Since $g^n \in H$ we have that $(g^n)^z \in H$ Thus, $a \in H$ Hence, $\langle g^n \rangle \subseteq H$

Let $h \in H$ Then $h \in G$ Let $x \in {\mathbb Z}$ with $h=g^x$ By the division algorithm, there exist $q,r \in {\mathbb Z}$ with $0 \le r<n$ such that $x=qn+r$ Thus, $h=g^x=g^{qn+r}=g^{qn}g^r=(g^n)^qg^r$ Therefore, $g^r=h(g^n)^{-q}$ Recall that $h,g^n \in H$ Hence, $g^r \in H$ By choice of $n$ $r$ cannot be positive. Thus, $r=0$ Therefore, $h=(g^n)^qg^0=(g^n)^qe_G=(g^n)^q \in \langle g^n \rangle$ Hence, $H \subseteq \langle g^n \rangle$

This proves the claim. It follows that every subgroup of $G$ is cyclic.