[proofplan] Choose a generator $g$ of $G$ and examine which powers of $g$ lie in the subgroup $H$. If $H$ is trivial, it is generated by the identity. Otherwise, the set of positive exponents $k$ with $g^k \in H$ is nonempty, so it has a least element $n$. We prove that $H=\langle g^n\rangle$ by dividing an arbitrary exponent by $n$ and using the minimality of $n$ to force the remainder to vanish. [/proofplan]

[step:Choose a generator and dispose of the trivial subgroup]Since $G$ is cyclic, choose an element $g \in G$ such that $G=\langle g\rangle$. If $H=\{e\}$, where $e$ denotes the identity element of $G$, then $H=\langle e\rangle$, so $H$ is cyclic. Henceforth assume $H \ne \{e\}$.[/step]

[guided]Because $G$ is cyclic, there is an element $g \in G$ with $G=\langle g\rangle$. This means that every element of $G$ is a power of $g$ with integer exponent. We first handle the case in which $H$ contains no element except the identity element $e$ of $G$. Then $H=\{e\}=\langle e\rangle$. Thus $H$ is cyclic in this case. This separate treatment matters because the later argument chooses a least positive exponent $n$ with $g^n \in H$, and such an exponent is not forced by the subgroup $H=\{e\}$ unless one already uses a positive order relation for $g$. We therefore assume from now on that $H \ne \{e\}$.[/guided]

[step:Choose the least positive exponent whose power lies in $H$] Define the set $S=\{k \in \mathbb{Z}_{>0}: g^k \in H\}$. The set $S$ is nonempty. Indeed, since $H \ne \{e\}$, choose $h \in H$ with $h \ne e$. Because $G=\langle g\rangle$, there exists $m \in \mathbb{Z}$ such that $h=g^m$. Since $h \ne e$, we have $m \ne 0$. If $m>0$, then $m \in S$. If $m<0$, then $h^{-1}=g^{-m} \in H$ because $H$ is a subgroup, and $-m \in \mathbb{N}$, so $-m \in S$. By the [well-ordering principle](/theorems/721) for $\mathbb{N}$, there exists a least element $n \in S$. [/step]

[step:Show that the subgroup generated by $g^n$ lies inside $H$] Since $n \in S$, we have $g^n \in H$. Because $H$ is a subgroup, it is closed under products and inverses. Hence, for every $q \in \mathbb{Z}$, \begin{align*} (g^n)^q=g^{nq}\in H. \end{align*} Every element of $\langle g^n\rangle$ is of this form, so $\langle g^n\rangle \subseteq H$. [/step]

[step:Divide an arbitrary exponent by the minimal positive exponent] Let $h \in H$. Since $G=\langle g\rangle$, choose $m \in \mathbb{Z}$ such that $h=g^m$. By the Euclidean division theorem, there exist $q \in \mathbb{Z}$ and $r \in \mathbb{Z}$ such that \begin{align*} m=qn+r \end{align*} and $0 \le r < n$. Using the exponent laws in the group $G$, we obtain \begin{align*} g^r=g^{m-qn}=g^m(g^n)^{-q}. \end{align*} Here $g^m=h \in H$, and $(g^n)^{-q} \in H$ because $g^n \in H$ and $H$ is a subgroup. Therefore $g^r \in H$. If $r>0$, then $r \in \mathbb{N}$ and $g^r \in H$, so $r \in S$. This contradicts the minimality of $n$, since $0<r<n$. Hence $r=0$. [/step]

[step:Conclude that every element of $H$ is a power of $g^n$] From $r=0$ we have $m=qn$, and therefore \begin{align*} h=g^m=g^{qn}=(g^n)^q. \end{align*} Thus $h \in \langle g^n\rangle$. Since $h \in H$ was arbitrary, $H \subseteq \langle g^n\rangle$. Together with $\langle g^n\rangle \subseteq H$, this gives $H=\langle g^n\rangle$. Therefore $H$ is cyclic. [/step]