Theorem. A subgroup of a cyclic group is cyclic.

Proof. Let $G$ be a cyclic group generated by $a$ and let $H\leq G$. If $H=\{e\}$ then $H=\langle e \rangle$ is cyclic. If $H\neq\{e\}$, then $a^n\in H$ for $n\in\mathbb{Z}^+$. Let $m$ be the smallest integer in $\mathbb{Z}^+$ such that $a^m\in H$.

I claim that $c=a^m$ generates $H$; that is, $H=\langle a^m \rangle = \langle c \rangle$. We must show that every $b\in H$ is a power of $c$. Let $b\in H$. Then since $H\leq G$, $b=a^n$ for some positive integer $n$. By the division algorithm for $\mathbb{Z}$, there exist $q, r \in \mathbb{Z}$ such that $n=mq+r$ where $0\leq r < m$. Then $a^n=a^{mq+r}=a^{mq}a^r$ so $a^r=a^na^{-mq}$. We must have that $a^r\in H$ because $H$ is a group and $a^n, a^{-mq} \in H$. Since $m$ is the smallest positive integer such that $a^m\in H$ and $0\leq r < m$, it must be the case that $r=0$. Thus, $n=mq$, $b=a^n=a^{mq}=c^q$, and so $b$ is a power of $c$. $\Box$