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