Cyclicity is subgroup-closed

Verbal statement
Every subgroup of a cyclic group is cyclic.

Facts used

 * 1) Every nontrivial subgroup of the group of integers is cyclic on its smallest element

For the infinite cyclic group
Any infinite cyclic group is isomorphic to the group of integers $$\mathbb{Z}$$, so we prove the result for $$\mathbb{Z}$$. The result follows from fact (1). Note that the trivial subgroup is cyclic anyway, and fact (1) states that every nontrivial subgroup is cyclic on its smallest element.

For a finite cyclic group
Any finite cyclic group is isomorphic to the group of integers modulo n, so it suffices to prove the result for those groups.

Suppose $$G = \mathbb{Z}/n\mathbb{Z}$$ is the group of integers modulo n, and suppose $$H$$ is a subgroup of $$G$$. Define $$K$$ as the subset of $$\mathbb{Z}$$ comprising those elements of $$\mathbb{Z}$$ in the congruence classes of $$H$$. In other words:

$$\{ K = a \in \mathbb{Z} \mid \exists c \in H, a$$ is in the congruence class $$c \}$$

Then, $$K$$ is clearly a subgroup of $$\mathbb{Z}$$, because congruences mod $$n$$ preserve addition, additive inverses and identity elements.

By fact (1), there exists a $$d \in \mathbb{Z}$$ such that $$K = d\mathbb{Z}$$. Clearly, $$n \in K$$, so $$d | n$$. Going back, we see that $$H$$ is cyclic on the congruence class of $$d$$.