# Cyclicity is subgroup-closed

## Statement

### Verbal statement

Every subgroup of a cyclic group is cyclic.

## Proof

### 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 $d$.