# Tour:Cyclicity is subgroup-closed

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General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part
WHAT YOU NEED TO DO:
• Read and understand the statement below.
• Understand the proof of the statement, and fill in missing details.
• Go to the previous page and review the survey article exploring cyclic groups.

## Statement

### 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

## 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 $d$.