Cyclicity is subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., cyclic group) satisfying a group metaproperty (i.e., subgroup-closed group property)
View all group metaproperty satisfactions | View all group metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for group properties
Get more facts about cyclic group |Get facts that use property satisfaction of cyclic group | Get facts that use property satisfaction of cyclic group|Get more facts about subgroup-closed group property

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 $d$.