Tour:Cyclicity is subgroup-closed

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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 $d$.

This page is part of the Groupprops guided tour for beginners. Make notes of any doubts, confusions or comments you have about this page before proceeding.
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