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Cyclicity is subgroup-closed

Statement

Verbal statement

Every subgroup of a cyclic group is cyclic.

Facts used

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.