Subgroup of Cyclic Group is Cyclic/Proof 2

Theorem

Let $G$ be a cyclic group.

Let $H$ be a subgroup of $G$.

Then $H$ is cyclic.

Proof

Let $G$ be a cyclic group generated by $a$.

Finite Group

Let $G$ be finite.

By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$.

As each one of these is cyclic by Subgroup of Finite Cyclic Group is Determined by Order‎, the result follows.

$\Box$

Infinite Group

By Morphism from Integers to Group, an infinite cyclic group is isomorphic to $\struct {\Z, +}$.

So all we need to do is show that any subgroup of $\struct {\Z, +}$ is cyclic.

Suppose $H$ is a subgroup of $\struct {\Z, +}$.

From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that:

$\exists m \in \Z_{> 0}: H = \ideal m$

where $\ideal m$ is the principal ideal of $\struct {\Z, +, \times}$ generated by $m$.

But $m$ is also a generator of the subgroup $\ideal m$ of $\struct {\Z, +}$, as:

$n \in \Z: n \circ m = n \cdot m \in \gen m$

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.8$ Corollary