# Subgroup of Cyclic Group is Cyclic

## Theorem

Let $G$ be a cyclic group.

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

Then $H$ is cyclic.

## Proof 1

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

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

If $H = \set e$, then $H$ is a cyclic group subgroup generated by $e$.

Let $H \ne \set e$.

By definition of cyclic group, every element of $G$ has the form $a^n$.

Then as $H$ is a subgroup of $G$, $a^n \in H$ for some $n \in \Z$.

Let $m$ be the smallest (strictly) positive integer such that $a^m \in H$.

Consider an arbitrary element $b$ of $H$.

As $H$ is a subgroup of $G$, $b = a^n$ for some $n$.

By the Division Theorem, it is possible to find integers $q$ and $r$ such that $n = m q + r$ with $0 \le r < m$.

It follows that:

- $a^n = a^{m q + r} = \paren {a^m}^q a^r$

and hence:

- $a^r = \paren {a^m}^{-q} a^n$

Since $a^m \in H$ so is its inverse $\paren {a^m}^{-1}$

By Group Axiom $G 0$: Closure, so are all powers of its inverse.

Now $a^n$ and $\paren {a^m}^{-q}$ are both in $H$, thus so is their product $a^r$, again by Group Axiom $G 0$: Closure.

However:

- $m$ was the smallest (strictly) positive integer such that $a^m \in H$

and:

- $0 \le r < m$

Therefore it follows that:

- $r = 0$

Therefore:

- $n = q m$

and:

- $b = a^n = \paren {a^m}^q$.

We conclude that any arbitrary element $b = a^n$ of $H$ is a power of $a^m$.

So, by definition, $H = \gen {a^m}$ is cyclic.

$\blacksquare$

## Proof 2

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 $\left({\Z, +}\right)$.

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

Suppose $H$ is a subgroup of $\left({\Z, +}\right)$.

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

- $\exists m \in \Z_{> 0}: H = \left({m}\right)$

where $\left({m}\right)$ is the principal ideal of $\left({\Z, +, \times}\right)$ generated by $m$.

But $m$ is also a generator of the subgroup $\left({m}\right)$ of $\left({\Z, +}\right)$, as:

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

Hence the result.

$\blacksquare$

## Proof 3

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

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

By Cyclic Group is Abelian, $G$ is abelian.

By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.

Let $G / H$ be the quotient group of $G$ by $H$.

Let $q_H: G \to G / H$ be the quotient epimorphism from $G$ to $G / H$:

- $\forall x \in G: \map {q_H} x = x N$

Then from Quotient Group Epimorphism is Epimorphism, $H$ is the kernel of $q_n$.

From Kernel of Homomorphism on Cyclic Group, $N$ is a cyclic group.

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Exercise $\text {Y}$ - 2002: John B. Fraleigh:
*A First Course in Abstract Algebra*(7th ed.)