# Subgroup of Cyclic Group is Cyclic/Proof 1

## 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 $a$.

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 $\text G 0$: 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.

$\blacksquare$