Theorem 6.6

Statement:

A subgroup of a cyclic group is cyclic.

Proof:

Let $G$ be a cyclic group generated by $a$ and let $H$ be a subgroup of $G$. If $H=\{e\}$, then $H=\langle e \rangle$ is cyclic. If $H \neq \{e\}$, then $a^{n} \in H$ for some $n \in \mathbb{Z^{+}}$. Let $m$ be the smallest integer in $\mathbb{Z^{+}}$ such that $a^{m} \in H$.

We claim that $c=a^{m}$ generates $H$; that is,

(1)
\begin{align} h = \langle a^{m} \rangle = \langle c \rangle \end{align}

We must show that every $b \in H$ is a power of $c$. Since $b \in H$ and $H \leq G$, we have $b=a^{n}$ for some $n$. Find $q$ and $r$ such that

(2)
\begin{align} n=mq + r \ \ \ \ for \ \ \ \ 0 \leq r < m \end{align}

in accord with the division algorithm. Then

(3)
\begin{equation} a^{n} = a^{mq+r} = (a^{m})^{q}a^{r}, \end{equation}

so

(4)
\begin{equation} a^{r} = (a^{m})^{-q}a^{n}. \end{equation}

Now since $a^{n} \in H$, $a^{m} \in H$, and $H$ is a group, both $(a^{m})^{-q}$ and $a^{n}$ are in $H$. Thus

(5)
\begin{align} (a^{m})^{-q}a^{n} \in H; \end{align}

that is,

(6)
\begin{align} a^{r} \in H. \end{align}

Since $m$ was the smallest positive integer such that $a^{m} \in H$ and $0 \leq r < m$, we must have $r=0$. Thus $n=qm$ and

(7)
\begin{equation} b = a^{n} = (a^{m})^{q} = c^{q}, \end{equation}

so $b$ is a power of $c$.