I'm having some trouble understanding the proof of the following theorem
A subgroup of a cyclic group is cyclic
I will list each step of the proof in my textbook and indicate the places that I'm confused and hopefully somewhere out there can clear some things up for me.
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 $H \neq \space {\{e\}}$, 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,
$$H = \langle a^m\rangle = \langle c\rangle.$$
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
$$ n = mq + r \space \space \space \space for \space \space \space 0 \leq r < m,$$
Alright this is the first part in the proof where I start to get confused. Where does the division algorithm come from and why are we using it? The proof continues as follows:
$$a^n = a^{mq + r} = (a^m)^q \cdot a^r,$$
so
$$ a^r = (a^m)^{-q} \cdot a^n.$$
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
$$(a^m)^{-q} \cdot a^n \in H; \space \space \space \text{that is,} \space \space a^r \in H.$$
This is another point at which I’m a little confused. What exactly about $a^n$ and $a^m$ being elements of $H$ allows us to accept that $(a^m)^{-q}$ and $a^{n}$ are in $H$? The proof continues:
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
$$b \space = \space a^n \space = \space (a^m)^q \space = \space c^q,$$
so $b$ is a power of $c.$
This final step is confusing as well, but I think its just because of the previous parts I was confused about. Any help in understanding this proof would be greatly appreciated
Best Answer
For the first question, the appearance of the division algorithm is best explained by its usefulness in the rest of the proof. You might think of it because you want $n=qm$, as you need to show that $a^n$ is a power of $a^m$, but the best you can do at that point is say $n=qm+r$ and then attempt to prove $r=0$.
For the second question, as $a^m\in H$, we have $(a^m)^{-1}\in H$, as subgroups are closed under taking inverses, and then $(a^m)^{-q}=((a^m)^{-1})^q\in H$, as subgroups are closed under multiplication.