About a case in a proof that every subgroup of a cyclic group is cyclic.

Question

In the following proof I do not understand why it is assumed that $\{e\}$ is cyclic and why we treat the case $h = \{e\}$ separately.

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

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

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

Let $H\neq\{e\}.$

By definition of cyclic group, every element of $G$ has the form $an$.

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

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

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

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

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

It follows that:

$$an=amq+r=(am)qar$$ and hence:

$$ar=(am)−qan$$

Since $am\in H$ so is its inverse $(am)^{−1}.$

By Group Axiom G0: Closure, so are all powers of its inverse.

Now $an$ and $(am)−q$ are both in $H$, thus so is their product $ar,$ again by Group Axiom G0: Closure.

However:

$m$ was the smallest (strictly) positive integer such that $am\in H$ and:

$$0\le r<m$$ Therefore it follows that:

$$r=0$$ Therefore:

$$n=qm$ and:

$$b=an=(am)q.$$ We conclude that any arbitrary element $b=an$ of $H$ is a power of $am$.

So, by definition, $H=\langle am\rangle$ is cyclic.

Answer 1

There's no real need to consider $\{e\}$ separately (as far as I am aware): the subgroups of $G=\langle a\rangle$ correspond to multiples (or "powers") of $a$, and we say, by convention, that $a^0=e$, the zero-th power of $a$; perhaps the additive nature of abelian groups (and hence cyclic groups) calls to question how one defines $0a$, which is the additive way of taking the zero-th power - the zero-th multiple of $a$ (which is still $e$ by convention) - and so, here, for pedagogical reasons, it is put to one side conceptually; and, besides, if $G$ is infinite, $\{e\}$ is its only finite subgroup.