The proof of the above theorem is similar to the proof of the above statement.

[Mr Davis 97]

We define a cyclic group to be one all of whose elements can be written as "powers" of a single element, so G is cyclic if $G= \{a^n ~|~ n \in \mathbb{Z} \}$ for some $a \in G$. Is it true that in this case, $G = \{ a^0, a^1, a^2, ... , a^{n-1} \}$? If so, why? And why do we write a cyclic group as $\{a^n ~|~ n \in \mathbb{Z} \}$, where n is allowed to be any integer, except just those 0 through n - 1?

 

[fresh_42]

We define a cyclic group to be one all of whose elements can be written as "powers" of a single element, so G is cyclic if $G= \{a^n ~|~ n \in \mathbb{Z} \}$ for some $a \in G$.

Yes. And until now, nothing is said about the order of $a$. It can be of finite order or of infinite order.
In the first case, say $a^m=1$ we get $(G,\cdot) \cong (\mathbb{Z}_m,+)$ and in the second case $(G,\cdot) \cong (\mathbb{Z},+)$

Is it true that in this case, $G = \{ a^0, a^1, a^2, ... , a^{n-1} \}$?

Where's the rest of the sentence? Is it true that in case $G \cong \mathbb{Z}_n$ ... what?

If so, why? And why do we write a cyclic group as $\{a^n ~|~ n \in \mathbb{Z} \}$, where n is allowed to be any integer, except just those 0 through n - 1?

This doesn't make sense. There is no restriction on $n$ as it runs through all integers. The only question is, whether $a^m=1$ for some integer $m$ or not. If $a^m=1$ then $m=0$ will get us $G=\{1\}~,$ if $m<0$ then we can exchange $a$ by $a^{-1}$ (which is in the group and also a generator) and get an $m > 0$, so that we may always assume $m \in \mathbb{N}$. If there is no such number $m$, then $G$ is a free Abelian group of rank $1$ which is isomorphic to $\mathbb{Z}~$.

 

[member 587159]

We define a cyclic group to be one all of whose elements can be written as "powers" of a single element, so G is cyclic if $G= \{a^n ~|~ n \in \mathbb{Z} \}$ for some $a \in G$. Is it true that in this case, $G = \{ a^0, a^1, a^2, ... , a^{n-1} \}$? If so, why? And why do we write a cyclic group as $\{a^n ~|~ n \in \mathbb{Z} \}$, where n is allowed to be any integer, except just those 0 through n - 1?

We can only write it that way when G is finite.

 

[Mr Davis 97]

Assume that G is a finite cyclic group. What I'm asking is whether writing $G = \{a^n ~ | ~ n \in \mathbb{Z} \}$ is the same as writing $\{a^0, a^1, ..., a^{n-1} \}$, and if so, why?

 

[member 587159]

Assume that G is a finite cyclic group. What I'm asking is whether writing $G = \{a^n ~ | ~ n \in \mathbb{Z} \}$ is the same as writing $\{a^0, a^1, ..., a^{n-1} \}$, and if so, why?

Yes: (I used o(g) to denote the order of the element g)

Proof:

If G is cyclic, we know there is an element $g \in G$ such that $G = \{g^k|k \in \mathbb{Z}\}$. We also know that $g^i = g^j \iff i \equiv j (mod \quad o(g))$.
Assume $o(g) = m$. As $G$ is finite, $m$ is finite as well $(m \in \mathbb{N}_0)$. Then, we see that $G = \{e,g,g^2,g^3, ..., g^{m-1}\}$, because the other elements were duplicates of those elements (as we can reduce the exponents modulo m; for example: $g^m = g^{2m}=g^{-m} = e, g^2 = g^{2-m} = g^{m+2}$). Every two elements that are still left are different (as they cannot be reduced mod m and give another element in the list).

I put the key part where you need that $G$ is finite in italics.

EDIT: Interesting to note:

Consider this theorem:

Let $G$ be finite of order $n$. Then:

$G$ cyclic $\iff \exists g \in G: o(g) = n$