Proof: Every Cyclic Group is Abelian Dr. Pinter's "A Book of Abstract Algebra"'s chapter on Cyclic Groups presents the exercise:

Prove that every cyclic group is abelian. 

Here's my attempt:
By Theorem 1 (of this chapter):

(i): For every positive integer $n$, every cyclic 
      group of order $n$ is isomorphic to $\mathbb{Z}_n$. 
      Thus, any two cyclic groups of orders $n$ are isomorphic.

Every cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$. Since $\mathbb{Z}_n$ is abelian under addition, so too then is the cyclic group.
Please let me know if is this a sound proof.
 A: Your proof works for finite cyclic groups, although it doesn't really get to the heart of why all cyclic groups, including infinite ones, are cyclic.
Suppose that $G = \langle g \rangle$ is a cyclic group, and that $a,b\in G$. Since $G$ is cyclic, we can write
$$a = g^n\\b=g^m$$
for some positive integers $n,m$. Can you use this to conclude that $ab = ba$?
A: For finite cyclic groups this is a very valid proof. The only addenum is the infinite cyclic group $\mathbb{Z}$. Of course, it can be proved directly (below) but, the proof you provide is much more revealing to the structure of abelian groups.
Now, consider $x,y\in G$ where $G$ is cyclic. Since $G$ is cyclic, it is generated by some element, say $a$. Then $xy=(a^m)(a^n)$ for some $m,n\in \mathbb{Z}$. Writing out this product, using the associativty, and then recollecting terms by definition of powers we see $xy=a^{m+n}$. Similarly, $yx=a^{m+n}$ so that $G$ is abelian.
A: Your proof is OK. But I think that it would be more interesting if you tried to do it without that theorem.
Hint for this:
Call $x$ the generator of the group. What would the other elements be?
A: $$\text{Let $G$ be a cyclic group.}$$
$$\therefore \exists n, m \in \mathbb{Z} \text{ such that}$$
$$a=g^{n} \text{ and } b = g^{m} \quad \forall a, b \in G$$
$$\text{and } G = \langle g \rangle \quad \forall g \in  G$$
$$\text{but $G$ is a group} \quad \therefore XY \in G \quad \forall X, Y \in G$$
$$\therefore ab = g^{n} g^{m} = g^{n+m} = g^{m+n} = g^{m}g^{n} = ba$$
$$\therefore ab = ba \quad \forall a, b \in G \quad \therefore G \text{ is a commutative group.}$$
