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.
Best Answer
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 $ab = ba$?