# 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.

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 $ab = ba$?

• I can show that $a^{m}a^{n}=a^{m+n}$, and vice-versa, by Chapter 10's Theorem 1: Law of exponents. Is that right? – Kevin Meredith Jun 8 '15 at 14:07
• Yes that's correct – Mathmo123 Jun 8 '15 at 14:08
• @Mathmo123 the above example works only if the operator (*) is 'product of and b' Can it be extended to other operators too? Is it a valid proof or am I missing something? – Anup Kumar Gupta Jul 12 '18 at 9:08
• @AnupKumarGupta I'm not sure what you mean. In a group, we have an operator $*$ which obeys certain axioms. The above proof works in any group. – Mathmo123 Jul 12 '18 at 20:44

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.

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?

$$let\ \ G \ \ is \ \ a \ cyclic\ group\\ \\ \therefore \exists \ \ n\ ,\ m\ \ \ \ \in Z \ \ such\ that\ \ \\ \\ a=g^{n}\ \ \ \ \ \ \ \ and\ \ \ b=g^{m}\ \ \ \ \ \forall \ a,\ b\ \ \in \ G \\ \\ and \ G=\ \ \ \ \ \ \forall \ \ g \ \in G \\ \\ but\ \ G\ is \ a \ group\ \ \ \therefore XY\ \in G\ \forall \ X\,\,, Y\ \in G\\ \\ \\$$ $$\therefore ab=g^{n}\ g^{m}=g^{n+m}=g^{m+n}=g^{m}g^{n}=ba\\ \\ \\ \therefore ab=ba\ \ \ \ \forall \ a\ ,\ b\ \in G \ \ \ \ \ \therefore G\ \ is\ commutative\ group$$