2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $ab = ba$?

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Kevin Meredith Jun 8 '15 at 14:07
  • $\begingroup$ Yes that's correct $\endgroup$ – Mathmo123 Jun 8 '15 at 14:08
  • $\begingroup$ @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? $\endgroup$ – Anup Kumar Gupta Jul 12 '18 at 9:08
  • $\begingroup$ @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. $\endgroup$ – Mathmo123 Jul 12 '18 at 20:44
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.