# Why is a group of order 3 abelian?

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• Mar 1st 2007, 04:34 PM
monsieur fatso
Why is a group of order 3 abelian?
I have some modern algebra homework due tomorrow (and a midterm in there, ) and I've done all of the homework except this problem. I can't for the life of me figure out why a group of order 3 is abelian. I don't see the correlation between a groups order and whether or not a group is communitive.

Do any of you guys have any idea how to show this? Or could you just give me some hints that will get me started?

Thanks.
• Mar 1st 2007, 04:42 PM
Plato
What is the order of any element is a group of order three?
Can any non-identity in the group generate the group?
Are cyclic groups Abelian?
• Mar 1st 2007, 04:55 PM
monsieur fatso
Quote:

Originally Posted by Plato
What is the order of any element is a group of order three?
Can any non-identity in the group generate the group?
Are cyclic groups Abelian?

I'm not sure I'm getting this.

Ok, the orders of any element in a group of order 3 is 3, except for 1?.
Yeah, in (z/3z,+) 1 is not the identity, and can generate the group, and it's abelian, but I'm not sure if all cyclic groups are abelian.
• Mar 1st 2007, 05:01 PM
Plato
Quote:

Originally Posted by monsieur fatso
I'm not sure if all cyclic groups are abelian.

Why don't you know that is true?
If not, there has been a major failure somewhere!
• Mar 1st 2007, 05:09 PM
topsquark
Try this: construct a group of order 3 with any properties you like. You will find that there is only one such group possible.

-Dan
• Mar 1st 2007, 05:10 PM
monsieur fatso
Quote:

Originally Posted by Plato
Why don't you know that is true?
If not, there has been a major failure somewhere!

I kind of figured it was, but I wasn't positive, as it hasn't really been introduced to us yet. I found some proof explaining why it is, however and convinced myself of it.

So, it's because every element in a group of order 3 has order 3, and the entire group can be generated by a non-identity element? Because it can be generated by a non-identity element, the group is cyclic, and every cyclic group is abelian?

Sorry if this seems like a really dumb question. I'm just really burning out on math. I have 2 math midterms tomorrow, and one next week, and I've been cramming hard core for a while now for it.
• Mar 1st 2007, 06:01 PM
ThePerfectHacker
Quote:

Originally Posted by monsieur fatso
I have some modern algebra homework due tomorrow (and a midterm in there, ) and I've done all of the homework except this problem. I can't for the life of me figure out why a group of order 3 is abelian. I don't see the correlation between a groups order and whether or not a group is communitive.
.

There exists only one such group (up to isomorphism of course) for any given prime number.
Since 3 is a prime, it must be a unique group. That is Z_3.

This group is cyclic, and hence abelian (as Plato said).
Q.E.D.
• Mar 1st 2007, 06:07 PM
Plato
Every cyclic group is Abelian.
Think about it. If x is in G then x=a^n for a in G.
If {x,y} is a subset of G, then x=a^n & y=a^m.
Then (xy)=[a^n][a^m]=a^(n+m) =a^(m+n)= =(a^m)(a^n)=yx
• Mar 1st 2007, 06:09 PM
monsieur fatso
So if the order is prime, the group z/pz is unique?

Makes sense now, that was what I was having trouble with I guess. Thanks a ton plato and perfect hacker.
• Mar 1st 2007, 06:56 PM
ThePerfectHacker
Quote:

Originally Posted by monsieur fatso
So if the order is prime, the group z/pz is unique?

Yes, I would say you learn that theorem 2 months into that course. It is called Lagrange's theorem (actually a consequence of it). If you never learned that theorem we can construct, as topsquark, suggests a table for group of order 3 and show that it must be unique.