Thread: 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.

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?

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.

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!

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

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.

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.

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

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.

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.