Abelian Group; what to do if the set is G=R-{1/3}?

[lostinmath08]

Homework Statement

On the set G=R-{1/3} the following operation is defined:
*G: GxG arrow G

(x,y) arrow x*y=x+y-3xy

Show that (G,*) is an abelian group.

Homework Equations

To proove something is an abelian group:

The Associative Law need to hold true x*(y*x)=(x*y)*x
Neutral Element needs to be true e*x=x*e=x for all or any x E G
Inverse Elements x*x'=x'*x=e where x' is called the inverse element of x
Commutativity x*y=y*x

The Attempt at a Solution

I don't know how to solve this, especially with a set like this. Any Hints or Advice would be greatly appreciated

 

[Dick]

You posted a problem like this before. Commutativity should be obvious. Try solving for e. Just get started with any property you want to start out with.

 

[lostinmath08]

I have already solved for the properties, but I thought the set matters.

 

[Dick]

The set does matter. Because once you have figured out that e=0, then 1/3 is the only element of R that doesn't have an inverse under the operation '*'. So if you include 1/3, it's not a group. Just like R-{0} is a multiplicative group and R isn't.