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

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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

    3. 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
     
  2. jcsd
  3. Feb 13, 2008 #2

    Dick

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    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.
     
  4. Feb 13, 2008 #3
    I have already solved for the properties, but I thought the set matters.
     
  5. Feb 13, 2008 #4

    Dick

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    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.
     
    Last edited: Feb 13, 2008
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