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Groups do not necessarily have to have only one operation

  1. Mar 28, 2009 #1
    To be sure of things, groups do not necessarily have to have only one operation; they may have more, but there is one and only one operation they are identified with. Am I right?
     
  2. jcsd
  3. Mar 28, 2009 #2

    CompuChip

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    Re: Groups

    There may be two operations o and * defined on a set G, such that (G, o) and (G, *) are both groups.
    When we say "let G be a group" we're sloppy, you should say what the operation is.
     
  4. Mar 28, 2009 #3
    Re: Groups

    Thus, if I were asked whether the set of all rational numbers Q with addition was a group, would it be valid to assume multiplication as an operation and state that a/b + c/d = (ad + cb)/(bd) ε Q (for a, b, c, d ε Z) in the case of proving that addition is closed with respect to Q.
     
  5. Mar 28, 2009 #4

    CRGreathouse

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    Re: Groups

    Yes, that's valid, but you're not multiplying group elements (which would be bad). You're just expanding the definition of rational addition.
     
  6. Mar 28, 2009 #5
    Re: Groups

    I see. Thanks for the replies.
     
  7. Mar 28, 2009 #6
    Re: Groups

    Well, in general there is only one operation that is defined in a group. That is there is only one operation which is applied berween the elements of a group.

    In your example, we could say let (Q,@) be a group, where Q is the set of rational numbers, and then we would say that @ is defined in this way:

    [tex]\frac{a}{b} @ \frac{c}{d}=\frac{a*d+c*b}{b*d}[/tex]

    where + is the natural addition symbol and * multiplication.

    But, like it was said above, here you are not multiplying the elements of Q, which we have assumed are of the form

    [tex]Q={ \frac{a}{b}: a,b \in Z }[/tex]
     
  8. Mar 29, 2009 #7

    HallsofIvy

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    Re: Groups

    But in any case, in
    [tex]\frac{a}{b}+ \frac{c}{d}= \frac{ad+ bc}{cd}[/tex]
    You are NOT multiplying rational number you are multiplying integers.
     
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