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Abelian group on the natural numbers (including 0) ?

  1. Apr 25, 2012 #1
    Is it possible to define an abelian group on the natural numbers (including 0)? It's just that for every binary operation I've tried, I can't find an inverse!
     
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  3. Apr 25, 2012 #2

    tiny-tim

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    addition? :wink:
     
  4. Apr 25, 2012 #3

    chiro

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    You can't do that.

    Inverses include negative numbers for a 0 identity element and these aren't part of the natural numbers.
     
  5. Apr 25, 2012 #4

    micromass

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    Not every element has an inverse in that case.

    To the OP: a group operation might be defined by "structure transport".
    For example, there exists a bijection [itex]f:\mathbb{N}\rightarrow \mathbb{Z}[/itex]. We have that [itex]\mathbb{Z}[/itex] is a group under addition.
    We can now define

    [tex]n*m = f^{-1}(f(n)+f(m))[/tex]

    The same thing works with bijections [itex]f:\mathbb{N}\rightarrow G[/itex] where G is a countable group.
     
  6. Apr 25, 2012 #5

    mathman

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    Addition mod n, where n is any integer.
     
  7. Apr 25, 2012 #6

    Hurkyl

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    Depending upon what precisely you mean, this is either not an operation on natural numbers, or it's obviously not a group operation on natural numbers.

    (e.g. on the latter point 0+n=0 and 0+0=0 would imply n = 0-0 = 0. Contradiction!)
     
  8. Apr 26, 2012 #7

    mathman

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    Did you mean ALL natural numbers had to be included?
    Otherwise it is a group on integers from 0 to n-1.
     
  9. Apr 26, 2012 #8

    Hurkyl

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    That was the original question, as far as I can tell.
     
  10. Apr 27, 2012 #9

    mathman

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    I believe the person who asked the original question should answer it. We are not mind readers.
     
  11. Apr 27, 2012 #10


    But we are readers. The OP explicitly says "Is it possible to define an abelian group on the natural numbers (including 0)?"

    DonAntonio
     
  12. Apr 28, 2012 #11
    Sorry for the late reply,guys!

    @mathman: As DonAntonio said, I clearly stated in the question what is required.

    I think that the answer given by micromass is the only possible way of achieving this, though it sort of leaves me unsatisfied. Is it possible to prove that there doesn't exist any other way of doing this?

    Thanks everyone, for the replies! :D
     
  13. Apr 28, 2012 #12

    chiro

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    I think mathman has made a good point.

    We know that from Fermat's and Eulers theorems in number, you can get inverses when trying to solve for exponentiation mod (n) which is used in a lot of the public key cryptosystems and inverses are guaranteed.

    The only thing though that I see with this kind of approach, is that in these kinds of situations, the n has to be bigger than the numbers involved that you are dealing with and if you had a bounded n for all of the natural numbers, then I don't think that a unique inverse would exist which means you could not form a group.

    If you can find a way to use the kind of behaviour you get using the Euler identity but having your n in your mod (n) part of the binary group operation where n is fixed and still maintain a unique inverse across all elements, then you could form a group.

    The reason why I don't think you can do this has to do with the behaviour of the mod function and it's behaviours with respect to bijections like you would get with normal sine or cosine functions over the whole real line.
     
  14. Apr 28, 2012 #13

    mathman

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    Without the word "all", there is some ambiguity, since my original reply gives groups on natural numbers, but not all natural numbers.
     
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