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Group axioms - Verify?

  1. Sep 25, 2006 #1
    It is known that "the integers under addition" form a group,
    that is (Z,+).
    I have always wondered how to actually proof that (Z,+) is a group?

    Definitions for a group from wikipedia:

    I'm especially interested in two things:
    1) Why does the associative law hold for (Z,+), that is
    a+(b+c) = (a+b)+c for a,b,c in Z.

    And moreover:
    2) Why is closure fulfilled?
    That is, if a and b in Z, then a+b is also in Z.
  2. jcsd
  3. Sep 25, 2006 #2
    In order to prove that addition on integers is commutative and that they are closed under addition, you need to use the definition of addition on the integers. However, in order to define addition you need to know how the integers are constructed.

    You can construct the natural numbers as a sequence of sets, and define addition. You can then construct the integers from the natural numbers, and define addition on the integers using the addition you've already defined on the natural numbers. The wikipedia pages on natural numbers and integers have some of the details.
  4. Sep 25, 2006 #3


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    Just to emphasize the importance of stating from which definition you're working -- in many contexts I would define the integers as "the free ring on zero generators"... in which case the additive group structure is trivial.
  5. Sep 25, 2006 #4
    How is that the case? Don't the integers have one generator? Either 1 or -1. Why isn't the ring with zero generators the trivial ring?
  6. Sep 25, 2006 #5


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    Because it's not free -- it satisfies a nontrivial relation amongst its elements. (in particular, 0 = 1)

    For a ring R to be freely generated by the empty set, that means:

    For any ring S, any function {} --> S extends uniquely to a homomorphism R --> S.

    (There is, of course, only one function {} --> S)

    If you plug in R = Z, you'll find the above is satisfied. If you plug in R = 0, you'll find it's not satisfied. (In fact, if 0 --> S is a homomorphism, then S = 0)
  7. Sep 25, 2006 #6
    So is this equivalent to saying that R has a basis? That was what I thought a free ring was.

    And how does Z have zero generators?
  8. Sep 25, 2006 #7


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    Because it's generated by the empty set. ({1} is also a generating set for Z, of course, but Z is not the free ring on one object)

    {} is clearly a subset of Z. What is the subring of Z generated by {}? Recall that it's the intersection of all subrings of Z that contain every element in {}. The only subring of Z is Z itself -- so {} generates Z.
  9. Sep 25, 2006 #8
    Okay, I think I see this. I was thinking of generating sets in terms of groups, and was trying to generate Z with addition. But that's not right. So the free ring on one generator would be Z[x], right?

    On the level of groups, 1 or -1 generate Z, correct? So Z is the free group on one generator.
  10. Sep 25, 2006 #9


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    Sounds right; I think things don't get annoying until you have two generators. (Unless you specify "free commutative ring" -- then everything remains nice. :smile:)

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