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Left Cosets of Z in Q

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    View [itex]Z[/itex] as a subgroup of the additive group of rational numbers [itex]Q[/itex]. Show that given an element [itex]\bar{x} \in Q/Z[/itex] there exists an integer [itex]n \geq 1[/itex] such that [itex]n \bar{x} = 0[/itex].


    2. Relevant equations



    3. The attempt at a solution
    As we are working in an additive group, it is clear that any left coset [itex]a+Z \in Q/Z[/itex] is constructed from [itex]a \in Q, 0 \leq a \ll 1[/itex]. This is because otherwise we get a nonzero intersection between cosets, and cosets either have a zero intersection or are completely identical.

    Now, the problem I am having understanding is the the existence of an integer [itex]n \geq 1[/itex] such that [itex]n \bar{x} = 0[/itex]. Because, the way I am interpreting it, for any coset in Q/Z there is a positive integer greater to or equal to 1 such that multiplying ANY element in that coset gives 0. I can't seem to picture this being possible, as some cosets will consist only of nonzero rational numbers, and there is no positive integer which when multiplied by any of them would result in 0.

    An alternate explanation would be that by 0, they mean 0 in [itex]Q/Z[/itex], which would be [itex]Z[/itex]. Here too I have issues, as [itex]Z[/itex] includes the integer 0, which again leads to the problem mentioned above.

    So, what am I misunderstanding about this problem?
     
  2. jcsd
  3. Oct 13, 2012 #2
    The problem you are having is that the number zero is not the zero in the quotient. The additive identity of Q/Z is the coset 0+Z. So 1/2+Z is a cose and if you multiply this coset by 2 you get 1+Z = Z.
     
  4. Oct 13, 2012 #3

    Dick

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    Actually, it's 2Z+1. It's certainly a subset of Z. But it's not really a coset.
     
  5. Oct 13, 2012 #4
    I was interpreting 2(1/2 + Z) as (1/2 + Z) + (1/2 + Z) = (1/2 + 1/2) + Z = 1 + Z. When the group is additive, isn't it pretty standard to write 2g for g+g? But the idea is that we add a rational number to itself a certain number of times (ie, multiply it by some integer) and we get an integer, z, so that z+Z = Z which is 0 in the quotient space. So if q is a rational number and q=z_1/z_2 then z_2q = z_1 and z_1 + Z = Z. Also, (z_2)(q+Z) = (q + q + ... + q)+Z = z_1 + Z = Z.
     
  6. Oct 13, 2012 #5

    Dick

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    Ah, right. g+g. Not 2 times every element of g. Thought I might be missing something...
     
  7. Oct 14, 2012 #6
    I had thought that, but then it seemed as though multiplication weren't quite distribution with that since

    Z = ..., -2, -1, 0, 1, 2, ...
    1/2 + Z = ..., -1 1/2, -1/2, 1/2, 1 1/2, 2 1/2, ...
    2(1/2 + Z) = ..., -3, -1, 1, 3, 5, .... != Z which makes it seem as though distributivity does not apply when multiplying an additive coset by a constant.

    And at any rate it would seem impossible get 0 from 1/2 + Z regardless of what we multiply it by....but 0 is in Z so we have to retrieve that somehow.
     
  8. Oct 14, 2012 #7

    Dick

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    I made the same mistake, listen to what Robert1986 is saying. In this problem 2(Z+1/2) means (Z+1/2)+(Z+1/2) NOT 2Z+2(1/2). So yes, it's not a distributive 'multiplication'.
     
  9. Oct 14, 2012 #8
    Oh, so it's not implying multiplying every element of the set by a constant. Well then that's easy enough to show lol.

    I guess I just got confused by what the notation was referring to ~~
     
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