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Quotient Groups

  1. Nov 13, 2008 #1
    1. Show that every element of the quotient group G = Q/Z has finite
    order. Does G have finite order?
    he problem statement, all variables and given/known data




    2. This is the proof

    The cosets that make up Q/Z have the form Z + q,
    where q belongs to Q. For example, there is a coset Z + 1/2, which is
    the set of all numbers of the form {n + 1/2}, where n is an integer.
    And there is a coset Z + 2/5, which consists of all numbers of the
    form n + 2/5, where n is an integer. The cosets form a group if you
    define the sum of A and B to be the set of all sums of an element in A
    and an element in B, and this group is the quotient group. The
    identity of Q/Z is just Z.

    Now if you take a rational number r/s, where r and s are integers,
    then

    s (r/s) = r

    which is an integer. Now anything in the coset Z + r/s is an integer
    plus r/s, so if you multiply anything in that coset by s, you get an
    integer. So if you multiply the coset by s (i.e. add it to itself s
    times) you get a coset consisting of all integers, but that's just Z
    itself. That is, the coset is of finite order s (or a divisor of s).



    3.I just need it to be explained especially the part with s(r/s)
     
  2. jcsd
  3. Nov 14, 2008 #2

    morphism

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    What you said is perfectly correct. To put it in precise terms, if r/s+Z is an element of Q/Z (where we may assume that s is a positive integer), then s(r/s+Z)=r+Z=Z. So every element of Q/Z has finite order.
     
  4. Nov 15, 2008 #3
    ok... I can see by example why this proof make sense with s in r/s being a prime number because s here is the order of the factor group, how about when s is not prime, does the proof still apply?
     
  5. Nov 16, 2008 #4

    morphism

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    Why would s have to be a prime number? The order of what factor group? Q/Z is definitely not of order s, or any other finite number!
     
  6. Nov 16, 2008 #5
    ok...
     
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