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Homework Help: Quotient groups Q/Z and R/Q

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that every element of the quotient group [tex]\mathbb{Q}/\mathbb{Z}[/tex] has finite order but that only the identity element of [tex]\mathbb{R}/\mathbb{Q}[/tex] has finite order.

    3. The attempt at a solution
    The first part of the question I solved. Since each element of [tex]\mathbb{Q}/\mathbb{Z}[/tex] is of the form [tex]\mathbb{Z}+\frac{r}{s}[/tex] if we add this element s times to itself, we get all [tex]\mathbb{Z}[/tex] back, since [tex]s\frac{r}{s}=r[/tex]. But for the second part of the question I have no clue... Can anyone hint me in the right direction?
     
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  3. Jan 14, 2010 #2

    Dick

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    Ok, so a nonidentity element of R/Q has the form Q+r, where r is irrational. What would be wrong with n*(Q+r)=Q?
     
  4. Jan 14, 2010 #3
    n*(Q+r)=n*r+Q since Q is normal. has it got something to do with the fact that for p=prime 1/p cannot be written as the sum of two rationals?
     
  5. Jan 14, 2010 #4
    oh no... 1/7=1/14+1/14
     
  6. Jan 14, 2010 #5

    Dick

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    If x+Q=Q what can you say about x? This has nothing to do with primes.
     
  7. Jan 14, 2010 #6
    If x is not Q, then this can never be true, since the sum of a non-rational and a rational number is non-rational.

    If x is in Q then x+Q = Q since for each element q in Q there exists an element q-x in Q which gives x+q-x=q, and each element in x+Q is in Q.

    x must be a rational.
     
  8. Jan 14, 2010 #7
    The only thing confusing me a little from the beginning is that if we have a nonidentity element of R/Q it has to have the form Q+r where r is irrational. Does r have to be irrational because Q is the identity of R/Q?
     
  9. Jan 14, 2010 #8
    I get it now! If x+Q=Q, x must be rational. Therefore, a nonidentity element of R/Q has the form Q+r where r is irrational. Now if n*(Q+r)=Q, n*r+Q=Q and n*r is rational. But this cannot be the case since r is irrational, and we have a contradiction, and a nonidentity element of R/Q does not have finite order.
    Thanks for the hints ;)
     
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