# Quotient groups Q/Z and R/Q

1. Jan 14, 2010

### 3029298

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

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

2. Jan 14, 2010

### Dick

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?

3. Jan 14, 2010

### 3029298

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?

4. Jan 14, 2010

### 3029298

oh no... 1/7=1/14+1/14

5. Jan 14, 2010

### Dick

If x+Q=Q what can you say about x? This has nothing to do with primes.

6. Jan 14, 2010

### 3029298

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.

7. Jan 14, 2010

### 3029298

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?

8. Jan 14, 2010

### 3029298

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 ;)