Finite Order in Quotient Groups: Q/Z and R/Q

Click For Summary

Homework Help Overview

The problem involves analyzing the orders of elements in the quotient groups \(\mathbb{Q}/\mathbb{Z}\) and \(\mathbb{R}/\mathbb{Q}\). The original poster attempts to demonstrate that every element of \(\mathbb{Q}/\mathbb{Z}\) has finite order, while only the identity element of \(\mathbb{R}/\mathbb{Q}\) possesses finite order.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forms of elements in \(\mathbb{R}/\mathbb{Q}\) and question the implications of irrational numbers in relation to finite order. There are attempts to clarify the reasoning behind the properties of these quotient groups and the nature of their elements.

Discussion Status

The discussion has progressed with participants providing hints and exploring the implications of their reasoning. Some have clarified that if \(x + Q = Q\), then \(x\) must be rational, leading to insights about the nature of nonidentity elements in \(\mathbb{R}/\mathbb{Q}\). There is ongoing exploration of the relationship between rational and irrational numbers in this context.

Contextual Notes

Participants are navigating the definitions and properties of quotient groups, particularly focusing on the implications of elements being rational or irrational. The original poster expresses uncertainty about the second part of the problem, indicating a need for further clarification on the nature of elements in \(\mathbb{R}/\mathbb{Q}\).

3029298
Messages
56
Reaction score
0

Homework Statement


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.

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?
 
Physics news on Phys.org
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?
 
Dick said:
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?

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?
 
oh no... 1/7=1/14+1/14
 
3029298 said:
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?

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

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.
 
Dick said:
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?

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?
 
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 ;)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K