1
$\begingroup$

If we start with the group of rational numbers $\mathbb{Q}$ and the subgroup of $\mathbb{Q}$; $\mathbb{Z}$ the integers, and then form the quotient group $\mathbb{Q}$/$\mathbb{Z}$ we have that this quotient group consists of all the cosets of $\mathbb{Z}$ in $\mathbb{Q}$.

These are of the form $a + \mathbb{Z}$, $a \in \mathbb{Q}$.

E.g. $...\frac{2}{3}, \frac{5}{3}, \frac{8}{3}...$ i.e. the coset $\frac{2}{3} + \mathbb{Z}$.

These cosets are the elements of the quotient group $\mathbb{Q}$/$\mathbb{Z}$ and is a partiton of $\mathbb{Q}$.

Then we have that $\mathbb{Q}$/$\mathbb{Z}$ is homomorphic to the circle $S^1$.

$\mathbb{Q}$/$\mathbb{Z} \:\cong \:S^1$

This is because the cosets are parameterized by elements belonging to the interval $[0,1]$. Every coset has exactly one element in it (as long as we make the identification that $0 = 1$, because it belongs to the original subgroup $\mathbb{Z}$)

$\mathbb{Q}$/$\mathbb{Z}\: \cong \:S^1$ ($\cong [0,1]$ with $0=1$)

I think these things are rather difficult and i wonder if my intuition, understanding, of them and the connection between them are ok, or if there are some flaws?


Clarification from the discussion in the comments: The statement $\mathbb{Q}$/$\mathbb{Z}\: \cong \:S^1$ comes from Wildberger's video lectures and what he probably means is that these are homeomorphic as spaces. But even this is wrong.

Conclusion: Do not watch his videos :)

$\endgroup$
  • 3
    $\begingroup$ $\mathbb Q/\mathbb Z$ is not isomorphic to $\mathbb S^1$. You confuse with $\mathbb R/\mathbb Z\cong\mathbb S^1$. $\endgroup$ – Surb Feb 16 '16 at 12:35
  • $\begingroup$ Usually $S^1$ denotes the unit circle in the real plane, which is far larger than this group. $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:35
  • $\begingroup$ math.stackexchange.com/questions/1202243/… $\endgroup$ – Taylor Ted Feb 16 '16 at 12:38
  • 2
    $\begingroup$ What does it even mean to be homomorphic? $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:44
  • 1
    $\begingroup$ Yes, it is also clearly false that they are homeomorphic, as one is countable while the other is not. But given the view of Wildberger on anything infinite, who knows? $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:53
2
$\begingroup$

As stated in the comments $\mathbb Q/\mathbb Z$ is not isomorphic to $\mathbb S^1$ which is the unit circle. What we have is the following isomorphism with the multiplicative group of roots of unit

$$\begin{align}\varphi: \mathbb Q/\mathbb Z &\to C^{\times}\\\frac{p}{q} + \mathbb Z &\mapsto e^{\frac{2\pi i p}{q}}\end{align}$$

$\endgroup$
  • $\begingroup$ And its not true that they are homomorphic as spaces? $\endgroup$ – JKnecht Feb 16 '16 at 12:49
  • $\begingroup$ @JKnecht What does it mean to be homomorphic? $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:49
  • $\begingroup$ I think Jknecht want to say isomorphic, no ? @TobiasKildetoft $\endgroup$ – Surb Feb 16 '16 at 12:50
  • $\begingroup$ Do you mean homeomorphic? $\endgroup$ – Aaron Maroja Feb 16 '16 at 12:51
  • $\begingroup$ @Surb That would be odd since he has already been told that this is not the case (and makes the "as spaces" odd too). $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.