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 :)

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    $\begingroup$ $\mathbb Q/\mathbb Z$ 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
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    $\begingroup$ What does it even mean to be homomorphic? $\endgroup$ – Tobias Kildetoft Feb 16 '16 at 12:44
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    $\begingroup$ 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

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

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

  • $\begingroup$ 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

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