# Intuition: groups, quotient groups, cosets, homomorphisms.

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 :)

• $\mathbb Q/\mathbb Z$ is not isomorphic to $\mathbb S^1$. You confuse with $\mathbb R/\mathbb Z\cong\mathbb S^1$. – Surb Feb 16 '16 at 12:35
• Usually $S^1$ denotes the unit circle in the real plane, which is far larger than this group. – Tobias Kildetoft Feb 16 '16 at 12:35
• math.stackexchange.com/questions/1202243/… – Taylor Ted Feb 16 '16 at 12:38
• What does it even mean to be homomorphic? – Tobias Kildetoft Feb 16 '16 at 12:44
• 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? – 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}