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I need to consider the additive cosets of $\mathbb Q/\mathbb Z$. However, I'm having a hard time visualizing what those cosets would look like? What do the cosets look like?

I know $q+\mathbb Z$ where $q\in\mathbb Z$. But I'm not sure how to interpret this.

What would be a concrete example of one? Are the only members of $q+\mathbb Z$ integers, or is there rationals in $q+\mathbb Z$ as well?

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The coset $q + \mathbb{Z}$ is literally the set $\{q + n : n \in \mathbb{Z}\}$. So for instance, if $q = 1/2$, then $1/2 + \mathbb{Z} = \{\ldots, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, \ldots\}$. So you can picture the coset $q + \mathbb{Z}$ as all the points on a line that are an integer distance away from $q$: each element is at distance $1$ from the previous one.

As for visualizing the cosets, here's a more geometric answer. As you may know, $\mathbb{R}/\mathbb{Z}$ is isomorphic to the unit circle $S^1$ via the map \begin{align*} \varphi: \mathbb{R}/\mathbb{Z} &\overset{\sim}{\longrightarrow} S^1\\ t + \mathbb{Z} &\longmapsto (\cos(2\pi t), \sin(2\pi t)) \, . \end{align*} Thus a coset $t + \mathbb{Z}$ is mapped to the point on the circle obtained by rotating $(1,0)$ through an angle of $2\pi t$ about the origin. $\mathbb{Q}/\mathbb{Z}$ is a subgroup of $\mathbb{R}/\mathbb{Z}$, and its image under $\varphi$ consists of all points obtained by rotating the point $(1,0)$ a rational multiple of $2\pi$. So you can picture $\mathbb{Q}/\mathbb{Z}$ as all points on the unit circle that have an angle that is a rational multiple of $2\pi$. (This subgroup happens to be the torsion subgroup of $\mathbb{R}/\mathbb{Z}$.)

This picture agrees with the answers using a $\mod{1}$ interpretation: under this equivalence relation $0$ and $1$ are identified (or "glued together"), which turns the interval $[0,1]$ into a circle.

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Cosets are in bijection with the rational numbers in $[0,1)$, and addition is performed modulo $1$. To see this, note that by the division algorithm, given any rational number $p/q$ there exists integers $n,r$ such that $p/q = n + r/q$ and $0<r<q$. Thus the coset of $p/q$ and $r/q$ is the same. Moreover, if $0\leqslant r,r'<1$ are rational, then $r-r'$ is lest than $1$, and in particular cannot be an integer unless it is zero. Thus $r$ and $r'$ are in different cosets.

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As I understand the quotient $\Bbb Q/\Bbb Z$ it is the partition of $\Bbb Q$ under the equivalence relation

$$r\sim s\iff r=s+z$$

where $r,s\in\Bbb Q$ and $z\in\Bbb Z$. Then if we rewrite $r=\frac{p_1}{q_1}$ and $s=\frac{p_2}{q_2}$ then we have that

$$\frac{p_1}{q_1}=\frac{p_2}{q_2}+z=\frac{p_2+zq_2}{q_2}$$

By example $1/2\sim 3/2$, etc... As someone said above $r\equiv s\pmod 1$. If we set as the representation of an equivalence class

$$x=\min\{|y|:y\in [r]\}$$

and we set the isomorphism $[r]\mapsto x$ then $\Bbb Q/\Bbb Z\cong[0,1)\cap\Bbb Q$.

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For general groups $A, B$ with $B$ a subgroup of $A$, the quotient group $A/B$ is defined to be the elements of $A$ "modulo" the elements of $B$. In other words, the elements of $B$ all behave like a zero element in $A/B$.

So in $\Bbb Q / \Bbb Z$, which I'll call $G$ for simplicity, the integers behave like zero in $G$. So, for example, $\frac12$ and $\frac32$ are actually the "same" in $G$ because $\frac12 + 1 = \frac32$, and $1$ behaves like $0$ in $G$.

Technically speaking, the elements of $G$ (and quotient groups in general) are equivalence classes, and not simply numbers. This is why I put "same" in quotations above. While $\frac12$ and $\frac32$ are different numbers, they're part of the same equivalence class in $G$, because they differ by an integer in $\Bbb Q$ and all integers behave like zero in $G$, and so $\frac12$ and $\frac32$ "differ by zero" in $G$. Therefore for all intents and purposes we can treat them as being equal to each other in $G$. And we can denote the equivalence class by $[\frac12]$.

Why use $\frac12$ as our representative to denote the equivalence class? Why not $\frac32$? Or $\frac72$, or $-\frac92$? It's because it's best to choose the "simplest" representative.

Note that for every rational number $q$, we can find some integer $n$ such that $q-n \in [0,1)$. Therefore every rational number belongs to the same equivalence class as some rational number in $[0,1)$. And most would agree that the rationals in $[0,1)$ are the simplest to use as our equivalence class representatives.

Therefore $G = \Bbb Q/\Bbb Z$ consists of equivalence classes of the form $[q]$, where $q \in [0,1) \cap \Bbb Q$. And for each such $q$, the elements of $[q]$ look like $q+k$, where $k$ is any integer.

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Well, they look like ... the rationals included inside $[0,1)$ with operations $\mod 1$

P.S.: You can intuitively imagine the quotient $\mathbb{Q}/\mathbb{Z}$, as setting the integers equal to zero inside the rationals. Then all intervals $[a,a+1)\cap \mathbb{Q}$, with $a$ integer, become identified since their rational values are equivalent. This is the exact meaning of $q+\mathbb{Z}$; for example: $$ ...\equiv-2+\frac{3}{8}\equiv -1+\frac{3}{8}\equiv\frac{3}{8}\equiv 1+\frac{3}{8}\equiv 2+\frac{3}{8}\equiv ... \mod1 $$

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  • $\begingroup$ But how did you get that? I don't understand how to get from $q+Z$ being an element of $Q/Z$ to what you have above. $\endgroup$ – Nate Oct 12 '16 at 23:37

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