I am having trouble understanding the factor group, $\mathbb{R}$/$\mathbb{Z}$, or maybe i'm not. Here's what I am thinking.

Okay, so i have a group $G=(\mathbb{R},+)$, and I have a subgroup $N=(\mathbb{Z},+)$. Then I form $G/N$. So this thing identifies any real number $x$ with the integers that are exactly 1 unit step away. So if $x=\frac{3}{4}$, then $[x]=({...,\frac{-5}{4},\frac{-1}{4},\frac{3}{4},\frac{7}{4},...})$ and i can do this for any real number. So therefore, my cosets are unit intervals $[0,1)+k$, for integers $k$. Herstein calls this thing a circle and I was not sure why, but here's my intuition. The unit interval is essentially closed and since every real number plus an integer identifies with itself, these "circles" keep piling up on top of each other as if its one closed interval. Since it's closed it is a circle. Does that make sense?

Now how do I extend this intuition to this?
$G'=[(a,b)|a,b\in{\mathbb{R}}], N'=[(a,b)|a,b\in{\mathbb{Z}}].$ What is $G'/N'$? How is this a torus? I can't get an intuitive picture in my head...

EDIT: Actually, are the cosets just simply $[x]=[x\in{\mathbb{R}}|x+k,k\in{\mathbb{Z}}]?$

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    $\begingroup$ The cosets are $[a]=\{x\in\mathbb{R}\vert x=a+k,k\in\mathbb{Z}\}$. For your second question: math.stackexchange.com/questions/360533/… $\endgroup$
    – Jared
    Commented Apr 27, 2013 at 14:51
  • $\begingroup$ Okay, so identification is this process of creating a factor group? $\endgroup$ Commented Apr 27, 2013 at 14:58
  • $\begingroup$ I'm not sure I understand your question. The 'elements' of a factor group are cosets of a normal subgroup, and the operation on cosets is inherited from the group operation: $aN\cdot bN=ab\cdot N$. The identification of elements giving cosets here is $a\sim b$ iff $a-b\in\mathbb{Z}$ $\endgroup$
    – Jared
    Commented Apr 27, 2013 at 15:11
  • $\begingroup$ In the link you provided it spoke of identifying the unit square's boundaries. I was trying to understand identification mathematically, since by creating the factor group, you've identified the upper and lower bounds, right? $\endgroup$ Commented Apr 27, 2013 at 15:15

3 Answers 3


One proves that $\mathbb R/\mathbb Z$ is isomorphic to the group of unit-modulus complex numbers (let's call it $G$), which is a circle, isn't it?

Let's prove the isomorphism. Take $\varphi : \mathbb R \rightarrow G$ defined by $\varphi(\theta) = e^{2\pi i\theta}$. We have $\varphi(\theta + \theta') = e^{2\pi i(\theta+\theta')} = e^{2\pi i\theta}e^{2\pi i\theta'} = \varphi(\theta)\varphi(\theta')$ so this is indeed a homomorphism. $\varphi$ is surjective, and $\varphi(\theta) = 1 \Leftrightarrow 2\pi\theta = 2k\pi (k\in\mathbb Z)$, so $ker(\phi) = \mathbb Z$.

By the first isomorphism theorem, $\mathbb R/\mathbb Z \simeq G$.

As for your second question, try to picture this: take a 1x1 square sheet, and join the opposite edges so as to get a torus. $G'/N'$ is exactly the same construction: you identify the 'points' $(+\infty,0)$ with $(-\infty,0)$ and $(0,+\infty)$ with $(0,-\infty)$.

I hope this clarifies a bit!

  • $\begingroup$ Okay, the isormorphism example helped a lot! I took Abstract a year ago as a summer course and i find the small finite groups are easy to visualize (probably because of Cayley tables), but i've noticed I'm having (and had) trouble with infinite groups. This helped a lot, so thank you! $\endgroup$ Commented Apr 27, 2013 at 15:18
  • $\begingroup$ Note that this is an isomorphism of topological groups. $\endgroup$ Commented May 2, 2013 at 17:06
  • $\begingroup$ As someone who wants to learn topology I like that I can actually wrap my head around this! Thanks guys! $\endgroup$ Commented May 5, 2013 at 16:50

You can also use the following nice facts. I hope you are inspired by them.

$$\mathbb R/\mathbb Z\cong T\cong\prod_p\mathbb Z(p^{\infty})\cong\mathbb R\oplus(\mathbb Q/\mathbb Z)\cong\mathbb C^{\times}$$

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    $\begingroup$ I'm inspired ;-) +1 $\endgroup$
    – amWhy
    Commented May 3, 2013 at 0:10
  • 2
    $\begingroup$ What is $T$ and $\product_{p}{\mathbb{Z}(p^{\infty})}$ $\endgroup$ Commented May 5, 2013 at 17:01
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    $\begingroup$ @ChristopherErnst: $T$ is circle group. And the other is the products of prufer groups. $\endgroup$
    – Mikasa
    Commented May 5, 2013 at 17:26
  • $\begingroup$ @BabakS.: very nice facts! =1 $\endgroup$
    – Amzoti
    Commented May 11, 2013 at 0:37
  • $\begingroup$ @Eleven-Eleven This isomorphism is proved in answers to this question. $\endgroup$ Commented Nov 13, 2023 at 20:39

Just try to add some intuitive answers:

  1. The essential intuition that $\mathbb{R}/\mathbb{Z}$ is isomorphic to unit circle in the complex plane is that we can map a real number $x$ to the radian so we get $e^{2\pi i x}$, and so if two number differ by integers they map to the same radian (we circle back).

  2. The same intuition carries over to the isomorphism that $\mathbb{R}^2/\mathbb{Z^2}$ is isomorphic to the torus: We just need to tweak how to generate the coordinates: It turns out that any coordinate on a torus can be indexed by two radians. See this video at 0:33, there is one horizontal circle, and one vertical circle, so you can think of the first real number as a radian that maps you to some position in the horizontal circle, and the second number maps you to some position in the vertical circle. Again, differing by integers does not matter any more.


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