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My modern algebra needs some work. Am I right in thinking that $\mathbb{Z}/2\mathbb{Z}$ refers to the two sets $$\{\pm0, \pm2, \pm4, \pm6, \ldots\}$$ and $$\{\pm1, \pm3, \pm5, \pm7\}~~?$$ What about $\mathbb{R}/2\mathbb{Z}$ if that makes sense to write? Would that mean $$\{,\ldots,[0,1),[2,3),[4,5),\ldots\}$$ and $$\{\ldots,[1,2),[3,4),[5,6),\ldots\}~~?$$ I haven't got round to looking at these "quotients" as I think they're called. It's on my list of things to do. I believe they're to do with equivalence classes? Are there some more "exotic" examples with concrete examples of the sets produced as above? Just asking to see if I'm thinking on the right track. So my original understanding of $\mathbb{R}/2\mathbb{Z}$ was incorrect (as pointed out in answer(s) below).

EDIT

Not sure if this is a good way to visualise what's going on with e.g. $\mathbb{R}/2\mathbb{Z}$, but imagine the Cartesian plane with $x$ and $y$ axes. For $\mathbb{R}/2\mathbb{Z}$ I can see the left hand side ($\mathbb{R}$) corresponding to the $y$-axis and the right hand side ($2\mathbb{Z}$) corresponding to the integers on the $x$-axis. If I imagine $2\mathbb{Z}$ on the $x$-axis slicing the plane vertically then what I'm left with is an infinite number of slices each of width $2$. The quotient kind of takes all these slices and stacks them on top of each other so that the only information available to me belongs to $[0,2)$. Positioning along the $x$-axis is lost.

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First of all, $\mathbb Z$ contains negative integers, so $\mathbb Z/2\mathbb Z$ has two elements, $0+2\mathbb Z=\{\dots,-4,-2,0,2,4,\dots\}$ and $1+2\mathbb Z=\{\dots,-5,-3,-1,1,3,5,\dots\}$. – Thomas Andrews Jan 6 at 20:40
    
Ah yes of course ... will update. – poirot Jan 6 at 20:41
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As a group, $\mathbb R/2\mathbb Z$ "looks like" the group of rotations of a circle, where $a+2\mathbb Z$ corresponds to rotating by the angle $a\pi$. – Thomas Andrews Jan 6 at 20:45
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About your (incorect) guess about the nature of $\Bbb R/\Bbb Z$: pay close attention to the second comment of @ThomasAndrews. – Lubin Jan 7 at 6:01
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A related thread. – Jyrki Lahtonen Jan 7 at 6:04
up vote 10 down vote accepted

A good way to think about quotients is to pretend that nothing has changed except your concept of equality.

You can think of $\mathbb{Z}/2\mathbb{Z}$ as just the integers (under addition) but multiples of 2 (i.e. elements of $2\mathbb{Z}$) are eaten up like they're zero. So in this quotient world $1=3=-5=41$ etc. and $0=6=104=-58$ etc.

What is $1+1$? Well, $1+1=2$. But in this quotient group $2=0$ so $1+1=0$. Notice that $1=-3=99$ so $1+1 =-3+99=96$ (which also $=0$). Equivalent "representatives" give equivalent answers.

Formally, yes, $\mathbb{Z}/2\mathbb{Z} = \{0+2\mathbb{Z}, 1+2\mathbb{Z}\}$ where $0+2\mathbb{Z}=2\mathbb{Z}=$ even integers and $1+2\mathbb{Z}=$ odd integers.

A more formal version of my previous calculation: $(1+2\mathbb{Z})+(1+2\mathbb{Z}) = (-3+2\mathbb{Z})+(99+2\mathbb{Z}) = (-3+99)+2\mathbb{Z} = 96+2\mathbb{Z}=0+2\mathbb{Z}$.

If we move to $\mathbb{R}/2\mathbb{Z}$, then elements are equivalence classes: $x+2\mathbb{Z} = \{x+2k \;|\; k \in \mathbb{Z}\} = \{\dots,x-4,x-2,x,x+2,x+4,\dots\}$. Addition works exactly the same as it does in $\mathbb{R}$ (except we have enlarged what "equals" means). So $((3+\sqrt{2})+2\mathbb{Z})+((-10+\pi)+2\mathbb{Z}) = (7+\sqrt{2}+\pi)+2\mathbb{Z}$. Of course, here, $7+\sqrt{2}+\pi$ could be replaced by something like $-3+\sqrt{2}+\pi$.

In fact, every $x+2\mathbb{Z}$ is equal to $x'+2\mathbb{Z}$ where $x' \in [0,2)$ (add an appropriate even integer to $x$ to get within the interval $[0,2)$). So as a set $\mathbb{R}/2\mathbb{Z}$ is essentially $[0,2)$ (each equivalence class in the quotient can be uniquely represented by a real number in $[0,2)$).

Alternatively, think of this group like $[0,2]$ with $0=2$. Take the interval $[0,2]$ and glue the ends together. It's a circle group. Basically $\mathbb{R}/2\mathbb{Z}$ as a group is just like adding angles (but $2=0$ not $2\pi=0$). :)

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Taking quotients means considering equality up to something you are not interested in. This always corresponds to building equivalence classes, as you correctly noted. However, in most practical uses, one is interested in cases where the mapping that sends elements to their equivalence classes furthermore respects the underlying structure.

For example, the fact that you can only take quotients of groups by normal subgroups, not just any subgroup, comes from the fact that only normal subgroups are those that are the kernels of group homomorphisms. Indeed, if the quotient map sending $a$ to $aN$ is a group homomorphism (implying that the equivalence classes can be turned into a group with the obvious construction. Here N is the normal subgroup with respect to which you take quotients) then the kernel of this map is $N$. To obtain the usual classification of normal subgroups note that for any kernel $N$ of a group homomorphism it holds that $aN=Na$ for all elements $a$ of the group.

In the example above, note that the sets $aN$, with $a\in G$ going through the elements of your group, form a partition of $G$ even when $N$ is not a normal subgroup (not trivial, but also not hard to show). Therefore, you can take quotients w.r.t. any subgroup, and this quotient taking will identify elements up to something you are not interested in. However, if your subgroup is not normal, this partitioning is quite useless because you cannot do any further algebra with it.

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The elements of $\Bbb R/2\Bbb Z$ are of the form: $$\{x,x\pm2,x\pm4,\ldots,x\pm 2n,\ldots\},$$ where $0\le x<2$.

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It is often helpful to realize how the cosets are formed. You're really talking about the collection $$A/B = \{ a + B : a\in A\}$$ Now it may happen that $a_1+B$ and $a_2+B$ are the same set for two (or more) different $a_i$'s. In the case of $\mathbb Z / 2\mathbb Z$, there are actually only two different such cosets, $0+2\mathbb Z$ and $1+2\mathbb Z$. All of the other $n+2\mathbb Z$ are one or the other of these. – MPW Jan 6 at 20:49
    
Yes. I was referring specifically to the original case. Perhaps I should add more about the other case – MPW Jan 6 at 20:51
    
No, leave it! It is instructive and correct. Oh, too late :( – MPW Jan 6 at 20:52

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