Basic understanding of quotients of "things"?

Question

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.

Answer 1

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 $2\pi=0$). :)

Answer 2

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

Answer 3

You are correct that quotients always involve a partition into equivalence classes, which provides a coarser or zoomed out view by identifying elements that only differ by some detail you choose to ignore.

However, this partition is usually not useful unless it respects the original mathematical structure, so that you can keep doing maths at the coarser level in a way that's compatible with, and informative about, the original structure. This structural compatibility requirement is what distinguishes quotients from arbitrary partitions.

In algebraic structures, compatibility means that the results of algebraic operations stay in the same equivalence class if you exchange operands by other members of their respective equivalence classes. Wikipedia calls this a congruence relation

In the special case of groups, this manifests in the fact that you can only take quotients of groups by normal subgroups.

Indeed, for any subgroup $N$ of a group $G$, the relation $a\sim b:\Leftrightarrow ab^{-1}\in N$ on $G$ is an equivalence relation and therefore forms a partition $H$ into disjoint subsets, or equivalence classes (proof below). However, the map $\phi\colon G\to H$ that sends elements to their equivalence classes respects the group structure of $G$ if and only if $N$ is normal. Furthermore, if $N$ is normal, then there is a natural element-wise group structure on $H$.

Proposition: The following are equivalent

1. For all $a\in G$ and $n \in N$, we have $ana^{-1}\in N$ (usual normality definition)
2. For all $a_1,a_2 \in G$ such that $a_1\sim a_2$, we have $a_1^{-1}\sim a_2^{-1}$. For all $a_1,a_2, b_1, b_2 \in G$ such that $a_1\sim a_2$ and $b_1\sim b_2$, we have $a_1b_1 \sim a_2b_2$ (congruence conditions)
3. $h_1\star h_2 := \{g_1g_2 : g_1\in h_1, g_2\in h_2\}$ defines a group structure on $H$ ("element-wise group structure")
4. There is a group structure on $H$ such that $\phi$ is a group homomorphism ("structure on equivalence classes compatible with original structure")

Furthermore, if any and therefore all of the conditions above are satisfied, then $\phi: G\to (H, \star) =: G / N$ is a homomorphism and $G / N$ is called the quotient of $G$ by $N$.

Summary: For any subgroup $N$, you can form equivalence classes by identifying elements that are equal up to left multiplication by an element of $N$. However, unless $N$ is normal, this partition is useless because your notion of equivalence is not invariant under group operations, you cannot define a natural group structure on the equivalence classes, and you cannot define any group structure on the equivalence classes that is compatible with the original group structure.

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Appendix:

Proof that $\sim$ is an equivalence relation: reflexivity, symmetry, and transitivity are directly equivalent to the axioms of subgroups, $e\in N$, $n^{-1}\in N$ and $n_1n_2\in N$ respectively.

Proof of proposition:

Some trivial facts that we'll be using that are true for all subgroups $N$:

• $N \in H$
• $g \in \phi(g)$ for all $g\in G$.
• $\phi^{-1}(N)=N$. In particular, $\phi(n)=N$ for all $n\in N$. In particular, $\phi(e_G)=N$.

(1)$\Rightarrow$(2): We first prove the second part: If $a_1\sim a_2$ and $b_1\sim b_2$, then $a_1a_2^{-1}\in N$ and $b_1b_2^{-1}\in N$ and therefore $$ (a_1b_1)(a_2b_2)^{-1} = (a_1(b_1b_2^{-1})a_1^{-1})(a_1a_2^{-1}) \in N, $$ that is, $a_1b_1\sim a_2b_2$. Next, to prove the first part, apply the second part twice to combine the three equivalences $a_1\sim a_2$, $a_1^{-1}\sim a_1^{-1}$ , and $a_2^{-1}\sim a_2^{-1}$ into $a_2^{-1} \sim a_1^{-1}$.

(2)$\Rightarrow$(3): First, $h_1\star h_2$ is an element of $H$: In fact, for any two elements $g_1g_2,\tilde{g}_1\tilde{g}_2$ of $h_1\star h_2$, we have $g_1\sim \tilde{g}_1$ and $g_2\sim \tilde{g}_2$ and by the second part of (1) we obtain $g_1g_2 \sim \tilde{g}_1\tilde{g}_2$ (it's annoying to show that $h_1\star h_2$ is not just a subset of an equivalence class but in fact all of one -- so lets just say we define $h_1\star h_2$ as the containing equivalence class). Second, it's easy to show that $N=e_H$. Third, inverses exist. In fact, for $h\in H$ define $h^{ -1}:=\{ g^{-1} : g\in h \}$. Any two elements $g^{-1}$ and $\tilde{g}^{-1}$ of $h^{-1}$ satisfy $g^{-1}\sim \tilde{g}^{-1}$ by the first part of (1); thus, $h^{\vee}\in H$ (again, annoying to show it's the full equivalence class). Since, $e_G\in h\star h^{-1}$ and $e_G\in N$, we further have $h^{-1}h = N = e_{H}$. Fourth and finally, associativity is trivial.

(3)$\Rightarrow$(4): If $(H, \star)$ is a group, then $\phi\colon G\to (H, \star)$ is a homomorphism of groups. In fact, since $ab$ is contained in both $\phi(ab)$ and $\phi(a)\star\phi(b)$ we have $\phi(ab) = \phi(a)\star\phi(b)$.

(4) $\Rightarrow$ (1): If $H$ has a group structure such that $\phi$ is a homomorphism, then $e_H = \phi(e_G) = N$. For $a\in G$ and $n\in N$, we therefore have $$ \phi(ana^{-1})=\phi(a)\phi(n)\phi(a^{-1})\\ =\phi(a)N\phi(a)^{-1}\\ =\phi(a)e_H\phi(a)^{-1}\\ =e_H\\ =N,$$ from which we conclude $ana^{-1}\in \phi^{-1}(N)=N$.

Additional fun facts that really don't have much to do with the original question anymore: the equivalence relation $\sim$ defined above identifies elements that are equal up to left multiplication by elements of $N$. In other word, the equivalence class of a given element $a\in G$ is the right coset $Na:=\{ na : n\in N \}$. That's arbitrary! And it's only the same as "equal up to right multiplication by elements of $N$" if -- you guessed it -- $N$ is normal. In that case, the right coset $Na$ is the same as the left coset $aN$, and the group structure on $H$ that we defined in part (3) of the proposition above can equivalently be defined via $aN \star bN := (ab) N$.