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The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group, I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question. My Question is :

  1. Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$

    So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:

    Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??

I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?

Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.

IDOK
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    You're finding how many cosets of $B$ are present in $A$. The quotient group is the group of cosets. – Michael Hardy Oct 01 '11 at 16:54
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    One should consider the fact that the quotient group $A/B$ was originally sometimes called difference group and denoted $A-B$ (although in the abelian case only, I guess) – Mariano Suárez-Álvarez Oct 01 '11 at 16:56
  • @MarianoSuárez-Alvarez,@Michael hardy:fantastic sir – IDOK Oct 01 '11 at 17:08
  • Sir suppose can you tell me ,how does it work in the case of some other groups(other than cyclic groups),I mean consider $E(\mathbb{Q})/2E(\mathbb{Q})$ ,where $E(\mathbb{Q})$ is the Group of rational points on the curve,so how can you explain that Quotient group,and also the multiplicative group ${\mathbb{Q}}^{\times}/({\mathbb{Q}}^{\times})^2$,how can you explain the above groups with similar interpretation ,i mean tell me the intuition,the Second one was the Multiplicative group ,which means that the $3$ and $\frac{12}{25}$ are similar – IDOK Oct 01 '11 at 17:10
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    @iyengar, a minor pro tip: This is a fairly egalitarian place, and sir'ing everyone you speak to will tend to make them slightly twitchy. We all try our best not to let that show, because it's really irrelevant, but doing so takes some mental energy that could be used to provide a better answer instead. (Using some full stops now and then rather than commas wouldn't hurt either, but that's easier to ignore than obsequious mannerisms). – hmakholm left over Monica Oct 01 '11 at 18:56
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    @Henning Alas, you are not the first (and not even the second) to tell him this. – Alex B. Oct 04 '11 at 01:10
  • but i cant refrain myself form calling so,as i want to show respect for all those who spend their time for answering questions,formatting them providing large and good answers@AlexB. – IDOK Oct 04 '11 at 06:06
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    If the people you want to respect are feeling uncomfortable with "sir", you are doing them a favor by not using "sir" to address them. –  Oct 05 '11 at 05:41
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    @iyengar You might also be interested to know that your obstinate insistence on ignoring spacing rules had the effect that I wasn't notified of your comment, AlexB not being my nick, nor part of my nick on math.stackexchange. – Alex B. Oct 07 '11 at 06:39
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    @Iyengar I see that you are very interested in number theory and the Birch and Swinnerton-Dyer Conjecture. However I think the following piece of advice is beneficial to both you and me: Before attacking such advanced topics in mathematics, one should first learn basic ideas in abstract algebra, number theory, analysis etc. Did you know that quotient groups are very useful in proving things about groups of order $p^n$? One often passes to a quotient and then applies the inductive hypothesis to it.

    In fact after having understood the quotient group you can pass to the quotient ring, and then

    –  Feb 09 '12 at 21:28
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    start learning commutative algebra, etc. This is how one slowly builds up one's base. You claim in the past that you have little access to computers, material, live in an isolated place, etc. Why not start learning the foundations these incredibly important things first (before tackling BSD or any or the Millenium Prize Problems)? Regards. –  Feb 09 '12 at 21:31
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    @BenjaminLim : Sir, thanks a lot for your beautiful comments, its true that I was completely ignoramus in such matters in the beginning, but I slowly started reading basic things and also some part of graduate things, its then I slowly started understanding the things posted here. But due to the lack of teachers to advice me, and also lack of basic libraries, I have been dependent on these communities, and became a target for thousands of criticisms and jokes. Any-way thanks a lot sir, for your advice. – IDOK Feb 10 '12 at 03:02
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    @Iyengar Even though I am not as experienced as many of the high-rep users here on this forum, I can try to help you whenever I can. My email is on my profile you can send me one anytime. –  Feb 10 '12 at 03:19
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    @BenjaminLim : Its not the experience that counts, but its the good mind that counts. And its not a matter of reputation here. You just appear to me as a cloud with silver lining. I again have to remind myself that, not everyone in the world are Brusque, There are always people like you who are ethical, but the proportion is microscopic ( only a quite few ). But I salute your good and kind heart, and thank you once again sir. – IDOK Feb 10 '12 at 08:47
  • @Iyengar I just noticed that your reputation has dropped by about 400 points. Did you award a bounty or something? –  Feb 10 '12 at 10:42
  • @BenjaminLim : Yes sir, to one question which is here . Thanks a lot for your concern. But would you mind in answering it if you know something about it, and if you find a small chunk of free time in your valuable schedule. And to add something, I have seen that the web-site you have kept routed me to the web-site of "Prof.Alexander". I think system on my side has got some error, to route it somewhere else. – IDOK Feb 10 '12 at 11:04
  • @Iyengar No that is the website of my mentor. I have put it there intentionally. –  Feb 10 '12 at 11:13
  • @BenjaminLim : But I am sorry to say that I have sent an email to that address . – IDOK Feb 10 '12 at 12:56

7 Answers7

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Often I think of a quotient group in terms of (loss of) information. When we move from a group to its quotient group we lose some information about the identity of the elements. For example, when we map an element of the additive group of integers $\mathbf{Z}$ to the quotient group $\mathbf{Z}/10\mathbf{Z}$ we lose the information of all the other digits save the least significant one. In other words after moving down to the quotient group we can no longer tell the difference between 9, 999, or 314159. In this sense we then equate 9 with 99 et cetera

Why would we want do this, as it amounts to loss of information? Well, there are several reasons. Sometimes we are only really interested in the residual information. For example, when we study the set of numbers of the form $a+b\root 3\of 2+c\root 3\of 4$, where $a,b,c$ are integers, and we want to start adding, subtracting and multiplying them, we quickly notice that those operations are very similar to the corresponding operations involving polynomials $a+bx+cx^2$. The difference is that we are only interested in the value of the polynomial at a single point $x=\root 3\of 2$. This shows in the multiplication rule, because the polynomial $x^3$ takes the value $2$. In order to make this correspondence between polynomials and numbers more accurate we are forced to equate the polynomial $x^3-2$ with the polynomial $0$. This time we get a quotient ring instead of a quotient group (see algebra textbooks for such details), but the idea is that some things we have learned about polynomial algebra will carry over to our set of numbers, and that gives us the benefit of economy of thinking. We don't need to relearn everything from scratch, if the next time we are interested in $\root 3\of 3$ instead.

Sometimes quotient groups are forced upon us. We are not in possession of all the information. A simple example is the following. Assume that somebody is counting coins, but the only counting aid available to him is a light switch. Every time he tallies one more coin he will toggle the light switch: lit, dark, lit, dark,... He may or may not be able to keep track of the actual tally, but if somebody else comes to the room, or the tallyman gets confused, the status of the light switch will only tell whether an odd or an even number of coins have been counted, i.e. we have moved from the group $\mathbf{Z}$ to the quotient group $\mathbf{Z}/2\mathbf{Z}$. Another very common quotient group in mathematics is used to decribe an angle of rotation. Let's say that we are studying a planar object spinning about its center of mass. It may have completed God knows how many full revolutions, but when we enter the room and observe its position, we have no way of knowing anything else but the current direction pointed at by, say, a small arrow somebody painted on the object for this purpose. A full revolution corresponds to an angle of rotation $2\pi$, so the total angle of rotation will have an uncertainty that can be any integer multiple of $2\pi$. In other words, we can only see an element of the quotient group $\mathbf{R}/2\pi\mathbf{Z}$, not an element of $\mathbf{R}$.

J. W. Tanner
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Jyrki Lahtonen
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  • very precise and good and deep explanation,thank you sir, – IDOK Oct 01 '11 at 17:51
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    Also, rather than "lose information", one may also sometimes view it as "ignoring information which is not very relevant for some purposes". Moding out by the squares/multiples of $2$ keeps information about parity, which may be all that is needed in one particular instance (so that we are ignoring information which is not relevant but may be obscuring what is "really going on" by sheer volume). – Arturo Magidin Oct 01 '11 at 20:47
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    @Arturo A good way of phrasing it! I was trying to include something like that when writing the paragraph on the theme only interested in the residual information, but forgot some of it and didn't quite manage to convey it all. The answer is long enough as it is. – Jyrki Lahtonen Oct 01 '11 at 21:04
  • @JyrkiLahtonen: suppose can you tell me ,how does it work in the case of some other groups(other than cyclic groups),I mean consider $E(\mathbb{Q})/2E(\mathbb{Q})$ ,where $E(\mathbb{Q})$ is the Group of rational points on the curve,so how can you explain that Quotient group,and also the multiplicative group ${\mathbb{Q}}^{\times}/({\mathbb{Q}}^{\times})^2$,how can you explain the above groups with similar interpretation ,i mean tell me the intuition,the Second one was the Multiplicative group ,which means that the $3$ and $\frac{12}{25}$ are similar" – IDOK Oct 02 '11 at 07:31
  • @JyrkiLahtonen:can you interpret the Above groups in the same way sir – IDOK Oct 02 '11 at 07:31
  • @iyengar: I have very little to say about those that would fit here. The quotient group $\mathbf{Q}^{\times}/(\mathbf{Q}^{\times})^2$ can be best understood starting from the unique factorization of the integers. There will be one component modulo $2$ for each prime $p$ because we can reduce the exponent of any prime number modulo two by dividing it with a square. Also the units $\pm1$ keep their identity in the quotient. If we moded out cubes instead of squares, then $\pm1$ would become equated, because they are both their own cubes. – Jyrki Lahtonen Oct 02 '11 at 18:21
  • Similarly in the quotient group of the rational points of an elliptice curve 2-torsion and the free abelian part each leave their mark in the quotient group, but information about odd order torsion is lost, because those points will also be among the doubled points. I'm not very familiar with that theory, so I don't know, why we would want to equate doubled points with the point at infinity. – Jyrki Lahtonen Oct 02 '11 at 18:22
  • That was indeed a beautiful try,thanks a lot,can you try my previous one sir which is here,thank you much,@JyrkiLahtonen – IDOK Oct 03 '11 at 06:31
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    @iyengar I won't. I'm not familiar with the Tate-Shafarevich group at all. You should spend a couple of years becoming fully conversant with the examples on this thread and other absolutely necessary background material before you even look at elliptic curves. Then you should study the elliptic curves for a couple of years: recognize the holes in your theoretical background and fill them. Only after that is there any chance that you would understand what the Tate-Shafarevich group is measuring. I don't, but my 30+ years of math have been mostly targeting other stuff. – Jyrki Lahtonen Oct 03 '11 at 07:02
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    @iyengar Sorry, if my previous comment sounded harsh. It was meant as a genuine advice. I really think that you should concentrate on studying and understanding these simple concepts, and not obsess about Birch - Swinnerton-Dyer. – Jyrki Lahtonen Oct 03 '11 at 09:22
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Quotient groups $A/B$ count cosets of $B$ inside $A$. The counting even works well with addition.

The Cartesian plane forms a group A, and a line through the origin is a subgroup B. The cosets of B inside A are all the parallel lines. How many are there?

Suppose B is the line: $$ B = \{ (x,y) : y = 2x \}$$ or just B is y = 2x for short. Parallel lines are parameterised by their "intercept" b, the coset b + B is the line with slope 2 and intercept b. $$b+B = \{(x,y) : y = 2x+b\}$$ This means that there is exactly one coset for every real number. In some sense we have counted the parallel lines.

The way we count them even keeps track of addition. If I took a point on the line B, say (2,4), and added it to a point on 7 + B, say (3,13), then I get the point (5,17), which is on the line $$(0 + B) + (7 + B) = 7 + B.$$ If I add the point (3,7) on 1 + B to the point (8,20) on 4 + B, then I get the point (11,27) on $$(1+B) + (4+B) = 5 + B$$ This is just because if x = 11, then 2x is 22, and 27 is 2x+5.

If one wanted to be more precise, I suppose one should say (0,5) + B, since it should be an element of $A$ plus $B$, but just like $A/B \cong \mathbb{R}$, we only need one number here too.

Jack Schmidt
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In a sense, the quotient group is indeed a measurement of how many copies of your normal subgroup are within the larger group. In the simple example of $\mathbb{Z}/3\mathbb{Z}$, the group has three elements: one for the subgroup $3\mathbb{Z}$ itself and one for each of its two cosets, which, if you were to plot them on a number line, for example, "look" the same as the original subgroup. And if you put the subgroup and its two cosets together, you get the whole group $\mathbb{Z}$. So in a sense the quotienting in this case tells you how many subsets resembling $3\mathbb{Z}$ are needed to break down $\mathbb{Z}$. The thing that makes this different from arithmetic division, of course, is the fact that the quotient is also a group--the group structure just comes from the way the "copies" of the subgroup interact with each other.

Will Dana
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  • "but Sir,can you give me a different example rather than the congruence modulo,suppose can you tell me ,how does it work in the case of some other groups(other than cyclic groups),I mean consider $E(\mathbb{Q})/2E(\mathbb{Q})$ ,where $E(\mathbb{Q})$ is the Group of rational points on the curve,so how can you explain that Quotient group,and also the multiplicative group ${\mathbb{Q}}^{\times}/({\mathbb{Q}}^{\times})^2$,how can you explain the above groups ,i mean tell me the intuition,the Second one was the Multiplicative group ,which means that the $3$ and $\frac{12}{25}$ are similar – IDOK Oct 01 '11 at 17:05
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Two of the most basic concepts in mathematics, namely "Sets" and "Relations" are much useful to create new things from old. For example, "integers" can be constructed from natural numbers by putting an equivalence relation on the set $\mathbb{N}\times \mathbb{N}$ (see Concrete Abstract Algebra: Niels Lauritzen), rationals can be constructed from integers by putting an equivalence relation on $\mathbb{Z}\times \mathbb{Z}$; in some way reals, complex numbers can also be constructed. The only things we have used are "Sets and Equivalence Relation".

In this way, we can think of quotient group of a group $G$ as a set of equivalence classes of group $G$ with a natural binary operation, induced from that of $G$:

If $H$ is a subgroup of $G$, define an equivalence relation on $G$ by $x\sim y$ iff $x^{-1}y\in H$, denote equivalence class of $a$ by $[a]$. If $K$ is the set of equivalence classes of $G$, a natural way to define a binary operation on $K$ is to define $*\colon K \times K\rightarrow K$ by $[a]*[b]=[ab]$.

(So we have constructed a new set using equivalence relation and trying to put a binary on it) The necessary (and sufficient) for this map to be well defined (hence binary operation) is that $H$ should be normal subgroup of $G$. Once we convince for it, it is easy to verify other conditions in the definition of group. The new group is called as quotient group of $G$ by $H$ (denoted $G/H$)

Therefore, the quotient group of a group by normal subgroup is a set of equivalence of classes of $G$ with a natural binary operation, induced from $G$

Beginner
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I modify Jack Schmidt's exquisite answer for more details.

Quotient groups $A/B$ count cosets of $B$ inside $A$. The counting even works well with addition.

The Cartesian plane forms a group A, and a line through the origin is a subgroup B. The cosets of B inside A are all the parallel lines. How many are there?

Suppose B is the line: $ B = \{ (x,y) : y = 2x \}$ or just $y = 2x$ for short. Parallel lines are parameterised by their "intercept" b. Hence the coset b + B is the line with slope 2 and intercept b. $$b+B = \{(x,y) : y = 2x+b\}$$ This means there is exactly one coset for every real number $x$. In some sense we've counted the parallel lines.

The way we count them even keeps track of addition.
Take a point on the line $\color{blue}{B = 0 + B}$, say (2,4).
Add it to a point on $\color{blue}{7 + B}$, say (3,13).
Then I get the point $(2,4) + (3,13) = (5,17)$,
which is on the line $\color{blue}{(0 + B) + (7 + B)} = 7 + B.$

If I add the point (3,7) on 1 + B to the point (8,20) on 4 + B, then I get the point (11,27) on $$(1+B) + (4+B) = 5 + B$$ This is just because if x = 11, then 2x is 22, and 27 is 2x+5. To boot, $y = 2(11) + 5$.

If one wanted to be more precise, one should say (0,5) + B, since a coset should be an element of $A$ plus $B$. But just like $A/B \cong \mathbb{R}$, we only need one number here too.

0

The quotient group is the result of a simplification done by an homomorphism

https://math.stackexchange.com/a/69063/53203 mentions that the quotient subgroup is a type of subgroup but "with less information".

This is also where my intuition lies, but I would like to make that a bit more precise by uttering the key missing keyword: "homomorphism". This is going to be a more focused subset of this other answer: Intuition behind normal subgroups

An isomorphism is a bijective function between two groups (of the same size since it's a bijection) and means that they are the exact same as far as the group structure is concerned. Pretty boring.

An homomorphism however does not have to be a bijection, only surjection: it can take a larger group and transform it into a smaller image group. Notably, several distinct inputs can map to the same output.

The tradeoff is that this smaller group contains a "coarser" group structure than the original group, as it ignores some finer part of the original group (preview: that finer part is the normal subgroup structure). This image structure is simpler because the homomorphism can map multiple input elements to a single output element.

Now, as I have explained in more detail at: Intuition behind normal subgroups:

by the fundamental theorem on homomorphisms, there is a one to one relation between homomorphisms quotient groups (or normal groups):

  • for every homomorphism, the image is a quotient group
  • for every quotient group, there is a corresponding homomorphism

Therefore, the quotient group is always the result of a simplification done by an homomorhpism.

I like this intuition, because it is very easy to understand what an homomorphism is: it is just a function that keeps group structure.

And now we've just seen that every quotient group maps one to one to this thing that is easy to understand.

By the fundamental theorem on homomorphisms, this is how every homomorphism looks like.

From this it is clear how the structure of the quotient G/N is simpler than the original G: we collapsed the structure of the entire normal group N to a single point! Therefore, an homomorphism is basically a simplification function that ignores the structure of the normal group while doing its transformation.

This also makes it clear why the name "quotient" is used: it is because we are splitting the larger group G into two smaller groups:

  • the normal group N
  • the quotient group G/N, which is specified by N

which gives us a clear analogy to integer multiplication and division.

It is important to note however that this intuition is only valid in the "division sense": because groups multiplication is more complex than integer multiplication (notably, non-abelian), there is no simple known way of "multiplying two smaller groups to recover a larger group".

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$$ \begin{array}{c|c|c} A & B & C \\ D & E & F \\ G & H & I \\ J & K & L \end{array} $$

$12\div 4 = 3$ because $3$ is how many sets of $4$ it takes to make a set of $12.$

A normal subgroup of order $4$ in a group of order $12$ has $3$ cosets; thus the quotient group has order $3$.