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The background for this question is that I know that many students starting to learn algebra focus on the standard construction of the quotient group $G/N$ instead of working with the universal property. Therefore it would be nice to give another construction, thus avoiding cosets, but so that the universal property is clear. This idea came to me years ago, but I've never found something.

For another example, there is a very nice and intuitive construction of the localization of a ring: $S^{-1} A := A[\{X_s\}_{s \in S}] / (s X_s = 1)$. The idea is: Invent new elements, and force them to be inverses to your elements of $S$. The universal property follows from the universal property of quotient and free algebra.

Perhaps quotient groups are a bit too elementary so that there could be another nice construction (please don't answer when you've just got this to say) ... anyway, any ideas are welcome :). I'm pretty sure that in this case there is no better one, but perhaps another one. I play around with the Cayley representation of $G$ ...

Sorry if this is too elementary for you ;-)

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Uh.. You realize that the construction is necessary to prove that the category has quotients. It doesn't just fall out of thin air. – Harry Gindi Feb 7 '10 at 16:06
the question is not about how one should develop the theory of the category of groups ... – Martin Brandenburg Feb 7 '10 at 16:10
Martin- I really have no idea what you're asking for in this question. The standard construction does make the universal property clear (obviously a map sending N to the identity send each coset a single point). – Ben Webster Feb 7 '10 at 16:13
The question seems clear, and is a reasonable one. I doubt there is any good answer, but I'd love to see one if there is. – Tom Church Feb 7 '10 at 16:19
Please note that there is a discussion on this subject on pp. 57-58 of Mac Lane. In particular, on the top of p. 58 it is stated that the universal property of the quotient group can be obtained without cosets at all, using the AFT. (I did not check the details; if I do, I will send a complete answer. In any case, I suspect that it is the Representabilty Theorem of p. 122 of Mac Lane that is really required.) – user2734 Feb 7 '10 at 16:48
up vote 3 down vote accepted

In order to make the construction of the quotient, it seems that it would be helpful to have at hand some quotients of the group $G$, described in some way other than formally via cosets.

Here is one possible approach:

Generalizing the regular action of $G$ on itself, we have the action of $G$ on its power set (given by translating a set).

If $S \subset G$ is any subset, we can look at the orbit of $S$ under $G$, which is some subset $\mathcal O_S$ of the power set of $G$, and we are given a homomorphism $G \to Perm(\mathcal O_S)$ (the group of permutations of $\mathcal O_S$). Let $G'$ be the image of $G$ under this homomorphism.

If we now take $S$ to be a normal subgroup $N$ of $G$, then $G' = G/N$. I wonder how hard this is to show directly, in terms of the universal property?

It is easy to see at least that $G \rightarrow G'$ has the property that $N$ is in its kernel. Can one show directly in this set-up that any map $G \rightarrow G''$ having $N$ in its kernel factors through the explicitly constructed quotient $G'$?

I don't see a pithy argument straight away, but it doesn't seem too unfeasible to me.

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this is exactly what I've been looking for! I don't see how the universal property can be verified directly from the construction (perhaps anybody else?), but for me it's sufficient that $p : G \to G'$ is a surjektive homomorphism with kernel exactly $N$. – Martin Brandenburg Feb 8 '10 at 0:59
Restatement: Let G' be the image of G acting by left translation on the set of left cosets of N in G. (Excuse my use of the C-word!) Amusingly, when N is a non-normal subgroup, this produces the quotient by the largest normal subgroup contained in N, whereas (as remarked by Tom Church) the unknown/Brandenberg/Poonen construction produces the quotient by the smallest normal subgroup containing N. – Bjorn Poonen Feb 8 '10 at 5:25
Dear Bjorn, Yes! – Emerton Feb 8 '10 at 5:43

(This was too long to fit in a comment.) Unknown's suggestion in the comments can be made just about as explicit as your construction of localization. It's just that the construction of the polynomial ring and the ideal generated by a set of elements and the quotient ideal are so familiar that one doesn't notice how much is involved in the construction of localization.

Let $M$ be the set of pairs $(H,f)$ where $H$ is a group whose underlying set is a subset of $G$ and $f \colon G \to H$ is a group homomorphism sending $N$ to the identity. Then $G/N$ is the image of $G \to \prod_{(H,f) \in M} H$.

The reason for requiring the underlying set of $H$ to be a subset of $G$ is to ensure that $M$ is a set. It is clear that the result has the universal property. It is also clear that this construction uses very little about the variety of groups.

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@BP: Even if I had found the solution set, I would not think of this nice explicit construction. Thank you for this answer. – user2734 Feb 7 '10 at 18:45
My goodness, this actually works. But from this definition you don't know what the kernel of $G\to "G/N"$ is. Reminds me of a joke. Q: What is 2+3? A: I don't know but it is the same as 3+2 because the addition is commutative. – VA. Feb 7 '10 at 20:14
Surjectivity of $G\to G/N$ implies nothing. Please prove that the kernel is $N$, without using cosets. No, please. P.S. The joke is due to Arnold. – VA. Feb 7 '10 at 20:22
To understand VA's point, consider the above construction applied to a non-normal subgroup of G, in which case the kernel is the normal closure of your subgroup. Without explicitly constructing the quotient, how do you know that for N normal in G and x not contained in N, there is some homomorphism G -> H vanishing on N but not on x? – Tom Church Feb 7 '10 at 21:33
By the way, I agree with Martin that the construction of G/N given in this "answer" is not a handy one. I think its pedagogical value lies mainly in the fact that a similar argument can sometimes be used in other situations to prove the existence of universal objects. – Bjorn Poonen Feb 7 '10 at 23:30

I have tried a method with a combinatorial group theory class, which is first to look at the universal quotient property relative to equivalence relations on a set. You explore this idea, and show that the set of equivalence classes is the answer you want and you derive the `first iso theorem' in that context. Exploring it to death if you like, with trivial do-able examples and constructions that students know. (e.g. different pairs (a,b) of integers giving the same rational numbers etc.) A function defines an equivalence relation and the equivalence relation defines a function and they are related. NB. no algebra at all in sight. This is strictly constructions on sets and functions.

The idea of the universal property is now centre stage i.e. where it should be. You can now ask if something similar is going to happen for groups and homomorphisms. The point is that it is not dividing out by a (normal) subgroup that is at stake here but by an equivalence relation (and that will not work of course). Equivalence relations seem to be more basic perhaps.

We all know the problem and the answer but the student has then something nearer to their previous experience. The equivalence relation does not work since although the quotient does seem to have an 'obvious' multiplication, that does not work and simple examples show it not to be well-defined (NB. 'Well definition' is a simple concept but students can (and do) find it very hard to grasp. This way, it is met as 'ill definition' first. The notion of well definition is perhaps not understood because it is usually presented as being a solution to a problem that the student has not ever met themselves.)

You now examine what goes wrong and get to the idea that you need a congruence not simply an equivalence relation. (Here I have used a categorical approach without the use of the term category, and really looked as a congruence as being an equivalence relation 'internal' to the category of groups... and no need to mention categories unless you feel like it. You can make the transition from equivalence relation to congruence in any way that seems appropriate for the background of the students.) This is the hard bit, not cosets, and it is hard, but sort of avoided in the usual treatment.

Now normal subgroups can be shown to be a reformulation of congruences and you can flip from one to the other and back again with no loss of info.

I like this treatment as the cosets only appear at the last moment, and then are the group theory analogue of equivalence classes. (If we have not got those across before we do cosets then there is no hope!)

This treatment also concentrates on the algebraic details that really need understanding. I do not mean group theory details as they are more specialised. We have equivalence classes, well definition and a universal property, three big ideas.

Normal subgroups come out as being natural.

Did this approach work? Not all the students could handle the idea of cosets, but they did seem happier with equivalence classes, well definition etc, and also seemed to like the universal property idea, which they met in various other contexts (product groups, free product, free group, etc.) in that course.

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Students don't understand well-definition because they don't understand the concept of a function. – Harry Gindi Feb 7 '10 at 17:55
That is a possible reason. The problem in that case would be to construct examples that attack both lacks of understanding. I think my point is that a possible analysis of the coset problem is that there are too many different things going on when students first meet them. (I do not know if the maths eduction and psychology people have looked at these areas, has anyone any idea?) – Tim Porter Feb 7 '10 at 18:10
I also think that it's very important to introduce equivalence relations before cosets. before you collapse groups, you should have collapsed some sets ;-). – Martin Brandenburg Feb 7 '10 at 20:25

What one really wants is to define a homomorphism with kernel equal to N. One way of doing it would be to define a function from G to the free group generated by the elements of G and then to identify everything you need to identify (in the free group) to make this map a homomorphism that vanishes on N. For instance, f(x) would be identified with the identity for every x in N, f(x)f(y) would be identified with f(xy), and so on. But proving that something like this works would presumably require you to go through more or less exactly the same steps you have to go through to prove that multiplication of cosets is well-defined, so it might end up being an unnecessarily complicated way of presenting essentially the same proof.

This is of course the usual experience with universal constructions: you rarely manage to get something for nothing.

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"it might end up being an unnecessarily complicated way of presenting essentially the same proof." unfortunately, yes. besides you also use quotients (by identifying things) in this construction. – Martin Brandenburg Feb 7 '10 at 16:41
I don't claim it's a good way of doing things, but one could (in desperation) argue that here one is defining quotients in just one situation (what you need to define a group in terms of generators and relations) and getting all quotients out of it. But I wouldn't want to go to the wall on this one ... – gowers Feb 7 '10 at 17:48

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