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 ;-)



alternative construction of the quotient group
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.


