15 June 2016

A few days ago, one of the viewers in my stream was commenting about how they were having trouble understanding quotient groups. I tried to give some examples and intuition on the fly, and more or less completely failed. So that got me thinking: what is a good way to motivate quotient groups?

## Groups are determined by how they act on sets.

Group constructions, and especially the quotient group construction, can look like very arbitrary symbol-pushing. Rather than trying to think about the group operation directly, it can make more sense to think about how the elements act on sets.

A group action of a group $$G$$ on a set $$S$$ is a function $$G \times S \to S$$, usually written in a similar manner to multiplication, so the result of $$g$$ acting on $$x$$ is $$gx$$. The function needs to be compatible with the group operation, so $$ex = x$$, where $$e$$ is the identity element, and $$g_1(g_2x) = (g_1g_2)x$$.

The group operation itself is a function $$G \times G \to G$$ and it satisfies these constraints, so any group can be seen as acting on itself via multiplication, and if you had this group action then you could determine the multiplication operation. So that indicates that actions determine the group. I believe that the correspondence can be made rigorous, but it is probably rather technical to avoid circular definitions, and so instead let's just use it as a guide toward the usual definitions.

## Quotients are a way to "undo" products.

With numbers, if $$c$$ is the product of $$a$$ and $$b$$, then $$a$$ is the quotient of $$c$$ by $$b$$, and $$b$$ is the quotient of $$c$$ by $$a$$. Our goal is to make something similar for groups.

Imagine we have $$G = A \times B$$, so an element of $$G$$ is a pair $$(a, b)$$. We want to be able to recover $$B$$, or something like it, from knowledge about $$G$$ and $$A$$. An action of $$B$$ on a set $$S$$ can always be thought of as an action of $$G$$ on $$S$$, by having $$A$$ act trivially. In other words, define $$(a, b)x = bx$$, where the right side of the equation comes from the action of $$B$$.

We can also go the other way. Starting from all of the actions of $$G$$, let's consider only those where $$A$$ acts trivially, i.e. $$ax = x$$ for any $$a \in A$$. Here we're identifying $$A$$ with its image in $$G$$.

What should an element of the quotient look like? It should come from an element of $$G$$, but some elements of $$G$$ should map to the same element in the quotient. That should happen exactly when those two elements always act the same. Since we're considering actions where $$A$$ always acts trivially, if $$a \in A$$ and $$g \in G$$, $$ag$$ and $$g$$ always act the same, and therefore should map to the same quotient group element.

So we can define an equivalence relation on $$G$$ by saying $$g \sim g'$$ if $$gg'^{-1} \in A$$. If two elements are in the same equivalence class, they should map to the same element in the quotient group. We'll use a little trick here, and define the elements of the quotient are actually these equivalence classes themselves. With a little bit of thought, you can see that these equivalence classes are the cosets $$Ag = \{ ag\ |\ a \in A \}$$.

## The quotient definition.

We know what the elements of the quotient $$G/A$$ should be. Now we need the group action. The map $$\varphi : G \to G/A$$ taking an element of $$g$$ to the corresponding coset $$Ag$$ should be a group homomorphism, so we want $$(Ag)(Ag') = A(gg')$$. Up until this point, we haven't required $$A$$ to be a normal subgroup. This is where that condition becomes important.

In order for the equation $$(Ag)(Ag') = A(gg')$$ to be a valid group operation, we need $$A(gg')$$ to be the same regardless of which $$g$$ and $$g'$$ we pick from the coset. In other words, for any $$a, a' \in A$$, there must be some $$a'' \in A$$ such that $$aga'g' = a''gg'$$. Rearranging this gives $$ga'g^{-1} = a^{-1}a''$$, so for any $$a' \in A, g \in G$$, we need $$ga'g^{-1} \in A$$, which is exactly the condition that $$A$$ is normal.

With that condition, defining $$(Ag)(Ag') = A(gg')$$ does, in fact, give a valid group operation, which is the usual definition of the quotient. You might also do this definition with left cosets ($$gA$$ rather than $$Ag$$), but $$A$$ being normal ensures that the result is the same.

So we started with the goal of trying to undo a product operation, and came to a reasonably general construction that actually does a lot more. Exploring all of the applications of quotients is not something that I can do in a single blog post, so this will have to suffice.