coset space

(group theory)

Let H be a subgroup of a group G. Define Ω := {Hg | g in G} - the set of right cosets of H in G. We can then define an action by G itself on Ω thusly :

for ω = Hg in Ω and x in G, define ωx := Hgx

This clearly satisfies the axioms of an action. Ω equipped with this action is called the coset space of H in G, denoted (G : H). Note that (G : H) is transitive - that is, its one orbit consists of the whole space.

Now, the essential reason that coset spaces are interesting is contained in the following theorem -

Theorem : Every transitive G-space is isomorphic to a coset space.

Here "G-space" means a set acted on by a group G, and "isomorphic" means there is an isomorphism between the space and the coset space - in the context of group actions, an isomorphism between the G-spaces X and Y is a bijective map T : X->Y s.t. for all x in X and g in G, T(xg) = T(x)g. This definition means that isomorphic G-spaces have the same "structure" under G. So this theorem shows that we can reduce any problem concerning transitive G-spaces of any kind to one concerning coset spaces on G, and hence show a whole host of situations to be essentially the same. And, since a non-transitive G-space is partitioned by its orbits, each of which can be considered as a transitive G-space, the entire of the theory of group actions can be considered in terms of coset spaces via this theorem.

Proof of the theorem : In particular, we show that if we pick an ω in Ω and let H := Stab(ω), then Ω is isomorphic to (G : H). Define T : (G : H) -> Ω by setting T(Hg) := ωg. T is onto (surjective) by the transitivity of &Omega, but we need to show it is 1-1 (injective) and, since we defined it indirectly, well-defined.

Now, for g₁, g₂ in G :
Hg₁ = Hg₂
<==> g₁g₂^-1 in H
<==> g₁g₂^-1 in Stab(ω) (H = Stab(ω))
<==> ω(g₁g₂^-1) = ω (by definition of stabilizer)
<==> ωg₁ = ωg₂ (by definition of an action)

Reading down this chain of deduction we have a proof that T is well-defined, and upwards that it is 1-1. So T is bijective. So to show it is an isomorphism, we need only the following :
T(Hxg) = ω(xg) = (ωx)g = (T(Hx))g

Which completes the proof.

One simple and important corollary - for a non-transitive G-space, we have that for each g in G, Orb(g) is a transitive G-space. Applying the theorem, Orb(g) is isomorphic to (G : Stab(g)).

To the soft-linker who linked this to In three years, after i've taken three classes on group theory, i'll probably understand this. For now, though, I guess I'll just vote it up, and to anyone who concurs - the idea of putting this on E2, with proper links to other write-ups on the concepts it uses, is that even someone with no idea about mathematics should be able to understand it by following the links down to things they can understand (probably basic set theory and notation) and then reconstructing upwards. I'm not saying that would be easy, but it is the idea. If anyone tries this, and finds any required concept that isn't properly dealt with here, please /msg me and I'll do my best to node it. (See Noether's w/u on I don't know what the hell this means, but I'm voting it up anyway. for a better expressed version of what I'm saying.)