Simple Characterization of the Hopf Fiberation as Cosets of the Circle Group (Stabilizer of a Point) A few months, ago before I took my first Algebra class, I asked a naive question  about the formula for one of the hopf fibers here 
A community member gave a really good answer, so I would recommend checking it out. My question is which version of the orbit-stabilizer theorem was used in making this characterization of the fibers? (Again, please refer to the answer provided in the link above).
A stabilizer of a point in $\mathbb{S}^3$ is the circle group $\mathbb{S}^1$. On way of written $\mathbb{S^1}$ is as $\cos(\theta)+i\sin(\theta)$. We can take cosets of $\mathbb{S^1}$ which will have the form $q(\cos(\theta)+i\sin(\theta))$ where $q\in\mathbb{S}^3$. Since quaternion multiplication is an isometrey (distance preserving linear transformation), the cosets of $\mathbb{S}^1$ are also unit circles. These are disjoint circles that fill up all of $\mathbb{S}^3$.
It follows from the orbit-stabilizer theorem that there is a bijection between the left cosets of the stabilizer and the orbit of a point. This is the version of the theorem I learned in my class and we only applied it to finite permutation groups. What is the special case of the orbit-stabilizer used in the answer I linked above? 
Is the orbit stabilizer theorem a version of the first isomorphism theorem? If so, does that make $\mathbb{S}^1$ the "kernel" in some sense. Sorry if last part of the question doesn't make any sense. I am just trying to understand this more concretely.        
 A: A $G$-set is a set $X$ equipped with a group action of $G$ on it. A morphism $X\to Y$ of $G$-sets is a function $f$ satisfying $f(gx)=gf(x)$ for all $(g,x)\in G\times X$. (It is variously called "intwertwining," "equivariant," or sometimes a $G$-morphism in various fields). If the function is invertible, then its inverse is also a $G$-morphism, and it may be called a $G$-isomorphism. The left coset space $G/H=\{ g'H\mid g'\in G\}$ is naturally a $G$-set with action $g(g'H):=(gg')H$ (which one must check is well-defined, BTW). If $H\le K\le G$ are subgroups then there is a projection map $G/H\to G/K$ defined by $gH\mapsto gK$ (which, again, is well-defined) which is an onto $G$-morphism. (Note if $H$ is trivial, then $G/\{e\}\cong G$ is itself a $G$-set, called the regular action, whose definition $G\times G\to G$ is just the group operation.) If $X$ is a $G$-set and $x\in X$, then the inclusion $\mathrm{Orb}(x)\to X$ is a one-to-one $G$-morphism.
Orbit-Stabilizer Theorem (algebraic version). The map $g\mapsto gx$ is a $G$-equivariant map $G\to\mathrm{Orb}(x)$ whose fibers are the cosets of $\mathrm{Stab}(x)$. Thus, $G/\mathrm{Stab}(x)\cong\mathrm{Orb}(x)$ are isomorphic as $G$-sets via the one-to-one correspondence $g\mathrm{Stab}(x)\leftrightarrow gx$.
By taking cardinalities, using the fact that cosets are equal size and partition the group, and Largrange's theorem, we may derive a corollary which is useful in combinatorics:
Orbit-Stabilizer Theorem (numerical version). $|\mathrm{Orb}(x)|=|G|/|\mathrm{Stab}(x)|$.
Note the division is undefined when $|G|$ and $|\mathrm{Stab}(x)|$ are the same infinite cardinal, although the equation $|\mathrm{Orb}(x)||\mathrm{Stab}(x)|=|G|$, or better, $|\mathrm{Orb}(x)|=[G:\mathrm{Stab}(x)]$ are true always.
In general, of course, if $H$ is not normal then $G/H$ does not have a sensible group operation defined on it (and writing "$gHg'H:=gg'H$" is actually ill-defined). Even if $H$ is normal, the group operation on $G/H$ is not the same as the group action of $G$ on $G/H$. (However, it is possible to relate the two: $H$ is normal iff the kernel of the action $G\curvearrowright G/H$ is $H$, so the action "factors through" $G/H$ by a so-called "universal property," which gives an action $G/H\times G/H\to G/H$, which then turns out, one can check, to be a group operation on $G/H$). For this reason, the first isomorphism is like an upgraded version of a special case of orbit stabilizer. ("Spiritually," though, I consider it the first isomorphism theorem in the category of $G$-sets, just as one can state first iso theorems in the categories of groups, modules, rings and various algebras.)

A (real or complex) Lie group $G$ is a (real or complex) manifold whose group operation is a continuous map $G\times G\to G$ (where $G\times G$ has the product topology) and inversion $G\to G$ (i.e. $g\mapsto g^{-1}$) is also a continuous map. It turns out, Lie groups admit unique differentiable structures, so without loss of generality they may be considered differentiable manifolds. There is also a difficult theorem that differentiable manifolds admit unique smooth and even analytic structures, so we could also assume analytic manifolds with analytic group operation and inversion (analytic means expressible with multivariable power series).
If $H$ is a closed subgroup of a Lie group $G$, then the coset space $G/H$ admits a manifold structure as well, and if there is a nice inner product on the lie algebra $\mathfrak{g}$ associated to $G$ ("nice" meaning left $G$-invariant and right $H$-invariant ... I think) which makes $G/H$ into a $G$-invariant metric space. (That is, the metric $d$ satisfies $d(gx,gy)=d(x,y)$ for all $x,y\in G/H$.) Note that in the context of actions of Lie groups $G\curvearrowright X$, one wants the action to be "nice," the nicest being a smooth map $G\times X\to X$, which will be the case here. In these cases, or else if we pick nice points $x\in X$, the orbit $\mathrm{Orb}(x)$ is a submanifold of $X$.
Orbit-Stabilizer Theorem (topological version). If everything is nice, the $G$-set isomorphism $G/\mathrm{Stab}(x)\cong\mathrm{Orb}(x)$ is a homeomorphism, and the aforementioned projection $G\to\mathrm{Orb}(x)$ is in fact a fiber bundle $ \mathrm{Stab}(x)\to G\to\mathrm{Orb}(x)$ in which the fibers are the cosets of $\mathrm{Orb}(x)$.
(Intuitively, saying $F\to E\to B$ is a fiber bundle means $E$ is a bunch of $F$s arranged in the shape of a $B$. If this sentence seems cryptic, replace the letters with words as necessary in the cases when the total space $E$ is a cylinder, Mobius band, torus, or Klein bottle.) Note we can upgrade everything to diffeomorphism, or
Orbit-Stabilizer Theorem (geometric upgrade). If everything is even nicer, $G/\mathrm{Stab}(x)$ is a Riemannian manifold (in particular, a metric space) and $G/\mathrm{Stab}(x)\cong\mathrm{Orb}(x)$ is an isometry, the action of $G$ on $G/\mathrm{Stab}(x)$ is by isometries, and the projection $G\to\mathrm{Orb}(x)$'s fibers are all isometric, i.e. same shape and size (moreover the element $g$ which takes $g'H$ to $gg'H$ restricts to an isometry between them).

Note the algebraic version of OST is stronger than the numerical version, since the former implies the latter but not conversely. There are can be same-size orbits of a group action $G\curvearrowright X$ which nonetheless correspond to distinct conjugacy classes of subgroups (the stabilizers of elements in an orbit form a conjugacy class of subgroup), i.e. the orbits are nonisomorphic as $G$-sets despite being same-size, so the algebraic version of OST says more than the numerical version can. For example, pick two same-size nonconjugate subgroups $H,K\le G$ of a finite group $G$ and consider the disjoint union $X=G/H\sqcup G/K$. For a small instance, pick $H=\langle(1234)\rangle$ and $K=\langle(12)(34)\rangle$ within the symmetric group $G=S_4$.
When $G$ and $X$ are topological spaces or smooth manifolds or Riemannian manifolds (in particular metric spaces), neither algebraic nor numerical versions of OST say anything about topology or geometry (shape and size), so the next two versions add to the first two. Unfortunately I don't know/remember all the adjectives that supplant the adjective "nice" nor do I have citations, but references should be scattered across standard advanced/comprehensive sources on differential geometry / Lie theory, and I know they apply for $S^3\curvearrowright S^2$.

In the case of the unit quaternions $S^3$, there is a group action $S^3\curvearrowright S^2$. The map $S^3\times S^2\to S^2$ is given by conjugation, i.e. $(r,\mathbf{v})\mapsto r\mathbf{v}r^{-1}$. Since this is just rotation, and we can rotate between any two vectors, the action is transitive, so in particular $\mathrm{Orb}(\mathbf{i})=S^2$, and so our various OSTs above say $S^3/S^1\cong S^2$ are isomorphic $S^3$-sets, diffeomorphic smooth manifolds, and isometric metric spaces, all via the same one-to-one correspondence.
