TITLE: Proof behind $S^n\cong SO(n+1)/SO(n)$
QUESTION [12 upvotes]: I have been trying to understand the fact that $S^n \cong SO(n+1)/SO(n)$. I believe I have the intuition correct at this point; consider the case when $n=2$ as we have $S^2 \cong SO(3)/SO(2)$.:
We are trying to find the correspondence between rotations in $\mathbb{R}^3$ and points on $S^2$. At first I incorrectly thought that these spaces were isomorphic, however one then realizes that there are more rotations than points on a sphere, in the following sense:
Consider the point $p$ at the "north pole" of the sphere $S^2$. We can correspond $p$ with any point on the sphere by rotating the sphere, that is, by applying elements on $SO(3)$ so that $p$ ends up at any place upon it. However, we can first apply any rotation around the $z$-axis (through $p$). So, by "modding out" these rotations, which are exactly the elements of $SO(2)$, we have our isomorphism.
I am looking for someone to help me formalize this into a proof. Thank you!

REPLY [15 votes]: It is basically the orbit–stabilizer theorem.
$SO(3)$ acts by rotations on $\mathbb R^3$. This action restricts to a transitive action on $S^2$. Fix a vector in $S^2$, say $e_1 = (1,0,0)$. One has a continuous map $SO(3) \to S^2$ given by $A \mapsto Ae_1$. The subgroup of $SO(3)$ stabilizing $e_1$, the "kernel" of this map, to abuse language, is the block-diagonal subgroup $H = \{1\} \times SO(2)$. It follows that the quotient $SO(3)/H$ is in continuous bijection with $S^2$. Because both spaces are compact Hausdorff, it is a homeomorphism.
As you seem to have noticed, there is nothing special about $n=3$ in this result.