Let $G=SO(3)$ and $K$ be the subgroup of $G$ .Let $K$ be the rotations around $Z$ axis .
$$K = {k(ϕ) : 0 ≤ ϕ < 2π}$$
$$K(ϕ) =\begin{pmatrix}cos ϕ & − sin ϕ &0\\sin ϕ &cos ϕ &0\\ 0 &0& 1 \end{pmatrix}$$ $K$ is a subgroup of $G$, and in fact $K = S_1$ .Introduce the coset of K as $$gK = \{gk : k ∈ K\}$$
The coset of $G/K$ is given by $$G/K = \{gK : g ∈ G\}.$$
What this coset intutively means can any one please explain ??
The wiki saays that let G be the group and H be the subgroup of G then the left coset is defined as $gH =\{gH :h$ an element of $H\}$ is the left coset of H in G with respect to g.
How it can be said that $SO(3)/SO(2)~= S^2$
$K$ consists of rotations in the $xy$-plane.
Consider what happens when you perform one of these, and then follow it with some other (general) rotation $g$ (e.g., rotate the $y$ axis 20 degrees towards the $z$ axis). The resulting rotations constitute the coset $gK$. In general if $g \notin K$, this will not contain the identity rotation.
What is $G/K$? (BTW, it's not a "coset", it's a "quotient".) It's the set of rotations where two rots $g, g'$ are considered equivalent if there's a rotation $h$ of the $xy$-plane with the property that $g = g'h$. To put it differently, they're equivalent if $g'^{-1}g$ is an $xy$-plane rotation. That means that they're equivalent if $g'^{-1}g$ leaves the north pole $N = (0,0,1)$ fixed... or you could say that $g$ and $g'$ both send $N$ to the same place.