Can I get some help reading a proof for classification of finite subgroups of $SO(3)$? There's one line I can't understand. The proof that I'm reading.
Here, we have $$\left|G \right|-1=\frac{1}{2}\sum_{p \in P} \left ( \frac{\left| G\right|}{\left|\text{Orb}(p) \right|}-1 \right )=\frac{1}{2}\sum_{O}\left ( \left| G\right|-\left| O\right| \right ) $$
It says this is done by rearranging to sum over orbits, but I don't understand what it means.
What is $O$ here?
Edit:
Thanks to ancient mathematician, I understand what $O$ is.
However, I'm still curious how RHS turned out; could you shed some insight on this?
 A: Each non-zero element of $SO(3)$ fixes a set of antipodal points on $S^2$ and no other points of $S^2$.
There is a natural action of $G$ on $P$.
Under this action of $G$, the set $P$ can be partitioned as a disjoint union of orbits of $G$.
Moreover, there is a map
$$f: G \setminus \{ 1 \} \to P/\{\pm 1\}$$
which sends each $g \in G$, with $g \neq 1$, to its fixed point set, which is a set of antipodal points.
We can thus write $G \setminus \{ 1 \}$ as the disjoint union of subsets of the form $f^{-1}(c)$, as $c \in P / \{ \pm 1 \}$. Hence
$$ |G| - 1 = \sum_{c \in P / \{ \pm 1 \}} |f^{-1}(c)| = \frac{1}{2} \sum_{p \in P} |f^{-1}([p])|, $$
where $[p]$ is the equivalence class of $p$ in $P / \{ \pm 1 \}$.
Note that $f^{-1}([p])$ is the stabilizer of $p$ in $G$ from which we have removed the identity.
Moreover, it is known that $G/Stab_G(p) \simeq \mathcal{O}_p$. This basically explains the first equality.
As for the second equality, well you combine terms which belong to the same orbit in $P$ under the action of $G$. If you take a particular $p \in P$, there are $|\mathcal{O}_p|$ such terms. This cancels with the denominator of the first term in the sum on the right-most side which then becomes simply $|G|$. The second terms also gets multiplied by $|\mathcal{O}_p|$ and this explains the second equality.
Edit 1: to make the second equality clearer, let us first start with the first equality:
$$ |G| - 1 = \frac{1}{2} \sum_{p \in P} \left( \frac{|G|}{|\mathcal{O}_p|} - 1 \right).$$
Suppose the orbit $\mathcal{O}_p = \{ p_1, p_2, \ldots, p_k \}$, say, where $p_1 = p$. When you are summing over $P$, such a sum contains a sum over $\mathcal{O}_p$, which contains $k = |\mathcal{O}_p|$ equal terms.
So, if $\Omega$ denotes the set of all orbits $\mathcal{O} \subseteq P$ under the action of $G$, we note that $\Omega$ is a partition of $P$ into orbits. We can then write
\begin{align*} |G| - 1 &= \frac{1}{2} \sum_{\mathcal{O} \in \Omega} |\mathcal{O}| \left( \frac{|G|}{|\mathcal{O}|} - 1 \right) \\
&= \frac{1}{2} \sum_{\mathcal{O} \in \Omega} \left( |G| - |\mathcal{O}| \right). \end{align*}.
