I'm trying to understand a proof of the 2nd and 3rd parts of Sylow's Theorem. In some preliminary work, my book establishes that $P$ is a Sylow p-subgroup of $G$.
Then it defines $\{P_1,P_2,...,P_r\}=S$
to be the set of all conjugates of $P$, i.e. $S=\{gPg^{-1} | g \in G\}$.
Then, Let $Q$ be any p-subgroup of $G$. Then, by definition of $S$, $G$ acts by conjugation on $S$.
My first question is this: I understand what it means for a group to act by conjugation on itself. We have $g.a = gag^{-1}$ where $a,g \in G$. However, in this case, $S$ is a set whose elements are also sets. So then, if $s \in S$ then $g.s$ looks like $gsg^{-1}$ but this doesn't actually make sense, since $s$ is not actually an element in $G$ but a set of elements in $G$ so it's unclear to me what the operation looks like. Is it reasonable to assume that I'm supposed to use the group operation iteratively here? I.e. $g.s = \{gs_0g^{-1} | s_0 \in s\}$?
My next question is about orbits. My book states that $S = O_1 \cup O_2 \cup ... \cup O_s$ where $O_i$ is one of the orbits of $G$ on $S$ which I believe. It then goes on to say that $r = |O_1| + ... + |O_s|$ where $r$ is the number of conjugates used earlier. I don't know why this would be true. Thanks in Advance.
$S$ is a set of subsets. You can conjugate a subset: $$gXg^{-1} = \{gxg^{-1} \mid x \in X\}$$ So the action on $S$ is by conjugation of the subsets. If $gPg^{-1}$ is a conjugate then the action of $h$ sends $gPg^{-1}$ to $hgPg^{-1}h^{-1} = (hg)P(hg)^{-1}$ which is another conjugate of $P$.
As for why $r$ is the sum of the sizes of the orbits. This is because the orbits partition $S$. Every conjugate in $S$ is in exactly one orbit and every orbit contains only elements of $S$. So counting all the elements in $S$ is the same as counting all the elements in all the orbits.