I try to understand the proof of the First Sylow theorem:
A finite group whose order is divisible by a prime $p$ contains a Slow $p-$subgroup.
The proof is that
Let $S$ be the set of all subsets of $G$ of order $p^e$. One of the subsets is a Sylow subgroup. We will show that the one of the subsets $[U]$ of order $p^e$ has a stabilizer of order $p^e$. That stabilizer will be the subgroup we are looking for.
We decompose $S$ into orbits for the operation of left multiplication, obtaining an equation: $$ N=|S|=\sum_{orbits O} |O| $$
Question 1: for left muplication, there is only one orbit (transitive operation). Why can we get this decomposition for $S$? I mean there is just only one orbit $O$ right?
According to Lemma Why can we say that `This is a transitive operation'?, $p$ does not divide $N$. So at least one orbit has an order that is not divisible by $p$, say the orbit $O_{[U]}$ of the subset $[U]$. Lemma tells us that the order of $H$ divides the order of $U$, which is $p^e$. (Question 2: why can we say the order of $H$ is $p^e$?). So $|H|$ is a power of $p$. We have $|H|\cdot |O_{[U]}|=|G|=p^em$, and $|O_{[U]}|$ is not divisible by $p$ (Question 3: Is this one by the counting formula: $|G|=|G_s||O_s|$ where $G_s$ is the order of stabilizer?). Hence, $|O_{[U]}|=m$ and $|H|=p^e$. So $H$ is a Sylow $p$-subgroup.