Lemma: $G$ is a finite group, $p$ a prime number with $|G|=p^lm$ with $l\geq 1 and p\nmid m$. Let $s$ denote the amount of sylow groups, then $s|m$.
Proof: Is once more pretty straight forward. Let $M=(S_1,S_2,...,S_s)$ be the set of all sylow-groups of G. Because of the second Sylow-theorem the conjugation-operation $$ G\times M \rightarrow M\, , (a,S)\mapsto aSa^{-1} $$ which is transitive and also for the orbit $G(S)$ of $S$ it holds that $G(S)=M$ for all $S\in M$ so that $G(S_1)=G(S_2)...=G(S_s)$ so basicly we only have one representation of the orbits. With the orbit-equality we get $$ s:=|M|=|G(S)|=|G:\text{Sta}_G(S)| $$ with $\text{Sta}_G(S)$ denoting the stabilizer. Now because the Stabilizer is a Centralizer/Normalizer under conjugation, we get $$ S \lhd \text{Sta}_G(S)=\text{Nor}_G(S)<G $$ and with Lagrange $$ m=|G:S|=|G:\text{Nor}_G(S)||\text{Nor}_G(S):S|=s \cdot |\text{Nor}_G(S):S| $$
What I don't understand is why we can write $$ |G:S|=|G:\text{Nor}_G(S)||\text{Nor}_G(S):S| $$ Any ideas?