Lemma: $|G|=p^lm \,, p\nmid m,$. Let $S<G$ be a p-Sylowgroup and $H<G$ a subgroup with $p|\, |H|$. Then there exists a to $S$ conjugate p-Sylowgroup $S'=aSa^{-1}$ with: $$ H\cap S' < H \text{ is a p-Sylowgroup in H} $$
Proof: The proof is pretty straight forward, I understand the steps but my problem is understanding why the Lemma follows from these steps. It is clear that every conjugate subgroup $S'=aSa^{-1}$ is a p-Sylowgroup because the orders are the same. Here is the proof.
We look at the Set $M:=G/S$. Then $p\nmid |M|$, which follows from the definition of p-Sylowgroups and Lagrange-theorem. $G$ operates on $M$ trough $G\times M \rightarrow M, \, (b,aS)\mapsto baS$. This operation is transitive and for the stabilazor it holds that $$ \text{Sta}_G(S)=S, \, \text{Sta}_G(aS)=aSa^{-1}=:S' $$ for all $a\in G$. Now if we let the above operate only on $H$, we get $\text{Sta}_H(aS)=H\cap S'$. We look at the operations from $H$ to $M$. Because of $p \nmid |M|$ there is a least one orbit $H(aS)\subset M$ with $p \nmid |H(aS)|$. This follows from the orbit-equality. But because $$ |H(aS)|= \text{ind}(H:\text{Sta}_H(aS))=\text{ind}(H:H\cap S') $$ it follows that $H\cap S'<H$ is a $p$-Sylowgroup for the chosen $a\in G$.
It is the last point I don't understand. I understand that because of $p \nmid |H(aS)|$ it follows that $\text{ind}(H:H\cap S')=1$ right? But I don't understand how they get to the fact that it is a $p$-Sylowgroup.
Edit//: I mean maybe it is this: Because $p | H$ it follows that $|H|=p^rm$ for $r>1,m\geq 1,p\nmid m$. If $p\nmid (H:H\cap S') \Rightarrow p | H\cap S'$ so with Lagrange $|(H:H\cap S')|=m$ and $(H\cap S')=p^r$ then?
What you get from the end of the argument is: $p$ does not divide $|H(aS)| = [H : H \cap S']$. But then the highest power of $p$ that divides $|H|$ must also divide $|H \cap S'|$. Thus it is a Sylow subgroup.
By the way, there is a way simpler proof but I am not sure if you are allowed to use all Sylow theorems. Choose a Sylow subgroup of $H$, this will be a $p$-subgroup of $G$, and so contained in a Sylow subgroup $S'$ of $G$. Then conclude by using the second Sylow theorem: all Sylow subgroups are conjugate.