relationship between $Syl_p(G)$ and $Syl_p(N_G(P))$ for a $p$-subgroup Let $P$ be a $p$-subgroup of a group $G$. I need to show that $P\in Syl_p(G)$ if and only if $P\in Syl_p(N_G(P))$.
I have a lemma that states that if $P\in Syl_p(G)$, and $Q$ is any $p$-subgroup of $G$, then $Q\cap N_G(P)=Q\cap P$. I also know that $P$ would be normal in $G$. I feel like I need to put these items together to get the forward direction. I feel like the backwards direction would be easy but I'm not really seeing it either. Thanks.
 A: Let $P$ be a $p$-subgroup of a (finite) group $G$. Then the following holds.
Proposition $P \in Syl_p(G) \iff P \in Syl_p(N_G(P))$. 
Proof Assume that $P \in Syl_p(G)$. Since $P \subseteq N_G(P)$, we have $|G:P|=|G:N_G(P)| \cdot |N_G(P):P|$, hence $p$ does not divide $|N_G(P):P|$, meaning $P$ is a $p$-Sylow subgroup of $N_G(P)$.
Conversely, suppose $P \in Syl_p(N_G(P))$. Now $P$ is a $p$-subgroup, hence applying Sylow Theory in $G$, there must be some $Q \in Syl_p(G)$ with $P \subseteq Q$. Of course, $P \subseteq Q \cap N_G(P)$. But $Q \cap N_G(P)$ is a $p$-subgroup of $N_G(P)$, hence, $P$ being Sylow (and thus a maximal $p$-subgroup), we must have $P=Q \cap N_G(P)=N_Q(P)$. If we would have $P \lt Q$, then by the "normalizers grow" principle in $p$-groups, we would have $N_Q(P) \gt P$, a contradiction. So in fact $P=Q$ and we are done.

Note There is yet another proof that depends on a number theoretic argument.
Lemma Let $P$ be a $p$-subgroup of a (finite) group $G$. Then the following holds.
$|G:P| \equiv |N_G(P):P|$ mod $p$

Proof (sketch) Let $P$ act on its left cosets in $G$. The fixed points are exactly those cosets, which have a representative in $N_G(P)$. Now apply the Cauchy-Frobenius (sometimes erroneously called the Burnside) counting formula.$\square$ 
From the Lemma it is crystal clear that the Proposition holds.
