Sylow subgroup of the normalizer of itself So I was given a finite group $G$, with $P \leq G $, P is a p subgroup of $G$, and $P\in Syl_p(N_G(P))$. I want to show that P is a Sylow p subgp of $G$. 
So I attempted a contradiction, supposing $P$ is not a Sylow p subgp of $G$, then by Sylow's 2nd thm, there exists a $P_1\in Syl_p(G)$ such that $P\leq P_1$. 
From the hypothesis, I can also get that $P$ is proper subgroup of $P_1$
So am I on the right track, what should I do next? 
 A: Hint 1: If $P \leq G$ is a $p$-group of order $p^\alpha$ and not a Sylow $p$-subgroup, then $P$ is contained in a $p$-group of order $p^{\alpha+1}$.
Hint 2: In a $p$-group, subgroups of index $p$ are normal.
A: There is nothing to do if $P$ is the identity subgroup, so suppose that $|P| = p^{n} >1.$ Suppose (towards a contradiction) that $[G:P]$ is divisible by $p.$ Now $P$ permutes the right cosets of $P$ in $G$ via the action $Px \to Pxy$ for all $x \in G$ and $y \in P.$ 
Now the coset $P = P1_{G}$ is fixed by $P.$ Since $P$ is a non-trivial $p$-group, the number of cosets fixed by $P$ in the given action must be divisible by $p,$ as the totals number of cosets is $[G:P]$, which is divisible by $p.$ Hence $P$ must fix a coset $Px$ with $x \not \in P.$ Then $Pxy =Px$ for all $y \in P,$ so that $xPx^{-1} \leq P$, and in fact $xPx^{-1} = P,$ by order considerations. Then $x \in N_{G}(P).$ We have shown that the only cosets $Px$ which are fixed by $P$ are those with $x \in N_{G}(P),$ (and all such cosets are indeed fixed. Hence the number of fixed cosets is $[N_{G}(P):P],$ which we know is prime to $p$, a contradiction, as we have seen that the number of fixed cosets is divisible by $p.$
A: Hint: 
If we put $\,|G|=p^nm\;,\;\;(m,p)=1\,$ , and assuming 
$$\,|P|=p^k\;,\;k<n\;,\;\;|N_G(P)|=p^nm'\;,\;\;m=xm'\,$$ 
then
$$p^{n-k}m=[G:P]=[G:N_G(P)][N_G(P):P]=xm'\implies p\mid xm'$$
and this is a contradiction (why?)
