Show that $P$ is a Sylow $p$-subgroup of $G$ if and only if $P$ is a Sylow $p$-subgroup of $N_G(P).$

Question

Question: Let $G$ be a finite group and $P$ be a $p$-subgroup of $G.$ Show that $P$ is a Sylow $p$-subgroup of $G$ if and only if $P$ is a Sylow $p$-subgroup of $N_G(P).$

My attempt: Suppose that $|P| = p^r$

$(\Rightarrow)$ Assume that $|G| = p^rm$ where $r$ does not divide $m.$ Since $P\leq N_G(P)\leq G,$ by Lagrange's Theorem, we have $p^r$ divides $|N_G(P)|$ and $|N_G(P)|$ divides $p^rm.$ It follows that $|N_G(P)| = p^r n$ where $p$ does not divide $n.$ Therefore, $P$ is a Sylow $p$-subgroup of $|N_G(P)|.$

$(\Leftarrow):$ Suppose that $|N_G(P)| = p^r n$ where $p$ doe not divide $n.$ I do not know how to proceed from here.

Any hint would be appreciated.

Answer 1

The answer to the question follows immediately from the following. In general, if $P$ is a $p$-subgroup of $G$ (so not necessarily Sylow) then $|G:P| \equiv |N_G(P):P|$ mod $p$. See for instance http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowpf.pdf for a proof. The proof is not difficult and depends on the action of $P$ on the left coset space $G/P$ by left multplication.