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I don't understand the last part of this proof:

http://www.proofwiki.org/wiki/Intersection_of_Normal_Subgroup_with_Sylow_P-Subgroup

where they say: $p \nmid \left[{N : P \cap N}\right]$, thus, $P \cap N$ is a Sylow p-subgroup of $N$. I don't see why this implicaction is true. On the other hand, I understand that $P$ being a Sylow p-subgroup of $G$ implies that $p \nmid [G : P]$, for $[G:P]=[G:N_G(P)][N_G(P):P]$ and $p$ does not divide any of these two factors. So, what I don't understand why is true is the inverse implication, that is, if $p \nmid [G : P]$ then $P$ is a Sylow p-subgroup of $G$.

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We know that $|P\cap N|=p^k$ for some $k\in\mathbb{N}$ by Lagrange's Theorem (since $P\cap N\leq P$ and $|P|$ is a power of $p$), so since $[N:P\cap N]=\frac{|N|}{|P\cap N|}$, $p\not\mid\frac{|N|}{p^k}$. Hence, $p^k$ is the biggest power of $p$ that divides $|N|$, so $P\cap N$ is a Sylow $p$-subgroup of $N$.

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There is a theorem that says that all subgroups of maximal prime power size $p^k$ are Sylow $p$-subgroups and they are all conjugate to one another.

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