# p divides the index $[G: N_G(P)]$?

Suppose G is a finite group with order $p^at, p\nmid t$, where $p$ is a prime number, and suppose $P$ is any $p-$subgroup of G, not necessarily to be Sylow $p-$subgroup, let $N_G(P)$ be its normalizer and asume $P$ is a sylow $p-$subgroup in $N_G(P)$, then is $p\nmid [G:N_{G}(P)]$ always true?

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Good question, and yes it is true. Because in general if $P$ is a $p$-subgroup of $G$ then $[G:P] \equiv [N_G(P):P]$ mod $p$. This can be proved by looking at the action of $P$ on the left coset space $G\backslash P$ by left multiplication. See for example here, chapter 2. In other words, $P$ is a Sylow $p$-subgroup in $G$ iff it is a Sylow $p$-subgroup in its normalizer.