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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.

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  • $\begingroup$ Your first implication is really an "if and only if". A subgroup is defined to be a $p$-Sylow subgroup if it is a group of order $p^r$, where $p^r$ is the largest power of $p$ dividing $|G|$. Since $P \leq N_G(P) \leq G$, you have that $p^r$ is the largest power of $p$ dividing both $|G|$ and $|N_G(P)|$. $\endgroup$ – Joe Johnson 126 Oct 5 '18 at 0:35
  • $\begingroup$ @JoeJohnson126 I understand that $p^r$ is the largest power of $p$ dividing $|N_G(P).|$ But I do not see how does $p^r$ is the largest power of $p$ divide $|G|.$ $\endgroup$ – Idonknow Oct 5 '18 at 1:04
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    $\begingroup$ For the other direction you need two things: 1. Any $p$-subgroup is contained in a Sylow $p$-subgroup. 2. In a $p$-group, all proper subgroups are properly contained in their normalizer. $\endgroup$ – Tobias Kildetoft Oct 5 '18 at 3:51
  • $\begingroup$ You may want to add an assumption that $G$ is finite. It might still be true otherwise, for all I know, but there may be more to the proof than intended. $\endgroup$ – C Monsour Oct 5 '18 at 17:42
  • $\begingroup$ @CMonsour Edited. Thanks. $\endgroup$ – Idonknow Oct 7 '18 at 2:41
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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.

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  • $\begingroup$ Yes, I manage to prove that $|G:P| \cong |N_G(P):P|$ if $P$ is a $p$-subgroup of $G.$ However, I do not see how the result can solve my problem. Any hint would be appreciated. $\endgroup$ – Idonknow Oct 7 '18 at 2:36
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    $\begingroup$ Assume that $P$ is a Sylow $p$-subgroup of $N_G(P)$, but not of $G$. Then $[N_G(P) : P]$ is not divisible by $p$, while $[G : P]$ is. But this is impossible, since $[G : P] \equiv [N_G(P) : P] \pmod{p}$. $\endgroup$ – D_S Oct 7 '18 at 4:20

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