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So the question is as follow:

$G$ is a finite group and $P$ is its sylow-p subgroup. Prove that if $N_G(P)\triangleleft G$ then $P\triangleleft G$

So what I'm trying to prove is that $N_G(P)=G$ somehow. my proof goes like this:

$N_G(P)\triangleleft G$ so let $h\in G \backslash N_G(P)$ and $g\in N_G(P)$ so we know that $h^{-1}gh \in N_G(P)$ with that we know that $h^{-1}g^{-1}hPh^{-1}gh=P$. I'm stuck from here I"m trying to show that somehow $P^h=P$ and that will result in $h\in N_G(P)$.

Thank you for your help!

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$P$ is characteristic in its normalizer. It is unmoved by any automophism of $N_G(P)$. In particular conjugation. Hence if $N_G(P) \unlhd G$, then $P \unlhd G$. Alternatively you can use the Frattini argument giving $N_G(P)=G$.

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  • $\begingroup$ why is it characteristic in its normalizer? is it because it's its sole sylow subgroup? is there a general theorem or just for sylows...? $\endgroup$ – Kim Seel Mar 21 '17 at 18:39
  • $\begingroup$ There is a general theorem: if $N \unlhd G$, and gcd($|N|,|G:N|)=1$, then $N$ is characteristic subgroup. That is, for every $\psi \in Aut(G)$, $\psi[N]=N$. It is not difficult to prove. So if a Sylow subgroup is normal, then it MUST be characteristic. $\endgroup$ – Nicky Hekster Mar 21 '17 at 20:26

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