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Let $G$ be a finite group and let $P$ be a Sylow p-subgroup with $N_G(P)=P$. Prove that $P$ is not contained in any proper normal subgroup of $G$.

We can suppose $P \subseteq K\lhd G$. Since $P$ is a p-group, it is contained in some Sylow p-subgroup $K ⊂ G$. Since $K$ is a solvable group. and we have $\forall g \in G, gPg^{-1} \subseteq gKg^{-1}$ which then is K since it is normal.

There is theorem that says, If P is a Sylow p-subgroup of a finite group G, then $N_G(N_G(P))=N_G(P)$

Now, I am supposing that I must then find a contradiction to this???

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    $\begingroup$ The second sentence of the second paragraph makes no sense at all. $P$ is already a Sylow $p$-subgroup of $G$ and therefore also of $K$. $\endgroup$ – Derek Holt Apr 10 '15 at 21:28
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    $\begingroup$ Here is a hint. Since $gPg^{-1}$ and $P$ are both Sylow $p$-subgroups of $K$, they must be conjugate by an element $k \in K$. So $kgPg^{-1}k^{-1} = P$. $\endgroup$ – Derek Holt Apr 10 '15 at 21:31
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    $\begingroup$ Perhaps another hint, from a different perspective: google "Frattini's Argument" to see what huge contradiction you'd get if there were a proper normal subgroup containing $\;P\;$ . Use this to reach a contradiction or, following Derek's hint, use the argument's proof to prove your problem. $\endgroup$ – Timbuc Apr 10 '15 at 22:33
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As Timbuc indicates, you better use the classical Frattini Argument: if $N \unlhd G$, $P \in Syl_p(N)$, then $G=NN_G(P)$.

In your case, $S \in Syl_p(G)$ with $S \subset N \subsetneq G$ and $N_G(S)=S$, then certainly $S \in Syl_p(N)$. Hence the Frattini Argument yields $G=NN_G(S)=NS=N$, a contradiction.

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