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Let $G$ be a finite group, let $p$ be a prime dividing $|G|$, and let $K$ be a $p$-Sylow subgroup of $G$. Show that $$N_G(N_G(K)) = N_G(K).$$

I can come up with $K \trianglelefteq N_G(K)$ and $N_G(K) \trianglelefteq N_G(N_G(K))$, but I don't know what I can do after these. Thank you very much!

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  • $\begingroup$ Doesn't $N_G(K)$ have a unique Sylow $p$-subgroup. $\endgroup$ – Lord Shark the Unknown Dec 7 '19 at 17:30
  • $\begingroup$ Yes, K is the unique Sylow p-subgroup of $N_G(K)$ $\endgroup$ – Shu Hu Dec 7 '19 at 17:33
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If $g\in G-N_G(K) $, then $gKg^{-1}\neq K$ is a Sylow subgroup of $G$. Since $K$ is normal in $N_G(K) $, it is the unique subgroup of the same order as $K$ in $N_G(K) $. Thus $N_G(K) $ cannot contain $gKg^{-1}$, so $g$ does not normalize $N_G(K) $.

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  • $\begingroup$ Thank you for answering! May I ask why gKg−1≠K is a Sylow subgroup of G when $g∈G−N_G(K)$? $\endgroup$ – Shu Hu Dec 7 '19 at 17:42
  • $\begingroup$ @Shu It's not equal to $K$ because $g$ is outside the normalizer, and it is a Sylow subgroup because it has the same order (Sylow subgroups are $p$-subgroups of maximal order). $\endgroup$ – Matt Samuel Dec 7 '19 at 17:44
  • $\begingroup$ I see, thank you so much! $\endgroup$ – Shu Hu Dec 7 '19 at 17:48
  • $\begingroup$ @ShuHu No problem. $\endgroup$ – Matt Samuel Dec 7 '19 at 17:48
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Let $g\in N_G(N_G(K))$ then $K$ and $K^g$ are Sylow p-subgroups of $N_G(K)$ and therefore $K=K^g$ i.e. $g\in N_G(K)$.

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Here are two different proofs. $K \in Syl_p(G)$ and put $H=N_G(N_G(K))$.

$(1)$ Since $N_G(K) \unlhd H$ and $K \in Syl_p(G)$, whence $K \in Syl_p(N_G(K))$, we can apply the Frattini Argument, which gives $H=N_H(K)N_G(K)=(H \cap N_G(K))N_G(K)=N_G(K)N_G(K)=N_G(K).$

The second proof relies on a counting argument.

$(2)$ Note that $K \ char \ N_G(K) \unlhd H$, it follows that $K \unlhd H$. So $H$ has a unique Sylow $p$-subgroup. Hence $n_p(H)=|H:N_H(K)|=|N_G(N_G(K)):N_G(K)|=1$ and again the result follows.

And here is a bonus:

Proposition Let $G$ be a finite group, $P \in Syl_p(G)$, then every subgroup between $N_G(P)$ and $G$ is self-normalizing, that is, if $H$ is a subgroup of $G$ with $N_G(P) \subseteq H \subseteq G$, then $N_G(H)=H$.

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