# 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 $N_G(N_G(K)) = N_G(K)$.

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!

• Doesn't $N_G(K)$ have a unique Sylow $p$-subgroup. – Lord Shark the Unknown Dec 7 '19 at 17:30
• Yes, K is the unique Sylow p-subgroup of $N_G(K)$ – Shu Hu Dec 7 '19 at 17:33

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 $$N_G(K)$$.

• Thank you for answering! May I ask why gKg−1≠K is a Sylow subgroup of G when $g∈G−N_G(K)$? – Shu Hu Dec 7 '19 at 17:42
• @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). – Matt Samuel Dec 7 '19 at 17:44
• I see, thank you so much! – Shu Hu Dec 7 '19 at 17:48
• @ShuHu No problem. – Matt Samuel Dec 7 '19 at 17:48

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. $$g\in N_G(K)$$.

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 $$N_G(H)=H$$.