Exercise 4.5.32 (Normalizers of Sylow $p$-subgroups are self-normalizing)

Question

Let $P$ be a Sylow $p$-subgroup of $H$ and let $H$ be a subgroup of $K$. If $P \unlhd H$ and $H \unlhd K$, prove that $P$ is normal in $K$. Deduce that if $P \in Syl_p(G)$ and $H = N_G(P)$, then $N_G(H) = H$ (in words: normalizers of Sylow $p$-subgroups are self-normalizing).

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $P$ be a Sylow $p$-subgroup of $H$, where $P \unlhd H$ and $H \unlhd K$. Write $|H| = p^k m$, where $p 
mid m$, so that $|P| = p^k$.

Since $P \unlhd H$, $P$ is the unique subgroup of $H$ of order $p^k$, thus $P$ is a characteristic subgroup of $H$. Moreover, $H \unlhd K$, thus $P\unlhd K$ by Section 4.4 (p. 135):
$$(P \unlhd^{\mathrm{char}} H, \ H \unlhd K) \Rightarrow P \unlhd K.$$

\bigskip
Suppose now that $P \in Syl_p(G)$ and $H = N_G(P)$. Then $P \unlhd H$ by definition of the normalizer, and $H \unlhd N_G(H)$ is always true. By the first part of this proof, $P \unlhd N_G(H)$, thus $N_G(H) \leq N_G(P) = H$: if $g \in N_G(H)$, then $gPg^{-1} = P$ because $P \unlhd N_G(H)$, so $g \in N_G(P)$. Moreover $H \leq N_G(H)$. This show that
$$H = N_G(H).$$
\end{proof}

Exercise 4.5.32 (Normalizers of Sylow $p$-subgroups are self-normalizing)

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Exercise 4.5.32 (Normalizers of Sylow $p$-subgroups are self-normalizing)

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