----- > [!proposition] Proposition. ([[normalization of p-Sylow subgroups is idempotent]]) > Let $H$ be a [[p-Sylow subgroup]] of a finite [[group]] $G$. Then $N_{G}(N_{G}(H))=N_{G}(H)$ > where $N_{G}(S)$ denotes the [[normalizer of a subgroup|normalizer]] of a [[subgroup]] $S$ in $G$. > [!proof]- Proof. ([[normalization of p-Sylow subgroups is idempotent]]) > Every [[subgroup]] is a [[normal subgroup]] of itself. Hence $N_{G}(H) \subset N_{G}(N_{G}(H))$. > > We first note that $H$ is a [[p-Sylow subgroup]] of $N_{G}(H)$. This is because by [[Lagrange's Theorem]] the [[order of a group|order]] of $N_{G}(H)$ [[divides]] the order of $|G|=p^{r}m$ where $r$ is maximal (i.e., $p \not{|}m$). By [[Euclid's Lemma]], we must have $|N_{G}(H)| \ \b | \ p^{r} \text{ or } |N_{G}(H)| \ \b| \ m.$ > If the first case, $|N_{G}(H)|= p^{i} \cdot 1$ for some $0<i \leq r$ and $k \in \mathbb{N}$, so $N_{G}(H)$ ... (think about it) > > By definition, $H$ is invariant under [[conjugate|conjugation]] by members of $N_{G}(H)$, implying by the [[the Sylow theorems|second Sylow theorem]] that $H$ is the *only* [[p-Sylow subgroup]] of $N_{G}(H)$. > > If $H$ and $H'$ are two distinct [[p-Sylow subgroup]]s of $G$, then $(*)$ $\begin{align} > N_{G}(H) = & \{ g \in G: g H g^{-1}= H \} \\ > \neq & \{ g' \in G : g'H'g'^{-1} = H' \} \\ > = & N_{G}(H'), > \end{align}$ > since if the sets were equal there would exist some $g^{*} \in G$ for which $H=g^{*}Hg^{*^{-1}}=g^{*^{}}Hg^{*^{^{-1}}}=H'$. Moreover, $N_{G}(H)$ is [[conjugate]] to $N_{G}(H')$, because by the [[the Sylow theorems|second Sylow theorem]] there exists $a \in G$ such that $aHa^{-1}=H'$, and so $\begin{align} > N_{G}(H')= & \{ g' \in G: g' a H a^{-1} g'^{-1}=aHa^{-1} \} \\ > = & \{ g' \in G : (g'a) H (g'a)^{-1} = aHa^{-1} \} \\ > = & \{ g' \in G : (g'a)^{-1}H(g'a) \}= aHa^{-1} \\ > = & \{ g' \in G : a^{-1}g'^{-1}Hg'a = aHa^{-1} \} \\ > = & \{ ag'a^{-1} \in G : g'^{-1} Hg' = H \} \\ > = & a\{ g' \in G: g'^{-1}Hg' = H\}a^{-1} \\ > = & aN_{G}(H)a^{-1}. > \end{align}$ > Now suppose $H'=aHa^{-1}$ is a [[p-Sylow subgroup]] of $N_{G}(N_{G}(H))$. Then $N_{G}(H')=aN_{G}(H)a^{-1}=N_{G}(H)$, since $N_{G}(H) \trianglelefteq N_{G}(N_{G}(H))$. By $(*)$, this implies $H'=H$. > > We thus conclude that $H$ is the only [[p-Sylow subgroup]] of $N_{G}N_{G}(H)$, and thus is a [[normal subgroup]] of $N_{G}N_{G}(H)$ by the [[the Sylow theorems|second Sylow theorem]]. But since $N_{G}(H)$ is the *largest* [[normal subgroup]] of $G$ containing $H$, it must be the case that $N_{G}N_{G}(H) \subset N_{G}(H)$. > > Hence the two-way inclusion is shown. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```