---- > [!definition] Definition. ([[normalizer of a subgroup]]) > Let $H$ be a [[subgroup]] of a [[group]] $G$. Define the **normalizer of $H$ in $G$** to be $N_{G}(H):=\{ g \in G: g H g ^{-1} = H \}. (\text{equivalently: }gH=Hg)$ > \ > $N_{G}(H)$ is the *largest* [[subgroup]] of $G$ containing $H$ in which $H$ is [[normal subgroup|normal]]. > [!intuition] (Per [this source](https://www.math.clemson.edu/~macaule/classes/m19_math4120/slides/math4120_lecture-3-06_h.pdf)) > The basic idea: if $H \leq G$ but $H$ is *not* [[normal subgroup|normal]] in $G$, can we measure "how far" $H$ is from being [[normal subgroup|normal]]? > \ > Recall that $H \trianglelefteq G$ iff $gH=Hg$ for all $g \in G$. So a natural way to answer this question would be to check *how many* $g$ satisfy this requirement. Imagine each $g \in G$ votes regarding, from their perspective, if $H$ is [[normal subgroup|normal]]: $gH=Hg \ \ \textcolor{LimeGreen}{\text{``yes"}} \ \ \ \ gH \neq Hg \ \ \textcolor{Apricot}{\text{``no"}}$ > \ > At a minimum, every $g \in H$ votes $\textcolor{LimeGreen}{\text{yes}}$. > \ > At a maximum, every $g \in G$ votes $\textcolor{LimeGreen}{\text{yes}}$, but this happens only when $H$ really is [[normal subgroup|normal]]. > \ > There can be levels *between* these two extremes— we the set of elements which vote $\textcolor{LimeGreen}{\text{yes}}$ in favor of $Hs [[normal subgroup|normality]] is precisely normalizer of $H$ in $G$ as defined above. > > [!justification] > We have to confirm that $(1)$ $N_{G}(H)$ is a [[subgroup]] of $G$ containing $H$, $(2)$ that $H$ is [[normal subgroup|normal]] in $N_{G}(H)$, and $(3)$ that $N_{G}(H)$ is the **largest** [[subgroup]] of $G$ containing $H$ in which $H$ is [[normal subgroup|normal]]. > > ## 1 ($N_{G}(H)$ is a [[subgroup]] of $G$ with $H \subset N_{G}(H)$) The identity element $e$ is inherited from $G$, since $eH=H=He$ trivially. Let $g_{1}, g_{2} \in N_{G}(H)$, so that $g_{1}H=Hg_{1}$ and $g_{2}H=Hg_{2}$. Then $(g_{1}g_{2})H=g_{1}(g_{2}H)=g_{1}(Hg_{2})=(g_{1}H)g_{2}=H(g_{1}g_{2}),$ so $g_{1}g_{2} \in N_{G}(H)$ as well. Finally, if $g \in N_{G}H$, then since $H$ is invariant under [[conjugate|conjugation]] by $g$, $gHg^{-1}=H,$ we can left-multiply both sides by $g^{-1}$ to obtain $Hg^{-1}=g^{-1}H,$ which shows that $g^{-1} \in N_{G}(H)$ as well. Thus, $N_{G}(H)$ is a [[subgroup]] of $G$. To see $H \in N_{G}(H)$, let $h \in H$ and observe that $hH=H=Hh$. >## 2 ($H$ is [[normal subgroup|normal]] in $N_{G}(H)$) This is pretty much by construction. Let $g \in N_{G}(H)$, then by definition $gH=Hg$; since $g$ was arbitrary we conclude $gH=Hg$ for all $g \in H$ — i.e., $H$ is [[normal subgroup|normal]] in $N_{G}(H)$. >## 3 $N_{G}(H)$ is the **largest** [[subgroup]] of $G$ containing $H$ in which $H$ is [[normal subgroup|normal]]. >Let $S \leq G$ be a [[subgroup]] of $g$ containing $H$ in which $H$ is [[normal subgroup|normal]]: for all $s \in S$, we have $sH=Hs$. Then clearly $s \in N_{G}(H)$. Since $s$ was arbitrary we conclude $S \subset N_{G}(H)$. ^9e23a4 ---- #### ---- #### 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 > ```