Conjugate subgroups lead naturally to the question of which elements leave a subgroup fixed under conjugation. The answer is the normalizer. The normalizer collects exactly those elements under which a subgroup is invariant, and it is the largest subgroup in which the given subgroup becomes normal. This is the subgroup-level analogue of the centralizer, and it plays the same counting role: centralizers count conjugates of elements, while normalizers count conjugates of subgroups. :::definition[Normalizer of a Subgroup in Another Subgroup] Let $G$ be a group and let $H,K$ be subgroups of $G$. The normalizer of $H$ in $K$ is the set $$\begin{equation*} N_K(H)=\{k \in K \mid kHk^{-1}=H\}. \end{equation*}$$ ::: :::theorem Let $G$ be a group and let $H,K$ be subgroups of $G$. Then $$\begin{equation*} N_K(H)=N_G(H)\cap K. \end{equation*}$$ ::: :::proof Given that $G$ is a group and $H,K$ are subgroups of $G$. To prove that $N_K(H)=N_G(H)\cap K$. Let $x \in N_K(H)$. Then $x \in K$ and $xHx^{-1}=H$. Therefore $x \in N_G(H)$ and $x \in K$. Hence $x \in N_G(H)\cap K$. Thus $$\begin{equation*} N_K(H)\subseteq N_G(H)\cap K. \end{equation*}$$ Conversely, let $x \in N_G(H)\cap K$. Then $x \in K$ and $xHx^{-1}=H$. Therefore $x \in N_K(H)$. Thus $$\begin{equation*} N_G(H)\cap K\subseteq N_K(H). \end{equation*}$$ Hence $N_K(H)=N_G(H)\cap K$. $\square$ ::: :::theorem Let $G$ be a group and let $H,K$ be subgroups of $G$. Then $N_K(H)$ is a subgroup of $K$. ::: :::proof Given that $G$ is a group and $H,K$ are subgroups of $G$. To prove that $N_K(H)$ is a subgroup of $K$. Since $e \in K$ and $eHe^{-1}=H$, we have $e \in N_K(H)$. Therefore $N_K(H)$ is non-empty. Let $k_1,k_2 \in N_K(H)$. Then $$\begin{equation*} k_1Hk_1^{-1}=H \end{equation*}$$ and $$\begin{equation*} k_2Hk_2^{-1}=H. \end{equation*}$$ From $k_2Hk_2^{-1}=H$, multiplying on the left by $k_2^{-1}$ and on the right by $k_2$ gives $$\begin{equation*} k_2^{-1}Hk_2=H. \end{equation*}$$ Therefore $$\begin{align*} (k_1k_2^{-1})H(k_1k_2^{-1})^{-1} &= k_1(k_2^{-1}Hk_2)k_1^{-1} \\ &= k_1Hk_1^{-1} \\ &= H. \end{align*}$$ Also $k_1k_2^{-1}\in K$ because $K$ is a subgroup of $G$. Hence $k_1k_2^{-1}\in N_K(H)$. Hence $N_K(H)$ is a subgroup of $K$. $\square$ ::: :::definition[Normalizer of a Subgroup] Let $G$ be a group and let $H$ be a subgroup of $G$. The normalizer of $H$ in $G$ is $$\begin{equation*} N_G(H)=\{g \in G \mid gHg^{-1}=H\}. \end{equation*}$$ When the ambient group is understood, it may be denoted by $N(H)$. ::: :::theorem Let $G$ be a group and let $H$ be a subgroup of $G$. Then $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal. ::: :::proof Given that $G$ is a group and $H$ is a subgroup of $G$. To prove that $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal. By definition, $$\begin{equation*} N_G(H)=\{g \in G \mid gHg^{-1}=H\}. \end{equation*}$$ Therefore $gHg^{-1}=H$ for every $g \in N_G(H)$. Hence $H$ is normal in $N_G(H)$. Let $L$ be a subgroup of $G$ such that $H$ is normal in $L$. Then $lHl^{-1}=H$ for every $l \in L$. Therefore $l \in N_G(H)$ for every $l \in L$. Thus $L\subseteq N_G(H)$. Hence $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal. $\square$ ::: :::corollary Let $G$ be a group and let $H$ be a subgroup of $G$. If $H$ is normal in $G$, then $N_G(H)=G$. If $G$ is abelian, then $N_G(H)=G$. ::: :::proof Given that $G$ is a group and $H$ is a subgroup of $G$. To prove that the two stated conclusions hold. [1] Suppose that $H$ is normal in $G$. Then $gHg^{-1}=H$ for every $g \in G$. Therefore every $g \in G$ belongs to $N_G(H)$. Hence $N_G(H)=G$. [2] Suppose that $G$ is abelian. For $g \in G$ and $h \in H$, $$\begin{align*} ghg^{-1} &= hgg^{-1} \\ &= h. \end{align*}$$ Therefore $gHg^{-1}=H$ for every $g \in G$. Hence $N_G(H)=G$. $\square$ ::: :::theorem[Counting Conjugate Subgroups] Let $G$ be a group and let $H,K$ be finite subgroups of $G$. Then the number of distinct conjugates of $H$ induced by elements of $K$ is $$\begin{equation*} [K:N_K(H)]. \end{equation*}$$ ::: :::proof Given that $G$ is a group and $H,K$ are finite subgroups of $G$. To prove that the number of distinct conjugates of $H$ induced by elements of $K$ is $[K:N_K(H)]$. Let $$\begin{equation*} \mathcal{T}=\{kHk^{-1}\mid k \in K\} \end{equation*}$$ be the set of distinct conjugates of $H$ induced by $K$. Let $$\begin{equation*} \mathcal{S}=\{kN_K(H)\mid k \in K\} \end{equation*}$$ be the set of left cosets of $N_K(H)$ in $K$. Define $f:\mathcal{S}\to\mathcal{T}$ by $$\begin{equation*} f(kN_K(H))=kHk^{-1}. \end{equation*}$$ To prove that $f$ is well-defined, let $k_1N_K(H)=k_2N_K(H)$. Then $k_2^{-1}k_1\in N_K(H)$. Therefore $$\begin{equation*} (k_2^{-1}k_1)H(k_2^{-1}k_1)^{-1}=H. \end{equation*}$$ Multiplying on the left by $k_2$ and on the right by $k_2^{-1}$ gives $k_1Hk_1^{-1}=k_2Hk_2^{-1}$. Therefore $f(k_1N_K(H))=f(k_2N_K(H))$. To prove that $f$ is one-one, suppose that $f(k_1N_K(H))=f(k_2N_K(H))$. Then $k_1Hk_1^{-1}=k_2Hk_2^{-1}$. Therefore $$\begin{equation*} (k_2^{-1}k_1)H(k_2^{-1}k_1)^{-1}=H. \end{equation*}$$ Thus $k_2^{-1}k_1\in N_K(H)$, so $k_1N_K(H)=k_2N_K(H)$. By definition of $\mathcal{T}$, every element of $\mathcal{T}$ has the form $kHk^{-1}$ for some $k \in K$. Therefore $f$ is onto. Therefore $f$ is a bijection. Hence $|\mathcal{T}|=|\mathcal{S}|=[K:N_K(H)]$. $\square$ ::: :::corollary Let $G$ be a group and let $H,K$ be finite subgroups of $G$. If $H$ is invariant under exactly $n$ elements of $K$, then the number of conjugates of $H$ induced by $K$ is $$\begin{equation*} \frac{|K|}{n}. \end{equation*}$$ ::: :::proof Given that $G$ is a group, $H,K$ are finite subgroups of $G$, and $H$ is invariant under exactly $n$ elements of $K$. To prove that the number of conjugates of $H$ induced by $K$ is $\frac{|K|}{n}$. The elements of $K$ under which $H$ is invariant are precisely the elements of $N_K(H)$. Therefore $|N_K(H)|=n$. By the theorem on counting conjugate subgroups, the number of conjugates induced by $K$ is $$\begin{equation*} [K:N_K(H)]. \end{equation*}$$ Since $K$ is finite, $$\begin{equation*} [K:N_K(H)]=\frac{|K|}{|N_K(H)|}=\frac{|K|}{n}. \end{equation*}$$ Hence the number of conjugates of $H$ induced by $K$ is $\frac{|K|}{n}$. $\square$ ::: :::example Let $G=S_3$ and let $H=\{e,(12)\}$. The conjugates of $H$ are $$\begin{equation*} \{e,(12)\},\quad \{e,(13)\},\quad \{e,(23)\}. \end{equation*}$$ Thus there are three conjugates. Since $|S_3|=6$, the counting formula gives $$\begin{equation*} 3=[S_3:N_{S_3}(H)]. \end{equation*}$$ Therefore $|N_{S_3}(H)|=2$. Since $H\subseteq N_{S_3}(H)$ and $|H|=2$, we get $N_{S_3}(H)=H$. ::: This preview shows the subgroup version of the orbit-stabilizer idea. The orbit consists of all conjugates of $H$, and the stabilizer is the normalizer $N_K(H)$, the elements of $K$ that leave $H$ unchanged. Choose a model and compare the size of $K$, the size of the normalizer, and the number of distinct conjugates. Start with the $S_3$ transposition subgroup, then compare the normal subgroup $A_3$ where every element normalizes it. :::scientific-preview[Normalizer Orbit Counting Explorer] ::: The normalizer is the stabilizer of the subgroup under conjugation. A larger normalizer means more elements fix the subgroup, so fewer distinct conjugates appear. If the normalizer is the whole group, the subgroup has only one conjugate and is normal. This calculator checks the formula for counting conjugate subgroups induced by a finite subgroup $K$. Enter $|K|$ and $|N_K(H)|$. The calculator validates divisibility and then computes $[K:N_K(H)]$. Use the preset from the exercise, $|K|=24$ and $|N_K(H)|=6$, and compare it with the $S_3$ example. :::calculator[Normalizer Index Calculator] ::: The index counts how many different positions the subgroup can occupy under conjugation by elements of $K$. This is the subgroup analogue of counting conjugates of an element by using the centralizer index. :::solved-problem Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. Prove that $$\begin{equation*} N_G(aHa^{-1})=aN_G(H)a^{-1}. \end{equation*}$$ ::: :::solution Let $G$ be a group, let $H$ be a subgroup of $G$, and let $a \in G$. Let $x \in N_G(aHa^{-1})$. Then $$\begin{equation*} x(aHa^{-1})x^{-1}=aHa^{-1}. \end{equation*}$$ Multiplying on the left by $a^{-1}$ and on the right by $a$ gives $$\begin{equation*} (a^{-1}xa)H(a^{-1}x^{-1}a)=H. \end{equation*}$$ Since $a^{-1}x^{-1}a=(a^{-1}xa)^{-1}$, we have $a^{-1}xa \in N_G(H)$. Therefore $x \in aN_G(H)a^{-1}$. Thus $$\begin{equation*} N_G(aHa^{-1})\subseteq aN_G(H)a^{-1}. \end{equation*}$$ Conversely, let $x \in aN_G(H)a^{-1}$. Then $x=aya^{-1}$ for some $y \in N_G(H)$. Since $yHy^{-1}=H$, $$\begin{align*} x(aHa^{-1})x^{-1} &= (aya^{-1})(aHa^{-1})(ay^{-1}a^{-1}) \\ &= ayHy^{-1}a^{-1} \\ &= aHa^{-1}. \end{align*}$$ Therefore $x \in N_G(aHa^{-1})$. Thus $$\begin{equation*} aN_G(H)a^{-1}\subseteq N_G(aHa^{-1}). \end{equation*}$$ Hence $N_G(aHa^{-1})=aN_G(H)a^{-1}$. $\square$ ::: :::exercise [1] Let $H$ be a subgroup of $G$. Prove that $H\subseteq N_G(H)$. [2] Let $H$ be a subgroup of $G$. Prove that $H$ is normal in $G$ if and only if $N_G(H)=G$. [3] Let $H,K$ be finite subgroups of $G$ and suppose $|K|=24$ and $|N_K(H)|=6$. How many conjugates of $H$ are induced by $K$? ::: :::answer [1] If $h \in H$, then $hHh^{-1}=H$ because $H$ is a subgroup and is closed under conjugation by its own elements. Hence $h \in N_G(H)$, so $H\subseteq N_G(H)$. [2] If $H$ is normal in $G$, then $gHg^{-1}=H$ for every $g \in G$, so $N_G(H)=G$. Conversely, if $N_G(H)=G$, then every $g \in G$ satisfies $gHg^{-1}=H$, so $H$ is normal in $G$. [3] The number is $[K:N_K(H)]=24/6=4$. ::: :::faq[Frequently Asked Questions] Q: How is a normalizer different from a centralizer? A: A centralizer fixes an element by commuting with it, while a normalizer fixes a subgroup as a set under conjugation. Q: Why is the normalizer important? A: It is the largest subgroup in which the given subgroup is normal, and it counts conjugate subgroups through its index. Q: What does $[K:N_K(H)]$ count? A: It counts the distinct subgroups of the form $kHk^{-1}$ with $k \in K$. ::: :::call-to-action[Section Complete] subtitle: The section has developed conjugacy classes, the class equation, conjugate subgroups, and normalizers for Sylow theory. button: Review First Page | published/abstract-algebra/sylow-theorems/conjugacy-classes/centralizers-and-conjugacy-of-elements button-ghost: Revisit Normalizers | published/abstract-algebra/sylow-theorems/conjugacy-classes/normalizers-and-counting-conjugate-subgroups :::