Normality depends on the ambient group, so it is natural to ask for the largest part of $G$ in which a given subgroup behaves normally. This leads to the normaliser. The normaliser collects exactly those elements of $G$ whose conjugation leaves the subgroup fixed. It turns a failed normality condition into a successful one by shrinking the ambient group from $G$ to $N(H)$. This concept is used frequently in group actions, Sylow theory, conjugacy arguments, and the study of internal symmetry. :::definition[Normaliser] Let $(G,\circ)$ be a group and let $H$ be a non-empty subset of $G$. The normaliser of $H$ in $G$ is the subset $$\begin{equation*} N_G(H)=\{x\in G:xHx^{-1}=H\}. \end{equation*}$$ When the ambient group is clear, it may be denoted by $N(H)$. ::: The normaliser is not defined by asking whether each element of $H$ commutes with $x$. It asks whether conjugation by $x$ sends the whole subset $H$ back to itself. If $H$ is a subgroup, then $N_G(H)$ is the largest subgroup of $G$ in which $H$ becomes normal. Thus normalisers measure how much normality survives when the whole group is too large. :::theorem Let $(G,\circ)$ be a group and let $H$ be a non-empty subset of $G$. Then $N_G(H)$ is a subgroup of $G$. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a non-empty subset of $G$. To prove that $N_G(H)$ is a subgroup of $G$. Since $$\begin{equation*} eHe^{-1}=H, \end{equation*}$$ we get $e\in N_G(H)$. Therefore $N_G(H)$ is non-empty. Let $x,y\in N_G(H)$. Then $$\begin{equation*} xHx^{-1}=H \end{equation*}$$ and $$\begin{equation*} yHy^{-1}=H. \end{equation*}$$ Since $yHy^{-1}=H$, we also get $$\begin{equation*} y^{-1}Hy=H. \end{equation*}$$ Now $$\begin{align*} (x\circ y^{-1})H(x\circ y^{-1})^{-1} &=x\circ y^{-1}Hy\circ x^{-1}\\ &=xHx^{-1}\\ &=H. \end{align*}$$ Therefore $$\begin{equation*} x\circ y^{-1}\in N_G(H). \end{equation*}$$ By the one-step subgroup criterion, $N_G(H)$ is a subgroup of $G$. Hence, $N_G(H)\leq G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. If $H$ is normal in $G$, then $$\begin{equation*} N_G(H)=G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $G$, and $H$ is normal in $G$. To prove that $N_G(H)=G$. Since $H$ is normal in $G$, $$\begin{equation*} xHx^{-1}=H \quad \forall x\in G. \end{equation*}$$ Therefore every $x\in G$ belongs to $N_G(H)$. Thus $G\subseteq N_G(H)$. Since $N_G(H)\subseteq G$ by definition, we get $$\begin{equation*} N_G(H)=G. \end{equation*}$$ Hence, $N_G(H)=G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $G$. Then $H$ is normal in $N_G(H)$. ::: :::proof Given that $(G,\circ)$ is a group and $H$ is a subgroup of $G$. To prove that $H$ is normal in $N_G(H)$. Let $x\in N_G(H)$. By the definition of the normaliser, $$\begin{equation*} xHx^{-1}=H. \end{equation*}$$ Therefore $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in N_G(H). \end{equation*}$$ Hence, $H$ is normal in $N_G(H)$. $\square$ ::: :::theorem Let $(G,\circ)$ 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 $(G,\circ)$ 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 the previous theorem, $$\begin{equation*} H\trianglelefteq N_G(H). \end{equation*}$$ Let $L$ be a subgroup of $G$ such that $H\trianglelefteq L$. To prove that $L\subseteq N_G(H)$. Let $x\in L$. Since $H$ is normal in $L$, $$\begin{equation*} xHx^{-1}=H. \end{equation*}$$ Therefore $x\in N_G(H)$. Thus $$\begin{equation*} L\subseteq N_G(H). \end{equation*}$$ Hence, $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal. $\square$ ::: :::example Let $G=S_3$ and let $$\begin{equation*} H=\langle(12)\rangle=\{e,(12)\}. \end{equation*}$$ The elements that normalise $H$ must conjugate $(12)$ to an element of $H$. In $S_3$, conjugation sends transpositions to transpositions, and only elements preserving the transposition $(12)$ keep $H$ fixed. Thus $$\begin{equation*} N_{S_3}(H)=H. \end{equation*}$$ So $H$ is normal in its normaliser, although $H$ is not normal in all of $S_3$. $\square$ ::: :::example Let $G=S_3$ and let $H=A_3$. Since $A_3\trianglelefteq S_3$, the normaliser theorem gives $$\begin{equation*} N_{S_3}(A_3)=S_3. \end{equation*}$$ This example should be compared with the previous one. The normaliser of a normal subgroup is the whole group, while the normaliser of a non-normal subgroup can be strictly smaller. $\square$ ::: :::solved-problem Let $H\leq G$. Prove that $H\trianglelefteq G$ if and only if $N_G(H)=G$. ::: :::solution Let $H\leq G$. If $H\trianglelefteq G$, then $xHx^{-1}=H$ for every $x\in G$. Therefore $N_G(H)=G$. Conversely, if $N_G(H)=G$, then every $x\in G$ satisfies $$\begin{equation*} xHx^{-1}=H. \end{equation*}$$ Therefore $H\trianglelefteq G$ by the conjugation equality criterion. ::: We observe the chain $H\leq N_G(H)\leq G$. Choose one of the standard $S_3$ examples and compare how large the normaliser becomes. For $A_3$, the normaliser is all of $S_3$ because $A_3$ is normal. For the subgroup generated by $(12)$, the normaliser is only the subgroup itself, so normality is recovered only after shrinking the ambient group. :::scientific-preview[Normaliser Ladder Explorer] ::: We use orders to check the subgroup chain $H\leq N_G(H)\leq G$. Enter the orders of $G$, $N_G(H)$, and $H$. The calculator verifies the divisibility conditions and computes the indices. When $|N_G(H)|=|G|$, the normaliser criterion says that $H$ is normal in $G$. :::calculator[Normaliser Index Calculator] ::: :::exercise 1. Define the normaliser of a non-empty subset $H$ of $G$. 2. Prove that $N_G(H)$ is a subgroup of $G$. 3. Prove that if $H\trianglelefteq G$, then $N_G(H)=G$. 4. Prove that $H\trianglelefteq N_G(H)$ when $H$ is a subgroup of $G$. 5. Find $N_{S_3}(A_3)$. ::: :::answer 1. $N_G(H)=\{x\in G:xHx^{-1}=H\}$. 2. The identity belongs to $N_G(H)$, and if $x,y\in N_G(H)$, then $(xy^{-1})H(xy^{-1})^{-1}=H$. 3. If $H$ is normal in $G$, then every element of $G$ fixes $H$ under conjugation, so every element of $G$ lies in $N_G(H)$. 4. If $x\in N_G(H)$, then $xHx^{-1}=H$, which is the normality condition inside $N_G(H)$. 5. Since $A_3\trianglelefteq S_3$, $N_{S_3}(A_3)=S_3$. ::: :::faq[Frequently Asked Questions] Q: What does the normaliser measure? A: It measures the largest part of $G$ that keeps $H$ fixed under conjugation. Q: Is the normaliser always a subgroup? A: Yes. The normaliser of any non-empty subset of a group is a subgroup. Q: When is the normaliser equal to the whole group? A: For a subgroup $H$, this happens exactly when $H$ is normal in $G$. Q: Why is $H$ normal in $N_G(H)$? A: Every element of $N_G(H)$ is defined to conjugate $H$ onto itself. ::: :::call-to-action[Continue to Quotient Groups] subtitle: Normal subgroups now lead to the construction of quotient groups. button: Next Section | published/abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/construction-of-quotient-groups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/normality-inside-related-subgroups.md :::