Some normal subgroups arise from special locations inside the group. The centre is the most important source, because elements of the centre commute with every element of the group. Cyclic subgroups also require careful treatment: a cyclic subgroup need not be normal merely because it is cyclic, but it is normal when its generator lies in the centre. In this lesson we separate valid central arguments from invalid cyclic shortcuts. This distinction is important because students often overgeneralise from abelian groups and assume that cyclic subgroups are automatically normal. :::theorem Let $(G,\circ)$ be a group. Then every subgroup of $Z(G)$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group. To prove that every subgroup of $Z(G)$ is normal in $G$. Let $H$ be a subgroup of $Z(G)$. Let $x\in G$ and let $h\in H$. Since $H\subseteq Z(G)$, we get $$\begin{equation*} h\circ x=x\circ h. \end{equation*}$$ Therefore $$\begin{align*} x\circ h\circ x^{-1} &=h\circ x\circ x^{-1}\\ &=h. \end{align*}$$ Thus $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $a\in Z(G)$. Then $\langle a\rangle$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group and $a\in Z(G)$. To prove that $\langle a\rangle$ is normal in $G$. Since $a\in Z(G)$, every power of $a$ also belongs to $Z(G)$. Therefore $$\begin{equation*} \langle a\rangle\subseteq Z(G). \end{equation*}$$ By the theorem that every subgroup of $Z(G)$ is normal in $G$, we get $$\begin{equation*} \langle a\rangle\trianglelefteq G. \end{equation*}$$ Hence, $\langle a\rangle$ is normal in $G$. $\square$ ::: :::example Let $G$ be the quaternion group $$\begin{equation*} Q_8=\{1,-1,i,-i,j,-j,k,-k\}. \end{equation*}$$ Its centre is $$\begin{equation*} Z(Q_8)=\{1,-1\}. \end{equation*}$$ Therefore $\{1,-1\}$ is a normal subgroup of $Q_8$. This example shows a typical use of the central source: once a subgroup lies in the centre, no separate coset comparison is needed. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $a\in G$. The statement that $\langle a\rangle$ is normal in $G$ is not true in general. ::: :::proof Given that the statement is that $\langle a\rangle$ is normal in $G$ for every $a\in G$. To prove that the statement is not true in general. Let $G=S_3$ and let $$\begin{equation*} a=(12). \end{equation*}$$ Then $$\begin{equation*} \langle a\rangle=\{e,(12)\}. \end{equation*}$$ Let $$\begin{equation*} x=(123). \end{equation*}$$ Then $$\begin{align*} x\circ (12)\circ x^{-1} &=(123)\circ(12)\circ(132)\\ &=(23). \end{align*}$$ But $$\begin{equation*} (23)\notin \langle(12)\rangle. \end{equation*}$$ Therefore $$\begin{equation*} x\langle a\rangle x^{-1}\nsubseteq \langle a\rangle. \end{equation*}$$ Hence, $\langle a\rangle$ need not be normal in $G$. $\square$ ::: :::lemma Let $C$ be a cyclic group. Then every subgroup of $C$ is fixed by every automorphism of $C$. ::: :::proof Given that $C$ is a cyclic group. To prove that every subgroup of $C$ is fixed by every automorphism of $C$. Let $C=\langle a\rangle$ and let $K$ be a subgroup of $C$. Since every subgroup of a cyclic group is cyclic, there exists an integer $m$ such that $$\begin{equation*} K=\langle a^m\rangle. \end{equation*}$$ Let $\varphi$ be an automorphism of $C$. Then $\varphi(a)$ is a generator of $C$. Therefore $$\begin{align*} \varphi(K) &=\varphi(\langle a^m\rangle)\\ &=\langle \varphi(a^m)\rangle\\ &=\langle \varphi(a)^m\rangle\\ &=\langle a^m\rangle\\ &=K. \end{align*}$$ Hence, every subgroup of $C$ is fixed by every automorphism of $C$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a cyclic subgroup of $G$. If $H$ is normal in $G$, then every subgroup of $H$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a cyclic subgroup of $G$, and $H$ is normal in $G$. To prove that every subgroup of $H$ is normal in $G$. Let $K$ be a subgroup of $H$. Let $x\in G$. Since $H$ is normal in $G$, the map $\varphi_x:H\to H$ defined by $$\begin{equation*} \varphi_x(h)=x\circ h\circ x^{-1} \end{equation*}$$ is an automorphism of $H$. Since $H$ is cyclic, every subgroup of $H$ is fixed by every automorphism of $H$. Therefore $$\begin{equation*} \varphi_x(K)=K. \end{equation*}$$ Thus $$\begin{equation*} xKx^{-1}=K \quad \forall x\in G. \end{equation*}$$ Hence, $K$ is normal in $G$. $\square$ ::: :::theorem Let $(G,\circ)$ be a group. If every cyclic subgroup of $G$ is normal in $G$, then every subgroup of $G$ is normal in $G$. ::: :::proof Given that $(G,\circ)$ is a group and every cyclic subgroup of $G$ is normal in $G$. To prove that every subgroup of $G$ is normal in $G$. Let $H$ be a subgroup of $G$. Let $x\in G$ and let $h\in H$. Since $\langle h\rangle$ is a cyclic subgroup of $G$, it is normal in $G$. Therefore $$\begin{equation*} x\circ h\circ x^{-1}\in \langle h\rangle. \end{equation*}$$ Since $\langle h\rangle\subseteq H$, we get $$\begin{equation*} x\circ h\circ x^{-1}\in H. \end{equation*}$$ Therefore $$\begin{equation*} xHx^{-1}\subseteq H \quad \forall x\in G. \end{equation*}$$ Hence, $H$ is normal in $G$. $\square$ ::: :::solved-problem Let $G$ be a group and let $a\in Z(G)$. Prove directly that $\langle a\rangle\trianglelefteq G$. ::: :::solution Let $x\in G$ and let $a^n\in \langle a\rangle$, where $n\in \mathbb{Z}$. Since $a\in Z(G)$, we get $$\begin{equation*} x\circ a=a\circ x. \end{equation*}$$ Therefore $$\begin{align*} x\circ a^n\circ x^{-1} &=a^n\circ x\circ x^{-1}\\ &=a^n. \end{align*}$$ Thus $$\begin{equation*} x\langle a\rangle x^{-1}\subseteq \langle a\rangle. \end{equation*}$$ Hence $$\begin{equation*} \langle a\rangle\trianglelefteq G. \end{equation*}$$ ::: We compare three different reasons a cyclic subgroup might be normal. A central generator gives a guaranteed normal subgroup, while a cyclic subgroup with a noncentral generator may fail normality. If a cyclic subgroup is already known to be normal, then all of its subgroups inherit normality. Use the choices to keep the central argument distinct from the invalid shortcut that every cyclic subgroup must be normal. :::scientific-preview[Central Cyclic Normality Explorer] ::: The safe conclusion depends on the hypothesis. A central generator proves normality immediately, while a merely cyclic subgroup still requires a separate normality check. We can test whether a proposed subgroup order can fit inside the centre before using the central normality theorem. Enter the orders of $G$, $Z(G)$, and $H$. If $|H|$ divides $|Z(G)|$ and $H$ is known to lie in the centre, the theorem proves $H\trianglelefteq G$. If the containment is not known, divisibility alone does not prove normality. :::calculator[Central Source Calculator] ::: :::exercise 1. Prove that every subgroup of $Z(G)$ is normal in $G$. 2. Give a cyclic subgroup that is not normal in $S_3$. 3. If $a\in Z(G)$, prove that $\langle a\rangle\subseteq Z(G)$. 4. Suppose $H$ is cyclic and normal in $G$. Explain why subgroups of $H$ are normal in $G$. ::: :::answer 1. If $H\leq Z(G)$, then $xhx^{-1}=h$ for every $x\in G$ and $h\in H$, so $H\trianglelefteq G$. 2. The subgroup $\langle(12)\rangle=\{e,(12)\}$ is not normal in $S_3$. 3. Since $a$ commutes with every element of $G$, every power of $a$ also commutes with every element of $G$. 4. Conjugation by any element of $G$ gives an automorphism of $H$, and every subgroup of a cyclic group is fixed by every automorphism. ::: :::faq[Frequently Asked Questions] Q: Is every cyclic subgroup normal? A: No. The subgroup generated by a transposition in $S_3$ is cyclic but not normal. Q: What extra condition makes a cyclic subgroup normal? A: One useful condition is that the generator lies in the centre of the group. Q: Why do subgroups of a normal cyclic subgroup become normal? A: Conjugation acts as an automorphism of the cyclic subgroup, and cyclic subgroups have automorphism-invariant subgroups. ::: :::call-to-action[Continue to Products and Intersections] subtitle: Learn how normal subgroups behave under intersections and products. button: Next Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/products-and-intersections-of-normal-subgroups.md button-ghost: Previous Lesson | published/abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/index-two-and-coset-criteria-for-normality.md :::