After using commutators to recognise abelian quotients, we now focus on quotients by central subgroups. The centre $Z(G)$ is always normal, so $G/Z(G)$ is always a quotient group. Surprisingly, if this quotient is cyclic, then the original group must be abelian. More generally, if $H$ lies inside the centre and $G/H$ is cyclic, then $G$ is abelian. These results are useful because they turn information about a quotient into information about the original group. :::theorem Let $(G,\circ)$ be a group. If $G/Z(G)$ is cyclic, then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group and $G/Z(G)$ is cyclic. To prove that $G$ is abelian. Since $G/Z(G)$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G/Z(G)=\langle aZ(G)\rangle. \end{equation*}$$ Let $x,y\in G$. Then there exist integers $m,n$ and elements $z_1,z_2\in Z(G)$ such that $$\begin{equation*} x=a^m\circ z_1 \end{equation*}$$ and $$\begin{equation*} y=a^n\circ z_2. \end{equation*}$$ Since $z_1,z_2\in Z(G)$, they commute with every element of $G$. Therefore $$\begin{align*} x\circ y &=(a^m\circ z_1)\circ(a^n\circ z_2)\\ &=a^m\circ a^n\circ z_1\circ z_2\\ &=a^{m+n}\circ z_1\circ z_2\\ &=a^{n+m}\circ z_2\circ z_1\\ &=a^n\circ a^m\circ z_2\circ z_1\\ &=(a^n\circ z_2)\circ(a^m\circ z_1)\\ &=y\circ x. \end{align*}$$ Thus $$\begin{equation*} x\circ y=y\circ x \quad \forall x,y\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: :::corollary Let $(G,\circ)$ be a non-abelian group. Then $G/Z(G)$ is not cyclic. ::: :::proof Given that $(G,\circ)$ is a non-abelian group. To prove that $G/Z(G)$ is not cyclic. If possible let $G/Z(G)$ be cyclic. Then, by the previous theorem, $G$ is abelian. This contradicts the given fact that $G$ is non-abelian. A contradiction. Hence, $G/Z(G)$ is not cyclic. $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $H$ be a subgroup of $Z(G)$. If $G/H$ is cyclic, then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group, $H$ is a subgroup of $Z(G)$, and $G/H$ is cyclic. To prove that $G$ is abelian. Since $H\subseteq Z(G)$, $H$ is normal in $G$. Since $G/H$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G/H=\langle aH\rangle. \end{equation*}$$ Let $x,y\in G$. Then there exist integers $m,n$ and elements $h_1,h_2\in H$ such that $$\begin{equation*} x=a^m\circ h_1 \end{equation*}$$ and $$\begin{equation*} y=a^n\circ h_2. \end{equation*}$$ Since $H\subseteq Z(G)$, we have $h_1,h_2\in Z(G)$. Therefore $$\begin{align*} x\circ y &=(a^m\circ h_1)\circ(a^n\circ h_2)\\ &=a^m\circ a^n\circ h_1\circ h_2\\ &=a^{m+n}\circ h_1\circ h_2\\ &=a^{n+m}\circ h_2\circ h_1\\ &=a^n\circ a^m\circ h_2\circ h_1\\ &=(a^n\circ h_2)\circ(a^m\circ h_1)\\ &=y\circ x. \end{align*}$$ Thus $$\begin{equation*} x\circ y=y\circ x \quad \forall x,y\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: :::example Let $G$ be a group such that $G/Z(G)$ has prime order. Since every group of prime order is cyclic, $G/Z(G)$ is cyclic. By the theorem, $G$ is abelian. Therefore a non-abelian group cannot have $G/Z(G)$ of prime order. Hence, the central quotient of a non-abelian group cannot be cyclic of prime order. $\square$ ::: :::example Let $G=Q_8$, the quaternion group. The group $Q_8$ is non-abelian, and therefore $$\begin{equation*} Q_8/Z(Q_8) \end{equation*}$$ is not cyclic. In fact, this quotient has four elements and is isomorphic in structure to the Klein four group. This example shows why the theorem is restrictive: a non-abelian group cannot have a cyclic quotient by its centre. $\square$ ::: We observe the implication from a cyclic central quotient back to the original group. Choose whether the subgroup being factored out is central and whether the quotient is cyclic. When both facts hold, every element of $G$ can be written as a power of one representative times a central element, forcing commutativity. If $G$ is known to be non-abelian, the same implication rules out a cyclic quotient by the centre. :::scientific-preview[Central Quotient Consequence Explorer] ::: Centrality lets the lifted representatives commute after rearranging the central factors. That is why cyclicity of the quotient becomes strong enough to force the whole group to be abelian. :::solved-problem Let $G$ be a group such that $|G/Z(G)|=p$, where $p$ is prime. Prove that $G$ is abelian. ::: :::solution Let $|G/Z(G)|=p$, where $p$ is prime. A group of prime order is cyclic. Therefore $G/Z(G)$ is cyclic. By the theorem that cyclicity of $G/Z(G)$ forces $G$ to be abelian, we get $$\begin{equation*} G \text{ is abelian}. \end{equation*}$$ ::: :::solved-problem Let $H\leq Z(G)$ and suppose $G/H=\langle aH\rangle$. Show directly that every element of $G$ has the form $a^n\circ h$ for some $n\in\mathbb{Z}$ and $h\in H$. ::: :::solution Let $x\in G$. Since $G/H=\langle aH\rangle$, there exists $n\in\mathbb{Z}$ such that $$\begin{equation*} xH=(aH)^n. \end{equation*}$$ Therefore $$\begin{equation*} xH=a^nH. \end{equation*}$$ Thus $$\begin{equation*} (a^n)^{-1}\circ x\in H. \end{equation*}$$ Let $$\begin{equation*} h=(a^n)^{-1}\circ x. \end{equation*}$$ Then $h\in H$, and $$\begin{equation*} x=a^n\circ h. \end{equation*}$$ ::: We combine quotient order and cyclicity facts. Enter the order of $G/Z(G)$ if it is finite. If the order is prime, then the quotient is cyclic, and the theorem forces $G$ to be abelian. The calculator also warns when a declared non-abelian group is paired with a cyclic central quotient claim. :::calculator[Central Quotient Test Calculator] ::: :::exercise 1. Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian. 2. Prove that if $G$ is non-abelian, then $G/Z(G)$ is not cyclic. 3. Let $H\leq Z(G)$. Prove that if $G/H$ is cyclic, then $G$ is abelian. 4. If $|G/Z(G)|=p$ for a prime $p$, prove that $G$ is abelian. 5. Explain why $Q_8/Z(Q_8)$ cannot be cyclic. ::: :::answer 1. If $G/Z(G)=\langle aZ(G)\rangle$, then every $x\in G$ has the form $a^m z$ with $z\in Z(G)$, and such elements commute. 2. This is the contrapositive of the theorem that cyclic $G/Z(G)$ implies abelian $G$. 3. If $G/H=\langle aH\rangle$ and $H\leq Z(G)$, then every element has the form $a^n h$ with $h\in H$, and such elements commute. 4. Since every group of prime order is cyclic, $G/Z(G)$ is cyclic, so $G$ is abelian. 5. Since $Q_8$ is non-abelian, its quotient by the centre cannot be cyclic. ::: :::faq[Frequently Asked Questions] Q: Why is $G/Z(G)$ always a quotient group? A: The centre $Z(G)$ is always a normal subgroup of $G$. Q: Does $G/Z(G)$ cyclic mean $G$ is cyclic? A: It means more strongly that $G$ is abelian, but it does not necessarily say that $G$ itself is cyclic. Q: Why does a central subgroup matter? A: Elements of a central subgroup commute with everything, so representatives lifted from a cyclic quotient can be rearranged safely. Q: Can a non-abelian group have a cyclic quotient? A: Yes, but not when the quotient is by the centre in the form $G/Z(G)$. ::: :::call-to-action[Review Quotient Group Construction] subtitle: Return to the beginning of quotient groups before moving to the next major topic. button: Review Quotient Groups | 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/quotient-groups/abelian-quotients-and-commutators.md :::