After studying subgroups formed from intersections, unions, and products, the next natural step is to study subgroups determined by commutativity conditions. The first such special subgroup is the centre of a group. The focus keyword for this lecture is centre of a group. The centre collects exactly those elements that commute with every element of the group. In this lecture, students will learn the definition of the centre, prove that it is a subgroup, and understand why an abelian group is precisely a group whose centre is the whole group. :::definition[Centre of a Group] Let $(G,\circ)$ be a group. The centre of the group $G$ is the set of all elements of $G$ which commute with every element of $G$, that is, $$\begin{equation*} Z(G)=\{a\in G:a\circ x=x\circ a\quad \forall\,x\in G\}. \end{equation*}$$ ::: An element $a\in G$ belongs to $Z(G)$ only when it commutes with every element of $G$, not merely with some elements of $G$. Thus the centre measures how much commutativity is present inside a possibly non-abelian group. If $G$ is abelian, then every element commutes with every other element, so the centre should be all of $G$. If $G$ is highly non-abelian, the centre may be very small. A common mistake is to check $a\circ x=x\circ a$ for one convenient element $x$ instead of checking it for all $x\in G$. :::theorem[Centre Is a Subgroup] Let $(G,\circ)$ be a group. Then $Z(G)$ is a subgroup of $G$. ::: :::proof[Proof of Centre Is a Subgroup] Given that $(G,\circ)$ is a group. To prove that $Z(G)$ is a subgroup of $G$. Since $e\circ x=x\circ e=x$ for all $x\in G$, we get $e\in Z(G)$. Therefore $Z(G)$ is non-empty. Let $a,b\in Z(G)$. To prove that $a\circ b^{-1}\in Z(G)$. Let $x\in G$. Since $a\in Z(G)$, we get $$\begin{equation*} a\circ x=x\circ a. \end{equation*}$$ Since $b\in Z(G)$, we get $$\begin{equation*} b\circ x=x\circ b. \end{equation*}$$ Therefore $$\begin{align*} b\circ x=x\circ b &\implies b\circ x\circ b^{-1}=x\\ &\implies x\circ b^{-1}=b^{-1}\circ x. \end{align*}$$ Thus $$\begin{equation*} b^{-1}\circ x=x\circ b^{-1}. \end{equation*}$$ Now $$\begin{align*} (a\circ b^{-1})\circ x &=a\circ(b^{-1}\circ x)\\ &=a\circ(x\circ b^{-1}) \quad [\because b^{-1}\circ x=x\circ b^{-1}]\\ &=(a\circ x)\circ b^{-1}\\ &=(x\circ a)\circ b^{-1} \quad [\because a\circ x=x\circ a]\\ &=x\circ(a\circ b^{-1}). \end{align*}$$ Therefore $a\circ b^{-1}\in Z(G)$. By the one-step subgroup criterion, $Z(G)$ is a subgroup of $G$. Hence, $Z(G)\leq G$. $\square$ ::: :::theorem[Centre of an Abelian Group] Let $(G,\circ)$ be a group. If $G$ is abelian, then $$\begin{equation*} Z(G)=G. \end{equation*}$$ ::: :::proof[Proof of Centre of an Abelian Group] Given that $(G,\circ)$ is a group. Let $G$ be abelian. To prove that $Z(G)=G$. Let $a\in G$. Since $G$ is abelian, we get $$\begin{equation*} a\circ x=x\circ a\quad \forall\,x\in G. \end{equation*}$$ Therefore $a\in Z(G)$. Thus $$\begin{equation*} G\subseteq Z(G). \end{equation*}$$ Also, by definition of $Z(G)$, we have $$\begin{equation*} Z(G)\subseteq G. \end{equation*}$$ Therefore $$\begin{align*} G\subseteq Z(G)\quad \text{and}\quad Z(G)\subseteq G &\implies Z(G)=G. \end{align*}$$ Hence, $Z(G)=G$. $\square$ ::: :::corollary[Abelian Centre Criterion] Let $(G,\circ)$ be a group. Then $G$ is abelian if and only if $$\begin{equation*} Z(G)=G. \end{equation*}$$ ::: :::proof[Proof of Abelian Centre Criterion] Given that $(G,\circ)$ is a group. To prove that $G$ is abelian if and only if $Z(G)=G$. [1] Let $G$ be abelian. By the theorem on the centre of an abelian group, we get $$\begin{equation*} Z(G)=G. \end{equation*}$$ [2] Let $Z(G)=G$. To prove that $G$ is abelian. Let $a,b\in G$. Since $Z(G)=G$, we get $a\in Z(G)$. Therefore $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ Thus every pair of elements of $G$ commutes. Hence, $G$ is abelian if and only if $Z(G)=G$. $\square$ ::: :::example[Centre of an Additive Group] Let $(\mathbb{Z},+)$ be the group of integers under addition. Since addition of integers is commutative, for all $a,x\in\mathbb{Z}$ we have $$\begin{equation*} a+x=x+a. \end{equation*}$$ Therefore every integer belongs to the centre. Hence, $$\begin{equation*} Z(\mathbb{Z})=\mathbb{Z}. \end{equation*}$$ ::: :::example[Centre May Be Proper] Let $(G,\circ)$ be a non-abelian group. Then at least one pair of elements $a,b\in G$ satisfies $$\begin{equation*} a\circ b\neq b\circ a. \end{equation*}$$ In that case, $a$ cannot belong to $Z(G)$ or $b$ cannot belong to $Z(G)$, and hence the centre is not necessarily the whole group. This shows why the centre is a special subgroup measuring the central part of a group rather than the whole group in every case. ::: We observe centrality in the concrete group $S_3$. Choose an element and test whether it commutes with every element of $S_3$. The identity passes every test, while a transposition or a three-cycle fails with at least one element. This shows why checking one convenient product is not enough to prove membership in the centre. :::scientific-preview[Centre Membership Explorer] ::: The centre collects elements that pass every commutativity test. In $S_3$, only the identity passes, which is why the centre of $S_3$ is trivial. :::solved-problem[Centre of a Two Element Group] Let $G=\{e,a\}$ be a group of order $2$. Prove that $Z(G)=G$. ::: :::solution[Solution of Centre of a Two Element Group] Let $G=\{e,a\}$ be a group of order $2$. To prove that $Z(G)=G$. Since $e$ is the identity element, we get $$\begin{equation*} e\circ x=x\circ e=x\quad \forall\,x\in G. \end{equation*}$$ Therefore $e\in Z(G)$. Since $a\neq e$ and $G$ has exactly two elements, the inverse of $a$ is $a$. Thus $$\begin{equation*} a\circ a=e. \end{equation*}$$ Now $a$ commutes with $e$ because $e$ is the identity element, and $a$ commutes with itself because $$\begin{equation*} a\circ a=a\circ a. \end{equation*}$$ Therefore $$\begin{equation*} a\circ x=x\circ a\quad \forall\,x\in G. \end{equation*}$$ Thus $a\in Z(G)$. Hence, $Z(G)=G$. $\square$ ::: We can use the size of the centre to test whether a finite group could be abelian. Enter the order of $G$ and the order of $Z(G)$. If the two orders are equal, the abelian centre criterion proves that $G$ is abelian. If the centre is smaller, the group is not abelian because an abelian group has centre equal to the whole group. :::calculator[Centre Size Calculator] ::: :::exercise[Exercises on Centre of a Group] 1. Prove that $e\in Z(G)$ for every group $(G,\circ)$. 2. Let $G$ be abelian. Prove directly from the definition that $Z(G)=G$. 3. Let $G$ be a group. Prove that if $Z(G)=\{e\}$, then no non-identity element of $G$ commutes with every element of $G$. ::: :::answer[Answers on Centre of a Group] 1. Since $e\circ x=x\circ e=x$ for all $x\in G$, we get $e\in Z(G)$. 2. If $a\in G$, then $a\circ x=x\circ a$ for all $x\in G$ because $G$ is abelian. Hence $a\in Z(G)$, so $G\subseteq Z(G)$. Also $Z(G)\subseteq G$, and therefore $Z(G)=G$. 3. If a non-identity element $a$ commuted with every element of $G$, then $a\in Z(G)$. This would contradict $Z(G)=\{e\}$. ::: :::faq[Frequently Asked Questions] Q: Is the centre always a subgroup? A: Yes. The centre is closed under the one-step subgroup test and contains the identity. Q: Is the centre always the whole group? A: No. The centre is the whole group exactly when the group is abelian. Q: Can the centre be the trivial subgroup? A: Yes. Some non-abelian groups have centre equal to $\{e\}$. ::: :::call-to-action[Continue to Centralisers] subtitle: Study the subgroup of elements that commute with one fixed element. button: Study Centralisers | published/abstract-algebra/subgroups-and-normal-subgroups/special-subgroups/centraliser-of-an-element.md button-ghost: Review Products of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/products-of-subgroups.md :::