After defining centralisers, we now connect them with the centre of a group. The focus keyword for this lecture is centre and centralisers. The centre consists of elements that commute with all elements, while a centraliser consists of elements that commute with one fixed element. This difference immediately gives an important inclusion: the centre is contained in every centraliser. In this lecture, students will prove this inclusion, express the centre as an intersection of centralisers, and study the condition $C(a)=Z(G)$. :::theorem[Centre Lies in Every Centraliser] Let $(G,\circ)$ be a group and let $a\in G$. Then $Z(G)$ is a subgroup of $C(a)$. ::: :::proof[Proof of Centre Lies in Every Centraliser] Given that $(G,\circ)$ is a group and $a\in G$. To prove that $Z(G)$ is a subgroup of $C(a)$. Since $Z(G)$ is a subgroup of $G$, it is enough to prove that $$\begin{equation*} Z(G)\subseteq C(a). \end{equation*}$$ Let $x\in Z(G)$. Then $$\begin{equation*} x\circ g=g\circ x\quad \forall\,g\in G. \end{equation*}$$ Since $a\in G$, we get $$\begin{equation*} x\circ a=a\circ x. \end{equation*}$$ Therefore $x\in C(a)$. Thus $$\begin{equation*} Z(G)\subseteq C(a). \end{equation*}$$ Since $Z(G)\leq G$ and $C(a)\leq G$, the subset relation gives $Z(G)\leq C(a)$. Hence, $Z(G)$ is a subgroup of $C(a)$. $\square$ ::: :::theorem[Centre as an Intersection of Centralisers] Let $(G,\circ)$ be a group. Then $$\begin{equation*} Z(G)=\bigcap_{a\in G}C(a). \end{equation*}$$ ::: :::proof[Proof of Centre as an Intersection of Centralisers] Given that $(G,\circ)$ is a group. To prove that $Z(G)=\bigcap_{a\in G}C(a)$. Let $x\in Z(G)$. Then $$\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}$$ Therefore $$\begin{equation*} x\in C(a)\quad \forall\,a\in G. \end{equation*}$$ Thus $$\begin{equation*} x\in \bigcap_{a\in G}C(a). \end{equation*}$$ Hence $$\begin{equation*} Z(G)\subseteq \bigcap_{a\in G}C(a). \end{equation*}$$ Conversely, let $x\in \bigcap_{a\in G}C(a)$. Then $$\begin{equation*} x\in C(a)\quad \forall\,a\in G. \end{equation*}$$ Therefore $$\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}$$ Thus $x\in Z(G)$. Hence $$\begin{equation*} \bigcap_{a\in G}C(a)\subseteq Z(G). \end{equation*}$$ Therefore $$\begin{align*} Z(G)\subseteq \bigcap_{a\in G}C(a)\quad \text{and}\quad \bigcap_{a\in G}C(a)\subseteq Z(G) &\implies Z(G)=\bigcap_{a\in G}C(a). \end{align*}$$ Hence, $$\begin{equation*} Z(G)=\bigcap_{a\in G}C(a). \end{equation*}$$ $\square$ ::: :::theorem[Centraliser Equal to Centre] Let $(G,\circ)$ be a group. If there exists an element $a\in G$ such that $$\begin{equation*} C(a)=Z(G), \end{equation*}$$ then $G$ is abelian. ::: :::proof[Proof of Centraliser Equal to Centre] Given that $(G,\circ)$ is a group. Let there exist an element $a\in G$ such that $$\begin{equation*} C(a)=Z(G). \end{equation*}$$ To prove that $G$ is abelian. Since $$\begin{equation*} a\circ a=a\circ a, \end{equation*}$$ we get $a\in C(a)$. Using $C(a)=Z(G)$, we get $a\in Z(G)$. Since $a\in Z(G)$, we get $$\begin{equation*} a\circ x=x\circ a\quad \forall\,x\in G. \end{equation*}$$ Therefore every element of $G$ commutes with $a$. Thus $$\begin{equation*} G\subseteq C(a). \end{equation*}$$ Also $$\begin{equation*} C(a)\subseteq G. \end{equation*}$$ Therefore $$\begin{equation*} C(a)=G. \end{equation*}$$ Using $C(a)=Z(G)$, we get $$\begin{equation*} Z(G)=G. \end{equation*}$$ By the abelian centre criterion, $G$ is abelian. Hence, $G$ is abelian. $\square$ ::: The condition $C(a)=Z(G)$ is stronger than it may first appear. Since $a$ always belongs to $C(a)$, equality forces $a$ to lie in the centre. But once $a$ is central, every element of the group commutes with $a$, so $C(a)=G$. Therefore $C(a)=Z(G)$ forces $Z(G)=G$, and this makes the whole group abelian. This argument is a useful example of how a local commutativity condition can become a global commutativity conclusion. :::example[Abelian Case] Let $(G,\circ)$ be an abelian group and let $a\in G$. Then $$\begin{equation*} Z(G)=G \end{equation*}$$ and $$\begin{equation*} C(a)=G. \end{equation*}$$ Thus $$\begin{equation*} C(a)=Z(G)=G. \end{equation*}$$ This example shows that the equality $C(a)=Z(G)$ occurs naturally in abelian groups. ::: :::solved-problem[Finding the Centre from Centralisers] Let $(G,\circ)$ be a group and suppose $$\begin{equation*} C(a)=G \end{equation*}$$ for every $a\in G$. Prove that $G$ is abelian. ::: :::solution[Solution of Finding the Centre from Centralisers] Let $(G,\circ)$ be a group and suppose $C(a)=G$ for every $a\in G$. To prove that $G$ is abelian. Let $x,y\in G$. Since $C(y)=G$, we get $x\in C(y)$. Therefore $$\begin{equation*} x\circ y=y\circ x. \end{equation*}$$ Thus every pair of elements of $G$ commutes. Hence, $G$ is abelian. $\square$ ::: :::solved-problem[Using the Intersection Formula] Let $(G,\circ)$ be a group. Prove that if $x$ belongs to every centraliser $C(a)$ with $a\in G$, then $x\in Z(G)$. ::: :::solution[Solution of Using the Intersection Formula] Let $(G,\circ)$ be a group. Suppose that $x\in C(a)$ for every $a\in G$. To prove that $x\in Z(G)$. Since $x\in C(a)$ for every $a\in G$, we get $$\begin{equation*} x\circ a=a\circ x\quad \forall\,a\in G. \end{equation*}$$ By the definition of the centre, this means $x\in Z(G)$. Hence, $x\in Z(G)$. $\square$ ::: We observe the formula $Z(G)=\bigcap_{a\in G}C(a)$ by tracking whether one candidate element belongs to several centralisers. A candidate is central only when it belongs to every centraliser. The display is deliberately logical rather than computational: missing even one centraliser means the element fails to commute with at least one group element. This helps separate a local centraliser condition from the global centre condition. :::scientific-preview[Centraliser Intersection Explorer] ::: The centre is the common part of all centralisers. Belonging to one centraliser only proves commutativity with one fixed element; belonging to all centralisers proves commutativity with every element. We use orders to test the special condition $C(a)=Z(G)$. Since $Z(G)\subseteq C(a)$, equal finite orders force equality. The theorem then says that if such an equality occurs for some element $a$, the whole group must be abelian. The calculator reports this conclusion and warns when the entered orders contradict the required containment. :::calculator[Centre Centraliser Equality Calculator] ::: :::exercise[Exercises on Centre and Centralisers] 1. Prove that $Z(G)\subseteq C(a)$ for every $a\in G$. 2. Prove that if $C(a)=G$ for some $a\in G$, then $a\in Z(G)$. 3. Prove that $G$ is abelian if and only if $C(a)=G$ for every $a\in G$. ::: :::answer[Answers on Centre and Centralisers] 1. If $x\in Z(G)$, then $x$ commutes with every element of $G$, so in particular $x\circ a=a\circ x$. Hence $x\in C(a)$. 2. Since $C(a)=G$, every $x\in G$ belongs to $C(a)$. Thus $x\circ a=a\circ x$ for every $x\in G$, and hence $a\in Z(G)$. 3. If $G$ is abelian, then every element commutes with every $a\in G$, so $C(a)=G$. Conversely, if $C(a)=G$ for every $a\in G$, then every pair of elements commutes, so $G$ is abelian. ::: :::faq[Frequently Asked Questions] Q: Why is $Z(G)$ contained in every $C(a)$? A: A central element commutes with every element, so it commutes with the particular element $a$. Q: What does $Z(G)=\bigcap_{a\in G}C(a)$ mean? A: It means that an element is central exactly when it belongs to the centraliser of every element of the group. Q: Why does $C(a)=Z(G)$ imply that $G$ is abelian? A: Since $a\in C(a)$, equality gives $a\in Z(G)$. Then $C(a)=G$, so $Z(G)=G$, and therefore $G$ is abelian. ::: :::call-to-action[Continue to Cosets of Subgroups] subtitle: Move from special subgroups to cosets, index, and the divisibility theorem for finite groups. button: Study Cosets | published/abstract-algebra/subgroups-and-normal-subgroups/cosets-of-subgroups/left-and-right-cosets.md button-ghost: Review Centralisers | published/abstract-algebra/subgroups-and-normal-subgroups/special-subgroups/centraliser-of-an-element.md :::