After studying the centre of a group, we now weaken the commutativity requirement. The centre asks an element to commute with every element of the group, while the centraliser asks which elements commute with one fixed element. The focus keyword for this lecture is centraliser of an element. This subgroup is important because it isolates the part of the group that behaves commutatively around a chosen element. In this lecture, students will define the centraliser, prove that it is a subgroup, and compute centralisers in standard situations. :::definition[Centraliser of an Element] Let $(G,\circ)$ be a group and let $a\in G$. The centraliser of the element $a$ in $G$ is the set of all elements of $G$ which commute with $a$, that is, $$\begin{equation*} C(a)=\{x\in G:x\circ a=a\circ x\}. \end{equation*}$$ ::: The centraliser $C(a)$ depends on the chosen element $a$. If $a$ changes, the subgroup $C(a)$ may also change. The centre $Z(G)$ is stricter because an element in the centre must commute with every element of $G$. Thus every central element belongs to every centraliser, but an element may belong to one centraliser without being central. A common mistake is to think that $x\in C(a)$ means $x$ commutes with all elements of $G$; it means only that $x$ commutes with the fixed element $a$. :::theorem[Centraliser Is a Subgroup] Let $(G,\circ)$ be a group and let $a\in G$. Then $C(a)$ is a subgroup of $G$. ::: :::proof[Proof of Centraliser Is a Subgroup] Given that $(G,\circ)$ is a group and $a\in G$. To prove that $C(a)$ is a subgroup of $G$. Since $$\begin{equation*} e\circ a=a\circ e=a, \end{equation*}$$ we get $e\in C(a)$. Therefore $C(a)$ is non-empty. Let $x,y\in C(a)$. Then $$\begin{equation*} x\circ a=a\circ x \end{equation*}$$ and $$\begin{equation*} y\circ a=a\circ y. \end{equation*}$$ From $y\circ a=a\circ y$, we get $$\begin{align*} y\circ a=a\circ y &\implies y\circ a\circ y^{-1}=a\\ &\implies a\circ y^{-1}=y^{-1}\circ a. \end{align*}$$ Thus $$\begin{equation*} y^{-1}\circ a=a\circ y^{-1}. \end{equation*}$$ Now $$\begin{align*} (x\circ y^{-1})\circ a &=x\circ(y^{-1}\circ a)\\ &=x\circ(a\circ y^{-1}) \quad [\because y^{-1}\circ a=a\circ y^{-1}]\\ &=(x\circ a)\circ y^{-1}\\ &=(a\circ x)\circ y^{-1} \quad [\because x\circ a=a\circ x]\\ &=a\circ(x\circ y^{-1}). \end{align*}$$ Therefore $x\circ y^{-1}\in C(a)$. By the one-step subgroup criterion, $C(a)$ is a subgroup of $G$. Hence, $C(a)\leq G$. $\square$ ::: :::example[Centraliser in an Abelian Group] Let $(G,\circ)$ be an abelian group and let $a\in G$. Since every element of $G$ commutes with every other element of $G$, we have $$\begin{equation*} x\circ a=a\circ x\quad \forall\,x\in G. \end{equation*}$$ Therefore every element of $G$ belongs to $C(a)$. Hence, $$\begin{equation*} C(a)=G. \end{equation*}$$ ::: :::example[Centraliser of the Identity] Let $(G,\circ)$ be a group with identity element $e$. For every $x\in G$, $$\begin{equation*} x\circ e=e\circ x=x. \end{equation*}$$ Therefore every element of $G$ commutes with $e$. Hence, $$\begin{equation*} C(e)=G. \end{equation*}$$ ::: :::solved-problem[Centraliser Contains Powers] Let $(G,\circ)$ be a group and let $a\in G$. Prove that $\langle a\rangle\subseteq C(a)$. ::: :::solution[Solution of Centraliser Contains Powers] Let $(G,\circ)$ be a group and let $a\in G$. To prove that $\langle a\rangle\subseteq C(a)$. Let $x\in\langle a\rangle$. Then there exists $n\in\mathbb{Z}$ such that $$\begin{equation*} x=a^n. \end{equation*}$$ Now $$\begin{align*} x\circ a &=a^n\circ a\\ &=a^{n+1}\\ &=a\circ a^n\\ &=a\circ x. \end{align*}$$ Therefore $x\in C(a)$. Thus $\langle a\rangle\subseteq C(a)$. Hence, every power of $a$ belongs to the centraliser of $a$. $\square$ ::: :::solved-problem[Centraliser of an Element of Order Two] Let $(G,\circ)$ be a group and let $a\in G$ such that $a^2=e$. Prove that $a\in C(a)$ and $a^{-1}\in C(a)$. ::: :::solution[Solution of Centraliser of an Element of Order Two] Let $(G,\circ)$ be a group and let $a\in G$ such that $a^2=e$. To prove that $a\in C(a)$ and $a^{-1}\in C(a)$. Since $$\begin{equation*} a\circ a=a\circ a, \end{equation*}$$ we get $a\in C(a)$. Since $a^2=e$, we get $a^{-1}=a$. Therefore $$\begin{equation*} a^{-1}=a\in C(a). \end{equation*}$$ Hence, $a\in C(a)$ and $a^{-1}\in C(a)$. $\square$ ::: The centraliser should be read as a local commutativity subgroup. It does not claim that the whole group is abelian. It only records those elements that commute with one fixed element. In many later topics, especially conjugacy and class equations, the size of $C(a)$ tells us how much freedom remains after fixing the element $a$. This is why centralisers appear naturally in structural questions about non-abelian groups. We observe centralisers in $S_3$. Choose the fixed element $a$, and the display lists exactly the elements $x$ satisfying $xa=ax$. The identity has the whole group as its centraliser, while a transposition has a smaller centraliser containing only itself and the identity. This makes the difference between the centre and a centraliser visible: a centraliser is local to one element. :::scientific-preview[Centraliser in S3 Explorer] ::: The centraliser always contains the fixed element and its powers. It may be much larger, but it does not have to be the whole group unless the fixed element commutes with every element. In a finite group, $C(a)$ is a subgroup, so its order must divide $|G|$. It also contains $\langle a\rangle$, so $|\langle a\rangle|$ must divide $|C(a)|$. Enter the three orders to test these necessary conditions. Passing the divisibility checks does not compute the centraliser, but it confirms that the proposed sizes are compatible with the subgroup facts in this lesson. :::calculator[Centraliser Order Calculator] ::: :::exercise[Exercises on Centralisers] 1. Prove that $a\in C(a)$ for every $a\in G$. 2. Prove that if $G$ is abelian, then $C(a)=G$ for every $a\in G$. 3. Let $x\in C(a)$. Prove that $x^{-1}\in C(a)$. ::: :::answer[Answers on Centralisers] 1. Since $a\circ a=a\circ a$, we get $a\in C(a)$. 2. If $G$ is abelian, then $x\circ a=a\circ x$ for every $x\in G$. Hence $C(a)=G$. 3. Since $x\in C(a)$, we have $x\circ a=a\circ x$. This gives $x^{-1}\circ a=a\circ x^{-1}$, and therefore $x^{-1}\in C(a)$. ::: :::faq[Frequently Asked Questions] Q: Does $C(a)$ always contain $a$? A: Yes. Every element commutes with itself, so $a\in C(a)$. Q: Is $C(a)$ always a subgroup? A: Yes. It satisfies the one-step subgroup criterion. Q: Is $C(a)$ the same as $Z(G)$? A: Not always. The centre contains elements that commute with every element, while $C(a)$ contains elements that commute with the fixed element $a$. ::: :::call-to-action[Continue to Centre and Centralisers] subtitle: Learn how the centre sits inside every centraliser and what happens when a centraliser equals the centre. button: Study Centre and Centralisers | published/abstract-algebra/subgroups-and-normal-subgroups/special-subgroups/centre-and-centralisers.md button-ghost: Review Centre of a Group | published/abstract-algebra/subgroups-and-normal-subgroups/special-subgroups/centre-of-a-group.md :::