Centralizers and conjugacy give the first language for measuring how elements behave under inner change of coordinates in a group. The centralizers and conjugacy viewpoint asks two connected questions: which elements commute with a fixed element, and which elements can be obtained from it by conjugation? These ideas are later used in the class equation, normalizers, and Sylow theory, where counting conjugates becomes a structural tool. Students often confuse commuting with conjugating; commuting keeps an element unchanged under a particular rearrangement, while conjugating moves it within its natural symmetry family. :::definition[Centralizer of an Element] Let $G$ be a group and let $a \in G$. The centralizer of $a$ in $G$ is the set of all elements of $G$ that commute with $a$. It is denoted by $C_G(a)$ and is defined by $$\begin{equation*} C_G(a)=\{g \in G \mid ga=ag\}. \end{equation*}$$ ::: :::theorem Let $G$ be a group and let $a \in G$. Then $$\begin{equation*} C_G(a)=G \iff a \in Z(G). \end{equation*}$$ ::: :::proof Given that $G$ is a group and $a \in G$. To prove that $C_G(a)=G \iff a \in Z(G)$. Suppose first that $C_G(a)=G$. Then every $g \in G$ belongs to $C_G(a)$. Therefore $ga=ag$ for every $g \in G$. Hence $a \in Z(G)$. Conversely, suppose that $a \in Z(G)$. Then $ga=ag$ for every $g \in G$. Therefore every $g \in G$ belongs to $C_G(a)$. Hence $C_G(a)=G$. $\square$ ::: :::theorem Let $G$ be a group and let $a \in G$. Then $C_G(a)$ is a subgroup of $G$. ::: :::proof Given that $G$ is a group and $a \in G$. To prove that $C_G(a)$ is a subgroup of $G$. Since $ea=ae=a$, we have $e \in C_G(a)$. Therefore $C_G(a)$ is non-empty. Let $x,y \in C_G(a)$. Then $xa=ax$ and $ya=ay$. From $ya=ay$, multiplying on the left by $y^{-1}$ and on the right by $y^{-1}$ gives $$\begin{equation*} y^{-1}a=ay^{-1}. \end{equation*}$$ Now $$\begin{align*} (xy^{-1})a &= x(y^{-1}a) \\ &= x(ay^{-1}) \\ &= (xa)y^{-1} \\ &= (ax)y^{-1} \\ &= a(xy^{-1}). \end{align*}$$ Therefore $xy^{-1} \in C_G(a)$. Hence $C_G(a)$ is a subgroup of $G$. $\square$ ::: :::definition[Conjugate Elements] Let $G$ be a group and let $a,b \in G$. The element $b$ is called a conjugate of $a$ in $G$ if there exists $g \in G$ such that $$\begin{equation*} b=gag^{-1}. \end{equation*}$$ ::: :::theorem Let $G$ be a group. Define a relation $\rho$ on $G$ by declaring $a \rho b$ if and only if $b=gag^{-1}$ for some $g \in G$. Then $\rho$ is an equivalence relation on $G$. ::: :::proof Given that $G$ is a group and $\rho$ is the relation on $G$ defined by $a \rho b$ if and only if $b=gag^{-1}$ for some $g \in G$. To prove that $\rho$ is an equivalence relation on $G$. [1] Let $a \in G$. Since $a=eae^{-1}$, we have $a \rho a$. Therefore $\rho$ is reflexive. [2] Let $a \rho b$. Then $b=gag^{-1}$ for some $g \in G$. Multiplying on the left by $g^{-1}$ and on the right by $g$ gives $$\begin{equation*} a=g^{-1}bg. \end{equation*}$$ Since $g^{-1} \in G$, the element $a$ is a conjugate of $b$. Therefore $b \rho a$. Hence $\rho$ is symmetric. [3] Let $a \rho b$ and $b \rho c$. Then there exist $g,h \in G$ such that $$\begin{equation*} b=gag^{-1} \end{equation*}$$ and $$\begin{equation*} c=hbh^{-1}. \end{equation*}$$ Therefore $$\begin{align*} c &= hbh^{-1} \\ &= h(gag^{-1})h^{-1} \\ &= (hg)a(hg)^{-1}. \end{align*}$$ Since $hg \in G$, the element $c$ is a conjugate of $a$. Therefore $a \rho c$. Hence $\rho$ is transitive. Hence $\rho$ is an equivalence relation on $G$. $\square$ ::: :::definition[Conjugacy Class of an Element] Let $G$ be a group and let $a \in G$. The conjugacy class of $a$ in $G$ is the equivalence class of $a$ under conjugacy. It is denoted by $Cl_G(a)$ and is given by $$\begin{equation*} Cl_G(a)=\{gag^{-1} \mid g \in G\}. \end{equation*}$$ ::: :::theorem[Conjugacy Class Formula] Let $G$ be a finite group and let $a \in G$. Then the number of conjugates of $a$ in $G$ is $$\begin{equation*} |Cl_G(a)|=[G:C_G(a)]. \end{equation*}$$ ::: :::proof Given that $G$ is a finite group and $a \in G$. To prove that $|Cl_G(a)|=[G:C_G(a)]$. Let $\mathcal{L}$ be the set of all left cosets of $C_G(a)$ in $G$. Define $f:\mathcal{L}\to Cl_G(a)$ by $$\begin{equation*} f(gC_G(a))=gag^{-1}. \end{equation*}$$ To prove that $f$ is well-defined, let $gC_G(a)=hC_G(a)$. Then $h^{-1}g \in C_G(a)$. Therefore $$\begin{equation*} (h^{-1}g)a=a(h^{-1}g). \end{equation*}$$ Multiplying on the left by $h$ and on the right by $g^{-1}$ gives $gag^{-1}=hah^{-1}$. Therefore $f(gC_G(a))=f(hC_G(a))$. To prove that $f$ is one-one, suppose that $f(gC_G(a))=f(hC_G(a))$. Then $gag^{-1}=hah^{-1}$. Therefore $$\begin{equation*} (h^{-1}g)a=a(h^{-1}g). \end{equation*}$$ Thus $h^{-1}g \in C_G(a)$, and so $gC_G(a)=hC_G(a)$. To prove that $f$ is onto, let $x \in Cl_G(a)$. Then $x=gag^{-1}$ for some $g \in G$. Therefore $x=f(gC_G(a))$. Therefore $f$ is a bijection. Hence $|Cl_G(a)|=[G:C_G(a)]$. $\square$ ::: :::example Let $G=S_3$ and let $a=(12)$. The conjugates of a transposition in $S_3$ are precisely the transpositions. Hence $$\begin{equation*} Cl_{S_3}((12))=\{(12),(13),(23)\}. \end{equation*}$$ Thus $|Cl_{S_3}((12))|=3$. Since $|S_3|=6$, the conjugacy class formula gives $$\begin{equation*} 3=[S_3:C_{S_3}((12))]. \end{equation*}$$ Therefore $|C_{S_3}((12))|=2$. In fact, $$\begin{equation*} C_{S_3}((12))=\{e,(12)\}. \end{equation*}$$ ::: This preview uses $S_3$ to connect conjugacy classes with centralizers. Choose an element and compare the elements that commute with it against the elements obtained by conjugation. The displayed formula checks $|Cl_G(a)|=[G:C_G(a)]$ for the selected element. Try the identity, a transposition, and a three-cycle, and observe that central elements have singleton conjugacy classes while noncentral elements move inside larger classes. :::scientific-preview[S3 Conjugacy and Centralizer Explorer] ::: The calculator shows that a conjugacy class size is not arbitrary; it is an index of a subgroup. This divisibility fact is the reason conjugacy becomes powerful in later counting arguments. :::solved-problem Let $G$ be a group and let $a \in G$. Prove that $a \in Z(G)$ if and only if $Cl_G(a)=\{a\}$. ::: :::solution Let $G$ be a group and let $a \in G$. Suppose first that $a \in Z(G)$. Then $ga=ag$ for every $g \in G$. Therefore $$\begin{align*} gag^{-1} &= agg^{-1} \\ &= a. \end{align*}$$ Thus every conjugate of $a$ is equal to $a$. Hence $Cl_G(a)=\{a\}$. Conversely, suppose that $Cl_G(a)=\{a\}$. Then $gag^{-1}=a$ for every $g \in G$. Multiplying on the right by $g$ gives $ga=ag$ for every $g \in G$. Therefore $a \in Z(G)$. Hence $a \in Z(G)$ if and only if $Cl_G(a)=\{a\}$. $\square$ ::: :::exercise [1] Let $G$ be an abelian group. Find $C_G(a)$ and $Cl_G(a)$ for $a \in G$. [2] Let $G$ be a finite group and let $a \in G$ have exactly two conjugates. Prove that $C_G(a)$ is normal in $G$. [3] In $S_3$, find the conjugacy class of $(123)$. ::: :::answer [1] Since every element of $G$ commutes with $a$, $C_G(a)=G$. Also $gag^{-1}=a$ for every $g \in G$, so $Cl_G(a)=\{a\}$. [2] Since $|Cl_G(a)|=[G:C_G(a)]$ and $a$ has exactly two conjugates, $[G:C_G(a)]=2$. Every subgroup of index $2$ is normal, so $C_G(a)$ is normal in $G$. [3] The conjugacy class is $\{(123),(132)\}$. ::: :::faq[Frequently Asked Questions] Q: Is the centralizer of an element always a subgroup? A: Yes. The centralizer contains the identity and is closed under the one-step subgroup test. Q: Does conjugacy mean equality? A: Not usually. Conjugate elements may be different, but they represent the same element after an inner change by an element of the group. Q: Why does the index of the centralizer count conjugates? A: Two elements $g,h \in G$ produce the same conjugate of $a$ exactly when they lie in the same left coset of $C_G(a)$. ::: :::call-to-action[Next Step] subtitle: Use centralizers and conjugacy classes to build the class equation for finite groups. button: Study Class Equation | published/abstract-algebra/sylow-theorems/conjugacy-classes/class-equation-and-conjugacy-class-decomposition button-ghost: Review Centralizers | published/abstract-algebra/sylow-theorems/conjugacy-classes/centralizers-and-conjugacy-of-elements :::