Conjugacy begins with a nuisance that turns into a method: the same group element can look different after the group renames itself from the inside. In a symmetry group, this is not a cosmetic relabelling imposed from outside. It is the operation of taking one symmetry, moving the whole system by another symmetry, performing the first symmetry, and moving back. In $S_n$, this internal relabelling preserves the cycle pattern of a permutation. In matrix groups, it is the algebraic version of changing basis. In every group, it divides elements into families whose members have the same internal type. Conjugacy classes are those families. [example: Relabelling a Transposition in $S_4$] Let $G = S_4$, let $g = (1\,2)$, and let $h = (1\,3\,4)$. Then $h^{-1} = (1\,4\,3)$, and since permutations act on the left, the composite $hgh^{-1}$ is evaluated from right to left. We compute its value on each symbol: \begin{align*} (hgh^{-1})(1) &= h(g(h^{-1}(1))) = h(g(4)) = h(4) = 1, \\ (hgh^{-1})(2) &= h(g(h^{-1}(2))) = h(g(2)) = h(1) = 3, \\ (hgh^{-1})(3) &= h(g(h^{-1}(3))) = h(g(1)) = h(2) = 2, \\ (hgh^{-1})(4) &= h(g(h^{-1}(4))) = h(g(3)) = h(3) = 4. \end{align*} Thus $hgh^{-1}$ fixes $1$ and $4$, sends $2$ to $3$, and sends $3$ to $2$, so \begin{align*} h(1\,2)h^{-1} = (2\,3). \end{align*} Equivalently, conjugation has transported the two moved letters $1$ and $2$ to $h(1)=3$ and $h(2)=2$. The result is still a transposition, showing in this case that conjugacy in $S_n$ preserves the shape of a permutation while renaming its symbols. [/example] This relabelling principle is not limited to permutations. It is the basic way a group acts on itself while respecting its multiplication. The chapter develops conjugacy classes as orbits, counts them by stabilisers, computes them in important groups, and explains why they control normal subgroups and character theory. ## Definition The first object to isolate is the motion itself. If $h$ is allowed to change the internal viewpoint, then $g$ is transported to $hgh^{-1}$. This operation is designed so that multiplication relations are preserved while the element being studied is moved to a different representative of the same internal type. Let $G$ be a group. Elements $g,k \in G$ are called conjugate in $G$ if there exists $h \in G$ such that $k = hgh^{-1}$. The ambient group is part of the data: two elements may become conjugate after enlarging the group, because the larger group supplies more elements $h$ with which to perform the transport. Once a single transported element is understood, the natural question is how far a fixed element can move. The answer is not a number at first, but a subset of the group: the whole family of elements obtained by internal relabelling. [definition: Conjugacy Class] Let $G$ be a group and let $g \in G$. The conjugacy class of $g$ in $G$ is the subset \begin{align*} \operatorname{Cl}_G(g) = \{hgh^{-1} : h \in G\} \subset G. \end{align*} [/definition] The definition gives the object, but it does not yet tell us whether conjugacy really partitions the group into coherent pieces. The next theme supplies that structure. ## Conjugacy as an Equivalence Relation A classification by internal type should have three basic features: every element should have its own type, the order of comparison should not matter, and two comparisons through an intermediate representative should compose. These are exactly the axioms of an [equivalence relation](/page/Equivalence%20Relation). [quotetheorem:5005] This theorem says that the conjugacy classes are not overlapping fragments. They form a partition of $G$, so every element belongs to exactly one conjugacy class. [example: The Conjugacy Classes of $S_3$] Let $G=S_3$. For every $h \in S_3$, \begin{align*} heh^{-1}(x) &= h(e(h^{-1}(x))) \\ &= h(h^{-1}(x)) \\ &= x \end{align*} for each $x \in \{1,2,3\}$, so $heh^{-1}=e$ and $\operatorname{Cl}_{S_3}(e)=\{e\}$. For a transposition $(a\,b)$ and any $h \in S_3$, the conjugate $h(a\,b)h^{-1}$ sends $h(a)$ to $h(b)$, sends $h(b)$ to $h(a)$, and fixes the remaining symbol. Hence \begin{align*} h(a\,b)h^{-1}=(h(a)\,h(b)). \end{align*} Applying this to $(1\,2)$ gives \begin{align*} e(1\,2)e^{-1} &= (1\,2), \\ (1\,3)(1\,2)(1\,3)^{-1} &= ((1\,3)(1)\,(1\,3)(2)) = (3\,2) = (2\,3), \\ (2\,3)(1\,2)(2\,3)^{-1} &= ((2\,3)(1)\,(2\,3)(2)) = (1\,3). \end{align*} Thus the conjugacy class of $(1\,2)$ is \begin{align*} \{(1\,2),(1\,3),(2\,3)\}. \end{align*} For the $3$-cycle $(1\,2\,3)$, the same relabelling rule gives \begin{align*} h(1\,2\,3)h^{-1}=(h(1)\,h(2)\,h(3)). \end{align*} Taking $h=e$ gives $(1\,2\,3)$, while taking $h=(1\,2)$ gives \begin{align*} (1\,2)(1\,2\,3)(1\,2)^{-1} &= ((1\,2)(1)\,(1\,2)(2)\,(1\,2)(3)) \\ &= (2\,1\,3) \\ &= (1\,3\,2). \end{align*} These are the only two $3$-cycles in $S_3$, so their class is \begin{align*} \{(1\,2\,3),(1\,3\,2)\}. \end{align*} Therefore $S_3$ is partitioned into conjugacy classes of sizes $1$, $3$, and $2$. [/example] The partition of $S_3$ already hints that conjugacy classes measure non-commutativity. The identity behaves like a central element, while transpositions and $3$-cycles move in families under relabelling. The equivalence relation viewpoint tells us that conjugacy classes partition $G$, but it does not explain their sizes. For counting, the orbit viewpoint is stronger. ## Orbits, Stabilisers, and the Class Equation ### Centralisers as Stabilisers Before counting the elements that fail to move a chosen $g$, it is useful to isolate the elements that are never moved by conjugation. These are precisely the elements that commute with everything in the group, and they form the fixed part of the conjugation action. [definition: Centre of a Group] Let $G$ be a group. The centre of $G$ is \begin{align*} Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}. \end{align*} [/definition] Elements in the centre have singleton conjugacy classes. They are invisible to conjugation because moving them by any $h \in G$ gives the same element back. [example: Conjugacy in an Abelian Group] Let $G$ be an abelian group and let $g \in G$. For every $h \in G$, commutativity gives $hg = gh$, so \begin{align*} hgh^{-1} &= (hg)h^{-1} \\ &= (gh)h^{-1} \\ &= g(hh^{-1}) \\ &= ge \\ &= g. \end{align*} Thus every conjugate of $g$ is equal to $g$, and taking $h=e$ also gives $ege^{-1}=g$, so \begin{align*} \operatorname{Cl}_G(g)=\{g\}. \end{align*} In an abelian group, conjugation does not move any element, so every conjugacy class has size $1$. [/example] The centre describes elements with no conjugation motion at all, but counting a class requires more local information. A noncentral element may still be fixed by many conjugating elements, and those failed motions are exactly what orbit-stabiliser needs as input. For a chosen $g$, the stabiliser under conjugation consists of the elements $h \in G$ satisfying $hgh^{-1} = g$, equivalently $hg = gh$. This subgroup deserves its own name because it measures how much of $G$ commutes with $g$, and therefore how many distinct conjugates of $g$ remain after the redundant relabellings have been identified. [definition: Centraliser] Let $G$ be a group and let $g \in G$. The centraliser of $g$ in $G$ is \begin{align*} C_G(g) = \{h \in G : hg = gh\}. \end{align*} [/definition] The centraliser lies between the cyclic subgroup generated by $g$ and the whole group. Before using its index, we need the algebraic fact that this set of commuting elements is itself a subgroup of $G$. [quotetheorem:5006] This subgroup result is what makes the notation $[G : C_G(g)]$ legitimate. Since the centraliser is the stabiliser of $g$ for the conjugation action, orbit-stabiliser should convert it into a formula for the size of $\operatorname{Cl}_G(g)$. The counting problem is that many elements $h \in G$ can produce the same conjugate $hgh^{-1}$. The duplication is exactly controlled by the elements that commute with $g$: changing $h$ by an element of $C_G(g)$ does not change the resulting conjugate. The needed formula identifies the conjugacy class with the cosets of this stabiliser. [quotetheorem:2404] This formula turns the geometric idea of moving an element into a precise count. To know the size of the conjugacy class, it suffices to count how many elements commute with $g$. [example: Centraliser of a Transposition in $S_4$] Let $G=S_4$ and let $g=(1\,2)$. For any $h \in S_4$, conjugation relabels the two moved symbols of the transposition: \begin{align*} h(1\,2)h^{-1}(h(1)) &= h(1\,2)(1) = h(2), \\ h(1\,2)h^{-1}(h(2)) &= h(1\,2)(2) = h(1), \end{align*} and if $x \notin \{h(1),h(2)\}$, then $h^{-1}(x) \notin \{1,2\}$, so \begin{align*} h(1\,2)h^{-1}(x) &= h(h^{-1}(x)) = x. \end{align*} Hence \begin{align*} h(1\,2)h^{-1}=(h(1)\,h(2)). \end{align*} Now $h$ commutes with $(1\,2)$ exactly when \begin{align*} h(1\,2)h^{-1}=(1\,2), \end{align*} which by the previous computation is equivalent to \begin{align*} (h(1)\,h(2))=(1\,2). \end{align*} Thus $\{h(1),h(2)\}=\{1,2\}$. Since $h$ is a bijection of $\{1,2,3,4\}$, this also forces \begin{align*} \{h(3),h(4)\}=\{3,4\}. \end{align*} There are two possible actions on $\{1,2\}$ and two possible actions on $\{3,4\}$, giving \begin{align*} C_{S_4}((1\,2))=\{e,(1\,2),(3\,4),(1\,2)(3\,4)\}. \end{align*} By *[Conjugacy Class Size Formula](/theorems/2404)*, \begin{align*} |\operatorname{Cl}_{S_4}((1\,2))| &= [S_4:C_{S_4}((1\,2))] \\ &= \frac{|S_4|}{|C_{S_4}((1\,2))|} \\ &= \frac{4!}{4} \\ &= \frac{24}{4} \\ &= 6. \end{align*} There are also exactly $\binom{4}{2}=6$ transpositions in $S_4$, one for each two-element subset of $\{1,2,3,4\}$, so this class is precisely the class of all transpositions in $S_4$. [/example] ### Global Counting The same count can be summed over all classes. Once the singleton classes from the centre have been separated, every remaining class contributes an index of a proper centraliser. This global count matters because it turns a partition of $G$ into arithmetic information about $|G|$. The class equation is the standard form of that count. [quotetheorem:3242] What does the class equation force when the group order is a power of a prime? In that setting, noncentral conjugacy classes have sizes divisible by $p$, so the centre must absorb the remaining congruence information. [quotetheorem:799] This result is one of the first places where conjugacy classes do real work. The theorem is not about a particular conjugacy class, but the proof depends on the fact that every noncentral class has size divisible by $p$. ## Computation in Symmetric and Matrix Groups ### Cycle Type in Symmetric Groups Conjugacy becomes concrete when the group has a familiar model. In symmetric groups, conjugation renames symbols. In matrix groups, conjugation changes basis. These two examples explain why conjugacy classes are so often treated as internal types. For permutations, the feature preserved by relabelling is cycle length. A transposition remains a transposition, a $3$-cycle remains a $3$-cycle, and a product of disjoint cycles keeps the same list of cycle lengths. [definition: Cycle Type] Let $\mathcal{P}(n)$ denote the set of partitions of $n$. The cycle type map is \begin{align*} \operatorname{type}: S_n &\to \mathcal{P}(n) \\ \sigma &\mapsto \text{the partition of } n \text{ whose parts are the lengths of the disjoint cycles of } \sigma, \end{align*} including parts of size $1$ for fixed points. [/definition] Cycle type is a compact invariant of a permutation, but invariance alone would not classify conjugacy classes: two permutations could conceivably have the same cycle lengths while still failing to be conjugate. The key issue is whether relabelling the underlying symbols can turn any permutation of a fixed cycle type into any other one of that same type. This is the point where cycle type must do more than describe an individual permutation. To use it for conjugacy classes, we need the converse direction: matching cycle types should guarantee that the two permutations are conjugate in $S_n$. [quotetheorem:798] This theorem turns a group-theoretic question into a combinatorial one. The conjugacy classes of $S_n$ are indexed by partitions of $n$. [example: Conjugacy Classes in $S_4$ by Cycle Type] By *Conjugacy Classes in $S_n$*, conjugacy classes in $S_4$ are indexed by cycle type. The partitions of $4$ are \begin{align*} 4, \quad 3+1, \quad 2+2, \quad 2+1+1, \quad 1+1+1+1. \end{align*} These correspond respectively to $4$-cycles, $3$-cycles with one fixed point, products of two disjoint transpositions, single transpositions, and the identity. For type $4$, a $4$-cycle uses all four symbols. There are $4!$ ways to write the four symbols in a row, but \begin{align*} (a\,b\,c\,d)=(b\,c\,d\,a)=(c\,d\,a\,b)=(d\,a\,b\,c), \end{align*} so each $4$-cycle is counted $4$ times. Hence the number of $4$-cycles is \begin{align*} \frac{4!}{4}=\frac{24}{4}=6. \end{align*} For type $3+1$, choose the fixed symbol in $4$ ways. The remaining three symbols form a $3$-cycle, and \begin{align*} (a\,b\,c)=(b\,c\,a)=(c\,a\,b), \end{align*} so the number of $3$-cycles on those three symbols is \begin{align*} \frac{3!}{3}=\frac{6}{3}=2. \end{align*} Thus the number of elements of type $3+1$ is \begin{align*} 4 \cdot 2 = 8. \end{align*} For type $2+2$, the elements are exactly \begin{align*} (1\,2)(3\,4), \quad (1\,3)(2\,4), \quad (1\,4)(2\,3), \end{align*} so there are $3$ of them. For type $2+1+1$, choose the two moved symbols: \begin{align*} \binom{4}{2}=\frac{4\cdot 3}{2}=6, \end{align*} so there are $6$ transpositions. For type $1+1+1+1$, the only element is the identity, so the class has size $1$. Therefore the five conjugacy class sizes in $S_4$ are \begin{align*} 6, \quad 8, \quad 3, \quad 6, \quad 1, \end{align*} and their sum is \begin{align*} 6+8+3+6+1=24=|S_4|. \end{align*} Thus the cycle-type classes account for every element of $S_4$ exactly once. [/example] ### Change of Basis in Matrix Groups The symmetric group example shows conjugacy as relabelling finite symbols. The matrix setting asks for the corresponding idea when the labels are basis vectors and the object being studied is a linear transformation. Matrices supply a different kind of internal relabelling. If a [linear map](/page/Linear%20Map) is represented by a matrix, changing basis replaces the matrix $A$ by $PAP^{-1}$. Conjugacy is therefore the algebraic relation behind normal forms. [definition: Similar Matrices] Let $k$ be a field, let $M_n(k)$ denote the set of $n \times n$ matrices over $k$, and let $GL_n(k)$ denote the group of invertible elements of $M_n(k)$. For $A, B \in M_n(k)$, the matrices $A$ and $B$ are similar over $k$ if there exists $P \in GL_n(k)$ such that \begin{align*} B = PAP^{-1}. \end{align*} [/definition] Similarity is conjugacy inside the group $GL_n(k)$ when $A$ and $B$ are invertible. It also makes sense for all matrices in the ring $M_n(k)$, where the same formula describes change of basis for linear maps. A tempting first invariant is the list of eigenvalues, but change of basis preserves more than the spectrum: it also preserves how eigenspaces and generalized eigenspaces fit together. The next example shows exactly where the naive eigenvalue test breaks. [example: A Jordan Block Conjugacy Invariant] Let $k=\mathbb{C}$ and consider \begin{align*} A &= \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}, & B &= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}. \end{align*} Their characteristic polynomials are the same, since \begin{align*} \det(tI-A) &= \det\begin{pmatrix}t-1 & -1 \\ 0 & t-1\end{pmatrix} \\ &= (t-1)(t-1)-(-1)\cdot 0 \\ &= (t-1)^2, \end{align*} and \begin{align*} \det(tI-B) &= \det\begin{pmatrix}t-1 & 0 \\ 0 & t-1\end{pmatrix} \\ &= (t-1)(t-1)-0\cdot 0 \\ &= (t-1)^2. \end{align*} Nevertheless, $A$ and $B$ are not similar. Subtracting the identity gives \begin{align*} A-I &= \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix} - \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \\ B-I &= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} - \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}. \end{align*} The columns of $A-I$ are \begin{align*} \begin{pmatrix}0 \\ 0\end{pmatrix}, \quad \begin{pmatrix}1 \\ 0\end{pmatrix}, \end{align*} so its column space is $\operatorname{span}\{(1,0)^\top\}$ and $\operatorname{rank}(A-I)=1$. The columns of $B-I$ are both zero, so $\operatorname{rank}(B-I)=0$. If $B=PAP^{-1}$ for some $P\in GL_2(\mathbb{C})$, then \begin{align*} B-I &= PAP^{-1}-I \\ &= PAP^{-1}-PP^{-1} \\ &= P(A-I)P^{-1}. \end{align*} Multiplication by $P$ and $P^{-1}$ is bijective on $\mathbb{C}^2$, so $P(A-I)P^{-1}$ has the same image dimension as $A-I$. Hence $\operatorname{rank}(B-I)=\operatorname{rank}(A-I)$, contradicting $0=1$. Thus equal eigenvalues, even with the same algebraic multiplicity, do not determine the conjugacy class of a matrix. [/example] The symmetric and matrix examples have the same pattern: conjugacy preserves the structure that survives an internal change of coordinates. In $S_n$ that structure is cycle type; for matrices it is encoded by rational or [Jordan normal form](/page/Jordan%20Normal%20Form). ## Normal Subgroups and Unions of Classes ### Stability under Internal Relabelling Conjugacy classes also explain why normal subgroups are the right subgroups for quotient groups. A subgroup is normal exactly when it is stable under every internal relabelling by elements of the ambient group. To state this stability as a reusable object, it is helpful to name subsets that are closed under conjugation. Such subsets are not necessarily subgroups, but they are the subsets visible from the viewpoint of conjugacy alone. [definition: Conjugation-Invariant Subset] Let $G$ be a group. A subset $A \subset G$ is conjugation-invariant if for every $a \in A$ and every $g \in G$, \begin{align*} gag^{-1} \in A. \end{align*} [/definition] A conjugation-invariant subset is built from whole conjugacy classes. To use that principle without checking each element separately, we need the precise equivalence between conjugation-invariance and being a union of classes. The characterisation below turns a stability condition into a set-theoretic description. It is the basic tool for passing between individual conjugates and whole subsets, and it gives the language needed for normal subgroups. [quotetheorem:5007] This theorem is often used without being named. It is the reason normal subgroups can be studied by listing conjugacy classes and checking which unions are closed under multiplication. [quotetheorem:802] The identity class must lie in any subgroup, and closure under products and inverses adds further restrictions. For small finite groups, this gives a practical method for finding all normal subgroups. [example: Normal Subgroups of $S_3$ from Classes] The conjugacy classes of $S_3$ are \begin{align*} \{e\}, \quad \{(1\,2), (1\,3), (2\,3)\}, \quad \{(1\,2\,3), (1\,3\,2)\}. \end{align*} By *Normal Subgroups Are Unions of Conjugacy Classes*, a [normal subgroup](/page/Normal%20Subgroup) of $S_3$ must be a union of these classes, and because every subgroup contains the identity, it must contain the class $\{e\}$. Thus the only possible unions are \begin{align*} \{e\}, \quad \{e,(1\,2\,3),(1\,3\,2)\}, \quad \{e,(1\,2),(1\,3),(2\,3)\}, \quad S_3. \end{align*} Their sizes are \begin{align*} 1, \quad 1+2=3, \quad 1+3=4, \quad 1+3+2=6. \end{align*} The union of size $4$ cannot be a subgroup, since by *[Lagrange's Theorem](/theorems/841)* the order of any subgroup of $S_3$ must divide \begin{align*} |S_3| = 3! = 6, \end{align*} and $4 \nmid 6$. The size $3$ union is \begin{align*} \{e,(1\,2\,3),(1\,3\,2)\}. \end{align*} Writing $r=(1\,2\,3)$, we have \begin{align*} r^2 &= (1\,3\,2), \\ r^3 &= e, \end{align*} so this union is \begin{align*} \{e,r,r^2\}=\langle r\rangle=A_3. \end{align*} It is a union of conjugacy classes, so it is normal in $S_3$ by *Normal Subgroups Are Unions of Conjugacy Classes*. The remaining two valid unions, $\{e\}$ and $S_3$, are normal in every group. Hence the normal subgroups of $S_3$ are \begin{align*} \{e\}, \quad A_3, \quad S_3. \end{align*} [/example] ### Conjugating Subgroups Conjugacy is not only an operation on elements. Many constructions in group theory depend on a subgroup only up to its position inside the ambient group, and moving every element of a subgroup by the same internal relabelling gives the corresponding operation on subgroups. [definition: Conjugate Subgroups] Let $G$ be a group and let $H,K \le G$. The subgroups $H$ and $K$ are conjugate in $G$ if there exists $g \in G$ such that \begin{align*} K = gHg^{-1}. \end{align*} [/definition] Normal subgroups are exactly the subgroups that do not move under this operation. For an arbitrary subgroup, the more local question is which elements of $G$ preserve that particular subgroup. This is the subgroup analogue of the centraliser. [definition: Normaliser] Let $G$ be a group and let $H \le G$. The normaliser of $H$ in $G$ is \begin{align*} N_G(H) = \{g \in G : gHg^{-1} = H\}. \end{align*} [/definition] The normaliser is the stabiliser of $H$ for the conjugation action on subgroups. Thus $H \trianglelefteq G$ exactly when $N_G(H) = G$, while $C_G(g)$ is the element-level stabiliser for the same kind of action. Conjugacy also explains why quotient groups remember normal subgroups but not arbitrary subgroups. A quotient identifies elements in cosets of $N$, and this is compatible with multiplication only when $N$ is stable under conjugation by all of $G$. ## Class Functions and Characters Some functions on a group are meant to measure internal type rather than a chosen representative. Such functions must be constant on conjugacy classes. This condition is the starting point for character theory. The definition below names these functions. It is useful even before representation theory enters the story, because it captures exactly which functions are blind to internal relabelling. [definition: Class Function] Let $G$ be a group and let $S$ be a set. A function $f: G \to S$ is a class function if for every $g, h \in G$, \begin{align*} f(hgh^{-1}) = f(g). \end{align*} [/definition] A class function is constant on each conjugacy class. For finite groups, giving a class function is the same data as assigning one value to each conjugacy class. [example: Trace as a Class Function] Let $V$ be a finite-dimensional [vector space](/page/Vector%20Space) over $\mathbb{C}$, let $\rho: G \to GL(V)$ be a [group homomorphism](/page/Group%20Homomorphism), and define \begin{align*} \chi_\rho: G &\to \mathbb{C} \\ g &\mapsto \operatorname{tr}(\rho(g)). \end{align*} We show that $\chi_\rho$ takes the same value on conjugate elements. For $g,h \in G$, the homomorphism property gives \begin{align*} \rho(hgh^{-1}) &= \rho(h)\rho(g)\rho(h^{-1}) \\ &= \rho(h)\rho(g)\rho(h)^{-1}. \end{align*} Write $A=\rho(h)$ and $B=\rho(g)$. Since $A$ is invertible, choose any basis of $V$ and compute trace in the corresponding matrices. If $A=(a_{ij})$, $A^{-1}=(b_{ij})$, and $B=(c_{ij})$, then \begin{align*} \operatorname{tr}(ABA^{-1}) &= \sum_i (ABA^{-1})_{ii} \\ &= \sum_i \sum_j \sum_k a_{ij}c_{jk}b_{ki} \\ &= \sum_j \sum_k c_{jk}\sum_i b_{ki}a_{ij} \\ &= \sum_j \sum_k c_{jk}(A^{-1}A)_{kj} \\ &= \sum_j \sum_k c_{jk}\delta_{kj} \\ &= \sum_j c_{jj} \\ &= \operatorname{tr}(B). \end{align*} Therefore \begin{align*} \chi_\rho(hgh^{-1}) &= \operatorname{tr}(\rho(hgh^{-1})) \\ &= \operatorname{tr}(\rho(h)\rho(g)\rho(h)^{-1}) \\ &= \operatorname{tr}(\rho(g)) \\ &= \chi_\rho(g). \end{align*} Thus the trace of a representation is unchanged by conjugating the group element, so $\chi_\rho$ is constant on conjugacy classes. [/example] The example suggests the fundamental representation-theoretic source of class functions. Whenever a group acts linearly, the trace of the acting matrix depends only on the conjugacy class of the group element. The point that needs a theorem is not the single computation itself, but its representation-theoretic stability. A representation sends conjugate group elements to conjugate linear operators, and trace is insensitive to conjugating a linear operator. This is the mechanism that makes characters depend on conjugacy classes rather than on chosen representatives. [quotetheorem:5008] Characters turn conjugacy from a classification of elements into a coordinate system for representation theory. Once functions are constant on conjugacy classes, the whole space of such functions should have one independent coordinate for each class. For a finite group, this coordinate count is exact. The theorem below gives the linear-algebraic reason that conjugacy classes index the columns of character tables and the coordinates of arbitrary class functions. [quotetheorem:5009] This theorem is formal but important. It explains why conjugacy classes are the indexing set for many tables, sums, and orthogonality relations in finite representation theory. ## Beyond and Connected Topics Conjugacy classes sit near the centre of finite group theory. The [Cambridge IA Groups](/page/Cambridge%20IA%20Groups) notes are the natural place to connect this chapter with group actions, orbit-stabiliser, normal subgroups, quotient groups, and Sylow theory. The [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) notes develop the algebraic structures that use conjugacy as a background tool. Normality, modules over group rings, and homomorphism kernels all interact with the principle that internal relabelling should preserve structure. The [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology) notes give a topological setting where conjugacy appears in fundamental groups. Changing basepoints in a path-[connected space](/page/Connected%20Space) identifies fundamental groups up to isomorphism, and loops related by suitable changes of basepoint produce conjugacy phenomena. The [Cambridge II Galois Theory](/page/Cambridge%20II%20Galois%20Theory) notes connect conjugacy with field symmetries. Galois groups act on roots, and conjugacy classes in these groups control splitting behaviour, cycle types of Frobenius elements, and arithmetic information about polynomials. The same idea applies to subgroups. Instead of moving a single element to $ghg^{-1}$, one moves a subgroup $H \le G$ to $gHg^{-1}$; normal subgroups are the fixed points of this operation, while normalisers are the stabilisers. Inner automorphisms package all these relabellings as homomorphisms $G \to \operatorname{Aut}(G)$. In representation theory, conjugacy classes become the columns of character tables. Irreducible characters form a basis for class functions, so the number of conjugacy classes matches the number of irreducible complex representations of a finite group. In Lie theory, conjugacy classes are geometric submanifolds in Lie groups. Their tangent spaces, centralisers, and adjoint orbits connect group theory with differential geometry and symplectic geometry. ## References Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge II Galois Theory](/page/Cambridge%20II%20Galois%20Theory). Dummit and Foote, *Abstract Algebra* (2004). Artin, *Algebra* (1991). Serre, *Linear Representations of Finite Groups* (1977).