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For a Finite Galois extension of $K(T)$ with algebraically closed characteristic-zero constants, the local decomposition and inertia groups coincide and are cyclic. A fixed compatible choice of primitive roots of unity selects a generator through its uniformizer character of inertia. Changing the prime above a base point conjugates this generator, and changing the uniformizer does not change its character. Without the roots-of-unity convention, only the inertia-subgroup conjugacy class is intrinsic. Over $\mathbb{C}$, positive local orientation gives the standard convention.

A group of automorphisms of a finite group permutes both its conjugacy classes and its irreducible characters. The two permutation representations have the same character: each automorphism fixes equally many conjugacy classes and irreducible characters. The invertible character table intertwines the two permutation matrices, which have equal traces. The claim concerns fixed classes, not merely fixed elements.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

An element of a group is central when it commutes with every element of that group. The set of all such elements is the center of a group. Equivalently, its conjugacy class is a singleton: $xg{x}^{-1}=g$ for every $x$ is exactly the commutation condition. This lets centrality identify one-element conjugacy classes without enumerating every conjugation.

The centralizer of an element $x$ of a group $G$ consists of all elements that commute with $x$:It is the stabilizer subgroup of $x$ under the conjugation action, hence a subgroup of $G$. In conjugacy classes of the symmetric group on three letters, the three representative computations areFor the transposition $(12)$, a commuting permutation must preserve the exchanged pair $\{1,2\}$ and fix $3$; only $e$ and $(12)$ do so. For $r=(123)$, a commuting permutation is determined by where it sends $1$: the equation $g(r(i))=r(g(i))$ forces the images of $2$ and $3$, leaving exactly $e,r,r^2$. For $e$, every permutation commutes. These centralizer sizes $6,2,3$ give conjugacy class sizes $1,3,2$.

The centralizer is useful because it measures how much freedom remains after fixing one operation. Its index of a subgroup counts the distinct results of conjugation, and the intersection of all element centralizers is the center of a group. A centralizer need not be an abelian group: $C_G(e)=G$ even when $G$ is a nonabelian group. In an abelian group, all element centralizers equal $G$. These observations also explain why the surrounding group matters: the centralizer of $(123)$ inside the alternating group $A_3$ is the whole $A_3$, whereas its centralizer inside $S_3$ is a proper subgroup. See McKernan, the centralizer definition preceding lemma 23.6.

For an arbitrary subset $A\subseteq G$, the centralizer of a subset fixes every element of $A$ individually under the conjugation action:This is a pointwise stabilizer of the elements of $A$, not of the vertices of a polygon in a geometric realization. In particular, $r=(123)$ belongs to $C_{S_3}(r)$ although its rotation moves every vertex of a polygon of the equilateral triangle. If $A$ generates a subgroup $H$, commuting with every element of $A$ implies commuting with every product of those elements and their inverse elements; hence $C_G(A)=C_G(H)$. For the triangle rotation subgroup $H=⟨r⟩$, this gives $C_{S_3}(H)=C_{S_3}(r)=H$.

To see the commutation test directly on the equilateral triangle, compare a rotation $r$ with the reflection $s=(12)$. Their permutations satisfyThe two orders send vertex of a polygon $1$ to different places: $rs(1)=3$, but $sr(1)=1$. Thus $s\notin C_{S_3}(r)$. By contrast $r$ and $r^2$ commute, with $r{r}^2=r^{2}r=e$. The reflection reverses the chosen rotation while preserving the subgroup of all triangle rotations, so $s\in N_{S_3}(H)$. The distinction is visible before any counting: commuting preserves an operation itself; membership in the normalizer permits exchanging it with another operation in the same subgroup.

A group of order thirty-six with a normal $C_3^2$ and a cyclic quotient of order four acting freely on its eight nonzero elements has six conjugacy classes of sizes $1,4,4,9,9,9$. Its four linear characters are lifted from the cyclic quotient. The remaining irreducible degrees are four and four, by the sum of squared degrees. Column orthogonality forces the last three values of both degree-four rows to vanish. The square of the order-four action is inversion on $C_3^2$, making the first two nontrivial class values real. Orthogonality then gives sum minus one and product minus two, so the rows are $(4,1,-2,0,0,0)$ and $(4,-2,1,0,0,0)$. This illustrates why reality must be justified when reconstructing a character table from orthogonality.

In the conjugacy class order $1,(123),(12)(34),(12345),(13452)$, the five complex irreducible characters of the alternating group $A_5$ areThe class sizes are $1,20,15,12,12$. Restricting the degree-four and degree-five irreducible characters of $S_5$ gives rows of squared character inner product one. The degree-six restriction has squared norm two and is orthogonal to those rows and the trivial row, so it splits into the two remaining irreducibles. The sum of squared degrees forces their degrees to be three. Orthogonality forces their last two values to have sum one and product minus one, producing the displayed roots. Since a normal subgroup is a union of classes and its order divides sixty, these class sizes also prove the Simplicity of the alternating group A5.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

For the nontrivial action of $C_3$ on $C_7$, the conjugacy class sizes are $1,7,7,3,3$. Three linear characters factor through the quotient $C_3$, and the remaining two irreducible characters have degree three. Put $\omega =e^{2\pi i/3}$ and let $\alpha ,\beta =(-1\pm i\sqrt{7})/2$. With suitable class labels the table isOne can derive the last two rows by inducing nontrivial characters of $C_7$ along the two orbits of the order-three automorphism on their exponents. Their degree-three orbit sums vanish off $C_7$. On a generator they give the two sums of seventh roots over quadratic residues and nonresidues. Their sum is $-1$ and their product is $2$: the nine pairwise exponent sums between the two triples contain zero three times and every nonzero residue once. Thus they are the roots of $X^2+X+2$. Alternatively column orthogonality forces the zeros on the centralizers of order three, and orthogonality with the identity column interchanges $\alpha$ and $\beta$ in the last row. This finite table illustrates how centralizer sizes and a small number of given values determine missing character values.

The quaternion group has conjugacy classes $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$, with sizes $1,1,2,2,2$. Over the complex numbers, the number of classes and the sum of squares of irreducible representation dimensions force four one-dimensional representations and one two-dimensional representation. For $ϵ,\eta \in \{\pm 1\}$, the linear character rows on these classes are $(1,1,ϵ,\eta ,ϵ\eta )$. Orthogonality then gives the remaining row $(2,-2,0,0,0)$. A representation with character $(4,-4,0,0,0)$ contains two copies of the two-dimensional irrep and none of the linear irreps, by the character inner product. The choice of scalar field matters: the corresponding quaternionic four-dimensional real representation is irreducible over the reals.

For a finite Galois extension $L/K$ and a conjugacy class $C$ in $G=\mathrm{Gal}(L/K)$, the unramified primes with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|G|$. In particular,One proof uses Brauer induction, the Artin L-function order at one, and character orthogonality. The trivial Artin L-function is the Dedekind zeta function of $K$. The full theorem also gives the natural density, but this logarithmic estimate alone establishes the Dirichlet density assertion.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

A class function on a group is a function constant on every conjugacy class. Equivalently, $f(hg{h}^{-1})=f(g)$ for all group elements $g,h$.

The cocenter of a finite-dimensional algebra is the quotient by the vector subspace spanned by additive commutators $ab-ba$; this subspace need not be an ideal. For a group algebra, its basis is indexed by conjugacy classes: every commutator of basis elements is a difference of conjugates, and every difference of conjugates is a commutator. For $M_n(k)$, off-diagonal matrix units and $E_{ii}-E_{jj}$ show that the quotient is one-dimensional, identified by the trace.

For a finite group, the sum of products of irreducible character values is zero when $g,h$ are not conjugate, and equals $|C_G(g)|$ when $g,h$ are conjugate. To prove this from row orthogonality relations for a character table, choose representatives $g_j$ of its conjugacy classes and form $U_{ij}=\sqrt{|g_j^G|/|G|}{\chi }_i(g_j)$. Row orthogonality says $U{U}^{\ast }=I$. There are equally many irreducible characters and conjugacy classes, so this is a square unitary matrix and $U^{\ast }U=I$ as well. Its column equations give the result, using $|G|/|g_j^G|=|C_G(g_j)|$.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

For a finite group $G$, write $k(G)$ for its number of conjugacy classes. Exactly the elements of the center of a group $Z(G)$ have singleton conjugacy classes; every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, $k(G)-|Z(G)|\le (|G|-|Z(G)|)/2$. If $G$ is a nonabelian group, the cyclic quotient by the center criterion and the fact that every group of group order at most three is a cyclic group give $[G:Z(G)]\ge 4$, hence $k(G)\le 5|G|/8$. The quaternion group attains this bound, with five conjugacy classes among eight elements.

Use $D_n$ for the group of order $2n$ with rotation $R$ and reflection $m$. Every element is $R^j$ or $R^{j}m$, $0\le j<n$. The rotational conjugacy classes are the sets $\{R^j,R^{-j}\}$, reduced to singletons for central rotations. For odd $n$, all reflections form one class, giving $(n+3)/2$ classes in total. For even $n$, the identity and $R^{n/2}$ are central singletons, the other rotations form pairs, and the reflections split into even and odd powers of $R$, giving $n/2+3$ classes. Indeed conjugation by $R^s$ changes a reflection exponent by $2s$, while conjugation by $m$ negates it. For odd prime $n$, a normal subgroup containing a reflection must contain all reflections and hence the whole group; the proper normal subgroups are therefore only the identity subgroup and the rotation subgroup.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.