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For an affine variety with coordinate ring $A$ acted on by a finite group $G$, the affine quotient is $\mathrm{Spec}{A}^G$. Elements of $A$ satisfy monic orbit polynomials over $A^G$, so the quotient map is a finite morphism. Over an algebraically closed field, invariant functions separate finite orbits; in characteristic not dividing $|G|$, interpolate and average a function separating two orbits. The quotient fibres are consequently the orbits.

Cayley theorem embeds a finite group of order $n$ into $S_n$. Extend $\sigma \in S_n$ to $n+2$ letters and multiply it by the disjoint transposition $(n+1n+2)$ exactly when $\sigma$ is odd. Additivity of parity makes this an injective group homomorphism into the alternating group $A_{n+2}$. Thus every finite group is isomorphic to a subgroup of some alternating group.

Every complex character of a representation of a finite group $G$ is a rational linear combination of characters of induced representations from one-dimensional group representations of cyclic subgroups $C_j$. One-dimensionality follows because the irreducible representations of a finite cyclic group over the complex numbers are one-dimensional. The rational coefficients distinguish this statement from Brauer induction, which permits integer coefficients by allowing more general subgroups. For a rational-valued character, another form of Artin induction uses rational combinations of induced representations of trivial group representations of cyclic subgroups.

Mathematische Annalen 20, 301–329. The opening finite group definition is reproduced in Oxford's exam and discussed in The abstract group concept by the MacTutor History of Mathematics Archive. The publisher page establishes the article's title, author, pages and issue date; the commentary does not assume access to its subscription facsimile.

Online source: doi.org/10.1007/BF01443599

For a finite group generated by $a,b$, take the regular covering of a thrice-punctured sphere with monodromy $a,b,(ab{)}^{-1}$ and fill each lifted puncture by a disk. Locally the projection is $z\mapsto z^m$, where $m$ is the order of its monodromy. The result is a closed oriented topological surface with an orientation-preserving group action. The nontrivial stabiliser subgroups are the conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$ at the three types of ramification points. Counting lifted cells gives the displayed Euler characteristic. A subgroup acts freely exactly when it meets every conjugate of these three cyclic groups trivially. Its quotient is then a regular covering of degree equal to the subgroup order, and quotient Euler characteristic is divided by that order.

The branching of a group representation $V$ to a subgroup $H$ is its decomposition into irreducible representations of $H$, when that restriction of a representation is a semisimple representation: $V{|}_H={⨁}_{\lambda }{m}_{\lambda }{W}_{\lambda }$. The nonnegative integers $m_{\lambda }$ are the branching multiplicities. For finite groups over the complex numbers, they are the character inner products of the restricted character with the irreducible characters of $H$. Restriction to an arbitrary subgroup need not be completely reducible.

Every complex character of a representation of a finite group $G$ is an integer linear combination of characters of induced representations from one-dimensional group representations of subgroups $H_j\le G$. Negative coefficients are allowed. Unlike Artin induction, the subgroups need not be cyclic groups. Applied to an Artin L-function, compatibility with induced representations turns this character identity into a product of one-dimensional Artin L-functions with integer exponents, reducing the behavior at $s=1$ to abelian extensions.

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.

Every finite group whose order is divisible by at most two distinct primes is a solvable group. A nonabelian simple group must therefore have at least three distinct prime divisors.

Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]

For a complex irreducible representation $D$ of a finite group $G$, the sum $A={\sum }_{g}D(g^2)$ commutes with every $D(h)$: conjugation sends its summands to $D((hg{h}^{-1}{)}^2)$, merely permuting the summation index. Schur's lemma gives $A=cI$. Taking the trace yields $c\dim D={\sum }_g\chi (g^2)=|G|\nu (D)$, proving the identity. Here $\nu (D)$ is the Frobenius-Schur indicator. For a reducible representation the sum is still central in its image, but need not be one scalar on all irreducible constituents.

For a finite group, the regular representation contains $d_{\rho }$ copies of each irreducible representation $\rho$. Its character of a representation is $|G|$ at the identity and zero elsewhere. Thus ${\sum }_{\rho }{d}_{\rho }{\chi }_{\rho }(x)=|G|1_{\{x=e\}}$, including the trivial representation. This formula converts independent uniform expectations into constrained configuration counts.

The character group of a topological group that is an abelian group consists of its continuous group homomorphisms to the circle group, with pointwise multiplication. For a finite group, all characters are continuous and their values are roots of unity, so this agrees with the character group of a finite abelian group.

For complex class functions on a finite group $G$, the character inner product isIf $\psi$ is irreducible, this inner product is its multiplicity in the representation with character $\chi$.

For a complex representation of a finite group, the eigenvalues of each represented element are roots of unity. They have absolute value one, so inversion equals complex conjugation on each eigenvalue. Taking traces gives the identity.

For a linear character $\chi$ of a finite group $K$,Indeed, if $\chi (h)\ne 1$, the bijection $k\mapsto hk$ gives ${\sum }_k\chi (k)=\chi (h){\sum }_k\chi (k)$, forcing the sum to vanish. The trivial character gives $|K|$ directly. The same argument applies to the restriction of a linear character of a larger group to a subgroup.

At an element $g$ of a finite group, a complex character value is the sum of the roots of unity which are the eigenvalues of its representing matrix. It is therefore an algebraic integer, and every algebraic conjugate has modulus at most the degree.

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 Clebsch-Gordan decomposition expresses a tensor product of group representations as a direct sum of irreducible representations, whenever complete reducibility of compact-group representations or Maschke's theorem applies. For a finite group over the complex numbers, the multiplicity of $V_{\lambda }$ is the character inner product $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V{\chi }_W⟩$. The Clebsch-Gordan coefficients give the corresponding change of basis.

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 $|G|/|g_j^G|=|C_G(g_j)|$.