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For strictly decreasing integer tuples $\kappa ,\eta$, let $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ on the diagonal maximal torus of the unitary group. Expanding both determinants, Fourier orthogonality makes the integral zero unless the exponent tuples have the same permutation orbit. Strict decreasing order makes that equivalent to $\kappa =\eta$; the $n!$ matching permutation terms then each contribute one. An alternating Laurent polynomial is a finite linear combination of these alternants. For an irreducible character ${\chi }_{\lambda }$, its highest monomial gives coefficient one for $A_{\lambda +\delta }$ in ${\chi }_{\lambda }{A}_{\delta }$, with $\delta =(n-1,\dots ,0)$. The Weyl integration formula and Schur orthogonality relations force the sum of squared coefficients to be one, so all other coefficients vanish. This proves the unitary Weyl character formula directly.

A finite group is ambivalent when every element is conjugate to its inverse. Since every complex character satisfies $\chi (g^{-1})=\overline{\chi (g)}$, an ambivalent group has real values for every irreducible character. Conversely, if every irreducible value is real, irreducible characters separate conjugacy classes and force every $g$ to be conjugate to $g^{-1}$.

Assume that every nontrivial one-dimensional Artin L-function over the intermediate fields is holomorphic and nonzero at one, while the trivial one is the base Dedekind zeta function with a simple pole. By Brauer induction, $\chi ={\sum }_j{n}_j{\mathrm{Ind}}_{H_j}^G{\psi }_j$ with integers $n_j$ and linear characters ${\psi }_j$. Compatibility of Artin L-functions with induced representations expresses $L_K(\chi ,s)$ as the corresponding product of $L_{L^{H_j}}({\psi }_j,s{)}^{n_j}$. Its order at one is $-{\sum }_j{n}_j⟨{\psi }_j,1_{H_j}⟩$. Frobenius reciprocity turns this sum into $-⟨\chi ,1_G⟩$. Hence every nontrivial irreducible character gives order zero and a holomorphic nonzero function near one. This local conclusion does not assert global Artin holomorphy.

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.

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 $\chi$.

If $\chi$ is an irreducible character of a finite group $G$ and $H\le G$, thenEquality holds exactly when $\chi$ vanishes on $G\setminus H$.

Normalized Haar measure makes this averaging a norm-one projection onto continuous class functions. Averaging a trace coefficient replaces its coefficient endomorphism by $\int \theta (h{)}^{-1}\alpha \theta (h)dh$. On an irreducible summand this average is $(\mathrm{tr}\alpha /\dim V)I$ by Schur lemma. It follows that averaging finite-dimensional matrix coefficients yields finite linear combinations of irreducible characters. Together with the Peter-Weyl theorem, this proves uniform density of those characters among continuous class functions.

For any partition of an integer $\lambda ⊢n$, the number $f^{\lambda }$ of standard Young tableaux of shape $\lambda$ isThe numerator is a factorial, and the denominator is the hook product of a partition, using every cell's hook length once. The empty shape has one empty standard Young tableau, and the formula still holds with $0!=1$ and an empty product equal to one. By the standard polytabloid basis, $f^{\lambda }$ is also the dimension of the Specht module $S^{\lambda }$ over the complex numbers, or equivalently the degree ${\chi }^{\lambda }(1)$ of its irreducible character. For a teaching derivation of this dimension formula, see Etingof et al., Introduction to Representation Theory, §5.17, Theorem 5.17.1.

In the Young diagram of $(3,2)$, the hook length array $[[4,3,1],[2,1]]$ givesThis agrees with the five explicit fillings in standard Young tableau. Its conjugate partition $(2,2,1)$ has the same hook lengths after transposition, hence also gives dimension five. At the extremes, for $n\ge 1$, a single row $(n)$ and a single column $(1^n)$ each have hook product $n!$ and exactly one standard Young tableau; their Specht modules over the complex numbers are the trivial representation and sign representation, respectively.

Why the formula is beautiful: global compatibility of all the row and column inequalities reduces to a product of local hook lengths. It is useful when enumeration is already expensive: computing one hook length per cell replaces testing up to $n!$ numbered Young tableaux. For $(3,2)$, the proportion of the $120$ bijective fillings that are standard Young tableaux is $5/120=1/24$, the reciprocal of the hook product. This product does not justify treating overlapping hook conditions as independent; Grinberg explains that tempting but invalid argument in §1.3 of his teaching slides.

Historically, the modern hook-product statement was published by J. S. Frame, G. de B. Robinson and R. M. Thrall in 1954: ^[ref]. Their manuscript was received on 1 September 1953, and their proof reorganizes an earlier irreducible character degree formula into the product of the hook lengths. The theorem unifies the counting and representation viewpoints developed earlier by Young. For proof routes within this wiki, see trace computation of Specht module dimension, determinant formula for Specht module dimension, and Greene–Nijenhuis–Wilf hook walk.

For $N◃G$ and $\theta \in \mathrm{Irr}(N)$, set ${\theta }^g(n)=\theta (g^{-1}ng)$. Its inertia group is its stabilizer under this conjugation action and contains $N$. Write $\mathrm{Irr}(G\mid \theta )$ for the irreducible characters of $G$ whose restriction to $N$ contains $\theta$.

For a finite group $G$ and prime number $p$, the notationdenotes the irreducible characters whose irreducible character degree is coprime to $p$.

The irreducible characters of a finite group form a basis of its complex class functions. Equality of every irreducible character value therefore forces equality of every class-function value, including indicators of conjugacy classes, and hence conjugacy. The converse follows from the class-function property.

A monomial character is a character induced from a linear character of a subgroup. When discussing monomial finite groups, this property is required for every irreducible character.

A finite group is monomial if every irreducible character is a monomial character. Finite metabelian groups have this property.

For a real-valued irreducible character of a finite odd-order group, pairing $g$ with $g^{-1}$ gives $⟨\chi ,1_G⟩=(\chi (1)+2A)/|G|$, with $A$ an algebraic integer. If this inner product were zero, $\chi (1)$ would be even, since a rational algebraic integer is an integer. But an irreducible character degree divides the group order, so it is odd. Therefore the character is trivial.

Let $\mathrm{Irr}(G)$ denote the irreducible characters of finite-dimensional complex group representations of $G$. A generalized character, or virtual character, is an integral linear combination of these irreducible characters. A class function is a function constant on conjugacy classes. For class functions use the character inner productA p-elementary group is a direct product of groups $P\times C$, where $P$ is a p-group and $C$ is a cyclic group of order coprime to $p$. An elementary subgroup is a subgroup of this form for some prime $p$. Either factor may be trivial. This notion is different from an elementary abelian group.

Brauer's characterization of characters says that a complex class function $f$ on $G$ is a generalized character exactly when its restriction to every elementary subgroup is a generalized character there. Equivalently,The character orthogonality relations make these inner products the coefficients of the restrictions in their irreducible bases. The theorem detects the integral character lattice. An ordinary character of a representation additionally requires every global irreducible coefficient $⟨f,\chi {⟩}_G$ to be nonnegative; integrality alone characterizes virtual characters, not necessarily actual representations.

A Frobenius group is a finite transitive group action on more than one point in which a nonidentity element fixes at most one point and a point stabilizer is nontrivial. Equivalently it has a nontrivial proper subgroup $H$ with $H\cap H^g=1$ for every $g\notin H$. The Frobenius complement is such a point stabilizer $H$.

The Frobenius kernel is the identity together with all fixed-point-free elements in this action, equivalentlyThe Frobenius kernel theorem asserts that this set is a normal subgroup, acts regularly on the points, and gives $G=N⋊H$. In particular $C_N(h)=1$ for $1\ne h\in H$.

Brauer's permutation lemma states that an automorphism of a finite group fixes equally many conjugacy classes and irreducible characters. More generally a group of automorphisms has equal permutation characters on those two sets. Indeed, the character table is invertible and intertwines the two permutation matrices: simultaneous permutation of rows and columns preserves character evaluation. Similar matrices have equal traces, and a permutation-matrix trace counts its fixed points. The fixed objects on the group side are classes, not individual elements.

Suppose first that $N$ is the Frobenius kernel. An element $g\notin N$ fixes a point in the Frobenius action. If $n\in C_N(g)$, then $n$ maps that point to another point fixed by $g$. Since $g$ has only one fixed point and $N$ acts regularly, $n=1$. ThusNo nonidentity conjugacy class of $N$ can be fixed by such a $g$. If the class of $n\ne 1$ were fixed, $gn{g}^{-1}=vn{v}^{-1}$ for some $v\in N$, so $v^{-1}g\notin N$ would centralize $n$, a contradiction. Conjugation by $g$ consequently fixes just the identity class. By Brauer's permutation lemma it fixes just one irreducible character, necessarily $1_N$.

For every nonprincipal $\phi \in \mathrm{Irr}(N)$, its inertia group of a character is therefore $I_G(\phi )=N$. Since $N$ is normal, the restriction of the induced character is the sum of its conjugates, counted over $G/N$:Frobenius reciprocity and character orthogonality giveAn actual character of norm one is irreducible. Every nonprincipal irreducible character of $N$ therefore induces irreducibly to $G$.

A finite monomial group, also called an M-group, is a finite group for which every irreducible character is a monomial character: for every $\chi \in \mathrm{Irr}(G)$ there are a subgroup $H\le G$ and a linear character $\lambda$ of $H$ such that $\chi ={\mathrm{Ind}}_H^G\lambda$. The inducing subgroup and linear character may depend on $\chi$.

Taketa's theorem states that every finite monomial group is solvable. Here being a solvable group means that the iterated derived subgroups eventually reach the identity.

Dade's embedding theorem for monomial groups states that every finite solvable group embeds as a subgroup of a finite monomial group. It does not say that every solvable group is itself monomial, nor that the embedding can always be taken normal. Together the results identify solvability as a necessary property of monomial groups without making monomiality closed under taking arbitrary subgroups.

Write $o(\theta )$ for the order of the determinant character of an irreducible character $\theta$. An extension has the same degree and restricts exactly to $\theta$, rather than just containing it as a constituent.

Extension theorem. Thompson's coprime character extension theorem says that if $N◃G$, $\theta \in \mathrm{Irr}(N)$ is $G$-invariant, and $\gcd (\theta (1),[G:N])=1$, then $\theta$ extends to $G$ exactly when its determinant character extends to a linear character of $G$. In particular, if also $\gcd (o(\theta ),[G:N])=1$, there is a unique extension with determinant order coprime to the index. For a p-group quotient, the version used below gives a canonical extension whenever both the degree and the determinant order are prime to $p$. This theorem is an assertion of extendibility, not the false statement that every invariant irreducible character extends.

The p-nilpotency lemma. Definewhere $m_p$ is the largest power of $p$ dividing an integer $m$. Isaacs' character-degree criterion for p-nilpotency assertsEquivalently, $s_p(G{)}_p=h_p(G)$. We prove both the divisibility and the criterion.

Let $N=O^p(G)$ be the p-residual of a finite group, the least normal subgroup with p-group quotient. It can also be described as the subgroup generated by all p-prime-order elements: every homomorphism to a p-group kills those elements, and after quotienting by their generated subgroup the image of every element has p-power order, so the quotient is a p-group by Cauchy's theorem.

The residual is p-perfect, meaning $O^p(N)=N$. Indeed $O^p(N)$ is characteristic in $N$, hence normal in $G$, and $G/O^p(N)$ is an extension of the two p-groups $N/O^p(N)$ and $G/N$. Minimality of $N$ forces equality. Consequently $N/N^{\prime }$ has order prime to $p$, since otherwise it would have a quotient of order $p$. Every linear character of $N$, in particular every determinant character of $N$, therefore has order prime to $p$.

Let $\mathcal{T}$ consist of the $G$-invariant irreducible characters $\theta$ of $N$ of degree prime to $p$. Each extends to an irreducible character $\hat{\theta }$ of $G$ by the extension theorem. The irreducible characters of $G$ lying over $\theta$ are exactlyHere is the representation argument for this description. If a representation lies over the invariant $\theta$, its $N$-module is copies of the module for $\theta$. Identify it with that module tensored with its multiplicity space. Dividing the $G$-action by the fixed extending action on the first factor leaves an action on the multiplicity space, trivial on $N$. Irreducibility is equivalent to irreducibility of that quotient action, proving the displayed parametrization.

Every irreducible degree of the p-group $G/N$ is a power of $p$, since it divides the group order. Thus the p-prime-degree characters lying over $\theta$ are exactly those with $\nu$ linear, and there areof them. For the last equality, $G/N{G}^{\prime }$ is the largest abelian p-group quotient of $G$, hence the p-primary part of its abelianization. In particular all these characters have degree $\theta (1)$.

Conversely, for any $\chi \in \mathrm{Irr}(G)$ of p-prime degree, Clifford theory writes its restriction to $N$ as $e$ times the sum of a single orbit of irreducible characters. The orbit size divides $[G:N]$ and also divides $\chi (1)$, so it is one; the constituent $\theta$ is invariant and has p-prime degree. Hence $\theta \in \mathcal{T}$. The preceding extension parametrization now forces the quotient character to be linear. We have therefore obtained the exact identityThis proves the required divisibility without assuming p-nilpotency.

The p-group $G/N$ acts on $\mathrm{Irr}(N)$ by conjugation. Every nonfixed orbit has size divisible by $p$, and degrees are constant on each orbit. Fixed characters whose degrees are divisible by $p$ also contribute zero modulo $p$. The sum of squares of irreducible degrees consequently givesThe quotient is prime to $p$ exactly when $N$ has order prime to $p$. Since $G/N$ is a p-group, that is exactly the assertion that $N$ is a normal p-complement. Conversely, any normal p-complement equals $O^p(G)$ because it contains all p-prime-order elements and has p-group quotient. This finishes the lemma.

Detection from the group algebra. For finite groups $G$ and $H$ with isomorphic complex group algebras, the Artin–Wedderburn theorem givesThe simple matrix-block sizes and their multiplicities are determined by the algebra isomorphism. Hence $G$ and $H$ have identical irreducible degree multisets. They have the same order, from the degree-square sum, and the same number of linear characters, which equals $|G:G^{\prime }|$. Thus $s_p$ and $h_p$ coincide for every prime $p$. The lemma shows that one group is p-nilpotent exactly when the other is.

A finite group is nilpotent exactly when it has a normal p-complement for every prime dividing its order. For completeness, if all those complements exist, set $L_q={\bigcap }_{p\ne q}{K}_p$, where $K_p$ is the normal p-complement. The diagonal map $G/L_q\to {\prod }_{p\ne q}G/K_p$ is injective, so $[G:L_q]$ is prime to $q$. At the same time, containment in each $K_p$ removes every prime other than $q$ from $|L_q|$. Thus $L_q$ is a normal Sylow q-subgroup. Distinct normal Sylow subgroups commute, giving the direct-product decomposition. Conversely, the direct-product decomposition of a finite nilpotent group provides all the complements. We concludeThis is the assertion that the complex group algebra detects p-nilpotency and hence nilpotency. It needs the multiplicities of the degrees, not just their distinct values.

Normal-complement consequence. If $p$ divides every nonlinear irreducible degree, the p-prime-degree characters are exactly the linear characters. ThereforeThe lemma yields a normal p-complement immediately:This proves Thompson's character-degree normal complement theorem, including the abelian case where the hypothesis on nonlinear characters is vacuous.

Frobenius reciprocity is the adjointness of character induction and restriction: for class functions $f$ on $H$ and $u$ on $G$,For characters this says that the multiplicity of an irreducible $G$-character in an induced character equals the corresponding multiplicity after restriction. To prove the identity directly, insert the induced class function formula and interchange the two finite sums. For each $x$, substitute $g=xh{x}^{-1}$ with $h\in H$. Since $u$ is constant on conjugacy classes,This proves the theorem for all class functions, not just irreducible characters.

For the group-theoretic conclusion, suppose that $p$ divides $|G^{\prime }\cap Z(G)|$. A central element of order $p$ lies in every Sylow p-subgroup: adjoining it to a Sylow subgroup gives a p-subgroup, so maximality forces it into that subgroup. Hence there is a subgroup $Q\le P\cap G^{\prime }\cap Z(G)$ of order $p$. Choose a nontrivial linear character $\lambda$ of $Q$.

Since $P$ is abelian, $\lambda$ extends to a linear character $\mu$ of $P$. One direct justification is to extend from an abelian subgroup $A$ to $⟨A,x⟩$ one generator at a time: if $d$ is the least positive exponent with $x^d\in A$, choose a complex $d$th root of $\lambda (x^d)$ as the value at $x$. The formula on $a{x}^j$ is then well defined and multiplicative. Iteration yields the claimed extension to $P$.

The induced character ${\mu }^G$ has degree $[G:P]$, which is prime to $p$. In its decomposition into irreducible characters, at least one constituent $\chi$ must have degree prime to $p$; otherwise the total degree would be divisible by $p$. Frobenius reciprocity also gives $⟨{\chi }_P,\mu {⟩}_P>0$ for this constituent. Restriction to the central subgroup $Q$ forcesIndeed, the induced representation itself restricts to $Q$ as copies of $\lambda$, because conjugation by every coset representative is trivial on $Q$; alternatively this follows from Schur's lemma and the positive multiplicity after restriction.

Each $q\in Q$ consequently acts in the representation affording $\chi$ as the scalar $\lambda (q)$. Taking its determinant character givesThe left side is trivial since $Q\le G^{\prime }$ and every linear character kills $G^{\prime }$. The right side is nontrivial since $\lambda$ has order $p$ and $p\nmid \chi (1)$. This contradiction provesThe argument is the central commutator has no torsion from an abelian Sylow subgroup determinant obstruction.