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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.

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.

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.

The graded abelian group ${⨁}_{n\ge 0}R(S_n)$ of ordinary virtual characters has product $\chi \star \psi ={\mathrm{Ind}}_{S_m\times S_n}^{S_{m+n}}(\chi ⊠\psi )$ for degrees $m,n$. Its unit is the trivial degree-zero character. Associativity follows by transitivity of induced representations through the three-factor Young subgroup; commutativity follows by conjugating the block positions. The Frobenius characteristic map identifies this ring with the ring of symmetric functions, sending irreducible Specht characters to Schur functions. This product is different from pointwise multiplication of characters of the same symmetric group.

Every uniform pro-p group embeds continuously in ${\mathrm{GL}}_n({\mathbb{Z}}_p)$ for some finite $n$. Apply the Ado theorem to ${\mathbb{Q}}_p\otimes L_G$ and choose a deep power lattice whose representing matrices lie in $p\mathrm{End}(\Lambda )$, or $4\mathrm{End}(\Lambda )$ at $p=2$. The matrix exponential and logarithm give a faithful representation of the corresponding open normal power subgroup. Induce that representation to the whole group using its finitely many cosets. An element outside the open subgroup permutes the coset blocks nontrivially; an element inside it is detected by the original representation. Thus the induced representation is faithful. All block matrices preserve a finite free ${\mathbb{Z}}_p$-module, giving the integral embedding.

For a future null standard momentum $k$, choose Lorentz transformations $L(p)k=p$ and a unitary irreducible representation $d$ of its little group. The Wigner rotation $R(\Lambda ,p)=L(\Lambda p{)}^{-1}\Lambda L(p)$ acts on the internal state. The resulting induced representation acts on wavefunctions with invariant measure $d^{3}p/(2p^0)$. Trivial little-group translations give fixed helicity; nontrivial translations give a continuous-spin representation with infinitely many polarization states.

Let $K$ be the intersection under consideration. The family is nonempty since it contains $G$, and is invariant under conjugation: conjugating an irreducible inducing pair $(H,\phi )$ gives another such pair. Hence $K$ is a normal subgroup of $G$.

For any $\chi \in \mathrm{Irr}(G)$, monomiality gives $\chi ={\lambda }^G$ with $\lambda$ a linear character of some $H$. This $H$ belongs to the family, so $K\le H$. Because $K$ is normal, $K\le tH{t}^{-1}$ for every $t\in G$. In the induced representation oneach coset line $t\otimes {\mathbb{C}}_{\lambda }$ is therefore invariant under $K$: for $k\in K$,Thus the restriction to $K$ is a direct sum of one-dimensional group representations. All representing matrices of $K$ commute, and the derived subgroup $K^{\prime }$ acts as the identity. Equivalently,The intersection of the character kernels is trivial: the regular representation is faithful and is a direct sum of irreducible representations. It follows that $K^{\prime }=1$.

This proves the abelian intersection of irreducible inducing subgroups without assuming that every subgroup of an M-group is itself monomial.

Let $V$ afford the given group representation of $H$. Its induced representation is the left $G$-moduleChoose representatives $t$ of the left cosets $G/H$; then it is the direct sum of the spaces $t\otimes V$. If $gt=t^{\prime }h$ with $h\in H$, the action sends $t\otimes v$ to $t^{\prime }\otimes \Psi (h)v$. This is independent of the representatives up to equivalence, and its dimension is $[G:H]\dim V$.

Only cosets fixed by $g$ contribute to its trace. Their contribution is $\psi (t^{-1}gt)$, giving the induced character formulaFor an arbitrary class function $f$ on $H$, the same averaging formula defines its induced class function on $G$. Equivalently, extend $f$ by zero outside $H$ and average $|H{|}^{-1}{\sum }_{x}f(x^{-1}gx)$. This definition is linear and does not require $f$ to be an actual character.

For a complex character $\psi$ of a finite group $G$ afforded by a representation $\rho$, its determinant character isMultiplicativity of the determinant makes this a linear character; it does not mean the determinant of the scalar value $\psi (g)$. The definition depends only on the character because characters determine the equivalence class of a finite-dimensional complex representation. In particular the determinant character is trivial on the derived subgroup $G^{\prime }$.

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.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.

Use the metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and first consider the connected Poincaré group, ${\mathbb{R}}^{1,3}⋊S{O}^{+}(1,3)$. Physical half-integer spins require unitary representations of its double cover ${\mathbb{R}}^{1,3}⋊SL(2,\mathbb{C})$; genuine representations of the group without this cover permit only integer spin. Write $P^{\mu }=d(P^{\mu })$ and $M^{\mu \nu }=d(M^{\mu \nu })$. Strong continuity gives self-adjoint generators, with commutators understood on a common invariant dense domain.

The commuting translation generators have a joint spectral resolution. Lorentz transformations carry the spectral four-momentum $p$ to $\Lambda p$, so an irreducible representation has its momentum spectral measure concentrated on a single Lorentz orbit. For positive-energy particles the orbits of interest are the future mass shells $p^2=m^2>0$, or $p^2=0$ with $p\ne 0$ and $p^0>0$. The zero-momentum orbit and spacelike or negative-energy orbits are outside these two particle cases.

Choose ${ϵ}^{0123}=1$ and define the Pauli-Lubanski pseudovector with the overall sign conventionThis sign makes the spatial rest-frame vector equal to $m\mathbf{J}$. The generators' Hermiticity and the antisymmetric epsilon contraction make $W^{\mu }$ Hermitian: the ordering correction from commuting $P$ through $M$ contracts metric tensors into epsilon and vanishes. The same antisymmetry proves Pauli-Lubanski orthogonality, $W^{\mu }{P}_{\mu }=0$.

The assumed commutators say that $P^{\mu }$ and $W^{\mu }$ transform as four-vectors and that $W$ commutes with all translations. Consequentlycommute with the entire Poincare algebra. For instance the two terms in the Lorentz commutator with a contracted vector square cancel. The Casimir operators therefore have fixed spectral values in a unitary irreducible representation, by the Schur lemma applied to their spectral projections. A second ingredient, the little group, distinguishes the internal states.

Fix a standard momentum $k$ on the orbit. Its little group is the subgroup of Lorentz transformations leaving $k$ fixed. Choose $L(p)$ with $L(p)k=p$. A Lorentz transformation acts on the internal fibre by the Wigner rotationIf $d_K$ is a unitary irreducible representation of this stabilizer, the induced representation acts on wavefunctions asThe measure $d^{3}p/(2p^0)$ on either future mass shell is Lorentz invariant, so this action is unitary. The identityproves the group multiplication law. Conversely, covariance of the translation spectral resolution supplies precisely these fibres and their stabilizer action. A commuting operator acts fibrewise; transitivity relates its fibre values, and irreducibility of $d_K$ makes it scalar. This explains why the construction gives, and classifies, the unitary irreducible representations on each orbit.

For a timelike momentum choose $k=(m,0,0,0)$. A Lorentz transformation fixing it acts only as a rotation of its spatial orthogonal complement, so the massive particle little group is $SO(3)$, or $SU(2)$ in the double cover. PutThe Poincare algebra gives $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The unitary irreducible representations of $SU(2)$ are labelled by $j=0,\frac{1}{2},1,\dots$, with ${\mathbf{J}}^2=j(j+1)$ and $J_3$ eigenvalues $-j,-j+1,\dots ,j$. Evaluation of the Pauli-Lubanski pseudovector at $k$ gives $W^0=0$, $\mathbf{W}=m\mathbf{J}$. Thus the massive Pauli-Lubanski Casimir and particle labels areThe full massive induced representation of the Poincare double cover is infinite-dimensional because momentum ranges continuously over the orbit; its finite internal multiplicity is $2j+1$.

For a null momentum choose $k=(E,0,0,E)$ with $E>0$. The infinitesimal stabilizer generators areTheir commutators with momentum, evaluated at $k$, vanish. From the rotation and boost commutators of the Poincare algebra one obtainsThese are the massless little-group generators of $ISO(2)$: a planar rotation and two commuting null rotations, which act like translations in the little-group algebra. Hence the massless particle little group is the Euclidean plane group, with the appropriate double cover for half-integer spin.

The definition of $W$ gives $W^0=\mathbf{P}\cdot \mathbf{J}$ and $\mathbf{W}=P^0\mathbf{J}-\mathbf{P}\times \mathbf{K}$. At the standard null momentum,Suppose the little-group representation has finite dimension, as for particles with finitely many polarizations. The commuting Hermitian $N_1,N_2$ have a finite joint spectrum. Conjugation by $e^{-i\varphi R}$ rotates this spectrum through every planar angle. The only finite set invariant under all these rotations is the origin. Thus both $N_i$ vanish, proving that finite-dimensional unitary representations of the massless little group have trivial translations. Irreducibility then leaves a one-dimensional rotation character, $R=hI$. The massless longitudinal Pauli-Lubanski eigenvalues arefor the double cover, with $h\in \mathbb{Z}$ for a genuine $SO(2)$ character. The number $h$ is helicity. Different helicities have the same two Casimir eigenvalues, so the little group is indispensable for the classification. A connected-group irreducible representation has one helicity; if parity is also implemented, it exchanges $h$ and $-h$ and normally requires their pair.

Unitarity alone also permits nontrivial null-rotation generators. The rotation-invariant operator $N_1^2+N_2^2$ is scalar in an irreducible little-group representation; if it has value $(\rho /E{)}^2>0$, one obtains a continuous-spin representation. A concrete realization on angular wavefunctions isThese operators satisfy the same commutators. Periodic or antiperiodic angular wavefunctions give integer or half-integer rotation modes, and the translation generators connect infinitely many such modes. The invariant labels areThese are the additional infinite-polarization massless unitary irreducible representations. Restricting to finitely many particle polarizations excludes them; they must not be ruled out merely by assuming unitarity. This completes the timelike and null cases of the Wigner classification.

Let $D,E$ be module vertices of $M$, and choose a module $T$ such that $M\mid {\mathrm{Ind}}_E^{G}T$. Relative $D$-projectivity then givesApply Mackey decomposition to the inner restriction and use induction transitivity. The resulting module is a direct sum of modules induced to $G$ from subgroups $D\cap {}^{g}E$, for representatives $g\in D\backslash G/E$. By Krull–Schmidt theorem, the indecomposable module $M$ is a summand of one of these induced representations. It is therefore relatively projective for one intersection contained in $D$. Minimality of $D$ forces $D=D\cap {}^{g}E$, so $D\le {}^{g}E$.

Interchanging $D,E$ gives $E\le {}^{h}D$ for some $h$. The two inclusions force equal orders, and henceThus all vertices of an indecomposable module are conjugate in $G$.

Let $B=kGe$ have defect group $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every indecomposable module in $B$ has a module vertex conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=ga{h}^{-1}$, is an indecomposable permutation-module summand with module vertex $\Delta D=\{(d,d):d\in D\}$ and trivial module source. The second fact is the bimodule characterization of a defect group of a block; the first follows by applying its relative-$\Delta D$ induction splitting to $B{\otimes }_{kG}M=M$.

If $D$ is cyclic, every module source for every module vertex $Q\le D$ occurs among the finitely many indecomposable modules for a cyclic p-group. Each module in the block is a summand of an induction of one of these finitely many module sources. There are finitely many subgroups $Q$ and each induced representation has finitely many indecomposable summands by Krull–Schmidt theorem. Thus cyclic defect implies finite representation type of the block.

For the converse, the restriction $B{\downarrow }_{D\times D}$ has the regular $kD$-bimodule as a direct summand. To see this from module vertices and module sources, its Brauer quotient of a module at $\Delta D$ is nonzero. Its summands are permutation modules for the $p$-group $D\times D$ and have module vertices of order at most $|D|$. A summand detected at $\Delta D$ therefore has module vertex $\Delta D$ and trivial module source, hence isHere a transitive permutation module for a $p$-group is indecomposable: it has one-dimensional head of a module over the local group algebra and cannot split into two nonzero modules. This also explains the Brauer-quotient detection used in the argument.

For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It giveswhere $\mid$ denotes being a direct summand. If $B$ has only finitely many indecomposable modules $M_1,\dots ,M_s$, each $B{\otimes }_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i{\downarrow }_D$. Thus $kD$ has finite representation type.

A noncyclic $p$-group has a quotient $C_p\times C_p$: its Frattini quotient has rank at least two, since a one-generator Frattini quotient forces the group itself to be cyclic. Inflation along this quotient embeds all the indecomposable modules of that elementary abelian group, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.

The conclusion remains valid if the splitting field $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ useAn endomorphism commuting with these has form $\left(\begin{array}{cc}A& 0\\ C& A\end{array}\right)$ with $A{T}_f=T_{f}A$. Its endomorphism ring has a square-zero ideal of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a field, so it is local and the module is indecomposable. Finite fields have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,

Let $P=L⋉U$ and $Q=M⋉V$ be rational parabolic subgroups with rational Levi subgroups. Work with complex representations of their finite groups of Lie type. Here Harish-Chandra induction and Harish-Chandra restriction meanThe latter is an $M^F$-module, and it is adjoint to the former by Frobenius reciprocity.

DefineFor $x\in \mathcal{S}(M,L{)}^F$, put $H_x=M\cap {}^{x}L$ and let $\mathrm{ad}x$ transport an $L^F$-module to a $({}^{x}L{)}^F$-module. The Mackey formula for Harish-Chandra induction is the natural functor isomorphismOn characters the direct sum becomes a sum. The intersection parabolics displayed in the subscripts are important: they specify exactly which Harish-Chandra induction and Harish-Chandra restriction occur.

We use the following parabolic-intersection lemmas, stating explicitly the geometric input to the proof. First, representatives of $M^F\backslash \mathcal{S}(M,L{)}^F/L^F$ represent exactly the double cosets $Q^F\backslash G^F/P^F$. Second, for such $x$, the groups $M\cap {}^{x}P$ and $Q\cap {}^{x}L$ are parabolic subgroups of $M$ and ${}^{x}L$, respectively, with common Levi subgroup $H_x$ and respective unipotent radicals $M\cap {}^{x}U$ and $V\cap {}^{x}L$. If $K=Q\cap {}^{x}P$, projection $Q\to M$ maps $K$ onto $M\cap {}^{x}P$ with kernel $K\cap V$. Projection ${}^{x}P\to {}^{x}L$ maps $K\cap V$ onto $V\cap {}^{x}L$ with kernel $V\cap {}^{x}U$. These assertions hold on rational points as well. The surjectivity on rational points uses connectedness of the unipotent kernels and the Lang theorem for algebraic groups. These are the compatible-Levi and rational double-coset lemmas; they follow by applying the root-subgroup decomposition after choosing the common maximal algebraic torus.

For completeness, the representation-theoretic reduction is as follows. The group algebra realization of induced representation splits according to the double cosets:as a $(Q^F,P^F)$-bimodule. Tensoring on the right with an inflated $L^F$-module $E$ identifies each summand with induction from $K^F=(Q\cap {}^{x}P{)}^F$. This proves the finite-group Mackey restriction formula in the formHere ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$.

We next record the elementary normal-subgroup identity underlying Harish-Chandra restriction. If $N◃D$ are finite groups, with $N$ a normal subgroup, and $C\subset D$, then for a complex $C$-module $A$,To see this, replace invariants by coinvariants using the averaging idempotent $|N{|}^{-1}{\sum }_{n\in N}n$. Taking coinvariants in $\mathbb{C}[D]{\otimes }_{\mathbb{C}[C]}A$ first replaces $\mathbb{C}[D]$ by $\mathbb{C}[D/N]$, and forces $C\cap N$ to act trivially on $A$. Thus it yields $\mathbb{C}[D/N]{\otimes }_{\mathbb{C}[CN/N]}{A}_{C\cap N}$; averaging over $C\cap N$ identifies this last module with the one displayed. This also shows why characteristic zero makes the argument exact.

Apply this identity to $D=Q^F$, $N=V^F$, $C=K^F$. By the intersection lemmas, $CN/N=(M\cap {}^{x}P{)}^F$. Since ${}^x\tilde{E}$ is trivial on $({}^{x}U{)}^F$, its $(K\cap V{)}^F$-invariants are exactlyMoreover $(M\cap {}^{x}U{)}^F$ acts trivially on this module, so induction from $(M\cap {}^{x}P{)}^F$ to $M^F$ is precisely Harish-Chandra induction from $H_x$. Substituting into each double-coset summand gives the boxed formula. This proves the formula for rational parabolics; it does not assume a general Mackey formula for nonrational Deligne-Lusztig induction.

Work over the complex numbers. By Maschke's theorem, restriction to the normal subgroup $A$ is a direct sum of its isotypic components. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.

If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:g{V}_W=V_W\}$, a proper subgroup. This is an irreducible representation of $H$: if $0\ne U⊊V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.

The natural map from the induced representation sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. ConsequentlyThis proves the dichotomy directly, including the irreducibility of the representation being induced.

We use complex representations. By Maschke's theorem, restriction to any finite subgroup is a semisimple module, so it has a canonical isotypic decomposition. In particular, the $\theta$-isotypic component $W_{\theta }$ is the sum of all copies of the irreducible $N$-module with character $\theta$. The central character idempotentprojects onto this component. These projections preserve every $N$-submodule, which consequently decomposes as the direct sum of its intersections with the isotypic components.

Let $W$ afford $\xi \in \mathrm{Irr}(T\mid \theta )$. The component $W_{\theta }$ is nonzero. Since $T=I_G(\theta )$ fixes $\theta$, it preserves $W_{\theta }$. Irreducibility of $W$ as a $T$-module therefore gives $W=W_{\theta }$: its restriction to $N$ consists entirely of copies of $\theta$.

Form the induced representationwhere $\mathcal{R}$ represents the left cosets $G/T$. Because $N◃G$, for $n\in N$ we haveThus $g\otimes W$ is $N$-isotypic of type ${\theta }^g$, where ${\theta }^g(n)=\theta (g^{-1}ng)$. These types are distinct for distinct cosets $gT$, precisely by the definition of the inertia group of a character $T$.

Let $Y$ be a nonzero $G$-submodule of $V$. Its isotypic decomposition as an $N$-module shows that it meets some $g\otimes W$ nontrivially. Acting by $g^{-1}$ gives $Y\cap (1\otimes W)\ne 0$. This intersection is a $T$-submodule of the irreducible $W$, hence equals $1\otimes W$. The $G$-translates of that component span $V$, so $Y=V$. The induced representation is therefore irreducible, and its restriction contains $\theta$. Its induced character satisfiesThis proves the irreducibility assertion directly from isotypic components; no form of the correspondence being proved has been assumed.

Let $V$ be an irreducible $G$-module with character $\chi \in \mathrm{Irr}(G\mid \theta )$, and let $W=V_{\theta }\ne 0$ be its $\theta$-isotypic component. It is invariant under the inertia group of a character $T$. The sum of its $G$-translates is a nonzero $G$-submodule, so irreducibility makes it all of $V$. Translates associated with different cosets $gT$ have distinct $N$-types and hence form a direct sum:To see that $W$ is irreducible over $T$, let $0\ne W_0\subseteq W$ be a $T$-submodule. Then ${\sum }_{g\in \mathcal{R}}g{W}_0$ is a nonzero $G$-submodule of $V$, hence equals $V$. Its $\theta$-isotypic part is exactly $W_0$, whereas the corresponding part of $V$ is $W$. Thus $W_0=W$.

Let $\xi$ be the irreducible character of $W$ as a $T$-module. It lies over $\theta$, and the natural map of induced representationsis an isomorphism: it identifies each summand with the corresponding $gW$, and these summands are direct and exhaust $V$. Consequently $\chi ={\xi }^G$, proving surjectivity of induction on the stated sets.

For injectivity, the construction in (a) shows that the $\theta$-isotypic component of ${\mathrm{Ind}}_T^{G}W$ is exactly $1\otimes W$, with its original $T$-action. Isomorphic induced $G$-modules therefore have isomorphic intrinsic $\theta$-components as $T$-modules. Since complex representations with the same character are isomorphic, ${\xi }_1^G={\xi }_2^G$ forces ${\xi }_1={\xi }_2$. Hence the Clifford correspondence is the bijectionwhose inverse takes the character of the $\theta$-isotypic component.

The induced character construction extends any class function $\psi$ on $H$ byFrobenius reciprocity for class functions is $⟨{\mathrm{Ind}}_H^G\psi ,\chi {⟩}_G=⟨\psi ,{\mathrm{Res}}_H^G\chi {⟩}_H$. It follows directly by substituting this sum into the left inner product, writing $g=xh{x}^{-1}$, and using the conjugacy invariance of $\chi$. If $\psi$ is a character, its inner product with every restricted irreducible character is a nonnegative integer, a multiplicity in its complete reducible decomposition. Expanding in the orthonormal irreducible-character basis therefore makes $\mathrm{Ind}\psi$ a nonnegative integer sum of irreducible characters: it is a character of $G$.

If $U$ affords $\psi$, a concrete induced representation is $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}U$, with $G$ acting by left multiplication on the first factor. Choosing coset representatives decomposes it into $[G:H]$ copies of $U$. An element $g$ permutes these copies; only fixed cosets contribute trace, and the contribution from $xH$ is $\psi (x^{-1}gx)$. Summing fixed cosets gives exactly the displayed induced class function.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

If a group representation of $G$ has character $\chi$, its restriction of a representation to $H$ has character $\chi {|}_H$. For an $H$-representation $V$ with character $\psi$, the induced representation is $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}V$, with $G$ acting on the first factor. Decomposing into coset copies of $V$ shows that its character isIndeed only cosets fixed by $g$ contribute to the trace, and each coset has $|H|$ representatives.

The Frobenius reciprocity says $⟨{\mathrm{Ind}}_H^G\psi ,\chi {⟩}_G=⟨\psi ,{\mathrm{Res}}_H^G\chi {⟩}_H$. Directly, with the usual character inner product,The first equality substitutes $g=xh{x}^{-1}$ in the finite sum, and the second uses conjugacy invariance. This proves the theorem.