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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.

Use the classes $(E,2C_3,3{\sigma }_v)$ of the C3v point group. For the three hydrogen $1s$ atomic orbitals, the character of a representation counts orbitals fixed by each symmetry operation. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. ThereforeThe character inner product gives multiplicitiesThus ${\Gamma }_{\mathrm{H}}=A_1\oplus E$, with the $E$ contribution accounting for two independent symmetry orbitals.

The C2v molecular point group has four one-dimensional irreducible representations. In the operation order used above, the character table rows are $A_1=(1,1,1,1)$, $A_2=(1,1,-1,-1)$, $B_1=(1,-1,1,-1)$ and $B_2=(1,-1,-1,1)$. Their character inner products with ${\chi }_{\Gamma }=(6,0,2,0)$ giveHence $\Gamma =2A_1\oplus A_2\oplus 2B_1\oplus B_2$, whose dimensions sum to six. The labels $B_1,B_2$ refer to the source axes; swapping $x$ and $y$ swaps these names.

The drawn $x,y$ axes pass through opposite sites, rather than along the square edges. In the D4h molecular point group class order $(E,2C_4,C_2,2C_2^{\prime },2C_2^{\prime \prime },i,2S_4,{\sigma }_h,2{\sigma }_v,2{\sigma }_d)$, an s orbital contributes $+1$ whenever its centre is fixed and zero otherwise. ThusFor example, a transverse twofold rotation about the printed $x$ axis fixes sites 1 and 3, while its diagonal counterpart exchanges all four sites. Reflection in the square plane fixes every site. Taking character inner products, with class sizes $(1,2,1,2,2,1,2,1,2,2)$, givesThese have dimensions $1+1+2=4$. The $B_{1g}$ label is tied to the printed diagonal-axis convention; rotating the transverse coordinate labels by $45^{\circ }$ would interchange $B_{1g}$ and $B_{2g}$.

Use the trapezium's printed axes, with principal axis $z$ in its plane and $y$ perpendicular to the plane. For either site pair $(1,2)$ or $(3,4)$, the group representation has characters of a representation $(2,0,2,0)$ in the order $(E,C_2(z),\sigma (xz),\sigma (yz))$: identity and the molecular-plane mirror fix both s orbitals, while the other two operations exchange them. The character inner products give ${\Gamma }_{12}={\Gamma }_{34}=A_1\oplus B_1$ in the C2v molecular point group.

For a finite group action $G\curvearrowright X$, Burnside's lemma statesCount pairs $(g,x)$ with $gx=x$ in two ways. Counting first by $g$ gives the displayed sum. Counting first by $x$ gives ${\sum }_x|G_x|$. For an orbit $\mathcal{O}$, the orbit-stabilizer theorem gives $|\mathcal{O}||G_x|=|G|$, so each orbit contributes exactly $|G|$. Division proves the formula.

Use colors $1,\dots ,m$, with variables $x_1,\dots ,x_m$, on the underlying finite set $S$. The weight of a coloring $f:S\to \{1,\dots ,m\}$ is $w(f)={\prod }_{s\in S}{x}_{f(s)}$. It is constant on each coloring orbit. The pattern inventory is $F_G(x)={\sum }_{\mathcal{O}}w(\mathcal{O})$, one weight for each orbit, including its multiplicity when different orbits have the same content.

If $c_r(g)$ is the number of length-$r$ cycles of $g$ on $S$, define the cycle indexIn the symmetric-function cycle-indicator convention, write $Z_G(x)$ for its substitution $t_r=p_r(x)={\sum }_{i=1}^m{x}_i^r$. A fixed coloring must be constant on every cycle of $g$. Coloring one length-$r$ cycle therefore has weight sum $p_r(x)$, independently of other cycles. The weighted Burnside lemma, proved by exactly the same orbit-stabilizer double count with weights, yieldsThis defines both sides in the same alphabet; the raw cycle index in the independent variables $t_r$ is specialized before comparison. The identity holds for all finite $m$ and hence stably for arbitrarily many colors.

For the last assertion, the permutation character on $T$ satisfies $\chi (g)=|{\mathrm{Fix}}_T(g)|$. A pair $(s,t)$ is fixed in the diagonal action on $T\times T$ exactly when both components are fixed. ThereforeThe character is real, so its character inner product with itself and another application of Burnside's lemma giveThis is the rank of a permutation action formula, and in fact does not require transitivity. If the action is transitive and $|T|\ge 2$, the diagonal is one orbit; rank two means exactly one further orbit on distinct ordered pairs, which is double transitivity.

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.

The character of a permutation representation at $g$ equals the number of points fixed by $g$, because a permuted basis vector contributes one to the trace exactly when it is fixed. Thus the character inner product isTo identify the last average, count pairs $(g,z)$ with $gz=z$. Each orbit contributes $|G|$, since its number of elements times their common stabilizer size is $|G|$ by the orbit-stabilizer theorem. ThereforeThe conjugation in the usual character inner product has no effect here because the permutation characters are real-valued.

For the special linear group $G={\mathrm{SL}}_3({\overline{\mathbb{F}}}_q)$, let $N=q-1$, and choose a generator $\chi$ of the character group of ${\mathbb{F}}_q^{\times }$. Every linear character of the diagonal rational maximal torus has a unique expressionThe upper triangular Borel subgroup $B=TU$ is rational. Consequently its Deligne-Lusztig induction is the actual principal series of a finite reductive groupwhere $\tilde{\theta }$ is the inflation of a group representation across $B^F\to T^F$. The geometric reason for this split case is $X(1)=G^F/B^F$: the associated torus covering supplies $\theta$, and the auxiliary unipotent fibres contribute a single even cohomological degree, hence no minus sign.

The rational Bruhat decomposition indexes $B^F\backslash G^F/B^F$ by $W=S_3$. Applying the Mackey restriction formula and Frobenius reciprocity givesEach intersection contains $T^F$, and the two linear characters are trivial on its unipotent part. Its contribution is therefore one exactly when $w$ fixes $\theta$, and zero otherwise. Complete reducibility over $\mathbb{C}$ makes this character inner product the sum of squares of irreducible representation multiplicities. Thus the principal series of a finite reductive group is irreducible exactly when the Weyl group stabilizer is trivial.

To compute that stabilizer, represent $\theta$ by the triple $({\chi }^r,{\chi }^s,1)$. Two triples give the same linear character on determinant-one diagonal matrices exactly when they differ by a common factor $(\nu ,\nu ,\nu )$. For example, triviality of $({\eta }_1,{\eta }_2,{\eta }_3)$ on all such matrices first gives ${\eta }_1={\eta }_3$ by varying $a$, and then ${\eta }_2={\eta }_3$ by varying $b$. The Weyl group permutes the three entries.

A transposition fixes $\theta$ precisely when its two interchanged entries coincide: the third, fixed entry forces the common factor to be one. These three obstructions are $r=0$, $s=0$, and $r=s$. A three-cycle can also fix $\theta$ without any two entries coinciding. Its common factor must satisfy ${\nu }^3=1$; in the nontrivial case the entries, up to order and common factor, are $(1,\nu ,{\nu }^2)$. This happens exactly when $3\mid N$ and $\{r,s\}=\{N/3,2N/3\}$.

The complete condition is thereforeIn particular pairwise distinct coordinate characters alone are insufficient. For $q=4$ the only distinct triples have the order-three symmetry, and no character of this split rational maximal torus gives an irreducible principal series of a finite reductive group.

Let $V$ afford the irreducible character $\chi$. The positive character inner product $⟨{\chi }_H,1_H{⟩}_H$ is the multiplicity of the trivial representation in $V$ restricted to $H$, so there is a nonzero vector $v$ fixed by every element of $H$. Since $N\subseteq \ker \chi$, every element of $N$ acts as the identity on all of $V$. By $G=NH$, every element of $G$ therefore fixes $v$.

The line $\mathbb{C}v$ is a nonzero invariant subspace of the irreducible representation $V$, so it must equal $V$. The action on this line is trivial, provingThe hypothesis on the kernel of a character is what makes an $H$-fixed vector into a $G$-fixed vector.

The preceding parts give a simple Specht module for each partition and show that different partitions give nonisomorphic simples. The conjugacy classes of $S_n$ are indexed by partitions of $n$, namely cycle types. We use the standard finite-group character theorem: the number of irreducible complex characters equals the number of conjugacy classes, and they form an orthonormal basis of complex class functions. Consequently the ${\chi }^{\lambda }$ are the complete list of irreducible characters of $S_n$.

By Maschke's theorem, the Young permutation module $M^{\mu }$ decomposes as a direct sum of these simple modules. For a semisimple complex module, Schur lemma makes the multiplicity of a simple module equal to the dimension of the homomorphism space from that simple; character orthogonality identifies the same number with the character inner product. HencePart (ii) makes $a_{\lambda \mu }=0$ unless $\lambda ⊵\mu$, and gives $a_{\mu \mu }=1$. Taking characters provesThus the permutation-character decomposition is triangular in the dominance order on partitions, with diagonal multiplicity one. These are the dominance and diagonal-multiplicity consequences of Young's rule, obtained here without assuming its full multiplicity formula.

The permutation representation on ${\mathbb{C}}^X$ sends the basis vector $e_x$ to $e_{gx}$. Its character is ${\pi }_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. Burnside lemma says that the number of orbits is $|G{|}^{-1}{\sum }_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is ${\sum }_g|X^g|={\sum }_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the orbit-stabilizer theorem.

A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,by Burnside lemma on the product.

The general linear group ${\mathrm{GL}}_2(q)$ acts on the $q+1$ lines of ${\mathbb{F}}_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The character inner product of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one irreducible representation, of dimension $(q+1)-1$. Its dimension is $\boxed{q}$.

For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\ge 4$, and a permutation carries any pair of a given type to any other. Hence $⟨{\pi }_Z,{\pi }_Z⟩=3$. The natural point permutation representation is the trivial representation plus the standard representation of the symmetric group, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.

Embed this point representation into ${\mathbb{C}}^Z$ by $e_i\mapsto {\sum }_{j\ne i}{e}_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\ge 3$. Thus the trivial and standard representations each occur in ${\mathbb{C}}^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying ${\sum }_{j\ne i}{a}_{ij}=0$ for every $i$. Its dimension isThe constituent dimensions are $\boxed{1,n-1,n(n-3)/2}$, each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.

The restriction of a representation $\rho :G\to \mathrm{GL}(V)$ to $H$ is the same vector space and operators with the domain limited to $H$; its character is ${\chi }_H(h)=\chi (h)$. Averaging a positive Hermitian inner product over $H$ makes it invariant, so invariant subspaces have invariant orthogonal complements. Repeating gives a direct sum of irreducible $H$-spaces and hence ${\chi }_H={\sum }_i{d}_i{\psi }_i$ with nonnegative integer multiplicities.

By character inner product orthogonality,since $\chi$ is irreducible and has normalized squared norm one on $G$. Equality holds precisely when the omitted nonnegative terms vanish, namely $\boxed{\chi (g)=0\text{ for every }g\notin H}$. This is the restriction multiplicity bound.

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.

We prove the Dirichlet density form of the Chebotarev density theorem. Let $L/K$ be a finite Galois extension with Galois group $G$, let $C$ be a conjugacy class in $G$, and let $S_C$ consist of the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. The assertion isThe same result holds for any union of conjugacy classes, by addition. The following proof uses exactly the allowed one-dimensional Artin L-function behavior at $1$ and the permitted induction and Euler-product properties.

First determine the order of vanishing at $1$ for an arbitrary character of a representation $\chi$ of $G$. The Brauer induction theorem supplies subgroups $H_j\le G$, linear characters ${\psi }_j$ of $H_j$, and integers $n_j$, possibly negative, such thatPut $K_j=L^{H_j}$. Compatibility of Artin L-functions with induced representations and direct sums gives, initially for $\mathrm{Re}s>1$,Each ${\psi }_j$ is a one-dimensional Galois representation for $L/K_j$, so the stated analytic assumptions apply to every factor. A trivial ${\psi }_j$ contributes a simple pole; a nontrivial ${\psi }_j$ is holomorphic and nonzero at one. Thus the product is meromorphic near one andThe character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_{g\in G}\alpha (g)\overline{\beta (g)}$. Applying Frobenius reciprocity to each induced term givesIn particular, a nontrivial irreducible character has order zero, so its Artin L-function is holomorphic and nonzero near one. For the trivial character, $L_K(1_G,s)={\zeta }_K(s)$, the Dedekind zeta function of the base field $K$, and its pole is simple. This argument establishes the behavior needed at one, not holomorphy everywhere for arbitrary Artin L-functions.

Next convert that order calculation into a prime sum. For an irreducible character $\chi$, choose a unitary representation of dimension $d_{\chi }$ affording it, and choose a Frobenius automorphism $g_{\mathfrak{p}}$ at each unramified prime. The logarithm specified by the absolutely convergent Euler product gives, for real $s>1$,Only finitely many ramified primes occur. Their Euler-factor logarithms $R_{\chi }(s)$ stay bounded near one, since the relevant eigenvalues have modulus one and $(N\mathfrak{p}{)}^{-1}<1$. Also $|\chi (g)|\le d_{\chi }$, so the contribution with $m\ge 2$ is uniformly bounded for $s\ge 1$ byConvergence follows, for example, from the convergent ideal series defining ${\zeta }_K(2)$. ThereforeWriting $a_{\chi }=⟨\chi ,1_G{⟩}_G$, the order computation means $L_K(\chi ,s)=(s-1{)}^{-a_{\chi }}{U}_{\chi }(s)$, where $U_{\chi }$ is holomorphic and nonzero near one. Choose its local logarithm. Along the real interval approaching one, this and the Euler-product logarithm differ only by a fixed multiple of $2\pi i$. ConsequentlyOnly the trivial irreducible character has $a_{\chi }=1$; all the others have $a_{\chi }=0$.

Finally, character orthogonality makes the irreducible characters an orthonormal basis of class functions. For $c\in C$, the coefficient of $\chi$ in the class indicator is $⟨1_C,\chi {⟩}_G=(|C|/|G|)\overline{\chi (c)}$. HenceSum this identity over the unramified primes with weight $(N\mathfrak{p}{)}^{-s}$. The finite character sum and the boxed estimates giveDividing by $\log (1/(s-1))$ proves the asserted Dirichlet density. The same Euler-product calculation with ${\zeta }_K$ gives ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-s}=\log (1/(s-1))+O(1)$, so using all primes as the denominator is equivalent. Finitely many ramified primes do not affect either limit. No prime-counting asymptotic or natural-density Tauberian step is required by this proof.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

The Specht module construction gives a nonzero inclusion $V^{\lambda }\hookrightarrow M^{\lambda }$ over the complex numbers: a polytabloid has coefficient one at its original tabloid, because the row and column stabilizers have trivial intersection, and its span is the corresponding irreducible representation. This establishes $a=M(\lambda ,\lambda )\ge 1$ without using the multiplicity-one conclusion. Similarly $a^{\prime }=M({\lambda }^{\prime },{\lambda }^{\prime })\ge 1$.

By the corrected tensor identity for induced representations and the conjugate-partition sign twist, the multiplicity of $V^{\lambda }$ in ${\tilde{M}}^{{\lambda }^{\prime }}$ is $a^{\prime }$. The character inner product already computed is the sum of products of multiplicities of common irreducible representations. ThusBoth are positive integers, so $a=a^{\prime }=1$. In particularThis deduction avoids assuming the full Young's rule, which would already contain the answer. Alternatively, the permitted one-dimensional R-module homomorphism space gives the same character inner product immediately.

The Young permutation module $M^{(n-k,k)}$ is the symmetric-group subset permutation representation on the $k$-element subsets of $\{1,\dots ,n\}$: the second row of a tabloid specifies the subset. For $k\ge 1$, restriction of a character to $S_{n-1}$ fixing $n$ splits these subsets into two orbits of a group action, according as they contain $n$.

The corresponding permutation representations are on the $k$-subsets and the $(k-1)$-subsets of $\{1,\dots ,n-1\}$. Their Young permutation modules have two row sizes $n-1-k,k$ and $n-k,k-1$, respectively. Sort each pair into decreasing order and omit any zero part; swapping the two row labels gives an isomorphic permutation representation. This sorting matters at $n=2k$.

We use Young's rule: over the complex numbers, the multiplicity of $S^{\rho }$ in $M^{\eta }$ is the Kostka number $K_{\rho \eta }$, and it can be nonzero only if $\rho$ dominates $\eta$ in the dominance order on partitions. Since each $\eta$ above has at most two parts and size $n-1$, dominance givesThus $\rho$ has at most two nonzero parts. The character inner product measures this constituent multiplicity, soIf $k=0$, the original and restricted modules are the trivial representations, giving the same conclusion. For $n=1$, there is no partition of $n-1=0$ with three parts, so the assertion is vacuous.

By character orthogonality, the squared multiplicities are the character inner product norm of the restriction of a character:Since $\chi$ is an irreducible character, its norm on $G$ is one, and henceThe omitted summands are nonnegative, so the character restriction norm bound is an equality precisely when