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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.

Take $g=(12\cdots n)$. Since $n\equiv 3(\mathrm{mod}4)$ is odd, its sign is $(-1{)}^{n-1}=1$, so $g\in A_n$. Define $s$ to fix $1$ and reverse the other cyclic labels: it interchanges $2$ with $n$, $3$ with $n-1$, and so on. Then $sg{s}^{-1}=g^{-1}$, whileThe centralizer of an $n$-cycle in $S_n$ is $⟨g⟩$: a commuting permutation is determined by the image of one point, and that image fixes the corresponding power of the cycle. All its elements are even, because $g$ is even. Any other conjugator taking $g$ to $g^{-1}$ differs from $s$ by a centralizer element and therefore remains odd. Thus $g$ and $g^{-1}$ are not conjugate in $A_n$. This is the parity obstruction described by inversion of an odd cycle in an alternating group.

For any complex character of a finite group, a unitary realization givesIf every $\theta \in \mathrm{Irr}(A_n)$ had real value at $g$, all these characters would take equal values on $g$ and $g^{-1}$. But irreducible characters separate conjugacy classes: they form a basis of class functions, so equality of every irreducible value would also give equality of the class indicator functions. The two distinct conjugacy classes cannot have that property. ConsequentlyIn other words, these alternating groups are not ambivalent groups, even though all symmetric-group irreducible characters are integer-valued.

In character theory, a p-elementary group is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An elementary group is p-elementary for at least one prime $p$; it need not be an elementary abelian group.

A generalised character means an integer linear combination of irreducible complex characters. Brauer's characterisation of characters states that a complex class function $f$ on a finite group $G$ is a generalised character if and only if $f_E$ is a generalised character for every elementary subgroup $E\le G$. Equivalently,An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.

For the extension construction, set $d=\theta (1)$. The determinant character $\det \theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a linear character. For any $g\in G$, the subgroup $J_g=N⟨g⟩$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and ${\mu }_{J_g}$ extends its determinant. The permitted coprime-degree determinant extension theorem therefore gives a unique character ${\chi }_g\in \mathrm{Irr}(J_g)$ withDefine a function on $G$ by $f(g)={\chi }_g(g)$. We next verify that these individually defined values form a compatible class function.

If $x\in G$, conjugation carries $J_g$ to $J_{xg{x}^{-1}}$. Transporting ${\chi }_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a class function. Uniqueness identifies the transported character with ${\chi }_{xg{x}^{-1}}$. Consequently $f(xg{x}^{-1})=f(g)$.

Now take an elementary subgroup $E\le G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of solvable groups is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives ${\chi }_J\in \mathrm{Irr}(J)$ extending $\theta$ with determinant ${\mu }_J$.

For $g\in E$, we have $J_g\le J$. The restriction $({\chi }_J{)}_{J_g}$ is irreducible: an invariant subspace for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is ${\mu }_{J_g}$. By uniqueness on $J_g$, it equals ${\chi }_g$. Thereforean ordinary character and hence a generalised character of $E$. Brauer's characterisation of characters now makes $f$ a generalised character of $G$. If $n\in N$, then $J_n=N$ and ${\chi }_n=\theta$, so $f(n)=\theta (n)$. We have proved, by gluing determinant-normalized character extensions,Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.

For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_{\pi }\times C_{{\pi }^{\prime }}$, where each factor has the indicated prime divisors in its order. If $p\in \pi$, put $E_{\pi }=P\times C_{\pi }$ and $E_{{\pi }^{\prime }}=C_{{\pi }^{\prime }}$. If $p\notin \pi$, put $E_{\pi }=C_{\pi }$ and $E_{{\pi }^{\prime }}=P\times C_{{\pi }^{\prime }}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the prime-set decomposition of elementary groupsHere a pi-group has order divisible only by primes in $\pi$, and a pi-element has such an element order; the identity qualifies for both complementary prime sets.

Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of conjugacy classes and are disjoint, because membership in both would force every prime divisor of the element order into $\pi \cap {\pi }^{\prime }=\varnothing$, hence force order $1$. Let $E\le K$ be elementary and use $E=E_{\pi }\times E_{{\pi }^{\prime }}$. If both factors were nontrivial, choose $a\ne 1$ in the first and $b\ne 1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and ${\pi }^{\prime }$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a pi-group or entirely a complementary-prime group.

Write $m=|K{|}_{\pi }$ and $r=|K{|}_{{\pi }^{\prime }}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd (m,r)=1$. The Chinese remainder theorem supplies an integer $D$ satisfyingDefine the class functionFor an elementary $\pi$-subgroup $E$, its order divides $m$. Let ${\rho }_E$ be the character of its regular representation, equal to $|E|$ at $1$ and zero elsewhere. Thenwith integral coefficient because $D\equiv 1(\mathrm{mod}|E|)$. For an elementary ${\pi }^{\prime }$-subgroup, its order divides $r$, andagain has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a generalised character. A second application of Brauer's characterisation of characters provesThe freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi =\varnothing$, ${\pi }^{\prime }=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.

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.

Sunada's theorem states that if a finite group $T$ acts isometrically on a closed Riemannian manifold $X$, and two Gassmann equivalent subgroups $H_1,H_2$ act freely, then $H_1\backslash X$ and $H_2\backslash X$ are isospectral for the function Laplace-Beltrami operator. Gassmann equivalence means $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$ of $T$. For each Laplace-Beltrami operator eigenspace $E_{\lambda }$, its invariant subspace has dimensionThe character of a representation is a class function, so the two multiplicities agree. A necessary condition for nonisometry is that the subgroups are not conjugate in $T$: conjugation by an ambient deck transformation would induce an isometry. This is not sufficient when the common covering has additional geometric symmetries.

Use the oriented hyperbolic triangle groupand the epimorphism $\rho :\Gamma \to T$ given by $a\mapsto s$, $b\mapsto d$. Its base orbifold is the doubled hyperbolic triangle with angles $\pi /3,\pi /3,\pi /6$, having signature $(0;3,3,6)$. Let $K=\ker \rho$ and $K_i={\rho }^{-1}(H_i)$.

The groups $K_i$ are torsion free. Indeed, torsion elements in an oriented triangle group are conjugate to nontrivial powers of its three vertex rotations. Such powers map either to elements of order three, which cannot lie in a subgroup of order $96/12=8$, or to the order-two cube $z=(sd{)}^3$. The latter is central and absent from both $H_i$, so it is absent from every conjugate as well. Thus the degree-twelve orbifold coversare smooth, closed, connected orientable surfaces. The common cover $X=K\backslash {\mathbb{H}}^2$ is smooth too; $H_i$ act freely on it. Orbifold Euler characteristic givesFor example, this formula follows by triangulating the base and weighting a vertex of order $r$ by $1/r$ before multiplying by the cover degree. Equivalently, each doubled triangle has hyperbolic area $\pi /3$, so $N_i$ has area $4\pi$ and Gauss-Bonnet theorem gives the same genus.

There is one extra issue in obtaining the same hyperbolic limit on a single marked surface: the two limiting hyperbolic covers must themselves be isometric. Nonconjugacy and the displayed element orders alone do not prove this. An explicit realization of the supplied example resolves the issue. The group seed is recorded in the original genus-two construction; the following coset actions and reflection identities give a direct verification. Label the twelve sheets so that the two right-coset actions of $s,d$ areThese are concrete permutation data for an order-$96$ realization with the required two subgroups. The permutationsatisfies, by direct substitution,Reflection of the doubled triangle reverses both vertex rotations. Apply this reflection on every fundamental domain and relabel sheets by $P$. The two identities ensure that identified edges go to identified edges, so this descends to a hyperbolic isometry $F_0:N_1\to N_2$. Fix $N=N_1$, let $g$ be its hyperbolic metric, and use $F_0$ to mark $N_2$ by the same smooth surface. Thus both pulled-back limiting Riemannian metrics are exactly $g$. The explicit reversed-action check is what makes the common limit justified, rather than treating it as a consequence of Sunada's theorem.

Choose smooth orbifold Riemannian metrics $h_k$ on the base converging to its hyperbolic metric $h_0$, with no nonidentity local isometry on the regular part. The density assertion in Sunada's local isometry lemma applies in orbifold charts as well: use equivariant perturbations at the finitely many cone points and ordinary local perturbations elsewhere. Smoothness here is orbifold smoothness, so the lifted Riemannian metrics on $X$ and the two smooth covers are ordinary smooth Riemannian metrics. Such choices can be made in successively smaller smooth neighbourhoods of $h_0$; cone orders and the topological coverings stay fixed.

For area separation of convergent Sunada families, adjust each $h_k$ by a constant positive factor so thatExplicitly, multiply the initially chosen Riemannian metric by the ratio of the desired area to its actual area. In dimension two this ratio is exactly the area multiplier. The ratios tend to one, so smooth convergence persists, and constant scaling preserves the absence of local isometries. Lift these adjusted Riemannian metrics to $N_i$ and pull them back by the fixed markings to obtain $g_{ik}$, for $i=1,2$. The three required properties follow below.

We use the Dirichlet density form of the Chebotarev density theorem, the version directly proved by the permitted information at $s=1$. If $L/K$ is a finite Galois extension of number fields with group $G$ and $C$ is a conjugacy class, let $S_C$ be the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. ThenThe same conclusion holds for a union of conjugacy classes, replacing $|C|$ by the size of that union. The denominator may equivalently be the sum over all primes of $K$. Here and below $s>1$ is real. This specifies the density being proved; an unweighted prime-counting asymptotic is a stronger formulation and is not being inferred merely from a limit at one.

First establish the necessary pole for a general base field. The Dedekind zeta function as a permutation Artin L-function gives it using exactly the allowed assumptions. Take a finite normal closure $E$ of $K/\mathbb{Q}$, and let $W$ be the permutation representation on ${\mathrm{Hom}}_{\mathbb{Q}}(K,E)$. Its trivial constituent has multiplicity one: an invariant function on the transitive set of embeddings is constant. At a rational prime, a basis of $W^I$ consists of the sums over inertia orbits of embeddings. Frobenius automorphism permutes these orbit sums. Within the decomposition-group orbit belonging to a prime $\mathfrak{p}$ of $K$, the resulting cycle has length $f_{\mathfrak{p}}$, because the residue embeddings are permuted cyclically by finite-field Frobenius. A cycle of length $f$ has determinant $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Consequently the local factor of an Artin L-function givesincluding at ramified primes. Decomposing $W=1\oplus {⨁}_{\psi \ne 1}{m}_{\psi }\psi$ givesAll factors after $\zeta$ are holomorphic and nonzero at one by the allowed hypothesis. Thus ${\zeta }_K$ has a simple pole there andThis argument also explains why the trivial relative Artin L-function is ${\zeta }_K$, rather than always the Riemann zeta function.

Now let $\chi$ be an irreducible character of $G$, of degree $d_{\chi }$. At an unramified prime the Euler product has local factor $\det (1-{\rho }_{\chi }(g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$, where $g_{\mathfrak{p}}$ is a Frobenius representative. The eigenvalues of ${\rho }_{\chi }(g)$ are roots of unity, since $g$ has finite order. Expanding the logarithm for $s>1$ therefore giveswhere the finitely many ramified local factors contribute a function bounded as $s\downarrow 1$. They are nonzero and finite near one because their eigenvalues have modulus one while $(N\mathfrak{p}{)}^{-1}<1$.

The terms $m\ge 2$ are uniformly bounded. Indeed $|\chi (g)|\le d_{\chi }$, and for $N\mathfrak{p}\ge 2$ and $s\ge 1$,There are at most $[K:\mathbb{Q}]$ primes of $K$ above each rational prime and each has norm at least that prime. Hence ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. Isolating $m=1$ proves the character prime sums in the Chebotarev proof identityFor nontrivial $\chi$, the allowed holomorphy and nonvanishing at one give a local holomorphic logarithm and a bounded right side. The logarithm specified by the Euler series differs from that local branch by a constant multiple of $2\pi i$ on the real interval near one, so it too is bounded. For the trivial character, the right side is $\log (1/(s-1))+O(1)$ by the pole of ${\zeta }_K$. Thus

Expand the conjugacy-class indicator by character orthogonality. For $c\in C$,To see the coefficient, take its inner product with $\chi$: $|G{|}^{-1}{\sum }_{g\in C}\overline{\chi (g)}=|C|\overline{\chi (c)}/|G|$. The irreducible characters form an orthonormal basis of class functions, so this is the full expansion. Substituting the prime-sum estimates leaves only the trivial character's logarithmic term:Division by the diverging logarithm proves the theorem. The same trivial-character calculation shows that all primes of $K$ have density one; adding or removing finitely many primes does not change a Dirichlet density.

There is a necessary qualification for the final polynomial application: the printed claim is false for degree one. For example, $X$ is monic and irreducible in $\mathbb{Z}[X]$, yet has the root zero modulo every prime. We prove the intended assertion for degree $n\ge 2$.

Let $L$ be the splitting field of the monic irreducible polynomial over $\mathbb{Q}$, and let $\Omega$ be its $n$ roots. The Galois group $G$ acts transitively on $\Omega$. There is a derangement in a transitive group action: count the pairs $(g,\alpha )$ satisfying $g\alpha =\alpha$. The orbit-stabilizer theorem givesIf every $g$ fixed at least one root, the identity, which fixes $n\ge 2$, would make this sum at least $|G|+n-1$, a contradiction. Therefore some $g$ fixes no root, and neither does any conjugate of $g$.

For completeness, relate this group action to reduction without assuming the desired splitting result. All roots are algebraic integers, since the polynomial is monic. Exclude the finitely many primes ramifying in $L$ or dividing the nonzero polynomial discriminant. At a prime $\mathfrak{q}$ over any remaining $p$, the roots reduce to distinct elements of $k_{\mathfrak{q}}$ and give all roots of the reduced polynomial. Arithmetic Frobenius permutes them by $b\mapsto b^p$. A reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed by this map. Because reduction is injective on the original root set, this happens exactly when the Frobenius permutation fixes an original root. This proves the Frobenius permutation and roots modulo a prime criterion.

The Chebotarev density theorem assigns the conjugacy class of our fixed-point-free $g$ the positive Dirichlet density $|C_g|/|G|$. All but finitely many of these primes have rootless reduction, so there are infinitely many. More precisely, applying the same argument to the union of all fixed-point-free classes gives

Take $N$ to be a closed compact Riemannian manifold and use the positive Laplace-Beltrami operator. Its Riemannian heat kernel $K_N(t,x,y)$ is the integral kernel of $e^{-t{\Delta }_N}$:It solves $({\partial }_t+{\Delta }_x)K_N=0$ for $t>0$ and tends to the identity kernel as $t\downarrow 0$. For an orthonormal eigenbasis, the spectral expansion of the Riemannian heat kernel isThe bar is required for a complex basis. With boundary, a specified invariant boundary condition must also be included in this definition.

For the quotient formula use the covering-space interpretation: $U$ acts freely and properly discontinuously by Riemannian isometries. On compact $N$ such a discrete group is finite. Freeness alone, without this covering hypothesis, does not justify the image sum: an infinite dense subgroup of circle rotations, for example, acts freely but has no manifold quotient. Positive-dimensional group actions require a different quotient analysis.

For lifts $x,y$ of $\bar{x},\bar{y}\in M$, the heat kernel on a finite isometric quotient isInvariance of $K_N$ under simultaneous Riemannian isometries and reindexing the sum show that this is independent of both lifts. A local isometry commutes with the Laplacian, so the sum satisfies the quotient heat equation. For its initial condition, integrate over a fundamental domain $F\subset N$ against a lifted function. The terms combine into the integral over all of $N$, whose initial limit is $f(x)$. Uniqueness of the heat evolution proves the formula. There is no factor $1/|U|$ in this kernel formula.

On the diagonal the sum is $U$-invariant. Its integral over $F$ is therefore $1/|U|$ times its integral over $N$, giving the heat traceThus the normalization factor appears in the trace, not the kernel. Put $F_t(g)={\int }_N{K}_N(t,x,gx)dV(x)$. For any finite isometry group $T$, changing variables $x=hy$ proves $F_t(hg{h}^{-1})=F_t(g)$: it is a class function on $T$.

For Gassmann equivalent subgroups $U_1,U_2\le T$, their intersections with each conjugacy class have equal size, and their orders are equal. If both act freely, the heat traces of their quotients satisfyso they agree for every $t>0$. Since $Z_M(t)={\sum }_{\lambda }m(\lambda )e^{-t\lambda }$, equality determines the spectrum with multiplicities: take $t\to \infty$ to recover the smallest eigenvalue and its multiplicity, subtract that term, and repeat. This proves Sunada theorem. Equivalently, projection onto $U_i$-invariant functions averages the group action, and the quotient eigenvalue multiplicity is $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, with ${\chi }_{\lambda }$ the eigenspace character of a representation. Almost conjugacy equalizes these averages.

For an integer tuple $\ell =({\ell }_1,\dots ,{\ell }_m)$, the monomial alternant isNegative exponents require nonzero coordinates; all exponents in the character expansion below are nonnegative. The power-sum symmetric polynomial is $s_k(x)={\sum }_{i=1}^m{x}_i^k$. Put $\delta =(m-1,m-2,\dots ,0)$, so $A_{\delta }(x)={\prod }_{i<j}(x_i-x_j)$ is the Vandermonde determinant.

Let a conjugacy class of $S_n$ have ${\alpha }_k$ cycles of length $k$, with $\sum k{\alpha }_k=n$, and put $p_{\alpha }(x)={\prod }_k{s}_k(x{)}^{{\alpha }_k}$. The product $p_{\alpha }{A}_{\delta }$ is an alternating polynomial homogeneous of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. In an alternating polynomial, a monomial with two equal exponents has zero coefficient, since interchanging those variables fixes the monomial and reverses its sign. Grouping the remaining monomials by their permutation orbits gives a unique expansion in alternants $A_{\ell }$ with ${\ell }_1>\cdots >{\ell }_m\ge 0$.

Such tuples of the indicated total degree are exactly ${\ell }_i={\lambda }_i+m-i$ for partitions of an integer $\lambda$ of $n$ with at most $m$ parts. Define the class functionThe displayed monomial occurs with coefficient $1$ in $A_{\lambda +\delta }$ and in no other ordered alternant, soThis proves the expansion and explicitly defines its coefficients. They depend only on the cycle counts and hence are class functions. Identifying these coefficients with Specht module characters is the Frobenius alternant character formula, which the remaining parts allow us to assume.

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

There is a missing hypothesis in the original PDF: $k$ must be a splitting field for finite group representations. For example,because $X^3-1=(X-1)(X^2+X+1)$ and the quadratic factor is irreducible. This algebra has two simple modules, whereas $C_3$ has three conjugacy classes, all $2$-regular. Thus the printed assertion for an arbitrary field is false.

Under the intended splitting hypothesis, fix compatible lifts defining Brauer characters. Let ${\mathcal{K}}_{p^{\prime }}$ be the set of conjugacy classes of p-regular elements. Character additivity and the tensor-product formula define the unital algebra homomorphismWe use the Brauer character basis theorem: over a splitting field the irreducible Brauer characters form a complex basis of the class functions on the p-regular conjugacy classes. Therefore $\Phi$ maps the basis $1\otimes [S]$ bijectively to a basis and is an algebra isomorphism. Comparing dimensions givesIsomorphism classes are understood in the count.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

For a finite subgroup $U$ of a group $T$ of Riemannian isometries of a closed manifold $N$, assume $U$ acts freely and define $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$. Invariance of the Riemannian heat kernel and change of variables show that $I_t$ is a class function on $T$. For $U\le T$, the heat kernel on a finite isometric quotient and integration over a fundamental domain give the displayed heat trace formula. Grouping by conjugacy classes proves equality for Gassmann equivalent subgroups. More generally the same argument allows fixed points of $T$ outside the freely acting subgroups; only the quotient under consideration must be smooth.

Let $G$ be a connected compact Lie group, $T$ a maximal torus, and $W$ its Weyl group. With normalized Haar measures and $D(t)={\prod }_{\alpha >0}(1-e^{-\alpha }(t))$, every continuous class function satisfies ${\int }_{G}f(g)dg=|W{|}^{-1}{\int }_{T}f(t)|D(t){|}^{2}dt$. For a general continuous function, replace $f(t)$ by its average ${\int }_{G/T}f(gt{g}^{-1})d(gT)$. Conjugation from $G/T\times T$ covers the regular elements $|W|$ times, and its real Jacobian determinant is $|D(t){|}^2$.

For a class function on SU(2) and normalized Haar measure, the displayed formula reduces a group integral to its conjugacy-class angle. The weight follows by identifying SU(2) with the unit sphere in four real dimensions: writing a unit quaternion as $\cos \theta +\mathbf{n}\sin \theta$ gives spherical volume proportional to ${\sin }^2\theta d\theta d\mathbf{n}$. Since ${\int }_0^{\pi }{\sin }^2\theta d\theta =\pi /2$, normalization gives $2/\pi$.