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First realize the free isometric action. Choose generators $t_1,\dots ,t_r$ of the finitely presented group $T$. In every dimension $d\ge 3$, the connected sumhas fundamental group the free group $F_r$. Map its generators onto the $t_j$ and take the connected regular covering $N\to B_d$ corresponding to the kernel. Its deck transformation group is $F_r/\ker \pi \cong T$. Lift any Riemannian metric on $B_d$ to $N$. Deck transformations are then Riemannian isometries, and a deck transformation fixing one point is the identity by uniqueness of lifts. This gives the free isometric realization of a finitely generated group. For the trivial group use $S^d$. If $T$ is infinite the covering manifold need not be compact, which is allowed in this first construction. Simple connectivity of $N$ is not being asserted here.

The requested nonhomeomorphism conclusion needs an extra hypothesis. For a counterexample even with distinct subgroups, take $T=S_3$, $U_1=⟨(12)⟩$ and $U_2=⟨(23)⟩$. They are conjugate and therefore Gassman equivalent. For every free isometric action of $T$, the conjugating element induces a Riemannian isometry $U_1\backslash N\to U_2\backslash N$. These quotients cannot be nonhomeomorphic. Identical subgroups give an even simpler obstruction. Thus the stated equivalence condition alone cannot imply the claimed topology.

A sufficient version is: for a finite $T$ with Gassmann equivalent, nonisomorphic subgroups $U_1,U_2$, there are closed isospectral manifolds with fundamental groups $U_1,U_2$, and hence they are not homeomorphic. To prove it, realize $T$ as the fundamental group of a closed smooth manifold $B$ of dimension four. Here is the needed closed-manifold realization of a finitely presented fundamental group. Start with a four-dimensional zero-handle and one one-handle for each generator. Attach two-handles along disjoint embedded boundary circles representing the relators, with arbitrary framings. The boundary is three-dimensional, so finitely many such loops can be chosen disjoint. The resulting compact handle manifold $W$ has ${\pi }_1(W)=T$ by the Seifert-van Kampen theorem.

The inclusion $\partial W\to W$ is surjective on fundamental groups: relative to the boundary the dual handle decomposition uses only handles of indices two, three and four, none of which introduces a fundamental-group generator. Double $W$ along its boundary. In the amalgamated product for the double, both boundary maps are the same surjection onto $T$, so the two copies of $T$ are identified completely and ${\pi }_1(B)=T$. The double is closed, connected and smooth after smoothing its collar. This construction works in any dimension at least four; it does not claim that every finitely presented group is a closed three-manifold group.

Let $N$ now be the universal cover of $B$ with the lifted metric. Since $T$ is finite, $N$ is compact and simply connected, with a free isometric deck action of $T$. By Sunada theorem, $M_i=U_i\backslash N$ are isospectral. Because $N$ is their universal cover, ${\pi }_1(M_i)\cong U_i$. Nonisomorphic fundamental groups rule out a homeomorphism. For an infinite $T$ the same qualified argument applies to finite-index subgroups with equal coset characters: divide first by the intersection $K={\mathrm{Core}}_T(U_1)\cap {\mathrm{Core}}_T(U_2)$ of their finite-index subgroup cores, producing a compact intermediate cover and a finite isometry group to which Sunada applies.

For explicit nonhomeomorphic examples, let $p$ be an odd prime. Let $H_p=(\mathbb{Z}/p\mathbb{Z}{)}^3$, and let $K_p$ be the Heisenberg group over a prime field, whose elements are triples with multiplicationBoth have order $p^3$, and every nonidentity element has order $p$. For $K_p$, induction givesso the assertion follows at $k=p$, since $p$ is odd. But $H_p$ is abelian and $K_p$ is not: $(1,0,0)$ and $(0,1,0)$ do not commute.

Embed both groups regularly in $S_{p^3}$. Every nonidentity element in either regular representation has cycle type $p^{p^2}$. Hence both embedded subgroups meet the identity class once, the class of that cycle type $p^3-1$ times, and all other classes zero times. They are nonisomorphic Gassmann equivalent regular subgroups. Taking $p=3$ and using the closed four-manifold construction with $T=S_{27}$ gives an explicit pair of isospectral nonhomeomorphic quotients, completing the intended construction under a sufficient hypothesis.

Finally choose distinct odd primes $p_1,\dots ,p_n$ and takeConjugacy classes in a direct product are products of classes, so their intersection counts factor. Thus all $2^n$ subgroups are Gassmann equivalent. They are pairwise nonisomorphic: the unique Sylow $p_j$-subgroup of $U_{ϵ}$ is its $j$th factor, and whether it is abelian records ${ϵ}_j$. Realize $T_n$ as the fundamental group of a closed four-manifold and quotient its universal cover by these subgroups. The binary family of nonhomeomorphic Sunada quotients givesThis existence statement is valid; the earlier universal nonhomeomorphism assertion from Gassmann equivalence alone is not.

A partition of an integer $n$ is a finite sequence ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }>0$ with sum $n$; append zeros when comparing lengths. Its Young diagram has cells $(i,j)$ with $1\le j\le {\lambda }_i$. A Young tableau is a bijective filling of these cells by $1,\dots ,n$. Write $R_t,C_t$ for its row and column stabilizers, and fix the conventionPermutations act on entries on the left, with the rightmost factor acting first. This defines the Young symmetrizer used throughout. In the dictionary order on integer partitions, $\lambda >\mu$ means that at the first differing part ${\lambda }_i>{\mu }_i$; equality is allowed in $\ge$.

Suppose there is no row-column collision between $t$ of shape $\lambda$ and $u$ of shape $\mu$. Each column of $u$ contains at most one entry from each row of $t$. The first $k$ rows of $t$ therefore contain at mostentries, where ${\mu }_j^{\prime }$ is a column length. Thus ${\sum }_{i\le k}{\lambda }_i\le {\sum }_{i\le k}{\mu }_i$ for every $k$: $\mu$ dominates $\lambda$ in dominance order on partitions. If $\lambda \ge \mu$ in dictionary order on integer partitions, a first strictly larger part would contradict the corresponding prefix inequality. Hence $\lambda =\mu$.

All the bounds must now be equalities. Every column $j$ of $u$ contains exactly one entry from row $i$ of $t$ whenever $i\le {\lambda }_j^{\prime }$. Choose $r_0\in R_t$ sending that entry to the entry of $t$ in cell $(i,j)$. These prescriptions are bijections within the rows. The columns of $r_{0}u$ then have exactly the same sets of entries as the columns of $t$, so $r_{0}u=c_{0}t$ for some $c_0\in C_t$. ConsequentlyIf a collision was present instead, its two entries supply the first alternative. This proves the row-column collision lemma, including the prescribed order of the two stabilizer factors.

We next prove the Specht module classification. By Maschke's theorem, the group algebra $A=\mathbb{C}{S}_n$ is a semisimple algebra. Use the permitted basic quasi-idempotence of a Young symmetrizer,and put $e_t=h_t/H_{\lambda }$. The coefficient of $1$ in $h_t$ is one, because $R_t\cap C_t=\{1\}$, so $e_t\ne 0$.

For $\pi \in S_n$, consider $r_t\pi c_t$. A collision between the rows of $t$ and the columns of $\pi t$ gives a transposition $\tau \in R_t\cap \pi C_t{\pi }^{-1}$. Row symmetrization fixes $\tau$, whereas column antisymmetrization changes its sign, so $r_t\pi c_t=0$. If there is no collision, the proved lemma gives $\pi =rc$ with $r\in R_t,c\in C_t$, andIt follows that $h_t\pi h_t$ is zero or $\mathrm{sgn}(c)H_{\lambda }{h}_t$. Since permutations span $A$, $e_{t}A{e}_t=\mathbb{C}{e}_t$. In a semisimple algebra this means that $e_t$ is a primitive idempotent, so $A{e}_t=A{h}_t$ is an irreducible left module.

If $\lambda >\mu$ in dictionary order on integer partitions, the collision lemma applied to every $\pi u$ gives $r_{t}A{c}_u=0$, and therefore $h_{t}A{h}_u=0$. Sincethe two simple modules cannot be isomorphic. Conversely, Young tableaux of the same shape are related by a permutation, which conjugates their Young symmetrizers and gives isomorphic left ideals. Finally, the center of $\mathbb{C}{S}_n$ has the conjugacy-class sums as a basis. Conjugacy classes are indexed by cycle-type partitions, so its dimension is the number of partitions of $n$. The Artin–Wedderburn theorem gives exactly that many simple-module isomorphism classes. We have already produced one for each partition. Thus the $A{h}_{\lambda }$ form a complete set of pairwise nonisomorphic irreducible modules.

For a partition $\nu$ with at most $m$ parts, let $D_{\nu }(V)$ be the Schur module constructed in Question 3, equivalently the image of a Young symmetrizer on $V^{\otimes |\nu |}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max (0,-{\lambda }_m)$ and $\nu =\lambda +k(1,\dots ,1)$. Define the rational Schur module byHere ${\det }^{-k}$ is the one-dimensional representation $g\mapsto \det (g{)}^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original polynomial Schur module. Larger shifts give the same module, as will also follow from the character formula below.

Write $s_r(x)={\sum }_{i=1}^m{x}_i^r$ and $p_{\alpha }(x)={\prod }_{r=1}^n{s}_r(x{)}^{{\alpha }_r}$. For a permutation $\sigma$ of the given cycle type, the trace of a permuted tensor power isIn a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\mathrm{tr}({\xi }^r)$. This proof applies to nondiagonalizable endomorphisms as well. The eigenvalues give $\mathrm{tr}({\xi }^r)=s_r(x)$.

The Schur–Weyl duality decomposition, on which $\sigma$ and ${\xi }^{\otimes n}$ act on the two respective factors, gives the same trace asFor a partition, the character extends polynomially to all endomorphisms because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.

We now derive the alternant character formula for the general linear group. Put $\delta =(m-1,m-2,\dots ,0)$ andUse the permitted symmetric-group character result, the Frobenius alternant character formula:It concerns characters of $S_n$, rather than assuming the character formula we seek for the general linear group. Substitute the proved trace identity and setThen ${\chi }^{\lambda }(\alpha )={\sum }_{\mu }{c}_{\lambda \mu }{\chi }^{\mu }(\alpha )$ for every conjugacy class. Independence of the irreducible symmetric-group characters forces $c_{\lambda \mu }={\delta }_{\lambda \mu }$.

Each character ${\varphi }_{\mu }$ on diagonal matrices is a symmetric homogeneous polynomial of degree $n$, since conjugation by permutation matrices permutes its arguments. Thus $a_{\delta }{\varphi }_{\mu }$ is alternating of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Every alternating polynomial of this degree has a unique expansion in alternants $a_{\lambda +\delta }$: a monomial with repeated exponents has zero coefficient, while each strictly decreasing nonnegative exponent vector is uniquely $\lambda +\delta$ for a partition $\lambda$ of $n$ with at most $m$ parts. Its coefficient at $x^{\lambda +\delta }$ is the coefficient of that alternant. The identities for $c_{\lambda \mu }$ therefore give $a_{\delta }{\varphi }_{\mu }=a_{\mu +\delta }$. We have derived the Weyl character formulaFor partitions the quotient is the Schur polynomial, with removable apparent singularities when eigenvalues coincide. For arbitrary dominant integer tuples, multiply the formula for $\nu$ by $(x_1\cdots x_m{)}^{-k}$; this shifts every numerator exponent by $-k$ and yields the same boxed formula. The variables must then be nonzero. The expression also shows independence of the shift used to define the determinant twist. Equality of characters identifies these irreducible modules: the group-algebra image on a direct sum of two irreducibles is finite-dimensional and semisimple, and its span of group operators detects the traces on every simple block.

Finally, the character of a dual representation evaluates the original character at $x_i^{-1}$. Reverse the numerator's columns after making this substitution. The exponents become $-{\lambda }_{m+1-j}+1-j$. Factoring $(x_1\cdots x_m{)}^{-(m-1)}$ converts these into ${\mu }_j+m-j$ for ${\mu }_j=-{\lambda }_{m+1-j}$. The denominator undergoes the identical column reversal and factor, so both signs and factors cancel. Hence ${\varphi }_{\lambda }(x^{-1})={\varphi }_{\mu }(x)$, andThis tuple is again weakly decreasing, so it is precisely the required dominant label.

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Its full symmetry group of a cube consists of signed permutation matrices. To see completeness, a cube symmetry fixes its center of a group and permutes the six face normals, so it sends coordinate axes to signed coordinate axes; conversely every such matrix preserves the cube. Coordinate permutations and sign changes take any edge to any other edge, so this is a transitive group action on edges.

Choose the edge $e=\{(1,1,z):-1\le z\le 1\}$. A symmetry stabilizing it must preserve its axis direction and its midpoint $(1,1,0)$. It can swap the first two coordinates and independently reverse the third coordinate, but cannot change the signs of the two fixed coordinates. HenceThere are twelve edges and four elements of the stabilizer subgroup. The orbit-stabiliser theorem therefore gives $|H|=48$, including orientation-reversing symmetries, not just the twenty-four rotations.

The action defines a group homomorphism $H\to S_{12}$. If an element fixes every edge as a set, it fixes every vertex, since each vertex is the unique common point of its three incident edges. A cube isometry fixing all vertices is the identity. Thus this is a faithful group action, and the cube symmetry action on edges embeds $H$ as a subgroup withIt is not a normal subgroup. Central inversion $\mathbf{x}\mapsto -\mathbf{x}$ belongs to $H$ and swaps the twelve edges in six opposite pairs. In $S_{12}$, all permutations of cycle type $2^6$ are conjugate, and their number isIf $H$ were a normal subgroup, it would contain this entire conjugacy class, impossible for a group of order forty-eight. This avoids relying on a classification of normal subgroups of the symmetric group.

Use the dihedral group presentation $G=⟨r,s:r^{2n}=s^2=1,srs=r^{-1}⟩$. Every element is $r^j$ or $s{r}^j$. A nonidentity rotation has order two only when its exponent is $n$, while $(s{r}^j{)}^2=1$ for every $j$. Thus the complete list isThe half-turn $r^n$ is central, so its conjugacy class is the singleton $\{r^n\}$ and its normal closure is $\{1,r^n\}$.

For reflections, $r^k(s{r}^j)r^{-k}=s{r}^{j-2k}$ and $s(s{r}^j)s^{-1}=s{r}^{-j}$. Hence the conjugacy class is exactly the $n$ reflections whose exponents have the same parity as $j$:The subgroup generated by this class contains $r^2$, as the product of two consecutive listed reflections, and contains $s{r}^j$. Conversely those two generators contain every listed reflection. Their subgroup is invariant under conjugation by both $r$ and $s$, so the smallest normal subgroup containing $s{r}^j$ isof order $2n$ and index two. This involutions in an even dihedral group classification distinguishes the two reflection classes from the central half-turn.

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

The conjugacy class of $x$ is its orbit under the conjugation action, and its stabilizer is the centralizer $C_G(x)$. The orbit-stabilizer theorem givesBy Lagrange's theorem, the order of the centralizer divides $p^n=|G|$, so the index is a power of $p$. A conjugacy class has size one exactly when its element belongs to the center of a group.

Separate these singleton classes in the class equation:where the sum takes one representative of each nonsingleton conjugacy class. Every summand is divisible by $p$. Thus $p\mid |Z(G)|$, and the identity ensures $|Z(G)|>0$. Consequently $|Z(G)|\ge p$. This is the nontrivial center of a finite p-group property.

Use the unique curvature $-1$ hyperbolic metric on a closed Riemann surface of genus $g\ge 2$, supplied by the uniformization theorem. This is the implicit setting for the partition and spectral theorems; a genus-zero or genus-one surface does not have this hyperbolic metric.

A partition is a pants decomposition by disjoint essential simple closed geodesics, whose complement is a union of pairs of pants. A pair of pants is a sphere with three disks removed, furnished here with geodesic boundary. Its Euler characteristic is $-1$, so there must be $2g-2$ pants. Every pant has three boundaries and every cutting curve occurs twice, giving $3g-3$ cutting curves. The Bers pants decomposition theorem asserts that a constant $B_g$ depending only on $g$ bounds the lengths of all cuffs in some such pants decomposition of every closed genus-$g$ hyperbolic surface.

To see what geometry the cuffs determine, cut each pair of pants along its three perpendicular seams into two congruent right-angled hyperbolic hexagons. The three alternate sides are ${\ell }_1/2,{\ell }_2/2,{\ell }_3/2$, where the ${\ell }_i>0$ are its boundary lengths. The right-angled hyperbolic hexagon identity determines the seam $s_i$ opposite the half-cuff ${\ell }_i/2$ byWe use the standard existence and uniqueness theorem for a right-angled hyperbolic hexagon with prescribed positive alternating side lengths. Thus the three boundary lengths determine the pair of pants up to isometry.

Fix a topological pants decomposition, label its cuffs, and choose reference seam endpoints and orientations. Gluing two boundaries of the same length requires a translation along that boundary; its signed distance is a twist ${\tau }_i$. The gluing graph, the $3g-3$ positive cuff lengths, and the $3g-3$ twists therefore determine the unmarked hyperbolic surface up to isometry. A full boundary translation has period ${\ell }_i$, so for an unmarked gluing one may take $0\le {\tau }_i<{\ell }_i$, with endpoints identified. For a marked surface the whole real twist records the number of Dehn twists, and the Fenchel–Nielsen coordinates areThe topological gluing data are necessary when the decomposition is not fixed. Different markings or decomposition graphs can describe the same unmarked surface.

Teichmüller space ${\mathcal{T}}_g$ consists of pairs $(X,f)$, where $X$ is a closed genus-$g$ Riemann surface and $f:{\Sigma }_g\to X$ is an orientation-preserving marking, modulo the equivalence $(X,f)\sim (Y,h)$ when a conformal orientation-preserving map $a:X\to Y$ has $a\circ f$ homotopic to $h$. The Fenchel–Nielsen coordinates identify it with ${\mathbb{R}}_{>0}^{3g-3}\times {\mathbb{R}}^{3g-3}$, of real dimension $6g-6$. The moduli space of Riemann surfaces forgets the marking by quotienting by the mapping class group.

The Wolpert generic spectral rigidity theorem says that the locus of closed genus-$g$ hyperbolic surfaces possessing an isospectral but nonisometric partner lies in a locally real-analytic exceptional locus of lower dimension in Teichmüller space. Thus a generic surface is determined, up to isometry, by its unmarked length spectrum, equivalently by its Laplacian spectrum. Here an isometry may reverse orientation; the length spectrum cannot distinguish the two orientations of the same metric. The equivalence of the two spectra is the compact hyperbolic Selberg trace formula.

The simplifying result is the Buser finite length spectrum theorem: for every $g\ge 2$ and $\epsilon >0$ there is $L=L(g,\epsilon )$ such that two closed genus-$g$ hyperbolic surfaces with hyperbolic systoles at least $\epsilon$ have the same entire unmarked length spectrum if their length multisets up to $L$ agree. Multiplicities are included. We will count primitive unoriented closed geodesics; counting all iterates is equivalent by successively removing shorter iterates. The theorem is uniform on the thick part of moduli space, not merely a cutoff chosen separately for a particular pair.

Here is a proof using compactness and polynomial trace equations. We state the subsidiary facts and explain their role. First, Mumford's compactness theorem makes the genus-$g$ thick moduli space compact. It can also be seen from the Bers pants decomposition theorem: cuffs in a Bers decomposition lie in $[\epsilon ,B_g]$, there are finitely many pants decomposition graphs, and unmarked twists may be reduced modulo their cuff lengths. These data lie in finitely many compact boxes of Fenchel–Nielsen coordinates and cover the thick moduli space.

Second, these finite compact boxes give a compact family $K$ of marked hyperbolic structures on a fixed smooth surface ${\Sigma }_g$ representing every member of the thick moduli space. We use the standard smooth dependence of the glued metrics and their hyperbolic holonomy representations on Fenchel–Nielsen coordinates. Choose smooth representatives over finitely many parameter neighborhoods and conjugate the hyperbolic holonomy representation using a fixed lifted Riemannian orthonormal frame. Passing to finite compact subboxes gives a compact set of representations, together with a compact family of representative smooth metrics. Changing the markings between boxes causes no problem: all are markings from the same fixed ${\Sigma }_g$, and the union is finite.

Third, uniformization theorem identifies each marked hyperbolic surface with ${\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$ for a faithful discrete hyperbolic holonomy representation into ${\mathrm{PSL}}_2(\mathbb{R})$. We use the standard lifting fact that a closed orientable hyperbolic holonomy representation admits a lift to ${\mathrm{SL}}_2(\mathbb{R})$, and lifts may be chosen continuously on small parameter neighborhoods. Choose lifts on finitely many such neighborhoods, allowing the finitely many possible sign choices. A representation is described by the matrices of $2g$ standard fundamental group generators. The polynomial encoding of hyperbolic geodesic lengths uses, for each fixed group word $w$, its squared matrix traceis a polynomial in their entries: products are polynomial, and the inverse of a determinant-one matrix is its polynomial adjugate. For a nontrivial hyperbolic word, the hyperbolic translation length obeysThus equality of the positive lengths is exactly equality of these trace polynomials, independent of lift signs.

Fourth, the length spectrum of a closed hyperbolic surface is locally finite with finite multiplicities. We need the uniform finiteness of short geodesic classes across the compact marked family. More uniformly, for each finite $b$, only finitely many unoriented primitive conjugacy classes can have length at most $b$ in any metric of $K$. To justify uniformity, compactness of the smooth metrics gives a common bilipschitz equivalence comparison with one fixed metric $h_0$ on ${\Sigma }_g$. If ${\ell }_h(\alpha )\le b$, the minimizing geodesic has $h_0$ length at most $Cb$, so ${\ell }_{h_0}(\alpha )\le Cb$. Local finiteness for $h_0$ then gives a finite list. Local finiteness itself follows from proper discontinuity of the cocompact group of the hyperbolic holonomy representation: conjugate a geodesic axis to meet a fixed compact fundamental set, and a bounded-length axis gives a group element moving that compact set a bounded distance, of which there are only finitely many.

Finally, the Hilbert basis theorem says that a polynomial ring in finitely many real variables is Noetherian. Consequently a decreasing sequence of sets cut out by polynomial equations eventually stabilizes. Indeed, their vanishing ideals form an increasing chain and stabilize. Finite unions of such sets are still algebraic: one can take all products of one defining polynomial from each component. This finiteness fact is the engine that turns arbitrarily many spectral comparisons into finitely many.

Enumerate the primitive unoriented conjugacy classes as ${\alpha }_1,{\alpha }_2,\dots$. For each $n$ letand let $F_n$ be the finite set of primitive classes that can have length at most $b_n$ anywhere in $K$. Define $Z_n$ in the finite-dimensional space of pairs of generator matrices as follows: it consists of pairs $(\rho ,\sigma )$ for which there exists an injectionand also an injection satisfying the analogous equations with $\rho$ and $\sigma$ interchanged. There are only finitely many injections. Each choice imposes finitely many polynomial equations; taking their finite unions and then intersecting the two directions proves that $Z_n$ is a real affine algebraic set. On determinant-one representations these equations are exactly the desired matching of lengths. Defining the polynomials on the ambient matrix-entry space is harmless; the argument will only apply them to representations in $K$.

Put $W_n=Z_1\cap \cdots \cap Z_n$. The sets $W_n$ form a decreasing chain of affine algebraic sets, hence for some $N$,For representations in $K$, membership in this intersection is equivalent to equality of the full length spectra. One direction follows by matching equal lengths, including their multiplicities. For the other, fix a length $\ell$. All primitive classes of length $\ell$ on the first surface occur in some finite initial segment. Its injective matching shows that the second surface has at least that multiplicity at $\ell$; the reverse injection gives the reverse inequality. Local finiteness makes both multiplicities finite. Doing this for every $\ell$ proves multiset equality.

TakeIf two surfaces in $K$ have equal length multisets up to $L$, each initial segment through $n\le N$ can be matched injectively on the other surface. Its matching classes lie in $F_n$, since the matched lengths are at most $b_n$. The reverse matching is available as well. Thus the pair belongs to $W_N$, hence to every $W_n$, and the preceding paragraph yieldsAll choices of the compact family were made using only $g$ and $\epsilon$, so the resulting cutoff has the asserted dependence. If the thick family is empty the assertion is vacuous. This proves Buser's finite length spectrum theorem without claiming that equal full length spectra always force isometry; exceptional isospectral pairs are compatible with the theorem and are precisely why Wolpert's theorem is a generic statement.

Maschke's theorem makes the group algebra $\mathbb{C}G$ semisimple, since $G$ is a finite group and the ground field has characteristic zero. By the Artin–Wedderburn theorem, write $\mathbb{C}G\cong {\prod }_{i=1}^r{M}_{d_i}(\mathbb{C})$. Its simple modules give exactly the $r$ nonisomorphic irreducible complex group representations.

The center of each matrix algebra consists of scalar matrices, so ${\dim }_{\mathbb{C}}Z(\mathbb{C}G)=r$. On the other hand, an element ${\sum }_g{a}_{g}g$ is central precisely when its coefficients are constant on conjugacy classes. The sums of the elements in the separate conjugacy classes are therefore a basis of the center. Hence the number of irreducible complex representations equals the number of conjugacy classes.

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.

For a finite group, the conjugacy classes partition the group, and a class is a singleton exactly when its element belongs to the centre of a group. Since the centre is trivial, the class equation becomeswhere the ${\mathcal{C}}_j$ are all the nonidentity conjugacy classes. If the prime number $p$ divided every $|{\mathcal{C}}_j|$, reduction modulo $p$ would give $0=1$, because $p\mid |G|$. Therefore at least one of these class sizes is not divisible by $p$. Primality now givesIts size is greater than one because the centre is trivial. This proves the prime-to-p conjugacy class lemma.

The conclusion requires $p$ to be prime. The printed question does not explicitly impose this hypothesis. If arbitrary composite divisors are allowed, take the symmetric group $S_3$ and $p=6$: its centre is trivial, its two nonidentity conjugacy classes have sizes $3$ and $2$, and neither is coprime to $6$. Thus the argument proves the intended prime case and also identifies why the unrestricted literal reading is false.