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For a finite group generated by $a,b$, take the regular covering of a thrice-punctured sphere with monodromy $a,b,(ab{)}^{-1}$ and fill each lifted puncture by a disk. Locally the projection is $z\mapsto z^m$, where $m$ is the order of its monodromy. The result is a closed oriented topological surface with an orientation-preserving group action. The nontrivial stabiliser subgroups are the conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$ at the three types of ramification points. Counting lifted cells gives the displayed Euler characteristic. A subgroup acts freely exactly when it meets every conjugate of these three cyclic groups trivially. Its quotient is then a regular covering of degree equal to the subgroup order, and quotient Euler characteristic is divided by that order.

Every finitely generated group acts freely and properly discontinuously by Riemannian isometries on a connected manifold in any dimension $d\ge 3$. Choose a surjection $F_r\to T$ and the regular covering of ${\#}_r(S^1\times S^{d-1})$ associated with its kernel. Pulling back a metric makes the deck transformation group isometric. Uniqueness of lifts makes its action free. The cover can be noncompact and need not be simply connected; no claim about realizing every group as a closed three-manifold fundamental group is involved.

Subgroups $U,V$ of a finite group $T$ have Gassmann equivalence whenSumming over the conjugacy classes gives $|U|=|V|$. Equivalently, their coset permutation representations have the same character of a representation. In particular, almost conjugate subgroups need not be conjugate.

Put $p=\mathrm{ord}A$, $q=\mathrm{ord}B$, and $r=\mathrm{ord}(AB)$. Begin with the sphere punctured at three points. Its fundamental group is $⟨a,b,c\mid abc=1⟩$, and the epimorphism sending $a,b,c$ to $A,B,(AB{)}^{-1}$ defines a connected regular $|T|$-sheeted covering space. Around the punctures the monodromies have cycles of lengths $p,q,r$. Fill each lifted puncture by a disk, using the local branched covering models $z\mapsto z^p$, $z\mapsto z^q$ and $z\mapsto z^r$. This produces a closed oriented surface $M$, with $T$ acting by homeomorphisms that are smooth in these local models. This construction works even if one of $p,q,r$ is one.

Triangulate the base sphere with the three marked points as vertices and two faces. Lifting this triangulation givesso the Euler characteristic isIn the hyperbolic case $1/p+1/q+1/r<1$, this is equivalently the triangle cover construction for Sunada surfaces. Take the hyperbolic triangle groupand its epimorphism $\theta :\Delta \to T$ with the same images. All finite-order elements of $\Delta$ are conjugate to powers of its three elliptic generators. Because their images have the same exact orders, $K=\ker \theta$ is torsion-free and $M=K\backslash {\mathbb{H}}^2$. It can be assembled from $2|T|$ hyperbolic triangles of angles $\pi /p,\pi /q,\pi /r$. Both examples below belong to this hyperbolic case.

For $M_1=U\backslash M$ and $M_2=V\backslash M$ to be smooth surfaces normally covered by $M$, the subgroup actions must be free. The precise condition is that neither subgroup meet any conjugate in $T$ of a nonidentity power of $A$, $B$ or $AB$. These are exactly the possible point stabilizers. Then $M\to M_1$ and $M\to M_2$ are regular coverings, with deck transformation groups $U,V$ respectively. No normality of $U,V$ in $T$ is needed: normality would concern the further covering of the base orbifold, not these coverings from $M$. For a free subgroup $W$,

Choose any $T$-invariant smooth Riemannian metric on $M$, for example by averaging an arbitrary metric. It descends to both quotients. On functions let $R_t$ be the unitary action of $t\in T$, and let $E=e^{-s{\Delta }_M}$ for $s>0$. The averaging operator $P_W=|W{|}^{-1}{\sum }_{w\in W}{R}_w$ projects orthogonally onto the $W$-invariant functions, which identify with functions on $W\backslash M$. The trace formula underlying the Sunada theorem isThe expression $\mathrm{Tr}(R_w{e}^{-s{\Delta }_M})$ is a class function, because ${\Delta }_M$ commutes with $T$. Thus Gassmann equivalence makes the two sums identical. Alternatively, apply the same averaging formula to each finite-dimensional Laplacian eigenspace: its invariant dimension is $|W{|}^{-1}{\sum }_{w\in W}{\chi }_{\lambda }(w)$. This proves equality of every eigenvalue multiplicity and hence isospectrality of the two quotient metrics.

If $V=tU{t}^{-1}$ for some $t\in T$, the action of $t$ gives an isometry between the quotients for every such metric. Nonconjugacy is therefore necessary. To make it sufficient, choose the quotient orbifold metric generically so that its regular locus has no nonidentity local isometries; this is the equivariant, or orbifold, version of the Sunada local isometry lemma. Smooth perturbations on regular coordinate balls provide this genericity while their lifts remain smooth on $M$. Any isometry between $M_1$ and $M_2$, when composed locally with their projections, would then identify the same points of the base regular locus. By continuation it is an isomorphism of their coverings of that locus, and the classification of connected covering spaces forces $U,V$ to be conjugate in $T$. Thus nonconjugate subgroups give non-isometric isospectral quotients for a suitable generic invariant metric. The genericity belongs to the base regular locus; the covering metric on $M$ itself necessarily has the prescribed deck symmetries.

For genus two, an explicit choice is the following permutation group on twelve labels, with permutation products composed from right to left:These parameters are those in Kang's genus-two construction. They admit a direct finite verification. Starting with $T_0=\{1\}$ and iterating $T_{j+1}=T_j\cup T_{j}A\cup T_{j}B$, the successive sizes are $1,3,7,13,23,39,61,81,93,96$, after which the set is unchanged. Thus $|T|=96$, and direct multiplication gives $(p,q,r)=(3,3,6)$ and $|U|=|V|=8$. The conjugacy-class triples $(|C|,|C\cap U|,|C\cap V|)$, obtained as $C=\{tx{t}^{-1}:t\in T\}$, areThis proves Gassmann equivalence. The subgroup $V$ fixes the labels $5,11$, whereas $U$ has no common fixed label, so they cannot be conjugate even in the full symmetric group. Both are groups of order eight, so they meet no nonidentity order-three element. The only order-two power of $AB$ iswhich is central in $T$ and lies in neither subgroup. This also excludes its conjugates and proves freeness. The quotient Euler characteristics and genera are thereforeThe covering surface $M$ has genus nine.

For genus three, use $T={\mathrm{GL}}_3({\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. Let $U$ stabilize a nonzero vector and let $V$ stabilize a two-dimensional subspace. Both have index seven and order twenty-four. An element $t$ fixes $2^{\dim \ker (t-I)}-1$ nonzero vectors. Hyperplanes correspond to nonzero dual vectors over ${\mathbb{F}}_2$, so the number fixed in the dual action is $2^{\dim \ker (t^T-I)}-1$, the same number. Equality of these permutation characters proves that the point and hyperplane stabilizers are almost conjugate.

They are not conjugate. In coordinates where $U$ fixes $e_1$, its elements include all matrices $\left(\begin{array}{cc}1& \alpha \\ 0& C\end{array}\right)$ with arbitrary row $\alpha$ and $C\in {\mathrm{GL}}_2({\mathbb{F}}_2)$. An invariant plane containing $e_1$ would project to a line invariant under every $C$, which is impossible. A plane not containing $e_1$ would be the graph of a linear functional on the quotient, and arbitrary $\alpha$ would move that graph. Thus $U$ stabilizes no plane, whereas every conjugate of $V$ does.

Choose the concrete generatorsOver ${\mathbb{F}}_2$, each of $A,B,AB$ has order seven. The same closure iteration has sizes $1,3,7,15,31,59,103,144,168$, verifying that $A,B$ generate $T$. Since seven does not divide $24$, neither stabilizer contains a nonidentity power of these generators, or any conjugate of such a power. We may therefore use $(p,q,r)=(7,7,7)$ in the triangle cover construction for Sunada surfaces. The quotient calculation isHere the covering surface $M$ has genus forty-nine. Choosing the generic invariant metrics described above completes both requested pairs.

Let $p=|a|$, $q=|b|$ and $r=|ab|$ denote element orders. Remove three points from an oriented sphere. Its fundamental group has peripheral generators ${\gamma }_a,{\gamma }_b,{\gamma }_c$ with ${\gamma }_a{\gamma }_b{\gamma }_c=1$. Send them to $a,b,(ab{)}^{-1}$. Since $a,b$ generate $T$, the kernel defines a connected regular covering of the punctured sphere with degree $|T|$ and deck transformation group $T$.

Complete the lifted punctures by disks. Above the three removed points the monodromy cycles have lengths $p,q,r$, respectively, and the disk projection has the local form $z\mapsto z^p$, $z^q$ or $z^r$. This gives a closed connected oriented topological surface $M$ and an orientation-preserving $T$-action. At the three kinds of filled-in points the stabiliser subgroups are conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$, respectively. Every other point has trivial stabiliser subgroup. Orders equal to one simply correspond to unbranched filled points.

The punctured sphere has Euler characteristic $-1$, so its cover has Euler characteristic $-|T|$. Filling the ends adds $|T|/p+|T|/q+|T|/r$ disks. This proves the formula for the branched surface action from two generators:

For a subgroup $U$, its quotient map is always regular as a branched quotient. It is an ordinary unbranched normal covering map exactly when $U$ acts freely, that is,Necessity follows from the listed stabiliser subgroups. Conversely, these conditions eliminate every possible nontrivial point stabiliser. A free finite action gives a covering space, and its deck transformations are $U$, acting transitively on each fibre; hence the cover is normal. In this caseThe condition $U⊴T$ concerns regularity of the different map $U\backslash M\to T\backslash M$; it is not needed for $M\to U\backslash M$ when the action of $U$ is free.

For the specified order-$96$ group, $p=q=3$. Set $c=ab$ and $z=c^3$. Since $z$ has order two, $c^6=1$ and $c^3\ne 1$, so $|c|$ is either two or six. The order-two alternative would make the displayed formula give $\chi (M)=96(1/3+1/3+1/2-1)=16$. That is impossible for a connected closed oriented topological surface, whose Euler characteristic is $2-2g\le 2$. Thus $|c|=6$, and

Each $U_i$ has order eight. By Lagrange's theorem, it contains no element of order three or six, so it meets all conjugates of $⟨a⟩$ and $⟨b⟩$ trivially. In $⟨c⟩$, the only nonidentity element whose order divides eight is $z=c^3$. Since $z$ is central, it is also the unique such element in every conjugate of $⟨c⟩$, and it is excluded from both $U_i$ by hypothesis. Both subgroup actions are therefore free. The quotients are closed oriented surfaces with

The branched-cover construction supplies a smooth structure, with the stabilisers acting by rotations in the disk charts. Average a smooth Riemannian metric over $T$ to make the whole action isometric. The Gassmann equivalence of $U_1,U_2$ then gives isospectral manifolds: for every upstairs eigenspace with character ${\chi }_{\lambda }$, its downstairs multiplicities are $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, and these are equal because the two subgroups have equal intersection sizes with every conjugacy class. This proves the spectral assertion for any such invariant metric.

To guarantee nonisometry, choose the invariant metric with curvature markers distinguishing finite-cover quotients. Here are the geometric details, since nonconjugacy alone would not guarantee nonisometry for an arbitrary metric. Choose a small disk in the free locus of the $T$-action, with all its translates disjoint. In that disk choose three noncollinear points in a geodesically convex neighbourhood. Supported conformal smooth bump functions can give them three distinct high values as strict Gaussian curvature maxima, occurring nowhere else in the unmodified metric. Arrange these three maxima as one tight cluster, much closer to each other than to any translated cluster, and copy the same modification to every translate. Curvature and its first few derivatives can be adjusted by the second and higher derivatives of the conformal factor; narrow bumps give high distinguished curvature with small change in metric distances. Further small supported perturbations remove accidental maxima of these exact heights. Thus the resulting metric remains smooth and $T$-invariant, and its marked clusters are intrinsic and uniquely labelled by their three curvature values.

Any proposed quotient isometry $F:U_1\backslash M\to U_2\backslash M$ must carry a whole labelled cluster to another. Lift its restriction near that cluster to the corresponding two marked disks in $M$, and identify these disks using the appropriate group element $t\in T$. The resulting local isometry fixes all three marked points. Its differential at the first point fixes the two independent directions of the unique geodesics to the other two; hence it is the identity, and the local isometry itself is the identity near that point. If $p_i:M\to U_i\backslash M$ are the quotient maps, we therefore haveon an open set. Both sides are local isometries on the connected complete Riemannian surface $M$. A local isometry is determined along each geodesic by its value and differential at the initial point, so the equality propagates to all of $M$.

For $u\in U_1$, this equality implies $p_2\circ tu=p_2\circ t$. Evaluating at a free point of the $T$-action shows $tu{t}^{-1}\in U_2$; equivalently these maps differ by a deck transformation of $p_2$. Thus $t{U}_1{t}^{-1}\subseteq U_2$, and equality of subgroup orders gives equality. This contradicts their assumed nonconjugacy. For this marked invariant metric the two quotients are nonisometric, isospectral Riemannian surfaces of genus two. No constant-curvature requirement is imposed or used.

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.

Let $G$ be generated by $r$ elements and choose $h\ge \max \{2,r\}$. The fundamental group of a closed orientable surface has the presentationMapping $a_1,\dots ,a_r$ to generators of $G$ and every $b_i$ and remaining $a_i$ to the identity defines a surjective group homomorphism ${\pi }_1(S_h)\twoheadrightarrow G$.

Its kernel determines a connected regular covering $S\to S_h$ with deck transformation group $G$. The action is free and orientation preserving. The covering surface has genus at least two, and the lifted hyperbolic metric makes every deck transformation an isometry. Part a now gives $G\hookrightarrow \mathrm{Mod}(S)$. This is the realization of a finite group as a surface deck group.