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Take a finite presentation $T=⟨x_1,\dots ,x_r\mid R_1,\dots ,R_s⟩$. For a closed-manifold realization of a finitely presented fundamental group, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with fundamental group the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ hasby the Seifert-van Kampen theorem. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this handle decomposition upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the fundamental group: there are no new one-handles or two-handle relations. Thus ${\pi }_1(B)\to {\pi }_1(W)$ is an isomorphism andThese are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed fundamental group.

Now assume $T$ finite. Choose any smooth Riemannian metric on $B$ and lift it to its universal cover $\tilde{B}$. This cover is compact, since its degree is $|T|$, and simply connected. Its deck transformations give a free isometric $T$-action. For the two Gassmann equivalent subgroups putEach is a closed Riemannian manifold, and $\tilde{B}$ is its simply connected universal cover, with deck transformation group $U_i$. Consequently ${\pi }_1(M_i)\cong U_i$.

Here is the complete spectral argument behind the Sunada theorem in this situation. Each finite-dimensional upstairs eigenspace $E_{\lambda }$ is a representation of $T$, because the action is isometric and commutes with the Laplace-Beltrami operator. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_{\lambda }^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its trace givesThe character ${\chi }_{\lambda }$ is constant on each conjugacy class of $T$. Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. The quotients are isospectral, with ${\pi }_1(M_i)\cong U_i$.

If $U_2=t{U}_1{t}^{-1}$, the map $[x{]}_{U_1}\mapsto [tx{]}_{U_2}$ is well defined, invertible and isometric because $t$ acts by an isometry upstairs. Conjugate subgroups give isometric quotients.

For an explicit example with different fundamental groups, use the order-sixteen Gassmann pairEmbed each in $T=S_{16}$ by left multiplication on its own sixteen elements, identified with the same sixteen-point set. These regular representations are faithful. Both groups have one identity element, three elements of order two, and twelve of order four: in $C_4^2$ the elements of order dividing two are precisely its four even-coordinate elements; in $Q_8\times C_2$ they are the four elements in $\{1,-1\}\times C_2$. Every remaining element has order four.

In a regular action, an order-$k$ element is a product of $16/k$ disjoint $k$-cycles. Thus each subgroup intersects the identity conjugacy class of $S_{16}$ in one element, the class of type $2^8$ in three, the class of type $4^4$ in twelve, and every other class in zero. They are Gassmann equivalent. However $H_1$ is abelian and $H_2$ is nonabelian, so they are not isomorphic. Apply the preceding construction with fundamental group $S_{16}$ and these subgroups. It producesIn particular they cannot even be homeomorphic, and therefore cannot be isometric.