Gassmann equivalence

Subgroups $H_1,H_2$ of a finite group $G$ are Gassmann equivalent if $|H_1\cap C|=|H_2\cap C|$ for every conjugacy class $C$. Equivalently their coset permutation representations have the same character of a representation. They need not be conjugate subgroups. The Sunada theorem turns this equality into equality of Laplacian eigenfunction multiplicities.

In a finite general linear group, the permutation representations on nonzero vectors and on nonzero dual vectors have equal characters: an element fixes $q^{\dim \ker (t-I)}-1$ vectors in each action because a matrix and its transpose have the same rank. In characteristic two the stabilizers of a nonzero vector and a hyperplane are therefore almost conjugate subgroups. In dimension three over ${\mathbb{F}}_2$, both have order $24$ and index $7$ in a group of order $168$, but they are not conjugate: a point stabilizer preserves no plane. A triangle cover construction for Sunada surfaces with generator orders $(7,7,7)$ gives quotient surfaces of genus three.

Embed $C_4\times C_4$ and $Q_8\times C_2$ in $S_{16}$ by their regular representations. Both have one element of order one, three of order two, and twelve of order four. An order-$r$ element in a regular action has cycle type $r^{16/r}$. Their intersections with every conjugacy class of $S_{16}$ consequently have the same size, so they are Gassmann equivalent. The first group is abelian and the second is not, so their underlying groups are not isomorphic. Finite-cover constructions from the Sunada theorem therefore give isospectral manifolds with different fundamental groups.

For an odd prime $p$, embed the elementary abelian group $(C_p{)}^3$ and the Heisenberg group over a prime field regularly in $S_{p^3}$. Every nonidentity element has order $p$ and hence regular permutation cycle type $p^{p^2}$. The two subgroups have equal intersection counts with every symmetric-group conjugacy class, so they are Gassmann equivalent, but they are not isomorphic because one is abelian and the other is not. Using a compact simply connected isometric cover with ambient deck group $S_{p^3}$ gives quotients with these nonisomorphic fundamental groups; Sunada theorem makes them isospectral while their topology differs.