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Use the coordinate conventionIn this convention the Heisenberg group over a prime field has the productThe subset $H=\{A(t,b,t):t,b\in {\mathbb{F}}_p\}$ is an Abelian subgroup of order $p^2$: the middle coordinate in a product is $b+b^{\prime }+t{t}^{\prime }$, which is symmetric in the factors. Also $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ lies in $H$. The map $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_p$, with kernel of a group homomorphism $H$. Hence $H$ is a normal subgroup of subgroup index $p$, and its left cosets are exactly the level sets of $a-c$.

For odd $p$, an explicit group isomorphism from $H$ to the additive group ${\mathbb{F}}_p^2$ isThe term $t{t}^{\prime }$ in matrix multiplication cancels the cross term in $(t+t^{\prime }{)}^2/2$. Thus $H$ is an elementary abelian group, isomorphic to $C_p\times C_p$.

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.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is

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.

Write $A(a,b,c)$ with $a$ in entry $(1,2)$, $b$ in entry $(1,3)$ and $c$ in entry $(2,3)$, and perform all arithmetic in the finite field ${\mathbb{F}}_5$. Direct multiplication givesThus the set is closed. Associativity comes from matrix multiplication. The identity element is $A(0,0,0)$, and solving the product formula for it givesBoth orders of multiplication with this inverse give the identity. Hence these matrices form a finite group, an instance of the Heisenberg group over a prime field. Each coordinate has five independent possibilities, with distinct triples giving distinct matrices, soThe explicit coordinate convention matters because the central coordinate is $b$ here; other descriptions of the Heisenberg group over a prime field may place the central coordinate last.