= Solution
For the constant-curvature constructions, the three data sets lead respectively to genera two, three and four. The first two pairs are isometric through a lifted reflection; the third gives nonisometric pairs under the length-identification hypothesis stated below. Nonconjugacy of the two subgroups inside the finite group is not by itself a test for nonisometry.
Replace each $T_i$, if necessary, by the joint image of its two coset actions, generated by the corresponding pair of generator permutations. The joint kernel lies in both subgroups, so passing to this finite quotient leaves the two coset actions unchanged. We use these faithful joint images in the constructions below.
First verify the <Gassmann equivalence> needed for the <Sunada theorem> for all group elements, not just the displayed generators. Write $P_{A,U},P_{B,U},P_{A,V},P_{B,V}$ for their <permutation matrices>. Invertible matrices $C_i$ satisfying
$$
P_{A,U}C_i=C_iP_{A,V},\qquad P_{B,U}C_i=C_iP_{B,V}
$$
intertwine every word in the generators, hence give equal <permutation characters>. Here are explicit certificates calculated from the coset actions. Define $C_{kj}=1$ when column label $j$ belongs to the $k$th support below, and zero otherwise. For $i=1$, labels run from one to twelve; the row supports in their order are
$$
\begin{gathered}
\{5,8,10,11\},\ \{6,9,11,12\},\ \{4,7,10,12\},\ \{1,8,9,12\},\\
\{2,7,9,10\},\ \{3,7,8,11\},\ \{2,3,5,12\},\ \{1,3,6,10\},\\
\{1,2,4,11\},\ \{3,4,5,9\},\ \{1,5,6,7\},\ \{2,4,6,8\}.
\end{gathered}
$$
For $i=2$ and $i=3$, labels run from zero to six, and the row supports are respectively
$$
\begin{gathered}
\{0,2,3,4\},\ \{1,3,4,5\},\ \{2,4,5,6\},\ \{0,3,5,6\},\\
\{0,1,4,6\},\ \{0,1,2,5\},\ \{1,2,3,6\};\\[3pt]
\{0,2,5,6\},\ \{0,1,3,6\},\ \{0,1,2,4\},\ \{1,2,3,5\},\\
\{2,3,4,6\},\ \{0,3,4,5\},\ \{1,4,5,6\}.
\end{gathered}
$$
Direct substitution gives the two intertwining identities, and elimination gives $\det C_1=-1024$, $\det C_2=\det C_3=32$. These nonzero determinants certify equivalent linear <permutation representations>. Equal characters imply almost conjugacy: for a subgroup $U$,
$$
\chi_{T/U}(g)=\frac{|C_T(g)|}{|U|}\,|[g]\cap U|,
$$
because a fixed coset corresponds to $t^{-1}gt\in U$, and each specified conjugate has $|C_T(g)|$ conjugating elements. The subgroup orders agree since their indices agree, so equality of these characters gives equal intersection counts with every conjugacy class.
Nonconjugacy can also be checked directly: a bijective equivariant sheet relabelling would be determined by the image of one label, since the actions are transitive. Propagating each of the twelve or seven possible images by the two generators gives no consistent bijection in any of the three cases.
For genus two, use $i=1$. The generator orders are $3,3$, and their product has order six; on both twelve-sheet coset spaces their cycles are respectively $3^4,3^4,6^2$. Map the <hyperbolic triangle group> $\Delta(3,3,6)$ onto the joint finite permutation image $T_1$. Its kernel is torsion-free because these orders are exact, giving a smooth normal cover $M$ of the sphere orbifold with cone orders $3,3,6$. The subgroups act freely by the <smoothness criterion from cone-monodromy cycles>, since all coset cycles have their full orders. The two quotient surfaces therefore have
$$
\chi=12\left(-1+\frac13+\frac13+\frac16\right)=-2,\qquad\boxed{g=2}.
$$
Their equal <permutation characters> give <isospectrality> by the <Sunada theorem>.
There is nevertheless an <isometry>. The sheet relabelling
$$
R_1=(1\ 2)(4\ 6)(7\ 10)(8\ 12)(9\ 11)
$$
fixes labels three and five and satisfies
$$
R_1A_U R_1^{-1}=A_V^{-1},\qquad R_1B_U R_1^{-1}=B_V^{-1}.
$$
To see the geometric significance, write the triangle-group generators as $a=s_1s_2$, $b=s_2s_3$, where $s_j$ reflect in triangle sides. Conjugation by $s_2$ sends both $a,b$ to their inverses. The two displayed permutation identities say exactly that reflecting the triangles and relabelling the sheets preserves their gluings. This <reflection intertwining of triangle-cover coset actions> lifts the base reflection to an <orientation-reversing isometry> of the quotient surfaces. Thus \b[the genus-two pair from these hyperbolic triangles is isometric], despite the subgroups being nonconjugate in the orientation-preserving finite group.
For genus three, use $i=2$. Both generators and their product act in seven-cycles, so the same construction with $\Delta(7,7,7)$ gives smooth seven-sheet quotients and
$$
\chi=7\left(-1+\frac17+\frac17+\frac17\right)=-4,\qquad\boxed{g=3}.
$$
Their <isospectrality> follows from the certificate $C_2$. The relabelling $R_2(j)=4-j\pmod7$, namely $(0\ 4)(1\ 3)(5\ 6)$ with label two fixed, again satisfies $R_2A_U R_2^{-1}=A_V^{-1}$ and $R_2B_U R_2^{-1}=B_V^{-1}$. Therefore \b[this constant-curvature genus-three pair is also isometric] by the same lifted triangle reflection. This assertion concerns the constructed triangle metrics, rather than arbitrary metrics placed independently on the two topological surfaces.
For genus four, use $i=3$ and a different base: a hyperbolic torus with one cone point of order seven. Its <orbifold fundamental group> has presentation
$$
\langle a,b\mid[a,b]^7=1\rangle.
$$
Map $a,b$ to $A_3,B_3$. Multiplying the given permutations shows that their commutator is a seven-cycle in both coset actions, so this map has torsion-free kernel and both subgroup quotients are smooth. The base <orbifold Euler characteristic> is $-(1-1/7)=-6/7$, giving
$$
\chi=7(-6/7)=-6,\qquad\boxed{g=4}.
$$
The certificate $C_3$ and the <Sunada theorem> give an isospectral pair for each base metric. This is the <cone-torus construction of genus-four Sunada surfaces>.
For nonisometry, choose simple closed base <geodesics> $\alpha,\beta$ representing $a,b$, avoiding the cone point and meeting once. We use the following additional geometric hypothesis, as permitted in this question: choose a cone-torus metric such that, in each cover, the length $4\ell(\alpha)$ identifies its unique degree-four lift, and the length $3\ell(\beta)$ identifies precisely its two degree-three lifts, among all primitive closed geodesics. This is a generic length-separation choice; it excludes coincidences with other geodesic lengths. No claim about an arbitrary symmetric choice of base metric is needed.
A lift component of a base <geodesic> corresponds to a cycle of its <monodromy permutation>. The degree-four $A_3$ cycle is $\{2,6,4,5\}$ on the $U_3$ sheets and $\{0,1,6,3\}$ on the $V_3$ sheets. The degree-three $B_3$ cycles are $\{1,2,5\},\{3,6,4\}$ on the first cover and $\{1,4,3\},\{2,5,6\}$ on the second. Since the base curves meet once, their lifted geometric intersection numbers are the sizes of the intersections of these sheet-label sets. The <cycle intersections count intersections of lifted curves> therefore gives the two unordered multisets
$$
\{2,2\}\quad\text{on the }U_3\text{ cover},\qquad\{2,1\}\quad\text{on the }V_3\text{ cover}.
$$
Negative curvature ensures that the geodesic representatives realize geometric intersection number. An <isometry> would preserve the length-identified degree-four lift and the set of the two length-identified degree-three lifts, even if it exchanged the latter, so it would preserve this unordered multiset. The multisets differ. The <intersection test for nonisometric finite covers> thus proves \b[the genus-four pair is nonisometric for this choice of constant-curvature metric]. Without a condition excluding relevant accidental length coincidences, subgroup nonconjugacy alone would not establish that conclusion.
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