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\section{The groups \texorpdfstring{$\tg{2}{2n}$}{Z/2Z x Z/2nZ}}\label{sec:2x2n}
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The groups $\tg{2}{2n}$ treated here are those with $2n\in\{32,34,36,38,40,42,50\}$, and we show that none
of them is a subgroup of the torsion of an elliptic curve over a number field of degree $10$. By
\Cref{lem:isogeny} this also excludes the cyclic groups $\torz{64}$, $\torz{68}$, $\torz{72}$, $\torz{76}$,
$\torz{80}$, $\torz{84}$ and $\torz{100}$. The curves $X_1(2,2n)$ with $2n\in\{32,34,36,40,42\}$ were used
in \cite[Section 11]{N69} over number fields of degree at most $9$, and we follow the method used there. In all these cases the
Jacobian of $X_1(2,2n)$ has rank $0$ over $\Q$, and we study the reduction of the trace divisor modulo $p=3$ or
$p=5$. The entirely cuspidal reductions are excluded by \Cref{prop:main}, which needs $\gon_\Q X_1(2,2n)>10$.
For $2n\in\{32,34,36\}$ Abramovich's bound does not give this, and we prove it in
\Cref{lem:2x2n-32gon,lem:2x2n-34gon,lem:2x2n-36gon}: for $2n=32$ with the Castelnuovo-Severi inequality, and
for $2n=34,36$ by specializing a hypothetical pencil of degree $10$ modulo $p$ (\Cref{lem:2x2n-pencil}) and
computing the short vectors in the lattice of divisors of modular units. The other reductions are excluded by
pushing them forward to curves $X_1(m)$ and $X_1(2,m)$ with $m\mid 2n$, to $X_0(42)$ and to quotients of
$X_1(2,2n)$ by groups of level automorphisms (\Cref{lem:classtest} and \eqref{eq:pushclass}), by the Hecke
sieve, and by showing that a level automorphism which fixes the reduction of the trace divisor also fixes the
trace divisor, or the cuspidal part of its reduction (\Cref{lem:section}).

Let $X=X_1(2,N)$ with $N=2n$. It parametrizes triples $(E,P,Q)$ with $P$ of order $2$ and $Q$ of order $N$ such
that $P\neq nQ$. If $n$ is even, the maps
\begin{equation}\label{eq:2x2n-aut}
(E,P,Q)\longmapsto\bigl(E,P+n\varepsilon_1Q,\ \varepsilon_2P+aQ\bigr),\qquad \varepsilon_1,\varepsilon_2\in\{0,1\},\
a\in(\torz{N})^\times,
\end{equation}
are level automorphisms, and taken modulo $-1$ they form a group of $2\varphi(N)$ automorphisms of $X$
\cite[Section 11]{N69}. If $n$ is odd, put $P_1=P$, $P_2=nQ$ and $P_3=P+nQ$. Then the maps
$(E,P,Q)\mapsto(E,P_i,P_j+2aQ)$, with $i\neq j$ and $a\in(\torz{n})^\times$, are level automorphisms, and taken
modulo $-1$ they form a group of $3\varphi(n)$ automorphisms of $X$, isomorphic to $\SL_2(\F_2)\times\Delta(n)$.
For $N=32,34,36,38,40,42,50$ these groups have $32,48,24,54,32,36,60$ elements. These level automorphisms are
defined over $\Q$, so they map an effective rational divisor $D$ of degree $10$ whose support contains no cusp to
another such divisor. Hence in the sieves below we may replace $D$ by its image under a level automorphism, and, as
in \Cref{subsec:sieve}, it suffices to test one representative of each orbit of the first closed point occurring in
the reduction of $D$, together with all the possibilities for the other closed points.

We also use the four maps $\pi_1,\dots,\pi_4:X\to X_1(N)$ which keep $Q$ or $P+Q$, or divide $E$ by $\dm P$ or
$\dm{P+nQ}$ and keep the image of $Q$, which has order $N$ in all cases. Sending $(E,P,Q)$ to $(E/\dm P,R)$,
where $R$ is one of the two points with $\widehat\phi(R)=Q$ for the dual $\widehat\phi$ of the quotient map
$\phi:E\to E/\dm P$, identifies $X$ over $\Q$ with the modular curve $X_\Delta(2N)$ of the group
$\Delta=\{\pm1,\pm(N+1)\}\subseteq(\torz{2N})^\times$. Here $R$ has order $2N$ as $2R=\phi(Q)$ has order $N$,
the two choices of $R$ differ by $NR$, and $\widehat\phi$ recovers $(E,P,Q)$. We describe points of $X$ over
finite fields by the model
$$y^2=x^3+\gamma x^2+\beta(1-\beta-\gamma)x,\qquad P=(0,0),\quad Q=(\beta,\beta).$$
All the class groups of curves over finite fields that we use are computed with Magma's proven class group
algorithm.

By \cite[Theorem 3.1(3)]{DEHMZ}, $\rk J_1(2,N)(\Q)=0$ for all the $N$ above. For $N=38,40,42,50$ we have
$i(2,N)=1080,1152,1152,1800$, so $\gon_\C X\geq\frac{325}{2^{15}}\,i(2,N)>10.7$, and hence
$\gon_\Q X\geq\gon_\C X\geq11$. For $N=32,34,36$ we have $i(2,N)=768,864,864$, and we prove $\gon_\Q X\geq12$,
$11$, $11$, respectively, in \Cref{lem:2x2n-32gon,lem:2x2n-34gon,lem:2x2n-36gon}. So $\gon_\Q X>10$ in all
cases, and by \Cref{prop:main}~a) the reduction of the trace divisor modulo an odd prime $p\nmid N$ is not
entirely cuspidal. In \Cref{tab:2x2n-points} we list, for each curve, its genus $g$, the prime $p$ that we use, the
degrees of the closed cusps of $X_{\F_p}$, and the numbers of non-cuspidal closed points of $X_{\F_p}$ of degree at
most $10$. We find these points by running through all the elliptic curves over $\F_{p^k}$, $k\leq10$, and all
their level structures, and we check the numbers against the zeta functions of the Jacobians. The scripts which
carry out these computations are linked in the proofs below.

\begin{table}[H]
\centering
\caption{Closed cusps and non-cuspidal closed points of $X_1(2,2n)_{\F_p}$.}
\label{tab:2x2n-points}
\footnotesize
\begin{tabular}{lrrll}
\toprule
$X$ & $g$ & $p$ & closed cusps & non-cuspidal closed points of degree $1,\dots,10$\\
\midrule
$X_1(2,32)$ & $37$ & $3$ & $1^{16},2^8,4^2,8^2$ & $0,0,0,0,0,192,320,960,1824,6208$\\
$X_1(2,34)$ & $49$ & $3$ & $1^{24},8^3$ & $0,0,0,36,48,48,144,738,2160,6504$\\
$X_1(2,36)$ & $41$ & $5$ & $1^{24},2^8,6^4$ & $0,0,48,120,720,2904,10416,50334,218080,979512$\\
$X_1(2,38)$ & $64$ & $3$ & $1^{27},9^3$ & $0,0,0,0,54,63,162,756,2270,6129$\\
$X_1(2,40)$ & $57$ & $3$ & $1^{16},2^{12},4^{10}$ & $0,0,0,32,96,64,480,504,2048,6000$\\
$X_1(2,42)$ & $61$ & $5$ & $1^{18},2^9,3^6,6^3$ & $0,0,0,126,648,2080,11376,48816,211776,983196$\\
$X_1(2,50)$ & $109$ & $3$ & $1^{30},4^6,10^3$ & $0,0,0,0,0,0,60,1170,1860,5040$\\
\bottomrule
\end{tabular}
\end{table}

We will use the following lemma, which turns a lower bound for the gonality over $\F_p$ into a lower bound over
$\Q$.

\begin{lemma}\label{lem:2x2n-pencil}
Let $X$ be a curve over $\Q$ with good reduction at a prime $p$, let $\Sigma$ be a finite set of closed points
of $X$, and let $d\geq1$. Assume that
\begin{enumerate}
\item[a)] $\gon_{\F_p}X_{\F_p}\geq d$,
\item[b)] the reduction modulo $p$ of every effective rational divisor of degree $d$ on $X$ whose support is
      disjoint from $\Sigma$ is supported on the reduction of $\Sigma$, and
\item[c)] every morphism $X_{\F_p}\to\PP^1$ of degree $d$ has a fibre above a point of $\PP^1(\F_p)$ which is
      not supported on the reduction of $\Sigma$.
\end{enumerate}
Then $\gon_\Q X>d$.
\end{lemma}

\begin{proof}
Let $\mathcal X$ be the smooth proper model of $X$ over $\Z_{(p)}$, and suppose that $f\in\Q(X)$ has degree
$e\leq d$. Let $\mathcal L$ be the line bundle on $\mathcal X$ which extends $\mathcal O_X(f^*\infty)$, and let
$V$ be the intersection of the span of $1$ and $f$ with $H^0(\mathcal X,\mathcal L)$. Then $V$ is a saturated
submodule of rank $2$ of the free $\Z_{(p)}$-module $H^0(\mathcal X,\mathcal L)$, so its reduction is a
$2$-dimensional subspace of $H^0(X_{\F_p},\mathcal L_{\F_p})$. It defines a morphism $X_{\F_p}\to\PP^1$ of degree
$e-b$, where $b$ is the degree of its base locus, and by a) we get $e-b\geq d$. Hence $e=d$ and $b=0$, so $V$
generates $\mathcal L$, and it defines a morphism $F:\mathcal X\to\PP^1_{\Z_{(p)}}$ which extends $f$ and whose
special fibre $\overline f$ has degree $d$. By c) there is $\overline t\in\PP^1(\F_p)$ such that
$\overline f^*(\overline t)$ is not supported on the reduction of $\Sigma$. Choose $t\in\PP^1(\Q)$ which reduces
to $\overline t$ and does not lie in the finite set $f(\Sigma)$. Then $f^*(t)$ is an effective rational divisor of
degree $d$ whose support is disjoint from $\Sigma$, and its reduction is $\overline f^*(\overline t)$, as the
pull-back by $F$ of the closure of $t$ in $\PP^1_{\Z_{(p)}}$ is the closure of $f^*(t)$ in $\mathcal X$. This
contradicts b).
\end{proof}


\subsection{The curve \texorpdfstring{$X_1(2,32)$}{X1(2,32)}}\label{subsec:2x2n-32}

The curve $X_1(2,32)$ has genus $37$, and we work at $p=3$. We use the level automorphism
$h_{32}:(E,P,Q)\mapsto(E,P,15Q)$, of order $2$, and the quotient $Y_{32}=X_1(2,32)/\dm{h_{32}}$, with quotient map
$\rho_{32}:X_1(2,32)\to Y_{32}$. The diamond operators commute with $h_{32}$, so they act on $Y_{32}$.

\begin{lemma}\label{lem:2x2n-32gon}
Let $X=X_1(2,32)$, let $h_{32}$ be the level automorphism $(E,P,Q)\mapsto(E,P,15Q)$ and let
$Y_{32}=X/\dm{h_{32}}$. Then $Y_{32}$ has genus $13$, $\gon_\Q Y_{32}\geq6$ and $\gon_\Q X\geq12$.
\end{lemma}

\begin{proof}
Under the identification of $X$ with $X_{\Delta_0}(64)$, $\Delta_0=\{\pm1,\pm31\}$, described at the beginning
of the section, $h_{32}$ corresponds to the diamond operator $\dm{15}$, as $\widehat\phi(15R)=15Q$. Hence
$Y_{32}\simeq X_{\Delta_1}(64)$ with $\Delta_1=\{\pm1,\pm15,\pm17,\pm31\}$, and $\gon_\Q Y_{32}\geq6$ by
\cite[Proposition 2.7 and Table 2]{Orlic}. We \gitlink{computations/next232_gonality_boundary.py}{compute} that the
groups belonging to $\Delta_0$ and $\Delta_1$ have index $768$ and $384$ in $\PSL_2(\Z)$, and that the curves
$X$ and $Y_{32}$ have genus $37$ and $13$. Suppose that $f\in\Q(X)$ has degree $d\leq11$. If $f$ is the pull-back
of a function on $Y_{32}$, this function has degree $d/2\leq5$, which is impossible. Otherwise $\Q(X)$ is
generated by $f$ and $\Q(Y_{32})$, as $[\Q(X):\Q(Y_{32})]=2$, and the Castelnuovo-Severi inequality
\cite[Theorem 3.11.3]{stichtenoth} gives
$$37=g(X)\leq2g(Y_{32})+(2-1)(d-1)=25+d\leq36,$$
which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-32Y}
Let $X=X_1(2,32)$, let $h_{32}$ and $Y_{32}$ be as in \Cref{lem:2x2n-32gon}, and let
$\rho_{32}:X\to Y_{32}$ be the quotient map.
\begin{enumerate}
\item[a)] The closed cusps of $X_{\F_3}$ of degree $1$ are the reductions of the $16$ rational cusps of $X$, and
      $h_{32}$ fixes every geometric point of the closed cusps of $X_{\F_3}$ of degree $2$ and $4$. The closed cusps
      of $Y_{32,\F_3}$ have degrees $1^8,2^8,4^4$, and those of degree $1$ are the images of the closed cusps of
      degree $1$ of $X_{\F_3}$.
\item[b)] Let $H_\Q\subseteq J_{Y_{32}}(\Q)$ be the subgroup generated by the classes of the rational cuspidal
      divisors of degree $0$. Then $J_{Y_{32}}(\Q)$ is finite and $[J_{Y_{32}}(\Q):H_\Q]\leq2$.
\item[c)] We have $J_{Y_{32}}(\F_3)\simeq(\torz2)^4\times\torz8\times(\torz{40})^2\times\torz{80}$, and the
      reduction $H_{32}$ of $H_\Q$ is isomorphic to $(\torz2)^2\times\torz4\times\torz{20}\times(\torz{40})^2$. It
      has order $512000$ and index $32$.
\end{enumerate}
\end{lemma}

\begin{proof}
a) We \gitlink{computations/next232_cusp_signatures.m}{compute} the Galois orbits of cusps of $X$ and their
reductions, using the valuations of integral modular units, and we
\gitlink{computations/next232_final_replay_v3.py}{check} the action of $h_{32}$ on the cusps, using their
description by Tate curves. We find $24$ geometric cusps of $X$ fixed by $h_{32}$, namely those in the closed cusps
of $X_{\F_3}$ of degree $2$ and $4$, and we \gitlink{computations/next232_quotient_cusps.py}{compute} the cusps of
$Y_{32}$ from this.

b) The Jacobian of $Y_{32}$ is a quotient of $J_X$, so $J_{Y_{32}}(\Q)$ is finite. For the index we use the
method of \cite[Section 4.3]{DEHMZ}. We work with the model of $Y_{32}$ given by $q$-expansions, on which the
operators $A_q$ of \Cref{prop:tools}~b) become $T_q-\dm q-q$ (see \Cref{subsec:hecke}). Let $M$ be the subgroup of
$J_{Y_{32}}(\Qbar)_\tors$ killed by $T_3-\dm3-3$, $T_5-\dm5-5$ and $\tau-1$, where $\tau$ is complex conjugation,
and let $\mathcal C\subseteq J_{Y_{32}}(\Qbar)$ be the subgroup generated by the classes of the cuspidal divisors
of degree $0$. By \Cref{prop:tools}~b), $J_{Y_{32}}(\Q)\subseteq M$. Using modular symbols, we
\gitlink{computations/next232_quotient13_integral_certificate.py}{compute} that $\#M=32768000$, that
$[M:M\cap\mathcal C]=2$, and that the subgroup of $\mathcal C$ fixed by $\Gal(\Qbar/\Q)$ is $H_\Q$, of order
$512000$. Hence $J_{Y_{32}}(\Q)\cap\mathcal C=H_\Q$, and $[J_{Y_{32}}(\Q):H_\Q]\leq[M:M\cap\mathcal C]=2$.

c) The curve $Y_{32}$ is the double cover of $X_1(2,16)$ obtained by adjoining $w$ with
$w^2=x(Q_{16})-x(8Q_{16})$, where $(E,P,Q_{16})$ is the universal triple over $X_1(2,16)$ in the model above, and
we \gitlink{computations/next232_quotient13_v2.m}{compute} $J_{Y_{32}}(\F_3)$ from this model. The $20$
closed cusps of $Y_{32,\F_3}$ come from $17$ Galois orbits of cusps of $Y_{32}$ over $\Q$, which we identify using
the valuations of integral modular units, and we \gitlink{computations/next232_quotient13_filter.m}{compute} the
subgroup $H_{32}$ generated by the reductions of these orbits.
\end{proof}

\begin{lemma}\label{lem:2x2n-32sieve}
Let $X=X_1(2,32)$ and let $D$ be an effective rational divisor of degree $10$ on $X$ whose support contains no
cusp. Suppose that the reduction of $D$ modulo $3$ is not entirely cuspidal. Then there is a level automorphism
$g$ of $X$ such that the reduction of $gD$ is $x_{32}+C'$, where $C'$ is an effective cuspidal divisor of degree
$4$ with no closed cusp of degree $1$ in its support, and $x_{32}$ is the closed point of degree $6$ of $X_{\F_3}$
represented by
$$\beta=\theta^5,\qquad\gamma=1+\theta^2+\theta^3+2\theta^4+\theta^5,\qquad\text{where }
\theta^6+2\theta^4+\theta^2+2\theta+2=0.$$
\end{lemma}

\begin{proof}
Write the reduction of $D$ as $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$ of degree $e$. By
\Cref{tab:2x2n-points}, $X_{\F_3}$ has no non-cuspidal closed points of degree at most $5$, so $D'$ is a single
closed point of degree $e\in\{6,\dots,10\}$. Up to the $32$ level automorphisms there are $8,10,32,57,198$ such
points of degree $6,7,8,9,10$, which we \gitlink{computations/next232_source_common.m}{compute}, and we replace $D$
by its image under a level automorphism to make $D'$ one of these $305$ representatives. The group
$J_1(32)(\Q)\simeq(\torz2)^3\times\torz4\times\torz{120}\times\torz{11640}$ is generated by rational cuspidal
divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and we
\gitlink{computations/next232_target_setup_v2.m}{compute} that its reduction $G$ has index $68$ in
$J_1(32)(\F_3)\simeq(\torz2)^3\times(\torz4)^2\times\torz{120}\times\torz{197880}$. We apply \eqref{eq:pushclass} to
the four maps $\pi_1,\dots,\pi_4$, allowing all the effective cuspidal divisors on $X_1(32)_{\F_3}$, and we
\gitlink{computations/next232_filter_common.m}{compute} that only one of the $305$ cases passes all four tests,
the point $x_{32}$. So $D'=x_{32}$ and $C'$ has degree $4$. We also
\gitlink{computations/next232_final_replay_v3.py}{compute} that, in the notation of \eqref{eq:pushclass} with a
rational cusp $y_0$ of $X_1(32)$, $[\pi_{i*}x_{32}-6\overline{y_0}]\notin G+\mathcal S_3$ for $i=1,\dots,4$.

Suppose that $C'$ contains a closed cusp of degree $1$. By \Cref{lem:2x2n-32Y}~a) it is the reduction
$\overline c$ of a rational cusp $c$ of $X$. The class $[\pi_{1*}(D-c)-9y_0]$ lies in $J_1(32)(\Q)$, and its
reduction is $[\pi_{1*}x_{32}+\pi_{1*}(C'-\overline c)-9\overline{y_0}]$, where $\pi_{1*}(C'-\overline c)$ is an
effective cuspidal divisor of degree $3$. Hence $[\pi_{1*}x_{32}-6\overline{y_0}]\in G+\mathcal S_3$, which is a
contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-32cosets}
Let $X$, $h_{32}$, $Y_{32}$, $\rho_{32}$ and $H_{32}$ be as in \Cref{lem:2x2n-32Y}, let $x_{32}$ be the closed
point of $X_{\F_3}$ of \Cref{lem:2x2n-32sieve}, and let $\overline{c_\ast}$ be a closed cusp of degree $1$ of
$Y_{32,\F_3}$. Then $h_{32}x_{32}=x_{32}$, and $y_{32}:=\rho_{32}(x_{32})$ is a closed point of degree $3$ with
$\rho_{32}^*y_{32}=x_{32}$. For $a\in\{1,3\}$ let $\mathcal T_a$ be the set of the elements $z$ of
$J_{Y_{32}}(\F_3)/H_{32}$ with $2z=0$ which are the classes of the divisors $\dm a y_{32}+c-5\overline{c_\ast}$,
with $c$ a closed cusp of degree $2$ of $Y_{32,\F_3}$. Then $\mathcal T_1\cap\mathcal T_3=\emptyset$.
\end{lemma}

\begin{proof}
We \gitlink{computations/next232_final_replay_v3.py}{check} that $h_{32}$ acts on the geometric points of
$x_{32}$ as the third power of the Frobenius, which proves the first two claims. In a basis of
$J_{Y_{32}}(\F_3)/H_{32}\simeq\torz2\times\torz2\times\torz8$ we
\gitlink{computations/next232_quotient13_orbit_filter.m}{compute}
$$\mathcal T_1=\{(1,1,0),(1,0,4)\}\qquad\text{and}\qquad\mathcal T_3=\{(1,1,4),(1,0,0)\}.$$
\end{proof}

\begin{proposition}\label{prop:2x32}
The groups $\tg{2}{32}$ and $\torz{64}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Let $X=X_1(2,32)$, and let $D$ be the trace divisor of a point of $Y_1(2,32)$ over a number field of degree $10$.
As $J_X(\Q)$ is finite and $\gon_\Q X\geq12$ by \Cref{lem:2x2n-32gon}, the reduction of $D$ modulo $3$ is not
entirely cuspidal by \Cref{prop:main}~a). By \Cref{lem:2x2n-32sieve}, after replacing $D$ by its image under a
level automorphism, the reduction of $D$ is $x_{32}+C'$, where $C'$ is an effective cuspidal divisor of degree
$4$ supported on closed cusps of degree $2$ and $4$. By \Cref{lem:2x2n-32Y}~a) and \Cref{lem:2x2n-32cosets},
$h_{32}$ fixes $C'$ and $x_{32}$, so it fixes the reduction of $D$, and \Cref{lem:stab}~a), with $C=0$ and
$\gon_\Q X>10$, gives $h_{32}D=D$. By \Cref{lem:CM}, the points in the support of $D$ correspond to elliptic curves
without complex multiplication, so by \Cref{lem:orbits} the geometric points in the support of $D$ form orbits of
length $2$ under $h_{32}$, on which the coefficients of $D$ are constant. So $D=\rho_{32}^*B$ for an effective rational
divisor $B$ of degree $5$ on $Y_{32}$. Its reduction
$\overline B$ satisfies $\rho_{32}^*\overline B=x_{32}+C'$, and as $\rho_{32}^*y_{32}=x_{32}$, we get
$\overline B=y_{32}+C_Y$ for an effective cuspidal divisor $C_Y$ of degree $2$ with $\rho_{32}^*C_Y=C'$. If $C_Y$
contained a closed cusp of degree $1$, then so would $C'$, by \Cref{lem:2x2n-32Y}~a). So $C_Y$ is a closed cusp
of degree $2$.

By \Cref{lem:2x2n-32Y}~a), $\overline{c_\ast}$ is the reduction of a rational cusp $c_\ast$ of $Y_{32}$. Let
$H_\Q$ be as in \Cref{lem:2x2n-32Y}~b). Reduction modulo $3$ is injective on $J_{Y_{32}}(\Q)$ and maps $H_\Q$
onto $H_{32}$, and $[J_{Y_{32}}(\Q):H_\Q]\leq2$, so the class $z$ of $[\overline B-5\overline{c_\ast}]$ modulo
$H_{32}$ satisfies $2z=0$, and $z\in\mathcal T_1$. The diamond operator $\dm3$ is an automorphism of $Y_{32}$
defined over $\Q$ which maps cusps to cusps, so it acts on $J_{Y_{32}}(\Q)/H_\Q$, trivially as this group has
order at most $2$. As $[\dm3c_\ast-c_\ast]\in H_\Q$, it follows that $[\dm3B-5c_\ast]-[B-5c_\ast]\in H_\Q$, so $z$
is also the class of $[\dm3y_{32}+\dm3C_Y-5\overline{c_\ast}]$. As $\dm3C_Y$ is a closed cusp of degree $2$, we
get $z\in\mathcal T_3$, which contradicts \Cref{lem:2x2n-32cosets}. Finally, $\torz{64}$ is excluded by
\Cref{lem:isogeny}.
\end{proof}

\subsection{The curve \texorpdfstring{$X_1(2,34)$}{X1(2,34)}}\label{subsec:2x2n-34}

The curve $X_1(2,34)$ has genus $49$, and we work at $p=3$. We
\gitlink{computations/new234_cusp_lattice.py}{compute} that its cusps over $\Q$ form $24$ rational cusps and $3$
Galois orbits of length $8$. The closed cusps of $X_1(2,34)_{\F_3}$ have degrees $1^{24},8^3$, so those of degree
$1$ are the reductions of the rational cusps. Abramovich's bound gives only $\gon_\C X_1(2,34)\geq9$, and
we have $\gon_{\F_3}X_1(2,34)\geq10$ by \cite[Table 1]{NV}.

\begin{lemma}\label{lem:2x2n-34sieve}
Let $X=X_1(2,34)$. The reduction modulo $3$ of every effective rational divisor of degree $10$ on $X$ is
supported on the cusps.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction is $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$ of
degree $e$. By \Cref{tab:2x2n-points}, the type of $D'$ is one of $[4],\dots,[10]$, $[4,4]$, $[4,5]$, $[4,6]$ and
$[5,5]$. Up to the $48$ level automorphisms, the non-cuspidal closed points of degree $4,\dots,10$ form
$2,1,2,3,18,45,138$ orbits, which we \gitlink{computations/new234_source_common.m}{compute}. Normalizing the
first point of $D'$ and taking all the $36$, $48$, $48$ and $48$ possibilities for the second point in the types
$[4,4]$, $[4,5]$, $[4,6]$ and $[5,5]$, we get $209+72+96+96+48=521$ cases. The group
$J_1(34)(\Q)\simeq\torz{8760}\times\torz{595680}$ is generated by rational cuspidal divisor classes
\cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and we
\gitlink{computations/new234_target_setup_v2.m}{compute} that its reduction $G$ has index $128$ in
$J_1(34)(\F_3)\simeq(\torz4)^3\times\torz{8760}\times\torz{1191360}$ and contains the classes of all the cuspidal
divisors on $X_1(34)_{\F_3}$. So \Cref{lem:classtest}, applied to $\pi_{i*}D$ as in \eqref{eq:pushclass}, gives
$[\pi_{i*}D'-e\,\overline{y_0}]\in G$ for $i=1,\dots,4$, and we
\gitlink{computations/new234_filter_common.m}{compute} that each of the $521$ cases fails this for one of the
four maps.
\end{proof}

\begin{lemma}\label{lem:2x2n-34units}
Let $c_1,\dots,c_{24}$ be the rational cusps of $X=X_1(2,34)$. If $f\in\Q(X)$ is not constant and
$\ddiv(f)=\sum_ia_ic_i$, then $\sum_ia_i^2>60$.
\end{lemma}

\begin{proof}
We use the identification of $X$ with $X_\Delta(68)$, $\Delta=\{\pm1,\pm33\}$, described at the beginning of the
section. A cuspidal divisor of degree $0$ is principal if and only if its image under the Abel-Jacobi map, which
we compute with modular symbols, lies in the period lattice. This gives the lattice of the principal divisors
supported on the cusps, and we \gitlink{computations/new234_cusp_lattice.py}{compute} its intersection with the
group of divisors supported on $c_1,\dots,c_{24}$, a lattice of rank $23$ in $\Z^{24}$. We
\gitlink{computations/new234_short_cusp_units.m}{compute} that it contains no non-zero vector of squared norm at
most $60$. A function over $\Qbar$ whose divisor is supported on rational points is a constant multiple of a
function over $\Q$, so this proves the claim.
\end{proof}

\begin{lemma}\label{lem:2x2n-34fibres}
Let $X=X_1(2,34)$. Every morphism $X_{\F_3}\to\PP^1$ of degree $10$ has a fibre above a point of $\PP^1(\F_3)$
whose support contains a non-cuspidal point.
\end{lemma}

\begin{proof}
Suppose that $\overline f:X_{\F_3}\to\PP^1$ has degree $10$ and that its fibres $A_1,\dots,A_4$ above the four
points of $\PP^1(\F_3)$ are cuspidal. By \Cref{tab:2x2n-points}, $X(\F_3)$ consists of $24$ cusps, and each of them
lies in exactly one $A_i$. Let $n_i$ be the number of these cusps in $A_i$, and order the fibres with
$n_1\geq\dots\geq n_4$.
Then $\sum_in_i=24$ and $n_i\leq10$, so $n_1\geq6$ and $n_2\geq\lceil(24-n_1)/3\rceil\geq5$. The other closed cusps
have degree $8>10-5$, so $A_1$ and $A_2$ are supported on rational points. An effective divisor of degree $10$
supported on exactly $n$ rational points has squared norm $\sum_ia_i^2\leq(11-n)^2+n-1$, and checking the
possibilities $6\leq n_1\leq10$ and $\lceil(24-n_1)/3\rceil\leq n_2\leq n_1$, we find that the squared norm of
$A_1-A_2$ is at most $60$, as $A_1$ and $A_2$ have disjoint supports. The divisor $A_1-A_2$ is principal, and it
is the reduction of a rational cuspidal divisor $A$ of degree $0$, as the closed cusps of degree $1$ of
$X_{\F_3}$ are the reductions of the rational cusps. The class $[A]$ lies in the finite group $J_X(\Q)$ and reduces to $0$, so $[A]=0$ by
Katz's result, and $A$ is the divisor of a function as in \Cref{lem:2x2n-34units} with $\sum_ia_i^2\leq60$, which
is a contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-34gon}
We have $\gon_\Q X_1(2,34)\geq11$.
\end{lemma}

\begin{proof}
We apply \Cref{lem:2x2n-pencil} to $X=X_1(2,34)$ with $p=3$, $d=10$ and $\Sigma$ the set of cusps. Condition a)
holds as $\gon_{\F_3}X\geq10$ \cite[Table 1]{NV}, condition b) is \Cref{lem:2x2n-34sieve}, and condition c) is
\Cref{lem:2x2n-34fibres}.
\end{proof}

\begin{proposition}\label{prop:2x34}
The groups $\tg{2}{34}$ and $\torz{68}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be the trace divisor of a point of $Y_1(2,34)$ over a number field of degree $10$. As
$J_1(2,34)(\Q)$ is finite and $\gon_\Q X_1(2,34)>10$ by \Cref{lem:2x2n-34gon}, the reduction of $D$ modulo $3$ is
not entirely cuspidal by \Cref{prop:main}~a). This contradicts \Cref{lem:2x2n-34sieve}. Finally, $\torz{68}$ is
excluded by \Cref{lem:isogeny}.
\end{proof}

\subsection{The curve \texorpdfstring{$X_1(2,36)$}{X1(2,36)}}\label{subsec:2x2n-36}

The curve $X_1(2,36)$ has genus $41$, and we work at $p=5$. We use the two maps $X_1(2,36)\to X_1(2,18)$ given by
$(E,P,Q)\mapsto(E,P,2Q)$ and $(E,P,Q)\mapsto(E/\dm{18Q},\phi(P),\phi(Q))$, where $\phi:E\to E/\dm{18Q}$ is the
quotient map, the four maps $\pi_1,\dots,\pi_4$ to $X_1(36)$, and two quotients of $X_1(2,36)$ which were also
used in \cite[Section 11.5]{N69}. The first one is the following.

\begin{lemma}[{\cite[Lemma 11.8 and its proof]{N69}}]\label{lem:2x2n-36Z}
Let $X=X_1(2,36)$, let $h_{36}$ be the level automorphism $(E,P,Q)\mapsto(E,P,13Q)$, of order $3$, and let
$Z_{36}=X/\dm{h_{36}}$, a curve of genus $9$. The classes of the cuspidal divisors of degree $0$ on
$Z_{36,\F_5}$ generate a subgroup $H_{36}\simeq(\torz3)^3\times(\torz{24})^3$ of index $8$ of
$J_{Z_{36}}(\F_5)\simeq(\torz3)^3\times(\torz{24})^2\times\torz{192}$, and $H_{36}$ contains the reduction modulo
$5$ of $J_{Z_{36}}(\Q)$.
\end{lemma}

For the second one, let $g_{36}$ be the level automorphism $(E,P,Q)\mapsto(E,P+18Q,P+Q)$ of $X_1(2,36)$, of order
$4$, with $g_{36}^2(E,P,Q)=(E,P,19Q)$. By \cite[Section 11.5]{N69}, the group generated by $g_{36}$ acts freely
on the non-cuspidal points of $X_1(2,36)$ and acts on the cusps with orbits of length $4$, so the quotient
$W_{36}=X_1(2,36)/\dm{g_{36}}$ is a curve of genus $11$ and the quotient map $\rho_{36}:X_1(2,36)\to W_{36}$ is
unramified. Let $t_9$ be the parameter of $X_1(9)$ in Tate normal form,
$$r_9=t_9^2-t_9+1,\quad c_9=t_9(r_9-1),\quad b_9=r_9c_9,\quad E_9:y^2+(1-c_9)xy-b_9y=x^3-b_9x^2,$$
and $j_9(t_9)=j(E_9)$. By \cite[Section 11.5]{N69}, the function $t_9$ of $(E,4Q)$ and the function
$w_{36}=4(x(9Q)-x(18Q))/(x(P)-x(P+18Q))$ are invariant under $g_{36}$, and $\Q(W_{36})=\Q(t_9,w_{36})$ with
$16(w_{36}^2+16)^3=j_9(t_9)w_{36}^4$.

\begin{lemma}\label{lem:2x2n-36sieve}
Let $X=X_1(2,36)$ and let $D$ be an effective rational divisor of degree $10$ on $X$ whose support contains no
cusp. Suppose that the reduction of $D$ modulo $5$ is not entirely cuspidal. Then there is a level automorphism
$g$ of $X$ such that the reduction of $gD$ is $x_{36}+C'$, where $C'$ is an effective cuspidal divisor of degree
$2$ and $x_{36}$ is the closed point of degree $8$ of $X_{\F_5}$ represented by
$$\beta=2\theta^4+3\theta^5+4\theta^7,\qquad\gamma=2+\theta+4\theta^3+2\theta^5+\theta^6+2\theta^7,\qquad
\text{where }\theta^8+\theta^4+3\theta^2+4\theta+2=0.$$
\end{lemma}

\begin{proof}
Write the reduction of $D$ as $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$ of degree $e$. By
\Cref{tab:2x2n-points}, the type of $D'$ is one of $[3],\dots,[10]$, $[3,3]$, $[3,4]$, $[3,5]$, $[3,6]$,
$[3,7]$, $[4,4]$, $[4,5]$, $[4,6]$, $[5,5]$, $[3,3,3]$ and $[3,3,4]$. Up to the $24$ level automorphisms, the
non-cuspidal closed points of degree $3,\dots,10$ form $2,5,30,126,434,2115,9088,40874$ orbits, which we
\gitlink{computations/new236_source_common.m}{compute}. Normalizing one point of $D'$ and taking all the
possibilities for the other points, we get $135282$ cases. We apply the following three tests, each of them an
instance of \eqref{eq:pushclass}.
\begin{enumerate}
\item[a)] The group $J_1(2,18)(\Q)\simeq\torz2\times\torz{42}\times\torz{126}$ is generated by rational cuspidal
      divisor classes \cite[Proposition 4.16 and Table 3]{DEHMZ}, and we
      \gitlink{computations/new236_filter_target18_setup.m}{compute} that its reduction has index $12$ in
      $J_1(2,18)(\F_5)\simeq\torz6\times\torz{84}\times\torz{252}$ and contains the classes of all the cuspidal
      divisors. We \gitlink{computations/new236_filter18_common_v2.m}{compute} that only $1251$ cases pass the test
      with both maps to $X_1(2,18)$.
\item[b)] By \Cref{lem:2x2n-36Z}, the class of the push-forward of $D'$ to $Z_{36,\F_5}$, relative to a rational
      cusp, lies in $H_{36}$. We
      \gitlink{C236_resolution_20261030/geometry/genus9_stage2_common_v2.m}{compute} that only $208$ of the $1251$
      cases pass this test.
\item[c)] The group $J_1(36)(\Q)\simeq\torz{12}\times\torz{252}\times\torz{7812}$ is generated by rational
      cuspidal divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and we
      \gitlink{computations/new236_geom_cyclic36_setup.m}{compute} that its reduction $G$ has index $214272$ in
      $J_1(36)(\F_5)\simeq(\torz4)^2\times(\torz{12})^2\times\torz{36}\times(\torz{7812})^2$, and that the images
      of the sets $\mathcal S_0,\dots,\mathcal S_7$ of \eqref{eq:pushclass} in $J_1(36)(\F_5)/G$ have
      $1,7,27,77,181,273,279,279$ elements. We
      \gitlink{C236_resolution_20261030/geometry/cyclic36_stage3_common.m}{compute} that only one of the $208$
      cases passes the test with the four maps $\pi_1,\dots,\pi_4$, the point $x_{36}$ with a cuspidal part of
      degree $2$.
\end{enumerate}
Hence $D'=x_{36}$ and $\deg C'=2$.
\end{proof}

\begin{lemma}\label{lem:2x2n-36x}
Let $X=X_1(2,36)$, let $g_{36}$, $W_{36}$ and $\rho_{36}$ be as above, and let $x_{36}$ be the closed point of
$X_{\F_5}$ of \Cref{lem:2x2n-36sieve}. Then $g_{36}x_{36}=x_{36}$, and
$$\ell\Bigl(x_{36}+\sum_{j=0}^3g_{36}^jC'\Bigr)=1$$
on $X_{\F_5}$ for every effective cuspidal divisor $C'$ of degree $2$ on $X_{\F_5}$.
\end{lemma}

\begin{proof}
The automorphism $g_{36}$ acts on the geometric points of $x_{36}$ as the sixth power of the Frobenius, as we
\gitlink{computations/new236_geom_cusp2_numeric.m}{compute}, so it fixes
$x_{36}$, and $x_{36}=\rho_{36}^*z_{36}$ for a closed point $z_{36}$ of degree $2$ of $W_{36,\F_5}$. After
completing the square in $E_9$, write its equation as $y^2=x^3+a_2x^2+a_4x+a_6$. The cubic $x^3+a_2x^2+a_4x+a_6$ and
the polynomial $w_{36}^2(a_2+3x)^2-4(w_{36}^2+4)(a_4+2a_2x+3x^2)$ have a unique common root $\xi$. Put
$\kappa=a_4+2a_2\xi+3\xi^2$ and $\Phi_{36}=16\kappa^3(1+4/w_{36})^2$. Over $\F_5$ the cover $\rho_{36}$ is given by
$z^4=\Phi_{36}$, and we \gitlink{computations/new236_geom_cusp2_numeric.m}{check} that
$\ddiv(\Phi_{36})=4\Lambda_{36}$ with $[\Lambda_{36}]$ of order $4$. Hence
$$\ell_X(\rho_{36}^*\Theta)=\sum_{j=0}^3\ell_{W_{36}}(\Theta+j\Lambda_{36})$$
for every divisor $\Theta$ on $W_{36,\F_5}$.

The closed cusps of $X_{\F_5}$ of degree $1$ form $6$ orbits of length $4$ under $g_{36}$, which map to $6$
closed cusps $c_1,\dots,c_6$ of degree $1$ of $W_{36,\F_5}$, and the closed cusps of degree $2$ form $2$ orbits of
length $4$, which map to $2$ closed cusps $u_1,u_2$ of degree $2$. As $g_{36}$ acts freely on the cusps, we have
$\sum_jg_{36}^jC'=\rho_{36}^*\rho_{36*}C'$, and $\rho_{36*}C'$ is one of the $21$ divisors $c_i+c_k$ with
$1\leq i\leq k\leq6$ or one of $u_1,u_2$. In all these $23$ cases we
\gitlink{computations/new236_geom_cusp2_numeric.m}{compute} that
$\ell_{W_{36}}(z_{36}+\rho_{36*}C'+j\Lambda_{36})=1,0,0,0$ for $j=0,1,2,3$. Hence
$\ell_X(x_{36}+\sum_jg_{36}^jC')=\ell_X(\rho_{36}^*(z_{36}+\rho_{36*}C'))=1$.
\end{proof}

\begin{lemma}\label{lem:2x2n-36nc}
Let $X=X_1(2,36)$. The reduction modulo $5$ of every effective rational divisor of degree $10$ on $X$ whose
support contains no cusp is supported on the cusps.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction is not supported on the cusps. By
\Cref{lem:2x2n-36sieve}, after replacing $D$ by its image under a level automorphism, the reduction of $D$ is
$x_{36}+C'$ with $C'$ effective cuspidal of degree $2$. By \Cref{lem:2x2n-36x}, $g_{36}$ fixes $x_{36}$ and
$\ell(x_{36}+\sum_{j=0}^3g_{36}^jC')=1$, and $g_{36}$ acts on the geometric cusps with orbits of length $4$. As
$J_X(\Q)$ is finite, \Cref{lem:section} gives $4\mid\deg C'=2$, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-36fibres}
Let $X=X_1(2,36)$. Every morphism $X_{\F_5}\to\PP^1$ of degree $10$ has a fibre above a point of $\PP^1(\F_5)$
whose support contains a non-cuspidal point.
\end{lemma}

\begin{proof}
Suppose that $\overline f:X_{\F_5}\to\PP^1$ has degree $10$ and that its fibres $A_1,\dots,A_6$ above the six
points of $\PP^1(\F_5)$ are cuspidal. By \Cref{tab:2x2n-points}, $X(\F_5)$ consists of $24$ cusps, and each of them
lies in exactly one $A_i$. Let $n_i$ be the number of these cusps in $A_i$, and order the fibres with
$n_1\geq\dots\geq n_6$.
Then $\sum_in_i=24$, so $n_1\geq4$ and $n_2\geq\lceil(24-n_1)/5\rceil$. For a cuspidal divisor
$A=\sum_cm_cc$ on $X_{\F_5}$, the sum being over the closed cusps, put $\|A\|^2=\sum_c\deg(c)m_c^2$, which is the
squared norm of the corresponding divisor on the $64$ geometric cusps. If $A$ is effective of degree $10$ and
contains exactly $n$ closed cusps of degree $1$, then $\|A\|^2\leq(11-n)^2+n-1$. Checking the possibilities for
$4\leq n_1\leq10$ and $n_2$, we find $\|A_1-A_2\|^2\leq104$, as $A_1$ and $A_2$ have disjoint supports.

The closed cusps of $X_{\F_5}$ are the reductions of the orbits of the Frobenius automorphism $\sigma_5$ on the
geometric cusps of $X$, which are defined over $\Q(\zeta_{36})$. Let $A$ be the cuspidal divisor over
$\Q(\zeta_{36})$ which reduces to the principal divisor $A_1-A_2$ and has the same multiplicities on each
$\sigma_5$-orbit. Its class is torsion by the Manin-Drinfeld theorem and reduces to $0$ modulo a prime above $5$, so
$A$ is principal by Katz's result. So the vector of multiplicities of $A$ lies in the lattice of principal
divisors supported on the $64$ geometric cusps and fixed by $\sigma_5$. Using the identification of $X$ with
$X_\Delta(72)$, $\Delta=\{\pm1,\pm35\}$, and modular symbols as in the proof of \Cref{lem:2x2n-34units}, we
\gitlink{computations/new236_cusp_lattice.py}{compute} this lattice, which has rank $35$, and we
\gitlink{computations/new236_cusp_short.m}{compute} that it contains $24552$ pairs $\pm v$ of non-zero vectors of
squared norm at most $104$, none of which has positive part of degree $10$. As the positive part of $A$ is $A_1$,
of degree $10$, this is a contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-36gon}
We have $\gon_\Q X_1(2,36)\geq11$.
\end{lemma}

\begin{proof}
We apply \Cref{lem:2x2n-pencil} to $X=X_1(2,36)$ with $p=5$, $d=10$ and $\Sigma$ the set of cusps. Condition a)
holds as $\gon_{\F_5}X\geq10$ \cite[Table 1]{NV}, condition b) is \Cref{lem:2x2n-36nc}, and condition c) is
\Cref{lem:2x2n-36fibres}.
\end{proof}

\begin{proposition}\label{prop:2x36}
The groups $\tg{2}{36}$ and $\torz{72}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Let $D$ be the trace divisor of a point of $Y_1(2,36)$ over a number field of degree $10$. As $J_1(2,36)(\Q)$ is
finite and $\gon_\Q X_1(2,36)>10$ by \Cref{lem:2x2n-36gon}, the reduction of $D$ modulo $5$ is not entirely
cuspidal by \Cref{prop:main}~a). This contradicts \Cref{lem:2x2n-36nc}. Finally, $\torz{72}$ is excluded by
\Cref{lem:isogeny}.
\end{proof}

\subsection{The curves \texorpdfstring{$X_1(2,38)$}{X1(2,38)}, \texorpdfstring{$X_1(2,40)$}{X1(2,40)} and
\texorpdfstring{$X_1(2,50)$}{X1(2,50)}}\label{subsec:2x2n-38-40-50}

For these three curves we work at $p=3$, and the push-forwards to smaller levels, followed by the Hecke sieve on
$X_1(38)$ and $X_1(50)$, exclude all the reductions of the trace divisor which are not entirely cuspidal.

\begin{lemma}\label{lem:2x2n-38}
Let $X=X_1(2,38)$. The reduction modulo $3$ of every effective rational divisor of degree $10$ on $X$ whose
support contains no cusp is supported on the cusps.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction is $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$
of degree $e$. By \Cref{tab:2x2n-points}, the type of $D'$ is one of $[5],\dots,[10]$ and $[5,5]$. Up to the $54$
level automorphisms, the non-cuspidal closed points of degree $5,\dots,10$ form $1,2,3,14,43,116$ orbits, which we
\gitlink{computations/late238_source.m}{compute}, and normalizing the first point and taking all the $54$
possibilities for the second point in the type $[5,5]$, we get $179+54=233$ cases. Put $V=2Q$, of order $19$. We
use the four maps $X\to X_1(19)$ which keep $V$, or divide $E$ by $\dm P$, $\dm{19Q}$ or $\dm{P+19Q}$ and keep the
image of $V$. The group $J_1(19)(\Q)\simeq\torz{4383}$ is generated by rational cuspidal divisor classes
\cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and its reduction $G$ has index $2$ in
$J_1(19)(\F_3)\simeq\torz{8766}$ and contains the classes of all the cuspidal divisors. We
\gitlink{computations/late238_filter19_v2.m}{compute} that only $35$ cases satisfy \eqref{eq:pushclass} for all
four maps: one of type $[8]$, two of type $[9]$, fourteen of type $[10]$ and eighteen of type $[5,5]$. In all of
them $\deg C'\leq2$.

For these we use the map $\pi_1:X\to X_1(38)$, $(E,P,Q)\mapsto(E,Q)$. We
\gitlink{computations/late_cusp_tate.py}{compute} that the closed cusps of $X_1(38)_{\F_3}$ have degrees
$1^{18},9^2$, and that those of degree $1$ are the reductions of rational cusps. So $\pi_{1*}C'$ is the reduction
of a rational cuspidal divisor, and $A_5(\pi_{1*}C')=0$ by \Cref{prop:tools}~c). As $J_1(38)(\Q)$ is finite
\cite[Theorem 3.1(2)]{DEHMZ}, \Cref{prop:tools}~d) shows that $A_5(\pi_{1*}D)$ is principal, and hence so is
$A_5(\pi_{1*}D')$. We \gitlink{continuation_20261029b/C238/hecke_template.m}{compute} that $A_5(\pi_{1*}D')$ is not
principal in all $35$ cases, which is a contradiction.
\end{proof}

\begin{lemma}\label{lem:2x2n-40}
Let $X=X_1(2,40)$. The reduction modulo $3$ of every effective rational divisor of degree $10$ on $X$ whose
support contains no cusp is supported on the cusps.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction is $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$
of degree $e$. By \Cref{tab:2x2n-points}, the type of $D'$ is one of $[4],\dots,[10]$, $[4,4]$, $[4,5]$, $[4,6]$
and $[5,5]$. Up to the $32$ level automorphisms \eqref{eq:2x2n-aut}, the non-cuspidal closed points of degree
$4,\dots,10$ form $1,3,2,15,21,64,195$ orbits, which we
\gitlink{continuation_20261029b/C240/enumerate_direct.m}{compute}. Normalizing the first point and taking all the
possibilities for the second one, we get $301$ cases with one point and $32+96+64+288=480$ cases with two points,
$781$ in total. We use the maps $X\to X_1(2,20)$ given by $(E,P,Q)\mapsto(E,P,2Q)$ and by
$(E,P,Q)\mapsto(E/\dm{20Q},\phi(P),\phi(Q))$, where $\phi:E\to E/\dm{20Q}$ is the quotient map. The group
$J_1(2,20)(\Q)\simeq\torz4\times\torz{60}\times\torz{120}$ is generated by rational cuspidal divisor classes
\cite[Proposition 4.16 and Table 3]{DEHMZ}, and its reduction has index $4$ in
$J_1(2,20)(\F_3)\simeq\torz8\times(\torz{120})^2$ and contains the classes of all the cuspidal divisors. We
\gitlink{continuation_20261029b/C240/filter_quotient20_linear.m}{compute} that only $288$ of the $781$ cases
satisfy \eqref{eq:pushclass} for the first map, and we
\gitlink{continuation_20261029b/C240/filter_quotient20_isogeny.m}{compute} that only $118$ of these $288$ cases
satisfy it for the second map. For
these we use the four maps $\pi_1,\dots,\pi_4$ to $X_1(40)$. The group
$J_1(40)(\Q)\simeq(\torz2)^3\times\torz8\times\torz{120}\times\torz{895440}$ is generated by rational cuspidal
divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, its reduction $G$ has index $4592$ in
$J_1(40)(\F_3)\simeq(\torz2)^5\times\torz{16}\times\torz{68880}\times\torz{895440}$, and we
\gitlink{continuation_20261029b/C240/filter_cyclic40_portable.m}{compute} that each of the $118$ cases fails
\eqref{eq:pushclass}, with all the effective cuspidal divisors on $X_1(40)_{\F_3}$ allowed, for one of the four
maps.
\end{proof}

\begin{lemma}\label{lem:2x2n-50}
Let $X=X_1(2,50)$. The reduction modulo $3$ of every effective rational divisor of degree $10$ on $X$ whose
support contains no cusp is supported on the cusps.
\end{lemma}

\begin{proof}
Let $D$ be such a divisor, and suppose that its reduction is $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$
of degree $e$. By \Cref{tab:2x2n-points}, $D'$ is a single closed point of degree $7,\dots,10$, and up to the $60$
level automorphisms there are $1,21,31,84$ of them, which we \gitlink{computations/late250_source.m}{compute},
$137$ cases in total. We use the four maps $X\to X_1(25)$ which keep $2Q$ on $E$ or on its quotients by its three
subgroups of order $2$. The group $J_1(25)(\Q)\simeq\torz{227555}$ is generated by rational cuspidal divisor
classes \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and its reduction has index $11$ in
$J_1(25)(\F_3)\simeq\torz{2503105}$ and contains the classes of all the cuspidal divisors. We
\gitlink{computations/late250_filter25.m}{compute} that only $3$ cases satisfy \eqref{eq:pushclass} for all four
maps, each of them a closed point of degree $8$, and hence $\deg C'=2$.

For these we use the map $\pi_1:X\to X_1(50)$, $(E,P,Q)\mapsto(E,Q)$. We
\gitlink{computations/late_cusp_tate.py}{compute} that the closed cusps of $X_1(50)_{\F_3}$ have degrees
$1^{20},4^4,10^2$, and that those of degree $1$ are the reductions of rational cusps. So $\pi_{1*}C'$ is the
reduction of a rational cuspidal divisor, and $A_7(\pi_{1*}C')=0$ by \Cref{prop:tools}~c). As $J_1(50)(\Q)$ is
finite \cite[Theorem 3.1(2)]{DEHMZ}, $A_7(\pi_{1*}D')$ is principal by \Cref{prop:tools}~d). We
\gitlink{continuation_20261029b/C250/hecke_template.m}{compute} that $A_7(\pi_{1*}D')$ is not principal in all
$3$ cases, which is a contradiction.
\end{proof}

\begin{proposition}\label{prop:2x38-2x40-2x50}
The groups $\tg{2}{38}$, $\tg{2}{40}$, $\tg{2}{50}$, $\torz{76}$, $\torz{80}$ and $\torz{100}$ are not subgroups of
the torsion of an elliptic curve over a number field of degree $10$.
\end{proposition}

\begin{proof}
Let $N\in\{38,40,50\}$ and let $D$ be the trace divisor of a point of $Y_1(2,N)$ over a number field of degree
$10$. As $J_1(2,N)(\Q)$ is finite and $\gon_\Q X_1(2,N)\geq11$, as explained at the beginning of the section, the
reduction of $D$ modulo $3$ is not entirely cuspidal by \Cref{prop:main}~a). This contradicts
\Cref{lem:2x2n-38,lem:2x2n-40,lem:2x2n-50}. Finally, the cyclic groups are excluded by \Cref{lem:isogeny}.
\end{proof}

\subsection{The curve \texorpdfstring{$X_1(2,42)$}{X1(2,42)}}\label{subsec:2x2n-42}

The curve $X_1(2,42)$ has genus $61$, we work at $p=5$, and by Abramovich's bound $\gon_\C X_1(2,42)\geq12$. The
cusps of $X_1(2,42)$ have degrees $1^{18},2^9,6^6$ over $\Q$ \cite[Section 11.6]{N69}, and the closed cusps of
$X_1(2,42)_{\F_5}$ have degrees $1^{18},2^9,3^6,6^3$ (\Cref{tab:2x2n-points}), so those of degree $1$ and $2$ are the reductions of Galois orbits of cusps of the same degree. We use
the two maps $X_1(2,42)\to X_1(2,14)$ given by $(E,P,Q)\mapsto(E,P,3Q)$ and by
$(E,P,Q)\mapsto(E/\dm{14Q},\phi(P),\phi(Q))$, where $\phi:E\to E/\dm{14Q}$ is the quotient map, the four maps
$X_1(2,42)\to X_1(21)$ which keep $2Q$ on $E$ or on its quotients by its three subgroups of order $2$, and six
maps $X_1(2,42)\to X_0(42)$. These are the compositions of the natural map $X_1(42)\to X_0(42)$ with the maps
$X_1(2,42)\to X_1(42)$ which keep $Q$, $P+Q$ or $P+22Q$, or divide $E$ by $\dm P$ or $\dm{P+21Q}$ and keep the
image of $Q$, or divide $E$ by $\dm{21Q}$ and keep the image of $P+Q$. All these points have order $42$. We denote
by $\pi_0:X_1(2,42)\to X_0(42)$ the first of these six maps, $(E,P,Q)\mapsto(E,\dm Q)$. We also use the following quotient.

\begin{lemma}[{\cite[Lemma 11.12 and its proof]{N69}}]\label{lem:2x2n-42Y}
Let $X=X_1(2,42)$, let $G_9$ be the group of level automorphisms generated by $s_1:(E,P,Q)\mapsto(E,21Q,P+Q)$ and
$s_2:(E,P,Q)\mapsto(E,P,25Q)$, and let $Y_{42}=X/G_9$. Let $v$ be the parameter of $X_1(7)$ in Tate normal form,
evaluated at $(E,6Q)$, put $t=(v^3-3v+1)/(v(v-1))-8$ and
$$f_1=t^2+13t+49,\qquad f_2=t^2+5t+1,\qquad f_3=t^4+14t^3+63t^2+70t-7.$$
Then $Y_{42}$ has genus $5$, and its function field is obtained from $\Q(u)$, where $u^2=t$, by adjoining the
coordinates of a point of order $3$ on
$$E_u:y^2=x^3-\frac{f_2}{48f_1}x+\frac{f_3}{864f_1^2}.$$
The reduction of $J_{Y_{42}}(\Q)$ modulo $5$, taken relative to a rational cusp, is contained in a subgroup
$H_{42}\simeq(\torz2)^3\times\torz{24}$ of index $18$ of $J_{Y_{42}}(\F_5)\simeq\torz2\times(\torz6)^2\times\torz{48}$.
\end{lemma}

\begin{lemma}\label{lem:2x2n-42X0}
Let $\vartheta:X_1(42)\to X_0(42)$ be the natural map, let $y_0$ be the image of a rational cusp of $X_1(42)$,
and let $H_0\subseteq J_0(42)(\F_5)$ be the reduction modulo $5$ of $\vartheta_*J_1(42)(\Q)$. Then
$J_0(42)(\F_5)\simeq\torz2\times\torz4\times(\torz{48})^2$, $H_0\simeq\torz2\times\torz{48}$ has index $192$, and
for $j=0,\dots,6$ the images in $J_0(42)(\F_5)/H_0$ of the sets $\mathcal S_j$ of the classes $[j\overline{y_0}-B]$,
with $B$ effective cuspidal of degree $j$ on $X_0(42)_{\F_5}$, have $1,6,12,16,20,24,24$ elements.
\end{lemma}

\begin{proof}
The group $J_1(42)(\Q)$ is generated by rational cuspidal divisor classes \cite[Theorem 3.1(2), Corollary 4.14 and
Table 2]{DEHMZ}, so $\vartheta_*J_1(42)(\Q)$ is generated by the push-forwards of the rational cuspidal divisors of
degree $0$. All the $8$ cusps of $X_0(42)$ are rational, and we
\gitlink{computations/next242_geom_x0_42_setup.m}{compute} the push-forwards of the Galois orbits of cusps of
$X_1(42)$, the group $J_0(42)(\F_5)$ and its subgroup $H_0$, and the sets $\mathcal S_j$.
\end{proof}

\begin{lemma}\label{lem:2x2n-42sieve}
Let $X=X_1(2,42)$ and let $D$ be an effective rational divisor of degree $10$ on $X$ whose support contains no
cusp. Suppose that the reduction of $D$ modulo $5$ is not entirely cuspidal. Then there is a level automorphism
$g$ of $X$ such that the reduction of $gD$ is $D'+C'$, with $C'$ effective cuspidal, where
\begin{enumerate}
\item[a)] either $D'$ is a closed point of degree $9$ which is fixed by a level automorphism of order $3$, and
      $\deg C'=1$,
\item[b)] or $D'=x_{42}$, the closed point of degree $6$ of $X_{\F_5}$ represented by
      $$\beta=2\theta^2+4\theta^3+2\theta^4+3\theta^5,\qquad\gamma=4+\theta+4\theta^2+\theta^5,\qquad\text{where }
      \theta^6+\theta^4+4\theta^3+\theta^2+2=0,$$
      and $\deg C'=4$.
\end{enumerate}
\end{lemma}

\begin{proof}
Write the reduction of $D$ as $D'+C'$ as in \Cref{subsec:sieve}, with $D'\neq0$ of degree $e$. By
\Cref{tab:2x2n-points}, the type of $D'$ is one of $[4],\dots,[10]$, $[4,4]$, $[4,5]$, $[4,6]$ and $[5,5]$. Up to
the $36$ level automorphisms, the non-cuspidal closed points of degree $4,\dots,10$ form $5$, $18$, $64$, $316$,
$1372$, $5888$ and $27374$ orbits, which we \gitlink{computations/next242_source_common.m}{compute}.
Normalizing the first point and taking all the $126$, $648$, $2080$ and $648$ possibilities for the second point
in the types $[4,4]$, $[4,5]$, $[4,6]$ and $[5,5]$, we get $35037+630+3240+10400+11664=60971$ cases. We apply the
following tests. In i), ii) and iii) we use \eqref{eq:pushclass}, with all the effective cuspidal divisors on the
target allowed.
\begin{enumerate}
\item[i)] The group $J_1(2,14)(\Q)\simeq(\torz2)^2\times\torz6\times\torz{18}$ is generated by rational cuspidal
      divisor classes \cite[Proposition 4.16 and Table 3]{DEHMZ}, and we
      \gitlink{computations/next242_target14_setup.m}{compute} that its reduction has index $4$ in
      $J_1(2,14)(\F_5)\simeq(\torz2)^2\times\torz{12}\times\torz{36}$ and contains the classes of all the
      cuspidal divisors. Only $4647$ cases pass the test with the two maps to $X_1(2,14)$, as we
      \gitlink{computations/next242_filter14_common.m}{compute}.
\item[ii)] With the subgroup $H_{42}$ of \Cref{lem:2x2n-42Y}, only $752$ of them pass the test with the
      quotient map $X\to Y_{42}$, as we
      \gitlink{next_cases_20261030/C242/geometry/quotient5_filter_common_v2.m}{compute}.
\item[iii)] The group $J_1(21)(\Q)\simeq\torz{364}$ is generated by rational cuspidal divisor classes
      \cite[Theorem 3.1(2), Corollary 4.14 and Table 2]{DEHMZ}, and its reduction has index $6$ in
      $J_1(21)(\F_5)\simeq\torz{2184}$. Only $111$ of the $752$ cases pass the test with the four maps to
      $X_1(21)$, as we \gitlink{next_cases_20261030/C242/geometry/cyclic21_filter_common.m}{compute}. Exactly $3$
      of them are single closed points of degree $9$ (with a cuspidal part of degree $1$), and we
      \gitlink{computations/next242_cheap_stabilizers.py}{check} that each of them is fixed by one of the level
      automorphisms $(E,P,Q)\mapsto(E,21Q,P+25Q)$ and $(E,P,Q)\mapsto(E,P+21Q,P+4Q)$, which have order $3$. These
      $3$ cases are the cases in a).
\item[iv)] For each of the six maps $\pi:X\to X_0(42)$ described above, the class of $\pi_*D-10y_0$ lies in
      $\vartheta_*J_1(42)(\Q)$, so \Cref{lem:classtest} gives $[\pi_*D'-e\,\overline{y_0}]\in H_0+\mathcal
      S_{10-e}$, with the notation of \Cref{lem:2x2n-42X0}. Only $10$ of the remaining $108$ cases pass this test
      for all six maps, as we \gitlink{next_cases_20261030/C242/geometry/x0_42_filter_common.m}{compute}: one of
      type $[6]$, namely $x_{42}$ with $\deg C'=4$, one of type $[10]$, four of type $[4,4]$, two of type
      $[4,6]$ and two of type $[5,5]$.
\item[v)] Let $\pi'$ be one of the maps $(E,P,Q)\mapsto(E,Q)$ and $(E,P,Q)\mapsto(E,P+Q)$ from $X$ to $X_1(42)$.
      As $J_1(42)(\Q)$ is finite, $A_q(\pi'_*D)$ is principal for $q=11$ and $q=17$ by \Cref{prop:tools}~d), and
      hence so is its reduction $A_q(\pi'_*D')+A_q(\pi'_*C')$. If $\deg C'=0$, then $A_{11}(\pi'_*D')$ is
      principal. As $17\equiv5^5\pmod{42}$,
      we have $A_{17}C=0$ for every cuspidal divisor $C$ on $X_1(42)_{\F_5}$ by \Cref{lem:cuspHecke}, so
      $A_{17}(\pi'_*D')$ is principal in all cases. For the $5$ cases of degree $10$ we
      \gitlink{computations/next242_direct_hecke_common_v2.m}{compute} that $\ell(A_{11}(\pi'_*D'))=0$, and for the
      $4$ cases of type $[4,4]$ that $\ell(A_{17}(\pi'_*D'))=0$, with $\pi'(E,P,Q)=(E,Q)$ except for one case of
      type $[4,4]$, where we use $\pi'(E,P,Q)=(E,P+Q)$.
\end{enumerate}
So only the case b) remains.
\end{proof}

\begin{lemma}\label{lem:2x2n-42cubic}
Let $X=X_1(2,42)$ and let $D$ be an effective rational divisor of degree $10$ on $X$ whose support contains no
cusp. Suppose that the reduction of $D$ modulo $5$ is $x_{42}+C'$, where $x_{42}$ is the closed point of
\Cref{lem:2x2n-42sieve} and $C'$ is an effective cuspidal divisor of degree $4$. Then $C'$ consists of a closed
cusp of degree $3$ and a closed cusp of degree $1$.
\end{lemma}

\begin{proof}
Let $\pi_0$, $\vartheta$, $y_0$ and $H_0$ be as above and in \Cref{lem:2x2n-42X0}. We
\gitlink{next_cases_20261030/C242/geometry/x0_42_filter_common.m}{compute} that
$[\pi_{0*}x_{42}-6\overline{y_0}]\notin H_0$. Suppose that $C'$ contains no closed cusp of degree $3$. As $C'$ has
degree $4$, it is then supported on closed cusps of degree $1$ and $2$, so it is the reduction of a rational
cuspidal divisor $C$. The class of $\pi_{0*}(D-C)-6y_0$ lies in $\vartheta_*J_1(42)(\Q)$ and reduces to
$[\pi_{0*}x_{42}-6\overline{y_0}]$, which is a contradiction. So $C'$ contains a closed cusp of degree $3$, and the
remaining part of $C'$ is a closed cusp of degree $1$.
\end{proof}

\begin{lemma}\label{lem:2x2n-42W}
Let $X=X_1(2,42)$, let $h_{42}$ be the level automorphism $(E,P,Q)\mapsto(E,P+21Q,P+4Q)$, and let $x_{42}$ be the
closed point of \Cref{lem:2x2n-42sieve}. Then $h_{42}$ has order $3$, it fixes $x_{42}$, it acts freely on the $72$
geometric cusps of $X$, and
$$\ell\bigl(x_{42}+C'+h_{42}C'+h_{42}^2C'\bigr)=1$$
on $X_{\F_5}$ for every effective cuspidal divisor $C'$ on $X_{\F_5}$ consisting of a closed cusp of degree $3$
and a closed cusp of degree $1$.
\end{lemma}

\begin{proof}
One checks directly that $h_{42}$ has order $3$, and that $h_{42}=s_2s_1^2$ with $s_1,s_2$ as in
\Cref{lem:2x2n-42Y}. We \gitlink{computations/next242_case26_phases.m}{check} that $h_{42}$ acts on the geometric
points of $x_{42}$ as the square of the Frobenius, so it fixes $x_{42}$. The automorphism $h_{42}$ permutes the
three points of order $2$ cyclically, so it fixes no cusp, as at a cusp one of the points of order $2$ is
distinguished (see the proof of \cite[Proposition 11.13]{N69}). As $h_{42}$ has order $3$, it acts freely on the
$72$ geometric cusps.

Let $W_{42}=X/\dm{h_{42}}$, which is the curve $V$ of genus $17$ of \cite[proof of Proposition 11.13]{N69}, as
$\dm{h_{42}}=\dm{s_1s_2^2}$, and let $\rho_{42}:X\to W_{42}$ be the quotient map. In the notation of
\Cref{lem:2x2n-42Y}, the function field of $W_{42}$ is obtained from that of $Y_{42}$ by adjoining a root
$w_{42}$ of
$$w_{42}^3-\frac{f_2}{48}w_{42}+\frac{f_3(2t+13)}{1728f_1}-\frac{u}{8f_1}=0.$$
Here $w_{42}=r_0v_0+r_1v_1+r_2v_2$, where $r_0,r_1,r_2$ are the $x$-coordinates of the points of order $2$ in a
normalized model of $E$, ordered by the level structure, and $v_0=v$, $v_1=(v-1)/v$, $v_2=1/(1-v)$.
This is the model of the curve $V$ of genus $17$ in \cite[proof of Proposition 11.13]{N69}: the data
\githubbare{next_cases_20261030/C242/geometry/diagonal17_normalized_model_data.m} and
\githubbare{next_cases_20261030/C242/geometry/diagonal17_parameter_data.m} are those exported by the script
\href{https://github.com/F-Najman/nonic-torsion/blob/14d4464ea4e46cd90c251b2ca9485c7ea52e72c1/C2x42/diagonal_genus17_lattice/export.m}{\path{C2x42/diagonal_genus17_lattice/export.m}}
of the repository of \cite{N69} for degree $9$.
The function $u$ is regular at $x_{42}$ and satisfies $u^2+4u+2=0$ there, and $j=0$ at $x_{42}$. We
\gitlink{computations/next242_geom_case26_sections.m}{find} exactly $2$ closed points of degree $2$ of
$W_{42,\F_5}$ at which $u^2+4u+2=0$ and $j=0$, and both are unramified in $\rho_{42}$, with a closed point of degree
$6$ above them. So $x_{42}=\rho_{42}^*z_{42}$ for one of them, $z_{42}$ say, and we treat both possibilities.

The closed cusps of $W_{42,\F_5}$ have degrees $1^6,2^3,3^2,6$. As $h_{42}$ acts freely on the cusps,
$C'+h_{42}C'+h_{42}^2C'=\rho_{42}^*\rho_{42*}C'$, and $\rho_{42*}C'$ is the sum of one of the $2$ closed cusps of
degree $3$ and one of the $6$ closed cusps of degree $1$ of $W_{42,\F_5}$, which gives $12$ possibilities. To
compute $\ell(\rho_{42}^*\Theta)$ for divisors $\Theta$ on $W_{42}$, we extend the constants to $\F_{25}$. Put
$\tilde t=t+8$, $\delta_1=\tilde t^2-3\tilde t+9$ and $\delta_2=2\tilde t^3-9\tilde t^2+27\tilde t-27$, and note
that $\delta_2^2-4\delta_1^3=-27\delta_1^2$. The parameter $v$ satisfies $v^3-\tilde tv^2+(\tilde t-3)v+1=0$, and
generates the function field of $X$ over that of $W_{42}$. For a primitive cube root of unity $\zeta\in\F_{25}$,
the cover $X_{\F_{25}}\to W_{42,\F_{25}}$ is the Kummer cover $y^3=\Phi_{42}$ with
$\Phi_{42}=(\delta_2+3(\zeta-\zeta^2)\delta_1)/2$, as $3v-\tilde t=y+\delta_1/y$. With
$\Lambda_j=\lfloor j\ddiv(\Phi_{42})/3\rfloor$ for $j=0,1,2$, the integral bases of this tame Kummer cover give
$$\ell_X(\rho_{42}^*\Theta)=\sum_{j=0}^2\ell_{W_{42}}(\Theta+\Lambda_j)$$
for every divisor $\Theta$ on $W_{42,\F_{25}}$. For both choices of $z_{42}$ and all $12$ possibilities for
$\rho_{42*}C'$ we \gitlink{computations/next242_geom_case26_sections.m}{compute} that
$\ell_{W_{42}}(z_{42}+\rho_{42*}C'+\Lambda_j)=1,0,0$ for $j=0,1,2$. Hence
$\ell_X(x_{42}+\sum_jh_{42}^jC')=\ell_X(\rho_{42}^*(z_{42}+\rho_{42*}C'))=1$.
\end{proof}

\begin{proposition}\label{prop:2x42}
The groups $\tg{2}{42}$ and $\torz{84}$ are not subgroups of the torsion of an elliptic curve over a number field
of degree $10$.
\end{proposition}

\begin{proof}
Let $X=X_1(2,42)$ and let $D$ be the trace divisor of a point of $Y_1(2,42)$ over a number field of degree $10$. As
$J_X(\Q)$ is finite and $\gon_\Q X\geq\gon_\C X\geq12$, the reduction of $D$ modulo $5$ is not entirely cuspidal by
\Cref{prop:main}~a). By \Cref{lem:2x2n-42sieve}, after replacing $D$ by its image under a level automorphism, one
of the cases a) and b) of that lemma holds.

Suppose first that a) holds, so the reduction of $D$ is $D'+\overline c$, where $D'$ is a closed point of degree
$9$ fixed by a level automorphism $h$ of order $3$, and $\overline c$ is a closed cusp of degree $1$. Let $c$ be a
cusp of $X$ which reduces to $\overline c$, which is defined over $\Q(\zeta_{42})$. Then $h$ fixes the reduction
$D'$ of $D-c$, and \Cref{lem:stab}~b), with $d=10$, $e=1$ and $\gon_\C X\geq12>11$, gives $hD=D$. By \Cref{lem:CM},
the points in the support of $D$ correspond to elliptic curves without complex multiplication, so \Cref{lem:orbits},
applied to the group of order $3$ generated by $h$, shows that $3\mid\deg D=10$, which is a contradiction.

Suppose now that b) holds, so the reduction of $D$ is $x_{42}+C'$ with $C'$ effective cuspidal of degree $4$. By
\Cref{lem:2x2n-42cubic}, $C'$ consists of a closed cusp of degree $3$ and one of degree $1$, so by
\Cref{lem:2x2n-42W} the automorphism $h_{42}$ fixes $x_{42}$, acts freely on the geometric cusps, and satisfies
$\ell(x_{42}+\sum_{j=0}^2h_{42}^jC')=1$. \Cref{lem:section} gives $3\mid\deg C'=4$, which is a
contradiction. Finally, $\torz{84}$ is excluded by \Cref{lem:isogeny}.
\end{proof}