diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 7f67d04..05fafad 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -810,3 +810,627 @@ Choosing $\ell$ smaller would shrink $B$'s fraction too much; choosing
 $\ell$ larger would not annihilate $B$ at all (the top-degree columns
 would survive). The exponent $\ell$ is exactly the right size to
 simultaneously annihilate on both sides while keeping $B$ fat.
+
+\subsection{Choosing a compatible projection}
+\label{subsec:admissibility}
+
+Our ultimate idempotent $e$ will be a block-diagonal sum, one block per
+character of~$A_0$. Each block is a projection matrix onto the image
+of~$B$. The two opposite annihilations we want follow automatically
+from Lemma~\ref{lem:B-annihilation} once $e$ is a projection onto
+$\im B$. But $e$ must live inside $K[H]$, and there the two shifts
+$u(s)$ and $v(s)$ act through matrix representations of different
+shapes. This subsection shows that for most characters the
+corresponding block projection exists and is nonsingular.
+
+\paragraph{Characters of $A_0$.}
+$A_0=(\mathbb Z/M\mathbb Z)^P$ is finite abelian of order
+$|A_0|=M^{p^3}$. A \emph{character} is a homomorphism
+$\lambda:A_0\to K^*$. It is uniquely determined on the coordinate
+generators $c_x$ (the function with value~$1$ at $x\in P$ and $0$
+elsewhere): $\lambda(c_x)=\zeta^{a_x}$ for some
+$a_x\in\{0,\ldots,M-1\}$. Because $\zeta$ has order $M$ and is an
+element of~$K$, every character has values in~$K$.
+
+The shift group $P\subseteq H$ acts on characters by
+$(s\lambda)(a)=\lambda((-s)\cdot a)$. Conjugating the Fourier
+idempotent $p_\chi^{A_0}$ of a character by $u(s)$ gives
+\[
+ u(s)\,p_\chi^{A_0}\,u(s)^{-1}=p_{s\chi}^{A_0}.
+\]
+
+\paragraph{The diagonal matrix and the Gram form.}
+For $\lambda\in\widehat{A_0}$ abbreviate $\chi_x=x\lambda$ (the
+translate of $\lambda$ by $x$) and define the diagonal
+$p^3\times p^3$ matrix
+\begin{equation}\label{eq:D-lambda}
+ D_\lambda=\diag\!\bigl(\chi_x(c):x\in P\bigr).
+\end{equation}
+Every entry lies in $K^*$ because $\chi_x$ is a character and $c$ is
+a group element. So $D_\lambda^{-1}$ exists and is diagonal.
+
+The matrix
+\begin{equation}\label{eq:N-lambda}
+ N_\lambda=B^{\mathsf T}\,D_\lambda^{-1}\,B
+\end{equation}
+is a symmetric $h\times h$ matrix over~$K$.
+
+Call $\lambda$ \emph{admissible} if
+\begin{enumerate}
+\item its $P$-stabilizer is trivial, i.e., the orbit $P\lambda$ has
+      size exactly $p^3$; and
+\item $N_\lambda$ is invertible.
+\end{enumerate}
+Both conditions are checkable in finite time, and the check is
+deterministic: enumerate the $p^3$ translates and compute one
+$h\times h$ determinant.
+
+\paragraph{Most characters are admissible.}
+The two modes of failure each account for less than half of the
+characters. Their union is then smaller than half, so more than half
+succeed. This count is the key estimate of the whole construction.
+
+\begin{lemma}\label{lem:good-count}
+More than half of the $|A_0|$ characters are admissible.
+\end{lemma}
+\begin{proof}
+\emph{Stabilizers.} Fix a nonzero $s\in P$. Translation by $s$
+partitions $P$ (of size $p^3$) into $p^2$ orbits of size~$p$. A
+character $\lambda$ is fixed by $s$ if and only if $\lambda(c_x)$ is
+constant on every such orbit, so exactly $M^{p^2}$ characters are
+fixed. The union over the $p^3-1$ nonzero $s$ therefore has size at
+most $(p^3-1)M^{p^2}$, i.e. a fraction at most
+\[
+ (p^3-1)\,M^{-p^2(p-1)}
+ \le (p^3-1)/M < 1/4 .
+\]
+The last inequality uses $M=p^4-1>4p^3$, which holds for $p\ge5$;
+we have $p>600$.
+
+\emph{Singular $N_\lambda$.} The entries of $D_\lambda^{-1}$ range
+independently over $K^*$ as $\lambda$ ranges over all characters of
+$A_0$, because $\chi_x(c)$ for distinct $x$ are independent characters.
+Introduce independent variables $\tau_x$ and Cauchy--Binet gives
+\begin{equation}\label{eq:weighted-gram}
+ Q(\tau):=\det\!\bigl(B^{\mathsf T}\diag(\tau_x)_{x\in P}B\bigr)
+ =\sum_{\substack{J\subseteq P\\|J|=h}}\det(B_J)^2\prod_{x\in J}\tau_x ,
+\end{equation}
+where $B_J$ is the $h\times h$ submatrix of $B$ consisting of the
+rows indexed by~$J$, taken in the fixed increasing order.
+
+The right side is a sum of distinct monomials (different $J$ give
+different exponent vectors), so it is a nonzero polynomial. Its total
+degree is exactly~$h$. Because $B$ has full column rank by
+Lemma~\ref{lem:monomial-basis}, some maximal minor $B_J$ is nonzero.
+The polynomial $Q$ therefore has a nonzero leading monomial and does
+not cancel.
+
+Now apply Schwartz's grid bound~\cite{Schwartz1980} (see also
+Zippel~\cite{Zippel1979}): a nonzero polynomial of total degree at
+most~$a$ vanishes on at most an $a/M$ fraction of a product grid whose
+coordinates each range over $M$ values. Here $Q$ has degree $h$ and
+the grid is $(K^*)^{p^3}$ of side $M$. Thus $Q$ vanishes on at most
+the fraction $h/M$ of characters. Since $h\le p^3$ (it counts a
+subset of $K^P$) and $M=p^4-1$, this is at most
+\[
+ \frac{h}{M}\le\frac{p^3}{p^4-1}<\frac14 ,
+\]
+because $p^4-1>4p^3$ for $p\ge5$.
+
+Combining: fewer than $1/4+1/4=1/2$ of characters fail either
+condition. So more than half admit a trivial stabilizer and an
+invertible Gram form simultaneously.
+\end{proof}
+
+\subsection{Realizing the projection in $K[H]$}
+\label{subsec:idempotent}
+
+Each admissible character $\lambda$ gives a projection $T_\lambda$ on
+the $p^3$-dimensional block attached to~$\lambda$. The task is to
+write that projection as an element of $K[H]$ whose regular rank we
+can compute. The Fourier idempotents do this job: for each character
+$\lambda$, the products $p_{x\lambda}u(s)$ (for $x,s\in P$) form a
+$K$-basis of one matrix block of $K[H]$. Then we sum over $\lambda$.
+
+\paragraph{One character, one block.}
+Fix $\lambda\in\widehat{A_0}$ with trivial $P$-stabilizer. For
+$x,y\in P$ set
+\begin{equation}\label{eq:E-xy}
+ E_{xy}^{\lambda}=p_{x\lambda}^{A_0}\,u(x-y).
+\end{equation}
+Because $u(s)\,p_{x\lambda}u(s)^{-1}=p_{(x+s)\lambda}$, the row
+indices line up with the shift. Fourier orthogonality gives
+\begin{equation}\label{eq:E-mult}
+ E_{xy}^{\lambda}E_{zt}^{\lambda}=\delta_{yz}\,E_{xt}^{\lambda}.
+\end{equation}
+Indeed, the Fourier product survives when $(x-y+z)\lambda=x\lambda$,
+i.e. $(y-z)\lambda=0$, which by the trivial stabilizer forces $y=z$.
+
+Put $q_\lambda=\sum_{x\in P}p_{x\lambda}^{A_0}$. Because $q_\lambda$
+commutes with $A_0$ and $P$ permutes its summands, $q_\lambda$ is a
+central idempotent. The basis $\{p_{x\lambda}u(s)\}_{x,s\in P}$
+identifies the ideal $q_\lambda K[H]$ with the matrix algebra
+$\Mat_{p^3}(K)$, via $u(s)\mapsto L_s$ and $c\mapsto D_\lambda$.
+
+\paragraph{Explicit projection.}
+Define
+\begin{equation}\label{eq:T-lambda}
+ T_\lambda=B\,N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
+ \in\Mat_{p^3}(K).
+\end{equation}
+Because $N_\lambda=B^{\mathsf T}D_\lambda^{-1}B$, we have
+\[
+ T_\lambda B=B,\qquad
+ T_\lambda^2=T_\lambda.
+\]
+So $T_\lambda$ is a projection. It fixes each column of $B$, so
+$\im B\subseteq\im T_\lambda$; and it factors through $B^{\mathsf T}$,
+so $\im T_\lambda\subseteq\im B$. Hence $\im T_\lambda=\im B$ and
+$\operatorname{rank}T_\lambda=h$.
+
+Define an element of the $\lambda$-block of $K[H]$:
+\begin{equation}\label{eq:e-lambda}
+ e_\lambda=\sum_{x,y\in P}(T_\lambda)_{xy}\,E_{xy}^{\lambda}.
+\end{equation}
+By the matrix-unit relation~\eqref{eq:E-mult},
+$e_\lambda^2=e_\lambda$. The action of $H$ on this block is the
+matrix action of $T_\lambda$.
+
+\paragraph{The two annihilations.}
+Left multiplication by $u(s)$ in this block is $L_s$. Therefore
+\[
+ (L_s-1)^\ell\,T_\lambda
+ =(L_s-1)^\ell B\;N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
+ =0,
+\]
+using $(L_s-1)^\ell B=0$ from Lemma~\ref{lem:B-annihilation}. The
+conjugate $v(s)$ acts by $D_\lambda L_sD_\lambda^{-1}$, so
+\[
+ D_\lambda L_s D_\lambda^{-1}-1=D_\lambda(L_s-1)D_\lambda^{-1}.
+\]
+Combining with the transpose identity $B^{\mathsf T}(L_s-1)^\ell=0$:
+\[
+ B^{\mathsf T}D_\lambda^{-1}(D_\lambda L_s D_\lambda^{-1}-1)^\ell
+ =B^{\mathsf T}(L_s-1)^\ell D_\lambda^{-1}=0.
+\]
+Hence $(u(s)-1)^\ell e_\lambda=0$ and
+$e_\lambda(v(s)-1)^\ell=0$ in $K[H]$.
+
+\paragraph{Summing over characters.}
+Blocks for distinct orbits are orthogonal: if $P\lambda\ne P\lambda'$
+then $e_\lambda e_{\lambda'}=0$. Choose the least admissible character
+in each $P$-orbit that contains one, and let $\mathscr R$ be the set
+of chosen characters. Define
+\begin{equation}\label{eq:e-full}
+ e=\sum_{\lambda\in\mathscr R}e_\lambda \in K[H].
+\end{equation}
+By orthogonality the sum is an idempotent: $e^2=e$. The two
+annihilations persist on~$e$.
+
+\paragraph{The regular rank.}
+Left multiplication by $T_\lambda$ on $\Mat_{p^3}(K)$ has image the
+matrices whose $p^3$ columns all lie in~$\im B$, of dimension $p^3h$.
+Summing over the blocks:
+\[
+ d:=\dim_K(eK[H])=p^3\,h\,|\mathscr R|.
+\]
+The admissible characters form more than half of $\widehat{A_0}$
+(Lemma~\ref{lem:good-count}), and each $P$-orbit has size $p^3$, so
+\[
+ |\mathscr R|>\frac{|A_0|}{2p^3}.
+\]
+Recall $w=|H|=|A_0|\,p^3$. Therefore
+\[
+ \frac dw=\frac{p^3h|\mathscr R|}{|A_0|p^3}
+ =\frac{h|\mathscr R|}{|A_0|}
+ >\frac{h}{2p^3}>\frac1{96},
+\]
+using $h/p^3>1/48$ from Lemma~\ref{lem:h}.
+
+\begin{proposition}\label{prop:modular}
+There is a specified idempotent $e\in K[H]$ and positive integers
+$d,w$ with $w=|H|$, $d=\dim_KeK[H]$, $d/w>1/96$, satisfying for
+every $s\in P$
+\[
+ (u(s)-1)^\ell\,e=0,\qquad
+ e\,(v(s)-1)^\ell=0 .
+\]
+Its coefficients are obtained by finite operations in the specified
+field $K$ of Lemma~\ref{lem:field}.
+\end{proposition}
+
+Remark on $\ell$. Choosing $\ell$ smaller would shrink $B$'s fraction
+below what we need; choosing $\ell$ larger would not annihilate the
+top-degree columns of $B$, so the identities would fail. The exponent
+$\ell=(p-1)/2$ is exactly the right size.
+
+\subsection{Ordered annihilations at edge level}
+\label{subsec:edge-zero}
+
+Now take $m$ copies $H_1,\ldots,H_m$ of $H$, one per vertex of
+$\Gamma$. For each edge $i<j$ identify the cyclic subgroup generated
+by $v(1,\bar j,\bar j^2)$ in $H_i$ with the one generated by
+$u(1,\bar i,\bar i^2)$ in $H_j$, where the bar denotes residues
+in~$\F_p$. Let $G_0$ be the resulting free-product quotient. We prove
+in Section~\ref{sec:gluing} that $H_i\to G_0$ is injective; here we
+only need the algebraic fact that the edge products vanish.
+
+Denote by $e_i\in K[G_0]$ the image of the idempotent $e$ under the
+map $H_i\to G_0$.
+
+\begin{corollary}\label{op:edge-zero}
+For every edge $i<j$ of $\Gamma$, $e_ie_j=0$ in $K[G_0]$.
+\end{corollary}
+\begin{proof}
+Let $y\in K[H_i]\subseteq K[G_0]$ denote the common generator of
+order $p$ across the edge, and put $X=y-1$. On the $H_i$ side we
+have $e_iX^\ell=0$ (as $y=v(s)$ is the conjugating shift); on the
+$H_j$ side we have $X^\ell e_j=0$ (as $y=u(s)$ is the incoming
+shift).
+
+Work inside $K[H_i]$, which is a free right $K[\langle y\rangle]$-module
+of rank $w/p$: pick representatives $a_1,\ldots,a_{w/p}$ of the left
+cosets of $\langle y\rangle$ in $H_i$. Then
+$e_i=\sum_a a\,q_a$ with $q_a\in K[\langle y\rangle]$.
+
+The cyclic group algebra is $K[\langle y\rangle]=K[X]/(X^p)$ because
+$(1+X)^p=1+X^p$ in characteristic $p$. Inside this quotient the right
+annihilator of $X^\ell$ is precisely the principal ideal
+$(X^{p-\ell})$, since $X^{\ell+j}=0$ in $K[X]/(X^p)$ iff
+$\ell+j\ge p$.
+
+Thus $q_aX^\ell=0$ forces $q_a=q_a'X^{p-\ell}$, and
+\[
+ e_i=q\,X^{p-\ell}\quad\text{for}\quad q=\sum_a a\,q_a'\in K[H_i].
+\]
+
+Because $p-\ell=\ell+1$,
+\[
+ e_ie_j=q\,X^{\ell+1}e_j=q\,X\,(X^\ell e_j)=0.
+\]
+No factor is commuted past $X$, no symmetry is used, and no reverse
+product $e_je_i$ is asserted.
+\end{proof}
+
+\paragraph{Summary so far.}
+We have a minimal exceptional spherical map: connected, on the sphere,
+with exactly one face of nonidentity value, and minimal among all such.
+We now perform local reductions. Each reduction either produces a
+strictly smaller exceptional map (contradiction to minimality) or
+forces a local restriction on the configuration. After the reductions,
+the restrictions combine with Euler's formula to give a final
+contradiction.
+
+\paragraph{Reduction 1: every edge is nonneutral.}
+Suppose first that a nonloop edge is neutral. Contract it: edges with
+value $1$ and their darts disappear from the boundary-word description,
+and the graph stays connected. Face evaluations are unchanged, since
+the two neutral darts contributed nothing. Minimality fails.
+
+If a neutral edge is a loop, keep the side containing the exceptional
+face; the retained side is connected, all its faces retain their
+evaluations, and the discarded loop disappears. Minimality fails.
+
+Hence every edge carries a nontrivial label.
+
+\paragraph{Reduction 2: every edge is a nonbridge.}
+Suppose an edge is a bridge with value $s$. Deleting it splits the map
+into two connected components, and the unique boundary walk of each
+adjacent face splits into two walks, one per component. After suitable
+choice of starting point and orientation, the old walk is
+$W_1\,s\,W_2\,s^{-1}$, where $W_1,W_2$ are the two split walks.
+
+If some $W_a$ has nonidentity value, keep the component containing it,
+with the same face type as before. This component is connected and has
+exactly one exceptional face (that of $W_a$). Minimality fails.
+
+If both $W_a$ are trivial, the original face was either ordinary or
+exceptional. If it was ordinary, the exceptional face is on one of the
+two components; keep that one. If the original face was the unique
+exceptional one, both $W_a$ are trivial (this was assumed), so $sWs^{-1}$
+was the exceptional word; keep the component whose walk is the whole
+$sWs^{-1}$ from the other side. In every case, a smaller exceptional map
+results.
+
+So every edge is a nonbridge.
+
+\paragraph{Reduction 3: the two faces across an edge have distinct types.}
+Suppose an edge separates two faces $\Phi,\Phi'$ of the same type $i\in
+V(\Delta)$. Delete the edge and merge the two faces into one of type
+$i$. If the boundary walk of $\Phi$ near the edge reads $sD$ and that
+of $\Phi'$ reads $s^{-1}E$ (with the shared edge contributing $s$ on
+one side and $s^{-1}$ on the other, and $D,E$ the remaining letters),
+the merged walk reads $DE$. Direct computation gives
+\begin{equation}\label{eq:gluing-merge}
+ DE=s^{-1}(sD)s\,(s^{-1}E).
+\end{equation}
+If $sD$ evaluates to $1$, then $s^{-1}(sD)s$ also evaluates to $1$; if
+$sD$ evaluates to $g\ne1$, then its conjugate $s^{-1}gs$ also has value
+$\ne1$. Conjugation preserves identity-versus-nonidentity. Hence:
+
+\begin{itemize}
+\item ordinary + ordinary $\to$ ordinary;
+\item ordinary + exceptional (in either order) $\to$ exceptional.
+\end{itemize}
+
+We are assuming the map has \emph{exactly one} exceptional face, so
+never both-exceptional. In every case, the merge produces a smaller
+map with at most one exceptional face, and thus a smaller exceptional
+map. Minimality fails.
+
+Therefore the two face types across an edge labelled $\gamma$ are
+exactly the two endpoints of~$\gamma$.
+
+\paragraph{Reduction 4: at most one loop, bounding the exceptional monogon.}
+Suppose there is a loop. Its two sides are faces (possibly with other
+stuff in between, but the loop has just its two immediate face-sides).
+Consider the side \emph{not} containing the exceptional face. Retain
+the loop and everything on that side; erase everything on the side
+that contains the exceptional face. Retained faces are all ordinary
+(they were ordinary before). The erased side becomes a single monogon
+whose walk is the loop alone, and its value is the loop's value,
+nonidentity by Reduction 1. So this monogon is the unique exceptional
+face of the new, smaller map.
+
+Minimality fails, unless the erased side consisted only of that
+isolated monogon originally. Hence every loop bounds the exceptional
+monogon in the original map. Since there is only one exceptional face
+(it bounds only one walk), \emph{there is at most one loop}, and if it
+exists it bounds exactly the exceptional monogon.
+
+\paragraph{Reduced configuration.}
+Combining the four reductions, a minimal exceptional map satisfies:
+
+\begin{itemize}
+\item every edge is nonneutral;
+\item every edge is a nonbridge;
+\item the two faces across an edge labelled $\gamma$ have the two
+      distinct endpoint types of $\gamma$;
+\item at most one loop exists, and it bounds the exceptional monogon.
+\end{itemize}
+
+We continue in the next reduction. The one remaining configuration to
+rule out is that a face has two consecutive nonloop darts with the
+same label. Doing so will give the final contradiction with Euler's
+formula.
+
+\paragraph{Reduction 5: no two consecutive nonloop darts with the same label.}
+Suppose a face $\Phi$ has consecutive nonloop darts $e:a\to b$ and
+$f:b\to d$ with a common label $\gamma\in E(\Delta)$ and values
+$\epsilon,\varphi\in C_\gamma\setminus\{1\}$. The edges $e$ and $f$
+are distinct: otherwise the same edge would occur on both sides of
+$\Phi$, making it a bridge, contradicted by Reduction 2.
+
+Let $X$ be the face on the opposite side of $e$, $Y$ the face on the
+opposite side of $f$. By Reduction 3, both have type $j\in V(\Delta)$,
+where $j$ is the other endpoint of $\gamma$ from $\Phi$'s type~$i$.
+The faces $X$ and $Y$ might coincide.
+
+Remove $e,f$. Insert a new edge $g:a\to d$ along their arcs, smoothed
+through the empty corner of $\Phi$ at~$b$. Give $g$ value
+$\epsilon\varphi$ and label $\gamma$ (or make it neutral if this
+equals $1$; Reduction 1 then applies if needed). The walk of $\Phi$
+replaces $ef$ by $g$: same face type $i$, same evaluation. Only $X$
+and $Y$ change; every other face keeps its type and value.
+
+\emph{Case $X\ne Y$.} Write $X=\bar e D$, $Y=\bar f E$, with $D:a\to b$
+and $E:b\to d$. The merged face has walk $\bar gDE$. Its evaluation is
+\begin{equation}\label{eq:gluing-smooth-merge}
+ (\epsilon\varphi)^{-1}DE
+ =\varphi^{-1}(\epsilon^{-1}D)\,\varphi\,(\varphi^{-1}E).
+\end{equation}
+Let $p=\epsilon^{-1}D$ (old $X$'s value) and $q=\varphi^{-1}E$ (old
+$Y$'s value). The merged value is $\varphi^{-1}p\varphi\cdot q$:
+
+\begin{itemize}
+\item $p=q=1$: merged value $\varphi^{-1}\cdot 1\cdot\varphi\cdot 1=1$;
+\item $p\ne1$, $q=1$: merged value $\varphi^{-1}p\varphi\ne1$ (conjugate
+      of nonidentity);
+\item $p=1$, $q\ne1$: merged value $q\ne1$;
+\item $p\ne1$, $q\ne1$: excluded (at most one exceptional face).
+\end{itemize}
+
+So the merged face is ordinary when both old faces were ordinary, and
+exceptional when exactly one was. A smaller exceptional map results,
+contradicting minimality.
+
+\emph{Case $X=Y$.} Now $X$ is a single face whose boundary walk
+encounters $e$ and $f$ on their opposite sides. Reading from the dart
+$\bar e$: the walk is $\bar eD\bar fE$, with $D:a\to d$ and
+$E:b\to b$.
+
+After smoothing, the single walk becomes \emph{two} walks
+$\bar gD$ and $E$. The rules are easiest to check by slots: at $a$
+and $d$ the new darts replace the old slots of $e,\bar f$; at $b$ the
+consecutive slots $\bar e,f$ are deleted. Counting $v$, $e$, $F$:
+$v$ unchanged, $e$ decreased by one, $F$ increased by one. Euler's
+formula gives $2c_{\text{new}}=v-(e-1)+(F+1)=2+2=4$, so the new
+drawing has \emph{exactly two} components.
+
+Every old face except $X$ survives in one component; $X$ was the only
+face shared between the two. One component contains $a,d$ (with $g$),
+the other contains $b$ (isolated if $E$ is empty, otherwise with the
+walk $E$). Both new faces have type $j$.
+
+Let $A=(\epsilon\varphi)^{-1}D$ be the value of walk $\bar gD$, and
+$B=E$ the value of walk $E$. The old walk of $X$ evaluated to
+$\epsilon^{-1}D\varphi^{-1}E$, whose cyclic shift starting at $\bar f$
+is $(\varphi^{-1}B\varphi)A$:
+
+\begin{itemize}
+\item $X$ exceptional: the old value was $\ne1$, so at least one of
+      $A,B$ is $\ne1$. Keep the component whose walk is that one.
+\item $X$ ordinary with $A=B=1$: both new walks are ordinary, so
+      the exceptional face is elsewhere. Keep the component
+      containing it.
+\item $X$ ordinary with $A\ne1$ and $B\ne1$: both new walks are
+      exceptional. The exceptional case sits either in the
+      $\bar gD$-component (if that also contains the original
+      exception), or in the $E$-component. Keep the component
+      without the face $\Phi$, whose unique exception is its
+      sole nonidentity boundary.
+\end{itemize}
+
+In every case, a smaller exceptional map results, contradicting
+minimality. So the reduced configuration forbids consecutive nonloop
+darts with a common label, on any face.
+
+\subsection{Euler deficit and the final contradiction}
+\label{subsec:gluing-euler-deficit}
+
+Recall the reduced configuration: no neutral edges, no bridges,
+distinct endpoint types across every edge, at most one loop (bounding
+the exceptional monogon), and no face has two consecutive nonloop
+darts with the same label.
+
+\paragraph{Vertices away from the loop.}
+Let $u$ be a vertex that is not the base of any loop. Read the face
+types around $u$ cyclically; call the resulting sequence
+$i_1,i_2,\ldots,i_r$, where $i_a$ is the type of the $a$-th face,
+and $r=\deg u$. Successive $i_a$ differ: the edge between them has
+endpoints $i_{a}$ and $i_{a+1}$, by Reduction 3. No immediate
+reversal by Reduction 5 (which says the labels, hence the ordered
+endpoints, are distinct).
+
+So we get a closed walk $i_1,i_2,\ldots,i_r,i_1$ in $\Delta$ with
+successive edges distinct and length $r\ge1$. Since $\Delta$ is
+simple and triangle-free, no such closed walk has $r<4$: length $1$
+or $2$ would be a loop or a doubling of an edge in a simple graph;
+length $3$ would be a triangle. Thus $\deg u\ge4$.
+
+\paragraph{Faces away from the loop.}
+Let $F$ be an ordinary face not incident to the loop. Its boundary is
+nonempty: any connected graph with at least one edge has nonempty
+dart cycles. Successive labels differ by Reduction 5, so the closed
+walk in $\Delta$ has length at least $4$. Thus $|\partial F|\ge4$.
+
+\paragraph{Euler deficit.}
+For a connected spherical drawing,
+\begin{equation}\label{eq:gluing-deficit}
+ \sum_{u}(4-\deg u)+\sum_{F}(4-|\partial F|)=8,
+\end{equation}
+which follows from Euler's formula and $\sum_u\deg u=\sum_F|\partial
+F|=2e$ (every edge contributes two to each side).
+
+\emph{No loop present.} Every vertex satisfies $\deg u\ge4$, so
+contributes $\le0$. Every ordinary face has $|\partial F|\ge4$, so
+contributes $\le0$. Only the exceptional face can contribute
+positively: its walk is nonempty, so $|\partial F|\ge1$ and the
+contribution is at most $3$. Total $\le3<8$, contradicting
+\eqref{eq:gluing-deficit}.
+
+\emph{Loop present.} The base vertex $u_0$ of the loop has $\deg u_0
+\ge2$ (the loop contributes two), so contributes at most $2$. The
+exceptional monogon has walk of length $1$, so contributes $3$. The
+ordinary face on the other side of the loop sees the loop's dart with
+value $\ne1$ (by Reduction 1), so its walk is nonempty and cannot be
+a monogon (a monogon would have just the loop's dart, giving value
+$\ne1$, contradicting ordinari-ness); hence its length is at least
+$2$, contributing at most $2$. All other vertices and faces
+contribute at most zero. Total $\le 2+3+2=7<8$, contradicting
+\eqref{eq:gluing-deficit}.
+
+Hence no minimal exceptional spherical map exists. So $L\to D_i^0$
+is injective, i.e., $D_i^0\to Q$ is injective, for every vertex~$i$.
+
+\subsection{Extending to the full vertex group}
+\label{subsec:gluing-extension}
+
+We now extend $D_i^0\to Q$ to $D_i\to Q$. The argument is a
+permutation-action construction.
+
+Fix $i$. Let $\Omega=L\times\mathbb{N}$, with $L$ acting by left
+multiplication on the first coordinate. The action of $D_i^0$ on
+$\Omega$, transported through $D_i^0\hookrightarrow L$, is free on
+countably infinitely many orbits. The same is true of the left
+$D_i^0$-action on $D_i\times\mathbb{N}$.
+
+Pair the $D_i^0$-orbits bijectively between the two sets and choose
+one point in each pair. Extending $D_i^0$-equivariantly gives a
+$D_i^0$-equivariant bijection $\phi_i:D_i\times\mathbb{N}\to\Omega$.
+Transport the $D_i$ left-multiplication along $\phi_i$: this gives a
+$D_i$-action on $\Omega$ whose restriction to $D_i^0$ is exactly the
+original $L$-action.
+
+Do this independently for each $i$. On a shared edge $\gamma=\{i,j\}$,
+both copies of the generator $x_\gamma\in L$ act on $\Omega$ by the
+same element, so the resulting actions glue to a $Q$-action on
+$\Omega$. The action restricted to each $D_i$ is faithful, since
+the original left action on $D_i\times\mathbb{N}$ is faithful and
+$\phi_i$ is a bijection.
+
+Therefore $D_i\to Q$ is injective for every vertex $i$. This
+completes the proof of Lemma~\ref{lem:triangle-free-gluing}.
+
+\subsection{Applying the criterion to $H_i$}
+\label{subsec:gluing-application}
+
+We now apply the lemma to our concrete setting. Recall $H=A_0\rtimes
+P$ with $A_0=(\mathbb Z/M\mathbb Z)^P$, $P=\F_p^3$, $U=u(P)$,
+$V=cUc^{-1}$, and $U\cap V=\{1\}$ from Proposition~\ref{prop:UV-triv}.
+For each edge $i<j$ of $\Gamma$ we identify the cyclic subgroup
+generated by $v(1,\bar j,\bar j^2)$ in $H_i$ with the one generated
+by $u(1,\bar i,\bar i^2)$ in $H_j$. Here the bars denote residues in
+$\F_p$, and both elements have exact order $p$.
+
+\begin{proposition}\label{prop:first-gluing-injective}
+Every natural homomorphism $H_i\to G_0$ is injective. Consequently
+the induced maps $K[H_i]\to K[G_0]$ on group algebras are injective.
+\end{proposition}
+
+\begin{proof}
+Apply Lemma~\ref{lem:triangle-free-gluing} with $\Delta=\Gamma$ and
+$D_i=H$. The vertex groups $H_i$ are finite, the cyclic edge groups
+have order $p$, and the edge embeddings are given by the displayed
+vectors. We must verify the short-word condition.
+
+Fix a vertex $i$. The incident edges split into two kinds:
+
+\begin{itemize}
+\item $j<i$: the syllable from edge $\{i,j\}$ in $H_i$ lies in~$U$ and
+      is $u(\lambda\cdot(1,\bar i,\bar i^2))$ for some scalar
+      $\lambda\in\F_p^*$; the vector $(1,\bar i,\bar i^2)$ is attached
+      via the $H_j$ side.
+\item $j>i$: the syllable from edge $\{i,j\}$ in $H_i$ lies in~$V$ and
+      is $v(\lambda\cdot(1,\bar j,\bar j^2))$ for some
+      $\lambda\in\F_p^*$.
+\end{itemize}
+
+Within each of $U$ or $V$, distinct neighbors $j$ correspond to the
+vectors $(1,\bar j,\bar j^2)\in\F_p^3$. Three of these vectors with
+three distinct residues $\bar j_1,\bar j_2,\bar j_3$ are linearly
+independent over $\F_p$: the determinant
+\[
+\det\begin{pmatrix}
+ 1 & \bar j_1 & \bar j_1^2\\
+ 1 & \bar j_2 & \bar j_2^2\\
+ 1 & \bar j_3 & \bar j_3^2
+\end{pmatrix}
+= (\bar j_2-\bar j_1)(\bar j_3-\bar j_1)(\bar j_3-\bar j_2)
+\ne 0,
+\]
+because $p>m>3$, so the three residues are distinct in $\F_p$. The
+same argument handles two vectors (a $2\times 2$ minor of the above)
+and one vector (nonzero).
+
+So a word of at most $3$ syllables, all in $U$ or all in $V$, with
+distinct labels, evaluates nontrivially in $F_p^3$ and hence in $H_i$.
+
+For a mixed word of length $2$ or $3$: exactly one of $U$ or $V$
+supplies a single syllable, and the other supplies one or two. After
+a cyclic permutation, a putative relation would equate a nonidentity
+element of $U$ to a product of elements of~$V$. But $U\cap V=\{1\}$
+and the $U$-element is nonidentity, contradiction. For lengths $2$
+and $3$, cyclically distinct labels are just distinct labels, so
+the check is exhaustive.
+
+All hypotheses of Lemma~\ref{lem:triangle-free-gluing} hold, giving
+the injection $H_i\to G_0$. The induced map $K[H_i]\to K[G_0]$ is
+injective because distinct elements of an embedded subgroup remain
+distinct basis elements of the group algebra.
+\end{proof}
+
+Consequently $G_0$ is a single group, finitely generated (by the
+finite sets of generators of the finitely many $H_i$), containing
+each $H_i$ as a subgroup, and the $e_i\in K[G_0]$ satisfy
+Corollary~\ref{op:edge-zero}.