\section{Local CRT box presentations for
\texorpdfstring{$C_{2pq}$}{C2pq}}\label{sec:boxes}

Fix distinct odd primes $p<q$ and identify
$C_{2pq}=C_2\times C_p\times C_q$.  In the fixed order
\[
 (E,P,Q,P1,Q1,F,G)
 =\bigl(\Orb2,\Orb p,\Orb q,\Orb{2p},\Orb{2q},\Orb{pq},
        \Orb{2pq}\bigr),                                    \tag{6.1}
\]
these are the seven nonzero automorphism orbits.  The orbit $\Orb1=\{0\}$
will be added only after the seven-box classification.

The name ``box'' records the nonzero CRT coordinates.  With
$a\in C_p^\times$ and $b\in C_q^\times$, the complete list is
\[
\begin{array}{c|c|c|c}
\text{box}&\text{points}&\text{order}&\text{cardinality}\\
\midrule
E&(1,0,0)&2&1\\
P&(0,a,0)&p&p-1\\
Q&(0,0,b)&q&q-1\\
P1&(1,a,0)&2p&p-1\\
Q1&(1,0,b)&2q&q-1\\
F&(0,a,b)&pq&(p-1)(q-1)\\
G&(1,a,b)&2pq&(p-1)(q-1).
\end{array}                                                   \tag{6.1a}
\]
Thus a stable support is a choice of boxes and possibly $\Orb1$.  There are
$2^7-1=127$ nonempty base masks and $2(127)+1=255$ nonempty stable supports.

The quotient $A_d(B)$ is universal for functional equations: a map
$h:B\to H$ into any abelian group kills every zero-sum row of length at most
$d$ exactly when it factors through $A_d(B)$.  The proof below solves these
equations and then uses coordinate characters to show that the resulting
parameters are independent.  This identifies the quotient, not merely an
upper bound on it.

For a single box $B$, the cumulative quotients through degree four are as
follows; ``same'' repeats the preceding entry, and the $p=3$ rows have
$q\ge5$.

\begin{table}[ht]
\centering
\small
\begin{tabular}{@{}lccc@{}}
\toprule
$B$ & $A_2(B)$ & $A_3(B)$ & $A_4(B)$\\
\midrule
$E$ & $C_2$ & $C_2$ & $C_2$\\
$P$ & $\Z^{(p-1)/2}$ & $C_p$ & $C_p$\\
$Q$ & $\Z^{(q-1)/2}$ & $C_q$ & $C_q$\\
$P1$, $p=3$ & $\Z$ & $\Z$ & $\Z$\\
$P1$, $p\ge5$ & $\Z^{(p-1)/2}$ & same & $C_{2p}$\\
$Q1$ & $\Z^{(q-1)/2}$ & same & $C_{2q}$\\
$F$ & $\Z^{(p-1)(q-1)/2}$ & $C_p\oplus C_q$ & same\\
$G$, $p=3$ & $\Z^{q-1}$ & same & $\Z\oplus C_q$\\
$G$, $p\ge5$ & $\Z^{(p-1)(q-1)/2}$ & same
    & $C_2\oplus C_p\oplus C_q$\\
\bottomrule
\end{tabular}
\caption{Cumulative local box presentations.}\label{tab:local}
\end{table}

Equivalently, at degree four a homomorphism from a selected box quotient to
an arbitrary abelian target has the universal normal forms
\[
\begin{array}{c|l}
E&c,\ 2c=0,\\
P&au,\ pu=0,\\
Q&bv,\ qv=0,\\
P1\ (p\ge5)&c+au,\ 2c=pu=0,\\
P1\ (p=3)&au,\ u\text{ free},\\
Q1&c+bv,\ 2c=qv=0,\\
F&au+bv,\ pu=qv=0,\\
G\ (p\ge5)&c+au+bv,\ 2c=pu=qv=0,\\
G\ (p=3)&au+bv,\ u\text{ free},\ qv=0.
\end{array}                                                   \tag{6.2}
\]

\begin{proposition}[Local box theorem]\label{prop:local-boxes}
Table~\ref{tab:local} and the normal forms in (6.2) hold uniformly for all
admissible $p,q$.
\end{proposition}

\begin{proof}
Every nontrivial box is stable under negation without fixed points.  Pair
rows identify opposite generators and give the free ranks in the $A_2$
column; on $E$, the row $2e_E$ gives $C_2$.

On a parity-zero punctured prime line, pairs and triples give partial
additivity $f(x+y)=f(x)+f(y)$ whenever the three terms are nonzero.  Moving
along the punctured line yields $f(x)=xu$, and the full prime cycle gives
$ru=0$.  This proves the $P,Q$ rows.

To see the punctured-line step explicitly, fix $x\ne0$.  Partial additivity
gives $f(kx)=kf(x)$ until the additive cycle returns to zero; the final step
gives $rf(x)=0$.  Every nonzero point is a multiple of $x$, so with
$u=x^{-1}f(x)$ one obtains $f(a)=au$.  Conversely this formula kills every
zero-sum row, proving that the $C_r$ factor is exact.

On a parity-one punctured line there are no odd rows.  For $r\ge5$, the
quadruples $(x,1,1,-x-2)$ give $f(x+2)=f(x)+2f(1)$.  Removing zero from the
translation-by-two cycle leaves a connected path, so $f(x)=c+xu$.
An opposite pair gives $2c=0$.  For $r\ge7$, the valid row
$(2,2,2,r-6)$ gives $4c+ru=0$, hence $ru=0$.  At the endpoint $r=5$, the
separate zero-sum row $(1,1,1,2)$ gives $4c+5u=0$, hence $5u=0$.
The coordinate characters realize both factors.  For $r=3$, every
quadruple contains two signs of each kind and gives nothing beyond the pair
relation, leaving one free generator.  This proves both $P1$ branches and the
always-generic $Q1$ row, including the endpoint $q=5$.

The exception at three is structural.  When $r\ge5$, deleting zero from the
translation-by-two cycle leaves enough overlapping edges to compare all
values.  For $r=3$, the two remaining vertices are opposites and every even
zero-sum relation uses the two signs equally.  The generic $C_2$ constant and
$C_3$ slope therefore merge into one unrestricted integral sign parameter.

For $F$ with $p,q\ge5$, fix a horizontal displacement $s$ and compare
\[
 (x+s,y)+(c,d)=(x,y)+(c+s,d).                               \tag{6.3}
\]
An auxiliary $(c,d)$ can avoid at most four $p$-classes and three
$q$-classes; at $p=5$ the endpoint is handled by using the already fixed
nonzero endpoints before choosing the auxiliary class.  Thus horizontal
differences are independent of the base point.  A second comparison makes
them additive, and the vertical argument is identical.  Hence
$f(a,b)=c+au+bv$ with $pu=qv=0$.  Pairs give $2c=0$ and a valid triple gives
$3c=0$, so $c=0$.

When $p=3$, write $h(b)=f(1,b)$ and recover the negative sheet by pairs.
Same-sheet triples make pair sums with a fixed nonzero total have the same
$h$-value.  A bridge point avoiding at most four $q$-classes covers zero
totals, already for $q=5$.  The difference equation gives
$h(b)=u+bv$, $3u=0$, $qv=0$.  Therefore $A_3(F)=C_3\oplus C_q$.

For $G$ with $p,q\ge5$, a zero-sum quadruple made from two opposite-sheet
pairs makes every horizontal and vertical difference base-point independent.
The same avoidance argument gives $f(a,b)=c+au+bv$.  Pairs impose $2c=0$,
and the three coordinate characters prove independence of the
$C_2,C_p,C_q$ factors.

Finally let $p=3$.  Again write $h(b)=f(1,b)$.  Every zero-sum quadruple has
two signs of each kind and is a Freiman relation on $C_q^\times$.  Its
universal solutions are $h(b)=u+bv$ with $u$ free and $qv=0$.  This gives
$A_4(G)=\Z\oplus C_q$ and completes the table.
\end{proof}

The avoidance arguments are uniform at the small endpoints.  The two points
being compared are fixed before the auxiliary point is selected.  At most
four $p$-classes and three $q$-classes are then unavailable.  At $p=5$ the
explicit endpoint relations leave the fifth class; no asymptotic density
estimate or division by $p-3$ is used.  The separate two-sheet proof handles
$p=3$, precisely where the generic auxiliary point need not exist.

Finally, the three CRT coordinate characters restrict to the parity constant,
the $p$-slope, and the $q$-slope in (6.2).  They certify independence of the
three primary factors in every generic row.  In the exceptional $P1,G$ rows,
the integral sign map has infinite image and proves that no local row through
degree four can kill the free generator.

The free sign in the last row is the unique local mechanism capable of
surviving through degree five.  It is the local form of the exceptional unit
support in Section~\ref{sec:units}.

\begin{corollary}[Compatibility on the unit box]\label{cor:unit-compat}
For $G=\Orb{2pq}=\units{2pq}$,
\[
 \depth(G)=6\quad(p=3),\qquad \depth(G)=4\quad(p,q\ge5).
\]
\end{corollary}

\begin{proof}
If $p=3$, the local form is $\Z\oplus C_q$; the critical-family argument in
Section~\ref{sec:profiles} shows exact depth six.  Independently,
Proposition~\ref{prop:unit-lattice} and Corollary~\ref{cor:unit-close} give
the same conclusion for $n=6q$.  If $p,q\ge5$, (6.2) is the ambient CRT
character group and degree three has not yet imposed the generic quartic
relations, so the exact depth is four; this also follows from
$3\nmid2pq$ in Proposition~\ref{prop:unit-lattice}.
\end{proof}