\section{The unit-support lattice for every even modulus}\label{sec:units}

We now compute the integral quotient in Theorem~\ref{thm:unit-main}.  The
main input is a universal ambient group for a sifted arithmetic progression.
For $q\ge1$ and $c\in\Z$, put
\[
 T_{q,c}=\{t\in C_q:\gcd(1+ct,q)=1\}.
\]
For $T\subset C_q$, let $\UAG(T)$ be the quotient of $\Z^T$ by the Freiman
rows
\[
 e_a+e_b-e_{a'}-e_{b'}\qquad(a+b=a'+b'\text{ in }C_q).
\]
This is the universal target for Freiman homomorphisms on $T$, in the sense of
Konyagin--Lev and the structural additive literature
\cite{KonyaginLev2000,Grynkiewicz2013}.

Call a prime divisor of $q$ \emph{restricting} if it does not divide $c$.
Let $d$ be the product of the restricting primes below five.  In the
application $d\in\{1,2,3,6\}$.

\begin{theorem}[Sifted universal ambient group]\label{thm:sifted-uag}
Assume every restricting prime not dividing $d$ is at least five.  Then
\[
 \UAG(T_{q,c})\cong \Z^{\varphi(d)}\oplus C_{q/d}.            \tag{4.1}
\]
\end{theorem}

\begin{proof}
Write $T=T_{q,c}$.  If $s,s'\in dC_q$, the Chinese remainder theorem supplies
$x$ such that
\[
 x,x+s\in T,\qquad x,x+s,x+s+s'\in T.                       \tag{4.2}
\]
Indeed, at a prime dividing $d$ these points occupy one residue class and we
avoid its unique forbidden class.  At every other restricting prime there
are at most three forbidden classes, while the prime has size at least five.
The choices lift through prime powers and then combine by CRT.

At a restricting prime power $\ell^a\Vert q$, membership in $T$ excludes one
residue class modulo $\ell$, not one class modulo $\ell^a$.  We first choose
$x$ modulo $\ell$ outside the at most three forbidden classes and then lift it
arbitrarily modulo $\ell^a$.  At a nonrestricting prime, $\ell\mid c$, so
$1+cx$ is never zero modulo $\ell$ and no avoidance is needed.  This explains
why $d$ records small primes but not their exponents.

Let $W$ be the attained admissible classes modulo $d$; then $|W|=\varphi(d)$.
Every Freiman relation preserves the multiset of its two classes in $W$.
This point deserves an explicit local check.  Modulo two there is only one
admissible class, so preservation is automatic; this uniqueness is what makes
the cases $d=2,6$ work.  Modulo three there are two admissible classes, and
the three unordered two-element multisets have distinct sums.  Combining
these statements by CRT proves the assertion modulo $d$.  Consequently the
class-count functions descend to $\UAG(T)$.

The multiset conclusion is stronger than preservation of the total class
sum.  Modulo three, translate the admissible classes to $\{0,1\}$: the three
multisets $\{0,0\},\{0,1\},\{1,1\}$ have distinct sums $0,1,2$.  Modulo two,
there is only one admissible class.  The latter uniqueness, rather than a
generic two-class argument, is the load-bearing fact in the $d=2,6$ cases.

For $s\in dC_q$, define
\[
 \Delta(s)=[e_{x+s}]-[e_x],
\]
where $x,x+s\in T$.  The first part of (4.2) and
$(x+s)+y=x+(y+s)$ make this independent of $x$.  The three-point part gives
$\Delta(s+s')=\Delta(s)+\Delta(s')$.  Thus $D=\Delta(d)$ has order dividing
$q/d$.

Choose $t_w\in T$ in each $w\in W$.  Every generator has the normal form
\[
 [e_t]=[e_{t_w}]+\Delta(t-t_w).
\]
Define
\[
 \Psi(e_t)=\left(e_w,\frac{t-t_w}{d}\right)
       \in\Z^W\oplus C_{q/d}.                               \tag{4.3}
\]
Class-multiset preservation cancels the first coordinates of a Freiman row,
and equality of the original sums cancels the second.  Hence $\Psi$ descends.
It sends the chosen representatives onto the free basis and $D$ to $1$, so it
is surjective.  Conversely the displayed normal forms generate $\UAG(T)$;
an element in that form with zero image has every free coefficient and the
slope coefficient equal to zero.  Thus $\Psi$ is injective.
\end{proof}

Equivalently, a Freiman homomorphism $F:T_{q,c}\to A$ is determined by one
base value for every admissible class $w$ modulo $d$ and one additive slope
$D$ of order dividing $q/d$:
\[
 F(t)=F(t_w)+\frac{t-t_w}{d}D.                              \tag{4.4a}
\]
There is no relation among the $\varphi(d)$ base values before inverse pairs
are imposed.  In the unit application with $d=3$, the difference of the two
base values is precisely the exceptional free direction.

\subsection{Fourth-order dichotomy}

Write $n=2N$.  The bijection $t\mapsto u(t)=1+2t$ identifies
$T_{N,2}$ with $\units n$.  Negation corresponds to the involution
$\iota(t)=-1-t$.  There are no odd-length zero-sum rows on $\units n$:
an odd number of odd residues cannot sum to zero modulo the even integer $n$.

Modulo the inverse-pair rows, every zero-sum quadruple becomes a Freiman
relation.  More precisely, replacing the last two entries by their inverse
partners turns
\[
 u(a)+u(b)+u(c)+u(d)=0
\]
into $a+b=\iota(c)+\iota(d)$ in $C_N$, and the converse construction is
immediate.  Therefore
\[
 \Z^{\units n}/\calL_{\le4}(\units n)
 \cong\UAG(T_{N,2})/\angles{\text{inverse rows}}.             \tag{4.4}
\]

\begin{proposition}[All-even unit quotient]\label{prop:unit-lattice}
For every even $n$,
\[
 \Z^{\units n}/\calL_{\le4}(\units n)\cong
 \begin{cases}
 C_n,&3\nmid n,\\
 \Z\oplus C_{n/6},&6\mid n.
 \end{cases}                                                 \tag{4.5}
\]
Moreover $\calL_{\le5}(\units n)=\calL_{\le4}(\units n)$.
\end{proposition}

\begin{proof}
If $3\nmid n$, Theorem~\ref{thm:sifted-uag} gives
$\Z\oplus C_N$, with $e_t\mapsto(1,t)$.  An inverse row maps to $(2,-1)$.
Writing $A$ for the free generator and $B$ for the torsion generator, the
quotient has $B=2A$ and $2NA=0$, hence is $C_{2N}=C_n$.

If $n=6m$, the same theorem gives $\Z^2\oplus C_m$.  The free coordinates
count the admissible classes $0,2$ modulo three, which the involution swaps.
Every inverse row therefore has image $(1,1,\kappa)$ for one fixed
$\kappa\in C_m$.  The vector $(1,1)$ is primitive.  Quotienting eliminates
exactly one free generator, and its torsion component can be absorbed into
that eliminated generator.  The result is $\Z\oplus C_m$.  Finally the
parity observation preceding (4.4) excludes every odd row, proving the last
assertion.
\end{proof}

In the second branch, the torsion absorption can be made explicit.  Let
$A_0,A_2$ be the free basis vectors for the two admissible classes modulo
three and let $B$ generate $C_m$.  Replace $A_2$ by $A_2+\kappa B$.  This is
an automorphism of $\Z^2\oplus C_m$, and it changes the inverse row to the
primitive vector $A_0+A_2$.  Quotienting leaves
$\Z(A_0-A_2)\oplus C_m$.  Thus the apparently nonzero slope component of the
inverse row neither kills nor enlarges any torsion factor.

The target map supplies a useful check.  In the generic branch the surviving
generator maps to $1\in C_n$, so its order is exactly $n$.  In the exceptional
branch the surviving free difference maps to zero in the finite target and is
therefore a genuine obstruction, while the $C_m$ coordinate is the residual
translation coordinate.

\subsection{Degree-six closure}

Suppose $n=6m$ and set
\[
 v_t=1+6t,\qquad
 T_m=\{t\in C_m:\gcd(1+6t,m)=1\}.
\]
Then $\units{6m}=\{\pm v_t:t\in T_m\}$.  After the inverse rows, send
\[
 e_{v_t}\longmapsto(1,t),\qquad e_{-v_t}\longmapsto(-1,-t)
       \quad\text{in }\Z\oplus C_m.                         \tag{4.6}
\]
Every row of length at most five maps to zero.  Its first coordinate is the
difference between the numbers of residues congruent to $1$ and $-1$ modulo
six; it is divisible by six and has absolute value at most five.  The
zero-sum congruence then kills the second coordinate.  The translation kernel
inside (4.6) is generated by $(6,-1)$.

\begin{lemma}[Sifted two-sum lemma]\label{lem:sumset}
$T_m+T_m=C_m$.
\end{lemma}

\begin{proof}
For prescribed $y\in C_m$, choose $a$ prime by prime so that both $a$ and
$y-a$ lie in $T_m$.  A prime restricting these conditions divides neither
six nor the relevant coefficient, so it is at least five; at most two residue
classes are forbidden.  Primes dividing six impose no restriction.  Lifting
through prime powers and applying CRT gives the required $a$.
\end{proof}

Prime by prime, the two excluded classes are explicitly $-6^{-1}$ and
$y+6^{-1}$ modulo a prime $\ell\nmid6$; they may coincide, but there are never
more than two.  Since then $\ell\ge5$, a class remains.  If $\ell=2$ or $3$,
$1+6a$ is automatically a unit modulo $\ell$, even when $\ell$ divides $m$.

Choose three pairs in $T_m$ with sums $-1,0,0$.  The corresponding six
positive units sum to $6+6(-1)=0$ modulo $6m$, and their row maps to $(6,-1)$.
Thus degree six kills the last kernel generator.

Every row of length at most five killed the entire presentation in (4.6),
whereas $(6,-1)$ is nonzero there.  The constructed sixth-order row is
therefore a generator of the complete translation kernel, not merely an
arbitrary relation appearing at degree six.  This supplies both the strict
lower bound and the closing upper bound.

\begin{corollary}\label{cor:unit-close}
If $6\mid n$, then $\calL_{\le6}(\units n)=K(\units n)$, while the free
summand in (4.5) proves nonclosure through degree five.
\end{corollary}

Proposition~\ref{prop:unit-lattice} and
Corollary~\ref{cor:unit-close} prove the lattice part of
Theorem~\ref{thm:unit-main}.  The global order-six reconstruction theorem of
Gr\"unbaum--Moore \cite{GrunbaumMoore1995} is compatible with this bound;
the explicit support quotient and its closing row are the calculations above.