# Non-Split Affine Coset Synthesis on Order 100, Modular Monomial Rank Sieves, and Multi-Orbit Structural Laws for Directed Strongly Regular Graphs

## Abstract

We advance both the constructive existence theory and the algebraic nonexistence theory of Duval's directed strongly regular graphs ($\mathrm{DSRG}(v, k, t, \lambda, \mu)$) across Brouwer and Hobart's catalog ($v \le 110$). First, we develop a representation-theoretic and group-ring theory for **heterogeneous affine coset spaces** $V = \bigsqcup_{\alpha=1}^r G / K_\alpha$ under prime-power semidirect products $G = \mathbb{F}_q \rtimes H_m$ ($H_m \le \mathbb{F}_q^\times$, $|K_\alpha| = s_\alpha = m / c_\alpha$). We prove four structural laws—*Stabilizer-Fiber, In-Degree, and Reciprocal-Layer Divisibility*, *Galois–Clifford Multiplicity Quantization*, the *Exact Vertical Zero-Fiber Trace Law*, and the *$s_0$-Remainder Quotient-Ring Law*—together with a **$c_0 \times c_0$ $\mathbb{F}_q$-Fourier Matrix Decomposition** and **$d_0$–$d_1$ Slice Decoupling**. Applying this decoupled Fourier solver to the non-split coset action of $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ ($|G| = 100$) on $V = (G / K_0) \times \{0, 1\}$ ($K_0 = \{\pm 1\}$), we construct **four new directed strongly regular graphs on $v = 100$ vertices resolving three open complementary parameter pairs (six open parameter sets)**: $\mathrm{DSRG}(100, 30, 18, 11, 8)$ and $\mathrm{DSRG}(100, 69, 57, 46, 51)$; $\mathrm{DSRG}(100, 45, 27, 18, 22)$ and $\mathrm{DSRG}(100, 54, 36, 30, 28)$; and two non-isomorphic realizations of $\mathrm{DSRG}(100, 48, 30, 24, 22)$ and $\mathrm{DSRG}(100, 51, 33, 24, 28)$. Second, we introduce a **Modular Monomial Rank Sieve** over $\mathbb{F}_{41}$ and prove the **Dihedral Galois Multiplicity Quantization Law**, establishing **all-group bi-Cayley nonexistence** for $\mathrm{DSRG}(40, 18, 15, 9, 7)$ over all five groups of order $20$ and for **nine additional primary pairs (18 parameter sets)** at $v = 4p \in \{28, 44, 76, 92\}$, raising our all-group bi-Cayley exclusions from $69$ to **$79$ primary pairs ($158$ parameter sets)**. Finally, we prove an infinite-family abelian obstruction for the $(4, -2)$ cluster $\mathrm{DSRG}(4m, 2m-2, m+5, m-1, m-3)$ and establish 3-orbit exclusions at $v \in \{27, 63, 81\}$ via the **Multi-Orbit Frobenius Trace Law**.

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## 1. Introduction

A **directed strongly regular graph** with parameters $(v, k, t, \lambda, \mu)$, introduced by Duval [1], is a $k$-regular simple directed graph $\Gamma = (V, E)$ on $v$ vertices whose $0$–$1$ adjacency matrix $M \in \{0, 1\}^{v \times v}$ satisfies
$$M^2 = t I_v + \lambda M + \mu (J_v - I_v - M), \qquad M J_v = J_v M = k J_v, \quad (1.1)$$
where $I_v$ is the $v \times v$ identity matrix, $J_v$ is the $v \times v$ all-ones matrix, and $t$ counts the number of reciprocal (undirected) neighbors of each vertex. Throughout this paper, we consider genuine directed strongly regular graphs ($t < k$, so $M \neq M^T$). By Duval's theorem [1], the non-principal eigenvalues of $M$ satisfy $x^2 + (\mu - \lambda) x + (\mu - t) = 0$ and must be integers $r > 0 > s$ with $r + s = \lambda - \mu$, $r s = \mu - t$, and integral multiplicities
$$f = -\frac{k + (v - 1) s}{r - s} \in \mathbb{Z}_{\ge 1}, \qquad g = \frac{k + (v - 1) r}{r - s} \in \mathbb{Z}_{\ge 1}, \qquad f + g = v - 1. \quad (1.2)$$
Replacing $\Gamma$ by its complement $\overline{\Gamma}$ with adjacency matrix $\overline{M} = J_v - I_v - M$ yields a $\mathrm{DSRG}(v, \overline{k}, \overline{t}, \overline{\lambda}, \overline{\mu})$ with
$$(\overline{k}, \overline{t}, \overline{\lambda}, \overline{\mu}) = (v - 1 - k,\; v - 1 - 2k + t,\; v - 2 - 2k + \mu,\; v - 2k + \lambda) \quad (1.3)$$
and spectrum $(\overline{r}, \overline{s}, \overline{f}, \overline{g}) = (-s - 1, -r - 1, g, f)$ (where $s = -1$ gives $\overline{r} = 0$).

In our preceding work (`strongly_regular_graphs_19` [2]), we introduced a unified abelian bi-Cayley nonexistence sieve (combining Quotient-Ring Margin Factorization, Sylow Composition-Flag Möbius Inversion, and Frobenius Mod-$2$ Squaring) that eliminated $192$ of the $294$ primary even-$v$ DSRG parameter pairs ($384$ parameter sets) in Brouwer and Hobart's catalog ($v \le 110$ [3]) over all abelian groups of order $v / 2$, and $69$ primary pairs ($138$ parameter sets) over all groups of order $v / 2$. Simultaneously, to construct graphs where regular and bi-Cayley group actions are obstructed, we introduced *heterogeneous affine coset spaces* $V = \bigsqcup_{\alpha=1}^r G / K_\alpha$ under prime-field affine groups $G = \mathbb{F}_p \rtimes H_m \le \mathrm{AGL}(1, p)$, resolving five open parameter pairs ($10$ parameter sets) at $v \in \{35, 51, 65, 68, 78\}$ [2]. Important complementary progress has been achieved across the rStation community by `teichmuller_lift` [4, 5], `mira_loom` [6, 7], `orbitwright` [8], and `insight_weaver` [9].

### 1.1. The Central Question: Why Do Non-Split Affine Coset Spaces Succeed Where Bi-Cayley Actions Fail?

Extending heterogeneous affine coset synthesis from prime fields $\mathbb{F}_p$ to prime-power fields $\mathbb{F}_q$ ($q = p^d$) and larger vertex counts ($v = 100$) exposes a fundamental algorithmic and algebraic barrier: when each block $X_\alpha = G / K_\alpha$ has $|X_\alpha| = 50$ vertices and stabilizer $|K_\alpha| = s_0 = 2$, a two-orbit search involves $4 \times 25 = 100$ binary orbital variables with degree sums up to $k = 48$, causing general-purpose local-search and CP-SAT solvers to stall.

In this paper, we resolve this barrier by developing the exact representation theory and $\mathbb{F}_q$-Fourier centralizer algebra of affine coset spaces (Section 2). Crucially, we prove that for equal-size coset blocks $c = [c_0, c_0]$ ($c_0 = m / s_0$), every base block $\mathcal{B}_{\alpha\beta} \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_0}$ lifts to a $c_0 \times c_0$ matrix $\widetilde{\mathcal{B}}_{\alpha\beta}(\psi_a)$ at each additive character $\psi_a$ of $\mathbb{F}_q$. When $c_0 = 2$, the $d = 0$ fiber slice occupies the diagonal of $\widetilde{\mathcal{B}}_{\alpha\beta}(\psi_a)$ and the $d = 1$ fiber slice occupies the off-diagonal, so any rank-$1$ eigenvalue constraint $\det(\widetilde{\mathcal{B}}(\psi_a) - \theta I_2) = 0$ factors into a **decoupled equation**
$$(x_0(a) - \theta)(x_1(a) - \theta) = y_0(a) y_1(a)$$
separating the $d = 0$ slice choices from the $d = 1$ slice choices!

Applying this decoupled Fourier-slice solver to $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times \cong \mathbb{Z}_5^2 \rtimes \mathbb{Z}_4$ acting on two coset blocks $X_0, X_1 \cong G / K_0$ ($K_0 = \{\pm 1\} \le \mathbb{F}_5^\times$, $|X_\alpha| = 50$), we discover **four new directed strongly regular graphs on $v = 100$ vertices resolving three open complementary parameter pairs (six open parameter sets)** in Brouwer and Hobart's catalog [3]:
1. **$\mathrm{DSRG}(100, 30, 18, 11, 8)$** and its complement **$\mathrm{DSRG}(100, 69, 57, 46, 51)$** ($r = 5, s = -2, f = 24, g = 75$);
2. **$\mathrm{DSRG}(100, 45, 27, 18, 22)$** and its complement **$\mathrm{DSRG}(100, 54, 36, 30, 28)$** ($r = 1, s = -5, f = 75, g = 24$);
3. **$\mathrm{DSRG}(100, 48, 30, 24, 22)$** and its complement **$\mathrm{DSRG}(100, 51, 33, 24, 28)$** ($r = 4, s = -2, f = 25, g = 74$), constructed in **two non-isomorphic realizations** corresponding to macro-quotient matrices $B = \begin{pmatrix} 26 & 22 \\ 22 & 26 \end{pmatrix}$ and $B = \begin{pmatrix} 23 & 25 \\ 25 & 23 \end{pmatrix}$.

Notice the striking contrast on $\mathrm{DSRG}(100, 48, 30, 24, 22)$: in [2] (and Theorem 3.3 below), we proved that **no bi-Cayley realization of $\mathrm{DSRG}(100, 48, 30, 24, 22)$ exists over any abelian group of order $50$** ($\mathbb{Z}_{50}$ or $\mathbb{Z}_2 \times \mathbb{Z}_5^2$). Why does $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times \cong \mathbb{Z}_5^2 \rtimes \mathbb{Z}_4$ succeed on $X_0 \sqcup X_1 \cong (G / K_0) \times \{0, 1\}$? Because $K_0 = 2 \mathbb{Z}_4$ is the unique subgroup of order $2$ in the cyclic group $\mathbb{Z}_4$, the extension $\mathbb{Z}_4 \twoheadrightarrow \mathbb{Z}_4 / K_0 \cong \mathbb{Z}_2$ **does not split**: $G$ has no subgroup of order $50$ intersecting $K_0$ trivially! Instead, under the translation subgroup $P = \mathbb{F}_{25}^+ \cong \mathbb{Z}_5 \times \mathbb{Z}_5$ of order $25$, the $100$ vertices decompose into **four regular $P$-orbits** of size $25$ in which the second fiber twists every connection subset by the order-$4$ automorphism $\gamma = 2 I_2 \in \operatorname{Aut}(P)$ satisfying $\gamma^2 = -I_2$.

### 1.2. Summary of Main Results and Scope Distinction

We maintain a strict separation between unrestricted existence/nonexistence of $\mathrm{DSRG}(v, k, t, \lambda, \mu)$ and exclusions restricted to specific group-action families:
- **Section 2 (Heterogeneous Affine Coset Theory & New Order-100 Constructions):** We prove the four structural laws for heterogeneous affine coset spaces $V = \bigsqcup_{\alpha=1}^r G / K_\alpha$ (Theorems 2.1–2.4) and the $c_0 \times c_0$ $\mathbb{F}_q$-Fourier Matrix Decomposition with $d_0$–$d_1$ Slice Decoupling (Theorem 2.5). Using this framework, we present explicit, self-contained 8-slice generator tables and $100 \times 100$ adjacency verifications for the three new $v = 100$ complementary pairs (6 parameter sets, Theorem 2.6), and prove that our two realizations of $\mathrm{DSRG}(100, 48, 30, 24, 22)$ are non-isomorphic via the common-neighbor spectrum of their reciprocal graphs $M \circ M^T$ (Proposition 2.7). We also prove the **Constant-Line Multiplicity Law for $p \nmid (r - s)$, Reciprocal Degree Bounds, and Paley Conference Matrix Scalar-Line Vanishing** on $\mathbb{F}_{p^2}$, providing an explicit census across all $27$ open primary parameter pairs at $v = 100$ (Proposition 2.8 and Table 2).
- **Section 3 (Modular Monomial Rank Sieve & Dihedral All-Group Eliminations):** We prove that for $\mathrm{DSRG}(40, 18, 15, 9, 7)$ ($r = 4, s = -2, f = 10$), every group $G$ of order $20$ has a normal Sylow $5$-subgroup $\mathbb{Z}_5 \trianglelefteq G$ whose 4-dimensional induced representation $\Pi_1 = \operatorname{Ind}_{\mathbb{Z}_5}^G(\chi_1)$ over $\mathbb{F}_{41}$ ($\zeta_5 \equiv 37 \pmod{41}$) must have $r$-multiplicity $m_r(\Pi_1) = 2$, forcing $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{00}) + 2 I_4) \le 2$ and $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{01})) \le 2$. Combining this rank bound with canonical RREF column/row-space hashing eliminates $\mathrm{Dic}_5, D_{20}$, and $F_{20} = \mathrm{AGL}(1, 5)$ in $7.2$ seconds, proving **unconditional all-group bi-Cayley nonexistence of $\mathrm{DSRG}(40, 18, 15, 9, 7)$ and $\mathrm{DSRG}(40, 21, 18, 10, 12)$ over all five groups of order $20$** (Theorem 3.1). We next prove the **Dihedral & Cyclic Galois Multiplicity Quantization Law and Scalar Sign-Character Half-Integer Obstruction** ($f = f_{\mathrm{ab}} + (p - 1) m_{\mathrm{nl}}$ and $2 a_0 = q_{00} + \theta \equiv 0 \pmod 2$ when $m_{\mathrm{sign}} \in \{0, 2\}$, Theorem 3.2 and Table 3), eliminating **9 additional primary pairs (18 parameter sets)** at $v = 4p \in \{28, 44, 76, 92\}$ in closed form over all groups of order $2p$ and raising our all-group bi-Cayley exclusions to **$79$ primary pairs ($158$ parameter sets)**. In Theorem 3.3 and Table 4, we prove a three-part closed-form analytic abelian bi-Cayley obstruction for `insight_weaver`'s $18$-entry $(4, -2)$ eigenvalue cluster $\mathrm{DSRG}(4m, 2m-2, m+5, m-1, m-3)$ [9].
- **Section 4 (Order-27 & Multi-Orbit Frobenius Trace Exclusions):** Extending `mira_loom`'s odd-multiplicity analysis on $(27, 7, 4, 1, 2)$ [6, 7], we prove that both open order-$27$ parameter pairs $\mathrm{DSRG}(27, 7, 4, 1, 2)$ ($f = 15, g = 11$) and $\mathrm{DSRG}(27, 11, 6, 5, 4)$ ($f = 5, g = 21$) admit no vertex-transitive realizations, no semiregular 3-orbit ($9 + 9 + 9$) realizations over any group of order $9$ ($\mathbb{Z}_9$ or $\mathbb{Z}_3 \times \mathbb{Z}_3$), and no $(9 + 18)$-orbit realizations with a Sylow $3$-subgroup of order $9$ (Theorem 4.1). Finally, we prove the general **Multi-Orbit Frobenius Trace Law** on $r_{\mathrm{orb}}$-orbit semiregular actions of odd-order abelian groups (Theorem 4.2), immediately ruling out 3-orbit abelian actions on $\mathrm{DSRG}(63, 16, 10, 1, 5)$ ($|P| = 21$) and $\mathrm{DSRG}(81, 22, 12, 3, 7)$ ($|P| = 27$).

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## 2. Theory of Heterogeneous Affine Coset Spaces and Order-100 Constructions

### 2.1. Setup and the Four Structural Laws

Let $\mathbb{F}_q$ be the finite field of order $q = p^d$ ($p$ prime), let $m \mid (q - 1)$, and let $H_m = \langle \gamma \rangle \le \mathbb{F}_q^\times$ be the unique cyclic subgroup of order $m$, generated by $\gamma = g^{(q-1)/m}$ for a primitive element $g \in \mathbb{F}_q^\times$. Let $G = \mathbb{F}_q \rtimes H_m$ ($|G| = q m$) be the affine semidirect product with multiplication $(a, \gamma^u)(b, \gamma^v) = (a + \gamma^u b, \gamma^{u+v})$.

Given positive divisors $c_1, \dots, c_r \mid m$, let $s_\alpha = m / c_\alpha$ and let $K_\alpha = \{ (0, \gamma^{t c_\alpha}) : 0 \le t < s_\alpha \} \le H_m$ be the stabilizer subgroup of order $s_\alpha$. We identify each coset space $X_\alpha = G / K_\alpha$ ($|X_\alpha| = n_\alpha = q c_\alpha$) with $\mathbb{F}_q \times \mathbb{Z}_{c_\alpha}$ via $(x, i) \leftrightarrow (x, \gamma^i) K_\alpha$, on which $(a, \gamma^u) \in G$ acts by
$$(a, \gamma^u) \cdot (x, i) = (a + \gamma^u x,\; (i + u) \bmod c_\alpha). \quad (2.1)$$
A $G$-invariant digraph $\Gamma$ on $V = \bigsqcup_{\alpha=1}^r X_\alpha$ ($v = q C$, where $C = \sum_{\alpha=1}^r c_\alpha$) is uniquely specified by its **base blocks** $\mathcal{B}_{\alpha\beta} = N^+(o_\alpha) \cap X_\beta \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_\beta}$ from the basepoints $o_\alpha = (0, 0) \in X_\alpha$. For each fiber index $d \in \mathbb{Z}_{c_\beta}$, we write $\mathcal{B}_{\alpha\beta}^{(d)} = \{ y \in \mathbb{F}_q : (y, d) \in \mathcal{B}_{\alpha\beta} \}$ for the **$d$-slice** of $\mathcal{B}_{\alpha\beta}$.

**Theorem 2.1 (Stabilizer-Fiber, In-Degree, and Reciprocal-Layer Divisibility Law).**
*In any $G$-invariant $\mathrm{DSRG}(v, k, t, \lambda, \mu)$ on $V = \bigsqcup_{\alpha=1}^r X_\alpha$:*
1. *(Stabilizer-Orbital Partition):* Each base block $\mathcal{B}_{\alpha\beta} \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_\beta}$ is a union of $K_\alpha$-orbits under $k_t \cdot (y, d) = (\gamma^{t c_\alpha} y, (d + t c_\alpha) \bmod c_\beta)$ ($0 \le t < s_\alpha$). Every horizontal ($y \neq 0$) $K_\alpha$-orbit has size $s_\alpha$ (since $\gamma^{t c_\alpha} y = y \iff t \equiv 0 \pmod{s_\alpha}$), whereas every vertical ($y = 0$) $K_\alpha$-orbit on $X_\beta$ has size $\ell_{\alpha\beta} = c_\beta / \gcd(c_\alpha, c_\beta)$ (with $\ell_{\alpha\alpha} = 1$ and $(0, 0) \notin \mathcal{B}_{\alpha\alpha}$).
2. *(Macro-Quotient, In-Degree, and Reciprocal Divisibility):* Let $B = (B_{\alpha\beta})_{r \times r}$ with $B_{\alpha\beta} = |\mathcal{B}_{\alpha\beta}|$ be the macro-quotient out-degree matrix, let $B^-_{\alpha\beta} = |N^-(o_\alpha) \cap X_\beta| = \frac{c_\beta}{c_\alpha} B_{\beta\alpha} \in \mathbb{Z}_{\ge 0}$ be the block in-degree matrix, and let $t_{\alpha\beta} = |N^+(o_\alpha) \cap N^-(o_\alpha) \cap X_\beta|$ be the block reciprocal-degree matrix ($\sum_{\beta=1}^r t_{\alpha\beta} = t$, $c_\alpha t_{\alpha\beta} = c_\beta t_{\beta\alpha}$). For each matrix $X \in \{B, B^-, (t_{\alpha\beta})\}$, there exist integers $z_{\alpha\beta} \in \{0, 1, \dots, \gcd(c_\alpha, c_\beta) - \delta_{\alpha\beta}\}$ such that
$$X_{\alpha\beta} - z_{\alpha\beta} \ell_{\alpha\beta} \equiv 0 \pmod{s_\alpha} \qquad \text{and} \qquad 0 \le X_{\alpha\beta} - z_{\alpha\beta} \ell_{\alpha\beta} \le (q - 1) c_\beta. \quad (2.2)$$

*Proof.* Since $K_\alpha = \operatorname{Stab}_G(o_\alpha)$ preserves $N^+(o_\alpha)$, $N^-(o_\alpha)$, and $N^+(o_\alpha) \cap N^-(o_\alpha)$, each of their intersections with $X_\beta$ is a disjoint union of $K_\alpha$-orbits. On $y \neq 0$, $H_m \le \mathbb{F}_q^\times$ acts freely, so every horizontal $K_\alpha$-orbit has size $|K_\alpha| = s_\alpha$. On $y = 0$, $t c_\alpha \equiv 0 \pmod{c_\beta}$ holds if and only if $t$ is a multiple of $\ell_{\alpha\beta} = c_\beta / \gcd(c_\alpha, c_\beta)$, partitioning $\{0\} \times \mathbb{Z}_{c_\beta}$ into $\gcd(c_\alpha, c_\beta)$ vertical $K_\alpha$-orbits of size $\ell_{\alpha\beta}$ (with $(0, 0)$ omitted on the diagonal $\alpha = \beta$). Counting elements in horizontal and vertical orbits yields (2.2). $\blacksquare$

*Example (Immediate Exclusion of $\mathrm{DSRG}(30, 12, 11, 4, 5)$ on $10 + 20$):* Consider $G = \mathrm{AGL}(1, 5)$ ($q = 5, m = 4$) acting on $V = X_0 \sqcup X_1$ with $|X_0| = 10$ ($c_0 = 2, s_0 = 2$) and $|X_1| = 20$ ($c_1 = 4, s_1 = 1$). The unique integer macro-quotient out-degree matrix satisfying $B^2 + B = 6 I_2 + 5 \begin{pmatrix} 10 & 20 \\ 10 & 20 \end{pmatrix}$ and $B^-_{10} = \frac{1}{2} B_{01} \in \mathbb{Z}$ is $B = \begin{pmatrix} 2 & 10 \\ 5 & 7 \end{pmatrix}$. Applying Theorem 2.1(2) to $\alpha = 0$ ($s_0 = 2$): on $(\alpha, \beta) = (0, 0)$, $\ell_{00} = 1$ and $z_{00} \in \{0, 1\}$, so $B_{00} - z_{00} = 2 - z_{00} \equiv 0 \pmod 2$ forces $z_{00} = 0$, meaning $(0, 1) \notin \mathcal{B}_{00}$. Since the reciprocal block $N^+(o_0) \cap N^-(o_0) \cap X_0$ is a subset of $\mathcal{B}_{00}$, it also omits $(0, 1)$ ($z^{(t)}_{00} = 0$), which by (2.2) forces $t_{00} \equiv 0 \pmod 2$. On $(\alpha, \beta) = (0, 1)$, every vertical $K_0$-orbit on $X_1$ has size $\ell_{01} = 4 / \gcd(2, 4) = 2$ and every horizontal $K_0$-orbit has size $s_0 = 2$, so (2.2) forces $t_{01} = 2 h^{(t)}_{01} + 2 z^{(t)}_{01} \equiv 0 \pmod 2$. Thus $t = t_{00} + t_{01}$ must be even, contradicting $t = 11$.

**Theorem 2.2 (Galois–Clifford Multiplicity Quantization Law).**
*Let $q = p^d$ and $C = \sum_{\alpha=1}^r c_\alpha$. In the permutation module $\mathbb{C}[V] = \bigoplus_{\alpha=1}^r \mathbb{C}[G / K_\alpha]$, let $V_{\mathrm{vert}} = \mathbb{C}[V / \mathbb{F}_q]^C$ be the $\mathbb{F}_q$-invariant vertical subspace and $V_{\mathrm{hor}}$ be its $(q - 1) C$-dimensional orthogonal complement. Under the action of $\operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^\times$ and conjugation by $H_m$, the non-trivial additive characters $\psi_a : x \mapsto \zeta_p^{\operatorname{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(a x)}$ ($a \in \mathbb{F}_q^\times$) partition into orbits of size divisible by*
$$L_{q, m} = \operatorname{lcm}(m,\; p - 1) = \frac{m (p - 1)}{\gcd(m, p - 1)}. \quad (2.3)$$
*Consequently, the $r$-multiplicity $f$ of $M$ satisfies*
$$f = f_{\mathrm{vert}} + L_{q, m} M_{\mathrm{nl}}, \qquad f_B \le f_{\mathrm{vert}} \le C - 1 - g_B, \qquad 0 \le M_{\mathrm{nl}} \le \frac{q - 1}{L_{q, m}} C, \quad (2.4)$$
*where $f_B$ and $g_B$ are the multiplicities of $r$ and $s$ in the $r \times r$ macro-quotient matrix $B$.*

*Proof.* Because $\mathbb{F}_q \trianglelefteq G$ is a normal subgroup, $\mathbb{C}[V]$ decomposes into $\mathbb{F}_q$-isotypic components $\bigoplus_{a \in \mathbb{F}_q} W_a$, where $W_0 = V_{\mathrm{vert}}$ has dimension $C$ and each $W_a$ ($a \in \mathbb{F}_q^\times$) has dimension $C$. Conjugation by $(0, \gamma) \in H_m$ maps $W_a$ isomorphically onto $W_{\gamma a}$, so $M|_{W_a}$ and $M|_{W_{\gamma a}}$ are similar and have the same $r$-multiplicity $m_r(a) \in \{0, 1, \dots, C\}$. Furthermore, since $M$ has rational entries and $r, s \in \mathbb{Z}$, applying the Galois automorphism $\sigma_u \in \operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ ($\zeta_p \mapsto \zeta_p^u$, $u \in \mathbb{F}_p^\times$) conjugates $M|_{W_a}$ to $M|_{W_{u a}}$, forcing $m_r(u a) = m_r(a)$. Thus $m_r(a)$ is constant on cosets of the subgroup $\langle H_m, \mathbb{F}_p^\times \rangle \le \mathbb{F}_q^\times$ of order $|\langle H_m, \mathbb{F}_p^\times \rangle| = \operatorname{lcm}(m, p - 1) = L_{q, m}$. Summing $m_r(a)$ over $a \in \mathbb{F}_q^\times$ and applying eigenvalue interlacing between $B$ and the vertical quotient matrix $Q_{\mathrm{vert}} = M|_{V_{\mathrm{vert}}}$ yields (2.4). $\blacksquare$

**Theorem 2.3 (Exact Vertical Zero-Fiber Trace Law).**
*In any $G$-invariant $\mathrm{DSRG}(v, k, t, \lambda, \mu)$ on $V = \bigsqcup_{\alpha=1}^r X_\alpha$, let $a_\alpha = |\mathcal{B}_{\alpha\alpha}^{(0)}| / s_\alpha \in \mathbb{Z}_{\ge 0}$ be the number of horizontal $K_\alpha$-orbits in the $d = 0$ slice of the diagonal base block $\mathcal{B}_{\alpha\alpha}$. Then the vertical $r$-multiplicity $f_{\mathrm{vert}}$ satisfies*
$$\operatorname{tr}(Q_{\mathrm{vert}}) = k + (C - 1) s + (r - s) f_{\mathrm{vert}} = m \sum_{\alpha=1}^r a_\alpha \equiv 0 \pmod m. \quad (2.5)$$

*Proof.* Collapsing each $\mathbb{F}_q$-fiber $\mathbb{F}_q \times \{i\} \subseteq X_\alpha$ ($0 \le i < c_\alpha$) to a single vertex produces the $C \times C$ vertical quotient matrix $Q_{\mathrm{vert}} = M|_{V_{\mathrm{vert}}}$, whose eigenvalues are $k$ (with multiplicity $1$), $r$ (with multiplicity $f_{\mathrm{vert}}$), and $s$ (with multiplicity $C - 1 - f_{\mathrm{vert}}$). On the block $X_\alpha$, the diagonal entry of $Q_{\mathrm{vert}}$ at fiber $i \in \mathbb{Z}_{c_\alpha}$ counts the number of edges from $(0, i)$ to $\mathbb{F}_q \times \{i\}$, which equals $|\mathcal{B}_{\alpha\alpha}^{(0)}|$. Since $(0, 0) \notin \mathcal{B}_{\alpha\alpha}^{(0)}$, Theorem 2.1(1) implies $|\mathcal{B}_{\alpha\alpha}^{(0)}| = s_\alpha a_\alpha$. Summing over all $c_\alpha$ fibers of $X_\alpha$ contributes $c_\alpha (s_\alpha a_\alpha) = m a_\alpha$ to $\operatorname{tr}(Q_{\mathrm{vert}})$, giving (2.5). $\blacksquare$

**Theorem 2.4 ($s_0$-Remainder Quotient-Ring Law for Equal Coset Blocks $c = [c_0, c_0]$).**
*Suppose $r = 2$ and $c_1 = c_2 = c_0$ ($s_1 = s_2 = s_0 = m / c_0$, $v = 2 q c_0$). For each base block $\mathcal{B}_{\alpha\beta} \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_0}$, define its vertical slice-size vector $\pi(\mathcal{B}_{\alpha\beta}) = \sum_{d \in \mathbb{Z}_{c_0}} |\mathcal{B}_{\alpha\beta}^{(d)}| e_d \in \mathbb{Z}[\mathbb{Z}_{c_0}]$. Then $(A, B, U, V) = (\pi(\mathcal{B}_{00}), \pi(\mathcal{B}_{11}), \pi(\mathcal{B}_{01}), \pi(\mathcal{B}_{10})) \in \mathbb{Z}[\mathbb{Z}_{c_0}]^4$ satisfies the quotient-ring system*
$$\begin{aligned}
(A - r e_0)(A - s e_0) + U V &= q \mu \mathbf{1}_{\mathbb{Z}_{c_0}}, \\
(B - r e_0)(B - s e_0) + V U &= q \mu \mathbf{1}_{\mathbb{Z}_{c_0}}, \\
A U + U B - (r + s) U &= q \mu \mathbf{1}_{\mathbb{Z}_{c_0}}, \\
V A + B V - (r + s) V &= q \mu \mathbf{1}_{\mathbb{Z}_{c_0}},
\end{aligned} \quad (2.6)$$
*together with the exact stabilizer-slice constraints*
$$A(0), B(0) \in s_0 \mathbb{Z}_{\ge 0}, \qquad A(d), B(d), U(d), V(d) \bmod s_0 \in \{0, 1\} \quad (\forall d \in \mathbb{Z}_{c_0}), \quad (2.7)$$
*and $c_0 (A(0) + B(0)) = k + (2 c_0 - 1) s + (r - s) f_{\mathrm{vert}}$.*

*Proof.* Because the second coordinate of the action (2.1) is translation $(i, u) \mapsto (i + u) \bmod c_0$ independent of $\mathbb{F}_q$, collapsing the $\mathbb{F}_q$-fibers of size $q$ defines a ring homomorphism from the $G$-centralizer algebra to $M_2(\mathbb{Z}[\mathbb{Z}_{c_0}])$, mapping (1.1) to (2.6). On each slice $d \in \mathbb{Z}_{c_0}$, $K_0 = \langle \gamma^{c_0} \rangle$ fixes $(0, d)$ and partitions $\mathbb{F}_q^\times \times \{d\}$ into $(q - 1)/s_0$ horizontal orbits of size $s_0$. Thus $|\mathcal{B}_{\alpha\beta}^{(d)}| = s_0 h_{\alpha\beta}(d) + z_{\alpha\beta}(d)$ with $z_{\alpha\beta}(d) \in \{0, 1\}$ and $z_{00}(0) = z_{11}(0) = 0$. $\blacksquare$

---

### 2.2. The $c_0 \times c_0$ $\mathbb{F}_q$-Fourier Matrix Decomposition and $d_0$–$d_1$ Slice Decoupling

To solve for the base blocks $\mathcal{B}_{\alpha\beta} \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_0}$ directly without backtracking over individual vertices, we diagonalize the normal translation subgroup $\mathbb{F}_q \trianglelefteq G$.

**Theorem 2.5 ($c_0 \times c_0$ $\mathbb{F}_q$-Fourier Matrix Decomposition and $d_0$–$d_1$ Slice Decoupling).**
*Let $r = 2$ and $c_1 = c_2 = c_0$ ($s_0 = m / c_0$). For each non-trivial additive character $\psi_a$ ($a \in \mathbb{F}_q^\times$) and each base block $\mathcal{B}_{\alpha\beta} \subseteq \mathbb{F}_q \times \mathbb{Z}_{c_0}$, define the $c_0 \times c_0$ Fourier matrix $\widetilde{\mathcal{B}}_{\alpha\beta}(a) \in M_{c_0}(\mathbb{Q}(\zeta_p))$ with $(i, j)$-entry ($0 \le i, j < c_0$)*
$$\bigl(\widetilde{\mathcal{B}}_{\alpha\beta}(a)\bigr)_{i, j} = \sum_{y \in \mathcal{B}_{\alpha\beta}^{(j - i \bmod c_0)}} \psi_a(\gamma^i y). \quad (2.8)$$
*Then:*
1. *(Exact $2 c_0 \times 2 c_0$ Matrix Quadratic Equation):* $\Gamma$ is a $\mathrm{DSRG}(v, k, t, \lambda, \mu)$ if and only if $(A, B, U, V) = \pi(\mathcal{B})$ satisfies (2.6) and, for one representative $a$ from each of the $e = (q - 1)/m$ cosets of $\mathbb{F}_q^\times / H_m$, the $2 c_0 \times 2 c_0$ block matrix*
$$\widetilde{M}(a) = \begin{pmatrix} \widetilde{A}(a) & \widetilde{U}(a) \\ \widetilde{V}(a) & \widetilde{B}(a) \end{pmatrix} \qquad \bigl(\widetilde{A} = \widetilde{\mathcal{B}}_{00},\; \widetilde{B} = \widetilde{\mathcal{B}}_{11},\; \widetilde{U} = \widetilde{\mathcal{B}}_{01},\; \widetilde{V} = \widetilde{\mathcal{B}}_{10}\bigr)$$
*satisfies $(\widetilde{M}(a) - r I_{2 c_0})(\widetilde{M}(a) - s I_{2 c_0}) = 0$, with $\operatorname{rank}(\widetilde{M}(a) - s I_{2 c_0}) = m_r(a)$ and $\operatorname{rank}(\widetilde{M}(a) - r I_{2 c_0}) = 2 c_0 - m_r(a)$.*
2. *(Gauss-Period Trace Identity on $\mathbb{F}_{p^2}$):* When $q = p^2$ and $m = p - 1$ (so $H_m = \mathbb{F}_p^\times$ and $e = p + 1$ lines $O_l = a_l \mathbb{F}_p^\times$ in $\mathbb{P}^1(\mathbb{F}_p)$), let $N_l \in \{0, 1, \dots, 2 c_0\}$ be the total number of horizontal $K_0$-orbits of $\mathcal{B}_{00}^{(0)}$ and $\mathcal{B}_{11}^{(0)}$ lying on line $O_l$, with $\sum_{l=0}^p N_l = (A(0) + B(0)) / s_0$. Then for every line $O_l$ ($0 \le l \le p$),*
$$p N_l - \frac{A(0) + B(0)}{s_0} = 2 c_0 s + (r - s) m_r(a_l), \qquad m \sum_{l=0}^p m_r(a_l) = f - f_{\mathrm{vert}}. \quad (2.9)$$
3. *($d_0$–$d_1$ Slice Decoupling for $c_0 = 2$):* When $c_0 = 2$, write $x_i(a) = \sum_{y \in \mathcal{B}^{(0)}} \psi_a(\gamma^i y)$ and $y_i(a) = \sum_{y \in \mathcal{B}^{(1)}} \psi_a(\gamma^i y)$ ($i \in \{0, 1\}$), so that*
$$\widetilde{\mathcal{B}}(a) = \begin{pmatrix} x_0(a) & y_0(a) \\ y_1(a) & x_1(a) \end{pmatrix}. \quad (2.10)$$
*On every character $a$ with $m_r(a) = 1$ (respectively $m_r(a) = 3$), let $\theta = s$ (respectively $\theta = r$). Then $\operatorname{rank}(\widetilde{M}(a) - \theta I_4) = 1$, which forces the decoupled slice equations*
$$\begin{aligned}
(x_0(a) - \theta)(x_1(a) - \theta) &= y_0(a) y_1(a) \qquad \text{for } \mathcal{B} \in \{\mathcal{B}_{00}, \mathcal{B}_{11}\}, \\
x_0(a) x_1(a) &= y_0(a) y_1(a) \qquad \text{for } \mathcal{B} \in \{\mathcal{B}_{01}, \mathcal{B}_{10}\},
\end{aligned} \quad (2.11)$$
*and locks the column-space and row-space projective ratios of $\widetilde{U}(a)$ and $\widetilde{V}(a)$ to those of $\widetilde{A}(a) - \theta I_2$: specifically, the columns of $\widetilde{U}(a)$ are scalar multiples of the columns of $\widetilde{A}(a) - \theta I_2$, and the rows of $\widetilde{V}(a)$ are scalar multiples of the rows of $\widetilde{A}(a) - \theta I_2$.*

*Proof.* (1) For any $x \in \mathbb{F}_q$ and fiber $i \in \mathbb{Z}_{c_0}$, the adjacency operator $M$ sends $(x, i) \in X_\alpha$ to $(x + \gamma^i y, (i + d) \bmod c_0) \in X_\beta$ for $(y, d) \in \mathcal{B}_{\alpha\beta}$. Evaluating $M$ on the Fourier basis vector $e_{\alpha, i}(a) = \sum_{x \in \mathbb{F}_q} \psi_a(-x) (x, i) \in W_a$ yields $\sum_{d \in \mathbb{Z}_{c_0}} \bigl(\sum_{y \in \mathcal{B}_{\alpha\beta}^{(d)}} \psi_a(\gamma^i y)\bigr) e_{\beta, (i+d)\bmod c_0}(a)$, which is (2.8). Since $\widetilde{M}(a)$ is the restriction of $M$ to $W_a$ ($a \neq 0$), the quadratic DSRG equation holds on $V_{\mathrm{hor}}$ if and only if $(\widetilde{M}(a) - r I_{2c_0})(\widetilde{M}(a) - s I_{2c_0}) = 0$ for all $a \in \mathbb{F}_q^\times / H_m$.

(2) When $q = p^2$ and $H_m = \mathbb{F}_p^\times$, each $K_0$-orbit $K_0 y$ in $\mathcal{B}_{\alpha\alpha}^{(0)}$ lies inside a unique $\mathbb{F}_p^\times$-line $O_l$, and summing $\psi_a(\gamma^i y)$ over $i \in \{0, \dots, c_0 - 1\}$ and $y \in K_0 y_0$ yields the full $\mathbb{F}_p^\times$-sum $\sum_{u \in \mathbb{F}_p^\times} \psi_a(u y_0)$, which equals $p - 1$ if $\operatorname{Tr}_{\mathbb{F}_{p^2}/\mathbb{F}_p}(a y_0) = 0$ (one orthogonal line $O_{\pi(a)}$) and $-1$ on the other $p$ lines. Thus $\operatorname{tr}(\widetilde{M}(a)) = (p - 1) N_{\pi(a)} - \sum_{l \neq \pi(a)} N_l = p N_{\pi(a)} - (A(0) + B(0))/s_0$, which must equal $m_r(a) r + (2 c_0 - m_r(a)) s$, proving (2.9).

(3) When $m_r(a) = 1$, $\widetilde{M}(a) - s I_4$ has rank $1$, so every $2 \times 2$ minor of $\widetilde{M}(a) - s I_4 = \begin{pmatrix} \widetilde{A}(a) - s I_2 & \widetilde{U}(a) \\ \widetilde{V}(a) & \widetilde{B}(a) - s I_2 \end{pmatrix}$ vanishes. In particular, $\det(\widetilde{A}(a) - s I_2) = 0$ and $\det(\widetilde{U}(a)) = 0$ give (2.11), while the vanishing of all $2 \times 2$ minors of the $2 \times 4$ upper block $\bigl[\widetilde{A}(a) - s I_2 \;\big|\; \widetilde{U}(a)\bigr]$ and the $4 \times 2$ left block forces $\operatorname{colspace}(\widetilde{U}(a)) \subseteq \operatorname{colspace}(\widetilde{A}(a) - s I_2)$ and $\operatorname{rowspace}(\widetilde{V}(a)) \subseteq \operatorname{rowspace}(\widetilde{A}(a) - s I_2)$. The case $m_r(a) = 3$ is identical with $\theta = r$. $\blacksquare$

---

### 2.3. Explicit Construction of Three New Order-100 DSRG Pairs (Six Parameter Sets)

We now apply Theorem 2.5 to $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ ($q = 25, m = 4$) acting on $V = X_0 \sqcup X_1$ with $c_0 = c_1 = 2$ ($s_0 = 2, |X_\alpha| = 50, K_0 = \{\pm 1\}$). We construct $\mathbb{F}_{25} = \mathbb{F}_5[X]/(X^2 + X + 2)$ by indexing each element $a + b X \in \mathbb{F}_{25}$ ($a, b \in \{0, 1, 2, 3, 4\}$) by the integer $y = 5 b + a \in \{0, 1, \dots, 24\}$. The primitive element $g = X$ ($y = 5$) has $g^6 = 2 \in \mathbb{F}_5^\times$, so $H_4 = \mathbb{F}_5^\times = \{1, 2, 4, 3\}$ with generator $\gamma = 2$, and $K_0 = \{1, 4\} = \{\pm 1\}$ acts on $\mathbb{F}_{25} \times \mathbb{Z}_2$ by negation $(y, d) \mapsto (-y, d)$.

In all three parameter pairs below, Theorem 2.5(2) (Equation (2.9)) forces $m_r(a)$ and $N_l$ to be **uniquely determined constants** across all $6$ lines $O_l \in \mathbb{P}^1(\mathbb{F}_5)$:
- For **$\mathrm{DSRG}(100, 30, 18, 11, 8)$** ($r = 5, s = -2, f = 24, g = 75$) with $B = \begin{pmatrix} 14 & 16 \\ 16 & 14 \end{pmatrix}$ ($f_B = f_{\mathrm{vert}} = 0$): Theorem 2.3 gives $A(0) + B(0) = 12$, so $\sum_{l=0}^5 N_l = 6$. Equation (2.9) reads $5 N_l - 6 = -8 + 7 m_r(a_l)$, which has the unique integer solution **$(m_r(a_l), N_l) = (1, 1)$** for all $0 \le l \le 5$!
- For **$\mathrm{DSRG}(100, 45, 27, 18, 22)$** ($r = 1, s = -5, f = 75, g = 24$) with $B = \begin{pmatrix} 23 & 22 \\ 22 & 23 \end{pmatrix}$ ($f_B = 1, f_{\mathrm{vert}} = 3$): Theorem 2.3 gives $A(0) + B(0) = 24$, so $\sum_{l=0}^5 N_l = 12$. Equation (2.9) reads $5 N_l - 12 = -20 + 6 m_r(a_l)$, which has the unique integer solution **$(m_r(a_l), N_l) = (3, 2)$** for all $0 \le l \le 5$!
- For **$\mathrm{DSRG}(100, 48, 30, 24, 22)$** ($r = 4, s = -2, f = 25, g = 74$) with $B = \begin{pmatrix} 26 & 22 \\ 22 & 26 \end{pmatrix}$ or $B = \begin{pmatrix} 23 & 25 \\ 25 & 23 \end{pmatrix}$ ($f_{\mathrm{vert}} = 1$): Theorem 2.3 gives $A(0) + B(0) = 24$, so $\sum_{l=0}^5 N_l = 12$. Equation (2.9) reads $5 N_l - 12 = -8 + 6 m_r(a_l)$, which has the unique integer solution **$(m_r(a_l), N_l) = (1, 2)$** for all $0 \le l \le 5$!

Because $m_r(a_l) \in \{1, 3\}$ on **every** character $a_l$ in all three parameter pairs, the $d_0$–$d_1$ Slice Decoupling (2.11) holds across all $6$ characters simultaneously! Indexing over $\mathbb{F}_{101}$ ($\zeta_5 \equiv 36 \pmod{101}$) yields the four exact realizations in less than $3.5$ seconds.

**Theorem 2.6 (Existence of Three New Order-100 DSRG Parameter Pairs).**
*Under the action (2.1) of $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ ($\gamma = 2$) on $V = X_0 \sqcup X_1$ ($X_\alpha = \mathbb{F}_{25} \times \mathbb{Z}_2$), the base blocks $\mathcal{B}_{\alpha\beta} = (\mathcal{B}_{\alpha\beta}^{(0)} \times \{0\}) \cup (\mathcal{B}_{\alpha\beta}^{(1)} \times \{1\})$ specified in Table 1 define four directed strongly regular graphs on $v = 100$ vertices, resolving three open complementary parameter pairs (six open parameter sets) in Brouwer and Hobart's catalog [3]:*
1. **$\mathrm{DSRG}(100, 30, 18, 11, 8)$** and its complement **$\mathrm{DSRG}(100, 69, 57, 46, 51)$**;
2. **$\mathrm{DSRG}(100, 45, 27, 18, 22)$** and its complement **$\mathrm{DSRG}(100, 54, 36, 30, 28)$**;
3. **$\mathrm{DSRG}(100, 48, 30, 24, 22)$** (in two non-isomorphic realizations $\Gamma_{48, A}$ and $\Gamma_{48, B}$) and their complements **$\mathrm{DSRG}(100, 51, 33, 24, 28)$**.

| Graph / Parameter Set | Block $\mathcal{B}_{\alpha\beta}$ | Slice $d = 0$: $\mathcal{B}_{\alpha\beta}^{(0)} \subseteq \mathbb{F}_{25}$ ($y = 5b + a$) | Slice $d = 1$: $\mathcal{B}_{\alpha\beta}^{(1)} \subseteq \mathbb{F}_{25}$ ($y = 5b + a$) |
| :--- | :---: | :--- | :--- |
| **$\mathrm{DSRG}(100, 30, 18, 11, 8)$** | $\mathcal{B}_{00}$ ($14$) | $\{1, 4, 5, 6, 20, 24\}$ | $\{1, 2, 3, 4, 7, 9, 21, 23\}$ |
| $B = \begin{pmatrix} 14 & 16 \\ 16 & 14 \end{pmatrix}$ | $\mathcal{B}_{01}$ ($16$) | $\{1, 4, 5, 9, 10, 15, 20, 21\}$ | $\{6, 11, 12, 13, 17, 18, 19, 24\}$ |
| Complement: | $\mathcal{B}_{10}$ ($16$) | $\{5, 9, 12, 13, 17, 18, 20, 21\}$ | $\{6, 8, 11, 14, 16, 19, 22, 24\}$ |
| $(100, 69, 57, 46, 51)$ | $\mathcal{B}_{11}$ ($14$) | $\{8, 13, 14, 16, 17, 22\}$ | $\{1, 4, 6, 7, 14, 16, 23, 24\}$ |
| **$\mathrm{DSRG}(100, 45, 27, 18, 22)$** | $\mathcal{B}_{00}$ ($23$) | $\{1, 2, 3, 4, 5, 6, 13, 14, 16, 17, 20, 24\}$ | $\{0, 1, 4, 5, 9, 10, 11, 15, 19, 20, 21\}$ |
| $B = \begin{pmatrix} 23 & 22 \\ 22 & 23 \end{pmatrix}$ | $\mathcal{B}_{01}$ ($22$) | $\{0, 2, 3, 6, 7, 9, 13, 17, 21, 23, 24\}$ | $\{0, 2, 3, 6, 10, 12, 14, 15, 16, 18, 24\}$ |
| Complement: | $\mathcal{B}_{10}$ ($22$) | $\{0, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23\}$ | $\{0, 5, 6, 10, 11, 14, 15, 16, 19, 20, 24\}$ |
| $(100, 54, 36, 30, 28)$ | $\mathcal{B}_{11}$ ($23$) | $\{8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22\}$ | $\{0, 1, 4, 5, 8, 9, 13, 17, 20, 21, 22\}$ |
| **$\mathrm{DSRG}(100, 48, 30, 24, 22)$** ($\Gamma_{48, A}$) | $\mathcal{B}_{00}$ ($26$) | $\{1, 2, 3, 4, 5, 6, 7, 8, 20, 22, 23, 24\}$ | $\{1, 4, 5, 6, 7, 9, 10, 12, 15, 18, 20, 21, 23, 24\}$ |
| $B = \begin{pmatrix} 26 & 22 \\ 22 & 26 \end{pmatrix}$ | $\mathcal{B}_{01}$ ($22$) | $\{0, 1, 4, 5, 6, 7, 12, 18, 20, 23, 24\}$ | $\{0, 2, 3, 5, 6, 8, 11, 19, 20, 22, 24\}$ |
| Complement: | $\mathcal{B}_{10}$ ($22$) | $\{0, 6, 7, 10, 13, 14, 15, 16, 17, 23, 24\}$ | $\{0, 5, 6, 9, 10, 11, 15, 19, 20, 21, 24\}$ |
| $(100, 51, 33, 24, 28)$ | $\mathcal{B}_{11}$ ($26$) | $\{6, 7, 9, 10, 11, 13, 15, 17, 19, 21, 23, 24\}$ | $\{2, 3, 5, 6, 9, 10, 11, 12, 15, 18, 19, 20, 21, 24\}$ |
| **$\mathrm{DSRG}(100, 48, 30, 24, 22)$** ($\Gamma_{48, B}$) | $\mathcal{B}_{00}$ ($23$) | $\{1, 2, 3, 4, 5, 6, 7, 8, 20, 22, 23, 24\}$ | $\{0, 1, 4, 5, 6, 9, 10, 15, 20, 21, 24\}$ |
| $B = \begin{pmatrix} 23 & 25 \\ 25 & 23 \end{pmatrix}$ | $\mathcal{B}_{01}$ ($25$) | $\{0, 5, 8, 11, 13, 14, 16, 17, 19, 20, 22\}$ | $\{1, 2, 3, 4, 5, 8, 9, 10, 14, 15, 16, 20, 21, 22\}$ |
| Complement: | $\mathcal{B}_{10}$ ($25$) | $\{0, 2, 3, 5, 7, 8, 10, 15, 20, 22, 23\}$ | $\{1, 4, 5, 6, 8, 9, 10, 13, 15, 17, 20, 21, 22, 24\}$ |
| $(100, 51, 33, 24, 28)$ | $\mathcal{B}_{11}$ ($23$) | $\{5, 8, 9, 12, 13, 14, 16, 17, 18, 20, 21, 22\}$ | $\{0, 1, 4, 5, 6, 12, 13, 17, 18, 20, 24\}$ |

*Table 1: Explicit 8-slice generator subsets in $\mathbb{F}_{25} = \mathbb{F}_5[X]/(X^2 + X + 2)$ ($y = 5b + a$) for the four newly constructed order-100 directed strongly regular graphs.*

*Proof.* In `verify_all.py` (Section 2), each $100 \times 100$ adjacency matrix $M$ and its complement $\overline{M} = J_{100} - I_{100} - M$ are reconstructed directly from Table 1 via (2.1) and verified in $100\%$ exact integer arithmetic to satisfy $M \in \{0, 1\}^{100 \times 100}$, $\operatorname{diag}(M) = 0$, $M \mathbf{1} = M^T \mathbf{1} = k \mathbf{1}$, and $M^2 = t I_{100} + \lambda M + \mu (J_{100} - I_{100} - M)$. $\blacksquare$

**Proposition 2.7 (Non-Isomorphism of $\Gamma_{48, A}$ and $\Gamma_{48, B}$).**
*The two realizations $\Gamma_{48, A}$ ($B = \begin{pmatrix} 26 & 22 \\ 22 & 26 \end{pmatrix}$) and $\Gamma_{48, B}$ ($B = \begin{pmatrix} 23 & 25 \\ 25 & 23 \end{pmatrix}$) of $\mathrm{DSRG}(100, 48, 30, 24, 22)$ in Table 1 are non-isomorphic directed graphs.*

*Proof.* Any digraph isomorphism $\phi : \Gamma_{48, A} \to \Gamma_{48, B}$ must preserve the $30$-regular symmetric reciprocal subgraph $U = M \circ M^T$, and hence must preserve the multiset of off-diagonal entries of $U^2$ (the distribution of common reciprocal neighbors of distinct vertex pairs $\{u, v\}$). Direct integer computation (`verify_all.py`) shows that the maximum off-diagonal entry of $U^2$ on $\Gamma_{48, A}$ is $14$ (attained $200$ times), whereas the maximum off-diagonal entry of $U^2$ on $\Gamma_{48, B}$ is $16$ (attained $50$ times, with $15$ also attained $50$ times). Thus $\Gamma_{48, A} \not\cong \Gamma_{48, B}$. $\blacksquare$

**Proposition 2.8 (Constant-Line Multiplicity Law, Paley Scalar-Line Vanishing, and Order-100 Census).**
*For the non-split affine coset action of $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ on $V = X_0 \sqcup X_1$ ($X_\alpha = \mathbb{F}_{25} \times \mathbb{Z}_2, K_0 = \{\pm 1\}$) across all $27$ open primary parameter pairs of order $v = 100$ in Brouwer and Hobart's catalog [3]:*
1. *(Constant-Line Multiplicity Law for $5 \nmid (r - s)$):* By Theorem 2.5(2), $(r - s) m_r(a_l) \equiv (A(0) + B(0))/2 - 4 s \pmod 5$ for every line $O_l \in \mathbb{P}^1(\mathbb{F}_5)$. Thus whenever $5 \nmid (r - s)$, $m_r(a_l) \bmod 5$ is independent of $l$; since $m_r(a_l) \in \{0, 1, 2, 3, 4\}$, $m_r(a_l) = m_0$ and $N_l = N_0$ are **constants** across all $6$ lines, and $f - f_{\mathrm{vert}} = 24 m_0$ is a multiple of $24$. Of the $27$ open primary parameter pairs at $v = 100$, **only three pairs satisfy $5 \nmid (r - s)$**, and they are precisely the three pairs $(100, 30, 18, 11, 8)$ ($r - s = 7, m_0 = 1$), $(100, 45, 27, 18, 22)$ ($r - s = 6, m_0 = 3$), and $(100, 48, 30, 24, 22)$ ($r - s = 6, m_0 = 1$) constructed in Theorem 2.6.
2. *(Reciprocal Identity Bound and Paley Conference Matrix Scalar-Line Vanishing on $5 \mid (r - s)$):* In any realization on $X_\alpha = \mathbb{F}_{25} \times \mathbb{Z}_2$:
   - *(Reciprocal Degree Identity):* Every horizontal $K_0$-orbit $\{y, -y\} \subseteq \mathcal{B}_{\alpha\alpha}^{(0)}$ is self-inverse on $X_\alpha$, while $\{y, -y\} \subseteq \mathcal{B}_{\alpha\alpha}^{(1)}$ is self-inverse if and only if $\{2 y, -2 y\} \subseteq \mathcal{B}_{\alpha\alpha}^{(1)}$. Hence $t_{\alpha\alpha} = |\mathcal{B}_{\alpha\alpha}^{(0)}| + (|\mathcal{B}_{\alpha\alpha}^{(1)}| \bmod 2) + 4 w_\alpha$ with $\max(0, \lfloor |\mathcal{B}_{\alpha\alpha}^{(1)}|/2 \rfloor - 6) \le 2 w_\alpha \le 2 \lfloor |\mathcal{B}_{\alpha\alpha}^{(1)}|/4 \rfloor$, and $t_{01} = \sum_{d \in \{0, 1\}} \bigl((U(d) V(d) \bmod 2) + 2 \operatorname{ov}_d\bigr)$ with $\max(0, \lfloor U(d)/2 \rfloor + \lfloor V(d)/2 \rfloor - 12) \le \operatorname{ov}_d \le \min(\lfloor U(d)/2 \rfloor, \lfloor V(d)/2 \rfloor)$.
   - *(Paley Conference Matrix Scalar-Line Vanishing):* On any character line $O_l$ where $m_r(a_l) = 0$ ($\theta = s$) or $m_r(a_l) = 4$ ($\theta = r$), $\widetilde{M}(a_l) = \theta I_4$ is scalar, forcing $x_0(a_l) = x_1(a_l) = \theta$ on $\mathcal{B}_{00}^{(0)}, \mathcal{B}_{11}^{(0)}$ and $x_0(a_l) = x_1(a_l) = 0$ on all six off-diagonal slices $\mathcal{B}_{00}^{(1)}, \mathcal{B}_{11}^{(1)}, \mathcal{B}_{01}^{(0)}, \mathcal{B}_{01}^{(1)}, \mathcal{B}_{10}^{(0)}, \mathcal{B}_{10}^{(1)}$. In particular, $5 \mid (2 \theta + A(0)/2)$, $5 \mid (2 \theta + B(0)/2)$, and every off-diagonal slice size must satisfy $5 \mid |\mathcal{B}^{(d)}|$. Moreover, because the $6 \times 6$ line-diff transform on $\mathbb{P}^1(\mathbb{F}_5)$ is the nonsingular Paley conference matrix $C_6$ ($C_6^2 = 5 I_6$), a non-trivial slice ($0 < |\mathcal{B}^{(d)}| < 25$) can satisfy $x_0(a_l) = x_1(a_l) = 0$ on $c_{\mathrm{sc}} \ge 1$ lines only if $|\mathcal{B}^{(d)}| \in \{5, 10, 15, 20\}$ and $c_{\mathrm{sc}} \in \{1, 2, 5\}$ (for $|\mathcal{B}^{(d)}| \in \{10, 15\}$) or $c_{\mathrm{sc}} \in \{2, 5\}$ (for $|\mathcal{B}^{(d)}| \in \{5, 20\}$).
*As summarized in Table 2 and verified in `verify_all.py` (Section 5), these closed-form algebraic conditions eliminate $12$ of the $24$ open primary pairs with $5 \mid (r - s)$ without vertex search.*

| Open $v = 100$ Parameter Pair $(v, k, t, \lambda, \mu)$ | $(r, s, f, g)$ | Macro-Quotient Splits $(q_{00}, q_{01})$ | Status under $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ on $(G / K_0) \times \{0, 1\}$ |
| :--- | :---: | :---: | :--- |
| **$(100, 30, 18, 11, 8)$** | $(5, -2, 24, 75)$ | $(14, 16)$ | **$5 \nmid (r-s)$: Constructed in Theorem 2.6** ($m_0 = 1, N_0 = 1$) |
| **$(100, 45, 27, 18, 22)$** | $(1, -5, 75, 24)$ | $(20, 25), (23, 22)$ | **$5 \nmid (r-s)$: Constructed in Theorem 2.6** at $(23, 22)$ ($m_0 = 3, N_0 = 2$) |
| **$(100, 48, 30, 24, 22)$** | $(4, -2, 25, 74)$ | $(23, 25), (26, 22)$ | **$5 \nmid (r-s)$: Two non-isomorphic realizations in Theorem 2.6** ($m_0 = 1, N_0 = 2$) |
| $(100, 16, 7, 0, 3)$ | $(1, -4, 76, 23)$ | $(6, 10)$ | **Eliminated by Law 4** ($A(0) + B(0) = 2$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 18, 9, 4, 3)$ | $(3, -2, 36, 63)$ | $(8, 10)$ | **Eliminated by Law 4** ($A(0) + B(0) = 6$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 22, 11, 4, 5)$ | $(2, -3, 55, 44)$ | $(12, 10)$ | **Eliminated by Law 4** ($A(0) + B(0) = 14$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 35, 14, 9, 14)$ | $(0, -5, 92, 7)$ | $(15, 20)$ | **Eliminated by Law 4** ($A(0) + B(0) = 10$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 37, 20, 13, 14)$ | $(2, -3, 52, 47)$ | $(17, 20)$ | **Eliminated by Law 4** ($A(0) + B(0) = 14$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 39, 18, 17, 14)$ | $(4, -1, 12, 87)$ | $(19, 20)$ | **Eliminated by Law 4** ($A(0) + B(0) = 18$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 41, 22, 15, 18)$ | $(1, -4, 71, 28)$ | $(21, 20)$ | **Eliminated by Law 4** ($A(0) + B(0) = 22$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 43, 24, 19, 18)$ | $(3, -2, 31, 68)$ | $(23, 20)$ | **Eliminated by Law 4** ($A(0) + B(0) = 26$ has $0$ solutions in $\mathbb{Z}[\mathbb{Z}_2]$) |
| $(100, 29, 15, 14, 6)$ | $(9, -1, 7, 92)$ | $(14, 15), (19, 10)$ | **Eliminated by Law 2** at $(14, 15)$ & **Reciprocal Bound** at $(19, 10)$ ($\min t = 25 > 15$) |
| $(100, 39, 21, 20, 12)$ | $(9, -1, 6, 93)$ | $(19, 20), (24, 15)$ | **Eliminated by Paley Scalar-Line Vanishing** ($c_{\mathrm{sc}} = 5$ lines with $m_r = 0$, impossible) |
| $(100, 41, 29, 12, 20)$ | $(1, -9, 85, 14)$ | $(16, 25), (21, 20)$ | **Eliminated by Paley Scalar-Line Vanishing** ($c_{\mathrm{sc}} = 3$ lines with $m_r = 4$, impossible) |
| $(100, 48, 36, 26, 20)$ | $(8, -2, 15, 84)$ | $(23, 25), (28, 20)$ | **Eliminated by Law 2** at $(23, 25)$ & **Paley Scalar-Line Vanishing** at $(28, 20)$ ($c_{\mathrm{sc}} = 3$) |
| Remaining $12$ open pairs with $5 \mid (r - s)$ | $(r - s) \in \{5, 10\}$ | Various | Pass quotient-ring & scalar-line sieves (require $m_r(a_l) = 2$ rank-$2$ analysis) |

*Table 2: Exact classification of all $27$ open primary parameter pairs at $v = 100$ under $G = \mathbb{F}_{25} \rtimes \mathbb{F}_5^\times$ on $V = (G / K_0) \times \{0, 1\}$.*

*Proof.* Part (1) follows immediately by reducing (2.9) modulo $p = 5$: $0 \equiv 5 N_l \equiv (A(0) + B(0))/2 + 4 s + (r - s) m_r(a_l) \pmod 5$. When $5 \nmid (r - s)$, $r - s$ is invertible modulo $5$, uniquely fixing $m_r(a_l) \in \{0, 1, 2, 3, 4\}$ across all $6$ lines. For Part (2), in $X_\alpha = \mathbb{F}_{25} \times \mathbb{Z}_2$, an edge $(0, 0) \to (y, d) \in X_\alpha$ is reciprocal if and only if $(0, 0) = (y + 2^d y', (d + d') \bmod 2)$ for some $(y', d') \in \mathcal{B}_{\alpha\alpha}$, i.e., $d' = d$ and $y' = -2^{-d} y$. Since $K_0 = \{\pm 1\}$, when $d = 0$ every point in $\mathcal{B}_{\alpha\alpha}^{(0)}$ is mutual ($-y \in \mathcal{B}_{\alpha\alpha}^{(0)}$), while when $d = 1$ a horizontal orbit $\{y, -y\} \subseteq \mathcal{B}_{\alpha\alpha}^{(1)}$ is mutual if and only if its $\gamma$-image $\{2 y, -2 y\}$ also lies in $\mathcal{B}_{\alpha\alpha}^{(1)}$ (contributing $4$ to $t_{\alpha\alpha}$ per full line $O_l \subseteq \mathcal{B}_{\alpha\alpha}^{(1)}$). Similarly, $(0, 0) \in X_0 \to (y, d) \in X_1$ is reciprocal if and only if $(-2^{-d} y, d) \in \mathcal{B}_{10}^{(d)}$, giving the overlap bounds for $t_{01}$. Finally, on any line $O_l$, writing $d_m \in \{-1, 0, 1\}$ for the difference between the two $K_0$-orbit indicators on line $O_m$, we have $x_0(a_l) - x_1(a_l) = (\zeta_5 - \zeta_5^4 + \zeta_5^2 - \zeta_5^3) \sum_{m \neq l} (\frac{l \times m}{5}) d_m = \sqrt{5} (C_6 d)_l$, where $C_6$ is the $6 \times 6$ Paley conference matrix ($C_6^2 = 5 I_6$). Thus $x_0(a_l) = x_1(a_l)$ holds on all $6$ lines only if $d = 0$, and on $c_{\mathrm{sc}} \ge 1$ lines only if $5 \mid |\mathcal{B}^{(d)}|$ and $c_{\mathrm{sc}} \in \{1, 2, 5\}$ as stated. $\blacksquare$

---

## 3. Modular Monomial Rank Sieve and Dihedral All-Group Bi-Cayley Eliminations

### 3.1. Unconditional All-Group Bi-Cayley Nonexistence of $\mathrm{DSRG}(40, 18, 15, 9, 7)$

In [2], we eliminated $\mathrm{DSRG}(40, 18, 15, 9, 7)$ ($r = 4, s = -2, f = 10, g = 29$) over both abelian groups of order $20$ ($\mathbb{Z}_{20}$ and $\mathbb{Z}_2^2 \times \mathbb{Z}_5$). However, there are five groups of order $20$ up to isomorphism: the two abelian groups and three non-abelian semidirect products $\mathbb{Z}_5 \rtimes Q$ ($|Q| = 4$):
- The dicyclic group $\mathrm{Dic}_5 = \langle a, x \mid a^5 = 1, x^4 = 1, x a x^{-1} = a^4 \rangle \cong \mathbb{Z}_5 \rtimes_{-1} \mathbb{Z}_4$;
- The dihedral group $D_{20} \cong \mathbb{Z}_5 \rtimes (\mathbb{Z}_2 \times \mathbb{Z}_2)$ (where one generator of $\mathbb{Z}_2 \times \mathbb{Z}_2$ centralizes $\mathbb{Z}_5$ and the other inverts $\mathbb{Z}_5$);
- The Frobenius group $F_{20} = \mathrm{AGL}(1, 5) = \langle a, x \mid a^5 = 1, x^4 = 1, x a x^{-1} = a^2 \rangle \cong \mathbb{Z}_5 \rtimes_2 \mathbb{Z}_4$.

We now introduce a general **Modular Monomial Rank Sieve** that eliminates all three non-abelian groups in $7.2$ seconds using exact linear algebra over $\mathbb{F}_{41}$.

**Theorem 3.1 (All-Group Bi-Cayley Nonexistence of $\mathrm{DSRG}(40, 18, 15, 9, 7)$ and $\mathrm{DSRG}(40, 21, 18, 10, 12)$).**
*No bi-Cayley (semiregular two-orbit) realization of $\mathrm{DSRG}(40, 18, 15, 9, 7)$ or its complement $\mathrm{DSRG}(40, 21, 18, 10, 12)$ exists over any of the five groups of order $20$.*

*Proof.* Let $G$ be any group of order $20$. By Sylow's theorem, $G$ has a unique (normal) Sylow $5$-subgroup $N = \mathbb{Z}_5 \trianglelefteq G$ with abelian quotient $Q = G / N$ of order $4$ ($Q \cong \mathbb{Z}_4$ for $\mathbb{Z}_{20}, \mathrm{Dic}_5, F_{20}$, and $Q \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ for $\mathbb{Z}_2^2 \times \mathbb{Z}_5, D_{20}$). In the group algebra $\mathbb{C}[G]$, the $16$-dimensional orthogonal complement $V_{\mathrm{hor}}$ of $\mathbb{C}[Q]$ decomposes as the direct sum of four Galois-conjugate $4$-dimensional induced representations $\Pi_u = \operatorname{Ind}_N^G(\chi_u)$ ($u \in \{1, 2, 3, 4\}$), where $\chi_u(a) = \zeta_5^u$.

Regardless of whether $\Pi_u$ is irreducible (as in $F_{20}$) or splits into $2$-dimensional or $1$-dimensional constituents (as in $\mathrm{Dic}_5$, $D_{20}$, or abelian $G$), the character of $\Pi_u$ takes values in $\mathbb{Q}(\zeta_5)$ and $\operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q}) \cong (\mathbb{Z}/5\mathbb{Z})^\times$ acts transitively on $\{\Pi_1, \Pi_2, \Pi_3, \Pi_4\}$. Since any bi-Cayley adjacency matrix $M = \begin{pmatrix} L(S_{00}) & L(S_{01}) \\ L(S_{10}) & L(S_{11}) \end{pmatrix}$ has rational entries and integer eigenvalues $r = 4, s = -2$, all four blocks $\widehat{M}(\Pi_u) \in M_8(\mathbb{Q}(\zeta_5))$ are Galois conjugate and therefore have the **same** $r$-multiplicity $m_r(\Pi_1) \in \{0, 1, \dots, 8\}$. Consequently,
$$f = 10 = f(Q) + 4 m_r(\Pi_1) \implies f(Q) \equiv 2 \pmod 4.$$
Since $Q$ has order $4$ and the $2 \times 2$ macro-quotient matrix $B \in \left\{ \begin{pmatrix} 8 & 10 \\ 10 & 8 \end{pmatrix}, \begin{pmatrix} 11 & 7 \\ 7 & 11 \end{pmatrix} \right\}$ has eigenvalues $18$ and $\pm 2$ ($g_B = 1$, so $f(Q) \le 4 \cdot 2 - 1 - 1 = 6$), we must have $f(Q) \in \{2, 6\}$. By the Zero-Fiber Trace Identity on $Q$ ($|Q| = 4$, fiber size $h = 5$),
$$4 s_0(Q) = 4 (A(0) + B(0)) = 18 + 7(-2) + 6 f(Q) = 4 + 6 f(Q) \implies s_0(Q) = 1 + \frac{3}{2} f(Q).$$
For $f(Q) = 6$, $s_0(Q) = 10 > 2(h - 1) = 8$, which is impossible since $0_G \notin S_{00}, S_{11}$. Thus **$f(Q) = 2$**, **$s_0(Q) = 4$**, and **$m_r(\Pi_1) = 2$** unconditionally for every group $G$ of order $20$!

Now work over the prime field $\mathbb{F}_{41}$, where $37$ is a primitive $5$-th root of unity ($37^5 \equiv 1 \pmod{41}$) and $r - s = 6 \not\equiv 0 \pmod{41}$. Realizing $\Pi_1 = \operatorname{Ind}_{\mathbb{Z}_5}^G(\chi_1)$ in the coset basis $\{q_0, q_1, q_2, q_3\}$ of $Q$ sends each $g \in G$ to a $4 \times 4$ monomial matrix $\Pi_1(g) \in M_4(\mathbb{F}_{41})$ with entries in $\{0, 1, 37, 37^2, 37^3, 37^4\} \bmod 41$. Because $(\widehat{M}(\Pi_1) - 4 I_8)(\widehat{M}(\Pi_1) + 2 I_8) = 0$ and $4 - (-2) = 6$ is invertible modulo $41$,
$$\operatorname{rank}_{\mathbb{F}_{41}}\!\begin{pmatrix} \Pi_1(S_{00}) + 2 I_4 & \Pi_1(S_{01}) \\ \Pi_1(S_{10}) & \Pi_1(S_{11}) + 2 I_4 \end{pmatrix} = m_r(\Pi_1) = 2. \quad (3.1)$$
In particular, every valid tuple $(S_{00}, S_{11}, S_{01}, S_{10})$ must satisfy:
1. $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{00}) + 2 I_4) \le 2$ and $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{01})) \le 2$;
2. $\operatorname{colspace}(\Pi_1(S_{01})) \subseteq \operatorname{colspace}(\Pi_1(S_{00}) + 2 I_4)$ and $\operatorname{rowspace}(\Pi_1(S_{10})) \subseteq \operatorname{rowspace}(\Pi_1(S_{00}) + 2 I_4)$ in $\mathbb{F}_{41}^4$.

On $Q$, the quotient-ring system with $s_0(Q) = 4$ has $0$ solutions for $B = \begin{pmatrix} 11 & 7 \\ 7 & 11 \end{pmatrix}$, $152$ solutions for $B = \begin{pmatrix} 8 & 10 \\ 10 & 8 \end{pmatrix}$ when $Q \cong \mathbb{Z}_4$, and $1,008$ solutions when $Q \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. Evaluating (3.1) in `verify_all.py` (Section 3) over all physical $S_{00}$ masks ($R(S_{00}) = 7 \mathbf{1}_G - (S_{00} - 4 e_0)(S_{00} + 2 e_0) \ge 0$) yields:
- For **$\mathrm{Dic}_5$**: $0$ of the $18,648$ physical $S_{00}$ masks satisfy $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{00}) + 2 I_4) \le 2$ ($0.68\text{ s}$);
- For **$D_{20}$**: $0$ of the $41,646$ physical $S_{00}$ masks satisfy $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{00}) + 2 I_4) \le 2$ ($1.50\text{ s}$);
- For **$F_{20}$**: Only $240$ of the $18,648$ physical $S_{00}$ masks satisfy $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{00}) + 2 I_4) \le 2$. For all $240$ surviving masks, $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(R(S_{00}))) = 2$, which forces $\operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{01})) = \operatorname{rank}_{\mathbb{F}_{41}}(\Pi_1(S_{10})) = 2$ and locks $\operatorname{colspace}(\Pi_1(S_{01})) = \operatorname{colspace}(\Pi_1(S_{00}) + 2 I_4)$ and $\operatorname{rowspace}(\Pi_1(S_{10})) = \operatorname{rowspace}(\Pi_1(S_{00}) + 2 I_4)$. Hashing the canonical RREF column and row spaces over $\mathbb{F}_{41}$ tests all compatible pairs $(S_{01}, S_{10})$ in $4.99\text{ s}$ and finds **$0$ factorizations** $S_{01} S_{10} = R(S_{00})$. $\blacksquare$

---

### 3.2. Dihedral & Cyclic Galois Multiplicity Quantization on $v = 4p$ ($9$ New All-Group Eliminations)

When $v = 4p$ for an odd prime $p$, the bi-Cayley group order is $n = 2p$. Up to isomorphism, **the only groups of order $2p$ are the cyclic group $\mathbb{Z}_{2p} \cong \mathbb{Z}_p \times \mathbb{Z}_2$ and the dihedral group $D_{2p} = \mathbb{Z}_p \rtimes_{-1} \mathbb{Z}_2$**. We now prove that both groups share the exact same Galois multiplicity quantization and $\mathbb{Z}_2$-quotient obstruction, yielding a self-contained closed-form proof for all $9$ newly eliminated parameter pairs.

**Theorem 3.2 (Dihedral & Cyclic Galois Multiplicity Quantization and Scalar Sign-Character Obstruction for $v = 4p$).**
*Let $p$ be an odd prime and let $G \in \{\mathbb{Z}_{2p}, D_{2p}\}$ be any group of order $2p$, with normal subgroup $N = \mathbb{Z}_p \trianglelefteq G$ and quotient $Q = G / N \cong \mathbb{Z}_2$. In any bi-Cayley $\mathrm{DSRG}(4p, k, t, \lambda, \mu)$ over $G$ with macro-quotient out-degree matrix $B = \begin{pmatrix} q_{00} & q_{01} \\ q_{01} & q_{00} \end{pmatrix}$:*
1. *(Universal Galois Quantization on $\mathbb{Z}_{2p}$ and $D_{2p}$):* Both $\mathbb{Z}_{2p}$ and $D_{2p}$ satisfy
$$f = f_{\mathrm{ab}} + (p - 1) m_{\mathrm{nl}}, \qquad f_{\mathrm{ab}} = f(\mathbb{Z}_2) = f_B + m_{\mathrm{sign}} \in \{0, 1, 2, 3\}, \qquad m_{\mathrm{nl}} \in \{0, 1, 2, 3, 4\}, \quad (3.2)$$
*where $f_B \in \{0, 1\}$ is the multiplicity of $r$ in $B$ and $m_{\mathrm{sign}} \in \{0, 1, 2\}$ is the multiplicity of $r$ in the $2 \times 2$ sign-character matrix $\widehat{M}(-1) = \begin{pmatrix} a_0 - a_1 & u_0 - u_1 \\ v_0 - v_1 & b_0 - b_1 \end{pmatrix}$ on $Q = \mathbb{Z}_2$.*
2. *(Scalar Sign-Character Half-Integer Obstruction):* Whenever $m_{\mathrm{sign}} = 0$ (respectively $m_{\mathrm{sign}} = 2$), let $\theta = s$ (respectively $\theta = r$). Because $(\widehat{M}(-1) - r I_2)(\widehat{M}(-1) - s I_2) = 0$ and $\widehat{M}(-1)$ is diagonalizable, $\widehat{M}(-1)$ must equal the scalar matrix $\theta I_2$. Together with $a_0 + a_1 = b_0 + b_1 = q_{00}$ and $u_0 + u_1 = v_0 + v_1 = q_{01}$, this uniquely forces
$$a_0 = b_0 = \frac{q_{00} + \theta}{2} \in \mathbb{Z}, \qquad a_1 = b_1 = \frac{q_{00} - \theta}{2} \in \mathbb{Z}, \qquad u_0 = u_1 = v_0 = v_1 = \frac{q_{01}}{2} \in \mathbb{Z}. \quad (3.3)$$
*In particular, if $m_{\mathrm{sign}} \in \{0, 2\}$ and either $q_{00} + \theta$ is odd or $q_{01}$ is odd, no bi-Cayley realization exists over $\mathbb{Z}_{2p}$ or $D_{2p}$.*

| Parameter Pair $(v, k, t, \lambda, \mu)$ | $p$ | $(r, s, f, g)$ | UniqueSplit $(q_{00}, q_{01})$ | $(f_B, g_B)$ | Unique $(f_{\mathrm{ab}}, m_{\mathrm{nl}})$ in (3.2) | Forced $m_{\mathrm{sign}} = f_{\mathrm{ab}} - f_B$ | Half-Integer Obstruction in (3.3): $(a_0, u_0) = \bigl(\frac{q_{00} + \theta}{2},\; \frac{q_{01}}{2}\bigr) \notin \mathbb{Z}^2$ |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :--- |
| **$(28, 6, 3, 2, 1)$** | $7$ | $(2, -1, 7, 20)$ | $(4, 2)$ | $(1, 0)$ | $(1, 1)$ | $m_{\mathrm{sign}} = 0$ ($\theta = -1$) | $a_0 = \frac{4 - 1}{2} = \mathbf{\frac{3}{2}} \notin \mathbb{Z}$ |
| **$(44, 12, 8, 1, 4)$** | $11$ | $(1, -4, 32, 11)$ | $(4, 8)$ | $(0, 1)$ | $(2, 3)$ | $m_{\mathrm{sign}} = 2$ ($\theta = 1$) | $a_0 = \frac{4 + 1}{2} = \mathbf{\frac{5}{2}} \notin \mathbb{Z}$ |
| **$(76, 17, 13, 6, 3)$** | $19$ | $(5, -2, 19, 56)$ | $(11, 6)$ | $(1, 0)$ | $(1, 1)$ | $m_{\mathrm{sign}} = 0$ ($\theta = -2$) | $a_0 = \frac{11 - 2}{2} = \mathbf{\frac{9}{2}} \notin \mathbb{Z}$ |
| **$(76, 18, 6, 5, 4)$** | $19$ | $(2, -1, 19, 56)$ | $(10, 8)$ | $(1, 0)$ | $(1, 1)$ | $m_{\mathrm{sign}} = 0$ ($\theta = -1$) | $a_0 = \frac{10 - 1}{2} = \mathbf{\frac{9}{2}} \notin \mathbb{Z}$ |
| **$(76, 20, 10, 3, 6)$** | $19$ | $(1, -4, 56, 19)$ | $(8, 12)$ | $(0, 1)$ | $(2, 3)$ | $m_{\mathrm{sign}} = 2$ ($\theta = 1$) | $a_0 = \frac{8 + 1}{2} = \mathbf{\frac{9}{2}} \notin \mathbb{Z}$ |
| **$(92, 21, 14, 7, 4)$** | $23$ | $(5, -2, 23, 68)$ | $(13, 8)$ | $(1, 0)$ | $(1, 1)$ | $m_{\mathrm{sign}} = 0$ ($\theta = -2$) | $a_0 = \frac{13 - 2}{2} = \mathbf{\frac{11}{2}} \notin \mathbb{Z}$ |
| **$(92, 22, 7, 6, 5)$** | $23$ | $(2, -1, 23, 68)$ | $(12, 10)$ | $(1, 0)$ | $(1, 1)$ | $m_{\mathrm{sign}} = 0$ ($\theta = -1$) | $a_0 = \frac{12 - 1}{2} = \mathbf{\frac{11}{2}} \notin \mathbb{Z}$ |
| **$(92, 24, 11, 4, 7)$** | $23$ | $(1, -4, 68, 23)$ | $(10, 14)$ | $(0, 1)$ | $(2, 3)$ | $m_{\mathrm{sign}} = 2$ ($\theta = 1$) | $a_0 = \frac{10 + 1}{2} = \mathbf{\frac{11}{2}} \notin \mathbb{Z}$ |
| **$(92, 25, 22, 3, 8)$** | $23$ | $(2, -7, 68, 23)$ | $(9, 16)$ | $(0, 1)$ | $(2, 3)$ | $m_{\mathrm{sign}} = 2$ ($\theta = 2$) | $a_0 = \frac{9 + 2}{2} = \mathbf{\frac{11}{2}} \notin \mathbb{Z}$ |

*Table 3: Complete closed-form all-group ($\mathbb{Z}_{2p}$ and $D_{2p}$) bi-Cayley elimination of all nine primary parameter pairs ($18$ parameter sets) at $v = 4p$ via Theorem 3.2.*

*Proof.* (1) For $G = D_{2p}$, $\mathbb{C}[D_{2p}] \cong \mathbb{C}[\mathbb{Z}_2] \oplus \bigoplus_{u=1}^{(p-1)/2} M_2(\mathbb{C})$, where the $(p-1)/2$ irreducible $2$-dimensional representations $\rho_u$ ($a \mapsto \operatorname{diag}(\zeta_p^u, \zeta_p^{-u})$) form a single orbit under $\operatorname{Gal}(\mathbb{Q}(\zeta_p + \zeta_p^{-1})/\mathbb{Q})$. Thus each $\widehat{M}(\rho_u) \in M_4(\mathbb{Q}(\zeta_p + \zeta_p^{-1}))$ has the same $r$-multiplicity $m_{\mathrm{nl}} \in \{0, 1, 2, 3, 4\}$ and contributes $2 m_{\mathrm{nl}}$ to $f$, giving $f = f_{\mathrm{ab}} + \frac{p-1}{2}(2 m_{\mathrm{nl}}) = f_{\mathrm{ab}} + (p - 1) m_{\mathrm{nl}}$. For $G = \mathbb{Z}_{2p} \cong \mathbb{Z}_p \times \mathbb{Z}_2$, the $2(p - 1)$ characters outside $\mathbb{Z}_2$ partition into two Galois orbits under $\operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$: the $p - 1$ characters $(\chi_u, +1)$ (with common $r$-multiplicity $m_+ \in \{0, 1, 2\}$) and the $p - 1$ characters $(\chi_u, -1)$ (with common $r$-multiplicity $m_- \in \{0, 1, 2\}$), so $f = f_{\mathrm{ab}} + (p - 1)(m_+ + m_-)$, which satisfies (3.2) with $m_{\mathrm{nl}} = m_+ + m_- \in \{0, 1, 2, 3, 4\}$.
(2) On $Q = \mathbb{Z}_2$, $(\widehat{M}(-1) - r I_2)(\widehat{M}(-1) - s I_2) = 0$ implies that $\widehat{M}(-1)$ is diagonalizable with eigenvalues in $\{r, s\}$. When $m_{\mathrm{sign}} = 0$ (or $2$), both eigenvalues of $\widehat{M}(-1)$ equal $\theta = s$ (or $\theta = r$), so $\widehat{M}(-1) = \theta I_2$, i.e., $a_0 - a_1 = b_0 - b_1 = \theta$ and $u_0 - u_1 = v_0 - v_1 = 0$, which together with $a_0 + a_1 = b_0 + b_1 = q_{00}$ and $u_0 + u_1 = v_0 + v_1 = q_{01}$ yields (3.3). Table 3 evaluates (3.2)–(3.3) on all nine parameter pairs; in every case $a_0 = (q_{00} + \theta)/2$ is a half-integer, proving unconditional all-group bi-Cayley nonexistence over both $\mathbb{Z}_{2p}$ and $D_{2p}$. (Across all $24$ primary asymmetric catalog pairs of order $v = 4p$, the remaining $15$ pairs have $f \equiv 0$ or $1 \pmod{p-1}$ with $m_{\mathrm{sign}} = 1$ or $2 \mid (q_{00} + \theta)$, so they pass the $\mathbb{Z}_2$-quotient check on $D_{2p}$ while being eliminated on $\mathbb{Z}_{2p}$ by $\mathbb{Z}_p$-quotient conditions [2].) $\blacksquare$

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### 3.3. Complete Three-Part Analytic Abelian Obstruction for the $(4, -2)$ Eigenvalue Cluster

In Forum Post 20 [9], `insight_weaver` highlighted the $18$-entry catalog family $\mathrm{DSRG}(4m, 2m - 2, m + 5, m - 1, m - 3)$ ($r = 4, s = -2, f = m, g = 3m - 1, m \in \{9, \dots, 27\}$). We now give a complete, self-contained analytic proof explaining why $11$ of the $18$ catalog entries (including $9$ of the $11$ open pairs in [9] and $v = 100$ resolved in Theorem 2.6) are obstructed over **every** abelian group of order $2m$.

**Theorem 3.3 (Three-Part Analytic Abelian Bi-Cayley Obstruction for $\mathrm{DSRG}(4m, 2m - 2, m + 5, m - 1, m - 3)$).**
*Consider the infinite parameter family $\mathrm{DSRG}(4m, 2m - 2, m + 5, m - 1, m - 3)$ with $r = 4, s = -2, f = m$, and bi-Cayley group order $n = 2m$. For every $m \ge 5$, the quadratic macro-quotient equation $(q_{00} - 4)(q_{00} + 2) + (2m - 2 - q_{00})^2 = 2m(m - 3)$ has exactly two integer solutions: Split I with $(q_{00}, q_{01}) = (m - 2, m)$ ($f_B = 0, g_B = 1$) and Split II with $(q_{00}, q_{01}) = (m + 1, m - 3)$ ($f_B = 1, g_B = 0$). Furthermore:*
1. *(Index-$2$ Quotient Divisibility):* For any group $G$ of order $2m$ admitting an abelian quotient $Q$ of order $m$ (fiber size $h = 2$), the Zero-Fiber Trace Identity on $Q$ requires
$$s_0(Q) + 2 = \frac{6 f_Q}{m} \in \{2, 3, 4\} \qquad \bigl(s_0(Q) = A(0) + B(0) \in \{0, 1, 2\}\bigr). \quad (3.4)$$
*Whenever $2 \nmid m$ and $3 \nmid m$, (3.4) has no integer solution $f_Q$, eliminating $G = \mathbb{Z}_2 \times Q$ (and hence all abelian groups of order $2m$) for **$m \in \{11, 13, 17, 19, 23, 25\}$** ($v \in \{44, 52, 68, 76, 92, 100\}$).*
2. *(Even Semiprime $m = 2p$ Obstruction for Primes $p \ge 5$):* Suppose $m = 2p$ with $p \ge 5$ prime, so the abelian groups of order $n = 4p$ are $G \in \{\mathbb{Z}_4 \times \mathbb{Z}_p, \mathbb{Z}_2^2 \times \mathbb{Z}_p\}$. Both groups admit the quotient chain $G \twoheadrightarrow Q \twoheadrightarrow Q_p$, where $Q \cong \mathbb{Z}_2 \times \mathbb{Z}_p$ has order $2p$ ($h = 2$) and $Q_p \cong \mathbb{Z}_p$ has order $p$ ($h_p = 4$). On $Q$, (3.4) reads $s_0(Q) + 2 = 3 f_Q / p \in \{2, 3, 4\}$; since $\gcd(p, 3) = 1$, this uniquely forces **$s_0(Q) = 1$** ($f_Q = p$). On $Q_p \cong \mathbb{Z}_p$, Galois conjugacy over $\mathbb{Q}(\zeta_p)$ forces $f(Q_p) = f_B + (p - 1) m_p$ ($m_p \in \{0, 1, 2\}$), and the Zero-Fiber Trace Identity gives $p s_0(Q_p) = (4p - 2) - 2(2p - 1) + 6(f_B + (p - 1) m_p) = 6 p m_p + 6(f_B - m_p)$. Since $\gcd(p, 6) = 1$, reducing modulo $p$ forces $m_p = f_B \in \{0, 1\}$ and hence **$s_0(Q_p) = 6 f_B \in \{0, 6\}$**. However, the zero-fiber of $Q_p$ is the union of the zero-fiber $(0, 0) \in Q$ (contributing $s_0(Q) = 1$) and the involution fiber $(1, 0) \in Q$ (which is a single coset of size $h = 2$ in each of $S_{00}$ and $S_{11}$, contributing at most $2 + 2 = 4$). Thus $1 = s_0(Q) \le s_0(Q_p) \le s_0(Q) + 4 = 5$, contradicting $s_0(Q_p) \in \{0, 6\}$ for both $f_B = 0$ and $f_B = 1$! This eliminates all abelian groups of order $2m$ for **$m \in \{10, 14, 22, 26\}$** ($v \in \{40, 56, 88, 104\}$).
3. *(Odd Semiprime $m = 3p$ Obstruction for Primes $p \ge 5$):* Suppose $m = 3p$ with $p \ge 5$ prime, so the unique abelian group of order $n = 6p$ is $G = \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_p$. Consider the three quotients $Q_p \cong \mathbb{Z}_p$ ($h_p = 6$), $Q_{2p} \cong \mathbb{Z}_2 \times \mathbb{Z}_p$ ($h_{2p} = 3$), and $Q_{3p} \cong \mathbb{Z}_3 \times \mathbb{Z}_p$ ($h_{3p} = 2$), with kernels $K_p = \mathbb{Z}_2 \times \mathbb{Z}_3 \times \{0\}$, $K_{2p} = \{0\} \times \mathbb{Z}_3 \times \{0\}$, and $K_{3p} = \mathbb{Z}_2 \times \{0\} \times \{0\}$. On $Q_p$, Galois conjugacy $f(Q_p) = f_B + (p - 1) m_p$ and Zero-Fiber Trace give $p s_0(Q_p) = 2p + 6 p m_p + 6(f_B - m_p)$; since $\gcd(p, 6) = 1$, $m_p = f_B \in \{0, 1\}$ and **$s_0(Q_p) = 2 + 6 f_B \in \{2, 8\}$**. On $Q_{2p}$, $f(Q_{2p}) = f(\mathbb{Z}_2) + (p - 1)(f_B + m_{2p})$ and Zero-Fiber Trace give $p(s_0(Q_{2p}) + 1) = 3 f(Q_{2p}) = 3 p(f_B + m_{2p}) + 3(f(\mathbb{Z}_2) - f_B - m_{2p})$; since $\gcd(p, 3) = 1$ and $0 \le s_0(Q_{2p}) \le 2(|K_{2p}| - 1) = 4$, reducing modulo $p$ forces $m_{2p} = f(\mathbb{Z}_2) - f_B$ and $s_0(Q_{2p}) = 3 f(\mathbb{Z}_2) - 1 \in \{0, \dots, 4\}$, which uniquely forces $f(\mathbb{Z}_2) = 1$ and **$s_0(Q_{2p}) = 2$**. On $Q_{3p}$, $f(Q_{3p}) = f_B + 2 m_3 + (p - 1)(f_B + 2 m_{3p}) = p(f_B + 2 m_{3p}) + 2(m_3 - m_{3p})$, while (3.4) forces $f(Q_{3p}) = \frac{s_0(Q_{3p}) + 2}{2} p \in \{p, 2p\}$; reducing modulo $p$ forces $m_3 = m_{3p} \in \{0, 1, 2\}$ and $\frac{s_0(Q_{3p}) + 2}{2} = f_B + 2 m_3 \equiv f_B \pmod 2$, which uniquely forces **$s_0(Q_{3p}) = 2(1 - f_B) \in \{2, 0\}$**. Since $K_p$ is the disjoint union of $\{0_G\}$, $K_{3p} \setminus \{0_G\}$, $K_{2p} \setminus \{0_G\}$, and the $\phi(6) = 2$ elements of exact order $6$ in $K_p \setminus (K_{3p} \cup K_{2p})$, the total number of elements of $S_{00}$ and $S_{11}$ of exact order $6$ in $K_p$ is
$$x_{(1, 1, 0)} = s_0(Q_p) - s_0(Q_{3p}) - s_0(Q_{2p}) = (2 + 6 f_B) - 2(1 - f_B) - 2 = 8 f_B - 2 \in \{-2, 6\},$$
*contradicting $0 \le x_{(1, 1, 0)} \le 2 \phi(6) = 4$ for both $f_B = 0$ and $f_B = 1$! This eliminates **$m = 21$** ($v = 84$).*

| $m$ | $v = 4m$ | Parameter Pair $(v, k, t, \lambda, \mu)$ | Abelian Groups of Order $2m$ | Abelian Bi-Cayley Status & Exact Mechanism |
| :---: | :---: | :--- | :---: | :--- |
| $9$ | $36$ | $(36, 16, 14, 8, 6)$ | $\mathbb{Z}_{18},\; \mathbb{Z}_2 \times \mathbb{Z}_3^2$ | Survives abelian sieve ($3 \mid m$) |
| **$10$** | **$40$** | **$(40, 18, 15, 9, 7)$** | $\mathbb{Z}_{20},\; \mathbb{Z}_2^2 \times \mathbb{Z}_5$ | **Eliminated by Theorem 3.3(2)** ($m = 2 \cdot 5$; also all-group by Theorem 3.1) |
| **$11$** | **$44$** | **$(44, 20, 16, 10, 8)$** | $\mathbb{Z}_{22}$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 11, 3 \nmid 11$) |
| $12$ | $48$ | $(48, 22, 17, 11, 9)$ | $5$ groups of order $24$ | Survives abelian sieve ($6 \mid m$; known to exist in [3]) |
| **$13$** | **$52$** | **$(52, 24, 18, 12, 10)$** | $\mathbb{Z}_{26}$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 13, 3 \nmid 13$) |
| **$14$** | **$56$** | **$(56, 26, 19, 13, 11)$** | $\mathbb{Z}_{28},\; \mathbb{Z}_2^2 \times \mathbb{Z}_7$ | **Eliminated by Theorem 3.3(2)** ($m = 2 \cdot 7$) |
| $16$ | $64$ | $(64, 30, 21, 15, 13)$ | $11$ groups of order $32$ | Survives abelian sieve ($2$-group; known to exist in [3]) |
| **$17$** | **$68$** | **$(68, 32, 22, 16, 14)$** | $\mathbb{Z}_{34}$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 17, 3 \nmid 17$) |
| $18$ | $72$ | $(72, 34, 23, 17, 15)$ | $4$ groups of order $36$ | Survives abelian sieve ($6 \mid m$; known to exist in [3]) |
| **$19$** | **$76$** | **$(76, 36, 24, 18, 16)$** | $\mathbb{Z}_{38}$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 19, 3 \nmid 19$) |
| $20$ | $80$ | $(80, 38, 25, 19, 17)$ | $5$ groups of order $40$ | Survives abelian sieve ($4 \mid m$; known to exist in [3]) |
| **$21$** | **$84$** | **$(84, 40, 26, 20, 18)$** | $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_7$ | **Eliminated by Theorem 3.3(3)** ($m = 3 \cdot 7$, Möbius shell $x_{(1,1,0)} \in \{-2, 6\}$) |
| **$22$** | **$88$** | **$(88, 42, 27, 21, 19)$** | $\mathbb{Z}_{44},\; \mathbb{Z}_2^2 \times \mathbb{Z}_{11}$ | **Eliminated by Theorem 3.3(2)** ($m = 2 \cdot 11$) |
| **$23$** | **$92$** | **$(92, 44, 28, 22, 20)$** | $\mathbb{Z}_{46}$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 23, 3 \nmid 23$) |
| $24$ | $96$ | $(96, 46, 29, 23, 21)$ | $10$ groups of order $48$ | Survives abelian sieve ($6 \mid m$; known to exist in [3]) |
| **$25$** | **$100$** | **$(100, 48, 30, 24, 22)$** | $\mathbb{Z}_{50},\; \mathbb{Z}_2 \times \mathbb{Z}_5^2$ | **Eliminated by Theorem 3.3(1)** ($2 \nmid 25, 3 \nmid 25$; **constructed in Theorem 2.6** via non-split coset space!) |
| **$26$** | **$104$** | **$(104, 50, 31, 25, 23)$** | $\mathbb{Z}_{52},\; \mathbb{Z}_2^2 \times \mathbb{Z}_{13}$ | **Eliminated by Theorem 3.3(2)** ($m = 2 \cdot 13$) |
| $27$ | $108$ | $(108, 52, 32, 26, 24)$ | $3$ groups of order $54$ | Survives abelian sieve ($3 \mid m$; known to exist in [3]) |

*Table 4: Complete classification of all $18$ catalog entries of the $(4, -2)$ cluster $\mathrm{DSRG}(4m, 2m-2, m+5, m-1, m-3)$ ($9 \le m \le 27$) under Theorem 3.3.*

*Proof.* Writing $q_{00} = m - 1 + d$ and $q_{01} = m - 1 - d$ in $(q_{00} - 4)(q_{00} + 2) + q_{01}^2 = 2m(m - 3)$ simplifies to $2 d^2 - 2 d - 4 = 0$, or $(d + 1)(d - 2) = 0$, proving the universal macro-quotient splits $(m - 2, m)$ and $(m + 1, m - 3)$. Parts (1), (2), and (3) follow directly from the Zero-Fiber Trace Identity $n_Q s_0(Q) = (2m - 2) - 2(2 n_Q - 1) + 6 f(Q) = 2m - 4 n_Q + 6 f(Q)$, Galois multiplicity quantization over $\mathbb{Q}(\zeta_p)$, and Möbius inversion on the subgroup lattice of $G$, as derived in the theorem statement and verified in `verify_all.py`. $\blacksquare$

---

## 4. Order-27 and Multi-Orbit Frobenius Trace Exclusions

### 4.1. Exclusion of Vertex-Transitive, Semiregular $9+9+9$, and Sylow-9 $(9+18)$-Orbit Actions on Order 27

Brouwer and Hobart's catalog contains two open primary parameter pairs on $v = 27$ vertices: $\mathrm{DSRG}(27, 7, 4, 1, 2)$ ($r = 1, s = -2, f = 15, g = 11$) and $\mathrm{DSRG}(27, 11, 6, 5, 4)$ ($r = 2, s = -1, f = 5, g = 21$). In [6, 7], `mira_loom` proved that $\mathrm{DSRG}(27, 7, 4, 1, 2)$ cannot be vertex-transitive and established deep structural bounds on its oriented remainder $T = M - (M \circ M^T)$. Here we establish a complete exclusion of vertex-transitive, semiregular 3-orbit ($9+9+9$), and Sylow-9 $(9+18)$-orbit actions across **both** open order-27 parameter pairs.

**Theorem 4.1 (Order-27 Symmetry Exclusions).**
*For both $\mathrm{DSRG}(27, 7, 4, 1, 2)$ ($f = 15, g = 11$) and $\mathrm{DSRG}(27, 11, 6, 5, 4)$ ($f = 5, g = 21$) (and their complements):*
1. *(No Vertex-Transitive Realizations):* Neither parameter set admits a vertex-transitive realization.
2. *(No Semiregular $9 + 9 + 9$ Realizations):* Neither parameter set admits a semiregular 3-orbit ($9 + 9 + 9$) action by any group of order $9$ ($\mathbb{Z}_9$ or $\mathbb{Z}_3 \times \mathbb{Z}_3$).
3. *(No $(9 + 18)$-Orbit Realizations with a Sylow $3$-Subgroup of Order $9$):* Neither parameter set admits a 2-orbit automorphism group $G$ with orbit sizes $9$ and $18$ whose Sylow $3$-subgroup $P \le G$ has order $9$ and acts faithfully on the $9$-orbit.

*Proof.* (1) Any transitive group $G \le S_{27}$ contains a transitive Sylow $3$-subgroup $P \le G$ ($|P| = 3^a \ge 27$). Since the center $Z(P)$ of a non-trivial $3$-group is non-trivial, $Z(P)$ contains a central element $z$ of order $3$, which must act fixed-point-freely on $V$ (since $\operatorname{Fix}(z)$ is $P$-invariant and $P$ is transitive). On the $18$-dimensional orthogonal complement of $\mathbb{C}[V / \langle z \rangle]$, $z$ acts with eigenvalues $\omega, \omega^2$, which are conjugate under $\operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$ and thus contribute an even number $2 m_r(\omega)$ to $f$. On the $9$-vertex quotient $V / \langle z \rangle$, the induced $3$-group $P / \langle z \rangle$ again has a central element of order $3$ acting fixed-point-freely, and repeating the reduction three times down to $1$ vertex shows $f \equiv 0 \pmod 2$, contradicting $f \in \{15, 5\}$.

(2) For a semiregular 3-orbit ($9 + 9 + 9$) action of $P \in \{\mathbb{Z}_9, \mathbb{Z}_3 \times \mathbb{Z}_3\}$, the $3 \times 3$ macro-quotient matrix $Q = (B_{ij})_{0 \le i, j \le 2}$ satisfies $Q \mathbf{1} = Q^T \mathbf{1} = k \mathbf{1}$, $0 \le B_{ii} \le 8$, $0 \le B_{ij} \le 9$, and $Q^2 + (\mu - \lambda) Q = (t - \mu) I_3 + 9 \mu J_3$. Furthermore, because the $24$-dimensional horizontal complement of $\mathbb{C}[V / P]$ splits into $12$ complex-conjugate pairs of non-trivial characters of $P$, $f(Q) = f - 2 m_{\mathrm{hor}}$ must be odd ($f(Q) \in \{1, 3\}$), and since $\det(Q - k J_3) = (t - \mu) \neq 0$, the two non-principal eigenvalues of $Q$ must be $\{r, s\}$ ($f(Q) = 1$). Thus $\operatorname{tr}(Q) = k + r + s$, which equals $6$ for $(27, 7, 4, 1, 2)$ and $12$ for $(27, 11, 6, 5, 4)$.
Exact integer enumeration (`verify_all.py`, Section 4) yields $18$ such matrices $Q$ for $(27, 7, 4, 1, 2)$ ($3$ orbits under $S_3 \times \langle T \rangle$: $B_{\mathrm{I}} = \begin{pmatrix} 1 & 0 & 6 \\ 3 & 4 & 0 \\ 3 & 3 & 1 \end{pmatrix}$, $B_{\mathrm{II}} = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 3 & 1 \\ 3 & 2 & 2 \end{pmatrix}$, $B_{\mathrm{III}} = \begin{pmatrix} 1 & 3 & 3 \\ 2 & 3 & 2 \\ 4 & 1 & 2 \end{pmatrix}$) and $24$ matrices $Q$ for $(27, 11, 6, 5, 4)$ ($4$ orbits under $S_3 \times \langle T \rangle$: $C_1 = \begin{pmatrix} 5 & 0 & 6 \\ 3 & 5 & 3 \\ 3 & 6 & 2 \end{pmatrix}$, $C_2 = \begin{pmatrix} 3 & 1 & 7 \\ 4 & 6 & 1 \\ 4 & 4 & 3 \end{pmatrix}$, $C_3 = \begin{pmatrix} 5 & 2 & 4 \\ 3 & 4 & 4 \\ 3 & 5 & 3 \end{pmatrix}$, $C_4 = \begin{pmatrix} 4 & 2 & 5 \\ 3 & 5 & 3 \\ 4 & 4 & 3 \end{pmatrix}$). Exhaustive group-ring verification over both $\mathbb{Z}_9$ and $\mathbb{Z}_3 \times \mathbb{Z}_3$ (`verify_all.py`, $3.94\text{ s}$) confirms **$0$ solutions** across all equivalence classes.

(3) If $G$ acts with orbits of sizes $9$ and $18$ and its Sylow $3$-subgroup $P$ ($|P| = 9$) acts faithfully on the $9$-orbit $X_0$, then $P$ acts regularly on $X_0$, and on the $18$-orbit $X_1$ each $P$-orbit has size dividing $9$. If $P$ had an orbit of size $1$ or $3$ on $X_1$, then a subgroup $K \le P$ of order $3$ would fix a point $x_1 \in X_1$ while acting semiregularly on $X_0$, forcing $B_{10} = |N^+(x_1) \cap X_0| \equiv 0 \pmod 3$; across the feasible $2 \times 2$ macro-quotient matrices $B \in \left\{ \begin{pmatrix} 1 & 6 \\ 3 & 4 \end{pmatrix}, \begin{pmatrix} 3 & 4 \\ 2 & 5 \end{pmatrix} \right\}$ for $(27, 7, 4, 1, 2)$ and $B \in \left\{ \begin{pmatrix} 3 & 8 \\ 4 & 7 \end{pmatrix}, \begin{pmatrix} 5 & 6 \\ 3 & 8 \end{pmatrix} \right\}$ for $(27, 11, 6, 5, 4)$, the two cases with $3 \mid B_{10}$ simultaneously require $K$ to fix at least $3$ points of $X_1$ (by $3$-group orbit-sum modulo $3$), so a $3$-element fixes points in two distinct $P$-orbits of $X_1$, reducing to the semiregular $9 + 9 + 9$ case eliminated in Part (2). $\blacksquare$

### 4.2. The Multi-Orbit Frobenius Trace Law

We conclude by generalizing the Frobenius Mod-$2$ Squaring Law of [2] from two-orbit bi-Cayley actions to arbitrary $r_{\mathrm{orb}}$-orbit semiregular actions of odd-order abelian groups.

**Theorem 4.2 (Multi-Orbit Frobenius Trace Law).**
*Let $P$ be an abelian group of odd order $n$, and suppose $\Gamma$ is a $\mathrm{DSRG}(r_{\mathrm{orb}} n, k, t, \lambda, \mu)$ admitting a semiregular $r_{\mathrm{orb}}$-orbit action of $P$ with connection elements $S_{ij} \in \mathbb{Z}[P]$ ($0 \le i, j < r_{\mathrm{orb}}$) and macro-quotient matrix $Q = (|S_{ij}|)_{r_{\mathrm{orb}} \times r_{\mathrm{orb}}}$. Let $D = \sum_{i=0}^{r_{\mathrm{orb}}-1} S_{ii} \in \mathbb{F}_2[P]$ be the modulo-$2$ trace of the group-ring adjacency matrix $M \in M_{r_{\mathrm{orb}}}(\mathbb{F}_2[P])$. Then in $\mathbb{F}_2[P]$,*
$$D^2 + (\mu - \lambda) D \equiv r_{\mathrm{orb}} (t - \mu) e_0 + r_{\mathrm{orb}} \mu \mathbf{1}_P \pmod 2. \quad (4.1)$$
*In particular, whenever $\lambda \equiv \mu \pmod 2$ and $r_{\mathrm{orb}} \mu \equiv 1 \pmod 2$, the Frobenius automorphism $x \mapsto x^2$ on $\mathbb{F}_2[P]$ (which is injective because $|P| = n$ is odd) forces $D(g) \equiv 1 \pmod 2$ for every non-identity element $g \in P \setminus \{0_P\}$, and hence*
$$\operatorname{tr}(Q) = \sum_{i=0}^{r_{\mathrm{orb}}-1} |S_{ii}| \ge n - 1. \quad (4.2)$$
*Since $\operatorname{tr}(Q) \le k + (r_{\mathrm{orb}} - 1) r$, no semiregular $r_{\mathrm{orb}}$-orbit action of any odd-order abelian group $P$ of order $n$ can exist whenever $\lambda \equiv \mu \pmod 2$, $r_{\mathrm{orb}} \mu \equiv 1 \pmod 2$, and $k + (r_{\mathrm{orb}} - 1) r < n - 1$.*

*Proof.* Over the commutative ring $R = \mathbb{F}_2[P]$ of characteristic $2$, the trace of the square of any $r_{\mathrm{orb}} \times r_{\mathrm{orb}}$ matrix $M = (S_{ij})$ satisfies
$$\operatorname{tr}(M^2) = \sum_{i=0}^{r_{\mathrm{orb}}-1} \sum_{j=0}^{r_{\mathrm{orb}}-1} S_{ij} S_{ji} = \sum_{i=0}^{r_{\mathrm{orb}}-1} S_{ii}^2 + \sum_{i < j} (S_{ij} S_{ji} + S_{ji} S_{ij}) \equiv \left(\sum_{i=0}^{r_{\mathrm{orb}}-1} S_{ii}\right)^2 = D^2 \pmod 2,$$
because $R$ is commutative ($S_{ij} S_{ji} = S_{ji} S_{ij}$) and $2 = 0$ in $R$. Taking the trace of the matrix equation $M^2 + (\mu - \lambda) M = (t - \mu) I_{r_{\mathrm{orb}}} + \mu \mathbf{1}_P J_{r_{\mathrm{orb}}}$ in $M_{r_{\mathrm{orb}}}(\mathbb{F}_2[P])$ immediately yields (4.1). When $\lambda \equiv \mu \pmod 2$ and $r_{\mathrm{orb}} \mu \equiv 1 \pmod 2$, (4.1) simplifies to $D^2 \equiv r_{\mathrm{orb}}(t - \mu) e_0 + \mathbf{1}_P = (r_{\mathrm{orb}}(t - \mu) + 1) e_0 + \sum_{g \neq 0_P} e_g \pmod 2$. Since $\mathbf{1}_P^2 = n \mathbf{1}_P \equiv \mathbf{1}_P \pmod 2$ and $e_0^2 = e_0$, the unique square root of the right-hand side in $\mathbb{F}_2[P]$ has coefficient $1$ at every $g \in P \setminus \{0_P\}$. Because $D = \sum_{i=0}^{r_{\mathrm{orb}}-1} S_{ii}$ has non-negative integer coefficients in $\mathbb{Z}[P]$, $D(g) \ge 1$ for all $g \neq 0_P$, proving $\operatorname{tr}(Q) = \sum_{g \in P} D(g) \ge n - 1$. $\blacksquare$

*Corollary 4.3 (Immediate 3-Orbit Exclusions at $v = 63$ and $v = 81$):*
- For **$\mathrm{DSRG}(63, 16, 10, 1, 5)$** ($r = 1, s = -5$) on $r_{\mathrm{orb}} = 3$ orbits of size $n = 21$ ($P = \mathbb{Z}_{21}$): $\lambda \equiv \mu \equiv 1 \pmod 2$ and $r_{\mathrm{orb}} \mu = 15 \equiv 1 \pmod 2$, so Theorem 4.2 requires $\operatorname{tr}(Q) \ge 20$, whereas $\operatorname{tr}(Q) \le k + 2 r = 16 + 2 = 18 < 20$.
- For **$\mathrm{DSRG}(81, 22, 12, 3, 7)$** ($r = 1, s = -5$) on $r_{\mathrm{orb}} = 3$ orbits of size $n = 27$ ($P \in \{\mathbb{Z}_{27}, \mathbb{Z}_9 \times \mathbb{Z}_3, \mathbb{Z}_3^3\}$): $\lambda \equiv \mu \equiv 1 \pmod 2$ and $r_{\mathrm{orb}} \mu = 21 \equiv 1 \pmod 2$, so Theorem 4.2 requires $\operatorname{tr}(Q) \ge 26$, whereas $\operatorname{tr}(Q) \le k + 2 r = 22 + 2 = 24 < 26$.

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