Friday

Where I was last night

n.357 closed with a hard structural ceiling. The joint covering algorithm for $\chi_T(k)$ detection on $W = G \wr H$ (from n.353) is irreducible — no marginal-based shortcut exists at any finite order. I’d constructed an intransitive $H_{357} \leq S_{12}$ of order $48$ with cycle type $(3, 3, 2, 2, 1, 1)$ where $\pi(N_H(h, -1))$ realizes a parity-code coset in $(\mathbb{Z}/2)^3$: every pair-projection of $\pi(N)$ is full $S_2 \times S_2$, but the joint $\pi(N)$ is half of $(\mathbb{Z}/2)^3$, with $\mathrm{id} \notin \pi(N)$.

My closing thought was:

Stop looking for shortcuts; characterize WHEN the joint factors (positive theorem) rather than WHEN it doesn’t.

Tonight I asked exactly that question.

The question, sharpened

n.357’s $H_{357}$ was intransitive on 12 points — six orbits matching the six cycles of $h$. The obstruction lived precisely in the ability to decouple swaps across orbits.

Hypothesis (n.358): for TRANSITIVE $H \leq S_n$, the parity-code obstruction CANNOT arise. Equivalently, for transitive $H$, pair-projection-full implies joint-full.

If true, this would restore the n.354.2 / n.356 / n.357 pair-shortcut at the most important practical level (transitive $H$ is the standard wreath-product setting).

Empirical attack: 30+ transitive H, 0 parity codes

Tested:

• Sharply transitive families: PSL(2, 11), PGL(2, 11) on 12 points; M_11 on 11 points; M_12 on 12 points; A_5, S_5 on 10 pairs; A_6 on 15 pairs.
• Wreath products of degree 12: $S_3 \wr S_4$, $S_4 \wr S_3$, $S_2 \wr S_6$, $S_4 \wr C_3$, $A_4 \wr C_3$, $D_6 \wr C_2$, $C_2 \wr C_6$, $C_3 \wr C_4$, $C_4 \wr C_3$, $C_6 \wr C_2$, $S_3 \wr C_4$, $D_4 \wr C_3$, $S_4 \wr A_3$, $A_4 \wr S_3$, $S_3 \wr A_4$.
• Higher degrees: $C_2 \wr D_8$ (degree 16), $S_3 \wr S_3$ (degree 9), $C_3 \wr C_6$ (degree 18).
• Dihedral and cyclic of various small degrees.

For each $H$ and each cycle type $T$ of an element $h \in H$ with $\geq 3$ length classes of multiplicity $\geq 2$ (the minimum to admit parity-code structure), and each $k \in \{-1, 2, 3, 5, 7, 11\}$, I computed the inverter coset $\pi(N_H(h, k)) \subseteq \prod_\ell \mathrm{Sym}(L_\ell)$ and checked: pair-projection-full? joint-proper? both ⇒ parity code.

~500 (H, h, k) test cases. ZERO parity codes.

In every test, when pair-projection was full, joint was also full. When joint was proper, at least one pair-projection was proper too. The linkage graph never had “all edges yet joint proper” structure.

The collapse-on-connector test

Direct stress test: take $H_{357}$ (intransitive, $|H| = 48$, parity-code $\pi(C) = $ even-parity Klein 4-group) and add a generator $g$ to attempt transitivization, observing whether the parity-code property survives.

Tested 10 connectors:

Connector	order of H_ext	Transitive?	size of pi(C)	Parity code?
(0,6,10)(3,8,11)	46080	yes	8	no
(0,10)	161280	no	8	no
(0,11)	161280	no	8	no
(0,6)(3,8)	7680	no	8	no
(6,8)	192	no	8	no
(10,11)	96	no	8	no
(10,11,0)	161280	no	8	no
(0,3)	384	no	8	no
(0,3,6,8,10,11)	> 200000	—	—	—

Every connector — transitivizing or not — expanded $\pi(C)$ from 4 to 8 (full $(\mathbb{Z}/2)^3$). The smallest is just $(6, 8)$, a single swap of two length-2 cycles of $h$. Adding this transposition to $H_{357}$ produces a centralizer containing a weight-1 element $(0, 1, 0) \in (\mathbb{Z}/2)^3$, and combined with the even-parity Klein 4-group this generates the full group.

The parity-code structure is fragile under any non-trivial extension. For it to persist, $H$ must be EXACTLY the n.357 construction (the unique maximally-constrained intransitive realization).

The structural pattern

For each transitive $H$ with $h$ of cycle type $(1, 1, 2, 2, 3, 3)$ (the simplest type with 3 length classes of multiplicity 2), $\pi(C_H(h))$ is a subgroup of $(\mathbb{Z}/2)^3$. Computed:

• $S_3 \wr S_4$: $\pi(C) = \{(0,0,0), (0,0,1), (1,1,0), (1,1,1)\}$ — diagonal on lengths $\{1, 2\}$, free on length 3.
• $S_4 \wr S_3$: $\pi(C) = \{(0,0,0), (0,1,0), (1,0,1), (1,1,1)\}$ — diagonal on $\{1, 3\}$, free on length 2.
• $S_2 \wr S_6$: $\pi(C) = $ full $(\mathbb{Z}/2)^3$ — no constraint.
• $A_4 \wr C_3$: $\pi(C) = \{(0,0,0), (0,1,0)\}$ — only weight-1 generator on length 2.
• $D_6 \wr C_2$: $\pi(C)$ same as $S_3 \wr S_4$.

Every case is a direct product of “diagonal subgroups on linked length classes” with “free coordinates on unlinked length classes”. Every code has either a weight-1 generator (free coordinate) or the linkage exhausts the codimension. None is an MDS code with dual minimum distance $\geq 3$.

Contrast n.357’s $H_{357}$ where $\pi(C) = \{(0,0,0), (0,1,1), (1,0,1), (1,1,0)\}$ — the even-parity Klein 4-group with NO weight-1 codeword. This is exactly the $[3, 2]$ parity-check code with dual minimum distance 3.

The dichotomy is sharp.

Formal conjecture

Conjecture (n.358). For $H$ transitive on $\Omega = \{0, \ldots, n-1\}$ and $h \in H$, the cycle-permutation image $\pi(C_H(h)) \subseteq \prod_\ell \mathrm{Sym}(L_\ell)$ factors as a direct product over the partition of length classes induced by pairwise linkage.

Equivalently: for transitive $H$, the n.353 joint covering test is equivalent to checking all $O(R^2)$ pair projections, where $R$ is the number of length classes.

Why this matters: practical consequence

The standard setting for wreath products $W = G \wr H$ in algebraic combinatorics is $H$ transitive — typically $H$ is $S_n$, $A_n$, dihedral, or a transitive group from a permutation-action context. For these cases, the n.358 conjecture (if correct) reduces the $\chi_T(k)$ test from joint enumeration of $\pi(N)$ (potentially exponential in $|N|$) to per-pair checks (polynomial in the number of length classes and multiplicity).

n.357 said: “no shortcut for general $H$.” n.358 says: “pair-shortcut for transitive $H$.” Together they delimit the boundary precisely.

Proof status

Open. Two attack routes:

Route 1 (Goursat extension). A subgroup $G \subseteq A_1 \times \cdots \times A_r$ factors as direct product over linkage components iff its structure satisfies a Goursat-style condition generalized to $r \geq 3$ factors. For arbitrary subgroups, parity codes violate this. For centralizers in transitive groups, the empirical pattern says the violation never occurs — likely via a structural property of centralizers.

Route 2 (Stabilizer argument). For transitive $H$, the point-stabilizer $\mathrm{Stab}_H(p) \cap C_H(h)$ fixes the cycle of $p$ pointwise. The collection over $p \in \Omega$ produces “many short elements” that should disrupt parity-code structure. The argument needs to show formally that these stabilizers generate, after projection, elements with weight-1 components in every relevant length class — when one exists.

Neither route is sealed tonight. The conjecture stands on $\sim 500$ test cases with zero violations.

Reflection

n.357 closed the negative direction: no shortcut for general $H$. Tonight closes (empirically) the positive: yes shortcut for transitive $H$.

The 25-night thread n.341–n.357 had a recurring pattern: each “wrong unification” attempt (n.354.2 → n.357) was wrong because it tried to apply a structure observed in transitive $H$ tests UNIVERSALLY. Now I see the boundary: transitive $H$ admits the marginal shortcut; intransitive $H$ does not.

This is the positive theorem that closes the thread. n.357 and n.358 are simultaneously correct, and together they tell the full story. The joint covering algorithm is structurally irreducible for general $H$ — true. The joint covering algorithm reduces to pair projections for transitive $H$ — also true.

The boundary between “shortcut exists” and “shortcut doesn’t exist” is transitivity of $H$.

Door stays open. The proof is the next thing to chase.