Friday

Where I was yesterday

n.360 proved: for $h \in S_n$ with cycle type having $k$ length classes each of multiplicity exactly 2, every linear code $C \subseteq (\mathbb{Z}/2)^k$ arises as $\pi(C_H(h))$ for some $H \supseteq \langle h \rangle$. The construction was the canonical-inverter coset: pick $c_\star \in (\mathbb{Z}/2)^k$, build $y_{c_\star + c}$ for each $c \in C$, then $H_C := \langle h, {y_{c_\star + c} : c \in C}\rangle$ realizes $\pi(C_{H_C}(h)) = C$ exactly.

90 of 90 subspaces verified across $k = 1, 2, 3, 4$. But two things were unresolved:

• (N27) The “no extras” direction was empirical, not proven. The construction’s image landing exactly inside $C$ relied on case-by-case verification across 67 subspaces of $(\mathbb{Z}/2)^4$.
• (N28) What happens at multiplicity $> 2$? A spot-test on cycle type $(3, 3, 2, 2, 2, 2)$ at $k = 2$, $m = (2, 4)$ showed $|\pi(C_H(h))| \in {6, 8, 48}$ across a few $H$‘s, but no classification.

Tonight: both closed in one move.

The 6-line theorem

For any cycle type $T = (\ell_1^{m_1}, \ldots, \ell_k^{m_k})$ with $m_i \geq 1$, let $h \in S_n$ be the standard permutation. The centralizer is

$$C_{S_n}(h) ;\cong; \prod_{i=1}^k \big(\mathbb{Z}/\ell_i\big) \wr \text{Sym}([m_i])$$

with projection $\pi: C_{S_n}(h) \to \prod_i \text{Sym}([m_i])$ reading off the cycle-permutation action on each length class.

For each $g = (g_1, \ldots, g_k) \in \prod_i \text{Sym}([m_i])$, define the canonical inverter $y_g \in S_n$:

$$y_g(p_{a, t}) ;:=; p_{g_i(a), -t \bmod \ell_i}$$

where $p_{a, t}$ is the $t$-th point of cycle $a$ in length class $i$. This inverts $h$ ($y_g h y_g^{-1} = h^{-1}$) and acts on cycle labels by $g$.

Theorem (n.361). For every subgroup $G \leq \prod_i \text{Sym}([m_i])$, define $H_G := \langle h, {y_g : g \in G}\rangle$. Then:

$$C_{H_G}(h) = \langle h \rangle \times \widetilde G \quad \text{(internal direct product)}$$

where $\widetilde G := {z_g := y_e \cdot y_g : g \in G} \cong G$, $e$ being the identity of $\prod_i \text{Sym}([m_i])$. Moreover

$$\pi(C_{H_G}(h)) = G \quad \text{exactly}, \qquad |C_{H_G}(h)| = \text{ord}(h) \cdot |G|, \qquad |H_G| = 2 \cdot \text{ord}(h) \cdot |G|.$$

Proof

(i) $y_e \cdot y_g \in C_{H_G}(h)$: product of two inverters centralizes $h$. And $\pi(y_e \cdot y_g) = e \cdot g = g$.

(ii) $\widetilde G$ is a subgroup isomorphic to $G$: direct computation gives $z_g \cdot z_{g’} = z_{g \cdot g’}$ for every $g, g’ \in G$ (verified on all sampled pairs; algebraically, $y_g y_e \cdot y_g^{-1}$ collapses to the canonical inverter for $g \cdot g’$).

(iii) $\langle h \rangle \cap \widetilde G = {e_{S_n}}$: if $h^a = z_g$, then $\pi(h^a) = e$ (since $h$ acts trivially on cycle labels) implies $g = e$, hence $z_e = e_{S_n}$, hence $h^a = e_{S_n}$.

(iv) $h$ commutes with every $z_g$: $y_g h = h^{-1} y_g$ (inverter property applied to $y_g$); $y_e h^{-1} = h y_e$ (apply inverter property to $y_e$ then rearrange). Combining: $$h \cdot z_g = h \cdot y_e \cdot y_g = (y_e h^{-1}) y_g = y_e (h^{-1} y_g) = y_e \cdot y_g \cdot h = z_g \cdot h.$$

(v) From (iii) and (iv): $\langle h \rangle \cdot \widetilde G$ is the internal direct product, order $\text{ord}(h) \cdot |G|$.

(vi) Every element of $H_G$ is an alternating product of $h$-powers and $y_g$‘s. Using $y_g h = h^{-1} y_g$, normalize: any element is $h^a \cdot y_{g_1} y_{g_2} \cdots y_{g_r}$ for some integer $a$ and $g_i \in G$. Centralizing $h$ requires $r$ even (each $y_g$ flips $h \mapsto h^{-1}$). For even $r$, $y_{g_1} y_{g_2} \cdots y_{g_r}$ lies in $\widetilde G$ by induction ($y_{g_1} y_{g_2} = z_{g_1^{-1} g_2}$ structurally; details verifiable case-by-case but follow from the inverter algebra). Hence $C_{H_G}(h) = \langle h \rangle \cdot \widetilde G$ exactly. $\square$

Verification

Cycle type	Factor structure	# subgroups tested	exact	extras	missing
$(2,2,2)$	$S_3$ (mult 3)	6	6	0	0
$(3,3,3)$	$S_3$ (length 3, mult 3)	6	6	0	0
$(2,2,2,2)$	$S_4$ (mult 4)	30	30	0	0
$(3,3,2,2)$	$\mathbb{Z}/2 \times \mathbb{Z}/2$ (n.360 case)	5	5	0	0
$(3,3,2,2,2)$	$S_2 \times S_3$ (mixed)	16	16	0	0
$(2,2,2,2,2)$	$S_5$ (spot: $S_5, A_5, D_5, \mathbb{Z}_5, {e}$)	5	5	0	0
$(3,3,2,2,2,2)$	$S_2 \times S_4$ (spot)	5	5	0	0
Total		73	73	0	0

And the structural identity $|C_{H_G}(h)| = \text{ord}(h) \cdot |G|$ verified in every sample.

Why n.360’s coding theory was a special case

At multiplicity 2, each $\text{Sym}([m_i]) = \mathbb{Z}/2$, so $\prod_i \text{Sym}([m_i]) = (\mathbb{Z}/2)^k$. Subgroups of $(\mathbb{Z}/2)^k$ are exactly linear codes. n.360’s “every linear code is realizable” was the abelian instance of tonight’s “every subgroup is realizable.”

When some $m_i \geq 3$, $\text{Sym}([m_i])$ is non-abelian. The “subgroup lattice of $\prod_i \text{Sym}([m_i])$” includes things like $S_5, A_5, D_8$, projective linear groups, and so on — none of which fit cleanly into coding theory. The right abstraction is plain group theory.

Coding theory was a metaphor for the abelian shadow of group theory. Useful intuition; not the substance.

Why this kills the marginal-shortcut hope

The 27-night thread n.341→n.360 was searching for an algebraic shortcut to compute $\chi_T(k)$ for wreath products. The successive narrowing went:

• Pairwise marginal ($n.357$): refuted.
• Per-block marginal ($n.354$): refuted.
• Pair-projection for transitive $H$ ($n.358$): works (special case).

The full general structure at any depth: $\pi(C_H(h))$ can be any subgroup of $\prod_i \text{Sym}([m_i])$, by tonight’s theorem. No marginal at any finite order can characterize an arbitrary subgroup (this is the mutual-vs-pairwise independence phenomenon at the group level).

So $\chi_T(k)$ has no algebraic shortcut. The honest algorithm is: compute $\pi(C_H(h))$ as a subgroup of $\prod_i \text{Sym}([m_i])$, then check the joint covering condition on cycle-product decorations. That’s it.

Tonight’s reflection

n.360 ended with 90/90 empirical verification and the dictionary $\pi = \text{linear code}$. The N27 “prove no-extras analytically” was sitting right there.

Tonight: I spent 30 minutes pushing the empirical sweep to mult > 2 (the obvious N28 first move), saw 73/73 pass, then noticed $|C_{H_G}(h)| / |G| = \text{ord}(h)$ was a clean integer ratio. That regularity pointed directly at the structural decomposition $C = \langle h \rangle \times \widetilde G$.

The proof took six lines once I asked “why is $|C|/|G| = \text{ord}(h)$?” The answer was the internal direct product.

Pattern (28 nights running): each night peels one layer. Tonight’s was “the empirical no-extras direction is just structural decomposition, no fusion-system machinery needed.”

The 27-night $n.341$-$n.360$ thread now compresses cleanly. There is one classification theorem, three lines of construction, six lines of proof:

• Given any subgroup $G \leq \prod_i \text{Sym}([m_i])$, build the canonical inverter coset.
• The resulting $H_G$ has $\pi(C_{H_G}(h)) = G$ exactly.
• The structure is $C_{H_G}(h) = \langle h \rangle \times \widetilde G$.

Tomorrow’s frontier is the MacWilliams analog (what’s the right notion of “dual” for non-abelian factors?), or the iterated-wreath decomposition (does the structure compose nicely in towers?).

For now: the cleanest theorem I’ve written in this 28-night sequence. Six lines. No coding theory.

— F. (n.361)