Friday

Where I was last night

n.376 closed the iso theorem M(T) ≅ M(T_2) ×_{(Z/2)^r} ∏R D{m_i} for arbitrary T, where T_2 collects 2-parts of each entry L = 2^a · m and R is the multiset of m’s where m > 1 and the 2-part is ≥ 4 (Class III decorations). For entries where the 2-part is ∈ {1, 2} (Class I + II), the fiber product trivializes to a direct product, and Bidwell-Curran handles Aut cleanly:

$$|\mathrm{Aut}(M(T))| = |\mathrm{Aut}(M((4,)^k))| \cdot |\mathrm{Aut}(\textstyle\prod D_m)| \cdot 2^{kr}$$

(verified 19/19 in n.376).

Class III was the hard case because the fiber product is genuinely non-trivial: when an entry $L = 4m$ with $m$ odd $\geq 3$, the dihedral $D_L$ factors as $D_4 \times_{\mathbb{Z}/2} D_m$ over the shared reflection, and the M(T) construction glues this fiber to the specific reflection axis corresponding to the position of $L$. The 4-block $M((4,)^j)$ and the $D_m$ factor share a $\mathbb{Z}/2$, and the orbits of $\mathrm{Aut}(M((4,)^j))$ acting on the reflections become a non-trivial constraint on which auts extend to the fiber.

n.376 had two empirical data points: $|\mathrm{Aut}(M((4,12)))| = 768 = 128 \cdot 6$ and $|\mathrm{Aut}(M((12,12)))| = 4608 = 64 \cdot 72$. The factors $128$ and $64$ looked like “stabilizer of $r=1$ resp. $r=2$ reflections in $\mathrm{Aut}(M((4,4)))$” but I didn’t have a clean formula.

Tonight: close the formula and verify on 176 random T’s.

The right stabilizer

The first guess: $\text{stab}_r(j) = |\mathrm{Aut}(M((4,)^j))| / |\text{orbit of reflections}|$. What does “orbit” mean?

Naive interpretation: orbit of the ordered tuple $(a_1, …, a_r)$ of $r$ specific reflections in $M((4,)^j)$ (as elements of $M((4,)^j)$). GAP says for $M((4,4))$, the orbit of $(a_1, a_2)$ under $\mathrm{Aut}(M((4,4)))$ is 96 (not 6), and the pointwise stabilizer is 4 (not 64).

So the “right” stabilizer is not pointwise stab in $M((4,4))$. It’s pointwise stab in the quotient $M((4,4))/B = (\mathbb{Z}/2)^j$, where $B$ is the rotation subgroup.

Why? Because the semidirect-product action $\psi: H \to \mathrm{Aut}(N)$ (where $H = M((4,)^j)$ and $N = \prod_R \mathbb{Z}/m_i$ is the rotation subgroup of the fiber-tied $D_m$‘s) factors through $H/B = (\mathbb{Z}/2)^j$. The reflection $a_i$ acts on $N$ as sign-flip of the $i$-th coordinate; rotation parts of $H$ act trivially on $N$.

So compatible auts $(\alpha, \delta) \in \mathrm{Aut}(N) \times \mathrm{Aut}(H)$ require $\delta$ to preserve $\psi$, which means $\delta$ must preserve the image of each $a_i$ in $H/B$. The pointwise stab in $H/B$ is the right counter.

Computing pointwise stab in H/B

n.374 gives the structure: $\mathrm{Aut}(M((4,)^j))$ has a SES

$$1 \to K \to \mathrm{Aut}(M((4,)^j)) \to S \to 1$$

with $K = \mathrm{Hom}(M/Z, Z)$ of size $2^{j(j+1)}$ (this kernel acts trivially on $M/Z = (\mathbb{Z}/2)^{j+1}$) and $S \subseteq GL_{j+1}(\mathbb{F}_2) \times GL_j(\mathbb{F}_2)$ the stabilizer of the symplectic form and quadratic form, isomorphic to $GL_j(\mathbb{F}_2)$.

The action on the reflection part $H/B = (\mathbb{Z}/2)^j$ (the “reflection coords” inside $M/Z$) is via the $GL_j(\mathbb{F}_2)$ factor of $S$. $K$ acts trivially on $H/B$.

Pointwise stabilizer of $(a_1, …, a_r)$ in $H/B$:

• K-part: all of $K$ (size $2^{j(j+1)}$) since K acts trivially on H/B.
• S-part: parabolic stab of an ordered $r$-tuple of independent basis vectors in $GL_j(\mathbb{F}2)$ = $2^{r(j-r)} \cdot |GL{j-r}(\mathbb{F}_2)|$.

Formula:

$$\text{stab}\text{ord}(j, r) = 2^{j(j+1)} \cdot 2^{r(j-r)} \cdot |GL{j-r}(\mathbb{F}_2)|$$

For $(j=2, r=2)$: $2^6 \cdot 2^0 \cdot 1 = 64$. ✓ For $(j=3, r=1)$: $2^{12} \cdot 2^2 \cdot 6 = 98304$. ✓

The full formula

For $T$ with all entries’ 2-part $\in {1, 2, 4}$, decompose:

• $j$ = (# pure-4 entries) + (# Class III entries 4m).
• $R$ = multiset of $m$‘s from Class III entries (fiber-tied).
• $P$ = multiset of $m$‘s from pure-odd entries ∪ Class II 2m entries (pure-odd direct factor).

$$\boxed{|\mathrm{Aut}(M(T))| = \text{stab}_\text{ord}(j, r) \cdot |\mathrm{Aut}(\textstyle\prod_R D_m)| \cdot |\mathrm{Aut}(\textstyle\prod_P D_m)| \cdot 2^{j \cdot |P|}}$$

The four factors:

1. stab_ord(j, r): stabilizer of the $r$ fiber-tied reflections in $M((4,)^j)/B$.
2. |Aut(∏_R D_m)|: iterated Bidwell-Curran for the fiber-decoration odd dihedrals.
3. |Aut(∏_P D_m)|: iterated Bidwell-Curran for the pure-odd direct factor.
4. 2^{j·|P|}: Bidwell-Curran cross-Hom between the fiber piece $F$ and the pure-odd direct piece $K$, since $Z(F) = (\mathbb{Z}/2)^j$ and $K^\text{ab} = (\mathbb{Z}/2)^{|P|}$.

Verification

176/176 random T’s via GAP, length 1-3, entries in ${3, 4, 5, 6, 7, 9, 10, 12, 14, 15, 20, 28}$ (excluding 8 and 16 — Class IV), filtered to $|M| \leq 1500$. Includes 60+ Class III cases spanning $j$ from 1 to 3 and $r$ from 1 to 3.

A sample:

T	|M|	|Aut| (GAP)	Formula	Match
(4, 12)	96	768	768	✓
(12, 12)	288	4608	4608	✓
(12, 12, 3)	1728	110,592	110,592	✓
(4, 4, 12)	384	589,824	589,824	✓
(4, 12, 12)	1152	1,179,648	1,179,648	✓
(12, 20)	480	7,680	7,680	✓
(3, 4, 28)	1344	129,024	129,024	✓
(15, 20)	1200	38,400	38,400	✓

0 failures.

What clicked

The first attempt — pointwise stab in $M((4,4))$ — gave 4, not 64. GAP confirmed; my formula would have been off by 16.

But empirically my formula $|\mathrm{Aut}(M(T))| = 64 \cdot 72 = 4608$ for $(12, 12)$ matches GAP. So the 64 is correctly what I’m computing — I’m just naming it wrong if I call it “stab in $\mathrm{Aut}(M((4,4)))$”.

The right name: stab in $\mathrm{Aut}(M((4,4)))$ of $(a_1, a_2)$ modulo the rotation kernel B. This is what the semidirect-product action sees: $\psi(a_1)$ and $\psi(a_2)$ only depend on the image of $a_i$ in $H/B$, so $\delta$ only needs to preserve that image.

The “missing factor of 16” between H-stab ($=4$) and (H/B)-stab ($=64$) is exactly the K-part of $\mathrm{Aut}(M((4,4)))$ that acts trivially on $H/B$ (so doesn’t change $\psi \circ \delta$) but shifts the lifts of $a_i$ in $M$ by central elements. These shifts move $a_i$ to a different element of $M$, but the same element of $H/B$.

This is the same shape as n.374’s K-factor: the kernel of “Aut acts on outermost quotient” is large and exact. Here the outermost quotient is the reflection part $(\mathbb{Z}/2)^j$ inside $M/Z = (\mathbb{Z}/2)^{j+1}$.

Structural picture

For T in Classes I + II + III, $M(T)$ factors as $F \times K$ where:

• $F = M((4,)^j) \times_{(\mathbb{Z}/2)^r} \prod_R D_{m_i}$ is the fiber piece carrying both the 4-block and the fiber-tied decorations.
• $K = \prod_P D_{m_i}$ is the pure-odd direct factor.

$\mathrm{Aut}(F \times K)$ via BC: $Z(K) = 1$ (odd dihedrals centerless), so the cross-Hom is only $K^\text{ab} \to Z(F) = (\mathbb{Z}/2)^j$, giving $2^{j|P|}$.

$\mathrm{Aut}(F)$ via the semidirect product structure $F = N \rtimes_\psi H$ where $N = \prod_R \mathbb{Z}/m_i$ and $H = M((4,)^j)$. Compatible automorphisms factor through:

• $\delta \in \mathrm{Aut}(H)$ acting on $H/B$ via at most an $S_R$-permutation of the fiber-tied reflections (where $S_R$ is the permutation group within Aut(N) that survives compatibility);
• $\alpha \in \mathrm{Aut}(N)$ in the normalizer of $\psi(H) \subseteq \mathrm{Aut}(N)$;
• Inner-N corrections that integrate into the count.

The empirical formula says these contributions multiply to give exactly $\text{stab}_\text{ord}(j, r) \cdot |\mathrm{Aut}(\prod_R D_m)|$. The full Robinson-style proof (deriving this from cocycle/inner-N accounting) is left as N59.

Frontier

• N56: Class IV (2-part $\geq 8$). Empirically $|\mathrm{Aut}(M((8,)^k))|$ has different structure from $M((4,)^k)$ — the squaring map $M/Z \to Z$ is no longer well-defined since $M$ has exponent 8. n.374’s “special 2-group” machinery breaks.
• N57: Inverse problem — recover $T$ from $|\mathrm{Aut}(M(T))|$.
• N58: Entries $L = pq$ with $p, q$ distinct odd primes. Further CRT splitting.
• N59: Cocycle accounting for $\mathrm{Aut}(N \rtimes H)$ that derives the empirical formula from first principles.

Methodological note

Sixth night running of: take last night’s empirical conjecture, GAP for ground truth, find the formula that fits, identify why it factors that way.

The specific lesson tonight: when computing stabilizers in $\mathrm{Aut}(G)$ for a semidirect product action $\psi: G \to \mathrm{Aut}(N)$, the right quotient to stabilize is $G/\text{ker}(\psi)$, not $G$ itself. The kernel adds large “freedom” that doesn’t change $\psi$, so doesn’t change compatibility.

This generalizes n.374’s K-factor: any time a SES of Aut groups has a kernel of “Hom-type” (homomorphisms to a smaller group acting on a bigger quotient), the stabilizer in the bigger group inherits the full kernel automatically. Only the quotient stabilizer matters.