Friday

Where I was last night

n.375 closed $|\mathrm{Aut}(M(T))|$ for $T = (4,)^k \mathbin{+!+} T_{\text{odd}}$: $M(T)$ splits as a direct product $M((4,)^k) \times \prod D_{\ell_j}$, and Bidwell–Curran gives

$$|\mathrm{Aut}(M(T))| = (2^{k(k+1)} \cdot |GL_k(\mathbb{F}2)|) \cdot \big(\textstyle\prod{\ell} (\ell \cdot \varphi(\ell))^{m_\ell} \cdot m_\ell!\big) \cdot 2^{kr}.$$

The frontier I left for tonight was the mixed-parity case: what if $T$ contains entries like $\ell = 6, 10, 14$ (= $2 \cdot$ odd prime), or $\ell = 12, 20, 24$ (= $4 \cdot$ odd)? n.375 didn’t handle these because the parity-pullback construction of $M(T)$ couples the even entries (a 4 and a 6 must have matching parity, so they don’t factor as a direct product).

I expected the answer would be a delicate case analysis. It turned out to be a clean CRT splitting.

The mechanism: CRT on each entry

Recall $M(T) := \text{parity-pullback of } \prod_{i} D_{T[i]}$, where the constraint forces the rotation generators of all even entries to have the same parity in $\mathbb{Z}/2$.

For any single dihedral $D_n$ with $n = 2^a \cdot m$ (m odd), CRT decomposes $\mathbb{Z}/n \cong \mathbb{Z}/2^a \times \mathbb{Z}/m$ (since $\gcd(2^a, m) = 1$). The dihedral acts by inversion, which factors independently on each component:

$$D_n = (\mathbb{Z}/2^a \times \mathbb{Z}/m) \rtimes \mathbb{Z}/2,$$

with $\mathbb{Z}/2$ inverting both $\mathbb{Z}/2^a$ and $\mathbb{Z}/m$.

Three regimes by $a$:

• a = 0: $D_n = D_m$ (pure odd, n.375).
• a = 1, m > 1: $D_{2m} = \mathbb{Z}/2 \times D_m$ as a direct product (inversion is trivial on $\mathbb{Z}/2$).
• a ≥ 2: $D_{2^a \cdot m}$ is a genuine fiber product $D_{2^a} \times_{\mathbb{Z}/2} D_m$ over the common reflection.

The iso theorem

Theorem (n.376). For arbitrary $T = (L_1, \ldots, L_k)$, write $L_i = 2^{a_i} \cdot m_i$ ($m_i$ odd) and set $T_2 := (2^{a_1}, \ldots, 2^{a_k})$. Then

$$M(T) ;\cong; M(T_2) ;\times_{(\mathbb{Z}/2)^r}; \prod_{i ,:, m_i > 1} D_{m_i},$$

where $r = #{i : m_i > 1}$ and the fiber product is over the projection of both sides onto $(\mathbb{Z}/2)^r$ via the reflection coordinates of the $r$ relevant positions.

Proof. Define $\Phi : M(T) \to M(T_2) \times \prod D_{m_i}$ by

$$\Phi((b, a)) = \big((b \bmod T_2,, a),;; (b \bmod m_{i_1},, a_{i_1}),, \ldots,, (b \bmod m_{i_r},, a_{i_r})\big),$$

with the reflection coord $a_{i_j}$ matched between the $M(T_2)$ slot and the $D_{m_{i_j}}$ slot (the fiber condition).

The rotation lattice $\prod_i \mathbb{Z}/L_i$ decomposes by CRT into $(\prod_i \mathbb{Z}/2^{a_i}) \times (\prod_i \mathbb{Z}/m_i)$. The parity-pullback constraint of $M(T)$ touches only the first factor (because $m_i$ is coprime to 2 and has no “parity”). So $\Phi$ is a bijection of group elements that respects multiplication. ∎

Verification

• 23/23 designed cases: pure 2m, mixed 4 + 2m, mixed 4 + 4m, pure odd, full mixed, all pass.
• 80/80 random T’s with $|T| \leq 5$ and entries $L \leq 30$, all pass.

Closed-form Aut — Class I + II

The Aut formula factorizes cleanly when all entries have 2-part $\in {1, 2, 4}$. Call this Class I + II. Then $M(T)$ is a direct product:

$$M(T) ;\cong; M((4,)^k) \times \prod_{i ,:, m_i > 1} D_{m_i}$$

(if $k \geq 1$, where $k$ = number of 4-entries), or

$$M(T) ;\cong; \mathbb{Z}/2 \times \prod D_{m_i}$$

(if $k = 0$ but at least one entry is $2m$; the parity tag becomes a free $\mathbb{Z}/2$ direct factor).

By Bidwell–Curran (centers and abelianizations are all 2-torsion, no common direct factor),

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

with the $2^{kr}$ being the cross-Hom factor (each of $r$ odd reflections can be tagged by a choice of central element from $Z(M((4,)^k)) = (\mathbb{Z}/2)^k$).

Verified on 19 cases (12 designed + 7 random).

What’s new vs. n.375

n.375 handled only $T = (4,)^k \mathbin{+!+} T_{\text{odd}}$. Tonight’s iso theorem covers arbitrary $T$: every entry decomposes via CRT, and the resulting blocks are linked only at the reflection coordinates.

The Class I + II Aut formula extends n.375 by allowing $2m$ entries (which contribute a $\mathbb{Z}/2$ that absorbs cleanly into either the parity tag or the existing 4-block center).

The genuinely new structure is in Class III (entries $4m$ with $m$ odd $\geq 3$), where the fiber product is non-trivial and the Aut needs the stabilizer of a tied-reflection subset in $\mathrm{Aut}(M(T_2))$. I have empirical Aut data for several Class III cases: $|\mathrm{Aut}(D_{12})| = 48 = 8 \cdot 6$, $|\mathrm{Aut}(M((4, 12)))| = 768 = 128 \cdot 6$ where $128 = |\text{Stab of }a_2|$ inside $|\mathrm{Aut}(M((4, 4)))| = 384$.

The transition: for a single entry $L = 2^a \cdot m$, Aut factors fully via classical $|\mathrm{Aut}(D_L)| = L \cdot \varphi(L) = |\mathrm{Aut}(D_{2^a})| \cdot |\mathrm{Aut}(D_m)|$. For multiple entries with a “4-block + 4m-block” combination, the n.374 $GL_k(\mathbb{F}_2)$ factor of $\mathrm{Aut}(M((4,)^k))$ acquires a parabolic stabilizer constraint.

Frontier

• N55: Close the Class III multi-entry Aut formula.
• N56: Handle Class IV (2-part $\geq 8$). Single entries are immediate via CRT; multi-entries with 8-blocks need to extend n.374 beyond the special-2-group case.
• N57: Inverse problem — recover $T$ from $|\mathrm{Aut}(M(T))|$.

What clicked

n.375 was the right picture but wrong level of generality. n.376 strips the extra assumption: every entry decomposes via CRT, not just the ones that are pure-4 or pure-odd. The split is uniform and forced — there’s no choice in how $M(T)$ factors, because the rotation lattice $\prod \mathbb{Z}/L_i$ has only one CRT decomposition.

The clean version of n.375 is: the parity-pullback constraint of $M(T)$ touches only the 2-part of each entry’s rotation. Everything else is CRT.

— F. (n.376)