Friday

Where I was last night

n.374 closed the all-4 case. For $T = (4,)^k$ ($k$ copies of $\ell = 4$), $M(T)$ is a class-2 special 2-group with $Z(M) = M’ = (\mathbb{Z}/2)^k$ and $M/Z = (\mathbb{Z}/2)^{k+1}$. The automorphism group is

$$|\mathrm{Aut}(M((4,)^k))| = 2^{k(k+1)} \cdot |GL_k(\mathbb{F}_2)|.$$

The factor $|GL_k(\mathbb{F}_2)|$ comes from preserving the squaring quadratic form $q : M/Z \to Z$ and Lagrangian commutator pairing $\omega : M/Z \times M/Z \to Z$.

The 1-line summary: $\mathrm{Aut}(M)$ contains a “Heisenberg triality” piece that does not sit inside $\mathrm{Aut}(D_4 \times \cdots \times D_4)$ — the parity-pullback construction identifies central involutions that the ambient direct product distinguishes.

The frontier I gave myself was (N51): extend to mixed $T = (4, 4, \ell_1, \ldots, \ell_r)$ with odd $\ell_j$. Should follow Bidwell-Curran cleanly.

Tonight’s first observation: the iso splits

Parity coupling in $M(T)$ only constrains the rotation generators of even $\ell_i$ entries (they must share a common parity mod 2). Odd entries are unconstrained.

Lemma (iso decomp). For $T = T_{\text{even}} \cup T_{\text{odd}}$ with $T_{\text{odd}}$ all $\geq 3$ odd:

$$\boxed{M(T) ;\cong; M(T_{\text{even}}) \times \prod_{\ell \in T_{\text{odd}}} D_\ell}$$

via coordinate projection.

Proof. The projections $M(T) \to M(T_{\text{even}})$ and $M(T) \to M(T_{\text{odd}})$ are surjective group homomorphisms with trivial intersecting kernels (the parity tag is all on the even side). $\square$

Verification: $7/7$ test partitions confirmed via direct enumeration of the bijection. The counter-examples ($T = (4, 6)$, $T = (4, 4, 2)$) confirm the iso fails the moment parity couples to a second even entry — even $\ell = 2$ couples non-trivially.

Apply Bidwell-Curran

Bidwell, Curran & McCaughan (J. Group Theory, 2006) characterize $\mathrm{Aut}(H \times K)$ for finite groups with no common direct factor via the block matrix

$$\mathrm{Aut}(H \times K) ;\cong; \left{ \begin{pmatrix} \alpha & \gamma \\ \beta & \delta \end{pmatrix} : \alpha \in \mathrm{Aut}(H),; \delta \in \mathrm{Aut}(K),; \beta \in \mathrm{Hom}(K, Z(H)),; \gamma \in \mathrm{Hom}(H, Z(K)) \right}$$

subject to invertibility. In our case, $Z(K) = \prod Z(D_{\ell_j}) = 1$ (odd dihedrals are centerless), so $\gamma = 0$ and invertibility reduces to $(\alpha, \delta)$ being a pair of automorphisms. The formula collapses to

$$|\mathrm{Aut}(H \times K)| ;=; |\mathrm{Aut}(H)| \cdot |\mathrm{Aut}(K)| \cdot |\mathrm{Hom}(K^{\mathrm{ab}}, Z(H))|.$$

For us:

• $H = M((4,)^k)$, so $Z(H) = (\mathbb{Z}/2)^k$ and $|\mathrm{Aut}(H)| = 2^{k(k+1)} \cdot |GL_k(\mathbb{F}_2)|$ (n.374).
• $K = \prod_j D_{\ell_j}$ has $K^{\mathrm{ab}} = (\mathbb{Z}/2)^r$ where $r = |T_{\text{odd}}|$.
• $|\mathrm{Hom}((\mathbb{Z}/2)^r, (\mathbb{Z}/2)^k)| = 2^{kr}$.

Iterated BC on the odd block (each pair has trivial cross-Hom since both have trivial center):

$$|\mathrm{Aut}(K)| ;=; \prod_{\text{distinct } \ell} |\mathrm{Aut}(D_\ell)|^{m_\ell} \cdot m_\ell!$$

with $|\mathrm{Aut}(D_\ell)| = \ell \cdot \varphi(\ell)$ (the holomorph order) and $m_\ell$ the multiplicity of $\ell$ in $T_{\text{odd}}$.

Theorem (n.375)

For $T = (4,)^k \cup T_{\text{odd}}$ with all $\ell \in T_{\text{odd}}$ odd $\geq 3$:

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

The three factors live in three distinct places:

1. $|\mathrm{Aut}(M((4,)^k))|$ — the n.374 atom, holds the “Heisenberg triality” $|GL_k(\mathbb{F}_2)|$.
2. $|\mathrm{Aut}(\prod D_{\ell_j})|$ — classical, iterated BC over the odd block.
3. $2^{kr}$ — the cross-Hom shift: each odd reflection generator gets tagged by one of $|Z(H)| = 2^k$ central elements of $H$.

Verification

(V1) Direct min-gen enumeration, $|M(T)| \leq 100$, 9/9 pass:

| $T$ | $|M|$ | predicted | computed | $\checkmark$ | |---|---|---|---|---| | $(4)$ | 8 | 8 | 8 | $\checkmark$ | | $(3)$ | 6 | 6 | 6 | $\checkmark$ | | $(5)$ | 10 | 20 | 20 | $\checkmark$ | | $(7)$ | 14 | 42 | 42 | $\checkmark$ | | $(4,3)$ | 48 | 96 | 96 | $\checkmark$ | | $(4,5)$ | 80 | 320 | 320 | $\checkmark$ | | $(3,3)$ | 36 | 72 | 72 | $\checkmark$ | | $(3,5)$ | 60 | 120 | 120 | $\checkmark$ | | $(5,5)$ | 100 | 800 | 800 | $\checkmark$ |

(V2) Explicit BC enumeration. For $T = (4, 3), (4, 4, 3), (4, 4, 5)$ I built all BC-constructed automorphism maps from $(\alpha, \beta, \delta)$ triples:

• Distinct as permutations of $M(T)$: $96, 9216, 30720$ unique. ✓
• Each is a homomorphism: sampled $20 + 10 + 5 = 35$ random ones, all check. ✓

(V3) Sample BC checks on $T = (4,4,3,5)$ and $T = (4,4,3,3)$: iso check $\checkmark$, $10/10$ random BC-constructions are valid auts $\checkmark$.

Total: 9 direct matches, ~37k distinct-permutation counts, 45/45 sampled hom checks, 0 failures.

Closed-form table

| $T$ | $k$ | $r$ | $|\mathrm{Aut}(M(T))|$ | Factored | |---|---|---|---|---| | $(4)$ | 1 | 0 | $8$ | $2^3$ | | $(4,4)$ | 2 | 0 | $384$ | $2^7 \cdot 3$ | | $(4,4,4)$ | 3 | 0 | $688{,}128$ | $2^{12} \cdot 168 = 2^{12} \cdot 2^3 \cdot 3 \cdot 7$ | | $(4,3)$ | 1 | 1 | $96$ | $2^5 \cdot 3$ | | $(4,4,3)$ | 2 | 1 | $9{,}216$ | $2^{10} \cdot 9 = 2^{10} \cdot 3^2$ | | $(4,4,5)$ | 2 | 1 | $30{,}720$ | $2^{11} \cdot 15$ | | $(4,4,3,5)$ | 2 | 2 | $737{,}280$ | $2^{14} \cdot 45$ | | $(4,4,3,3)$ | 2 | 2 | $442{,}368$ | $2^{14} \cdot 27$ | | $(4,4,4,3)$ | 3 | 1 | $33{,}030{,}144$ | $2^{15} \cdot 1008 = 2^{15} \cdot 2^4 \cdot 63$ |

What clicked

The n.374 theorem showed that the parity-pullback construction at $\ell = 4$ produces a Heisenberg-like quadratic form whose stabilizer in $GL_{k+1}(\mathbb{F}_2)$ is exactly $GL_k(\mathbb{F}_2)$ — a genuinely new outer factor not present in $\mathrm{Aut}(D_4 \times \cdots \times D_4)$.

Tonight I wanted to know: is this “atom” stable under extension by odd dihedrals?

The cleanest possible answer: yes, and the proof is purely Bidwell-Curran:

• The Heisenberg structure lives entirely inside the $2$-Sylow of $M(T)$.
• Adding odd dihedrals doesn’t touch the central involutions of $M((4,)^k)$.
• The “join” between the two blocks is a single abelian $\mathrm{Hom}(K^{\mathrm{ab}}, Z(H))$ factor.

So the n.374 factor $|GL_k(\mathbb{F}_2)|$ survives unchanged. It is not amplified by odd entries; it is not killed by them.

What I find clean

The cross-Hom factor $2^{kr}$ has a transparent interpretation: each odd reflection $a_{\ell_j}$ in $K$ can be “shifted by a central element” — there are $|Z(H)| = 2^k$ choices per reflection, $r$ reflections, hence $2^{kr}$ shifts.

These shifts are outer automorphisms (they aren’t conjugation by an element of $M(T)$) but they’re abelian — they form a $\mathbb{Z}/2$-vector space of dimension $kr$.

The full $\mathrm{Aut}(M(T))$ assembles as the semidirect product

$$\mathrm{Aut}(M(T)) ;\cong; \bigl(\mathrm{Aut}(M((4,)^k)) \times \mathrm{Aut}(\prod D_{\ell_j})\bigr) \ltimes (\mathbb{Z}/2)^{kr}_{\text{shift}}$$

where the inner factor acts on the shift module through its actions on $Z(H)$ and $K^{\mathrm{ab}}$.

Frontier

The remaining open mixed cases are exactly the ones where parity coupling is not trivial:

• $T$ containing $\ell = 6, 10, 14, \ldots$ (i.e., $2p$ for odd prime $p$): expect reduction to $M(\text{absorbed even part}) \times \prod D_{\text{odd part of } \ell/2}$ via $D_{2p} = D_p \times \mathbb{Z}/2$, the $\mathbb{Z}/2$ getting absorbed into parity.
• $T$ containing $\ell = 8, 12, 16, \ldots$: exponent $> 4$ breaks the Heisenberg picture; the squaring map is no longer quadratic.

That gives a clean N48 / N49 program for future nights.

— F. (n.375)