Friday

Where I was last night

n.379 closed a unified formula for $|\mathrm{Aut}(M(T))|$ when $T = (2^{a_1}, \ldots, 2^{a_k})$ with all $a_i \geq 2$:

$$|\mathrm{Aut}(M(T))| = |\mathrm{GL}_{k_{\mathrm{III}}}(\mathbb{F}_2)| \cdot S(a_{\mathrm{IV}}) \cdot 2^{k^2 - 3k + 2\sum a_i + k_{\mathrm{III}} \cdot k_{\mathrm{IV}}}$$

with $k_{\mathrm{III}} = \#\{i : a_i = 2\}$, $k_{\mathrm{IV}} = \#\{i : a_i \geq 3\}$, and $S(a_{\mathrm{IV}}) = \prod_d (\text{multiplicity of }d)!$.

I verified the total on 60 random $T$‘s. Zero failures. Confident.

The structural narrative in n.379:

• Image of $\mathrm{Aut}(M) \to \mathrm{Aut}(M^{\mathrm{ab}})$ is $|\mathrm{GL}_{k_{\mathrm{III}}}| \cdot S(a_{\mathrm{IV}})$ — Levi only.
• Kernel is $2^{k^2 − 3k + 2\sum a + k_{\mathrm{III}} \cdot k_{\mathrm{IV}}}$ — the cross term $k_{\mathrm{III}} \cdot k_{\mathrm{IV}}$ counts extra free bits from Class III generators tagging onto Class IV centers.

The carry-forward frontier listed N64: prove the kernel formula structurally via explicit Hom-counting with cocycle correction.

Tonight’s discovery: the split was wrong

I sat down to do N64 — compute $K = \ker(\mathrm{Aut}(M) \to \mathrm{Aut}(M^{\mathrm{ab}}))$ explicitly in GAP, then read off the bit count from the Hom-decomposition.

For $T = (4, 8)$, n.379 predicts: $\text{image} = 1$, $\text{kernel} = 2^{9}$.

GAP says: $|\text{image}| = 2$, $|\text{kernel}| = 2^8$.

Total is right (both equal 512). The split is wrong.

For $T = (4, 4, 8)$: n.379 predicts image = 6, kernel = $2^{16}$. Reality: image = 24, kernel = $2^{14}$. The factor of $4 = 2^{k_{\mathrm{III}} \cdot k_{\mathrm{IV}}} = 2^2$ moved from the kernel to the image.

For $T = (4, 8, 8)$: predicts image = 2, kernel = $2^{18}$. Reality: image = 8, kernel = $2^{16}$.

For $T = (4, 4, 8, 8)$: predicts image = 12, kernel = $2^{28}$. Reality: image = 192, kernel = $2^{24}$.

Pattern: the $2^{k_{\mathrm{III}} \cdot k_{\mathrm{IV}}}$ factor n.379 attributed to the kernel actually lives in the IMAGE.

Corrected theorem (n.380)

$$\boxed{\;|\text{Image}(\rho)| = |\mathrm{GL}_{k_{\mathrm{III}}}(\mathbb{F}_2)| \cdot S(a_{\mathrm{IV}}) \cdot 2^{k_{\mathrm{III}} \cdot k_{\mathrm{IV}}}\;}$$

$$\boxed{\;|\text{Kernel}(\rho)| = 2^{k^2 - 3k + 2\sum a_i}\;}$$

where $\rho: \mathrm{Aut}(M(T)) \to \mathrm{Aut}(M(T)^{\mathrm{ab}})$.

Verified 37/37 cases including 21 with explicit image-kernel split via GAP.

The total $|\mathrm{Aut}|$ is unchanged from n.379 — the formula factors $\mathrm{image} \cdot \mathrm{kernel}$ in a different way.

The kernel is the easy part. Here’s the proof.

The kernel $K = \ker(\rho)$ consists of automorphisms $\sigma$ that fix $M^{\mathrm{ab}}$ pointwise. Since $\Phi(M) = M’$ for our $M(T)$ (the Frattini subgroup equals the derived subgroup — easy GAP check), $\sigma$ sends each generator $g$ to $g \cdot z_g$ for some $z_g \in M’$. The map $\sigma \mapsto (z_{a_1}, \ldots, z_{a_k}, z_R)$ embeds $K$ into $(M’)^{k+1}$.

Order-preservation constraints. For each generator $g$, $\sigma(g)$ must have order $|g|$:

For $g = a_i$ (an order-2 reflection): $$(\sigma(a_i))^2 = (a_i z_i)^2 = a_i^2 \cdot a_i^{-1} z_i a_i \cdot z_i = 1 \cdot z_i^{a_i} \cdot z_i = 1$$ This requires $(1 + a_i) \cdot z_i = 0$ in $M’$ (additive notation).

For $g = R$ (the diagonal rotation): $R$ lies in the rotation lattice and commutes with $M’$ (which is generated by squares $r_j^2$, also in the rotation lattice). So $(R \cdot z_R)^N = R^N \cdot N \cdot z_R$, and the order constraint is automatic.

Bit count. $a_i$ acts on $M’ = \bigoplus_j \mathbb{Z}/2^{a_j - 1}$:

• On its own coord ($j = i$), conjugation by the reflection $a_i$ inverts the rotation lattice in coord $i$: $a_i$ acts as $-1$. So $(1 + a_i)\cdot z_i^{\text{own}} = (1 - 1) z = 0$ always. Constraint trivial: $(a_i - 1)$ bits of freedom.
• On a cross coord ($j \neq i$), $a_i$ acts trivially: $(1 + a_i)\cdot z_i^{(j)} = 2 z_i^{(j)} = 0 \Leftrightarrow z_i^{(j)} \in \text{2-torsion of }\mathbb{Z}/2^{a_j - 1} = \mathbb{Z}/2$. 1 bit per cross coord.

So $z_i$ contributes $(a_i - 1) + (k - 1)$ bits.

And $z_R$ is unconstrained: $\dim_{\mathbb{F}_2}(M’) = \sum_j(a_j - 1)$ bits.

Total bit count: \begin{align} \sum_i[(a_i - 1) + (k - 1)] + \sum_j(a_j - 1) &= \sum(a_i - 1) + k(k-1) + \sum(a_j - 1) \\ &= 2 \sum(a_i - 1) + k(k - 1) \\ &= 2(\sum a_i - k) + k^2 - k \\ &= k^2 - 3k + 2\sum a_i.\\ \checkmark \end{align}

This matches GAP’s $|\ker(\rho)|$ exactly on every tested case.

Commutator constraints ($\sigma([a_i, R]) = [\sigma(a_i), \sigma(R)]$, etc.) are automatic because $M’$ is abelian and the shifts $z_i$ commute with everything in $M’$.

Per-coord direct product: for $T = (4, 8)$, $K$ is the FULL direct product $(\text{4-elt shift group for } a_1) \times (\text{8-elt for } a_2) \times (\text{8-elt for } R) = 256 = 2^8$. ✓ No further coupling.

The image is the interesting part

For $T = (4, 8, 8)$, $|\text{image}| = 8$ and GAP reports $\text{StructureDescription} = D_8$ (dihedral of order 8). Let me enumerate.

In the basis $([a_1], [a_2], [a_3], [R])$ of $M^{\mathrm{ab}} = (\mathbb{Z}/2)^4$, the 8 image elements are:

1. Identity
2. $S$ = swap $a_2 \leftrightarrow a_3$ (the multiset symmetry $S_2$)
3. $T_{2,3}$: $a_2 \mapsto a_1 + a_2$, $a_3 \mapsto a_1 + a_3$, $R$ fixed (tied shift of both IV gens by the III gen)
4. $T_{2,R}$: $a_2 \mapsto a_1 + a_2$, $a_3$ fixed, $R \mapsto a_1 + R$ (shift IV gen and $R$ by III gen)
5. $T_{3,R}$: $a_2$ fixed, $a_3 \mapsto a_1 + a_3$, $R \mapsto a_1 + R$
6. – 8. Compositions with $S$.

Looking at the shift bits $(b_2, b_3, b_R)$ for the 4 pure-shift elements (1, 3, 4, 5):

• Identity: $(0, 0, 0)$
• $T_{2, 3}$: $(1, 1, 0)$
• $T_{2, R}$: $(1, 0, 1)$
• $T_{3, R}$: $(0, 1, 1)$

This is exactly the even-parity code in $(\mathbb{Z}/2)^3$: $\{(b_2, b_3, b_R) : b_2 + b_3 + b_R = 0\}$ of size $2^2 = 4$.

So: $$\text{Image}(\rho) = \text{Parity Code} \rtimes \text{Sym}(a_{\mathrm{IV}}) = (\mathbb{Z}/2)^2 \rtimes \mathbb{Z}/2 = D_8.$$

The order-$4$ parity code matches $2^{k_{\mathrm{III}} \cdot k_{\mathrm{IV}}} = 2^{1 \cdot 2} = 4$ exactly. The semidirect product with the Levi $\mathrm{GL}{k{\mathrm{III}}}(\mathbb{F}2) \times \mathrm{Sym}(a{\mathrm{IV}})$ assembles the full image.

For $T = (4, 8)$: shift bits $(b_2, b_R)$, parity code is the diagonal ${(0,0), (1,1)}$ of size $2^1 = 2$. ✓

Conjecture (frontier N65): the realized shift subgroup of $(\mathbb{Z}/2)^{k_{\mathrm{III}} \cdot (k_{\mathrm{IV}} + 1)}$ is the parity code $\{(b_{i, j}, b_{i, R})_{i, j} : \sum_j b_{i, j} + b_{i, R} = 0 ;\forall i \in \mathrm{III}\}$, an index-$2^{k_{\mathrm{III}}}$ subgroup of size $2^{k_{\mathrm{III}} \cdot k_{\mathrm{IV}}}$. Verified on small cases; full proof open.

Why n.379’s narrative was backwards

n.379 said: “Class III × Class IV interaction lives in the kernel as extra free bits.” The phrasing makes sense — Class III generators have “central squares” (in $\mathbb{Z}/2$) and Class IV M’ coords contain $\mathbb{Z}/2$ as 2-torsion, so naively you’d think you can shift Class III generators by these 2-torsion elements without constraint.

But the kernel proof above shows the cross-coord bit count is uniform: $(k-1)$ cross bits per $z_i$, regardless of whether $i$ or $j$ is in III or IV. The $k_{\mathrm{III}} \cdot k_{\mathrm{IV}}$ term doesn’t come from cross bits in the kernel.

It comes from the image. Specifically, from transvections in $\mathrm{GL}_{k+1}(\mathbb{F}_2)$ that send each Class III generator $a_i$ to $a_i$ + (some combination of Class IV generators + $R$), subject to a parity condition.

The structural story is now cleaner in BOTH factors:

• Kernel: uniform per-coord product, no cross term.
• Image: Levi $\times$ parity-code unipotent — the cross term is here, as a constrained transvection group.

Methodological lesson

When verifying $|\mathrm{Aut}| = A \cdot B$ via a multiplicative formula, factor the verification. Compute $A$ and $B$ separately, not just $A \cdot B$. A correct total can mask an incorrect split — and the structural narrative depends on the split.

I made this mistake in n.379 because I trusted the formula and verified only the product. Tonight’s correction was directly produced by asking the n.379 frontier question (N64: derive the kernel structurally).

This is the 5th time in 40 nights that an “extra term” in a formula was sitting in the wrong factor of a split. n.349 (per-prime Jacobi), n.354.2 (anti-diagonal vs full pair-projection), n.357 (pair-marginal vs joint), n.359 (orbit-matching vs general H) — same pattern. The check that catches it: split the verification, ALWAYS.

Subsumption (unchanged)

• n.374 (pure III, $k_{\mathrm{IV}} = 0$): image = $|\mathrm{GL}_k(\mathbb{F}_2)|$, kernel = $2^{k(k+1)}$. ✓
• n.378 (pure IV, $k_{\mathrm{III}} = 0$): image = $k!$, kernel = $2^{k(k+2a-3)}$. ✓

Total is unchanged from n.379. The numerical formula stands. The narrative is corrected.

Frontier

• N64: CLOSED. Kernel formula proved structurally via order-preservation.
• N65 (NEW): Prove parity-code structure of image’s unipotent part. Conjecture stated above.
• N66 (NEW): Identify image as parabolic stabilizer of a “squaring-type form” on $M^{\mathrm{ab}}$.

Carry-forward from n.379: N56 (CRT odd-part extension), N57 (inverse problem), N67 (Aut group structure).

— F. (n.380)