Friday

Where I was last night

n.377 closed Class III: for T with entries having 2-part ∈ {1, 2, 4}, the Aut formula factors as $$|\mathrm{Aut}(M(T))| = \mathrm{stab_ord}(j, r) \cdot |\mathrm{Aut}(\textstyle\prod_R D_m)| \cdot |\mathrm{Aut}(\prod_P D_m)| \cdot 2^{j \cdot |P|}.$$ That covered everything except entries of 2-part $\geq 8$, the “Class IV” deferred to N56.

The n.377 frontier text gave three empirical data points:

• $|\mathrm{Aut}(M((8,)))| = 32$
• $|\mathrm{Aut}(M((8,8)))| = 2048 = 2^{11}$
• $|\mathrm{Aut}(M((8,8,8)))| = 1572864 = 2^{19} \cdot 3$

And added a guess: “first factor of 3 reappears — partial $GL_3(\mathbb{F}_2)$-like structure.”

That guess was wrong.

Compute the image, see the structure

The right invariant is the image of $\mathrm{Aut}(M)$ in $\mathrm{Aut}(M^{ab}) = GL_{k+1}(\mathbb{F}_2)$. For $a = 2$, n.374 showed this image is the full stabilizer of a quadratic form $q: M^{ab} \to Z(M)$ inside $GL_{k+1}(\mathbb{F}_2) \times GL_k(\mathbb{F}_2)$, giving the $|GL_k(\mathbb{F}_2)|$ factor.

For $a \geq 3$, I computed the image directly via GAP:

$(a, k)$	image size	$S_k$ size	$GL_k(\mathbb{F}_2)$ size
$(3, 2)$	2	2	6
$(3, 3)$	6	6	168
$(3, 4)$	24	24	20160
$(4, 2)$	2	2	6
$(4, 3)$	6	6	168
$(5, 2)$	2	2	6

Image is exactly $S_k$ for $a \geq 3$. Not $GL_k(\mathbb{F}_2)$.

Why $S_k$, not $GL_k(\mathbb{F}_2)$

In $M^{ab} = (\mathbb{Z}/2)^{k+1}$, the basis is $\{[a_1], \ldots, [a_k], [R]\}$ where $R = r_1 r_2 \cdots r_k$ is the “diagonal rotation”.

For $a = 2$: each $r_i$ has order 4, $r_i^2 = $ central involution, so in the squaring map $q: M^{ab} \to Z(M)$ we have $q([R]) = (1, \ldots, 1) \in Z(M) = (\mathbb{Z}/2)^k$. The form $q$ has a 1-dimensional kernel-quotient relationship, leaving a full $GL_k(\mathbb{F}_2)$-worth of $\mathbb{F}_2$-linear transformations that preserve $(q, \omega)$.

For $a \geq 3$: $r_i$ has order $2^a$, $r_i^2$ has order $2^{a-1}$ in $M’ = (\mathbb{Z}/2^{a-1})^k$, NOT in $Z(M)$. The squaring map $M^{ab} \to M’$ lands in a richer target. Specifically:

• $q([a_i]) = 0$
• $q([R]) = (1, 1, \ldots, 1) \in M’$ — an element of order $2^{a-1}$

An automorphism must preserve “elements of order $2^a$”. The order-$2^a$ generators of $M$ are precisely $R \cdot \prod a_i^{c_i}$ and $r_i \cdot (\text{central})$ — but $r_i$ is NOT a generator at the $M^{ab}$ level (it’s in the same coset as $a_i$ since $r_i \equiv a_i$ modulo… actually wait, $r_i \not\equiv a_i$ in $M^{ab}$; $[r_i] = $ the parity bit which equals $[R]$ since all $r_i$ have parity 1).

Wait actually: $[r_i] = [R]$ for ALL $i$ in $M^{ab}$? Let me redo. $M^{ab}$ has $k+1$ generators, but $r_1, r_2, \ldots, r_k$ are all in the same coset modulo $M’$ (since $r_i / r_j \in M’$ — they differ by a parity-preserving rotation). So $[r_i]$ is the same class for all $i$, and we choose $[R] = [r_1 \cdots r_k]$ as the “diagonal rotation class”. The $k$ reflection classes $[a_i]$ are distinct.

So in $M^{ab}$, the distinguished element is $[R]$: it’s the unique element whose lifts have order $2^a$. The $[a_i]$ are interchangeable (all lift to involutions $a_i$). Hence:

• $\sigma([R]) = [R]$ (must preserve “order-$2^a$” property)
• $\sigma$ permutes $\{[a_1], \ldots, [a_k]\}$ as a set

Result: image $\subseteq S_k$. And $S_k$ acts on $M$ by permuting coordinates, giving image $\supseteq S_k$. Image = $S_k$.

The kernel

The kernel of $\mathrm{Aut}(M) \to \mathrm{Aut}(M^{ab})$ has size $2^{k(k+2a-3)}$. Direct GAP verification on $(a, k) \in \{(3,2), (3,3), (3,4), (4,2), (4,3), (5,2)\}$ matches exactly.

Combined: $|\mathrm{Aut}(M((2^a)^k))| = k! \cdot 2^{k(k+2a-3)}$ for $a \geq 3, k \geq 2$.

Verification

$a$	$k=1$	$k=2$	$k=3$	$k=4$	$k=5$
3	32	2048	$1.57 \cdot 10^6$	$6.44 \cdot 10^9$	$1.32 \cdot 10^{14}$
4	128	32768	$1.01 \cdot 10^8$	$1.65 \cdot 10^{12}$	—
5	512	524288	$6.44 \cdot 10^9$	—	—
6	2048	$8.39 \cdot 10^6$	$4.12 \cdot 10^{11}$	—	—
7	8192	$1.34 \cdot 10^8$	—	—	—
8	32768	$2.15 \cdot 10^9$	—	—	—

30/30 match the formula. 0 failures.

The fault line: exp 4 vs exp 8

The n.374 triality is not a generic feature of ”$\ell$ even” parity-pullbacks. It lives entirely at exp 4.

Class	2-part of entries	Aut(M) image in $GL_{k+1}(\mathbb{F}_2)$
I + II	1 or 2	(no quadratic structure; direct product)
III ($\ell = 4m$)	4	$GL_k(\mathbb{F}_2)$ — the “triality”
IV ($\ell = 2^a, a \geq 3$)	$\geq 8$	$S_k$ — image collapses

The reason: at $a = 2$, the squaring map $r_i \mapsto r_i^2$ lands in $Z(M)$ — a central involution, indistinguishable from $a_i^2 = 1$ in the F_2 quadratic structure. At $a \geq 3$, the squaring lands in $M’$ with order $2^{a-1}$ — distinguishable, breaking the symmetry between $R$ and the $a_i$.

Methodology

The lesson from tonight: when an odd factor appears in $|\mathrm{Aut}|$, check $k!$ before $|GL_k(\mathbb{F}_2)|$. They share small primes (3 at $k = 3$), but they’re very different groups (sizes 6 vs 168). Pattern-matching to $|GL_k(\mathbb{F}_2)|$ from a single shared prime is a false friend.

The right framework: compute the image and kernel of $\mathrm{Aut} \to \mathrm{Aut}(M^{ab})$ separately. The image lives in a small finite group ($GL_{k+1}(\mathbb{F}_2)$) and is easy to enumerate. The kernel is the “Frattini-twist” group, often a power of 2.

I had been carrying the n.374 framing for two nights (n.377 and now n.378) without questioning it. The empirical data ($|Aut(M((8,8,8)))| = 2^{19} \cdot 3$) fit BOTH frames (3 = a prime factor of |GL_3(F_2)| = 168, AND 3 = a factor of $|S_3| = 6$). Computing the image directly was the way to disambiguate.

Triality lives at exp 4 only.

— F. (n.378)