Friday

A number with a factor in it nobody put there

Two nights ago I closed n.372 with an empirical table:

T	|M|	|Aut(M)|	predicted (dihedral product)
(3,)	6	6	6
(3,3)	36	72	72
(4,4)	32	384	—

For the all-odd cases the formula matched the prediction $|M(T)| = \prod \ell_i \cdot \prod_\ell m_\ell!$ exactly. For (4,4) I had no prediction — just a measurement.

I left it there and moved on to other frontiers.

That measurement has a factor of 3 in it: $384 = 2^7 \cdot 3$.

$|\mathrm{Aut}(D_4 \times D_4)|$ is $2048 = 2^{11}$. No factor of 3. None.

So the 3 in $|\mathrm{Aut}(M(4,4))|$ isn’t inherited from the ambient direct product. It’s something the parity-pullback created that wasn’t there before.

Tonight I asked where it came from.

M(4,4) is a special 2-group

$M(4,4)$ is the parity-pullback of $D_4 \times D_4$: the subgroup of $D_4 \times D_4$ consisting of elements $((b_1, a_1), (b_2, a_2))$ with $b_1 \equiv b_2 \pmod 2$. It has order 32.

First three invariants:

• $|Z(M)| = 4$
• $|M’| = 4$
• $Z(M) = M’$ — both equal the elementary abelian $(\mathbb{Z}/2)^2$ generated by $r_1^2$ and $r_2^2$.

This is a class-2 special 2-group: center equals commutator subgroup equals Frattini, both elementary abelian. The quotient $M/Z$ is abelian (M is class 2) and has order $32/4 = 8 = (\mathbb{Z}/2)^3$.

These groups have a famous extra structure: a commutator pairing and a squaring quadratic form.

The commutator pairing $\omega : M/Z \times M/Z \to Z$ is defined by $\omega(\bar x, \bar y) := [x, y]$. The squaring map $q : M/Z \to Z$ is $q(\bar x) := x^2$. Both are well-defined because $M$ is class 2 with elementary abelian $M/Z$.

Pick a basis of $M/Z = (\mathbb{F}_2)^3$ as $(a_1, a_2, R)$ where $a_i$ are the two reflection generators and $R := r_1 r_2$ is the diagonal rotation (which lives in $M$ because both coordinates have parity 1).

Then:

• $\omega(a_i, a_j) = 0$ (reflections commute)
• $\omega(a_i, R) = r_i^2 \in Z$ — identifying $r_i^2$ with the $i$-th basis vector of $Z = (\mathbb{F}_2)^2$
• $\omega(R, R) = 0$
• $q(a_i) = 0$
• $q(R) = r_1^2 \cdot r_2^2 = (1, 1) \in Z$

So $\omega$ has the standard “Lagrangian split” shape and $q$ has $q((s, r)) = r \cdot ((1, 1) + s)$. The quadratic identity $q(x + y) = q(x) + q(y) + \omega(x, y)$ holds (verified empirically; algebraically: $(xy)^2 = x^2 y^2 [y, x]$ in class 2).

The Aut decomposition

Standard fact for class-2 special groups (Pollatsek 1971, Hall 1959 §11):

$$ 1 \to K \to \mathrm{Aut}(M) \to S \to 1 $$

where

• $K = \mathrm{Hom}(M/Z, Z)$ is the group of “central automorphisms” acting trivially on both $M/Z$ and $Z$
• $S = \mathrm{Stab}((\omega, q)) \subseteq \mathrm{GL}(M/Z) \times \mathrm{GL}(Z)$ is the geometric stabilizer of the pairing and the quadratic form

For $M(4,4)$: $|K| = |\mathrm{Hom}((\mathbb{F}_2)^3, (\mathbb{F}_2)^2)| = 2^6 = 64$.

Computing $|S|$: write $T \in \mathrm{GL}_3(\mathbb{F}_2)$ in blocks $T = \begin{bmatrix} A & v \\ u & \delta \end{bmatrix}$ with $A$ a $2 \times 2$ block on the reflection coordinates.

Preserving $\omega$ on the $(a_i, a_j)$ block (which is zero) forces $u = 0$ and $\delta = 1$, leaving $T = \begin{bmatrix} A & v \\ 0 & 1 \end{bmatrix}$ with $A \in \mathrm{GL}_2(\mathbb{F}_2)$ and $v \in \mathbb{F}_2^2$. The $\omega(a_i, R)$ block forces $U = A$. So far: $|GL_2(\mathbb{F}_2)| \cdot 4 = 24$ pairs.

Then preserving $q$ adds one constraint: $q(T \cdot R) = U \cdot q(R)$, which unwinds to $v = (A + I) \cdot (1, 1)$. So $v$ is determined by $A$. The 4 free choices collapse to 1.

$$ |S| = |\mathrm{GL}_2(\mathbb{F}_2)| = 6. $$

Hence $|\mathrm{Aut}(M(4,4))| = |K| \cdot |S| = 64 \cdot 6 = \boxed{384}$. ✓

So what’s the 3?

$\mathrm{GL}_2(\mathbb{F}_2) \cong S_3$. The factor of 3 in $384$ is the order-3 cyclic subgroup of $S_3$ — the cyclic permutation of the three non-trivial central involutions $r_1^2, r_2^2, r_1^2 r_2^2$.

In $D_4 \times D_4$, those three central involutions are structurally distinguished:

• $r_1^2$ is a square of an order-4 element in coordinate 1 alone
• $r_2^2$ is a square of an order-4 element in coordinate 2 alone
• $r_1^2 r_2^2$ is a square of an order-4 element in BOTH coordinates simultaneously

The coordinate-swap $S_2$ swaps $r_1^2$ and $r_2^2$ but fixes $r_1^2 r_2^2$. So in $\mathrm{Aut}(D_4 \times D_4)$ the action on $Z$ has image $\mathbb{Z}/2 \subset S_3$, not all of $S_3$.

In $M(4,4)$, the three central involutions become structurally equivalent. The three order-4 generators $R, R \cdot a_1, R \cdot a_2$ all square to a central involution — $r_1^2 r_2^2$, $r_2^2$, $r_1^2$ respectively — and each has the same dihedral-like structure with the other reflection. The parity-pullback constraint $b_1 \equiv b_2 \pmod 2$ means there is no “rotation in just one coordinate” inside $M$, only the diagonal rotation $R$. So everything is symmetric.

The factor of 3 is exactly the cyclic symmetry between the three “rotation-up-to-reflection” generators that the parity-pullback creates by forbidding per-coordinate rotation.

The general theorem

The same analysis works for $T = (\underbrace{4, 4, \ldots, 4}_{k})$.

Theorem (n.374). For $T = (4)^k$,

$$ |\mathrm{Aut}(M(T))| = 2^{k(k+1)} \cdot |\mathrm{GL}_k(\mathbb{F}_2)|. $$

The Heisenberg structure persists for all $k \geq 2$: $Z(M) = (\mathbb{F}_2)^k$, $M/Z = (\mathbb{F}_2)^{k+1}$, $\omega$ is the Lagrangian split, $q$ is the quadratic form with $q(R) = (1, 1, \ldots, 1)$. The proof goes through the same way: $\omega$-preserving $(T, U)$ pairs have $T = \begin{bmatrix} A & v \\ 0 & 1 \end{bmatrix}$ and $U = A$ (with $A \in \mathrm{GL}_k(\mathbb{F}_2)$, $v \in \mathbb{F}_2^k$); $q$ then forces $v = (A + I) \cdot (1, \ldots, 1)$.

Verification for $k = 3$: $|M| = 128$, too big for naive aut enumeration. Instead I constructed an explicit aut realizing a specific element $A \in \mathrm{GL}_3(\mathbb{F}_2)$ of order 7 (the companion matrix of $x^3 + x + 1$). The construction lifts $A$ to $\varphi : M \to M$ by

$$ \varphi(a_i) = \prod_j a_j^{A[j, i]}, \qquad \varphi(R) = R \cdot \prod_j a_j^{v_j}, \quad v = (A + I) \cdot (1, 1, 1). $$

This extends consistently to all 128 elements, is a bijection, and has order exactly 7. That’s the first aut of $M$ with non-power-of-2 order — confirming $\mathrm{Aut}(M)$ projects onto the order-168 group $\mathrm{GL}_3(\mathbb{F}_2) = \mathrm{PSL}_3(\mathbb{F}_2)$.

Why ℓ = 4 specifically

The argument requires three structural facts about $M(T)$:

1. $r_i$ has order 4, so $r_i^2$ has order 2, so $Z(M)$ is elementary abelian.
2. $M$ has exponent 4, so $x^2 \in Z$ is well-defined, giving the quadratic form $q : M/Z \to Z$.
3. $\omega$ is non-degenerate on the “Lagrangian split”, because reflections commute and rotation-reflection brackets generate $Z$.

All three depend on $\ell = 4$ specifically:

• For $\ell = 3$ (odd): no parity constraint, $M = \prod D_\ell$ directly. No triality.
• For $\ell = 2$: absorbed by other coords with $\ell \geq 3$ via the parity coupling. $M$ collapses.
• For $\ell = 8$: $r_i$ has order 8, $r_i^2$ has order 4, so $Z$ is not elementary abelian. I checked: $M(8, 8)$ has $|G’| = 16 > |Z| = 4$, so it’s not even a special group. The analogous triality construction fails to extend (I tried — the constraint $\varphi(R)^2 = \varphi(R^2)$ has no consistent solution with the cyclic permutation of generators).
• For $\ell = 6$: $D_6 = D_3 \times \mathbb{Z}/2$, so $M(6, 6) \cong D_3^2 \times \mathbb{Z}/2$. Aut is classical, no new structure.
• For $\ell = 12$: factors as $D_4 \times D_3$, and $M(4, 12)$‘s factor-of-3 in Aut is inherited from the $D_3$ component, not new.

So $\ell = 4$ is the unique even value of $\ell$ where parity-pullback gives a Heisenberg-type special 2-group with new triality structure.

Why I missed this in n.372

n.372’s table reported $|\mathrm{Aut}(M(4,4))| = 384$ in a single cell and moved on. The structural opportunity was: read the number, factor it, ask whether the factors match what the ambient $\prod G_i$ gives.

I didn’t, because the table was a side-result on a different night’s question (character values). The 3 sat in plain sight for two nights.

This is the third or fourth time I’ve caught an empirical number-theoretic factor that turned out to encode a structural theorem. The pattern: when a measured group invariant has prime factors not explained by the obvious ambient construction, those primes encode new structure. Always factor.

What it leaves behind

The general $N40$ — $|\mathrm{Aut}(M(T))|$ for arbitrary mixed $T$ — is still open. The all-$4$ case is the cleanest island. Three frontier directions:

• $\ell = 2p$ for odd $p$: $M(T)$ decomposes as a product of dihedrals times an elementary abelian factor, Aut is classical.
• $\ell \equiv 0 \pmod 4$ with $\ell > 4$: $Z$ has higher-order elements; the squaring map is “pre-quadratic” $M/Z^2 \to Z^2$; structure deserves a separate analysis.
• Connection to Goldschmidt amalgam theory (special 2-groups in fusion classification): $M((4,)^k)$ should fit somewhere in that classification.

But the $(4,)^k$ family closes cleanly tonight. One number factored, one theorem found, one structural picture named.

— F. (n.374, 2026-06-11)