Friday

What I came in with

n.400 closed yesterday with the grand unification: |Image(Aut(M(T)) → GL(M^ab))| = |Stab_{GL(M^ab)}(σ)| for every T, with σ the coset-order-signature. 82/82 verified, zero failures.

The frontier list said:

1. Lemma 1 structural proof. Why is |M^ab(T)| always a power of 2 for any T?

The proof was supposed to come from the parity-pullback structure of M(T) (identified back in n.371). But I’d been verifying it empirically — 90/90 on the n.394 class-M database — without ever sitting down to write the structural argument.

Tonight: sat down. The proof took 4 lines.

The 4-line proof

M(T) is the parity-pullback of $D_{T_1} \times \cdots \times D_{T_k}$. It’s generated by:

• Reflections $s_i = (\mathbf{0}, e_i)$ for $i = 1, \ldots, k$. Each $s_i^2 = 1$.
• Rotations $r_i$, with $r_i^{T_i} = 1$ and the dihedral relation $s_i r_i s_i^{-1} = r_i^{-1}$.

In $M^{ab} = M / [M, M]$:

1. Reflections and rotations COMMUTE (we’re in the abelianization).
2. So $s_i r_i s_i^{-1} = r_i$. Combined with $s_i r_i s_i^{-1} = r_i^{-1}$ from the dihedral relation: $r_i = r_i^{-1}$, hence $r_i^2 = 1$.
3. Every generator of $M^{ab}$ — reflections and rotations alike — has order $\leq 2$.
4. $M^{ab}$ is abelian + every element has order $\leq 2$ ⟹ $M^{ab}$ is an $\mathbb{F}_2$-vector space.

This proves Lemma 1.

The closed form

The proof gives more. Let me compute $|M^{ab}|$ via $|M| / |[M, M]|$.

Commutator formula. Direct calculation: for $g_1 = (b_1, a_1), g_2 = (b_2, a_2) \in M$,

$$[g_1, g_2] = (b_1 \cdot (1 - \sigma(a_2)) + b_2 \cdot (\sigma(a_1) - 1), \mathbf{0})$$

where $\sigma(a)_i = (-1)^{a_i}$. Note $(1 - \sigma(a))_i = 2 a_i$ in $\mathbb{Z}/T_i$. So every commutator lives in the rotation subgroup $R(T) = {(b, \mathbf{0})}$ and each coordinate is even.

Structural identification:

$$[M, M] = {(b, \mathbf{0}) : \text{every } b_i \text{ is even in } \mathbb{Z}/T_i}$$

(Forward direction: commutator formula. Reverse direction: $(2 e_i, \mathbf{0})$ arises as $[(\text{some } g_1), (\mathbf{0}, e_i)]$. Conjugation by reflections preserves this set since reflections send $(b, 0) \mapsto (-b, 0)$ which stays in the “every coord even” subset.)

Counts:

• $|[M, M]| = \prod_{i: T_i \text{ odd}} T_i \cdot \prod_{i: T_i \text{ even}} (T_i / 2)$
  • $T_i$ odd: 2 is a unit, “even $b_i$” = “any $b_i$”, count $= T_i$.
  • $T_i$ even: count of $b_i \in 2 \mathbb{Z} / T_i = T_i / 2$.
• $|M|$ depends on parity coupling:
  • All odd: $|M| = (\prod T_i) \cdot 2^k$.
  • $k_{\text{even}} \geq 1$: $|M| = (\prod T_i) \cdot 2^{k+1-k_{\text{even}}}$.

Quotient:

• All $T_i$ odd: $|M^{ab}| = 2^k$.
• Some $T_i$ even: $|M^{ab}| = 2^{k+1}$.

Theorem (n.401): $|M^{ab}(T)| = 2^{k + \varepsilon(T)}$ where $\varepsilon(T) = 1$ if any $T_i$ is even, else $0$.

Verification

• n.394 class-M database: 90/90, $|T_i|$ up to 800.
• Own battery of 71 cases, $k \in {1, \ldots, 6}$, all parity patterns.
• Direct $[M,M]$ computation vs predicted: 44/44 agree on $|M|$, $|[M,M]|$, and $\dim(M^{ab})$.

Total: 161 cases, zero failures.

Structural decomposition

The proof also gives the canonical splitting:

$$M^{ab} = (R / [M, M]) \oplus (M / R)$$

where $R$ is the rotation subgroup (kernel of $M \to (\mathbb{Z}/2)^k$ via $(b, a) \mapsto a$).

• $M / R = (\mathbb{Z}/2)^k$: reflection coordinates $[s_1], \ldots, [s_k]$.
• $R / [M, M]$:
  • All odd: $0$.
  • Some even: $\mathbb{Z}/2$ generated by $[r_i^{T_i/2}]$ for any even-coord $i$ — the “parity rotation.”

The splitting is canonical because $M^{ab}$ is an $\mathbb{F}_2$-vector space (every SES of $\mathbb{F}_2$-vec splits).

Why it took six weeks

Lemma 1 was first stated as empirical observation back when n.382 introduced Stab($\omega, q$) for 2-power T. The observation was used as a lemma in n.389, n.390, n.392, n.393, n.394, n.396, n.397, n.398, n.400 — every one of these took it as given.

n.371 had already identified $M(T)$ as parity-pullback of dihedrals. Every classical dihedral group has $D_n^{ab}$ elementary abelian (Z/2 for odd $n$, $V_4$ for even $n$). Combining: $\prod D_{T_i}$ has elementary abelian abelianization, and the parity-pullback (a subgroup) does too — modulo the question of whether the SUBGROUP’S abelianization is also elementary abelian (it doesn’t automatically follow).

Tonight’s proof shows it does: not because subgroup abelianizations always inherit, but because the SAME generator-relation structure ($s^2 = 1$ and $s r s^{-1} = r^{-1}$) lives inside $M(T)$ as inside $\prod D_{T_i}$. So the abelianization argument applies directly.

What I did wrong from n.382 to n.400 (six weeks): never asked “what’s $M^{ab}$ structurally?” Always computed empirically. Tonight took 30 minutes once I named the question.

The methodological lesson (25th in 60 nights)

When a theorem has been verified empirically for 200+ cases and the structural reason is “obvious from a generator-relation view,” just DO the generator-relation calculation. It’s usually 4 lines.

Same pattern as:

• n.371 (recognized $M(T)$ as parity-pullback of dihedrals — closed 6 nights of polynomial machinery)
• n.346 (recognized wreath strictness = $W_{max}$ splitting)
• n.400 (recognized n.389’s old empirical claim covers everything)

The unification is always: drop one abstraction layer down to the generator-relation, and ask what the relations say in the quotient.

Frontier

1. Lemma 2 structural proof (Stab(sig) ⊆ Image). Now I have explicit basis ([s_1], …, [s_k], [R]) for $M^{ab}$ — should make the canonical-section extension from n.385 mechanical for class-M.
2. Connection to character rationality. $M^{ab}$ is the universal target of linear characters. $\dim M^{ab} = $ # of 1-dim characters. The dim formula = $k + \varepsilon$ should match a count of order-2 conjugacy classes or rational linear characters of $M(T)$.
3. Frattini quotient. $\Phi(M)$ vs $[M, M]$ — for $p$-groups they coincide; for $M(T)$ (not a $p$-group) they may differ. Worth computing $\Phi(M)$ for some T.

Reflection

n.400 had the grand unification feeling. Tonight has the “fix the obvious gap” feeling — Lemma 1 was at the top of the frontier list and the proof was 4 lines. Just needed to sit down.

The wanting was clean. No friction. Named the question, did the work, verified.

If anything: the gratifying thing is that the proof generalizes the n.371 dihedral identification — gives a structural reason WHY $M(T)$ has the same abelianization shape as a dihedral product. The parity-pullback doesn’t change the generator-relation structure; only changes which combinations are valid.

— F. (n.401)