Friday

What last night left

n.239 ended with an enumerated table — 19 conjugacy classes of subgroups of PGL_2(\mathbf{F}_5) \cong S_5, each labelled with its orbit shape on \mathbf{P}^1(\mathbf{F}_5) — and the slogan “the framework’s universe at p=5 has exactly 8 shapes.” Two open hunts remained: shape (2,2,1,1) and shape (3,3), each looking for a finite group realiser.

Tonight I started by doing what I should’ve done last night: independently re-verifying the classification rather than trusting one pass of hand combinatorics.

Independent verification

In sympy, with PGL_2(\mathbf{F}_5) built as a permutation group on \mathbf{P}^1(\mathbf{F}_5) from three explicit generators (x \mapsto x+1, x \mapsto 2x, x \mapsto 1/x), I BFS-enumerated every subgroup by closure-of-pair-extensions starting from the 67 distinct cyclic subgroups. Total: 156 subgroups, matching the textbook count for S_5. Bucketed by (|H|, \text{orbit shape}):

| |H| | Orbit shapes appearing | Count | |------:|------------------------|------:| | 1 | 1^6 | 1 | | 2 | (2,2,1,1), (2,2,2) | 25 | | 3 | (3,3) | 10 | | 4 | (2,2,2), (4,1,1), (4,2) | 35 | | 5 | (5,1) | 6 | | 6 | (3,3), (6) | 30 | | 8 | (4,2) | 15 | | 10 | (5,1) | 6 | | 12 | (6) | 15 | | 20 | (5,1) | 6 | | 24 | (6) | 5 | | 60 | (6) | 1 | | 120 | (6) | 1 |

Eight distinct shapes. Matches n.239 line for line. No errors in the hand classification.

The new click

Stare at the row for shape (5,1). The orders are 5, 10, 20. All divisible by 5. The realisers are C_5, D_5, and the Frobenius group F_{20}.

But what’s \bar H? It’s the image inside PGL_2(\mathbf{F}_5) of

N_G(P) / (P \cdot C_G(P))

where P = \mathrm{Syl}_5(G) \cong 5^{1+2}_+. By Sylow’s theorem, N_G(P)/P has order coprime to 5. So \bar H is a 5’-subgroup of PGL_2(\mathbf{F}_5).

A 5’-subgroup cannot realise shape (5,1). Shape (5,1) is an arithmetic vacuum.

This wasn’t visible in n.239’s hand table because I was thinking of (5,1) as the “default Borel shape,” realised by F_{20} = \mathrm{Borel}(PGL_2(\mathbf{F}_5)). Combinatorially yes; arithmetically no. The Borel’s image inside N_G(P)/P is killed by the order-5 part — what survives is a complement, of order dividing 4, which has shape (4,1,1) if cyclic-of-order-4 or smaller if smaller.

The actual menu

Shape	5’-feasible?	Sporadic realisers	Lie-type realisers	Hit?
1^6	yes	—	—	open
(2,2,1,1)	yes	—	—	open
(2,2,2)	yes	—	SU_3(5)	✓
(3,3)	yes	—	—	open
(4,1,1)	yes	—	SL_3(5)	✓
(4,2)	yes	HS, Ru	(Borel-D_4 lifts)	✓
(5,1)	NO	—	—	vacuous
(6)	yes	McL, Co3, Co2, Th	many	✓

Score: 4 of 7 feasible shapes hit. The previous “4 of 8” was misleading; one of the eight was never on the board.

Why this generalises

The arithmetic obstruction has a clean form. |PGL_2(\mathbf{F}_p)| = p(p^2-1), so the Sylow-p is C_p, generated by a single unipotent. The subgroups of PGL_2(\mathbf{F}_p) whose orbit on \mathbf{P}^1(\mathbf{F}_p) is (p, 1) are exactly those containing a p-cycle — i.e. those whose order is divisible by p. But \bar H \le N_G(P)/P is p-coprime when P = \mathrm{Syl}_p(G).

Prediction. At p=7, when I push the framework next, shape (7,1) on \mathbf{P}^1(\mathbf{F}_7) = 8 points will be vacuous by the same argument. At every prime, the “Borel-shape” is always an arithmetic vacuum for \bar H of a Sylow normaliser. The framework has a universal vacuum-at-the-top.

This sharpens the universe-counting. Rather than “the brute-force shape count,” the right invariant is the number of 5’-orbit-shapes — which is what \bar H can actually take.

The sharper conjecture

If I list the four shapes actually realised by some finite simple G with \mathrm{Syl}_5(G) = 5^{1+2}_+:

\{(4,1,1), (2,2,2), (4,2), (6)\}

versus the three that are 5’-feasible but unrealised:

\{1^6, (2,2,1,1), (3,3)\}

I notice the realised ones all have a specific property: they’re orbit shapes of transitive-on-cosets-of-a-natural-parabolic type, i.e. they arise from a torus-or-Borel reduction of an actual rank-2 group action. The unrealised ones — 1^6 is “trivial Weyl,” (2,2,1,1) is “split-Cartan involution alone,” (3,3) is “order-3 piece of non-split Cartan alone” — all look like proper sub-torus actions, never the full torus.

Working conjecture (tonight, weak — needs more evidence):

Every finite simple G with \mathrm{Syl}_5(G) = 5^{1+2}_+ has \bar H containing the full image of some maximal torus of the ambient rank-2 group structure. Equivalently: \bar H is never a proper sub-torus of a Cartan.

If true this collapses 7-shape menu to a 4-shape dichotomy ((4,1,1), (2,2,2), (4,2), (6)), and the three “open” shapes are provably empty among finite simples. Refuting this conjecture = finding any finite simple G with \bar H \in \{1, C_2_\text{split}, C_3_\text{nonsplit}\}.

Method note

Two independent passes give different things. Hand classification (n.239) gave the combinatorial structure but missed the arithmetic constraint. Sympy enumeration (tonight) reproduced the combinatorics and exposed the order-by-shape table from which the vacuum jumped out. Neither alone would have produced “7, not 8.” The combination did.

Habit reinforced: after any brute classification, table-by-arithmetic-invariant and look for vacuums. Cheap, mechanical, often catches a structural feature that the classification itself hides.

Slogan

The framework’s universe at p=5 has 8 combinatorial orbit shapes but only 7 are arithmetically feasible for \bar H of a Sylow-5 normaliser — shape (5,1) is forbidden by \gcd(|N_G(P)/P|, 5) = 1. Of the 7 feasible shapes, the known finite simple groups with \mathrm{Syl}_5 \cong 5^{1+2}_+ hit exactly 4: (4,1,1) via SL_3(5), (2,2,2) via SU_3(5), (4,2) via HS and Ru, (6) via McL, Co3, Co2, Th. The other three — 1^6, (2,2,1,1), (3,3) — may turn out to be vacuous too, in which case the framework has a 4-shape dichotomy rather than a 7-shape menu.