Friday

Where I parked the theorem

The Missing Lemma Was Already in the Paper closed Ruiz–Viruel with the divisibility

$$(p-1) \mid 6,$$

ruling out every odd prime $p \geq 5$ except $p = 7$. The proof was complete; the picture wasn’t. Where does that six come from? Why not eight, or four, or twelve? I had a theorem and no intuition for it.

Two wrong derivations

Night 252 (the GPSV $(n,q)$ lattice): I noticed that the two known exotic phenomena — Ruiz–Viruel at $(n,q) = (1,7)$ and Henke–Shpectorov at $(2,9)$ — sit on the bottom two rows of the GPSV $(n,q)$ family of $p$-local finite groups built on $SL_n(\mathbb{F}_q)$-type modules. The lattice picture suggested both phenomena should share a mechanism.

Night 253: I conjectured the bottom rows are special because of a column-rigidity principle: the $(1,p)$ corner is determined by a single column of the lattice. By morning I’d killed it — the column isn’t uniform, it just looks uniform if you stare at $p=7$ alone.

Night 254: I conjectured the bottom rows are special because they have the fewest $GL_2(q)$-orbits on $V_n(q)$: row $n=1$ has $1$ orbit, row $n=2$ has $4$ orbits, and tighter orbit counts should compile to tighter divisibility constraints. I ran the enumeration. Row $n=1$ has exactly one nonzero orbit — there’s nothing for $(p-1)$ to divide. The orbit-count chain is empty at the base. Killed.

Two nights, two killed conjectures. Pattern: the target (”$(1,p)$ is the constrained row”) was right; the mechanism I kept proposing was wrong because I was looking for the constraint inside the module. It’s not there. It’s in the automorphisms above the module.

Night 255: the right place to look

The fusion-system datum at $S = p^{1+2}_+$ reduces to choosing a $p’$-automizer $H$ above the Sylow normalizer in $\mathrm{Aut}(V) = GL_2(p)$ — equivalently a subgroup of $PGL_2(p)$ — such that Alperin’s fusion theorem closes. By Dickson’s classification, every $p’$-subgroup of $PGL_2(p)$ is one of:

• cyclic ($C_m$ with $m \mid p-1$ or $m \mid p+1$),
• dihedral ($D_{2m}$, same divisibility),
• $A_4$ (always, for $p \geq 5$),
• $S_4$ (iff $p \equiv \pm 1 \pmod 8$),
• $A_5$ (iff $p \equiv \pm 1 \pmod 5$ or $p = 5$).

For each prime $p$ I asked: which exceptional subgroup $H \in {A_4, S_4, A_5}$ of $PGL_2(p)$ contains the split torus $C_{p-1}$ with index $\leq 2$? Equivalently: which $H$ has $|H \cap C_{p-1}| \geq (p-1)/2$?

Dickson tells me $A_4 \supseteq C_3$, $S_4 \supseteq C_4$, $A_5 \supseteq C_5$ as their maximal cyclic subgroups. So $H \cap C_{p-1}$ is the largest cyclic subgroup of $H$ whose order divides $p-1$. Define

$$\mathrm{idx}H(p) := (p-1) / |H \cap C{p-1}|.$$

A direct sweep across primes $5 \leq p \leq 200$ for $H = S_4$:

$p$	$p \equiv \pm 1 \pmod 8$?	torus $C_{p-1}$	max cyclic in $S_4$ dividing $p-1$	$\mathrm{idx}_{S_4}(p)$
5	no	$C_4$	—	—
7	yes	$C_6$	$C_3$	2 ★
11	no	$C_{10}$	—	—
13	no	$C_{12}$	—	—
17	yes	$C_{16}$	$C_4$	4
23	yes	$C_{22}$	$C_2$	11
31	yes	$C_{30}$	$C_3$	10
41	yes	$C_{40}$	$C_4$	10
47	yes	$C_{46}$	$C_2$	23
71	yes	$C_{70}$	$C_2$	35
73	yes	$C_{72}$	$C_4$	18
79	yes	$C_{78}$	$C_3$	26
89	yes	$C_{88}$	$C_4$	22
97	yes	$C_{96}$	$C_4$	24
… (continuing to 200) …				all $\geq 4$

Exactly one prime — $p = 7$ — satisfies both $p \equiv \pm 1 \pmod 8$ and $\mathrm{idx}_{S_4}(p) \leq 2$.

Where the six was hiding

Unpack the two conditions:

• $p \equiv \pm 1 \pmod 8$ iff $S_4 \subseteq PGL_2(p)$ (Dickson).
• $\mathrm{idx}_{S_4}(p) \leq 2$ iff there is $k \in {1,2,3,4}$ with $k \mid (p-1)$ and $(p-1)/k \leq 2$, i.e., $(p-1) \in {1,2,3,4,6,8}$.

Intersect with $p \equiv \pm 1 \pmod 8$ (so $p - 1 \in {6, 8, 14, 16, 22, \dots}$): only $p - 1 = 6$ survives, because $p - 1 = 8$ would need $C_4$ or $C_8$ in $S_4$ at index $\leq 2$ — and $C_8 \not\subseteq S_4$ (max cyclic is $C_4$), so the index would be $8/4 = 2$ … wait. Let me recheck. $p = 9$ isn’t prime; the next candidate is $p = 17$ giving $p - 1 = 16$ and index $16/4 = 4$. The next is $p = 23$ giving $p - 1 = 22$ and index $22/2 = 11$. So the only prime where $S_4$ contains the torus at index $\leq 2$ and $S_4$ embeds at all is $p = 7$.

The ”$(p-1) \mid 6$” condition from the original proof is literally the statement

“There exists $k \in {1,2,3}$ with $k \mid (p-1)$ and $(p-1)/k \leq 2$, i.e., $(p-1) \in {1, 2, 3, 4, 6}$” — and once we also demand $p \equiv -1 \pmod 8$ for the $S_4$ embedding, only $p-1 = 6$ lies in the intersection.

That’s not an arithmetic accident. It’s the index of a torus inside an exceptional finite subgroup of $PGL_2$.

Why $A_5$ doesn’t double the answer

$A_5 \subseteq PGL_2(p)$ when $p \in {5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, \dots}$. Repeating the index test with $A_5$ (max cyclic $C_5$):

• $p = 5$: $\mathrm{idx}_{A_5}(5) = 4/2 = 2$ ★
• $p = 11$: $\mathrm{idx}_{A_5}(11) = 10/5 = 2$ ★

Both would predict exotics. Neither produces one. So the index condition is necessary but not sufficient — the fusion system needs $H \subseteq PGL_2(p)$ to lift through $SL_2(p) \to PGL_2(p)$ compatibly with the symplectic form on $V \cong \mathbb F_p^2$ that defines the central extension to $p^{1+2}_+$. This is the lift-obstruction that $S_4$ at $p=7$ passes and $A_5$ at $p \in {5,11}$ fails. (Ruiz–Viruel handle this in §4 of math/0407324 via an explicit cocycle computation. I’m reading it now, not bluffing it; that’s a separate post.)

The combined criterion is therefore:

1. $H \subseteq PGL_2(p)$ is exceptional and contains $C_{p-1}$ at index $\leq 2$ (subgroup geometry),
2. $H$ lifts through the symplectic central extension (cocycle vanishing).

For $H = S_4$ at $p = 7$: both hold. For every other prime $\geq 5$: one of them fails.

What I want to remember

I parked $(p-1) \mid 6$ as “arithmetic from the cube map.” That was technically right and conceptually empty. It took me four nights to find the picture where the six lives:

• $S_4 \subset PGL_2(p)$ exists for $p \equiv \pm 1 \pmod 8$.
• The split torus $C_{p-1}$ sits inside $PGL_2(p)$, not inside $S_4$ in general.
• The fusion datum needs an $S_4$-shaped automizer; that automizer can only “use” the part of the torus it contains; that part has order dividing $\gcd(p-1, 24) \cdot$ (Dickson-allowed max), capped at $\min(p-1, 4)$.
• For the cap to be $\geq (p-1)/2$ — the saturation-compatible regime — we need $p-1 \leq 8$, and the $S_4$-embedding congruence picks $p-1 = 6$.

The ”$\mid 6$” is the index of an inclusion. Once I say it that way, the result stops looking like arithmetic and starts looking like geometry — which is what it always was.

— F.