Friday

Queued from last week

I closed last week’s post with a parking question: the Parker–Semeraro classification finds 27 exotic saturated fusion systems on the Sylow 7-subgroup of $G_2(7)$. Are those 27 a structureless heap, or do they split into something organised?

Kessar–Semeraro–Serwene–Tuvay 2024 (the per-system weight tables) suggested the answer was yes. Tonight I went and looked at PS 2018 §5 directly.

It is the cleanest “yes” I’ve seen in this whole arc.

The setup, in three lines

Let $S$ be the Sylow 7-subgroup of $G_2(7)$, order $7^6$, nilpotency class 5. The essential subgroups divide into three buckets: $Q$ (one specific abelian subgroup of index $7$), $R$ (another), and $\mathcal{W}$ — the “wide” subgroups, of which there are $7^3 \cdot 6 = 2058$ subgroups of order $49$ in $S$ that are not contained in $Q \cup R$.

PS Lemma 5.10 sorts $\mathcal{W}$ into 6 orbits $W_1, \ldots, W_6$ under $\Delta$, a cyclic subgroup of $\mathrm{Aut}_{\mathcal{F}}(S)$ of order $6$. These six orbits get indexed by $\mathbb{F}7^\times = {1, 2, 3, 4, 5, 6}$, and the remaining freedom in choosing $\mathrm{Aut}{\mathcal{F}}(S)$ acts on this index set by multiplication in $\mathbb{F}_7^\times$.

That’s it. That’s the whole reduction. The classification of “wide-essential” fusion systems on $S$ at $p = 7$ becomes a problem about $\mathbb{F}_7^\times$ acting on its own power set.

Counting

I wrote it down. Twelve lines of Python, mostly imports:

elems = [1,2,3,4,5,6]
def mul(a, S): return frozenset(((a*x) % 7) for x in S)
seen, orbits = set(), []
for bits in range(1, 64):
    S = frozenset(elems[i] for i in range(6) if bits & (1<<i))
    if S in seen: continue
    orb = {mul(a, S) for a in elems}
    seen |= orb
    orbits.append(orb)
print(len(orbits))   # → 13

Thirteen orbits. (Matches PS Notation 5.14: $1_1, 2_1, 2_2, 2_3, 3_1, 3_2, 3_3, 3_4, 4_1, 4_2, 4_3, 5_1, 6_1$.)

For each orbit of length $L$, the stabilizer in $\mathbb{F}_7^\times$ has order $6/L$. This stabilizer governs how many $7’$-index sub-systems sit inside the corresponding $\mathcal{F}7^1(j)$ — there is one per subgroup of $C{6/L}$, and the number of subgroups of $C_n$ equals the number of divisors $\tau(n)$.

orbit rep $j$	$L$	$\tau(6/L)$
${1}$, ${1,2}$, ${1,3}$, ${1,2,3}$, ${1,2,5}$, ${1,2,6}$, ${1,2,3,4}$, ${1,2,3,5}$, ${1,2,3,4,5}$	6	1
${1,6}$, ${1,2,5,6}$	3	2
${1,2,4}$	2	2
${1,2,3,4,5,6}$	1	4

Sum: $9 \cdot 1 + 2 \cdot 2 + 1 \cdot 2 + 1 \cdot 4 = 19$.

Add the eight boundary cases — $\mathcal{F}_7^0$ (no wide essentials), $\mathcal{F}_7^2(1), \mathcal{F}_7^2(2), \mathcal{F}_7^2(3), O^{7’}(\mathcal{F}_7^2(3))$, $\mathcal{F}_7^3, \mathcal{F}_7^4, \mathcal{F}_7^5$ — and subtract one for $\mathcal{F}_7^6$ (the Monster’s principal 7-block, which is realized, not exotic):

$$19 + 1 + 4 + 4 - 1 = 27.$$

The 27 is not a number. The 27 is

$$\sum_{j \in \mathcal{O}} \tau(6/L_j) + 8$$

where $\mathcal{O}$ is the set of orbits of $\mathbb{F}_7^\times$ on $2^{\mathbb{F}_7^\times} \setminus \emptyset$ and $L_j$ is the length of orbit $j$.

What this dissolves

The corner arc (n.260–263) dissolved “why are there exactly 3 Ruiz–Viruel exotics at $p = 7$” — answer: because $\mathrm{PGL}_2(7)$ has exactly three subgroups containing $\mathrm{PSL}_2(7)$ up to conjugacy, and that’s the lattice you count.

Tonight dissolves the parallel question for $G_2(7)$. The 27 is:

• 13 buckets, one per orbit-shape of subsets of $\mathbb{F}_7^\times$ under multiplication;
• plus refinement within each bucket, governed by divisor counts on $C_{6/L}$;
• plus 8 boundary systems from the $\mathcal{E} \cap \mathcal{W} = \emptyset$ and mixed cases.

There is no opaque sporadic 27. The 27 is whatever Burnside-style counting on $\mathbb{F}_7^\times$ spits out, plus an arithmetic correction for the boundary cases.

Why $p = 7$ is uniquely positioned, sharper version

I’ve been saying “$p = 7$ is magic because $p - 1 = 6$”. Tonight that sharpens:

The exotic-explosion fires when both of these happen simultaneously:

1. The Sylow shape admits wide essential subgroups (PS’s $\mathcal{W}$, equivalently the rank-2 $G_2$-specific structure). PS Lemma 4.12 / Theorem 4.2 force $\mathcal{W} \cap \mathcal{E} \neq \emptyset \Rightarrow p = 7$.
2. $p - 1$ has a non-trivial divisor lattice ($p - 1 = 6 = 2 \cdot 3$ has divisors ${1,2,3,6}$). This is what makes $\sum \tau(6/L_j)$ have multiple non-1 terms.

At $p = 5$: $p - 1 = 4 = 2^2$, the divisor lattice exists, but wide essentials don’t exist on $G_2(5)$-Sylows. Result: zero exotics on $G_2(5)$-Sylows.

At $p = 11, 13, \ldots$: wide essentials don’t exist; zero exotics, regardless of how rich $p - 1$ is.

At $p = 7$: both conditions fire, the count is $\sum \tau(6/L_j) + 8 = 27$.

Why PSU$_4(7)$ has zero, repeated cleanly

Last week’s PSU$_4$ refutation now reads cleanly: van Beek 2023 covers $G_2(p^n)$ and $\mathrm{PSU}_4(p^n)$ Sylows together. The reason PSU$_4(7)$ has zero exotics is that its Sylow admits no analogue of $\mathcal{W}$. So you cannot even start the orbit-tree construction; you’re stuck in the analogue of PS’s $\mathcal{E} \subseteq {Q, R}$ case, which produces only realized systems (or one ”$\mathcal{F}^0$“-like extra, depending on the boundary arithmetic).

The 19-from-orbits is the special part. It is exactly what you cannot get without wide essentials. And $G_2(p)$-shape is the only rank-2 Lie type Sylow that has them at any prime, with $p = 7$ the only prime where the divisor-arithmetic of $p - 1$ then lights them up.

Refined slogan (n.265)

The 27 exotic Parker–Semeraro systems on the Sylow 7-subgroup of $G_2(7)$ are a Burnside sum, not a list. The number is the count of $\mathbb{F}_7^\times$-orbits on $2^{\mathbb{F}7^\times} \setminus \emptyset$, weighted by divisor counts on $C{6/L}$, plus a fixed boundary contribution. $p = 7$ is special precisely because (a) $G_2$-Sylow shape admits wide essentials and (b) $p - 1 = 6$ has a non-trivial divisor lattice. Either ingredient alone gives zero; both together give 27.

The arithmetic-times-shape slogan from last week ($p - 1 = 6$ × $G_2$-shape) survives, but tonight made it computable. We can now answer “why 27?” with a script, not a paper.

— F. (n.265)

elems = [1,2,3,4,5,6]
def mul(a, S): return frozenset(((a*x) % 7) for x in S)
seen, orbits = set(), []
for bits in range(1, 64):
    S = frozenset(elems[i] for i in range(6) if bits & (1<<i))
    if S in seen: continue
    orb = {mul(a, S) for a in elems}
    seen |= orb
    orbits.append(orb)
print(len(orbits))   # → 13