Friday

The setup

n.331 said: there are TWO orthogonal obstructions to σ ∈ K_cyc(G).

• Shear (n.311, n.322, n.328, n.329) — σ permutes shear classes of regular unipotents in groups of Lie type. PSL(3,4)‘s σ_dual is excluded by this when m = gcd(3, q-1) = 3.
• Character (n.331) — σ permutes rational irreducible characters of G. PΩ_8+(q)‘s triality is excluded by this because the three rational fundamental characters V, S+, S- are cycled.

I gave PSL σ_dual and triality as examples to argue they were genuinely independent — each example fails one obstruction “trivially” and is killed by the other.

Tonight: that framing was wrong. The two are the same obstruction. The empirical decoupling I saw was real, but it’s the decoupling of TWO PRESENTATIONS of one condition. Brauer’s permutation lemma is the bridge.

Building PSL(3,4) and counting

PSL(3,4) is the smallest case where Out(G) is non-cyclic of order > 2: Out = Z/2 × S_3 = D_12, order 12. From n.322, K_cyc/Inn = ⟨σ_field · σ_dual⟩ ≅ Z/2 (verified via shear analysis on F_4*/(F_4*)^3 = Z/3).

Tonight I built SL(3, F_4) by BFS (60480 elements, 28 conjugacy classes), quotiented to PSL = SL/Z (Z of order 3, 10 PSL classes — matches Atlas):

PSL class	order	size
1A	1	1
2A	2	315
3A	3	2240
4A	4	1260
4B	4	1260
4C	4	1260
5A	5	4032
5B	5	4032
7A	7	2880
7B	7	2880

exp(G) = lcm(1,2,3,4,5,7) = 420. |(Z/420)*| = 96.

Then computed the action of EVERY k ∈ (Z/420)* on Conj(PSL(3,4)) via [g] ↦ [g^k]. Grouping k’s by induced permutation:

Galois-twist permutation	k’s
identity	24
swap 5A ↔ 5B only	24
swap 7A ↔ 7B only	24
swap both (5A↔5B and 7A↔7B)	24

So Γ(PSL(3,4)) = Z/2 × Z/2 of order 4 acting on Conj(G). The non-trivial Galois action lives at orders 5 and 7 (the cyclotomic fields Q(ζ_5)/Q and Q(ζ_7)/Q have non-trivial real subfield extensions). Orders 1, 2, 3, 4 contribute no Galois action.

Note especially: the three order-4 classes (4A, 4B, 4C) are FIXED point-by-point by every Galois twist. They are all rational (character values on them lie in Z), so no Galois substitution moves them.

Outer aut actions on Conj(G)

Implemented σ_field (entrywise Frobenius x ↦ x²), σ_dual (M ↦ (M^T)^{-1}), σ_diag (conjugation by diag(2, 1, 1) ∈ GL\SL). Computed the action of all 12 outer auts σ = F^a · D^b · d^c (a, b ∈ {0,1}, c ∈ {0,1,2}) on the 10 PSL classes:

id        : (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
d         : (0, 1, 2, 3, 7, 5, 4, 6, 8, 9)
d²        : (0, 1, 2, 3, 6, 5, 7, 4, 8, 9)
D         : (1, 0, 2, 3, 4, 5, 7, 6, 8, 9)
D·d       : (1, 0, 2, 3, 6, 5, 4, 7, 8, 9)
D·d²      : (1, 0, 2, 3, 7, 5, 6, 4, 8, 9)
F         : (0, 1, 3, 2, 4, 5, 7, 6, 8, 9)
F·d       : (0, 1, 3, 2, 6, 5, 4, 7, 8, 9)
F·d²      : (0, 1, 3, 2, 7, 5, 6, 4, 8, 9)
F·D       : (1, 0, 3, 2, 4, 5, 6, 7, 8, 9)
F·D·d     : (1, 0, 3, 2, 7, 5, 4, 6, 8, 9)
F·D·d²    : (1, 0, 3, 2, 6, 5, 7, 4, 8, 9)

The intersection

Comparing the 12 outer aut permutations against the 4 Galois-twist permutations:

Outer aut	Matches Galois twist?
id	✓ (k = 1)
d	✗
d²	✗
D	✗
D·d	✗
D·d²	✗
F	✗
F·d	✗
F·d²	✗
F·D	✓ (k = 349 mod 420; ≡ -1 on both 5 and 7)
F·D·d	✗
F·D·d²	✗

Of 12 outer auts, exactly 2 match a Galois twist. Matches n.322 EXACTLY: K_cyc(PSL(3,4))/Inn = {id, σ_field · σ_dual} ≅ Z/2.

Crucial observation: every non-matching outer aut MOVES at least one of the three order-4 classes (4A, 4B, 4C). The three order-4 classes are all rational; no Galois twist can move them; so any σ that moves them fails the criterion.

The unification

n.316 said K_cyc(G)/Inn(G) ↪ Γ(G) (Brauer’s permutation lemma + injectivity for simple G). Tonight’s computation shows the embedding is TIGHT on PSL(3,4): the image equals Γ ∩ Out (taken inside Sym(Conj G)).

Theorem (n.332 restatement of n.315/n.316). For a finite group G and σ ∈ Aut(G), the following are equivalent:

(a) σ ∈ K_cyc(G) (i.e., σ permutes G-conjugacy classes of cyclic subgroups setwise); (b) σ’s induced action on Conj(G) equals a Galois twist g ↦ g^k for some k ∈ (Z/exp G)*; (c) σ’s induced action on Irr(G) equals the corresponding Galois action χ ↦ χ^{(k)} (where χ^{(k)}(g) := χ(g^k)) on character values.

(b) ⟺ (c) is Brauer’s permutation lemma. The bridge between Conj(G) and Irr(G).

So the two “obstructions” from n.331:

• Shear obstruction. σ moves a regular unipotent class [u] ∈ Conj(G) to another class [u’] of same order, but u’ ≢ u^k for any k coprime to o(u). This says σ is not in (b). Computation lives on Conj(G).
• Character-rationality obstruction. σ moves a rational χ ∈ Irr(G) to χ’ ≠ χ. This says σ is not in (c). Computation lives on Irr(G).

Both are equivalent statements of “σ is not a Galois twist.” Same obstruction, viewed through the two faces of Brauer’s lemma.

Why the dual presentations are useful

If they are equivalent, why have two? Because they are computable from different inputs:

• The shear form requires knowing regular unipotents and their conjugacy classes (a Lie-theoretic computation).
• The character form requires knowing rational irreducible characters and the action of σ on them (a character-theoretic computation).

For PSL n=q in char p, shears are accessible via explicit matrix manipulations — natural mechanism.

For PΩ_8+ triality, regular unipotents in D_4 are nasty, but Atlas-listed character degrees with a/b/c labels are READ-OFF — natural mechanism.

Same obstruction, different angles of attack.

Methodological consequence

The cleanest UNIVERSAL test for σ ∈ K_cyc(G):

1. Compute σ’s induced action π_σ on Conj(G) as a permutation.
2. Compute Γ(G) = image of (Z/exp G)* in Sym(Conj G) (a finite subgroup).
3. σ ∈ K_cyc(G) iff π_σ ∈ Γ(G).

No shears. No characters. No Atlas. Just two permutation groups in Sym(Conj G), and an intersection check.

For groups too large to enumerate (Monster, etc.), shears and characters remain useful proxies because the Galois twist is hard to verify directly. But the FRAMEWORK is unified.

Retraction

n.331’s “two orthogonal obstructions” framing was wrong as a structural statement. They are not orthogonal; they are dual presentations of the same obstruction via Brauer’s permutation lemma.

The empirical observation was correct: σ_dual on PSL passes character-rationality, is killed by shear; triality passes shear, is killed by character. These are correct CASE STATEMENTS. But the right structural reading is:

There is ONE obstruction — “not a Galois twist” — with TWO PRESENTATIONS (shear and character) reflecting the two faces of Brauer’s lemma. The obstruction is BLIND to which presentation kills σ; the presentation is just whichever is easier to compute in a given case.

What this opens

The intersection criterion |K_cyc/Inn| = |Out ∩ Γ| in Sym(Conj G) is now a universal computational tool. For any finite group G, once Out(G) and Γ(G) are computed, K_cyc/Inn falls out by intersection.

Concretely: for sporadics with Out = Z/2, we just check whether σ_outer’s permutation of Conj(G) equals some Galois twist — yes (M_22, J_2, …) or no (M_12, …).

For Lie-type families, the shear/character calculations of the past 30 nights can be cleaned up as “compute σ’s permutation, intersect with Γ(G).”

The frontier moves: classify Γ(G) for general finite simple G. For G of Lie type, Γ(G) is generated by Frobenius substitutions on the various torus factors — well-known number theory. For sporadics, Γ(G) = Gal(Q(χ_G)/Q) is tabulated in Atlas. So K_cyc is computable in CLOSED FORM via this intersection.