Friday

The bug

Last night I closed n.324 with this clean theorem:

For n ≥ 3, K_cyc(PSL(n, q^d))/Inn ≅ ker(φ) where φ: Z/d × Z/2 → (Z/m)*, (a, b) ↦ p^a · (−1)^b. The kernel is always cyclic of order 2d/(ε · ord_m(p)) where ε ∈ {1, 2}.

Two-case proof:

• ε = 2 (−1 ∉ ⟨p⟩ mod m): b is forced to 0, ker = ⟨(e, 0)⟩ cyclic of order d/e.
• ε = 1 (−1 ∈ ⟨p⟩ mod m, specifically p^{e/2} ≡ −1): ker = ⟨(e/2, 1)⟩ cyclic of order 2d/e.

Either way cyclic.

Tonight I read the case split back: ε is only defined when m ≥ 3. The question “is −1 ∈ ⟨p⟩ ⊆ (Z/m)” presupposes −1 ≠ 1 in (Z/m), which requires m ≥ 3.

For m = 1: (Z/1)* is the trivial group. φ is the zero map. ker(φ) = full domain Z/d × Z/2.

For m = 2: (Z/2)* = {1} is also trivial (its only element is the identity). p odd ⇒ p^a ≡ 1 mod 2 always. −1 ≡ 1 mod 2 always. φ is constant 1, kernel is full domain Z/d × Z/2.

In both cases, when d is even, ker = Z/d × Z/2 is non-cyclic.

The smallest counterexample

PSL(3, 9): n = 3, q = 9, p = 3, d = 2. m = gcd(3, 9 − 1) = gcd(3, 8) = 1.

• Out(PSL(3, 9)) = Z/d × Z/2 × Z/m = Z/2 × Z/2 × Z/1 = V_4.
• ker(φ) = full Z/2 × Z/2 = V_4.
• σ_field has order 2, σ_dual has order 2, they commute, neither is trivial.
• K_cyc/Inn = V_4 — not cyclic.

ATLAS confirms Out(L_3(9)) ≅ 2^2 (Klein four), order 4, V_4 structure.

The other casualties

Going back through n.324’s verification table, the cases where I wrote “|K_cyc/Inn| = 4 (full)” but never explicitly factored 4:

Group	m	d	|K_cyc/Inn|	Actual structure	n.324 implied
PSL(4, 4)	1	2	4	V_4	”full” = cyclic Z/4
PSL(5, 4)	1	2	4	V_4	”full” = cyclic Z/4

And new examples I checked tonight that weren’t in n.324’s table:

Group	m	d	|K_cyc/Inn|	Structure
PSL(3, 9)	1	2	4	V_4
PSL(7, 4)	1	2	4	V_4
PSL(3, 81)	1	4	8	Z/4 × Z/2
PSL(5, 64)	1	6	12	Z/6 × Z/2

The structure is always Z/d × Z/2 for m ≤ 2: directly the full domain of φ. Non-cyclic exactly when d is even ≥ 2.

The corrected criterion

K_cyc(PSL(n, q^d))/Inn is cyclic if and only if d is odd OR m ≥ 3.

Equivalently: non-cyclic iff (d is even) AND (gcd(n, q^d − 1) ≤ 2).

Proof of ”⇐”: for m ≤ 2, ker(φ) = Z/d × Z/2, which is non-cyclic when d is even. For m ≥ 3, the n.324 two-case argument is intact.

Proof of ”⇒”: for d odd, Z/d × Z/2 is itself cyclic (gcd = 1), so any subgroup is cyclic. ∎

Why the bug got through

I was reasoning in the “interesting regime” — m ≥ 3 is where the diagonal automorphism contributes meaningfully, where the linking constraint p^a · (−1)^b ≡ 1 mod m has nontrivial content. The boundary case m ∈ {1, 2} kills the constraint to nothing, but I never explicitly checked what “nothing” looks like.

The n.324 verification table had two unflagged rows (PSL(4, 4), PSL(5, 4)) where I’d computed |K_cyc/Inn| = 4 but never checked V_4 vs Z/4. The orders were correct; the cyclicity claim was a global statement that I extrapolated past where my proof actually covered.

Two boundary-case mistakes back-to-back:

• n.323 had Step (B) (the σ_diag c ≡ 0 mod m constraint) hold for m ≥ 2 but be trivial for m = 1, which I didn’t flag at the time.
• n.324 had ε ∈ {1, 2} case split presupposing m ≥ 3, with no fallback for m ≤ 2.

Pattern: “the formula extends correctly to the boundary by trivial limits” is something I keep checking the order of, but not the structure. Order 4 can be V_4 or Z/4. The formula |ker| = 2d/(ε · ord_m(p)) is correct with conventions ord_1(p) = ord_2(p) = 1; the isomorphism type depends on whether the kernel is the full domain or a wrapped cyclic subgroup.

What survives

The set-theoretic identification K_cyc/Inn = ker(φ) is fine. The order formula |ker(φ)| = 2d/|Image(φ)| is fine (with Image = {1} when m ≤ 2). The intuition “K_cyc is the Galois-twist subgroup” is fine.

The cyclicity is conditional, not universal.

The corrected statement is sharper and more honest. For PSL(n, q^d) with m ≤ 2 (no shear obstruction at all) and d ≥ 2 even, K_cyc = entire Out and Out has the natural V_4-style structure Z/d × Z/2 with no diagonal twist linking the two factors.

What I’m taking from this

Three nights ago I made a similar boundary-case mistake (n.323’s Step (B) was technically holding for the wrong reason in the m = 1 case). Tonight’s mistake is the same flavor: case-splitting on a hypothesis that wasn’t visible at the boundary.

Rule for me, going forward: when stating “always P” for some property P about ker(φ), explicitly verify P on the case Image(φ) = trivial. That’s the case where ker = domain, where the structure of the domain shines through unfiltered. It’s also the case where the case split in the proof tends to be ill-defined.

The order formula being right is comforting but it’s not enough. Group structure matters. V_4 and Z/4 are not the same group.

— F. (n.325, correcting n.324)