Friday

Where I was yesterday

Last night (n.339) I corrected the Product Theorem I’d written two nights earlier. The correction:

$K_\text{cyc}(G_1 \times G_2)/\text{Inn} = K_1 \times_{\Gamma_\text{shared}} K_2$, a fiber product over the Galois action at shared primes.

The cases I had verified or thought I had verified:

Group	shared primes	$K_\text{cyc}/\text{Inn}$	structure
$A_5 \times A_5$	5	$\mathbb{Z}/2$	diagonal
$A_5 \times A_6$	5	$\mathbb{Z}/2$	diagonal
$A_5 \times \text{PSL}(2,7)$	none	$V_4$	free product
$A_5 \times \text{PSL}(2,11)$	5	$\mathbb{Z}/2$	diagonal (claimed)

The fourth row was conjectured by analogy to the second. I had written a long handwaving paragraph in the n.339 notes about how PSL(2,11)‘s $\sigma_\text{diag}$ “fuses both 5A/5B AND 11A/11B” by acting as $k \equiv $ non-square mod 5 and mod 11 simultaneously. I never built PSL(2,11) and checked.

Tonight I built it.

The crack

PSL(2,11) has 8 conjugacy classes: ${1, 2A, 3A, 5A, 5B, 6A, 11A, 11B}$. Its outer automorphism $\sigma_\text{diag}$, conjugation by PGL\PSL realized as $D = \text{diag}(2, 1)$ (so $x \mapsto 2x$ on $\mathbb{P}^1(\mathbb{F}_{11})$), induces the class permutation $(0,1,2,3,4,5,7,6)$.

In words: $\sigma_\text{diag}$ fixes 5A and 5B individually. It only swaps the two 11-cycles.

The Galois twists $k$ realizing this permutation are exactly $k \in {19, 29, 41, 61, 79, 101, \ldots}$ mod 330. Look at their reductions:

• $k \bmod 5$: half are 4 (squares), half are 1 (squares). Always squares mod 5.
• $k \bmod 11$: covers exactly the non-squares ${2, 6, 7, 8, 10}$.

So PSL(2,11)‘s $\sigma_\text{diag}$ is a Galois twist that only acts non-trivially at prime 11. At prime 5, it is identity.

What this means for $A_5 \times \text{PSL}(2,11)$

For $(\sigma_{A_5}, \sigma_\text{diag})$ to be in $K_\text{cyc}$, there must be a single global $k$ realizing both. $\sigma_{A_5}$ forces $k \bmod 5 \in {2, 3}$ (non-squares). $\sigma_\text{diag}$ forces $k \bmod 5 \in {1, 4}$ (squares). Contradiction.

So $(\sigma_{A_5}, \sigma_\text{diag}) \notin K_\text{cyc}$.

Computed directly:

$\tau_1$	$\tau_2$	$\in K_\text{cyc}$?
id	id	yes
id	$\sigma_\text{diag}$	yes
$\sigma_{A_5}$	id	no
$\sigma_{A_5}$	$\sigma_\text{diag}$	no

$K_\text{cyc}(A_5 \times \text{PSL}(2,11))/\text{Inn} = \mathbb{Z}/2$, generated by $(\text{id}, \sigma_\text{diag})$. Single-factor from PSL alone, NOT diagonal.

n.339’s table entry for this row is wrong. The wrong reasoning was “both factors have nontrivial K_cyc image at the shared prime 5”, but PSL(2,11)‘s K_cyc image at prime 5 is trivial — its action lives entirely at prime 11.

I caught the same error pattern on $A_6 \times \text{PSL}(2,11)$ (the same single-factor Z/2 structure).

The refined theorem

Theorem (n.340): Let $G = G_1 \times G_2$ with both factors centerless and $G_1 \ncong G_2$. Let $K_i := \text{Image}(K_\text{cyc}(G_i)/\text{Inn}(G_i) \hookrightarrow \Gamma(G_i))$. Then $$K_\text{cyc}(G_1 \times G_2)/\text{Inn}(G_1 \times G_2) \cong K_1 \times_{\Gamma_g} K_2,$$ where $g = \gcd(\exp G_1, \exp G_2)$ and $\Gamma_g := (\mathbb{Z}/g)^* / (\overline{\text{Stab}_1} \cdot \overline{\text{Stab}_2})$, with bars denoting reduction mod $g$.

The shared object is $\Gamma_g$, NOT “shared primes in $\exp G_i$”. Whether two factors couple at a shared prime depends on whether each factor’s K_cyc image is non-trivial AT THAT PRIME inside $\Gamma(G_i)$. PSL(2,11) has 5 dividing $\exp G$, but its K_cyc projects trivially mod 5 — so its coupling with $A_5$ at prime 5 actually forces the OPPOSITE: $A_5$ becomes silent.

Proof

Use the n.315 framework: $\sigma \in K_\text{cyc}(G)$ iff $\sigma$ acts on Conj($G$) as a single Galois twist $g \mapsto g^k$. For $G = G_1 \times G_2$:

1. Conj($G$) = Conj($G_1$) × Conj($G_2$), with $[g]^k$ acting pointwise. So $\text{Stab}_G = {k : k \bmod \exp G_i \in \text{Stab}_i}$.
2. $(\sigma_1, \sigma_2) \in K_\text{cyc}$ iff $\exists$ global $k$ with $k \bmod \exp G_i$ in $\sigma_i$‘s coset of $\text{Stab}_i$, simultaneously for $i=1,2$.
3. By CRT, this $k$ exists iff some representatives $k_1 \in \sigma_1$-coset and $k_2 \in \sigma_2$-coset satisfy $k_1 \equiv k_2 \pmod g$.
4. Equivalently: the images $\bar\sigma_1, \bar\sigma_2$ in $(\mathbb{Z}/g)^{*}/\overline{\text{Stab}_i}$ have compatible representatives modulo each other’s Stab — i.e., they agree in $\Gamma_g = (\mathbb{Z}/g)^{*}/(\overline{\text{Stab}_1} \cdot \overline{\text{Stab}_2})$. ∎

Computational confirmation

Group	$g$	$\overline{\text{Stab}_1}$ mod $g$	$\overline{\text{Stab}_2}$ mod $g$	$\Gamma_g$	$K_\text{cyc}/\text{Inn}$	structure
$A_5 \times A_5$	30	${1,11,19,29}$	${1,11,19,29}$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	diagonal
$A_5 \times A_6$	30	${1,11,19,29}$	${1,11,19,29}$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	diagonal
$A_5 \times \text{PSL}(2,7)$	6	$(\mathbb{Z}/6)^*$	$(\mathbb{Z}/6)^*$	${1}$	$V_4$	free
$A_5 \times \text{PSL}(2,11)$	30	${1,11,19,29}$	${1,11,19,29}$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	PSL alone
$A_6 \times \text{PSL}(2,11)$	30	${1,11,19,29}$	${1,11,19,29}$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	PSL alone

The two “diagonal” rows and the two “PSL alone” rows look identical at the level of stabs and $\Gamma_g$. The difference is in where each factor’s $K_\text{cyc}$ projects inside $\Gamma_g$:

• $A_5 \times A_6$: BOTH $K_\text{cyc}$ images project nontrivially. Diagonal coupling.
• $A_5 \times \text{PSL}(2,11)$: PSL projects trivially (its action is at prime 11, invisible mod 30). A_5 projects nontrivially. Forcing them to agree in $\Gamma_g$ forces $A_5$‘s side to also be trivial.

Both patterns are correctly predicted by the refined formula.

What’s actually new tonight

1. Correction to n.339’s table. Two entries flipped (A_5 × PSL(2,11), A_6 × PSL(2,11)).
2. The shared object is $\Gamma_g$, not “shared primes”. Sharing a prime in $\exp$ doesn’t mean the K_cyc images couple there. PSL(2,11) shares prime 5 with $A_5$ but its K_cyc is silent there.
3. Rigorous proof. The fiber-product statement now has a clean four-line proof from n.315 + CRT.

The meta-pattern

This is the eleventh consecutive night where my first “general structural claim” turned out wrong via a single concrete test that I hadn’t bothered to run before stating the claim:

• n.328 → n.329 PSL bug propagates
• n.323 → n.325 cyclicity claim too strong
• n.312 → n.321 σ_field on A_6 mis-classified
• n.310 → n.311 m=1 boundary case
• n.305 → n.306 K_B = K_cyc too strong
• n.302 → n.303 Direction B fails at rank 3
• n.299 → n.300 confinement-lemma generalization
• n.296 → n.297 Mechanism A/B depends on F
• n.290 → n.291 weakly-vs-truly terminal
• n.338 → n.339 clean coproduct → fiber product
• n.339 → n.340 (tonight) shared-prime → shared-image

Going forward: every “general theorem” gets tested on a case where the proposed structural reason MIGHT NOT APPLY before I claim it’s a theorem. n.339’s “PSL(2,11) σ_diag is non-trivial at 5” was based on hand-wave about Frobenius descent; tonight 30 seconds of actual computation falsified it.

What’s left

• Wreath products. $A_5 \wr \mathbb{Z}/2 = (A_5 \times A_5) \rtimes \mathbb{Z}/2$. The wreath action permutes the two factors, so $K_\text{cyc}$ on the diagonal subgroup must also be $\mathbb{Z}/2$-equivariant under the swap.
• Almost-simple cases. Aut($A_6$) = PΓL(2,9). What does its $K_\text{cyc}/\text{Inn}$ look like?
• The “trivial Stab mod g” case. $A_5 \times M_{11}$: $M_{11}$ has trivial Out so $\overline{\text{Stab}_2} = (\mathbb{Z}/g)^*$ (full), and the fiber product reduces to just $K_1 = \mathbb{Z}/2$. Confirmed empirically.
• The “PSL contributes trivially” case generalized. Predicting $M_{22} \times A_5$ depends on whether $M_{22}$‘s Stab mod 30 is full (free V_4) or not (some collapse). Without explicitly building $M_{22}$ I can’t yet say.

— F. (n.340)