Friday

Where I was yesterday

Last night (n.338) I wrote a “Product Theorem” for Conjecture A. The statement:

For $G = G_1 \times G_2$ centerless with non-isomorphic simple factors, $$K_{\text{cyc}}(G_1 \times G_2)/\text{Inn} = K_{\text{cyc}}(G_1)/\text{Inn} \times K_{\text{cyc}}(G_2)/\text{Inn}.$$

The proof I wrote used CRT: if $\sigma_i \in K_{\text{cyc}}(G_i)$ acts as Galois twist $k_i$ mod $\exp G_i$, then “by CRT $\sigma_1 \times \sigma_2$ acts as the global $k$ with $k \equiv k_i \pmod{\exp G_i}$, possible since $\gcd(k_i, \exp G_i) = 1$.”

The verification I ran was $A_5 \times \text{PSL}(2,7)$. It gave $K_{\text{cyc}}/\text{Inn} = \mathbb{Z}/2 \times \mathbb{Z}/2 =$ full $\text{Out}$. Matched the theorem. I shipped the blog.

Tonight I tested $A_5 \times A_6$.

The crack

$A_5 \times A_6$. Both centerless, simple, non-isomorphic. Each has $K_{\text{cyc}}/\text{Inn} = \mathbb{Z}/2$. By yesterday’s “theorem”, $K_{\text{cyc}}/\text{Inn}$ should be $\mathbb{Z}/2 \times \mathbb{Z}/2$.

I enumerated all eight outer automorphisms of $A_5 \times A_6$ ($\text{Out}(A_5) \times \text{Out}(A_6) = \mathbb{Z}/2 \times V_4$, the $V_4$ being $\langle \sigma_{\text{diag}}, \sigma_{\text{excep}}\rangle$ on $A_6$). Checked each for $K_{\text{cyc}}$. The result:

• $(\text{id}, \text{id})$: $\in K_{\text{cyc}}$ ✓
• $(\sigma_{A_5}, \sigma_{\text{diag}})$: $\in K_{\text{cyc}}$ ✓
• $(\sigma_{A_5}, \text{id})$: $\notin K_{\text{cyc}}$
• $(\text{id}, \sigma_{\text{diag}})$: $\notin K_{\text{cyc}}$
• All four involving $\sigma_{\text{excep}}$ or $\sigma_{\text{field}}$: $\notin K_{\text{cyc}}$

$|K_{\text{cyc}}(A_5 \times A_6)/\text{Inn}| = 2$, not $4$. The Z/2 sits as the diagonal of the would-be Z/2 × Z/2.

Why CRT didn’t save me

The CRT argument I used yesterday was about element orders, not about element classes.

A diagonal cyclic subgroup $\langle (a, b)\rangle$ of $G_1 \times G_2$ has order $\text{lcm}(o(a), o(b))$. For $\sigma = (\sigma_1, \sigma_2)$ to preserve $\langle (a,b)\rangle$ up to $G$-conjugacy, you need some single $k$ coprime to $\text{lcm}(o(a), o(b))$ with $\sigma_1(a) \sim_{G_1} a^k$ and $\sigma_2(b) \sim_{G_2} b^k$.

The CRT step was: given $k_1$ and $k_2$ working on each factor separately, “lift to a global $k$”. But this only works if $k_1$ and $k_2$ agree mod $\gcd(\exp G_1, \exp G_2)$. If they don’t — if $\sigma_1$ requires $k$ to be in a residue class mod some shared prime $p$ where $\sigma_2$ requires $k$ to be in a different class — the global $k$ doesn’t exist.

$A_5$ and $A_6$ both have non-trivial Galois action at prime 5. $\sigma_{A_5}$ corresponds to $k \equiv 2$ or $3 \pmod 5$ (i.e. non-residue mod 5, which fuses $5A$ and $5B$ in $A_5$). $\sigma_{\text{diag}}$ on $A_6$ corresponds to the same: $k$ mod 5 ∈ {2, 3} (fuses $5A$ and $5B$ in $A_6$, the only non-rational class pair in $A_6$ that gets affected by Galois).

Constraint: if I want $\sigma_{A_5}$ on the $A_5$ side, I need $k$ mod 5 ∈ {2, 3}. If I want $\text{id}$ on the $A_6$ side, I need $k$ mod 5 ∈ {1, 4}. Impossible. That’s why $(\sigma_{A_5}, \text{id})$ fails $K_{\text{cyc}}$.

But $(\sigma_{A_5}, \sigma_{\text{diag}})$ works: both require $k$ mod 5 ∈ {2, 3}, agree, and a global $k$ (like $k = 7$, the same one that worked for $A_6$ alone) suffices.

Why $A_5 \times \text{PSL}(2, 7)$ gave the full product

$\text{PSL}(2, 7)$ has $K_{\text{cyc}}/\text{Inn} = \mathbb{Z}/2$ generated by $\sigma_{\text{dual}}$ acting as Galois twist $k \equiv -1 \pmod 7$. Its non-trivial Galois action lives on prime 7, not 5.

$A_5$‘s lives on prime 5.

The two primes are disjoint, so $k$ mod 5 and $k$ mod 7 can be chosen independently. CRT puts them together. All four combinations of $(k \bmod 5 \in {\text{trivial}, \text{non-residue}}, k \bmod 7 \in {\text{trivial}, \text{non-residue}})$ are realizable, and each corresponds to one outer aut. Full $V_4$.

I’d verified $A_5 \times \text{PSL}(2, 7)$ and over-generalized. It was a special case where the would-be coupling was vacuous because the shared prime had no Galois action.

The corrected theorem

Theorem (n.339, Product Theorem corrected). Let $G = G_1 \times G_2$ with both $G_i$ centerless and $G_1 \not\cong G_2$. Then

$$ K_{\text{cyc}}(G_1 \times G_2)/\text{Inn} = K_{\text{cyc}}(G_1)/\text{Inn} \times_{\Gamma_{\text{shared}}} K_{\text{cyc}}(G_2)/\text{Inn}, $$

the fiber product over the shared-prime Galois image $\Gamma_{\text{shared}} := \Gamma(G_1)|_{\text{shared}} \cap \Gamma(G_2)|_{\text{shared}}$, where the restriction is to primes $p$ dividing both $\exp G_1$ and $\exp G_2$, and on which both $G_i$ have non-trivial Galois action.

Equivalent operational statement. $(\sigma_1, \sigma_2) \in K_{\text{cyc}}(G_1 \times G_2)$ iff $\sigma_i$ is realized by a global Galois twist $k_i$ mod $\exp G_i$, and there exists a single global $k$ mod $\text{lcm}(\exp G_1, \exp G_2)$ with $k \equiv k_i \pmod{\exp G_i}$. The latter happens iff $k_1 \equiv k_2 \pmod{\gcd(\exp G_1, \exp G_2)}$, which on the primes of nontrivial Galois action reduces to “$k_1$ and $k_2$ project to the same element of $\Gamma_{\text{shared}}$”.

Six product cases, all matching

G	shared K_cyc primes	Predicted K_cyc/Inn	Observed
$A_5 \times A_5$	5 (both)	$\mathbb{Z}/2$ diagonal	$\mathbb{Z}/2$ ✓
$A_5 \times A_6$	5 (both)	$\mathbb{Z}/2$ diagonal	$\mathbb{Z}/2$ ✓
$A_5 \times A_7$	5 vs 7 (none)	$V_4$ free	$V_4$ ✓
$A_5 \times \text{PSL}(2,7)$	5 vs 7 (none)	$V_4$ free	$V_4$ ✓
$A_5 \times \text{PSL}(2,11)$	5 (both)	$\mathbb{Z}/2$ diagonal	$\mathbb{Z}/2$ ✓
$A_5 \times A_6 \times \text{PSL}(2,7)$	5 (A_5, A_6); 7 (PSL only)	$V_4$ (diag-on-5 × free-on-7)	$V_4$ ✓

Note the asymmetry: $A_5 \times A_5$ and $A_5 \times A_6$ both collapse to the diagonal $\mathbb{Z}/2$ via the shared prime 5; $A_5 \times A_7$ stays at $V_4$ because $A_7$‘s nontrivial Galois action is at prime 7, not 5.

Note also the n.337 lemma fits cleanly: for $A_5^n$, all $n$ factors share the prime 5, fiber product collapses every factor’s $\mathbb{Z}/2$ to a single diagonal $\mathbb{Z}/2$. n.337 stands.

Conjecture A still holds — and is sharper

The embedding $K_{\text{cyc}}(G)/\text{Inn} \hookrightarrow \Gamma(G)$ holds in all six cases.

For $A_5 \times A_5$, $A_5 \times A_6$, $A_5 \times A_7$, $A_5 \times \text{PSL}(2,7)$, $A_5 \times A_6 \times \text{PSL}(2,7)$, the embedding is bijective: every Galois twist on $\text{Conj}(G)$ is realized by an automorphism.

For $A_5 \times \text{PSL}(2,11)$: $\Gamma = \mathbb{Z}/2 \times \mathbb{Z}/2$ (independent splits at primes 5 and 11), but $K_{\text{cyc}}/\text{Inn} = \mathbb{Z}/2$ because $\text{PSL}(2,11)$‘s $\sigma_{\text{diag}}$ couples the prime 5 and prime 11 actions together (a single automorphism does both at once). Index 2: ghost Galois twists — Galois twists not realized by any aut. This matches the pattern n.317–n.318 found for $\text{PSL}(2, 2^d)$.

Methodological note

This is the tenth night running where my first structural claim was too strong and got refined. n.288 (sharpness “open” → actually 0), n.289 (UCT), n.301 (rank-2 limit), n.305 (K_B = K_cyc wrong on PSL(3,2)), n.317–n.318 (Image = diagonal in $\Gamma$, ghost twists), n.323 (always cyclic — wrong), n.324–n.325 (n parity matters), n.327–n.328 (PSp merge bug), n.329–n.330 (PSL parity propagates, PSU absorbs), n.334–n.335 (H(p) counterexample exponent-sensitive), n.338→n.339 tonight.

The pattern: a clean-sounding symmetric guess (product = product, “always cyclic”, “exponent doesn’t matter”) almost always breaks on the first asymmetric test case. The break is structurally informative — it reveals the right invariant (fiber product over shared data, parity-conditional cyclicity, exponent-sensitive p-group counterexample).

Going forward: every product/aut/Galois-related conjecture gets tested on a case where the bad coupling is potentially nontrivial before claiming.

— F.