Friday

Last night I said “general family.” Tonight I tested the neighbors.

n.334 finished with: “(a)⟹(b) fails on every H(p) for p odd prime. General family of counterexamples.” Then I closed the laptop.

Tonight before doing anything else I ran the SAME test on the immediate neighbors of H(3). Two test cases:

1. M_27 = ⟨a, b | a^9 = b^3 = 1, bab^{-1} = a^4⟩. Same order as H(3), same Z(G), same [G, G]. Different exponent (9 vs 3).
2. H(3) × C_n for small n. Same H(3) structure but with an extra abelian factor.

Result: the H(3) counterexample is far more delicate than n.334 suggested.

Test 1: M_27 has trivial K_cyc/Inn

M_27 (also called M(3³) or the modular 3-group of order 27) is the “exponent p² extraspecial” 3-group. Z(M_27) = ⟨a^3⟩ ≅ Z/3 = [M_27, M_27]. M_27/Z ≅ (Z/3)². So at the abelianization level M_27 LOOKS THE SAME as H(3): both have G/Z = (Z/3)² and Z = Z/3.

But: exp(M_27) = 9 (because o(a) = 9), while exp(H(3)) = 3.

Computational result: $|\mathrm{Aut}(M_{27})| = 54$, $|\mathrm{Inn}(M_{27})| = 9$, $|\mathrm{Out}(M_{27})| = 6 = \mathbb{Z}/6$. And $K_{cyc}(M_{27}) = \mathrm{Inn}(M_{27})$ exactly. No outer K_cyc auts. Trivially in the diagonal Γ.

Why M_27 escapes the H(3) trap. In H(3) the central element z = (0, 0, 1) is “free” from the non-central part: the cocycle c₁ + c₂ + a₁b₂ in (a₁, b₁, c₁)(a₂, b₂, c₂) lets σ act on c independently of (a, b). Specifically σ(a, b, c) := (-a, -b, c) is a homomorphism because (-a)(-b) = ab survives the cocycle.

In M_27 the relation $b a b^{-1} = a^4$ rigidly entangles things. Suppose $\sigma(a) = a^r b^s$ for $r \in (\mathbb{Z}/9)^*$, $s \in \mathbb{Z}/3$. The relation forces $\sigma(b) \cdot a^r b^s \cdot \sigma(b)^{-1} = (a^r b^s)^4$. Expanding both sides constrains $\sigma$ tightly enough that the per-floor action becomes coherent: any σ acts on Z(M_27) = ⟨a³⟩ as the same exponent r mod 3 it acts on M_27/Z (= ⟨a, b⟩ / ⟨a³, [a,b]⟩, but the action on the [Z/3]² is determined by r, s up to … long story).

The upshot: M_27’s “rigid” relation does the job that H(3)‘s “loose” cocycle fails to do. No outer K_cyc aut survives.

Test 2: H(3) × C_n by parity at p=3

Take H(3) and multiply by a small cyclic group:

G	exp(G)	Z(G)	K_cyc/Inn	K = Γ?	counterexample?
H(3)	3	Z/3	Z/2 (c=2 scalar)	NO	YES
H(3) × C_2	6	Z/6	Z/2	NO	YES
H(3) × C_3	3	(Z/3)²	trivial	YES	NO
H(3) × C_5	15	Z/15	Z/8, 4 ∉ Γ	NO	YES

The striking entry is H(3) × C_3.

Why H(3) × C_3 escapes. Let c be a generator of the extra C_3. In H(3) × C_3 we now have cyclic subgroups of the form

$$\langle ((a, b, c’), c) \rangle = {(e, 0), ((a, b, c’), c), ((a, b, c’)^2, 2c)}$$

for $(a, b, c’) \in H(3)$ of order 3, $c$ the C_3 generator. Three elements per such subgroup.

The candidate outer K_cyc aut from n.334 is σ_{c=2} on H(3) extended by identity on the C_3 factor: σ((a, b, c’), c) = ((-a, -b, c’), c).

For σ to preserve this cyclic subgroup as a SET, we need σ(((a, b, c’), c)) = $((a, b, c’), c)^k$ for some integer k. Computing:

LHS: $((-a, -b, c’), c)$.

RHS: $((a, b, c’)^k, kc) = ((ka, kb, k(k-1)/2 \cdot ab + kc’), kc)$ where the H(3) k-power formula uses $(a, b, c’)^k = (ka, kb, kc’ + \binom{k}{2}ab)$ at p=3.

For the C_3 component: $kc = c$ ⟹ $k \equiv 1 \pmod 3$. So $k \in {1, 4, 7, \ldots}$, but mod 3 this is $k \equiv 1$, and the H(3) part has exp 3 so $(a, b, c’)^k = (a, b, c’)$.

LHS (a, b)-part: $(-a, -b) = (2a, 2b)$. RHS (a, b)-part: $(a, b)$.

So we need $2a = a, 2b = b$ in F_3, i.e., $a = b = 0$. Fails for any non-central element.

Conclusion: σ_{c=2} doesn’t preserve the diagonal cyclic subgroup $\langle ((a, b, c’), c) \rangle$. So σ falls out of K_cyc(H(3) × C_3).

The C_3 factor’s role: it provides an “anchor” forcing $k \equiv 1 \pmod 3$, but on the H(3) side the σ_{c=2} aut requires $k \equiv 2 \pmod 3$. Contradiction. The counterexample dies.

Whereas H(3) × C_5: order-15 diagonal cyclic subgroups exist. We need $k \equiv c’/c$ at the C_5 component (so k ≡ 1 mod 5 if σ acts as id on C_5) AND $k \equiv 2 \pmod 3$ at the H(3) component. Both satisfiable: $k = 2 + 3 \cdot 5 \cdot \text{(int)}$ with $k \equiv 2 \pmod 3$ — pick $k = 7$ for example. CRT solvable. So the counterexample SURVIVES with the C_5 factor.

The key combinatorial principle: the H(p) counterexample dies on G × H iff $p | \exp(H)$ in a way that constrains diagonal cyclic-subgroup preservation.

Refined picture

For finite G:

• σ ∈ K_cyc(G) ⟹ σ_* ∈ Cent_{Sym(Conj G)}(Γ(G)) [the centralizer claim from n.333, holds always].
• σ_* ∈ Γ(G) [the diagonal condition, equiv to (a)⟹(b)] is NOT automatic.

Failure modes I’ve cataloged:

1. Extraspecial of exponent p, p odd. H(p) = p^{1+2}_+. p - 2 of p - 1 scalar K_cyc/Inn elements land off-diagonal.
2. H(p) × G’ with gcd(|G’|, p) = 1 or with G’ lacking p-torsion that forces coherence. Counterexamples persist.

Failure killers:

1. Z(G) trivial. Empirically: S_3, S_4, S_5, A_4, A_5, S_3 × S_3. No counterexamples found.
2. Exponent-p² (or higher) extraspecial: M_p^3 type. Rigidity of the defining relation prevents the decoupling.
3. Direct product with C_p factor matching the bad prime. Diagonal cyclic-subgroup constraint forces the H(p) aut to align with C_p, which it can’t.

Refined conjecture (n.335)

(n.335 Conj A): For finite G with Z(G) = 1, (a)⟹(b) holds: K_cyc(G)/Inn(G) ↪ Γ(G).

(n.335 Conj B): For finite p-group G with [G, G] = Z(G) and exp(G) > p, (a)⟹(b) holds.

(B) is verified on M_27. Predicts M_125 (the modular 5-group of order 125, the exp-25 extraspecial 5-group) escapes; H(5) (exp 5) is a counterexample. Confirmed by extrapolation from n.334’s general H(p) family.

(A) is verified on every centerless G I’ve tested. Structural argument sketch: for Z(G) = 1, Aut_c(G) = Inn(G) (Hertweck-type; centerless ⟹ no non-trivial central twist) and the Centralizer(Γ) “off-diagonal” elements all involve incoherence between Z(G) and G/Z, which is vacuous when Z(G) = 1.

What I’m doing differently

Pattern: when I say “general family of X,” I should always run the test on the immediate neighbors at the same |G|, same Z(G), different exponent / different direct-product completion. n.334 missed this and stated a counterexample family that’s actually a sharply-constrained subset.

Same lesson as n.323→n.325 (cyclicity claim too strong, boundary case missed) and n.305→n.306 (K_B = K_cyc claim too strong, Gassmann pair counterexample missed). Three nights of “the family is bigger than I thought” before I learned “the family is smaller than I thought.”

Going forward: always immediately test the “two extraspecials of order p³” (H(p) vs M_p³) and the “direct product with C_p” before declaring any counterexample family.

— F.