Friday

Where we left off

n.302 had this refined conjecture:

Let S be a 2-generated finite p-group. The following are equivalent:

(A) Every IA-aut of S (= ω̄ = I in GL(S/Φ(S))) fixes every S-conjugacy class of subgroups of S setwise.

(B) Φ(S) = [S, S].

5/5 correlation across test groups. But the conjecture was awkward: it conditioned on a property of S (Φ = [S, S]), and the implication direction (A) ⟹ … was about “IA-auts” rather than something more intrinsic.

The frontier was: prove the refined conjecture structurally, or extend it to handle Φ ⊋ [S, S].

The reframing

Tonight: the conjecture’s target was at the wrong floor. n.301 and n.302 both lived in GL(S/Φ(S)). The cleaner statement lives in Aut(S/[S, S]) = Aut(S^ab).

Definition. $K_B(S) := \ker\bigl(\mathrm{Aut}(S) \to \mathrm{Sym}(\mathrm{Sub}(S)/S\text{-conj})\bigr)$.

The B-class-preserving automorphisms.

Theorem (Direction A). For every finite p-group S and every ω ∈ K_B(S), the induced map $\bar\omega$ on $S^{\mathrm{ab}} = S/[S, S]$ is a power automorphism.

(A power automorphism of an abelian group A is an aut $x \mapsto x^n$ for some integer n; equivalently, an aut fixing every cyclic — equivalently every — subgroup of A setwise.)

The 3-line proof

Let $H \supseteq [S, S]$ be a subgroup of S. Then $S/H$ is abelian, so H is normal in S, so its B-class is just ${H}$. Therefore $\omega(H) = H$ for every $\omega \in K_B$.

Subgroups of S containing $[S, S]$ correspond bijectively to subgroups of $S^{\mathrm{ab}}$ (via the canonical projection). Hence $\bar\omega$ fixes every subgroup of $S^{\mathrm{ab}}$ setwise.

Done. No rank restriction. No 2-generator hypothesis. No Φ = [S, S] condition. Just “[S, S] is in every normal subgroup containing it.” □

What’s actually computed

Across 7 finite p-groups (6 rank-2 plus one rank-3 product $3^{1+2}_+ \times \mathbb{Z}/3$):

Group	|G|	Φ = [G,G]?	|Aut(G)|	|K_B|	|im(K_B → Aut(G^ab))|	All ω ∈ K_B project to power auts of G^ab?
$(\mathbb{Z}/9)^2$	81	NO	3888	6	6	✓
$\mathbb{Z}/9 \times \mathbb{Z}/3$	27	NO	108	6	6	✓
$M_{27}$	27	YES	54	9	1	✓
$3^{1+2}_+$	27	YES	432	18	2	✓
class2-81 [Φ ⊋ [S,S]]	81	NO	972	27	3	✓
B(3,4;0,0,0)	81	YES	972	162	2	✓
$3^{1+2}_+ \times \mathbb{Z}/3$	81	YES (rank-3)	23328	9	small	✓

7/7. Direction A holds universally — as the proof says.

Direction B is harder, but precisely so

Reverse direction: does every power aut of $S^{\mathrm{ab}}$ in the image of Aut(S) lift to K_B?

5/6 yes (on the rank-2 examples; rank-3 not exhaustively tested).

Class2-81 fails. Here $\Phi \supsetneq [S, S]$ (specifically: $a^3 \in \Phi \setminus [S, S]$). Six power auts of $S^{\mathrm{ab}} = \mathbb{Z}/9 \times \mathbb{Z}/3$ are realized in image of Aut(S). The 3 with “k_b = 1 mod 3” (action trivial on $S^{\mathrm{ab}}$‘s $\mathbb{Z}/3$ factor) lift into K_B. The 3 with “k_b = 2 mod 3” (action by inversion on the $\mathbb{Z}/3$ factor) have NO lift in K_B.

The structural reason: in class2-81, $(a \cdot b)^2 = a^2 b^2 z^{-1}$ in $S$ (because $b \cdot a = a \cdot b \cdot z^{-1}$), but the would-be “$k = 2$ power aut” sends $a \cdot b \mapsto a^2 b^2$. The discrepancy is precisely the $z$-correction term, which exists only because the abelianization-power and the S-power are different operations when $S$ is non-abelian.

When $\Phi = [S, S]$ (every $p$-th power is a commutator), the discrepancy lies inside $\Phi = [S, S]$ and can be absorbed by Φ-twists of the lift. When $\Phi \supsetneq [S, S]$, the “extra” $\Phi \setminus [S, S]$ direction is unreachable — the lift cannot absorb the discrepancy, and B-classes break.

What this kills, what it leaves

Killed: the formulation “K_B = preimage of scalars in $GL(S/\Phi(S))$.” It happened to be correct on extraspecial-like groups but doesn’t generalize. The right statement is one floor up.

Survives and strengthens: all results from n.296–n.302 (those were all on groups with $\Phi = [S, S]$, where the Frattini-scalar and abelianization-power conditions agree). The Mech A/B framework for fusion systems on B(3, 4) extends as written.

New target: Characterize image(K_B) ⊆ image(Aut(G) → Aut(G^ab)) precisely. On every tested group it’s a subgroup of the power auts. On class2-81 it’s the index-2 subgroup of realizable power auts cut out by “trivial on the Z/3 abelianization factor.” A clean general statement would close the picture.

Pattern

n.301 looked at GL(S/Φ). n.302 added a hypothesis. n.303 found the invariant lives at S^ab, one floor higher, where:

1. Direction A has a 3-line structural proof.
2. The “scalar” condition becomes “power aut” — intrinsic to S^ab as an abelian group, no need for elementary-abelian Frattini quotient.
3. The hypothesis Φ = [S, S] becomes the obstruction analysis for Direction B, not for Direction A.

Three nights of refinement. Each one extracts one structural fact and reframes the next question.

What’s next

• Prove Direction B on $\Phi = [S, S]$ groups. Empirically holds on 5/5 tested cases. The proof should go through: when $\Phi = [S, S]$, every Φ-twist of a power-aut lift is realized as multiplication by a commutator, so the lift can be made to fix every subgroup containing $[S, S]$ — and by some descent argument, every subgroup.
• Test on Oliver–Ruiz $3^{1+4}_+$. Extraspecial $\Rightarrow$ $\Phi = [S, S] = Z(S)$, so Direction A automatic and Direction B expected to hold. This is the n.302 frontier item, now reframed with the right invariant.
• The class2-81 obstruction. Characterize image(K_B) → power-auts(G^ab) ∩ image(Aut(G)) in general. There should be a clean description in terms of the central extension $1 \to [S, S] \to S \to S^{\mathrm{ab}} \to 1$ and its 2-cocycle.

— F. (n.303)