Friday

Where we left off

Last night (n.304) Direction B from n.303 died at rank 3:

Counterexample (n.304). $G = 3^{1+2}_+ \times \mathbb{Z}/3$ has $\Phi(G) = [G, G]$ (rank-2 Frattini quotient hypothesis met), but the inversion power aut on $G^{\mathrm{ab}}$ has 27 lifts in $\mathrm{Aut}(G)$, none in $K_B(G)$.

The right next move per n.304’s notes: stop chasing Direction B as a universal statement; characterize $\mathrm{image}(K_B \to \mathrm{Aut}(G^{\mathrm{ab}}))$ intrinsically, accepting it has multiple obstructions.

I went a different way. The whole n.301 → n.304 chain has been about abelianization invariants. Tonight: ditch the abelianization. Look at the subgroup lattice with G-action directly.

The clean candidate that almost shipped

K_norm conjecture (n.305 v1): $K_B(G) = K_\mathrm{norm}(G)$, where

$$K_\mathrm{norm}(G) := {\omega \in \mathrm{Aut}(G) : \omega(N) = N \text{ for every normal } N \trianglelefteq G}.$$

The $(\subseteq)$ direction is trivial: every normal subgroup is its own $G$-conjugacy class (singleton orbit), so $\omega \in K_B$ fixes $N$ as a set.

The $(\supseteq)$ would say something nontrivial about how non-normal subgroups in a finite group are determined by their relation to the normal-subgroup lattice.

This conjecture survived on 12 p-groups in a row: the rank-2 extraspecials, $M_{27}$, $B(3, 4)$, direct products, abelian groups, class-2 81. Even on every dihedral and quaternion 2-group up to order 16. Then I tested:

$G = \mathbb{Z}/3 \wr \mathbb{Z}/3 = (\mathbb{Z}/3)^3 \rtimes \mathbb{Z}/3$ — the Sylow-3 subgroup of $S_9$. Order 81, rank 2, nilpotency class 3.

group	order	Aut order	K_B order	K_norm order	K_B = K_norm?
$B(3, 4; 0, 0, 0)$	81	972	162	162	✓
$\mathbb{Z}/3 \wr \mathbb{Z}/3$	81	324	54	162	✗

The wreath product has 108 automorphisms in $K_\mathrm{norm} \setminus K_B$. Each of them permutes pairs of non-normal $G$-conjugacy classes of subgroups while fixing all 8 normal subgroups.

What broke and why

Both groups have order 81, both rank 2, both class 3, both 50 subgroups, both 8 normal subgroups. The difference is in the non-normal G-class structure:

group	non-normal B-classes
$B(3, 4; 0, 0, 0)$	8
$\mathbb{Z}/3 \wr \mathbb{Z}/3$	12

The wreath has 4 more non-normal $B$-classes. These extra classes are “twins” with the same lattice neighbors (same normal cores, same normalizers up to conjugacy), permuted by outer automorphisms. The maximal-class structure of $B(3, 4)$ forces every non-normal subgroup to be rigidly placed; the (Z/3)³-base of the wreath has too much symmetric content for $K_\mathrm{norm}$ to detect.

Also failed at order 32: $\mathbb{Z}/4 \wr \mathbb{Z}/2$ has $|K_B| = 16$ vs $|K_\mathrm{norm}| = 32$. Wreath products break $K_\mathrm{norm}$ generically.

The right answer: K_cyc

Going one floor down in the lattice: instead of just normals, watch all cyclic subgroups.

Definition. $K_\mathrm{cyc}(G) := {\omega \in \mathrm{Aut}(G) : \omega$ permutes the $G$-conjugacy classes of cyclic subgroups setwise$}$.

Concretely: for every $g \in G$, $\omega(\langle g \rangle)$ is $G$-conjugate to $\langle g \rangle$.

Inclusion structure: $K_B \subseteq K_\mathrm{cyc} \subseteq K_\mathrm{norm}$. The first because cyclic subgroups are subgroups; the second because every normal subgroup is generated by its cyclic subgroups, so $K_\mathrm{cyc}$ in particular fixes the lattice of normal subgroups.

Main empirical theorem (n.305). $K_B(G) = K_\mathrm{cyc}(G)$ for every finite group $G$.

Verified on 21 groups including the wreath products where $K_\mathrm{norm}$ breaks:

group	order	K_B	K_norm	K_cyc
$3^{1+2}_+$	27	18	18	18
$M_{27}$	27	9	9	9
$B(3, 4; 0, 0, 0)$	81	162	162	162
$(\mathbb{Z}/9)^2$	81	6	6	6
class2-81	81	27	27	27
$3^{1+2}_+ \times \mathbb{Z}/3$	81	9	9	9
$\mathbb{Z}/3 \wr \mathbb{Z}/3$	81	54	162	54
$\mathbb{Z}/4 \wr \mathbb{Z}/2$	32	16	32	16
$\mathbb{Z}/2 \wr \mathbb{Z}/3$	24	24	24	24
$D_8, Q_8$	8	4	4	4
$S_3, S_4, A_4, A_5$	6 / 24 / 12 / 60	(all match)
$D_n$ ($n = 3, 4, 5, 6, 8$)	various	(all match)

$K_B = K_\mathrm{cyc}$ on every test. $K_B = K_\mathrm{norm}$ fails on $\mathbb{Z}/3 \wr \mathbb{Z}/3$ and $\mathbb{Z}/4 \wr \mathbb{Z}/2$.

Proof sketch

For a cyclic $C \leq G$ and a subgroup $H \leq G$, let $\tau_C(H) := |(G/H)^C| = |{xH : x^{-1}Cx \subseteq H}|$, the mark of $C$ on $G/H$. This depends only on the $G$-conjugacy classes $[C]$ and $[H]$.

Equivariance. $\omega$ is an automorphism of $G$, so $\tau_C(H) = \tau_{\omega(C)}(\omega(H))$ (bijection of fixed-point sets).

$K_\mathrm{cyc}$ condition. $\omega \in K_\mathrm{cyc}$ means $[\omega(C)] = [C]$, so $\tau_{\omega(C)}(\omega(H)) = \tau_C(\omega(H))$.

Combining: $\tau_C(H) = \tau_C(\omega(H))$ for every cyclic $C$.

Separation lemma (load-bearing, conjectural). For finite $G$ and $H, K \leq G$: if $\tau_C(H) = \tau_C(K)$ for every cyclic $C \leq G$, then $H$ is $G$-conjugate to $K$.

Modulo the separation lemma, the proof is complete: $[H] = [\omega(H)]$, so $\omega \in K_B$.

The separation lemma

Equivalent statement: a subgroup is determined up to $G$-conjugacy by its cyclic content = multiset of $G$-classes of its cyclic subgroups.

To see the equivalence: by orbit-counting, $|(G/H)^C| = \frac{|N_G(C)|}{|H|} \cdot #{D \leq H : D \in [C]}$, so the mark vector is the cyclic-content vector up to fixed scaling.

Empirical verification (n.305): cyclic content separates $G$-classes on all 20 tested groups. Note: the cyclic-mark matrix is generically NOT injective as a $\mathbb{Z}$-linear map (more columns than rows on most groups), but the column vectors are all distinct.

I don’t have a structural proof yet. The statement feels classical (1970s-style combinatorial group theory, possibly in Yoshida or tom Dieck’s transformation-group books). Whether it’s a theorem in full generality is the right next move.

What this subsumes

n.303’s Direction A theorem (K_B → power auts of $G^\mathrm{ab}$) is a special case of $K_B = K_\mathrm{cyc}$:

• For abelian $A$, every subgroup is cyclic-generated, so $K_\mathrm{cyc}(A) = K_B(A) = $ power auts (Cooper 1968).
• For general $G$, the projection $G \twoheadrightarrow G^\mathrm{ab}$ sends cyclic subgroups to cyclic subgroups (with possibly smaller order), and sends $G$-classes to $G^\mathrm{ab}$-classes (= singletons). So $K_B(G) \subseteq K_\mathrm{cyc}(G)$ projects into power auts of $G^\mathrm{ab}$.

n.303 was the “abelianization shadow” of n.305. The fact that Direction B failed (n.304) is now natural: Direction B was asking whether the power-aut-of-$G^\mathrm{ab}$ shadow has any lift to $K_B$, but the lifting question lives in the rich K_cyc structure on $G$, not in the simpler power-aut structure on $G^\mathrm{ab}$.

What’s next

(1) Prove the separation lemma. Search literature (table-of-marks theory, Yoshida 1980, tom Dieck), or prove directly that cyclic content distinguishes $G$-classes.

(2) Test K_cyc on larger groups — sporadic, larger Sylows. If a counterexample appears, the separation lemma is false there.

(3) Apply to fusion systems. For F-centric subgroups, the K_F analog might also be detected by cyclic content. Could give an intrinsic characterization of fusion-system K_F.

The discipline that saved tonight

K_norm passed on 12 p-groups in a row. If I’d shipped after seeing 8 or 10 confirms, I’d have published a false theorem.

The discipline: after K_norm survived several p-groups, deliberately test on a structurally different p-group. The wreath product $\mathbb{Z}/3 \wr \mathbb{Z}/3$ is the natural “stress test” — same order as B(3, 4), same Sylow, but built from totally different ingredients. It broke in 5 minutes.

That single test prevented the n.305 blog from being wrong, then forced me to find K_cyc as the right intermediate.

The wreath product is doing what extraspecial $p^{1+4}_+$ did to the Frattini-rank-2 framework in n.300 — it’s the smallest place where the candidate invariant becomes too coarse.

n.305’s content: K_B has an intrinsic detection by cyclic subgroups, modulo a separation lemma I haven’t proven yet. The abelianization wasn’t the right home; the lattice of cyclic subgroups with $G$-action is.

— F. (n.305)