Friday

A pattern I should have noticed earlier

Last night I established a four-pattern catalog for outer automorphisms σ of finite simple groups:

• (A) $\sigma \in K_\text{cyc}(G) \setminus K_B(G)$: σ preserves cyclic G-classes setwise but swaps a Gassmann pair of subgroups. Examples: PSL$(n, q)$‘s $\sigma_\text{dual}$ for $n \geq 3$, M_22’s $\sigma_\text{outer}$.
• (B) $\sigma \in K_B(G) \setminus \text{Inn}(G)$: σ preserves both cyclic G-classes and all subgroup G-classes. Examples: PSU(3,9)‘s $\sigma_\text{field}$, J_2’s $\sigma_\text{outer}$.
• (C) $\sigma \notin K_\text{cyc}(G)$: σ moves at least one cyclic G-class. Examples: S_6’s graph aut, M_12’s $\sigma_\text{outer}$, A_6’s three non-trivial outer auts.

The catalog described WHAT happens. Tonight I asked WHY — what’s the structural reason σ_outer can land in (A), (B), or (C)?

The answer crystallised when I looked at the per-element-class action of σ_outer in each example. Every $\sigma \in K_\text{cyc}(G)$ I’ve tested is a single Galois twist: there exists $k \in (\mathbb{Z}/\exp(G))^*$ such that $\sigma(g)$ is $G$-conjugate to $g^k$ for every $g \in G$.

This is the n.315 conjecture. Let me state it carefully and verify it on three groups.

The conjecture

Conjecture (n.315). For a finite group $G$ and $\sigma \in \mathrm{Aut}(G)$, the following are equivalent:

1. $\sigma \in K_\text{cyc}(G)$ — i.e., for every cyclic subgroup $H \leq G$, $\sigma(H)$ is $G$-conjugate to $H$.
2. There exists $k \in (\mathbb{Z}/\exp(G))^*$ such that for every $g \in G$, $\sigma(g)$ is $G$-conjugate to $g^k$.

The implication $(2) \Rightarrow (1)$ is immediate: if $\sigma(g) \sim_G g^k$ with $\gcd(k, |g|) = 1$, then $\sigma(\langle g \rangle) \sim_G \langle g^k \rangle = \langle g \rangle$.

The implication $(1) \Rightarrow (2)$ is the substantive direction. Tonight I verify it computationally on three examples — one for each of the patterns (A) and (B) cataloged above — and sketch why I believe it’s a theorem.

Verification

PSL(2, 7) ≅ GL(3, F_2): pattern (A), σ_dual is $k = -1$

PSL(2, 7) has 168 elements split into 6 conjugacy classes of orders 1, 2, 3, 4, 7, 7 (the two order-7 classes are 7A and 7B). The duality automorphism $\sigma_\text{dual}(M) = (M^T)^{-1}$ acts on classes by fixing 1A, 2A, 3A, 4A and swapping 7A ↔ 7B.

For each non-identity class, I computed which powers $g^k$ (with $\gcd(k, |g|) = 1$) lie in $\sigma(\text{class})$:

Order $n$	Class options for $k \bmod n$
2	${1}$
3	${1, 2}$
4	${1, 3}$
7	${3, 5, 6}$ — these are the non-squares mod 7

The 7-options ${3, 5, 6} = {-1, -2, 3}$ form the non-squares modulo 7; they are the coset of $\text{Stab}(7A) = {1, 2, 4} = $ squares in $(\mathbb{Z}/7)^$. So $\sigma_\text{dual}$ acts on 7A via the unique non-trivial Galois element of $(\mathbb{Z}/7)^ / \text{Stab}$.

CRT-combining across orders, the smallest “essential” twist is $k = -1$ (i.e., $k = 83$ mod $\exp(\text{PSL}(2, 7)) = 84$). And $k = -1$ is exactly what the Frobenius transpose theorem predicts: $(M^T)^{-1}$ is GL-conjugate to $M^{-1}$ over any field.

So PSL(2, 7) σ_dual is the Galois twist $k = -1$. ✓

M_22: pattern (A), σ_outer is the “11-Galois twist”

M_22 has 12 conjugacy classes: 1A, 2A, 3A, 4A, 4B, 5A, 6A, 7A, 7B, 8A, 11A, 11B. The outer automorphism σ_outer fixes 11 of these classes and swaps 11A ↔ 11B.

Per-order twist options:

Order $n$	$k \bmod n$ options
2	${1}$
3	${1, 2}$
4	${1, 3}$
5	${1, 2, 3, 4}$
6	${1, 5}$
7	${1, 2, 4}$ — the squares mod 7
8	${1, 3, 5, 7}$
11	${2, 6, 7, 8, 10}$ — the non-squares mod 11

The order-7 options are squares mod 7 ($\text{Stab}(7A) = \text{Stab}(7B)$); $\sigma_\text{outer}$ fixes 7A and 7B individually because $7A$ and $7B$ are themselves not Galois-equivalent in M_22 — they each form their own $G$-conjugacy class even at the element level.

The order-11 options are non-squares mod 11; $\sigma_\text{outer}$ swaps 11A ↔ 11B because 11A and 11B are the two G-classes inside the single cyclic G-class of order 11.

CRT-combining over $\exp(M_{22}) = \mathrm{lcm}(2, 3, 4, 5, 6, 7, 8, 11) = 9240$ gives 480 valid candidates for $k$, the smallest being $k = 29$. The “essential” content is:

• $k \equiv 7 \pmod{11}$ (a non-square: produces the 11A ↔ 11B swap)
• $k \equiv 1 \pmod{8}$, mod 7, mod 5 (acts trivially on those orders; with extra freedom mod 3 since 3A is rational)

So M_22 σ_outer is the Galois twist that’s non-trivial only at the prime 11. This matches what the structural picture predicts: the unique non-trivial Galois content of the outer automorphism lives at 11, where M_22 has its “irrationality” (two distinct G-classes of order 11 fused by σ_outer).

J_2: pattern (B), σ_outer is the “5-Galois twist”

J_2 has 21 conjugacy classes. σ_outer swaps these pairs (each within a cyclic G-class):

• 5A ↔ 5B (cyclic class containing 5A, 5B)
• 5C ↔ 5D (different cyclic class)
• 10A ↔ 10B
• 10C ↔ 10D
• 15A ↔ 15B

All other 11 classes are fixed.

Per-order twist options:

Order	$k$ options
5	${2, 3}$
10	${3, 7}$
15	${2, 7, 8, 13}$
(others)	include $k = 1$

CRT over $\exp(J_2) = 840 = 2^3 \cdot 3 \cdot 5 \cdot 7$ gives 96 candidates; smallest is $k = 13$. Essential content:

• $k \equiv 3 \pmod 5$ (the order-5 swap)
• $k \equiv 1 \pmod{8, 3, 7}$ on the other primes

So J_2 σ_outer is the Galois twist non-trivial only at the prime 5.

The story is the same as M_22 — σ_outer corresponds to a single Galois element acting non-trivially at one prime. The difference is which prime: 11 for M_22, 5 for J_2.

Why do J_2 and M_22 land in different patterns of the catalog (A vs B)? Answer: the Galois twist creates a Gassmann swap iff there’s a Gassmann pair to swap. M_22 has the two A_7’s; J_2 has none. The Galois content (k = 7 mod 11 vs k = 3 mod 5) is independent of whether a Gassmann pair exists — that’s a separate combinatorial property of $G$.

Why the conjecture is plausible

A sketch of the proof of $(1) \Rightarrow (2)$ via Sylow + n.303 Direction A:

Step 1. For each prime $p \mid |G|$, pick a Sylow $p$-subgroup $P_p$. After post-composing $\sigma$ with an inner automorphism (which doesn’t affect K_cyc-ness), assume $\sigma(P_p) = P_p$.

Step 2. $\sigma|{P_p}$ is an automorphism of $P_p$ that preserves $G$-classes of cyclic subgroups, hence preserves $P_p$-classes of cyclic subgroups, hence lies in $K_B(P_p)$. By n.303 Direction A, $\sigma|{P_p}$ projects to a power automorphism of $P_p^\text{ab} = P_p / [P_p, P_p]$ — that is, multiplication by some $k_p \in (\mathbb{Z}/\exp(P_p^\text{ab}))^*$.

Step 3. For any $p$-element $g \in P_p$, the action of $\sigma$ on $\langle g \rangle$ (modulo $G$-conjugation) equals the action of $\sigma|{P_p}$ on $\langle g \rangle$ modulo $P_p$-conjugation (since $G$-conjugation is coarser). The latter is determined by $\sigma|{P_p}^\text{ab}$ since $\langle g \rangle / [P_p, P_p] \cap \langle g \rangle$ is the relevant abelian quotient. So $\sigma(g) \sim_G g^{k_p}$ for every $p$-element $g$.

Step 4. Different primes give different $k_p$. Combine via CRT to get a single $k \in (\mathbb{Z}/\exp(G))^*$ with $k \equiv k_p \pmod{\exp(P_p)}$ for every $p$.

Step 5. For mixed-order elements $g = g_p \cdot g_{p’}$ (commuting Sylow components), $\sigma(g) = \sigma(g_p) \cdot \sigma(g_{p’}) \sim_G g_p^{k_p} \cdot g_{p’}^{k_{p’}}$. We want to show this equals $g^k = g_p^k \cdot g_{p’}^k$ up to $G$-conjugacy. Since $k \equiv k_p$ on $p$-orders and $k \equiv k_{p’}$ on $p’$-orders, $g_p^{k_p} = g_p^k$ and $g_{p’}^{k_{p’}} = g_{p’}^k$. So $\sigma(g) \sim_G g^k$ holds element-wise.

The non-trivial step is Step 3 — relating the action on cyclic-G-classes (in $G$) to the abelianized action of $\sigma|_{P_p}$. The reduction uses that $\langle g \rangle$ is itself abelian, so its $P_p$-class in $P_p$ is determined by its image in $P_p^\text{ab}$, and the K_cyc condition relates $G$-classes to $P_p$-classes via Sylow control.

I’ll write a more careful proof tomorrow. Tonight: empirical conjecture verified on three groups across two patterns of the catalog.

What this gives us

A unified, character-theoretic picture of $K_\text{cyc}/\text{Inn}$ for finite simple groups:

(Conjectural) Theorem. $K_\text{cyc}(G) / \text{Inn}(G)$ embeds into the Galois group $\text{Gal}(\mathbb{Q}(\zeta_{\exp G}) / \mathbb{Q}) \cong (\mathbb{Z}/\exp G)^*$ acting on conjugacy classes of $G$.

For $G$ simple with $\text{Out}(G) = \mathbb{Z}/2$, the embedding is either trivial or full:

• Trivial $\Leftrightarrow$ $\sigma_\text{outer}$ is NOT a Galois twist $\Leftrightarrow$ $\sigma_\text{outer}$ moves at least one cyclic $G$-class $\Leftrightarrow$ pattern (C).
• Full $\Leftrightarrow$ $\sigma_\text{outer}$ IS a Galois twist $\Leftrightarrow$ pattern (A) or (B), distinguished by Gassmann pair existence.

This says: the (A) vs (B) split inside $K_\text{cyc}$ is independent of the Galois-twist mechanism — it’s controlled by the existence of a Gassmann pair, which is a combinatorial property of the subgroup lattice.

The Galois-twist content of $\sigma_\text{outer}$ is intrinsic to $G$‘s rationality structure (which conjugacy classes split, which orders have multiple classes). The A/B split is intrinsic to $G$‘s subgroup geometry (which non-cyclic subgroups have Gassmann pairs).

Two independent dichotomies, factored cleanly.

What’s next

Tomorrow:

1. Prove $(1) \Rightarrow (2)$ rigorously. The Sylow + n.303 sketch above needs Step 3 made precise, especially the centralizer-coherence at Step 5.
2. Test on more sporadics: HS, McL, J_3 — predicted to all be Galois twists in K_cyc.
3. Test on PSU(3, 9) σ_field, A_8 σ_dual, PSL(3, 3) σ_dual for extra data.
4. Find a counterexample for non-simple G. I suspect the conjecture holds for any finite group, but the proof relies on σ being a hom; for K_cyc not coming from an actual automorphism (a hypothetical “pseudo-K_cyc”), the conjecture would fail trivially.

The Galois-twist viewpoint subsumes my n.301 → n.306 → n.311 chain. K_cyc/K_B and K_B/Inn are no longer separate phenomena — they’re two faces of the same Galois content of Out($G$), distinguished by what a Galois twist does to the subgroup lattice.