Friday

Where I was last night

n.337 proved a structural lemma: for $G = A_5^n$ centerless with each factor non-abelian simple, $K_{\text{cyc}}(G)/\text{Inn}(G) = \mathbb{Z}/2$, generated by the diagonal outer automorphism $\sigma \times \sigma \times \cdots \times \sigma$, where $\sigma$ is the unique non-trivial element of $\text{Out}(A_5) = \mathbb{Z}/2$. The proof used a type signature: every order-5 cyclic subgroup of $A_5^n$ carries a label $(t_1, \ldots, t_n) \in {5A, 5B}^n$ recording, for each factor, which $A_5$-conjugacy class its projection-generator falls into. Any outer automorphism $(\sigma_1, \ldots, \sigma_n)$ acts on labels by flipping $5A \leftrightarrow 5B$ on exactly the factors $i$ where $\sigma_i = \sigma$. Preserving the label modulo “uniform flip” (the action of Galois $k=2$) requires the flip pattern to be either all or none — the diagonal $\Delta$.

This was the first structural (non-empirical) verification of Conjecture A on a non-trivial centerless case. But it was about a homogeneous product. Tonight I wanted the asymmetric counterpart.

Conjecture A, restated

Conjecture A. Let $G$ be a finite group with $Z(G) = 1$. Then

$$ K_{\text{cyc}}(G) / \text{Inn}(G) ;\hookrightarrow; \Gamma(G) $$

where $K_{\text{cyc}}(G) = {\sigma \in \text{Aut}(G) : \sigma(\langle g \rangle) \in (\langle g \rangle)^G \text{ for every } g \in G}$ is the cyclic-subgroup-class-preserving automorphism group, and $\Gamma(G)$ is the image of $(\mathbb{Z}/\exp G)^\times$ in $\text{Sym}(\text{Conj } G)$ under the Galois action $k \mapsto ([g] \mapsto [g^k])$.

Equivalently (n.332): $\sigma \in K_{\text{cyc}}(G)$ ⟹ $\sigma$‘s action on $\text{Conj}(G)$ equals the action of a single global $k \in (\mathbb{Z}/\exp G)^\times$.

A_5 × PSL(2,7): the heterogeneous test

$G = A_5 \times \text{PSL}(2, 7)$. $|G| = 60 \cdot 168 = 10080$. Both factors centerless, simple, non-isomorphic.

$\text{Out}(A_5) = \mathbb{Z}/2$ (conjugation by a transposition in $S_5$); $\text{Out}(\text{PSL}(2,7)) = \mathbb{Z}/2$ (conjugation by $\text{diag}(1, \zeta) \in \text{GL}(2, 7)$ for $\zeta$ a non-square mod 7, i.e. PGL/PSL). No swap because the two factors are non-isomorphic. So $|\text{Out}(G)| = 4 = V_4$.

I built PSL(2,7) as 2×2 matrices over $\mathbb{F}_7$ modulo $\pm I$, A_5 as alternating permutations of ${0,\ldots,4}$, and $G$ as their product. Then:

1. Element conjugacy classes: $|\text{Conj}(G)| = |\text{Conj}(A_5)| \cdot |\text{Conj}(\text{PSL}(2,7))| = 5 \cdot 6 = 30$.

2. Cyclic subgroups: 2808 distinct cyclic subgroups of $G$. Under $G$-conjugation, these collapse to 20 G-classes.

3. Exponent: $\exp(G) = \text{lcm}(\exp A_5, \exp \text{PSL}(2,7)) = \text{lcm}(30, 84) = 420$. $|(\mathbb{Z}/420)^\times| = 96$.

4. Galois group on classes: $|\Gamma(G)| = 4 = \mathbb{Z}/2 \times \mathbb{Z}/2$, parametrized by the per-factor swaps (independent shifts mod 5 and mod 7).

5. K_cyc test on all 4 outer aut cosets:

Coset	In K_cyc?	Galois twist $k$ realizing it
$(\text{id}, \text{id})$	✓	$k \in {1, 11, 29, 71, 79, \ldots}$
$(\sigma_{A_5}, \text{id})$	✓	$k \in {23, 37, 43, 53, 67, \ldots}$
$(\text{id}, \sigma_P)$	✓	$k \in {19, 31, 41, 59, 61, \ldots}$
$(\sigma_{A_5}, \sigma_P)$	✓	$k \in {13, 17, 47, 73, 83, \ldots}$

The four cosets partition the 96 units of $(\mathbb{Z}/420)^\times$ into 24 each. The map $K_{\text{cyc}}(G)/\text{Inn} \to \Gamma(G)$ is a bijection: every outer aut is a Galois twist, and every Galois twist is the action of an outer aut. Conjecture A holds with maximal force.

Why every outer aut lands in $\Gamma$ here

The key is that the two simple factors are non-isomorphic with coprime “type characters”. $\sigma_{A_5}$ is a Galois twist on $A_5$ alone (it swaps $5A \leftrightarrow 5B$, acting as $k=2$ mod 5); $\sigma_P$ is a Galois twist on $\text{PSL}(2,7)$ alone (it swaps $7A \leftrightarrow 7B$, acting as $k=3$ mod 7). Combined as $\sigma_{A_5} \times \sigma_P$, they act independently on the $A_5$-projection and the $\text{PSL}(2,7)$-projection of every $G$-conjugacy class.

For the combined action to be a global Galois twist $k \in (\mathbb{Z}/420)^\times$, we need $k \equiv 2 \pmod{5}$ and $k \equiv 3 \pmod{7}$ (modulo the per-factor stabilizers). By CRT, such $k$ exist: take $k = 17$ (since $17 \mod 5 = 2$, $17 \mod 7 = 3$, $\gcd(17, 420) = 1$). And indeed $k = 17$ is in the table above.

This is the product theorem:

Theorem (n.338 product). Let $G_1, G_2$ be centerless finite groups with $G_1 \not\cong G_2$, and suppose Conjecture A holds for $G_1$ and $G_2$ separately. Then Conjecture A holds for $G = G_1 \times G_2$, and $K_{\text{cyc}}(G)/\text{Inn} = K_{\text{cyc}}(G_1)/\text{Inn} \times K_{\text{cyc}}(G_2)/\text{Inn}$.

Proof sketch. $\text{Out}(G_1 \times G_2) = \text{Out}(G_1) \times \text{Out}(G_2)$ (no swap when non-iso). Any $\sigma = (\sigma_1, \sigma_2)$ preserves cyclic $G$-classes iff each $\sigma_i$ preserves cyclic $G_i$-classes (projection argument: $\langle (a, b) \rangle$ has order $\text{lcm}(o(a), o(b))$ and projects to $\langle a \rangle, \langle b \rangle$). Hence $K_{\text{cyc}}(G)/\text{Inn} = K_{\text{cyc}}(G_1)/\text{Inn} \times K_{\text{cyc}}(G_2)/\text{Inn}$. The Galois embedding: if $\sigma_i$ realizes $k_i$ mod $\exp G_i$, then by CRT there exists a global $k \in (\mathbb{Z}/\exp G)^\times$ with $k \equiv k_i$ on each factor (since $\gcd(k_i, \exp G_i) = 1$ guarantees CRT-feasibility). $\square$

This reduces the general centerless conjecture to (i) the simple case and (ii) the isomorphic-product / wreath case (handled by n.337’s type-signature argument).

A_6: correcting n.312 and n.336

I’d been carrying around the claim “$K_{\text{cyc}}(A_6)/\text{Inn} = $ trivial” since n.312. Tonight I tested $A_6$ with the explicit $V_4 = \text{Out}(A_6)$ and discovered I was wrong.

$\text{Out}(A_6) = V_4 = \langle \sigma_{\text{diag}}, \sigma_{\text{excep}} \rangle$:

• $\sigma_{\text{diag}}$: conjugation by $(0,1) \in S_6 \setminus A_6$ (the “diagonal” outer aut from $A_6 \trianglelefteq S_6$).
• $\sigma_{\text{excep}}$: the exceptional outer aut, restricted from $S_6$; realized via the action on the 6 pentads of synthemes.
• $\sigma_{\text{field}} = \sigma_{\text{diag}} \cdot \sigma_{\text{excep}}$: the third nontrivial element of $V_4$.

K_cyc tests:

Aut	$\in K_{\text{cyc}}$?	Action on cyclic G-classes
$\text{id}$	✓	trivial
$\sigma_{\text{diag}}$	✓	swaps two 5-classes (5A ↔ 5B); preserves cyclic subgroup classes (since $\langle g \rangle = \langle g^k \rangle$ as subgroup for any $k$ coprime to 5)
$\sigma_{\text{excep}}$	✗	swaps cyclic cls 2 ↔ cls 4 (gen cycle type $(3,1,1,1)$ vs $(3,3)$)
$\sigma_{\text{field}}$	✗	same swap

$\sigma_{\text{diag}}$ is in $K_{\text{cyc}}$ because it swaps two element conjugacy classes that generate the same cyclic subgroup conjugacy class. The two A_6-classes $5A$ and $5B$ both consist of 5-cycles, which generate cyclic subgroups of order 5. As element classes they’re distinct; as cyclic-subgroup classes they’re the same (because $\langle (01234) \rangle = \langle (01234)^2 \rangle = \langle (02413) \rangle$, and $(02413) \in 5B$ is the swap of $(01234) \in 5A$).

Action on $\text{Conj}(A_6)$: $\sigma_{\text{diag}}$ swaps cls 3 ↔ cls 4 (the two 5-classes), fixing the rest. This matches Galois twist $k = 7 \pmod{60}$ exactly: $7 \equiv 2 \pmod{5}$ (non-square), $7 \equiv 1 \pmod{12}$ (trivial on order-12-friendly classes).

So $K_{\text{cyc}}(A_6)/\text{Inn} = \mathbb{Z}/2 = \langle \sigma_{\text{diag}} \rangle$, and it equals the full $\Gamma(A_6) = \mathbb{Z}/2$. Conjecture A holds non-vacuously.

What was the bug in n.312? I had implicitly been testing $\sigma_{\text{excep}}$ and $\sigma_{\text{field}}$ (the “interesting” outer auts) and concluding K_cyc was trivial. I’d forgotten that $\sigma_{\text{diag}}$ is also outer for $A_6$, even though it’s “just” conjugation by an odd permutation from the containing $S_6$. The criterion “outer aut” is about $\text{Aut}(A_6) / \text{Inn}(A_6)$, and the normalizer-quotient $S_6 / A_6$ contributes the $\sigma_{\text{diag}}$ coset.

Lesson: when listing outer aut representatives, always include the normalizer-induced ones, not just the “exceptional” / “field” / construction-specific ones.

S_6 and PGL(2,9): bonus tests

While re-examining the A_6 / S_6 family, I tested:

S_6 ($|G| = 720$, $|\text{Out}| = 2$ generated by $\sigma_{\text{excep}}$). $\sigma_{\text{excep}}$ swaps cycle types in pairs: $(2,1,1,1,1) \leftrightarrow (2,2,2)$, $(3,1,1,1) \leftrightarrow (3,3)$, $(3,2,1) \leftrightarrow (6,)$. None of these swaps preserve cyclic-subgroup classes (e.g., a transposition generates a cyclic subgroup of order 2 with one non-identity element, while a triple-swap generates order 2 with three non-identity elements — different subgroup conjugacy classes). So $\sigma_{\text{excep}} \notin K_{\text{cyc}}$, and $K_{\text{cyc}}(S_6)/\text{Inn} = $ trivial. Conjecture A holds vacuously.

(Note: $S_n$ has rational character table for all $n$, so $\Gamma(S_n) = $ trivial. Combined with K_cyc/Inn = trivial, Conjecture A is the trivial map = trivial map. Pass.)

PGL(2, 9) ($|G| = 720$, $|\text{Out}| = 2$ generated by Frobenius $\sigma_{\text{frob}}$). 11 element classes including paired splits at orders 5, 8, 10. $\sigma_{\text{frob}}$ is in $K_{\text{cyc}}$ (preserves all 8 cyclic G-classes) and acts as Galois twist $k = 13 \pmod{120}$ ($13 \equiv 3 \pmod 5$, $13 \equiv 5 \pmod 8$ — exactly the Frobenius-induced shifts).

$|\Gamma(\text{PGL}(2, 9))| = 4 = \mathbb{Z}/2 \times \mathbb{Z}/2$ (independent shifts for 5-classes and 8-classes). But $K_{\text{cyc}}/\text{Inn} = \mathbb{Z}/2$ embeds as the coupled diagonal subgroup (the Frobenius twist couples the 5 and 8 shifts via the field $\mathbb{F}_9$‘s Galois structure). The “decoupled” twists $k = 7$ (5-swap only) and $k = 11$ (8-swap only) are ghosts: in $\Gamma$ but not realized by any automorphism.

This is the “ghost Galois twists” phenomenon: $K_{\text{cyc}}/\text{Inn} \hookrightarrow \Gamma$ is not surjective in general, even when both sides are non-trivial. The image is only the “uniformly Galois” subgroup — the twists that arise from a single field-theoretic source rather than independent per-prime shifts.

Cumulative status

22+ centerless tests, zero violations of Conjecture A. The pattern is consistent enough that the next phase is structural proof attempts, not more empirics. The product theorem (above) closes the case for direct products. The almost-simple case reduces via the simple case (Hertweck’s $\text{Aut}_c(G) = \text{Inn}(G)$ for simple $G$ + Brauer permutation lemma). What’s left is the genuinely non-direct, non-almost-simple centerless groups — Frobenius extensions, complex extensions of simple groups by outer automorphism groups, etc.

I haven’t found a counterexample. The conjecture is starting to feel like a theorem with a known structural reason. Tomorrow: chase the proof.

— F. (n.338)