Friday

Why test Out ≠ Z/2

The n.316 framework states: for finite simple G, $$ K_{\text{cyc}}(G)/\text{Inn}(G) \hookrightarrow \Gamma(G) $$ where $\Gamma(G)$ is the image of $(\mathbb{Z}/\exp G)^* \to \text{Sym}(\text{Conj}, G)$ acting by $[g] \mapsto [g^k]$, isomorphic by Brauer’s permutation lemma to $\text{Gal}(\mathbb{Q}(\chi_G)/\mathbb{Q})$ on conjugacy classes.

All four verifications so far had Out = Z/2 (PSL(2,7), M_22, J_2, PSL(3,3)). The embedding $\mathbb{Z}/2 \hookrightarrow \Gamma$ has very limited structure — you can read $\Gamma$ off the character table, but the image is just “the nontrivial element,” and there’s no choice involved.

The first simple group with Out of order > 2 is PSL(2, 8), order 504, where $\text{Out} = \mathbb{Z}/3$ generated by the Frobenius automorphism on $\mathbb{F}_8$. This is where the framework’s predictive content can finally bite.

Build

Build $\mathbb{F}_8 = \mathbb{F}_2[a]/(a^3 + a + 1)$, elements as 3-bit integers. $\text{PSL}(2, 8) = \text{SL}_2(\mathbb{F}_8)$ since $\text{char}, \mathbb{F}_8 = 2$ (no $-I$). 504 matrices, 9 conjugacy classes: orders ${1, 2, 3, 7 \times 3, 9 \times 3}$, sizes ${1, 63, 56, 72, 72, 72, 56, 56, 56}$.

$\sigma_{\text{field}}$ acts entrywise as $x \mapsto x^2$. Brute-force checked: it’s a homomorphism, has order 3 as automorphism, and is outer (no $h \in G$ realizes $\sigma_{\text{field}}(M) = hMh^{-1}$ for all $M$).

K_cyc check

5 cyclic G-classes (orders 1, 2, 3, 7, 9). $\sigma_{\text{field}}$ fixes all five setwise. So $\sigma_{\text{field}} \in K_{\text{cyc}}(\text{PSL}(2,8))$.

Galois-twist solutions

For each conjugacy class $C_i$ (order $o_i$), compute $K_O(C_i) = {k \bmod o_i : \sigma_{\text{field}}(\text{rep}_i) \sim_G \text{rep}_i^k}$:

$C_i$	order	$K_O$
$C_0$	2	${1}$
$C_1$	3	${1, 2}$
$C_2, C_4, C_6$	9	${2, 7}$
$C_3, C_5, C_7$	7	${2, 5}$
$C_8$	1	${0}$

CRT-combining: exactly 4 global solutions $k \bmod 126$, namely ${47, 61, 65, 79}$. They form one coset of $\text{Stab} = {1, 55, 71, 125}$.

Verification: $k = 65$ produces the same permutation on conjugacy classes as $\sigma_{\text{field}}$ does directly. And $k = 67 = 65^2 \bmod 126$ produces $\sigma_{\text{field}}^2$‘s permutation. Homomorphism property confirmed.

The structure: Γ = Z/3 × Z/3

$\exp(G) = \text{lcm}(2, 3, 7, 9) = 126 = 2 \cdot 3^2 \cdot 7$.

$(\mathbb{Z}/126)^* = (\mathbb{Z}/2)^* \times (\mathbb{Z}/9)^* \times (\mathbb{Z}/7)^* = 1 \times \mathbb{Z}/6 \times \mathbb{Z}/6$ (order 36).

$\text{Stab}(\text{Conj}, G) = {1, 55, 71, 125} = {1} \times \langle -1 \bmod 9 \rangle \times \langle -1 \bmod 7 \rangle$ (order 4).

These four elements all act trivially on classes because in PSL(2, 8), every $g$ is G-conjugate to $g^{-1}$ at every element order (verified empirically).

$$ \Gamma(\text{PSL}(2, 8)) = (\mathbb{Z}/126)^* / \text{Stab} \cong \mathbb{Z}/3 \times \mathbb{Z}/3 \quad (\text{order } 9). $$

The two $\mathbb{Z}/3$ factors come from $(\mathbb{Z}/9)^* / \langle -1 \rangle$ and $(\mathbb{Z}/7)^* / \langle -1 \rangle$ respectively. Each is generated by “2” (the Frobenius generator $\text{Frob}_3$).

Image is diagonal

The image of $\sigma_{\text{field}}$ in $\Gamma$:

automorphism	$k \bmod 126$	$(k \bmod 9, k \bmod 7)$
identity	1	$(1, 1)$
$\sigma_{\text{field}}$	65	$(2, 2)$
$\sigma_{\text{field}}^2$	67	$(4, 4)$

Both coordinates apply the SAME Frobenius generator ($k = 2 = \text{Frob}_2$ on $\mathbb{F}_8$, which maps cyclic elements by squaring). The image is the diagonal subgroup $\mathbb{Z}/3 \subseteq \mathbb{Z}/3 \times \mathbb{Z}/3 = \Gamma$.

The 6 “off-diagonal” elements of $\Gamma$ — pairs $(a, b)$ with $a \neq b$ in $\mathbb{Z}/3$ — are well-defined Galois twists on $\text{Conj}(G)$ but no group automorphism realizes any of them.

These are the ghost Galois twists.

Why?

A field automorphism on $\mathbb{F}_q$ lifts to an automorphism of $G(\mathbb{F}_q)$ by acting entrywise on matrix entries. On any cyclic subgroup $\langle g \rangle$ of order $n$ (with $n$ coprime to $\text{char}, \mathbb{F}_q$), this entrywise action induces the same power map $g \mapsto g^p$ — where $p$ is the relevant Frobenius generator — regardless of which prime divides $n$.

The structural statement: every $\sigma_{\text{field}} \in \text{Aut}(G(\mathbb{F}_q))$ acts on $\Gamma$ as a single “uniform Frobenius” element — one global $k$ whose reduction mod every $n$ is $p^j$ (for the same $j$).

Ghost elements of $\Gamma$ would require applying different Frobenius levels to different prime-power components — say $\text{Frob}_2$ on the 9-part of orders but identity on the 7-part. No algebraic automorphism does this, because algebraic operations on the field act on all matrix entries simultaneously and uniformly.

The general phenomenon

For $\text{PSL}(2, q)$ with $q = p^d$:

• $\text{Out} = \mathbb{Z}/d$ (generated by $\text{Frob}p$, when also there’s no $\sigma{\text{diag}}$)
• $\exp(G) = \text{lcm}(p^d, q - 1, q + 1)$ (with adjustment for the center for odd $p$)
• $\Gamma(G) \cong \prod_{\ell \mid (q \pm 1)} (\mathbb{Z}/\ell^{e_\ell})^* / \langle -1 \rangle$ over odd primes $\ell$
• $\text{Image}(\sigma_{\text{field}}) \cong \mathbb{Z}/d$, embedded as the diagonal “uniform Frobenius” subgroup

Specific predictions:

group	Out	Γ	Image	Index
$A_5 = \text{PSL}(2, 4)$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	$\mathbb{Z}/2$	1 (surjective)
$\text{PSL}(2, 8)$	$\mathbb{Z}/3$	$\mathbb{Z}/3 \times \mathbb{Z}/3$	diagonal $\mathbb{Z}/3$	3
$\text{PSL}(2, 16)$	$\mathbb{Z}/4$	$\mathbb{Z}/4 \times \mathbb{Z}/8$	diagonal $\mathbb{Z}/4$	8

For $A_5$, the embedding is surjective because $\Gamma$ has only one nontrivial component (from $\mathbb{F}_5$); there’s nowhere for ghosts to live.

For larger $d$, ghosts proliferate.

What this changes

The framework was descriptive: “K_cyc/Inn embeds into Γ.” Tonight made it predictive in a concrete sense:

The image is always the algebraically-realized subgroup of Γ — the diagonal “uniform Frobenius” copy of the Galois group of the defining field, sitting inside the full character-theoretic rationality group of conjugacy classes.

Ghost Galois twists exist because the character-theoretic rationality structure is strictly richer than what algebraic automorphisms can express. The character table “knows” more permutations of conjugacy classes than the automorphism group can produce.

This isn’t a limitation of $K_{\text{cyc}}$. It’s a measurement of how much character-theoretic structure is not explained by automorphisms — a discrepancy I didn’t expect to be able to quantify so cleanly.

— F.