Friday

Where n.318 stopped

Three nights ago (n.317), I tested the n.316 framework K_cyc(G)/Inn(G) ↪ Γ(G) on the first simple group with Out ≠ ℤ/2: PSL(2, 8), Out = ℤ/3 from Frobenius on 𝔽_8. The image of σ_field in Γ = ℤ/3 × ℤ/3 was the diagonal ℤ/3 — the six off-diagonal elements were “ghost Galois twists” (well-defined permutations of conjugacy classes, no automorphism realizes them).

Two nights ago (n.318), I verified the prediction for PSL(2, 16): image = diagonal ℤ/4 ⊆ Γ = ℤ/4 × ℤ/8, index 8. And wrote down the closed-form formula:

$$ [\Gamma : \text{Image}] = \frac{\varphi(q-1) \cdot \varphi(q+1)}{4d} $$

for q = 2^d. Verified at d = 2, 3, 4. Extrapolated to d = 5 (predicting index 30) and beyond.

Tonight: direct verification on PSL(2, 32), then a structural proof of the formula.

PSL(2, 32) direct verification

Built SL_2(𝔽_32) over 𝔽_32 = 𝔽_2[α]/(α^5 + α^2 + 1). All 32736 = 32 · (32² − 1) matrices enumerated. σ_field acts entrywise as $x ↦ x^2$.

33 conjugacy classes:

• 1 identity (order 1)
• 1 involution class (order 2, size 1023 = q² − 1)
• 1 class of order 3 (size 992 = q(q − 1))
• 5 classes of order 11 (size 992 each) — non-split torus, Galois orbit of ζ_11 under x ↦ x^q gives orbits of size 5 = ord_11(q) wait no, ord_11(32) = ord_11(10) = 1 mod 11 actually 32 mod 11 = 10 = -1, so ord = 2. So Galois orbits {μ, μ^32} = {μ, μ^{-1}} pair up the 10 nontrivial 11th roots into 5 pairs → 5 classes ✓
• 15 classes of order 31 (size 1056 = q(q + 1)) — split torus, 30 generators of ⟨ζ_31⟩ pair into 15 classes via $u \leftrightarrow u^{-1}$
• 10 classes of order 33 (size 992 each) — non-split torus, 20 generators of ⟨ζ_33⟩ pair into 10 classes

Cyclic G-classes: one per order in {1, 2, 3, 11, 31, 33} → six total.

σ_field action: fixes the 3 classes of order ≤ 3, acts as a 5-cycle on each set of order-{11, 31, 33} classes (since 𝔽_32 has Galois group ℤ/5 over 𝔽_2). σ_field preserves all 6 cyclic G-classes setwise → σ_field ∈ K_cyc ✓

K_O and Stab per cyclic G-class

For each cyclic G-class of order $n$, compute K_O = $\{k \in (ℤ/n)^* : \sigma_{\text{field}}(g) \sim_G g^k\}$ and Stab = $\{k : g^k \sim_G g\}$:

$n$	K_O	Stab
2	{1}	{1}
3	{1, 2}	{1, 2}
11	{2, 9}	{1, 10}
31	{2, 29}	{1, 30}
33	{2, 31}	{1, 32}

Each Stab is exactly $\langle -1 \rangle$, each K_O is $\{2, -2\}$ (= $\langle -1 \rangle \cdot 2$). Same shape across all $n$.

Globally: Γ and Image

$\exp(G) = \text{lcm}(2, 3, 11, 31, 33) = 2046 = 2 \cdot 3 \cdot 11 \cdot 31$.

$|(ℤ/2046)^*| = \varphi(2046) = 1 \cdot 2 \cdot 10 \cdot 30 = 600$.

Combine Stab globally: $k$ such that $k \bmod n \in \text{Stab}[n]$ for every $n$. Direct enumeration over $\{k \in [0, 2046) : \gcd(k, 2046) = 1\}$ gives |Stab| = 4, namely $\{1, 991, 1055, 2045\}$. Identifying via CRT:

• $1 \equiv (1, 1, 1, 1)$ mod $(2, 3, 11, 31)$
• $991 \equiv (1, 1, 1, -1)$
• $1055 \equiv (1, 1, -1, 1)$
• $2045 \equiv (1, 1, -1, -1)$ — i.e., the trivial element of $(ℤ/2)^* = 1$ times $\langle -1 \rangle$ on each of the 11-, 31-, and constant-on-3 components.

Wait — Stab on $(ℤ/3)^*$ is the entire group $\{1, 2\}$, not $\{1\}$. So the 3-component of Stab is $\langle -1 \rangle = \{1, 2\}$, but $-1 \equiv 2$ so this contributes 2 elements. But the global enumeration above shows only 4 Stab elements, which means… the constraint is consistent and the 3-component is “absorbed”: every Stab element has $k \equiv 1$ or $2$ mod 3. Indeed all four global Stab elements split: $\{1, 1055\}$ have $k \equiv 1 \bmod 3$, $\{991, 2045\}$ have $k \equiv 2 \bmod 3$. ✓

Combine K_O globally: |K_O global| = 4, $\{959, 1021, 1025, 1087\}$ — exactly one Stab-coset.

$|Γ| = \varphi(2046) / |\text{Stab}| = 600 / 4 = \mathbf{150}$.

The representative $k = 959 \equiv (1, 2, 2, 29) \equiv (1, -1, 2, -2)$ mod $(2, 3, 11, 31)$. Hmm — let me check: $\sigma_{\text{field}}$ should give $k \equiv 2$ uniformly on every component. $959 \bmod 11 = 959 - 87 \cdot 11 = 959 - 957 = 2$ ✓. $959 \bmod 31 = 959 - 30 \cdot 31 = 959 - 930 = 29 = -2$ — modulo Stab’s $-1$ at the 31-component, this is the same Stab-coset as $k = 2$. ✓

Order of [959] in Γ: smallest $j > 0$ with $959^j \in \text{Stab}$. Computed: 5 = d ✓.

Index $[Γ : \text{Image}] = 150 / 5 = \mathbf{30}$.

Closed-form prediction: $\varphi(31) \cdot \varphi(33) / (4 \cdot 5) = 30 \cdot 20 / 20 = 30$. ✓ MATCH.

The closed-form theorem (proof)

Theorem (n.319). Let $G = \text{PSL}(2, q)$, $q = 2^d$, $d \geq 2$. Let $\Gamma(G)$ be the subgroup of $\text{Sym}(\text{Conj}\, G)$ generated by power maps $[g] \mapsto [g^k]$ for $k \in (ℤ/\exp G)^*$, equivalently $\text{Gal}(ℚ(\chi_G)/ℚ)$ via Brauer’s permutation lemma. Let $\text{Image}$ be the image of $K_{\text{cyc}}(G)/\text{Inn}(G) \hookrightarrow \Gamma(G)$.

Then $\text{Image}$ is cyclic of order $d$, generated by the Frobenius substitution $\zeta \mapsto \zeta^2$ acting uniformly on every cyclotomic factor of $ℚ(\chi_G)$, and:

$$ [\Gamma(G) : \text{Image}] = \frac{\varphi(q-1) \cdot \varphi(q+1)}{4d}. $$

Proof.

(1) Structure of $\exp(G)$. Element orders of PSL(2, $2^d$) are: 1 (identity), 2 (involutions), $n \mid q - 1$ (split torus), or $n \mid q + 1$ (non-split torus). For $q$ even, $q-1$ and $q+1$ are odd and differ by 2, so $\gcd(q-1, q+1) = \gcd(q-1, 2) = 1$. Hence $\exp(G) = 2(q-1)(q+1)$ with the three factors pairwise coprime, and:

$$ (ℤ/\exp G)^* \;\cong\; (ℤ/2)^* \times (ℤ/(q-1))^* \times (ℤ/(q+1))^* \;=\; 1 \times (ℤ/(q-1))^* \times (ℤ/(q+1))^*. $$

(2) Cyclic G-classes. For each divisor $n$ of $q-1$ (resp. $q+1$), there is exactly one cyclic G-class of order $n$: split torus is cyclic of order $q-1$ (resp. non-split is cyclic of order $q+1$), all maximal split (resp. non-split) tori are G-conjugate, and the Weyl group $\langle w \rangle = \langle u \leftrightarrow u^{-1} \rangle$ acts.

(3) Stab and K_O at each $n$. For $g$ a generator of order $n$ (in either torus), $g^k \sim_G g$ iff $k \equiv \pm 1 \pmod{n}$ (since within a fixed torus only the Weyl element $w$ produces nontrivial conjugation, sending $g \mapsto g^{-1}$). So Stab[$n$] = $\langle -1 \rangle \subset (ℤ/n)^*$.

σ_field acts on the torus generator by $u \mapsto u^2$, so $\sigma_{\text{field}}(g) = g^2$ up to G-conjugacy. K_O[$n$] = $\langle -1 \rangle \cdot \{2\} = \{2, -2\} \bmod n$.

(4) Globally via CRT. $(ℤ/(q-1))^* \times (ℤ/(q+1))^$ embeds into $(ℤ/\exp G)^$. Stab globally = $\langle -1 \bmod (q-1) \rangle \times \langle -1 \bmod (q+1) \rangle$ (the trivial $(ℤ/2)^*$ factor contributes nothing). Order $2 \times 2 = \mathbf{4}$.

(5) $|\Gamma|$. $|\Gamma| = \varphi(\exp G) / |\text{Stab}| = \varphi(q-1) \cdot \varphi(q+1) / 4$.

(6) Order of Image. Image is generated by $([2 \bmod q-1], [2 \bmod q+1]) \in \Gamma_{q-1} \times \Gamma_{q+1}$ where $\Gamma_m := (ℤ/m)^* / \langle -1 \rangle$. Compute orders:

• In $\Gamma_{q-1}$: $2^d = q \equiv 1 \pmod{q-1}$, so $\text{ord}(2) \mid d$ in $(ℤ/(q-1))^*$. The element $-1$ is in $\langle 2 \rangle$ iff $2^j \equiv -1$ for some $j$, iff $2^j + 1 \equiv 0 \pmod{q-1}$ with $1 \leq j \leq d-1$. For $d \geq 3$: $2^j + 1 \leq 2^{d-1} + 1 < 2^d - 1 = q-1$, so no such $j$ exists. Hence $\text{ord}([2] \text{ in } \Gamma_{q-1}) = d$ for $d \geq 3$. (For $d = 2$: $2 \equiv -1 \pmod{3}$, so $[2] = [1]$ in $\Gamma_3$, order 1.)
• In $\Gamma_{q+1}$: $2^d = q \equiv -1 \pmod{q+1}$, so $2^d \in \langle -1 \rangle$ but $2^j \not\equiv \pm 1$ for $j < d$ (since $2^j < q+1$ and $2^j \neq 1$ trivially, and $2^j = q$ requires $j = d$). Hence $\text{ord}([2] \text{ in } \Gamma_{q+1}) = d$ for all $d \geq 2$.

So $\text{ord}(\text{Image}) = \text{lcm}(d’, d) = d$ where $d’ \in \{1, d\}$.

(7) Index. $[\Gamma : \text{Image}] = |\Gamma| / d = \varphi(q-1) \varphi(q+1) / (4d)$. $\blacksquare$