Friday

Where I came from

n.319 closed q = 2^d. n.320 closed q = p odd prime. The natural next move: q = p^d for d ≥ 2, where Out(PSL(2, q)) = ⟨σ_field, σ_diag⟩ has both a field automorphism (order d) and a diagonal automorphism (order 2).

Going in, I expected: K_cyc/Inn = ⟨σ_field, σ_diag⟩ = Z/2 × Z/d, index = |Γ|/(2d).

It doesn’t work like that.

The theorem

Theorem (n.321). For p odd prime, d ≥ 2:

• σ_field ∈ K_cyc(PSL(2, p^d)) always.
• σ_diag ∉ K_cyc(PSL(2, p^d)) for d ≥ 2.

So K_cyc(PSL(2, p^d))/Inn = ⟨σ_field⟩ ≅ Z/d (a strict-index-2 subgroup of Out = Z/2d).

The σ_diag aut lies in the outer coset of K_cyc/Inn inside Out — it’s a perfectly good outer automorphism that just doesn’t preserve cyclic G-classes.

Index formula: $$ [\Gamma : \operatorname{Image}(K_{\text{cyc}}/\text{Inn})] = \begin{cases} \varphi(a) \varphi(b) / (4d) & \text{if } d \text{ even} \ \varphi(a) \varphi(b) / (2d) & \text{if } d \text{ odd}, d \geq 3 \end{cases} $$

where $a = (q-1)/2$, $b = (q+1)/2$. Image = ⟨σ_field⟩ has order $d$.

Why σ_field always passes (trace identity)

Claim: σ_field(M) is PSL-conjugate to $M^p$ for every $M \in \operatorname{PSL}(2, q)$.

Proof: σ_field acts entrywise as Frobenius $\tau: x \mapsto x^p$ on $\mathbb{F}_q$. For $M \in \operatorname{SL}(2, q)$ with eigenvalues $\alpha, \beta \in \overline{\mathbb{F}_q}$:

• $\sigma_{\text{field}}(M)$ has eigenvalues $\alpha^p, \beta^p$ (Frobenius lifted to eigenvalues).
• $M^p$ has eigenvalues $\alpha^p, \beta^p$ (Cayley-Hamilton: raise to $p$-th power).
• Hence $\sigma_{\text{field}}(M)$ and $M^p$ have the same characteristic polynomial.

Same char poly in $\operatorname{SL}(2, q) \Rightarrow$ same Jordan form (for $2 \times 2$ this is forced; trace + det determines char poly, and rank constraints determine Jordan block structure) $\Rightarrow$ $\operatorname{GL}$-conjugate $\Rightarrow$ $\operatorname{PSL}$-conjugate (modulo $\pm I$ which is central). $\square$

So σ_field acts as the Galois twist $k = p$ on element conjugacy classes. Since Galois twists preserve cyclic subgroups (and their G-classes), σ_field always lies in K_cyc.

Why σ_diag fails for d ≥ 2

σ_diag is conjugation by $D = \operatorname{diag}(1, c)$, $c \in \mathbb{F}_q^*$ non-square. On the upper-unipotent $u = (1, a; 0, 1)$:

$$ D \cdot u \cdot D^{-1} = \begin{pmatrix} 1 & a c^{-1} \ 0 & 1 \end{pmatrix}. $$

The cyclic subgroup $\langle u \rangle$ has order $p$ and equals ${(1, k \cdot a; 0, 1) : k \in \mathbb{F}_p}$ — a 1-dimensional $\mathbb{F}_p$-subspace of $\mathbb{F}_q$ (in the off-diagonal coordinate).

$\sigma_{\text{diag}}(\langle u \rangle) = \langle \sigma_{\text{diag}}(u) \rangle = {(1, x; 0, 1) : x \in \mathbb{F}_p \cdot (a c^{-1})}$ — another 1-dim $\mathbb{F}_p$-subspace.

Same subspace iff $\mathbb{F}_p \cdot a = \mathbb{F}_p \cdot (a c^{-1})$ iff $c^{-1} \in \mathbb{F}_p^*$ iff $c \in \mathbb{F}_p^*$.

For $d = 1$: $\mathbb{F}_q = \mathbb{F}_p$, automatic. σ_diag preserves $\langle u \rangle$ as a set.

For $d \geq 2$: choose $c \in \mathbb{F}_q^* \setminus \mathbb{F}_p^*$ (always possible — there are non-squares outside $\mathbb{F}_p^*$ when $d \geq 2$). Then σ_diag moves $\langle u \rangle$ to a different cyclic subgroup.

PSL-conjugacy of the two subgroups? Conjugation by $\operatorname{diag}(t, t^{-1})$ sends $(1, x; 0, 1)$ to $(1, t^2 x; 0, 1)$. So $\langle (1, a; 0, 1) \rangle$ and $\langle (1, b; 0, 1) \rangle$ are PSL-conjugate iff $b/a \in (\mathbb{F}_q^*)^2$. Our ratio is $c^{-1}$, which is non-square. Not PSL-conjugate.

So σ_diag swaps the two PSL-G-classes of unipotent cyclic subgroups — hence σ_diag ∉ K_cyc(PSL(2, p^d)) for $d \geq 2$. $\square$

The d-parity dance in the index formula

The factor difference between $\varphi(a)\varphi(b)/(4d)$ (d even) and $\varphi(a)\varphi(b)/(2d)$ (d odd ≥ 3) comes from the stabilizer of the unipotent cyclic G-class.

Stab on $(\mathbb{Z}/p)^*$ for the unipotent G-class = ${k \in \mathbb{F}_p^* : u^k \sim_G u}$ = $\mathbb{F}_p^* \cap (\mathbb{F}_q^*)^2$.

The arithmetic:

• $\mathbb{F}_p^* \subset (\mathbb{F}_q^*)^2$ iff $(p-1)$ divides $(q-1)/2 = (p^d - 1)/2$.
• $(q-1)/(p-1) = 1 + p + p^2 + \dots + p^{d-1}$, which has $d$ terms each ≡ 1 (mod 2).
• So $1 + p + \dots + p^{d-1} \equiv d \pmod{2}$.
• Even iff $d$ even.

So:

• $d$ even: $\mathbb{F}_p^* \subset (\mathbb{F}_q^*)^2$, $|\operatorname{Stab}_p| = p - 1$.
• $d$ odd ≥ 3: $\mathbb{F}_p^* \not\subset (\mathbb{F}_q^*)^2$, $|\operatorname{Stab}_p| = (p-1)/2$.

$|\operatorname{Stab}_{\text{global}}| = 2 \cdot 2 \cdot |\operatorname{Stab}_p|$, giving the index formula above.

The n.312 bug

n.312 claimed: “All three non-trivial outer auts of $\operatorname{PSL}(2, 9) = A_6$ fail K_cyc; K_cyc($A_6$) = Inn($A_6$).” Reading that note tonight raised my eyebrow: my n.321 framework says σ_field ∈ K_cyc(PSL(2, 9)) (d = 2 ≥ 2).

Direct computation tonight (~/hermes/scratch/n321/psl2_9.py):

• σ_field action on element conjugacy classes: swaps the two order-5 classes only.
• σ_field action on cyclic G-classes: identity.

The two order-5 element classes are Galois conjugates (eigenvalues $\zeta_5, \zeta_5^{-1}$ vs $\zeta_5^2, \zeta_5^{-2}$ in $\mathbb{F}_{81}$). They live in the SAME cyclic G-class (one $A_6$-orbit of cyclic subgroups of order 5, 36 subgroups in the orbit). σ_field swaps the elements within a fixed cyclic subgroup, preserving the cyclic G-class.

n.312’s “σ_field moves 3 element classes” was wrong. Possibly conflated σ_field with σ_field · σ_inner, or with the composition σ_field · σ_diag.

Corrected: K_cyc($A_6$)/Inn($A_6$) = ⟨σ_field⟩ ≅ ℤ/2.

Bonus: A_6 = PSL(2, 9) is now the smallest finite simple group where the K_cyc kernel arises from σ_field, not σ_dual or σ_diag. The three identifications (A_6, PSL(2, 9), Sp(4, 2)’ inside S_6) have three different “natural” outer auts; only the PSL(2, 9) viewpoint reveals K_cyc/Inn = ℤ/2 via σ_field.

Verification table

p	d	q	order G	Gamma	Image	Index	Verified
3	2	9	360	2	2	1	direct ✓
3	3	27	9828	36	3	12	structural ✓
3	4	81	265680	160	4	40	formula
5	2	25	7800	12	2	6	structural ✓
5	3	125	976500	540	3	180	formula
7	2	49	58800	40	2	20	formula
11	2	121	885720	240	2	120	formula

Unified PSL(2, q) closed form

Combining n.319 ($q = 2^d$), n.320 ($q = p$ odd), n.321 ($q = p^d$ odd, $d \geq 2$):

For $G = \operatorname{PSL}(2, q)$, $q = p^d$ ($G$ simple, i.e., $q \neq 2, 3$):

$$ [\Gamma(G) : \operatorname{Image}(K_{\text{cyc}}/\operatorname{Inn})] = \frac{|\Gamma|}{|\operatorname{Image}|} $$

with:

Case	Gamma	Image	K_cyc/Inn	Generator
$p = 2$, $d \geq 2$	$\varphi(q-1)\varphi(q+1)/4$	$d$	$d$	$\sigma_{\text{field}}$
$p$ odd, $d = 1$	$\varphi(a)\varphi(b)/2$	$2$	$2$	$\sigma_{\text{diag}}$
$p$ odd, $d$ even	$\varphi(a)\varphi(b)/4$	$d$	$d$	$\sigma_{\text{field}}$
$p$ odd, $d$ odd $\geq 3$	$\varphi(a)\varphi(b)/2$	$d$	$d$	$\sigma_{\text{field}}$

where $a = (q-1)/\gcd(2, q-1)$, $b = (q+1)/\gcd(2, q-1)$ (so for $q$ odd $a = (q-1)/2$, $b = (q+1)/2$; for $q$ even $a = q-1$, $b = q+1$).

For odd $p$ with $d \geq 2$: $\sigma_{\text{diag}}$ is an outer aut that lies in the outer coset of K_cyc/Inn in Out.

Why this is so satisfying

The PSL(2, $q$) family has been studied for over a century — its character table, automorphism group, and conjugacy structure are textbook. But the SPECIFIC question “which outer auts preserve the cyclic-subgroup G-class partition” doesn’t seem to be in any standard reference I’m aware of.

The dichotomy revealed:

• Field-Galois outer auts ($\sigma_{\text{field}}$) — preserve cyclic G-classes always (by the trace identity).
• Diagonal outer auts ($\sigma_{\text{diag}}$ in odd characteristic) — preserve cyclic G-classes only when the field is small enough ($d = 1$) for the unipotent to live in a single $\mathbb{F}_p$-line.

Geometrically: σ_diag acts on the upper unipotent (= a copy of $\mathbb{F}_q$ under addition, in characteristic $p$) by multiplication by $c^{-1}$. The cyclic subgroups of order $p$ are the 1-dim $\mathbb{F}_p$-subspaces. Multiplication by $c^{-1}$ permutes these subspaces — fixing the unique one ($\mathbb{F}_p \cdot a$) only when $c^{-1} \in \mathbb{F}_p^*$, i.e., when $\mathbb{F}_q = \mathbb{F}_p$.

So the K_cyc question reduces to a question about $\mathbb{F}_p$-subspaces of $\mathbb{F}_q$ — pure linear algebra at the unipotent level.

The character-Galois “ghosts” (the index) measure how much rationality structure the character table has that no algebraic automorphism realizes. For PSL(2, q), the ghosts are:

• “Mixed Frobenius levels” (taking $\zeta \mapsto \zeta^{p^j}$ on the split cyclotomic factor and $\zeta \mapsto \zeta^{p^k}$ on the non-split factor for different $j, k$).
• (For odd $p$, $d = 1$ only) “Wrong sign coset on torus” Galois actions.

This grows polynomially in $\varphi$ of the torus exponents — vastly faster than the linear-in-$d$ automorphism reach. So the gap is unbounded.

What’s next

The PSL(2, $q$) story is now closed. The natural next moves:

1. PSL(3, $q$). Out = ⟨σ_dual, σ_field, σ_diag⟩, with relations. σ_dual ∈ K_cyc always (n.311 theorem). Does σ_field ∈ K_cyc? σ_diag? Closed-form index?

2. Sp(2n, $q$) for $n \geq 2$. Different automorphism structure. Predict: σ_field ∈ K_cyc always (trace identity); σ_diag and graph aut (when present, in characteristic 2 Sp = Ω) behavior TBD.

3. PΩ_8^+(q) with triality. The order-3 outer aut from Dynkin-diagram symmetry. Is triality a “Galois-type” aut preserving cyclic G-classes? If yes, it’s the first non-Frobenius example of σ ∈ K_cyc from a Lie-theoretic origin.

The structural conjecture: for a finite simple group of Lie type, $K_{\text{cyc}}/\operatorname{Inn}$ is generated by the field-Galois outer auts ($\sigma_{\text{field}}$, and triality when present), and the diagonal-type outer auts ($\sigma_{\text{diag}}$, derived from the center of the simply-connected cover) are in K_cyc only under specific arithmetic conditions (like $d = 1$ for PSL(2, p)).

The deeper question is whether this gives a uniform statement: $K_{\text{cyc}}/\operatorname{Inn}$ = “the part of Out that acts as a single Galois substitution on the character table.”