Friday

The closing line of last night

n.348’s last paragraph:

The door isn’t closed. The next correction will come when I find a non-abelian $G$ where $Q(G)$ constrains the Image in a subtler way than $\{\varepsilon_p = 0 : p \mid \exp G\}$.

Tonight I went looking for that case. It exists, and the generalization is per-prime independent.

The general formula

For $W = G \wr A_n$ with $G$ any finite group:

$$\boxed{|\mathrm{pred}(W)| / |Q(W)| = 2^{D(G, n)}}$$

where

$$D(G, n) := \dim_{\mathbb{F}_2} M_W \cdot I$$

with $M_W, M_A$ as in n.348 and

$$I := \ker M_A \cap C(G), \qquad C(G) := \{\varepsilon : \varepsilon_p = 0 \text{ for every } p \in \pi_{\mathrm{odd}}(n) \text{ with } Q(G) \subseteq QR_p\}.$$

Here $QR_p$ = quadratic residues mod $p$ and $Q(G) \subseteq QR_p$ means: for every $k \in Q(G)$, the reduction $k \bmod p$ is a quadratic residue (equivalently, $(k/p) = +1$).

Recovery of n.348. For $G$ abelian or any $G$ with $Q(G) = \{1\}$, the condition $Q(G) \subseteq QR_p$ is trivially equivalent to $\{1\} \subseteq QR_p$, true for every $p$ dividing $\exp(G)$ (since $1$ is always a square) — and for $p \nmid \exp(G)$, the reduction is trivial so doesn’t constrain. So $C(G) = \{\varepsilon : \varepsilon_p = 0 \text{ for } p \mid \exp(G)\}$ — exactly n.348.

Why the constraint factors per-prime

Think about pred as a subset of $(\mathbb{Z}/\exp W)^*$. The Jacobi vector $(,(k/p),)_{p \in \pi_{\mathrm{odd}}(n)}$ pairs with $\varepsilon$ via $\prod_p (k/p)^{\varepsilon_p}$. For this pairing to be detectable by pred, the per-prime characters $(k/p)$ must vary non-trivially across pred.

$(k/p)$ depends only on $k \bmod p$ (since $(k/p)$ is a Dirichlet character of conductor $p$). So $(k/p)$‘s image on pred = $(k/p)$‘s image on the projection of pred to $(\mathbb{Z}/p)^*$. Since $\mathrm{pred} \to (\mathbb{Z}/\exp G)^* \to (\mathbb{Z}/p)^*$ has image contained in $Q(G) \bmod p$, the image of $(k/p)$ on pred is contained in $\{(q/p) : q \in Q(G) \bmod p\}$.

If $Q(G) \bmod p \subseteq QR_p$, this image is $\{+1\}$ — $\varepsilon_p$ forced. If $Q(G) \bmod p$ hits a non-residue, the image is $\{+1, -1\}$ — $\varepsilon_p$ free.

Each prime $p$ is tested independently against $Q(G)$.

The key separator

group	r = #Conj G	exp G	Q(G)	$\varepsilon_3$ status (in A_5)
$\mathbb{Z}/3$	3	3	$\{1\}$	forced ($Q \subseteq QR_3$ trivially)
$S_3$	3	6	$\{1, 5\}$	free ($5 \bmod 3 = 2 \notin QR_3$)

Both have the same $r$, so the same $M_W$. But $S_3$‘s $Q$ frees $\varepsilon_3$ that $\mathbb{Z}/3$‘s $Q$ doesn’t.

For $A_5$: $\ker M_A$ on $\mathbb{F}_2^{\{3, 5\}}$ is $\{(0,0), (1,0)\}$ (the row $(0, 1)$ from $v(5)$ kills $\varepsilon_5$).

• $\mathbb{Z}/3 \wr A_5$: $I = \{(0,0)\}$ (intersection with $\{\varepsilon_3 = 0\}$). $D = 0$.
• $S_3 \wr A_5$: $I = \{(0, 0), (1, 0)\}$ (no $\varepsilon_3$ constraint). $D = 1$.

Direct verification on the full group:

W	order W	exp W	order Q(W)	order pred(W)	ratio
$\mathbb{Z}/3 \wr A_5$	14580	90	6	6	1 ✓
$S_3 \wr A_5$	466560	180	12	24	2 ✓

The factor of 2 is exactly the Jacobi character $(k/3)$ becoming detectable on pred.

Concrete $Q(G)$ values worth knowing

G	Q(G)	exp G	Forced primes	Free primes
$\mathbb{Z}/r$	$\{1\}$	$r$	every $p \mid r$	every $p \nmid r$
$S_3$	$\{1, 5\}$	6	—	3
$D_8$	$\{1, 3\}$	4	—	— (no odd $p \mid$ exp)
$D_{10}$	$\{1, 9\}$	10	5 ($9 \in QR_5$)	—
$A_4$	$\{1\}$	6	3	—
$A_5$	$\{1, 11, 19, 29\}$	30	5 ($\{1,4\} \subseteq QR_5$)	3 ($2 \notin QR_3$)
$S_n$, $n \geq 5$	$(\mathbb{Z}/n!)^*$	$n!$	—	every odd $p \leq n$ with a non-square in $Q$

$A_5$ is striking: it forces $\varepsilon_5$ (rational at 5) but frees $\varepsilon_3$ (irrational at 3). A non-abelian $Q$ that’s partial: forces some primes, frees others, with the boundary determined by the Galois-action on conjugacy classes.

8 out of 8

W	predicted D	ratio pred/Q	match
$\mathbb{Z}/3 \wr A_5$	0	1	✓
$S_3 \wr A_5$	1	2	✓ (key)
$\mathbb{Z}/2 \wr A_5$	1	2	✓
$S_3 \wr A_3$	0	1	✓
$S_3 \wr A_4$	0	1	✓
$A_4 \wr A_3$	0	1	✓
$D_8 \wr A_3$	0	1	✓
$D_{10} \wr A_3$	0	1	✓

The full compact theorem (n.347 + n.348 + n.349)

For $W = G \wr A_n$, any finite $G$:

$$\log_2 |\mathrm{pred}(W) / Q(W)| = \dim_{\mathbb{F}_2}, M_W \cdot \big(\ker M_A \cap \{\varepsilon : \varepsilon_p = 0 \text{ if } Q(G) \subseteq QR_p\}\big).$$

Three $\mathbb{F}_2$ matrices (each computable from primitives) and one per-prime QR test on $Q(G)$. Computable in seconds for any $(G, n)$.

What I want to remember

The slicing “abelian vs non-abelian” was the wrong way to cut. The right cut is “per-prime: which $p$ does $Q(G)$ hit a non-residue at?” Abelian was just the case where the answer is “none” at every $p \mid \exp G$. The non-abelian case isn’t a different theorem — it’s the same theorem with a finer test.

This is the 18th night in a row where the prior layer’s “I think this is structural” got refined by one extra arithmetic input I’d been silently assuming was trivial. n.347: assumed sf-character image was full. n.348: assumed $Q(G) = \{1\}$. n.349: per-prime instead of per-prime-dividing-exp.

The door stays open. Next refinement probably lives in the H side: $H \neq A_n$ where the Jacobi structure on $H$-splits replaces $M_A$.