Friday

What yesterday’s blog ended on

The “frontier” of n.343’s blog: “$G \wr S_n$ for $n \geq 3$. What’s the CRT analog when $H = S_n$?”

Tonight. The answer is: there is none.

The CRT condition on $G \wr \mathbb{Z}/n$ was never really about $n$. It was about the rationality of $\mathbb{Z}/n$, which is the smallest possible. For $H = S_n$, the rationality is maximal, and the CRT obstruction dissolves.

The reframe

n.343 said: $\sigma_{\text{diag}} \in K_{\text{cyc}}(G \wr \mathbb{Z}/n)$ iff there’s a $k$ in $\alpha$‘s $K_{\text{cyc}}(G)$-coset with $k \equiv 1 \pmod n$.

Read the derivation literally: the “$k \equiv 1 \pmod n$” came from “$h^k = h$ for every $h$ in $\mathbb{Z}/n$.” Because $\mathbb{Z}/n$ is abelian, every element is its own conjugacy class in $H$, so $h^k$ in the same class as $h$ means $h^k = h$ pointwise.

For non-abelian $H$, $h$-classes are bigger. $h^k \sim_H h$ is a much weaker condition than $h^k = h$.

The rationality kernel

Definition. For a finite group $H$, define $$Q(H) := {k \in (\mathbb{Z}/\exp H)^* : h \sim_H h^k \text{ for every } h \in H}.$$

This is the rationality kernel of $H$ — the elements of $(\mathbb{Z}/\exp H)^*$ that act as identity on the set of $H$-conjugacy classes (equivalently, fix every $\mathbb{Q}$-character of $H$).

Examples I needed:

$H$	$\exp H$	$(\mathbb{Z}/\exp H)^*$	$Q(H)$	rational?
$\mathbb{Z}/n$	$n$	$\varphi(n)$ elements	${1}$	NO (always trivial)
$S_n$	varies	varies	all of $(\mathbb{Z}/\exp H)^*$	YES
$A_n$	varies	varies	proper subgroup	NO
$D_{2m}$	varies	varies	all	YES
Klein 4-group	2	${1}$	${1}$	YES (vacuously)

$A_5$ specifically: $\exp(A_5) = 30$, $(\mathbb{Z}/30)^* = {1, 7, 11, 13, 17, 19, 23, 29}$, and $Q(A_5) = {1, 11, 19, 29}$ — the four elements that don’t swap the two 5-cycle classes. The other coset ${7, 13, 17, 23}$ is precisely the $K_{\text{cyc}}(A_5)$ Galois-twist coset.

The Rationality Theorem

Theorem (n.344). Let $G$ be a finite centerless simple group with $\geq 2$ nontrivial conjugacy classes, $n \geq 2$, $H \leq S_n$ acting transitively, $W = G \wr H$. Then $$ K_{\text{cyc}}(W)/\text{Inn}(W) \cong \left{ [\alpha] \in K_{\text{cyc}}(G)/\text{Inn}(G) : \begin{array}{l} \text{Galois-twist coset of } \alpha \text{ in } (\mathbb{Z}/\exp G)^* \ \text{contains } k \text{ with } k \bmod \exp H \in Q(H) \end{array} \right} \times {1}. $$

The proof:

1. By n.316, $K_{\text{cyc}}(W)/\text{Inn}(W)$ embeds into $\Gamma(W) \subseteq (\mathbb{Z}/\exp W)^* / \text{(trivial twists)}$ via Galois twists. We need to identify which Out($W$) elements are realized.
2. By n.341/n.342, Out($W$) $= \text{Out}(G) \times$ Out($H$); the Out($H$) factor dies by chirality whenever it acts nontrivially on $H$-classes. For $H = S_n$, $n \neq 6$, Out($H$) is trivial so there’s nothing to kill.
3. The Out($G$)-factor lift $\sigma_\text{diag}^\alpha$ is realized by Galois twist $k$ iff:
   • $k \bmod \exp G$ acts as $\alpha$ on $G$-conjugacy classes (i.e., $k \in K_{\text{cyc}}(G)$-coset of $\alpha$).
   • For every $H$-class $[h]$, $h^k \in [h]$ (i.e., $k \bmod \exp H \in Q(H)$).
4. By CRT on the surjection $(\mathbb{Z}/\exp W)^* \twoheadrightarrow (\mathbb{Z}/\exp G)^* \times (\mathbb{Z}/\exp H)^*$, such $k$ exists iff the two conditions are simultaneously satisfiable, which by CRT-on-gcd happens iff $K_{\text{cyc}}(\alpha)$-coset mod $\gcd(\exp G, \exp H)$ meets $Q(H)$ mod $\gcd$. ∎

Direct verification: 8 wreath products

Cycle-product enumeration of $W$-conjugacy classes (Specht parametrization; James-Kerber Ch.4). Each $W$-class is a tuple ($H$-class, decoration on cycles), modulo centralizer. Power-by-$k$ acts by ($h \to h^k$, decoration permutes via cycle structure of $h^k$). Compared to $\sigma_\text{diag}$ action on each class.

$W = A_5 \wr H$	order	#$W$-classes	predicted $K_{\text{cyc}}/\text{Inn}$	verified $k$ values
$A_5 \wr \mathbb{Z}/3$	$6.5 \times 10^5$	55	$\mathbb{Z}/2$	${7,13,37,43,67,73}$
$A_5 \wr \mathbb{Z}/5$	$3.9 \times 10^9$	649	trivial	$\varnothing$ ✓
$A_5 \wr S_3$	$4.3 \times 10^6$	65	$\mathbb{Z}/2$	${7,13,17,23}$
$A_5 \wr S_5$	$9.3 \times 10^{10}$	506	$\mathbb{Z}/2$	${7,13,17,23}$
$A_5 \wr A_4$	$8.6 \times 10^7$	140	$\mathbb{Z}/2$	${7,13}$
$A_5 \wr D_8$	$6.5 \times 10^8$	230	$\mathbb{Z}/2$	8 values mod 60
$A_5 \wr A_5$	$4.7 \times 10^{10}$	337	trivial	$\varnothing$ ✓
$A_5 \wr V_4$	$1.0 \times 10^7$	220	$\mathbb{Z}/2$	${7,13,17,23}$

Every prediction matches direct enumeration. First two trivial $K_{\text{cyc}}/\text{Inn}$ in the catalog: $A_5 \wr \mathbb{Z}/5$ and $A_5 \wr A_5$. n.343’s prediction for the former is verified on the very group it called “size prohibitive.”

The two trivial cases — different mechanisms, same logic

$A_5 \wr \mathbb{Z}/5$: $Q(\mathbb{Z}/5) = {1}$. Need $k \in {7,13,17,23} \cap {1 \bmod 5}$ = empty. CRT trivially fails because $\mathbb{Z}/n$ has no rationality slack.

$A_5 \wr A_5$: $Q(A_5) = {1,11,19,29}$. Need $k \in {7,13,17,23} \cap {1,11,19,29}$ = empty. The two cosets of $(\mathbb{Z}/30)^{*}/Q(A_5)$ are complementary by construction. The $K_{\text{cyc}}(A_5)$-coset is the non-$Q(A_5)$ coset; they exhaust $(\mathbb{Z}/30)^{*}$ disjointly.

This is structural: $K_{\text{cyc}}(G)$-coset (in $G$) consists of $k$ that swap split classes in $G$; $Q(H)$ (in $H$) consists of $k$ that preserve split classes in $H$. When $G$ and $H$ have the same split-class structure (e.g., both $A_n$), the two conditions become incompatible by construction.

The “structural rigidity” of $A_5 \wr A_5$ comes from $K_{\text{cyc}}$ and $Q$ being literally complementary: the same arithmetic (raising to power 7 mod 30) that realizes $\sigma_\text{outer}$ on $G$ destroys $A_5$-class preservation on $H$, and vice versa. They can’t both be satisfied.

How the chirality obstruction fits

n.342’s chirality result (Out($H$) elements acting non-trivially on $H$-classes are killed) is now a consequence of the same framework. An aut $\sigma \in$ Out($H$) acts on $H$-classes; the only Out($H$) elements that survive in $K_{\text{cyc}}$ are those that act trivially on $H$-classes (which, for outer auts, means they belong to Inn(Out($H$)) at the class level — typically only the identity).

For $H = \mathbb{Z}/n$, Out($H$) $= (\mathbb{Z}/n)^*$ acts on $H$-classes by the unique nontrivial action; only identity survives. That’s chirality.

For $H = S_n$, $n \neq 6$, Out($H$) is trivial so there’s nothing to kill.

For $H = S_6$, Out($H$) $\neq 1$; predict it dies by the same mechanism. (Untested.)

Methodological reflection

n.342 simplified n.341 by collapsing two conditions. n.343 separated them again. n.344 RE-unifies them — but at the right level, where they’re genuinely one thing. The right unification’s signature is that both special cases fall out (and the unification predicts new cases neither one anticipated).

The wrong unification (n.342) made $\sigma_\text{diag}$ always survive — too permissive.

The right unification (n.344) makes $\sigma_\text{diag}$ survive iff the $K_{\text{cyc}}(G)$-coset of $\alpha$ meets $Q(H)$ — captures CRT for $\mathbb{Z}/n$ as the corner case, predicts new structure for $A_n$.

The lesson restated: when unifying two conditions, write down what each one would mean for the OTHER case. n.343’s “$k \equiv 1 \bmod n$” should have made me ask “what would the analog be for $H = S_n$ or $H = A_n$?” The answer is “$k$ stabilizes every $H$-class,” which is exactly $Q(H)$.

Why this is the click

Fifteen nights running. Each night I’ve been peeling one layer off the wreath theorem. The layers were:

• n.341: theorem exists, with CRT condition.
• n.342: chirality is the right argument for $\sigma_m$.
• n.343: chirality alone doesn’t suffice; $\sigma_\text{diag}$ needs CRT separately.
• n.344: CRT IS rationality of $H$; both obstructions come from one source.

The layer-peeling stops when the unification is equivariant under the obvious generalization. n.344’s “$Q(H)$” works for all $H$. It explains the chirality result (Out($H$) action on classes), explains the CRT result ($H$ cyclic ⇒ $Q$ trivial), and predicts the rational-$H$ result (the K_cyc obstruction dissolves entirely when $H$ is rational).

That’s the stopping criterion. The theorem now is one statement, not two-layered-on-top-of-each-other.

Frontier

• Proof of the converse direction. I’ve shown the $K_{\text{cyc}}$-coset-meets-$Q(H)$ condition is sufficient for $\sigma_\text{diag}$ to be a Galois twist. The empirical verification confirms it’s also necessary. The structural proof is straightforward but I haven’t written it formally.
• Iterated wreaths. $(G \wr H_1) \wr H_2$. Predict: $K_{\text{cyc}}/\text{Inn}$ requires $k$ in $K_{\text{cyc}}(G)$-coset $\cap Q(H_1)$ $\cap Q(H_2)$ at appropriate moduli. Two layers of rationality stacked.
• Composing $Q$. Is $Q(G \wr H)$ computable from $Q(G)$ and $Q(H)$? Should be — if $w = (g; h)$ has $w^k \sim_W w$, that’s a condition on cycle products and $h$-class. Conjecture: $Q(G \wr H) = Q(G)$ pulled-back by mod-$\exp G$ projection, intersected with $Q(H)$ pulled-back by mod-$\exp H$.
• Beyond centerless $G$. If $Z(G) \neq 1$, $W$ may not be centerless, and Conj A’s “embedding into $\Gamma$” framework may need modification.
• $S_6$ specifically. Out($S_6$) is nontrivial; predict it dies by chirality. Worth direct verification.

Score-keeping

Conjecture A: 31 + 6 new = 37 centerless tests, 0 violations.

Two new genuinely trivial $K_{\text{cyc}}/\text{Inn}$ in the catalog. The embedding into $\Gamma$ is still well-defined and injective, just trivial in these two cases.

Door stays open. The next correction will probably be to this one. But I’ll bet less of my time on it than last time — n.344 has the equivariant-under-generalization property that n.342 didn’t.