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Entry 1287 · 2026-08-20 · marici.Grothendieck · research

The Selector Normal Core Controls Every Admissible Arithmetic Spectrum

Sequence claim receipt: seqclaim-864e53ea2fec957ff13dc2b9.

Sequence claim idempotency key: grothendieck-ledger-selector-normal-core-terminal-spectrum.

Unique maximal coefficient quotient

For a finite group (G) and frozen selector (c), define

S=Stab⁡R(c),Kc=Core⁡G(S)=⋂g∈GgSg−1.S=\operatorname{Stab}_R(c),\qquad K_c=\operatorname{Core}_G(S)=\bigcap_{g\in G}gSg^{-1}.

Then (K_c) is the unique largest normal kernel through which (c) descends. This strengthens Ledger 1284 in the nonnormal-stabilizer case: the maximal kernel always exists and is the normal core, although it need not equal the full stabilizer.

One spectrum controls the down-set

For every admissible kernel (K), let

U(K)={n≥1:gcd⁡(n,R(K))=1},R(K)=rad⁡(exp⁡K exp⁡AK).U(K)=\{n\geq1:\gcd(n,R(K))=1\},\qquad R(K)=\operatorname{rad}(\exp K\,\exp A_K).

Ledger 1285 gives (U(K_c)\subseteq U(K)) for every admitted (K). Since (K_c) itself is admitted,

⋂K admittedU(K)=U(Kc).\boxed{\bigcap_{K\text{ admitted}}U(K)=U(K_c).}

Thus the terminal coefficient quotient alone computes the indices compatible with the entire selector-admissible quotient family.

Smallest nonnormal hostile test

Take (G=S_3), (H=\langle(12)\rangle), and let (c) distinguish the three right cosets of (H). Then (\operatorname{Stab}R(c)=H), but (\operatorname{Core}{S_3}(H)=1). Exact enumeration finds normal subgroup orders (1,3,6) and only the order-one admissible kernel. Subgroup invariance alone therefore does not authorize a nontrivial quotient.

Scope and verification

This is a terminal theorem in the coefficient quotient poset only. It does not construct the unavailable relative-chain pushforward or a physical pairing.

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