LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every normalising element factors through the root-permuting subgroup

Statement

Let $(W, S)$ be an arbitrary-rank Coxeter system in its standard real root representation, with simple roots $Π = {α_{i} : i \in Γ_{0}}$ and length $ℓ$ . Fix $J \subseteq Γ_{0}$ , put $A_{J} = {α_{j} : j \in J}$ and $W_{J} = ⟨ s_{j} : j \in J ⟩$ , and write $M = {Norm}_{W} (W_{J})$ and $N_{J} = {n \in W : n A_{J} = A_{J}}$ . These are the conventions of Root-permuting normaliser complement; $N_{J}$ is a subgroup of $M$ .

$M = W_{J} N_{J}$ .

Facts & Assumptions

[F1]

The declared suppliers provide the subgroup and previously established assertions:

[F2]

For a subgroup $H$ of a group $G$ and $g \in G$ , the left coset is $g H = {g h : h \in H}$ . Left and right cosets gH and Hg of a subgroup

Proof

Given: The data in the statement. Goal: Prove its conclusion.

Root-permuting normaliser complement

Minimum representatives of normaliser cosets preserve the simple roots

1.1F1F2algebra

Let $g \in M$ . By [F1], the minimum $d$ in $g W_{J}$ belongs to $N_{J}$ . By [F2], write $d = g v$ with $v \in W_{J}$ ; then $g = d v^{- 1}$ . Put $u = v^{- 1} \in W_{J}$ . Since $d$ normalises $W_{J}$ , the element $h = d u d^{- 1}$ lies in $W_{J}$ and $g = h d$ . This proves $M \subseteq W_{J} N_{J}$ .

2.1F1step 1.1∎

Both $W_{J}$ and $N_{J}$ lie in the normaliser subgroup $M$ by [F1], so every product of an element of the former and an element of the latter belongs to $M$ . This proves the reverse inclusion.

Depends on

Used by · 1

Coxeter normaliser semidirect product

Dependency tree · 2 levels

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