Lemma 1
Tag 0ATJ, permanent. Cite it as [Bourbaki, Tag 0ATJ].
Lie Groups and Lie Algebras, chapter V, § 6. The Coxeter transformation
Let X be a finite forest, and a map from X to a group such that and are conjugate whenever and are not linked in X. Let be the set of total orderings on X. For all , let be the product in of the sequence defined by . Then the elements are conjugate in .
1) We proceed by induction on . The case is immediate, so assume that . There exists in a terminal vertex (Chap. IV, Appendix, no. 3, Prop. 2). Let be a vertex linked to if one exists; if is not linked to any vertex in , let in be arbitrary. In all cases, commutes with for . Let be such that is the largest element of and the largest element of ; we let and prove that are conjugate.
2) Assume first that, for , is the largest element of and the largest element of . Let be the full subgraph , which is a forest. Define a map from to by putting if , . Let be the restrictions of to . The induction hypothesis applies, so and are conjugate. But it is clear that , proving the lemma in this case.
3) Assume that is the largest element of for . Let (resp. ) be the set of elements of strictly larger (resp. smaller) than ; let be the restriction of to . Then
and this element is conjugate to . We are thus reduced to case 2).
4) In the general case, let (resp. ) be the set of elements of strictly larger (resp. smaller) than ; let be the restriction of to . Then , and this element is conjugate to . We are thus reduced to case 3).
It follows from Prop. 1 that all the Coxeter transformations are of the same order . This number is called the Coxeter number of .
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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