Bourbaki

Theorem 2

Tag 0AMF, permanent. Cite it as [Bourbaki, Tag 0AMF].

Lie Groups and Lie Algebras, chapter IV, § 1. Coxeter Groups

(i) For any subset XX of SS, the pair (WX,X)(W_X, X) is a Coxeter system.

(ii) Let (Xi)i∈I(X_i)_{i \in I} be a family of subsets of SS. If X=⋂i∈IXiX = \bigcap_{i \in I} X_i, then WX=⋂i∈IWXiW_X = \bigcap_{i \in I} W_{X_i}.

(iii) Let XX and X′X' be two subsets of SS. Then WX⊂WX′W_X \subset W_{X'} (resp. WX=WX′W_X = W_{X'}) if and only if X⊂X′X \subset X' (resp. X=X′X = X').

Every element of XX is of order 2 and XX generates WXW_X. Let x∈Xx \in X and w∈WXw \in W_X with lX(xw)≤lX(w)=ql_X(xw) \leq l_X(w) = q. By Cor. 4 of Prop. 7, we have

lS(xw)≤lS(w)=q. l_S(xw) \leq l_S(w) = q.
Let x1,…,xqx_1, \ldots, x_q be elements of XX such that w=x1…xqw = x_1 \ldots x_q. Since (W,S)(W, S) satisfies the exchange condition (Th. 1 of no. 6), there exists an integer jj such that 1≤j≤q1 \leq j \leq q and xx1…xj−1=x1…xj−1xjxx_1 \ldots x_{j-1} = x_1 \ldots x_{j-1} x_j. Thus, (WX,X)(W_X, X) satisfies the exchange condition and is therefore a Coxeter system (Th. 1 of no. 6). This proves (i).

Assertions (ii) and (iii) follow immediately from Cor. 1 of Prop. 7.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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