Remark 2

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

Lie Groups and Lie Algebras, chapter IV, § 2. Tits Systems

Let $(G, B, N, S)$ be a Tits system, and let $Z$ be a normal subgroup of $G$ contained in $B$ . Let $G' = G/Z, B' = B/Z, N' = N/(Z \cap N)$ , and let $S'$ be the image of $S$ in $N'/(B' \cap N')$ . Then one sees immediately that $(G', B', N', S')$ is a Tits system.

Throughout this paragraph, with $(G, B, N, S)$ denoting a Tits system, we set $T = B \cap N$ and $W = N/T$ . A double coset means a double coset of $G$ with respect to $B$ . For any $w \in W$ , we set $C(w) = BwB$ ; this is a double coset.

We are going to deduce several elementary consequences of the axioms (T1) to (T4). We denote by $w, w', \ldots$ elements of $W$ and by $s, s', \ldots$ elements of $S$ . The following relations are clear:

C(1) = B, \quad C(ww') \subset C(w).C(w'), \quad C(w^{-1}) = C(w)^{-1}.

Axiom (T3) can also be written in the form

C(s).C(w) \subset C(w) \cup C(sw).

Moreover, since $C(sw) \subset C(s).C(w)$ by (1) and since $C(s).C(w)$ is a union of double cosets, there are only two possibilities:

C(s).C(w) = \begin{cases} C(sw) & \text{if } C(w) \not\subset C(s).C(w) \\ C(w) \cup C(sw) & \text{if } C(w) \subset C(s).C(w). \end{cases}

By (T4), $B \neq C(s).C(s)$ ; putting $w = s$ in (3) and using the relation $s^2 = 1$ , we obtain

C(s).C(s) = B \cup C(s).

This formula shows that $B \cup C(s)$ is a subgroup of $G$ . Multiplying both sides of (4) on the right by $C(w)$ , and using formula (3) and the relation

B.C(w) = C(w),

we obtain

C(s).C(s).C(w) = C(w) \cup C(sw).

Taking the inverses of the sets entering into formulas (2), (3) and (5) and then replacing $w$ by $w^{-1}$ , we obtain the formulas

C(w).C(s) \subset C(w) \cup C(ws)

C(w).C(s) = \begin{cases} C(ws) & \text{if } C(w) \not\subset C(w).C(s) \\ C(w) \cup C(ws) & \text{if } C(w) \subset C(w).C(s) \end{cases}

C(w).C(s).C(s) = C(w) \cup C(ws).

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

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