Theorem 3

a) For any subset X of S, the set G_X is a subgroup of G, generated by $\bigcup_{s \in X} C(s)$.

b) The map X ↦ G_X is a bijection from $\mathscr{P}(S)$ to the set of subgroups of G containing B.

c) Let (X_i)_{i \in I} be a family of subsets of X. If $X = \bigcap_{i \in I} X_i$, then $G_X = \bigcap_{i \in I} G_{X_i}$.

d) Let X and Y be two subsets of S. Then $G_X \subset G_Y$ (resp. $G_X = G_Y$) if and only if $X \subset Y$ (resp. $X = Y$).

It is clear that $G_X = (G_X)^{-1}$; Lemma 1 of no. 1 shows that $G_X . G_X \subset G_X$; and hence a) follows, taking into account Cor. 1 of Th. 2.

The injectivity of X ↦ G_X follows from that of X ↦ W_X (§1, no. 8, Th. 2). Conversely, let H be a subgroup of G containing B. Let U be the set of $w \in W$ such that $C(w) \subset H$. We have $H = BUB$ since H is a union of double cosets. Let $X = U \cap S$; we show that $H = G_X$. Clearly, $G_X \subset H$. On the other hand, let $u \in U$ and let $(s_1, \ldots, s_q)$ be a reduced decomposition of u. Cor. 3 of Th. 2 implies that $C(s_j) \subset H$, and hence that $s_j \in X$ for $1 \leq j \leq q$. Thus, $u \in W_X$, and since H is the union of the $C(u)$ for $u \in U$, we have $H \subset G_X$, which proves b).

Assertions c) and d) follow from analogous properties of W_X (§ 1, no. 8, Th. 2).