We argue by induction on $q$, the case $q = 0$ being trivial. If $q \geqslant 1$, we have $C(s_1 \ldots s_q).C(w) \subset C(s_1).C(s_2 \ldots s_q).C(w)$. By the induction hypothesis, $C(s_2 \ldots s_q).C(w)$ is contained in the union of the $C(s_{j_1} \ldots s_{j_p} w)$, where

By (T3), the set $C(s_1).C(s_{j_1} \ldots s_{j_p} w)$ is contained in the union of the sets $C(s_1 s_{j_1} \ldots s_{j_p} w)$ and $C(s_{j_1} \ldots s_{j_p} w)$. This proves the lemma.

It is clear that $\mathrm{BWB}$ is stable under $x \mapsto x^{-1}$, and Lemma 1 shows that it is stable under the product. Since it contains $B$ and $N$, it is equal to $G$.

It remains to prove that $C(w) \neq C(w')$ if $w \neq w'$. For this, we shall prove by induction on the integer $q$ the following assertion:

(A_q) If $w$ and $w'$ are distinct elements of $W$ such that $l_S(w) \geq l_S(w') = q$, then $C(w) \neq C(w')$.

(For the definition of $l_S(w)$, see § 1, no. 1.)

This assertion is clear for $q = 0$, since then $w' = 1$ and $w \neq 1$, hence $C(w') = B$ and $C(w) \neq B$.

Assume that $q \geq 1$ and that $w, w'$ satisfy the hypotheses of (A_q). There exists $s \in S$ such that $sw'$ is of length $q - 1$. We have

hence $w \neq sw'$. Moreover, $sw \neq sw'$; by formula (3) of § 1, no. 1, we have

By the induction hypothesis, $C(sw')$ is distinct from $C(w)$ and from $C(sw)$; from formula (2) it follows that

Since $C(sw') \subset C(s).C(w')$, we have finally that $C(w) \neq C(w')$.

For any $s \in S$, let $P_s$ be the set of elements $w \in W$ such that

We are going to verify that the $P_s$ satisfy conditions (A'), (B') and (C) of § 1, no. 7; the two assertions of the theorem will then follow from Prop. 6 of § 1, no. 7.

Condition (A') is clear.

We verify (B'). If $P_s$ and $sP_s$ had an element $w$ in common, we would have $w \in P_s$ and $sw \in P_s$, and hence

It would follow that $C(s).C(s).C(w) = C(w)$ and, by formula (5), this would imply that $C(w) = C(sw)$, which would contradict Th. 1.

We verify (C). Let $s, s' \in S$ and $w, w' \in W$ with $w' = ws'$. The assumption that $w \in P_s$ and $w' \notin P_s$ implies that

by (3).

From (9) and the relation $w = w's'$, it follows that

By formula (2'), $C(w').C(s') \subset C(w') \cup C(w's')$, which immediately implies that Since $C(w')$ is a union of left cosets $gB$ and since formula (10) shows that $C(s)w'$ meets $C(w')$ and a fortiori that $C(s)w's'B$ meets $C(w')s'B$. It follows from formulas (11) and (12) that the double coset $C(sw)$ is equal to one of the double cosets $C(ws')$ and $C(w)$; since $sw \neq w$, Th. 1 allows us to conclude that $sw = ws'$.

On taking reduced decompositions of the $w_i$, one is reduced to the case of a reduced decomposition

If $u = s_2 \ldots s_q$, then $w = s_1 u$ and $l_S(s_1 u) > l_S(u)$, so $C(w) = C(s_1).C(u)$ by the theorem. The required formula follows from this by induction on $q$.

If $t \in T_w$, there exist by definition elements $w', w'' \in W$ and $s \in S$ such that

By Cor. 1,

Hence,

By (4), $C(s) \subset C(s).C(s)$. Hence,

a) For any reduced decomposition $(s_1, \ldots, s_q)$ of $w$,

b) The group $H_w$ contains $C(w)$ and is generated by $C(w)$.

We prove a) by induction on $j$. Assume that $C(s_k)$ is contained in $H_w$ for $k < j$. Let

The element $t$ belongs to the subset $T_w$ of $W$ defined in Lemma 2 of § 1, no. 4. By Cor. 2, $C(t) \subset H_w$, and hence $C(s_j) \subset H_w$.

Since $C(w) = C(s_1) \ldots C(s_q)$, cf. Cor. 1, we have $C(w) \subset H_w$, and b) follows.

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).

The elements $w \in W$ such that $B \cup C(w)$ is a subgroup of G are those for which there exists $X \subset S$ with $W_X = \{1, w\}$. Moreover, if $w \neq 1$, we necessarily have Card(X) = 1 , i.e. $w \in S$.

We have $G_X = BW_XB = BN'B$, which shows that $G_X$ is generated by $B \cup N'$. The verification of the axioms (T1) to (T4) is now immediate.

Let $s_1, \ldots, s_q \in X$ and $t_1, \ldots, t_q \in Y$. Lemma 1 shows that

and hence that The opposite inclusion is obvious.

Let $w \in W$ be such that $g \in C(w)$. Since $B$ is a subgroup of $G_X$, the hypothesis $g B g^{-1} \subset G_X$ implies that $C(w).C(w^{-1}) \subset G_X$, and hence that $C(w) \subset G_X$ by Cor. 3 of Th. 2, so $g$ belongs to $G_X$.

a) $P$ is parabolic if and only if there exists a subset $X$ of $S$ such that $P$ is conjugate to $G_X$ (cf. no. 5 for the definition of $G_X$).

b) Let $X, X' \subset S$ and $g, g' \in G$ be such that $P = g G_X g^{-1} = g' G_{X'} {g'}^{-1}$. Then, $X = X'$ and $g' g^{-1} \in P$.

Assertion a) follows from Th. 3, b). Under the hypotheses of b), we have

and Prop. 3 shows that $g^{-1} g' \in G_X$. Hence, $G_{X'} = G_X$ and $X' = X$ by Th. 3, b). Finally, which proves b).

If the parabolic subgroup $P$ is conjugate to $G_X$, where $X \subset S$, then $P$ is said to be of type X.

(ii) Two parabolic subgroups whose intersection is parabolic are not conjugate.

(iii) Let $Q_1$ and $Q_2$ be two parabolic subgroups of $G$ contained in a subgroup $Q$ of $G$. Then any $g \in G$ such that $gQ_1g^{-1} = Q_2$ belongs to $Q$.

(iv) Every parabolic subgroup is its own normaliser$^6$.

Assertion (i) follows from Props. 3 and 4, and implies (ii). Under the hypotheses of (iii), we have $gQ_1g^{-1} \subset Q$, which implies that $g \in Q$ by (i). Finally, (iv) follows from (iii) by taking $Q_1 = Q_2 = Q$.

By first transforming $P_1$ and $P_2$ by an inner automorphism of $G$, we may assume that $B \subset P_1$. Let $g \in G$ be such that $gB_2g^{-1} \subset P_2$. By Th. 1, there exist $n \in N$ and $b, b' \in B$ such that $g = bn b'$. Since $T$ is normal in $N$,

and

which proves the proposition.

Since $BH$ is a subgroup of $G$ containing $B$, there exists a unique subset $X$ of $S$ such that $BH = G_X$ (Th. 3).

Let $s_1 \in X$ and $s_2 \in S - X$; let $n_1$ and $n_2$ be representatives in $N$ of $s_1$ and $s_2$, respectively. Then $n_1 \in G_X = BH$ and there exists $b \in B$ such that $bn_1 \in H$. Since $H$ is normal in $G$, the element $h = n_2 b n_1 n_2^{-1}$ of $G$ belongs to $H$. This means that

If the length of $s_2 s_1 s_2$ is equal to 3, Cor. 1 of Th. 2 implies that

(6) If $H$ is a subgroup of a group $G$, the normaliser of $H$ in $G$ is the subgroup $\mathfrak{N}(H)$ consisting of the elements $g$ of $G$ such that $gHg^{-1} = H$. A subgroup $H'$ is said to normalise $H$ if $H' \subset \mathfrak{N}(H)$, in which case $HH' = H'H$ is a subgroup of $G$ in which $H$ is normal.

and hence that $h \in H \cap C(s_2 s_1 s_2)$. Since $H \cap C(s_2 s_1 s_2)$ is non-empty, $s_2 s_1 s_2 \in W_X$. As $(s_2, s_1, s_2)$ is a reduced decomposition, it follows that $s_2 \in X$, contrary to our assumption.

Thus $l_S(s_2 s_1 s_2) \leq 2$; if $l_S(s_2 s_1 s_2) = 1$, then $s_1 s_2 \in S$ and so $(s_1 s_2)^2 = 1$, or $s_1 s_2 = s_2 s_1$. If $l_S(s_2 s_1 s_2) = 2$, property (E) of §1, no. 5 implies that $s_2 s_1 = s_1 s_2$, since $s_1 \neq s_2$. Q.E.D.

The following property of a group U enters into Th. 5 below:

(R) For any normal subgroup V of U distinct from U, the commutator subgroup (cf. Algebra, Chap. I, §6, no. 8) of U/V is distinct from U/V.

Every soluble group satisfies (R); in particular, every abelian group satisfies (R). It can be shown that the symmetric group $\mathfrak{S}_n$ satisfies (R) (cf. Exerc. 29).

(1) U is normal in B and B = UT. (2) U has property (R). (3) G1 is equal to its commutator subgroup*. (4) The Coxeter system (W, S) is irreducible (cf. §1, no. 9).

Then every subgroup H of G normalised by G1 is either contained in Z or contains G_1*.

First we show that G = G1 T*. The group $G_1 T$ contains B and hence is its own normaliser (Th. 4); but as N normalises $G_1$ and T, it also normalises $G_1 T$, so $N \subset G_1 T$. Since G is generated by B and N, it follows that $G = G_1 T$.

Next, set

We have $G = G'T$ since $G'$ contains $G_1$, and hence $N = N'T$. The inclusion of $N'$ into N thus defines, on passing to the quotient, an isomorphism $\alpha : W' \to W$. Let $S' = \alpha^{-1}(S)$.

We now show that (G', B', N', S') is a Tits system. Since $G = BNB$ and $B = TU = UT$, we have $G = UNU$. Since U is a subgroup of $G'$, it follows that $G' = UN'U$; since $U \subset B'$, this proves (T1). Axiom (T2) is satisfied since $\alpha$ is an isomorphism. Let $w \in W$ and let $w' = \alpha^{-1}(w)$ be the corresponding element of $W'$. We have

From this we conclude that $G' \cap BwB = B'w'B'$, which means that the inclusion of $G'$ into G defines on passing to the quotient a bijection from

B'\setminus G'/B' to B\setminus G/B. Axiom (T3) follows immediately. Axiom (T4) follows from B = B'T.

The subgroup H is normal in G'. By Lemma 2 applied to (G', B', N', S'), there exists a subset X' of S' such that B'H = G'_X', and every element of S' - X' commutes with every element of X'. In view of assumption (4), there are only two possibilities:

a) X' = \varnothing, i.e. B'H = B', so H \subset B' \subset B. If g \in G, then g = g_1 t with g_1 \in G_1, t \in T, and H \subset g_1 B g_1^{-1} since G_1 normalises H. Thus H \subset g B g^{-1}, and since Z is the intersection of the g B g^{-1}, we have H \subset Z.

b) X' = S', i.e. B'H = G'. Since G = G'T, we have

As B normalises U, every conjugate of U is of the form hUh^{-1} with h \in H. Such a subgroup is contained in the group UH, hence G_1 \subset UH by the definition of G_1. Thus, we have the isomorphisms

By assumption (3), G_1/(G_1 \cap H) is equal to its commutator subgroup. Assumption (2) now shows that the group U/(U \cap H), which is isomorphic to G_1/(G_1 \cap H), reduces to the identity element. Hence G_1 \cap H = G_1 and G_1 \subset H, which completes the proof.

Th. 5 shows that G_1/(G_1 \cap Z) is simple or reduces to the identity element. On the other hand, assumption (3) implies that it is equal to its commutator subgroup. Hence the corollary.