Theorem 5

Let Z be the intersection of the conjugates of B, let U be a subgroup of B and let G1 be the subgroup generated by the conjugates of U in G. We make the following assumptions:*

(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.