We know $w^{-1}\alpha <0$ by the minimality of the expression (see Appendix II.19 and II.22). Hence if $\beta =-w^{-1}\alpha >0,$ then $G_1\supseteq w{𝔛}_{\beta }{w}^{-1}={𝔛}_{w\beta }={𝔛}_{-\alpha }\text{.}$ Thus, $w_{\alpha }\in G_1\text{.}$ Since $w_{\alpha }wB{w}^{-1}{w}_{\alpha }^{-1}\subseteq G_1$ and since $\text{length}\hspace{0.17em}{w}_{\alpha }w<\text{length}\hspace{0.17em}w,$ we may complete the proof by induction.

$\square$

Part (a) follows from $BwB·B{w}_{\alpha }B\subseteq Bw{w}_{\alpha }B\cup BwB\text{.}$ (b) Suppose $\pi ,\pi \prime$ are distinct subsets of the set of simple roots, say $\alpha \in \pi \prime ,\alpha \notin \pi \text{.}$ Now $w_{\alpha }\alpha =-\alpha$ and $w\alpha =\alpha +\underset{\beta \in \pi }{\Sigma }{C}_{\beta }\beta$ if $w\in W_{\pi }\text{.}$ Thus $w_{\alpha }\alpha \ne w\alpha ,$ since simple roots are linearly independent. Hence, $w_{\alpha }\notin W_{\pi },$ $W_{\pi \prime }\ne W_{\pi },$ and $G_{\pi \prime }\ne G_{\pi }$ since distinct elements of the Weyl group correspond to distinct double cosets. (c) Let $A$ be any subgroup containing $B\text{.}$ Set $\pi =\left\{\alpha \hspace{0.17em}\right|\hspace{0.17em}\alpha \hspace{0.17em}\text{simple,}\hspace{0.17em}{w}_{\alpha }\in A\}\text{.}$ We shall show $A=G_{\pi }\text{.}$ Clearly, $A\supseteq G_{\pi }\text{.}$ Since $G=\underset{w\in W}{\cup }BwB$ and $A\supseteq B,$ we need only show $w\in A$ implies $w\in G_{\pi }$ to get $A\subseteq G_{\pi }\text{.}$ Let $w\in A,$ $w=w_{\alpha }{w}_{\beta }\dots ,$ a minimal expression of $w$ as a product of simple reflections. By Lemma 29, $w_{\alpha },w_{\beta },\dots \in A\text{.}$ Hence, $\alpha ,\beta ,\dots \in \pi ,$ $w\in W_{\pi },$ and $w\in G_{\pi }\text{.}$

$\square$

We first show $N\not\subset B\text{.}$ Suppose $N\subseteq B$ and $1\ne x\in N,$ $x=uh,$ $u\in U,$ $h\in H\text{.}$ If $u\ne 1,$ then for some $w\in W,$ $wx{w}^{-1}\notin B,$ a contradiction. If $u=1,$ then $h\ne 1\text{.}$ Since $G$ is adjoint, it has center 1, and $h{x}_{\alpha }\left(t\right)h^{-1}=x_{\alpha }\left(t\prime \right)$ with $t\prime \ne t$ for some $t,t\prime \in k,$ $\alpha \in \Sigma \text{.}$ Hence $(h,x_{\alpha }\left(t\right))=x_{\alpha }(t\prime -t)\in N,$ $x_{\alpha }(t\prime -t)\ne 1,$ and we are back in the first case.

We now prove the lemma. By Lemma 30(c), $NB=G_{\pi }$ for some $\pi \text{.}$ We must show $\pi$ contains all simple roots. Suppose it does not. Since $N⊈B,$ we see $\pi \ne \varnothing \text{.}$ Also since $\Sigma$ is indecomposable, we can find simple roots $\alpha ,\beta$ with $\alpha \in \pi ,$ $\beta \notin \pi$ and $\alpha$ not orthogonal to $\beta \text{.}$ Let $b_1{w}_{\alpha }{b}_2\in N,$ $b_i\in B,$ then $b{w}_{\alpha }\in N$ with $b=b_2{b}_1\in B\text{.}$ Then $w_{\beta }b{w}_{\alpha }{w}_{\beta }^{-1}\in N\cap (B{w}_{\alpha }{w}_{\beta }B\cup B{w}_{\beta }{w}_{\alpha }{w}_{\beta }B)$ by Lemma 25(b). Hence either $w_{\alpha }{w}_{\beta }\in W_{\pi }$ or $w_{\beta }{w}_{\alpha }{w}_{\beta }\in W_{\pi }\text{.}$ Now $w_{\beta }{w}_{\alpha }{w}_{\beta }=w\gamma ,$ where $\gamma =w_{\beta }\alpha =\alpha -⟨\alpha ,\beta ⟩\beta \text{.}$ Since $⟨\alpha ,\beta ⟩\ne 0,\gamma$ is not a simple root and $N\left(w_{\beta }{w}_{\alpha }{w}_{\beta }\right)\ne 1,$ so that $N\left(w_{\beta }{w}_{\alpha }{w}_{\beta }\right)\ge 3$ by Appendix II.20. Hence $w_{\alpha }{w}_{\beta }$ and $w_{\alpha }{w}_{\beta }{w}_{\alpha }$ are both expressions of minimal length. By Lemma 29, $w_{\beta }\in W_{\pi },$ a contradiction. Thus, $\pi$ is the set of all simple roots and $NB=G_{\pi }=G\text{.}$

$\square$

Since $G$ is generated by the ${𝔛}_{\alpha }\text{'s}$ we must show that every ${𝔛}_{\alpha }\subseteq G\prime \text{.}$ We will do this in several steps, excluding as we proceed the cases already treated. The first step takes us almost all the way.

(a) Assume $\left|k\right|\ge 4\text{.}$ We may choose $t\in k^{*},$ $t^2\ne 1\text{.}$ Then $(h_{\alpha }\left(t\right),x_{\alpha }\left(u\right))=x_{\alpha }\left((t^2-1)u\right)\text{.}$ Since $\alpha$ and $u$ are arbitrary, every ${𝔛}_{\alpha }\subseteq G\prime \text{.}$

By (a) we may henceforth assume that the rank $\ell$ is at least 2 and that $\left|k\right|=2$ or 3. By the corollary to Lemma 15, we may write the right side of (A) as $x_{\beta +\gamma }\left(N_{\beta ,\gamma }tu\right)·\Pi \prime ,$ the factor with $i=j=1$ having been isolated. We will use the fact $(*)$ that $N_{\beta ,\gamma }=\pm (r+1)$ with $r=r(\beta ,\gamma )$ as in Theorem 1, the maximum number of times one can subtract $\gamma$ from $\beta$ and still have a root.

(b) Assume that $\alpha$ is a root which can be written $\beta +\gamma$ so that no other positive integral combination of $\beta$ and $\gamma$ is a root and $N_{\beta ,\gamma }\ne 0\text{.}$ Then ${𝔛}_{\alpha }\subseteq G\prime ,$ as follows at once from (A) with $\Pi \prime =1\text{.}$ This covers the following cases:

To see this we use the fact that all roots of the same length are congruent under the Weyl group, imbed $a$ in an appropriate root system based on a pair of simple roots, and use $(*)\text{.}$ In all cases but the second cases in (2) and (3) this system can be chosen of type $A_2$ with $\beta$ and $\gamma$ roots of the same length as $\alpha ,$ while in those cases it can be chosen of type $B_2$ with $\beta$ and $\gamma$ short roots.

$\square$

By Lemma 14, we have

Here $N_{\gamma ,\beta }=\pm 1$ and $N_{\gamma ,\beta +\gamma }=\pm 2$ since $\beta -\gamma$ is not a root. If we multiply this equation by $-t,$ exponentiate, observe that the three factors on the right side commute, and then shift the first of them to the left, we get Lemma 33.

$\square$

Since the center of a Chevalley group is always finite by Lemma 28(d), we need only prove the first statement. Also we may assume $G=G_0,$ the adjoint group, since by Corollary 5 to Theorem 4', there is a homomorphism $\phi$ of $G$ onto $G_0$ with $\text{ker}\hspace{0.17em}\phi \subseteq \text{center of}\hspace{0.17em}G$ and $G_0$ has center 1. Now we may write $G=G_1·G_2\dots G_r$ where $G_i$ $i=1,2,\dots ,r$ is the adjoint group corresponding to an indecomposable subsystem of $\Sigma \text{.}$ By Theorem 5, each $G_i$ is simple. Thus any normal subgroup of $G$ is a product of some of the $G_i\text{'s.}$ If it also is solvable, the product is empty and the subgroup is 1.

$\square$