Using (A) and (B) we can reduce every element of ${𝔛}_s^{\prime }$ to the form $\prod _{\alpha \in S}{x}_{\alpha }^{\prime }\left(t_{\alpha }\right)$ $(t_{\alpha }\in k),$ and we know by Lemma 17 that every element of ${𝔛}_s$ can be written uniquely in this form.

$\square$

(a) Assume $\alpha \ne \pm \beta \text{.}$ Let $S$ be the set of roots of the form $i\alpha +j\beta$ where $i$ and $j$ are integers and $j>0\text{.}$ By (B) ${𝔛}_s^{\prime }$ is normalized by ${𝔛}_{\alpha }^{\prime }$ and ${𝔛}_{-\alpha }^{\prime }$ and hence by $w_{\alpha }^{\prime }\left(t\right)\text{.}$ Thus $w_{\alpha }^{\prime }\left(t\right)x_{\beta }^{\prime }\left(u\right)w_{\alpha }^{\prime }(-t)\in {𝔛}_s^{\prime }\text{.}$ By Lemma 36 we need only prove that relation (a) holds in $G\text{.}$ But in $G$ (a) follows from the relations (R). Now assume $\alpha =\beta$ and $\text{rank}\hspace{0.17em}\Sigma \ge 2\text{.}$ In this case we use the fact (see the Corollary to Lemma 33)

$(*)$ There exist roots $\delta$ and $\eta$ and a positive integer $j$ such that $\alpha =\delta +j\eta$ and $(x_{\delta }^{\prime }\left(t\right),x_{\eta }^{\prime }\left(u\right))=\prod _{m,n>0}{x}_{m\delta +n\eta }^{\prime }\left(c_{m,n}{t}^m{u}^n\right)$ and $c_{1j}\ne 0\text{.}$ Set $T=\{m{w}_{\alpha }\delta +n{w}_{\alpha }\eta \hspace{0.17em}|\hspace{0.17em}m,n\hspace{0.17em}\text{positive integers}\}\text{.}$ Transforming both sides of the above equation by $w_{\alpha }^{\prime }\left(t\right)$ and applying the case of (a) already proved we see that the transform of every term except $x_{\delta +j\eta }^{\prime }\left(c_{1j}t{u}^j\right)\in {𝔛}_T^{\prime }\text{.}$ Hence $w_{\alpha }^{\prime }{x}_{\delta +j\eta }^{\prime }\left(c_{1j}t{u}^j\right)w_{\alpha }^{\prime }(-t)\in {𝔛}_T^{\prime },$ so by the earlier argument, with $T$ in place of $S,$ (a) holds. If $\alpha =\beta$ and $\text{rank}\hspace{0.17em}\Sigma =1$ then (a) holds by (B'). Since $w_{\alpha }^{\prime }{\left(t\right)}^{-1}=w_{\alpha }^{\prime }(-t),$ the case $\alpha =-\beta$ follows from the case $\alpha =\beta \text{.}$

(b)-(f) follow from (a) and the definitions of $w_{\alpha }^{\prime }\left(t\right)$ and $h_{\alpha }^{\prime }\left(t\right)\text{.}$

Part (a) of Theorem 8 follows from parts (a)-(d) of Lemma 37.

$\square$

(a) follows from Lemma 37 (f).

(b) Let $\beta$ be any root, and write $\beta =w{\alpha }_i$ with ${\alpha }_i$ simple and $w\in W\text{.}$ Let $w=w_{\alpha }\dots$ be a minimal expression for $w$ as a product of simple reflections. Set $\gamma =w_{\alpha }\beta ,$ Then $h_{\beta }^{\prime }\left(t\right)=w_{\alpha }^{\prime }\left(1\right)h_{\gamma }^{\prime }\left(c{(-1)}^{-⟨\beta ,\alpha ⟩}t\right)h_{\gamma }^{\prime }{\left(c{(-1)}^{-⟨\beta ,\alpha ⟩}\right)}^{-1}{w}_{\alpha }^{\prime }(-1)$ by Lemma 37 (c), and hence by Lemma 37 (e) $h_{\beta }^{\prime }\left(t\right)=h_{\gamma }^{\prime }\left(c{(-1)}^{-⟨\beta ,\alpha ⟩}t\right)h_{\gamma }^{\prime }{\left(c{(-1)}^{-⟨\beta ,\alpha ⟩}\right)}^{-1}{w}_{\alpha }^{\prime }\left(t^{-⟨\beta ,\alpha ⟩}\right)w_{\alpha }^{\prime }(-1)\in {𝔥}_{\gamma }^{\prime }{𝔥}_{\alpha }^{\prime }\text{.}$ By induction on the length of $w,$ (b) follows.

$\square$

Let $G\prime \prime$ be the group generated by $\left\{x_{\alpha }^{\prime \prime }\left(t\right)\hspace{0.17em}\right|\hspace{0.17em}\alpha \in \Sigma ,t\in k\}$ subject to the relations (A), (B) if $\text{rank}\hspace{0.17em}\Sigma >1$ or (B') if $\text{rank}\hspace{0.17em}\Sigma =1,$ and (C). Let $w_{\alpha }^{\prime \prime }\left(t\right),h_{\alpha }^{\prime \prime }\left(t\right),\dots$ be defined as usual in terms of the $x_{\alpha }^{\prime \prime }\left(t\right)\text{.}$ We wish to prove that $\pi \prime \prime :G\prime \prime \to G$ is an isomorphism. Let $x\in \text{ker}\hspace{0.17em}\pi \prime \prime \text{.}$ By Corollary 1 of the proposition in §3, $x\in H\prime \prime \text{.}$ By Lemma 38 and (C) $x=\prod h_i^{\prime \prime }\left(t_i\right)$ $(t_i\in k^{*})\text{.}$ Applying $\pi \prime \prime$ we obtain $1=\prod h_i\left(t_i\right)\text{.}$ Since $G$ is universal each $t_i=1,$ so $x=1\text{.}$

$\square$

Since $f(t,u)\in \text{ker}\hspace{0.17em}\pi ,$ $f(t,u)\in$ center of $G\prime \text{.}$ Set $h_{\alpha }\left(t\right)=h\left(t\right)\text{.}$

(a)

(b) Let $v=v^m,$ $u=v^n$ with $m,n\in \mathbb{Z}\text{.}$ Then $h\left(t\right)=h{\left(v\right)}^{m}c,$ $h\left(u\right)=h{\left(v\right)}^{n}d$ with $c,d\in$ center $G\prime ,$ since $G\prime$ is a central extension of $G\text{.}$ Thus $h\left(t\right),h\left(u\right)$ commute and $f(t,u)=f(u,t)\text{.}$

(c) $h\left(t\right)=h\left(u\right)h\left(t\right)h{\left(u\right)}^{-1}=h\left(t{u}^2\right)h{\left(u^2\right)}^{-1}$ (by Lemma 37 (f)), so that $f(t,u^2)=1\text{.}$

(d) Abbreviate $x_{\alpha },x_{-\alpha },w_{\alpha },h_{\alpha }$ to $x,y,w,h,$ respectively. We have:

Then

(by (A) for $y(-t^{-1}{u}^{-1})y\left(t^{-1}{u}^{-2}\right)=y\left(t^{-1}{u}^{-2}(1-u)\right)=y\left(u^{-2}\right)\text{)}$

proving (d).

$\square$

If $\left|k\right|=q$ there are $(q+1)/2$ squares. Since $((q+1)/2)\nmid q$ the squares do not form an additive group, so we can find $a,b,c$ so that $a+b=c$ where $a$ and $b$ are squares and $c$ is not. Then take $t=a/c,$ $u=b/c\text{.}$

$\square$

Let $t,u\in k^{*}\text{.}$ We must show $f(t,u)=1$ where $f$ is as in Lemma 39. By Lemma 39 (b and c) if either $t$ or $u$ is a square $f(t,u)=1\text{.}$ Assume that both are not squares. By Lemma 40 (applied to the finite field generated by $t$ and $u\text{)}$ $t=r^2{t}_1,$ $u=s^2{u}_1$ with $r,s\in k^{*},$ $t_1+u_1=1,$ $t_1$ and $u_1$ not squares. Then $f(t,u)=f(t,s^2{u}_1)=f(t,u_1)=f(r^2{t}_1,u_1)=f(t_1,u_1)=1$ by Lemma 39(a and d).

$\square$