If $(\pi ,E)$ and $(\pi \prime ,E\prime )$ are u.c.e. of $G,$ let $\phi :E\to E\prime$ and $\phi \prime :E\prime \to E$ be such that $\pi \prime \phi =\pi$ and $\pi \phi \prime =\pi \prime \text{.}$ Now $\phi \prime \phi :E\to E$ and $\pi \left(\phi \prime \phi \right)=\pi \text{.}$ Hence $\phi \prime \phi$ is the identity on $E$ by the uniqueness of $\phi$ in the definition of a u.c.e. Similarly $\phi \phi \prime$ is the identity on $E\prime \text{.}$

$\square$

Consider the central extension $(\pi \prime ,E\prime )$ where $E\prime =E\times E/𝒟E$ and $\pi \prime (a,b)=\pi \left(a\right),$ $a\in E,$ $b\in E/𝒟E\text{.}$ Now if ${\phi }_1\left(a\right)=(a,1)$ and ${\phi }_2\left(a\right)=(a,a+𝒟E),$ then $\pi \prime {\phi }_i=\pi ,$ $i=1,2,$ and hence ${\phi }_1={\phi }_2\text{.}$ Thus, $E/𝒟E=1$ and $E=𝒟E\text{.}$

$\square$

We have $\pi 𝒟E=𝒟\left(\pi E\right)=𝒟G=G\text{.}$ Hence, $E=C𝒟E$ where $C=\text{ker}\hspace{0.17em}\pi \text{.}$ Also, $𝒟E=𝒟\left(C𝒟E\right)={𝒟}^{2}E\text{.}$

$\square$

For each $x\in G$ we introduce a symbol $e\left(x\right)\text{.}$ Let $F$ be the group generated by $\{e\left(x\right),x\in G\}$ subject to the condition that $e\left(x\right)e\left(y\right)e{\left(xy\right)}^{-1}$ commutes with $e\left(z\right)$ for all $x,y,z\in G\text{.}$ If $\pi :e\left(x\right)\to x,$ then by using induction on the length of an expression in $F,$ we see that $\pi$ extends to a central homomorphism of $F$ onto $G\text{.}$

(a) $(\pi ,F)$ covers all central extensions of $G\text{.}$ To see this, let $(E\prime ,\pi \prime )$ be any central extension of $G\text{.}$ Choose $e\prime \left(x\right)\in E\prime$ such that $\pi \prime e\prime \left(x\right)=x\text{.}$ Since $\pi \prime$ is central, the $e\prime \left(x\right)\text{'s}$ satisfy the condition on the $e\left(x\right)\text{'s.}$ Hence, there is a homomorphism $\phi :F\to E\prime$ such that $\phi e\left(x\right)=e\prime \left(x\right),$ and
thus $\pi \prime \phi =\pi \text{.}$

(b) If $E=𝒟F$ and $\pi$ also denotes the restriction of $\pi$ to $E,$ then $(\pi ,E)$ covers all central extensions uniquely. By (iii), we have that $(\pi ,E)$ covers all central extensions. If $(\pi ,E\prime )$ is a central extension of $G$ and if $\pi \prime \phi =\pi =\pi \prime \phi \prime ,$ then $\phi \left(x\right)\phi \prime {\left(x\right)}^{-1}\in$ center of $E\text{.}$ Thus, $\psi :x\to \phi \left(x\right)\phi \prime {\left(x\right)}^{-1}$ is a homomorphism of $E$ into an Abelian group. Since $E=𝒟E,$ $\psi$ is trivial and $\phi =\phi \prime \text{.}$

$\square$

If $a\in \text{ker}\hspace{0.17em}\psi \pi ,$ let $\phi$ be the map $\phi :x\to (a,x)=ax{a}^{-1}{x}^{-1},x\in E\text{.}$ Now $\phi \left(x\right)\in$ center of $E,$ since $\pi (a,x)=(\pi a,\pi x)=1$ because $\pi a\in$ center of $F\text{.}$ Now $\phi$ is a homomorphism, so $\phi$ is trivial, and $a\in$ center of $E\text{.}$

$\square$

One direction is immediate by taking $G\prime =G$ and $\rho =\text{id}\text{.}$ Conversely, suppose the diagram is given and $(\psi ,E)$ is a u.c.e. Let $H$ be the subgroup of $G\times E\prime ,$ $H=\left\{(x,e\prime )\hspace{0.17em}\right|\hspace{0.17em}\rho x=\pi \prime e\prime \}\text{.}$ If $\psi :H\to G$ is given by $\psi (x,e\prime )=x,$ then $\psi$ is central. Since $(\pi ,E)$ is a u.c.e. of $G,$ there is a unique homomorphism $\theta :E\to H$ such that $\psi \theta =\pi \text{.}$ Now the homomorphism $\phi :E\to E\prime ,$ $\phi =\psi \prime \theta ,$ where $\psi \prime :H\to E\prime$ is given by $\psi \prime (x,e\prime )=e\prime ,$ satisfies $\rho \pi =\rho \psi \theta =\pi \prime \psi \prime \theta =\pi \prime \phi \text{.}$ If $\rho \pi =\pi \prime \phi \prime ,$ then let $\theta \prime :E\to H$ be given by $\theta \prime \left(e\right)=(\pi \left(e\right),\phi \prime \left(e\right))\text{.}$ Since $\psi \theta \prime =\pi ,$ we have $\theta \prime =\theta$ and $\phi \prime =\psi \prime \theta \prime =\psi \prime \theta =\phi ,$ proving the uniqueness of $\phi \text{.}$

$\square$

Since $\left|k\right|>4,$ $G=𝒟G,$ $G\prime =𝒟G\prime ,$ and u.c.e. exist for both $G$ and $G\prime \text{.}$ The conclusion becomes $G\prime$ is centrally closed, by the above remarks. We need only show that every central extension $(\psi ,E)$ of $G\prime$ splits, i.e., there exists $\theta :G\prime \to E$ so that $\psi \theta ={\text{id}}_{G\prime }\text{;}$ i.e., the relations defining $G\prime$ can be lifted to $E\text{.}$

We may assume $E=𝒟E\text{;}$ but then $(\pi \psi ,E)$ is a central extension of $G$ by (v). We need only show

(1) If $(\psi ,E)$ is a central extension of $G,$ then the relations (A),(B),(B') can be lifted to $E\text{.}$

Let $C=\text{ker}\hspace{0.17em}\psi ,$ a central subgroup of $E\text{.}$ We have:

(2) A commutator $(x,y)$ with $x,y\in E$ depends only on
the classes mod $C$ to which $x$ and $y$ belong.

Choose $a\in k^{*}$ so that $c=a^2-1\ne 0\text{.}$ Then in $G$ $(h_{\alpha }\left(a\right),x_{\alpha }\left(t\right))=x_{\alpha }\left(ct\right)$ for all $\alpha \in \Sigma ,t\in K\text{.}$ We define $\phi x_{\alpha }\left(t\right)\in E$ $(\alpha \in \Sigma ,t\in k)$ so that $\psi \phi x_{\alpha }\left(t\right)=x_{\alpha }\left(t\right)$ and so that

(3) $(\phi h_{\alpha }\left(a\right),\phi x_{\alpha }\left(t\right))=\phi x_{\alpha }\left(ct\right)$ and then $\phi h\text{'s}$ (and later $\phi w\text{'s)}$ in terms of the $\phi x\text{'s}$ by the same formulas which define the $h\text{'s}$ and the $x\text{'s.}$ in terms of the $x\text{'s.}$ Note that this choice is not circular because of (2). We shall show that the relations (A),(B),(B') hold with $\phi x\text{'s}$ in place of $x\text{'s.}$

(4) $\phi h\phi x_{\alpha }\left(t\right){\left(\phi h\right)}^{-1}=\phi \left(h{x}_{\alpha }\left(t\right)h^{-1}\right)$ for all $h\in H,$ $\alpha \in \Sigma ,$ $t\in k\text{.}$

Set $h{x}_{\alpha }\left(t\right)h^{-1}=x_{\alpha }\left(dt\right)$ with $d\in k^{*}\text{.}$ Conjugating (3) by $\phi h,$ we get $(\phi h_{\alpha }\left(a\right),\phi x_{\alpha }\left(dt\right))=\phi h\phi x_{\alpha }\left(ct\right)\phi {\left(h\right)}^{-1},$ and the left side equals $\phi x_{\alpha }\left(cdt\right)=\phi \left(h{x}_{\alpha }\left(ct\right)h^{-1}\right)$ by (3). Similarly we have

(4') $\phi n\phi x_{\alpha }\left(t\right){\left(\phi n\right)}^{-1}=\phi \left(n{x}_{\alpha }\left(t\right)$ n^{-1}$\right)$ for all $n\in N,$ $\alpha \in \Sigma ,$ $t\in k\text{.}$

(5) If $\alpha$ and $\beta$ are roots, $\alpha +\beta \ne 0,$ and $\alpha +\beta$ is not a root, then $\phi x_{\alpha }\left(t\right)$ and $\phi x_{\beta }\left(u\right)$ commute for all $t,u\in k\text{.}$ Set $\phi x_{\alpha }\left(t\right)\phi x_{\beta }\left(u\right)\phi x_{\alpha }{\left(t\right)}^{-1}=f(t,u)\phi x_{\beta }\left(u\right),$ $t,u\in k,$ $f(t,u)\in C\text{.}$ We must show $f(t,u)=1\text{.}$ Clearly, from the definitions we have

(6) $f$ is additive in both positions.

(5a) Assume $\alpha \ne \beta \text{.}$ If $(\alpha ,\beta )=0,$ then $f(t,u)=f(t{v}^2,u)$ $(v\ne 0)$ by Lemma 20(c) and by (4) with $h=h_{\alpha }\left(v\right)\text{.}$ If $(\alpha ,\beta )>0,$ then $f(t,u)=f(t{v}^d,u)$ where $d=4-⟨\alpha ,\beta ⟩⟨\beta ,\alpha ⟩$ by Lemma 20(c) and by (4) with $h=h_{\alpha }\left(v^2\right)h_{\beta }\left(v^{-⟨\beta ,\alpha ⟩}\right)\text{.}$ In both cases, $f(t(1-v^d),u)=1$ by (6) for some $d=1,2,$ or $3\text{.}$ Choose $v$ so $v^d-1\ne 0\text{.}$ Then we get $f\equiv 1\text{.}$

(5b) Assume $\alpha =\beta$ and $\text{rank}\hspace{0.17em}\Sigma >1\text{.}$ If there is a root $\gamma$ so that $⟨\alpha ,\gamma ⟩=1,$ set $h=h_{\gamma }\left(v\right)$ in the preceding argument and obtain $(*)$ $f(t,u)=f(tv,uv)\text{.}$ Choose $v$ so that $v-v^2\ne 0$ and $1-v+v^2\ne 0\text{.}$ By $(*)$ and (6), $f(t(v-v^2),u)=f(t,u/(v-v^2))=f(t,u/v)f(t,u/(1-v))=f(vt,u)f((1-v)t,u)=f(t,u),$ whence $f(t(1-v+v^2),u)=1$ and $f\equiv 1\text{.}$ If there is no such $\gamma ,$ then $\Sigma$ is of type $C_n$ and $\alpha$ is a long root. In this case, however, $\alpha =\beta +2\gamma$ with $\beta$ and $\gamma$ roots. Thus, $(\phi x_{\beta }\left(t\right),\phi x_{\gamma }\left(u\right))=g\phi x_{\beta +\gamma }(\pm tu)\phi x_{\beta +2\gamma }(\pm t{u}^2)$ with $g\in C,$ by Lemma 33. Since $\phi x_{\alpha }\left(v\right)$ commutes with all factors but the last by (5a), it also commutes with the last.

(5c) Assume $\alpha =\beta$ and $\text{rank}\hspace{0.17em}\Sigma =1\text{.}$ At least we have $f(t,u)=f(t{v}^2,u{v}^2),$ $t,u\in k,$ $v\in k^{*},$ using $h=h_{\alpha }\left(v\right)$ in the argument above. We may also assume that $\left|k\right|$ is not a prime. If it were, then $x_{\alpha }\left(t\right)$ and $x_{\alpha }\left(u\right)$ would be powers of $x_{\alpha }\left(1\right)$ and (5) would be immediate. Referring to the proof of (5b), we see it will suffice to be able to choose $v$ so that $v,1-v$ are squares and $v-v^2\ne 0,$ $1-v+v^2\ne 0\text{.}$ If $k$ is finite of characteristic 2, this is possible since all elements of $k$ are squares. Otherwise, set $v={(2w/(1+w^2))}^2\text{.}$ Then $1-v={((1-w^2)/(1+w^2))}^2,$ and we need only choose $w$ so that $1+w^2\ne 0,$ $v-v^2\ne 0,$ and $1-v+v^2\ne 0\text{.}$ Since at most $13$ values of $w$ are to be avoided and $\left|k\right|\ge 25$ in the present case, this too is possible. T his completes the proof of (5).

(7) $\phi$ preserves the relations (A). The element $x=\phi x_{\alpha }\left(t{c}^{-1}\right)\phi x_{\alpha }\left(u{c}^{-1}\right){\left(\phi x_{\alpha }\left((t+u)c^{-1}\right)\right)}^{-1}$ is in $C,$ and hence the transform of $x$ by $h_{\alpha }\left(a\right)$ is $x$ itself. However, by (3), (4), (5) this transform is also $x\phi x_{\alpha }\left(t\right)\phi x_{\alpha }\left(u\right){\left(\phi x_{\alpha }(t+u)\right)}^{-1}\text{.}$

(8) $\phi$ preserves the relations (B). We have $\phi x_{\alpha }\left(t\right)\phi x_{\beta }\left(u\right)\phi x_{\alpha }{\left(t\right)}^{-1}=f(t,u)\prod \phi x_{i\alpha +j\beta }\left(c_{ij}{t}^i{u}^j\right)\phi x_{\beta }\left(u\right),$ where $f(t,u)\in C\text{.}$ One proves $f=1$ by induction on $n,$ the number of roots of the form $i\alpha +j\beta$ $i,j\in {\mathbb{Z}}^{*}\text{.}$ If $n=0,$ this is just (5). If $n>0,$ the inductive hypothesis and (7) imply $f$ satisfies (6), and then the argument in (5a) may be used.

(9) $\phi$ preserves the relations (B'). This follows from (4').

This completes the proof of the theorem.

$\square$

This follows from various of the generalities at the beginning of this section.

E.g., if $k$ is finite, $\left|k\right|>4$ and ${SL}_2\left(9\right)$ is excluded,
then ${SL}_2\left(k\right),$ ${Sp}_n\left(k\right),$ and ${\text{Spin}}_n\left(k\right)$ all have trivial Schur multipliers, and the natural central extensions ${SL}_n\to {PSL}_n,$ ${Sp}_n\to P{Sp}_n,$ ${\text{Spin}}_n\to {SO}_n$ are all universal.

$\square$

Elements of the form $f(t,u)=h_{\alpha }\left(t\right)h_{\alpha }\left(u\right)h_{\alpha }{\left(tu\right)}^{-1}$ in $G\prime$ $\alpha \in \Sigma$ generate $C\text{.}$ We have $f(t,v{w}^2)=f(t,v)f(t,w^2)$ by Lemma 39(a). By induction, we get $f(t,w^{2n})=f{(t,w^2)}^n$ for arbitrary $n\text{.}$ Since for $u\in k^{*}$ we can find $w\in k^{*}$ such that $u=w^{2n},$ the proof is complete.

$\square$

Let $(\psi ,E)$ be a central extension of $G$ with $\text{ker}\hspace{0.17em}\psi$ reduced. Since $(\pi ,G\prime )$ is u.c.e. we have $\phi :G\prime \to E$ so that $\psi \phi =\pi \text{.}$ Since $C=\text{ker}\hspace{0.17em}\pi$ is divisible, so is $\phi C\subseteq \text{ker}\hspace{0.17em}\psi \text{.}$ Hence $\phi C=1$ and $\text{ker}\hspace{0.17em}\phi \supseteq \text{ker}\hspace{0.17em}\pi \text{.}$ Thus, there is a homomorphism $\theta :G\to E$ so that $\theta \pi =\phi \text{.}$ Therefore, $\psi \theta \pi =\pi$ on $G\prime$ and $\psi \theta =1$ on $G\text{.}$

$\square$

Assume $\sigma :G\to PGL\left(V\right)\text{.}$ Since $G=𝒟G,$ we have $\sigma :G\to PSL\left(V\right)\text{.}$ Let $\rho :SL\left(V\right)\to PSL\left(V\right)$ be the natural projection. Since $\text{dim}\hspace{0.17em}V$ is finite, we have $\text{ker}\hspace{0.17em}\rho$ is finite and thus $\text{ker}\hspace{0.17em}\rho$ is reduced. Consider the central extension $(\psi ,E)$ of $G$ where $E=\left\{(x,y)\hspace{0.17em}\right|\hspace{0.17em}\sigma x=\rho y,x\in G,y\in SL\left(V\right)\}\subseteq G\times SL\left(V\right)$ and $\psi (x,y)=x,$ $(x,y)\in E\text{.}$ Now $\text{ker}\hspace{0.17em}\psi =1\times \text{ker}\hspace{0.17em}\rho$ is reduced, so by Corollary 3b, we have $\theta :G\to E$ with $\psi \theta =1$ on $G\text{.}$ If $\psi \prime (x,y)=y,$ $(x,y)\in E,$ then $\sigma \prime =\psi \prime \theta :G\to SL\left(V\right)\subset GL\left(V\right)$ with $\rho \sigma \prime =\rho \psi \prime \theta =\sigma \psi \theta =\sigma \text{.}$

$\square$

By Theorem 10, we could assume $\left|k\right|\le 4$ or $\left|k\right|=9\text{.}$ However, the proof does not use this assumption or Theorem 10. If $(\psi ,E)$ is a central extension of $G$ such that $C=\text{ker}\hspace{0.17em}\psi$ has no $p\text{-torsion,}$ then we wish to show (A), (B), and (B') can be lifted to $E\text{.}$

(1) Assume $C$ is divisible by $p\text{.}$ Choose $\phi x_{\alpha }\left(t\right)\in E$ so that $\psi \phi x_{\alpha }\left(t\right)=x_{\alpha }\left(t\right)$ and ${\left(\phi x_{\alpha }\left(t\right)\right)}^p=1,$ $\alpha \in \Sigma ,$ $t\in k\text{.}$ We claim relations (A), (B) and (B') hold on the $\phi x\text{'s.}$

(1a) If $\alpha ,\beta$ are roots, $\alpha +\beta$ not a root, and $\alpha +\beta \ne 0,$ then $\phi x_{\alpha }\left(t\right)$ and $\phi x_{\beta }\left(u\right)$ commute, $t,u\in k\text{.}$ We have $\phi x_{\alpha }\left(t\right)\phi x_{\beta }\left(u\right)\phi x_{\alpha }{\left(t\right)}^{-1}=f\phi x_{\alpha }\left(u\right)$ with $f\in C\text{.}$ Taking $p^{\text{th}}$ powers, we get $1=f^p$ which implies $f=1$ since $C$ has no $p\text{-torsion.}$

(1b) The relations (A) hold. Taking $p^{\text{th}}$ powers of $\phi x_{\alpha }\left(t\right)\phi x_{\alpha }\left(u\right)=f{\phi }_{\alpha }(t+u),$ $f\in C,$ we get $f=1$ as before.

(1c) Exercise: Relations (B) and (B') also hold.

(2) General case.

(2a) $C$ can be embedded in a group $C\prime$ which is divisible by $p$ and has no $p\text{-torsion.}$ We have a homomorphism $\theta$ of a free Abelian group $F$ onto $C\text{.}$ Now $F{\otimes }_{\mathbb{Z}}Q$ is a divisible group, and we can identify $F$ with $F{\otimes }_{\mathbb{Z}}\mathbb{Z}\subset F{\otimes }_{\mathbb{Z}}Q\text{.}$ Hence $C=F/\text{ker}\hspace{0.17em}\theta \subset F{\otimes }_{\mathbb{Z}}Q/\text{ker}\hspace{0.17em}\theta =D,$ say, and $D$ is a divisible group. Moreover, since $C$ has no $p\text{-torsion,}$ $C\cap D_p=1$ where $D_p$ is the $p\text{-component}$ of $D\text{.}$ Thus, $C$ projects faithfully into $D/D_p=C\prime$ which is divisible and has no $p\text{-torsion.}$

(2b) Conclusion of proof. Form $E\prime =EC\prime ,$ the direct product of $E$ and $C\prime$ with $C$ amalgamated, and define $\psi \prime :E\prime \to G$ by $\psi \prime \left(ec\prime \right)=\psi \left(e\right),$ $e\in E,$ $c\prime \in C\prime \text{.}$

Now $(\psi \prime ,E\prime )$ satisfies the assumptions of (1) so the relations (A), (B), (B') can be lifted to $E\prime \text{.}$ However, by Lemma 32', the lifted group is its own derived group and hence contained in $E\text{.}$

$\square$

These are easy exercises.

$\square$

(1) If ${𝔥}_{\alpha }$ is the group generated (in G') by all $h_{\alpha }\left(t\right),$ $\alpha$ a fixed long root, then $C\subseteq {𝔥}_{\alpha }\text{.}$ We know that $h_{\alpha }\left(t\right)h_{\alpha }\left(u\right)h_{\alpha }{\left(tu\right)}^{-1},$ $\alpha \in \Sigma ,$ $t\in k^{*},$ form a generating set for $C\text{.}$ Using the Weyl group we can narrow the situation to at most two roots $\alpha ,\beta$ with $\alpha$ long, $\beta$ short, and $(\alpha ,\beta )>0\text{.}$ Hence, $⟨\beta ,\alpha ⟩=1$ and $(h_{\alpha }\left(t\right),h_{\beta }\left(u\right))=h_{\beta }\left(tu\right)h_{\beta }{\left(t\right)}^{-1}{h}_{\beta }{\left(u\right)}^{-1}=h_{\alpha }\left(t\right)h_{\alpha }\left(u^{⟨\alpha ,\beta ⟩}\right)h_{\alpha }{\left(t{u}^{⟨\alpha ,\beta ⟩}\right)}^{-1}$ by Lemma 37(f). This shows a will suffice.

(2) $\phi$ is a mapping onto $C\text{.}$ This follows from (1).

(3) $\phi$ is a homomorphism. We must show that the relations hold if $f(t,u)$ is replaced by $h_{\alpha }\left(t\right)h_{\alpha }\left(u\right)h_{\alpha }{\left(tu\right)}^{-1}\text{.}$ The relations (a) are obvious. A special case $(u=1)$ of (e) has been shown in Lemma 39(d). The other relations (b), (c), (d) follow from the commutator relations connecting the $h\text{'s}$ and the $w\text{'s.}$

(3') Assume $\Sigma$ is not of type $C_n\text{.}$ In this case there is a root $\gamma$ so that $⟨\alpha ,\gamma ⟩=1\text{.}$ Thus $f(t,v)=h_{\gamma }\left(u\right)f(t,v)h_{\gamma }{\left(u\right)}^{-1}=h_{\alpha }\left(tu\right)h_{\alpha }{\left(u\right)}^{-1}{h}_{\alpha }\left(uv\right)h_{\alpha }{\left(tuv\right)}^{-1}=f{(t,u)}^{-1}f(t,uv)$ or $f(t,uv)=f(t,u)f(t,v)\text{.}$ By relation (c), $f(uv,t)=f(u,t)f(v,t)\text{.}$

(4) $\phi$ is an isomorphism. This is done by constructing an explicit model for $G\prime \text{.}$

$\square$

Let $(\pi ,E)$ be a covering of $G\text{.}$ Locally $\pi$ is invertible so we may set $\phi ={\pi }^{-1}$ on some neighborhood of 1 in $G\text{.}$ We shall show that $\phi$ can be extended to a homomorphism of $G$ onto $E,$ i.e., $(\text{id},G)$ covers $(\pi ,E)\text{.}$ It suffices to show that $\phi$ can be extended to all of $G$ so that the relations (A), (B), (B'), (C) hold on the $\phi x\text{'s.}$

Consider the relations

Since $\phi$ is locally an isomorphism, there is $\epsilon >0$ such that (A) holds for $\left|t\right|<\epsilon ,$ $\left|u\right|<\epsilon \text{.}$ If $t\in k,$ $t=\Sigma t_i,$ $\left|t_i\right|<\epsilon ,$ then set $\phi x_{\alpha }\left(t\right)=\prod \phi x_{\alpha }\left(t_i\right)\text{.}$ Using induction and Lemma 41, we see that $\phi x_{\alpha }\left(t\right)$ is well defined. Clearly, (A) then holds for all $t,u\in k\text{.}$ Alternatively, we could note that ${𝔛}_{\alpha }$ is topologically equivalent to $ℂ$ and hence simply connected. Thus, $\phi$ extends to a homomorphism of ${𝔛}_{\alpha }$ into $E$ and (A) holds. Clearly the extension of $\phi$ to ${𝔛}_{\alpha }$ is unique.

To obtain the relations (B), let $\alpha ,\beta$ be roots $\alpha \ne \pm \beta ,$ let $S$ be the set of roots of the form $i\alpha +j\beta$ $(i,j\in {\mathbb{Z}}^{+}),$ and let ${𝔛}_S$ be the corresponding unipotent subgroup of $G\text{.}$ Topologically, ${𝔛}_S$ is equivalent to ${ℂ}^n$ for some $n,$ and is hence simply connected. As before $\phi$ can be extended to a homomorphism of ${𝔛}_S$ into $E,$ and the relations (B) hold. This extension is consistent with those above, by the uniqueness of the latter.

We now consider $h_{\alpha }\left(t\right)=x_{\alpha }\left(t\right)x_{-\alpha }(-t^{-1})x_{\alpha }\left(t\right)\text{.}$ $w_{\alpha }(-1)=x_{\alpha }(t-1)x_{-\alpha }{(1-t^{-1})}^{x_{\alpha }(-1)}{x}_{\alpha }{(t-1)}^{w_{\alpha }(-1)}$ where $x^y=y^{-1}xy\text{.}$ Hence, if $t$ is near 1 in $ℂ,$ then $\phi h_{\alpha }\left(t\right)$ is near 1 in $E\text{.}$ Thus, $\phi h_{\alpha }\left(t\right)$ is multiplicative near 1 and hence Abelian everywhere (recall that $\psi$ is central). We then have $\phi h_{\alpha }\left(u\right)=\phi h_{\alpha }\left(t\right)\phi h_{\alpha }\left(u\right)\phi h_{\alpha }{\left(t\right)}^{-1}=\phi h_{\alpha }\left(t^{2}u\right)\phi h_{\alpha }{\left(t^2\right)}^{-1}$ by Lemma 37(f). Since $ℂ$ has square roots, we have $\phi h_{\alpha }$ is multiplicative, i.e., (C) holds.

$\square$

The relation $(1+t{E}_{ik},1+E_{kj})=1+t{E}_{ij}$ shows (b) and hence also $E\left(R\right)\subseteq 𝒟GL\left(R\right)\text{.}$ If $x,y\in {GL}_n\left(R\right),$ then $xy{x}^{-1}{y}^{-1}\in E\left(R\right)$ because in ${GL}_{2n}\left(R\right)$ we have

$\square$

(a) $\pi$ is central. If $x\in \text{ker}\hspace{0.17em}\pi ,$ choose $n$ large enough so that $x$ is a product of $x_{ij}\text{'s}$ with $i,j<n\text{.}$ Let $F_n$ be the subgroup of $U\left(R\right)$ generated by the $x_{kn}\text{'s}$ $(k\ne n,k=1,2,\dots )\text{.}$ Now by (A) and (B), any element of $P_n$ can be expressed as ${\prod }_k{x}_{kn}\left(t_k\right)\text{.}$ Since in $E\left(R\right)$ this form is unique, $\pi \mid P_n$ is an isomorphism. Also by (A) and (B), $x_{ij}\left(t\right)P_n{x}_{ij}{\left(t\right)}^{-1}\subseteq P_n$ if $i,j<n\text{.}$ Thus, $x{P}_n{x}^{-1}\in P_n\text{.}$ If $y\in P_n,$ then $\pi (x,y)=1,$ and since $\pi \mid P_n$ is an isomorphism we have $(x,y)=1\text{.}$ In particular, $x$ commutes with all $x_{kn}\left(t\right)\text{.}$ Similarly, $x$ commutes with all $x_{nk}\left(t\right)$ and hence with all $x_{ij}\left(t\right)=(x_{in}\left(t\right),x_{nj}\left(1\right))\text{.}$ Thus, $x$ is in the center of $U\left(R\right)\text{.}$