Last update: 4 July 2013
Our object is to prove that if $\pi ,G\prime ,$ and $G$ are as in §6, then $(\pi ,G\prime )$ is a universal central extension of $G$ in a sense to be defined. The reader is referred to the lecturer's paper in Colloque sur la théorie des groupes algébriques, Bruxelles, 1962 and for generalities to Schur's papers in J. Reine Angew. Math. 1904, 1907, 1911.
Definition: A central extension of a group $G$ is a couple $(\pi ,G\prime )$ where $G\prime $ is a group, $\pi $ is a homomorphism of $G\prime $ onto $G,$ and $\text{ker}\hspace{0.17em}\pi \subseteq $ center of $G\prime \text{.}$
Examples:
(a)  $\pi ,G\prime ,G$ as in §6. 
(b)  $\pi :G\prime \to G$ the natural homomorphism of one Chevalley group onto another constructed from a smaller weight lattice. E.g., $\pi :{SL}_{n}\to {PSp}_{n},$ and $\pi :{\text{Spin}}_{n}\to {SO}_{n}\text{.}$ 
(c)  $\pi :G\prime \to G$ a topological covering of a connected topological group; i.e., $\pi $ is a local isomorphism, carrying a neighborhood of $1$ isomorphically onto one of $G\text{.}$ We note that $\pi $ is central since a discrete normal subgroup of a connected group is necessarily central. To see this, let $N$ be a discrete normal subgroup of a connected group $G$ and let $n\in N\text{.}$ Since the map $G\to N$ given by $g\to gn{g}^{1},$ $g\in G,$ has a discrete and connected image, $gn{g}^{1}=n$ for all $g\in G\text{.}$ 
Definition: A central extension $(\pi ,E)$ of a group $G$ is universal if for any central extension $(\pi \prime ,E\prime )$ of $G$ there exists a unique homomorphism $\phi :E\to E\prime $ such that $\pi \prime \phi =\pi ,$ i.e., the following diagram is commutative:
$$\begin{array}{ccccc}E& \multicolumn{3}{c}{\stackrel{\phi}{\u27f6}}& E\prime \\ & \pi \searrow & & \swarrow \pi \prime & \\ & & G& & \end{array}$$We abbreviate universal central extension by u.c.e. We develop this property in a sequence of statements.
(i) If a u.c.e. exists, it is unique up to isomorphism.
Proof.  
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 $ 
(ii) If $(\pi ,E)$ is a u.c.e. of $G$ then $E=\mathcal{D}E$ and hence $G=\mathcal{D}G,$ where $\mathcal{D}H$ is the derived group of $H\text{.}$
Proof.  
Consider the central extension $(\pi \prime ,E\prime )$ where $E\prime =E\times E/\mathcal{D}E$ and $\pi \prime (a,b)=\pi \left(a\right),$ $a\in E,$ $b\in E/\mathcal{D}E\text{.}$ Now if ${\phi}_{1}\left(a\right)=(a,1)$ and ${\phi}_{2}\left(a\right)=(a,a+\mathcal{D}E),$ then $\pi \prime {\phi}_{i}=\pi ,$ $i=1,2,$ and hence ${\phi}_{1}={\phi}_{2}\text{.}$ Thus, $E/\mathcal{D}E=1$ and $E=\mathcal{D}E\text{.}$ $\square $ 
(iii) If $G=\mathcal{D}G$ and $(\pi ,E)$ is a central extension of $G,$ then $E=C\xb7\mathcal{D}E$ where $C$ is a central subgroup of $E$ on which $\pi $ is trivial. Moreover, $\mathcal{D}E={\mathcal{D}}^{2}E\text{.}$
Proof.  
We have $\pi \mathcal{D}E=\mathcal{D}\left(\pi E\right)=\mathcal{D}G=G\text{.}$ Hence, $E=C\mathcal{D}E$ where $C=\text{ker}\hspace{0.17em}\pi \text{.}$ Also, $\mathcal{D}E=\mathcal{D}\left(C\mathcal{D}E\right)={\mathcal{D}}^{2}E\text{.}$ $\square $ 
(iv) If $G=\mathcal{D}G,$ then $G$ possesses a u.c.e.
Proof.  
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 $E=\mathcal{D}F$ 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 $E=\mathcal{D}E,$ $\psi $ is trivial and $\phi =\phi \prime \text{.}$ $\square $ 
Remark: Part (a) shows that if $G$ is any group then there is a central extension covering all others.
(iv') If $(\pi ,E)$ is a central extension of $G$ which covers all others and if $E=\mathcal{D}E,$ then $(\pi ,E)$ is a u.c.e.
(v) If $\pi :E\to F$ and $\psi :F\to G$ are central extensions, then so is $\psi \pi :E\to G,$ provided $E=\mathcal{D}E\text{.}$
Proof.  
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 $ 
(vi) Exercise: In (v), $(\pi ,E)$ is a u.c.e. of $F$ if and only if $(\psi \pi ,E)$ is a u.c.e. of $G\text{.}$
Definition: A group $G$ is said to be centrally closed if $(\text{id},G)$ is a u.c.e. of $G\text{.}$
(vii) Corollary: If $(\pi ,E)$ is a u.c.e. of $G,$ then $E$ is centrally closed.
(viii) If $E$ is centrally closed, then every central extension $\psi :F\to E$ of $E$ splits; i.e, there exists a homomorphism $\phi :E\to F$ such that $\psi \phi =\text{id}\text{.}$
(ix) $(\pi ,E)$ is a u.c.e. of $G$ if and only if every diagram of the form
$$\begin{array}{ccc}E& \u290f& E\prime \\ \pi \begin{array}{c}\downarrow \end{array}& & \begin{array}{c}\downarrow \end{array}\pi \prime \\ G& \underset{\rho}{\u27f6}& G\prime \end{array}$$can be uniquely completed, where $(\pi \prime ,E\prime )$ is a central extension of $G\prime $ and $\rho $ is a homomorphism.
Proof.  
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 $ 
Definition: A linear (respectively projective) representation of a group $G$ is a homomorphism of $G$ into some $GL\left(V\right)$ (respectively $PGL\left(V\right)\text{).}$
Since $GL\left(V\right)$ is a central extension of $PGL\left(V\right)$ we have the following result:
(x) Corollary: If $(\pi ,E)$ is a u.c.e. of $G,$ then every projective representation of $G$ can be lifted uniquely to a linear representation of $E\text{.}$
(xi) Topological situation: If $G$ is a topological group one can replace the condition $G=\mathcal{D}G$ by $G$ is connected, the condition $(\pi ,E)$ is a u.c.e. by $(\pi ,E)$ is a universal covering group in the topological sense, and the condition $G$ is centrally closed by $G$ is simply connected in the above discussion and obtain similar results.
Definition: If $(\pi ,E)$ is a u.c.e of the group $G,$ then we call $\text{ker}\hspace{0.17em}\pi $ the Schur multiplier of $G\text{.}$
If we write $\text{ker}\hspace{0.17em}\pi =M\left(G\right)$ to indicate the dependence on $G,$ then a homomorphism $\phi :G\to G\prime $ leads to a corresponding one $M\left(\phi \right):M\left(G\right)\to M\left(G\prime \right)$ by (ix). Thus $M$ is a functor from the category of groups $G$ such that $G=\mathcal{D}G$ to the category of Abelian groups with the following property: if $\phi $ is onto, then so is $M\left(\phi \right)\text{.}$
Remark: Schur used different definitions (and terminology) since he considered only finite groups but did not require that $G=\mathcal{D}G\text{.}$ If $G=\mathcal{D}G$ our definitions are equivalent to his. One of Schur's results, which we shall not use, is that if $G$ is finite then so is $M\left(G\right)\text{.}$
Theorem 10: Let $\Sigma $ be an indecomposable root system and $k$ a field such that $\leftk\right>4$ and, if $\text{rank}\hspace{0.17em}\Sigma =1,$ then $\leftk\right\ne 9\text{.}$ If $G$ is the corresponding universal Chevalley group (abstractly defined by the relations (A),(B),(B'),(C) of §6), if $G\prime $ is the group defined by the relations (A),(B),(B') (we use (B') only if $\text{rank}\hspace{0.17em}\Sigma =1\text{),}$ and if $\pi $ is the natural homomorphism from $G\prime $ to $G,$ then $(\pi ,G\prime )$ is a u.c.e. of $G\text{.}$
Remark: There are exceptions to the conclusion. E.g. ${SL}_{2}\left(4\right)$ and ${SL}_{2}\left(9\right)$ are such. Indeed ${SL}_{2}\left(4\right)\cong {PSL}_{2}\left(5\right)$ and ${SL}_{2}\left(5\right)$ is a central extension of ${PSL}_{2}\left(5\right)\text{.}$ For ${SL}_{2}\left(9\right)$ see Schur. It can be shown that the number of couples $(\Sigma ,k)$ for which the conclusion fails is finite.
Proof.  
Since $\leftk\right>4,$ $G=\mathcal{D}G,$ $G\prime =\mathcal{D}G\prime ,$ 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 $E=\mathcal{D}E\text{;}$ 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 $d=4\u27e8\alpha ,\beta \u27e9\u27e8\beta ,\alpha \u27e9$ by Lemma 20(c) and by (4) with $h={h}_{\alpha}\left({v}^{2}\right){h}_{\beta}\left({v}^{\u27e8\beta ,\alpha \u27e9}\right)\text{.}$ 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 $\u27e8\alpha ,\gamma \u27e9=1,$ 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 $1v+{v}^{2}\ne 0\text{.}$ By $(*)$ and (6), $f(t(v{v}^{2}),u)=f(t,u/(v{v}^{2}))=f(t,u/v)f(t,u/(1v))=f(vt,u)f((1v)t,u)=f(t,u),$ whence $f(t(1v+{v}^{2}),u)=1$ 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 $\leftk\right$ 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 $v,1v$ are squares and $v{v}^{2}\ne 0,$ $1v+{v}^{2}\ne 0\text{.}$ 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 $1v={((1{w}^{2})/(1+{w}^{2}))}^{2},$ and we need only choose $w$ so that $1+{w}^{2}\ne 0,$ $v{v}^{2}\ne 0,$ and $1v+{v}^{2}\ne 0\text{.}$ Since at most $13$ values of $w$ are to be avoided and $\leftk\right\ge 25$ 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 $ 
Exercise: Assume $\mathcal{L}$ is the original Lie algebra with coefficients transferred by means of a Chevalley basis to a field $k$ whose characteristic does not divide any ${N}_{\alpha ,\beta}\ne 0\text{.}$ Also assume $\Sigma $ is indecomposable of $\text{rank}>1\text{.}$ Prove:
(a)  The relations $[{X}_{\alpha},{X}_{\beta}]={N}_{\alpha ,\beta}{X}_{\alpha +\beta},\alpha +\beta \ne 0,$ form a defining set for $\mathcal{L}\text{.}$ Hint: define ${H}_{\alpha}=[{X}_{\alpha},{X}_{\alpha}]$ and show that the relations of Theorem 1 hold. 
(b)  $\mathcal{L}=\mathcal{D}\mathcal{L},$ the derived algebra of $\mathcal{L}\text{.}$ 
(c)  Every central extension of $\mathcal{L}$ splits. Hint: parallel the proof of Theorem 10. 
Corollary 1: The relations (A),(B),(B') can be lifted to any central extension of $G\text{.}$
Corollary 2:
(a)  $G\prime $ is centrally closed. Each of its central extensions splits. Its Schur multiplier is trivial. It yields the u.c.e. of all the Chevalley groups of the given type, and covers linearly all of the projective representations of these groups. 
(b)  If $k$ is finite or more generally an algebraic extension of a finite field, then (a) holds with $G\prime $ replaced by $G\text{.}$ 
Proof.  
This follows from various of the generalities at the beginning of this section. E.g., if $k$ is finite, $\leftk\right>4$ 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 $ 
Corollary 3: Assume $G,G\prime ,$ and $\pi $ are as above. If ${k}^{*}$ is infinite and divisible $\text{(}u\in {k}^{*},n\in \mathbb{Z}$ implies there exists $v\in {k}^{*}$ with ${v}^{n}=u\text{),}$ then the Schur multiplier of $G\text{;}$ i.e., $C=\text{ker}\hspace{0.17em}\pi ,$ is also divisible.
Proof.  
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 $ 
Corollary 3a: If ${k}^{*}$ is infinite and divisible by a set of primes including 2, then $C$ is also divisible by these primes.
Corollary 3b: If ${k}^{*}$ is infinite and divisible, then any central extension of $G$ by a kernel which is a reduced group (no divisible subgroups other than 1) is trivial; i.e. it splits.
Proof.  
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 $ 
Corollary 3c: If ${k}^{*}$ is infinite and divisible, then any finite dimensional projective representation of $G$ can be lifted uniquely to a linear representation.
Proof.  
Assume $\sigma :G\to PGL\left(V\right)\text{.}$ Since $G=\mathcal{D}G,$ 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 $ 
Example: Corollary 3c says, for example, that every finite dimensional representation of ${SL}_{n}\left(\u2102\right)$ can be lifted to a linear one. (The novelty is that the representation is not assumed to be continuous.)
Theorem 11: If $\Sigma $ is an indecomposable root system, if $\text{char}\hspace{0.17em}k=p\ne 0,$ and if $G=\mathcal{D}G$ (i.e. we exclude $\leftk\right=2,$ $\Sigma $ of type ${A}_{1},{B}_{2},$ or ${G}_{2}$ and $\leftk\right=3,$ $\Sigma $ of type ${A}_{1}\text{),}$ then $(\pi ,G\prime )$ uniquely covers all central extensions of $G$ for which the kernel has no $p\text{torsion.}$
Proof.  
By Theorem 10, we could assume $\leftk\right\le 4$ or $\leftk\right=9\text{.}$ 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 $ 
Corollary 1: Every projective representation of $G\prime $ in a field of characteristic $p$ can be lifted to a linear one.
Corollary 2: The Schur multiplier of $G\prime $ is a $p\text{group.}$
Proof.  
These are easy exercises. $\square $ 
Since the kernel of the map $\pi :G\prime \to G$ above turns out to be the Schur multiplier of $G,$ its structure for $k$ arbitrary is of some interest. The result is:
Theorem 12: (Moore, Matsumoto) Assume $\Sigma $ is an indecomposable root system and $k$ a field with $\leftk\right>4\text{.}$ If $G$ is the universal Chevalley group based on $\Sigma $ and $k,$ if $G\prime $ is the group defined by (A), (B), (B'), and if $\pi $ is the natural map from $G\prime $ to $G$ with $C=\text{ker}\hspace{0.17em}\pi ,$ the Schur multiplier of $G,$ then $C$ is isomorphic to the abstract group $A$ generated by the symbols $f(t,u)$ $(t,u\in {k}^{*})$ subject to the relations:
(a)  $f(t,u)f(tu,v)=f(t,uv)f(u,v),$ $f(1,u)=f(u,1)=1$ 
(b)  $f(t,u)f(t,{u}^{1})=f(t,1)$ 
(c)  $f(t,u)=f({u}^{1},t)$ 
(d)  $f(t,u)=f(t,tu)$ 
(e)  $f(t,u)=f(t,(1t)u)$ 
(ab')  $f$ is bimultiplicative. 
(c')  $f$ is skew 
(d')  $f(t,t)=1$ 
(e')  $f(t,1t)=1\text{.}$ 
Remark: These relations are satisfied by the norm residue symbol in class field theory, which is a significant aspect of Moore's work.
Partial Proof.  
(1) If ${\U0001d525}_{\alpha}$ is the group generated (in G') by all ${h}_{\alpha}\left(t\right),$ $\alpha $ a fixed long root, then $C\subseteq {\U0001d525}_{\alpha}\text{.}$ 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, $\u27e8\beta ,\alpha \u27e9=1$ and $({h}_{\alpha}\left(t\right),{h}_{\beta}\left(u\right))={h}_{\beta}\left(tu\right){h}_{\beta}{\left(t\right)}^{1}{h}_{\beta}{\left(u\right)}^{1}={h}_{\alpha}\left(t\right){h}_{\alpha}\left({u}^{\u27e8\alpha ,\beta \u27e9}\right){h}_{\alpha}{\left(t{u}^{\u27e8\alpha ,\beta \u27e9}\right)}^{1}$ 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 $\u27e8\alpha ,\gamma \u27e9=1\text{.}$ 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 $ 
Now let $G$ be a connected topological group. A covering of $G$ is a couple $(\pi ,E)$ such that $E$ is a connected topological group and $\pi $ is a homomorphism of $E$ onto $G$ which maps a neighborhood of 1 in $E$ homeomorphically onto a neighborhood of 1 in $G\text{;}$ i.e., which is a local isomorphism. A covering is universal if it covers all other covering groups. If $(\text{id},G)$ is a universal covering, we say that $G$ is simply connected.
Remarks:
(a)  A covering $(\pi ,E)$ of a connected group is necessarily central as was noted at the beginning of this section. 
(b)  If a universal covering exists, then it is unique and each of its coverings of other covering groups is unique. This follows from the fact that a connected group is generated by any neighborhood of 1. 
(c)  If $G$ is a Lie group, then a universal covering for $G$ exists and simple connectedness is equivalent to the property that every continuous loop can be shrunk to a point (See Chevalley, Lie Groups or Cohn, Lie Groups.) 
Theorem 13: If $G$ is a universal Chevalley group over $\u2102$ viewed as a Lie group, then $G$ is simply connected.
Before proving Theorem 13, we shall first state a lemma whose proof we leave as an exercise.
Lemma 41: If ${t}_{1},{t}_{2},\dots ,{t}_{n}$ are complex numbers such that $\left{t}_{1}\right<\epsilon ,i=1,2,\dots ,n$ and $\sum _{i=1}^{n}{t}_{i}=0,$ there exist ${t}_{i}$ and ${t}_{j}$ such that ${t}_{i}+{t}_{j}<\epsilon \text{.}$
Proof of Theorem 13.  
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 $$\begin{array}{cc}\text{(A)}& \phi {x}_{\alpha}\left(t\right)\phi {x}_{\alpha}\left(u\right)=\phi {x}_{\alpha}(t+u)\phantom{\rule{2em}{0ex}}\alpha \in \Sigma \text{.}\end{array}$$Since $\phi $ is locally an isomorphism, there is $\epsilon >0$ such that (A) holds for $\leftt\right<\epsilon ,$ $\leftu\right<\epsilon \text{.}$ 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 ${\U0001d51b}_{\alpha}$ is topologically equivalent to $\u2102$ and hence simply connected. Thus, $\phi $ extends to a homomorphism of ${\U0001d51b}_{\alpha}$ into $E$ and (A) holds. Clearly the extension of $\phi $ to ${\U0001d51b}_{\alpha}$ 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 ${\U0001d51b}_{S}$ be the corresponding unipotent subgroup of $G\text{.}$ Topologically, ${\U0001d51b}_{S}$ is equivalent to ${\u2102}^{n}$ for some $n,$ and is hence simply connected. As before $\phi $ can be extended to a homomorphism of ${\U0001d51b}_{S}$ into $E,$ and the relations (B) hold. This extension is consistent with those above, by the uniqueness of the latter. We now consider ${h}_{\alpha}\left(t\right)={x}_{\alpha}\left(t\right){x}_{\alpha}({t}^{1}){x}_{\alpha}\left(t\right)\text{.}$ ${w}_{\alpha}(1)={x}_{\alpha}(t1){x}_{\alpha}{(1{t}^{1})}^{{x}_{\alpha}(1)}{x}_{\alpha}{(t1)}^{{w}_{\alpha}(1)}$ where ${x}^{y}={y}^{1}xy\text{.}$ Hence, if $t$ is near 1 in $\u2102,$ 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 $\u2102$ has square roots, we have $\phi {h}_{\alpha}$ is multiplicative, i.e., (C) holds. $\square $ 
Examples: ${SL}_{n}\left(\u2102\right),$ ${Sp}_{n}\left(\u2102\right),$ and ${\text{Spin}}_{n}\left(\u2102\right)$ are simply connected. These cases can also be proved by induction on $n\text{.}$ (See Chevalley, Lie Groups, Chapter II.)
Remarks:
(a) If $\u2102$ is replaced by $\mathbb{R}$ in the preceding discussion, then relations (A), (B) can be lifted exactly as before. Also $\phi {h}_{\alpha}$ is still multiplicative if one of the two arguments is positive. Further $\text{ker}\hspace{0.17em}\pi $ is generated by $\phi {h}_{\alpha}{(1)}^{2},$ $\alpha $ a fixed long root, and, if type ${C}_{n}$ $(n\ge 1)$ is excluded, then ${h}_{\alpha}{(1)}^{4}=1$ or ${w}_{\alpha}{(1)}^{8}=1\text{.}$ Prove all of this.
(b) Moore has constructed a universal covering of $G$ and has determined the fundamental group in case $k$ is a $p\text{adic}$ field, using appropriate modifications of the definitions (here $G$ is totally disconnected.)
(c) Let $G$ be a Chevalley group over $k,G\prime $ the corresponding universal group, and $\pi :G\prime \to G$ the natural homomorphism. If $k$ is algebraically closed and if only appropriate coverings are allowed, then $(\pi ,G\prime )$ is a universal covering of $G$ in the sense of algebraic groups.
We close this section with a result in which the coefficients may come from any ring (associative with 1). The development is based in part on a letter from J. Milnor. Let $R$ be the ring, and let $GL\left(R\right)$ be the group of infinite matrices which are equal to the identity everywhere except for a finite invertible block in the upper left hand corner. Thus, ${GL}_{n}\left(R\right)\subset GL\left(R\right),$ $n=1,2,\dots \text{.}$ Let $E\left(R\right)$ be the subgroup of $GL\left(R\right)$ generated by the elementary matrices $1+t{E}_{ij}$ $(t\in R,i\ne j,i,j=1,2,\dots ),$ where ${E}_{ij}$ is the usual matrix unit. For example, if $R$ is a field, then $E\left(R\right)=SL\left(R\right),$ a simple group whose double coset decomposition involves the infinite symmetric group. Indeed, if $R$ is a Euclidean domain, then $E\left(R\right)=SL\left(R\right)\text{.}$
Lemma 42:
(a)  $E\left(R\right)=\mathcal{D}GL\left(R\right)$ 
(b)  $E\left(R\right)=\mathcal{D}E\left(R\right)$ 
Proof.  
The relation $(1+t{E}_{ik},1+{E}_{kj})=1+t{E}_{ij}$ shows (b) and hence also $E\left(R\right)\subseteq \mathcal{D}GL\left(R\right)\text{.}$ 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 $ 
We call ${K}_{1}\left(R\right)=GL\left(R\right)/E\left(R\right)$ the Whitehead group of $R\text{.}$ This concept is used in topology. The case in which $R=\mathbb{Z}\left[G\right]$ is of particular interest.
Example: If $R$ is a Euclidean domain, then ${K}_{1}\left(R\right)={R}^{*},$ the group of units. (See Milnor, Whitehead Torsion.)
By Lemma 42 and (iv), $E\left(R\right)$ has a u.c.e. $(\pi ,U\left(R\right))\text{.}$ Set ${K}_{2}\left(R\right)=\text{ker}\hspace{0.17em}\pi \text{.}$ This notation is partly motived by the following exact sequence.
$$1\u27f6{K}_{2}\left(R\right)\u27f6U\left(R\right)\u27f6GL\left(R\right)\u27f6{K}_{1}\left(R\right)\u27f61\text{.}$$${K}_{2}$ is a functor from rings to Abelian groups with the following property: if $R\to R\prime $ is onto, then so is the associated map ${K}_{2}\left(R\right)\to {K}_{2}\left(R\prime \right)\text{.}$
Remark: ${K}_{2}$ is known to the lecturer in the following cases:
(a)  If $R$ is a finite field (or an algebraic extension of a finite field), then ${K}_{2}=1\text{.}$ 
(b)  If $R$ is any field, see Theorem 12. 
(c)  If $R=\mathbb{Z},$ then $\left{K}_{2}\right=2\text{.}$ 
Theorem 14: Let $U\left(R\right)$ be the abstract group generated by the symbols ${x}_{ij}\left(t\right)$ $(t\in R,i\ne j,i,j=1,2,\dots )$ subject to the relations
(A)  ${x}_{ij}\left(t\right)$ is additive in $t\text{.}$ 
(B)  $({x}_{ik}\left(t\right),{x}_{\ell j}\left(u\right))=\{\begin{array}{cc}{x}_{ij}\left(tu\right)& \text{if}\hspace{0.17em}k=\ell ,i\ne j\text{.}\\ 1& \text{if}\hspace{0.17em}k\ne \ell ,i\ne j\text{.}\end{array}$ 
Proof.  
(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 $U\left(R\right)\text{.}$ (b) $\pi $ is universal. From (B), it follows that $U\left(R\right)=\mathcal{D}U\left(R\right)\text{.}$ Hence it suffices to show it covers all central extensions. Let $(\psi ,A)$ be a central extension of $E\left(R\right)$ and let $C$ be the center of $A\text{.}$ We must show that we can lift the relations (A) and (B) to $A\text{.}$ Fix $i,j,i\ne j$ and choose $p\ne i,j\text{.}$ Choose ${y}_{ij}\left(t\right)\in {\psi}^{1}{x}_{ij}\left(t\right)$ so that $(*)$ $({y}_{ip}\left(t\right),{y}_{pj}\left(1\right))={y}_{ij}\left(t\right)\text{.}$ We will prove that the $y\text{'s}$ satisfy the equations (A) and (B). (b1) If $i\ne j,k\ne \ell ,$ then ${y}_{ik}\left(t\right)$ and ${y}_{\ell j}\left(u\right)$ commute. Choose $q\ne i,j,k,\ell $ and write ${y}_{\ell j}\left(u\right)=c({y}_{\ell q}\left(u\right),{y}_{qj}\left(1\right)),c\in C\text{.}$ Since ${y}_{ik}\left(t\right)$ commutes up to an element of $C$ with ${y}_{\ell q}\left(u\right)$ and ${y}_{qj}\left(1\right),$ it commutes with ${y}_{\ell j}\left(u\right)\text{.}$ Hence (b2) $\left\{{y}_{ij}\left(t\right)\right\},$ $i,j$ fixed, is Abelian. (b3) The relations (A) hold. The proof is exactly the same as that of statement (7) in the proof of Theorem 10. (b4) ${y}_{ij}\left(t\right)$ in $\left(*\right)$ is independent of the choice of $p\text{.}$ If $q\ne p,i,j,$ set $w={y}_{qp}\left(1\right){y}_{pq}(1){y}_{qp}\left(1\right)\text{.}$ Transforming $(*)$ by $w$ and using (b1) we get $(*)$ with $q$ in place of $p\text{.}$ (b5) The relations (B) hold. We will use: $(**)$ If, $a,b,c$ are elements of a group such that $a$ commutes with $c$ and such that $(b,c)$ commutes with $(a,b)$ and $c,$ then $(a,(b,c))=((a,b),c)\text{.}$ Since a commutes with $c,(a,(b,c))=((a,b),(b,c),c)\text{.}$ The other conditions insure $((a,b),(b,c)c)=((a,b),c)\text{.}$ Now assume $i,j,k$ are distinct. Choose $q\ne i,j,k,$ so that $({y}_{ik}\left(t\right),{y}_{kj}\left(u\right))=({y}_{ik}\left(t\right),({y}_{kq}\left(u\right),{y}_{qj}\left(1\right)))=(({y}_{ik}\left(t\right),{y}_{kq}\left(u\right)),{y}_{qj}\left(1\right))=({y}_{iq}\left(tu\right),{y}_{qj}\left(1\right))={y}_{ij}\left(tu\right)$ by $(*),$ $(**),$ and (b4). This completes the proof of the theorem. $\square $ 
Let ${U}_{n}\left(R\right)$ denote the subgroup of $U\left(R\right)$ generated by ${y}_{ij}\left(t\right)$ with $i,j\le n\text{.}$
Corollary 1: If $n\ge 5,$ then ${U}_{n}\left(R\right)$ is centrally closed.
Corollary 2: If $R$ is a finite field and $n\ge 5$ then ${SL}_{n}\left(R\right)$ is centrally closed.
Proof.  
This follows from Corollary 1 and the equations ${E}_{n}\left(R\right)={SL}_{n}\left(R\right)={U}_{n}\left(R\right)\text{.}$ $\square $ 
Remarks: (a) It follows that if $R$ is a finite field and if ${SL}_{n}\left(R\right)$ is not centrally closed, then either $\leftR\right=9,$ $n=2$ or $\leftR\right\le 4$ and $n\le 4\text{.}$ The exact set of exceptions is: ${SL}_{2}\left(4\right),$ ${SL}_{2}\left(9\right),$ ${SL}_{3}\left(2\right),$ ${SL}_{3}\left(4\right),$ ${SL}_{4}\left(2\right)\text{.}$
$*$ Exercise: Prove this.
(b) The argument above can be phrased in terms of roots, etc. As such, it carries over very easily to the case in which all roots have one length. The only other exception is ${D}_{4}\left(2\right)\text{.}$
(c) By a more complicated extension of the argument, it can also be shown that the universal Chevalley group of type ${B}_{n}$ or ${C}_{n}$ over a finite field (or an algebraic extension of a finite field) is centrally closed if $n$ is large enough. Hence, only a finite number of universal Chevalley groups with $\Sigma $ indecomposable and $k$ finite fail to be centrally closed.
Now we sketch a proof that ${K}_{2}\left(\mathbb{Z}\right)$ is a group of order $2\text{.}$ The notation $U,{U}_{n},\dots $ above will be used. The proof depends on the following result:
(1) For $n\ge 3,$ ${SL}_{n}\left(\mathbb{Z}\right)$ is generated by symbols ${x}_{ij}(i,j=1,2,\dots ,n;i\ne j)$ subject to the relations
(B)  $({x}_{ik},{x}_{\ell j})=\{\begin{array}{cc}{x}_{ij}& \text{if}\hspace{0.17em}k=\ell ,i\ne j,\\ 1& \text{if}\hspace{0.17em}k\ne \ell ,i\ne j\text{.}\end{array}$ 
(C)  If ${w}_{ij}={x}_{ij}{x}_{ij}^{1}{x}_{ij},$ ${h}_{ij}={w}_{ij}^{2},$ then ${h}_{ij}^{2}=1\text{.}$ 
Identifying ${x}_{ij}$ with the usual ${x}_{ij}\left(1\right)$ and using ${x}_{ij}\left(t\right)={x}_{ij}{\left(1\right)}^{t},$ we see that the relations (B) here imply those of Theorem 14. Since the last relation may be written ${h}_{ij}{(1)}^{2}={h}_{ij}\left(1\right)$ and $\pm 1$ are the only units of $\mathbb{Z},$ we have ${SL}_{n}\left(\mathbb{Z}\right)$ defined by the usual relations (A), (B), (C) of §6. Perhaps there are other rings, e.g., the $p\text{adic}$ integers, for which this result holds. For the proof of (1) see W. Magnus, Acta Math. 64 (1934), which gives the reference to Nielsen, who proved the key case $n=3$ (it takes some work to cast Nielsen's result into the above form). The case $n=2,$ with (B) replaced by (B') ${w}_{12}{x}_{12}{w}_{12}^{1}={x}_{21}^{1},$ is simpler and is proved in an appendix to Kurosh, Theory of Groups.
Now let ${x}_{ij},{w}_{ij},{h}_{ij}$ refer to elements of ${U}_{n}\left(\mathbb{Z}\right)\text{.}$
(2) If ${C}_{n}$ is the kernel of ${\pi}_{n}:{U}_{n}\left(\mathbb{Z}\right)\to {SL}_{n}\left(\mathbb{Z}\right)$ and $n\ge 3,$ then ${C}_{n}$ is generated by ${h}_{12}^{2}m$ and ${\left({h}_{12}^{2}\right)}^{2}=1\text{.}$
As usual, we only require (C) when $i,j=1,2,$ and ${h}_{12}^{2}\in {C}_{n}$ Setting ${h}_{12}^{2}=1$ amounts to dividing by $\u27e8{h}_{12}^{2}\u27e9,$ which thus equals ${C}_{n}\text{.}$ The relation ${h}_{23}{h}_{12}{h}_{23}^{1}={h}_{12}^{1},$ which may be deduced from (B) as in the proof of Lemma 37, then yields ${h}_{12}^{2}={h}_{12}^{2}\text{.}$
(3) ${h}_{12}^{2}\ne 1$ if $n\ge 3\text{.}$
Assume not. There is a natural map ${U}_{n}\left(\mathbb{Z}\right)\to {U}_{n}\left(\mathbb{R}\right),$ ${x}_{ij}\to {x}_{ij}\left(1\right)\text{.}$ This maps ${h}_{12}^{2}$ onto ${h}_{12}{(1)}^{2},$ which (see Remark (a) after the proof of Theorem 13) generates the kernel of ${U}_{n}\left(\mathbb{R}\right)\to {SL}_{n}\left(\mathbb{R}\right)\text{.}$ Thus ${SL}_{n}\left(\mathbb{R}\right)$ is centrally closed, hence simply connected. Since ${SL}_{n}\left(\mathbb{R}\right)$ can be contracted to ${SO}_{n}$ (by the polar decomposition, which will be proved in the next section) which is not simply connected since ${\text{Spin}}_{n}\to {SO}_{n}$ is a nontrivial covering, we have a contradiction.
It now follows from (2), (3) and Theorem 14 that $\left{K}_{2}\left(\mathbb{Z}\right)\right=2\text{.}$
By Corollary 1 above the same conclusion holds with $SL\left(\mathbb{Z}\right)$ replaced by any ${SL}_{n}\left(\mathbb{Z}\right)$ with $n\ge 5\text{.}$
Exercise: Let ${SA}_{n}$ be ${SL}_{n}\hspace{0.17em}x$ translations of the underlying space, ${k}_{n}$ with $k$ again a field. I.e., ${SA}_{n}$ is the group of all $(n+1)\times (n+1)$ matrices of the form $\left[\begin{array}{cc}x& y\\ 0& 1\end{array}\right]$ where $x\in {SL}_{n},$ $y\in {k}^{n}\text{.}$ ${SA}_{n}$ is generated by ${x}_{ij}\left(t\right),$ $t\in k,$ $i\ne j,$ $i=1,2,\dots ,n,$ $j=1,2,\dots ,n+1\text{.}$ Prove:
(1) 
If the relation


(2)  If $k$ is finite, (C) may be omitted.  
(3)  If $n$ is large enough, the group defined by (A) and (B) is a u.c.e. for ${SA}_{n}\text{.}$  
(4)  Other analogues of results for ${SL}_{n}\text{.}$ 
This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract AROD33682303143033.