Last update: 1 August 2013

The main purpose of this section is to prove the following theorem:

*Theorem 5* (Chevalley, Dickson): Let $G$ be an adjoint group and assume $\mathcal{L}$ is simple
$\text{(}\Sigma $ indecomposable). If
$\left|k\right|=2,$ assume $\mathcal{L}$ is not of type
${A}_{1},{B}_{2},$ or
${G}_{2}\text{.}$ If $\left|k\right|=3,$
assume $\mathcal{L}$ is not of type ${A}_{1}\text{.}$ Then $G$ is simple.

*Remark:* The cases excluded in Theorem 5 must be excluded. If $\left|k\right|=2,$
then $G$ has ${\mathcal{A}}_{3},{\mathcal{A}}_{6},{SU}_{3}\left(3\right)$
as a normal subgroup of index 2 if $\mathcal{L}$ is of type ${A}_{1},$
${B}_{2},$ ${G}_{2}$ respectively. If
$\left|k\right|=3$ and $\mathcal{L}$ is of type
${A}_{1},$ then ${\mathcal{A}}_{4}$ is a normal subgroup of
$G$ of index 2. Here $\mathcal{A}$ denotes the alternating group.

A proof of Theorem 5 essentially due to Iwasawa and Tits will be given here in a sequence of lemmas.

*Lemma 29:* Let $G$ be a Chevalley group. If $w\in W,$
$w={w}_{\alpha}{w}_{\beta}\dots $
is a minimal expression as a product of simple reflections, then
${w}_{\alpha},{w}_{\beta},\dots \in {G}_{1},$
the group generated by $B$ and $wB{w}^{-1}\text{.}$

Proof. | |

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 ${G}_{1}\supseteq w{\U0001d51b}_{\beta}{w}^{-1}={\U0001d51b}_{w\beta}={\U0001d51b}_{-\alpha}\text{.}$ 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 $ |

Lemma 30: If $G$ again is any Chevalley group, if $\pi $ is a subset of the set of simple roots, if ${W}_{\pi}$ is the group generated by all ${w}_{\alpha},\alpha \in \pi ,$ and if ${G}_{\pi}=\underset{w\in {W}_{\pi}}{\cup}BwB,$ then

(a) | ${G}_{\pi}$ is a group. |

(b) | The ${2}^{\ell}$ groups so obtained are all distinct. |

(c) | Every subgroup of $G$ containing $B$ is equal to one of them. |

Proof. | |

Part (a) follows from $BwB\xb7B{w}_{\alpha}B\subseteq Bw{w}_{\alpha}B\cup BwB\text{.}$ (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 $ |

A group conjugate to some $G$ is called a *parabolic subgroup* of $G\text{.}$
We state without proof some further properties of parabolic subgroups which follow from Lemma 29.

(1) | No two ${G}_{\pi}\text{'s}$ are conjugate. |

(2) | Each parabolic subgroup is its own normalizer. |

(3) | ${G}_{\pi}\cap {G}_{\pi \prime}={G}_{\pi \cap \pi \prime}\text{.}$ |

(4) | $B\cup BwB$ $(w\in W)$ is a group if and only if $w=1$ or $w$ is a simple reflection. |

*Example:* If $G={SL}_{n},$ then $\pi $
corresponds to a partition of the $n\times n$ matrices into blocks with the diagonal blocks being
square matrices. Clearly, there are ${2}^{n-1}$ possibilities for such partitions.
${G}_{\pi}$ is then the subset of ${SL}_{n}$
of matrices whose subdiagonal blocks are zero.

*Lemma 31:* Let $\mathcal{L}$ be simple and let $G$ be the adjoint Chevalley group. If
$N\ne 1$ is a normal subgroup of $G,$ then
$NB=G\text{.}$

Proof. | |

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 $N\u2288B,$ 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 $\gamma ={w}_{\beta}\alpha =\alpha -\u27e8\alpha ,\beta \u27e9\beta \text{.}$ Since $\u27e8\alpha ,\beta \u27e9\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 $ |

*Lemma 32:* If $\mathcal{L}$ and $G$ are as in Theorem 5, then $G=G\prime ,$ the
derived group of $G\text{.}$

Before proving Lemma 32, we first show that Theorem 5 follows from Lemmas 31 and 32. Let $N\ne 1$ be a a normal subgroup of $G\text{.}$ By Lemma 31, $NB=G$ so $G/N\simeq B/B\cap N\text{.}$ Now $G/N$ equals its derived group and $B/B\cap N$ is solvable. Hence $G/N=1$ and $N=G\text{.}$

Instead of proving Lemma 32 directly, we prove the following stronger statement:

*Lemma 32':* If $\mathcal{L}$ is as in Theorem 5 then $G\prime =G$
holds in any group $G$ in which the relations (R) hold, in fact in which the relations:

(A) | $({x}_{\beta}\left(t\right),{x}_{\gamma}\left(u\right))=\Pi {x}_{i\beta +j\gamma}\left({c}_{ij}{t}^{i}{u}^{j}\right)$ |

(B) | ${h}_{\alpha}\left(t\right){x}_{\alpha}\left(u\right){h}_{\alpha}{\left(t\right)}^{-1}={x}_{\alpha}\left({t}^{2}u\right)$ |

Proof. | |||||||||||

Since $G$ is generated by the ${\U0001d51b}_{\alpha}\text{'s}$ we must show that every ${\U0001d51b}_{\alpha}\subseteq G\prime \text{.}$ 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 ${\U0001d51b}_{\alpha}\subseteq G\prime \text{.}$ 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 ${x}_{\beta +\gamma}\left({N}_{\beta ,\gamma}tu\right)\xb7\Pi \prime ,$ 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 ${\U0001d51b}_{\alpha}\subseteq G\prime ,$ 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 $ |

Because of the exclusions in the theorem, this leaves the following cases:

(6) | ${B}_{\ell}$ $(\ell \ge 2),$ $\alpha $ short. |

(7) | ${G}_{2},$ $\alpha $ short, $\left|k\right|=$ 3\text{.}$$ |

(8) | ${C}_{\ell}$ $(\ell \ge 3),$ $\alpha $ long, $\left|k\right|=2\text{.}$ |

(c) If (6) or (7) holds, then ${\U0001d51b}_{\alpha}\subseteq G\prime \text{.}$ in both of these cases we can find roots $\beta ,\gamma $ so that $\alpha =\beta +\gamma ,$ all other roots $i\beta +j\gamma $ $\text{(}i,j$ positive integers) are long, and ${N}_{\beta \gamma}\ne 0\text{:}$ in (6) we can choose $\beta $ long and $\gamma $ short, in (7) both short. Then $\Pi \prime $ belongs to $G\prime $ by cases already treated, hence so does ${\U0001d51b}_{\alpha},$ by (A).

(d) If (8) holds, then ${\U0001d51b}_{\alpha}\subseteq G\prime \text{.}$ Choose roots $\beta ,\gamma $ with $\beta $ long, $\gamma $ short, and $\alpha =\beta +2\gamma \text{.}$ Since ${\U0001d51b}_{\beta +\gamma}\subseteq G\prime $ because $\beta +\gamma $ is short, our assertion will follow from ${C}_{12}\ne 0$ in (A), hence from the next lemma.

*Lemma 33:* If $\beta $ and $\gamma $ form a simple system of type ${B}_{2}$ with
$\beta $ long and $\gamma $ short, then
$({x}_{\beta}\left(t\right),{x}_{\gamma}\left(u\right))={x}_{\beta +\gamma}(\pm tu){x}_{\beta +2\gamma}(\pm t{u}^{2})\text{.}$

Proof. | |

By Lemma 14, we have $${x}_{\gamma}\left(u\right){X}_{\beta}{x}_{\gamma}{\left(u\right)}^{-1}=\text{exp}\hspace{0.17em}\left(\text{ad}\hspace{0.17em}u{X}_{\gamma}\right){X}_{\beta}={X}_{\beta}+u{N}_{\gamma ,\beta}{X}_{\beta +\gamma}+{u}^{2}{N}_{\gamma ,\beta}{N}_{\gamma ,\beta +\gamma}/2{X}_{\beta +2\gamma}\text{.}$$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 $ |

The proof of Theorem 5 is now complete.

In the course of this discussion, we have established the following result.

*Corollary:* If $\Sigma $ is indecomposable and of rank $>1$ and if
$\alpha $ is any root, then there exist roots $\beta $ and $\gamma $ and a positive integer $n$
such that $\alpha =\beta +n\gamma $ and
${c}_{1n}\ne 0$ in the relations (A) of Lemma 32'.

*Corollary* (To Theorem 5): If $\left|k\right|\ge 4$ and $G$
is a Chevalley group based on $k,$ then every solvable normal subgroup of $G$ is central and hence finite.

Proof. | |

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 $G={G}_{1}\xb7{G}_{2}\dots {G}_{r}$ 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 $ |

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 ARO-D-336-8230-31-43033.