Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 1 August 2013

§4. Simplicity of G

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

Theorem 5 (Chevalley, Dickson): Let G be an adjoint group and assume is simple (Σ indecomposable). If |k|=2, assume is not of type A1,B2, or G2. If |k|=3, assume is not of type A1. Then G is simple.

Remark: The cases excluded in Theorem 5 must be excluded. If |k|=2, then G has 𝒜3,𝒜6,SU3(3) as a normal subgroup of index 2 if is of type A1, B2, G2 respectively. If |k|=3 and is of type A1, then 𝒜4 is a normal subgroup of G of index 2. Here 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 wW, w=wαwβ is a minimal expression as a product of simple reflections, then wα,wβ,G1, the group generated by B and wBw-1.


We know w-1α<0 by the minimality of the expression (see Appendix II.19 and II.22). Hence if β=-w-1α>0, then G1w𝔛βw-1=𝔛wβ=𝔛-α. Thus, wαG1. Since wαwBw-1wα-1G1 and since lengthwαw<lengthw, we may complete the proof by induction.

Lemma 30: If G again is any Chevalley group, if π is a subset of the set of simple roots, if Wπ is the group generated by all wα,απ, and if Gπ=wWπBwB, then

(a) Gπ is a group.
(b) The 2 groups so obtained are all distinct.
(c) Every subgroup of G containing B is equal to one of them.


Part (a) follows from BwB·BwαBBwwαBBwB. (b) Suppose π,π are distinct subsets of the set of simple roots, say απ,απ. Now wαα=-α and wα=α+ΣβπCββ if wWπ. Thus wααwα, since simple roots are linearly independent. Hence, wαWπ, WπWπ, and GπGπ since distinct elements of the Weyl group correspond to distinct double cosets. (c) Let A be any subgroup containing B. Set π={α|αsimple,wαA}. We shall show A=Gπ. Clearly, AGπ. Since G=wWBwB and AB, we need only show wA implies wGπ to get AGπ. Let wA, w=wαwβ, a minimal expression of w as a product of simple reflections. By Lemma 29, wα,wβ,A. Hence, α,β,π, wWπ, and wGπ.

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

(1) No two Gπ's are conjugate.
(2) Each parabolic subgroup is its own normalizer.
(3) GπGπ=Gππ.
(4) BBwB (wW) is a group if and only if w=1 or w is a simple reflection.

Example: If G=SLn, then π corresponds to a partition of the n×n matrices into blocks with the diagonal blocks being square matrices. Clearly, there are 2n-1 possibilities for such partitions. Gπ is then the subset of SLn of matrices whose subdiagonal blocks are zero.

Lemma 31: Let be simple and let G be the adjoint Chevalley group. If N1 is a normal subgroup of G, then NB=G.


We first show NB. Suppose NB and 1xN, x=uh, uU, hH. If u1, then for some wW, wxw-1B, a contradiction. If u=1, then h1. Since G is adjoint, it has center 1, and hxα(t)h-1 =xα(t) with tt for some t,tk, αΣ. Hence (h,xα(t))= xα(t-t) N, xα(t-t)1, and we are back in the first case.

We now prove the lemma. By Lemma 30(c), NB=Gπ for some π. We must show π contains all simple roots. Suppose it does not. Since NB, we see π. Also since Σ is indecomposable, we can find simple roots α,β with απ, βπ and α not orthogonal to β. Let b1wαb2N, biB, then bwαN with b=b2b1B. Then wβbwαwβ-1 N(BwαwβBBwβwαwβB) by Lemma 25(b). Hence either wαwβWπ or wβwαwβWπ. Now wβwαwβ=wγ, where γ=wβα=α-α,ββ. Since α,β0,γ is not a simple root and N(wβwαwβ)1, so that N(wβwαwβ)3 by Appendix II.20. Hence wαwβ and wαwβwα are both expressions of minimal length. By Lemma 29, wβWπ, a contradiction. Thus, π is the set of all simple roots and NB=Gπ=G.

Lemma 32: If and G are as in Theorem 5, then G=G, the derived group of G.

Before proving Lemma 32, we first show that Theorem 5 follows from Lemmas 31 and 32. Let N1 be a a normal subgroup of G. By Lemma 31, NB=G so G/NB/BN. Now G/N equals its derived group and B/BN is solvable. Hence G/N=1 and N=G.

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

Lemma 32': If is as in Theorem 5 then G=G holds in any group G in which the relations (R) hold, in fact in which the relations:

(A) (xβ(t),xγ(u))= Πxiβ+jγ(cijtiuj)
(B) hα(t) xα(u) hα(t)-1 =xα(t2u)


Since G is generated by the 𝔛α's we must show that every 𝔛αG. 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 |k|4. We may choose tk*, t21. Then (hα(t),xα(u)) =xα((t2-1)u). Since α and u are arbitrary, every 𝔛αG.

By (a) we may henceforth assume that the rank is at least 2 and that |k|=2 or 3. By the corollary to Lemma 15, we may write the right side of (A) as xβ+γ(Nβ,γtu)·Π, the factor with i=j=1 having been isolated. We will use the fact (*) that Nβ,γ=±(r+1) with r=r(β,γ) as in Theorem 1, the maximum number of times one can subtract γ from β and still have a root.

(b) Assume that α is a root which can be written β+γ so that no other positive integral combination of β and γ is a root and Nβ,γ0. Then 𝔛αG, as follows at once from (A) with Π=1. This covers the following cases:

(1) If all roots have the same length: types A, D, E.
(2) B (3), α long; B2, α long, |k|=3.
(3) C (3), α short; or α long and |k|=3.
(4) F4.
(5) G2, α long.

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 (*). In all cases but the second cases in (2) and (3) this system can be chosen of type A2 with β and γ roots of the same length as α, while in those cases it can be chosen of type B2 with β and γ short roots.

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

(6) B (2), α short.
(7) G2, α short, |k|=3.
(8) C (3), α long, |k|=2.

(c) If (6) or (7) holds, then 𝔛αG. in both of these cases we can find roots β,γ so that α=β+γ, all other roots iβ+jγ (i,j positive integers) are long, and Nβγ0: in (6) we can choose β long and γ short, in (7) both short. Then Π belongs to G by cases already treated, hence so does 𝔛α, by (A).

(d) If (8) holds, then 𝔛αG. Choose roots β,γ with β long, γ short, and α=β+2γ. Since 𝔛β+γG because β+γ is short, our assertion will follow from C120 in (A), hence from the next lemma.

Lemma 33: If β and γ form a simple system of type B2 with β long and γ short, then ( xβ(t), xγ(u) ) = xβ+γ(±tu) xβ+2γ(±tu2).


By Lemma 14, we have

xγ(u)Xβ xγ(u)-1= exp(aduXγ) Xβ=Xβ+u Nγ,βXβ+γ +u2Nγ,β Nγ,β+γ/2 Xβ+2γ.

Here Nγ,β=±1 and Nγ,β+γ=±2 since β-γ 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.

The proof of Theorem 5 is now complete.

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

Corollary: If Σ is indecomposable and of rank >1 and if α is any root, then there exist roots β and γ and a positive integer n such that α=β+nγ and c1n0 in the relations (A) of Lemma 32'.

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


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=G0, the adjoint group, since by Corollary 5 to Theorem 4', there is a homomorphism φ of G onto G0 with kerφcenter ofG and G0 has center 1. Now we may write G=G1·G2Gr where Gi i=1,2,,r is the adjoint group corresponding to an indecomposable subsystem of Σ. By Theorem 5, each Gi is simple. Thus any normal subgroup of G is a product of some of the Gi's. If it also is solvable, the product is empty and the subgroup is 1.

Notes and References

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.

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