2. MARCO TEÓRICO
2.2 COMPRENSIÓN DE LECTURA
2.2.2 LA IMAGEN, OTRO PROCESO DE LECTURA
Let G be a simple algebraic group with Frobenius map F . Then
K = (GFad)0 ∼= GFsc/Z(GFsc)
Theorem 9.4.1 Let K be as above. Then for any irreducible character χ of K there exists x ∈ K such that χ(1) divides |clK(x)|.
In fact we prove the following corollary.
Corollary 9.4.2 There exist two conjugacy classes in K such that the degree of any irreducible character of K divides the order one of them.
Lemma 9.4.3 Let x be any regular unipotent element in GF
ad. Then (|Z(G∗F∗)|, |CGF ad(x)|) = 1. Proof. By Lemma 8.6.1 |CGF ad(x)| = q l. The order of |Z(G∗F∗)| = |Z(GF sc)| is p-prime by [19, p.19]. Lemma 9.4.4 Let u ∈ GF ad be unipotent. Then u ∈ K.
Proof. A unipotent element in GF has order a power of p, see [16]. Therefore, since
|GF
ad : K| = |Z(GFsc)| has p-prime order we have the result.
Lemma 9.4.5 Let x be a regular unipotent element in GF. Then C
K(x) = CGF(x).
Proof. By Lemma 8.6.13, CG(x) = CU(x) where U is a maximal connected unipotent
subgroup of G containing x. Therefore all the elements in G which centralize x are unipotent and so the same must be true in GF. Hence CGF(x) 6 K, and we have
CK(x) = CGF(x).
Lemma 9.4.6 Let x be a regular unipotent element in GF. Then |clK(x)| =
|clGF(x)| |Z(G∗F ∗)| Proof. We have |clK(x)| = |C|K| K(x)| = |GF| |Z(G∗F ∗)||C GF(x)| = |clGF(x)| |Z(G∗F ∗)|.
Lemma 9.4.7 Let χ be an irreducible unipotent character of GF which is not the Stein- berg character, and let x be a regular unipotent element of K. Then χ(1) divides |clK(x)|.
Proof. If χ is a unipotent character of GF, then it is also a unipotent character of G∗F∗. So χ(1) divides |G∗F
∗
|
|Z(G∗F ∗)| by [24, Th. 6.5, p.68]. We also have χ(1) divides
|GF|
|CGF(x)|by Theorem
9.3.1. Therefore, since (|Z(G∗F∗)|, |CGF(x)|) = 1, we have χ(1) divides |G F|
|Z(G∗F ∗)||C GF(x)|
=
|clK(x)| by Lemma 9.4.6.
The following lemma is from [9, p.288].
Lemma 9.4.8 Let χs be a semisimple character of GF. Let s∗ be a semisimple element
in the conjugacy class of the dual group determined by χs, as in Lemma 9.1.2. Then
χ(1) = |G∗F∗ : CG∗F ∗(s∗)|p0.
Lemma 9.4.9 Let χ be an irreducible character of GF which is not unipotent and let x
be a regular unipotent element of K. Then χ(1) divides |clK(x)|.
Proof. By Lemma 9.1.9, we have χ(1) = χs(1)χu(1), where χs is a semisimple char-
acter of GF and χu a unipotent character of CG∗F ∗(s∗) for s∗ a semisimple element of
G∗F∗ in the conjugacy class determined by χs. Then χu(1) divides |C G∗F∗(s ∗)| |Z(C G∗F∗(s ∗))| by [24, Th. 6.5, p.68], and χs(1) divides |G∗F∗| |C G∗F∗(s ∗)| by Lemma 9.4.8. So χs(1)χu(1) divides |G∗F∗| |Z(C G∗F∗(s ∗))| and hence |G∗F∗|
|Z(G∗F ∗)|. By Lemma 9.1.10, χs(1)χu(1) divides |clGF(x)| and so,
since (|Z(G∗F∗)|, |CGF(x)|) = 1, χs(1)χu(1) divides |G F|
|Z(G∗F ∗)||C GF(x)|
= |clK(x)|.
Lemma 9.4.10 Let χ be the Steinberg character of GF
ad. The there exists x ∈ K such
that χ(1) divides |clK(x)|.
Proof. By Lemma 8.7.3 there exists a regular semisimple element x in K. Then |clK(x)|p = |K|p
|CK(x)|p = |K|p = |G|p since |Z(G
F
sc)| is p-prime. Therefore the result follows by Theorem
9.2.10.
Proof (Theorem 9.4.1). Since K is a normal subgroup of GF
ad, the degree of any irreducible
character of K must divide the degree of some irreducible character of GF
9.4.7, 9.4.9 and 9.4.10 we have shown the degree of any irreducible character of GFaddivides the size of some conjugacy class of K. Therefore the result follows.
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