2-Blocks and 2-Modular Characters of the Chevalley Groups G 2 (q)

The structure of blocks with a Klein four defect group

“…For ease of notation, we use the convention that a.X could be any central extension of X by a subgroup of order a, including the case of a direct product. We note that blocks with Klein four defect group do indeed exist for G 2 (q) for certain q by [25].…”

Irreducible restrictions of Brauer characters of the Chevalley group G2(q) to its proper subgroups

“…From the results about blocks and Brauer trees of G 2 (q) in [H1,HS1,HS2,S1,S2], we see that if 3 q then ψ = X 32 in all cases except when = 3 and q ≡ 1 (mod 3) where ψ = X 32 − 1 G . If 3 | q then ψ = X 22 except when = 2 where ψ = X 22 − 1 G .…”

“…The degrees of irreducible -Brauer characters of G 2 (q) when | |G| and q can be read off from [H1,HS1,HS2,S2], and [S3]. They are listed in Tables 2, 3 for = 2 and = 3.…”

“…Lastly, when p = 2, it is computed by Enomoto and Yamada in [EY]. In a series of papers [H1,HS1,HS2,S2,S3,S4], Hiss and Shamash have determined the blocks, Brauer trees and (almost completely) the decomposition numbers for G. In order to solve the problem, it turns out to be useful to know the small degrees of irreducible Brauer characters of G. The smallest degrees of nontrivial absolutely irreducible representations of many quasi-simple groups are collected in [T1] and the result for G is used in this paper. From the knowledge of the degrees of irreducible Brauer characters of G, we compute the second smallest such degree.…”

Uno's invariant conjecture for Chevalley groups G2(q) in nondefining characteristics

“…Thus Ψ is a required bijection of Irr 1 (b H ) onto Irr 1 (B 1 ).SinceIrr a+1 (b H ) = Y(b H , Z q 2 −1 ) and Irr a+1 (B 1 ) = Y(B 1 , Z q 2 −1 ), it follows that Ψ (ϕ y,1 ) = χ y,1 is a required bijection of Irr a+1 (b H ) onto Irr a+1 (B).Finally,Irr 2 (b H ) = Y(b H , T ), | Irr 2 (b H )| = 1 4 (2 2a − 2 a+1) and if ζ y,1 ∈ Irr 2 (b H ), we may suppose y = diag{y 1 , y 2 , y 3 } ∈ T such that y 2 = {y 1 , y −1 1 }. By[21], Irr 2 (B 1 ) = Y(B 1 , T ), | Irr 2 (B 1 )| = 1 12 (2 a − 4)(2 a − 2) and if χ y,1 ∈ Irr 2 (B 1 ), then C G (y) = T . Define Ψ : Irr 2 b(C) −→ Irr 2 (B 1 )…”