Irreducible A1 subgroups of exceptional algebraic groups

Abstract

A closed subgroup of a semisimple algebraic group is called irreducible if it lies in no proper parabolic subgroup. In this paper we classify all irreducible A 1 subgroups of exceptional algebraic groups G. Consequences are given concerning the representations of such subgroups on various G-modules: for example, the conjugacy classes of irreducible A 1 subgroups are determined by their composition factors on the adjoint module of G.

On the involution fixity of exceptional groups of Lie type

“…In the first case, the notation indicates that X 1 and X 2 are generated by long and short root subgroups of T , respectively. In the latter two cases, we assume that the final four A 1 factors ofH • are contained in a D 4 subsystem subgroup ofḠ (this is consistent with [60], where the composition factors of L(Ḡ)↓H • are given in Tables 59 and 60).…”

“…To do this, first note that X is constructed in [37, Lemma 1.5] as a diagonal subgroup of A 1 A 1 < A 4 A 4 , with each A 1 acting as V (4) on V A 4 (λ 1 ). In the notation of [60], we have X = E 8 (#17 {0} ), which is a subgroup of D 8 with…”

“…Writē G =Ĝ/Z andH =Ĥ/Z, whereĜ is the simply connected group and Z = Z(Ĝ). To be consistent with [60], we will order the factors ofH so thatĤ =Ĵ 1Ĵ2 andĴ 1 is simply connected (note that the composition factors of L(Ḡ)↓H are given in [60, Table 59]). It follows that H 0 = (Ĵ 1Ĵ2 ) σ /Z = (SL 2 (q) × PGL 2 (q))/Z = L 2 (q) × PGL 2 (q) and this explains why we can choose an involution of the form (1, t 1 ) in H 0 .…”

On the minimal dimension of a finite simple group

“…where L(X) denotes the Lie algebra of X and M i is the natural module forH i (see [36,Chapter 12], for example), it is easy to see that x ∈H 2 . Let P be the unique Sylow 3-subgroup of L. Then P is contained in a Sylow 3-subgroup of the second L 2 (q) factor of H, which is cyclic, so |P | = 3 and thus P Ḡ σ 2 .…”

Finite subgroups of simple algebraic groups with irreducible centralizers

“…By considering the 2 2 subgroups of F , all of which must be as in (i) of Proposition 2.6, we see that one of these, say e 1 , e 2 , is 2B-pure, so that C G (e 1 , e 2 ) 0 = D 2 4 . We have C G (e 1 )/ e 1 ∼ = P SO 16 , and consider the preimage of F/ e 1 in SO 16 . This preimage is elementary abelian by Proposition 2.5, so can be diagonalised, and we can take e 2 = (−1 8 , 1 8 ).…”