On splitting of the normalizer of a maximal torus in groups of Lie type

On splitting of the normalizer of a maximal torus in groups of Lie type Alexey Galt 07.08.2017

Example 1 Let G = SL 2 ( (F p ) be the ) special linear group of degree 2 over F p. λ 0 Then T = { 0 λ 1, λ F p} is a maximal torus of G. The normalizer N G (T ) is the group of all monomial matrices of G and N G (T )/T ( Sym 2. ) But G contains only one element of 1 0 order two:, and this element lies in T. Hence, 0 1 N G (T ) does not split over T.

Example 2 Let G = GL n (F p ) be the general linear group of degree n over F p. Then T = D n (F p ) is a maximal torus of G. The normalizer N G (T ) is the group of all monomial matrices of G and N G (T )/T Sym n. There is a canonical embedding of Sym n into the group of all monomial matrices of G. If H is an image of Sym n under this embedding, then H is a complement for T in N G (T ). Since the center Z(G) of G is contained in T, then a maximal torus of PGL n (F p ) also has a complement in their normalizer. Moreover, PGL n (F p ) PSL n (F p ) and the same is true for PSL n (F p ).

Problems Let G be a simple connected linear algebraic group over the algebraic closure F p of a finite field of positive characteristic p. Let σ be a Steinberg endomorphism and T a maximal σ-invariant torus of G. It s well known that all the maximal tori are conjugated in G and the quotient N G (T )/T is isomorphic to the Weyl group W of G. The following problem arises. Problem 1 Describe the groups G in which N G (T ) splits over T.

Problems A similar problem arises in finite groups G of Lie type. Let T = T G be a maximal torus in a finite group of Lie type G, N(G, T ) = N G (T ) G an algebraic normalizer of G. Notice that N(G, T ) N G (T ), but the equality is not true in general. Problem 2 Describe the groups G and their maximal tori T in which N(G, T ) splits over T.

History J.Tits Normalisateurs de tores I. Groupes de Coxeter Étendus // Journal of Algebra, 1966, V.4, 96 116. An answer to Problem 1 for simple Lie groups was given in M. Curtis, A. Wiederhold, B. Williams, Normalizers of maximal tori // Springer, Berlin, 1974, Lecture Notes in Math., V. 418, 31 47.

Algebraic groups The answer for Problem 1 is in the following table: Group Conditions of existence of a complement SL n (F p ) p = 2 or n is odd PSL n (F p ) No conditions Sp 2n (F p ) p = 2 PSp 2n (F p ) p = 2 or n 2 SO 2n+1 (F p ) No conditions SO 2n (F p ) No conditions PSO 2n (F p ) No conditions G 2 (F p ) No conditions F 4 (F p ) p = 2 E k (F p ) p = 2

Let O p (G σ ) G G σ be a finite group of Lie type. Two maximal tori in G are not necessary conjugate in G. Let W be a Weyl group of G, π a natural homomorphism from N = N G (T ) into W. Two elements w 1, w 2 are called σ-conjugate if w 1 = (w 1 ) σ w 2 w for some element w of W. Proposition There is a bijection between the G-classes of σ-stable maximal tori of G and the σ-conjugacy classes of W. Define C W,σ (w) = {x W (x 1 ) σ wx = w}. Proposition Let g σ g 1 N and π(g σ g 1 ) = w. Then (N G (T g )) σ /(T g ) σ C W,σ (w).

Linear groups In case of linear group W Sym n and the σ-conjugacy classes C W,σ (w) of W are coincide with ordinary conjugacy classes of symmetric group. Each such class corresponds to the cycle-type (n 1 )(n 2 )... (n m ). Let {n 1,..., n m } be a partition of n. We assume that n 1 =... = n l1 <... < n l1 +...+l r 1 +1 =... = n l1 +...+l r and a 1 = n l1 l 1, a 2 = n l1 +l 2 l 2,..., a r = n l1 +...+l r l r. Theorem Let T be a maximal torus of G = SL n (q) with the cycle-type (n 1 )(n 2 )... (n m ). Then T has a complement in N if and only if q is even or a i is odd for some 1 i r.

Symplectic and orthogonal groups Let n = n + n, {n 1,..., n k } and {n k+1,..., n m } be partitions of n and n, respectively. A set { n 1,..., n k, n k+1,..., n m } will be called a cycle-type and denoted by (n 1 )... (n k )(n k+1 )... (n m ). As above we assume that n 1 =... = n l1 <... < n l1 +...+l r 1 +1 =... = n l1 +...+l r Let a 1 = n l1 l 1, a 2 = n l1 +l 2 l 2,..., a r = n l1 +...+l r l r.

Theorem Let q be a power of a prime p. Let T a maximal σ-invariant torus of G, T a corresponding maximal torus of G with the cycle-type (n 1 )... (n k )(n k+1 )... (n m ) and m > 4. Then Group Conditions of existence of a complement PSp 2n (q) q is even q 1 (mod 4) Ω 2n+1 (q) a i is odd for some 1 i r q 3 (mod 4) and n i is even for all 1 i m q 1 (mod 4) PΩ + 2n (q) a i is odd for some 1 i r q 3 (mod 4) and n i is even for all 1 i m q is even PSL n (q) a i is odd for some 1 i r (n) 2 < (q 1) 2

The answer for Problem 2 for groups E 6 (q) is in the following table: No Representative ω ω C W (ω) Structure of C W (w) Torus T 1 1 1 51840 O 5 (3) : Z 2 (q 1) 6 2 ω 1 2 1440 S 2 S 6 (q 1) 4 (q 2 1) 3 ω 1 ω 2 2 192 D 8 S 4 (q 1) 2 (q 2 1) 2 4 ω 3 ω 1 3 216 Z 3 (S3 2 : Z 2) (q 1) 3 (q 3 1) + 5 ω 2 ω 3 ω 5 2 96 Z 2 Z 2 S 4 (q 2 1) 3 6 ω 1 ω 3 ω 5 6 36 Z 6 S 3 (q 1) (q 2 1) (q 3 1) + 7 ω 1 ω 3 ω 4 4 32 Z 4 D 8 (q 1) 2 (q 4 1) 8 ω 1 ω 4 ω 6 ω 36 2 1152 (q + 1) 2 (q 2 1) 2 9 ω 1 ω 2 ω 3 ω 5 6 24 Z 3 D 8 (q 2 1) (q + 1)(q 3 1) + 10 ω 1 ω 5 ω 3 ω 6 3 108 Z 3 S 3 S 3 (q 1) (q 2 + q + 1) (q 3 1) + 11 ω 1 ω 4 ω 6 ω 3 4 16 Z 4 Z 2 Z 2 (q 2 1) (q 4 1) 12 ω 1 ω 4 ω 3 ω 2 5 10 Z 2 Z 5 (q 1) (q 5 1) + 13 ω 3 ω 2 ω 5 ω 4 6 36 Z 6 S 3 (q 2 1) (q 1)(q 3 + 1) + 14 ω 3 ω 2 ω 4 ω 14 4 96 SL 2 (3) : Z 4 (q 1)(q 2 + 1) 2 15 ω 1 ω 5 ω 3 ω 6 ω 2 6 36 Z 6 S 3 (q 2 + q + 1) (q + 1)(q 3 1) + 16 ω 1 ω 4 ω 6 ω 3 ω 36 4 96 Z 4 S 4 (q + 1) 2 (q 4 1) 17 ω 1 ω 4 ω 5 ω 3 ω 36 10 10 Z 10 (q + 1)(q 5 1) + 18 ω 1 ω 4 ω 6 ω 3 ω 5 6 12 Z 6 Z 2 (q 2 + q + 1) (q 1)(q 3 + 1) + 19 ω 2 ω 5 ω 3 ω 4 ω 6 8 8 Z 8 (q 2 1)(q 4 + 1) + 20 ω 20 ω 5 ω 4 ω 3 ω 2 12 12 Z 12 (q 1)(q 2 + 1)(q 3 + 1) + 21 ω 1 ω 5 ω 2 ω 3 ω 6 ω 36 3 648 (((Z3 2 ) : Z 3) : Q 8 ) : Z 3 (q 2 + q + 1) 3 + 22 ω 1 ω 4 ω 6 ω 3 ω 5 ω 36 6 36 Z 6 S 3 (q + 1) (q 5 + q 4 + q 3 + q 2 + q + 1) + 23 ω 1 ω 4 ω 6 ω 3 ω 2 ω 5 12 12 Z 12 (q 2 + q + 1)(q 4 q 2 + 1) + 24 ω 1 ω 4 ω 14 ω 3 ω 2 ω 6 9 9 Z 9 (q 6 + q 3 + 1) + 25 ω 1 ω 4 ω 14 ω 3 ω 2 ω 31 6 72 Z 3 SL 2 (3) (q 2 q + 1) (q 4 + q 2 + 1) +