By Flaass D.G.

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Then one of the following holds: (i) q is of plus type and there is a unique partition {A, C} of P6 into two triples such that q(t) = 1 for t ∈ T if and only if t stabilizes the partition; (ii) q is of minus type and there is a unique element a ∈ P6 such that q(t) = 1 for t ∈ T if and only if t stabilizes a. 3 (ii)) will be denoted by Q(P6 , {A, C}) and Q(P6 , a), respectively. By elementary calculations one can see that there are ten forms of plus type, each having exactly six non-singular vectors and six forms of minus type, each having exactly ten non-singular vectors.

2). 1), so that Cγ = {cγ(c) | c ∈ C} is a complement to Q in G. Suppose further that γ(c) ∈ Z(Q) for every c ∈ C. Then dγ : cq → cγ(c)q is a deck automorphism of G which centralizes Q and maps C onto Cγ . We conclude this section a special case of Schur–Zassenhaus Theorem (cf. 10) in Aschbacher (1986)) which we present without a proof. 10 Suppose that C is a group of odd order and V is a GF (2)-module for C. Then H 1 (C, V ) is trivial, so that all complements to V in G = V : C are conjugate in G.

10 L2 (7) and L3 (2) Let P8 = {∞, 0, 1, . . , 6} be the projective line over the ﬁeld GF (7) of seven elements. The stabilizer of the projective line structure in the symmetric group of P8 is P GL2 (7) and the latter contains the group L2 (7) with index 2. 9 provides us with an embedding P GL2 (7) ≤ O6+ (2). In this section we reﬁne this embedding geometrically. 1 Let (V, f, q) = Q(P8 ) and let O6+ (2) be the corresponding orthogonal group. Then there is a pair I1 , I2 of disjoint maximal totally singular subspaces in V such that (i) L2 (7) is the joint stabilizer in O6+ (2) of I1 and I2 ; (ii) P GL2 (7) is the stabilizer in O6+ (2) of the unordered pair {I1 , I2 }.