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By William. Kerby
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Example text
17) implies that (E, +, ·) is a sub near-field of (F, +, ·). By definition E is the largest sub near-field of F. 15) we have D 1 n E = {1}, since Eisa near-field and D 1 C f1JJ. This implies DEn E = {1}, for DE = { da, e : a E F, e E E}, since da, e E E, e E E =? e- 1 da, e e E E =? de-• a, 1 E E =? de-• a 1 = 1, and we have da e = ede-•a 1 e-1 = 1. Now suppose E =I= F, and a E F\E. Fore, e E Ewe can now prove: , , J , da, e E*=da, e·E*$>e=e'. 1) (vi) de•, ada, e E E*. 1) (xii) we have de', ada, e = de· +a,_ e'+e = de• +a, e", for e" = -e' + e.
Thus 1 wehave:A-a,b CAa,b ~fAa,b =Aa,b and~EAa,b ~f(~)=(~t 1 ~o:- 1 ~- 1 = 1 (~t ~~=~a:, and (c) is verified. 15) fa:= o: 2 , for a: E Ca b• and hence a: = o:4 = To:T0: 2 = TO:T0:- 1 . a b· ' '· ' Here as in the case of sharply 2-transitive groups one can define pappian groups. We call a sharply 3-transitive group G pappian if the groups Ga are pappian. Hence G is pappian if and only if the yomplements Ga, b are commutative. As we shall see in § 13 Theorem (13,2), it is not necessary to define desarguesian groups in the sharply 3-transitive case.
4), a is the only fixed point of a. Therefore, G is of type (1, 1) or (1,0). From part (a) we conclude that G is of type (1, 1). 8) If G is a sharply 3-transitive group of type ( 1, 1), then Ji = (K 11 Ga) U {id}, for each a EM, and each 1; forms a subgroup of G. 9) If G is of type (1, 1), then for any three distinct points a, b, c EM, there exists exactly one a E K such that: ' a(a) =a, and a(b) =c. 9) for groups of type (2,n), for n E {0, 1}. It is, in fact, a generalization of a theorem of Tits [39].