By S. Payne, T. Thas
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4. 3; q/. Proof. 5; q/, and let x 2 Q. 5; q/ not containing x. 4; q/ and the tangent 4-space of Q at x. 1/. 4; q/ (from x). 3; q/ which contains a point of O; if L contains x, then LÂ is a point of O. O/. 5; q/. 5; q/. 5; q/ is regular. O/ is regular. O/, and suppose they define distinct points x0 and x1 of O. If fL0 ; L1 g? D fM0 ; M1 ; : : : ; Mq g and fL0 ; L1 g?? O/. 4; q/ Li is concurrent with Mj , i; j D 0; : : : ; q. Hence L0 ; : : : ; Lq ; M0 ; : : : ; Mq are contained in a three-dimensional space P , and moreover fL0 ; L1 g?
Are of type (i), it is not difficult to show that S 0 is a GQ of order q. Next we show that all points of S 0 are regular. 6 it is sufficient to prove that any triad of points of S 0 is centric. There are several cases according to the types of the points in the triad. 1/ the type IV point. There are many cases to consider, but several of them are easy. We present the details only for the least trivial of the cases. First of all we consider the case (IV,I,I). Let u and v be the points of type I.
D fx; y; zg?? Proof. s 2 C 1/. Let L be a line of B 0 . If L is incident with some point of X [ Y , then clearly L is of type uv, with u 2 X and v 2 Y . Then all points incident with L are in P 0 . 1 L is again incident with s C 1 points of P 0 . s; t 0 /. st 0 C 1/ we have t 0 D s, and so S 0 is a subquadrangle of order s. Since X [Y P 0 , jX j D jY j D s C 1, and each point of X is collinear with each point of 0 0 0 Y , we have fx; yg? D fx; y; zg? and fx; yg? D fx; y; zg?? s; t /, s > 1, t > 1.