By P. K. Jain, Ahmed Khalid

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P 0 ; B 0 ; I0 / in the following way. c; b/, and the points of P . The elements of B 0 are the lines of type (ii) and (iii) of B, the elements Œc, and Œ1. q/, u I0 L iff u I L for all u 2 P and all lines L of type (ii) or (iii) of B. It is easily checked that each point of P 0 is incident with q C 1 lines of B 0 , and each line of B 0 is incident with q C 1 points of P 0 . q/ the q lines of fL; M 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.

Q/. O/. For q odd, every oval O is an irreducible conic by B. 4; q/. , lines). Consequently for q odd there arises only one GQ of S. E. Payne. 2; q/. Now assume that q is even. O/ depends, naturally, on the nature of the oval O. 1/ and all lines incident with it are regular. O/ must belong to a completely determined list of examples (cf. 3 and Chapter 12 for the details). And for q D 2h > 8, there are examples of O for which there is a unique line L1 of regular points. O/; x/ was shown by S. E.

70]. Then a construction was found by S. E. Payne [121] which included all these examples and for q even produced some additional ones (cf. [120], [124]). These examples yield the only known cases (with s ¤ 1 and t ¤ 1) in which s and t are not powers of the same prime. 3 ([1], [70]). q 1; q C 1/, q D 2h . Proof. 3; q/ D P . 1. Description of the known GQ 33 for lines just those lines of P which are not contained in H and meet O (necessarily in a unique point). The incidence is that inherited from P .