2003
DOI: 10.1017/s0017089503001381
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Translation Generalized Quadrangles for Which the Translation Dual Arises From a Flock

Abstract: Abstract. It is shown that each finite translation generalized quadrangle (TGQ) S, which is the translation dual of the point-line dual of a flock generalized quadrangle, has a line [∞] each point of which is a translation point. This leads to the fact that the full group of automorphisms of S acts 2-transitively on the points of [∞], and the observation applies to the point-line duals of the Kantor flock generalized quadrangles, the Roman generalized quadrangles and the recently discovered Penttila-Williams g… Show more

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Cited by 16 publications
(16 citation statements)
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“…In [38], the converse of (iii) was (unexpectedly) obtained: (a) they have some line L each point of which is an elation point (in fact, each of these points is a translation point); (b) each line M of the GQ which meets L is a regular line (including L).…”
Section: Introduction and Statement Of The Main Resultsmentioning
confidence: 98%
See 1 more Smart Citation
“…In [38], the converse of (iii) was (unexpectedly) obtained: (a) they have some line L each point of which is an elation point (in fact, each of these points is a translation point); (b) each line M of the GQ which meets L is a regular line (including L).…”
Section: Introduction and Statement Of The Main Resultsmentioning
confidence: 98%
“…As there are s + 1 distinct -orbits in S\ , it immediately follows that L is contained in at least (and then precisely, by straightforward counting) s 3 + s 2 distinct subGQ's of S of order s, all isomorphic to Q(4, s). By [38] (see also, implicitly, [30]), it follows that each point incident with L is a translation point, and so by [37], S is the translation dual of the point-line dual of a flock GQ S(F). Now suppose that does not act semiregularly on S\ .…”
Section: The Cases (B)-(c)-(d) S < T ≤ Smentioning
confidence: 99%
“…r Remark 4.3. After the writing of this paper was completed, in [12] it was shown that there exists a symmetry about a line not on the point ðyÞ (of the form calculated in [5, 5.4]), from which it then follows that every point of the line EðyÞ is a translation point of TðEÞ, and hence that for a fixed subGQ S and a fixed point of type (i) z of TðEÞnS, there exists an automorphism of TðEÞ fixing S and mapping the point z to a point of type (ii). This implies that the ovoid of S subtended from z is isomorphic to an ovoid subtended from a point of type (ii).…”
Section: ] and Remark 22 Thenmentioning
confidence: 99%
“…This implies that the ovoid of S subtended from z is isomorphic to an ovoid subtended from a point of type (ii). In this way the solution of the isomorphism problem was completed in [12].…”
Section: ] and Remark 22 Thenmentioning
confidence: 99%
“…With the BLT-set of W ðsÞ there corresponds a BLT-set on Qð4; sÞ: Let wAE and project from w onto a PGð3; sÞCPGð4; sÞ with Qð4; sÞCPGð4; sÞ and wePGð3; sÞ: With Qð4; sÞ there corresponds a GQ T 2 ðCÞ in PGð3; sÞ; C a conic, and with E\fwg there corresponds a set E 0 of size s in PGð3; sÞ\g; with g the plane of C: If e 1 ; e 2 AE 0 ; e 1 ae 2 ; then /e 1 ; e 2 S-E 0 ¼ J has size s 0 and J,ð/e 1 ; e 2 S-gÞ is a subline over GFðs 0 Þ of /e 1 ; e 2 S: It follows that any flock corresponding to the BLTset is a semifield flock, that is, it follows that the point-line dual # S of S is a TGQ with any line of S incident with ðNÞ as base point; see [6,10,11,19]. Now, by Theorem 44 of Thas [24], the flocks corresponding to the BLT-set are Kantor semifield flocks, that is, S is a Kantor semifield flock GQ. Hence, by Theorem 4.1, S 0 DW ðsÞ and either SDHð3; s 2 Þ or S 0 is one of the s 3 þ s 2 subquadrangles of order s containing the point ðNÞ: & Proof.…”
Section: Remarks (A) Ifmentioning
confidence: 99%