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A 195,747,435 Vertex Graph Related to the Fischer Group Fi23, Part III

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A 195,747,435 Vertex Graph Related to the Fischer Group Fi23, Part III A 195,747,435 vertex graph related to the Fischer group Fi23, part III Rowley, Peter and Walker, Louise 2009 MIMS EPrint: 2009.18 Manchester Institute for Mathematical Sciences School of Mathematics The University of Manchester Reports available from: http://eprints.maths.manchester.ac.uk/ And by contacting: The MIMS Secretary School of Mathematics The University of Manchester Manchester, M13 9PL, UK ISSN 1749-9097 A 195,747,435 VERTEX GRAPH RELATED TO THE FISCHER GROUP F i23 ,III Peter Rowley Louise Walker School of Mathematics University of Manchester Oxford Road Manchester M13 9JL UK 1 Introduction This paper, together with our earlier work in [3] and [4], ¯nishes our anatomical description of the 195,747,435 vertex graph G. This graph is the point-line collinearity graph of a certain geometry ¡ associated with the Fischer group F i23 and is described in Section 2 of [3]. We shall continue the section numbering of [3] and [4]. Our main results are stated in Section 1 { also in Section 1 and the introduction of [4] there is a discussion of various aspects of this investigation of G. We direct the reader to Section 2 for our notation as well as the all important line orbit data. Again, a denotes a ¯xed point of G. Our objective here is to complete the proofs of Theorems 9, 10, 11, 12, 13, 14 and 15. So, for the most part, we are interested in dissecting parts of ¢4(a), the fourth disc of a. In particular 3 5 6 we shall concentrate on the Ga-orbits ¢4(a), ¢4(a) and ¢4(a). Sections 9, 11, 3 5 6 12 are mainly devoted to, respectively, ¢4(a), ¢4(a) and ¢4(a). As a result of analysing these orbits we are able to ¯ll in gaps in our knowledge of other 1 4 Ga-orbits in ¢4(a), namely ¢4(a) and ¢4(a). There are also some loose ends in 5 6 5 the third disc concerning ¢3(a) and ¢3(a). For ¢3(a) these are settled as easy 6 consequences of certain results on ¢4(a). While for ¢3(a) much more work is needed, and most of Section 10 is given over to examining this orbit. We give no further information here of where speci¯c results are stated. But remind the reader of Appendix B in which a detailed record of where the point distributions of particular line orbits are to be found. 5 9 A ¯rst look at ¢4(a) 5 We now ¯x a point x in ¢4(a) and begin to look at the points incident with lines in ¡1(x). 5 Firstly we summarize the basic properties of ¢4(a). 5 15 Lemma 9.1. (i) j¢4(a)j = 2 :11:23. 4 (ii) j¢1(x) \ ¢3(a)j = 1. » (iii) Gax = A7 and jGax \ Q(a)j = 1. 2 5 Proof. We prove (iii) ¯rst. By the de¯nition of ¢4(a) (see (2.15)(xiii)), there 4 exists d 2 ¢1(x) \ ¢3(a) with d + x 2 ®0(d; X(d; a)). Since ¿(X(d; a)) 2 Q(d)anGx by Lemma 3.2, (2.6) and Lemma 5.2 imply that ¡0(d + x)nfdg ⊆ 5 5 ¢4(a) and ¢1(d) \ ¢4(a) is a Gad-orbit of points. Hence [Gad : Gadx] = 352 » by (2.6) and using [1] together with Lemma 4.8(iv) yields Gadx = A7. Let g 2 Gax \ Q(a). Then g ¯xes X(d; a) and so g 2 Gadx by Lemma 5.3(i). Therefore Gax \ Q(a) = 1 because Gadx is simple. We now assume Gadx 6= Gax and argue for a contradiction. Since Q(a)x = 1, Gax is isomorphic to a subgroup of M23. Considering possible subgroups which 4 » contain A7 in [1] we conclude that Gax » 2 : A7, A8, M22 or M23. If Gax = M22 4 or M23, then j¢1(x)\¢3(a)j = 176 or 23£176 by Lemma 5.4, which is impossible 4 4 by Lemma 5.3(ii). If Gax » 2 : A7 or A8, then j¢1(x) \ ¢3(a)j = 16 or 8 4 respectively. Now ¢1(x) \ ¢3(a) is a Gax-orbit and by considering the possible orbit sizes of Gax on ¡1(x) given in (2.1) and (2.10) we see this is impossible. » Hence Gax = Gadx = A7, which proves (iii). An immediate consequence of 4 Gax = Gadx is that j¢1(x) \ ¢3(a)j = 1. 5 Since ¢4(a) is a Ga-orbit by Lemma 5.2, part (iii) gives 5 jGaj 15 j¢4(a)j = = 2 :11:23; jGaxj so completing the proof of the lemma. Therefore Gax has orbits on ¡1(x) as described in (2.13). For the rest of this 4 section we ¯x fdg := ¢1(x) \ ¢3(a). 1 2 ? Lemma 9.2. (i) Let c 2 ¢2(x) \ ¢2(a) \ ¢1(d) and fbg = fa; cg . Then 6 b 2 ¢3(x). 6 6 (ii) Let l 2 ®3(x; x + d; +). Then j¡0(l) \ ¢3(a)j = j¡0(l) \ ¢4(a)j = 1. 5 6 (iii) Let l 2 ®3(x; x + d; ¡). Then j¡0(l) \ ¢3(a)j = j¡0(l) \ ¢4(a)j = 1. 5 (2) Furthermore, if z 2 ¡0(l) \ ¢3(a), then z + x 2 ®3 (z; z + c; ¡) where 2 fcg = ¢1(z) \ ¢2(a). Proof. By Theorem 4.13(vi), for every l 2 ¡1(d; X(d; a)), we have ¡0(l) \ 2 ¢2(a) 6= ;. Examining the MOG in [2] we see there are 35 possibilities for 3 2 1 c 2 ¢1(d) \ ¢2(a) with c 2 ¢2(x). For any such c, T (c; x) is a triad not inci- ? dent with the hyperplane X(c; a) in ­c because ¡3(a; x) = ;. Let fbg = fa; cg . Then T (c; x) \ (c + b) = ; in ­c because (c + d) \ (c + b) = fX(c; a)g and so 6 b 2 ¢3(x) by Lemma 4.6(ii). This proves part(i). ? For each y 2 fc; xg nfdg, c + y2 = ¡1(X(c; a)), otherwise ¡3(a; x) 6= ; 2 by Lemma 3.6 because d 2 ¡0(X(c; a)). Furthermore d 2 ¢2(b), whence ? c+y 2 ®3;0(c; c+b; X(c; a)) for one point y 2 fc; xg nfdg and c+y 2 ®1;0(c; c+ b; X(c; a)) for three points y 2 fc; xg?nfdg using (2.3) and Lemma 3.11(ii). ? 6 ? Therefore (2.15)(vii),(viii) imply that jfc; xg \ ¢3(a)j = 1 and jfc; xg \ 5 ¢3(a)j = 3. ? 6 0 0 Let fyg = fc; xg \ ¢3(a), ¡0(c + d) = fc; d ; dg and ¡0(x + y) = fx; y ; yg: So we have ry r ´ ©XX ´ @ © b XX ´ @ © XX ´ 0 © XX´r @ry © c ©© @ @ ©r @ @ a @r @rx d0 @ ´´ ´ @ ´ @´r d 0 3 0 0 Then d 2 ¢3(a) by Lemma 4.15(iii) and y 2 ¢1(d ) by Lemma 3.10. Hence 0 4 6 y 2 ¢4(a)[¢4(a)[¢3(a), using (2.3), (2.15)(xii),(xiv) and x 2 ¢4(a). However 0 0 0 X(d ; a) = X(d; a) and so, since ¡3(a; x) = ; , y 2= ¡0(X(d ; a)). Appealing to 0 4 6 0 ? 0 0 0 0 Lemma 4.15 yields y 2 ¢4(a)[¢4(a). Since fy ; cg > 1, d +y 2 ®3(d ; d +c) 0 6 and we conclude that y 2 ¢4(a) by (2.15)(xiv). ? 5 0 Next ¯x z 2 fc; xg \ ¢3(a) and let ¡0(x + z) = fx; z ; zg. Using a similar 0 0 6 argument as for y we deduce that z 2 ¢4(a). We have x + y, x + z 2 ®3(x; x + d; +) [ ®3(x; x + d; ¡) by (2.13) because 1 2 1 y; z 2 ¢2(d). Let c vary through the 35 possibilities in ¢2(a) \ ¢2(x) \ ¢1(d). This generates 35 distinct possibilities for y and 105 distinct possibilities for z ? 5 (because jfc; xg \ ¢3(a)j = 3). Examining the Gax-orbit sizes in (2.13) we conclude that x + y 2 ®3(x; x + d; +) and x + z 2 ®3(x; x + d; ¡). 5 Finally, since T (c; x) \ (c + b) = ; in ­c, X(z; b) 2= ¡3(x) (where z 2 ¢3(a) \ 4 ? fc; xg ) and so by de¯nition (see (2.7)), z + x 2 ®3(z; z + c; ¡). Appealing to (2) Lemma 5.7 (and de¯nition of ®3 (z; z + c; ¡) following that lemma) we obtain the result. We can use Lemma 9.2 to extract information about certain lines incident 6 with a point in ¢3(a). 6 Lemma 9.3. Let l 2 ®3(x; x + d; +) and y 2 ¡0(l) \ ¢3(a) with fcg = ¢1(y) \ 1 1 ¢2(a) and fbg = ¢1(a)\¢2(y). Then y+x 2 ®1(y; y+c) and (y+x)\T (y; b) = ; in ­y. 0 ? Proof. There exists c 2 fd; yg \ ¡0(X(d; a)) by Lemma 3.11(ii) and then 0 2 0 ? c 2 ¢2(a) by Theorem 4.13(vi) and Lemma 4.15. Let fbg = fa; c g . By 6 ? Lemma 9.2(i) b 2 ¢3(x). Moreover b 2 fa; cg by Lemma 4.10. Thus we have »»r r r»» c@ a b@ @ @ @ @ @ry »»»PP @Pr» PP c0 P P PP »»Prx PPdr»» 0 Applying Lemma 4.14(iii) to b and x, we have that T (y; b) \ T (c ; x) = ; in ­y.
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