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

1 Introduction

This paper, together with our earlier work in [3] and [4], finishes our anatomical description of the 195,747,435 vertex graph G. This graph is the point-line collinearity graph of a certain geometry Γ 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 fixed 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 fill 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 specific 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 ﬁrst look at ∆4(a)

5 We now ﬁx a point x in ∆4(a) and begin to look at the points incident with lines in Γ1(x). 5 Firstly we summarize the basic properties of ∆4(a).

5 15 Lemma 9.1. (i) |∆4(a)| = 2 .11.23.

4 (ii) |∆1(x) ∩ ∆3(a)| = 1.

∼ (iii) Gax = A7 and |Gax ∩ Q(a)| = 1.

2 5 Proof. We prove (iii) ﬁrst. By the deﬁnition of ∆4(a) (see (2.15)(xiii)), there 4 exists d ∈ ∆1(x) ∩ ∆3(a) with d + x ∈ α0(d, X(d, a)). Since τ(X(d, a)) ∈

Q(d)a\Gx by Lemma 3.2, (2.6) and Lemma 5.2 imply that Γ0(d + x)\{d} ⊆ 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 ∈ Gax ∩ Q(a). Then g ﬁxes X(d, a) and so g ∈ 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 |∆1(x)∩∆3(a)| = 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 |∆1(x) ∩ ∆3(a)| = 16 or 8 4 respectively. Now ∆1(x) ∩ ∆3(a) is a Gax-orbit and by considering the possible orbit sizes of Gax on Γ1(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 |∆1(x) ∩ ∆3(a)| = 1. 5 Since ∆4(a) is a Ga-orbit by Lemma 5.2, part (iii) gives

5 |Ga| 15 |∆4(a)| = = 2 .11.23, |Gax| so completing the proof of the lemma.

Therefore Gax has orbits on Γ1(x) as described in (2.13). For the rest of this 4 section we ﬁx {d} := ∆1(x) ∩ ∆3(a).

1 2 ⊥ Lemma 9.2. (i) Let c ∈ ∆2(x) ∩ ∆2(a) ∩ ∆1(d) and {b} = {a, c} . Then 6 b ∈ ∆3(x).

6 6 (ii) Let l ∈ α3(x, x + d, +). Then |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

5 6 (iii) Let l ∈ α3(x, x + d, −). Then |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1. 5 (2) Furthermore, if z ∈ Γ0(l) ∩ ∆3(a), then z + x ∈ α3 (z, z + c, −) where 2 {c} = ∆1(z) ∩ ∆2(a).

Proof. By Theorem 4.13(vi), for every l ∈ Γ1(d, X(d, a)), we have Γ0(l) ∩ 2 ∆2(a) 6= ∅. Examining the MOG in [2] we see there are 35 possibilities for

3 2 1 c ∈ ∆1(d) ∩ ∆2(a) with c ∈ ∆2(x). For any such c, T (c, x) is a triad not inci- ⊥ dent with the hyperplane X(c, a) in Ωc because Γ3(a, x) = ∅. Let {b} = {a, c} .

Then T (c, x) ∩ (c + b) = ∅ in Ωc because (c + d) ∩ (c + b) = {X(c, a)} and so 6 b ∈ ∆3(x) by Lemma 4.6(ii). This proves part(i). ⊥ For each y ∈ {c, x} \{d}, c + y∈ / Γ1(X(c, a)), otherwise Γ3(a, x) 6= ∅ 2 by Lemma 3.6 because d ∈ Γ0(X(c, a)). Furthermore d ∈ ∆2(b), whence ⊥ c+y ∈ α3,0(c, c+b, X(c, a)) for one point y ∈ {c, x} \{d} and c+y ∈ α1,0(c, c+ b, X(c, a)) for three points y ∈ {c, x}⊥\{d} using (2.3) and Lemma 3.11(ii). ⊥ 6 ⊥ Therefore (2.15)(vii),(viii) imply that |{c, x} ∩ ∆3(a)| = 1 and |{c, x} ∩ 5 ∆3(a)| = 3. ⊥ 6 0 0 Let {y} = {c, x} ∩ ∆3(a), Γ0(c + d) = {c, d , d} and Γ0(x + y) = {x, y , y}. So we have ry r ¨XX @ ¨ b XX @ ¨ XX 0 ¨ XXr @ry ¨ c ¨¨ @ @ ¨r @ @ a @r @rx d0 @ @ @r d

0 3 0 0 Then d ∈ ∆3(a) by Lemma 4.15(iii) and y ∈ ∆1(d ) by Lemma 3.10. Hence 0 4 6 y ∈ ∆4(a)∪∆4(a)∪∆3(a), using (2.3), (2.15)(xii),(xiv) and x ∈ ∆4(a). However 0 0 0 X(d , a) = X(d, a) and so, since Γ3(a, x) = ∅ , y ∈/ Γ0(X(d , a)). Appealing to 0 4 6 0 ⊥ 0 0 0 0 Lemma 4.15 yields y ∈ ∆4(a)∪∆4(a). Since {y , c} > 1, d +y ∈ α3(d , d +c) 0 6 and we conclude that y ∈ ∆4(a) by (2.15)(xiv). ⊥ 5 0 Next ﬁx z ∈ {c, x} ∩ ∆3(a) and let Γ0(x + z) = {x, z , z}. Using a similar 0 0 6 argument as for y we deduce that z ∈ ∆4(a).

We have x + y, x + z ∈ α3(x, x + d, +) ∪ α3(x, x + d, −) by (2.13) because 1 2 1 y, z ∈ ∆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 |{c, x} ∩ ∆3(a)| = 3). Examining the Gax-orbit sizes in (2.13) we conclude that x + y ∈ α3(x, x + d, +) and x + z ∈ α3(x, x + d, −). 5 Finally, since T (c, x) ∩ (c + b) = ∅ in Ωc, X(z, b) ∈/ Γ3(x) (where z ∈ ∆3(a) ∩

4 ⊥ {c, x} ) and so by deﬁnition (see (2.7)), z + x ∈ α3(z, z + c, −). Appealing to (2) Lemma 5.7 (and deﬁnition of α3 (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 ∈ α3(x, x + d, +) and y ∈ Γ0(l) ∩ ∆3(a) with {c} = ∆1(y) ∩ 1 1 ∆2(a) and {b} = ∆1(a)∩∆2(y). Then y+x ∈ α1(y, y+c) and (y+x)∩T (y, b) = ∅ in Ωy.

0 ⊥ Proof. There exists c ∈ {d, y} ∩ Γ0(X(d, a)) by Lemma 3.11(ii) and then 0 2 0 ⊥ c ∈ ∆2(a) by Theorem 4.13(vi) and Lemma 4.15. Let {b} = {a, c } . By 6 ⊥ Lemma 9.2(i) b ∈ ∆3(x). Moreover b ∈ {a, c} by Lemma 4.10. Thus we have r r r c@ a b@ @ @ @ @ @ry PP @rP 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 Ωy.

Hence y + x ∈ α1(y, y + c) and (y + x) ∩ T (y, b) = ∅ in Ωy by Lemma 3.11(ii), as required.

At this stage we require the following symmetry result, which we now prove.

5 Lemma 9.4. a ∈ ∆4(x).

6 Proof. By Lemma 9.3 there exists y ∈ ∆1(x) ∩ ∆3(a) with y + x ∈ α1(y, y + c) 1 and (y + x) ∩ T (y, b) = ∅ in Ωy where {c} = ∆1(y) ∩ ∆2(a) and {b} = 1 2 ∆1(a) ∩ ∆2(y). By deﬁnition, c ∈ ∆2(x). Further, by Lemma 3.11(ii), there 0 ⊥ 0 exists b ∈ {a, c} ∩ Γ0(X(c, x)). We have b 6= b because (y + x) ∩ T (y, b) = ∅ in Ωy and so b 6∈ Γ0(X(c, x)). b0 Pr r PP aPP PP PP Prc PPbr @ @ @ @ @ (((rx @ y((( @r @r

5 0 2 Lemma 4.10 implies that b ∈ ∆2(y) and appealing to (2.15)(v),(vi) yields 0 3 4 b ∈ ∆3(x) ∪ ∆3(x). 0 3 4 6 Assume b ∈ ∆3(x). Then a ∈ ∆4(x) ∪ ∆4(x) by (2.15)(xii) and (xiv) 5 4 because Γ3(a, x) = ∅. Since |∆4(a)| 6= |∆4(a)| by Lemmas 8.7(ii) and 9.1(i) 6 6 6 5 15 we must have a ∈ ∆4(x) with |∆4(x)| = |∆4(a)| = |∆4(a)| = 2 .11.23. Since 0 6 0 6 |∆1(b ) ∩ ∆4(x)| = 160 and ∆1(b ) ∩ ∆4(x) is a Gxb0 -orbit by Lemma 8.4(v), using Lemmas 5.4 and 4.8(iii) we get that

(9.4.1) 210.7.11.23.160 |∆ (a) ∩ ∆3(x)| = = 35 1 3 215.11.23 and that

3 (9.4.2) ∆1(a) ∩ ∆3(x) is a Gxa-orbit.

5 6 We now switch our attention to x ∈ ∆4(a) and let z ∈ ∆1(x) ∩ ∆4(a) with x + z ∈ α3(x, x + d, +); by Lemma 9.2(i) such a z exists. Observe that, by 3 3 (9.4.2), ∆1(z) ∩ ∆3(a) is a Gaz-orbit. Hence {l ∈ Γ1(z) | Γ0(l) ∩ ∆3(a) 6= ∅} is a 5 6 6 5 Gaz-orbit of size 35 by (9.4.1). If ∆4(a) = ∆4(a), then a ∈ ∆4(x) = ∆4(x), and 5 6 we are done. So we may assume that ∆4(a) 6= ∆4(a). Therefore, by Lemma 5.5, 5 6 z + x is contained in a Gza-orbit of Γ1(z) of size 35 (since |∆4(a)| = |∆4(a)|). ∼ ∼ 6 Now Gaz = A7 (because Gax = A7 and a ∈ ∆4(x)) and so, by (2.13), Gaz has exactly one orbit on Γ1(z) of size 35. But, on the one hand, we have that the 5 6 6 three points of Γ0(z + x) are in either ∆4(a), ∆3(a) or ∆4(a) by Lemma 9.2(ii) 3 and on the other Γ0(z + x) ∩ ∆3(a) 6= ∅. From this contradiction we infer that 0 3 0 4 5 b ∈/ ∆3(x). Therefore b ∈ ∆3(x) and now a ∈ ∆4(x) by (2.15)(xiii) because

Γ3(a, x) = ∅.

5 In the next lemma we focus our attention on ∆3(a).

5 2 Lemma 9.5. Let y ∈ ∆3(a) and l ∈ α1(y, y + c, +) where {c} = ∆1(y) ∩ ∆2(a). 4 5 Then |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

0 Proof. Let Γ0(l) = {y, z, z }. By deﬁnition (see (2.7)), X(z, c) = X(y, b) where 2 {b} = ∆1(a)∩∆2(y). Therefore y +z ∈ α1,1(y, y +c, X(y, b)) and by Theorem 4 3 0 4 3 4 0 we may assume z ∈ ∆3(b) and z ∈ ∆3(b). By Lemma 4.12, b ∈ ∆3(z) ∩ ∆3(z ). 0 We have Γ3(a, z) = Γ3(a, z ) = ∅ by Lemmas 4.9(i), 4.15 and 5.1. This together

6 2 4 5 0 with the fact that a ∈ ∆2(c) yields a ∈ ∆4(z) ∩ ∆4(z ) (see (2.15)). The result now follows by Lemmas 8.14 and 9.4.

5 We now return to x ∈ ∆4(a).

Lemma 9.6. Let l ∈ α1(x, x + d, +).

5 4 (i) |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

5 2 (ii) If y ∈ Γ0(l) ∩ ∆3(a), then y + x ∈ α1(y, y + c, +)({c} = ∆1(y) ∩ ∆2(a)).

4 (iii) If z ∈ Γ0(l) ∩ ∆4(a), then z + x ∈ α5(z, END, +).

5 Proof. By Lemmas 5.2 and 9.5 there exists y ∈ ∆1(x) ∩ ∆3(a) with y + x ∈ 4 α1(y, y + c, +). Since Γ0(y + x)\{y, x} ⊆ ∆4(a) we can apply Lemma 5.5 to show 5 |α1(y, y + c, +)|.|∆3(a)| m = 5 |∆4(a)| where m is the size of the Gax-orbit on Γ1(x) containing x + y. Using (2.7), Lemmas 4.11(i) and 9.1(i) we get m = 42. Examining the possibilities in (2.13) we see that x + y ∈ α1(x, x + d, +). Parts (i) and (ii) follow by Lemmas 5.5 and

9.5 together with the fact that α1(x, x + d, +) is a Gax-orbit.

For (iii) we again appeal to Lemma 5.5 to show that z + x lies in a Gaz-orbit on Γ1(z) of size 5 |α1(y, y + c, +)|.|∆3(a)| 4 = 11 |∆4(a)| (by Lemmas 4.11(i) and 8.7(ii)). Therefore (2.12) implies that z+x ∈ α1(z, END, +)∪

α1(z, END, −) ∪ α5(z, END, −). The result now follows from Lemmas 8.8 and 8.13.

5 6 Lemma 9.7. ∆4(a) 6= ∆4(a).

5 6 5 Proof. Suppose ∆4(a) = ∆4(a), and argue for a contradiction. Let x ∈ ∆4(a). 0 3 0 By (2.15)(xiv) there exists d ∈ ∆1(x)∩∆3(a) and, using Lemma 8.4(v), ∆1(d )∩ 5 ∆4(a) is a Gax-orbit of size 160. Lemmas 4.8(iii), 5.4 and 9.1(i) imply that 3 3 ∆1(x) ∩ ∆3(a) is a Gax-orbit of size 35. So {l ∈ Γ1(x)|Γ0(l) ∩ ∆3(a) 6= ∅} is a

Gax-orbit of Γ1(x) of size 35. By (2.13) this orbit must be α3(x, x+d, +). Then 0 6 5 6 Lemma 9.2 forces d ∈ ∆3(a), a contradiction. Therefore ∆4(a) 6= ∆4(a).

7 6 10 Finishing ∆3(a)

6 6 1 In this section we reconsider ∆3(a). Fix x ∈ ∆3(a), {b} = ∆1(a) ∩ ∆2(x) and 1 {c} = ∆2(a) ∩ ∆1(x) for the whole of Section 10. Earlier in Lemma 4.11(ii) we 9 2 6 showed that |Gax| = 2 .3 . In order to make further headway with ∆3(a) we ∗x need to determine Gax. The ﬁrst result of this section does just that.

∗x ∗x ∼ Lemma 10.1. Gax is equal to the stabilizer in Gx (= M23) of the heptad ∗x 7 2 x + c and the three element subset T (x, b) of x + c. Moreover |Gax| = 2 .3 and 2 |Gax ∩ Q(x)| = 2 .

∗x Proof. From Lemmas 4.10 and 4.11(ii) we already know that Gax is contained ∗x 7 2 in H, the stabilizer in Gax of x+c and T (x, b). Thus, as |H| = 2 .3 , the lemma 2 9 2 will follow once we establish that |Gax ∩ Q(x)| = 2 . Since |Gax| = 2 .3 we 2 2 clearly have |Gax ∩ Q(x)| ≥ 2 . Assuming that |Gax ∩ Q(x)| > 2 we seek a 2 contradiction. By Lemma 7.5 there exists f1 ∈ ∆1(x) ∩ ∆4(a). From (2.10), 6 Corollary 7.4 and Lemma 8.12 we deduce that ∆1(f1) ∩ ∆3(a) is a Gaf1 -orbit 2 of size 70. Hence, by Lemmas 4.11(ii), 5.4 and 7.3(ii) ∆1(x) ∩ ∆4(a) is a Gax- 2 3 orbit of size 16. Now Gax ∩ Q(x) ≤ Gaxf1 (recall that ∆4(a) 6= ∆4(a) by ∼ (2.15)(xi) and Corollary 7.4) and Gaxf1 contains a Sylow 3-subgroup of Gaf1 (= 4 A8). Consequently, by our assumption and properties of A8, |Gax ∩ Q(x)| = 2 . 5 Appealing to Lemma 9.2(ii) yields the existence of f2 ∈ ∆1(x) ∩ ∆4(a). Since 5 6 ∆4(a) 6= ∆4(a) by Lemma 9.7, Lemma 5.5 implies that {x + f ∈ Γ1(x) | f ∈ 5 4 ∆4(a) and f + x ∈ α3(f, f + d, +) where {d} = ∆1(f) ∩ ∆3(a)} is a Gax-orbit of size 64. Hence ( ¯ ) ¯ ¯ x+f ∈ Γ1(x) and f +x ∈ α3(f, f +d, +) Γ (x + f) ∩ ∆5(a) ¯ 0 4 ¯ 4 where {d} = ∆1(f) ∩ ∆3(a)

5 is a Gax-orbit of ∆4(a) of size 64. But since Gax ∩ Q(x) ≤ Gaxf2 and [Gax : 5 2 5 2 Gax ∩ Q(a)] = 2 .3 we see such an orbit must have size dividing 2 .3 , a 2 contradiction. From this contradiction we infer that Gax ∩Q(x) = 2 , so proving the lemma.

With Lemma 10.1 to hand, (2.8) gives a listing of the Gax-orbits on Γ1(x) (setting TRI = T (x, b) and with c playing the role of b in (2.8)).

8 2 Lemma 10.2. Let y ∈ ∆1(x) ∩ ∆4(a). Then,

(i) y + x ∈ α4(y, O(y, a));

(ii) x + y ∈ α3,0(x, x + c, TRI); and

2 3 (iii) if l ∈ α3,0(x, x + c, TRI), then |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

Proof. By Corollary 7.4 and Lemma 7.5 and (2.10),

6 |∆1(y) ∩ ∆3(a)| = 70 + 168n for n = 0, 1 or 2

2 3 As a consequence of Corollary 7.4(ii) and (2.15) ∆4(a) 6= ∆4(a). Therefore Lemma 5.5 implies that

2 2 |∆4(a)| |∆1(x) ∩ ∆4(a)| = (70 + 168n). 6 |∆3(a)| 212.11.23 = (70 + 168n) 29.5.7.11.23 8 = (70 + 168n) 35 (using Lemmas 4.11(ii) and 7.3). Hence n = 0 and (2.10) yields y + x ∈

α4(x, O(x, a)), giving part (i). For part (ii) we may use Lemma 5.5 again to show 2 |∆4(a)||α4(x, O(x, a))| m = 6 |∆3(a)| where m is the size of the Gax-orbit on Γ1(d) containing x + y. So

212.11.23.70 m = = 16 29.5.7.11.23 by (2.10) and Lemmas 4.11(ii) and 7.3(ii). Perusing (2.8) we see that x + y ∈

α3,0(x, x + c, TRI) and so we have part (ii).

Part (iii) follows from Lemma 7.5 and part (ii) because α3,0(x, x + c, TRI) is a Gax-orbit of Γ1(x).

5 6 Lemma 10.3. If l ∈ α1,0(x, x+c, TRI), then |Γ0(l)∩∆4(a)| = |Γ0(l)∩∆4(a)| = 1.

9 6 5 5 Proof. Since ∆3(a) and ∆4(a) are Ga-orbits, there exists y ∈ ∆1(x) ∩ ∆4(a) 4 with y + x ∈ α3(y, y + d, +) by Lemma 9.2(ii) (where {d} = ∆1(y) ∩ ∆3(a)).

Using Lemma 9.3 and (2.8) we have that x + y ∈ α1,0(x, x + c, TRI). The result now follows from Lemma 9.2(ii) again because α1,0(x, x + c, TRI) is a Gax-orbit of Γ1(x).

Lemma 10.4. Let l ∈ α3,1(x, x + c, TRI). Then

1 (i) |Γ0(l) ∩ ∆4(a)| = 2; and

1 L (ii) if y ∈ Γ0(l) ∩ ∆4(a) it follows that y + x ∈ α3,0(y, y + d, DUAD) where 1 {d} = ∆1(y) ∩ ∆3(a).

Proof. Let TRI = {X1,X2,X3}. Then Xi ∈ Γ3(x, b) for i = 1, 2, 3. Since 6 a ∈ ∆3(x) by Lemma 4.12, a∈ / Γ0(Xi) for i = 1, 2, 3. Hence τ(Xi) interchanges a and the point in Γ0(a+b)\{a, b} for each i by Lemma 3.2. Appealing to Lemma 6 1 6.6(iii) and the fact that ∆3(a) is a Ga-orbit, there exists y ∈ ∆1(x) ∩ ∆4(a) L 1 with y + x ∈ α3,0(y, y + d, DUAD) where {d} = ∆1(y) ∩ ∆3(a). Using (2.8) together with Lemmas 4.10, 10.3 and 10.2(iii) we must have x + y ∈ α1,1(x, x + c, TRI) ∪ α3,2(x, x + c, TRI) ∪ α3,1(x, x + c, TRI). If |(x + y) ∩ TRI| = 1 in Ωx we may assume X1 ∈ Γ3(x + y) and X2,X3 ∈/ Γ3(x + y). If |(x + y) ∩ TRI| = 2 we may assume X1,X2 ∈ Γ3(x + y) and X3 ∈/ Γ3(x + y). In either case let 0 τ := τ(X1)τ(X3). Then τ ∈ Gax and τ interchanges y and y by Lemma 3.2, 0 where Γ0(x + y) = {x, y , y}. Therefore we may apply Lemma 5.6 to show that x + y lies in a Gax-orbit of Γ1(x) of size

1 L |∆4(a)||α3,0(y, y + d, DUAD)| m = 6 . 2|∆3(a)| By (2.9) and Lemmas 4.11(ii) and 6.2(i) we have m = 72. Now (2.8) implies 0 1 that x + y ∈ α3,1(x, x + c, TRI). Since τ ∈ Ga, y ∈ ∆4(a) and the proof of the lemma is complete.

3 Lemma 10.5. Let l ∈ α1,1(x, x + c, TRI). Then |Γ0(l) ∩ ∆4(a)| = 2.

0 ⊥ Proof. Let y ∈ Γ0(l)\{x}. By Lemma 3.11(ii) there exists c ∈ {b, x} with 0 0 1 0 x + c ∈ α3(x, l). Therefore c ∈ ∆2(y). Since l ∈ α1(x, x + c), c 6= c and so 0 2 Lemma 4.10 implies that c ∈ ∆2(a).

10 r 0 c@ @ r r y @ b@ @x r @ @ a r @rc @ @ @ @ r

Since (x + y) ∩ TRI = (x + y) ∩ (x + c) is a singleton subset of Ωx, there ex- 0 0 ists a unique hyperplane X ∈ Γ3(b, c , x, y). Moreover X 6= X(c , a) because 0 ⊥ 0 Γ3(a, x) = ∅. Let d ∈ {c , y} ∩ Γ0(X(c , a)) (such a point exists by Lemma 0 1 3.11(ii)). Then d ∈ Γ0(X) ∩ Γ0(X(c , a)), whence d ∈ ∆2(b) by Lemma 3.8(i). 2 2 Since ∆1(d) ∩ ∆2(a) 6= ∅ and Γ3(d, a) 6= ∅ we deduce that d ∈ ∆3(a) using The- 0 orem 4.13. We have d + c ∈ α4,1(d, O(d, a),X(d, a)) by Theorem 4.13(iv)(b). 1 ⊥ Also, since b ∈ ∆1(a), ∆2(a) ∩ {b, d} 6= ∅ by Lemma 3.11(ii), which, together with Theorem 4.13(iv)(a) implies that T (d, b) ∩ O(d, a) = ∅ in Ωd. We already know that T (d, x) ∩ T (d, b) = {X}, whence |T (d, x) ∩ O(d, a)| = 2 because 0 T (d, x) ∪ T (d, b) ⊆ d + c as subsets of Ωd. Hence, in Ωd, |(d + z) ∩ O(d, a)| = 2 for two points z in {d, x}⊥ and |(d + z) ∩ O(d, a)| = 4 for three points z in {d, x}⊥. Using (2.15)(ix) and (xi) we have

⊥ 0 1 3 (10.5.1) {d, x} consists of c , two points in ∆4(a) and two points in ∆4(a).

Since l ∈ α1,1(x, x + c, TRI) we have |T (x, d) ∩ T (x, b)| = 1 with T (x, b) ⊆ x + c in Ωx. Therefore (2.8) implies x + z ∈ α3,1(x, x + c, TRI) for two points ⊥ 0 ⊥ 0 z ∈ {x, d} \{c } and x+z ∈ α1,1(x, x+c, TRI) for two points z ∈ {x, d} \{c }. 1 ⊥ Appealing to Lemma 10.4 we have z ∈ ∆4(a) for each z ∈ {x, d} with x + z ∈ 3 ⊥ α3,1(x, x+c, TRI). Therefore (10.5.1) yields that z ∈ ∆4(a) for each z ∈ {x, d} with x + z ∈ α1,1(x, x + c, TRI). Since y is any point in Γ0(l)\{x}, the lemma now follows.

1 Lemma 10.6. Let l ∈ α3,2(x, x + c, TRI). Then |Γ0(l) ∩ ∆4(a)| = 2.

1 Proof. Let y ∈ Γ0(l)\{x}. Since TRI = T (x, b) (where {b} = ∆1(a) ∩ ∆2(x)), 1 by the deﬁnition of α3,2(x, x+c, TRI), |Γ3(y, b)| ≥ 2. Hence b ∈ ∆1(y)∪∆2(y)∪ 1 i ∆3(y) by Theorem 4.13. We have y∈ / ∆2(a) ∪ ∆3(a) for i = 1,..., 5 by (2.8)

11 1 and Lemmas 4.10 and 4.15. Therefore b∈ / ∆1(y). If b ∈ ∆2(y), then since 1 b ∈ ∆2(x), b is collinear with a point in Γ0(l) using Theorem 3. However we 1 then have l ∈ Γ1(TRI) which is impossible. Hence b ∈ ∆3(y). 1 1 Since Γ3(a, y) = ∅ , (2.15)(ix) implies that a ∈ ∆4(y). Then y ∈ ∆4(a) by

Lemma 6.4. As y was an arbitrary point in Γ0(l)\{x}, the lemma is proved.

Combining together all the results in this section with Lemma 4.10 and Theorem 4.13(vii),(viii) we see that Theorem 10 is now proven.

3 11 A ﬁrst look at ∆4(a)

3 We are now in a position to consider the Ga-orbit ∆4(a). Let x be a ﬁxed point 3 in ∆4(a). The ﬁrst result in this section summarizes some of the main properties of 3 ∆4(a).

6 2 Lemma 11.1. (i) |Gax| = 2 .3 and Q(x) ∩ Ga = 1;

3 12 (ii) |∆4(a)| = 2 .5.7.11.23;

2 (iii) |∆1(x) ∩ ∆4(a)| = 2;

2 (iv) |∆1(x) ∩ ∆3(a)| = 3; and

(v) Gax stabilizes a disjoint triad TRI and octad OCT in Ωx together with a partition of OCT into two 4-sets, each of whose union with TRI forms a

heptad in Γ1(x).

3 Proof. Since ∆4(a) is a Ga-orbit by Lemma 5.2, Lemma 7.5 implies there exists 6 2 l ∈ Γ1(x) where Γ0(l) = {x, y, z} with y ∈ ∆3(a) and z ∈ ∆4(a). Appealing 1 to Lemma 10.2(ii), we have l ∈ α3,0(y, y + c, TRI) where {c} = ∆1(y) ∩ ∆2(a). 1 Thus c ∈ ∆2(x). Since Γ3(a, x) = ∅ by Lemma 5.1, we have T (c, a)∩T (c, x) = ∅ ⊥ 1 1 in Ωc. For each d ∈ {c, x} , d∈ / ∆3(a) by Theorem 5 because x∈ / ∆4(a) by (2.15)(xi) and Lemma 6.3(ii). Therefore Lemma 3.11(ii) and Theorem 3 imply ⊥ 2 there exist d, d1, d2, y1 ∈ {c, x} \{y} with d, d1, d2 being distinct points in ∆3(a) 6 and y1 ∈ ∆3(a). Note that d + x ∈ α4,0(d, O(d, a),X(d, a)) by Lemma 6.3(ii) 0 and Corollary 7.4(ii). Let Γ0(d+c) = {d, d , c}. So the state of play is as follows.

12 dr @ @ @ d0 r dr1 @ ¨¨H @ ¨ HH Xr ¨ H @ XX ¨ H XX ¨ H@ XX ¨ H r X Xr H@¨ rx a cHXX ¨ @H XX ¨ @HH XX r ¨ H X ¨¨ @ H d2 H ¨¨ @ H¨r rz @ y1 @ @ @ r y 0 2 Then d ∈ ∆1(z) ∩ ∆3(a) by Theorem 3 and Lemma 3.10. Viewing l from z,

Lemma 10.2(i) implies l = z + x ∈ α4(x, O(x, a)). Set P := Q(x) ∩ Ga.

(11.1.1) Either P = 1 or P =∼ 22.

∼ Now P ﬁxes l and hence P ≤ Gazl. By Lemma 7.3(i) Gaz = A8. Using ∼ 2 (2.10) we note that P ≤ O2(Gazl) = 2 . Since |α4(x, O(x, a))| = 70, Gazl will contain a Sylow 3-subgroup S of Gaz. Because CGaz (S) = S we must have P = 1 or P =∼ 22.

(11.1.2) P ≤ Q(d).

3 Since x∈ / Γ0(X(d, a)), Γ0(x + d)\{d} ⊆ ∆4(a) by Lemma 3.2. So P fixes d and fixes X(d, a). Consequently P fixes the 8 elements of Ωd in Γ3(d + x) ∪ ∗d ∼ {X(d, a)}. Hence P ≤ Q(d) because involutions in Gd (= M23) fix exactly 7 elements of Ωd. This proves (11.1.2). 6 3 ∗d By (2.5) and Lemma 4.8(ii), |Gadx| = 2 .3 with |Q(d)a| = 2 and Gadx ∩ ∗d ∗a O2(Gad) = 1. So, by (11.1.2), Gadx ∩ Q(a) = P ∩ Q(a) = 1. Therefore |Gadx| = 6 ∗a ∗a ∗a 3 2 .3 and P E Gadx. Now |P | = |P | = 1, 2 or 2 by (2.5) which, together with (11.1.1), yields P = 1. 5 2 By (2.8) and Lemma 4.11(ii) we have |Gacyx| = 2 .3 because l ∈ α3,0(y, y + 0 1 0 c, TRI) and Gal = Gax. For each c ∈ ∆1(d)∩∆2(a), d+c ∈ α0,1(d, O(d, a),X(d, a)). 1 Since d + x ∈ α4,0(d, O(d, a),X(d, a)), c is the unique point in ∆2(a) ∩ ∆1(d) ∩ 1 6 ∆2(x). Hence Gacdx = Gadx and so |Gacdx| = 2 .3.

13 6 2 Put H := hGacyx,Gacdxi. Clearly 2 .3 | |H|.

In Ωx deﬁne TRI = (x + y) ∩ (x + y1) and OCT = (x + y) ⊕ (x + y1)

(⊕ is symmetric diﬀerence). Then |TRI| = 3 and so OCT is an octad in Ωx. Furthermore H ﬁxes (setwise) TRI and OCT and leaves invariant the partition of OCT into the tetrads OCT ∩ (x + y) and OCT ∩ (x + y1). Since H ≤ Gax and 6 2 ∗x ∗x Q(x) ∩ Ga = P = 1, 2 .3 | |H |, and therefore we conclude that H is the stabilizer in M23 of an octad and a partition of the octad into two tetrads and 6 2 that H has order 2 .3 . Furthermore the Gax-orbits on Γ1(x) are certain unions of the H-orbits described in (2.11). Note that α3,4|0(x, TRI, OCT) = {x+y, x+ 3 2 y1} and that each of these lines has one point in ∆4(a), one point in ∆4(a) and 6 2 one point in ∆3(a). For any k ∈ α3,0(x, TRI, OCT), |Γ0(k)∩∆3(a)| = 1 because x+d ∈ α3,0(x, TRI, OCT). This together with (2.11) and Lemma 5.3(ii) implies 2 2 that |∆1(x) ∩ ∆3(a)| = 3 or 15. If |∆1(x) ∩ ∆3(a)| = 15, then, using Lemma 4.8(ii),

|∆2(a)|.112 |∆3(a)| = 3 4 15 = 212.7.11.23.

3 4 4 16 Thus ∆4(a) 6= ∆4(a) since |∆4(a)| = 2 .3.7.23. Then Theorem 12 and Lemma 7.3(ii) yield that

2 2 |∆4(a)|.70 |∆1(x) ∩ ∆4(a)| = 3 |∆4(a)| = 10.

2 3 This contradicts (2.11) and so we must have |∆1(x)∩∆3(a)| = 3 with |∆4(a)| = 12 2 2 .5.7.11.23, |∆1(x) ∩ ∆4(a)| = 2 and Gax = H. This completes the proof of the lemma.

6 Lemma 11.2. (i) Let l ∈ α3,4|0(x, TRI, OCT). Then |Γ0(l) ∩ ∆3(a)| = 2 6 2 |Γ0(l) ∩ ∆4(a)| = 1. Moreover, if y ∈ Γ0(l) ∩ ∆3(a) and z ∈ Γ0(l) ∩ ∆4(a), 1 then l ∈ α3,0(y, y + c, TRI) ∩ α4(z, O(z, a)) (where {c} = ∆1(y) ∩ ∆2(a)).

2 3 (ii) Let l ∈ α3,0(x, TRI, OCT). Then |Γ0(l)∩∆3(a)| = 1 and |Γ0(l)∩∆4(a)| = 2.

14 6 3 (iii) Let l ∈ α1,0(x, TRI, OCT). Then |Γ0(l)∩∆3(a)| = 1 and |Γ0(l)∩∆4(a)| = 6 2. Moreover, if y ∈ Γ0(l) ∩ ∆3(a) then l ∈ α1,1(y, y + c, TRI) (where 1 {c} = ∆1(y) ∩ ∆2(a)).

5 1 (iv) Let l ∈ α1,2|2(x, TRI, OCT). Then |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1. 5 1 Moreover, if y ∈ Γ0(l) ∩ ∆3(a) and z ∈ Γ0(l) ∩ ∆4(a), then l ∈ α3(y, y + 2 c, +)∩α1,1(z, z+d, DUAD) (where {c} = ∆1(y)∩∆2(a) and {d} = ∆1(z)∩ 1 ∆3(a)).

Proof. Parts (i) and (ii) are a consequence of (2.5), (2.8), (2.10), (2.11), Lemmas 4.8(ii), 5.6, 7.3(ii), 7.5, 10.2 and 11.1(ii). 6 3 For part (iii), by Theorem 10, if y ∈ ∆1(x) ∩ ∆3(a), then |∆1(y) ∩ ∆4(a)| = 112. Therefore

6 6 |∆3(a)| |∆1(x) ∩ ∆3(a)| = 3 .112 |∆4(a)| 29.5.7.11.23.112 = 212.5.7.11.23 = 14 by Lemmas 4.11(ii) and 11.1(ii). Considering the orbit sizes in (2.11) we have x + y ∈ α3,4|0(x, TRI, OCT) ∪ α1,0(x, TRI, OCT). Therefore part (iii) follows from part (i) and Lemma 10.5.

Turning to (iv), we obtain the result from Lemma 5.5 using the orbit sizes in (2.7), (2.9), (2.11) and Lemmas 4.11(i), 6.2(i) and 11.1(ii), together with the information about α1,1(z, z + d, DUAD) given in Lemma 6.8(i).

5 Lemma 11.3. Let y ∈ ∆1(x) ∩ ∆3(a). Then

3 (i) |∆1(y) ∩ ∆4(a)| = 140.

(ii) x + y ∈ α1,2|2(x, TRI, OCT) ∪ α0,1|1(x, TRI, OCT) ∪ α2,1|1(x, TRI, OCT).

5 3 (iii) |Γ0(l)∩∆3(a)| = 1 and |Γ0(l)∩∆4(a)| = 2 for some line l ∈ α0,1|1(x, TRI, OCT)∪ 5 (1) α2,1|1(x, TRI, OCT). Moreover, if y ∈ Γ0(l) ∩ ∆3(a), then l ∈ α3 (y, y + 2 c, −) (where {c} = ∆1(y) ∩ ∆2(a)).

15 5 Proof. By Theorem 4.13(vii), Lemmas 5.7, 6.7 and 9.5, if y ∈ ∆1(x) ∩ ∆3(a) 3 (1) and k ∈ Γ1(y), then |Γ0(k) ∩ ∆4(a)| = 2 whenever k ∈ α3 (y, y + c, −); |Γ0(k) ∩ 3 3 ∆4(a)| = 1 whenever k ∈ α3(y, y + c, +); and Γ0(k) ∩ ∆4(a) = ∅ whenever k ∈ (2) 3 {y+c}∪α1(y, y+c, +)∪α3 (y, y+c, −). So |∆1(y)∩∆4(a)| = 80+60+96n where n = 0, 1 or 2 by (2.11), depending on the point distribution of α1(y, y + c, −). Using Lemmas 4.11(i) and 11.1(ii) we have

5 5 |∆3(a)|(140 + 96n) |∆1(x) ∩ ∆3(a)| = 3 |∆4(a)| 212.3.7.11.23 = (140 + 96n). 212.5.7.11.23

3 For this to be an integer we must have n = 0, whence |∆1(y) ∩ ∆4(a)| = 5 140. This proves part (i). Also we have |∆1(x) ∩ ∆3(a)| = 84. Using (2.11) 5 together with Lemma 11.2 yields that {l ∈ Γ1(x) | Γ0(l) ∩ ∆3(a) 6= ∅} =

α1,2|2(x, TRI, OCT) ∪ αi,1|1(x, TRI, OCT) for some i = 0 or 2. Therefore we have part (ii). (1) For part (iii) we know that y +x ∈ α3 (y, y +c, −)∪α3(y, y +c, +). Suppose 5 y + x ∈ α3(y, y + c, +). Then Lemma 6.8(i) implies |Γ0(y + x) ∩ ∆3(a)| = 1 3 |Γ0(y + x) ∩ ∆4(a)| = |Γ0(y + x) ∩ ∆4(a)| = 1. In this case Lemma 5.5 yields that x + y lies in a Gax-orbit of size 36. Using Lemma 5.5 again we have x + y ∈ α1,2|2(x, TRI, OCT) if and only if y + x ∈ α3(y, y + c, +). Thus if (1) y +x ∈ α3 (y, y +c, −), then x+y ∈ α0,1|1(x, TRI, OCT)∪α2,1|1(x, TRI, OCT). Now Lemma 5.7(i) (and the remark immediately following that lemma) gives the result.

6 12 A ﬁrst look at ∆4(a)

We now turn our attention to the final Ga-orbit in the fourth disc of G. So let 6 3 x be a fixed point in ∆4(a). By (2.15)(xiv), ∆1(x) ∩ ∆3(a) 6= ∅. Our first task, 3 achieved in Lemma 12.2, is to determine the size of ∆1(x) ∩ ∆3(a).

3 3 Lemma 12.1. Let d ∈ ∆1(x) ∩ ∆3(a) and put Ud := {z ∈ ∆1(x) ∩ ∆3(a) ∩ 1 ⊥ 1 ∆2(d)|{d, z} ∩ ∆3(a) 6= ∅} ∪ {d}. Then the following hold.

16 3 ⊥ (i) |Ud| = 5 and for each z ∈ Ud\{d}, ∆3(a) ∩ {e, x} = {d, z}, where {e} = ⊥ 1 {d, z} ∩ ∆3(a).

3 (ii) 5 | |∆1(x) ∩ ∆3(a)|.

3 Proof. Let d ∈ ∆1(x) ∩ ∆3(a). Then d + x ∈ α3,0(d, d + c, X(d, a)) by Lemma 2 8.4, where {c} = ∆1(d) ∩ ∆2(a) (see also (2.15)(xiv)). Using Appendix A × × × × × × with d + c = , d + x = and X(d, a) = × × × × × × × × ◦

in Ωd, we see that there are precisely 4 heptads in α1,1(d, d+

c, X(d, a))∩α3(d, d+x) (namely h4, h7, h11, h15 ). Lemma 4.15(i) together with 1 (2.15) then imply that there are precisely four points in Sd := ∆3(a) ∩ ∆1(d) ∩ 1 ∆2(x).

For any e ∈ Sd, T (e, x) ∩ D(e, a) = ∅ in Ωe because Γ3(a, x) = ∅ by Lemma ⊥ 3 5.1(ii). Hence {e, x} consists of d, one point z (6= d) in ∆3(a) and three points 1 ⊥ 1 in ∆4(a). Moreover, {d, z} ∩ ∆3(a) = {e} because Γ3(a, x) = ∅. Hence, letting 3 e vary throughout Sd we obtain four distinct points in (∆1(x) ∩ ∆3(a))\{d}. 3 ⊥ Therefore Ud = 5 and ∆3(a) ∩ {e, x} = {d, z}, so giving (i).

To prove part (ii) it is enough to show that if y1 and y2 are distinct points 3 in ∆1(x) ∩ ∆3(a) and y2 ∈ Uy1 , then Uy1 = Uy2 . Assume y2, z ∈ Uy1 with 1 ⊥ z 6= y2. By the deﬁnition of Uy1 there exist e1, e2 ∈ ∆3(a) with e1 ∈ {z, y1} ⊥ and e2 ∈ {y2, y1} . If e1 = e2, then z ∈ Uy2 by deﬁnition and we are done. So we may suppose that e1 6= e2, and then

17 r ¨¨zAZ ¨ Z e1 ¨r A Z @ A Z r H @ A Z H 0 Z H @ Are Z r H ¢ a H ((@r Zrx H ((( y1 HH(cr1 ((( ¢ ¢ ¢ r ¢ e2 H HH ¢ H¢r y2

2 0 where {c1} = ∆1(y1) ∩ ∆2(a). (We deﬁne e shortly.)

In Ωy1 , the heptads y1 + e1 and y1 + e2 intersect y1 + c1 in exactly X(y1, a) by Lemma 4.15(i). Moreover |(y1 + e1) ∩ (y1 + e2) ∩ (y1 + x)| = 2. Thus

|T (y1, z)∩T (y1, y2)| = 2 in Ωy1 , and so |Γ3(z, y1, y2)| = 2. Note that y2 ∈ ∆1(z) would imply D(a, z) = D(a, y2) by Lemma 4.16 which, by Lemma 3.6, yields 1 the impossible Γ3(a, x) 6= ∅. Therefore y2 ∈ ∆2(z) by Lemma 3.8(i). Appealing 0 ⊥ 0 to Lemma 3.11(ii) we can choose e ∈ {y2, z} with Γ3(a, e ) 6= ∅. Suppose 0 0 2 e ∈ ∆2(a). Then, by Theorem 7, e ∈ ∆2(a) and consequently Γ3(a, z, y2) 6= ∅ 0 which again, by Lemma 3.6, gives the untenable Γ3(a, x) 6= ∅. Hence e ∈ ∆3(a) 0 i 0 i and so e ∈ ∆3(a) for i = 1, 2, 3 or 4. If e ∈ ∆3(a) for i = 2, 3 or 4, then 0 0 Lemmas 4.15 and 4.16 imply that y2, z ∈ Γ0(X(e , a)) and thus x ∈ Γ0(X(e , a)) 0 1 by Lemma 3.6, a contradiction. Therefore e ∈ ∆3(a) and so z ∈ Uy2 . Thus

Uy1 = Uy2 , as required, and the proof of the lemma is complete.

3 Lemma 12.2. (i) |∆1(x) ∩ ∆3(a)| = 5.

6 15 (ii) |∆4(a)| = 2 .7.11.23.

∼ ∗x ∗x ∗x (iii) Gax = Gax ∼ (3 × A5)2 and Gax is the stabilizer in Gx of a triad TRI

and a 5-element subset FIX in Ωx with TRI ∩ FIX = ∅.

3 6 (iv) If l ∈ α0,4(x, TRI, FIX), then |Γ0(l) ∩ ∆3(a)| = 1 and |Γ0(l) ∩ ∆4(a)| = 2.

3 1 (v) If e1, e2 ∈ ∆1(x) ∩ ∆3(a) with e1 6= e2, then e1 ∈ ∆2(e2) and there exists ⊥ 1 e ∈ {e1, e2} with e ∈ ∆3(a).

18 3 Proof. Let d ∈ ∆3(a) ∩ ∆1(x). As in Lemma 12.1, set

1 1 Sd = ∆3(a) ∩ ∆1(d) ∩ ∆2(x) and 3 1 ⊥ 1 Ud = {z ∈ ∆1(x) ∩ ∆3(a) ∩ ∆2(d) | {d, z} ∩ ∆3(a) 6= ∅} ∪ {d}.

Recall that |Sd| = 4 and |Ud| = 5, and note that X(a, z1) 6= X(a, z2) for 3 5 ∗d 4 z1, z2 ∈ ∆1(x) ∩ ∆3(a), z1 6= z2. Since Gad ∼ 2 A6, Gad ∼ 2 A6 and |∆1(d) ∩ 6 3 2 ∼ ∆4(a)| = 160 by Lemmas 4.8(ii) and 8.4(v), Gadx ∼ 2 3 ≥ H = 3 × A4 ∼ ∗d ∗d with Gadx = Gadx being the stabilizer in Gad of the line d + x. Hence, by

(2.3), there exists T E Gadx with |T | = 3 and T fixing each of the four lines in 2 α3(d, d+x)∩α1,1(x, d+c, X(d, a)) ({c} = ∆1(d)∩∆2(a)). Consequently T fixes 3 ⊥ each point in Sd. Since by Lemma 12.1(i), for each z ∈ Ud\{d}, ∆3(a)∩{e, x} = ⊥ 1 {d, z} (e ∈ {d, z} ∩ ∆3(a)) we deduce that T fixes each point in Ud and hence

(12.2.1) T ﬁxes each of the 5 hyperplanes X(a, z), z ∈ Ud.

3 Put n = |∆1(x) ∩ ∆3(a)|.

(12.2.2) n = 5 or 20.

3 By Lemmas 5.3(ii) and 12.1(ii) n = 5, 10, 15 or 20. Now Ud ⊆ ∆1(x)∩∆3(a), 3 and {X(a, z) | z ∈ ∆1(x) ∩ ∆3(a)} is a T -invariant subset of Ωa of size n. Since ∗a ∼ an element of order 3 in Ga (= M23) ﬁxes exactly 5 elements of Ωa, by (12.2.1)

T ﬁxes each element in {X(a, z) | z ∈ Ud} and has orbits of length 3 on the remaining elements of Ωa. This rules out n = 10 or 15, so proving (12.2.2).

3 (12.2.3) {x + y | y ∈ ∆1(x) ∩ ∆3(a)} is a Gax-orbit of Γ1(x) of length n.

This follows from Lemmas 5.4 and 8.4(iii),(v).

Using (12.2.2) and Lemmas 4.8(iii) and 8.4(v) gives

3 2 5 2 (12.2.4) |Gax| = 2 .3 .5 (if n = 5) or 2 .3 .5 (if n = 20).

∗x Putting Gx = Gx and using the bar notation for subgroups of Gx, we next prove

(12.2.5) R E Gax where |R| = 3.

19 i 2 By (12.2.4) |Gax| = 2 .3 .5 where i ≤ 5. Let N be a minimal normal subgroup of Gax. First we consider the case when N is abelian. Since |CM23 (ϑ)| = 15 for ϑ an element of order 5, |N| = 3 and 24 are the only possibilities. Sup- 4 4 4 pose N = 2 were to hold. Then Gax ≤ 2 (3 × A5)2 or 2 A7, and hence

Gax/N ≤ (3 × A5)2 or A7. Now Gax/N must contain a Hall subgroup of order 2 2 3 .5. However neither (3 × A5)2 nor A7 contain a subgroup of order 3 .5, which rules out the possibility N = 24. So (12.2.5) holds with R = N. ∼ Now suppose N is not abelian. Then the order of Gax implies that N = A5 or A6. If the latter holds, then Gax/N is a 2-group and hence N contains ∼ ∼ 2 2 ∼ H = 3 × A4 (note that H = 3 would give 2 = H ∩ Q(x) ≤ Z(Gax), whereas ∼ Z(H) = 3). Since A6 contains no such subgroups we must have N = A5. Then, by centralizers of 5-elements in M and Out(A ) = 2 we obtain C (N) = 3. 23 5 Gax Taking R = C (N) in this case, proves (12.2.5). Gax

(12.2.6) (i) n = 5; and

∼ (ii) Gax = Gax ∼ (3 × A5)2.

By (12.2.5), G ≤ N (R) ∼ (3 × A )2. Since G intersects 3 × A in a ax Gx 5 ax 5 2 ∼ subgroup of order at least 3 .5, we deduce that Gax = 3 × A5 or (3 × A5)2. As a consequence, the size of a Gax-orbit on Γ1(x) will be either one of the sizes or half of one of the sizes given in (2.14). Consulting (2.14) reveals that Gax cannot have an orbit on Γ1(x) of length 20 and therefore (12.2.2) and (12.2.3) 3 2 3 2 force n = 5. Hence |Gax| = 2 .3 .5 by (12.2.4). Since Gadx ∼ 2 .3 has no ∼ central elements of order 2, we must have Gax = Gax ∼ (3 × A5)2, so yielding (12.2.6). 6 15 By (12.2.6)(ii) |∆4(a)| = [Ga : Gax] = 2 .7.11.23 and so we have proved 3 2 parts (i) and (ii). Because T E Gadx (∼ 2 .3 ) ≤ Gax (∼ (3 × A5)2) we see that ∗x T E G and so G = N ∗x (T ). Let FIX denote the set of (5 elements) of Ω ax ax Gax x ﬁxed by T and set [ TRI = Ωx\ {l ∈ Γ1(x) | |l ∩ FIX| = 4}.

∗x ∗x Then (see (2.14)) Gax is the stabilizer in Gx of TRI and FIX which proves part (iii).

20 We now prove parts (iv) and (v). Clearly, by part (i) and Lemma 12.1(i), 3 Ud = ∆1(x) ∩ ∆3(a). As previously observed in Lemma 12.1(i) for each z ∈ 3 ⊥ ⊥ 1 Ud\{d}, ∆3(a) ∩ {e, x} = {d, z} (e ∈ {d, z} ∩ ∆3(a)). Looking at the list in

(2.14), and noting that all the lines in α3,1(x, TRI, FIX) lie in the same diamond, 3 we deduce that {x + y | y ∈ ∆1(x) ∩ ∆3(a)} = α0,4(x, TRI, FIX), from which 3 part (iv) follows. Since Ue1 = ∆1(x) ∩ ∆3(a), Lemma 12.1(i) gives part (v).

1 Lc Lemma 12.3. Let y ∈ ∆1(x) ∩ ∆4(a) with y + x ∈ α3,0(y, y + d, DUAD) (where 1 {d} = ∆1(y) ∩ ∆3(a)). Suppose x + y lies in the Gax-orbit Ox on Γ1(x). Then

(i) Ox is α2,2(x, TRI, FIX) or α1,3(x, TRI, FIX).

1 4 6 (ii) For all l ∈ Ox, |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

Proof. The result is a consequence of Lemma 5.5, using the orbit information given in (2.9), (2.14) and Lemmas 6.2(i), 8.10(i) and 12.2(ii).

1 Lemma 12.4. Let y ∈ ∆4(a) and l ∈ α1,0(y, y+d, DUAD) where {d} = ∆1(y)∩ 1 ∆3(a). Then

6 (i) |Γ0(l) ∩ ∆4(a)| = 2.

6 (ii) If z ∈ Γ0(l) ∩ ∆4(a), then l ∈ α0,2(z, TRI, FIX).

6 (iii) For every m ∈ α0,2(x, TRI, FIX), |Γ0(m) ∩ ∆4(a)| = 2 and |Γ0(m) ∩ 1 ∆4(a)| = 1.

0 2 0 Proof. By Lemma 6.6(i) we may choose d ∈ ∆3(a)∩∆1(y) with y+d ∈ α3(y, l). For example, if y + d is the standard heptad, DUAD is the set given in (2.9) × × × × × × × × × × × × and l = we can choose y +d0 = . We have × ×

21 y r #@ r # @rz0 a # @ 0 # @ d# @ @ ¨r r rz @ ¨ @ # ¨ # @¨r @ # @ # @r# e 0 0 1 where Γ0(l) = {y, z , z} and d ∈ ∆2(z). By Lemma 3.11(ii) there exists 0 ⊥ 0 i e ∈ {d , z} ∩ Γ0(X(d , a)), whence e ∈ ∆3(a) ∩ ∆2(a) for i = 2, 3 by Lemma 2 0 4.15. Assume e ∈ ∆3(a) ∪ ∆2(a). Then z or z is collinear with a point in 0 0 ∆2(a) ∩ Γ0(e + d ) by Lemma 3.10. Therefore z or z ∈ ∆3(a) and Theorems 7,8 0 5 and 10 together with Lemmas 6.1, 6.3(ii) and 6.6 imply that z or z ∈ ∆3(a). We 5 0 5 may suppose z ∈ ∆3(a) (the argument is similar if z ∈ ∆3(a)). Appealing to 5 (2.9) and Lemmas 6.1, 6.6, 6.8(i) and 8.10 we have |∆1(y) ∩ ∆3(a)| = 42 + 56 or 1 42+112. This, together with Lemmas 4.11(i) and 6.2(i) gives |∆1(z)∩∆4(a)| = 140 or 220. Now (2.7) and Lemma 5.7 yield a contradiction. 3 0 0 Therefore e ∈ ∆3(a), and Lemma 4.15(ii) implies that Γ0(e + d )\{e, d } ⊆ 3 0 4 6 0 4 ∆3(a). Thus by (2.15)(xii),(xiv) we have z, z ∈ ∆4(a)∪∆4(a). If z, z ∈ ∆4(a), then Lemmas 6.2(i) and 8.7(ii) imply that z + y lies in a Gaz-orbit on Γ1(z) of size 110. From (2.12) we must have z + y ∈ α3(z, END, −). However Lemma

8.10(ii) now yields the contradiction l = y + z ∈ α3,1(y, y + c, DUAD). 6 0 4 6 Thus we may assume that z ∈ ∆4(a) (and z ∈ ∆4(a) ∪ ∆4(a)). Using Lemmas 4.11(i), 6.6, 6.8, 8.10 and 12.2(ii) we have

6 1 1 |∆1(y) ∩ ∆4(a)||∆4(a)| |∆1(z) ∩ ∆4(a)| = 6 |∆4(a)| (56 + 56n).213.3.5.11.23 = 215.7.11.23 6 where n = |Γ0(l) ∩ ∆4(a)| = 1 or 2. So

1 (12.4.1) |∆1(z) ∩ ∆4(a)| = 30(1 + n).

By Lemma 12.3 there exists k ∈ α2,2(z, TRI, FIX) ∪ α1,3(z, TRI, FIX) with 1 |Γ0(k) ∩ ∆4(a)| = 1. Therefore, considering the possibilities in (2.14), we conclude that z + y ∈ α2,2(z, TRI, FIX) ∪ α0,2(z, TRI, FIX) ∪ α1,3(z, TRI, FIX).

Assume TRI and FIX are the subsets of Ωz described in (2.14). Without loss

22 × × × × × of generality we may assume z + e = by Lemma 12.2(i),(iv). × ×

We now show z+y ∈ α2,2(z, TRI, FIX)∪α1,3(z, TRI, FIX) is untenable. Assume that z + y ∈ α2,2(z, TRI, FIX) ∪ α1,3(z, TRI, FIX).

0 ⊥ (12.4.2) {z + f | f ∈ {d , z} \{e}} consists of one line in α2,2(z, TRI, FIX), one line in α1,3(z, TRI, FIX) and two lines in α0,2(z, TRI, FIX).

0 First suppose z + y ∈ α2,2(z, TRI, FIX). If |T (z, d ) ∩ FIX| = 0, 1 or 3, then 0 none of the heptads in Γ1(T (z, d )) lie in α2,2(z, TRI, FIX). So we must have |T (z, d0) ∩ FIX| = 2 and then (12.4.2) follows from (2.14) in this case. (For × × × example if T (z, d0) = , then the four heptads in {z + f | f ∈

× × × × × × × × × × × × × × × {d0, z}⊥\{e}} are , , × × × × × × × ×

× 0 and .) Next assume z + y ∈ α1,3(z, TRI, FIX). So |T (z, d ) ∩ × × × × FIX| = 2 or 3. If T (z, d0) ⊆ FIX, then {d0, z}⊥ contains a point e0 (6= e) 3 ∈ ∆3(a) by Lemma 12.2(iv). However Lemma 12.2(v) now implies that there 0 ⊥ 1 0 exists d1 ∈ {e, e } ∩ ∆3(a). Thus Γ3(a, d1, d , e) 6= ∅ by Lemma 4.15, whence 0 Γ3(a, z) 6= ∅ . With this contradiction we conclude that |T (z, d ) ∩ FIX| = 2 and so (12.4.1) follows from (2.14) again. 0 ⊥ Let {d , z} = {e, y, y1, y2, y3} where z+y lies in one of the orbits α2,2(z, TRI, FIX) and α1,3(z, TRI, FIX), and z + y1 lies in the other orbit. Also z + y2, z + y3 ∈

α0,2(z, TRI, FIX) by (12.4.1). Since y + z ∈ α1,0(y, y + d, DUAD), Lemma 12.3 Lc 00 00 1 implies that y1 + z ∈ α3,0(y1, y1 + d , DUAD) (where {d } = ∆1(y1) ∩ ∆3(a))

23 1 4 0 2 1 and y1 ∈ ∆4(a) ∪ ∆4(a). We have d ∈ ∆3(a) and therefore y1 ∈ ∆4(a) by 0 0 0 0 Theorem 6. In Ωd0 , d + y, d + y1 and d + e intersect O(d , a) in two elements by Lemmas 4.15(ii) and 6.3(ii). So |T (d0, z) ∩ O(d0, a)| = 0, 1 or 2. If 0 0 0 ⊥ 2 T (d , z) ∩ O(d , a) = ∅, then {d , z} ∩ ∆4(a) 6= ∅ by (2.15)(x). This is impos- 2 0 0 sible because ∆1(z) ∩ ∆4(a) = ∅ by Theorem 12. Also |T (d , z) ∩ O(d , a)| 6= 2, 0 ⊥ 3 otherwise |{d , z} ∩ ∆4(a)| = 3 by (2.15)(xi) which contradicts the fact that 3 0 0 y, y1, e∈ / ∆4(a). Hence |T (d , z) ∩ O(d , a)| = 1 in Ωd0 . Thus, we may suppose 0 0 0 0 0 0 0 0 that d +y2 ∈ α2,0(d ,O(d , a),X(d , a)) and d +y3 ∈ α4,0(d ,O(d , a),X(d , a)). 1 3 Therefore y2 ∈ ∆4(a) and y3 ∈ ∆4(a) by (2.15)(xi) and Lemma 6.3(ii).

Since z + y2 and z + y3 both lie in α0,2(z, TRI, FIX) by (12.4.2), if m ∈ 1 3 6 α0,2(z, TRI, FIX), then |Γ0(m) ∩ ∆4(a)| = |Γ0(m) ∩ ∆4(a)| = |Γ0(m) ∩ ∆4(a)| = 3 1. However we have l ∈ α1,0(y, y + d, DUAD) with Γ0(l) ∩ ∆4(a) = ∅ . This, together with Lemmas 6.1, 6.6, 6.8 and 8.10, implies that there are no lines in 3 6 1 Γ1(y) incident with points in both ∆4(a) and ∆4(a). Since ∆4(a) is a Ga-orbit 1 3 6 there are no lines in Γ1 incident with points in each of ∆4(a), ∆4(a) and ∆4(a).

This contradicts the point distribution of Γ0(m).

Hence our assumption that z +y ∈ α2,2(z, TRI, FIX)∪α1,3(z, TRI, FIX) was 0 6 false and we have z+y ∈ α0,2(z, TRI, FIX) as required. To show that z ∈ ∆4(a), 1 6 notice that |∆1(z) ∩ ∆4(a)| = 90 by (2.14), whence n = |Γ0(l) ∩ ∆4(a)| = 2 by (12.4.1). 6 Finally part (iii) follows from (i) and (ii) because ∆4(a) is a Ga-orbit. This completes the proof of the lemma.

Theorem 11 has now been proved.

1 4 Lemma 12.5. (i) |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1 for all l in exactly

one of α2,2(x, TRI, FIX) and α1,3(x, TRI, FIX).

3 4 (ii) |Γ0(l)∩∆4(a)| = |Γ0(l)∩∆4(a)| = 1 for all l in exactly one of α2,2(x, TRI, FIX)

and α1,3(x, TRI, FIX).

1 Proof. Put O = α2,2(x, TRI, FIX) ∪ α1,3(x, TRI, FIX). Let y ∈ ∆1(x) ∩ ∆4(a) with x + y ∈ α0,2(x, TRI, FIX) and y + x ∈ α1,0(y, y + c, DUAD), where {c} = 1 ∆1(y) ∩ ∆3(a) (such a y exists by Lemma 12.4). Then we can choose e ∈

24 3 ∆1(x) ∩ ∆3(a) with x + e ∈ α3(x, x + y) and |T (e, y) ∩ FIX| = 2 in Ωx because every 2-element subset of FIX is contained in some heptad in α0,4(x, TRI, FIX). × × × For example if TRI and FIX are as in (2.14) and x + y = × × × × × × × × × then we can choose x + e to be . × × ⊥ There exists d ∈ {e, y} ∩ Γ0(X(e, a)) by Lemma 3.11(ii). Since y ∈ 1 2 ∆4(a), Theorem 11 implies that d ∈ ∆3(a) because y + x ∈ α1(y, y + c) ⊥ and so c 6= d. In Γ1(x) the set {x + z | z ∈ {x, d} } consists of x + e, two lines in α0,2(x, TRI, FIX) (one of which is x + y), one line in α1,3(x, TRI, FIX) and one line in α2,2(x, TRI, FIX). Let x + z1 ∈ α1,3(x, TRI, FIX), x + z2 ∈ ⊥ α2,2(x, TRI, FIX) and x+z3 ∈ α0,2(x, TRI, FIX) where {x, d} = {z1, z2, z3, e, y}. y r @ r @ ¡@ y0 r zr1 @ b ¨H 0 r ¡ @ ¨b¨ H z1@ aH ¨ b HHr HH ¡ @ ¨ b H@ H¡r @ ¨r b 0 @Hr H`` z2 ¨ x d @H``` z2 ! b r ¨ H `r ! ¨r @ H !! ¨0 !H ¨ z3 @0 !r H¨r e @ z3 @ @ r e

0 0 0 0 0 Below we introduce the points e , y , z1, z2 and z3. ⊥ We now consider {x, d} from the viewpoint of d. In Γd, d+y ∈ α2,0(d, O(d, a),X(d, a)) and d+e ∈ α2,1(d, O(d, a)X(d, a)) by Theorem 6. There are three possible cases.

In Ωd either

⊥ 1 ⊥ (1) |T (d, x) ∩ O(d, a)| = 0 , so we have |{d, x} ∩ ∆4(a)| = 3 and |{d, x} ∩ 2 ∆4(a)| = 1; or

⊥ 1 ⊥ (2) |T (d, x) ∩ O(d, a)| = 1, so we have |{d, x} ∩ ∆4(a)| = 3 and |{d, x} ∩

25 3 ∆4(a)| = 1; or

⊥ 1 ⊥ (3) |T (d, x) ∩ O(d, a)| = 2 , so we have |{d, x} ∩ ∆4(a)| = 1 and |{d, x} ∩ 3 ∆4(a)| = 3.

(using (2.15)(x),(xi) and Lemma 6.3(ii)). Case (1) is impossible by Theorem 12. Using Lemma 12.4(iii) and the fact that x + y, x + z3 ∈ α0,2(x, TRI, FIX) we see that case (3) cannot occur either. 1 Therefore case (2) holds, whence, using Lemma 12.4(iii), Γ0(l) ∩ ∆4(a) 6= ∅ for 3 all l in one of the orbits in O and Γ0(l) ∩ ∆4(a) 6= ∅ for all l in the other orbit in O. 0 0 0 Let Γ0(d+e) = {d, e , e},Γ0(y+d) = {y, y , d} and Γ0(x+zi) = {x, zi, zi} for 0 4 i = 1, 2, 3. To complete the proof we need to show that zi ∈ ∆4(a) for i = 1, 2. 0 3 By Lemma 4.15(ii) e ∈ ∆3(a). Appealing to (2.15)(xii),(xiv) and Lemma 3.10 0 4 6 1 yields that zi ∈ ∆4(a) ∪ ∆4(a) for i = 1, 2. We know zj ∈ ∆4(a) for some 0 Lc 0 j ∈ {1, 2}, whence zj + x ∈ α1,0(zj, zj + d , DUAD) ∪ α3,0(zj, zj + d , DUAD) by 0 1 Theorem 11 (where {d } = ∆1(zj)∩∆3(a)). However Lemma 12.4(ii) now yields Lc 0 zj + x ∈ α3,0(zj, zj + d , DUAD) because x + zj ∈/ α0,2(x, TRI, FIX). Therefore 0 4 1 zj ∈ ∆4(a) by Lemma 8.10(i). Using (2) and Lemma 12.4(iii), z3 ∈ ∆4(a) and 0 6 z3 ∈ ∆4(a). So by Lemma 3.10 we have

0 0 ⊥ i (12.5.1) {e , y } ∩ ∆4(a) 6= ∅ for i = 4 and 6.

0 0 0 0 0 0 We now consider T (e , y )∩(e +c ) as a subset of Ωe0 (where {c } = ∆1(e )∩ 2 0 0 0 0 0 0 ⊥ 6 ∆2(a)). If |T (e , y ) ∩ (e + c )| = 2 or 3 then {e , y } ⊆ ∆4(a) ∪ ∆3(a) by Theorem 7 which contradicts (12.5.1). Also if T (e0, y0) ∩ (e0 + c0) = ∅, then 0 0 ⊥ 4 0 0 {e , y } \{d} ⊆ ∆4(a) which is impossible by (12.5.1) again. Hence |T (e , y ) ∩ 0 0 0 0 ⊥ 4 (e + c )| = 1 and thus {e , y } \{d} consists of two points in each of ∆4(a) and 6 6 6 ∆4(a). Since |Γ0(x + y) ∩ ∆4(a)| = |Γ0(x + z3) ∩ ∆4(a)| = 2 by Lemma 12.4(iii) 0 0 4 we obtain the required result that z1, z2 ∈ ∆4(a).

4 3 Lemma 12.6. Let y ∈ ∆4(a) and l ∈ α5(y, END, −). Then |Γ0(l) ∩ ∆4(a)| = 6 6 |Γ0(l) ∩ ∆4(a)| = 1. Moreover if f ∈ Γ0(l) ∩ ∆4(a), then l ∈ α2,2(f, TRI, FIX) ∪

α1,3(f, TRI, FIX).

6 Proof. By Lemma 12.5 there exists f ∈ ∆1(y)∩∆4(a) with f+y ∈ α2,2(f, TRI, FIX)∪ 4 α1,3(f, TRI, FIX). Since the point in Γ0(f + y)\{f, y} does not lie in ∆4(a) ∪

26 6 ∆4(a) by Lemma 12.5 again, we may appeal to Lemma 5.5 to show that y + f lies in a Gay-orbit Oy of Γ1(y) of size

6 |∆4(a)| 4 .30 |∆4(a)|

(since |α2,2(f, TRI, FIX)| = |α1,3(f, TRI, FIX)| = 30 by (2.14)). Using Lemmas 8.7(ii) and 12.2(ii) we conclude that

215.7.11.23.30 |O | = = 55. y 216.3.7.23

So y + f ∈ α5(y, END, −) by (2.12). Hence Theorem 11 and Lemmas 8.10(i) 3 6 and 12.5 imply that |Γ0(y + f) ∩ ∆4(a)| = |Γ0(y + f) ∩ ∆4(a)| = 1. The lemma now follows using Lemma 5.5 because Gay is transitive on α5(y, END, −) by (2.12).

We remark that Theorem 14 has now been proved (see Lemmas 8.8, 8.9(i),(ii), 8.13, 9.6(i),(iii) and 12.6).

Lemma 12.7. Let l ∈ α1,3(x, TRI, FIX). Then

1 4 (i) |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

4 (ii) If y ∈ Γ0(l) ∩ ∆4(a) then l ∈ α3(y, END, +).

Proof. By Lemma 12.5 there exists l ∈ α1,3(x, TRI, FIX) ∪ α2,2(x, TRI, FIX) 1 4 with |Γ0(l)∩∆4(a)| = |Γ0(l)∩∆4(a)| = 1. We need to show that l ∈ α1,3(x, TRI, FIX). × × ×

Let TRI and FIX be the subsets of Ωx described in (2.14) and let T = .

Using the MOG in [2], three of the heptads in Γ1(x), incident with T are × × × × × × × × × × × × and which we denote by x + e1, x + e2 × × and x + f respectively. Then x + ei ∈ α0,4(x, TRI, FIX) (i = 1, 2) and x + f ∈ 3 α1,3(x, TRI, FIX). By Lemma 12.2(iv),(v) we may assume that ei ∈ ∆3(a) for

27 ⊥ 1 each i and that {e} = {e1, e2} ∩ ∆3(a). Since T (x, e) = T , we have that ⊥ 1 1 3 1 f ∈ {e, x} . Therefore f ∈ ∆2(a) ∪ ∆3(a) ∪ ∆3(a) ∪ ∆4(a) by Theorem 5. But 1 Lemma 12.5 implies that f ∈ ∆4(a) and part (i) follows by Lemma 12.5(i). For part (ii) we appeal to Lemmas 5.5, 8.7(ii) and 12.2(ii) to show that l lies in a Gay-orbit of Γ1(y) of size 55. Thus (2.12) and Lemma 12.6 imply that l ∈ α3(y, END, +) as required.

Lemma 12.8. Let l ∈ α2,2(x, TRI, FIX). Then

3 4 (i) |Γ0(l) ∩ ∆4(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

4 (ii) If y ∈ Γ0(l) ∩ ∆4(a), then l ∈ α5(y, END, −).

3 (iii) If z ∈ Γ0(l) ∩ ∆4(a), then l ∈ α2,1|1(z, TRI, OCT).

Proof. Parts (i) and (ii) follow immediately from Lemmas 12.5, 12.6 and 12.7. 4 3 We now look at part (iii). Let Γ0(l) = {x, y, z} where y ∈ ∆4(a) and z ∈ ∆4(a).

By Lemmas 5.5, 11.1(ii) and 12.2(ii) z + x lies in a Gaz-orbit of Γ1(z) of size 48.

So z + x ∈ α0,1|1(z, TRI, OCT) ∪ α2,1|1(z, TRI, OCT) by (2.11).

(12.8.1) (i) If z + x ∈ α0,1|1(z, TRI, OCT), then there exists a unique point 1 2 in ∆1(z) ∩ ∆2(x) ∩ ∆3(a).

(ii) If z + x ∈ α2,1|1(z, TRI, OCT), then there exist three points in ∆1(z) ∩ 1 2 ∆2(x) ∩ ∆3(a).

2 This follows because z + d ∈ α3,0(z, TRI, OCT) for all d ∈ ∆1(z) ∩ ∆3(a).

Since x + z ∈ α2,2(x, TRI, FIX), there exist three points in S := ∆1(x) ∩ 1 3 ∆2(z) ∩ ∆3(a) because three lines in α0,4(x, TRI, FIX) lie in α3(x, x + z). For ⊥ 2 each e ∈ S, there exists d ∈ {e, z} with d ∈ ∆3(a) because z∈ / ∆1(f) for 2 i any f ∈ ∆2(a) ∪ ∆3(a) ( i = 1,3,4) by Theorems 3,4 and 7. If, for two distinct 2 ⊥ ⊥ elements e1, e2 ∈ S, there exists d ∈ ∆3(a) with d ∈ {e1, z} ∩ {e2, z} , then

X(e1, a) = X(e2, a) by Lemma 4.16, whence Γ3(x, a) 6= ∅ by Lemma 3.6. This 1 2 is impossible and so we have three distinct points in ∆1(z) ∩ ∆2(x) ∩ ∆3(a).

Now (12.8.1) implies that z + x ∈ α2,1|1(z, TRI, OCT).

28 3 Lemma 12.9. Let z ∈ ∆4(a) and l ∈ α0,1|1(z, TRI, OCT). Then

5 3 (i) |Γ0(l) ∩ ∆3(a)| = 1 and |Γ0(l) ∩ ∆4(a)| = 2.

5 (1) 2 (ii) If y ∈ Γ0(l)∩∆3(a), then l ∈ α3 (y, y+c, −) ( where {c} = ∆1(y)∩∆2(a)).

5 3 Proof. Let y ∈ ∆1(z) ∩ ∆3(a). By Lemma 11.3(i) |∆1(y) ∩ ∆4(a)| = 140. Hence 5 Lemmas 4.11(i) and 11.1(ii) imply that |∆1(z) ∩ ∆3(a)| = 84. Thus Lemma 5 11.2 and (2.11) imply that {z + y | y ∈ ∆1(z) ∩ ∆3(a)} = α1,2|2(z, TRI, OCT) ∪

αi,1|1(z, TRI, OCT) for i = 0 or 2. However z + y∈ / α2,1|1(z, TRI, OCT) for all 5 5 y ∈ ∆1(z) ∩ ∆3(a) by Lemma 12.8(iii) and so {z + y | y ∈ ∆1(z) ∩ ∆3(a)} =

α1,2|2(z, TRI, OCT) ∪ α0,1|1(z, TRI, OCT). The result now follows by Lemma 11.3(iii).

We ﬁnish this section by proving two results which pull together information 5 6 5 about the orbits ∆3(a), ∆3(a) and ∆4(a).

Lemma 12.10. Let l ∈ α3,1(x, TRI, FIX). Then

6 5 (i) |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

6 (ii) If y ∈ Γ0(l) ∩ ∆3(a), then l ∈ α1,0(y, y + c, TRI) (where {c} = ∆1(y) ∩ 1 ∆2(a)).

5 4 (iii) If z ∈ Γ0(l)∩∆4(a), then l ∈ α3(z, z +d, +) (where {d} = ∆1(z)∩∆3(a)).

5 Proof. By Lemmas 5.2 and 9.2(ii) we can ﬁnd z ∈ ∆1(x) ∩ ∆4(a) with z + x ∈ 4 6 α3(z, z+d, +) ({d} = ∆1(z)∩∆3(a)). Let Γ0(z+x) = {x, y, z}. Then y ∈ ∆3(a) 6 by Lemma 9.2(ii) again. Lemma 5.5, together with the orbit sizes |∆3(a)|, 5 6 |∆4(a)| and |∆4(a)| given in Lemmas 4.11(ii), 9.1(i) and 12.2(ii) respectively, imply that y + z (= x + z) lies in a Gay-orbit of Γ1(y) of size 64 and a Gax-orbit of Γ1(x) of size 5. By (2.14) and Lemma 12.2(iv), x + z ∈ α3,1(x, TRI, FIX).

The result now follows from (2.8) and Lemma 5.5 because α3,1(x, TRI, FIX) is a Gax-orbit.

Lemma 12.11. Let l ∈ α0,0(x, TRI, FIX). Then

5 5 (i) |Γ0(l) ∩ ∆3(a)| = |Γ0(l) ∩ ∆4(a)| = 1.

5 (3) 2 (ii) If y ∈ Γ0(l)∩∆3(a), then l ∈ α2 (y, y+c, −) (where {c} = ∆1(y)∩∆2(a)).

29 5 4 (iii) If z ∈ Γ0(l)∩∆4(a), then l ∈ α3(z, z +d, −) (where {d} = ∆1(z)∩∆3(a)).

6 Proof. Let Γ0(l) = {x, y, z}. By Lemma 9.2(iii) and the fact that ∆4(a) is a 5 5 Ga-orbit, we may assume that y ∈ ∆3(a) and z ∈ ∆4(a) with y + x = l ∈ (2) 2 α3 (y, y + c, −) and z + x = l ∈ α3(z, z + d, −) (where {c} = ∆1(y) ∩ ∆2(a) and 4 {d} = ∆1(z) ∩ ∆3(a)). Appealing to Lemmas 5.5, 9.1(i) and 12.2(ii) we have that l = x + z lies in a Gax-orbit on Γ1(x) of size 15. Thus (2.14) and Lemma 9.2(iii) yield the result.

13 And ﬁnally. . .

In this, our last section, we stitch up a few loose ends so as to complete the proofs of Theorems 9, 13, 15 and 16 and thus, as a consequence, prove Theorem 5 1. Firstly we focus our attention on ∆4(a).

5 Lemma 13.1. Let x ∈ ∆4(a).

i (i) ∆1(x) ∩ ∆4(a) = ∅ for i = 1 and 2.

3 5 (ii) If l ∈ α1(x, x + d, −), then Γ0(l) ⊆ ∆4(a) ∪ ∆4(a) (where {d} = ∆1(x) ∩ 4 ∆3(a)).

Proof. Part (i) is a consequence of Theorems 11 and 12. Turning to (ii), let y ∈ Γ0(l)\{x}. First suppose y ∈ ∆3(a). By Theorems 5–8 and 10 and Lemmas 5 9.1 and 9.2(ii) we must have y ∈ ∆3(a). Then (2.13) and Lemmas 9.1(ii), 9.2(ii),(iii) and 9.6(i) imply that

5 |∆1(x) ∩ ∆3(a)| = 147 + 70 or 147 + 140.

Hence 5 5 |∆4(a)| |∆1(y) ∩ ∆4(a)| = 5 .n |∆3(a)| 5 5 with n = 217 or 287 because ∆3(a) and ∆4(a) are Ga-orbits. Appealing to Lemmas 4.11(i) and 9.1(i) yields

248 328 |∆ (y) ∩ ∆5(a)| = or 1 4 3 3

30 i which is clearly impossible. Therefore y∈ / ∆3(a). By part (i), y ∈ ∆4(a) for 4 i = 3, 4, 5 or 6. We have y∈ / ∆4(a), otherwise Theorem 14 and Lemma 9.5 6 imply that l ∈ α1(x, x + d, +). If y ∈ ∆4(a), then

6 |∆1(x) ∩ ∆4(a)| = 140 + 70 or 140 + 140 by (2.13) and Lemmas 9.1(ii), 9.2(ii),(iii) and 9.6(i). Using the orbit sizes given in Lemmas 9.1(i) and 12.2(ii) we obtain

5 |∆1(y) ∩ ∆4(a)| = 30 or 40.

However this contradicts the fact that

5 |∆1(y) ∩ ∆4(a)| = 20 + 18p + 90q for some p, q ∈ {0, 1, 2} by (2.14) and Lemmas 12.2(iv), 12.4(iii), 12.7(i), 12.8(i), 12.10(i) and 12.11(i). This completes the proof of part (ii) and the lemma.

5 Lemma 13.2. Let x ∈ ∆4(a) and l ∈ α1(x, x + d, −) (where {d} = ∆1(x) ∩ 4 ∆3(a)). Then

3 (i) |Γ0(l) ∩ ∆4(a)| = 2;

3 (ii) for y ∈ Γ0(l) ∩ ∆4(a) we have l ∈ α0,3|1(y, TRI, OCT); and

3 3 (iii) for all y ∈ ∆4(a) and l ∈ α0,3|1(y, TRI, OCT) we have |Γ0(l) ∩ ∆4(a)| = 2 5 and |Γ0(l) ∩ ∆4(a)| = 1.

Proof. Let x+d be the standard heptad and X be the element

6 of Ωx. By (2.13) and Lemma 9.2(ii) we may choose e ∈ ∆1(x) ∩ ∆3(a) with × × × x + e = ; so x + e ∈ α3(x, x + d, +). We have e + x ∈ × × × ×

31 1 α1,0(e, e + c, TRI) by (2.8) and Lemma 9.3 (where {c} = ∆1(e) ∩ ∆2(a)). Let × × × ×

1 × y ∈ ∆1(x)∩∆2(e) with x+y = . Then x+y ∈ α1(x, x+d, −). × × 0 0 3 5 Let Γ0(x + y) = {x, y, y }. By Lemma 13.1(ii), y, y ∈ ∆4(a) ∪ ∆4(a). 0 5 Suppose that y, y ∈ ∆4(a) holds. Then Lemma 3.11(ii) implies that there ⊥ 1 exists z ∈ {e, y} with z ∈ ∆2(c). Hence e + z ∈ α3,i(e, e + c, TRI) for some i = 0, 1, 2, 3 (see (2.8)). So z 6= x. Using Theorem 10 we must have y or y0 2 1 2 collinear with a point in ∆2(a)∪∆4(a)∪∆4(a). This contradicts Lemma 13.1(i) 0 0 3 and the fact that y, y ∈ ∆4(a). Therefore at least one of y and y lies in ∆4(a); 3 suppose that y ∈ ∆4(a). Lemmas 9.1(ii), 9.2(ii),(iii) and 9.6(i) together with (2.13) imply that

3 |∆1(x) ∩ ∆4(a)| = 70 or 140.

α0,3|1(y, TRI, OCT) are orbits under the action of Gax and Gay respectively.

We have now proved Theorem 15 (see Lemmas 9.1(ii), 9.2(ii),(iii), 9.6(i) and 13.2(i)).

3 Lemma 13.3. Let x ∈ ∆4(a) and l ∈ α1,2|0(x, TRI, OCT). Then

6 (i) |Γ0(l) ∩ ∆4(a)| = 2;

6 (ii) for y ∈ Γ0(l) ∩ ∆4(a), l ∈ α1,1(y, TRI, FIX); and

6 3 (iii) for all y ∈ ∆4(a) and l ∈ α1,1(y, TRI, FIX) we have |Γ0(l) ∩ ∆4(a)| = 1 6 and |Γ0(l) ∩ ∆4(a)| = 2.

32 0 2 Proof. Let Γ0(l) = {x, y, y } and d ∈ ∆1(x) ∩ ∆3(a). With TRI and OCT as in (2.11) we may suppose that

× × × × × l = × ×

× × × and, by Lemma 11.2(ii), that x + d = . × × × × By Theorems 5,6,7,8 and 10 and Lemmas 11.2 and 11.3(i) we have y, y0 ∈ ⊥ ∆4(a). Lemma 3.11(ii) implies that there exists e ∈ {d, y} with e ∈ Γ0(X(d, a)) 2 3 2 and thus e ∈ ∆2(a) ∪ ∆3(a) ∪ ∆3(a) by Theorem 6. If e ∈ ∆2(a) ∪ ∆3(a), then 0 by Theorem 6 again and Lemma 3.10, y or y is collinear with a point in ∆2(a). 0 3 0 Since y, y ∈ ∆4(a) this is impossible, whence e ∈ ∆3(a). If Γ0(d+e) = {d, e , e}, 0 3 then e ∈ ∆3(a) by Lemma 4.15(ii). Appealing to Theorem 7 and Lemma 3.10 0 4 6 4 yields y, y ∈ ∆4(a) ∪ ∆4(a). If, say, y ∈ ∆4(a), then Theorem 14 implies that y + x ∈ α5(y, END, −). But then Lemma 12.8(ii),(iii) forces x + y = l ∈ α2,1|1(x, TRI, OCT), whereas l ∈ α1,2|0(x, TRI, OCT) by assumption. Thus 4 0 4 0 6 y∈ / ∆4(a) and likewise y ∈/ ∆4(a). Hence y, y ∈ ∆4(a), so giving part (i). For part (ii), by (2.11) and Lemmas 11.2, 11.3(ii), 12.8(iii) and 13.2 together with part (i), 6 |∆1(x) ∩ ∆4(a)| = 192. Therefore 3 12 3 |∆4(a)|.192 2 .5.7.11.23.192 |∆1(y) ∩ ∆4(a)| = 6 = 15 = 120 |∆4(a)| 2 .7.11.23 by Lemmas 11.1(i) and 12.2(ii). Using Lemmas 12.2(iv), 12.4(iii), 12.7(i), 12.8(i), 12.10(i) and 12.11(i) together with (2.14) we must have

3 {y + z | z ∈ ∆4(a)} = α2,2(y, TRI, FIX) ∪ α1,1(y, TRI, FIX).

Since l∈ / α2,2(y, TRI, FIX) by Lemma 12.8(iii) we conclude that l ∈ α1,1(y, TRI, FIX). 6 Part (iii) follows from (i) and (ii) because ∆4(a) is a Ga-orbit.

33 We have now veriﬁed Theorem 13 ( see Lemmas 11.2, 12.8(i),(iii), 12.9(i), 13.2(iii) and 13.3(i) ). We turn our attention to the ﬁnal pair of line orbits still to be considered, 5 6 emanating from points in ∆3(a) and ∆4(a).

5 2 Lemma 13.4. Let x ∈ ∆3(a) and l ∈ α1(x, x+c, −) where {c} = ∆1(x)∩∆2(a). 5 6 Then Γ0(l) ⊆ ∆3(a) ∪ ∆4(a). Proof. This follows from the point distributions given in Theorems 2–8 and 10–15 together with Lemmas 6.8(ii), 9.2(iii), 9.6(ii) and 12.9(ii).

5 2 Lemma 13.5. Let x ∈ ∆3(a) and l ∈ α1(x, x+c, −) where {c} = ∆1(x)∩∆2(a). Then

6 (i) |Γ0(l) ∩ ∆4(a)| = 2; and

6 (ii) for y ∈ Γ0(l) ∩ ∆4(a), l ∈ α2,0(y, TRI, FIX).

6 1 Proof. We ﬁrst show that Γ0(l)∩∆4(a) 6= ∅. Let f ∈ ∆1(x)∩∆4(a) with x+f ∈

α3(x, x + c, +) (such a point f exists by Lemma 6.7). Then Lemmas 5.5 and 6.8 1 imply that f + x ∈ α1,1(f, f + d, DUAD) where {d} = ∆1(f) ∩ ∆3(a). Let X be the element of Ωf in DUAD ∩ (f + x). Then X ∈ Γ3(f + x)\Γ3(f + d) by (2.9). 2 0 1 0 Let {b} = ∆1(a) ∩ ∆2(x) and {b } = ∆1(a) ∩ ∆3(f). Since DUAD = D(f, b ), 0 we have X ∈ Γ3(b ). We now show that X 6= X(x, b). Assume X = X(x, b). 0 Then b = b otherwise Lemma 3.6 implies that a ∈ Γ0(X), contrary to the fact 1 2 that x ∈ ∆4(a). Therefore f ∈ ∆3(b) and x ∈ ∆1(f)∩ ∆2(b). This is impossible by Theorem 5 and so we conclude that X 6= X(x, b).

Thus, in Ωx we may choose a heptad k ∈ Γ1(x) with |k ∩ (x + f)| = 3,

|k ∩ (x + c)| = 1, X ∈ Γ3(k) and X(x, b) ∈/ Γ3(k). For example if x + f = × × × × ◦ × × × × × × , x+c = and X(x, b) = × × × × × × × × × × × × we may choose k to be or , depending on × × × × × ×

34 which element of (x+f)\{X(x, b)} is X. By (2.7), k ∈ α1(x, x+c, −). Let T be 0 the unique triad incident with x + f and k. Then T ∈ Γ2(X). Let k ∈ Γ1(f, T ) 0 0 be such that , in Ωf , DUAD lies in the heptad k (k exists by Lemma 3.11(ii) 0 because X lies in (DUAD∩T ) in Ωf ). So k ∈ αi,2(f, f +d, DUAD) for i = 1 or 3. 0 0 0 2 6 Appealing to Lemma 6.6(i),(iv), there exists z ∈ Γ0(k ) with z ∈ ∆3(a)∪∆3(a). 0 Furthermore z ∈ ∆1(z) for some z ∈ Γ0(k) by Lemma 3.10. Using Lemma 13.4 6 together with Lemma 4.15 yields z ∈ ∆4(a). Since α1(x, x + c, −) is a Gax-orbit 6 of Γ1(x), we have proved that Γ0(l) ∩ ∆4(a) 6= ∅. 6 Let y ∈ Γ0(l) ∩ ∆4(a). Then Lemmas 12.2(iv), 12.4(iii), 12.7(i), 12.8(i),

12.10(i), 12.11(i) and 13.3(iii) imply that l ∈ α2,0(y, TRI, FIX). We have 6 5 |∆1(x)∩∆4(a)| = 40+96n for n = 1 or 2 and |∆1(y)∩∆3(a)| = 15+18m for m = 5 1 or 2 by (2.7), (2.14), Lemmas 5.7, 12.11(i) and the fact that Γ0(l) ∩ ∆3(a) 6= ∅ 6 5 and Γ0(l) ∩ ∆4(a) 6= ∅. Taking this together with the orbit sizes of ∆3(a) and 6 ∆4(a) given in Lemmas 4.11(i) and 12.2(ii) yields n = 1 and m = 2. Therefore 5 6 |Γ0(l) ∩ ∆3(a)| = 2 and |Γ0(l) ∩ ∆4(a)| = 1 as required. This completes the proof of the lemma.

Thus we have now proved Theorems 9 and 16 (see Theorem 4.13(vii) and Lemmas 5.7, 6.7, 9.5 and 13.5(i) for Theorem 9; and Lemmas 12.2(iv), 12.4(iii), 12.7(i), 12.8(i), 12.10(i), 12.11(i), 13.3(iii) and 13.5(i),(ii) for Theorem 16).

Appendix A

× ◦ × × We list the 16 heptads meeing in :- × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

h1 h2 h3 h4

35 × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

h5 h6 h7 h8

× × × × × × × × × × × × × × × × × × × × × × × × × × × ×

h9 h10 h11 h12

× × × × × × × × × × × × × × × × × × × × × × × × × × × ×

h13 h14 h15 h16

Appendix B

Here we list the results in which we determine the point distribution of the

Gax-orbits on lines in Γ1(x) for x ∈ ∆3(a) ∪ ∆4(a).

1 (1) x ∈ ∆3(a)

α2(x, D(x, a)) Theorem 4.13(iii)

α1(x, D(x, a)) Lemma 4.15(i)

α0(x, D(x, a)) (2.15)(ix) and Lemma 3.2

2 (2) x ∈ ∆3(a)

α0,1(x, O(x, a),X(x, a)) Theorem 4.13(iv)

α0,0(x, O(x, a),X(x, a)) (2.15)(x) and Lemma 3.2

α4,1(x, O(x, a),X(x, a)) Theorem 4.13(iv)

α2,1(x, O(x, a),X(x, a)) Lemma 4.15(ii)

α4,0(x, O(x, a),X(x, a)) (2.15)(xi) and Lemma 3.2

α2,0(x, O(x, a),X(x, a)) Lemma 6.6(i),(ii)

36 3 (3) x ∈ ∆3(a) {x + b} Theorem 4.13(v)

α1,1(x, x + b, X(x, a)) Lemma 4.13(i)

α3,1(x, x + b, X(x, a)) Lemma 4.13(ii)

α3,0(x, x + b, X(x, a)) (2.15)(xiv) and Lemma 3.2

α1,0(x, x + b, X(x, a)) (2.15)(xii) and Lemma 3.2

4 (4) x ∈ ∆3(a)

α1(x, X(x, a)) Theorem 4.13(vi)

α0(x, X(x, a)) (2.15)(xiii) and Lemma 3.2

5 (5) x ∈ ∆3(a) {x + b} Theorem 4.13(vii)

α1(x, x + b, +) Lemma 9.5 (1) α3 (x, x + b, −) Lemma 5.7(i) (2) α3 (x, x + b, −) Lemma 5.7(ii)

α3(x, x + b, +) Lemma 6.7

α1(x, x + b, −) Lemma 13.5(i)

6 (6) x ∈ ∆3(a) {x + b} Theorem 4.13(viii)

α3,3(x, x + b, TRI) Lemma 4.10 and Theorem 4.13(viii)

α3,0(x, x + b, TRI) Lemma 10.2(ii),(iii)

α1,1(x, x + b, TRI) Lemma 10.5

α3,2(x, x + b, TRI) Lemma 10.6

α1,0(x, x + b, TRI) Lemma 10.3

α3,1(x, x + b, TRI) Lemma 10.4(i)

1 (7) x ∈ ∆4(a)

37 {x + b} Lemma 6.1

α3,2(x, x + b, DUAD) Lemma 6.6(i)

α1,2(x, x + b, DUAD) Lemma 6.6(iv) L α3,0(x, x + b, DUAD) Lemma 6.6(iii)

α1,1(x, x + b, DUAD) Lemma 6.8(i)

α1,0(x, x + b, DUAD) Lemma 12.4(i) Lc α3,0(x, x + b, DUAD) Lemma 8.10(i)

α3,1(x, x + b, DUAD) Lemma 8.10(ii)

2 (8) x ∈ ∆4(a)

α0(x, O(x, a)) Corollary 7.4

α4(x, O(x, a)) Lemma 7.5

α2(x, O(x, a)) Lemma 8.12

3 (9) x ∈ ∆4(a)

α3,4|0(x, TRI, OCT) Lemma 11.2(i)

α3,0(x, TRI, OCT) Lemma 11.2(ii)

α1,0(x, TRI, OCT) Lemma 11.2(iii)

α0,3|1(x, TRI, OCT) Lemma 13.2(iii)

α1,2|2(x, TRI, OCT) Lemma 11.2(iv)

α0,1|1(x, TRI, OCT) Lemma 12.9(i)

α2,1|1(x, TRI, OCT) Lemma 12.8(i),(iii)

α1,2|0(x, TRI, OCT) Lemma 13.3(i)

4 (10) x ∈ ∆4(a)

α1(x, END, +) Lemma 8.13

α1(x, END, −) Lemma 8.8

α5(x, END, +) Lemma 9.6(i),(iii)

α3(x, END, +) Lemma 8.9(ii)

α5(x, END, −) Lemma 12.6

α3(x, END, −) Lemma 8.9(i)

5 (11) x ∈ ∆4(a)

38 {x + b} Lemma 9.1(ii)

α3(x, x + b, +) Lemma 9.2(ii)

α1(x, x + b, +) Lemma 9.6(i)

α1(x, x + b, −) Lemma 13.2(i)

α3(x, x + b, −) Lemma 9.2(iii)

6 (12) x ∈ ∆4(a)

α0,4(x, TRI, FIX) Lemma 12.2(iv)

α3,1(x, TRI, FIX) Lemma 12.10(i)

α0,0(x, TRI, FIX) Lemma 12.11(i)

α2,0(x, TRI, FIX) Lemma 13.5(i),(ii)

α1,3(x, TRI, FIX) Lemma 12.7(i)

α2,2(x, TRI, FIX) Lemma 12.8(i)

α0,2(x, TRI, FIX) Lemma 12.4(iii)

α1,1(x, TRI, FIX) Lemma 13.3(iii)

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