2-ARC TRANSITIVE POLYGONAL GRAPHS OF LARGE GIRTH AND VALENCY

DISSERTATION

Presented in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy in the

Graduate School of The Ohio State University

By

Eric Allen Swartz, B.A.

*****

The Ohio State University

2009

Dissertation Committee: Approved by

Professor Akos´ Seress, Advisor Professor Boris Pittel Advisor Professor Neil Robertson Graduate Program In Mathematics c Copyright by

Eric Allen Swartz

2009 ABSTRACT

A near-polygonal graph is a graph Γ which has a set C of m-cycles for some positive integer m such that each 2-path of Γ is contained in exactly one cycle in C.

If m is the girth of Γ then the graph is called polygonal. Up until now, the only examples of 2-arc transitive polygonal graphs with arbitrarily large valency had girth no larger than seven, and the 2-arc transitive polygonal graph with largest girth had valency five and girth twenty-three (in fact, even with no restrictions on the automorphism group, there were no examples of polygonal graphs with odd girth greater than twenty-three). This thesis provides a construction of an infinite family of polygonal graphs of arbitrary girth m with 2-arc transitive automorphism groups, showing that there are 2-arc transitive polygonal graphs of arbitrarily large valency for each girth m. Furthermore, this thesis also provides a construction that, given a polygonal graph of valency r and girth m, produces a polygonal graph of valency r and girth 3m, and that the graphs constructed via this method will be 2-arc transitive if the original graph was 2-arc transitive. Finally, this thesis provides a construction of a new infinite family of near-polygonal graphs of valency 10 and a method for determining which graphs can have a given girth, which yields a few new examples of polygonal graphs.

ii To my parents

iii ACKNOWLEDGMENTS

I would first and foremost like to thank my advisor, Dr. Akos´ Seress. His direction, patience, and encouragement, as well as countless hours of his time, are what made this dissertation possible.

I would like to thank Dr. Boris Pittel and Dr. Neil Robertson for giving up valuable hours of their time to serve on my dissertation committee.

I am indebted to my all friends and family for the constant support throughout the years. Their positivity has made everything much easier, and they have always been able to bring me comfortably back to the real world no matter how far my work drags me away.

I would especially like to thank Jenn, whose support has allowed me to remain sane through this process. She was there to help at every bump in the road, offer- ing encouragement, and her computer skills made writing this dissertation infinitely easier.

Finally, I must thank my parents. They have always been there for me and supported me in every possible way. I have been given every opportunity in life to succeed, and I owe everything to them.

iv VITA

June 3, 1982 ...... Born - Parkersburg, WV, USA

2004 ...... B.A. Mathematics, Harvard University. 2005-present ...... Graduate Teaching Associate, The Ohio State University.

FIELDS OF STUDY

Major Field: Mathematics

v TABLE OF CONTENTS

Page

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita...... v

List of Tables ...... viii

Chapters:

1. INTRODUCTION ...... 1

1.1 Background ...... 1 1.2 Preliminary Results ...... 2 1.2.1 Trivalent Polygonal Graphs ...... 3 1.2.2 Polygonal Graphs with Valency at Least 4 ...... 3 1.3 Outline of Thesis ...... 8

2. A FAMILY OF NEAR-POLYGONAL GRAPHS OF VALENCY 10 . . . 12

2.1 Preliminary Results ...... 12 2.2 A New Family of Polygonal Graphs ...... 15 2.3 Determining the length of the special cycles ...... 18 2.4 Calculations ...... 23

3. A CONSTRUCTION OF AN INFINITE FAMILY OF 2-ARC TRANSI- TIVE POLYGONAL GRAPHS OF ARBITRARY ODD GIRTH . . . . . 25

3.1 Preliminary results ...... 25 3.2 A family of polygonal graphs with cycles of a fixed odd length . . . 27

vi 4. A CONSTRUCTION OF AN INFINITE FAMILY OF 2-ARC TRANSI- TIVE POLYGONAL GRAPHS OF ARBITRARY EVEN GIRTH . . . . 40

5. AN ANALYSIS OF THE GIRTH-DOUBLING CONSTRUCTION FOR POLYGONAL GRAPHS ...... 49

5.1 The construction ...... 49 5.2 Another construction ...... 52 5.3 Relating the two constructions ...... 58

Appendices:

A. Girth-finding Program ...... 62

B. Another Girth-finding Program ...... 68

C. Large Numbers ...... 75

Bibliography ...... 78

vii LIST OF TABLES

Table Page

1.1 2-arc transitive polygonal graphs constructed using Theorem 1.2.10 . 7

1.2 New 2-arc transitive polygonal graphs of valency 10 ...... 10

2.1 Results obtained from Theorem 2.3.5 ...... 23

2.2 New 2-arc transitive polygonal graphs of valency 10 ...... 24

viii CHAPTER 1

INTRODUCTION

1.1 Background

Let Γ be a graph. For a positive integer l, an l-walk of Γ is a sequence of vertices

(α0, α1, ..., αl) such that αi is adjacent to αi+1 for 0 ≤ i ≤ l − 1. If in addition

αi−1 6= αi+1 for 1 ≤ i ≤ l − 1, then an l-walk is called an l-arc; while if further all the

αi are distinct then the l-arc is called an l-dipath (directed path). The identification of an l-dipath (α0, α1, ..., αl) and its reverse (αl, ..., α1, α0) is called an l-path, and

denoted by [α0, α1, ..., αl]. An m-cycle is an (m-1)-path [α1, ..., αm] such that αm is

adjacent to α1 [6].

A near-polygonal graph is a graph Γ with a set of distinguished cycles C, all of length m for some m ∈ N, such that every 2-arc of Γ is contained in a unique member of C. Furthermore, if m is also the girth, or length of the shortest cycle, of Γ, then Γ is called a polygonal graph. If Γ is a polygonal graph such that the set C of distinguished m-cycles is in fact the set of all m-cycles of Γ, then Γ is called a strict polygonal graph.

Finally, if a graph Γ has an automorphism that maps a given 2-arc to any other 2-arc, then Γ is said to be a 2-arc transitive graph. If the element of the automorphism group mapping one 2-arc to another is always unique, the automorphism group is said to

1 be sharply 2-arc transitive. A graph that is both polygonal and 2-arc transitive is a

2-arc transitive polygonal graph.

Polygonal graphs are a natural generalization of the edge- and vertex-set of poly- gons and Platonic solids, and one immediately notes that these are themselves strict polygonal graphs, with the special set of cycles being the polygon itself or the faces of the solid, respectively. Kn, the complete graph on n points, is a strict polygonal

graph of girth 3, and the Petersen graph is a polygonal graph of girth 5 that is not a

strict polygonal graph [14].

Manley Perkel invented the notion of a polygonal graph in [8] and a near-polygonal

graph in [12]. Even though polygonal graphs have been around for a few decades and

occur both in topological and algebraic graph theory and in geometries related to

finite groups, the theory of the subject is still in its infancy. Polygonal graphs are far

from being classified; indeed, the amount of symmetry they must possess in unclear,

the exact structure of their automorphism groups is unknown, and, in fact, up until

now, only a scarce supply of these graphs could even be constructed. [14]

1.2 Preliminary Results

In this section, we detail some of the known results involving polygonal graphs.

First, a graph is said to be regular if all its vertices have the same valency, or number

of adjacent edges. We have the following lemma:

Lemma 1.2.1. [8, Lemma 2.1] A near-polygonal graph is regular.

Hence it makes sense to discuss the valency of a near-polygonal graph, since it is

the same at every vertex.

2 1.2.1 Trivalent Polygonal Graphs

As is noted in [14], many examples of trivalent polygonal graphs come from topo-

logical constructions. The reason for this is made clear by the following observation:

Observation 1.2.2. [14, p. 180] If a trivalent graph Γ of girth m can be embedded

on a (not necessarily orientable) surface such that each face is an m-gon, then the

set of faces form the distinguished cycle set C of a polygonal graph. Conversely, if Γ

is a trivalent polygonal graph with a distinguished cycle set C = {C1,C2, ..., Ck}, then

attaching a 2-cell to each Ci we get a 2-dimensional topological space which can be

embedded on some surface.

Proceeding in this fashion has yielded the following results:

Theorem 1.2.3. [10, p. 46] The only trivalent polygonal graphs of girth 3, 4, and 5

are, respectively, the edge- and vertex-sets of the tetrahedron, cube, and dodecahedron.

Theorem 1.2.4. [7] All trivalent strict polygonal graphs of girth 6 are duals of tri-

angulations of the torus or the Klein bottle (of which there are infinitely many).

Theorem 1.2.5. [11, Theorem 5] There exists a trivalent polygonal graph of girth 7

with 14n vertices for all n ∈ N.

Theorem 1.2.6. [1] For all n divisible by 4, there exist trivalent polygonal graphs of girth n.

1.2.2 Polygonal Graphs with Valency at Least 4

Obviously, considering the complete graphs Kn for n ≥ 3 shows that there are infinitely many polygonal graphs of girth 3, and each n ≥ 5 yields a polygonal graph

3 of valency at least 4. We obtain the following lemma by considering the points and

edges of the regular cube in k-dimensional space, k ≥ 2:

Lemma 1.2.7. [9, p. 1311] For all k ≥ 2 there is a polygonal graph with valency k and girth 4.

There are three main techniques for constructing polygonal graphs of valency greater than 3. The first is the “girth-doubling” construction outlined in [2]:

Theorem 1.2.8. [2, Theorem 1.1] If there exists a polygonal graph Γ of valency r and girth m, then there exists a polygonal graph Γ2 of valency r and girth 2m. Moreover,

if Γ is a strict polygonal graph, then so is Γ2.

By starting with the complete graph Kn, which has valency n − 1, and repeatedly

applying this theorem, we obtain:

Corollary 1.2.9. [2, Corollary 1.2] For all r ≥ 2, there exist polygonal graphs of

valency r and arbitrarily large girth.

This is a remarkable theorem that produces a large number of graphs. However,

we are forced to start with a specific polygonal graph, and, since there are very few

known graphs with odd girth, we are limited in the graphs we construct. It is also

not immediately clear what the automorphism of the newly constructed graph will

be; for example, if we start with a 2-arc transitive polygonal graph Γ of girth m,

does the graph Γ2 of girth 2m that we construct necessarily have a 2-arc transitive

automorphism group?

The other two primary techniques for constructing polygonal graphs of valency

greater than 4 will be discussed briefly here, but, because they play a large role in

4 the results of this thesis, will be discussed in much greater depth in the following chapters. Both techniques are algebraic in nature, and rely on the following group- theoretic construction of the coset graph. For a group G and a subgroup H < G, denote by [G : H] the set of right cosets of H in G. For an element g ∈ G\H with g2 ∈ H, the coset graph Γ := Cos(G, H, HgH) is defined as the graph with vertex set

[G : H] such that two vertices Hx and Hy are adjacent if and only if yx−1 ∈ HgH; alternatively we can view the edge set as {(Ht, Hgt): t ∈ G}. Observe that from the condition g2 ∈ H it follows that HgH = Hg−1H and Hx and Hy are adjacent if and only if Hy and Hx are adjacent, implying that Γ is undirected. Denote by α

g the vertex H of Γ and by β the vertex α = Hg, and let Gα1α2...αn be the stabilizer

g g2 g in G of the vertices α1, α2, ..., αn. Note that β = α = α, Gα = H,Gβ = H , and

g Gαβ = H ∩ H . The neighbor set NΓ(α) of α consists of the cosets in HgH, and the valency of α is the index |Gα : Gαβ| ([6]).

One technique relies on the following theorem:

Theorem 1.2.10. [12, Theorem 2.3]Let G be a group, and H a subgroup of G. Let g ∈ G\H with g2 ∈ H, and suppose that G = hH, gi. Let h ∈ H\Hg,K = H ∩ Hg, and L = H ∩ Hg ∩ Hgh. Suppose that

(i) H = K ∪ KhK,

(ii) if L ≤ Kk for some k ∈ H, then k ∈ K ∪ Kh, and

(iii) if t ∈ G with Lt ≤ K, then Lt = Ls for some s ∈ H.

Then either G = H ∪HgH (in which case the graph we have constructed is a complete graph) or there is a near-polygonal graph Γ with G ≤ Aut(Γ).

5 Observation 1.2.11. [14, p.183-184] The near-polygonal graphs constructed by The-

orem 1.2.10 are 2-arc transitive.

Carefully chosen families of groups will thus allow us to construct an infinite family

of 2-arc transitive near-polygonal graphs. The obvious limitation here is that in no

way does this theorem guarantee that the graph we have constructed is polygonal,

i.e. we have no idea what the girth of our constructed graph is without specifically

checking. Some of the polygonal graphs obtained using this construction have been

detailed in [12] and [13], and the results can be summarized in Table 1.1 from [14].

As is noted in [14], up until now, the graphs with girth 23 listed in Table 1.1

actually had the largest girth of any known 2-arc transitive polygonal graph.

The final technique, known as “spinning the cycle,” is based upon the following:

Theorem 1.2.12. [6, Corollary 1.2] Let Γ be a graph and G ≤ Aut(Γ) such that G

acts sharply 2-transitively on the 2-arcs of Γ. Assume that for an arc (α, β) of Γ there

exists an involution g ∈ G such that (α, β)g = (β, α). Then Γ is near-polygonal.

This method of construction yields the following:

Theorem 1.2.13. [6, Theorems 1.3, 1.4, 1.5, 1.6] There exist 2-arc transitive polyg- onal graphs of arbitrarily large valency of girth 5, 6, and 7, respectively.

As is noted in [14], up until now, the only examples of 2-arc transitive polygonal graphs with arbitrarily large valency had girth no larger than 7, and those graphs are the ones constructed above.

6 G valency girth strict polygonal? PSL(2, 32) 5 3 strict PSL(2, 31) 5 5 strict PSL(2, 41) 5 7 strict PSL(2, 132) 5 7 strict PSL(2, 431) 5 9 strict PSL(2, 199) 5 11 strict PSL(2, 432) 5 11 strict PSL(2, 1951) 5 13 strict PSL(2, 36209) 5 17 strict PSL(2, 522919) 5 19 strict PSL(2, 459649) 5 21 strict PSL(2, 17472) 5 23 strict PSL(2, 5336599) 5 23 strict PGL(2, 11) 5 4 strict PGL(2, 19) 5 6 non-strict PGL(2, 379) 5 14 non-strict PSL(2, 19) 6 5 non-strict PSL(2, 139) 6 7 strict PSL(2, 359) 6 9 strict PSL(2, 1321) 6 11 strict PSL(2, 16901) 6 13 strict PSL(2, 12239) 6 17 strict PSL(2, 74761) 6 21 non-strict PGL(2, 32) 6 4 non-strict PGL(2, 72) 6 8 non-strict PGL(2, 132) 6 12 non-strict

Table 1.1: 2-arc transitive polygonal graphs constructed using Theorem 1.2.10

7 1.3 Outline of Thesis

The rest of the thesis is organized as follows.

Chapter 2 provides a construction for an infinite family of near-polygonal graphs

of valency 10 via the method presented in Theorem 2.1.3. Section 2.2 proves the

following theorem:

Theorem 1.3.1. Let q be a prime power such that q ≡ 1, 19 (mod 30). Then there ex- ist H ≤ PSL(3, q) and g ∈ PSL(3, q) such that, if G := hH, gi, Γ := Cos(G, H, HgH) is a near-polygonal graph.

In Section 2.3 we determine which prime powers q yield near-polygonal graphs with cycles of length m for a given m. We first prove the following lemma:

Lemma 1.3.2. If Γ is a near-polygonal graph constructed as in Theorem 1.3.1, then the length of the special cycles in Γ must be even.

0 0 The next two lemmas construct sequences of numbers an, bn that are crucial to determining which prime powers will yield near-polygonal graphs with cycles of a given length. First, given that

√ √ √ 308052 + 239699 −3 + 82209 5 − 321908 −15 c2 := √ , (1.1) 719097 − 82209 −15 we have:

2 Lemma 1.3.3. If x is a root of x − (1 + c2)x − 1 = 0, then for any positive integer

n n we have x = anx + bn, where

(i) a2 = (1 + c2),

8 (ii) b2 = 1,

(iii) an+2 = (1 + c2)(an(1 + c2) + bn) + an,

(iv) bn+2 = an(1 + c2) + bn.

0 0 4n 0 0 Lemma 1.3.4. If an, bn are coefficients such that x = anx + bn, then

0 4 2 3 2 2 2 3 (i) an = an(1 + c2)((1 + c2) + 2) + 4anbn((1 + c2) + 1) + 6anbn(1 + c2) + 4anbn,

0 4 2 3 2 2 4 (ii) bn = an((1 + c2) + 1) + 4anbn(1 + c2) + 6anbn + bn.

Using these sequences, we prove the following:

Theorem 1.3.5. If q = p, p2 for some prime p 6= 2, 3, 5, the near-polygonal graph Γ constructed from PSL(3, q) as in Theorem 1.3.1 has special cycles of length m, and

0 0 am, bm are defined as above, then the equation

0 2 0 0 02 (1 − bm) − am(1 + c2)(1 − bm) − am = 0 (1.2) must hold in GF(q).

This provides us with a number that must be 0 in GF(q). The complex norm of the numerator of this number is an integer that must be 0 in GF(q), and hence the prime divisors of this integer (or their squares) are the only possibilities for q.

In Section 2.4, using the above method, we are able to provide some new examples of 2-arc transitive polygonal graphs, as summarized by Table 1.2.

In Chapter 3, we construct new 2-arc transitive polygonal graphs via the method presented in Theorem 1.2.12, culminating in the following theorem proved in Section

3.2:

9 Group Cycle Length Polygonal? Strict Polygonal? PSL(3, 49) 6 Yes Yes PSL(3, 121) 6 Yes Yes PSL(3, 79) 8 Yes Yes PSL(3, 19) 10 No No PSL(3, 592) 10 Yes Yes PSL(3, 2392) 12 Yes Yes PSL(3, 292) 14 No No PSL(3, 132) 14 Yes No PSL(3, 1009) 14 Yes Yes

Table 1.2: New 2-arc transitive polygonal graphs of valency 10

Theorem 1.3.6. There are 2-arc transitive polygonal graphs of arbitrarily large va- lency and girth m for all odd m ≥ 3.

In Chapter 4, we construct even more 2-arc transitive polygonal graphs via the

method presented in Theorem 1.2.12, this time culminating in the proof of the fol-

lowing:

Theorem 1.3.7. There are 2-arc transitive polygonal graphs of arbitrarily large va-

lency and girth m for all even m ≥ 4.

Corollary 1.3.8. There are infinitely many 2-arc transitive polygonal graphs of girth

m for all m ≥ 3, and there 2-arc transitive polygonal graphs of arbitrarily large valency

for each m.

Chapter 5 shows that the same construction used to prove Theorem 1.2.8 can be

extended slightly, leading to the proof of the following in Section 5.3:

10 Theorem 1.3.9. If there exists a polygonal graph Γ of valency r and girth m, then there exists a polygonal graph Γ3 of valency r and girth 3m. Moreover, if Γ is a strict polygonal graph, then so is Γ3.

Furthermore, we are able to show that many properties of the automorphism group are preserved by this construction, namely:

Theorem 1.3.10. If Γ is a vertex transitive, edge transitive, arc transitive, or 2-arc transitive polygonal graph, then so is Γk, the graph constructed via Theorem 1.2.8, for k = 2,3.

11 CHAPTER 2

A FAMILY OF NEAR-POLYGONAL GRAPHS OF VALENCY 10

2.1 Preliminary Results

Following Perkel ([12]), we will let Ω be the vertex set of a graph Γ. For a vertex

δ ∈ Ω, let W (r)(δ) denote the set of vertices reachable by an r-long walk from δ.

1 We will also denote W (δ) by NΓ(δ) and refer to it as the neighbor set or simply as the neighbors of δ. We denote by Aut(Γ) the set of automorphisms of the graph Γ.

Finally, for G ≤ Aut(Γ) and α, β, γ, ... elements of Ω,Gαβγ... denotes the subgroup of

G fixing each vertex α, β, γ, ..., and for a subgroup H ≤ Aut(Γ), Ω(H) is the set of vertices of Γ fixed by every element of H.

Our construction uses the following results of Perkel ([12]).

Lemma 2.1.1. [12, Lemma 2.1] Let Γ be a connected graph on a set Ω, and let

G ≤ Aut(Γ), with G transitive on Ω. let α, β, γ ∈ Ω with β, γ ∈ NΓ(α), and β 6= γ.

Suppose that

(i) Gα is 2-transitive on NΓ(α),

(ii) Gαβγ fixes no vertices of NΓ(α)\{β, γ}, and

12 (iii) Gαβγ is conjugate in Gα to every G-conjugate of Gαβγ which lies in Gαβ, i.e.

g g h for g ∈ G, (Gαβγ) < Gαβ ⇒ (Gαβγ) = (Gαβγ) for some h ∈ Gα. (Note in

particular that this condition holds if Gα contains just one conjugacy class of

subgroups isomorphic to Gαβγ.)

Then connected components of the subgraph of Γ induced by Ω(Gαβγ) are either iso-

lated vertices of cycles of the graph.

Theorem 2.1.2. [12, Theorem 2.2] Assume the hypothesis of Lemma 2.1.1 and as-

sume Γ is not the complete graph on Ω. Then we can find a set C of m-cycles of Γ such that Γ is a near-polygonal graph, and G ≤ Aut(Γ).

Using the coset graph as a basis, Perkel proves the following theorem which gives some explicit conditions when a group will satisfy Lemma 2.1.1.

Theorem 2.1.3. [12, Theorem 2.3] Let G be a group, and H a subgroup of G. Let g ∈ G\H with g2 ∈ H, and suppose that G = hH, gi. Let h ∈ H\Hg,K = H ∩ Hg, and L = H ∩ Hg ∩ Hgh. Suppose that

(i) H = K ∪ KhK,

(ii) if L ≤ Kk for some k ∈ H, then k ∈ K ∪ Kh, and

(iii) if t ∈ G with Lt ≤ K, then Lt = Ls for some s ∈ H.

Then either G = H ∪ HgH (in which case Γ is the complete graph on Ω) or there is a near-polygonal graph Γ with G ≤ Aut(Γ).

Using Theorem 2.1.3 as a basis, we can now prove the following Corollary:

13 Corollary 2.1.4. Let G be a group and H ≤ G isomorphic to PSL(2, r). Let K be

r−1 a maximal subgroup of H isomorphic to F r(r−1) , i.e. the Frobenius group r : 2 , and 2 r−1 let L be a cyclic subgroup of K of order 2 . Suppose that:

(i) There exists g ∈ G such that g∈ / H, g2 ∈ H,K = H ∩ Hg, and G = hH, gi.

(ii) There exists h ∈ H\Hg such that L = H ∩ Hg ∩ Hgh.

Then the coset graph Γ := Cos(G, H, HgH) is a near-polygonal graph.

Proof. We will show that our group G satisfies the conditions of Theorem 2.1.3. We

−1 h −1 first determine the size of KhK. Suppose k1hk2 = k3hk4. Then (k1 k3) = k2k4 ∈ K,

−1 h h r−1 i.e. (k1 k3) ∈ K ∩K = L. Given k1, there are only 2 possible choices for k3 (since

r−1 |L| = 2 ). If we are also given k2, then, once k3 is chosen, k4 is uniquely determined.

r−1 Thus for any given k1, k2, there are precisely 2 pairs k3, k4 such that k1hk2 = k3hk4. Hence |K| · |K| r(r − 1) r(r − 1) 2 r2(r − 1) |KhK| = r−1 = · · = . (2.1) 2 2 2 r − 1 2 Now, since h∈ / K,K ∩ KhK = ∅,

r(r − 1) r2(r − 1) r(r2 − 1) |K ∪ KhK| = + = = |H|, (2.2) 2 2 2

and so H = KhK as desired.

Next, by [4] we know that H contains precisely (r + 1) conjugates of K. We also

know that H contains (r + 1) distinct commutative groups of order r, one in each

subgroup conjugate to K. Furthermore, each group conjugate to K is the normalizer

in H of its respective commutative r-group. Each group conjugate to K contains r

r−1 conjugate subgroups of order 2 . Being that there are (r + 1) conjugates of K in H,

r−1 this counts r(r + 1) cyclic subgroups of order 2 . Since there actually are a total of 14 r(r+1) r−1 r−1 2 conjugate cyclic groups of order 2 in H, this means each cyclic 2 -subgroup normalizes exactly two commutative r-subgroups, i.e. L ≤ K, L ≤ Kh and that’s

all. Thus if L ≤ Kk for some k ∈ H, then either Kk = K or Kk = Kh, and since

NH (K) = K, we have k ∈ K ∪ Kh, as desired.

t t r−1 Finally, if L ≤ K for some t ∈ G, then L is another cyclic subgroup of order 2 in

r−1 K. However, again by [4], all the cyclic subgroups of order 2 in H are conjugate, so there is an s ∈ H such that Lt = Ls. We can now apply Theorem 2.1.3 to see that

Γ = Cos(G, H, HgH) is either near-polygonal or a complete graph.

Suppose now that Γ is a complete graph. Since Hg and Hgh are distinct vertices of Γ that are both adjacent to H, there must exist t ∈ G such that (Ht, Hgt) = (Hg, Hgh).

So, comparing Hgt and Hgh, there must be an element k ∈ K such that t = kh, but then Hg = Ht = H, a contradiction. Hence Γ is near-polygonal.

2.2 A New Family of Polygonal Graphs

In order to apply Corollary 2.1.4, we note that for q ≡ 1, 19 (mod 30), the group ∼ PSL(3, q) contains a subgroup isomorphic to PSL(2, 9) = A6 ([3]). Note also that for

all primes p 6= 2, 3, 5, we have p2 ≡ 1, 19 (mod 30). For calculation purposes, we view

PSL(3, q) as the image of SL(3, q) modulo scalar matrices.

Consider SL(3, q) where q is a prime power and q ≡ 1, 19 (mod 30). Then SL(3, q)

0 ∼ has a subgroup H = 3.A6 (see for example [3]). From [3], we see furthermore that

H0 = hT 0,M 0,A0i, where −1 0 0 0 T =  0 −1 0 , (2.3) 0 0 1

15  1 1  − 2 2 − t −t 0 1 1 M = t − 2 t − 2  , and (2.4) 1 1 t − 2 2 − t −1 0 0 0 A =  0 0 ω , (2.5) 0 −ω2 0 √ √ 3 2 −1+ −3 1+ 5 where ω = 1 6= ω and t satisfies 4t − 2t − 1 = 0 (e.g. ω = 2 , t = 4 ).  1 1  − 2 2 − t t 0 0 0 0 1 1 Let B = A M T =  ωt − 2 ω ω(t − 2 ) . 2 1 2 1 2 −ω ( 2 − t) ω t 2 ω 0 0 02 012 0 0 5 Then note that 3.A6 = hA ,B i with relations A = B = (A B ) = 1. If

φ : SL(3, q) → PSL(3, q) is the natural map with φ(A0) = A and φ(B0) = B, we have

∼ 2 4 5 H = A6 = φ(3.A6) = hA, B|A = B = (AB) = 1i.

0 0 0 0 Furthermore, we have K = 3.F36 ≤ 3.A6 with 3.F36 = hA ,C i, where C =

0 0 0 0 0 0 0 ∼ B A B A B B . Naturally, if φ(C ) = C, then K = F36 = hA, Ci.

0 0 0 0 0 0 0 0 0 0 0 0 0 Define S1 := C A C C C A and S2 := C C A . Note that S1 and S2 generate the

0 0 normal subgroup of order 27 in 3.F36 (and hence φ(S1) = S1 and φ(S2) = S2 generate the normal subgroup of order 9 in K = F36).

2 Lemma 2.2.1. There is an order 4 element g in PSL(3, q) such that g∈ / A6, g ∈ A6,

g and A6 ∩ A6 = F36.

Proof. We will explicitly construct g. First note that

 1 1  − 2 t − 2 −t 0B0−1A0B0−2 0C0−1A0 1 1 A = A = t − 2 −t − 2  (2.6) 1 1 −t − 2 t − 2

is an element of 3.F36 and 3.A6. This element has order 2. We introduce the following

change of basis matrix:

1 cω2 c  0 1 2 P := 1 c ω  , (2.7) ω 1 ω c 16 1+2ωt 0 where c = − 2(1+2ω2t) (and naturally P := φ(P )). Note that

1 + cω −c2ω2 −c2ω2  1 P 0−1 = · −c c(1 + cω) −c2ω . (2.8) (1 + 2cω)(1 − cω)   −cω2 −c2 cω2(1 + cω)

We now, unbelievably, have the following nice matrices:

1 0 0  0P 0−1 A = 0 0 −1 , (2.9) 0 −1 0

ω2 0 0 0P 0−1 S1 =  0 1 0 , (2.10) 0 0 ω

−1 0 0  0C0−1A0P 0−1 A =  0 0 −ω2 , and (2.11) 0 −ω 0

 0 0 d 0P 0−1 ω2 S2 =  d 0 0 , (2.12) 0 ω 0

c 2 where d = 2(1+2cω)(1−cω) (c (ω − 4ωt − 2t) + c(−4t − 2ω + 4ωt) + ω). Now look at 2  √1 −1 −ω  −3 3a 3a x0 =  a √ω √ω  , (2.13)  −3 −3  √1 √ω ωa −3 −3

2 0−1 0 0−1 √ω 02 0C A P 0 where a = d −3 . Note that x = A , so x has order 4 and is an element

P 0−1 0P 0−1x0 0P 0−1 0P 0−1x0 0P 0−1 2 0 of (3.F36) . Also, A = A and S1 = S2 · ω I, i.e. φ(x ) = x normalizes K = F36. Since x∈ / K, A6 is simple, and F36 is a maximal subgroup of

P A6, x∈ / H. Thus our desired g is simply x .

Now let h := φ(B0−1T 0B0). Note that h ∈ H\Hg and h has order 2. We can now prove the main theorem.

17 ∼ Proof of Theorem 1.3.1. We let H be a copy of A6 = PSL(2, 9) in PSL(3, q) and g the element provided by Lemma 2.2.1 above. Furthermore, we note that L :=

H ∩ Hg ∩ Hgh is a cyclic subgroup of order 4 (generated by the element ` := CCAC).

Therefore, by Corollary 2.1.4, Γ := Cos(G, H, HgH) is a near-polygonal graph.

2.3 Determining the length of the special cycles

We continue with the notation from the previous section. Explicit calculation

shows that gh`(`gh)−1 = I, i.e.

gh ∈ CG(L). (2.14)

Since, by construction, L is the stabilizer of the 2-arc (Hgh, H, Hg), L must also

stabilize the special cycle containing the 2-arc (Hgh, H, Hg). Also by construction,

we note that L fixes precisely the special cycle containing the 2-arc (Hgh, H, Hg) (see the proof of [12, Lemma 2.1]). On the other hand, (2.14) implies that L fixes H(gh)j for any integer j > 0. Thus the circuit (H, Hgh, Hghgh, Hghghgh, ...) must be the special cycle containing the 2-arc (Hgh, H, Hg), and finding the length of the special cycles is equivalent to finding the smallest m ∈ Z such that (gh)m ∈ H. Note further that if (gh)m ∈ H, then (gh)m stabilizes our circuit and hence (gh)m ∈ L.

In order to determine when a group PSL(3, q) will yield a near-polygonal graph

with cycles of length m for a given m, we will start working in the group SL(3, C) with matrices defined as in the above section (note that this reintroduces the scalar

matrices). The following computations were done with the help of Mathematica

[16]. Now we have

18   g1 g2 g3 g = g4 g5 g6 , (2.15) g7 g8 g9

√ √  −1− 5 1 −1+ 5  4 2√ 4√ h =  1 −1+ 5 1+ 5  , and (2.16)  2√ 4√ 4  −1+ 5 1+ 5 1 4 4 − 2 √ √ √ √  1+i 3 1 3+i 3+ 5−i 15  √ √4 √ √ √2 √ 8 ` =  3+i 3− 5+i 15 3−i 3+ 5+i 15 0  . (2.17)  8 8√ √ √ √  1 1−i 3 −3+i 3+ 5+i 15 2 4 8

The numbers g1, ..., g9 are listed in Appendix C. Note that g and h still have orders

4 and 2, respectively, but ` now has order 12. Further calculation shows that

  gh1 gh2 gh3 gh = gh4 gh5 gh6 . (2.18) 0 gh8 gh9

The numbers gh1, ..., gh9 are listed in Appendix C.

Some calculation shows that gh has characteristic polynomial

3 2 χgh(x) = x − c2x + c1x − 1, (2.19) where √ √ √ 308052 + 239699i 3 + 82209 5 − 321908i 15 c2 = Tr(gh) = √ , (2.20) 719097 − 82209i 15

√ √ √ −1 c11 + c12i 3 + c13 5 + c14i 15 c1 = Tr((gh) ) = √ √ √ , (2.21) c15 + c16i 3 + c17 5 + c18i 15 and the constant coefficient is -1 since det(gh) = 1. The numbers c11, ..., c18 are listed in Appendix C.

Calculation also shows that χgh(−1) = −1−c2 −c1 −1 = 0, meaning (x+1)|χgh(x).

We conclude that the remaining roots of the characteristic polynomial (and hence the remaining eigenvalues) must be roots of

19 χ (x) gh = x2 − (1 + c )x − 1. (2.22) x + 1 2

Lemma 2.3.1. If Γ is a near-polygonal graph constructed as in Theorem 1.3.1, then

the length of the special cycles in Γ must be even.

√ √ −1+i 3 −1 1−i 3 Proof. First, since Det(`) = 1, Tr(`) = 2 , and Tr(` ) = 2 , we see that √ √ −1 + i 3 1 − i 3 χ (x) = x3 − x2 + x − 1, (2.23) ` 2 2

√ √ √ −1+i 3 i+ 3 −i− 3 and the roots of χ`(x) (and hence the eigenvalues of `) are 2 , 2 , and 2 . Now suppose that Γ has special cycles of length m, m odd. Thus (gh)m ∈ L =

h`i. Since ` and gh commute (see (2.14)), as matrices ` and gh are simultaneously

diagonalizable. Mathematica [16] gives us the following eigenvectors of `:

v1 = (v11, v12, 1), (2.24)

v2 = (v21, v22, 1), (2.25)

v3 = (v31, v32, 1). (2.26)

The numbers v11, ..., v32 are listed in Appendix C. We now define a change of basis

matrix Q by Q = [v1 v2 v3] (vectors written in column form), and from this we see

that

−1 0 0  −1 Q ghQ =  0 r1 0  , (2.27) 0 0 r2

where r1 and r2 are the other two roots of χgh(x) (their exact values are unimportant

for this proof), and so the change of basis matrix Q puts the eigenvalue -1 in the first

20 row, first column. Applying the change of basis to `, we see that  √ √ √  i−√3+3i√5+√15 0 0 i+ 3−3i 5+ 15 √ √ √ −1  i(i− 3+3i 5+ 15)  Q `Q =  0 √ √ √ 0  . (2.28)  i+ 3−3i 5+ 15 √ √ √  0 0 1+i√ 3+3√5−i√ 15 i+ 3−3i 5+ 15 √ √ √ √ Hence we have i−√3+3i√5+√15 = −1+i 3 = ω in the first row, first column. i+ 3−3i 5+ 15 2 Now m odd implies that the first row, first column of Q−1(gh)mQ is -1. On the other hand, for any integer j, the first row, first column of Q−1`jQ must be one of

1, ω, ω2, none of which are -1, a contradiction. Hence the cycles of Γ must have even length.

Lemma 2.3.2. If we consider g, h, ` as elements of SL(C) and (gh)m = s ∈ h`i, then s4 = I.

Proof. As in the above proof, we note that the first row, first column of a diagonalized

(gh)m must be 1. If the first row, first column entry of `j is 1, then j = 3, 6, 9, 12.

Thus s = `3, `6, `9, `12 and, in any case, s4 = I.

2 Lemma 2.3.3. If x is a root of x − (1 + c2)x − 1 = 0, then for any positive integer

n n we have x = anx + bn, where

(i) a2 = (1 + c2),

(ii) b2 = 1,

(iii) an+2 = (1 + c2)(an(1 + c2) + bn) + an,

(iv) bn+2 = an(1 + c2) + bn.

2 Proof. The first two parts are clear from x = (1 + c2)x + 1. For the other two parts, we proceed by induction. We have our base case, so assume the conclusion is true for k = n. Then

21 xn+2 = x2 · xn (2.29)

= ((1 + c2)x + 1)(anx + bn) (2.30)

2 = (1 + c2)anx + bnx + anx + bn (2.31)

= (1 + c2)an((1 + c2)x + 1) + bnx + anx + bn (2.32)

= ((1 + c2)(an(1 + c2) + bn) + an)x + (an(1 + c2) + bn), (2.33)

as desired.

0 0 4n 0 0 Lemma 2.3.4. If an, bn are coefficients such that x = anx + bn, then

0 4 2 3 2 2 2 3 (i) an = an(1 + c2)((1 + c2) + 2) + 4anbn((1 + c2) + 1) + 6anbn(1 + c2) + 4anbn,

0 4 2 3 2 2 4 (ii) bn = an((1 + c2) + 1) + 4anbn(1 + c2) + 6anbn + bn.

4n n 4 4 Proof. We note that x = (x ) = (anx + bn) . Multiplying this out and repeatedly

2 substituting (1 + c2)x + 1 for x as in the proof of Lemma 2.3.3 yields the desired

result.

Theorem 2.3.5. If q = p, p2 for some prime p 6= 2, 3, 5, the near-polygonal graph Γ

constructed from PSL(3, q) as in Theorem 1.3.1 has special cycles of length m, and

0 0 am, bm are defined as above, then the equation

0 2 0 0 02 (1 − bm) − am(1 + c2)(1 − bm) − am = 0 (2.34)

must hold in GF(q).

Proof. Since Γ has cycles of length m, we know that (gh)m ∈ L. By Lemma 2.3.1, we

know that m is even, and by Lemma 2.3.2 we know that (gh)4m = I when we consider

22 m z · z n 4 21022 237 · 5 6 21566 251 · 5 · 72 · 112 8 22110 271 · 5 · 792 10 22654 283 · 53 · 192 · 592 12 23198 2101 · 5 · 72 · 112 · 2392 14 23742 2115 · 5 · 132 · 292 · 10092

Table 2.1: Results obtained from Theorem 2.3.5

2 gh as a matrix in SL(3, C). This means that the roots of x − (1 + c2)x − 1 = 0, the

4m 0 0 eigenvalues of gh that are not -1, must satisfy x = amx + bm = 1. Solving for x, we get 0 1 − bm x = 0 . (2.35) am

2 Substituting (2.35) into x − (1 + c2)x − 1 = 0 and clearing the denominator, we get

(2.34), as desired.

2.4 Calculations

Table 2.1 contains the results obtained from the above recursion using Mathe- matica ([16]), determining the primes p for which PSL(3, p) or PSL(3, p2) admits a near-polygonal graph of a given girth m. In each case, the numerator of the number obtained from (2.34) was of the form n · z, where n was an integer and z · z was a

power of 2.

Note that we were able to calculate and factor the numerator through m = 38,

and that we probably could factor many numbers beyond that. However, the limits

of modern computing makes it impossible to determine the girth of these graphs, and,

since we are searching for polygonal graphs, the results were omitted.

23 Group Cycle Length Polygonal? Strict Polygonal? PSL(3, 49) 6 Yes Yes PSL(3, 121) 6 Yes Yes PSL(3, 79) 8 Yes Yes PSL(3, 19) 10 No No PSL(3, 592) 10 Yes Yes PSL(3, 2392) 12 Yes Yes PSL(3, 292) 14 No No PSL(3, 132) 14 Yes No PSL(3, 1009) 14 Yes Yes

Table 2.2: New 2-arc transitive polygonal graphs of valency 10

In Table 2.2, we include the results obtained from analyzing some of these near- polygonal graphs using GAP ([15], see Appendices A and B).

Hence our construction has explicitly produced seven new examples of polygonal graphs, and, once we have access to computers with better memory, possibly more.

24 CHAPTER 3

A CONSTRUCTION OF AN INFINITE FAMILY OF 2-ARC TRANSITIVE POLYGONAL GRAPHS OF ARBITRARY ODD GIRTH

3.1 Preliminary results

As in [6], an (l,m)-path-cycle cover of a graph Γ is a set C of m-cycles such that

each l-path of Γ is covered by at least one cycle in C; sometimes an (l,m)-path-cycle

cover is simply called an (l,m)-cover.. If in addition every l-path of Γ lies in a constant

number λ of cycles of C, then C is called a regular λ-(l,m)-cover, or simply called a

λ-(l,m)-cover. Hence near-polygonal graphs are the graphs that have a 1-(2,m)-cover

for some m.

For a graph Γ and a group G ≤ Aut(Γ), an (l, m)-cover C is called G-symmetrical

if C = {C1,C2, ..., Cn} is such that

(i) the restriction G|Ci of G to each Ci contains all rotations of Ci;

(ii) G induces a transitive action on C.

25 ∼ There are two possibilities for G|Ci to contain all rotations of Ci, namely Ci = Zm ∼ or Ci = D2m. The corresponding symmetrical covers will be called G-rotary or G- dihedral, respectively. For a positive integer l, a graph Γ is called (G,l)-arc transi- tive, (G,l)-dipath transitive, or (G,l)-path transitive if G acts transitively on l-arcs, l-dipaths, or l-paths of Γ, respectively. In the case of dipath and path transitivity, we also require that l-dipaths or l-paths exist in Γ, respectively.

We now present the basis for the construction of near-polygonal graphs in [6].

Lemma 3.1.1. [6, Lemma 1.1] Let Γ be a regular graph of valency at least three, let

G ≤ Aut(Γ), and let l ≥ 1 be an integer. Then (a) ⇒ (b) ⇒ (c) ⇒ (d) holds for the following four statements (a)-(d).

(a) Γ has a G-dihedral (l, m)-cover for some m ≥ 3.

(b) Γ is (G, l)-dipath transitive.

(c) Γ has a G-rotary (l, m)-cover for some m ≥ 3.

(d) Γ is (G, l)-path transitive.

Moreover, if Γ has a G-dihedral (l, m)-cover C and G acts sharply transitively on the l-dipaths in Γ then C is a 1-(l, m)-cover.

We will use Lemma 3.1.1 in conjunction with the following method for construc- tion.

Construction 3.1.2. [6, Construction 2.1] Let (α0, ..., αl) and (α1, ..., αl, αl+1) be l-

dipaths in a graph Γ (allowing that α0 = αl+1), let G ≤ Aut(Γ), and suppose that there

g exists g ∈ G such that αi = αi+1 holds for 0 ≤ i ≤ l. Let C be the cycle generated by

hgi G the vertices α0 , and let C = C .

26 We shall refer to the method described in Construction 3.1.2 as spinning an l- dipath. In order to apply Lemma 3.1.1, we need to ensure that a target group G occurs as a group of automorphisms of some graph Γ. That goal can be achieved by defining Γ as a coset graph. [6]

For constructing near-polygonal graphs as in [6], we want to spin a 2-dipath

(γ, α, β). The spinning element can be described easily in terms of the coset graph.

Lemma 3.1.3. [6, Lemma 2.3] For a coset graph Γ = Cos(G, H, HgH) with g2 ∈ H,

g let α = H and β = α . Then an element f ∈ G maps α to β if and only if f ∈ Gαg.

Furthermore, for f ∈ G such that αf = β, we have that β 6= αf −1 if and only if f ∈ Gαg\Gαβg.

Finally, we will use the following well-known lemma as stated in [6].

Lemma 3.1.4. [6, Lemma 2.4] Let Γ be a graph, and let G ≤ Aut(Γ) be transitive on the vertex set of Γ. Then Γ is (G, 2)-dipath transitive if and only if Gα acts 2- transitively on NΓ(α); furthermore, Γ is sharply (G, 2)-dipath transitive if and only if

Gα acts sharply 2-transitively on NΓ(α).

3.2 A family of polygonal graphs with cycles of a fixed odd length

This is a generalization of the construction in [6, Section 3]. Let m = 2k+1, k ∈ N. We define G to be the direct product of k copies of PGL(2, q), where q is a

Qk prime power, i.e. G := i=1 PGL(2, q). The elements of PGL(2, q) can be iden- tified with equivalence classes of 2 × 2 invertible matrices over the field GF(q),

with two matrices equivalent if and only if they are scalar multiples of each other.

With a slight abuse of notation, we shall write that the matrices themselves are

27 elements of PGL(2, q), and that the elements of G are k-tuples of matrices. We

identify H ∼= AGL(1, q) with the set of (equivalence classes of) lower triangular ma- a 0  trices : a, b, c ∈ GF(q), ac 6= 0 and define H := Diag(Qk AGL(1, q)) = b c i=1 {(h, h, ..., h): h ∈ H} ≤ G. 1 0 For a ∈ GF(q), let p(a) := ∈ H and p(a) := (p(a), p(a), ..., p(a)) ∈ H. a 1 a 0 Moreover, if a 6= 0 then let d(a) := ∈ H and d(a) := (d(a), d(a), ..., d(a)) ∈ H. 0 1 Let P = {p(a): a ∈ GF(q)}, P = {p(a): a ∈ GF(q)},D = {d(a): a ∈ GF(q)∗}, and

∗ D = {d(a): a ∈ GF(q) }. Then H = PD, H = P D,P C H, and P C H.  0 y For y ∈ GF(q)∗, let g(y) = . Then, for −1 0

g = g(y1, y2, ..., yk) := (g(y1), g(y2), ..., g(yk)) ∈ G (3.1)

we have g2 = 1 ∈ H, so we can define the coset graph

Γ = Γ(y1, y2, ..., yk) := Cos(G, H, HgH). (3.2)

Let α denote the vertex H and let β denote the vertex Hg. First we determine the

number of vertices, the valency, and bound the number of components of Γ. The

number of vertices is

|G : H| = |G|/|H| = qk−1(q − 1)k−1(q + 1)k. (3.3)

g −1 g For any d ∈ D we have d = d , so Gαβ = H ∩ H ≥ D. Since D is a maximal

subgroup of H, we must have equality here, and so the valency of Γ is

|Gα : Gαβ| = |H : D| = q. (3.4)

For a vertex δ of Γ, let W (r)(δ) denote the set of vertices reachable by an r-long

walk from δ. Then we have the following, which has the same proof as [6, Lemma

3.1]:

28 Lemma 3.2.1. W (r)(α) = HgP gP ...gP (r iterations of gP ).

Now we will bound the number of components of Γ.

∗ Lemma 3.2.2. Let y1, y2, ..., yk be distinct elements of GF(q) . Then Γ has at most

2k−1 components.

Proof. The component containing α consists of the cosets reachable by some walk

S r ∗ in Γ, that is, the cosets in r≥0 H(gH) = hH, gi. If we define G := hH, gi, then

∗ ∗ the number of components of Γ will be |G : G |. Since the yi are all distinct, G

contains a nondiagonal subgroup G∗∗ that projects bijectively on each coordinate of

Qk i=1 PSL(2, q) with a different map for each coordinate, and this can only happen

∗∗ Qk when G = i=1 PSL(2, q). If q is even then PGL(2, q) = PSL(2, q) and so G = G∗ = G∗∗, implying that Γ is connected. If q is odd then |PGL(2, q) : PSL(2, q)| = 2

and G ≥ G∗ > G∗∗ (since each coordinate of G∗ surjects onto PGL(2, q), this last

containment is strict). |G : G∗∗| = 2k, so this implies that the number of components

is

|G : G∗| ≤ 2k−1, (3.5)

and its exact value depends on how many pairwise yiyj are squares in GF(q).

∗ Lemma 3.2.3. For all y1, y2, ..., yk ∈ GF(q) , the graph Γ is near-polygonal.

Proof. The group H = Gα acts sharply 2-transitively on the cosets of D = Gαβ so,

by Lemma 3.1.4, the graph Γ = Γ(y1, ..., yk) is sharply (G, 2)-dipath transitive. We will show that Γ has a G-dihedral (2, n)-cover for some n, which, by Lemma 3.1.1, will show that Γ is near-polygonal. To this end, we define

    0 y1 0 yk f = f(y1, y2, ..., yk) := p(1) · (g(y1), ..., g(yk)) = , ..., . (3.6) −1 y1 −1 yk

29 We have f ∈ Hg\Dg and so, by Lemma 3.1.3, the vertex

    −1 y −y y −y γ := αf = H 1 1 , ..., k k (3.7) 1 0 1 0

is different from β and (γ, α, β) is a 2-dipath. Spinning the dipath as in Construction

3.1.2, we obtain a cycle C = [δ0 = α, δ1 = β, δ2, ..., δn−1 = γ] for some n, and the

(2, n)-cover C = CG. We claim the group element

−1 0 −1 0 z := , ..., (3.8) −1 1 −1 1

z is an involution satisfying δi = δn−i for 1 ≤ i ≤ n − 1. Direct computation shows that

2 z −1 zf f −1z z z z f −1 z = 1 and f = f . So β = β = α = α, implying δ1 = β = α = γ = δn−1.

z fz zf −1 f −1 Then, by induction on i = 1, 2, ..., n − 1, it follows that δi+1 = δi = δi = δn−i =

δn−i−1. Hence hf, zi acts as the dihedral group on C and so C is a G-dihedral (2, m)- cover. Consequently, Lemma 3.1.1 implies that Γ is a near-polygonal graph.

Next, we describe how to choose the values y1, ..., yk such that Γ = Γ(y1, ..., yk)

has a 1-(2,m) cover. We define a sequence of polynomials un(y) by u0(y) := 0,

n X 2n − j u (y) := (−1)j yn−j, (3.9) 2n+1 j j=0

n X 2n + 1 − j u (y) := (−1)j yn−j for n ≥ 0. (3.10) 2n+2 j j=0 We now have the following, which as far as we know was first stated in [17]:

Lemma 3.2.4. Let j ≥ 1 be an integer.

−1 th (a) The roots of uj(y) in C are the numbers ξ + ξ + 2, where ξ is a j root of unity different from ±1.

30 (b) For all j ≥ 1, we have u2j(y) = u2j−1(y) − u2j−2(y) and u2j+1(y) = yu2j(y) −

u2j−1(y).

(c) For all j ≥ 0, we have

2j −1 X −j+i u2j+1(x + x + 2) = x , (3.11) i=0

j −1 X −j+2i u2j+2(x + x + 2) = x . (3.12) i=0

 0 y a b  (d) Let f(y) := . Then for all j ≥ 1, we have f(y)j = j j , where −1 y cj dj d j e d j+1 e b j c d j e aj = −y 2 uj−1(y), bj = y 2 uj(y), cj = −y 2 uj(y), and dj = y 2 uj+1(y).

We can use Lemma 3.2.4 to prove the following:

th Lemma 3.2.5. Let q ≡ ±1 (mod m), and let ζ = ζm be a primitive m root of unity

2 i m−i ∗ in GF(q ). Then, if yi := ζ + ζ + 2 ∈ GF(q) for 1 ≤ i ≤ k, all the yi are distinct

and the graph Γ = Γ(y1, ..., yk) has a 1-(2,m)-cover.

2 Proof. First we prove that yi ∈ GF(q). Indeed, computing in GF(q ) we have

q i m−i q qi q(m−i) i m−i yi = (ζ + ζ + 2) = ζ + ζ + 2 = ζ + ζ + 2 = yi (3.13)

since q ≡ ±1 (mod m), which implies that yi ∈ GF(q). Note further that yi = 0 if

i i ∗ and only if ζ = −1, and since m is odd, ζ 6= −1 and so for all 1 ≤ i ≤ k, yi ∈ GF(q) .

We also have yi 6= yj for i 6= j because

(ζi − ζj)(ζi+j − 1) y − y = 6= 0 (3.14) i j ζi+j

by our assumptions on i and j. By Lemma 3.2.4(c), um(yi) = 0 and then Lemma

3.2.4(b) implies um+1(yi) = −um−1(yi) for 1 ≤ i ≤ k. Hence, by Lemma 3.2.4(d), for

31 m the spinning element f = f(y1, ..., yk) as defined in Lemma 3.2.3 we have f = 1.

th Moreover, since ζ is a primitive m root of unity, we have un(y1) 6= 0 for 1 ≤ n < m

because ζ is not an nth root of unity. Therefore f n ∈/ H. This means that the cycle

spinned by f has length m and so Γ has a 1-(2,m)-cover.

We now proceed to show that the girth of the graph Γ we have just constructed

is at least m. We define l (l) Y r (y) := g(y)p(ai), (3.15) i=1 ∗ ∗ (l) where y ∈ GF(q) and for all i, ai ∈ GF(q) . Note that we view r (y) as a function

∗ of y given arbitrary fixed units a1, ..., al ∈ GF(q) .

" (l) (l) # (l) r11 (y) r12 (y) Lemma 3.2.6. If r (y) = (l) (l) , then for all l ≥ 2, r21 (y) r22 (y)

(l) (l−1) (a) r12 (y) = yr11 (y).

(l) (b) deg(r11 (y)) = l.

(l) l l d 2 e (c) r11 (y) = y sl(y), where sl(y) is a degree b 2 c polynomial in y with leading coef-

ficient a1a2...al. ( yalsl−1(y) − sl−2(y), l even; (d) sl(y) = alsl−1(y) − sl−2(y), l odd.

b l c−1 (e) For all l, the coefficient of y 2 in sl(y), denoted by Tl, is the opposite of the

l−2 sum of l − 1 terms, each of which is the product of l − 2 different ai, d 2 e of which are odd i.

l l−1 (f) The constant term of sl(y), denoted by Kl, is (−1) 2 for l even and (−1) 2 (a1 +

a3 + a5 + ··· al) for l odd.

32 (g) The linear term of sl(y) for l even has coefficient Sl, where Sl is the product of

l ( l )( l +1) 2 +1 2 2 (−1) and the sum of 2 terms of the form aiaj, where i is odd and j is even.

(l) (l−1) (h) r22 = yr21 (y).

(l) (i) deg(r21 (y)) = l − 1.

(l) l−1 l−1 d 2 e (j) r21 (y) = y tl(y), where tl(y) is a degree b 2 c polynomial in y with leading

coefficient −a2a3...al. ( yaltl−1(y) − tl−2(y), l odd; (k) tl(y) = altl−1(y) − tl−2(y), l even.

0 l+1 l (l) The constant term of tl(y), denoted by Kl, is (−1) 2 for l odd and (−1) 2 (a2 +

a4 + a6 + ··· al) for l even.

ya y Proof. Note first that g(y)p(a) = . Thus −1 0

ya y r(1)(y) = 1 (3.16) −1 0

and y(ya a − 1) y(ya ) r(2)(y) = 1 2 1 . (3.17) −ya2 −y

This will prove our inductive step in (b) and (c). Furthermore, we note that

ya y r(l)(y) = r(l−1)(y) · l −1 0 " # r(l−1) r(l−1) ya y = 11 12 · l (l−1) (l−1) −1 0 r21 r22 " (l−1) (l−1) (l−1) # yalr11 (y) − r12 (y) yr11 (y) = (l−1) (l−1) (l−1) . (3.18) yalr21 (y) − r22 (y) yr21 (y)

33 This proves (a), and this also gives us

(l) (l−1) (l−1) r11 (y) = yalr11 (y) − r12 (y)

(l−1) (l−2) = yalr11 (y) − yr11 (y), (3.19) immediately using the result from (a).

To prove (b), we proceed by induction and assume it is true for all n < l. By (3.19),

(l) r11 (y) is by inductive hypothesis the difference of a degree l polynomial in y and a degree l − 1 polynomial in y and hence is itself a degree l polynomial in y.

To prove (c), we again proceed by induction on l and assume it holds for all n < l.

Applying our inductive hypothesis,

(l) (l−1) (l−2) r11 (y) = yalr11 (y) − yr11 (y)

d l−1 e d l−2 e = yal · y 2 sl−1(y) − y · y 2 sl−2(y)

d l e  d l+1 e−d l e  = y 2 y 2 2 alsl−1(y) − sl−2(y) . (3.20)

Since sl−1(y) has leading coefficient a1a2...al−1, we now see that sl(y) has leading coefficient a1a2...al−1al, and

 l  deg(s (y)) = deg(r(l)(y)) − l 11 2  l   l  = l − = , (3.21) 2 2 as desired.

For (d), we note that s1(y) = a1, s2(y) = ya1a2 − 1 and proceed by induction. We

34 assume the result holds for n = l − 2, l − 1. Then

ya y r(l)(y) = r(l−1)(y) · l −1 0 " # r(l−1) r(l−1) ya y = 11 12 · l (l−1) (l−1) −1 0 r21 r22 " (l−1) (l−1) (l−1) # yalr11 (y) − r12 (y) yr11 (y) = (l−1) (l−1) (l−1) , (3.22) yalr21 (y) − r22 (y) yr21 (y)

and so using (a),(b) we find that

(l) l d 2 e r11 (y) = y sl(y) (3.23)

(l−1) (l−1) = yalr11 (y) − r12 (y) (3.24)

d l−1 e d l−2 e = yal(y 2 sl−1(y)) − y(y 2 sl−2(y)). (3.25)

(3.26)

Checking the cases for l odd and l even distinctly yields the desired result.

For (e), we proceed by induction. Given that s2(y) = ya1a2 − 1, s3(y) = ya1a2a3 −

(a1 + a3), we have our base case. Assume the result holds for n = l − 1. By (d) and

(c),

Tl = al(Tl−1) − a1a2...al−2, (3.27)

so we have (l − 2) + 1 = (l − 1) terms, each of which is the product of (l − 2) different

l−2 ai, d 2 e are odd i.

For (f), we proceed by induction with base cases s1(y) = a1, s2(y) = ya1a2 − 1,

s3(y) = ya1a2a3 − (a1 + a3), and assume the result is true for n = l − 2, l − 1. We note

by (d) that for l even,

Kl = −Kl−2, (3.28)

35 proving the even case, and for l odd,

Kl = −Kl−2 + alKl−1 (3.29)

= −(Kl−2 + al), (3.30) which proves the result.

For (g), we again proceed by induction. We have base case s2(y) = ya1a2 − 1. We now assume that the result is true for n = l − 2 even. By (d),

Sl = −Sl−2 + alKl−1. (3.31)

Using (f), this gives the desired result.

For (h), we again use induction and refer simply to (3.22).

For (i) and (j), we use induction and refer to (3.16) and (3.17) as our base cases.

(3.22) gives us

(l) (l−1) (l−1) r21 (y) = yalr21 (y) − r22 (y)

(l−1) (l−2) = yalr21 (y) − yr21 (y), (3.32) immediately using the result from (h).

To prove (i), we proceed by induction and assume it is true for all n < l. By (3.32),

(l) r21 (y) is by inductive hypothesis the difference of a degree l − 1 polynomial in y and a degree l − 2 polynomial in y and hence is itself a degree l polynomial in y.

To prove (j), we again proceed by induction on l and assume it holds for all n < l.

36 Applying our inductive hypothesis,

(l) (l−1) (l−2) r21 (y) = yalr21 (y) − yr21 (y)

d l−2 e d l−3 e = yal · y 2 tl−1(y) − y · y 2 tl−2(y)

d l−1 e  d l e−d l−1 e  = y 2 y 2 2 altl−1(y) − tl−2(y) . (3.33)

Since tl−1(y) has leading coefficient −a2a3...al−1, we now see that tl(y) has leading coefficient −a2a3...al−1al, and

l − 1 deg(t (y)) = deg(r(l)(y)) − l 21 2 l − 1 l − 1 = (l − 1) − = , (3.34) 2 2 as desired.

For (k), we again use induction with base cases t1(y) = −1, t2(y) = −a2, t3(y) = y(−a2a3) + 1. We assume the result is true for all n < l, and using (h),(i), and (3.22) we find that

(l) l−1 d 2 e r21 (y) = y tl(y) (3.35)

(l−1) (l−1) = yalr21 (y) − r22 (y) (3.36)

d l−2 e d l−3 e = yal(y 2 tl−1(y)) − y(y 2 tl−2(y). (3.37)

(3.38)

Checking the cases for l odd and l even distinctly yields the desired result.

For (l), we proceed by induction with base cases t1(y) = −1, t2(y) = −a2, t3(y) =

2 y(−a2a3) + 1, t4(y) = y (−a2a3a4) + (a2 + a4) and assume the result is true for n < l.

We note by (k) that for l odd,

0 0 Kl = −Kl−2, (3.39)

37 proving the odd case, and for l even,

0 0 0 Kl = −Kl−2 + alKl−1 (3.40)

0 = −(Kl−2 + al), (3.41) which proves the result and the lemma.

It should be noted that only parts (a)-(c) of the above lemma are applied in this

chapter. Parts (d)-(l) are needed for Chapter 4. We now can prove our main result.

Proof of Theorem 1.3.6. If m = 2k + 1 for some k ∈ N, q ≡ ±1 (mod m), ζ is a

th 2 i m−i primitive m root of unity in GF(q ), and yi := ζ + ζ + 2, then, by Lemma

3.2.5, Γ = Γ(y1, ..., yk) is a near-polygonal graph with cycles of length m. Assume

(n) Qn Γ has girth n < m. This means that α ∈ W (α) = H i=1 gP by Lemma 3.2.1 and that there were no returns along our n-walk, i.e. we have an (n − 1)-dipath

(δ0 = α, δ1, ..., δn−1) with δn−1 adjacent to α and no shorter dipaths from α to itself Qn in Γ. Thus H i=1 gP = H, which means that there exist a1, a2, ..., an ∈ GF(q) such Qn Qn that i=1 gp(ai) ∈ H, which in turn implies that i=1 g(yj)p(ai) is a lower diagonal

matrix for each 1 ≤ j ≤ k. Note that for 1 ≤ t ≤ (n − 1), at 6= 0, since otherwise

n Y gp(ai) = gp(a1)...gp(at−1)g1gp(at+1)...gp(an) i=1

= gp(a1)...gp(at−1 + at+1)...gp(an), (3.42)

38 which is at most a nontrivial (n − 1)-walk from α, contradicting our assumption of girth n. Note that

n Y (n−1) g(y)p(ai) = r (y) · g(y)p(an) i=1 " # r(n−1) r(n−1) ya y = 11 12 · n (n−1) (n−1) −1 0 r21 r22 " (n−1) (n−1) (n−1) # yanr11 (y) − r12 (y) yr11 (y) = (n−1) (n−1) (n−1) , (3.43) yanr21 (y) − r22 (y) yr21 (y) and so

n " (n−1) (n−1) (n−1) # Y y1anr11 (y1) − r12 (y1) y1r11 (y1) gp(ai) = (n−1) (n−1) (n−1) ,... i=1 y1anr21 (y1) − r22 (y1) y1r21 (y1) " (n−1) (n−1) (n−1) #! ykanr11 (yk) − r12 (yk) ykr11 (yk) (n−1) (n−1) (n−1) , (3.44) ykanr21 (yk) − r22 (yk) ykr21 (yk),

and by Lemma 3.2.6(c)

(n−1) n−1 d 2 e yr11 (y) = y · y sn−1(y). (3.45)

Qn Hence each yi is a root of the following equation in GF(q) since i=1 gp(ai) is lower diagonal:

d n−1 e y · y 2 sn−1(y) = 0. (3.46)

We know from Lemma 3.2.5 that for all i, yi 6= 0. Hence each yi for 1 ≤ i ≤ k must

be a root of sn−1(y) in GF(q) if the girth of Γ is n. The yi are all distinct, and so this

implies that deg(sn−1(y)) ≥ k. But, by Lemma 3.2.6(c),

n − 1 2k − 1 deg(s (y)) = ≤ = k − 1, (3.47) n−1 2 2

since n < m, a contradiction. Therefore the girth of Γ is at least m, and Γ is polygonal

of girth m.

39 CHAPTER 4

A CONSTRUCTION OF AN INFINITE FAMILY OF 2-ARC TRANSITIVE POLYGONAL GRAPHS OF ARBITRARY EVEN GIRTH

We proceed as in Section 3.2, except we let m = 2k + 2, k ∈ N, as opposed to letting m = 2k + 1. The results concerning the number of vertices, the valency, and

bound for the number of components of Γ is exactly the same as in Section 3.2, and

Lemmas 3.2.3 and 3.2.5 hold with the same proof verbatim.

We now proceed to show that the girth of the graph Γ we have just constructed

is at least m.

Lemma 4.0.7. Let m = 2k + 2 for some k ∈ N, q a prime power such that q ≡ ±1 (mod m) and let Γ be a near-polygonal graph with special cycles of length m as

constructed in Lemma 3.2.5. Then Γ does not have girth less than m − 1.

Proof. The proof is the same as the proof of Theorem 1.3.6.

Theorem 4.0.8. Let m = 2k + 2 for some k ∈ N, p a prime, q = pn for some n ∈ N, q ≡ ±1 (mod m) and let Γ be a near-polygonal graph with special cycles of length m as constructed in Lemma 3.2.5. Then, if Γ has girth (m − 1), either

(i) p|(k − 1) and p|(2k−1 − 1), or

40 (ii) p|(4k + 3) and p|(5k − 4 · 3k).

Proof. We must look precisely at the case when we have a cycle of length m−1. Since

we have a 2-arc transitive automorphism group and Γ is (G, 2)−dipath transitive, we

may assume that our cycle of length m − 1 includes [γ = Hf −1, α = H, β = Hg],

i.e. we may assume that γ ∈ W (2k−1)(β). Now, W (2k−1)(β) = Hg · P g ··· P g (2k − 1 copies of P g). Thus Hg · P g ··· P g = Hf −1, or

gP ··· gP gf ∈ H, (4.1)

or for some a1, a2, ..., a2k−1,

gp(a1)gp(a2) ··· gp(a2k−1)gf ∈ H. (4.2)

Note that the ai are nonzero since there can be no returns. So let’s look at

−y y2  g(y)p(a )...g(y)p(a )g(y)f(y) = r(2k−1)(y) · 1 2k−1 0 −y  k+1 k+1  y (−s2k−1(y)) y (ys2k−1(y) − s2k−2(y)) = k k+1 y (−t2k−1(y)) y (t2k−1(y) − t2k−2(y)) (4.3) based on Lemma 3.2.6(c),(j). We will use repeatedly throughout this proof the fact

that, since q > k, two polynomials of degree k or smaller differ by a constant if they

have the same roots. Noting that plugging in each value of yi in for y makes this,

coordinate-wise, an element of H, we can conclude the following:

1. The row 1, column 2 entry of (4.3) must be zero for each value of yi. Since

deg(ys2k−1(y) − s2k−2(y)) = k, this means that for some constant c1 we have

c1(ys2k−1(y) − s2k−2(y)) = u2k+2(y). (4.4)

41 k−1 Note that, by Lemma 3.2.6(f), the constant term on the left is −c1(−1) =

k k2k+1−k
k c1(−1) and on the right the constant term is (−1) k = (−1) (k + 1), and

so c1 = (k + 1), giving us:

(k + 1)(ys2k−1(y) − s2k−2(y)) = u2k+2(y). (4.5)

Note that comparing leading coefficients gives us

(k + 1)a1...a2k−1 = 1. (4.6)

2. The ratio of the row 1, column 1 entry over the row 2, column 2 entry in (4.3)

must be constant for all yi. Thus for come constant c2 we have

k+1 y (−s2k−1(y)) k+1 = c2. (4.7) y (t2k−1(y) − t2k−2(y))

Hence −s2k−1(y) = c2(t2k−1(y) − t2k−2(y)). Note that this is at most a degree k − 1

polynomial in y (s2k−1(y), t2k−1(y) are degree k − 1, t2k−2(y) is degree k − 2) with

at least k distinct roots, and so it must be identically 0. Furthermore, the leading

coefficient of −s2k−1(y) is −a1a2...a2k−1 and the leading coefficient of c2t2k−1(y) is

−c2a2a3...a2k−1. Since all the ai are nonzero (again, no returns and also (4.6)), we

have c2 = a1, and so

s2k−1(y) + a1t2k−1(y) − a1t2k−2(y) = 0. (4.8)

3. Finally, the ratio of the row 2, column 1 entry over the row 1, column 1 entry in

(4.3) must be constant for all yi. Thus for some constant c3 we have

42 k y (−t2k−1(y)) k = c3. (4.9) y (−ys2k−1(y))

So −c3ys2k−1(y)+t2k−1(y) = 0 must hold for y = y1, ..., yk. Noting that, by Lemma

3.2.6(c),(j), −c3ys2k−1(y)+t2k−1(y) is a degree k polynomial in y, for some constant

c4 we get that

c4(−c3ys2k−1(y) + t2k−1(y)) = u2k+2(y). (4.10)

k Since, by Lemma 3.2.6(l), the constant term of c4t2k−1(y) is c4(−1) and the con-

k stant term of u2k+2(y) is (k+1)(−1) , we get c4 = k+1. Now, the leading coefficient

of −(k+1)c3ys2k−1(y) is −(k+1)c3a1...a2k−1, and the leading coefficient of u2k+2(y)

is 1 = (k + 1)a1...a2k−1 (from (4.6)), so c3 = −1, finally giving us

(k + 1)(ys2k−1(y) + t2k−1(y)) = u2k+2(y). (4.11)

Comparing (4.5) and (4.10), we get that

s2k−2(y) = −t2k−1(y). (4.12)

Using Lemma 3.2.6(c),(j), the leading coefficient of t2k−1(y) is −a2...a2k−1 and the leading coefficient of −s2k−2(y) is −a1a2...a2k−2, which means that

a2k−1 = a1 (4.13)

Now, from (4.8), since t2k−1(y) = −s2k−2(y), we get

s2k−1(y) − a1s2k−2(y) = a1t2k−2. (4.14)

Noting (4.13), from part (d) of Lemma 3.2.6 we get:

s2k−1(y) − a1s2k−2(y) = −s2k−3(y). (4.15)

43 Combining these together, we get

s2k−3(y) = −a1t2k−2(y). (4.16)

The leading coefficient of −s2k−3(y) is −a1a2...a2k−3 and the leading coefficient of

a1t2k−2 is −a1a2...a2k−3a2k−2, and so we conclude that

a2k−2 = 1. (4.17)

We now claim that

a1 = a3 = a5 = ... = a2k−3 = a2k−1, (4.18)

a2 = a4 = a6 = ... = a2k−4 = a2k−2 = 1. (4.19)

We will proceed inductively. Assume that for some i > 0, we have that

a2k−1 = a2k−3 = ... = a2k+1−2i = a1, (4.20)

a2k−2 = a2k−4 = ... = a2k−2i = 1, (4.21) and that for all integers 1 ≤ j ≤ i we have

s2k−2j(y) = −t2k−2j+1(y), (4.22)

s2k−2j−1(y) = −a1t2k−2j. (4.23)

Again using part (d) of Lemma 3.2.6 and (4.21), we have

s2k−2i(y) = ya2k−2is2k−2i−1(y) − s2k−2i−2(y)

= ys2k−2i−1(y) − s2k−2i−2(y). (4.24)

44 From part (k) of Lemma 3.2.6, (4.22), (4.20), and (4.23), we have:

s2k−2i(y) = −t2k−2i+1(y)

= ya2k−2i+1(−t2k−2i(y)) + t2k−2i−1(y)

= ya1(−t2k−2i(y)) + t2k−2i−1(y)

= ys2k−2i−1(y) + t2k−2i−1(y). (4.25)

Comparing (4.24) and (4.25), we get

s2k−2i−2(y) = −t2k−2i−1(y). (4.26)

The leading coefficient of s2k−2i−2(y) is a1a2...a2k−2i−2 and the leading coefficient of

−t2k−2i−1 is a2...a2k−2i−2a2k−2i−1, and so we conclude that

a2k−2i−1 = a1. (4.27)

Again using part (d) of Lemma 3.2.6 and (4.27), we have

s2k−2i−1(y) = a2k−2i−1s2k−2i−2(y) − s2k−2i−3(y)

= a1s2k−2i−2(y) − s2k−2i−3(y). (4.28)

From part (k) of Lemma 3.2.6, (4.23), (4.21), and (4.26), we have:

s2k−2i−1(y) = a1(−t2k−2i(y))

= a1(a2k−2i(−t2k−2i−1(y)) + t2k−2i−2(y))

= a1(−t2k−2i−1(y)) + a1t2k−2i−2(y)

= a1s2k−2i−2(y) + a1t2k−2i−2(y). (4.29)

Comparing (4.28) and (4.29), we get

s2k−2i−3(y) = −a1t2k−2i−2(y). (4.30)

45 The leading coefficient of s2k−2i−3(y) is a1a2...a2k−2i−3 and the leading coefficient of

−a1t2k−2i−2 is a1a2...a2k−2i−3a2k−2i−2, and so we conclude that

a2k−2i−2 = 1. (4.31)

Therefore, by induction, we can continue this process indefinitely, and thus (4.18)

and (4.19) must hold. Going back to (4.6), we have (k + 1)a1...a2k−1 = 1, and so

k (k + 1)a1 = 1, or 1 ak = . (4.32) 1 k + 1

Now, we go back to (4.5), (k + 1)ys2k−1(y) − (k + 1)s2k−2(y) = u2k+2(y), and look

at the coefficient of yk−1. From part (e) of Lemma 3.2.6, the coefficient of yk−1 in

(k + 1)ys2k−1(y) is

k−1 −(k + 1)(2k − 2)a1 .

k−1 The coefficient of y in (k + 1)s2k−2(y) is just the leading coefficient, which is

k−1 −(k + 1)a1...a2k−2 = −(k + 1)a1 .

k−1 Finally, the coefficient of y in u2k+2(y) is

2k + 1 − 1 (−1)1 = 2k. 1

Noting that

k−1 1 a1 = , a1(k + 1) from (4.5) we have 2k − 2 1 + = 2k, a1 a1 or 2k − 1 a = . (4.33) 1 2k 46 Again, we go back to (4.5), but this time we look at the coefficient of y. From part

(f) of Lemma 3.2.6, the linear term of (k + 1)ys2k−1(y) is

k−1 k−1 (k + 1)(−1) (a1 + a3 + ... + a2k−1) = (k + 1)(−1) (ka1) 2k − 1 = (−1)k−1(k + 1) 2

From part (g) of Lemma 3.2.6, the linear term of (k + 1)s2k−2 is

(k − 1)k 2k − 1 (k + 1)(k − 1)(2k − 1) (−1)k(k + 1) = (−1)k . 2 2k 4

Finally, the linear term in u2k+2(y) is

2k + 1 − k + 1 (k + 2)(k + 1)k (−1)k−1 = (−1)k−1 . k − 1 6

Putting these together, from (4.5) we get

(k + 1)(2k − 1) (k + 1)(k − 1)(2k − 1) (k + 2)(k + 1)k + = , 2 4 6 which simplifies to

(4k + 3)(k − 1) = 0. (4.34)

CASE 1. If (k − 1) = 0, then k = 1 in GF(q). We know that

2k − 1 1 a = = 1 2k 2

in GF(q). Going back to (4.6), we get that

k (k + 1)a1 = 1

1 (2)( )k = 1 2

2k−1 − 1 = 0 (4.35)

47 Thus we must have p|(k − 1), p|(2k−1 − 1).

3 CASE 2. If (4k + 3) = 0, then k = − 4 in GF(q). We know that

2k − 1 5 a = = 1 2k 3

in GF(q). Going back to (4.6), we get that

k (k + 1)a1 = 1

1 5 ( )( )k = 1 4 3

5k − 4 · 3k = 0 (4.36)

Thus we must have p|(4k + 3), p|(5k − 4 · 3k).

We now can easily prove the main results.

Proof of Theorem 1.3.7. For a given even m = 2k+2, we construct our infinite family

of near-polygonal graphs of girth m as in Lemma 3.2.5. By Lemma 4.0.7 these graphs

all have girth at least (m − 1), and by Theorem 4.0.8 for only finitely many primes p

can the constructed graph have girth (m − 1). Hence there are infinitely many 2-arc

transitive polygonal graphs of girth m for all even m ≥ 4.

Proof of Corollary 1.3.8. Combining the results of Theorem 1.3.6 and Theorem 1.3.7

proves this result.

48 CHAPTER 5

AN ANALYSIS OF THE GIRTH-DOUBLING CONSTRUCTION FOR POLYGONAL GRAPHS

5.1 The construction

The following section is a (slight) generalization of [2, Section 2] which was cer-

tainly known to the authors of [2] but has not been published. We begin with a

polygonal graph Γ of valency r and set of special cycles C of girth m, and we let V and E be the vertex-set and edge-set of Γ, respectively. Let T be the edge set of a spanning tree in G. For a fixed integer k ∈ N, we define a group

Y H := Zk. (5.1) e∈E\T Group elements are thus n-tuples for n = |E\T |, where each coordinate is generated

by an element of order k. The coordinates are indexed by the edges in E\T, and

|H| = kn.

Next, to each edge e ∈ E we assign a group element of H, which is referred to as

the voltage of e [5]. We will denote this group element ge. If e ∈ T , then ge is the

identity of H, i.e. an n-tuple with 0 in every coordinate. If e ∈ E\T, then ge is the

49 generator of the cyclic group of order k in the coordinate indexed by e and 0 in every other coordinate.

We now use this assignment to define Γk. Its vertex set is Vk = V × H, and we define the edge set Ek in the following manner. For each edge e ∈ E, we fix an ordering

(α, β) of the two vertices incident to e. Then, for all g ∈ H, we add {(α, g), (β, g+ge)} to the edge set Ek. We say that the directed edge (α, β) has voltage ge = gαβ and the

n directed edge (β, α) has voltage −ge = gβα. Note that we have constructed k edges for each e ∈ E, one for every choice of g.

There is a natural covering map φ :Γk → Γ defined by supressing the group element. Note for each α ∈ V and e ∈ E we have

|φ−1(α)| = |φ−1(e)| = kn. (5.2)

Furthermore, if α has neighbors β1, β2, ...βr in Γ whose connecting directed edges are assigned voltages g1, ..., gr, going from α to each respective βi, then (α, g) ∈ Vk has exactly r neighbors. Since Γ is r-regular, so is Γk.

Now consider an m-cycle C ∈ C. Since C has m edges and is 2-regular, φ−1(C) must contain knm edges and must be 2-regular. Thus φ−1(C) is the disjoint union of cycles C1, ..., Cl for some l. Start at a vertex α in C and walk along the cycle. In ˜ Γk we start at (α, g) for some g ∈ H. Upon returning to α in C, the path in Γk is at (α, g + h), where h is the sum of the voltages along C. A second walk leads us to

(α, g + 2h), a third to (α, g + 3h), etc., finally returning to (α, g) after k trips around

C in Γ. Hence each cycle C˜ ∈ φ−1(C) is k times as long as C, and there are kn−1 disjoint cycles in φ−1(C).

50 Note that if P is a path of length 2 contained in C, any P˜ ∈ φ−1(P ) is covered by

a unique C˜ ∈ φ−1(C). It follows that

C˜ = {φ−1(C)|C ∈ C} (5.3)

covers each path of length two in Γk exactly once. Thus Γk is a near-polygonal graph

with special cycles of length km.

Lemma 5.1.1. If Γ is connected, then so is Γk.

Proof. Let α ∈ V, the vertex set of Γ. By construction, every vertex in Γk is connected

to a vertex of the form (α, g) for some g ∈ H. It thus suffices to show that (α, 0) is

Pl connected to (α, g) for all g ∈ H. Let g = i=1 gαiβi , where gαiβi is the voltage for

the edge going from vertex αi to vertex βi (note that the gαiβi are not necessarily

distinct). Begin at (α, 0) in Γk and start a walk. In Γ we can walk from α to α1 using

only edges in T , which do not contribute voltage. Walk on the edge from α1 to β1,

then walk from β1 back to α using only edges in T . In Γk we have walked from (α, 0)

to (α, gα1β1 ). We now repeat this process for each i. This is altogether a closed walk

in Γ with voltage g that begins and ends at α. Hence (α, 0) and (α, g) are connected

in Γk, and so Γk is connected.

Lemma 5.1.2. If Γ is a polygonal graph of girth m, k > 3 is an integer, and Γk is the near-polygonal graph constructed as above, then Γk is not polygonal.

Proof. Take two special cycles C1 (with total voltage h1) and C2 (with total voltage h2) in Γ containing an edge e with vertices α and β with voltage g going from α to

β. Traverse C1 starting with e, and then traverse C2 ending with e. Now traverse

C1 in the opposite direction to β and then C2 in the opposite direction to α. From

51 this, we have a closed walk W of length 4m − 2 in Γ with no returns. We lift W to a

walk in Γk, starting at (α, 0): we go to (β, g), then to (α, h1), to (β, h1 + h2 + g) to

(α, h1 + h2), to (β, h2 + g) and finally back to (α, 0). This is a closed walk in Γk, and

it has length 4m − 2. Hence the special cycles in Γk, which have length km ≥ 4m, are

longer than the girth, and Γk is not polygonal.

5.2 Another construction

In order to fully analyze the case k = 3 and the automorphism groups of our graphs, we need the following construction, which is a slight modification of the one in the previous section. We begin with a polygonal graph Γ of valency r and set of special cycles C of girth m, and we let V and E be the vertex-set and edge-set of Γ, respectively, as above. For a fixed integer k ∈ N, we define a group

Y G := Zk. (5.4) e∈E Group elements are thus N-tuples for N = |E|, where each coordinate is generated by

an element of order k. The coordinates are indexed by the edges in E, and |H| = kN .

Next, to each edge e ∈ E we assign a group element of H, which is referred to as

the voltage of e [5]. We will refer to this group element as ge. For all e ∈ E, ge is the

generator of the cyclic group of order k in the coordinate indexed by e and 0 in every

other coordinate. If e is incident with vertices α and β and directed from α to β, we

will keep the terminology of the previous section and let gαβ = ge and gβα = −ge.

∗ We define a graph Γk with vertex set V × G and edge set analagously as above.

∗ Note that Γk has valency r and is near-polygonal with cycles of length km. Also,

∗ ∗ there is still a natural covering projection φ :Γk → Γ defined by suppressing the

52 ∗ group element. However, as we will show later, Γk is not connected. We will now prove the following:

∗ Theorem 5.2.1. If Γ is a polygonal graph of valency r and girth m, then Γ3 is a polygonal graph of valency r and girth 3m. Moreover, if Γ is a strict polygonal graph,

∗ then so is Γ3.

Proof. First, given a polygonal graph Γ of valency r and girth m, we may use the

∗ construction above to produce a near-polygonal graph Γ3 of valency r and special

∗ cycles of length 3m. We must now show that Γ3 has girth at least 3m.

∗ ∗ Let C be a cycle of shortest possible length in Γ3. As above, we have a natural map

∗ ∗ ∗ ∗ φ :Γ3 → Γ from suppressing the group elements (i.e. voltages). Consider φ (C ) in Γ. We note first that φ∗(C∗) is a closed walk in Γ of the same length as C∗ where no edge is traversed consecutively in opposite directions, i.e. no returns (Theorem 2.1.1 of [5]). Since φ∗(C∗) is a closed walk in Γ, φ∗(C∗) must contain at least one cycle in

Γ. Since Γ has girth m, this walk must have length at least m.

We walk along φ∗(C∗) as we walk along C∗. If φ∗(C∗) = C for some cycle C in Γ, then once we start to walk along C∗ in one direction, we must walk along C in that direction. Since there are no returns, there is exactly one direction along C in Γ we are able to walk. Thus we walk in this one direction the entire way, never leaving the cycle C. Walking the cycle this way requires us to traverse the cycle three times in order to cancel the voltage, so we have walked at least 3m edges. Hence if C∗ has length less than 3m, then φ∗(C∗) must contain a cycle C and an edge e not contained in C.

We now make a couple of observations about our closed walk in Γ. First, if we “leave” a vertex α at least three times along our closed walk φ∗(C∗), then we must come back

53 to this vertex α as many times as we left. Each time we leave and come back, since there are no returns, we must have traversed a cycle in Γ before getting back to α.

This cycle has length at least m by assumption, so if we leave a vertex three times in

our closed walk, our walk is at least 3m edges long.

Next, we note that since every edge in Γ has a unique nonzero voltage, every edge

we traverse in φ∗(C∗) must be used at least twice, i.e. once forwards and once

backwards,three times forward, or three times backwards, in order to cancel out the

∗ ∗ voltage and return to our starting point in C in Γ3. Without a loss of generality, we may assume that we begin our walk at a vertex α, traverse a cycle C, and then leave α again along an edge e not in C. Note that since

this is a closed walk, we must eventually come to the vertex α again along our walk.

CASE 1. We come back to α along a walk that is edge-disjoint with C.

Note that C is of length at least m and this new walk beginning with edge e also must have length at least m (no returns). However, since we must cancel out the voltage of each edge we traversed, we now must walk each of these again, which is more than

3m edges.

CASE 2. We leave α along an edge e and come back to C at a vertex β, where we

then traverse an edge in C.

Suppose the path along C from α to β is i edges long. Then, since the girth of Γ is m, the other path along C from α to β and the path from α to β that is edge-disjoint

from C both have length at least m − i. Note that these two paths themselves form

a closed walk in Γ, and so we have:

2(m − i) ≥ m. (5.5)

54 All of these edges must be traversed at least twice, and:

2(2(m − i) + i) = 2m + 2(m − i) ≥ 3m, (5.6)

∗ ∗ ∗ ∗ so |φ (C )| ≥ 3m. Hence any cycle in Γ3 has length at least 3m and Γ3 is polygonal.

∗ Now suppose that Γ is strict polygonal. In order to show that Γ3 is also strict

∗ polygonal, we must show that all 3m-cycles in Γ3 are lifts of m-cycles of Γ. Assume

∗ ∗ ∗ ∗ ∗ ∗ that C is a cycle in Γ3 of length 3m with image φ (C ) such that φ (C ) contains contains branching vertices (i.e. vertices incident to at least three different edges of

φ∗(C∗)).

Let α be an arbitrary branching vertex. We may assume that the closed walk φ∗(C∗) starts at α, and so φ∗(C∗) can be written as the concatenation of k closed walks

W1, ..., Wk, where each Wi is from α to α, and α does not occur as an inner vertex in any Wi. Since the girth of Γ is m, each Wi has length at least m. Hence, if k ≥ 4 then the length of φ∗(C∗) is at least 4m, a contradiction. Moreover, since α is branching,

α is incident to at least three edges, and each of these edges must be traversed at least twice to cancel their voltage; hence k ≥ 3. Summarizing, we obtain that k = 3,

∗ ∗ each Wi has length m, α is incident to exactly three edges in φ (C ), and each of these three edges are traversed exactly twice, in opposite directions.

Let (α, β) be the first edge of W1. We have seen that (β, α) must be the last edge of

Wi for some i ≤ 3. If i = 1 then W1 contains a proper subwalk from β to β of length m − 2 that contains no returns, a contradiction to the fact that Γ has girth m. The case i = 3 is also impossible, since φ∗(C∗) cannot contain consecutive edges (β, α)

and (α, β). Hence i = 2.

We claim now that β is branching. Indeed, if β is not branching, then let {β, γ} be

the unique (undirected) edge different from from {α, β} and incident to β in φ∗(C∗).

55 st The second edge of W1 must be (β, γ) and the (m − 1) edge of W2 must be (γ, β).

Hence the 2-arc (α, β, γ) occurs in W1 and the 2-arc (γ, β, α) occurs in W2. Since W1

and W2 are closed walks of length m, in fact they must be cycles. Moreover, since Γ

is strict polygonal and both W1 and W2 contain the 2-path (α, β, γ), we must have

that W1 = W2, only they are traversed in opposite directions. This is a contradiction,

∗ ∗ because the last edge of W1 and the first edge of W2 are a return in φ (C ). Hence β

is branching.

So far, we have seen that each branching vertex must occur exactly three times

in φ∗(C∗) and must partition φ∗(C∗) into the concatenation of three closed walks of

length m. This is a final contradiction, because we have also shown that the branching

∗ vertex β has two occurrences of distance 2m − 2 from each other. Therefore, Γ3 must be strict polygonal, as desired.

∗ We now begin our analysis of the automorphism group of Γk.

∗ ∗ Theorem 5.2.2. Each σ ∈ Aut(Γ) lifts to σ ∈ Aut(Γk), i.e. there exists K ≤

∗ ∼ Aut(Γk) such that K = Aut(Γ).

∗ Proof. For σ ∈ Aut(Γ), we define a permutationσ ˜ of G, the voltage group of Γk the following way. Let {α, β} ∈ E (recall that E is the edge set of Γ), and let

σ σ σ˜ δ = α ,  = β . Then {δ, } ∈ E because σ ∈ Aut(Γ). For gαβ ∈ G, define gαβ = gδ. Since every edge in E has a voltage, and the edges are being permuted by σ,σ ˜ is a

permutation of the edge voltages. Every element g ∈ G can be written uniquely as

Pl a sum of the edge voltages, i.e. if g = i=1 gi, where each gi is an edge voltage (not

necessarily all distinct), then this expression is unique up to permuting the gi. Thus

σ˜ Pl σ˜ σ˜ extends naturally to the group G itself defined by g = i=1 gi .

56 ∗ ∗ ∗ We may now define a map σ of Γk as follows: for each vertex (α, g) of Γk, we define

σ∗ σ σ˜ ∗ (α, g) = (α , g ). This map clearly permutes the vertices of Γk. If we have two

∗ ∗ σ σ˜ adjacent vertices (α, g) and (β, g + gαβ) in Γk, then σ takes these to (α , g ) and

σ σ˜ σ˜ σ σ˜ σ∗ (β , g + gαβ) = (β , g + gασβσ ), preserving their adjacency. Furthermore, if (α, g) and (β, h)σ∗ are adjacent, we simply apply (σ−1)∗ to them to note that (α, g) and

∗ ∗ (β, h) must have been adjacent also. Therefore σ is an automorphism of Γk.

Finally, we note that two different automorphisms σ1 and σ2 of Γ must have different

∗ ∗ ∗ actions on V . Hence σ1 and σ2 have different actions on the vertices of Γk, and we

∗ ∼ see that there exists K ≤ Aut(Γk) such that K = Aut(Γ).

Corollary 5.2.3. If Γ is vertex transitive, edge transitive, arc transitive, or 2-arc

∗ transitive, then so is Γk.

Proof. The proof of each statement is analagous to the proofs of the others, so we will

only show that 2-arc transitivity is maintained under the construction. Suppose that Γ

∗ is 2-arc transitive, and look at two 2-arcs of Γk, namely ((α, g−gαβ), (β, g), (γ, g+gβγ))

and ((δ, h − gδ), (, h), (ρ, h + gρ). Since Γ is 2-arc transitive, there exists σ ∈ Aut(Γ)

σ ∗ ∗ such that (α, β, γ) = (δ, , ρ). By Lemma 5.2.2, there exists σ ∈ Aut(Γk) such that

σ∗ σ˜ σ˜ σ˜ ((α, g − gαβ), (β, g), (γ, g + gβγ)) = ((δ, g − gδ), (, g ), (ρ, g + gρ)). (5.7)

∗ ∗ Since Γk is a voltage graph, for all t ∈ G there exists an automorphism Φt of Γk defined

0 0 ∗ 0 0 by taking each vertex (α , g ) of Γk to (α , g + t) (Section 2.2.1 of [5]). Therefore,

∗ σ Φ σ˜ ((α, g − gαβ), (β, g), (γ, g + gβγ)) h−g = ((δ, h − gδ), (, h), (ρ, h + gρ)), (5.8)

∗ and Γk is 2-arc transitive, as desired.

57 5.3 Relating the two constructions

∗ Since the above constructions are so similar, it is natural to ask how Γk and Γk are related when constructed from the same graph Γ. Throughout this section we will assume that the initial graph Γ is connected (and hence, by Lemma 5.1.1, so is Γk).

0 ∗ Lemma 5.3.1. There exists a covering projection φ :Γk → Γk.

Proof. These is seen by suppressing the group elements corresponding to the voltages along the spanning tree T, giving us a natural covering projection like the ones that

∗ map Γk and Γk onto Γ.

This leads to the following immediate corollary:

∗ Corollary 5.3.2. Each component of Γk projects onto Γk.

Proof. This is obvious since we are assuming that Γ (and hence Γk) is connected.

We now note that there is an interesting embedding of the voltage group H of

∗ Γk into the the voltage group G of Γk. Let f = {β, γ} ∈ E\T, and suppose that we gave the orientation (β, γ) to f at the definition of the voltage gf in H and G. Let

Wf = (γ = γ0, γ1, ..., γl = β) be the unique path in Γ using only edges of T , and define l−1 ∗ X gf := gf + gγiγi+1 . (5.9) i=0 ∗ (The voltages gγiγi+1 are defined only in G.) The map taking gf to gf on the generating set of H extends naturally to a homomorphism ϕ : H → G. Clearly, ϕ is injective,

P ∗ P since f∈E\T af gf = 0 implies f∈E\T af gf = 0, and the latter equation holds if and

∗ only if all af = 0. We shall refer to ϕ(H) as H .

58 Let α be a distinguished vertex of Γ. For any edge f = {β, γ} ∈ E\T there are two unique paths Wαβ and Wαγ using only edges of T from α to β and γ, respectively.

Note that if we traverseWαβ, then walk across f, and then walk back to f using Wαγ

∗ backwards, the resulting walk Wα,f has voltage gf ; indeed, we only added edges from a path P from α to a vertex in Wf (if we added any at all) to Wf . Each edge in P is

∗ traversed exactly twice in Wα,f , giving us a total voltage of exactly gf for Wα,f . The

∗ following lemma makes the relationship between Γk and Γk more explicit by analyzing the embedding H∗ < G:

∗ Lemma 5.3.3. Vertices (α, g1) and (α, g2) are in the same component of Γk if and

∗ only if g1 − g2 ∈ H .

Proof. First, suppose that g −g ∈ H∗. Then we have g −g = Pk g∗ , where each 1 2 1 2 i=1 fi

∗ fi ∈ E\T (the fi are not necessarily distinct). We start at the vertex (α, g1) in Γk

∗ and begin a walk. Since Γk is a covering space of Γ, a closed walk in Γ will lift to a

∗ unique walk in Γk starting at the vertex (α, g1). Let W be the concatenation of the

walks Wα,fi as defined in the previous paragraph. Then W is a closed walk, and the

sum of the voltages along W is Pk g∗ = g − g . Since Γ∗ is a covering space of Γ, i=1 fi 1 2 k

∗ the walk W will lift to a walk in Γk starting at (α, g2) and ending at (α, g1).

∗ Conversely, suppose that (α, g1) and (α, g2) are in the same component of Γk. Let W

∗ be a walk in Γk from (α, g1) to (α, g2) with no returns and let ((β1, h1), (γ1, h1+ge1 )),...,

((βj, hj), (γj, hj + gej )) be the edges in W that project down to edges e1, ..., ej in Γ

that are in E\T, and suppose that we traverse them along W in precisely this order.

Hence every other edge in W projects down to an element of T . Note that since T is

a tree and there are no returns in W , there is a unique path Pi along edges of T that

goes from each γi to βi+1 without returns. We will now modify our walk as follows:

59 ∗ ∗ for i = 1, ..., j − 1, we replace Pi by Pi , where Pi is the unique walk in Γ that is the concatenation of two paths in T : one from γi to α and the other from α to βi+1, i.e.

the concatenation of Wαγi (as defined above) traversed in the opposite direction and

∗ Wαβi+1 . Note that there may be some returns along Pi , but it is unique. Note also

∗ that the voltage of Pi is the same as Pi since we only (possibly) modified Pi by going to α along some edges in T and then returning along those same edges in T. Thus

∗ W also realizes that (α, g1) and (α, g2) are in the same component and has the same voltage as W . However, the voltage of W ∗ is clearly

j X g∗ ∈ H∗, (5.10) ei i=1

∗ and hence g1 − g2 ∈ H .

Given this lemma, it is now very easy to prove the following:

∗ |T | Theorem 5.3.4. Γk is isomorphic to k disjoint copies of Γk.

∗ Proof. From Lemma 5.3.3, (α, g1) and (α, g2) are in the same component of Γk if and

∗ ∗ ∗ only if g1 −g2 ∈ H . Thus each coset of H in G determines a component of Γk. Since

|G : H∗| = |G : H| = k|T |, (5.11)

∗ |T | Γk has k connected components. Each of these connected components must project

|E|−|T | ∗ down onto Γk, so each component has at least k vertices. However, Γk has ex-

|E| |T | ∗ actly k vertices, and therefore each of the k components of Γk must be isomorphic to Γk.

We can now easily prove the main results.

60 Proof of Theorems 1.3.9, 1.3.10. We have already shown the corresponding theorems

∗ ∗ for Γk (Theorem 5.2.1, Corollary 5.2.3), and Theorem 5.3.4 tells us that Γk is isomor-

|T | phic to k disjoint copies of Γk. The results follow.

61 APPENDIX A

GIRTH-FINDING PROGRAM

The following is a program for GAP that determines whether a near-polygonal graph constructed using Theorem 1.3.1 is a polygonal graph and also whether it is a strict polygonal graph. girth10:= function(p, bound) local t, half, unity, r3, changebase, inv, d, c, niceg, invniceg, i1, x, y, cycle, A, B, C, g, h, H, K, rcosets, rprst, sign, baseset, buildup, oldbuildup, built1, built2, built, ii, count, w, z, totalcount;

if not (IsPrimePowerInt(p) and (p mod 30) in [1,19]) then return "not good prime power"; fi;

t:= (Z(p)^0 + SquareRoots(GF(p), 5*Z(p)^0)[1])/(4*Z(p)^0);

half:= 1/(2*Z(p)^0);

62 unity:= (-Z(p)^0 + SquareRoots(GF(p), (-3)*Z(p)^0)[1])/(2*Z(p)^0);

r3:= SquareRoots(GF(p), Z(p)^0*(-3))[1];

A:= Z(p)^0*[[-1,0,0],[0,0, -unity],[0, -unity^2,0]];

B:= Z(p)^0*A*[[-half, half - t, -t],[t - half, t, -half],

[t, -half, half - t]]*DiagonalMat([-1,-1,1]);

H:= Group(Z(p)^0*A, Z(p)^0*B); if ( not (Order(H) = 1080)) then

return "failed to make A_6"; fi;

C:= Z(p)^0*B*A*B*A*B*B;

K:= Group(Z(p)^0*A, Z(p)^0*C); if (not (Order(K) = 108)) then

return "failed to make F_36"; fi;

c:= -half*(Z(p)^0 + Z(p)^0*2*unity*t)/(Z(p)^0 + Z(p)^0*2*unity^2*t);

63 changebase:= Z(p)^0*[[1, c*unity^2, c], [1, 1/c, unity^2],

[1, unity, unity/c] ];

inv:= Inverse(changebase);

d:= Z(p)^0*c*(c^2*(unity - 4*unity*t - 2*t)

+ c*(-4*t - 2*unity + 4*unity*t) + unity)/

(2*(1+2*c*unity)*(1 - c*unity));

x:= unity^2/(d*r3); niceg:= Z(p)^0*[[1/r3, -1/(3*x), -unity^2/(3*x)],

[x, unity/r3, unity/r3], [unity*x, 1/r3, unity/r3]];

invniceg:= Inverse(niceg);

g:= inv*niceg*changebase;

h:= Inverse(B)*DiagonalMat([-Z(p)^0,-Z(p)^0,Z(p)^0])*B;

rcosets:= RightCosets(H,K);

rprst:= List(rcosets, z->Representative(z));

for i1 in [1..10] do

64 if rprst[i1] in K then sign:=i1;

fi;

od;

Remove(rprst,sign);

baseset:= [g*rprst[1], g*rprst[2], g*rprst[3], g*rprst[4], g*rprst[5], g*rprst[6], g*rprst[7], g*rprst[8], g*rprst[9]];

H:= AsSet(Elements(H));

buildup:=ShallowCopy(baseset);

ii:=3;

repeat

oldbuildup:=buildup;

buildup:=[];

for z in oldbuildup do

for y in baseset do

Add(buildup, z*y);

od;

od;

ii:=ii+1;

until ii=bound;

65 built1:= []; built1:= List(buildup, w->w*g); built1:= AsSet(built1);

built2:= []; built2:=List(oldbuildup, w->w*g*h); built2:=AsSet(built2);

built:=[]; for z in built2 do for w in H do

Add(built, w*z); od; od; built:= AsSet(built);

count:= Length(Intersection(built1, built)); if (not (count = 0)) then

return ["Girth at most ", 2*bound-1]; fi;

built2:= []; built2:=List(buildup, w->w*g*h); built2:=AsSet(built2);

66 built:=[]; for z in built2 do for w in H do

Add(built, w*z); od; od; built:= AsSet(built);

totalcount:= Length(Intersection(built1, built));

if (totalcount = 0) then return ["Girth at least ", 2*bound+1]; fi;

if (totalcount = 1) then return ["Strict polygonal of girth ", 2*bound]; fi;

if (not (totalcount in [0,1])) then return ["Non-strict polygonal of girth ", 2*bound]; fi;

end;

67 APPENDIX B

ANOTHER GIRTH-FINDING PROGRAM

The following is a program for GAP that determines whether a near-polygonal graph constructed using Theorem 1.3.1 is a polygonal graph also whether or not it is a strict polygonal graph. It is less time-efficient than the program contained in Appendix A but requires less memory, so it should be used when the program contained in Appendix A fails to run due to memory constraints. newgirth10:= function(p, bound) local t, half, unity, r3, changebase, inv, d, c, niceg, invniceg, i1, x, y, cycle, A, B, C, g, h, H, K, rcosets, rprst, sign, baseset, leveltwo, buildup, oldbuildup, built1, built2, built, ii, count, jj, w, z, totalcount;

if not (IsPrimePowerInt(p) and (p mod 30) in [1,19]) then return "not good prime power"; fi;

t:= (Z(p)^0 + SquareRoots(GF(p), 5*Z(p)^0)[1])/(4*Z(p)^0);

68 half:= 1/(2*Z(p)^0);

unity:= (-Z(p)^0 + SquareRoots(GF(p), (-3)*Z(p)^0)[1])/(2*Z(p)^0);

r3:= SquareRoots(GF(p), Z(p)^0*(-3))[1];

A:= Z(p)^0*[[-1,0,0],[0,0, -unity],[0, -unity^2,0]];

B:= Z(p)^0*A*[[-half, half - t, -t],[t - half, t, -half],

[t, -half, half - t]]*DiagonalMat([-1,-1,1]);

H:= Group(Z(p)^0*A, Z(p)^0*B); if ( not (Order(H) = 1080)) then

return "failed to make A_6"; fi;

C:= Z(p)^0*B*A*B*A*B*B;

K:= Group(Z(p)^0*A, Z(p)^0*C); if (not (Order(K) = 108)) then

return "failed to make F_36"; fi;

69 c:= -half*(Z(p)^0 + Z(p)^0*2*unity*t)/(Z(p)^0 + Z(p)^0*2*unity^2*t);

changebase:= Z(p)^0*[[1, c*unity^2, c], [1, 1/c, unity^2],

[1, unity, unity/c] ];

inv:= Inverse(changebase);

d:= Z(p)^0*c*(c^2*(unity - 4*unity*t - 2*t)

+ c*(-4*t - 2*unity + 4*unity*t) + unity)/

(2*(1+2*c*unity)*(1 - c*unity));

x:= unity^2/(d*r3); niceg:= Z(p)^0*[[1/r3, -1/(3*x), -unity^2/(3*x)],

[x, unity/r3, unity/r3], [unity*x, 1/r3, unity/r3]];

invniceg:= Inverse(niceg);

g:= inv*niceg*changebase;

h:= Inverse(B)*DiagonalMat([-Z(p)^0,-Z(p)^0,Z(p)^0])*B;

rcosets:= RightCosets(H,K);

rprst:= List(rcosets, z->Representative(z));

70 for i1 in [1..10] do

if rprst[i1] in K then sign:=i1;

fi;

od;

Remove(rprst,sign);

baseset:= [g*rprst[1], g*rprst[2], g*rprst[3], g*rprst[4], g*rprst[5], g*rprst[6], g*rprst[7], g*rprst[8], g*rprst[9]];

H:= AsSet(Elements(H));

buildup:=ShallowCopy(baseset);

ii:=3;

repeat

oldbuildup:=buildup;

buildup:=[];

for z in oldbuildup do

for y in baseset do

Add(buildup, z*y);

od;

od;

ii:=ii+1;

71 until ii=bound;

built2:= []; built2:=List(buildup, w->w*g*h); built2:=AsSet(built2);

for y in [1..Length(buildup)] do

built1:=[];

built1:= List(baseset, z -> buildup[y]*z*g);

built:= [];

for z in built1 do

for w in H do

Add(built, w*z);

od;

od;

built:= AsSet(built);

count:= Length(Intersection(built, built2));

if (not (count = 0)) then

return ["Girth at most ", 2*bound-1];

fi;

72 od;

leveltwo:= List([0..80], w-> baseset[QuoInt(w,9)+1]

*baseset[(w - QuoInt(w,9)*9)+1]); leveltwo:= AsSet(leveltwo);

totalcount:= 0; count:=0;

for y in [1..Length(buildup)] do

built1:=[];

built1:= List(leveltwo, z -> buildup[y]*z*g);

built:= [];

for z in built1 do

for w in H do

Add(built, w*z);

od;

od;

built:= AsSet(built);

count:= Length(Intersection(built, built2));

totalcount:= totalcount + count; od;

73 if (totalcount = 0) then return ["Girth at least ", 2*bound+1]; fi;

if (totalcount = 1) then return ["Strict polygonal of girth ", 2*bound]; fi;

if (not (totalcount in [0,1])) then return ["Non-strict polygonal of girth ", 2*bound]; fi;

end;

74 APPENDIX C

LARGE NUMBERS

The following is a list of large numbers that would not fit neatly onto the page in

Section 2.3:

√ √ √ g := −39+11i 3+9√5−i 15 1 −66+18i 15

√ √ √ g := 21−11i 3−9 5√−29i 15 2 −132+36i 15

√ √ √ g := 111+11√i 3+9 √5−7i √15 3 −66−66i 3−54 5+18i 15

√ √ √ g := −78−44i√ 3−36√ 5−2i √15 4 −51+51i 3−63 5−21i 15

√ √ √ g := −27−95i√3+27√5+19√i 15 5 6(17−17i 3+21 5+7i 15)

√ √ √ g := −129+61i√ 3+15√ 5−23√i 15 6 6(17+17i 3−21 5+7i 15)

√ √ √ g := −36−14i√ 3+6 √5+4i √15 7 −33+33i 3+27 5+9i 15

75 √ √ √ g := −39−61√i 3−15√ 5−i √15 8 −66+66i 3+54 5+18i 15

√ √ √ g := −3−8i √3−24 √5−5i √15 9 −33−33i 3−27 5+9i 15

√ √ gh := 86√i 3−√46 5 √ 1 −7−7i 3+99 5−33i 15

√ √ √ gh := 495−495√i 3−√21 5−7√i 15 2 6(7+7i 3−99 5+33i 15)

√ √ √ gh := −21+7√i 3−99√ 5−99√i 15 3 6(7+7i 3−99 5+33i 15)

√ √ gh := −93i 3+145√ 5 4 −223+119i 15

√ √ √ gh := −1785+1785i 3−669 √5−223i 15 5 −2676+1428i 15

√ √ √ gh := −669+223i 3+357 √5+357i 15 6 −2676+1428i 15

√ gh := −237+53√i 15 8 6(7+33i 15)

√ √ √ gh := −795−265i 3+237√ 5−237i 15 9 84+396i 15

c11 := −5371383613575

c12 := −1454745577717

76 c13 := 9416703784329

c14 := 1706532297823

c15 := −5119596893469

c16 := 5119596893469

c17 := −3222830168145

c18 := −1074276722715

√ √ 2( 3+i 5) v := − √ √ √ 11 5i+ 3+i 5+ 15

√ v := 1−√5 12 1+ 5

√ √ √ (1+i)(−3i+3i 3−3 5+ 15) v := − √ √ √ 21 (2+3i)+ 3+i 5−(1+2i) 15

√ √ √ (3+4i)−i 3+ 5+i 15 v := √ √ √ 22 (2+3i)+(1−2i) 3+i 5− 15

√ √ √ (1+i)(−3i−3i 3+3 5+ 15) v := − √ √ √ 31 (3+2i)−i 3+ 5+(2+i) 15

√ √ √ (3−4i)−i 3+ 5+i 15 v := √ √ √ 32 (2−3i)−(1+2i) 3−i 5+ 15

77 BIBLIOGRAPHY

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