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Linear Algebra and its Applications 432 (2010) 23812398

Contents lists available atScienceDirect

Linear Algebra and its Applications

j o u r n a l h o m e p a g e : w w w . e l s e v i e r . c o m / l o c a t e / l a a

Search for properties of the missing Moore graph

Martin Macaja ,, Jozef irnb,c

a Comenius University, Bratislava, Slovakiab Open University, Milton Keynes, UKc Slovak University of Technology, Bratislava, Slovakia

A R T I C L E I N F O A B S T R A C T

Article history:Received 30 April 2009Accepted 13 July 2009Available online 19 August 2009

Submitted by R.A. Brualdi

AMS classification:05C5005C2520B2520C15

Keywords:Moore graphsSpectral graph theoryRational representations

In the degree-diameter problem, the only extremal graph the ex-istence of which is still in doubt is the Moore graph of order 3250,degree 57 and diameter 2. It has been known that such a graph can-not be vertex-transitive. Also, certain restrictions on the structureof the automorphism group of such a graphhave been known in the

case when the order of the group is even. In this paper we furtherinvestigate symmetries and structural properties of the missingMoore(57, 2)-graph(s) with the help of a combination of spectral,group-theoretic, combinatorial, and computational methods. Oneof the consequences is that the order of the automorphism groupof such a graph is at most 375.

2009 Elsevier Inc. All rights reserved.

1. Introduction

In graphtheory thereare a number of problems linking graphs with linear algebra andgroup theory.A prominent example that has been around for five decades is thedegree-diameter problem. We recallthatthe degree ofavertexofagraphisthenumberofedgesincidentwiththevertex,whilethe diameterof a graph is the smallestksuch that any two vertices in the graph are connected by a path of lengthat mostk. In its broadest formulation the degree-diameter problem is to find, for any given positiveintegersdand k, the largest order of a graph of maximum degreed and diameterk and classify thecorrespondingextremalgraphs.ResearchinthisareawasinitiatedbythepioneeringpaperbyHoffmanand Singleton [9], prompted by Moore who suggested the problem. By now the bibliography of related

Corresponding author.E-mail addresses: [email protected] (M. Macaj), [email protected] (J. ira n).

0024-3795/$ - see front matter 2009 Elsevier Inc. All rights reserved.doi:10.1016/j.laa.2009.07.018
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2382 M. Macaj, J. iran / Linear Algebra and its Applications 432 (2010) 23812398

papers, including those addressing analogues of the problem for structures such as digraphs and finitegeometries, counts hundreds of papers. For history and details we refer to the relatively recent surveypaper by Miller and the second author [11]. In what follows we only summarize the facts essential forour contribution.

A simple spanning tree argument[9] shows that the order of a graph of maximum degreedanddiameterkcannot exceed 1+d+d(d1)+ +d(d1)k1. This quantity is theMoore boundand graphs of order equal to the Moore bound are known as the Moore (d, k)-graphs. Ifk= 1 the Moore(d, k)-graphs are the complete graphs of orderd+1 for anyd 1. The other trivial cases are d = 1andd = 2 where the Moore graphs are the complete graph of order two and cycles of length 2k+1for anyk 1. Existence of Moore(d, k)-graphs ford 2 andk 2 turned out to be a hard problemthat stimulates research until now. In[9] the authors proved that fork= 2, Moore(d, k)-graphs existonly ifk = 2 andd = 2, 3, 7, and possibly 57, and that there are no Moore(d, k)-graphs fork = 3. Inthe first three cases the Moore graphs are unique, namely, the 5-cycle, the Petersen graph, and theHoffman-Singleton graph. Non-existence of Moore (d, k)-graphs fork 4 was proved independentlyby Bannai and Ito [2] and Damerell[5]. All these papers were based on an extensive use of methods oflinear algebra and spectral methods in particular.

It is striking that all the known Moore (d, k)-graphs turn out to be vertex-transitive; in fact, theirautomorphism groups have rank 3. This has naturally led to investigation of symmetry properties ofthe hypothetical(57,2)-Moore graph(s). For brevity, letbe a Moore(57, 2)-graph and letGbe itsautomorphism group.The study ofG was initiated by Aschbacher [1] by proving that G cannotbearank3 group. Later in a series of lectures for his graduate students, Graham Higman showed that cannotbe vertex-transitive; see Camerons monograph[3]for an account of the proof. The same argumentshows that the order ofGis not divisible by 4. This was taken further by Makhnev and Paduchikh [10]by a closer investigation of the structure ofG, assuming thatGcontains an involution. A consequenceof their investigation is the bound |G| 550 ifGhas even order.

The aim of this contribution is to further investigate possible symmetries (in conjunction withother properties) of the Moore(57, 2)-graph(s) the existence of which is still in doubt. Using a blend

of spectral, group-theoretic, combinatorial, and computational methods we show that |G| can assumeonly a very restricted set of values. In particular, we obtain the inequality |G| 375 with no restrictionon the parity of|G|.

The structure of the paper is as follows. The known properties of a Moore (57,2)-graph , withemphasis on symmetries, arereviewedin Section 2. InSections 3 and 4 wepresentaselectionofmatrixmethodsusedininvestigationofautomorphismgroupsofgraphsandproveanumberofgeneralresultstailored to our needs. In the interest of clarity, Section5contains arithmetic restrictions on charactersand other related invariants of automorphisms of . In Sections6,7and8we derive upper boundson orders of Sylow p-subgroups of the automorphism groupG offor odd primes. Finally, usingsolvability ofG, in Section9we combine our results to determine possible orders ofG. We concludethe paper by a handful of remarks.

In our study we use a wide range of methods. Attempting to keep the length of the paper withinreasonable bounds we decided to omit definitions of some concepts that are considered generallyknown. We refer the reader to the monographs of Curtis and Reiner[4] for character theory, Godsil[8] for equitable partitions, Rotman [13]for general group theory, and Dixon and Mortimer [6] forpermutation groups.

Groups of small order appearing in our investigation are described in form of direct and semi-directproducts, except for groups of order 81 and 625 the description of which comes from the SmallGrouplibrary of GAP[7].

2. The missing Moore graph: State-of-the-art

Throughout the paper we let denote a Moore (57, 2)-graph with adjacency matrix A. Generalgraph-theoretic terms will, in most cases, be used with reference to ,andtoA whenever appropriate.We begin with listing a selection of basic properties ofthat can be extracted from [9].

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Proposition 1. The graph has3250vertices and girth5.Its adjacency matrix A satisfies the equationA2 +A56I=J,where J is the all-one matrix.Consequently,its eigenvalues are 57, 7,and 8,withmultiplicities1, 1729,and1520,respectively.

Any Moore graph of diameter 2 hasgirth 5 and therefore such graphs have the property that anytwovertices have 0 or exactly 1 common neighbor according as they are adjacent or not. That is, Moore graphsof diameter 2 are exactly the strongly regular (0, 1)-graphs. If the regularity condition is omitted, then,apart from the trivial cases of an empty graph, an isolated vertex, or an isolated edge, the only othergraphs with the above property are the starsK1,nforn 2.

ThestartingpointinthestudyofautomorphismsofMooregraphswasanobservationofAschbacher[1] regarding fixed-point subgraphs. For a groupXof automorphisms oflet Fix(X)be the subgraphinduced by the set of all fixed points ofX.

Lemma 1. Let X be a group of automorphisms of.Then, Fix(X)is empty,an isolated vertex,a pentagon,the Petersen graph, the Hoffman-Singleton graph,or a star K1,nfor some n 1.

Since has diameter 2, the imagevx of any vertex v under any automorphismx of is eitherv itself, or a neighbor ofv , or else a vertex at distance two from v. The following three numericalinvariants related to this observation have turned out to be extremely useful. For =0, 1, 2 letai(x)=|{v ; d(v, vx)= i}| where d standsforthedistance.ThemainreasonofsuccessofspectralmethodsinthestudyofMooregraphsis,infact,acloserelationbetweenthefunctions aiand spectral propertiesof automorphisms of, which we make explicit in Theorem1.

Every permutationxof the vertex set of our Moore (57, 2)-graphcan be represented in the formof a permutation matrixPx . As it is well known,xis an automorphism ofif and only if

PxA= APx, (1)

whereA is the adjacency matrix of. This induces a natural linear representation ofG = Aut()in the vector space generated by vertices of , of dimension 3250, as well as on the eigenspacescorresponding to the three eigenvalues of. By the fundamental observation of Higman, charactersof these representations are closely related to combinatorial properties of automorphisms of.

Theorem 1(Higman; see [3]).Letbe a Moore(57, 2)-graph with the adjacency matrix A.Let V0,V1,V2be the eigenspaces of A for eigenvalues57, 7,and 8,respectively.Let X be an automorphism group ofand let0,1and 2be characters of the restriction of X onto V0,V1and V2,respectively.As before,

for x X let ai(x)= |{v ; d(v, vx)= i}|,i = 0, 1, 2.Finally,let P=

1 1 157 7 8

3192 8 7

and Q =

13250 1 1 1

1729 637/3 13/31520 640/3 10/3

.Then,Q=P1 and(0(x), 1(x), 2(x))T =Q(a0(x), a1(x), a2(x))T.

We recall that Higman demonstrated power of this theorem by showing that it implies that cannot be vertex-transitive; see [3]. We will make use of the following two consequences.

Lemma2 [3]. Letxbeaninvolutoryautomorphismof. Then, a0(x)= 56 anda1(x)= 112. Consequently,the order of Aut()is not divisible by4.

Lemma 3. For any x X we have

1(x)= 1

15(8a0(x)+a1(x)65)

and therefore

a1(x) 7a0(x)+5 mod 15.

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Proof. The first statement follows from the fact thata0(x)+a1(x)+a2(x)= 3250. The congruenceis a consequence of the fact that the algebraic integer 1(x)is, according to the first statement, also arational number and therefore an integer.

Makhnev and Paduchikh[10] investigated possible groups of automorphisms ofof even orderand obtained the following result.

Theorem 2 [10, Theorem 1]. Letbe a Moore (57,2)-graph and let G = Aut().Assume that G containsan involution t.Then the following statements hold:

(a) G=Yt X for some subgroups X and Y of odd order,Y is inverted by t,and either|Y| divides5or57,or|Y| divides21,

(b)if X /= 1,thenFix(X)can be one of the following:a star(Y=1 and |X| =7);a pentagon,inwhich case|Y|divides5 and |X|divides55;the Petersen graph,in which case |Y|divides3 and|X| divides 27; and finallythe Hoffman-Singleton graph, in which case Y divides 5 or7 and X divides

25.

The proof of this result is divided into a series of lemmas dealing with properties of involutoryautomorphisms of . The series is preceded with the following more specific version of Lemma1, afurther extension of which is one of the cornerstones of our investigation.

Lemma 4. Let X be a group of automorphisms ofof odd prime order.Then,one of the following holds:

(1) Fix(X)is empty and|X| divides135;(2) Fix(X)is a singleton and|X| divides319;(3) Fix(X)is a star with|Fix(X)| =2 +7l and|X| divides7;(4) Fix(X)is a pentagon and|X| divides115;

(5) Fix(X)is the Petersen graph and |X| divides3;(6) Fix(X)is the Hoffman-Singleton graph and|X| divides5.

3. Equitable partitions

Equitable partitions are a useful tool in spectral analysis of graphs, see [8]. To the best of ourknowledge, this tool has not been used in connection with Moore (57,2)-graph(s). Since we are onlyinterested in such graphs we will introduce the concepts related to equitable partitions just for thisspecial case.

Letbe a Moore(57, 2)-graph and let S= {S1, S2, . . . , Sk}be a partition of the vertex set of .We say that Sis anequitable partitionofif there exist integersb

ijsuch that each vertex fromS

ihas

exactlybijneighbors inSj. We say that the matrixB= (bij)kkis theadjacency matrixofS.The most important but not the only instances of equitable partitions come from orbits of auto-

morphism groups. For example, a different kind of equitable partition ofis obtained by taking, for

any vertexv, the partition {{v}, N(v), \(Nv {v})} whose adjacency matrix is

0 57 01 0 560 1 56

.

Spectral properties of the adjacency matrix of an equitable partitionof a graph in general are closelyrelated to the spectral properties of the adjacency matrix of the graph. In our special case, the propertyof being a Moore graph also manifests in the properties of the equitable partitions.

Lemma 5. LetS= {S1, S2, . . . , Sk} beanequitablepartitionofaMoore (57, 2)graph suchthat|Si| =si

and let B= (bij)be the adjacency matrix ofS.Then(1) s ibi,j =sjbj,i;(2) 57is an eigenvalue of B with an eigenvector(1,1, . . . , 1)T;(3) the characteristic polynomial of B divides(x57)(x7)1729(x+8)1520;

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(4)coefficients bkijof Bk are numbers of k-walks from a vertex of Siinto Sj;

(5)B2 +B56I=(1,1, . . . , 1)T(s1, s2, . . . , sk).

Proof. Items 14 follow directly from the general theory, cf. [8], and item 5 can be derived from 4 ina similar way as the equationA2

+A

56I

=Jfor the adjacency matrixAof

.

LetXbe an automorphism group of a Moore(57,2)-graphand letB = (bi,j)be the adjacencymatrix of the equitable partition formed bythe orbits ofX.Forbrevitywewillsaythat B isthe adjacencymatrixofX. Moreover, for any matrix invariant ofBwe will say that the invariant is a property ofX.That is, we will be speaking about the traceofX,eigenvaluesofX, an so forth. Finally ifOiis theithorbit ofXwe will say that the numberbi,iis thetraceofOi. In general, we define the traceof a set ofverticesSto be the average degree of the subgraph induced byS.

Letxbe an element ofXand letObe an orbit ofX. We say thatx contributestoOif for some vertexv Othe vertexvx is adjacent tov; in symbols,vx v.

Lemma 6. Let O be an orbit of X and let x X contribute to O.Then(1)x1 contributes to O;(2) if|X| is odd,thenTr(X)is even;(3)if x is central in X,thenTr(O) 2;(4) Tr(O)2

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Proof. By[12,Lemma 4.1] we have

(577)|S|(3250 |S|)

3250 e(S, \S)(57+8)

|S|(3250 |S|)

3250 ,

wheree(U, V)is the number of edges between the setsUandV. The result now follows from the factthate(S, \S)= |S|(57Tr(S)).

Corollary 1. For any x X we have a1(x) 500.

Proof. LetS= {v ; v vx}. Clearlya1(x)= |S|. If for someu, v S we havevx /= u v /= ux ,

then we have ina quadrangleu v vx ux u. Therefore Tr(S) 2 anda1(x)= |S| 50(8+2)= 500.

4. Characters

Because of the fact thathas integral eigenvalues, all its eigenspaces have bases over the field ofrational numbers. Linear representations ofover these eigenspaces are therefore rational represen-tations. Although the field of rational numbers is not algebraically closed, thanks to the fact that itscharacteristic is zero onestill may use Maschkes Theorem (see [4]) on decompositions into irreduciblerational representations. We would like to emphasize that byirreducible rational representationswemean rational representations irreducible over the field of rational numbers.

Besides Maschkes Theorem we will be using the following properties. Let us recall that two ele-mentsx,yof a groupHwill be said to belong to the samerational classofHif and only if the cyclicgroupsx and y are conjugate subgroups ofH.

Theorem 3[4].Let H be a finite group.Then,any rational representation of H is constant on all rationalclasses of H and the number of irreducible rational representations of H is equal to the number of rationalclasses of H.

Proposition 2. Let H be a finite group and let x1,x2, . . . ,xube representatives of rational classes of H.LetR1, R2, . . . , Rube irreducibleQ-representations of X with characters r1, r2, . . . , ru.Then,for any rationalrepresentation R of H with character the system of linear equation with the matrix

r1(x1) r2(x1) . . . ru(x1) | (x1)r1(x2) r2(x2) . . . ru(x2) | (x2)

.

.....

.

..... | ...

r1(xu) r2(xu) . . . ru(xu) | (xu)

has a solution in non-negative integers.

Proof. It is sufficient to realize that if a decomposition ofRinto irreducible rational representationscontainsnicopiesRifor 1 i u, then =n1r1+n2r2+ +nuru.

Combining Theorems3and1we immediately obtain:

Lemma 11. Let X be an automorphism group of a Moore(57,2)-graph.Then,the functions a0

,a1

,anda2are constant on rational classes of X.

Proof. It is sufficient to observe that the values ai(x) are linear combinations ofj(x), which arecharacters of rational representations ofX.

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In the next section we will apply Proposition2to the representation ofXwith the character 1from Theorem1to specify values of the functiona1.

We conclude with an interesting observation. By Proposition2,decomposition of a rational repre-sentation into irreducibles is determinedby its character. The decomposition of the representation ofX

onis determined by the functiona0. This means that if one knows all permutation representationsofXtogether with their decompositions, then the values ofa0 impose stronger restrictions on thestructure of orbits ofXon than the ones impled by the orbit counting lemma. Although we did notuse this observation in our paper, we believe that it can be helpful in further research.

5. Tables

In previous sections we derived algebraic tools for obtaining restrictions on values ofa0,a1anda2. In this section we will apply the tools for cyclic groups. There are two reasons for this restriction.Firstly, results for cyclic groups are sufficient for our purposes. Secondly, non-commuting elementsinfluence each other to a much smaller extent than commuting ones, as one can observe already on

non-Abelian groups of orderpq.

Lemma 12. Let x be an automorphism of a Moore (57, 2)-graph of prime order p. Then, the values a1(x)and1(x)satisfy:

a0(x) p a1(x) 1(x)

0 5 50+75k 500 1+5k0 13 65+195k 500 13k1 3 27+45k= 0

1 19 57+285k

500 19k5 5 10+75k 500 1+5k5 11 55+165k 500 2+11k10 3 0 150 5 25+75k 350 24+5k56 2 112 332 7 49+105k 500 7k9 7 98+105k 500 7+7k16 7 42+105k 500 7+7k23 7 91+105k 500 14+7k30 7 35+105k 500 14+7k

37 7 84+105k 392 21+7k44 7 28+105k 260 21+7k51 7 77 2858 7 21+105k 0

In particular,the starred cases p = 3,a0(x)= 1and p = 7,a0 =58cannot occur.

Proof. By Lemma3and Theorem1, we havea1(x)= 7a0(x)+5+15lfor somel Z and1(x)=4+a0(x)+l. By Proposition 2 applied to the cyclic group x, the parameter l must be such that the

system

1 p1 | 17291 1 | 1(x)

hasasolutioninnon-negativeintegers.Valuesinthetablearepresented

in such a way thatkis non-negative. The general bounda1(x) 500 follows from Corollary1.

The value ofa1(x)for p = 2 is taken from [3]. Ifp = 3 anda1(x) >0 then would contain atriangle, a contradiction. Ifp= 7 and a0(x)= 2 +7m, then there are only 56(8m) orbits of size7 not connected with Fix(x)and x can contribute to at most one third of them. The arguments aresimilar for the casesp= 19,a0(x)= 1, andp= 5,a0(x)= 50.

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Lemma 13. Let x be an automorphism ofof order p2 where p= 3 or p= 5.Then,the values a1(x),a1(x

p),andTr(x)satisfy:

p a0(x) a0(xp) a1(x) a1(x

p) Tr(x)

5 0 0 50+75k 125+375l 60+60k+60l 2605 0 50 50+75k 100 56+60k 2765 5 50 10+75k 100 24+60k 2445 50 50 25+75k 100 36+60k 563 1 10 27+45k 0 18+30k3 10 10 45k 0 30k

Proof. By Lemma 3 and Theorem 1 we have a1(x)= 7a0(x)+5+15l and 1(x)= 4+a0(x)+l

for some l Z. By Proposition 2 applied to the cyclic group x, the system

1 p1 p2 p | 17291 p1 p | 1(x

p)

1 1 0 | 1(x)hasasolutioninnon-negativeintegers.Solvingthesystemweobtain

the table:

p a0(x) a0(xp) a1(x) a1(x

p) 1(x) 1(xp) Tr(x)

5 0 0 50+75k 125+375l 1+5k 4+25l 60+60k+60l5 0 5 50+75k 85+375l 1+5k 4+25l 33,6+60k+60l5 0 50 50+75k 100+375l 1+5k 29+25l 56+60k+60l5 5 5 10+75k 85+375l 1+5k 4+25l 21, 6+60k+60l5 5 50 10+75k 100+375l 1+5k 29+25l 24+60k+60l5 50 50 25+75k 100+375l 24+5k 29+25l 36+60k+60l3 1 10 27

+45k 0

2

+3k 1 18

+30k

3 10 10 45k 0 1+3k 1 30k

The lines marked by a star require non-integral traces and therefore cannot occur.If|Fix(x5)| =50, then 2500 vertices are adjacent to Fix(xp)and thereforea1(x

5) 700/2.Finally, every orbit of size 25 adjacent to Fix(x) has trace equal 0, which implies the upper bound

on Tr(x).

Remark.Later in Proposition3we will show that the casea0(x)= 50 cannot occur.

Lemma 14. Let X=PQ be an automorphism group of such that P (Q) acts semi-regularly on

\Fix(P)( \Fix(Q)) and(|P|, |Q|)= 1.Then,for any central element x X,

a1(x) b1(x)mod |X|, (2)

whereb1(x)= |{v Fix(P)Fix(Q);v vx}|. Moreover, if x= xPxQforxPP,xQ Q, thenb1(x)=

b1(xP)+b1(xQ).

Proof. AsXacts semi-regularly on \(Fix(P)Fix(Q)), the congruence (2)follows from Lemma6.SincethekerneloftheactionofXon Fix(P) (Fix(Q))isexactly P(Q),wehave b1(x)= b1(xP)+b1(xQ).

Lemma 15. Let x be an automorphism ofof order pq, where p q are primes.Assume that the values pq,a0(x),a0(x

p),and a0(xq)are as in the first four columns of the table below.Then the values a1(x),a1(x

p),a1(x

q),andTr(x)satisfy:

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pq a0(x) a0(xp) a0(x

q) a1(x) a1(xp) a1(x

q)Tr(x)

6 2 10 56 4+90k 0 112 20+30k10 1 5 56 102+150k 10+150l 112 56+60k+60l

10 6 50 56 62+150k 100+150l 112 76+60k+60l14 7 9 56 84+210k 98+210l 112 86+90k+90l 40014 14 16 56 28+210k 42+210l 112 38+90k+90l 34414 21 23 56 182+210k 196+210l 112 170+90k+90l 28814 28 30 56 126+210k 140+210l 112 122+90k+90l 23214 35 37 56 70+210k 84+210l 112 74+90k+90l 17614 42 44 56 14+210k 28+210l 112 26+90k+90l 12014 49 51 18222 1 5 56 222 220 112 20615 0 0 10 5+75k+225m 50+75k 0 16+60k+120m35 1 16 50 42+105k 147+105k 175 74+90k 18635 1 51 50 252 77 175 206 186

55 5 5 5 55 55 385 7865 0 0 0 650

Moreover,the cases marked by a star cannot occur.

Proof. We need to solve the system

1 p1 q1 (p1)(q1) | 17291 1 q1 1q | 1(x

q)1 p1 1 1p | 1(x

p)1 1 1 1 | 1(x)

in non-negative

integers. Existence of an integral solution is equivalent to the conditions

p|1(xp)1(x), (3)

q|1(xq

)1(x), (4)pq|17291(x

p)1(xq)+1(x). (5)

After solving the system with the help of Lemmas12and14and applying the arguments used inthe proof of Lemmas12and13we obtain the table. Upper bounds on traces follows from facts thatevery orbit in Abelian group has trace at most 2 and that every orbit connected to Fix(x)has traceequal to 0.

As an illustration we show details for the casepq = 35,a1(x)= 1,a1(x5)= 16 anda1(x

7)= 50.FromthepropertiesoftheautomorphismgroupoftheHoffman-Singletongraphweobtain b1(x

5)= 7,b1(x

7)= 0 andb1(x)= 7. Further, Lemmas12and 14yielda1(x5)= 42 +105l, 1(x

5)= 7 +7l,a1(x

7)= 175, and1(x7)= 34. Similarlya1(x)= 42 +105kand 1(x)= 1+7k. Moreover, the

condition (3)impliesl = 1 +k+5m, that is,a1(x5

)= 147+105k+525m(observe that the onlysolution withm /= 0 isk = 4 andm= 1).The groupX= xhas on the subgraph Fix(x5)Fix(x7)one orbit of size 1, 3 orbits of size

5 and 6 orbits of size 7. Out of these orbits exactly three, of size 7 each, have trace equal to 2 whilethe others have trace equal to 0. The remaining points ofform 91 orbits of size 35, one of which isadjacent to Fix(X). Therefore Tr(X) 6+902 = 186, which impliesm= 0 andk 1.

6. p-Groups of automorphisms

IntheirproofofLemma 4, Makhnev and Paduchikh exploited the observation thatnon-trivial orbitsof an automorphism of a prime orderphave lengthpand their argument is based on counting orbitsaround a suitably chosen fixed vertex. Realizing that the size of a non-trivial orbit of somep-group is apower ofp, one can not only extend Lemma4top-groups but also derive an upper bound on the sizeof ap-group with a given set of fixed points. Our findings are divided into four statements. Since theproof techniques are repetitious we only include a proof for Lemma18.

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Lemma 16. Let X be a group of automorphisms ofsuch that X is a p-group for some odd prime p.Then,one of the following holds:

(1) Fix(X)is empty and p {5, 13};(2) Fix(X)is a singleton and p {3, 19};

(3) Fix(X)is a star with|Fix(X)| =2 +7l and p = 7;(4) Fix(X)is a pentagon and p {5, 11};(5) Fix(X)is the Petersen graph and p = 3;(6) Fix(X)is the Hoffman-Singleton graph and p = 5.

Lemma 17. Let X be a an automorphism group ofof order3k.Then one of the following holds:

(1) Fix(X)is the Petersen graph and |X| divides27;(2) Fix(X)is a singleton and|X| divides81.

For any vertexvofwe letN(v)denote the set of the 57 neighbors ofv.

Lemma 18. Let X be a group of automorphisms such that|X|is a5-group.Then one of the followingholds:

(1) Fix(X)is the Hoffman-Singleton graph and|X| divides25;(2) Fix(X)is a pentagon and|X| divides125;(3) Fix(X)is empty and|X| divides56.

Proof. (1) Leta Fix(X). If there was an orbit O in N(a)\Fix(X)such that|O|< |X|, then for anyelemento O the point stabilizerXo ofo would be a proper subgroup ofXwith|Fix(Xo)| > 50, acontradiction. Therefore,Xact semi-regularly onN(a)\Fix(X)and |X| divides|N(a)\Fix(X)| =50.

(2) Leta Fix(X). Since|N(a)\Fix(X)| =55, the groupXhas onN(a)\Fix(X)an orbitOof size5. LetYbe a point stabilizer in this orbit. IfY /= X, then Fix(Y)is the Hoffman-Singleton graph and

by (1) we have |Y| 25 and|X| =5|Y| 125.(3) As 3250 125 mod 625, the smallest orbit ofXon has size at most 125. Let Ybe a point

stabilizer in this orbit. IfY /= X, then Fix(Y)is a pentagon or the Hoffman-Singleton graph and by (1)or (2) we have|Y| 125 and|X| =125|Y| 56.

Lemma 19. Let p> 5be a prime and let X be a group of automorphisms ofof order pk.Then one of thefollowing holds:

(1) Fix(X)= and X=Z13;(2) Fix(X)is a singleton and X=Z19;(3) Fix(X)is a pentagon and X=Z11;(4) Fix(X)is a star on2+7l vertices and X=Z7;

(5) Fix(X)is an edge and X=Z7 Z7.

With the help of further trickery, in the next section we will push down the bound on the order ofa 3-groupXacting onfrom Lemma17in the case when Fix(X)is a singleton. Likewise, in Section8we will decrease the upper bound on orders of 5-groups acting on for any possible fixed subgraphs.

Theorem 4. Let X be a group of automorphisms ofof order3k.Then,k 3.

Theorem 5. Let X be a group of automorphisms of oforderapowerof5. Then one of the following holds:

(1) Fix(X)is the Hoffman-Singleton graph and|X| divides5;

(2) Fix(X)is a pentagon and|X| divides25;(3) Fix(X)is empty and|X| divides125.

The proofs will require extension of our methods by considering a few general tools regarding groupactions on sets.

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8. Proof of Theorem5

We begin with auxiliary results some of which are of independent interest.

Proposition 3. Let X be a group of automorphisms of a Moore (57,2)-graph of order a power of5.IfFix(X)is the Hoffman-Singleton graph,then |X| 5.

Proof. Let us assume that |X| =25. The semi-regular action ofXon \Fix(X)gives 50 orbits of size1 and 128 orbits of size 25. In the neighborhood of any fixed point ofXthere are exactly two orbitsof size 25 and both have trace equal to 0. As Xis Abelian and has odd order, each of the remaining 28orbits of size 25 has trace equal to 0 or 2. The trace ofXtherefore does not exceed 56. Moreover, thetrace is even and congruent to 8168 6 mod 15. It follows that at least one of these 28 orbits hastrace equal to 0.

Let us order the orbits ofXin such a way thatO1, . . . , O50are the fixed points ofX, points inO50+iand O100+ilie in the neighborhood ofOi(i= 1,2, . . . , 50),andlet O178be an orbit with zero trace. Let

B= (bi,j)be the adjacency matrix ofX. By Lemma5,Bsatisfies the equalityB2 +B56I=(1,1, . . . , 1)T(s1, s2, . . . , sk). (6)

We will analyze the way the entry in the bottom right corner in the matrix on the right-hand side ofthis equality is obtained.

We begin with looking at the entries b178,j. Since O178is not connected to Fix(X),wehave b178,i =0for 1 i 50. Letibe a vertex of Fix(X), 1 i 50. Sinceis a Moore graph, every vertex inO178isadjacent to precisely one vertex in O50+i O100+i.Thismeansthat b178,50+i+ b178,100+i =1 for everyi= 1,2, . . . , 50. As every vertex inO178has degree 57 andb178,178 =0, we have

177

i=151

b178,i =7. (7)

By semi-regularity of the action ofXon \Fix(X)it further follows thatb178,j =bj,178. Substitutingall this information into (6)yields

177i=151

b2178,i =31. (8)

Eqs. (7) and (8), have, however, no simultaneous solution in non-negative integers.

This proves part (1) of Theorem5.Part (2) follows from (1) in much the same way as in Lemma18.The third part will be proved by demonstrating non-existence of a group of order 625. Details of

the proof depend on the size of the smallest orbit. Note that part (2) of Theorem5implies that thesmallest orbit size cannot be equal to 5.

Lemma 22. If X is a group of automorphisms ofof order625and with the smallest orbit size25,then Xis isomorphic to SmallGroup(625, 12)of the GAP library.

Proof. LetO be an orbit ofXof size 25 and let Xo be the vertex stabilizer of a vertex o O. Then|Xo| =25 and, by Proposition3, |Fix(Xo)| =5. Therefore Fix(Xo) O,Xohas five conjugates inXandOis unique orbit with stabilizerXo. In particularXis not Abelian.

Letp O \Fix(Xo). Then|Xop| =5, therefore Fix(Xop)is a copy of the Hoffman-Singleton graphandO Xop. The remaining points of Fix(Xop)either form another orbit of size 25, or there exists an

orbit of size 125 such thatXopis a stabilizer of a vertex in this orbit.IfCore(O)= Xop, thenXopis normal inXand cannot be the vertex stabilizer of an orbit of size 125.Therefore there exists a unique orbitO /= OwithCore(O)= Core(O)= Xop.

There are 10 non-Abelian groups of order 625, their indexes in the SmallGroup library being3, 4, 6, 7, 8, 9, 10, 12, 13, 14. Groups 6 and 14 are excluded since they have only normal subgroups of

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order 25. In groups 3, 4, 9, 10, 12, 14 all non-normal subgroups of order 25 have core of size 5. Thenumber of cores must be a multiple of 5, which is true only for the groups 3 and 12, in both cases thereare five cores. However in the group 3 one core lies in six classes of groups and remaining four coreslies in 1 class each, whereas we need every core to lie in two conjugacy classes of groups of order 25

(which holds in the group 12).It remains to exclude groups 7 and 8. In these groups there are six conjugacy classes of subgroupsof order 25 with core of size 5; moreover, here the core is always the centre of the group. There arealso five classes with trivial core. Therefore there are exactly five orbits of size 25 and the number oforbits with a non-trivial core is either 0 or 2.

In both cases the centre has order 5 and the Frattini subgroup is elementary Abelian of order 25and contains the centre. Iffis a non-central element from the Frattini subgroup, then every conjugacyclass of subgroups of order 25 with trivial core has a subgroup containing f, andfhas five conjugatesinX.

It follows that in every orbit O of size 25 with a trivial core there exists an element o fixed byfand |Fix(f)O| =5. Iffgenerates the stabilizer of a vertex in an orbit of size 125, then it fixes25 elements of this orbit. Hence the possibility that Xhas exactly three orbits of size 5 with a trivialcore cannot occur. Thus, there are five orbits of size 25, and all stabilizers have trivial core (and aredistinct).

LetO be an orbit of size 25 with a trivial core. The conjugacy classes of subgroups consisting ofstabilizers of vertices fromOcomprise five groups any two of which intersect non-trivially while anythree have trivial intersection. This implies that for any pair of pointso,p O such thatp /Xo, thesubgroupXopfixes a Hoffman-Singleton graph and Fix(Xop)contains exactly 10 elements ofO. AsXophas five conjugates inX, it cannotbe the vertexstabilizer of an orbit of size 125 (it would fix 25 points ofthis orbit and the equation 50 = 25a+10bhas no solutions inN). ThereforeXopfixes 10 points fromevery orbit of size 25. Consequently, the system of subgroupsXop, which is the system of intersectionsof pairs of vertex stabilizers of elements in O,donotdependonthechoiceofO. However, computationsin GAP shows that this is not the case.

Proposition 4. The groupSmallGroup(625, 12) cannot act as an automorphism group of a Moore (57, 2)-graphwith the smallest orbit of size 25.

Proof. Let us assume that X= SmallGroup(625, 12)acts as an automorphism group ofwith thesmallest orbit of size 25.

First we sum up some fact about the groupX, which has in GAP a polycyclic presentation in fourgeneratorsf1,f2,f3andf4. In this presentation, the centre ofXis generated byf3andf4and the Frattinisubgroup ofXis generated byf4. Every nonidentity element has order 5 and every non-central elementhas five conjugates. Consequently, any vertex stabilizer of an orbit of size 125 must fix 25 elements inthis orbit and 50 elements in total.

Thereare30conjugacyclassesofnon-normalsubgroupsoforder25.Onecanchooserepresentativesin such a way that each representative is generated by a pairu andv of elements whereu U=f1,f2,f1f2,f

21f2,f

31f2,f

41f2

and v Vf3,f3f4,f3f

24,f3f

34,f3f

44

. In all cases the core is generated byv .

Therefore we have 10 orbits of size 25 and, for every v V, two orbits have core generated by v. Bythe pigeonhole principle at least one point ofUlies in at least two representatives.

If some elementu Ulies in two representatives, then it fixes 50 elements, five in each orbit ofsize 25 and 25 in each orbit of size 125 in which it fixes a point. Thereforeulies in five representatives

and also fixes a point in exactly one orbit of size 125. The conjugates ofu are

u, uf4, uf24, uf

34, uf

44

,

therefore the elementsuf3,uf23,uf

33,uf

43 have the same property asu, i.e., they fix a point in five orbits

of size 25 and in one orbit of size 125. Thus Xhas at least five orbits of size 125.

Let O1be the orbit of size 125 in which u fixes some point and let Ojbe a different orbit. If|Oj| =25and u fixessomepointin Ojthen b1j =1, otherwise b1j =0.If|Oj| 125, then u fixesnopointin Ojand5|b1j. Moreover

jb1jbj1+b1156 = 125 and

jb1j =57. Computations in GAP show, however,

that such entriesb1jdo not exist.

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Lemma 23. Let X be an automorphism group ofof order625with the smallest orbit of size125.ThenX contains a subgroup Y of order5which is a vertex stabilizer in at least one and at most two orbits of size125.

Proof. LetO be an 125-element orbit ofXandYbe the vertex stabilizer of a vertex o O. The pair(|Fix(Y)|, |Fix(Y)O|)is one of(5, 5), (50,5), (50, 25). Depending on(|Fix(Y)|, |Fix(Y)O|),Yis avertex stabilizer of exactly 1, 10, and 2 orbits, respectively. The statement now follows from the factthat the number of orbits of size 125 is congruent to 1 mod 5.

Lemma 24. Let X have order625 and let the smallest orbit of X have size 125.Then X has at least twoorbits of size125.

Proof. Letx Xbe a central element of order 5. By Lemma12,a1(x) >0. Thereforexcontributes toat least one orbitO. Sincexis central, Lemma7shows that|O| a1(x), anda(x) 500 by Corollary1.The orbitOtherefore cannot have size greater than 500. Moreover,xandx2 contribute to different

orbits.

Proposition 5. The graph does not admit a group of automorphisms of order625 with the smallest orbitsize125.

Proof. Let us assume that|X| =625 and thatXhas a total ofkorbits numbered in such a way thatorbits O1, O2, . . . , Oihavesize625, Oi+1, . . . , Okhave size 125, and Okis an orbit whose vertex stabilizerYfixes elements from at most two orbits; its existence is guaranteed by Lemma23).

Letbk1, bk2, . . . , bkkbe thekth row of the adjacency matrixB ofX. Since each of the firsti orbitshas size 625 we have 125bkj =625bjkfor 1j i. IfOjis an orbit of size 125 such thatYdoes not fixa vertex inOjand ifo Okis fixed byY, thenYacts semi-regularly on the neighbors ofoin the orbit

Oj , that is,bkj is a multiple of 5. Hence, 5 divides all the entries bk1, bk2, . . . , bkiand all but at mostone entries amongbkjwherei+1 j k 1. Considering the bottom right entry in the equationB2 +B56I=(1,1, . . . , 1)T(s1, s2, . . . , sk)from Lemma5we obtain

b2k1+ . . .+b2ki

/5+

b2k,i+1+ +b

2kk

+bkk =181. (9)

Ifi 3, then computations shows that suchbs do not exist. Ifi = 4, then there are six unorderedpossibilities for a row of the adjacency matrix for an orbit of size 125 (in all cases on the diagonal is 2).

{b1, b2, b3, b4} {b5, b6, b7, b8, b9, b10}

{10,10,5,5} {5,5,5,5,5,2}{15,10,5,5} {5,5,5,5,2,0}{15,15,5,5} {5,5,5,2,0,0}{20, 10, 10, 5} {5,5,2,0,0,0}{20, 15, 10, 5} {5,2,0,0,0,0}{20, 15, 15, 5} {2,0,0,0,0,0}

According to the table we will say that orbit of type 125 is of type 1, 2, 3, 4, 5 or 6.By Lemma10,the trace of an orbit of size 625 is an even integer from {6,8,10,12,14}. The trace of

Xis congruent to 0 mod 60 and every orbit of size 125 has trace equal to 2. Therefore Tr(X)= 60 andsum of traces of the orbits of size 625 equals 48.

IfXcontains an orbit of type 1 or 6, then it is impossible to find four rows of adjacency matrixcorresponding to orbits of size 625. For type 2, no solution has sum of traces equal to 48. For types 3and 5, every solution with sum of traces equal to 48 requires another orbit of type 6. Finally, for type4, every solution with sum of traces equal to 48 requires at least one orbit of another type.

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9. Mixing the primes

Our goal specification of possible automorphism groups of a Moore (57,2)-graph of odd order is facilitated by two classical results in group theory. By the Odd Order Theorem of Feit and Thompson,

any groupXunder our consideration is solvable. Thus, by the result of Philip Hall (see [13]), whenever|X| =ab for relatively primea and b,Xhas a subgroup of order a. This enables us to approach thegroupsXfrom below in the sense of the number of prime divisors.

The next step in our investigation is considering groups of automorphisms ofof non-prime-power order, in particular, of orderpaqb for distinct odd primesp, qand fora, b 1. We recall that aSylowp-subgroup of a groupXof orderpau with (p, u)= 1 is any subgroup ofXof orderpa. By SylowsTheorem, the number of Sylowp-subgroups ofXis congruent to 1 modpand dividesu.

In what follows letXbe an automorphism group ofsuch that|X| =paqb wherep, qare distinctodd primes anda, b 1. Further, letPandQbe Sylowp- andq-subgroups ofX, respectively.

Lemma 25. If P is normal in X,then Q acts on orbits of P.In particular,Q acts as an automorphism group

ofFix(P).

Lemma 26. In the above notation,p 5or q 5.

Proof. Supposethatp, q {7, 11, 13, 19}. Recall that, by Sylows theorem, the number of distinct Sylowp-subgroups of a group of orderpaqb is a power ofqcongruent to 1 modp; a similar remark applies toSylowq-subgroups. Considering all possibilities listed in Lemma19it follows that bothPandQmustbe normal inX, that is,Xis the direct product ofPand Q. ThereforePacts on the set of fixed pointsofQ and vice versa. This is clearly possible only ifp = 7, q = 19,P=Z7, and|Fix(P)| =58, whichcontradicts Lemma12.

Proposition 6. Let p = 3and q= 5.Then QX.Moreover,

(1) if PX,then |Fix(P)| =10, |Fix(Q)| =0,and|Q| =5;(2) if PX,then |P| =3and Q {Z25,Z

35,Z

25 Z5}.

Proof. Since |P| divides 27, normality ofQfollows from Sylows Theorem.Now, assume thatPX, that is,X=PQ, and consider a cyclic subgroupYXof order 15. Let

P andQ be the Sylow subgroups ofY. Then, Fix(P)is the Petersen graph and is a union of orbits ofQ. Similarly, Fix(Q)is a union of orbits ofP. As an automorphism of order 5 of the Petersen graphacts semi-regularly on vertices, we have just two possibilities: either Fix(Q)is empty, or Fix(Q)isthe Hoffman-Singleton graph and Fix(P) Fix(Q). The second possibility, however, cannot occurbecause an automorphism of order 3 in the Hoffman-Singleton graph cannot have exactly 10 fixedpoints.

IfPX, thenPacts onQand the preceding arguments show that this action is faithful. From the5-groups of order at most 125 onlyZ25,Z

35, andZ

25 Z5have an automorphism of order 3. In addition,

in all three cases the Sylow 3-subgroup of the automorphism group has order 3.

Proposition 7. Let p= 3and q > 5.Then q /=11,QX,and PX.If q= 19,then |P| divides 9,and ifq {7, 13},then |P| =3.

Proof. Ifq /= 13, normality ofQ inXfollows from Sylows Theorem as in the previous proof. AsPcontains an element of order 3 fixing a copy of the Petersen graph with automorphism group of order

120,Pcannot be normal inX.Ifq = 13, thenPcannot be normal inXbecause Fix(P)does not have an automorphism of order13. IfQX, thenXcan only be the extension ofZ33by the automorphism of order 13. In this case,however,PX a contradiction.

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Consequently, for anyqthe groupPacts faithfully as a group of automorphisms onQ. IfQ = Z27,then every element ofPcan be identified with a 22 matrix over Z7. If|P| =9, thenPcontains a

matrix

1 00 2

which fixes the element (1, 0) Q. In other words,Xcontains a cyclic subgroup of

order 21, which is not possible.

Proposition 8. Letp= 5 andq> 5. Thenq {7, 11} andQX. If q= 11, then |P| divides 25. If q= 7,then PX and|X| =35.

Proof. Again, Sylows Theorem provides normality ofQ inXand also normality ofPin Xifq /= 11.Therefore, ifq /= 11 thenX=PQ.

Let q= 19. As Qfixes exactly one point, Fix(P) is non-empty and has an automorphism of order 19 a contradiction.

Let q= 13. SinceX=PQ,itcontainsanelementx of order 65. Such an automorphism, however,cannot exist by Lemma15.

Letq = 7. By Lemma19, Fix(Q)is a star and hence Fix(P)is non-empty. Since a pentagon does

not have an automorphism of order 7, Fix(P)must be the Hoffman-Singleton graph and|P| =5. As agroup of order 49 cannot fix a single vertex in the Hoffman-Singleton graph, we conclude that |Q| =7and, by Lemma15, |Fix(Q)| =16.

Finally letq= 11. AsQis normal inX,Pact as a group of automorphisms ofQ. LetP be the kernelof this action. It suffices to show that|P| divides 5.

First, we show that for any element y P we have Fix(y)= Fix(Q). Asy centralizesQwe haveFix(y) Fix(Q). If Fix(y)= , then the groupy Qhas on one orbit of size 5 equal to Fix(Q)and 59 orbits of size 55. As Fix(Q) is a pentagon, exactly one ofy andy2 contributes to the orbit Fix(Q).As other orbits have size 55 we obtaina1(y) /= a1(y

2)which is a contradiction with Lemma11.Let |P| > 1.ByLemma 13, P does not containan element of order 25.As every nonidentity element

ofP hasorder5wehaveFix(P)= Fix(Q) and P actssemi-regularlyon \Fix(Q).Therefore |P| =5.

Remark.The preceding results show that ifx is an automorphism ofof odd order pq, thenx iscovered by Lemma15.Automorphisms of order 2p,p odd, can be handled by similar arguments incombination with the results of Makhnev and Paduchikh.

Now we are ready to prove our main results.

Theorem 6. Letbe a Moore graph of degree 57 on 3250 vertices and G= Aut().If|G| is odd then |G|divides1932, 133, 52 11, 72 3, 75, 53 3,or33 5.

Proof. IfGhas at most two prime divisors, then its size lies in the list above. By the facts mentioned

earlier it suffices to show non-existence ofGwith|G|having three prime divisors. By Propositions6,7and8the only possibility for the odd part ofGis the direct product ofZ5and Z7 Z3. However, byLemma 15 the element of order 5 in such a group Gfixes simultaneously the Hoffman-Singleton graphand the empty set.

Finally, applying our methods to possible groups of automorphismsof even order we are able toimprove the earlier results of Makhnev and Paduchikh.

Theorem 7.Letbe a Moore graph of degree 57on3250vertices and G = Aut().If|G|is even then|G| divides1152, 52 2, 33 2,or2p,p {7, 11, 19}.

Proof. By Theorem 2 G= Yt Xwhere tis an involution andXand Yhave odd order. In their proof,Makhnev and Paduchikh showed that any automorphism of order 3 in Yfixes exactly one point whichis not possible. By Theorem5 if Fix(X)is the Hoffman-Singleton graph, then |X| /= 25. Finally, byLemma15G cannot contain Z55an d Z22with common subgroup of order 11 or Z10and Z35withcommon subgroup of order 5.

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Corollary 3. Letbe a Moore graph of degree 57 on 3250 vertices and G = Aut(). Then, |G| 375, andif|G| is even,then |G| 110.

10. Conclusion

By the time of writing this article, progress in the research into symmetries of the Moore (57, 2)-graph(s) was achieved mostly by studying properties of individual automorphisms. Higman (see[3])successfully used methods of linear algebra, whereas tools developed by Makhnev and Paduchikh[10]are of combinatorial and group-theoretical nature. In our investigation we have combined allthe previous methods, enriched by equitable partitions and supported by the use of computers. Wefeel that our main contribution lies in studying entire subgroups rather than single automorphisms,focusing on ways the automorphisms within a subgroup influence each other.

Thanks to Makhnev and Paduchikh [10] it was sufficient to restrict ourselves to groups of odd order,the Odd Order Theorem with results of Philip Hall thus enabled us to study possible automorphismgroups of the Moore (57,2)-graph(s) in terms of the number of prime divisors of their orders. Our

upper bounds on possible orders ofp-groups led to the finding that in groups of orderpaqb at leastone of the Sylow subgroups is normal. This was the main step towards completion of characterizationof possible orders of automorphism groups of the graph(s). As a matter of fact, with the exception of110, all possible orders have at most two distinct prime factors.

The choice of the form of presentation of our main results was determined by trying to avoidunnecessary technical details and preferring compactness of statements. In particular, restrictions onorders of groups acting on the Moore(57,2)-graph(s) are not the strongest pieces of information weare able to derive. Our methods enable one to substantially restrict the values of the functionsa0anda1, and hence the orbit structure of the actions. For example, from Lemma15it follows that there isno automorphism of order 110.

On the one hand, it is possible that some of the orders listed in our two main theorems could be

excluded by showing non-existence of suitable adjacency matrices for groups of automorphism, asdone for the order 625 in section8.On the other hand, in the study of possible actions of groups oforder 375 with 10 orbits we found hundreds of matrices satisfying conditions of Lemma 5which wewere not able to exclude by our techniques. An example of such a matrix is:

2 5 8 10 6 8 5 6 4 35 8 8 3 8 6 8 7 4 08 8 6 6 2 4 9 7 3 4

10 3 6 6 8 5 4 10 3 26 8 2 8 10 5 8 5 2 38 6 4 5 5 12 7 6 2 2

5 8 9 4 8 7 4 8 0 46 7 7 10 5 6 8 8 0 0

12 12 9 9 6 6 0 0 2 19 0 12 6 9 6 12 0 1 2

.

We therefore think that one of the challenges for future research is to develop methods to deal withsuch situations.

ForthosebelievingintheexistenceoftheMoore (57,2)-graph(s) ,ourmethodspointatpromisingplaces to look for candidates. Regarding symmetries, the theoretically best possible negative resultwould be to show that the automorphism group ofis trivial. In the course of our investigation wediscovered a number of configurations that can appear in . A study of such configurations couldbe of interest for researchers, even for those believing in the non-existence of . Examples of suchconfigurations are 15 mutually non-adjacent copies of the Hoffman-Singleton graph, and 11 copies ofthe Hoffman-Singleton graph sharing a pentagon.

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Acknowledgments

Research of the first author was supported by the APVV Research Grant No. 0111-07 and the VEGAResearch Grants No. 1/0588/09 and 1/0406/09. Research of the second author was supported from the

APVV Research Grants No. 040-06 and 0104-07, the LPP Research Grants No. 0145-06 and 0203-06,and the VEGA Research Grant No. 1/0489/08. The first author would like to thank the Open Universityfor hospitality during his research visit that initiated the work on this paper.

References

[1] M. Aschbacher, The nonexistence of rank three permutation groups of degree 3250 and subdegree 57, J. Algebra 19 (3)(1971) 538540.

[2] E. Bannai, T. Ito, On finite Moore graphs, J. Fac. Sci., Univ. Tokyo, Sect. I A 20 (1973) 191208.[3] P. Cameron, Permutation Groups, Cambridge University Press, 1999.[4] C.W. Curtis, I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley, 1988.[5] R.M. Damerell, On Moore graphs, Math. Proc. Cambridge Philos. Soc. 74 (1973) 227236.

[6] J.D. Dixon, B. Mortimer, Permutation Groups, GTM 163, Springer, 1996.[7] The GAP Group, GAP Groups, Algorithms, and Programming, Version 4.4.10, 2007,.[8] C.D. Godsil, Algebraic Combinatorics, Chapman and Hall, 1993.[9] A.J. Hoffman, R.R. Singleton, On Moore graphs with diameters 2 and 3, IBM J. Res. Dev. 4 (1960) 497504.

[10] A.A. Makhnev, D.V. Paduchikh, Automorphisms of Aschbacher graphs, Algebra Logic 40 (2) (2001) 6974.[11] M. Miller, J. irn, Moore graphs and beyond: a survey of the degree/diameter problem, Electron. J. Combin., Dynamic

Survey DS14 (2005), 61 pp.[12] B. Mohar, Graph Laplacians, in: Beineke, Wilson (Eds.), Topics in Algebraic Graph Theory, Cambridge University Press,

2004, pp. 113136.[13] J.J. Rotman, An Introduction to the Theory of Groups, GTM 148, fourth ed., Springer, 1999.
http://www.gap-system.org/http://www.gap-system.org/http://www.gap-system.org/

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