Primitive groups and synchronization
João Araújo, Wolfram Bentz, Peter J. Cameron, Gordon Royle, Artur Schaefer
aa r X i v : . [ m a t h . G R ] M a y Primitive groups, graph endomorphisms andsynchronization
João Araújo
Universidade Aberta, R. Escola Politécnica, 1471269-001 Lisboa, Portugal & CAUL/CEMAT, Universidade de Lisboa1649-003 Lisboa, [email protected]
Wolfram Bentz
CAUL/CEMAT, Universidade de Lisboa1649-003 Lisboa, [email protected]
Peter J. Cameron
Mathematical Institute, University of St AndrewsNorth Haugh, St Andrews KY16 9SS, [email protected]
Gordon Royle
Centre for the Mathematics of Symmetry and ComputationThe University of Western AustraliaCrawley, WA 6009, [email protected]
Artur Schaefer
Mathematical Institute, University of St AndrewsNorth Haugh, St Andrews KY16 9SS, [email protected]
Abstract
Let Ω be a set of cardinality n , G a permutation group on Ω , and f : → Ω a map which is not a permutation. We say that G synchronizes f ifthe transformation semigroup h G, f i contains a constant map, and that G isa synchronizing group if G synchronizes every non-permutation.A synchronizing group is necessarily primitive, but there are primitivegroups that are not synchronizing. Every non-synchronizing primitive groupfails to synchronize at least one uniform transformation (that is, transforma-tion whose kernel has parts of equal size), and it had previously been conjec-tured that this was essentially the only way in which a primitive group couldfail to be synchronizing — in other words, that a primitive group synchro-nizes every non-uniform transformation.The first goal of this paper is to prove that this conjecture is false, by ex-hibiting primitive groups that fail to synchronize specific non-uniform trans-formations of ranks and . As it has previously been shown that primi-tive groups synchronize every non-uniform transformation of rank at most , these examples are of the lowest possible rank. In addition we producegraphs with primitive automorphism groups that have approximately √ n non-synchronizing ranks , thus refuting another conjecture on the number ofnon-synchronizing ranks of a primitive group.The second goal of this paper is to extend the spectrum of ranks for whichit is known that primitive groups synchronize every non-uniform transforma-tion of that rank. It has previously been shown that a primitive group ofdegree n synchronizes every non-uniform transformation of rank n − and n − , and here this is extended to n − and n − .Determining the exact spectrum of ranks for which there exist non-uniformtransformations not synchronized by some primitive group is just one of sev-eral natural, but possibly difficult, problems on automata, primitive groups,graphs and computational algebra arising from this work; these are outlinedin the final section. Let Ω be a set of size n and let f be a transformation on Ω of rank (size of image)smaller than n (in other words, f is a non-permutation). A permutation group G of degree n on Ω synchronizes f if the transformation semigroup h G, f i containsa constant transformation. The kernel of f is the partition of Ω determined by theequivalence relation x ≡ y if and only if xf = yf . If the parts of the kernel allhave the same size, then f is called uniform ; it is non-uniform otherwise. A groupis called synchronizing if it synchronizes every non-permutation. A synchronizinggroup is necessarily primitive (see [13]) but the converse does not hold and con-siderable efforts have been made to determine exactly which primitive groups aresynchronizing.To show that a primitive group is not synchronizing it is necessary to find a2 itness , which is a transformation f such that h G, f i does not contain a constantmap. Neumann [39] proved that any non-synchronizing primitive group has a uni-form witness. This prompted the definition of a primitive group as being almostsynchronizing if it synchronizes every non-uniform transformation. In [4] the al-most synchronizing conjecture was stated. This asserts that primitive groups haveno non-uniform witnesses; that is, they are almost synchronizing.This conjecture has previously been proved for transformations of ranks , , , n − and n − or, in other words, for transformations of very low , or very high ,rank (see [6, 39, 43]). In this paper, we prove three main results, showing differentoutcomes for the “low rank” and the “high rank” cases. A graph admitting a vertex-primitive automorphism group will be called a primitive graph . Theorem 1.1
There are primitive graphs admitting non-uniform endomorphisms,and hence not every primitive group is almost synchronizing.
Section 3 describes a number of different graph constructions used to demon-strate this result. This theorem resolves the almost synchronizing conjecture in thenegative, but shows that the structure of endomorphisms of primitive graphs canbe more complex than previously suspected, prompting the difficult problem offinding a classification of the primitive almost synchronizing groups.Our second main theorem considers the high rank case and extends the spec-trum of ranks for which transformations of that rank are known to be synchronizedby every primitive group.
Theorem 1.2
A primitive group of degree n synchronizes every transformation ofrank n − or n − . Sections 4 and 6 detail the somewhat intricate arguments required to deal withthe considerable number of graphs that arise in proving this result.Our third main theorem considers groups of small permutation rank . Groupswith permutation rank are doubly transitive and easily seen to be synchronizing,so the first non-trivial case is for groups of rank . In this case, the group actsprimitively on a strongly regular graph and properties of such graphs can be usedto considerably extend the n − bound. Theorem 1.3
A primitive permutation group of degree n and permutation rank synchronizes any non-permutation with rank at least n − (1 + √ n − / . The original context of this research is automata theory, as we now outline. Ourautomata are always finite and deterministic. On reading a symbol, an automatonundergoes a change of state; so each symbol defines a transition , a transformation3n the set of states. The set of transformations realised by reading a word or se-quence of symbols is the semigroup generated by the transitions of the automaton.Thus, from an algebraic viewpoint, an automaton is a subsemigroup of the fulltransformation semigroup on a finite set with a prescribed set of generators.A deterministic finite-state automaton is said to be synchronizing if there is afinite word such that, after reading this word, the automaton is in a fixed state, in-dependently of its state before reading the word. In other terms, the word evaluatesto a transformation of rank , mapping the set of states to a single state. Such aword is called a reset word or synchronizing word .One of the oldest and most famous problems in automata theory, the well-known ˇCerný conjecture, states that if an automaton with n states has a synchro-nizing word, then there exists one of length ( n − . (For many references onthe growing bibliography on this problem please see the two websites [40, 44] andalso Volkov’s talk [48]; so far the best bound for the length of a reset word is cubic[41].) Solving this conjecture is equivalent to proving that given a set S of transfor-mations on a finite set of size n then, if the transformation semigroup h S i containsa constant transformation, then it contains one that can be expressed as a word oflength at most ( n − in the generators of S . This conjecture has been establishedfor aperiodic automata , that is, when h S i is a semigroup with no non-trivial sub-groups [45]. So it remains to prove the conjecture for semigroups that do containnon-trivial subgroups, and the case when the semigroup contains a permutationgroup is a particular instance of this general problem. Indeed, the known exam-ples witnessing the optimality of the ˇCerný bound contain a permutation amongthe given set of generators.We note that if a transformation semigroup S contains a transitive group G butnot a constant function, then the image I of a transformation f of minimum rank in S is a G -section for the kernel of f , in the sense that Ig is a section or transversal for ker( f ) , a set meeting every kernel class in a single element. In addition, thetransformation f has uniform kernel (see Neumann [39]).Although they did not use this terminology, results due to Rystsov [43] andNeumann [39] cover some cases of the almost sychronizing conjecture. In particu-lar, Rystsov [43] showed that a transitive permutation group of degree n is primitiveif and only if it synchronizes every transformation of rank n − , while Neumann[39] showed that a primitive permutation group synchronizes every transformationof rank .In earlier work, Araújo and Cameron [6] resolved some additional cases of theconjecture: Theorem 1.4
A primitive permutation group G of degree n synchronizes maps ofkernel type ( k, , . . . , (for k ≥ ) and maps of rank n − , as well as non-uniform aps of rank or . That paper, like the present one, uses a graph-theoretic approach due to thethird author [19]. In what follows, graphs are always simple and undirected. The clique number of a graph is the largest number of vertices in a complete subgraph,while the chromatic number is the smallest number of colours required for a propercolouring.
Theorem 1.5
A transformation semigroup does not contain a constant transfor-mation if and only if it is contained in the endomorphism monoid of a non-nullgraph. Moreover, we may assume that this graph has clique number equal to chro-matic number.
One direction of the theorem is clear, since a non-null graph has no rank endomorphisms. For the other direction, define a graph by joining two vertices ifno element of the semigroup maps them to the same place, and show that this graphhas the required property. We will elaborate further in the next section.This paper is in three main parts. In the first part, we show that the almostsynchronizing conjecture fails for maps of small rank, so the results in [6] are bestpossible. We construct four examples of primitive groups (with degrees , , and ) which fail to synchronize non-uniform maps of rank . In addition,we find infinitely many examples for rank , along with yet another sporadic ex-ample of rank of degree . Also, we provide a construction of primitive graphswhose automorphism groups have approximately √ n non-synchronizing ranks , re-futing a conjecture of the third author’s on the number of non-synchronising ranksof a primitive group.In the second part, we press forward with maps of large rank, showing that aprimitive group synchronizes all maps with kernel type ( p, , , . . . , or kerneltype ( p, , , . . . , for p ≥ , as well as all maps of rank n − and n − . We alsoshow that a primitive group synchronizes every map in which one non-singletonkernel class is sufficiently large compared to the other non-singleton kernel classes.In the third part, we consider the special situation where the primitive group G has permutation rank , in which case any graph with automorphism group contain-ing G is either trivial or strongly regular. In the latter case, we prove a general resultabout endomorphisms of strongly regular graphs, and deduce that G synchronizesevery non-permutation transformation of rank at least n − (1 + √ n − / .The paper ends with a number of natural but challenging problems related tosynchronization in primitive groups and in related combinatorial settings.Given the enormous progress made in the last three or four decades, permu-tation groups now has the tools to answer questions coming from the real world5hrough transformation semigroups; these questions translate into beautiful state-ments in the language of permutation groups and combinatorial structures, as shownin many recent investigations (as a small sample, please see [2, 4, 6, 7, 8, 12, 13,20, 29, 39, 42]). The critical idea underlying our study is a graph associated to a transformationsemigroup in the following way. If S is a transformation semigroup on Ω , thenform a graph, denoted Gr( S ) , with vertex set Ω where v and w are adjacent if andonly if there is no element f of S which maps v and w to the same point. Now thefollowing result is almost immediate ( cf. [21, 19]). Theorem 2.1 (See [6, 21]) Let S be a transformation semigroup on Ω and let Gr( S ) be defined as above. Then(a) S contains a map of rank if and only if Gr( S ) is null (i.e., edgeless).(b) S ≤ End(Gr( S )) , and Gr(End(Gr( S ))) = Gr( S ) .(c) The clique number and chromatic number of Gr( S ) are both equal to theminimum rank of an element of S . In particular, if S = h G, f i for some group G , then G ≤ Aut(Gr( S )) . So, forexample, if G is primitive and does not synchronize f , then Gr( S ) is non-null andhas a primitive automorphism group, and so is connected.In this situation, assume that f is an element of minimum rank in S ; then thekernel of f is a partition ρ of Ω , and its image A is a G -section for ρ (that is, Ag is a section for ρ , for all g ∈ G ). Neumann [39], analysing this situation, defineda graph ∆ on Ω whose edges are the images under G of the pairs of vertices in thesame ρ -class. Clearly ∆ is a subgraph of the complement of Gr( S ) , since edgesin ∆ can be collapsed by elements of S . Sometimes, but not always, ∆ is thecomplement of Gr( S ) .We now introduce a refinement of the previous graph Gr( S ) , which will allowus to obtain the results of the remaining cases more easily. The new graph isdenoted by Gr ′ ( S ) . The same construction was used in a different context in [20],where it was called the derived graph of Gr( S ) .Suppose that Gr( S ) has clique number and chromatic number r (where r isthe minimum rank of an element of S ). We define Gr ′ ( S ) to be the graph with thesame vertex set as Gr( S ) , and whose edges are all those edges of Gr( S ) which arecontained in r -cliques of Gr( S ) . Theorem 2.2
Let S be a transformation semigroup on Ω and let Gr( S ) and Gr ′ ( S ) be defined as above. Then a) S contains a map of rank if and only if Gr ′ ( S ) is null.(b) S ≤ End(Gr( S )) ≤ End(Gr ′ ( S )) .(c) The clique number and chromatic number of Gr ′ ( S ) are both equal to theminimum rank of an element of S .(d) Every edge of Gr ′ ( S ) is contained in a maximum clique.(e) If S = h G, f i , where G is a primitive permutation group and f a map whichis a non-permutation not synchronized by G , then Gr ′ ( S ) is neither completenor null. Proof
Elements of
End(Gr( S )) preserve Gr( S ) and map maximum cliques tomaximum cliques, so End(Gr( S )) ≤ End(Gr ′ ( S )) . The existence of an r -cliqueand an r -colouring of Gr ′ ( S ) are clear, and so (c) holds; then (a) follows. Part (d)is clear from the definition. For (e), the hypotheses guarantee that the minimumrank of an element of S is neither nor n . (cid:3) Note that strict inequality can hold in (b). If Γ is the disjoint union of completegraphs of different sizes, then Gr ′ (End(Γ)) consists only of the larger completegraph, and has more endomorphisms than Γ does.The next lemma is proved in [6], but since the techniques it introduces areimportant in subsequent arguments we provide its proof here. Lemma 2.3
Let X be a nontrivial graph and let G ≤ Aut( X ) be primitive. Thenno two vertices of X can have the same neighbourhood. Proof
For a ∈ X denote its neighbourhood by N ( a ) . Suppose that a, b ∈ X ,with a = b , and N ( a ) = N ( b ) . We are going to use two different techniques toprove that this leads to a contradiction. The first uses the fact that the graph has atleast one edge, while the second uses the fact that the graph is not complete.First. Define the following relation on the vertices of the graph: for all x, y ∈ X , x ≡ y ⇔ N ( x ) = N ( y ) . This is an equivalence relation and we claim that ≡ is neither the universal relationnor the identity. The latter follows from the fact that by assumption a and b aredifferent and N ( a ) = N ( b ) . Regarding the former, there exist adjacent vertices c and d (because X is non-null); now c ∈ N ( d ) but c / ∈ N ( c ) , so c d . As G is a group of automorphisms of X it follows that G preserves ≡ , a non-trivialequivalence relation, and hence G is imprimitive, a contradiction.Second. Assume as above that we have a, b ∈ X such that N ( a ) = N ( b ) .Then the transposition ( a b ) is an automorphism of the graph. A primitive groupcontaining a transposition is the symmetric group. (This well-known result from7he nineteenth century can be found, for example, in [32, p.241].) Hence X is thecomplete graph, a contradiction. (cid:3) We conclude this section recalling another result from [6] about primitive graphs . Lemma 2.4 ([6]) Let Γ be a non-null graph with primitive automorphism group G , and having chromatic number r . Then Γ does not contain a subgraph isomor-phic to the complete graph on r + 1 vertices with one edge removed. This lemma was important in [6] and it will be here too (please see the obser-vations after Lemma 6.5).
In this section, we discuss various counterexamples to the conjecture that primitivegroups are almost synchronizing. From the discussion above, it suffices to find anon-null graph Γ with a primitive automorphism group, and then exhibit a non-uniform proper endomorphism f of Γ . Such an endomorphism is then a witnessthat G is not almost synchronizing for any primitive group G ≤ Aut(Γ) .In the first subsection, we present a number of sporadic examples of vertex-primitive graphs, each with non-uniform proper endomorphisms of rank or .The smallest of these, on vertices, can be shown (by computer) to be the uniquesmallest counterexample to the almost-synchronizing conjecture; the details aregiven in Section 8. None of the graphs described here are Cayley graphs.In the second subsection, we present infinite families of primitive graphs withnon-uniform proper endomorphisms of rank and above. In this subsection, we give the first counterexample to the conjecture that primitivegroups are almost synchronizing. In particular, we construct a primitive group ofdegree that fails to synchronize a non-uniform map of rank with kernel type (5 , , , , . We give two proofs of this, which extend in different ways.Our primitive group is PΓL(2 , (also known as A : 2 ), acting on points(this is PrimitiveGroup(45,3) in both
GAP and M
AGMA ). This group hasa suborbit of length , and the orbital graph Γ has the property that any edge iscontained in a unique triangle: the closed neighbourhood of a vertex is a “butterfly”consisting of two triangles with a common vertex (see Figure 1). Indeed, this graphis the line graph of the celebrated Tutte–Coxeter graph on vertices, which in turn8s the incidence graph of the generalized quadrangle W (2) of order . The graphwas first found by Tutte [46] with a geometric interpretation by Coxeter [25, 47].Let D be a dihedral subgroup of order of the automorphism group of thegraph. It is clear that elements of order in D fix no vertices of the graph, and alittle thought shows that their cycles are independent sets in the graph.The full automorphism group of the graph is the automorphism group of S (that is, the group extended by its outer automorphism), and there are two conju-gacy classes of dihedral groups of order . It is important to take the right onehere: we want the D which is not contained in S .For this group D , we find that each orbit of D is an independent set in Γ ; sothere is a homomorphism of Γ in which each orbit is collapsed to a single vertex. Asmall calculation shows that the image of this homomorphism is the graph shownbelow: r r r rr r r ✔✔✔✔ ✔✔✔✔ ✔✔✔✔❚❚❚❚ ❚❚❚❚ ❚❚❚❚ Now this graph can be found as a subgraph of Γ , as the union of two butterfliessharing a triangle; therefore the homomorphism can be realised as an endomor-phism of Γ of rank , with kernel classes of sizes (10 , , , , , , . The endo-morphisms of ranks and can now be found by folding in one or both “wings”in the above figure.Our second approach uses the fact that the chromatic number and clique num-ber of this graph are each equal to ; thus, each triangle has one vertex in each ofthe three colour classes, each of size . So there is a uniform map of rank notsynchronized by G .We used GAP to construct the graph (the vertex numbering is determined bythe group), and its package
GRAPE to find all the independent sets of size in Γ up to the action of G . One of the two resulting sets is A = { , , , , , , , , , , , , , , } . The induced subgraph on the complement of this set has two connected compo-nents, a -cycle and a -cycle. If we let B and C be the bipartite blocks in the -cycle and D and E those in the -cycle, we see that A, B, C, D, E are all in-dependent sets, and the edges between them are shown in Figure 1. Thus there is aproper endomorphism mapping the graph to the closed neighbourhood of a vertex,with kernel classes
A, B, C, D, E .Using software developed at St Andrews (see Section 8 for details) we wereable to calculate all the proper endomorphisms of this graph: there are DECB
Figure 1: The butterflyof these, with ranks , and ; the numbers of endomorphisms of each of theseranks are , and respectively. Then, using GAP, we were ableto determine that the endomorphism monoid of this graph is given by End( X ) = h G, t i , where G is PΓL(2 , and t is the transformation t = Transformation ([1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . The endomorphisms of each possible rank form a single D-class. The structurefor the H-classes is S , D and D for the three classes respectively, where D is the dihedral group on points. (Note that these groups are the automorphismgroups of the induced subgraphs on the image of the maps.)A very similar example occurs in the line graph of the Biggs–Smith graph [16, 17], a graph on vertices whose automorphism group is isomorphic to
PSL(2 , ( PrimitiveGroup(153,1) in both
GAP and M
AGMA ). Thisgraph has an endomorphism of rank and kernel type (6 , , , , constructedin a virtually identical way.However, this particular construction gives no additional examples. A vertex-primitive -regular graph whose neighbourhood is a butterfly is necessarily thelinegraph of an edge- primitive cubic graph. These were classified by Weiss [49],who determined that the complete list is K , , the Heawood graph, the Tutte-Coxeter graph and the Biggs-Smith graph. From either a direct analysis, or simplyreferring to the small-case computations described in Section 8, it follows that thefirst two of these do not yield examples.However, we have found three additional examples with the help of the com-puter. Surprisingly all three of them are associated with the group Aut( M ) = M : 2 . This group has two inequivalent primitive actions of degree . Eachof them is the automorphism group of a graph of valency in which the closedneighbourhood of a vertex consists of three triangles with a common vertex, andin each case, the graph has chromatic number . In each case, there is a subgroup10igure 2: The graph K (cid:3) K of the automorphism group with orbits of sizes , , , , ; each orbit isan independent set and the connections between the orbits give a homomorphismonto the butterfly.The third example is associated with a different primitive action of M : 2 ,this time of degree . In this action, M : 2 is the full automorphism groupof a -regular graph where each open neighbourhood is the disjoint union of twotriangles. The group has a subgroup of order , which has equal-sized or-bits each inducing an independent set. These orbits can each be mapped toa single vertex in such a way that the entire graph is mapped onto the closedneighbourhood of a vertex, yielding an endomorphism of rank , with kernel type (220 , , , , , , . As the closed neighbourhood of a vertex consistsof two -cliques overlapping in a vertex, we may perhaps view this just as a butter-fly with bigger wings? While the constructions of the previous subsection seem to be sporadic examples,we can also find several infinite families of vertex-primitive graphs with propernon-uniform endomorphisms.Recall that the
Cartesian product X (cid:3) Y of two graphs X and Y is the graphwith vertex set V ( X (cid:3) Y ) = V ( X ) × V ( Y ) and where vertices ( x , y ) and ( x , y ) are adjacent if and only if they have equal entries in one coordinate position andadjacent entries (in X or Y accordingly) in the other. Figure 2 shows the graph K (cid:3) K both to illustrate the Cartesian product and because it plays a role later inthis section.If X is a vertex-primitive graph then the Cartesian product X (cid:3) X is alsovertex-primitive, with automorphism group Aut( X ) wr Sym(2). In addition, ifthe chromatic and clique number of X are both equal to k , then V ( X ) can bepartitioned into k colour classes of equal size — say V , V , . . . , V k , and there is11 Figure 3: Non-uniform homomorphisms from K (cid:3) K to its complementa surjective homomorphism X (cid:3) X → K k (cid:3) K k with kernel classes { V i × V j | ≤ i, j ≤ k } . Therefore if there is a homomorphism f : K k (cid:3) K k → X , then bycomposing homomorphisms X (cid:3) X −→ K k (cid:3) K k f −−→ X −→ X (cid:3) X, there is an endomorphism of X (cid:3) X . Moreover, if the homomorphism f is non-uniform, then the endomorphism is also non-uniform.Although at first sight, there appears to be little to be gained from this observa-tion, in practice it is much easier (both computationally and theoretically) to findhomomorphisms between the two relatively small graphs K k (cid:3) K k and X , thanworking directly with the larger graph X (cid:3) X . Although this finds only a restrictedsubset of endomorphisms, it turns out to be sufficient to find large numbers ofnon-uniform examples.We start by considering the case where X = K k (cid:3) K k , where we assume thatthe vertices of both graphs are labelled with pairs ( i, j ) , where ≤ i, j < k andthat in K k (cid:3) K k two distinct vertices are adjacent if and only if they agree in onecoordinate position, while in its complement, they are adjacent if and only if they disagree in both coordinate positions.The graph X has chromatic number and clique number equal to k . (A diagonalset { ( i, i ) : 0 ≤ i < k } is a k -clique, while using the first coordinate as colourgives a k -colouring.) The homomorphisms we seek are those from K k (cid:3) K k to itsown complement.In particular, Figure 3 exhibits three non-uniform homomorphisms of ranks , and from K (cid:3) K to its complement, where the diagrams show the image ofeach vertex, but using xy to represent ( x, y ) . Verifying that this function is a homo-morphism merely requires checking for each row that the four pairs assigned to ithave pairwise distinct first co-ordinates and pairwise distinct second co-ordinates,and similarly for each column. This happens if and only if the pairs are obtainedfrom the super-position of two Latin squares of order , one determining the first12o-ordinate and the other the second co-ordinate. The rank of the homomorphismis then just the total number of distinct pairs that occur — this number ranges froma minimum of k (when the two Latin squares are identical) to a maximum of k (when the two Latin squares are orthogonal). In the example of Figure 3 the kerneltypes of the homomorphisms are { , } , { , , } and { , } , which corre-spond to endomorphisms of X (cid:3) X of the same rank, but with kernel classes each times larger.This argument clearly generalises to all k ≥ (non-uniform homomorphismsdo not arise when k < ) and so any two Latin squares of order k ( not necessarilyorthogonal) will determine an endomorphism of K k (cid:3) K k . Two Latin squares aresaid to be r -orthogonal if r distinct pairs arise when they are superimposed. Thuswe find an endomorphism of rank r from any pair of r -orthogonal Latin squares.The following result, due to Colbourn & Zhu [24] and Zhu & Zhang [50] showsexactly which possible ranks arise in this fashion. Theorem 3.1
There are two r -orthogonal Latin squares of order k if and only if r ∈ { k, k } or k + 2 ≤ r ≤ k − , with the following exceptions:(a) k = 2 and r = 4 ;(b) k = 3 and r ∈ { , , } ;(c) k = 4 and r ∈ { , , , , } ;(d) k = 5 and r ∈ { , , , , } ;(e) k = 6 and r ∈ { , } . (cid:3) In particular, for any k ≥ , there is an endomorphism of rank k + 2 with imagetwo k -cliques overlapping in a ( k − -clique. As k increases, we get a sequenceof butterflies with increasingly fat bodies, but fixed-size wings.This construction also sheds some light on the possible non-synchronizingranks for a group. For a group G of degree n , a non-synchronizing rank is a value r satisfying ≤ r ≤ n − such that G fails to synchronize some transformationof rank r .A transitive imprimitive group of degree n , having m blocks of imprimitivityeach of size k , preserves both a disjoint union of m complete graphs of size k (which has endomorphisms of ranks all multiples of k ) and the complete m -partitegraph with parts of size k (which has endomorphisms of all ranks between k and n inclusive). From this, a short argument shows that such a group has at least (3 / − o (1)) n non-synchronizing ranks. It was suspected that a primitive grouphas many fewer non-synchronizing ranks, perhaps as few as O (log n ) . However, asthis construction provides approximately k non-synchronizing ranks for a groupof degree k , this cannot be the case.It is natural to wonder whether this construction can be used to find non-uniform endomorphisms for graphs other than X = K k × K k . Unsurprisingly,13 Figure 4: A homomorphism from K (cid:3) K to L ( K ) the answer to this question is yes, with the line graph of the complete graph L ( K n ) (also known as the triangular graph ) being a suitable candidate for X whenever n is even. For example L ( K ) is a -vertex graph with chromatic number and cliquenumber equal to . The vertices of L ( K ) can be identified with the endpoints ofthe corresponding edge in K , and thus each vertex of L ( K ) is represented by a -set of the form { x, y } which we will abbreviate to xy .Figure 4 depicts a surjective homomorphism from K (cid:3) K to L ( K ) by la-belling each of the vertices of K (cid:3) K with its image in L ( K ) under the ho-momorphism. For each of the horizontal or vertical lines — corresponding to thecliques of K (cid:3) K — it is easy to confirm that the images of the five vertices inthe line share a common element and thus are mapped a clique of L ( K ) . Thishomomorphism has rank and kernel type { , } and hence yields a non-uniform endomorphism of L ( K ) (cid:3) L ( K ) with kernel classes of times thesize. The pattern shown in Figure 4 can be generalised to all triangular graphs, bydefining a map f : K n − (cid:3) K n − −→ L ( K n ) by f (( a, b )) = { a + 1 , b + 1 } if a = b and f (( a, a )) = { , a + 1 } . This homomorphism has kernel type { n − , ( n − n − / } .We finish this section with yet another construction that provides an infinitefamily of rank non-uniform non-synchronizable transformations.Let p be a prime greater than , and let V be the vector space spanned by e , . . . , e p − (we think of the indices as elements of the integers mod p ) with thesingle relation that their sum is zero. Let Γ be the Cayley graph for V with con-nection set of size p consisting of the vectors e i and e i + e i +1 with i running overthe integers mod p . It is clear that the group V : D p acts as automorphisms ofthis graph, and is primitive provided that is a primitive root mod p (this is thecondition for V to be irreducible as a C p -module).Now let X be the subspace spanned by e i + e i +2 for i = 0 , , . . . , p − .14hese vectors are linearly independent and so span a space of codimension , with cosets. Check that this subspace contains no edge of the graph: no two of itsvectors differ by a single basis vector or a sum of two consecutive basis vectors.We can take coset representatives to be , e , e , e + e , e p − , e p − + e , e p − + e , e p − + e + e .Each coset contains no edges of the graph, and indeed the unions ( X + e ) ∪ ( X + e p − ) and ( X + e + e ) ∪ ( X + e + e p − ) also contain no edges. Collapsingthese two unions and the other four cosets to a vertex, we find by inspection thatthe graph is a “butterfly”: r r rr r r (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)❅❅❅❅❅❅❅❅❅❅ The two vertices forming the butterfly’s body are the “double cosets”.Now we can find a copy of the butterfly in the graph, using the vertices and e for the body, e and e + e for one wing, and e p − and e + e p − for the otherwing.So there is an endomorphism of rank , with two kernel classes of size p − and four of size p − . n − The goal of this section is to show that primitive groups synchronize maps of rank n − . Moreover, we will establish general properties of graphs with primitiveautomorphism groups that will also be used in the next sections; therefore theseresults will be stated and proved as generally as possible. At several points during the argument (in this and in the next sections), we willestablish that the (primitive) automorphism group of a graph under considerationcontains a permutation which is a product of three or four disjoint transpositions,and hence has support of size m ∈ { , } . As the automorphism group of a non-trivial graph is not -transitive (in particular, it does not contain A n ), it followsfrom [38] that it has degree less than ( m/ . Thus in the worst case, we needonly examine primitive groups of degree less than , which can easily be handledcomputationally. For convenience all of the computational results are collated inSection 8. 15 .2 A bound on the intersection of neighbourhoods The goal of this subsection is to prove the following result about
Gr( S ) , the graphintroduced in Section 2. (We note that Spiga and Verret [42] have proved someresults about neighbourhoods of vertex-primitive graphs that also imply this prop-erty.) Theorem 4.1
Let G be a group acting primitively on a set X , and suppose that f ∈ T ( X ) is not synchronized by G . Let S be the semigroup generated by G and f , and let k be the valency of the graph Γ = Gr( S ) . Then for all distinct x, y ∈ X ,their neighbourhoods N ( x ) , N ( y ) in Γ satisfy | N ( x ) ∩ N ( y ) | ≤ k − . The previous theorem is an important ingredient in the proof of the main resultsin this paper. Its proof is based on methods from [4, 6, 21] and is a consequence ofthe following sequence of lemmas.
Lemma 4.2
Let Γ be a non-null graph with primitive automorphism group G , andhaving chromatic number and clique number r . Let x be a vertex of Γ , and C an r -clique containing x . Then for every vertex y N ( x ) ∪ { x } , we have that ( N ( x ) \ N ( y )) ∩ C = ∅ . Proof
Assume instead that for some y N ( x ) ∪ { x } we have ( N ( x ) \ N ( y )) ∩ C = ∅ . Then every element of C , different from x , is a neighbour of y . Thus theset C ∪ { y } induces a subgraph that is isomorphic to the complete graph with oneedge removed. This contradicts Lemma 2.4. (cid:3) We next state an observation on primitive groups and quasiorders (reflexive andtransitive relations). The proof is an easy exercise.
Lemma 4.3
Let G be a permutation group on the finite set X . Then G is primitiveif and only if the only G -invariant quasiorders are the identity and the universalrelation. This immediately implies:
Lemma 4.4
Let Γ be a graph with primitive automorphism group G and cliquenumber r on the vertex set X . Assume that there are distinct elements a, b ∈ X satisfying the following property: every r -clique containing a also contains b . Then Γ is complete. Proof
The relation → , defined by x → y if every r -clique containing x also con-tains y , is easily seen to be a quasiorder, and so the result follows from Lemma 4.4.16he following corollary follows from Lemma 4.4, Lemma 4.2, and Lemma 2.3. Corollary 4.5
Let Γ be a non-complete, non-null graph with primitive automor-phism group G , clique number r equal to its chromatic number, and valency k .Then for any two distinct vertices x , y we have | N ( x ) ∩ N ( y ) | ≤ k − . Proof
By Lemma 2.3, no two vertices of Γ have the same neighbourhood. Henceit suffices to show that there are no distinct vertices x , y satisfying | N ( x ) ∩ N ( y ) | = k − . For a contradiction, suppose that x and y have this property.Assume first that x and y are not adjacent, and let z be the unique element in N ( x ) \ N ( y ) . Now let C be a clique of size r containing x . As y N ( x ) ∪ { x } , itfollows from Lemma 4.2 that ( N ( x ) \ N ( y )) ∩ C = ∅ , and as N ( x ) \ N ( y ) = { z } ,it follows that z ∈ C . Therefore every r -clique containing x also contains z , andso by Lemma 4.4, Γ is complete, contradicting the hypotheses on Γ .If on the other hand x and y are adjacent, then N ( x ) ∪{ x } = N ( y ) ∪{ y } , whichdefines a non-trivial G -invariant equivalence relation on X , and so Γ is empty orcomplete, once again contradicting the hypotheses on Γ .We conclude that | N ( x ) ∩ N ( y ) | 6 = k − , and so | N ( x ) ∩ N ( y ) | ≤ k − . (cid:3) Theorem 4.1 is now an immediate consequence of the previous corollary andTheorem 2.1. The results above apply in particular to the graph Gr ′ ( S ) , where S = h G, f i , since the automorphism group of this graph contains the primitivegroup G . Proposition 4.6
Suppose that G is primitive and does not synchronize f . Let S = h G, f i . Then Gr ′ ( S ) has the following properties:(a) If x = y , then | N ( x ) ∩ N ( y ) | ≤ k − , where k is the valency of Gr ′ ( S ) .(b) If x and y are distinct, there exists a maximum clique in Gr ′ ( S ) containing x but not y . Proof
The first claim follows from Lemma 4.4 and Corollary 4.5. The second isclear since Gr ′ ( S ) has the same maximum cliques as Gr( S ) . (cid:3) n − The aim of this subsection is to prove the following:
Theorem 4.7
Primitive groups synchronize maps of rank n − . It is proved in [6] that a primitive group G synchronizes every map of kerneltype (4 , , . . . , . Therefore, to cover all maps of rank n − , we have to considerthe maps of kernel type (3 , , , . . . , and (2 , , , , . . . , .17 ernel type ( p, , , . . . , It was shown in [6] that every primitive group G synchronizes every map f ofkernel type ( p, , , . . . , for p = 2 , and for idempotent maps in the case of p = 3 .We will show that every primitive group G synchronizes every map f of kernel type ( p, , , . . . , , for p ≥ . This result was recently proved independently by Spigaand Verret [42]. Theorem 4.8
Let X be a set with at least elements, p ≥ , G a primitive groupacting on X and f ∈ T ( X ) a map of kernel type ( p, , , . . . , , that is, f hasone kernel class of size p , one kernel class of size , and an arbitrary number ofsingleton kernel classes. Then G synchronizes f . Proof
Let S = h G, f i , and Γ = Gr( S ) , let k be the valency of Γ . Assume that G does not synchronize f ; then Γ is not null by Theorem 2.1, and it is not completeeither as f has non-singleton kernel classes.Let A = { a , a } be the two-element kernel class of f and let B be its largestkernel class. Let b , b be distinct elements in B , and K = A ∪ B .Now let N B be the set of all vertices in K , the complement of K , that areadjacent to at least one element of B . As f maps N B injectively into N ( b f ) \{ a f } , we get | N B | ≤ k − .By definition we have N ( b ) , N ( b ) ⊆ N B ∪ A , and | N B ∪ A | = | N B | + | A | ≤ ( k −
1) + 2 = k + 1 . As | N ( b ) | = | N ( b ) | = k , it follows that | N ( b ) ∩ N ( b ) | ≥ k − by thepigeonhole principle. This contradicts Theorem 4.1, and so G synchronizes f . (cid:3) Kernel type (2 , , , , . . . , We are going to prove the following result:
Theorem 4.9
Let G act primitively on X and let f ∈ T ( X ) have kernel type (2 , , , , . . . , . Then G synchronizes f . Let S = h G, f i and Γ = Gr( S ) . By Theorem 2.1, S is a set of endomorphismsof Γ . Assume that G does not synchronize f ; then Γ is not null, once again byTheorem 2.1. Moreover Γ has clique number equal to its chromatic number. Let k be the valency of Γ .Let A = { a , a } , B = { b , b } , C = { c , c } be the non-singleton kernelclasses of f , and let K = A ∪ B ∪ C . 18y [39], the smallest non-synchronizing group has degree ; therefore, everyprimitive group of degree at most synchronizes every singular transformation;hence we can assume that n ≥ and so f of kernel type (2 , , , , . . . ) has at least singletons classes so that K = ∅ . As Γ has primitive automorphism group, itis connected and hence there is at least one edge between K and K , say at some a i ∈ A . We claim that there is an edge between A and B ∪ C . For the sake ofcontradiction, assume otherwise. Then, as f maps K injectively, both N ( a ) and N ( a ) are mapped injectively to N ( a f ) and as all of these sets have size k , weget that N ( a ) = N ( a ) , contradicting Lemma 2.3. So there is an edge between A and, say B , and hence between a f and b f .Repeating the same argument for the remaining class C we get that there mustalso be at least one edge between C and one of A or B . Up to a renaming of theclasses, we have two situations: Case 1: there are no edges between A and C .We exclude this case with an argument already used. For there are no edgesbetween A and C , and so any neighbour of a or a must lie in B ∪ N A , where N A is the set of elements in K that is adjacent to at least one of a , a .Now | N A | ≤ k − , as its elements are mapped injectively to N ( a f ) \{ b f } by f , and so | B ∪ N A | ≤ k +1 . By the pigeonhole principle, | N ( a ) ∩ N ( a ) | ≤ k − ,contradicting Theorem 4.1. Case 2: there are edges between every pair from A , B , and C , and hencetheir images a f, b f, c f form a 3-cycle.Consider the induced subgraph on Xf . We will obtain upper and lower boundson the number of edges in Xf , using methods analogous to those used in [6].Let e be the number of vertices in Γ , and let l be the number of edges within K . As Xf is obtained by deleting three vertices of X , the induced graph on Xf contains at most e − k + 3 edges (a loss of k edges at each vertex not in the imageof f , with at most edges counted twice).For the lower bound we count how many edges are at most sent to a commonimage by f . Let r, s, t be the number of edges between A and B , B and C , C and A , respectively, hence l = r + s + t . Since the sets A , B and C each have twoelements, it follows that r, s, t ≤ . These l edges are collapsed onto edges, sowe loose l − edges from within K .For each c ∈ K , such that ( c, a ) , ( c, a ) are edges, we map two edges intoone (and hence lose one). Now r + t edges connect A to K \ A . These edges are19onnecting to just two vertices, namely a and a , so one of them connects to atleast ⌈ ( r + t ) / ⌉ edges from within K . Hence there are at most k − ⌈ ( r + t ) / ⌉ edges between one of the a i and K , and so this is the maximal number of values c ∈ K for which ( c, a ) and ( c, a ) are edges. Symmetric arguments yield thefollowing result. Lemma 4.10
The transformation f identifies at most k − ⌈ ( r + t ) / ⌉ of the edgesbetween K and A , at most k − ⌈ ( r + s ) / ⌉ of the edges between K and B , and atmost k − ⌈ ( s + t ) / ⌉ of the edges between K and C . Hence the number of edges in Xf is at least e − ( k − ⌈ ( r + t ) / ⌉ ) | {z } loss in K – A − ( k − ⌈ ( r + s ) / ⌉ ) | {z } loss in K – B − ( k − ⌈ ( s + t ) / ⌉ ) | {z } loss in K – C − ( l − | {z } loss within K ≥≥ e − k + ( r + t ) / r + s ) / s + t ) / − ( l −
3) = e − k + 3 . (1)This equals our upper bound. It follows that all estimates used in deriving ourbounds must be tight. We have proved half of the following result. Lemma 4.11
Under the conditions of Case 2, and with notation as above, f iden-tifies exactly k − ( r + t ) / pairs of edges between K and A , k − ( r + s ) / pairsof edges between K and B , and k − ( s + t ) / pairs of the edges between K and C . In addition, N ( a ) ∩ K = N ( a ) ∩ K , N ( b ) ∩ K = N ( b ) ∩ K , N ( c ) ∩ K = N ( c ) ∩ K . Proof
As our upper and lower bounds agree, the estimates from Lemma 4.10 mustbe tight. Moreover, the inequality in (1) must be tight, as well, which implies that ( r + t ) / , ( r + s ) / , ( s + t ) / are equal to their ceilings and hence integers. Thisproves the first claim.Thus k − ( r + t ) , the number of edges between A and K , is an even number.In addition, k − ( r + t ) / , the number of edges between A and K identified by f , is then exactly half of the number of edges between A and K . However f canonly map at most two such edges onto one, as A has only elements. It followsthat if ( c, a ) is an edge with c ∈ K , then ( c, a ) is an edge as well, and vice versa.Hence N ( a ) ∩ K = N ( a ) ∩ K , and the remaining claims follow by symmetry. (cid:3) Theorem 4.1 implies that N ( b ) ∪ N ( b ) must contain at least four vertices thatare in exactly one of N ( b ) , N ( b ) . By Lemma 4.11, N ( b ) ∩ K = N ( b ) ∩ K .So the four vertices that are in exactly one of N ( b ) , N ( b ) must be a , a , c , c ,with each of b and b connected to exactly two of them, and so b and b have20o common neighbour in K . The same holds for the pairs from A and C . Thisshows that the two vertices adjacent to b cannot both lie in A , for otherwise a and a would be both adjacent to b . By symmetry each vertex is adjacent toexactly one element of the other non-singleton kernel classes. Each vertex with itstwo neighbours form a transversal for { A, B, C } ; thus there are only two possibletype of configurations: either the edges form two disjoint -cycles, both of whichintersect all of A , B , C or the edges form a -cycle that transverses A , B , C in aperiodic order. In different words, we can assume without loss of generality thatwe have a − b − c and a − b − c . Thus, either a − c (and hence we havetwo -cycles a − b − c − a and a − b − c − a , see Figure 5), or a − c and we have one -cycle a − b − c − a − b − c − a (see Figure 6). • ••• • • Figure 5: One of the two possible induced subgraphs on K •• ••• • ❖❖❖❖❖❖❖❖❖❖❖❖♦♦♦♦♦♦♦♦♦♦♦♦ Figure 6: One of the two possible induced subgraphs on K In either case, the triple transposition g = ( a a )( b b )( c c ) is an automor-phism of the induced subgraph on K . In fact, since N ( a ) ∩ K = N ( a ) ∩ K , N ( b ) ∩ K = N ( b ) ∩ K , N ( c ) ∩ K = N ( c ) ∩ K , the trivial extension of g isan automorphism of Γ . As explained in Subsection 4.1, this is impossible. So wehave: Theorem 4.12
Let G act primitively on X , and let f ∈ T ( X ) have kernel type (2 , , , , . . . , . Then G synchronizes f . (3 , , , . . . , (taking p = 3 in Theorem 4.8) and the results from [6] ( k = 4 in Theorem 2)about transformations of kernel type (4 , , . . . , , we get Theorem 4.7. In this section, we will exploit sets of vertices that share large number of adjacentvertices to show that certain kernel types are always synchronized. A consequenceof the results in this section is that primitive groups synchronize every transforma-tion of kernel type ( p, , , . . . , , for p ≥ . We also introduce some notation thatwill be very important in the next section.Let Γ be a regular graph with valency k , and let A ⊆ V (Γ) . We say that A is a small neighbourhood set of defect d if | ∪ a ∈ A N ( a ) | ≤ k + d .We will assume throughout this section that Γ is a graph with primitive auto-morphism group and clique number equal to its chromatic number. Lemma 5.1
Assume that A is a small neighbourhood set of defect in Γ of size l ≥ . Set N A = ∪ a ∈ A N ( a ) . Let x, y, z ∈ A be distinct, and w ∈ N A . Then(a) | N ( x ) ∩ N ( y ) | = k − ,(b) N ( x ) ∪ N ( y ) = N A ,(c) N ( z ) ⊆ N ( x ) ∪ N ( y ) ,(d) | N ( w ) ∩ A | ≥ l − ,(e) the elements of N A \ N ( z ) are in N ( x ) ∩ N ( y ) ,(f) N A contains at least l elements that are not adjacent to all elements of A . Proof
The first two claims follow from | N ( x ) ∪ N ( y ) | ≤ k +2 in connection withthe pigeonhole principle. The third follows from the second. For the fourth, assumethat x, y ∈ A \ N ( w ) , x = y . Then N ( x ) ∪ N ( y ) = N A , as w / ∈ N ( x ) ∪ N ( y ) ,for a contradiction. For (e) notice that any counterexample w would contradict (d).The last claim now follows from (e). (cid:3) Define l = 2 and l d = l d − + d for d ≥ . Lemma 5.2
For d ≥ , Γ does not contain any small neighbourhood set A ofdefect d and size l d . Proof
The proof is by induction on d . For d = 1 , notice that a small neighbor-hood set A of defect and size contradicts Corollary 4.5 in connection with thepigeonhole principle.So let d ≥ and assume that the result holds for smaller values of d . By wayof contradiction let A be a small neighbourhood set of defect d with l d distinct22lements. We may assume that | N A | = k + d . Let w ∈ N A . We claim that | N ( w ) ∩ A | > l d − l d − . Indeed, if A ′ ⊆ A \ N ( w ) for some A ′ with | A ′ | = l d − ,then ∪ a ∈ A ′ N ( a ) ⊆ N A \ { w } , and A ′ would be a small neighbourhood set ofdefect at most d − and size l d − . Such an A ′ does not exist by our inductiveassumption, and so | N ( w ) ∩ A | > l d − l d − .Now let x, y ∈ A , and g ∈ G such that xg ∈ N A , yg / ∈ N A . Such g exists byprimitivity. As N ( x ) ∪ N ( y ) is contained in a set of size k + d , N ( y ) intersects anysubset of N ( x ) of size d + 1 . The same holds for N ( yg ) and N ( xg ) . As xg ∈ N A ,there are at least l d − l d − + 1 = d + 1 elements in A ∩ N ( xg ) . Hence yg isconnected to one of those elements, and hence in N A , for a contradiction. (cid:3) We have everything needed to prove the main theorem of this section.
Theorem 5.3
Let d , . . . , d j ≥ be integers and d = − j + Σ d i . Let l ≥ l d . Then G synchronizes every map f of kernel type ( l, d , d , . . . , d j , , . . . , . Proof
Assume otherwise, and let A be the kernel class of size l of f , B i be theother non-singleton kernel classes of f , and x ∈ A . In Γ , the elements of N A allmap to N ( xf ) of size k . This set has at most k +( d − d − · · · +( d j −
1) = k + d preimages. It follows that A is a small neighbourhood set of defect d andsize at least l d , contradicting Lemma 5.2. (cid:3) Theorem 5.3 is applicable if j = 1 and d = 3 , in which case l d = 4 . In Subsection6.1, we will show that a primitive permutation group synchronizes every map ofkernel type (3 , , , . . . , . Together, these results imply the following corollary(see also Theorem 4.8). Corollary 5.4
Let p ≥ , and G a primitive permutation group on X . Then G synchronizes every transformation on X of kernel type ( p, , , . . . , . n − The aim of this section is to prove the following:
Theorem 6.1
Let G be a primitive group acting on a set of vertices X with | X | = n ≥ . Then G synchronizes every map of rank n − . We will first prove various auxiliary lemmas and describe our general proof strat-egy. The actual proofs involve a large number of subcases and will be divided overthe next three subsections, each of which covers a particular kernel class.23hroughout our proof of Theorem 6.1, we assume that G is a primitive groupof degree n over a set X , f is a transformation of rank n − , and G does notsynchronize f . We let Γ ′ = Gr ′ ( S ) be the graph constructed earlier for S = h G ∪ f i ) , k be the valency of Γ ′ , and r its clique size.The five possible kernel classes for a map of rank n − are (5 , , . . . , , (4 , , , . . . , , (3 , , , . . . , , (3 , , , , . . . , , and (2 , , , , , . . . , . If f isone of the first two types, the result was shown in [6] and Theorem 4.8. The re-maining three cases are covered in the following subsections.For each kernel type, we will denote by K the union of the non-singleton kernelclasses of f . For any given non-singleton kernel class Z , we let N Z = ∪ z ∈ Z N ( z ) ,and let N ′ Z = N Z ∩ K . We repeat that for all such Z , | N Z | ≥ k + 2 , as neigh-bourhoods of distinct elements in Z may only have intersection of size at most k − .We will distinguish several cases by the induced subgraph on the set Kf . Let Z be a non-singleton kernel class of f with image z ′ . Let Y , . . . , Y m be thosenon-singleton kernel classes that map to neighbours of z ′ . We refer to the number p Z = | Y | + · · · + | Y m | + ( k − m ) as the number of potential neighbours of Z , andto p ′ Z = k − m as the number of potential singleton kernel class neighbours of Z . Lemma 6.2 | N Z | ≤ p Z , | N ′ Z | ≤ p ′ Z . Proof z ′ has m neighbours that are images of non-singleton kernel classes andhence k − m neighbours that are either images of singleton kernel classes or not inthe image of f . If z ∈ Z and y is such that z − y , then y must be a preimage ofa neighbour of z ′ , hence y ∈ Y i for some i or y is the singleton class preimage ofone of remaining k − m elements of N ( z ′ ) . The results follow. (cid:3) For z ∈ X , let [ z ] denote the kernel class of f containing z . Lemma 6.3
Let r be the number of edges in the induced subgraph of K . Then r ≥ Σ z ∈ K ( k − p ′ [ z ] ) . Proof
For any given z ∈ K , all neighbours of z that lie in singleton kernel classesare in N ′ [ z ] . By the previous lemma | N ′ [ z ] | ≤ p ′ [ z ] . Hence z has at least k − p ′ [ z ] neighbours in Z . Summing over all z ∈ K , we obtain a lower bound on thenumber of pairs in the adjacency relation on K . The result follows. (cid:3) Lemma 6.4
Suppose that there are s non-singleton kernel classes, and that theinduced subgraph on Kf has r ′ edges. Let r be the number of edges in K , then r ≤ sk − r ′ − Σ | N ′ Z | + 6 , where the sum is over the non-singleton kernel classes Z of f . roof Consider the two induced graphs on X and Xf . We will estimate thedifference in their number of edges in two ways. Xf is obtained from X by deleting vertices, namely the non-images of f .Each of these is a vertex of k edges. Hence we lose k edges minus the numberthat we count twice because both of their vertices are non-images of f . There areat most such edges between vertices. Hence we lose at least k − edges.We obtain another estimate by comparing various subsets of edges and theirimages under f . We start with those edges that are within K : here r edges aremapped onto r ′ edges for a loss of r − r ′ .For each non-singleton kernel class, Z let r Z be the number of edges between Z and K \ Z . Then there are | Z | k − r Z edges between Z and elements in singletonkernel classes. These edges map to the | N ′ Z | edges between the image of Z and theimages of N ′ Z . Hence we have an effective loss of | Z | k − r Z − | N ′ Z | of edges.Finally we note that all edges between singleton classes are mapped injectivelyto other edges, so we do not encounter any loss for them.Summing up, we obtain a loss of at most ( r − r ′ ) + Σ (cid:0) | Z | k − r Z − | N ′ Z | (cid:1) = r − r ′ + Σ | Z | k − Σ r Z − Σ | N ′ Z | = r − r ′ + | K | k − r − Σ | N ′ Z | = | K | k − r − r ′ − Σ | N ′ Z | edges, where the sums are over the set of non-singleton kernel classes indexed by Z . Comparing with the lower bound k − , we get that r ≤ ( | K | k − r ′ − Σ | N ′ Z | ) − (4 k − | K | − k − r ′ − Σ | N ′ Z | + 6= sk − r ′ − Σ | N ′ Z | + 6 . (cid:3) In the following, we will only be dealing with kernel classes Z that satisfy p Z ∈{ k + 2 , k + 3 } . As p Z ≥ | N Z | ≥ k + 2 , in cases where p Z = k + 2 , we getthat p Z = | N Z | . Hence every potential neighbour of Z is in fact a neighbour.In particular, every potential singleton class neighbour is also a neighbour, whichimplies that | N ′ Z | = p ′ Z = k − m Z , where m Z is the number of neighbours of theimage of Z in Kf . In case that p Z = k + 3 , one potential neighbour might not bea neighbour (or might not exist, if the image of Z has a neighbour that is not in theimage of f ). Hence in this case | N ′ Z | ∈ { p ′ Z , p ′ Z − } .25 emma 6.5 Under the conditions of Lemma 6.4, assume that for all non-singletonkernel classes Z , p Z ∈ { k + 2 , k + 3 } . Let d be the number of kernel classes forwhich p Z = k + 3 . Then r ≤ r ′ + d + 6 .Moreover, for each Z , let m Z be the number of neighbours of the image of Z that lie in Kf . If r = r ′ + i + 6 , for some ≤ i ≤ d , then there are at least i non-singleton kernel classes Z for which | N ′ Z | = p ′ Z − k − m Z − . Proof
By Lemma 6.4, r ≤ sk − r ′ − Σ | N ′ Z | +6 , and as pointed out after the lemma,we have | N ′ Z | = k − m Z , if p z = k + 2 , or | N ′ Z | ≥ k − m Z − , if p z = k + 3 .Assume that there are exactly j kernel classes Z for which | N ′ Z | = k − m Z − .Then r ≤ sk − r ′ − Σ | N ′ Z | + 6= sk − r ′ − (Σ( k − m Z ) − j ) + 6= sk − r ′ − (cid:0) sk − r ′ (cid:1) + j + 6= r ′ + j + 6 . | N ′ Z | = k − m Z − implies that p Z = k +3 , therefore j ≤ d , and the first statementof the lemma follows. Assuming r = r ′ + i + 6 , we obtain i ≤ j , which shows thesecond statement. (cid:3) Our proof of Theorem 6.1 proceeds by considering for each kernel class all poten-tial combinations of induced subgraphs on K and Kf . All configurations whosenumber of edges lie within the bounds of Lemmas 6.3 and 6.5 will be further re-stricted and eventually excluded.One of our most common arguments will be to construct a contradiction toLemma 2.4. As we will use this construction extensively, we will introduce somespecial notation for it. By a CME – standing for clique minus one edge – we mean aset of vertices of size r + 1 that contains at most one non-edge, i.e., a configurationthat violates either Lemma 2.4 or the fact that r is the clique number of Γ ′ .For distinct vertices x, y, z , with x − y , the expression CME ( x − y, z ) meansthat for any r -clique L that contains the edge from x to y (whose existence followsfrom the definition of Γ ′ ), the set L ∪ { z } is a CME. A typical application will bethat z is in the same kernel class as one of x or y , and adjacent to the other one.Often we will have that N ′ [ z ] ⊆ N ( z ) due to z having not enough neighbours in K to omit a vertex from N ′ [ z ] . It then just remains to check that all vertices in K adjacent to both x and y are also adjacent to z .Another tool is to utilize small neighbourhood sets of defect . We always havesuch a set of size at least available if we have a kernel class Z with p Z = k + 2 .By transitivity of G , every element is then part of such a set. The following lemmasdraw consequences in these cases. 26 emma 6.6 Let x, y ∈ X , such that in Γ , | N ( x ) ∩ N ( y ) | = k − .(a) x and y are non-adjacent.(b) Suppose that Z is a kernel class of f such that N ( x ) ∩ Z = ∅ 6 = N ( y ) ∩ Z ,but that Z ∩ N ( x ) = Z ∩ N ( y ) . Then xf = yf . Proof
Assume that x and y are adjacent. Then x ∈ N ( y ) \ N ( x ) . Let z bethe other element of N ( y ) \ N ( x ) . By Proposition 4.6, there exists an r -clique L containing y , but not containing x . Then L ∪ { x } is a CME, as it has r + 1 elementsand at most one non-edge between x and z . By contradiction, we obtain (a).Now in the situation of (b), say w.l.o.g. that z ∈ ( Z ∩ N ( y )) \ N ( x ) . Let L bean r -clique containing y and avoiding the unique element in N ( y ) \ ( N ( x ) ∪ { z } ) .Then z ∈ L for otherwise L ′ = L ∪ { x } is a CME. Hence z ∈ L ′ , and L ′ ismissing two edges, namely ( x, y ) and ( x, z ) . Now Z ∩ N ( x ) = ∅ , hence there isan edge from x to an element of Z , and hence the non-edge ( x, z ) maps to the edge ( xf, zf ) . It follows that L ′ f cannot have r + 1 elements, for otherwise it would bea CME. So f must identify two elements of L ′ . These cannot be any elements ofthe clique L . x is adjacent to all elements of L \ { y, z } , and ( xf, zf ) is an edge.Thus xf = yf by elimination. (cid:3) Lemma 6.7
Suppose that in Γ ′ we have a small neighbourhood set of defect andsize at least . Then there exist vertices x, y, z ∈ Γ ′ such that | N ( x ) ∩ N ( y ) | = k − , | N ( y ) ∩ N ( z ) | = k − , | N ( x ) ∩ N ( y ) ∩ N ( z ) | < k − . Moreover, suchtriples exist for any chosen vertex y . Proof
Let ∼ be the relation on Γ ′ defined by x ∼ y if either x = y or | N ( x ) ∩ N ( y ) | = k − . The relation ∼ is clearly reflexive, symmetric, and preserved by G . Assume that for all x, y, z ∈ Γ ′ , | N ( x ) ∩ N ( y ) | = k − | N ( y ) ∩ N ( z ) | implies that | N ( x ) ∩ N ( z ) | = k − . Our assumption means that ∼ is transitiveand hence a G -compatible equivalence relation on X . By primitivity of G , ∼ istrivial or universal. However, ∼ is non-trivial as we assumed that Γ ′ has a smallneighbourhood set of defect , and it is not universal, as adjacent elements of Γ ′ are not in ∼ by Lemma 6.6(a). By contradiction, there exist x, y, z ∈ Γ ′ , with | N ( x ) ∩ N ( y ) | = k − | N ( y ) ∩ N ( z ) | , and k − > | N ( x ) ∩ N ( z ) | ≥| N ( x ) ∩ N ( y ) ∩ N ( z ) | .The last assertion follows from the transitivity of G . (cid:3) Lemma 6.8
Suppose that in Γ ′ we have a small neighbourhood set of defect andsize at least . Let y ∈ Γ ′ , and y ′ , ¯ y ∈ N ( y ) , y ′ = ¯ y . Then there exists a w ∈ Γ ′ such that | N ( y ) ∩ N ( w ) | = k − and N ( w ) ∩ { y ′ , ¯ y } 6 = ∅ . roof Given y , let x, z be the elements constructed in Lemma 6.7. Then | N ( x ) ∩ N ( y ) | = | N ( z ) ∩ N ( y ) | = k − . We claim that one of x, z is adjacent to an elementof { y ′ , ¯ y } . For assume otherwise, then | N ( x ) ∩ N ( y ) ∩ N ( z ) | = | N ( y ) \{ y ′ , ¯ y }| = k − , contradicting Lemma 6.7. The result follows. (cid:3) Lemma 6.9
Let x, y ∈ Γ ′ , xf = yf , such that { x, y } is a small neighbourhoodset of defect 2. Let N = N ( x ) ∩ N ( y ) . If for every non-singleton kernel class Z of f , | N ∩ Z | ≤ , then xf and yf are non-adjacent. Proof As { x, y } is a small neighbourhood set of defect 2, | N | = k − . Consider N f . As | N ∩ Z | ≤ for all kernel classes Z , f maps N injectively, and so | N f | = k − . Moreover, x, y are adjacent to every element in N , and as xf = yf , N ∪ { x, y } is mapped injectively by f , as well. It follows that | N ( xf ) ∩ N ( yf ) | ≥| N f | = k − , which implies that { xf, yf } are also a small neighbourhood set ofdefect 2. The result now follows with Lemma 6.6 (a). (cid:3) Lemma 6.10
Let A , A be small neighbourhood sets of defect and size . If | A ∩ A | ≥ then A = A . Proof
Let A = { x, y, z } , A = { x, y, z } . Then k + 2 ≤ | N { x,y } | ≤ | N A | = k + 2 , and so N { x,y } = N A . Symmetrically, N A = N { x,y } = N A which implies that N A ∪ A = N A , and so | N A ∪ A | = k + 2 . By Lemma 5.2, there are no smallneighbourhood sets of defect and size , hence z = z and A = A . (cid:3) (3 , , , . . . , Let f be a map of kernel type (3 , , , . . . , , A = { a , a , a } , B = { b , b , b } the non-singleton kernel classes of f , and assume that Γ has r edges between A and B . In order for p A ≥ k + 2 , the images of A and B need to be connected andwe get k + 2 = p A = | N A | = p B = | N B | and | N ′ A | = | N ′ B | = k − . Hence A, B are small neighbourhood sets of defect .Our next goal is to bound r . By Lemma 5.1, every element of N A is adjacentto at least elements in A , hence r ≥ . Lemma 6.5 shows that r ≤ .We will treat the two cases r = 6 , simultaneously. If r = 6 , then everyelement of B is adjacent to exactly elements of A and vice versa. If r = 7 thenexactly one element of A is adjacent to all vertices in B , exactly one element of B is adjacent to all vertices in A , and the remaining elements of A ∪ B have exactly neighbours in K . Hence w.l.o.g., we may assume that all edges in A ∪ B lie on28he -cycle a − b − a − b − a − b − a , except for potentially an extra edgebetween a and b in case that r = 7 . These two configurations are depicted inFigures 7 and 8. ••• ••• ❚❚❚❚❚❚❚❚❚❚❚❥❥❥❥❥❥❥❥❥❥❥❚❚❚❚❚❚❚❚❚❚❚ ❥❥❥❥❥❥❥❥❥❥❥ a a a b b b Figure 7: The induced subgraph on K with edges ••• ••• ❚❚❚❚❚❚❚❚❚❚❚❥❥❥❥❥❥❥❥❥❥❥❚❚❚❚❚❚❚❚❚❚❚ ❥❥❥❥❥❥❥❥❥❥❥ a a a b b b Figure 8: The induced subgraph on K with edges Lemma 6.11
There exist unique elements z ∈ N ′ A , c ∈ N ′ B that are not adjacentto a , b , respectively. Moreover, c is adjacent to a . Proof
We have that | N ( b ) ∩ A | = 2 . It follows that | N ( b ) ∩ N ′ B | = k − . As | N ′ B | = k − , there is exactly one element c in N ′ B that is not connected to b .The existence and uniqueness of z follow symmetrically. By (d) of Lemma 5.1, wehave the edges b − c − b , and a − z − a .Now consider an r -clique L containing the edge a − b . We have that b − a and L \{ a , b } ⊆ N ′ B ⊆ N ( b ) ∪{ c } . It follows that c ∈ L for otherwise L ∪{ b } would be a CME, missing only an edge between b and b . Hence c − a .The construction from this lemma is depicted in Figure 9. Note that theremay be additional edges that are not depicted, except for the confirmed non-edges ( c, b ) , ( z, a ) . The dotted edge is the additional edge in the case r = 7 . (cid:3) Let g ∈ G be such that a g ∈ A , a g / ∈ A . Consider A ′ = Ag − . It is a smallneighbourhood set of defect , as A has this property. Moreover a ∈ A ∩ A ′ but A = A ′ , as a g / ∈ A . By Lemma 6.10, A ′ ∩ A = { a } . Let A ′ = { a , x, y } . b ∈ N ( a ) and hence by (d) of Lemma 5.1, one element of x, y , say x , must beadjacent to b . Hence x ∈ N B \ A . 29 •• z • ••• ❱❱❱❱❱❱❱❱ ❡❡❡❡❡❡❡❡ ❚❚❚❚❚❚❚❚❚❚❚❥❥❥❥❥❥❥❥❥❥❥❚❚❚❚❚❚❚❚❚❚❚ ❴❴❴❴❴❴ ❥❥❥❥❥❥❥❥❥❥❥ a a a b b b c • ✐✐✐✐✐✐✐✐❩❩❩❩❩❩❩❩ ★ Figure 9: The construction from Lemma 6.11.As xf = a f , by Lemma 6.6, N ( a ) ∩ B = N ( x ) ∩ B = { b , b } . As b / ∈ N ( x ) , x = c , where c is from Lemma 6.11. By the same lemma, we have that x − a . As x is not adjacent to a , we have that x = c = z once again by Lemma6.11.Now, consider the third element y of A ′ . If y would be adjacent to b or b ,repeating the argument from the previous paragraph yields y = c . As x = y , itfollows that y is not adjacent to b or b . As | N ( c ) ∪ N ( y ) | = | N ( x ) ∪ N ( y ) | = k + 2 , it follows that y is adjacent to every element in N ( c ) \ { b , b } . So y ∈ N ( a ) andhence y ∈ N A . As y / ∈ N ( a ) the uniqueness of z = x implies that y / ∈ N ′ A . So y ∈ B , and hence y = b , as b , b are adjacent to a . It follows that { a , b } is asmall neighbourhood set of defect .Consider N = N ( a ) ∩ N ( b ) of size k − . N has no elements in K , andhence | N ∩ Z | ≤ for all kernel classes Z of f . By Lemma 6.9, a f and b f arenon-adjacent; however, this is false in our construction.Our assumption was that G synchronizes the transformation f . Hence by con-tradiction, Theorem 6.1 holds for transformations of kernel type (3 , , , . . . , . (3 , , , , . . . , Let A = { a , a } , B = { b , b , b } , C = { c , c } be the non-singleton kernelclasses of f . The requirement that p Z ≥ k + 2 for all kernel classes Z implies that Kf is connected. Hence the induced graph on Kf is a -path or a triangle. The induced graph on Kf is a -path The requirement that p B ≥ k + 2 implies that there must be edges from B toboth A and C , hence a f − b f − c f . 30et r be the number of edges in K , by Lemma 6.5 we conclude that r ≤ .As | N ′ B | = k − , each element of B has at least neighbours in K . Together,these constraints imply that at least one element of B has exactly neighbours in K . If this holds for all elements of B , then for at least two distinct x, y ∈ B , N ( x ) ∩ N ( y ) ∩ K = ∅ . Otherwise, there are x, y ∈ B , with | N ( x ) ∩ K | = 2 , | N ( y ) ∩ K | ≥ . In both cases, x, y ∈ B satisfy | N ( x ) ∩ K | = 2 , and N ( x ) ∩ N ( y ) ∩ K = ∅ . Say w.l.o.g. that x = b , y = b , and b − c − b .We claim that we have a CME ( b − c , b ) . For let L be an r -clique containing b , c , then L \ { b , c } ⊆ N ′ B . However N ′ B ⊆ N ( b ) , as | N ′ B | = k − , and b has only two neighbours in K . Hence L ∪ { b } is only missing one edge between b and b , and is a CME.By contradiction, we can exclude the case that Kf is a -path. The images of the non-trivial kernel classes form a triangle
In this case, B is a small neighbourhood set of defect , and A and C aresmall neighbourhood sets of defect or . Lemma 6.5 shows that the number ofedges r in K satisfies r ≤ . Moreover, by the same lemma if r = 11 , then | N ′ A | = k − | N ′ C | , and if r = 10 then | N ′ A | = k − or | N ′ C | = k − . We willassume w.l.o.g. that | N ′ A | ≤ k − whenever r = 10 . Moreover, if r = 11 we willassume w.l.o.g. that there are at least as many edges from B to C as there are from B to A . Lemma 6.12
Each element of x ∈ A ∪ C is adjacent to at least 2 elements of B ,and there is at least one edge from A to C . Proof
The first part follows from property (d) of Lemma 5.1. If there would beno edges between A and C , then a , a could only be adjacent to the elements in B and the k − elements in N ′ A , leaving | N A | ≤ k + 1 , for a contradiction. (cid:3) Lemma 6.12 implies that r ≥ , hence K contains , , or edges. Lemma 6.13
There exists an element x ∈ C that is adjacent to exactly one elementof A . Proof
Lemma 6.12 together with the fact that r satisfies ≤ r ≤ implies thatthere are to edges from A to C . The statement of the Lemma is true unlessthere are exactly edges from A to C that share a vertex in C . Say w.l.o.g. thatthese are the edges a − c − a , so c / ∈ N A . Now, with the results of Lemma6.12, the edges between A and C require that r ≥ , and hence | N ′ A | ≤ k − by assumption. But then | N A | ≤ | N ′ A | + | B | + |{ c }| ≤ k + 1 , | N A | ≥ k + 2 . (cid:3) Hence, we may assume that c − a , and that c is non-adjacent to a . The followingfigure depicts the minimal amount of edges in K . •• a a ••• b b b · · ·· · ·· · ·· · ·· · ·· · · •• c c ❨❨❨❨❨❨❨❨❨❢❢❢❢❢❢❢❢❡❡❡❡❡❡❡❡❡❱❱❱❱❱❱❱❱❡❡❡❡❡❡❡❡❡ ❨❨❨❨❨❨❨❨❨ ❨❨❨❨❨❨❨❨❨ ❤❤❤❤❤❤❤❤❤ ❑ By transitivity of G , there exists a small neighbourhood set D (the image of B under some g ∈ G ) of defect and size with c ∈ D . As a − c , by Lemma5.1(c), there exists d ∈ D, d = c with a − d . Hence, d ∈ N ′ A ∪ B ∪ { c } . Thefollowing lemmas will examine these possibilities. Lemma 6.14 d / ∈ B. Proof
Assume otherwise, say that d = b . Consider the set N = N ( b ) ∩ N ( c ) with | N | = k − . Then a is the only element in N ∩ A , as c is not adjacent to a . The other elements of N may not be in B or C , as b and c are, and hence arein singleton classes. x(cid:127)y~z}{| N \ { a }•• a a ••• b b b · · ·· · ·· · ·· · · •• c c ♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞❤❤❤❤❤❤❙❙❙❙❙❙❙❨❨❨❨❨❨❨❨❨ ❜❜❜❜❜❜❜❜❜❜❜❜❜❜❜❜❜ ❨❨❨❨❨❨❨❨❨ ❤❤❤❤❤❤❤❤❤ ★★★★★★★★★★ ✛✛✛✛✛✛✛✛✛✛ ✾✾✾✾✾✾✾✾✾✾✾✾✾✾✾✾✾✾✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻❀❀❀❀❀❀❀❀❀❀❀❀❀❀❀❀❀❀❀❑ By Lemma 6.9, b f and c f are non-adjacent. However, this is false, for acontradiction. (cid:3) emma 6.15 d = c . Proof
Assume otherwise. Then c − a , and there are at least two edges between A and C . Together with at least edges from A to B , there are at most edgesfrom B to C . As p B = k + 2 , we have B ⊆ N C , and with at most availableedges, it follows that N ( c ) ∩ B = N ( c ) ∩ B .Now let e / ∈ { c , c } be the third element of D . We have that B ⊆ N ( c ) ∪ N ( c ) = N D . As | N D | = k + 2 and | B | = 3 , e ∈ N B . •• a a hoinjmkl N ( c ) ∩ B hoinjmkl N ( c ) ∩ B • e ✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱✱ ?? B · · ·· · ·· · ·· · · •• c c ❯❯❯❯❯❯ ❧❧❧❧❧❧ ❯❯❯❯❯❯ ❧❧❧❧❧❧ ✍❅❭❭❭❭❭❭❭❭❭❭❭❭❭❳❳❳❳❳❳❳❳❳❳❳❳❳ Now, N ( e ) ∩ B must differ from one of N ( c ) ∩ B , N ( c ) ∩ B . This contradictsLemma 6.6(b), for ef = c f = c f . (cid:3) Lemma 6.16 d / ∈ N ′ A . Proof
Assume otherwise. By Lemma 6.6(b), N ( d ) ∩ A = N ( c ) ∩ A = { a } , andso a / ∈ N ( d ) . This implies that a must have at least k − ( | N ′ A | − neighboursin K .Now, if r = 9 , then | N ′ A | ≤ k − , and so a requires at least neighbors in K .However, Lemma 6.12 accounts for all edges in K , showing that a has exactly neighbors in K (recall that the edge from A to C was assumed to be a − c ).This excludes the case r = 9 .If r ≥ , then | N ′ A | = k − , and so a requires at least neighbors in K ,which must be the elements of B ∪ { c } . With edges from a to B , a − c ,33 − c , edges from a to B , and edges between B and C , we see that r = 11 . ••• da a ••• b b b · · ·· · ·· · ·· · ·· · · •• c c ❨❨❨❨❨❨❨❨❨❢❢❢❢❢❢❢❢❡❡❡❡❡❡❡❡❡❱❱❱❱❱❱❱❱❡❡❡❡❡❡❡❡❡ ❨❨❨❨❨❨❨❨❨ ♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠♠ ❭❭❭❭❭❭❭❭❭❭❭❭❭❭❭❭❭ ❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞❞ ❑t However, for the case that r = 11 , we assumed that there are at least as manyedges from B to C as there are from B to A . Our final configuration violates thisassumption, for a contradiction. (cid:3) We have excluded every possible location for d . Therefore, Theorem 6.1 holds fortransformations f of kernel type (3 , , , , . . . , . (2 , , , , , . . . , Let
A, B, C, D be the non-singleton kernel classes of f , and let A = { a , a } , B = { b , b } , C = { c , c } , D = { d , d } .For each kernel class Z with image z ′ , p Z ≥ k + 2 implies that z ′ must be ad-jacent to at least other images of non-singleton kernel classes. Hence the inducedsubgraph on Kf must have , , or edges, and in the last case, these must form a -cycle.Throughout, g will denote the transformation ( a a )( b b )( c c )( d d ) . Asnoted in Subsection 4.1, we are done if we can establish that g is an automorphismof Γ ′ . The image of K has edges arranged in a cycle We may suppose that the images of the non-singleton kernel classes are a f − b f − c f − d f − a f . In this case each non-trivial kernel class Z satisfies p Z = k + 2 , and is hence a small neighbourhood set of defect . Hence Z ⊆ N Z for every pair ( Z , Z ) of adjacent kernel classes. In particular, there are at leasttwo edges between each such pair.Let r be the number of edges in K . By Lemma 6.3 and Lemma 6.5, we have ≤ r ≤ . 34 contains edges Here there are exactly two edges between each pair ( Z , Z ) of adjacent kernelclasses. Now Z ⊆ N Z and Z ⊆ N Z is only possible if the two edges between Z and Z have disjoint vertices. The only two possible configurations are depictedin Figures 10 and 11. •• •• ••• • Figure 10: One of the two configuration with edges •• •• ••• • ❖❖❖❖❖❖❖❖❖❖❖❖♦♦♦♦♦♦♦♦♦♦♦♦ Figure 11: One of the two configuration with edgesIt is now easy to check that g is an automorphism of Γ ′ , for a contradiction. K contains edges We may assume that A and B are the unique non-singleton kernel classes thathave edges between them, and that b − a − b − a . However in this case, wehave N ′ B ⊆ N ( b ) , which implies the CME ( a − b , b ) , for a contradiction. K contains edges Suppose first that we have two kernel classes that have only three edges be-tween them, say A and B with edges b − a − b − a . By the number of availableedges, at least one of b , a is a vertex of only two edges from within K . Henceeither N ′ B ⊆ N ( b ) or N ′ A ⊆ N ( a ) , and so we have the CME ( a − b , b ) orCME ( a − b , a ) , as in the case that K contains edges.35o assume instead that there are two kernel classes with edges between them,say A and B ; then we may assume that all edges within K are a − b − a − b − a , b − c − d − a , and b − c − d − a (see Figure 12). We have N ′ C ⊆ N ( c ) ∩ N ( c ) and N ′ D ⊆ N ( d ) ∩ N ( d ) . Moreover N ′ B consists of k − elements that areadjacent to both b and b , one element b ′ adjacent to b but not b , and one element b ′ adjacent to b but not b . •• d d •• a a •• b b •• c c ❖❖❖❖❖❖❖❖❖❖❖❖ ♦♦♦♦♦♦♦♦♦♦♦♦ t❑ Figure 12: The remaining configuration for K with edges Lemma 6.17 | N ( b ′ ) ∩ N ( c ) | = k − . Proof
By Lemma 6.8 applied to y = c , y ′ = b , ¯ y = d , there exist z ∈ Γ ′ with | N ( c ) ∩ N ( z ) | = k − , such that z is adjacent to one of b , d . We want to narrowthe location of z .As N ( z ) ∩ { b , d } 6 = ∅ , z ∈ { a , a } ∪ ( N ′ B \ { b ′ } ) ∪ N ′ D . If z ∈ { a , a } then N ( z ) ∩ B = B = { b } = N ( c ) ∩ B , contradicting Lemma 6.6. Similarly,if z ∈ N ′ D then N ( z ) ∩ D = D = { d } = N ( c ) ∩ D , and if z ∈ N ′ B \ { b ′ , b ′ } then N ( z ) ∩ B = B = { b } = N ( c ) ∩ B . Hence z = b ′ . (cid:3) Lemma 6.18 N ( b ′ ) = { a , a , b } ∪ H where H ⊆ N ′ C . Proof b ′ must be in every r -clique containing a − b , for otherwise we obtaina CME ( a − b , b ) . Hence a ∈ N ( b ′ ) . Similarly, a ∈ N ( b ′ ) to avoid aCME( a − b , b ) .Hence N ( b ′ ) \ N ( c ) = { a , a } . By Lemma 6.17, all remaining neighbours of b ′ are in N ( c ) . One of those elements is b . If d ∈ N ( b ′ ) , then N ( b ′ ) ∩ A = A = { a } = N ( d ) ∩ A , contradicting Lemma 6.6. Hence N ( b ) \ { a , a , b } ⊆ N ′ C . (cid:3) Lemma 6.19
There exists x ∈ Γ ′ such that | N ( b ) ∩ N ( x ) | = k − , x is adjacentto b ′ , and x is not adjacent to c . roof By Lemma 6.8 applied to y = b , y ′ = b ′ , ¯ y = c , there exist x ∈ Γ ′ with | N ( b ) ∩ N ( x ) | = k − , such that x is adjacent to one of b ′ , c .If x is adjacent to c then either x = d or x ∈ N ′ C . Now if x = d then N ( b ) ∩ A = A = { a } = N ( d ) ∩ A , contradicting Lemma 6.6. Similarly, if x ∈ N ′ C , then N ( x ) ∩ C = C = { c } = N ( b ) ∩ C . Hence x ∈ N ( b ′ ) \ N ( c ) . (cid:3) By Lemma 6.18, we get that x ∈ A . This implies that b − x , contradicting Lemma6.6(a). The image of K has edges We may assume that the edges in the image of K are a f − b f − c f − d f − a f − c f . Hence B and D are small neighbourhood classes of defect and A and C are small neighbourhood classes of defect or . As p B = p D = k − , thereare at least edges between each kernel class pair in { A, C } × {
B, D } .Let r be the number of edges in K ; by Lemma 6.3 and Lemma 6.5, we obtain ≤ r ≤ . Moreover, r = 13 implies that | N ′ A | = k − | N ′ C | , and r = 12 implies that | N ′ A | = k − or | N ′ C | = k − . Lemma 6.20
Let z ∈ B ∪ D . Suppose that | N ( z ) ∩ K | = 2 . Then | N ( z ) ∩ A | =1 = | N ( z ) ∩ C | . Proof
Suppose otherwise, say w.l.o.g that N ( b ) ∩ K = A .We claim that one element x ∈ A satisfies N ′ A ⊆ N ( x ) . If r ≤ , then atmost edges have a vertex in A , as at least edges lie between C, B and
C, D .Thus one of a , a must have k − | N ′ A | neighbours outside of K .If r ≥ , then | N ′ A | = k − or | N ′ C | = k − . However b / ∈ N C , and so N C ∩ K has at most elements. As | N C | ≥ k + 2 , it follows that | N ′ C | = k − ,and so | N ′ A | = k − . Because r ≤ at most edges have a vertex in A , so oneof a , a must have k − neighbours outside of K .In either case N ′ A ⊆ N ( x ) for some x ∈ A , say for a . However, we nowhave a CME( b − a , a ) , for a contradiction. So | N ( b ) ∩ A | = 1 , and thus | N ( b ) ∩ C | = 1 . (cid:3) Lemma 6.21 Γ ′ has at least 10 edges that lie between the pairs of kernel classesfrom { A, C } × {
B, D } . Proof
Assume to the contrary that there are at most edges between the pairs ofkernel classes from { A, C } × {
B, D } . We will construct a contradiction to Lemma6.9.As there are at least two edges between the pairs in { A, C } × {
B, D } , eachpair has either or edges between them, with at most one case of edges. We37ay assume that the exceptional pair in the case of edges is ( C, D ) . ApplyingLemma 6.20 to the or vertices z ∈ B ∪ D that have exactly two neighboursin K , we see that if there are two edges between any pair ( Y, Z ) of kernel classes,those edges have disjoint vertices.Hence, w.l.o.g. we may assume that we have the edges b − a − d and b − a − d . In case that there are edges between C and D , we may further assumethat d is the unique vertex in D with neighbours in K . Applying Lemma 6.8 with y = a , y ′ = b , ¯ y = d , we see that there is a z such that | N ( a ) ∩ N ( z ) | = k − ,with z adjacent to b or d .We claim that z ∈ C . As z ∈ N ( b ) ∪ N ( d ) , we have z ∈ C ∪ N ′ B ∪ N ′ D .Now for all w ∈ N ′ D , w ∈ N ( d ) as d has only two neighbours in K . Hence N ( w ) ∩ D = { d } = N ( a ) ∩ D , and so z / ∈ N ′ D by Lemma 6.6(b). An analogousargument show that z / ∈ N ′ B , and so z ∈ C .Let N = N ( a ) ∩ N ( z ) . We claim that for every non-singleton kernel class Z of f , | N ∩ Z | ≤ . N ∩ B ⊆ N ( a ) ∩ B = { b } and N ∩ D ⊆ N ( a ) ∩ D = { d } , sothe claim holds for Z = B and Z = D . Moreover, N does not have any elementsin A or C , as a ∈ A, z ∈ C .Hence Lemma 6.9 is applicable to N . By the lemma a f and zf are non-adjacent. However, we have that a f − c f = zf , as z ∈ C , for a contradiction. (cid:3) K contains or edges By Lemma 6.21, in these cases there is at most one edge between A and C .Our next Lemma shows that this is not possible, for a contradiction. Lemma 6.22 If r ≤ , there are at least two edges from A to C . Proof
At least one edge must cross from A to C , for otherwise not all elementsin A ∪ C could have neighbours in K .Assume that there is only one edge between A and C . As at least edges gofrom A to K \ A , there must be at least from A to B ∪ D , and by symmetry atleast edges from C to B ∪ D . This accounts for the maximum edges. Hencethere are exactly edges from A to B ∪ D .W.l.o.g. we may assume that there are edges from A to D , say a − d − a − d , and edge from A to B . The two edges from A to B must be adjacent todifferent elements of A as A ⊆ N B . This implies that the edge between A and C is adjacent to a , and hence N ( a ) ∩ C = ∅ . Moreover, N ( a ) ∩ ¯ K = N ′ A , as a has only three neighbours in K .However, we now obtain CME ( a − d , a ) for a contradiction. Hence thereare at least two edges between A and C . (cid:3) contains edges In this case | N ′ A | = k − or | N ′ C | = k − , say | N ′ A | = k − . Hence at least edges go from A to K \ A , while at least edges go from C to K \ C . With r = 12 this implies that at least edges lie between A and C . With Lemma 6.21, we seethat there are exactly edges between A and C . As C needs to be contained in N B and N D , there exactly edges each between ( C, B ) and ( C, D ) . This leaves edges between ( A, B ) and ( A, D ) , and all edges are accounted for. Hence a , a are both adjacent to exactly elements in K , and thus N ′ A ⊆ N ( a ) ∩ N ( a ) .Assume first that there are edges between each of these pairs, where we mayassume that b − a − b − a . We have CME ( a − b , a ) , unless there is anelement in C (which we may assume to be c ) such that b − c − a , and that a isnot adjacent to c . This implies that the second edge between B and C is b − c ,and so in particular c / ∈ N ( b ) . But then N ( a ) ∩ N ( b ) ∩ C = ∅ , and we obtainCME ( a − b , a ) , for a contradiction.Up to symmetry, the only remaining option is that there are edges between A and B , and edges between ( A, D ) . We obtain a CME ( a − b , a ) , unless oneelement of C , say c , satisfies a − c − b and c / ∈ N ( a ) . However, we nowobtain a CME ( a − b , a ) , unless there exists x ∈ C satisfying a − x − b and that x / ∈ N ( a ) . x = c , for otherwise c / ∈ N B , as there are only two edges from C to B . Hence x = c and N ( a ) ∩ C = ∅ . Finally, we obtain the CME ( a − b , a ) ,for a contradiction.Hence we can exclude the possibility that K has edges. K contains edges By Lemma 6.5, | N ′ A | = | N ′ C | = k − . Hence each element of A ∪ C has at least neighbours in K , and as r = 13 this is only possible if there are at least edgesfrom A to C . In fact, Lemma 6.21 show that there are exactly edges between A and C , which in turn implies that each x ∈ A ∪ C has exactly neighbours in K .This implies that N ′ A ⊆ N ( a ) ∩ N ( a ) .Up to symmetry, we may assume that there are edges from A to B , and edges from A to D , say that b − a − b − a . As there are 3 edges between A and C one of a , a is adjacent to both elements in C . This must be a , for otherwise a has neighbours in B ∪ C and could not be in N D . So c − a − c . But thenwe have a CME ( a − b , a ) for a final contradiction. The image of K has edges Now let r be the number of edges between the elements of K . By Lemmas 6.3and 6.5 we get ≤ r ≤ . Moreover, by Lemma 6.5, if p = r − , there are atleast p non-singleton kernel classes Z for which | N ′ Z | ≤ k − .39onversely, if there are p non-singleton kernel classes Z for which | N ′ Z | ≤ k − , there are at least edges from each such Z to K \ Z and at least edges fromany other class Y to K \ Y . This requires at least (8 p + 6(4 − p )) / p = r edges. Hence if there are
12 + p edges, there are exactly p kernel classes X forwhich | N ′ X | = k − , and exactly − p kernel classes with | N ′ Z | = k − . As thisaccounts for all edges, we have proved the following lemma. Lemma 6.23
Let Z be a non-singleton kernel class, and x ∈ Z . If | N ′ Z | = k − ,then x has exactly neighbours in K . If | N ′ Z | = k − , then x has exactly neighbours in K . In particular, N ′ Z ⊆ N ( x ) . Lemma 6.24
Let x ∈ Z , where Z is a kernel class with | N ′ Z | = k − . Then allthree neighbours of x in K lie in different kernel classes. Proof
Suppose otherwise, say w.l.o.g. that x = b , and that a − b − a . Thenwe have CME ( a − b , a ) , unless there exists x ∈ C ∪ D satisfying a − x − b and that x / ∈ N ( a ) . Hence a , a , x account for all neighbours of b in K . Butnow we have CME ( a − b , a ) , for a contradiction. (cid:3) Note that if | N ′ Z | = k − , then | N Z | = k +2 , and so Z is a small neighbourhoodset of defect . K contains edges Then | N ′ X | = k − for all X and by Lemma 6.24, every element of K hasexactly neighbours in K , all from different kernel classes. This implies that thereare exactly edges between each pair of kernel classes, and that these edges havedisjoint vertices. It follows that g is an automorphism, and the result follows. K contains , , or edges In this case, we have kernel classes
Y, Z such that | N ′ Y | = k − , | N ′ Z | = k − .Note that Y is a small neighbourhood set of defect .Assume w.l.o.g. that Z = A , then by Lemma 6.24, we may assume that N ( a ) ∩ K = { b , c , d } . | N ( a ) ∩ N ( a ) ∩ K | ≤ , for otherwise | N A | < k + 2 .Thus we may further assume that b − a − c . By Lemma 6.24, a has no additionalneighbours in B ∪ C .Now applying Lemma 6.8 with y = a , y ′ = b , ¯ y = c there exists z ∈ N ( b ) ∪ N ( c ) with | N ( z ) ∩ N ( a ) | = k − . z = a , as a is not adjacent to b or c . If w ∈ N ′ B , then N ( w ) ∩ B = B = { b } = N ( a ) ∩ B , and so w = z byLemma 6.6(b). Analog, we get that z / ∈ N ′ C . It follows that z ∈ B ∪ C ∪ D .Now consider N = N ( a ) ∩ N ( z ) . As N ( a ) intersects every kernel classin at most one point, the same holds for N . By Lemma 6.9, a f and zf are non-adjacent. However, as z ∈ B ∪ C ∪ D , this is false, for a contradiction.40 contains edges Here | N ′ Z | = k − for all non-singleton kernel classes Z . As | N Z | ≥ k + 2 ,this implies that K \ Z ⊆ N Z . It follows that if there are exactly two edges betweena pair of kernel classes, those edges have disjoint vertices. By Lemma 6.23, eachelement of K has exactly four neighbours in K . Up to symmetry, there are twopossibilities:(a) There are edges between A and B , edges between C and D , and edgeseach between the other pairs of kernel classes;(b) There are edges between A and B , edges between C and D , and edgeseach between the other pairs of kernel classes;In the first case, it is easy to see that g is a graph automorphism, as the edgesbetween pairs of classes other than ( A, B ) and ( C, D ) have disjoint vertices. Soassume we are in the situation (b).We may assume that the three edges between A and C are c − a − c − a .Hence c has two neighbours in A , one neighbour in D , and thus one neighbour in B , which we may assume to be b . Similarly, c has two neighbours in B , and weget the edges b − c − b − c between B and C . Continuing in this fashion, weget the edges d − b − d − b and a − d − a − d .Now we have the CME ( a − c , c ) , unless c − d . This implies c − d .Further, we get CME ( a − c , a ) unless a − b , which implies that a − b . Thisaccounts for all edges (see Figure 13). •• d d •• a a •• c c •• b b ❖❖❖❖❖❖❖❖❖❖❖❖ ❖❖❖❖❖❖❖❖❖❖❖❖ ❖❖❖❖❖❖❖❖❖❖❖❖ ❙❦❯ ◆✐ ✐ ❈ Figure 13: The final configurationBut now we have the CME ( b − d , b ) , as N ( b ) ∩ N ( d ) ∩ K = { c } ⊆ N ( b ) , and N ′ B ⊆ N ( b ) . This contradiction excludes the final case in the proofof Theorem 6.1.We have shown that every primitive group synchronizes every transformationof rank n − . 41 Primitive groups of permutation rank The arguments of the preceding sections apply in complete generality because theyuse only the fact that the groups involved are primitive. We can get stronger resultsby focussing on a restricted class of primitive groups, in particular, the primitivepermutation groups of rank . (Unfortunately the term “rank” is used in a differentsense by permutation group theorists!)More precisely, the (permutation) rank of a transitive permutation group G acting on a set X is the number of orbits of G on X × X , the set of ordered pairsof elements of X . Equivalently, it is the number of orbits on X of the stabiliser ofa point of X .If | X | > , then the rank of G is at least , because no permutation can map ( x, x ) to ( x, y ) . A primitive group of rank is doubly transitive (and hence syn-chronizing), and thus the first non-trivial cases are primitive groups of rank . Theaim of this section is to prove the following result. Theorem 7.1
A primitive permutation group of degree n and permutation rank synchronizes any map with rank strictly larger than n − (1 + √ n − / . Although a complete classification of the primitive groups of rank is known(see [35, 37, 36]), we do not use this, but use instead combinatorial properties ofstrongly regular graphs. (A graph is strongly regular if the numbers k , λ , µ ofneighbours of a vertex, an edge, and a non-edge respectively are independent ofthe chosen vertex, edge or non-edge. See [22] for the definition and properties ofstrongly regular graphs. It is well known that a group with permutation rank iscontained in the automorphism group of a strongly regular graph.)More precisely, we shall prove the following result for strongly regular graphs.In the statement of this result – and throughout this section – we call a stronglyregular graph non-trivial if it is connected and its complement is connected, whichis the same as requiring that µ > and k > µ (the word “primitive” is sometimesused to denote this property, but to avoid confusion with our many other uses ofprimitive, we will not use it in this sense). Theorem 7.2
Let Γ be a non-trivial strongly regular graph on n vertices and let f ∈ End(Γ) be an endomorphism of Γ of rank r . Then n − r ≥ √ n − / . The proof of this uses three simple lemmas:
Lemma 7.3 If Γ is a non-trivial strongly regular graph with parameters ( n, k, λ, µ ) ,and f is a proper endomorphism of Γ of rank r , then n − r ≥ ( k − µ + 4) / . roof Suppose that the kernel of f has t singleton classes, and therefore n − t vertices in non-singleton classes. As f is not an automorphism, it follows that n − t ≥ , and because the non-singleton classes each have size at least , we have r ≤ t + ( n − t ) / . By adding ( n − t ) / to each side of this last expression andrearranging, we conclude that n − t ≤ n − r ) .Let v and w be two vertices in the same kernel class of f and let V , W bethe neighbours of v and w respectively that lie in singleton kernel classes. As f identifies v and w , and maps the vertices of V ∪ W injectively to the neighboursof vf it follows that | V ∪ W | ≤ k . Vertices v and w are each adjacent to at most n − t − vertices lying in non-singleton kernel classes so | V | ≥ k − ( n − t − and similarly for | W | . Therefore | V ∩ W | = | V | + | W | − | V ∪ W |≥ k − ( n − t −
2) + k − ( n − t − − k = k − n − t ) + 4 ≥ k − n − r ) + 4 . Finally, as v and w are not adjacent, it follows that | V ∩ W | ≤ µ and the resultfollows by combining the two bounds for | V ∩ W | . (cid:3) Lemma 7.4 If Γ is a non-trivial strongly regular graph with parameters ( n, k, λ, µ ) ,then k − µ ≥
13 min( k, k ′ ) . where k ′ = n − k − is the valency of the complement of Γ . Proof If Γ is a conference graph, then n = 4 µ + 1 and k = 2 µ and so k − µ = k/ k ′ / , thereby satisfying the conclusion of the theorem. Otherwise the threeeigenvalues of Γ , which we denote k , r and s (with r > > s ), are all integers,and in particular r ≥ . (There is possible confusion with the use of r as the rankof an endomorphism; note that we only use r in the present sense within this proof,following the notation of [22], and endomorphisms will not occur here.)It is well-known that all the parameters of a strongly regular graph can beexpressed purely in terms of k , r and s (see [22, Chapter 2]) and from this it canbe deduced that kr ( k ′ + r + 1) k ( r + 1) + k ′ r = krs ( r + 1)( r − k ) k ( k − r )( r + 1) = − rs, by substituting k ′ = k ( k − λ − µ = − k ( r + 1)( s + 1) k + rs k − µ = − rs = k ( k ′ + r + 1) k (1 + r ) + k ′ ≥ k ′ k ′ k ≥ k ′ , for k ′ ≤ k.k kk ′ + 1 ≥ k, for k ≤ k ′ . where the final inequalities arise from dividing by either k or k ′ , and then using thefact that r ≥ . (cid:3) Lemma 7.5 If Γ is a non-trivial strongly regular graph with parameters ( n, k, λ, µ ) ,then min( k, k ′ ) ≥ √ n − . Proof As Γ and its complement are both connected graphs of diameter , theMoore bound implies that n ≤ k + 1 and n ≤ k ′ + 1 and the result followsimmediately. (cid:3) Thus combining the results of Lemmas 7.3, 7.4 and 7.5, we conclude that aproper endomorphism of rank r of a non-trivial strongly regular graph on n verticessatisfies n − r ≥ √ n − / , thereby completing the proof of Theorem 7.2. Remark
The constant / in this theorem is not best possible, and can be im-proved by using the classification of primitive permutation groups of rank men-tioned above. Details will appear elsewhere. Remark
No non-trivial strongly regular graphs are known that have any properendomorphisms other than colourings (i.e. endomorphisms whose image is a clique).
In this section we briefly describe the results of searching for endomorphisms insmall vertex-primitive graphs, namely those on (strictly) fewer than vertices. Inaddition to confirming that the linegraph of the Tutte-Coxeter graph is the smallestexample of a vertex-primitive graph admitting a non-uniform endomorphism, there44 Values of ( k, χ ) occurring , , , , (9 , , , (16 , , , (12 , , (16 , , , (8 , , (18 , , (20 , , , (12 , , (15 , , (18 , , (21 , , , , (25 , Table 1: ( k, χ ) for n -vertex primitive graphs with ω = χ are various points in the theoretical arguments that terminate by requiring that cer-tain small cases be checked, so for convenience, we gather all this information inone place.The primitive groups of small degree are easily available in both GAP andM
AGMA , though the reader is warned that these two computer algebra systems use different numbering systems so that, for example,
PrimitiveGroup(45,1) is PGL(2 , in GAP , but M in M AGMA . As we are only seeking vertex-primitivegraphs whose chromatic number and clique number are equal, we need not considerthe primitive groups of prime degree, which have a large number of orbitals. Theremaining groups have a much more modest number of orbitals and it is easy toconstruct all possible graphs stabilised by each group by taking every possiblesubset of the orbitals (ensuring that if a orbital that is not self-paired is chosen,then so is its partner).For the sizes we are considering (up to vertices), it is fairly easy to determinethe chromatic and clique numbers of the graphs and thus extract all possible graphswhose endomorphism monoids might contain non-uniform endomorphisms. Thereare only such graphs on fewer than vertices and in Table 1, we give summarydata listing just the order n , the valency k and the chromatic number χ of each ofthese graphs. For example, the entry (12 , in the row for n = 25 indicates thaton vertices, there are three -regular vertex-primitive graphs with ω = χ = 5 .There are no further examples on – vertices and so this list is complete for n < .The bottleneck in this process is not the construction of the graphs, nor the cal-culation of their chromatic or clique numbers, but rather the computation of theirendomorphisms. Apart from some obvious use of symmetry (for example, requir-45ng that a vertex be fixed), we know no substantially better method than to performwhat is essentially a naive back-track search. This finds an endomorphism by as-signing to each vertex in turn a candidate image, determines the consequences ofthat choice (in terms of reducing the possible choices for the images of other ver-tices), and then turns to the next vertex, until either a full endomorphism is found,or there are unmapped vertices for which no possible choice of image respects theproperty that edges are mapped to edges.Such a search can easily be programmed from scratch, but in this case weused the constraint satisfaction problem solver M INION . This software, whichwas developed at St Andrews, performs extremely well for certain types of searchproblem. Using M
INION , we confirmed that for all but two of the graphs listedin Table 1, every endomorphism is either an automorphism or a colouring. Thetwo exceptions are the - and -regular graphs on vertices which also have“in-between” endomorphisms whose image is the -vertex Paley graph P (9) . The -regular graph is the Cartesian product P (9) (cid:3) K = K (cid:3) K (cid:3) K , while the -regular graph is the direct product P (9) × K = K × K × K .On vertices, there are non-trivial vertex-primitive graphs with equal chro-matic and clique number, including the linegraph of the Tutte-Coxeter graph. Ofthe remaining graphs, some are sufficiently dense that we have been unable yet tocompletely determine all of their endomorphisms. However by a combination ofcomputation and theory, we at least know that none of the -vertex graphs otherthan the linegraph of the Tutte-Coxeter graph admit proper endomorphisms otherthan colourings. This paper started with the intention of providing further evidence that primitivegroups are almost synchronizing but, rather inconveniently, this turns out not tobe true. Therefore, faced with an unexpectedly complex situation, we pose thefollowing problem, although with the expectation that resolving it is likely to bedifficult:
Problem 9.1
Classify the almost synchronizing primitive groups.It might be more feasible to focus on the “large-rank” end of the spectrum,where we still believe that the following weaker version of the almost synchroniz-ing conjecture is true.
Conjecture 9.2
A primitive group of degree n synchronizes any map whose rank r satisfies n/ < r < n (all such maps are non-uniform).
46s we have seen, showing that primitive groups synchronize maps of rank n − required a long case analysis. Further progress will require a solution of thefollowing problem. Problem 9.3
Find new techniques to show that large-rank transformations are syn-chronized by primitive groups, and use them to extend the range below n − .The previous problems deal with the spectrum of ranks synchronized by prim-itive groups. An orthogonal approach is to investigate the kernel types that aresynchronized by primitive groups, along the line of the results in Section 5. Problem 9.4
Find new kernel types synchronised by a primitive group. In partic-ular, prove that all primitive groups synchronize maps with the following kerneltypes: (2 , . . . , , , . . . , or ( p, q, , . . . , , for all p, q > . Problem 9.5
Is there a “threshold” function f such that a transitive permutationgroup of degree n is imprimitive if and only if it has more than f ( n ) non-syn-chronizing ranks? (A positive answer to Conjecture 9.2 would show that f ( n ) = n/ would suffice.) In particular, is the number of non-synchronizing ranks of aprimitive group o ( n ) ?Theorem 7.1 concerns the synchronizing power of groups of permutation rank and maps of large rank . In the spirit of the remaining results of this paper, itwould be interesting to investigate what happens with maps of small rank . Problem 9.6
Find the largest natural number k such that groups of permutationrank synchronize every non-uniform map of rank l , for all l ≤ k .The previous problem is somehow connected to the next, the classification of aclass of groups lying strictly between primitive and synchronizing. Problem 9.7
Is it possible to classify the primitive groups which synchronize ev-ery rank map?The previous problem is equivalent to classifying the permutation groups G ,acting primitively on a set Ω , such that for every -partition P of Ω and everysection S for P , there exists g ∈ G such that Sg is not a section for P .Note that there are primitive groups that do not synchronize a rank map (seethe example immediately before Section 2 in [6]). And there are non-synchronizinggroups which synchronize every rank map. Take for example PGL(2 , of de-gree ; this group is non-synchronizing, but synchronizes every rank map since is not divisible by . 47here are very fast algorithms to decide if a given set of permutations generate aprimitive group, but is it possible that such an algorithm exists for synchronization? Problem 9.8
Find an efficient algorithm to decide if a given set of permutationsgenerates a synchronizing group or show that such an algorithm is unlikely to exist.
Problem 9.9
Formulate and prove analogues of our results for semigroups of lin-ear maps on a vector space. Note that linear maps cannot be non-uniform, butwe could ask for linear analogues of results expressed in terms of rank such asTheorem 4.7.
Problem 9.10
Solve the analogue of Problem 9.9 for independence algebras (fordefinitions and fundamental results see [3, 9, 10, 11, 5, 23, 27, 28, 30])Suppose the diameter of a group G (acting on a set Ω ) is at most n − (thatis, given any set S of generators of G , every element of G can be generated bythe elements of S in a word of length at most n ). Suppose, in addition, that G and a transformation t of Ω generate a constant map tg t . . . g n − t . Then we canreplace the g i by a word (on the elements of S ) of length at most n and hence wehave a constant written as a word of length meeting the ˇCerny bound. However,finding the diameters of primitive groups is a very demanding problem. Thereforewe suggest the following two problems. Problem 9.11
Let Ω be a set. Let G be a synchronizing group acting primitivelyon Ω and let S ⊆ G be a set of generators for G . Let X ⊆ Ω be a proper subset of Ω , and let P be a partition of Ω in | X | parts. Is it true that there exist two elementsin the set X that can be carried to the same part of P by a word (on the elementsof S ) of length at most n ?We consider the previous problem one of the most important by its implicationson the ˇCerny conjecture, in the case of transformation semigroups containing aprimitive synchronizing group.The previous problem admits also a general version for primitive groups. Problem 9.12
Let Ω be a set. Let G be a group acting primitively on Ω and let S ⊆ G be a set of generators for G . Let X ⊆ Ω be a proper subset of Ω , and let P be a partition of Ω in | X | parts. Let Q ⊆ X × X be the set of pairs ( x, y ) such thatfor some g ∈ G we have xg and yg belonging to the same part of P . Assuming Q = ∅ , is it true that there exists ( x , y ) ∈ Q and a word w (on the elements of S ), of length at most n , such that x w and y w belong to the same part of P ?48he computations in this paper were critical to prove our results; and the gener-alizations of our results will certainly require to push the limits of the computationsabove. Problem 9.13
Extend the computational results of Section 8.
Acknowledgements
The second author has received funding from the European Union Seventh Frame-work Programme (FP7/2007-2013) under grant agreement no. PCOFUND-GA-2009-246542 and from the Foundation for Science and Technology of Portugal un-der PCOFUND-GA-2009-246542 and SFRH/BCC/52684/2014, as well as throughthe CAUL / CEMAT project.The first author has been partially supported by an grant from the Foundationfor Science and Technology of Portugal.
References [1] D. S. Ananichev and M. V. Volkov, Some results on ˇCerný type problemsfor transformation semigroups,
Semigroups and languages , World Scientific,Singapore, 2004, 23–42.[2] J.M. André, J. Araújo and P.J. Cameron, The classification of par-tition homogeneous groups with applications to semigroup theory. http://arxiv.org/abs/1304.7391 [3] J. Araújo. Idempotent generated endomorphisms of an independence algebra.
Semigroup Forum (3) (2003), 464–467.[4] J. Araújo, W. Bentz and P. J. Cameron, Groups synchronizing a transforma-tion of non-uniform kernel. Theoret. Comput. Sci. , (2013), 1–9.[5] J. Araújo, W. Bentz and J. Konieczny, The largest subsemilattices of the en-domorphism monoid of an independence algebra. Linear Algebra and its Ap-plications (2014), 1: 60 – 79.[6] J. Araújo and P. J. Cameron, Primitive groups synchronize non-uniform mapsof extreme ranks,
Journal of Combinatorial Theory, Series B , (2014), 98–114. 497] J. Araújo and P.J. Cameron, Two Generalizations of Homogeneity in Groupswith Applications to Regular Semigroups . Transactions of the AmericanMathematical Society (to appear) .[8] J. Araújo, P.J. Cameron, J.D. Mitchell and M. Neunhöffer, The classificationof normalizing groups.
Journal of Algebra (2013), 1: 481 – 490.[9] J. Araújo, M. Edmundo and S. Givant. v ∗ -algebras, independence algebrasand logic. International Journal of Algebra and Computation (7) (2011),1237–1257.[10] J. Araújo and J. Fountain. The origins of independence algebras Proceed-ings of the Workshop on Semigroups and Languages (Lisbon 2002) , WorldScientific, (2004), 54–67[11] J. Araújo and J. D. Mitchell. Relative ranks in the monoid of endomorphismsof an independence algebra.
Monatsh. Math. (1) (2007), 1–10.[12] J. Araújo, J.D. Mitchell and C. Schneider, Groups that together with anytransformation generate regular semigroups or idempotent generated semi-groups.
Journal of Algebra (2011), 1: 93 – 106.[13] F. Arnold and B. Steinberg, Synchronizing groups and automata.
Theoret.Comput. Sci. (2006), no. 1-3, 101–110.[14] J. Bamberg, N. Gill, T. P. Hayes, H. A. Helfgott, A. Seress and P.Spiga. Bounds on the diameter of Cayley graphs of the symmet-ric group.
J. Algebraic Combinatorics , , (2014), 1–22. (Also at http://arxiv.org/abs/1205.1596 )[15] Y. Beneson, T. Paz-Elizur, R. Adar, E. Keinan, Z. Livneh, E. Shapiro, Pro-gramable and autonomous computing machine made of biomolecules. Nature (2001), no. 1, 430–434.[16] N. L. Biggs, Three remarkable graphs,
Canad. J. Math. (1973), 397–411.[17] N. L. Biggs and D. H. Smith, On trivalent graphs, Bull. London Math. Soc. (1971), 155–158.[18] P. J. Cameron, Projective and Polar Spaces , Queen Mary Maths Notes, QueenMary, University of London, 1991.[19] P. J. Cameron, Synchronization, London TaughtCourse Centre notes, 2010, available from .5020] P. J. Cameron, Dixon’s theorem and random synchronization,
Discrete Math. (2013), 1233–1236.[21] P. J. Cameron and A. Kazanidis, Cores of symmetric graphs.
J. AustralianMath. Soc. (2008), 145–154.[22] P. J. Cameron and J. H. van Lint, Designs, graphs, codes and their links , Lon-don Math. Soc. Student Texts , Cambridge University Press, Cambridge,1991.[23] P. J. Cameron and C. Szabó, Independence algebras, J. London Math. Soc. , (2000), 321–334.[24] Colbourn, Charles J. and Zhu, L. The spectrum of R -orthogonal Latinsquares, Combinatorics advances , Math. Appl, , 1995, 49–75.[25] H. S. M. Coxeter, The chords of the non-ruled quadric in PG(3,3),
Canad. J.Math. (1958), 484–488.[26] E.W. Dijkstra, Self-stabilizing systems in spite of distributed control. Com-munications of the ACM (11) (1974), 643–644.[27] J. Fountain and A. Lewin. Products of idempotent endomorphisms of an in-dependence algebra of finite rank. Proc. Edinburgh Math. Soc. (2) (1992),493–500.[28] J. Fountain and A. Lewin. Products of idempotent endomorphisms of anindependence algebra of infinite rank. Math. Proc. Cambridge Philos. Soc. (2) (1993), 303–319.[29] C. D. Godsil and G. F. Royle, Cores of geometric graphs,
Ann. Combinatorics (2011), 267–276.[30] V. Gould, Independence algebras. Algebra Universalis (1995), 294–318.[31] H. A. Helfgott and A. Seress, On the diameter of permutationgroups, Ann. of Math. (2014), 611–658. (Also available at http://arxiv.org/abs/1109.3550 )[32] I. M. Isaacs,
Finite Group Theory . Amer. Math. Soc., Providence, RI, 2008.[33] Cai Heng Li, Zai Ping Lu and Dragan Marušiˇc, On primitive permutationgroups with small suborbits and their orbital graphs,
J. Algebra (2004),749–770. 5134] H. Jørgensen, Synchronization.
Inform. and Comput. (2008), no. 9-10,1033–1044.[35] W. M. Kantor and R. A. Liebler, The rank 3 permutation representations ofthe finite classical groups,
Trans. Amer. Math. Soc. (1982), 1–71.[36] Martin W. Liebeck, The affine permutation groups of rank 3,
Proc. LondonMath. Soc. (3) (1987), 477–516.[37] Martin W. Liebeck and Jan Saxl, The finite primitive permutation groups ofrank 3, Bull. London Math. Soc. (1986), 165–172.[38] M. Liebeck and J. Saxl, Minimal degrees of primitive permutation groups,with an application to monodromy groups of covers of Riemann surfaces, Proc. London Math. Soc. (3) (1991), 266 – 314.[39] P. M. Neumann, Primitive permutation groups and their section-regular par-titions. Michigan Math. J. (2009), 309–322.[40] J.-E. Pin, ˇCerný’s conjecture. [41] J.-E. Pin, On two combinatorial problems arising from automata theory. An-nals of Discrete Mathematics (1983), 535–548.[42] P. Spiga and G. Verret, Vertex-primitive digraphs hav-ing vertices with almost equal neighbourhoods. Available at http://arxiv.org/abs/1501.05046 [43] I. Rystsov, Quasioptimal bound for the length of reset words for regular au-tomata. Acta Cybernet. (1995), no. 2, 145–152.[44] A. N. Trahtman, Bibliography, synchronization @ TESTAS [45] A. N. Trahtman, The ˇCerný conjecture for aperiodic automata, Discr. Math.& Theoret. Comput. Sci. (2007), (2) 3–10.[46] W. T. Tutte, A family of cubical graphs, Proc. Cambridge Philos. Soc. (1947), 459–474.[47] W. T. Tutte, The chords of the non-ruled quadric in PG(3,3), Canad. J. Math. (1958), 481–483.[48] M. Volkov, Synchronizing finite automata. http://csseminar.kadm.usu.ru/SLIDES/synchrolectures/lecture1.pdf Combinatorial TheorySer. B (1973), 269–288.[50] Zhu, L. and Zhang, Hantoa, Completing the spectrum of r -orthogonal Latinsquares, Discrete Math ,268