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## Material Information- Title:
- Bivariegated graphs and their isomorphisms
- Creator:
- Riddle, Fay Aycock, 1949-
- Copyright Date:
- 1978
- Language:
- English
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- v, 93 leaves : ill. ; 28 cm.
## Subjects- Subjects / Keywords:
- Automorphisms ( jstor )
Distance functions ( jstor ) Graduates ( jstor ) Graph theory ( jstor ) Isomorphism ( jstor ) Line graphs ( jstor ) Mathematics ( jstor ) Permutations ( jstor ) Semigroups ( jstor ) Vertices ( jstor ) Dissertations, Academic -- Mathematics -- UF Graphic methods ( lcsh ) Mathematics thesis Ph. D - Genre:
- bibliography ( marcgt )
non-fiction ( marcgt )
## Notes- Thesis:
- Thesis--University of Florida.
- Bibliography:
- Bibliography: leaves 91-92.
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- Typescript.
- General Note:
- Vita.
- Statement of Responsibility:
- by Fay Aycock Riddle.
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04361966 ( OCLC ) AAH0865 ( NOTIS )
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BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS By FAY AYCOCK RIDDLE A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1978 To Dennis ACKNOWLEDGEMENTS The author would like to express her sincere thanks for the leadership and guidance of Professor A. R. Bednarek, who has been most patient and under- standing in the long-range completion of this effort. His insight into the open problems in graph theory has provided the basic topics of this work. The author also wishes to acknowledge the remainder of her supervisory committee: Drs. J. E. Keesling, M. P. Hale, Jr., T. T. Bowman, and W. D. Hedges for their personal contributions to her academic training. The author would like to thank her family for allowing her the opportunity to attend the superior elementary, secondary, and undergraduate institutions that laid the foundation for higher academic pursuits. Finally, the author would like to thank her husband, Dennis, for his encouragement and support during her graduate program. TABLE OF CONTENTS Page ACKNOWLEDGEMENTS ............................... ABSTRACT ........................................ INTRODUCTION ................................... CHAPTERS I BASIC DEFINITIONS AND THEORY ........... II MATRIX REPRESENTATION AND BIVARIEGATED HUSIMI TREES ........................... III GRAPH ISOMORPHISMS ..................... IV AUTOMORPHISM GROUPS .................... Automorphism Groups of a Graph and Its Factors ........................ Permutation Graphs ................... V GRAPHS OF SEMIGROUPS AND SEMIGROUPS OF GRAPHS .............................. Graphs of Semigroups ................. Semigroups of Graphs ................. VI DISCUSSION ............................. APPENDIX I: BIVARIEGATED GRAPHS ............... APPENDIX II: BIVARIEGATED HUSIMI TREES ........ REFERENCES ..................................... BIOGRAPHICAL SKETCH .............. ............... Abstract of Dissertation Presented to the Graduate Council of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS By Fay Aycock Riddle June, 1978 Chairman: A. R. Bednarek Major Department: Mathematics Characterizations of bivariegated graphs, in particular, bivariegated Husimi trees, are presented. The relationship between graph isomorphisms and length of irreducible cycles in a bivariegated graph is investigated. Also investigated are automorphism groups of bivariegated graphs and the relationship between bivariegated graphs and semigroups. The largest class of graphs for which graph isomorphisms can be characterized by all irreducible cycles being of length four is the class of Husimi trees. INTRODUCTION A graph G is a finite nonempty set V of vertices together with a set E of unordered pairs of distinct vertices of V, called edges. A graph is bivariegated if the vertex set can be partitioned into two sets of equal size such that each vertex is adjacent to one and only one vertex in the set not containing it. This defines a bijection f between the two vertex sets. Consider the following question: What is the relationship between the structure of a bivariegated graph G and the bijection f? In particular, when is f an isomorphism between the two graphs into which G is partitioned? Bednarek and Sanders [2] prove that if these two graphs are trees, then f is an isomorphism if and only if every irreducible cycle in G has length four. This paper extends this theorem by proving it for a class of graphs larger than the class of trees. Furthermore, in studying the above question, other information about bivariegated graphs is obtained. Chapter I presents some basic definitions and elementary results of graph theory that are useful in later chapters. The definitions primarily follow the notation used by Harary [6 ]. Chapter II gives a representation of the adjacency matrix of a bivariegated graph and presents some characterizations of bivariegated Husimi trees. Chapter III answers the question posed above by extending to a larger class of graphs a theorem relating graph isomorphisms and bivariegated graphs. Chapter IV examines the relationship between the automorphism groups of a bivariegated graph and its factors and presents a result obtained by considering the case in which the factors are equal. Chapter V examines the relationship between bivariegated graphs and graphs of semigroups. Then, a theorem relating semigroups to endomorphisms of a graph is examined for the case in which the graph is bivariegated. The final chapter discusses the results and questions arising from them. In the appendices are listings of bivariegated graphs with 2, 4, 6, and 8 vertices and listings of bivariegated Husimi trees with 2, 4, 6, 8, 10, and 12 vertices. CHAPTER I BASIC DEFINITIONS AND THEORY Definition 1.1: A graph G is a nonempty finite set of points (or vertices), V, along with a prescribed set E of unordered pairs of distinct points of V, known as edges. We write G = (V,E). If two distinct points, x and y, of a graph are joined by an edge, they are said to be adjacent. We write e = {x,y} to denote the edge between x and y, and we say that x and e are incident with each other, as are y and e. A subgraph of G is a graph having all of its points and lines in G. A spanning subgraph contains all the points of G. Two graphs G and H are isomorphic, denoted G C H, if there exists a one-to-one correspondence, called an isomorphism, between their point sets which preserves adjacency. Definition 1.2: A walk of a graph G is a finite sequence of points such that each point of the walk is adjacent to the point of the walk immediately preceding it and to the point immediately following it. If the first and last points of a walk are the same point, we say the walk is closed, or is a cycle (provided there are three or more distinct points and all points are distinct except the initial and final points). A graph is acyclic if it contains no cycles. A walk is a path if all the points are distinct. A complete cycle has every pair of points adjacent. The complete graph K has every pair of its p points ------ p adjacent. A graph is connected if every pair of points is joined by a path. A maximal connected subgraph of G is called a connected component or simply component of G. Definition 1.3: The degree of a point v of G, denoted deg v, is the number of lines incident with v. The point v is isolated if deg v = 0; it is an end point if deg v = 1. A graph is regular of degree n if every vertex has degree n. Theorem 1.4: The sum of the degrees of the points of a graph G is twice the number of lines. The proof of Theorem 1.4 can be found in [ 6 ]. Definition 1.5: A cut point of a graph is one whose removal increases the number of connected components. A nonseparable graph is connected, nontrivial, and has no cut points. A block of a graph is a maximal non- separable subgraph. The notation V {v} means V minus the vertex v and all its incident edges. Theorem 1.6: Let v be a point of a connected graph G. The following statements are equivalent: (1) The point v is a cut point of G. (2) There exist points u and w distinct from v such that v is on every u-w path. (3) There exists a partition of the set of points of V {v} into subsets U and W such that for any points u e U and w E W, the point v is on every u-w path. Theorem 1.6 is proven in [6 ]. Definition 1.7: A tree is a connected acyclic graph. A Husimi tree is a connected graph in which every block of G is a complete graph. The length of a walk (v ,v ,... v ) is n, the number of occurrences of lines in it. The distance d(u,v) between two points u and v in G is the length of a shortest path connecting them, if any. A metric space M is called a graph metric space if there exists a graph G whose vertex set can be put in one-to-one correspondence with the points in M in such a way that the distance between every two points of M is equal to the distance between the corresponding vertices of G. A graph is Ptolemaic if its associated metric space M is such that for any x,y,z,w e M, the three numbers d(x,y)*d(z,w),d(x,z)*d(y,w), and d(x,w)- d(y,z) satisfy the triangle inequality. A graph G is said to be weakly geodetic if for every pair u,v of vertices of G such that d(u,v) < 2, there is exactly one shortest u-v path. Kay and Chartrand [ 9] prove the following: Theorem 1.8: A graph G is a Husimi tree if and only if G is weakly geodetic and Ptolemaic. Definition 1.9: A graph G = (V,E) is said to be bivariegated if G = G1 f G2 = (V 1 u V2, E u E2 u Ef), where G1 = (V1,E1), G2 = (V2,E2), VI n V2 = 0, f: V, V2 is a bijection, and Ef = {{x,f(x)}jx VI}, where we let {x,f(x)} denote the edge incident with x and f(x). Therefore, a graph G is bivariegated if and only if its vertex set can be partitioned into two disjoint equal sets such that each vertex is adjacent to one and only one vertex in the set not containing it. If G, = G2' we say that G is the permutation graph of G1 (or, equivalently, of G2). Definition 1.10: An n-factor of a graph G is a spanning subgraph of G which is not totally disconnected and is regular of degree n. In particular, G has a 1-factor if it has a spanning subgraph consisting of disjoint edges. Sanders [13] characterizes trees with 1-factors: Theorem 1.11: A tree is bivariegated if and only if it has a 1-factor. Theorem 1.12: A bivariegated graph with p vertices has at least p/2 and at most p2 /4 lines. Proof: Let G = G1 f G2. The minimum number of lines G can have is when both G1 and G2 are totally disconnected. In that case there exist p/2 pairs of points in G edged by lines in Ef. The maximum number of lines G can have is when both G1 and G2 are complete. In that case, each of G and G2 has n(n 1)/2 edges, where n = p/2. In addition, there are n edges in E f. Thus, G has a total of n(n 1) + n = n2 = p2/4 lines. Definition 1.13: A coloring of a graph is an assignment of colors to its points so that no two adjacent points have the same color. An n-coloring of a graph G uses n colors. The chromatic number X(G) is defined as the minimum n for which G has an n-coloring. Clearly, the chromatic number of a tree is 2. The chromatic number of the complete graph on p vertices is p. Thus, a Husimi tree which is not also a tree has as its chromatic number the length of its longest cycle. CHAPTER II MATRIX REPRESENTATION AND BIVARIEGATED HUSIMI TREES It is the purpose of this paper to determine some characterizations relating to bivariegated graphs. Sanders [13 ] considers a subclass of these, the bivariegated trees, and characterizes it. Thus, in this chapter we attempt to further generalize the properties of bivariegated graphs by considering a slightly larger subclass, the bivariegated Husimi trees, and attempt to characterize it. In the process of doing this, it became evident that it would be extremely helpful in making our general- izations if we were to have at our disposal a set of examples of all bivariegated graphs with p or fewer vertices, for some p. Thus, before we present our observations concerning bivariegated Husimi trees, we will first look at how the examples were generated. Adjacency Matrices of Bivariegated Graphs By definition, a bivariegated graph G = G1 f - G2 with 2n vertices can be factored into graphs G1 and G2 with n vertices where each vertex in G1 = (V1,E1) is edged with a unique vertex in G2 = (V2,E2). Thus, to generate examples of all bivariegated graphs with 2n vertices, one needs to consider all possible combinations of pairs of graphs with n vertices, connected by all possible bijections f: V1 V2. Let V1 = {ul,u2, ... ,un} and V2 = {11v 2,...v }. For any bijection f, the vertices of V2 can be renamed v1',V 2',...,v n' in such a way that f(u.) = v.', i = 1,2,...,n. In light of this, one can generate all bivariegated graphs having 2n vertices by first considering all possible combinations of pairs of graphs Gl and G2 with n vertices. Then, rather than considering bijections between G1 and G2, we merely consider all possible permutations of the vertices of G2. Because for rather small values of n this becomes a tedious process, it was desirable to write a computer program to perform the process above. The program is based on the representation of a graph by its adjacency matrix. Definition 2.1: The adjacency matrix A = [a..] of a labeled graph G with p points is the p x p matrix in which a.. = 1 if v. is adjacent to v. and a., = 0 other- 1] 1 ] 13 wise. Clearly there is a one-to-one correspondence between labeled graphs with p points and p x p symmetric binary matrices with zero diagonal. Now consider the adjacency matrix A of a bivariegated graph. Let the first n rows and columns of A correspond to ul,u2,...,un and the second n rows and columns correspond to vl' ,v2' ..., v as defined above. Partition A into 4 square sub- n matrices. Because f(u.) = v.', i = 1,2,...,n, it should be clear that each of the upper right and lower left submatrices is the identity matrix. Now we need to generate the upper left and lower right submatrices, corresponding to adjacency matrices of G1 and G2, respectively. Consider the adjacency matrix of G Because it is symmetric, we need only to consider possible entries above its zero diagonal. For an n x n matrix, there are 1 + 2 + ... + (n-1) = n(n-l)/2 entries above the diagonal. Then all possible graphs G1 can be found by generating all binary sequences of length m = n(n-l)/2, of which there are 2m. Similarly, there are 2 possible adjacency matrices for G2. The number of combinations of 2 different elements from a set of p is p:/(2(p-2)!), and the number of ways to choose two identical elements from a set of size p is p. The sum of these can be easily shown to be p(p+l)/2. Now, if p = 2 we have 2m- (2m+l) adjacency matrices for G1 f G2. However, since the representation of a graph by adjacency matrix is not unique, this merely gives us an upper bound on the number of bivariegated graphs with 2n vertices. Thus, we have proven the following: Theorem 2.2: Let G = G f G2 be a bivariegated graph with 2n vertices. Then the vertices of G can be parti- tioned in such a way that the adjacency matrix of G is the 2n x 2n symmetric matrix, I A2 with zeros on the diagonal, A1 and A2 the adjacency matrices of GI and G2, and I representing the n x n identity matrix. Corollary 2.3: There are at most 2 m-(2m+l) bivariegated graphs with 2n vertices, where m = n(n-l)/2. Because the method outlined above generates duplicates of the same graph, the computer program was written to group the adjacency matrices that it generates according to the number of non-zero entries in each row (which corresponds to the degree sequence of each graph). These groups were then checked by hand for duplicates. From this we obtain the list in Appendix I of all bivariegated graphs with at most 8 vertices. Bivariegated Husimi Trees Now we return to the purpose of this chapter, that of characterizing bivariegated Husimi trees. Let H be a Husimi tree. Obviously, if H is bivariegated, it has an even number of vertices. Sanders [13] proves that a tree is bivariegated if and only if it has a 1-factor. Clearly it is true that if H is bivariegated, then it has a 1-factor. However, the converse is not true, as shown in Figure II-1, below. v 2 v3 4 5 6 7 8 FIGURE II-1 This Husimi tree has a 1-factor, consisting of the edges {vl,v2}, {v3,v4, {v5 ,v6}, and {v7'v 8}, yet it is not bivariegated. One can state, however, that if a Husimi tree H has a 1-factor such that no edge in the 1-factor is contained in a cycle, then H is bivariegated. The proof is exactly as in the theorem by Sanders. The-above exemplifies the following: all of Sanders' characterizations of bivariegated trees cannot be extended to bivariegated Husimi trees. Because of this, an examination of the properties of the bivariegated Husimi trees that were generated earlier was begun. However, since there are so few bivariegated Husimi trees with 8 or fewer vertices, little could be generalized from these examples. Thus, it was necessary to derive a scheme for generating examples of bivariegated Husimi trees only. We will now present the results obtained while studying this problem. Proposition 2.4: Let G = G f G2 be a bivariegated Husimi tree. If a is a cycle in G, then a must lie completely within G1 = (Vl,E ) or G2 = (V2,E2). Proof: Let a = (vl,v 2'...v" ,v 1. Suppose that v1 s V1 and that f(v1) = V2 E V2 is in the cycle a. Now consider any other vertex, say v31 in a. Because G is a Husimi tree, a is complete. Assume v3 E Vi. Because a is complete, v3 and v2 are edged, contradicting G being bivariegated, since v2 is also edged with v1 E V A similar contradiction can be obtained if we assume v3 e V2' We also obtain the following result: Proposition 2.5: Let G = G f G2 be a bivariegated Husimi tree. Then both GI and G2 cannot be connected graphs. Proof: Suppose G = (VI,E ) is connected. Let ul,u2 E V1 and a = (u ,u ,ui ... ,ui+nU2) be a path in G1 between uI and u2. Now f maps ul and u2 to f(u ) = v1 and f(u2 = v2 in G2 = (V2,E ). Suppose there exists a path B = (vl,vj, vj+'....' j+m',v2) in G2 between v1 and v2. Then(ul,ui,..., ui+nu2'v2'v j+m* ..vjvl'1l) is a cycle in G, with vertices in both V1 and V2. This is complete because G is a Husimi tree, contradicting G being bivariegated (since, for example, u V1 is edged with both vI and v2 V2)' Corollary 2.6: Let G = G1 f G2 be a bivariegated Husimi tree. If there is a path in G connecting the vertices u1 and u2 in V1, then there is no path in G2 connecting the vertices f(u ) and f(u2) in V2. Using the information gathered up to this point to obtain examples of bivariegated Husimi trees, it was noticed that the portions of these graphs that are not cycles are bivariegated trees. This is for- malized in the following theorem: Theorem 2.7: Let G be a Husimi tree. Consider the graph G' which consists of G minus each edge which is contained in a cycle. Then G is bivariegated if and only if all the components of G' are bivariegated trees. Proof: Suppose one of the components of G' (call it T) is not bivariegated. Clearly each component of G' is a tree. Let W be the set of all vertices of T contained in cycles of G. The set W is not empty, or else G has no cycles, in which case G = G' is not bivariegated. If T is not bivariegated, the connection of the vertices in W to cycles in G can still not be bivariegated. This is because all vertices in a cycle of a graph G = G1 f G2, if G is to be bivariegated, must all be in either G or G2. Thus, a vertex in W which is in, say G1, and which has no "mate" in G2 contained in T does not get a mate by being connected to a cycle. Thus, the addition of any number of cycles to any number of vertices of components of G' will not permit G itself to be bivariegated. Now, suppose each component of G' is bivariegated. Partition the vertices of a component T into two sets, V1 and V2. corresponding to the bivariegation. Any cycle connected to a vertex v of T will have all of its vertices in the same set, say V1, as v. Further, each of the other vertices of such a cycle is connected to a bivariegated tree (otherwise, it has no mate in V2). Because the bivariegation of a tree is unique, once one of its vertices is assigned to V1 or V2, the remainder of its vertices are uniquely bivariegated. Apply this process to each cycle and bivariegated trees connected to it, and a bivariegated Husimi tree will be obtained. Sanders [13] proves that the bivariegation of a tree is unique. We prove an analogous result below. Theorem 2.8: The bivariegation of a Husimi tree is unique. Proof: If G is bivariegated, the components of G' are bivariegated trees, where G' is defined as in Theorem 2.7. Now consider any cycle of G. Suppose the bivariegation places its vertices into, say VI. Each vertex of that cycle must be connected to a bivariegated tree or both a bivariegated tree and a cycle, or else it cannot be adjacent to a vertex in V2. Now, once the bivariegation of a single vertex of a tree is determined, the bivariegation of the remainder of its vertices is uniquely determined. This tree may or may not be connected to other cycles, but if it is, the bivariegation of the cycles attached to it is uniquely determined by the bivariegation of the vertex of the tree to which it is attached. Therefore, because (i) a bivariegated Husimi tree consists of bivariegated trees and cycles, (ii) a tree is uniquely bivariegated, and (iii) each cycle is uniquely bivariegated according to the bivariegation of the vertex of the tree to which it is attached, the entire Husimi tree is uniquely bivariegated. We now see that if a bivariegated Husimi tree contains a cycle, then each vertex of that cycle is attached to a bivariegated tree. Further, if it contains more than one cycle, the block between any two cycles is also a bivariegated tree. This motivates the use of the following for constructing all bivariegated Husimi trees: Sanders [13] constructs all bivariegated trees by taking the smallest bivariegated tree (two vertices connected by an edge) and successively adding remote end vertices. A remote end vertex is defined as follows: Definition 2.9: Let e be an end point of graph G, and let x be the unique point adjacent to e. If deg x = 2, then e is called a remote end vertex. We can construct all bivariegated Husimi trees by taking the smallest bivariegated Husimi tree (which is also the smallest bivariegated tree) and attaching, one at a time, to a vertex a graph of one of the forms in Figure 11-2, below. FIGURE 11-2 That is, attach to the vertex either a remote end vertex or a complete n-cycle with n-1 end points, where the attachment is made at the circled vertex. Thus, we obtain the following theorem. Theorem 2.10: Let G be a bivariegated Husimi tree. Construct G' as follows: let v be any vertex of G. To v, attach either a remote end vertex or a complete cycle such that the vertices other than v are edged with an end vertex. Then G' is a bivariegated Husimi tree. Proof: Suppose G = G1 f G2 where GI = (V1,E2). Suppose v s V1 and a remote end vertex is attached at v. Then the remote end vertex can be added to V2 to form the set V2', and the vertex x (as in Definition 2.9) can be added to V1 to form the set Vl'. Clearly G' = G f G2', where GI' = (V1',E,') and G2' = (V2',E2') is a bivariegated Husimi tree. Now consider the case where a cycle of the type described above is added to v c VI. Add every other vertex in that cycle to V1 to form the set VI', and add the end vertices attached to this cycle to V2 to form V2'. Then G' = Gi' f G2' is a bivariegated Husimi tree. Appendix II gives the results of generating all bivariegated Husimi trees with 2, 4, 6, 8, 10, and 12 vertices by using the method in Theorem 2.10. As Sanders [13) points out, the technique of adding remote end vertices may produce duplicates of a particular graph. The same is true for the method outlined above, as shown in Figure 11-3. FIGURE 11-3 CHAPTER III GRAPH ISOMORPHISMS The contents of this chapter are the results of attempts to extend to a larger class of graphs a theorem relating graph isomorphisms and bivariegated graphs. Definition 3.1: Let a graph G have vertex set V and edge set E. A finite sequence, a = (v ,vlv 2..."' n, of distinct vertices of G is called a path provided {v ,v. } I E for i = 0,1,2,... ,n-l. A cycle is a sequence a = (v0,v, .... v n-_v n) such that v0 = vn and (v0,v1 ....,v nl ) and (vl,v2 ....,v n) are paths. A cycle is irreducible if it contains no diagonals; that is, for v.,v. vertices of a, {v.,v.}E E implies li-jl = 1 or n-l. If a = (v0,vI .... Vn), then the length of a is n. Bednarek [ 1 ] characterized tree isomorphisms in terms of bivariegated trees: Theorem 3.2: If T1 = (V1,E ) and T2 = (V2,E2) are trees and if f: V, V2 is a bijection, then f is 23 an isomorphism if and only if all irreducible cycles in T1 f T2 have length four. One may wonder whether or not we may replace "trees" in the preceding theorem by some larger class of graphs. Bednarek [ 1] gives an example to show that "trees" may not be replaced by "connected graphs." This is demonstrated by Figure III-1, in which G1 and G2 are squares and f the isomorphism given by yi = f(x ), i = 0,1,2,3, and in which a = (xO,xl,x2',2,y3,y ,x0) is an irreducible cycle of length six. xo x1 Y y yl y3 Y2 FIGURE III-1 It is one of the aims of this chapter to find the largest class of graphs for which "trees" in Theorem 3.2 may be replaced. It can be proven that "trees" may be replaced by "Husimi trees": Theorem 3.3: If T1 = (V1,E1) and T2 = (V2,E2) are Husimi trees and if f: V1 V2 is a bijection, then f is an isomorphism if and only if all irreducible cycles with edges in both T1 and T2 have length four. Proof: Suppose f is not an isomorphism. Then either there is an edge {x,y} E E for which {f(x),f(y)} / E2 or there exists an edge {s,t} c E2 such that {f (s), f- (t)} / E1. Considering the first case we see that because {f(x),f(y)} / E2, then f(x) and f(y) are not in the same block of V2 since every block in a Husimi tree is complete. So V2 is separable, and we can decompose it into blocks B1,B2,.. .,B k. Let B1 be the block such that f(x) E B1 and Bk the block such that f(y) E Bk. Then there is a path a = (f(x),tl,...,t , f(y)) from B1 to Bk between f(x) and f(y) where each ti, i = 1,2,...,n, is a cutpoint of a block through which a passes. Because each block is complete, there is only one edge between each cutpoint of a particular block. Now consider the path a' = (x,f(x),tl, .... tn,f(y),y,x). This is a cycle of length greater than or equal to 5 and irreducible because of the way a was chosen. To prove the necessity of the condition, suppose f is an isomorphism. Let a = (v0,v 2,v"..,v k) be the largest irreducible cycle in T1 f T2 with edges in both T1 and T2. Let v0 = vk V1 and vk-1 E V2. Thus f-1 (vkl) = vk = V0 e V, so f(vk) = f(v0) = Vkl. We claim that Vk-2 e V2. For if not, then Vk-2 C VVk-l C V2, and f(vk-2 = Vk-1 However, this is impossible since f(v0) vk-l' contradicting the one-to-oneness of f. We next claim that v1 e V For if not, then v1 E V2,v0 E V1, and f(v0) = vI. However, this contradicts the functionality of f, as f(v0) = vk- also. We also claim that Vk-3 C V2. For if not, then Vk-3 E Vlv k-2 C V2, and f(vk-3) -= Vk. However, since {Vk-r' Vk-2 is an edge, {f- 1 k- f-(vk-2) = {v0,v k-3 must also be an edge. Finally, we claim that v2 E V For if not, then v2 E V2 and f(v ) = v2. However, since {v0,v 1 is an edge, {f(v0) f(v )1 = {v v is also an edge, contradicting the irreducibility of a. Now, consider f-1 (vk-2 ),f 1 (vk-3) V1 and f(vl) ,f(v2) c V2. Construct the cycle a' = (v0,vk ,f(vl),f(v2),v2,v3,... ,v k-3, f-1(vk-3 ),f-1vk2),v0). The length of a' is equal to the length of a plus two. We claim that a' is irreducible. To prove this, we will check each vertex in a' and show that it is adjacent to no vertices in a' other than those immediately preceding and succeeding it in the definition of a'. Clearly v0 is not adjacent to either f(v1), f(v2) ,v2 .. ,vk-3. Also {v0,f (vk-3)} is not an edge, for if it were, {vk-lv k-3} would also be an edge, contradicting the irreducibility of a. Clearly, Vk-1 is not adjacent to either f(v2) (since that would imply {v0,v2} is an edge), v2,..., or vk-3. Referring to Figure III-2, it is clear that vk-1 is not adjacent to either f- k-3) or f- (vk-2 ' as these vertices are in V.l Also, f(vI) is not adjacent to either v2,v 3..., or vk-3 (otherwise, there would be a cycle in V2, forcing {vkl' ,v k-3} to be an edge, contradicting the irreducibility of a). Since f-1 (vk-3) and f- (vk-2) are in V1 f(v1 is not adjacent to either of these. As for f(v ) above, f(v2) is not adjacent to either v3 .. k-3' 1 (vk-3), or f- (vk-2). By definition of a, v2 is not adjacent to either v4,., or vk-3. Also, {v2,f (vk-3) is not an edge, for if it were, then 6 = (v0lv2' f- (vk3) ,f-(vk-2) -1 ) = vk = v0) is a cycle in V. Since V1 is the set of vertices of a Husimi tree, S is a complete cycle, forcing {v0,f-1 (v k-3)} to be an edge. However, we have shown earlier that this cannot be an edge. Vk-2 FIGURE 111-2 Similarly, {v2',f (vk-2)} is not an edge, for if it were, then V- = (v0vl,v2f (-1 k-2),f -1 Vk) = 0) would be a cycle in VI. Then, completeness of cycles in T1 = (V1,E-) forces v0,v 2} c E contradicting the irreducibility of a. By definition of a, it is clear that none of v3" ... vk-3 is adjacent to another. Also, none of v3 .... Vk-3 is adjacent to f- (vk-3) since f-1 (vk-3) is only adjacent to vk-3 and f -l(vk-2). By the same reasoning, none of v3,..., k-3 is adjacent to f-1 (v-2). Therefore, a' is irreducible, contradicting our definition of a as being the largest irreducible cycle. Then it must be true that all irreducible cycles in T1 f T2 have length four. In trying to extend the theorem to a larger class of graphs, it was noted that much of the proof above depends on the existence of diagonals, that is, edges {vi,v } in a cycle (v0,vI ... ,v = 0) such that either li-jl i 1 or n-l. There is a characterization of Ptolemaic graphs in terms of number of diagonals, proven by Howorka [ 8 ] : Theorem 3.4: A graph G is Ptolemaic if and only if every n-cycle of G where n = 2k + r for r = 0,1, has at least 3k +- 2r 5 diagonals. It was conjectured that "trees" in Theorem 3.2 could be replaced by "Ptolemaic." However, Figure III-3 shows a counter example. Both G1 with V1 = {xl,x2,x31x4} and G2 with V2 = {y1,y2,y3,y4} are Ptolemaic since their four cycles have 1 diagonal. However, a = (xl,x4,x3,y3, y2',y,x1) is an irreducible cycle of length six, and the map f(x ) = y., i = 1,2,3,4, is an isomorphism. The following theorem gives the additional requirement in order for Theorem 3.2 to hold for Ptolemaic graphs: Theorem 3.5: If G1 = (V1,E1) and G2 = (V2,E2) are Ptolemaic graphs such that every cycle of length four induces a complete subgraph, and if f: G, G2 is a FIGURE III-3 xl X3 Y3 Yf bijection, then f is an isomorphism if and only if all irreducible cycles in G1 f G2 with edges in both G1 and G2 have length four. Proof: The sufficiency of the condition is proved as in the proof of Theorem 3.2. To prove the necessity of the condition, we construct a cycle a' exactly as in the proof of Theorem 3.2. By appeals to (i) the irreducibility of a, and (ii) the one-to-oneness of f, we can prove that neither v0,vk-l f(v ), nor f(v2) is edged with another vertex in a' as to contradict the claim that a' is irreducible. Similarly, v2 is not edged with either v3,v4,..., or Vk-3' If {v2,f (v k-3)} were an edge, then (v0,vl, v2f1 (vk-3 ),f1 (vk-2 ),v0) is a five-cycle in G Since G1 is Ptolemaic, there must be at least three diagonals. Possible diagonals are (i) {vf- (vk3) i, (ii) {vi f-1 k-3), (iii) {vlf1 (vk2), or (iv) {v2, f1 (vk-2 However, case (i) cannot be a diagonal, or else it would force {vk-l ,k-3} to be an edge. Therefore, the other three cases must be the diagonals. This gives us a cycle (v,vv 2',f-1 (vk-2 ),v0) of length four, which by our hypothesis, must be complete. This is a contradiction because it would force {v',v 2 to be an edge. Also, {v2,f (Vk-2)} is not an edge, for if it were, we would have a cycle (v0,v,v 2,f (v k-2), f- (k- )) of length four, again with completeness forcing {v0,v21 to be an edge. Similarly, none of v3, .. vk-3 is adjacent to any vertex of a' such that there would be a contradiction to the irreducibility of a'. Thus, a' is irreducible, contradicting a being the largest such cycle. In view of this, the theorem is proved. It turns out that Theorem 3.3 and 3.5 are equivalent because of the following theorem: Theorem 3.6: G is a Husimi tree if and only if G is Ptolemaic and every cycle of length four induces a complete subgraph. Proof: If G is a Husimi tree, then every cycle is complete and satisfies the diagonal requirement for being Ptolemaic. Conversely, suppose G is Ptolemaic and every cycle of length four induces a complete subgraph. Let P (n) be the proposition that every cycle of length n induces a complete subgraph. Clearly, P(4) is true. Now suppose that P(m) is true for m = 5,...,n-l. Consider a cycle a = (vl,v 2,...,v ,v1) of length n. There is at least one diagonal, and we may assume without loss of generality that this is the edge {vlVk }. Divide the vertex set of a into two subsets: A = {v ,v2,...,vk} and B = {vk ... n'v 1} By the inductive hypothesis, each of these sets is complete. We claim that there is an edge between any v. E A and v. E B. For consider the cycle (v ,vl,v,vkv i) of length four. By completeness, we see that {v.,v.} is an edge. Thus P(n) is true, and therefore, P(m) is true for all m greater than or equal to 5. The following theorem proves that the largest class of graphs for which Theorem 3.2 holds is the class of Husimi trees: Theorem 3.7: Suppose G1 = (VI,E ) and G2 =(V2,E2) are connected graphs and f: V,1 V2 is a bijection. Further, suppose that f is an isomorphism if and only if all irreducible cycles in G1 f G2 with edges in both G1 and G2 have length four. Then G1 and G2 must be Husimi trees. Proof: We will show that if G1 and G2 are not Husimi trees, then f being an isomorphism is not equivalent to the requirement on the length of irreducible cycles. Suppose G1 is not a Husimi tree and that (a v ',v,. ,v n) is a cycle in G1 with at least one missing diagonal. Without loss of generality, assume that v0 is one of the vertices of a missing diagonal in a. Let j be the smallest index such that {v0,vj} is not an edge. A shortest path from v0 to v is (v0,vj-_,Vj), since clearly {v0,vj. _l} El or else a contradiction of the choice of j would exist. Now, consider a shortest path from f(v ) to f(vn) of increasing index. The length of 8 is at least two since {f(v ),f(vn)} Z E2. No diagonals exist, or else the shortness of the path would be contradicted. Consider the cycle ', = (v0,v j ,vj,f(vj) ,..,f(vn),vn where the portion of a' between f(v ) and f(v n) is exactly as in S. Because f is an isomorphism, v. 1 is edged with no vertex in 8, v0 is edged only with f(v) = f(0) in 8, and v. is edged only with f(v.) in B. Thus a' is an irreducible cycle, and its length is 6. So far, we have only proven variations of Theorem 3.2 for finite graphs. The theorem may be expanded to the case of infinite trees: Theorem 3.8: If TI = (V-,EI) and T2 = (V2,E2) are infinite trees and if f: V1 + V2 is a bijection, then f is an isomorphism if and only if all irreducible cycles in T1 f T2 have length four. Proof: The sufficiency of the condition is proven as in the proof of Theorem 3.2. The necessity of the condition is proven differently than in that proof because its use of a longest irreducible cycle is inapplicable in the case of infinite trees. To prove necessity, suppose f is an isomorphism. Let a = (v,v2,... ,v n) be an irreducible cycle in T1 f T2 of length greater than 4. Some of the vertices are in VI, and some are in V2. Let v1 = vn a V1 and f(vI) = vn-l. Write a as (V ...,vnl,vnl+ n2' .'. v 1,v n.+l -'-,vnk ,v ) v1 ~-1 "2 k-l nk 1 6 V C V2 6 V2 where n1 / 2 and f(vI) = v Consider the closed walk k a' in V2, where a' = (f(v ) ,...,. ,f(vn ),v nl+2 .... n2 f(vn2+2) ,...,vnk) 1 2- 2 where f(v ) = v n.+l for i odd, v = f (v n.+l) for 1 1 1 1 i even, and f(v1) = v nk Claim: The length of a' is greater than two. Note that v3 is not in V2 because if it were, then f(v2) = v3. However, since f is an isomorphism and {vlv 2} E1, it would imply that {f(v1),f(v2)} = {vn-' ,v3} E2, contradicting the irreducibility of a. Thus vl,v21V3 are all in VI, and a' = (f(vl),f(v2),f(v3) ... f(vl)) has length greater than 2 because f(v3) and f(v1) do not form an edge (by the one-to-oneness of f). Claim: The walk a' is a trail. For if not, there is an edge e = {x,y} that is repeated in a'. If e is an edge in a and there do not exist v. and v. in a such that f(v ) = x and f(v ) = y, then e is repeated in a, contradicting the fact that a is a cycle. So there exist v. and v. such that f(v.) = x and f(v.) = y. But this would imply that {v.,v.} is an edge in a, contradicting the irreducibility of a -1 since x and v. are in a and {x,f (x) = v.} is an edge in T1 f T2. Thus, e = {x,y} is an edge in a', not a. So there exist v. and v. in V such that f(v.) = x and f(v.) = y. If e is repeated, then either (i) {v.,v.} is repeated in a (but this would contradict a being a cycle), or (ii) two edges in T1 being mapped to e (contradicting f being one-to-one). Therefore, no edge in a' is repeated, and thus a' is a trail. Moreover, a' is a circuit. This means that a' contains a cycle, contradicting the fact that T2 is a tree. This completes the proof. One can define an equivalence relation R on vertices of bivariegated trees: Definition 3.9: Let x and y be vertices in T1 f T2' and define x = y modulo R if and only if x = f(y), y = f(x), or x = y. This equivalence relation identifies each vertex in T1 with its image in T2 such that one can consider T1 collapsing onto T2 to form a new graph T1 f T2/R. If we let E be the set of edges in T1 f T2/R we can define adjacency as follows: Definition 3.10: Let [x] and [y] be the equivalence classes of the vertices x and y, respectively, of T1 f T2 under R. Then {[x],[y]} E E if and only if there exists a p e [x] and a q [y] such that {p,ql E E16E2 where E1 and E2are the edge sets of T1 and T2, respectively. Using this equivalence relation, we can prove another version of Theorem 3.2: Theorem 3.11: If T1 = (VI,E ) and T2 = (V2,E2) are trees and if f: V,1 V2 is a bijection, then f is an isomorphism if and only if T1 f T2/R is a tree. Proof: Suppose f is an isomorphism. We want to show that T1 is isomorphic to T1 f T2/R. For every v.i VI, define g(v ) = [v ]. This is clearly a one- to-one and onto mapping. If {x,y} c El, then {g(x),g(y)} = {[x] [y] } E. Also, if {g(x),g(y) } E, there is a p e [x] and a q c [y] such that {p,q} c E 1E2. If {p,q} E then p = x and q = y, so {x,y} c E1. If {p,q} e E2, then since f is an isomorphism, {f-1 (p)},f (q)} = {x,yl} E1. Therefore, g is an isomorphism, and T1 f T2/R is a tree since it is isomorphic to one. Now suppose f is not an isomorphism. Assume that {x,y} E E, but {f(x),f(y)} / E2. In T1 f T2/R this corresponds to {[x],[y] } E and the existence of a path a from [f(x)] = [x) to [f(y)] = [y] of length greater than one, where a = ([f(x)],tl ... tn, [f(y)] [f(x)]). No t. = [x] or [y], so a is a cycle, and, therefore, T1 f T2/R is not a tree. CHAPTER IV AUTOMORPHISM GROUPS In this chapter we examine the relationship between the automorphism groups of G = G1 f G2 and its factors and, in conclusion, quote a result obtained by considering the case in which G1 = G2' Automorphism Groups of a Graph and Its Factors Definition 4.1: An automorphism of a graph G = (V,E) is a permutation h: V V such that if {u,v} E E, then {h(u),h(v)} c E. The set of all automorphisms of G forms a group, which will be denoted Aut(G). It is an interesting problem to look at the relationship between the automorphism groups of the bivariegated graph G = G f G2 and its factors, G1 = (V1,E1) and G2 = (V2,E2). For, suppose h is an automorphism of one of the factors of G, say G1. Then, unless h is the identity map, h maps at least one vertex, say u, to another. Because G is bivariegated, u is edged with some vertex v in G2. Now, in order for h to correspond to an automorphism on the entire graph G, h must similarly move v. This motivates the following theorem: Theorem 4.2: Let g be any automorphism of one of the factors of G = G1 f G2 (without loss of generality, the factor GI). If g is an automorphism on G that agrees with g on G1 (that is, g restricted to G1 is identically g), then it must be true that for any vertex v c V 2' g(v) = f(g(f-l(v))). Proof: If g is an automorphism of G, then it preserves edges in G. Then, because G = (V,E) is bivariegated, for any v c V2, there exists a u c V1 such that f(u) = v and {u,v} c E. So it must also be true that {g(u), g(v)} E However, since g restricted to G1 is iden- tically g, g(u) = g(u), and g(v) e V2 (since all vertices in V1 are already images of some vertex under the map g). But g(u) e V1 is edged with only one element of V2, namely f(g(u)), because of the bivariegation. Thus, g(v) f(g(u)) = f(g(u)) = f(g(f-1(v))). This theorem suggests that if we are trying to relate automorphism groups of the factors of G = G1 f G2 to Aut(G), then we can extend the auto- morphisms of the factors of G to ones on G itself and examine whether or not these extensions are automorphisms of G. Furthermore, as suggested by Theorem 4.2, the extension of an automorphism g on G1 (or G2) to a possible automorphism g of G is uniquely determined by the bivariegation. To obtain an understanding of this problem, it is helpful to look at a few examples. In all the examples, let V = {1, 2' ,v3} and V2 = {v4,v ,v 6. For a mapping g, the notation (iI i2 3 ... i n) means that g(v. ) = v. g(v )= v. ,..., and g(v. ) = v. . 1 2 2 3 n 1 If a particular subscript, say i., is omitted, then assume g(v. ) = v. Further, the notation id will 1. 1. refer to the identity map. Example 4.3: Consider the bivariegated graph in Figure IV-1. V2 vl 2 v 3 v4 v6 v 5 FIGURE IV-1 Aut(G1) = {id,(l 3)}. Clearly, the identity on G1 induces the identity map on G. Now let g be the automorphism (1 3). Theorem 4.2 forces g to be the map (1 3) (4 6). It should be clear that Aut(G) consists only of g as defined above and the identity. Thus, Aut(G ) induces all the automorphisms in Aut(G). Similarly, Aut(G2) induces Aut(G). Example 4.4: In the graph below, v 2 FIGURE IV-2 Aut(G1) = {id, (1 3)}. If g is the automorphism (1 3) on Gi, it forces the automorphism g on G to be (1 3) (4 6). This, together with the identity, is Aut(G). Thus, Aut(G) is completely determined by Aut(G ). However, Aut(G2) = {id, (5 6), (4 5), (4 5 6), (4 6 5), (4 6)} = S3, the symmetric group on three elements. Only one of these, (4 6), induces an automorphism on G. Example 4.5: Consider the graph in Figure IV-3. v V 2 v4 v6 v5 FIGURE IV-3 Aut(CG) = {id,(l 3)}. As in the previous examples, these induce the identity and the automorphism (1 3) (4 6) on G. However, Aut(G) = {id, (l 3)(4 6), (1 4) (2 5) (3 6), (1 6) (2 5) (3 4)}. Thus, although every automorphism on GI, and similarly G2, induces one on G, Aut(G ) does not completely determine Aut(G). Example 4.6: Consider the graph in Figure IV-4: v4 v6 vI v3 5 FIGURE IV-4 Aut(G2) = {id, (4 5)}. These induce the identity and the automorphism (4 5) (1 2) on G. However, Aut(G) = {id,(4 5) (1 2),(3 6), ( 2)(4 5)(3 6)}, so Aut(G2) does not completely determine Aut(G). Now consider Aut(GI) = S3. Only two of these six automorphisms, the identity along with (1 2), induce maps on G which are automorphisms. Example 4.7: Consider the graph in Figure IV-5: V2 v 5 FIGURE IV-5 Aut(Gl) = {id,(2 3)}. This forces the identity and the map (2 3)(5 6) on G. This is not an automorphism, however, since {v4,v5} 5 EE but {g(v4),g(v5)} = {v4,v6} / E. Thus, only the identity on G induces an automorphism on G. The examples above point out the following: in some cases, the automorphism aroup of a factor of G completely determines Aut(G); in some cases, the automorohism qroup of a factor determines some (but not all) of the automorphisms of Aut(G). In other instances, only the identity induces an automorphism on G. The following theorem generalizes when one of the above cases occurs. Theorem 4.8: Let G = G1 f G2 be a bivariegated graph with 2n vertices. If Aut(G2) (or equivalently Aut(G )) is Sn, then every automorphism of G induces a unique automorphism of G. Proof: It is clear that the automorphisms must be unique, for any map so induced according to Theorem 4.2 must be uniquely defined by the structure imposed on G by the bivariegation. So, let g c Aut(Gl). Extend g to g on G by defining for v e V2, g(v) = f(g(f (v))). Claim: The map g preserves all edges of G. Clearly it preserves edges in G1 since g restricted to G1 is defined to be g. Now, consider any u E V1 and v e V2 such that f(u) = v. Then since {u,v} s E, {g(u),g(v)} must be an edge. But g(v) = f(g(f-l(v))) = f(g(u)), and by definition of bivariegation {g(u),g(v)} = {g(u),f(g(u))} = {g(u),f(g(u))} E E. Thus, g preserves edges joining G1 and G2. Finally, g preserves edges in G2. For, consider {s,t} c V2. Since Aut(G2) = Sn', any permutation of elements in G2 preserves edges in G2. In particular, g does. Thus, {g(s),g(t)} E V2. It should be clear at this point that if Aut(G) = Sn and Aut(G ) 7 Sn, then every automorphism of G2 does not induce an automorphism of G. Example 4.4 exhibits this. It should also be clear that Aut(G ) (or equivalently Aut(G2)) is S if and only if GI is either complete or totally disconnected. If one is to attempt to answer the question of when an automorphism of G1 extended to G is not an automorphism of G, one can notice the following: when- ever there is an automorphism g of Gl such that for s,t e VI, {f(s),f(t)} s E but {f(g(s)),f(g(t)) } E, then g is not an automorphism of G. For example, in the graph in Example 4.4, let g be the automorphism of G2 corresponding to (4 5 6). Then for v5 and v6' {f-1 (v5 )f -(v6)} {v2,v3} E but {f-1 (g(v5)) f1 (g(v6)) = {f- 1 v 1f-l(14)} v {V3,V} 4 E. Figure IV-6 provides a table of all bivariegated graphs with six or less points and their automorphism groups. The automorphism group of G1 or G2 is of type I if only the identity induces an automorphism on G, type II if every automorphism induces an automorphism on G, and type III if the automorphisms induce all of the 50 automorphisms on G. Further, the automorphism group of G1 or G2 is of type IV if it contains at least one non-identity element that induces an automorphism on G and at least one element which does not induce an automorphism on G. For each graph, the top half represents G1 and the bottom half G2. Si, II Aut(G) = S2 S1, I S 2' III Aut(G) = S2[E2] S2, III I ] 3' II Aut (G) S3' II = group of order 48 S + El, I Aut(G) = S2[E4] S2 + El, I S2 + El, II Aut(G) = group of S2 + E, II order 16 S2 + El, I Aut(G) = S2 [E4] S2 + El, I S2 + El, II Aut(G) = K4 S2 + E II S2 + El, III Aut(G) = S2 + E4 S3' IV S3, II Aut(G) = D6 S3, II 0 0 0 S3, IV Aut(G) = K4 S2+ E1, II S3, IV Aut(G) = S2[E4] S2+E1 III S2 + El, III Aut(G) = S2[E4] S2 + El, III S III Aut(G) = group of S3, III order 6 S2 + El, I Aut(G) = K4 S2 + El, I S2 + E III Aut(G) = S2 [E4] S3, IV FIGURE V-6 S2' II Aut(G) = D4 S2, II S2 II Aut(G) = D4 S2, II w 0 Li- I' Permutation Graphs It appears that in order to obtain a feel for the automorphism groups of a bivariegated graph and its factors, one needs to examine this problem by looking at a smaller class of bivariegated graphs. We will conclude this chapter by looking at some results obtained by doing this. Rather than examine the automorphism groups of the general class G1 f G2 of bivariegated graphs, M. Borowiecki [ 3 1 considers the case where G1 = G = G. First, he makes the following definitions: Definition 4.9: Let G be a graph whose vertices are labeled vl,...,vn and let a be a permutation on the set {l,...,n1. Then by the a-permutation graph P, (G) of G is meant the graph consisting of two disjoint, identically labeled copies of G, say G and G', together with n additional edges e., 1 < i < n, where e. ioins the vertex labeled v. in G with v () in G'. For clarity, we label a vertex G' as v.'. Denote G = (V,E) and G' = (V',E'). Let a* be the permutation of V such a*(v.) = v. if and only if a(i) = j. We define the mapping a: V V' by a(v.) = v.' if and only if a(i) = j. Thus, P (G) corresponds to G f G, with a corresponding to f. Now, Borowiecki states the following problem of Frechen [5 ]: Consider the class H(P,G) of all permutations a* which preserve a property P of a graph G under transformation from G to Pa(G). Determine the properties P and graphs G for which (a) H(P,G) is a subgroup of S n, n = IVI (b) H(P,G) = Aut(G). Borowiecki considers the above problem for the chromatic number X (G) of a graph G and proves the following: Theorem 4.10: If X(G) > 2, then H (X,G) is a group. Corollary 4.11: If G is not totally disconnected, then Aut(G)c H(x,G)C S n n CHAPTER V GRAPHS OF SEMIGROUPS AND SEMIGROUPS OF GRAPHS A metric characterization of the graph of a semigroup prompted an examination of the relationship between bivariegated graphs and graphs of semigroups. Then, a theorem relating semigroups to endomorphisms of graphs motivated a search for a connection between this theorem and bivariegation. Graphs of Semigroups In the following, let S represent a semigroup, and S*the set of proper subsemigroups of S. Definition 5.1: The graph E(S) of a semigroup S is the graph whose vertices are the proper subsemigroups of S with two of these, say, A and B, edged if and only if AnB / 0. Example 5.2: Suppose S is the semigroup given by the multiplication table a b a a a b b b Then S* = {{a},{b}}, and E(S) is a graph consisting of two disconnected vertices. Example 5.3: Suppose S is the semigroup given by a b c a a a a b b b b C c c c Then S* = {{a},{b},{c},{a,b},{a,c},{b,c}}, and E (S) is the graph in Figure V-1. {b} {a,b} {b,cl {a} ac. f{c} FIGURE V-I Bosak [4 ] proved some results concerning graphs of semigroups, and posed two problems: (1) Does there exist a semigroup with more than two elements whose graph is disconnected? (2) Find a necessary and sufficient condition for a graph to be the graph of a semigroup. The first problem was answered in the negative in a theorem proven by Lin [10]: Theorem 5.4: Every semigroup with more than two elements has a connected graph. Pondelicek [12] also solved Bosak's first problem, and he further obtained a metric characterization of the graph of a semigroup: Theorem 5.5: The diameter of the graph of a semigroup does not exceed three. Because of this metric characterization, and the work in this paper with graphs that have distinct metric characterizations, it was thought that it might be interesting to look at the relationships (if any) between the graphs of semigroups and bivariegated graphs. In particular, it was hoped to partially solve the second Bosak problem by answering some of the following questions: Which finite semigroups have bivariegated graphs? Which connected bivariegated graphs are the graphs of semigroups? First, the graphs of all semigroups of order four or less were examined, using the scheme given by Petrich [11]. For any class C of semigroups, a possibly larger class may be constructed by performing one or both of the following types of operations on the members of C: (1) adjunction of an identity or zero, (2) inflation, and (3) forming direct products. Inflation may be defined as follows: Definition 5.6: To every element s of a semigroup S, associate a set Z such that: the sets of Z are pair- wise disjoint and Z nS = {s}. The set V = u Z , together with the multiplication x y = ab if x Za and y E Zb' is an inflation of S. The remaining semigroups that cannot be obtained by one of the above operations are listed by Petrich according to either their multiplication tables or the configuration of their greatest semilattice decompositions and number of elements in each class. Definition 5.7: A commutative semigroup in which all elements are idempotent is called a semilattice. Clearly, the trivial semigroup of order one has no graph since it has no proper subsemigroup. The semigroups of order two have as their graphs either one or two disconnected points. From the graphs of semigroups of order three, only two are bivariegated. First, consider G2, the cyclic group of order two, with identity a and multiplication table defined by a a = a = b b and a b = b a = b. Inflate about the non-identity element b, obtaining the addi- tional element c, with multiplication table a b c a a b b b b a a c b a a Clearly, {a} and {a,b} are proper subsemigroups, producing a graph consisting of two edged vertices which is a bivariegated graph. ___ ii In addition, the semigroup defined by the multiplication table a b c a a a a b a a a c a a b has {a} and {a,b} as its proper subsemigroups, producing the same graph as above. For the semigroups of order four, the same graph as above is obtained by inflating G3 about each of its non-identity elements. In addition, the same graph is obtained by forming the graph of G4. From these results, conjectures were made. It is clear that the graph of a cyclic group is complete because each subgroup contains the identity. The only bivariegated complete graph is the one consisting of two edged vertices. Thus, for the graph of a cyclic group to be bivariegated, it must have exactly two proper subsemigroups. Recalling that the number of subgroups of G is equal to the number of distinct divisors of n, it can be seen that the graph of G , where n has three distinct divisors, is bivariegated. Three distinct divisors are necessary since one divisor of n will be the number one, corresponding to the trivial subgroup, and one divisor of n will be n itself, corresponding to the improper subgroup Gn- That leaves one nontrivial proper subgroup to correspond to the second vertex of the bivariegated graph. Now, the integer n has three distinct divisors only when it is the square of a prime number. Thus, the following theorem has been proven: Theorem 5.8: If n is the square of a prime number, then the graph of Gn is bivariegated. It was stated above that the graphs of infla- tions of certain cyclic groups are bivariegated. By definition of inflation, multiplication by the element obtained by inflation is the same as for the element from which the inflation is generated. Thus, in the multiplication table, the rows and columns corresponding to the new element and the inflating element are identical. If n is prime, then Gn has no nontrivial proper subgroup. If the inflating element is not the identity of Gn, then no additional subgroups are obtained by adding the new element since its mul- tiplication table is identical to that of the inflating element. (This is not true if the inflating element is the identity of G since the new subgroup consisting of the identity and the new element is formed.) Clearly, Gn itself is a proper subsemigroup of the semigroup obtained by inflation about a non-identity of G n If n is prime, then this semigroup contains only one other proper subsemigroup, the trivial one. Thus, the following theorem has been proven: Theorem 5.9: If n is prime, then the graph of the semigroup obtained by inflating G about a non-identity n element is bivariegated. After examining a number of graphs of semi- groups, the following was conjectured and proven: Theorem 5.10: Suppose G is the graph of some semigroup S (i.e., I(S) 2 G). If G is bivariegated, then G has exactly two vertices. Corollary 5.11: If E(S) is bivariegated, then S has exactly two proper subsemigroups. Proof: Let S1 be a vertex of G = E(S). Since G = G f - G2 is bivariegated, there is another vertex S2 G2 with which S1 E G1 is edged. By definition of E(S), S1 and S2 are edged if and only if they have at least one element in common; call it xl. In order for S2 and S1 to be distinct, one of these must contain another element, say x2. Without loss of generality, let x2 e S2 and x2 / SI. If G has exactly two vertices, there is nothing to show. So, assume that G has more than two vertices. Let S3 be a third vertex which is connected to either S1 or S2. Without loss of generality, assume S2 and S3 are edged. It is clear that S3 C G2. Assume x2 E S2 n S3. Note that xI / S3. Now, if G is bivariegated, there exists another vertex S4 such that f(S4) = S3. If S4 and S3 are edged, they have an element in common. This element cannot be x1, or else S4 e G1 would be edged with both S2 and S3. Similarly, the element in common cannot be x2. So, both S4 and S3 contain some other element, say x3. Notice that S2 n S3 is a subsemigroup (as it is the intersection of two subsemigroups) containing x2 and possibly other elements. Thus we have a fifth sub- semigroup, and thus a fifth vertex. Call S5 = S2 n S3. Since x2 E S5, it must be in G2' or else S2 would be edged with two vertices in GI. Then there must be an S6 in G1 such that f(S6) = S5. S6 cannot contain xl, or else S2 e G2 would be edged with two vertices in GI. Similarly, x2 / S-, or else S6 would be edged with both S2 and S5 in G2. Also, x3 i S6, or else both S4 and S6 in G1 would be edged with S3 in G2. So, S6 must contain an additional element, say x4, that is also an element of S5. But S5 = S2 n S3. Thus, x4 must also be in both S2 and S3' which forces S6 e G1 to be connected to S2' S3, and S5, all of which are in G2. This contradicts the fact that G is bivariegated. Thus, G has no more than two vertices. This theorem partially answers Bosak's second problem, for it says that there are quite large classes of graphs that are not graphs of semigroups. That is, out of the entire class of bivariegated graphs, there is only one, the two-point bivariegated graph, that is the graph of a semigroup. Semigroups of Graphs Now we will examine another problem relating to graphs and semigroups. In an article written by Z. Hedrlin and A. Pultr [7 ] there is a discussion of relations with given finitely generated semigroups. Definition 5.12: Let X be a set and R a binary rela- tion of X, R C X x X. We write xRy if (x,y) s R, x,y e X. Further, R = (R,X) is used to explicitly show the set X. In graph theoretic terms, every relation (R,X) may be associated with a graph G, where X is the set of vertices of G and R is the set of all edges of G. A transformation f of X is called compatible with a relation (R,X) if xRy implies f(x)Rf(y) for all x,y e X. The set of all compatible transformations with a relation (R,X) is denoted by C(R,X). In graph theoretic terms, a transformation of X is called an endomorphism of G if it sends each edge into an edge. Evidently, f is an endomorphism of G if and only if f is compatible with (R,X). Further, the set C(R,X), the set of all endomorphisms of the graph associated with R, is a semigroup under composition of transformations. Hedrlin and Pultr prove the following theorems: Theorem 5.13: Let S be a finitely generated semi- group with unity element. Then there exists a relation RT on a set XT such that C(RT,XT) under composition of transformations is isomorphic with S. Moreover, there exist infinitely many non-isomorphic relations RT with this property. The above theorem can be translated into the language of graph theory as follows: Theorem 5.14: For each finitely generated semigroup S with unity element, there exist infinitely many non-isomorphic graphs G such that S is isomorphic with the semigroup of all endomorphisms of G. Earlier in this chapter the relationship between semigroups and bivariegated graphs was examined. Thus, it became an interesting problem to look at the following question: Of the infinitely many graphs that arise in connection with Theorem 5.14, how many of these are bivariegated? Further, what are the necessary and sufficient conditions on S to insure that G is bivariegated? In order to attempt to answer these questions, the construction of (RTXT ) in the proof of Theorem 5.13 must be examined. Hedrlin and Pultr [7] define {(R.,X), i = 1,2,...,n} to be a system of relations on X, n > 2, and T to be a sequence from the following lemma. Lemma 5.15: Let n be a natural number. Then there exist infinitely many sequences T = {tl,t, ..., t n} of natural numbers such that t. / t. for i / j t. + t. > tk for all i,j,k c {l,2,...,n}. The proof, stated in [7], is based on the definition t. = i + h(n + 1), where h is an arbitrary natural number. It is clear from the definition of t. 1 that if n > 2, then t. > 2 for i = 1,2,...,n. Continuing the description of the construction in [7], denote by Y the set of all quadruples (x,y,i,j) such that x R. y, 1 < j < t., where t. is the i-th member of 1 1 1 the sequence T. Let U = {uo,u }, V = {voVlv2} be sets. Define XT = XUYuUuV, the union of mutually disjoint sets. Let two elements in X be in the relation RT in the following cases: x RT(x,y,i,l), (x,y,i,j) R T(x,y,i,j + 1) for j = 1,2,... ,ti 1, (x,y,i,t ) R y, UoRTX,voRTX for all x e X, uoRTUlulRTuo' voRTV ,vlRTV2 ,V2R Tv The remainder of this chapter is a record of what was attempted in order to find some relation- ship between Theorem 5.14 and bivariegated graphs. First, we wanted to examine the following question: Given a finitely generated semigroup S with identity, what additional properties of S are necessary in order that there exist a bivariegated graph G such that S is isomorphic to the semigroup of endormorphisms of G? The first attempt at obtaining a hint to the nature of these properties was to choose examples of bivariegated graphs, form their semigroups of endomorphisms, and check the additional properties of these semigroups. This attempt proved futile because graphs of very small order normally have a very large number of endomorphisms, causing the corresponding semigroup to be too large to deal with efficiently. Next, the characteristics of the graph corresponding to (RTX T) were examined to determine at least necessary conditions for it to be bivariegated. In particular, since the minimum requirement for a graph to be bivariegated is that it have even number of vertices, the conditions that cause the set XT to have an even number of elements were sought. The following results were obtained: Lemma 5.16: Let m be the number of elements in S and let {R., i = 1,2,...,n) be a system of relations on S. Then XT contains m + mn(n + 1)(2h + 1)/2 + 5 elements. Proof: Recall XT = XUYUUUV, the union of mutually disjoint sets. The set X corresponds to S, having m elements. To determine the number of elements in Y = {(x,y,i,j): xR.y, 1 < j < ti.}, note that for each R. there are mt. members of Y because j takes on all values between 1 1 and t., of which there are t. values. Thus, the order 1 1 of Y is n mt1 + mt2 + ... + mt = mt . 1 n i=1 But by Lemma 5.15, t. = i + h(n + 1), so the above 1 equals n n n mZ(i + h(n + 1)) = m(Zi + Zh(n + 1)) i=l i=l i=l = m(n(n + 1)/2 + hn(n + 1)) = mn(n + 1) ( + h) = mn(n + 1) (2h + 1)/2. Now, clearly U = {u ou } has 2 elements and V = {v ,v,v 2 1 has 3. Therefore XT has m + mn(n + 1) (2h + 1)/2 + 2 + 3 elements. As a result of the above lemma, we obtain a necessary condition in order for the construction of (R T,X T) to produce a bivariegated graph: Theorem 5.17: A necessary condition for the graph corresponding to (RT ,XT) to be bivariegated is that S have odd order and the number n of relations on S be such that either n or n + 1 is divisible by 4. Proof: If m + mn(n + 1)(2h + 1)/2 + 5 must be even, then m + mn(n + 1)(2h + 1)/2 must be odd. Consider the following cases: Case 1: The number m is even. Then mn(n + 1)(2h + 1)/2 must be odd. Note that m/2 is a whole number. No matter what n is, either n or n + 1 is even. If one factor in a product is even, the product is even, so (m/2) (n) (n + 1) (2h + 1) is even. Thus if m is even, (RT,X ) does not correspond to a bivariegated graph. Case 2: The number m is odd. Then mn(n + 1)(2h + 1)/2 must be even. Suppose n is even. Then n + 1 is odd. Let p = n/2. Clearly, p is a whole number. Now, it is necessary for mp(n + 1) (2h + 1) to be even. For any h, 2h + 1 is odd. So each of m, n + 1, and 2h + 1 are odd. Thus we must have p = n/2 even, implying that n is divisible by 4. Now suppose that n is odd. Then n + 1 is even, and 2h + 1 is odd. Let p = (n + 1)/2, a whole number. It is necessary for mnp(2n + 1) to be even. Since each of m, n, and 2h + 1 is odd, p = (n + 1)/2 must be even. This forces n + 1 to be divisible by 4. Thus, for the graph corresponding to (RTXT) to be bivariegated, m must be odd and either n or n + 1 must be divisible by 4. Although the restriction that a graph have an even number of vertices is far from being a sufficient condition for bivariegation, it nevertheless imposes quite a restriction on the semigroup S. Thus, it became interesting to examine the restriction on S caused by the limitation on the number of lines of a bivariegated graph. Recall that in the construction of (RTXT) above, the first relations defined were of the form xRT(x,y,i,l). This produces mn lines because there are m elements in X and n values of i (each corresponding to {(R, i = 1, 2,...,n)). From the relations of the form (x,y,i,j)R (x,y,i,j + 1) for j = 1,2,...,ti 1, we obtain m(t. 1) relations for each i. This produces n n Xm(t. 1) = mEt. mn i=l 1 =1 which, from the proof of Lemma 5.16, is = mn (n + 1)(2h + 1)/2 mn. From the relations of the form (x,y,i,ti)R y there are clearly mn different relations. Further, from uoRTx,v oRT for all x e X there are 2m relations. Finally, from uoR lulR uo' VoRTVl TVlRTV2 V2RTVo there are 4 relations. Thus, there is a total of mn + mn(n + 1)(2h + 1)/2 mn + mn + 2m + 4 or mn + mn(n + 1)(2h + 1)/2 + 2m + 4 relations. Now the question is whether or not this lies between the minimum number p/2 and the maximum number p /4 of lines that a bivariegated graph may have. For sake of simplicity, let s = mn(n + 1) (2h + 1)/2. Then, the first question is whether or not the number s + mn + 2m + 4 of relations corresponding to the construction of (RTp,XT) is greater than or equal to half the number s + m + 5 of elements in X T; that is, it should be true that s + mn + 2m + 4 > (s + m + 5)/2. This is clearly true. In addition, the second question is whether or not the number of relations is less than one-fourth of the square of the number s + m + 5 of elements in XT; that is, it should be true that s + mn + 2m + 4 < (s + m + 5) 2/4. However, 4s + 4mn + 8m + 16 < s2 + 2sm + m2 + 10m + 10s + 25 4mn < s2 + 2sm + m2 + 2m + 6s + 9. At least one of the terms, 2sm, alone is larger than 4mn because 2sm = 2m 2n (n + 1)(2h + 1)/2 = m2n(n + 1)(2h + 1) > mn(2) (3) > 4mn. Thus, the bounds on the number of lines in a bivariegated graph places no additional restrictions on the semigroup S. Now, notice that in the definition of R , V0RTVI, VlRTV2, V2RTV0, and V0RTx for all x c X. Suppose (R ,X ) is to be bivariegated by partitioning some of the elements into G1 and some into G2 in such a way that (RT,XT) can be transformed into G1 f G2' Without loss of generality, let v0 e G Recall v0R v . Then suppose v1 E G2. Because v1R Tv2 v2 must be in G2 or else v would be related to two different elements in G However, V2RT 0, which forces v2 to be related to two different elements in GI. Thus, the assumption that v1 e G2 was incorrect. Now since v1 e G if v2 c G2, it would be related to two different elements in Gi, so v2 e G However, since v0 is the only element of V related to elements outside V and all of the elements of V are in GI, it must be true that neither v1 nor v2 is related to an element in G2. Thus, the construction (RT,XT) does not produce a bivariegated graph. Therefore, it has been shown that in Theorem 5.14, none of the infinitely many graphs G such that S is isomorphic to the semigroup of endomorphisms of G is bivariegated. Thus, there must be many graphs with the same semigroup of endomorphisms. Finally, with regard to the relationship between semigroups and bivariegated graphs, we have shown that given a semigroup, its graph, if bivariegated, can only be the connected tree on two vertices. Additionally, we have shown that for a given semigroup S, no bivariegated graph G can be constructed from it such that S is isomorphic to the endomorphism semigroup on G. CHAPTER VI DISCUSSION Although some of the results of Sanders [13] related to bivariegated trees were extended in this paper to the case of bivariegated Husimi trees, it would be desirable to obtain some further characterizations of bivariegated Husimi trees. Sanders proves the following: Theorem 6.1: A tree with 2n vertices has a 1-factor if and only if the largest maximal independent set of vertices of T contains n vertices. Theorem 6.2: A tree is bivariegated if and only if T has a 1-factor. Thus, a bivariegated tree is characterized in terms of independent sets of its vertices. There- fore, the question arises: What is the relationship between a bivariegated Husimi tree and independent sets of vertices? Another question arises due to the relationship between trees and Husimi trees: Is there a way to shrink the cycles of a bivariegated Husimi tree to obtain a tree, and, if so, when is this tree bivariegated? Or, a similar question is, can we expand the vertices of a tree (possibly bivariegated) in some way to obtain a bivariegated Husimi tree? In one part of this paper we looked at the relationship between Aut(G), Aut(G ), and Aut(G2), where G = G1 f G2. It would be desirable to have a better characterization of this relationship. One way to proceed in solving this problem is in the manner of Borowiecki [3 ], that is, by restricting G1 and G2 to be equal and the bijection f to be a part of a smaller class of bijections. Then one can attempt to answer some of the questions posed by Frechen [5 ]: What are the properties P of G (=G2) which are preserved in G1 f G2? What are the properties P and graphs G (=G2) for which the class of all bijections preserving P is a subgroup of the symmetric group of order n or is isomorphic to Aut(G1l)? Another topic presented in this paper is the relationship between semigroups and bivariegated graphs. It would be interesting to further study the restriction on the diameter of the graph of a semigroup with respect to some of the metric characterizations presented by Howorka [ 8 ], and establish any connection with some subclass of bivariegated graphs. Finally, the semigroup of endomorphisms of a bivariegated graph was studied. Hedrlin and Pultr [7 ] present a construction which produces a graph from a semigroup. A question which arises is: Can we derive directly from the bivariegated graph (from which we have obtained a semigroup) the graph produced by this construction? Further, what is the relationship between these graphs? APPENDIX 1: BIVARIEG7ATED GRAPHS p = 2 p 4 p = 6 p = 8 I J (h/Il I p = 8 (continued) p = 8 (continued) p = 8 (continued) FTI p = 8 (continued) rf h h l' N J Mzz \J/ \..1/ p = 8 (continued) / K 1\ \--^ /\ A\ fc^ C0 00( ~~\ /_ \x^ BJ/\ APPENDIX II: BIVARIEGATED HUSIIMI TREES p = 4 p = 2 I p = 6 p= 8 I A t II IWr I L 4, L -T p = 10 - i [Sd IE -M L -V m p = 12 r "'I p^ ^ - rm1 En 1 :1 21117l 31711 TVV1 ET1Vf I7\ TlT]L ! iiAn hi~i J\^T p = 12 (continued) iL A 1 7L AL mi_ 1TE E TL\ p = 12 (continued) A-\ R-A L L Err Kr- E -] -Ah AIL1 m zzc' Lv- Z7 p : 12 (continued) FIKZ IZE7 I^Z77 N^-n REFERENCES 1. A. R. Bednarek, A note on tree isomorphisms, J. Comb. Theory, (B) 16 (1974), 194-196. 2. A. R. Bednarek and E. L. Sanders, A characterization of bivariegated trees, J. Discrete Math., 5 (1973), 1-14. 3. M. Borowiecki, On the a-permutation graphs, Recent Advances in Graph Theory, (Proc. Second Czechoslovak Sympos., Prague, 1974), Academia, Prague, 1975, 89-92. 4. J. Bosak, The graphs of semigroups, Theory of Graphs and its Applications, (Proc. Sympos. Smolenice, June 1963), Academic Press, New York, 1965, 119-125. 5. J. B. Frechen, On the number of cycles in permutation graphs, The Many Facets of Graph Theory, Springer- Verlag, Berlin, 1970, 83-87. 6. F. Harary, Graph Theory, Addison-Wesley, Reading, 1969. 7. Z. Hedrlin and A. Pultr, Relations (graphs) with finitely generated semigroups, Monatshefte fur Math., 68 (1964), 213-217. 8. E. Howorka, On metric properties of certain clique graphs, to appear. 9. D. C.Kay and G. Chartrand, A characterization of certain ptolemaic graphs, Canad. J. Math., 17 (1965), 342-346. 10. Y.-F. Lin, A problem of Bosak concerning the graphs of semigroups, Proc. A.M.S., 21 (1969), 343- 346. 92 11. M. Petrich, Introduction to Semigroups, Charles E. Merrill Pub. Co., Columbus, 1973. 12. B. Pondelicek, Diameter of a graph of a semigroup, Casposis Pest. Mat., 92 (1967), 206-211. 13. E. L. Sanders, Independent sets and tree structure, Doctoral Dissertation, University of Florida, 1975. BIOGRAPHICAL SKETCH Leita Fay Aycock Riddle was born on December 22, 1949, to Thomas G. and Leita M. Aycock of Atlanta, Georgia. In June, 1967, she was valedictorian of her graduation class at The Arlington Schools in Atlanta. In the fall of 1967, she entered H. Sophie Newcomb College of Tulane University. On June 20, 1970, she married Dennis Lee Riddle of Madison, Wisconsin. While in attendance at Tulane, she was appointed to Tulane University Scholars and Chi Beta honorary society. In May, 1971, she graduated cum laude with the degree of Bachelor of Science in mathematics. The following fall she joined the mathematics faculty at Evans Junior High School in Spartanburg, South Carolina. In September, 1972, she entered the Graduate School of the University of Florida in mathematics. In August, 1974, she received the degree of Master of Science. From August, 1975, to date she has been a member of the faculty at the University of South Carolina at Spartanburg, where she teaches mathematics and computer science. At the same time, she has continued her study in pursuit of the degree of Doctor of Philosophy. Leita Fay Aycock Riddle is a member of the American Mathematical Society and the Honor Society of Phi Kappa Phi. I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. A. R. Bednarek, Chairman Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. J. E. Keesling Professor of Math matics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. M. P. Hale, Jr. Associate Professor of Mathematics II I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. T. T. Bowman Assistant Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. W. D. Hedges Professor of Education This dissertation was submitted to the Graduate Faculty of the Department of Mathematics in the College of Arts and Sciences and to the Graduate Council, and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. June, 1978 Dean, Graduate School |

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PAGE 1 BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS By FAY AYCOCK RIDDLE A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1978 PAGE 2 To Ve.nn4.-S PAGE 3 ACKNOWLEDGEMENTS The author would like to express her sincere thanks for the leadership and guidance of Professor A. R. Bednarek, who has been most patient and understanding in the long-range completion of this effort. His insight into the open problems in graph theory has provided the basic topics of this work. The author also wishes to acknowledge the remainder of her supervisory committee: Drs. J. E. Keesling, M. P. Hale, Jr., T. T. Bowman, and W. D. Hedges for their personal contributions to her academic training. The author would like to thank her family for allowing her the opportunity to attend the superior elementary, secondary, and undergraduate institutions that laid the foundation for higher academic pursuits. Finally, the author would like to thank her husband, Dennis, for his encouragement and support during her graduate program. PAGE 4 TABLE OF CONTENTS Page ACKNOWLEDGEMENTS iii ABSTRACT v INTRODUCTION 1 CHAPTERS I BASIC DEFINITIONS AND THEORY 3 II MATRIX REPRESENTATION AND BIVARIEGATED HUSIMI TREES 9 III GRAPH ISOMORPHISMS 22 IV AUTOMORPHISM GROUPS 41 Automorphism Groups of a Graph and Its Factors 41 Permutation Graphs 52 V GRAPHS OF SEMIGROUPS AND SEMIGROUPS OF GRAPHS 54 Graphs of Semigroups " 54 Semigroups of Graphs 64 VI DISCUSSION 76 APPENDIX I: BIVARIEGATED GRAPHS 79 APPENDIX II: BIVARIEGATED HUSIMI TREES 85 REFERENCES SIBIOGRAPHICAL SKETCH 93 PAGE 5 Abstract of Dissertation Presented to the Graduate Council of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy BIVARIEGATED GRAPHS AND THEIR ISOMORPHISMS By Fay Aycock Riddle June, 1978 Chairman: A. R. Bednarek Major Department: Mathematics Characterizations of bivariegated graphs, in particular, bivariegated Husimi trees, are presented. The relationship between graph isomorphisms and length of irreducible cycles in a bivariegated graph is investigated. Also investigated are automorphism groups of bivariegated graphs and the relationship between bivariegated graphs and semigroups. The largest class of graphs for which graph isomorphisms can be characterized by all irreducible cycles being of length four is the class of Husimi trees . PAGE 6 INTRODUCTION A graph G is a finite nonempty set V of vertices together with a set E of unordered pairs of distinct vertices of V, called edges. A graph is bivariegated if the vertex set can be partitioned into two sets of equal size such that each vertex is adjacent to one and only one vertex in the set not containing it. This defines a bijection f between the two vertex sets. Consider the following question: What is the relationship between the structure of a bivariegated graph G and the bijection f? In particular, when is f an isomorphism between the two graphs into which G is partitioned? Bednarek and Sanders [2] prove that if these two graphs are trees, then f is an isomorphism if and only if every irreducible cycle in G has length four. This paper extends this theorem by proving it for a class of graphs larger than the class of trees. Furthermore, in studying the above question, other information about bivariegated graphs is obtained. PAGE 7 Chapter I presents some basic definitions and elementary results of graph theory that are useful in later chapters. The definitions primarily follow the notation used by Harary [ 6 ] . Chapter II gives a representation of the adjacency matrix of a bivariegated graph and presents some characterizations of bivariegated Husimi trees. Chapter III answers the question posed above by extending to a larger class of graphs a theorem relating graph isomorphisms and bivariegated graphs. Chapter IV examines the relationship between the automorphism groups of a bivariegated graph and its factors and presents a result obtained by considering the case in which the factors are equal. Chapter V examines the relationship between bivariegated graphs and graphs of semigroups. Then, a theorem relating semigroups to endomorphisms of a graph is examined for the case in which the graph is bivariegated. The final chapter discusses the results and questions arising from them. In the appendices are listings of bivariegated graphs with 2, 4, 6, and 8 vertices and listings of bivariegated Husimi trees with 2, 4, 6, 8, 10, and 12 vertices. PAGE 8 CHAPTER I BASIC DEFINITIONS AND THEORY Definition 1.1: A graph G is a nonempty finite set of po ints (or vertices) , V, along with a prescribed set E of unordered pairs of distinct points of V, known as edges . We write G = (V,E). If two distinct points, x and y, of a graph are joined by an edge, they are said to be adjacent . We write e = {x,y} to denote the edge between x and y, and we say that x and e are incident with each other, as are y and e. A s ubgraph of G is a graph having all of its points and lines in G. A spanning subgraph contains all the points of G. Two graphs G and H are isomorphic , denoted G = H, if there exists a one-to-one correspondence, called an isomorphism , between their point sets which preserves adjacency. PAGE 9 Definition 1.2: A walk of a graph G is a finite sequence of points such that each point of the walk is adjacent to the point of the walk immediately preceding it and to the point immediately following it. If the first and last points of a walk are the same point, we say the walk is closed, or is a cycle (provided there are three or more distinct points and all points are distinct except the initial and final points). A graph is acyclic if it contains no cycles. A walk is a path if all the points are distinct. A complete cyc le has every pair of points adjacent. The comple te graph K p has every pair of its p points adjacent . A graph is c onnected if every pair of points is joined by a path. A maximal connected subgraph of G is called a conn ected componen t or _sjÂ£ip2y_^om Â£ onent of G. Definition 1.3: The degree of a point v of G, denoted deg v, is the number of lines incident with v. The point v is isolated if deg v = 0; it is an end point if deg v = 1. A graph is regular of degree n if every vertex has degree n. PAGE 10 Theorem 1.4: The sum of the degrees of the points of a graph G is twice the number of lines. The proof of Theorem 1.4 can be found in [ 6 ] . Definition 1 .5: A cut point of a graph is one whose removal increases the number of connected components. A nonse parable graph is connected, nontrivial, and has no cut points. A block of a graph is a maximal nonseparable subgraph. The notation V {v} means V minus the vertex v and all its incident edges. Theorem 1.6 : Let v be a point of a connected graph G. The following statements are equivalent: (1) The point v is a cut point of G. (2) There exist points u and w distinct from v such that v is on every u-w path. (3) There exists a partition of the set of points of V {v} into subsets U and W such that for any points u e U and w e W, the point v is on every u-w path. Theorem 1.6 is proven in [6 ]. Definition 1.7: A tree is a connected acyclic graph. PAGE 11 A Hu simi tree is a connected graph in which every block of G is a complete graph. The length of a walk 6r /V^... v R ) is n, the number of occurrences of lines in it. The d istance d(u,v) between two points u and v in G is the length of a shortest path connecting them, if any. A metric space M is called a graph metric spac e if there exists a graph G whose vertex set can be put in one-to-one correspondence with the points in M in such a way that the distance between every two points of M is equal to the distance between the corresponding vertices of G. A graph is Ptolemaic if its associated metric space M is such that for any x,y,z,w e M, the three numbers d (x ,y ) Â• d (z ,w) , d (x , z) -d (y ,w) , and d(x,w)d(y,z) satisfy the triangle inequality. A graph G is said to be weakly geodetic if for every pair u,v of vertices of G such that d(u,v) < 2, there is exactly one shortest u-v path. Kay and Chartrand [ q ] prove the following: Th eorem 1.8 : A graph G is a Husimi tree if and only if G is weakly geodetic and Ptolemaic. PAGE 12 De finition 1.9 : A graph G = (V,E) is said to be bivariegated if G = G, f G 2 = (V 1 u V 2 , E 1 u E 2 u E f ) , where G Â± = (V-^E^ , G 2 = (V 2 ,E 2 ), V ! n v 2 = 0, f: V, * V 2 is a bijection, and E f = {{x,f(x)}|x e V^}, where we let (x,f(x)} denote the edge incident with x and f(x). Therefore, a graph G is bivariegated if and only if its vertex set can be partitioned into two disjoint equal sets such that each vertex is adjacent to one and only one vertex in the set not containing it. If G = G,, we say that G is the permutation graph of G 1 (or, equivalently , of G 2 ) Â• Definition 1.10 : An n-factor of a graph G is a spanning subgraph of G which is not totally disconnected and is regular of degree n. In particular, G has a 1-f actor if it has a spanning subgraph consisting of disjoint edges . Sanders [13] characterizes trees with 1-factors: Theorem 1.11 : A tree is bivariegated if and only if it has a 1-factor. Theorem 1.12 : A bivariegated graph with p vertices has at least p/2 and at most p 2 /4 lines. PAGE 13 Proo fLet G = G f G 2 . The minimum number of lines G can have is when both G] _ and G 2 are totally disconnected, in that case there exist p/2 pairs of points in G edged by lines in E f . The maximum number of lines G can have is when both G Â± and G 2 are complete. In that case, each of G and G 2 has n (n l)/2 edges, where n = p/2. In addition, there are n edges in E f . Thus, G has a total of n(n 1) + n = n 2 = p 2 /4 lines. Definition 1.13 : A coloring of a graph is an assignment of colors to its points so that no two adjacent points have the same color. An n^coloring of a graph G uses n colors. The chromati^jiumber X (6) is defined as the minimum n for which G has an n-coloring. Clearly, the chromatic number of a tree is 2. The chromatic number of the complete graph on p vertices is p. Thus, a Husimi tree which is not also a tree has as its chromatic number the length of its longest cycle. PAGE 14 CHAPTER II MATRIX REPRESENTATION AND BIVARIEGATED HUSIMI TREES It is the purpose of this paper to determine some characterizations relating to bivariegated graphs. Sanders [13 ] considers a subclass of these, the bivariegated trees, and characterizes it. Thus, in this chapter we attempt to further generalize the properties of bivariegated graphs by considering a slightly larger subclass, the bivariegated Husimi trees, and attempt to characterize it. In the process of doing this, it became evident that it would be extremely helpful in making our generalizations if we were to have at our disposal a set of examples of all bivariegated graphs with p or fewer vertices, for some p. Thus, before we present our observations concerning bivariegated Husimi trees, we will first look at how the examples were generated. Adjacency Matrices of Bivariegated Graphs By definition, a bivariegated graph G = G^ f G ? with 2n vertices can be factored into graphs G^ and G 2 PAGE 15 :o 10 with n vertices where each vertex in G, = (V, ,E,) is edged with a unique vertex in G 2 = (V 2 ,E 2 ). Thus, tc generate examples of all bivariegated graphs with 2n vertices, one needs to consider all possible combinations of pairs of graphs with n vertices, connected by all possible bijections f: V Â± * V 2 . Let V 1 = {u 1 ,u 2 , . . . ,u n > and V 2 = {v ,vÂ„,...v }. For any bijection f, the vertices of V 2 can be renamed v, ' , v 2 ' , . . . , v n ' in such a way that f(u.) = v. 1 , i = l,2,...,n. In light of this, one can generate all bivariegated graphs having 2n vertices by first considering all possible combinations of pairs of graphs G, and G 2 with n vertices. Then, rather than considering bijections between G, and G 2 , we merely consider all possible permutations of the vertices of G 2 . Because for rather small values of n this becomes a tedious process, it was desirable to write a computer program to perform the process above. The program is based on the representation of a graph by its adjacency matrix. Definition 2.1 : The a djacency matrix A = ta^.] of a labeled graph G with p points is the p x p matrix in which a. .= 1 if v. is adjacent to v. and a . . = otherÂ±j i D 1 3 PAGE 16 11 Clearly there is a one-to-one correspondence between labeled graphs with p points and p x p symmetric binary matrices with zero diagonal. Now consider the adjacency matrix A of a bivariegated graph. Let the first n rows and columns of A correspond to u^i^,...,!.^ and the second n rows and columns correspond to v 1 , ,v 2 ',. v ' as defined above. Partition A into 4 square subn matrices. Because f(u.) = v.', i = l,2,...,n, it should be clear that each of the upper right and lower left submatrices is the identity matrix. Now we need to generate the upper left and lower right submatrices, corresponding to adjacency matrices of G, and G,, respectively. Consider the adjacency matrix of G,. Because it is symmetric, we need only to consider possible entries above its zero diagonal. For an n x n matrix, there are 1 + 2 + ... + (n-1) = n(n-l)/2 entries above the diagonal. Then all possible graphs G can be found by generating all binary sequences of length m = n(n-l)/2, of which there are 2 m . Similarly, there are 2Â™ possible adjacency matrices for G~. The number of combinations of 2 different elements from a set of p is p ! / (2 (p-2) !), and the number of ways to choose two identical elements from a set of PAGE 17 12 size p is p. The sum of these can be easily shown to be p(p+l)/2. Now, if p = 2 m , we have 2 m-1 (2 m +l) adjacency matrices for G, f G 2 > However, since the representation of a graph by adjacency matrix is not unique, this merely gives us an upper bound on the number of bivariegated graphs with 2n vertices. Thus, we have proven the following: Theorem 2.2 : Let G = G f G 2 be a bivariegated graph with 2n vertices. Then the vertices of G can be partitioned in such a way that the adjacency matrix of G is the 2n x 2n symmetric matrix, 1 with zeros on the diagonal, A, and A 2 the adjacency matrices of G, and G 9 , and I representing the n x n identity matrix. Corolla ry 2.3: There are at most 2 m_ (2 m +l) bivariegated graphs with 2n vertices, where m = n(n-l)/2. Because the method outlined above generates duplicates of the same graph, the computer program was PAGE 18 13 written to group the adjacency matrices that it generates according to the number of non-zero entries in each row (which corresponds to the degree sequence of each graph) . These groups were then checked by hand for duplicates. From this we obtain the list in Appendix I of all bivariegated graphs with at most 8 vertices. Bivariegated Husimi Trees Now we return to the. purpose of this chapter, that of characterizing bivariegated Kusimi trees. Let H be a Husimi tree. Obviously, if H is bivariegated, it has an even number of vertices. Sanders [13] proves that a tree is bivariegated if and only if it has a 1-factor. Clearly it is true that if H is bivariegated, then it has a 1-factor. However, the converse is not true, as shown in Figure II1, below. T v l v. V 3 V 4 V , V 6 V 7 V ! FIGURE II-l PAGE 19 14 This Husimi tree has a 1-factor, consisting of the edges {v 1# v 2 }, {v 3 ,v 4 }, {v 5 ,v 6 }, and {v 7 ,v g }, yet it is not bivariegated. One can state, however, that if a Husimi tree H has a 1-factor such that no edge in the 1-factor is contained in a cycle, then H is bivariegated. The proof is exactly as in the theorem by Sanders. The~above exemplifies the following: all of Sanders' characterizations of bivariegated trees cannot be extended to bivariegated Husimi trees. Because of this, an examination of the properties of the bivariegated Husimi trees that were generated earlier was begun. However, since there are so few bivariegated Husimi trees with 8 or fewer vertices, little could be generalized from these examples. Thus, it was necessary to derive a scheme for generating examples of bivariegated Husimi trees only. We will now present the results obtained while studying this problem. Proposition 2.4 : Let G = G, f G 2 be a bivariegated Husimi tree. If a is a cycle in G, then a must lie completely within G 1 = (V^E^ or G 2 = (V 2 ,E 2 ). PAGE 20 15 P roof : Let a = (v ][ , v 2 , . . . , v , v^ . Suppose that v 1 e V 1 and that f (v, ) = v~ e V~ is in the cycle a. Now consider any other vertex, say v->, in a. Because G is a Husimi tree, a is complete. Assume v, e V, . Because a is complete, v-. and vÂ„ are edged, contradicting G being bivariegated, since v 2 is also edged with v, e V, . A similar contradiction can be obtained if we assume V3 E V 2 . We also obtain the following result: Proposition 2.5 : Let G = G, f G 2 be a bivariegated Husimi tree. Then both G and G 2 cannot be connected graphs . Proof: Suppose G, = (V ,E.) is connected. Let u 1# u 2 Â£ V^ and a = (u, ,u . ,u . , . . . ,u . ,u ? ) be a path in G. between u, and u ? . Now f maps u, and u 2 to f (u ) = v, and f(u 2 ) = v 2 in G 2 = (V 2 ,E 2 ). Suppose there exists a path B = (v 1 ,v., v. , ,...,v. ,v_) in G, between v, and v . Then (u, ,u . , . . . , j+1 ' j+m 2 2 i z ii u. ,u ,vÂ„,v., ,...v.,v n ,u ) is a cycle in G, with l+n 2' 2 3+m j 1 1 vertices in both V and V 2 This is complete because G is a Husimi tree, contradicting G being bivariegated (since, for example, u e V is edged with both v^ and v 2 e V 2 ). PAGE 21 16 or T ^ r r f Go be a bivariegated Husimi tree. If there is a path in G Â± connecting the vertices u Â± and u 2 in V^ then there is no path in G 2 connecting the vertices f (u^ and f (u 2 ) in V,,. Using the information gathered up to this pcint to obtain examples of bivariegated Husimi trees, it was noticed that the portions of these graphs that are not cycles are bivariegated trees. This is formalized in the following theorem: ^eorem,!^: Let G be a Husimi tree. Consider the graph G . which consists of G minus each edge which is contained in a cycle. Then G is bivariegated if and only if all the components of G' are bivariegated trees. Proof: Suppose one of the components of G' (call it T) is not bivariegated. Clearly each component of G' is a tree. Let W be the set of all vertices of T contained iÂ„ Â„f G -he set VI is not empty, or else G has no cycles, in cycles or (j. xne s>Â«iin which case G = Gis not bivariegated. If T is not bivariegated, the connection of the vertices in W to cycles in G can still not be bivariegated. This is because all vertices in a cycle of a graph G = G 1 f " =2' lf '' is to be bivariegated, must all be in either G^ or G.,. Thus, a vertex in W which is in, say G,, and which has no PAGE 22 17 "mate" in G ? contained in T does not get a mate by being connected to a cycle. Thus, the addition of any number of cycles to any number of vertices of components of G' will not permit G itself to be bivariegated. Now, suppose each component of G 1 is bivariegated. Partition the vertices of a component T into two sets, V, and V 9 , corresponding to the bivariegation. Any cycle connected to a vertex v of T will have all of its vertices in the same set, say V^ as v. Further, each of the other vertices of such a cycle is connected to a bivariegated tree (otherwise, it has no mate in V" 2 ) Â• Because the bivariegation of a tree is unique, once one of its vertices is assigned to V or V 2 , the remainder of its vertices are uniquely bivariegated. Apply this process to each cycle and bivariegated trees connected to it, and a bivariegated Husimi tree will be obtained. Sanders [13] proves that the bivariegation of a tree is unique. We prove an analogous result below. Theorem 2.8: The bivariegation of a Husimi tree is unique. PAGE 23 13 Proof: If G is bivariegated , the components of G 1 are bivariegated trees, where G' is defined as in Theorem 2.7. Now consider any cycle of G. Suppose the bivariegation places its vertices into, say V, . Each vertex of that cycle must be connected to a bivariegated tree or both a bivariegated tree and a cycle, or else it cannot be adjacent to a vertex in V-. Now, once the bivariegation of a single vertex of a tree is determined, the bivariegation of the remainder of its vertices is uniquely determined. This tree may or may not be connected to other cycles, but if it is, the bivariegation of the cycles attached to it is uniquely determined by the bivariegation of the vertex of the tree to which it is attached. Therefore, because (i) a bivariegated Husimi tree consists of bivariegated trees and cycles, (ii) a tree is uniquely bivariegated, and (iii) each cycle is uniquely bivariegated according to the bivariegation of the vertex of the tree to which it is attached, the entire Husimi tree is uniquely bivariegated. We now see that if a bivariegated Husimi tree contains a cycle, then each vertex of that cycle is attached to a bivariegated tree. Further, if it contains PAGE 24 19 more than one cycle, the block between any two cycles is also a bivariegated tree. This motivates the use of the following for constructing all bivariegated Husimi trees: Sanders [13] constructs all bivariegated trees by taking the smallest bivariegated tree (two vertices connected by an edge) and successively adding remote end vertices. A remote end vertex is defined as follows : Definition 2.9 : Let e be an end point of graph G, and let x be the unique point adjacent to e . If deg x = 2, then e is called a remote end vertex . We can construct all bivariegated Husimi trees by taking the smallest bivariegated Husimi tree (which is also the smallest bivariegated tree) and attaching, one at a time, to a vertex a graph of one of the forms in Figure II-2, below. FIGURE II-2 PAGE 25 20 That is, attach to the vertex either a remote end vertex or a complete n-cycle with n-1 end points, where the attachment is made at the circled vertex. Thus, we obtain the following theorem. Theorem 2.10 : Let G be a bivariegated Husimi tree. Construct G' as follows: let v be any vertex of G. To v, attach either a remote end vertex or a complete cycle such that the vertices other than v are edged with an end vertex. Then G 1 is a bivariegated Husimi tree . Proof : Suppose G = G f G 2 where G = (V ,E 2 ). Suppose v e V and a remote end vertex is attached at v. Then the remote end vertex can be added to V 2 to form the set V ? ' , and the vertex x (as in Definition 2.9) can be added to V, to form the set V * . Clearly G' = G ' f G 2 ', where G' = (V 1 ',E 1 ') and G 2 ' = (V ',E ') is a bivariegated Husimi tree. Now consider the case where a cycle of the type described above is added to v e V, . Add every other vertex in that cycle to V to form the set V, ' , and add the end vertices attached to this cycle to V to form V 2 ' . Then G' = G, ' f G 2 ' is a bivariegated Husimi tree, PAGE 26 21 Appendix II gives the results of generating all bivariegated Husimi trees with 2, 4, 6, 8, 10, and 12 vertices by using the method in Theorem 2.10. As Sanders [13] points out, the technique of adding remote end vertices may produce duplicates of a particular graph. The same is true for the method outlined above, as shown in Figure II-3. -e* Â• -* Â• -Â©FIGURE II-3 PAGE 27 CHAPTER III GRAPH ISOMORPHISMS The contents of this chapter are the results of attempts to extend to a larger class of graphs a theorem relating graph isomorphisms and bivariegated graphs . Definition 3.1 : Let a graph G have vertex set V and edge set E. A finite sequence, a = (v Q , v^ , v 2 , . . . , v n ) , of distinct vertices of G is called a path provided {v.,v. , } e E f or i = 0,1,2, ... ,n-l. A cycle is a 1 l + l sequence a = ( v ' v x ' Â• Â• Â• ' v n -i ' v n ) SUch that V = V n and (v Q ,v 1 ,...,v n _ 1 ) and (v Â± , v 2 , . . . , vj are paths . A cycle is irreducible if it contains no diagonals; that is, for v., v. vertices of a, {v^v./e E implies |i-j| = 1 or n-1. If a = (v Q , v^ . . . ,v n ) , then the length of a is n. Bednarek [ 1 ] characterized tree isomorphisms in terms of bivariegated trees: Theorem 3.2 : If T Â± = (V^E^ and T 2 = (V 2 ,E 2 ) are trees and if f: V 1 Â•* V 2 is a bijection, then f is 22 PAGE 28 23 an isomorphism if and only if all irreducible cycles in T, f T~ have length four. One may wonder whether or not we may replace "trees" in the preceding theorem by some larger class of graphs. Bednarek [ 1] gives an example to show that "trees" may not be replaced by "connected graphs." This is demonstrated by Figure III-l, in which G^ and Gp are squares and f the isomorphism given by y. = f(x.), i = 0,1,2,3, and in which a = (x ,x 1 ,x 2 ,y 2 ,y 3 ,y Q ,x ) is an irreducible cycle of length six. FIGURE III-l It is one of the aims of this chapter to find the largest class of graphs for which "trees" in Theorem 3.2 may be replaced. PAGE 29 24 It can be proven that "trees" may be replaced by "Husimi trees": Theorem 3.3 : If T. = (V-^E^ and T 2 = (^ 2 ' E 2 ) are Husimi trees and if f: V, > V 2 is a bijection, then f is an isomorphism if and only if all irreducible cycles with edges in both T, and T 2 have length four. Proof: Suppose f is not an isomorphism. Then either there is an edge {x,y} e E, for which {f(x),f(y)} t E 2 or there exists an edge {s,t} e E 2 such that {f (s) , f -1 (t)} Â£ E, . Considering the first case we see that because (f(x),f(y)} i E 2 , then f(x) and f (y) are not in the same block of V" 2 since every block in a Husimi tree is complete. So V 2 is separable, and we can decompose it into blocks B lf B 2 , . . . ,B k Let B Â± be the block such that f (x) e BL and B k the block such that f(y) e B,. Then there is a path a = (f (x) , t 1 , . . . , t R , f(y)) from B, to B, between f(x) and f (y) where each t., i = 1,2,. ..,n, is a cutpoint of a block through which a passes. Because each block is complete, there is only one edge between each cutpoint of a particular PAGE 30 25 block. Now consider the path a' = (x,f (x) ,t lf . . . /t n /f (y) ,y,x) . This is a cycle of length greater than or equal to 5 and irreducible because of the way a was chosen. To prove the necessity of the condition, suppose f is an isomorphism. Let a = ( v q ' v i ' v 2 ' " * ' ' v k^ be the largest irreducible cycle in T, f T 2 with edges in both T, and T 2 . Let v Q = v R Â£ V Â± and v k-1 e V, Thus f" 1 (v k _ 1 ) = v k = v Q e V 1 , so f (v fc ) = Â£ (v Q ) = v^^ We claim that v,_ 2 e V 2 . For if not, then v k-2 e v r v k-i Â£ v 2' and f(v k-2 ) = v k-r However < this is impossible since f (v Q ) = v k _2_/ contradicting the one-to-oneness of f. We next claim that v, Â£ V... For if not, then v, e v ?' v n e V i ' and f ^ v 0^ = V l" However, this contradicts the functionality of f, as f (v Q ) = v k-1 also. We also claim that v, 3 Â£ V" 2 . For if not, then v k _ 3 Â£ V 1 ,v k _ 2 Â£ V 2 , and f(v k _ 3) = v k _ 2 However, PAGE 31 then 26 since ^ v k _ 1 ' v k _ 2 ^ is an ed 9 e ' ^ f ( v k -l ),f ^ v k _2^ = (v-.V, _} must also be an edqe. k-3 J Finally, we claim that vÂ» e V . For if not, vÂ„ e V_ and f(v,) = vÂ„ . However, since {v n ,v,} is an edge, {f(v ),f(v .)} = {v,_,,v_} is also an edge, contradicting the irreducibility of a. Now, consider f ( v k _o) /f ( v v_t) Â£ ^i and f(v,),f(vÂ„) e VÂ„ . Construct the cycle a ' = (v 0' V k-l' f ^ V l^ ,f (v 2^ ,v 2' v 3 ' ' ' ' ,v k-3 ' f" 1 (v k _ 3 ),f" 1 (v k _ 2 ),v (] ). The length of a' is equal to the length of a plus two. We claim that a' is irreducible. To prove this, we will check each vertex in a' and show that it is adjacent to no vertices in a' other than those immediately preceding and succeeding it in the definition of a ' . Clearly v is not adjacent to either f(v, ), f (v 2 ) /V 2 / Â• Â• . /V, ~. Also ^ v n' f ( v b-_i^ ^ s not an ed 9 f for if it were, (v,, ,v. _} would also be an edge, contradicting the irreducibility of a. PAGE 32 27 Clearly, v ,_. is not adjacent to either f (V2) (since that would imply {vÂ„,v 2 } is an edge), v 2 , . . . , or v -, . Referring to Figure III-2, it is clear that v k-1 is not adjacent to either f ^-3) or f ^ v k-2^ ' as these vertices are in V, . Also, f (v, ) is not adjacent to either v~,v 3 ,..., or v, (otherwise, there would be a cycle in V~ 1 forcing {v -,v, ,} to be an edge, contradicting the irreducibility K _L K -> of a). Since f (v,_-J and f (v,_ ? ) are in V\ , f(v,) is not adjacent to either of these. As for f (v ) above, f(v,) is not adjacent to either v 3 , . . . , v R _ 3 , f ^-3)' or f ^ v k-2^ ' By definition of a, vÂ„ is not adjacent to either v.,..., or v, _. Also, (v 2 ,f ( v k-3^ is not an edge, for if it were, then 6 = (v ,v 1 ,v 2 ,f1 (v k _ 3 ),f1 (v k _ 2 ),f1 (v k _ 1 ) = v k = v Q ) is a cycle in V.. . Since V 1 is the set of vertices of a Husimi tree, 8 is a complete cycle, forcing t v Q' f (v ) } to be an edge. However, we have shown earlier that K. j this cannot be an edge. PAGE 33 23 k-1 k-2 FIGURE III-2 PAGE 34 29 Similarly, {v-,f ( v i,_o) ^ i s not an edge, for if it were, then B' = (v ,v r v 2 ,f 1 (v k . 2 ),f" 1 (v k _ 1 ) = v Q ) would be a cycle in V . Then, completeness of cycles in T, = (V,,E:.) forces { v n / v 2 ^ G E i Â» contradicting the irreducibility of a. By definition of a, it is clear that none of v~,...,v, _ is adjacent to another. Also, none of v -,,... v, -. is adjacent to f (v. _) , since f (v, -,) 3 k-3 k3 k-3 is only adjacent to v, __ and f ( v k _ 2 ^ Â• B ^ the same reasoning, none of v_.,...,v, , is adjacent to f ^ v v-2^ ' Therefore, a' is irreducible, contradicting our definition of a as being the largest irreducible cycle. Then it must be true that all irreducible cycles in T, f Tp have length four. In trying to extend the theorem to a larger class of graphs, it was noted that much of the proof above depends on the existence of diagonals, that is, PAGE 35 30 edges {v.,v.} in a cycle (v Q , v 1 , . . . , v r = v Q ) such that either | ij | f 1 or n-1. There is a characterization of Ptolemaic graphs in terms of number of diagonals, proven by Howorka [ 8 ] : Theorem 3.4 : A graph G is Ptolemaic if and only if every n-cycle of G where n = 2k + r for r = 0,1, has at least 3k 42r 5 diagonals. It was conjectured that "trees" in Theorem 3.2 could be replaced by "Ptolemaic." However, Figure III3 shows a counter example. Both G with V, = (x 1 ,x 2 ,x 3 ,x 4 } and G 2 with V 2 = {y 1 ,Y 2 >Y 3 'Y4 } are Ptolemaic since their four cycles have 1 diagonal. However, a = (x 1 ,x 4 ,x 3 ,y 3 , yÂ„,y ,x ) is an irreducible cycle of length six, and the map f(x.) = y., i = 1,2,3,4, is an isomorphism. The following theorem gives the additional requirement in order for Theorem 3 . 2 to hold for Ptolemaic graphs : Theorem 3.5 : If G Â± = (V-^E^ and G 2 = ( V 2 ' E 2 ) are Ptolemaic graphs such that every cycle of length four induces a complete subgraph, and if f: G^ + G 2 is a PAGE 36 31 -4 FIGURE III-3 PAGE 37 32 bijection, then f is an isomorphism if and only if all irreducible cycles in GL f G 2 with edges in both G, and G~ have length four. Proof: The sufficiency of the condition is proved as in the proof of Theorem 3.2. To prove the necessity of the condition, we construct a cycle a' exactly as in the proof of Theorem 3.2. By appeals to (i) the irreducibility of a, and (ii) the one-to-oneness of f, we can prove that neither v v , f(v ), nor f(v ) is edged with another vertex ' k1 1 z in a' as to contradict the claim that a' is irreducible. Similarly, v 2 is not edged with either v 3 ,v 4 ,..., or V k-3' If {v 2 ,f _1 (v k _ 3 ) 1 were an edge, then (v Q ,v ;L , v 2 ,f _1 (v k _ 3 ) ,f _1 (v k _ 2 ) ,v Q ) is a five-cycle in G y Since G is Ptolemaic, there must be at least three diagonals. Possible diagonals are (i) {v Q ,f ^ v k-3^' (ll) ^ v l' f _1 (v k _ 3 )}, (iii) {v 1 ,f" 1 (v k _ 2 )), or (iv) {v 2 ,f~ (v R _ 2 )}. However, case (i) cannot be a diagonal, or else it would force {v, ,,v, , } to be an edge. Therefore, the other k-1 k-3 PAGE 38 33 three cases must be the diagonals. This gives us a cycle (v ,v..,v_,f ( v i;._;>) / v n ) Â°f length four, which by our hypothesis, must be complete. This is a contradiction because it would force {v n ,vÂ„} to be an edge. Also, {v~,f (v, _J } is not an edge, for if it were, we would have a cycle (v ,v ,v_,f ( v i,_o)' f ( v i,_-| ) ) Â°f. length four, again with completeness forcing {v ,v } to be an edge. Similarly, none of v,,...,v . _ is adjacent to any vertex of a ' such that there would be a contradiction to the irreducibility of a ' . Thus, a' is irreducible, contradicting a being the largest such cycle. In view of this, the theorem is proved. It turns out that Theorem 3.3 and 3.5 are equivalent because of the following theorem: Theorem 3.6 : G is a Husimi tree if and only if G is Ptolemaic and every cycle of length four induces a complete subgraph. Proof : If G is a Husimi tree, then every cycle is complete and satisfies the diagonal requirement for being Ptolemaic. PAGE 39 34 Conversely, suppose G is Ptolemaic and every cycle of length four induces a complete subgraph. Let P(n) be the proposition that every cycle of length n induces a complete subgraph. Clearly, P(4) is true. Now suppose that P (m) is true for m = 5,...,n-l. Consider a cycle a = ( v i' v 2 ' ' Â• * Â» v n' V l* Â° f length n " There is at least one diagonal, and we may assume without loss of generality that this is the edge {v ,v, }. Divide the vertex set of a into two subsets: _L K A = {v 1 ,v 2 ,...,V k } and B = {v R , . . . , v r , v Â± } By the inductive hypothesis, each of these sets is complete. We claim that there is an edge between any v. Â£ A and v. e B. For consider the cycle (v i , v 1 , v.. , v k , v i ) of length four. By completeness, we see that {v^v. } is an edge. Thus P (n) is true, and therefore, P (m) is true for all m greater than or equal to 5. The following theorem proves that the largest class of graphs for which Theorem 3.2 holds is the class of Husimi trees: PAGE 40 35 Theorem 3.7 : Suppose G, = (V, ,E ) and G 2 =(V 2 ,E ) are connected graphs and f: V, -> VÂ„ is a bijection. Further, suppose that f is an isomorphism if and only if all irreducible cycles in G f G with edges in both G.. and G~ have length four. Then G and G 2 must be Husimi trees. Proof : We will show that if G, and G 2 are not Husimi trees, then f being an isomorphism is not equivalent to the requirement on the length of irreducible cycles. Suppose G, is not a Husimi tree and that n = (vÂ„.v, ,...,v ) is a cycle in G, with at least one u 1 n 1 missing diagonal. Without loss of generality, assume that v~ is one of the vertices of a missing diagonal in a . Let j be the smallest index such that {v_,v.} is not an edge. A shortest path from v to v. is (v_ , v . _, , v . ) , since clearly (v ,v. . \ c E-. or else a contradiction of J l j-l J 1 the choice of j would exist. Now, consider a shortest path q from f(v.) to f(v ) of increasing index. The length of 3 is at least two since {f(v.),f(v )} Â£ E 2 No diagonals exist, or else the shortness of the path would be contradicted. Consider the cycle a' = (v ,v._ 1 ,v.,f(v.),...f(v n ),v n ) PAGE 41 36 where the portion of a 1 between f(v.) and f (v ) is D n exactly as in g. Because f is an isomorphism, v._, is edged with no vertex in 3, v_ is edged only with f (v ) = f (v ) in 3, and v. is edged only with f(v.) in 3. Thus a 1 is an irreducible cycle, and its length is > 6. So far, we have only proven variations of Theorem 3.2 for finite graphs. The theorem may be expanded to the case of infinite trees: Theorem 3.3 : If T = (V^E^ and T 2 = (V 2 ,E 2 ) are infinite trees and if f: V, + V. is a bijection, then f is an isomorphism if and only if all irreducible cycles in T, f TÂ« have length four. Proof: The sufficiency of the condition is proven as in the proof of Theorem 3.2. The necessity of the condition is proven differently than in that proof because its use of a longest irreducible cycle is inapplicable in the case of infinite trees. To prove necessity, suppose f is an isomorphism. Let a = (v,,v , ...,v ) be an irreducible cycle in T f T~ JL Z n i. *Â• of length greater than 4. Some of the vertices are in V., and some are in V" 2 . Let v, = v n e V^ and f (v^ = v n _]_- PAGE 42 37 Write a as (v, , . . . ,v ,v ,.,...,v ,...,v ,,,...,v ,v. ) 1 n x n 1+ l' n 2 ' V^+l' n R 1 e v 1 e V 2 e V 2 where n, ^ 2 and f (v, ) = v . Consider the closed walk k a ' in V 2 , where a' = (f (v ) , . . . ,f (v ) ,v ,...,v ,f(v .,)Â».. ./V_ ) 1 n 1 n 1+ 2 n 2 n 2 + 2 n R where f (v ) = v , , for i odd, v = f (v . , ) for n . n . +1 n . n . +1 11 11 i even, and f (v n ) = v 1 n. k Claim : The length of a' is greater than two. Note that v^ is not in Vbecause if it were, then f (v 2 ) = v, . However, since f is an isomorphism and {v, ,v^} e E, , it would imply that {f (v, ) , f (v 2 ) } = {v _wV 3 ) e E 2 , contradicting the irreducibility of a. Thus v, ,vÂ„,v-. are all in V , and a' = (f (v 1 ) ,f (v 2 ) ,f (v 3 ) ,. ..,f (v 1 )) has length greater than 2 because f (v,) and f (v. ) do not form an edge (by the one-to-oneness of f ) . Claim: The walk a' is a trail. For if not, there is an edge e = {x,y} that is repeated in a'. If e is an PAGE 43 38 edge in a and there do not exist v. and v . in a such that f(v.) = x and f (v . ) = y, then e is repeated in a, contradicting the fact that a is a cycle. So there exist v. and v. such that f (vj = x and f(v.) = y. But this would imply that (v^v. } is an edge in a, contradicting the irreducibility of a since x and v. are in a and {x,f (x) = v^^ } is an edge in T f T~. Thus, e = {x,y} is an edge in a' , not a. So there exist v. and v. in V such that f (v^ = x and f(v.) = y. If e is repeated, then either (i) {v., v.} is repeated in a (but this would contradict a i 1 being a cycle) , or (ii) two edges in T^ being mapped to e (contradicting f being one-to-one) . Therefore, no edge in a 1 is repeated, and thus a' is a trail. Moreover, a' is a circuit. This means that a' contains a cycle, contradicting the fact that T., is a tree. This completes the proof. One can define an equivalence relation R on vertices of bivariegated trees: PAGE 44 39 Definition 3.9 : Let x and y be vertices in T, f T~ , and define x = y modulo R if and only if x = f(y), y = f (x) , or x = y . This equivalence relation identifies each vertex in T, with its image in TÂ„ such that one can consider T, collapsing onto T 2 to form a new graph T, f T 2 /R. If we let E be the set of edges in T f T 2 /R we can define adjacency as follows: Definition 3.10 : Let [x] and [y] be the equivalence classes of the vertices x and y, respectively, of T, f T 2 under R. Then { [x] , [y] } e E if and only if there exists a p e [x] and a q e [y] such that (p,q) z E-,i>E 2 where E, and E 2 are the edge sets of T^ and T 2 , respectively. Using this equivalence relation, we can prove another version of Theorem 3.2: Theorem 3.11 : If T, = (V 1 ,E 1 ) and T 2 = ( V 2 ' E 2 ) are trees and if f: V, + V, is a bijection, then f is an isomorphism if and only if ^ f T 2 //R is a tree - PAGE 45 40 Proof: Suppose f is an isomorphism. We want to show that T is isomorphic to 1^ f T 2 /R. For every v. Â£ V n , define g(v.) = [v.]. This is clearly a onex 1 i i to-one and onto mapping. If {x,y} e E, , then {g(x),g(y)} = { [x] , [y] } Â£ E. Also, if (g(x),g(y)} Â£ E, there is a p e [x] and a q Â£ [y] such that {p,q} Â£ E,UE 2 . If (p,ql Â£ E 1 , then p = x and q = y, so {x,y} Â£ E, . If (p,q> Â£ E 2 , then since f is an isomorphism, {f _1 (p) },{f _1 (q) } = {x,y} e E Â± . Therefore, g is an isomorphism, and T, f T 2 /R is a tree since it is isomorphic to one. Now suppose f is not an isomorphism. Assume that {x,y} e E 1 , but {f(x),f(y)l i E 2 . In T Â± f T 2 /R this corresponds to { [x] , [y] } Â£ E and the existence of a path a from [f(x)] = [x] to [f (y) ] = [y] of length greater than one, where a = ([f (x)] ,t 1 ,...,t n , [f (y)] , [f (x)]) . No t. = [x] or [y] , so a is a cycle, and, therefore, T, f T 2 /R is not a tree. PAGE 46 CHAPTER IV AUTOMORPHISM GROUPS In this chapter we examine the relationship between the automorphism groups of G = G. f G 2 and its factors and, in conclusion, quote a result obtained by considering the case in which G^ = G^. Automorphism Groups of a Graph and Its Factors Definition 4.1 : An automorphism of a graph G = (V,E) is a permutation h: V -*Â• V such that if {u,v} e E, then {h(u),h(v)} e E. The set of all automorphisms of G forms a group, which will be denoted Aut (G) . It is an interesting problem to look at the relationship between the automorphism groups of the bivariegated graph G = G, f G 2 and its factors, G = (V ,E ) and G 2 = (V 2 ,E 2 ). For, suppose h is an automorphism of one of the factors of G, say G^. Then, unless h is the identity map, h maps at least one vertex, say u, to another. Because G is bivariegated, u is edged with some vertex v in G 2 . Now, in order 41 PAGE 47 42 for h to correspond to an automorphism on the entire araph G, h must similarly move v. This motivates the followinq theorem: Theorem 4.2 : Let g be any automorphism of one of the factors of G = G, f G 2 (without loss of generality, the factor G ) . If g is an automorphism on G that agrees with g on G (that is, g restricted to G, is identically g) , then it must be true that for any vertex v e V, g(v) = f (g(f _1 (v))) . Proof: If g is an automorphism of G, then it preserves edges in G. Then, because G = (V,E) is bivariegated , for any v e V 2 , there exists a u e V such that f(u) = v and {u,v} e E. So it must also be true that {g(u), g(v)} e E. However, since g restricted to G, is identically g, g(u) = g(u), and g (v) e V 2 (since all vertices in V, are already images of some vertex under the map g) . But g (u) e V, is edged with only one element of V" 2 , namely f (g (u) ) , because of the bivariegation . Thus, i(v) = f(g(u)) = f(g(u)) = f (g(f _1 (v))). PAGE 48 43 This theorem suggests that if we are trying to relate automorphism groups of the factors of G = G, f G 2 to Aut(G), then we can extend the automorphisms of the factors of G to ones on G itself and examine whether or not these extensions are automorphisms of G. Furthermore, as suggested by Theorem 4.2, the extension of an automorphism g on G (or G 2 ) to a possible automorphism g of G is uniquely determined by the bivariegation . To obtain an understanding of this problem, it is helpful to look at a few examples. In all the examples, let V = W^v^v-j} and V 2 = {v 4 ,v 5 ,v 6 h For a mapping g, the notation (i, i 2 i 3 ... i n ) means that g(v. ) = v. , g(v. ) = v. ,..., and g (v ) = v. . x l x 2 X 2 X 3 X n X l If a particular subscript, say i., is omitted, then assume g(v. ) = v. . Further, the notation id will refer to the identity map. Example 4.3 ; Consider the bivariegated graph in Figure IV1. PAGE 49 44 FIGURE IV1 Aut(G,) = {id,(l 3)}. Clearly, the identity on G, induces the identity map on G. Now let g be the automorphism (1 3). Theorem 4.2 forces g to be the map (1 3) (4 6). It should be clear that Aut(G) consists only of g as defined above and the identity. Thus, Aut(G,) induces all the automorphisms in Aut(G). Similarly, Aut(G2) induces Aut (G) . Example 4.4: In the graph below, FIGURE IV2 PAGE 50 45 Aut(G ) = (id, (1 3)}. If g is the automorphism (1 3) on G , it forces the automorphism g on G to be (1 3) (4 6). This, together with the identity, is Aut(G). Thus, Aut(G) is completely determined by Aut(G,). However, Aut(G 2 ) = {id, (5 6) ,(4 5), (4 5 6), (4 6 5), (4 6) } = S 3 , the symmetric group on three elements. Only one of these, (4 6) , induces an automorphism on G. Example 4.5: Consider the graph in Figure IV-3. FIGURE IV-3 Aut(G ) = {id, (1 3)}. As in the previous examples, these induce the identity and the automorphism (1 3) (4 6) on G. However, Aut (G) = {id,(l 3) (4 6),(1 4) (2 5) (3 6) , (1 6) (2 5) (3 4)}. Thus, although every automorphism on G,, and similarly G 2 , induces one on PAGE 51 46 G, Aut(G,) does not completely determine Aut (G) Example 4.6: Consider the graph in Figure IV4 FIGURE IV4 Aut(G-) = (id, (4 5)}. These induce the identity and the automorphism (4 5) (1 2) on G. However, Aut(G) = {id, (4 5) (1 2) , (3 6) , (1 2) (4 5) (3 6) }, so Aut(G 2 ) does not completely determine Aut (G) . Now consider Aut(G,) = S3. Only two of these six automorphisms, the identity along with (1 2), induce maps on G which are automorphisms. PAGE 52 47 Example 4.7: Consider the graph in Figure IV5: FIGURE IV5 Aut(G, ) = {id, (2 3)}. This forces the identity and the map (2 3) (5 6) on G. This is not an automorphism, however, since {v 4 ,v 5 > e E, but {g (v 4 ) , g (v 5 ) } = {v,,v c } f, E. Thus, only the identity on G. induces an automorphism on G. The examples above point out the followincr: in some cases, the automorphism group of a factor of G completely determines Aut(G); in some cases, the automorphism qroup of a factor determines some (but not all) of the automorphisms of Aut(G). In other instances, onlv the identity induces an automorphism on G. PAGE 53 48 The following theorem generalizes when one of the above cases occurs. Theorem 4.8 : Let G = G f GÂ„ be a bivariegated graph with 2n vertices. If AutfG,) (or equivalently Aut(G,)) is S , then every automorphism of G. induces 1 n 1 a unique automorphism of G. Proof: It is clear that the automorphisms must be unique, for any map so induced according to Theorem 4.2 must be uniquely defined by the structure imposed on G by the bivariegation. So, let g e Aut(G,). Extend g to g on G by defining for v e V 2 , g (v) = f(g(f (v) ) ) . Claim: The map g preserves all edges of G. Clearly it preserves edges in G, since g restricted to G is defined to be g. Now, consider any u e V, and v e V 2 such that f(u) = v. Then since {u,v} e E, {g(u),g(v)} must be an edge. But g(v) = f(g(f~ (v) ) ) = f(g(u)), and by definition of bivariegation (g(u),g(v)} = {g (u) , f (g (u) ) } = {g (u) , f (g (u) ) } e E. Thus, g preserves edges joining G, and G ? . Finally, g preserves edges in G 2 For, consider {s,t} e V 2 Since Aut(G 2 ) = S r , any permutation of elements in G 2 preserves edges in G 2 In particular, g does. Thus, {g(s),g(t)} e V" 2 . PAGE 54 49 It should be clear at this point that if Aut(G-) = S and Aut(G,) f S , then every automorphism of G does n In ^ not induce an automorphism of G. Example 4.4 exhibits this. It should also be clear that Aut (G ) (or equivalently Aut(G 2 )) is S if and only if G. is either complete or totally disconnected. If one is to attempt to answer the question of when an automorphism of G, extended to G is not an automorphism of G, one can notice the following: whenever there is an automorphism g of G, such that for S,t e V-, (f(s),f(t)} Â£ E but {f (g(s) ) ,f (g(t) ) } t E, then g is not an automorphism of G. For example, in the graph in Example 4.4, let g be the automorphism of G 9 corresponding to (4 5 6). Then for v^ and v , {f _1 (v 5 ) ,f _1 (v 6 ) } = {v 2 ,v 3 ) e E but {f _1 (g(v 5 )) ,f _1 (g(v 6 ))} = {f" 1 (v 6 ),f"" 1 (v 4 ) } = {v 3 ,v 1 ) t E. Figure IV6 provides a table of all bivariegated graphs with six or less points and their automorphism groups. The automorphism group of G, or G 2 is of type I if only the identity induces an automorphism on G, type II if every automorphism induces an automorphism on G, and type III if the automorphisms induce all of the PAGE 55 50 automorphisms on G. Further, the automorphism group of G, or G~ is of type IV if it contains at least one non-identity element that induces an automorphism on G and at least one element which does not induce an automorphism on G. For each graph, the top half represents G, and the bottom half G 2 Â« PAGE 56 51 S , II Aut(G) = S. S , II Ts 2 , ii Aut (G) = D Â•s 2 , II s 2 , III Aut (G) = S 2 [E 2 ] 4 s 2 , III 1S V II Aut(G) = D *s 2 , II S 3' J1 Aut(G) = group of c TT order 48 b 3' S 2 +~E lf I Aut (G) = S 2 [E 4 1 S 2 + E 1# I S 2 + E,, II ! Aut(G) = group of S 2 + E , II order 16 s 3 , iv Aut(G) = K 4 S 2 + E , II S , IV Aut(G) = S 2 [E 4 ] 32+E-j^ , III S 2 + E , III Aut(G) = S 2 [E 4 ] S 2 + E , III Aut(G) = S 2 [E 4 ] s 3 , III Aut(G) = group of s 3 , in order 6 S 2 + E,, I Aut(G) = K 4 S 2 + E , I S 2 + E , III Aut(G) = S 2 [E ] s 3 , IV FIGURE V-6 PAGE 57 52 Pe rmutation Graphs It appears that in order to obtain a feel for the automorphism groups of a bivariegated graph and its factors, one needs to examine this problem by looking at a smaller class of bivariegated graphs. We will conclude this chapter by looking at some results obtained by doing this. Rather than examine the automorphism groups of the general class G 1 f G_ of bivariegated graphs M. Borowiecki [ 3 ] considers the case where G, = GÂ„ = G. First, he makes the following definitions: Definition 4.9 : Let G be a graph whose vertices are labeled v 1 ,...,v n and let a be a permutation on the set U,...,nJ. Then by the g-permutation graph P.^ (G) of G is meant the graph consisting of two disjoint, identically labeled copies of G, say G and G* , together with n additional edqes e. , 1 < i < n, where e. ioins the vertex labeled v. in G with v , ., in G'. For i a ( i ) clarity, we label a vertex G' as v . ' . Denote G = (V,E) and G' = (V',E'). Let a* be the permutation of V such a*(v i ) = v if and only if a(i) = j. We define the PAGE 58 53 mapping a: V -* V by a(v.) = v. 1 if and only if a(i) = j . Thus, P (G) corresponds to G f G, with a corresponding to f. Now, Borowiecki states the following problem of Frechen [5 ]: Consider the class H(P,G) of all permutations a* which preserve a property P of a graph G under transformation from G to Pa(G). Determine the properties P and graphs G for which (a) H(P,G) is a subgroup of S n , n = |V| (b) H(P,G) = Aut(G) . Borowiecki considers the above problem for the chromatic number x (G) of a graph G and proves the following: Theorem 4.10 : If X (G) > 2, then H (x,G) is a group. Corollary 4.11 : If G is not totally disconnected, then Aut (G) <= H(x,G) <= S . PAGE 59 CHAPTER V GRAPHS OF SEMIGROUPS AND SEMIGROUPS OF GRAPHS A metric characterization of the graph of a semigroup prompted an examination of the relationship between bivariegated graphs and graphs of semigroups. Then, a theorem relating semigroups to endomorphisms of graphs motivated a search for a connection between this theorem and bivariegation . Graphs of Semigroups In the following, let S represent a semigroup, and s* the set of proper subsemigroups of S. Definition 5.1 : The graph E (S) of a semigroup S is the graph whose vertices are the proper subsemigroups of S with two of these, say, A and B, edged if and only if AnB ^ PAGE 60 55 * a b a a a b b b Then S* = {{a},{b}}, and E (S) is a graph consisting of two disconnected vertices. Example 5.3: Suppose S is the semigroup given by * a b c a a a a b b b b c c c c Then S* = { { a} , { b} , { c} , { a ,b} , { a , c} , { b, c} } , and Â£ (S) is the graph in Figure V-l. {a,b} {a} {b,c} FIGURE V-l PAGE 61 56 Bosak [ 4 ] proved some results concerning graphs of semigroups, and posed two problems: (1) Does there exist a semigroup with more than two elements whose graph is disconnected? (2) Find a necessary and sufficient condition for a graph to be the graph of a semigroup. The first problem was answered in the negative in a theorem proven by Lin [10] : Theorem 5.4 : Every semigroup with more than two elements has a connected graph. Pondelicek [12] also solved Bosak 1 s first problem, and he further obtained a metric characterization of the graph of a semigroup: T heorem 5.5 : The diameter of the graph of a semigroup does not exceed three. Because of this metric characterization, and the work in this paper with graphs that have distinct metric characterizations, it was thought that it might be interesting to look at the relationships (if any) between the graphs of semigroups and bivariegated graphs. PAGE 62 57 In particular, it was hoped to partially solve the second Bosak problem by answering some of the following questions: Which finite semigroups have bivariegated graphs? Which connected bivariegated graphs are the graphs of semigroups? First, the graphs of all semigroups of order four or less were examined, using the scheme given by Petrich [ll]. For any class C of semigroups, a possibly larger class may be constructed by performing one or both of the following types of operations on the members of C: (1) adjunction of an identity or zero, (2) inflation, and (3) forming direct products. Inflation may be defined as follows: Definition 5.6 : To every element s of a semigroup S, associate a set Z such that: the sets of Z are pairs s r wise disjoint and Z ns = {s}. The set V = U _Z , J s seS s together with the multiplication x * y = ab if x e Z and y e Z, , is an inflation of S. The remaining semigroups that cannot be obtained by one of the above operations are listed by Petrich according to either their multiplication tables PAGE 63 58 or the configuration of their greatest semilattice decompositions and number of elements in each class. Definition 5.7 : A commutative semigroup in which all elements are idempotent is called a semilattice. Clearly, the trivial semigroup of order one has no graph since it has no proper subsemigroup. The semigroups of order two have as their graphs either one or two disconnected points. From the graphs of semigroups of order three, only two are bivariegated, First, consider G 2 , the cyclic group of order two, with identity a and multiplication table defined by a*a=a=b*b and a* b = b * a = b. Inflate about the non-identity element b, obtaining the additional element c, with multiplication table * a b c a a b b b b a a c b a a Clearly, {a} and {a,b} are proper subsemigroups , producing a graph consisting of two edged vertices which is a bivariegated graph. PAGE 64 59 In addition, the semigroup defined by the multiplication table * PAGE 65 60 where n has three distinct divisors, is bivariegated . Three distinct divisors are necessary since one divisor of n will be the number one, corresponding to the trivial subgroup, and one divisor of n will be n itself, corresponding to the improper subgroup G . That leaves one nontrivial proper subgroup to correspond to the second vertex of the bivariegated graph. Now, the integer n has three distinct divisors only when it is the square of a prime number. Thus, the following theorem has been proven: Theorem 5.8 : If n is the square of a prime number, then the graph of G is bivariegated. It was stated above that the graphs of inflations of certain cyclic groups are bivariegated. By definition of inflation, multiplication by the element obtained by inflation is the same as for the element from which the inflation is generated. Thus, in the multiplication table, the rows and columns corresponding to the new element and the inflating element are identical. If n is prime, then G has no nontrivial proper subgroup. If the inflating element is not the identity of G , then no additional subgroups are PAGE 66 61 obtained by adding the new element since its multiplication table is identical to that of the inflating element. (This is not true if the inflating element is the identity of G since the new subgroup consisting of the identity and the new element is formed.) Clearly, G itself is a proper subsemigroup of the semigroup obtained by inflation about a non-identity of G . 2 J n If n is prime, then this semigroup contains only one other proper subsemigroup, the trivial one. Thus, the following theorem has been proven: Theorem 5.9 : If n is prime, then the graph of the semigroup obtained by inflating G about a non-identity element is bivariegated. After examining a number of graphs of semigroups, the following was conjectured and proven: Theorem 5.10 : Suppose G is the graph of some semigroup S ( i.e . , E(S) = G) . If G is bivariegated, then G has exactly two vertices. Corollary 5.11 : If E (S) is bivariegated, then S has exactly two proper subsemigroups . PAGE 67 6 2 Proof : Let S, be a vertex of G = Â£ (S) . Since G = G, f G 2 is bivariegated , there is another vertex S~ e GÂ„ with which S, e G, is edged. By definition of I(S), S, and S 2 are edged if and only if they have at least one element in common; call it x, . In order for S~ and S, to be distinct, one of these must contain another element, say x ? . Without loss of generality, let x 2 e S 2 and x ? Â£ S, . If G has exactly two vertices, there is nothing to show. So, assume that G has more than two vertices. Let S^ be a third vertex which is connected to either S.. or S 2 . Without loss of generality, assume S~ and S., are edged. It is clear that S-, e G 2 . Assume x 2 e S^ n S-,. Note that x-, Â£ S-.. Now, if G is bivariegated, there exists another vertex S 4 such that f(S.) = S 3 . If S. and S^ are edged, they have an element in common. This element cannot be X.. , or else S. . e G, would be edged with both S n and 14 1 * 2 S_. Similarly, the element in common cannot be x 2 . So, both S . and S contain some other element, say x-, . 43 J 3 Notice that S ? n S., is a subsemigroup (as it is the intersection of two subsemigroups) containing x~ and possibly other elements. Thus we have a fifth subsemigroup, and thus a fifth vertex. Call S^ = S^ n S-.. PAGE 68 63 !ince x_ e Sj-, it must be in G-, , or else SÂ„ would be edged with two vertices in G, . Then there must be an S g in G, such that f(Sg) = Sr-. S fi cannot contain x. , or else SÂ„ e G~ would be edged with two vertices in G, . Similarly, x~ / S fi , or else S, would be edged with both S 2 and S,in G~ . Also, x-, Â£ S,, or else both S, and S fi in G. would be edged with S-, in G-, . So, S_ must contain an additional element, z 6 say x,,, that is also an element of S c . But S c = 4 DO S~ n S,. Thus, x. must also be in both SÂ„ and S-., which forces S, f_ G n to be connected to S~, S-,, and 6 1 z J Sri all of which are in GÂ„ . This contradicts the fact that G is bivariegated . Thus, G has no more than two vertices. This theorem partially answers Bosak's second problem, for it says that there are quite large classes of graphs that are not graphs of semigroups. That is, out of the entire class of bivariegated graphs, there is only one, the two-point bivariegated graph, that is the graph of a semigroup. PAGE 69 64 Semigroups of Graphs Now we will examine another problem relating to graphs and semigroups. In an article written by Z. Hedrlin and A. Pultr [7 ] there is a discussion of relations with given finitely generated semigroups. Definition 5.12 : Let X be a set and R a binary relation of X, R <= x x X. We write xRy if (x,y) e R, x,y e X. Further, R = (R,X) is used to explicitly show the set X. In graph theoretic terms, every relation (R,X) may be associated with a graph G, where X is the set of vertices of G and R is the set of all edges of G. A transformation f of X is called compatible with a relation (R,X) if xRy implies f(x)Rf(y) for all x,y z X. The set of all compatible transformations with a relation (R,X) is denoted by C(R,X). In graph theoretic terms, a transformation of X is called an endomorphism of G if it sends each edge into an edge. Evidently, f is an endomorphism of G if and only if f is compatible with (R,X). Further, the set C(R,X), the set of all endomorphisms of the graph associated PAGE 70 65 with R, is a semigroup under composition of transformations . Hedrlin and Pultr prove the following theorems : Theorem 5.13 : Let S be a finitely generated semigroup with unity element. Then there exists a relation R_ on a set X T such that C(R T ,X T ) under composition of transformations is isomorphic with S. Moreover, there exist infinitely many non-isomorphic relations R with this property. The above theorem can be translated into the language of graph theory as follows: Theorem 5.14 : For each finitely generated semigroup S with unity element, there exist infinitely many non-isomorphic graphs G such that S is isomorphic with the semigroup of all endomorphisms of G. Earlier in this chapter the relationship between semigroups and bivariegated graphs was examined. Thus, it became an interesting problem to look at the following question: Of the infinitely PAGE 71 66 many graphs that arise in connection with Theorem 5.14, how many of these are bivariegated? Further, what are the necessary and sufficient conditions on S to insure that G is bivariegated? In order to attempt to answer these questions, the construction of (R_,X_) in the proof of Theorem 5.13 must be examined. Hedrlin and Pultr [7] define {(R.,X), i = l,2,...,n} to be a system of relations on X, n > 2, and T to be a sequence from the following lemma. Lemma 5.15 : Let n be a natural number. Then there exist infinitely many sequences T = {t,,tÂ„,...,t } of natural numbers such that t. ? t. for i ^ j t. + t. > t, for all i,j,k e {1,2,. ..,n}. 1 J K The proof, stated in [7] , is based on the definition t. = i + h(n + 1), where h is an arbitrary natural number. It is clear from the definition of t. l that if n > 2, then t. > 2 for i = l,2,...,n. Continuing the description of the construction in [7], denote by Y the set of all quadruples (x,y,i,j) such that x R. y, 1 < j < t. , where t. is the i-th member of PAGE 72 67 the sequence T. Let U = {u ^K V = { v ' v ]_' v 2^ be sets. Define XÂ„ = XuYuUuV, the union of mutually disjoint sets. Let two elements in X be in the relation R T in the following cases: x R T (x,y,i,l) , (x,y,i,j) R T (x,y,i,j + 1) for j = 1,2,. . . / t i 1, (x^i,^) R T y, u q R t x,v o R t x for all x e X, u oVl' u lVo' v o R T v 1 ,v 1 R T v 2 ,v 2 R T v o . The remainder of this chapter is a record of what was attempted in order to find some relationship between Theorem 5.14 and bivariegated graphs. First, we wanted to examine the following question: Given a finitely generated semigroup S with identity, what additional properties of S are necessary in order that there exist a bivariegated graph G such that S is isomorphic to the semigroup of endormorphisms of G? The first attempt at obtaining a hint to the nature of these properties was to choose examples of PAGE 73 68 bivariegated graphs, form their semigroups of endomorphisms , and check the additional properties of these semigroups. This attempt proved futile because graphs of very small order normally have a very large number of endomorphisms, causing the corresponding semigroup to be too large to deal with efficiently. Next, the characteristics of the graph corresponding to (RÂ„,X ) were examined to determine at least necessary conditions for it to be bivariegated. In particular, since the minimum requirement for a graph to be bivariegated is that it have even number of vertices, the conditions that cause the set X T to have an even number of elements were sought. The following results were obtained: Lemma 5.16 : Let m be the number of elements in S and let {R., i = l,2,...,n} be a system of relations on S. Then X contains m + mn (n + 1) (2h + l)/2 + 5 elements . Proof: Recall X = XuYuUUV, the union of mutually disjoint sets. The set X corresponds to S, having m elements, PAGE 74 69 To determine the number of elements in Y = {(x,y,i,j): xR.y, 1 < j < t.}. note that for each R. there are x x 1 mt . members of Y because j takes on all values between l 1 and t., of which there are t. values. Thus, the order l l of Y is n mt, + mt~ + . . . + mt = mSt. . 1 2 n i But by Lemma 5.15, t. = i + h(n +1), so the above equals n n n ml(i + h(n + 1)) = m(Zi + Eh (n + 1) ) i=l i=l i=l = m(n(n + l)/2 + hn (n + 1) ) = mn (n + 1) (h + h) = mn(n + 1) (2h + l)/2. Now, clearly U = (u ,u, } has 2 elements and V = { v ,v, ,v } has 3. Therefore XÂ„ has m + mn (n + 1) o 1 2 l (2h + l)/2 + 2 + 3 elements. As a result of the above lemma, we obtain a necessary condition in order for the construction of (R_,X _) to produce a bivariegated graph: Th eorem 5.17 : A necessary condition for the graph corresponding to (R ,X ) to be bivariegated is that S PAGE 75 70 have odd order and the number n of relations on S be such that either n or n + 1 is divisible by 4. Proof : If m + mn(n + 1) (2h + l)/2 + 5 must be even, then m + mn(n + 1) (2h + l)/2 must be odd. Consider the following cases: Case 1 : The number in is even. Then mn (n + l)(2h + l)/2 must be odd. Note that m/2 is a whole number. No matter what n is, either n or n + 1 is even. If one factor in a product is even, the product is even, so (m/2) (n) (n + 1) (2h + 1) is even. Thus if m is even, (R^X^ does not correspond to a bivariegated graph. Case 2: The number m is odd. Then mn (n + 1) (2h + l)/2 must be even. Suppose n is even. Then n + 1 is odd. Let p = n/2. Clearly, p is a whole number. Now, it is necessary for mp (n + 1) (2h + 1) to be even. For any h, 2h + 1 is odd. So each of m, n + 1, and 2h + 1 are odd. Thus we must have p = n/2 even, implying that n is divisible by 4. Now suppose that n is odd. Then n + 1 is even, and 2h + 1 is odd. Let p = (n + l)/2, a whole number. PAGE 76 71 It is necessary for mnp(2n + 1) to be even. Since each of m, n, and 2h + 1 is odd, p = (n + l)/2 must be even. This forces n + 1 to be divisible by 4. Thus, for the graph corresponding to (R T ,X T ) to be bivariegated, m must be odd and either n or n + 1 must be divisible by 4. Although the restriction that a graph have an even number of vertices is far from being a sufficient condition for bivariegation, it nevertheless imposes quite a restriction on the semigroup S. Thus, it became interesting to examine the restriction on S caused by the limitation on the number of lines of a bivariegated graph. Recall that in the construction of (R T ,X T ) ibove, the first relations defined were of the form xR T (x,y,i,l) . This produces mn lines because there are m elements in X and n values of i (each corresponding to (R^, i = 1, 2,...,n>). From the relations of the form (x,y,i,j)R (x,y,i,j + 1) for j 1,2,. ..,t i 1, PAGE 77 72 we obtain m(t. 1) relations for each i. This produces n n Em(t. 1) = mlt. ran i=l 1 i=l X which, from the proof of Lemma 5.16, is = mn(n + 1) (2h + l)/2 mn . From the relations of the form (x,y,i,t i )R T y there are clearly mn different relations. Further, from u R m x,v R m x for all x e X o T o T there are 2m relations. Finally, from u o R T u 1 ,u 1 R T u o , V oW V l R T V 2' V 2 R T V o there are 4 relations. Thus, there is a total of mn + mn(n + 1) (2h + l)/2 mn + mn + 2m + 4 or mn + mn(n + 1) (2h + l)/2 + 2m + 4 relations . PAGE 78 73 Now the question is whether or not this lies between the minimum number p/2 and the maximum number 2 p /4 of lines that a bivariegated graph may have. For sake of simplicity, let s = mn(n + 1) (2h + l)/2. Then, the first question is whether or not the number s + mn + 2m + 4 of relations corresponding to the construction of (R ,X ) is greater than or equal to half the number s + m + 5 of elements in X ; that is, it should be true that s + mn + 2m + 4 > (s + m + 5)/2. This is clearly true. In addition, the second question is whether or not the number of relations is less than one-fourth of the square of the number s + m + 5 of elements in X ; that is, it should be true that 2 s + mn + 2m + 4 < (s + m + 5) /4. However, 2 2 4s + 4mn + 8m + 16 < s + 2sm + m + 10m + 10s + 25 2 2 4mn < s + 2sm + m + 2m + 6s + 9. PAGE 79 74 At least one of the terms, 2sm, alone is larger than 4mn because 2sm = 2m 2 n(n + 1) (2h + l)/2 = m 2 n(n + 1) (2h + 1) > mn(2) (3) > 4mn. Thus, the bounds on the number of lines in a bivariegated graph places no additional restrictions on the semigroup S. Now, notice that in the definition of R T , V R T V 1' V 1 R T V 2' V 2 R T V 0' and V oV f Â° r a11 X Â£ X ' Suppose (R ,XÂ„) is to be bivariegated by partitioning some of the elements into G^^ and some into G 2 in such a way that (R , X ) can be transformed into G 1 f G 2 Without loss of generality, let v Q e G^ Recall VqR,^. Then suppose v e G 2 . Because v 1 R T v 2 , v 2 must be in G 9 or else v would be related to two different elements in G . However, v^Vq, which forces v 2 to be related to two different elements in G^ Thus, the assumption that v e G 2 was incorrect. Now since v^ e G^, if v e G-, it would be related to two different elements PAGE 80 75 in G, , so v., E G, . However, since v~ is the onlyelement of V related to elements outside V and all of the elements of V are in G, , it must be true that neither v, nor v-, is related to an element in G2 Â• Thus, the construction (R T ,X T ) does not produce a bivariegated graph. Therefore, it has been shown that in Theorem 5.14, none of the infinitely many graphs G such that S is isomorphic to the semigroup of endomorphisms of G is bivariegated. Thus, there must be many graphs with the same semigroup of endomorphisms. Finally, with regard to the relationship between semigroups and bivariegated graphs, we have shown that given a semigroup, its graph, if bivariegated, can only be the connected tree on two vertices. Additionally, we have shown that for a given semigroup S, no bivariegated graph G can be constructed from it such that S is isomorphic to the endomorphism semigroup on G. PAGE 81 CHAPTER VI DISCUSSION Although some of the results of Sanders [13] related to bivariegated trees were extended in this paper to the case of bivariegated Husimi trees, it would be desirable to obtain some further characterizations of bivariegated Husimi trees. Sanders proves the following: Theorem 6.1 : A tree with 2n vertices has a 1-f actor if and only if the largest maximal independent set of vertices of T contains n vertices. Theorem 6.2 : A tree is bivariegated if and only if T has a 1-factor. Thus, a bivariegated tree is characterized in terms of independent sets of its vertices. Therefore, the question arises: What is the relationship between a bivariegated Husimi tree and independent sets of vertices? Another question arises due to the relationship between trees and Husimi trees: Is there a way to 76 PAGE 82 77 shrink the cycles of a bivariegated Husimi tree to obtain a tree, and, if so, when is this tree bivariegated? Or, a similar question is, can we expand the vertices of a tree (possibly bivariegated) in some way to obtain a bivariegated Husimi tree? In one part of this paper we looked at the relationship between Aut (G) , Aut(G ), and Aut(G2), where G = G, f G~ . It would be desirable to have a better characterization of this relationship. One way to proceed in solving this problem is in the manner of Borowiecki [3 ], that is, by restricting G, and G 2 to be equal and the bijection f to be a part of a smaller class of bijections. Then one can attempt to answer some of the questions posed by Frechen [5 ] : What are the properties P of G, (=G 2 ) which are preserved in G, f G 2 ? What are the properties P and graphs G, (=G 2 ) for which the class of all bijections preserving P is a subgroup of the symmetric group of order n or is isomorphic to Aut (G, ) ? Another topic presented in this paper is the relationship between semigroups and bivariegated graphs. It would be interesting to further study the restriction PAGE 83 78 on the diameter of the graph of a semigroup with respect to some of the metric characterizations presented by Howorka [ 8 ] , and establish any connection with some subclass of bivariegated graphs. Finally, the semigroup of endomorphisms of a bivariegated graph was studied. Hedrlin and Pultr [7 ] present a construction which produces a graph from a semigroup. A question which arises is: Can we derive directly from the bivariegated graph (from which we have obtained a semigroup) the graph produced by this construction? Further, what is the relationship between these graphs? PAGE 84 APPENDIX I: BIVARIEGATED GRAPHS P = 2 p = 4 p 6 ' 1 f ^ 1^^ Â• L * Jw p = 8 li | f 1 f ' .. ./ ' ' IV ' ur 79 PAGE 85 3 P = 8 (continued) t Â« VLV * ^^ su M Â» Â« r t \lk ^n? A \lJ/ PAGE 86 p = 8 (continued) V U^ HJ SJ/ i Â— i X^7 p r ^^ ^^ A^\ , \Js/ \Jsf A "^^r ^lv m Ai , , t^k . M V ^W K PAGE 87 82 p = 3 (continued) ^r *\t I iL 4zz r=K , r^ M~K XB^ XQ7 ^p* ^^ . r^ , ns ^ . n^ . pk V ^3^ ^^ ^^ ^s M V M N V M /I N \u/ PAGE 88 p = 8 (continued) . m^ a A /TX \lV ^ A A N A K M^ ^^ ^3^ JTX AA , ^r z] X y| X Z. ^t^ ^^ ^t^ ^W" PAGE 89 84 p = 8 (continued) A N HJ/ ^t A K X2? ^^ V "q ^^ N^ ^^ ^^ ,4 IX ^^ V ^ ,4^K ^^ iTh /m ^ PAGE 90 APPENDIX II: BIVARIEGATED HUSIMI TREES P = 2 p = 4 p = 6 p = 8 Â• Â• Â• Â• i Â« Â— i Â» > i Â— Â» Â« t Â— t Â— Â• Â— * Â• Â— Â• * Uk Â» PAGE 91 8 6 p = 10 Â• 1 Â» Â» Â• Â•Â» > Â• Â• ii i Â• Â« > Â« Â« k k Â• Â•Â» Â» Â» Â• Â• -Â» Â• k Â» * Â» t Â• Â•Â• 1Â• Â•Â• a Â• Â• Â•- PAGE 92 p = 12 i > # i i * Â« Â» Â» i i I Â» t Â« Â» Â» Â» ii Â• Â• Â• Â• m .11 Â» Â» t Â« I I I < f * ioi T Â— T Â— * Â— * Â— 1 Â— Â» n 1 Â« Â» Â— i i Â— Â• Â— * Â— Â» Â« Â— Â« 1 4 Â• . Â• Â• Â« * Â» Â« ' * * 1 * 1 Â• Â• Â» Â• Â• 1 no uAn * Â— x 1 * ' Â• Â— t Â• 1 Â» Â» Â• t Â» t Â• TAO JUT JMTL jcn xrn cm Â• f 4 Â• Â• > i m 4 Â• 4 Â•-Â» Â• Â• * 1 Â• Â• 1 I J lit 1 Â• I i * 1 Â» j no /vm PAGE 93 p = 12 (continued) Â» Â» Â» Â• Â» Â» Â» < Â« m. Â• Â— < Â» f Â— Â» en Â« Â— * Â— Â• Â— Â• Â— Â• Â» Â» Â» Â« Â« Â— # Â— ^ ror nac E?r m et nn e E * Â— Â• Â» "Z2L rcE" PAGE 94 89 p = 12 (continued) 3T im Â• V Â• rrrr rTra k ~* Â— * Â— Â• Â• Â• 1C Â— Â• 1 Â• T T urz ur~3\ ^st ^5L UDL K 31 _. Â• f Â• Â» k Â» Â» i :snr ztm: PAGE 95 90 p = 12 (continued) _xx PAGE 96 REFERENCES 1. A. R. Bednarek, A note on tree isomorphisms, J. Comb. Theory, (B) 16 (1974), 194-196. 2. A. R. Bednarek and E. L. Sanders, A characterization of bivariegated trees, J. Discrete Math., 5 (1973) , 1-14. 3. M. Borowiecki, On the a-permutation graphs, Recent Advances in Graph Theory , (Proc. Second Czechoslovak Sympos., Prague, 1974) , Academia, Prague, 1975, 89-92. 4. J. Bosak, The graphs of semigroups, Theory of Graphs and its Applications , (Proc. Sympos. Smolenice, June 1963), Academic Press, New York, 1965, 119-125. 5. J. B. Frechen, On the number of cycles in permutation graphs, The Many Facets of Graph Theory , SpringerVerlag, Berlin, 1970, 83-87. 6. F. Harary, Graph Theory , Addison-Wesley , Reading, 1969, 7. Z. Hedrlin and A. Pultr, Relations (graphs) with finitely generated semigroups, Monatshefte fur Math., 68 (1964), 213-217. 8. E. Howorka, On metric properties of certain clique graphs, to appear. 9. D. C.Kay and G. Chartrand, A characterization of certain ptolemaic graphs, Canad. J. Math., 17 (1965) , 342-346. 10. Y.-F. Lin, A problem of Bosak concerning the graphs of semigroups, Proc. A. M.S., 21 (1969), 343346. 91 PAGE 97 92 11. M. Petrich, Introduction to Semigroups , Charles E. Merrill Pub. Co., Columbus, 1973." 12. B. Pondelicek, Diameter of a graph of a semigroup, Casposis Pest. Mat., 92 (1967), 206-211. 13. E. L. Sanders, Independent sets and tree structure, Doctoral Dissertation, University of Florida, 1975. PAGE 98 BIOGRAPHICAL SKETCH Leita Fay Aycock Riddle was born on December 22, 1949, to Thomas G. and Leita M. Aycock of Atlanta, Georgia. In June, 1967, she was valedictorian of her graduation class at The Arlington Schools in Atlanta. In the fall of 1967, she entered H. Sophie Newcomb College of Tulane University. On June 20, 1970, she married Dennis Lee Riddle of Madison, Wisconsin. While in attendance at Tulane, she was appointed to Tulane. University Scholars and Chi Beta honorary society. In May, 1971, she graduated cum laude with the degree of Bachelor of Science in mathematics. The following fall she joined the mathematics faculty at Evans Junior High School in Spartanburg, South Carolina. In September, 1972, she entered the Graduate School of the University of Florida in mathematics. In August, 1974, she received the degree of Master of Science. From August, 197 5, to date she has been a member of the faculty at the University of South Carolina at Spartanburg, where she teaches mathematics and computer science. At the same time, she has continued her study in pursuit of the degree of Doctor of Philosophy. Leita Fay Aycock Riddle is a member of the American Mathematical Society and the Honor Society of Phi Kappa Phi. 93 PAGE 99 I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. A. R. Bednarek, Chairman Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Â£ J. E. Keesling Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. \~UiL ^.^hcr< Hale, Jr. Associate Professor of Mathematics PAGE 100 I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly oresentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. Z ^ 8< T. T. Bowman Assistant Professor of Mathematics I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy. W. D. Hedges \; Professor of Education This dissertation was submitted to the Graduate Faculty of the Department of Mathematics in the College of Arts and Sciences and to the Graduate Council, and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. June, 1978 Dean, Graduate School PAGE 101 UNIVERSITY OF FLORIDA 3 1262 08553 2736 |