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 longrange 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 onetoone 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 uw 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 uw 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 onetoone 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 uv 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 nfactor 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 1factor
if it has a spanning subgraph consisting of disjoint
edges.
Sanders [13] characterizes trees with 1factors:
Theorem 1.11: A tree is bivariegated if and only if it
has a 1factor.
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 ncoloring of a graph G uses
n colors. The chromatic number X(G) is defined as the
minimum n for which G has an ncoloring.
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 onetoone 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 + ... +
(n1) = n(nl)/2 entries above the diagonal. Then all
possible graphs G1 can be found by generating all binary
sequences of length m = n(nl)/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(p2)!), 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(nl)/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 nonzero 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 1factor. Clearly it is true
that if H is bivariegated, then it has a 1factor.
However, the converse is not true, as shown in Figure
II1, below.
v 2
v3 4 5 6 7 8
FIGURE II1
This Husimi tree has a 1factor, 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 1factor such that no edge in the 1factor
is contained in a cycle, then H is bivariegated. The
proof is exactly as in the theorem by Sanders.
Theabove 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 112, below.
FIGURE 112
That is, attach to the vertex either a remote end vertex
or a complete ncycle with n1 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 113.
FIGURE 113
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,... ,nl. 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
lijl = 1 or nl. 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 III1, 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 III1
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 vk1 E V2.
Thus f1 (vkl) = vk = V0 e V, so f(vk) = f(v0) = Vkl.
We claim that Vk2 e V2. For if not, then
Vk2 C VVkl C V2, and f(vk2 = Vk1 However,
this is impossible since f(v0) vkl' contradicting
the onetooneness 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 Vk3 C V2. For if not,
then Vk3 E Vlv k2 C V2, and f(vk3) = Vk. However,
since {Vkr' Vk2 is an edge, {f 1 k f(vk2) =
{v0,v k3 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 f1 (vk2 ),f 1 (vk3) V1 and
f(vl) ,f(v2) c V2. Construct the cycle
a' = (v0,vk ,f(vl),f(v2),v2,v3,... ,v k3,
f1(vk3 ),f1vk2),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 .. ,vk3. Also {v0,f (vk3)} is not an edge,
for if it were, {vklv k3} would also be an edge,
contradicting the irreducibility of a.
Clearly, Vk1 is not adjacent to either f(v2)
(since that would imply {v0,v2} is an edge), v2,...,
or vk3. Referring to Figure III2, it is clear that
vk1 is not adjacent to either f k3) or f (vk2 '
as these vertices are in V.l
Also, f(vI) is not adjacent to either v2,v 3...,
or vk3 (otherwise, there would be a cycle in V2, forcing
{vkl' ,v k3} to be an edge, contradicting the irreducibility
of a). Since f1 (vk3) and f (vk2) 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 .. k3' 1 (vk3), or f (vk2).
By definition of a, v2 is not adjacent to
either v4,., or vk3. Also, {v2,f (vk3) is not
an edge, for if it were, then
6 = (v0lv2' f (vk3) ,f(vk2) 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,f1
(v k3)} to be an edge. However, we have shown earlier that
this cannot be an edge.
Vk2
FIGURE 1112
Similarly, {v2',f (vk2)} is not an edge, for
if it were, then
V = (v0vl,v2f (1 k2),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" ... vk3 is adjacent to another. Also, none of
v3 .... Vk3 is adjacent to f (vk3) since f1 (vk3)
is only adjacent to vk3 and f l(vk2). By the same
reasoning, none of v3,..., k3 is adjacent to f1 (v2).
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 lijl i 1 or nl. 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 ncycle 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 III3
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 III3
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 onetooneness of f, we can prove that neither
v0,vkl 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
Vk3'
If {v2,f (v k3)} were an edge, then (v0,vl,
v2f1 (vk3 ),f1 (vk2 ),v0) is a fivecycle in G Since
G1 is Ptolemaic, there must be at least three diagonals.
Possible diagonals are (i) {vf (vk3) i, (ii) {vi
f1 k3), (iii) {vlf1 (vk2), or (iv) {v2, f1 (vk2
However, case (i) cannot be a diagonal, or else it would
force {vkl ,k3} to be an edge. Therefore, the other
three cases must be the diagonals. This gives us a
cycle (v,vv 2',f1 (vk2 ),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 (Vk2)} is not an edge, for if
it were, we would have a cycle (v0,v,v 2,f (v k2),
f (k )) of length four, again with completeness
forcing {v0,v21 to be an edge.
Similarly, none of v3, .. vk3 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,...,nl.
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) = vnl.
Write a as
(V ...,vnl,vnl+ n2' .'. v 1,v n.+l ',vnk ,v )
v1 ~1 "2 kl 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 onetooneness 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 onetoone).
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
toone 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,
{f1 (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(fl(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(f1(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 IV1.
V2
vl 2 v 3
v4 v6
v 5
FIGURE IV1
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 IV2
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 IV3.
v
V
2
v4 v6
v5
FIGURE IV3
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 IV4:
v4 v6
vI v3
5
FIGURE IV4
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 IV5:
V2
v
5
FIGURE IV5
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(fl(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'
{f1 (v5 )f (v6)} {v2,v3} E but {f1 (g(v5)) f1
(g(v6)) = {f 1 v 1fl(14)} v {V3,V} 4 E.
Figure IV6 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
nonidentity 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 V6
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 apermutation 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 V1.
{b}
{a,b} {b,cl
{a} ac. f{c}
FIGURE VI
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 nonidentity 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 nonidentity 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 nonidentity 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 nonidentity
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 twopoint 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 nonisomorphic
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
nonisomorphic 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 ith 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
onefourth 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 1factor
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 1factor.
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\ RA L L
Err Kr E ]
Ah
AIL1
m
zzc'
Lv Z7
p : 12 (continued)
FIKZ
IZE7
I^Z77
N^n
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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), 206211.
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
