A partially ordered semigroup of Boolean spaces.

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A partially ordered semigroup of Boolean spaces

Hadida, Ahmed Mohamed, Ph.D.

The University of Arizona, 1988

D-M·! 300 N. Zeeb Rd. Ann Arbor, MI 48106

A PARTIALLY ORDERED SEMIGROUP OF BOOLEAN SPACES

by

Ahmed Mohamed Hadida

A Dissertation Submitted to the Faculty of the

DEPARTMENT OF MATHEMATICS

In Partial Fulfillment of the Requirements

For the Degree of

DOCTOR OF PHILOSOPHY

In the Graduate College

THE UNIVERSITY OF ARIZONA

1 988

THE UNIVERSITY OF ARIZONA GRADUATE COLLEGE

As members of the Final Examination Committee, we certify that we have read

the dissertation prepared by ____ ~A=h~m~e~d~M~.~H=a~d=i=d=a~ ________________________ __

entitled A PARTIALLY ORDERED SEMIGROUP OF BOOLEAN SPACES

and recommend that it be accepted as fulfilling the dissertation requirement

for the Degree of Doctor of Philosophy

4/12/88 Date

4/12/88 Date

4/12/88 Fred W. Stevenson Date

Bru~ 0 bkf 4/12/88 Date

4/12/88 Date

Final approval and acceptance of this dissertation is contingent upon the candidate's submission of the final copy of the dissertation to the Graduate College.

I hereby certify that I have read this dissertation prepared under my direction and recommend that it be accepted as fulfilling the dissertation requirement.

4/12/88 Dissertation Director Date

3

STATEMENT BY AUTHOR

This dissertation has been submitted in partial fulfillment of require­ments for the advanced degree at the University of Arizona and is deposited in the University Library to be made available to borrowers under rules of the Library.

Brief quotations from this dissertation are allowable without special permission, provided that accurate acknowledgment of source is made. Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or the Dean of the Graduate College when in his or her judgment the proposed use of the material is in the interests of scholarship. In all other instances, however, permission must be obtained from the author.

Signed: U ;t (, /bi-<-Jf

4

ACKNOWLEDGMENTS

I would like to express my appreciation to my advisor Professor Richard

S. Pierce for his kindness, assistance and wealth of knowledge. It is not unrea­

sonable to say that my dissertation would not have been completed without his

help.

My sincere thanks to the excellent teachers Larry Grove, John Lomont,

Fred Stevenson and Bruce Wood for helping me towards mathematical maturity.

I also thank Professor John Brillhart from whom I have learned much more than

he realizes and Professor William Velez for his sense of humor which eased the

tense times.

I thank my friend Pontelis Damanou, my family; my father for waiting

all these years, my wife and children for enduring the experience with me.

Finally, I thank Sarah Oordt for her terrific job of typing this manu-

script.

TABLE OF CONTENTS

LIST OF ILLUSTRATIONS

ABSTRACT ...... .

CHAPTER 1. BACKGROUND

1.1. Boolean Algebras and Boolean Spaces 1.2. Semiring of Isomorphism Types 1.3. Primitive Boolean Algebras . . . . . 1.4. Finitary Boolean Algebras . . . . . .

CHAPTER 2. THE ARITHMETIC OF THE PARTIALLY· ORDERED SEMIGROUP D

2.1. Axioms . . . . . . 2.2. Arithmetic Properties 2.3. Torsion Elements . . 2.4. Torsion Free Quotient

CHAPTER 3. CRITERIA FOR FINITE ORDER

3.1. Convex Semigroups . . . 3.2. Elements of Infinite Order . . . . . . 3.3. Applications . . . . . . . . . . . . 3.4. Sufficient Conditions for Infinite Order 3.5. The Condition B(if!,r) ..... 3.6. Reducing if! . . . . . . . . . . 3.7. Minimal elements of Infinite Order

REFERENCES

5

page

6

7

8

8 10 13 16

..:.. 21

21 25 'r ~I

34

44

44 49 56 67 72 77 81

88

6

LIST OF ILLUSTRATIONS

page

Figure l. The elements of height one I 19

Figure 2. Nonuniqueness I 27

Figure 3. The elements of height one I I 30

Figure 4. Idempotent and pseudo-indecomposable elements 31

Figure 5. Elements of order 2 32

Figure 6. An element of order 3 33

Figure 7. An element of given order m 35

Figure 8. p is not a homomorphism 40

Figure 9. Nonuniqueness II 43

Figure 10. Elements of infinite order I 54

Figure 11. Convex subsemigroup 55

Figure 12. An element of height 3 56

Figure 13. Elements of infini te order I I 57

7

ABSTRACT

In this thesis we are concerned with arithmetic in a certain partially

ordered, commutative semigroup D. The first chapter investigates the class of

countable Boolean algebras from which this semigroup arises. The elements of

D correspond to the isomorphism classes of the Boolean algebras under consid­

eration. In Chapter 2 we begin the study of the semigroup structure of D. D is

axiomatically described by three groups of axioms. It is proved that these ax­

ioms are categorical. The ordering of D is used to investigate the multiplication.

The set of T of torsion elements of D (elements with only finite many distinct

powers), form a subsemigroup whose structure is studied. There is a natural

torsion free quotient D IT whose structure is also investigated. In Chapter 3,

the axioms are used to characterize elements s of T in terms of the arithmetic

in the subsemigroup generated by the elements that are smaller than s. The

characterization is used to determine elements of T that cover a single element.

In the last part of Chapter 3, we obtain some sufficient, purely combinatorial

conditions for an element to have infinite order.

CHAPER 1

BACKGROUND

8

This thesis is concerned with the structure of a certain partially ordered,

commutative semi group that occurs in an investigation of the class of countable

Boolean algebras. The first. section is an exposition of the background from

which this semi group arises. The body of the thesis, which begins with Chapter

2, can be understood, though perhaps not appreciated, without reading Section

1.

1.1. Boolean Algebras and Boolean Spaces

Boolean algebras can be defined by various axiom systems. The two

most natural definitions are these: a Boolean algebra is a bounded distributive

lattice in which each element has a complement; a Boolean algebra is an asso­

ciative ring with unity element in which each element is idempotent, that is,

a2 = a. V\Te will adopt the lattice theoretical approach. Our standard reference

on the theory of Boolean algebras is [B - Nfl.

Notation 1.1.1. Let A be a Boolean algebra (henceforth abbreviated

B.A.). For elements of a, b E A, the join of a and b is a + b, the meet of a and b

is ab, and the complement of a is -a. The zero of A is 0 (or OA if the subscript

is needed for clarity, and the unit of A is 1 (or lA if necessary). If al,'" ,an

are elements of A, the expression al +a2+' .. +a n designates the join al + ... + an, subject to the hypothesis that these elements are pairwise disjoint, that is

ajaj = 0 for all i i- j.

A Boolean space (abbreviated B.S.) is a compact, zero-dimensional

Hausdorff space X. In particular, X has a neighborhood basis consisting of

clop en sets, that is, sets that are simultaneously closed and open in X.

9

The classes of Boolean algebras and Boolean spaces are categories in

which the morphisms are homomorphisms and continuous mappings respec­

tively. The algebraic theory of B.A.'s and the topological theory of B.S.'s are

essentially interchangeable by virtue of Stone's representation theory.

THEOREM 1.1.2. (M. H. Stone). For a Boolean algebra A the set ultA of all

ultrafilters in A is a Boolean space with the hull-kernel topology. For a Boolean

space X, the set clop X of all clopen subsets of X is a Boolean Algebra with

the operations of set union, intersection and complement. The correspondences

A -+ ultA and X -+ clopX

are object maps of contravariant functors between the categories of Boolean

algebras and Boolean spaces, and the compositions of these functors in each

order are naturally equivalent to the identity functors.

Thus, the categories of Boolean algebras and Boolean spaces are dually

equivalent. In particular, every Boolean algebra is isomorphic to the algebra of

clop en subsets of a Boolean space.

Countable Boolean Algebras 1.1.3. If X is a Boolean space, then it

IS easy to see that clop X is a countable Boolean algebra if and only if X is

second countable. By Urysohn's metrization theorem, a compact Hausdorff

space is second countable if and only if it is metrizable. Thus, Stone's theorem

establishes a dual equivalence between the full categories of countable Boolean

algebras and metrizable Boolean spaces.

It is not difficult to see that if F is a free Boolean algebra on a denumer­

ably infinite set of generators, then ultF is homeomorphic to the cantor middle

third set C. Plainly every countable Boolean algebra is a homomorphic image of

F. This fact translates via Stone's duality theorem to the classical theorem that

10

every metrizable Boolean space is homeomorphic to a closed subset of C. This

observation provides a concrete set of repersentatives of the class of metrizable

Boolean spaces.

1.2. Semiring of Isomorphism Types

The algebraic structure of the set of isomorphism classes of countable

Boolean algebras makes the study of these algebras more orderly.

Notation 1.2.1. For a Boolean algebra A, let [A] be the isomorphism

class of A that is, the class of all Boolean algebras B such that B '" A. Denote

the set of all isomorphism classes of countable Boolean algebras by BA. '"

For the corresponding topologicial construction, let [X] be the homeo­

morphism class of the Boolean space X, and denote by BS the set of all home-'"

omorphism classes of metrizable Boolean spaces. There is a one-to-one corre­

spondence between B",A and B,.§ defined by the inverse mappings [A] ~ [ultA.]

and [X] ~ [clopX].

The direct product of countable Boolean algebras A and B, denoted by

A x B, is a countable Boolean algebra. The coproduct, or tensor product, of A

and B is denoted by A EEl B. If A and B are countable, then so is A EEl B. If

A '" A' and B '" B' then A x B ~ A' X B' and A EEl B '" A' EEl B'. Therefore,

the direct product and coproJuct induce sum and multiplication operations on

B",A, given by the following definition:

Definition 1.2.2. For [A] and [B] in B",A define

(i) [A] + [B] = [A x B]

(ii) [A] . [B] = [A EEl B].

Denote by 0 the class of the one element B.A. and by 1 the class of the

two element B.A.

The sum in B A corresponds to the operation on B S that is defined by '" '"

11

[X]+[Y] = [XUY] where XUY is the disjoint union of X and Y. In the category

of Boolean spaces, the dual of AEBB is the topological product (ultA) X (ultB) of

the Stone spaces of A and B, so the product in BA corresponds to the operation '"

[X] . [Y] = [X x Y] in BS. '"

PROPOSITION 1.2.3. ([B-MJ). BA is a commutative semiring with identity; that '"

is (BA, +, 0) and (BA,·, 1) are commutative monoids; and product distributes '" '"

over the sum.

In fact, BA is a special kind of semiring, called an r-semiring. '"

Definition 1.2.4. (1) An r-monoid is a commutative monoid (S, +, 0)

that satisfies:

(a) x + y = 0 if and only if x = Y = OJ and

(b) the refinement property: if x = 2: Xi, Y = 2: Yj, then there exist Zij E S i<n j<m

such that Xi = 2: Zij and Yj = 2: Zij for all i and j. j<m i<n

(2) An r-semiring is a commutative semiring (S, +,·,0,1) with unity

such that (S, +, 0) is an r-monoid that satisfies:

(a) Integrability property: xy = 0 if and only if x = 0 or Y = 0, and

(b) the product decomposition property: if ab = 2: Ck then there are decompo­k<r

sitions a = 2: ai, b = 2: bi and a partition n x m = Go U G1 U ... U G r - l i<m i<n

such that Ck = 2: aibj. (i,j)EGk

THEOREM 1.2.5. (B",A, +,·,0,1) is an r-semiring.

Definition 1.2.6. (1) The natural ordering of the monoid (M, +, 0) IS

defined by a ~ b if b = a + C for some c EM.

(2) If a is an element of the monoid (.LVI, +,0) denote Mia = {b E AI :

b ~ a}. We call M locally countable if lvI i a is a countable set for all a E AI.

12

As usual, we will write a < b if a::; b and a i= b.

The relation::; is transitive and reflexive, but generally not anti-symmetric;

also, a ::; b implies a + c ::; b + c for all c.

If A is countable Boolean algebra, then BA i [A] = {[A i x] : x E A}, '"

where A i x denotes the principal ideal {y E A : y ::; x}, viewed as a Boolean

algebra. Since A is countable so is BA i [A]. Thus BA is locally countable. Of '" '"

course, B A itself is uncountable. '"

Definition 1.2.7. Let (11/1, +, 0) and (N, +,0) be monoids, a V-relation

between M and N is a set R ~ l'll X N such that:

(i) If aRb then a = 0 if and only if b = O.

(ii) If aRb then a = ao + al in ]vI is equivalent to b = bo + bI in N with aoRbo

and aIRbI .

If A1 = N then a V -relation between M and N is called V -relation on

]vI. If the V -relation R is a morphism from M to N then R is called V -morphism.

It is clear that any subset of the identity relation !:lM on AI is a V­

relation. We say that AI is V-simple if R ~ !:lu for all V-relations Ron M. An

r-semiring S is V -simple if (S, +, 0) is V -simple.

A submonoid N of M is called hereditary if a ::; bEN implies a E N. In

this case, a V-relation on N is also a V-relation on M. Hence if M is V.·simple,

so is N.

The following theorem, due to Dobbertin, characterizes the semi ring

B",A up to isomorphism.

THEOREM 1.2.8. (a) B",A is a locally countable, V-simple r-semiring.

(b) If S is a locally countable r-semiring then there is a unique

V - morpllism (): S ~ BA.

13

(c) B""A is determined to within isomorphism by the properties (a) and

(b).

Undoubtedly, the deepest result in the theory of Countable Boolean

algebras is a theorem that was proved by Ketonen in 1976.

KETONEN'S THEOREM 1.2.9. If M is a countable, commutative semigroup then

M is isomorphic to a subsemigroup of B A. ""

This result shows that the arithmetic structure of (B ...... A, +, 0) IS very

complicated. Ketonen's work was motivated by a question that Tarski posed: is

there a countable Boolean algebra A such that A ~ A x A x A and A ~ A x A?

Since the two element group can be embedded in B",A, according to Ketonen's

result, the answer to Tarski's question is an emphatic "yes".

1.3. Primitive Boolean Algebras

The following property of elements in a monoid will be very important

for us.

Definition 1.3.1. An element a in the commutative monoid 111 is pseudo­

indecomposable (in M) if a = b + c implies a = b or a = c. We denote the set

of non-zero pseudo-indecomposable elements in a monoid 111 by D(1I1).

The concept of a pseudo-indecomposable element in the realm of Boolean

algebras is important as well. A countable Boolean algebra A is (additively)

pseudo-indecomposable (or pi for short) if the isomorphism class [AJ is a pseudo­

indecomposable element of the monoid (BA, +,0). This means that if A = B x C ......

then A '" B or A ~ C.

The zero B.A. and the two element B.A. are clearly pseudo-indecompos­

able; so is the countably infinite free Boolean algebra F.

Pseudo-indecomposability can be formulated in the language of Boolean

spaces; a metrizable Boolean space X is pseudo-indecomposable if, for every

14

elopen subset W of X, either W '" X or X\W '" X. Clearly, a countable

Boolean algebra A is pi if and only if uitA is a pi space.

It is not hard to show that a metrizable Boolean space X is pi if and

only if there is a point p E X such that if W is a elopen neighborhood of p in

X, then W '" X.

Definition 1.3.2. A countable Boolean algebra A is primitive if every . .

e E A admits a decomposition e = fo + ... + fn-l such that for all i, A i fi is

pseudo-indecomposable.

A topological characterization of primitive Boolean algebras is given in

the following lemma whose proof can be found in [B-M].

LEMMA 1.3.3. If X is the Stone space of a countable Boolean algebra A, then

A is primitive if and only if the pi elopen subsets of X form a basis for the

topology of X.

The set of all isomorphism elasses of primitive Boolean algebras will

be denoted by P. By our earlier remark, 0, 1, q = [F] are in P. In fact P ~ ~ ~

is a large subset of BA; IPI = IBAI = 2No. However, P has nicer properties """J ""'-I ""'-I "-I

than BA. For example, by Ketonen's theorem the natural ordering of BA is not ~ ~

antisymmetric, that is, the Schroder-Bernstein property fails. Ho\\'ever, that

flaw does not occur in P.

THEOREM 1.3.4. fB-IvI}. (a) ~ is a hereditary subsemiring of B,.,:;4, that is if

a ~ b E P, then a E P. ~

(b) The natural ordering of ~ is a partial order.

(c) The set D(~) of all pi elements in ~, is a submonoid of the multi­

plicative monoid of BA.

15

We noted that a hereditary submonoid of a V -simple monoid is V­

simple; thus P is a V-simple, r-semiring that is additively generated by D(P). - -It is possible to reconstruct ~ from D(~). In fact, it is possible to prove

a more general structure theorem. The statement of this result involves some

new concepts.

Definition 1.3.5. A primitive monoid is an r-monoid M such that

(i) The natural ordering of M is partial order.

(ii) M is generated, as a monoid, by the set D(M) of all non-zero elements

that are pseudo-indecomposable in M.

The monoid (~, +,0) is an important example of a primitive monoid.

More generally any hereditary submonoid of ~ is a primitive monoid.

Let M be a primitive monoid, the natural ordering of 1\1 restricts to a

partial ordering of D(1\1), but a refinement of this ordering will provide more

information about the structure of D(M).

For elements e, f E D(M) write e <l f if and only if e + f = f. It is

not hard to see that e <l f if and only if either e < f or e = f and e + e = e. <l

is transitive and antisymmetric, but generally not reflexive. Vve denote the set

of elements of D(1\1) on which <l is reflexive by Dl(1\1) that is D1(1vI) = {e E

D (1\1) : e + e = e}.

The following lemma is the principal tool that is used in the proof of

the structure theorem 1.3.8 below.

LEMMA 1.3.6. ((B-MJ). If 1\1 is a primitive monoid, then every a E ]0.1 has

unique representation: a = L: ek in whidl ek E D(M) for all k < r and ek ;tiel k<r

for k i= I. Such a sum is called the reduced representation of a.

Now we introduce the relational systyem (D, <l).

16

Notation 1.3.7. Let D a be non-empty set and let <l be a transitive,

antisymmetric relation on D. Denote the free, commutative monoid on D by

F(D). As usual, the elements of F(D) will be viewed as formal sums of elements

from D:

where nk < wand the ek E D are distinct.

Let = be the monoid congruence that is generated by the relations

e + f = f if e <l f. Denote the quotient monoid F(D)/ = by fleD). Let 7r :

F(D) -t fl(D) be the natural projection homomorphim.

Note that if (D,<l) I"V (D',<l') as relational systems, then fleD) I"V fleD')

as monoids.

THEOREM 1.3.8. ([B-Mj). (a) fleD) is a primitive monoid and 7r is a bijective

mapping of D to D(fl(D)) such that e <l f if and only if 7re + 7r f = 7r f.

(b) If M is a primitive monoid, then 1v1 '" fl(D(M)).

Definition 1.3.9. For each countable Boolean algebra A, denote V(A) =

{e E D(.~) : e ::; [A]}, viewed as a relational systcm under the ordering e <l f if

f = e + f. We will call V(A) the diagram of A.

The diagrams of primitive Boolean algebras can be characterized ab­

stractly as relational systems that satisfy certain conditions, but we will not

use that characterization. Clearly, if A '" B, then V( A) = V( B). A stronger

converse result can be proved: if A and B are primitive Boolean algebras and

V(A) '" V(B) as relational systems, then A '" Band V(A) = V(B).

1.4. Finitary Boolean Algebras

Dcfinition 1.4.1. A countable Boolean algebra A is finitary if A is prim­

itive and the diagram V(A) of A is finite.

17

It is useful to have a topological characterization of finitary Boolean

algebras.

Topological Boolean Algebras 1.4.2. Definition. For any subset A of

Boolean space X, denote by acA the set of all accumulation points of A in X.

acA is closed subset of X such that

(i) ac(A U B) = acA U acB

(ii) ac(acA) ~ acA

(iii) the closure of A is A

A' = acAnA.

(iv) acA = A\(A\A').

acA U A, and the topological derivative of A is

Consider the mapping A ~ acA as unary operation on the set P(X)

of all subsets of X. In addition P(X) supports the Boolean algebra operations

of union, intersection, complementation; and the empty set cP and the universe

X can be viewed as distinguished elements in P(X). Such a system is called a

topological Boolean algebra.

Considered as a topological Boolean algebra P(X) contains a smallest

sub algebra U(X). The space X is said to be of finite type if U(X) is finite.

THEOREM 1.4.3. ({B-Mj). If X is a Stone space of countable Boolean algebra

A, then A is finitary if and only if X is a space of finite type.

Proposition 1.4.4. Let FA denote the set of isomorphism types of fin i-"'"

tary Boolean algebras, that is, FA = {[AJ : A is a finitary B.A.}; then FA is "'" "'"

hereditary submonoid of P. Moreover D(F A) is an order ideal of D(P) and a "'" "'" "'"

subsemigroup of F"",A that additively generates F...:4.

The multiplication in FA is uniquely determined by the semi group

structure of D(F"",A). Questions about the semiring structure of F",A can usu-

18

ally be translated into problems about D(F A). The natural ordering of FA '" '"

induces a partial ordering of D(F",A).

Notation 1.4.5. (1) We denote

DI(FA) = {e E D(FA): e <l e} '" '"

D2(FA) = D(FA)\DI(FA). '" '" '"

Define the mapping K : D(}~A) ~ {1,2} by K( e) = 1 if e E DI (.~A), K( e) = 2

if e E D 2(F",A).

(2) \Ve define the principal ideals J(e), JI(e). J2(e) for e E D(F",A) by:

J(e) = {f E D(F",A): f <l e}

JI(e) = {f E D(ZA): f::; e}

h(e) = {f E D(F",A) : f < e}.

By the definition of FA, the principal ideals in D(F A) are finite. If e, f E D(FA) '" '" ,...,

then the ideal:

is finite. The following notation will be useful

THEOREM 1.4.6. ([B-MJ). The binary relation <l on D(F A) (defined above) is '"

transitive and antisymmetric; hence e ::; fife <l f or e = f is a partial ordering

of D(F,...,A).

(i) All principal ideals in D(F A) are finite. '"

Oi) D(F A) bas two minimal elements 1 and q = [FJ, and 1 < e if e I- q, '"

moreover, ]{(1) = 2, K(q) = 1, that is, 1 til, q <l q.

19

(iii) Ife, f E D(FA) satisfy J2(e) = J2(1) and K(e) = K(f) then e = f. '"

(iv) If K(e) = 1 then JI(e) =I- J2(1) for all f E D(F",A).

(v) If E is a finite, non-empty antichain in D(F A) and E =I- {e} ~ DI (FA), '" '"

then there are elements el and e2 in D(F A) such that E = max h( eI) = '"

Tllere is a product operation under which D(F",A) is a commutative

semigroup with unity 1 and zero q. Moreover the following conditions are satis-

fied:

(vii) e<Jf implieseg<Jfg.

(viii) e <J fg if and only if e = fogo where either fo <l f and go :::; g, or fo :::; f and

go <J g.

(ix) Ifr(f,g) = {e} and K(e) = 1 then fg = e; ifr(l, g) = {e} and K(e) = 2 or

if IfCf,g)1 > 1, then r(f,g) = maxJ2 (1g) and K(fg) = min{K(f),K(g)}.

It is clear from (iv) that the product operation of D( F",A) is determined

by (ix).

By virtue of part (iv) in the theorem, an element e of D(F",A) is uniquely

represented by the Hasse diagram of the finite ordered set JI (e), provided some

device is used to distinguish the elements of DI(FA) from those of D2(Fo4). ~Te '" '"

will make this distinction by using small circles for the elements in DI (FA) and '"

small crosses for elements in D 2(F A). For example, '"

1 1 1 f 1 f 1 1

Figure 1. The elements of height one I.

20

are the diagrams of all elements of height one in D(F",A). We will often use such

diagrams to designate ideals in D(F A). '"

21

CHAPTER 2

THE ARITHMETIC OF THE PARTIALLY ORDERED SEMIGROUP D

In this chapter we begin the study of the partially ordered semi group

D(F",A). The exposition is an elaboration of parts of the papers [PI] and [P2] of

Pierce. Section 2.4 presents new material.

2.1. Axioms

For the sake of generality, and to simplify notation, D will be charac­

terized by three sets of axioms. They reflect properties of D that are introduced

in Chapter 1 of this thesis.

Axiom Group A 2.1.l.

AI. D = (D,·,::;) is a partially ordered, commutative semigroup;

A 2 . If xED then 8(x) = {y ED: y::; x} is finite;

A 3 • D contains an identity element 1 such that x . 1 = x for all xED and a zero

element 0 =1= 1 such that x . 0 = 0 for all xED. Also 0 and 1 are minimal

elements of D, and 1 ::; x for all x =1= 0 in D;

A4 . If x ::; YI • Y2 then there exists ZI ::; YI and Z2 ::; Y2 such that x = ZI . Z2·

Remarks and Notations 2.1.2. The following observations apply to any

partially ordered semi group that satisfies the axioms of group A.

(1) The product x . Y in D will be denoted by concatenation xV.

(2) The notation 8(x) = {y ED: Y ::; x} will be used throughout this section.

For 5 ~ D denote 8(5) = U 8(x). xES

(3) By A 2 , D satisfies the descending chain conditions; thus every non-empty

subset of D contains at least one minimal element.

22

(4) by Al and A3 every x=/:.o in D is positive, that is xy ~ y for all y E D. In

particular 1 :S x :s x2 :s x3 :s .... (5) A4 is analogous to the definition of distributivity in semilattices (see [P3] p.

117).

(6) By A4 and induction: if x :S YI Y2,· .. ,Yn, then there exist ZI :S Yl, Z2 <

Y2,··· ,Zn :S Yn such that x = ZI Z2 ... Zn·

(7) D* = {x ED: x =/:. OJ. By A3, X E D* if and only if x ~ 1. Thus by Remark

(4) x, Y E D* implies xy E D*.

(8) The height h(x) of xED is the number of elements in the longest chain

below x, that is, h(x) = max{n : :JXI < X2 < ... < Xn < x}. By axiom A 2 ,

h(x) is finite for all xED. Clearly, if x < Y then h(x) < h(y).

(9) x covers Y in D if x > y and {z ED: x > Z > y} = <p. Denote ,( x) = {y E

D : x covers y}.

(10) For S ~ D, let J-l(S) be the set of maximal elements of Sj x E J-l(S) if and

only if {y E 5 : y ~ x} = {x}.

(11) A set S ~ D is an antichain if 5 = 1£(5). Let A(D) denote the set of all

finite antichains of D. Plainly, the empty set and all one element subsets of

D belong to A( D). Moreover, it is clear from A2 that ,( x) E A( D) for all

xED.

Definition 2.1.3. For S E A(D) denote E(5) = {x ED: ,(x) = 5}. An

element y in D is of the first kind if E( {y } ) = <pj otherwise y is of the second kind.

Define K : D --+ {I, 2} by K(y) = 1 if Y is of the first kind and K(y) = 2 if y is

of the second kind. Set DI = {y ED: K(y) = I}, D2 = {y ED: K(y) = 2}.

Axiom Group B 2.1.4.

BI If 5 E A(D) and 151 =/:. 1 then E(5) =/:. <p.

B2 If 5 E A(D) and E(S) =/:. <p, then E(S) = {X8' Y8} where K(X8) = 1 and

23

Since ,(x) E A(D) for every xED, by A2 and B2 we have the following

uniqueness statement.

LEMMA 2.1.5. If x, y E D satisfy ,(x) = ,(y) and K(x) = K(y) then x = y.

Notation 2.1.6. For x, y E D, denote r(x,y) = p{x,(y) U y,(x)}. The

following lemma motivates the final group of axioms. Its proof uses only the

axioms of group A.

LEMMA 2.1.7 .. For x, y E D, either ,(xy) = rex, y), or rex, y) = {xy}.

PROOF: If z E ,(x) then zy ~ xy; and if w E ,(y) then wx ~ xy. Thus u ~ xy

for all u E r(x,y). If u < xy then by A4 , u = XlYI where either Xl < x and

YI ~ Y or Xl ~ x and YI < y. Consequently, u = Xl YI ~ zy for some z E ,(x)

or u = XlYI ~ xw for some w E ,(y). Since r(x,y) is an antichain, the lemma

follows.

Remarks 2.1.8. (1) If r( x, y) = {z} where K (z) = 1 then necessarily

z = xy.

(2) By A 3 , 1 ~ x for all x =f. o. It follows that K(O) = 1. Then since

,(I) = ,(0) = ¢ and 0 =f. 1, 2.1.5 yields E(1) = 2.

Axiom C 2.1.9. Let x, y E D. If either Ir(x,y)1 > 1, or r(x,y) = {z} ~

D2 then

,(x,y) = r(x,y) and E(xy) = min{E(x),K(y)}.

It follows from 1.4.6 that D(F A) satisfies all the axioms. The following '"

theorem proves that up to isomorphism, D(F,..A) is the only partially ordered

semigroup that satisfies all the axioms of groups A, B anc C.

24

THEOREM 2.1.10. If D and D are partially ordered semigroups that satisfy all

of the axioms of groups A, Band C, then there is a unique order preserving

semigroup isomorphism of D to D.

PROOF: For n < w denote D(n) = {x ED: h(x) :::; n}, and

D(n) = {x E D: h(x):::; n}.

Since 0 and 1 are minimal, h(O) = h(l) = 0, D(O) = D(O) = {O, I}, and there is

an order isomorphism Bo : D(O) --+ D(O) given by Bo(O) = 0, Bo(1) = 1. Assume

that Bo ~ Bl ~ B2 ~ ... ~ Bn- l have been constructed, so that Bn- l : D(n-l) --+

D( n -1) is an order isomorphism, and K( ()n-l (x)) = K( x) for all x E D( n -1).

Note that if x E D(n), then ,(x) E A(D(n - 1)), so that by axiom B 2, either

x = X8 or x = Y8 where 5 = ,(x). Let 5 = Bn- l (5) E A(D(n - 1)), and

extend Bn- 1 by defining Bn(x 8) = xa, Bn(Y8) = Ya' By 2.1.5 and the hypothesis

that K(Bn_l(x)) = K(x) for all x E D(n - 1), it follows that Bn is well defined

order isomorphism of D(n) onto D(n). By construction K(Bn(x)) = K(x) for

all x E D(n). Let B = U Bn. By axiom A2, B is an order isomorphism of D to n<w

D.

To show that B(xy) = B(x)B(y), we use induction on m = h(x) + h(y).

Ifm = 0 then h(x) = h(y) = 0, x, Y E {0,1}, and the desired result is clear.

Now suppose that the result is true for all x, y wit.h h(x) + h(y) < m. If

,(xv) = r(x,y) = p{x,(y) u y,(x)},

then

,(B(xy)) = p{B(x,(y)) U B(y,(x)}

since B is an order isomorphism. Since h( x) + h( z) < m for z E ,(y) and

h(y) + h(w) < m for w E ,(x), the induction hypothesis gives B(x,(y)) =

25

{B(xz) : z E ,(y)} = {B(x)B(z) : z E ,(y)} = B(x)r(B(y)) and B(y,(x)) =

B(y)r(B(x))). Thus ,(B(xy)) = ,(B(x)B(y)). Also K(B(x)) = K(x) for all xED.

It follows from 2.1.6 that B( xy) = B( x )B(y). On the other hand if ,( xy) f= r( x, y)

then xy E J.l{x,(y) U y,(x)} and K(xy) = 1. By the induction hypothesis,

J.l{B(x)r(B(y)) U B(y)r(B(x))} = B(xy) E D1 . Hence B(xy) = B(x)B(y) by axiom

The uniqueness of B is a direct consequence of the following lemma.

LEM M A 2.1.11. Let x and y be elements of D. Suppose that there is an order

isomorphism B of b(x) onto bey) such that K(B(z)) = K(z) for all z E 8(x).

Tilen x = y.

PROOF: We argue by induction on hex). If hex) = 0 then x = 0 and K(x) = 1

or x = 1 and K(x) = 2, and in this case the result is clear. Now let hex) > 0,

and assume that the lemma is true for all z with h(z) < hex). vVe note that,

Bb(x)) = ,(y) and h(z) < hex) for all z E ,(x). The induction hypothesis gives

,(x) = ,(y). Since K(x) = K(B(x)), we have x = y by 2.1.5. As in 1.4.5, this

last lemma shows that an element x of D is uniquely represented by the Hasse

diagram of 8 (x), labeling the elements of 8 ( x) n D 1 by small circles and elements

of 8( x) n D2 by small crosses.

2.2. Arithmetic Properties

In this subsection and most of the following two, only the proper­

ties of axioms group A will be used, assume that D is any partially ordered

commutative semigroup that satisfies axioms A 2 , A3 and A 4 • As in 2.1.2 (7)

D* = {x ED: x f= OJ.

Definition 2.2.1. An element p E D* is (multiplicatively) pseudo­

indecomposable (pi for short) if p f= 1 and p = xy implies p = x or p = y.

Denote W = {p E D* : p is pseudo-indecomposable}.

26

In the rest of this thesis, we will not refer to the additively pseudo­

indecomposable elements that were encountered in 1.3. Thus, it is safe to omit

the term "multiplicatively" in all that follows.

The next lemma follows by induction from A4 and the definition of

pseudo-indecomposable elements.

LEMMA 2.2.2. Let p E D* be pseudo-indecomposable. If p :::; Xl X2 ••• Xn then

p:::; Xi for some i.

THEOREM 2.2.3. [Pl}. Every element in D* can be written as a product of

pseudo-indecomposable elements.

PROOF: Let Q be the set of all products of pseudo-indecomposable elements

including 1, the product of the empty set. Then Q is closed under multiplication.

If Q =I D*, then there is an X in D* - Q of minimal height. Then X =11 and x is

not pseudo-indecomposable. Thus x = zy with y < x, z < x. Since h(y) < h(x),

h(z) < h(y). It follows that y, z E Q, by the minimality of h(x). Consequently

x E Q because Q is closed under multiplication. This contradiction proves the

theorem.

In general the representation of the elements of D* as pseudo­

indecomposable elements is not unique. Many instances of this non-uniqueness

occur in the semigroup D satisfying all the axioms of Section 2.1.

Example 2.2.4. Let p, rand q be the elements of D represented by the

Hasse diagram (Figure 2). To show the labeling of these diagrams is correct we

need axioms Band C. Plainly p and q are pseudo-indecomposable. r(p, p) =

J.l{JYY(p)} = {p}, I«p) = 1, so that p2 = p. r(q,q) = J.l(q,(q» = {q}, I«q) = 1,

thus q2 = q. r(p,q) = J.l{JYY(q) U q,(p)} = J.l{p,O,p,q} = {p,q}, so ,(pq) =

{p, q} and I«pq) = min{I«p), K(q)} = 1. Thus, r =1= pq, so that r is pseudo-

pq

p

1

Figure 2. Nonuniqueness I.

indecomposable. Now

2 r

r(r,q) = p{r,(q) U q,(r)} = p{r,O,q2,pq} = {r,pq}.

27

Hence, ,(rq) = {r,pq}; and r(r,p) = p{r,O,p2,pq} = {r,pq} implies ,(rp) = . {r,pq} = ,(rq). Since K(rq) = K(r'p) = min{K(r),K(p)} = 1, we have rp = rq. Finally, r(r,r) = p{r,(r')} = p{rq,rp} = {rp} and K(rp) = 1 implies

1'2 = rp = r·q. That is, 1'2 = rp = rq has three distinct representations as a

product of pseudo-indecomposable elements. Note that 1'3 = 1'1'2 = r(r'p) =

r2p = (rp)p = r'p = 1'2.

2.3. Torsion Elements

By axiom A 3 , for every x E D* we have 1 ::; x ::; x 2 ::; ;r3 .. '. If there is

some natural number n such that xll = x ll+1 then x ll+1 = xll+2 = x ll + 3 = ....

This observation leads to the following definition.

Definition 2.3.1. The order of the element x E D* is O(x) = inf{n < 1.0 :

xll = x ll+1 }, where inf ¢ = 00. By convention xO = 1. Denote T = {x E D* :

O(x) < oo}.

28

THEOREM 2.3.2. (P1J). T is an order convex subsemigroup of D*; that is T is

closed under multiplication, 1 E T, and y :::; x E T implies yET.

PROOF: It is clear that O(xy) :::; max{O(x),O(y)} for x, YET. This means

that T is closed under multiplication. Since 1° = 1, 1 E T, and if y :::; x,

O(x) = n < 00 then for all m 2: n we have ym :::; xm = xn. Thus for y =I 0,

O(y) :::; h(xn), so that yET.

Definition 2.3.3. An element e E T is said to be indempotent if e2 = e.

Denote

E={eET:e2 =e}.

Define the mapping e : T -+ E by e(x) = xO(x).

LEMMA 2.3.4. E is a subsemigroup of T with 1 E E. Moreover e is an order

preserving retractive homomorphism of T to E.

PROOF: First note that e(x) = xn for all n > O(x). In particular, e(x)

x 2.O(x) = e(x)2, so that e(x) E E. Also

e(xy) = (xy)max{o(x),O(y)} = e(x)e(y).

If x :::; y, then e(x) = xO(x) :::; yO(x) :::; yO(x)+O(y) = e(y)j that is e is order

preserving. Finally, if fEE, then 0(1) = 1 and e(1) = f. Thus, e is a

retraction.

Notation 2.3.5. Denote 'l1 = {p E D* : p is pseudo-indecomposable },

'l1(x) = 'l1 n b(x)

for all x E D*.

29

THEOREM 2.3.6. ({Pl}). The mapping I --t W(!) is an order isomorphism from

E onto the lattice of all finite hereditary subsets of T U W.

PROOF: By 2.3.2 and A 2 , W(I) is a finite, hereditary subset of Tn Wi that is,

P < q E W(!) implies p E W(f)· Moreover, if I ::; g, then W(f) ~ W(g). The

converse ofthis implication follows from the observation that I = n e(p). In pEIJ1(f)

fact, if p E W(f), then e(p) ::; e(f) = I, so that n e(p)::; IIIJ1(p)1 = I. On the pEIJ1(f)

other hand, by 2.2.3 there exist PI, P2,··· ,Pn E W(!) such that 1= PIP2··· Pn-

Thus, I = e(!) = e(PI )e(p2)··· e(Pn)::; IT e(p). To complete the proof, let pEIJ1(f)

H be a finite, hereditary subset of Tn w. Define I = n e(p). Then lEE pEH

and H ~ W(I). If q E W(!) then q::; n pO(p), so that by 2.2.2, q ::; P for some pEH

P E H. Hence q E H since H is hereditary. This shows that H = W(!).

COROLLARY 2.3.7. (1) E is a distributive lattice with ordering inl1Crited from

D.

(2) e is an order isomorphism of Tn W onto the partially ordered set of

all join irreducible elements of E.

(3) Each element lEE has unique representation in the form I =

e(pd ... e(Pn) where Pi E T n \II and Pi i Pj for i =1= j.

PROOF: (1) The result follows from 2.3.6, since if Hand J{ are hereditary

subsets of Tn W, then H U J{ and H n J{ are also hereditary.

(2) In the lattice of finite hereditary subsets of Tnw, the join irreducible

elements are precisely the principal ideals W(p), since n e(q) = e(p), the qEIJ1(p)

result follows from the proof of 2.3.6.

(3) It is a standard result of lattice theory (see [P3] p.39, for instance)

that in a distributive lattice with the chain condition, every element has a unique

representation as an irredundant join of join irreducible elements. Thus, (3) is

30

a consequence of (1) and (2).

Up to this point, the results of 2.3 depend only on the axioms of group

A. However, somewhat sharper statements can be made if the full axiom system

A-C is invoked. The following fact is a sample of such results.

LEMMA 2.3.8. If e E E and e =I 1 then K(e) = 1.

PROOF: By definition, r(e,e) = p{e,(e)}. The hypothesis that e =I 1 implies

that there exists x E ,(e) such that x 2: 1. Consequently, there is an element

Y E f(e,e) such that y 2: e. If Ir(e,e)1 > lor f(e,e) = {V} ~ D 2 , it would

follow from axiom C that e :S y < e2 = e. Hence, f( e, e) = {y} C D 1 , and

e2 = y. That is, K(e) = K(e2) = K(y) = 1.

Example 2.3.9. (1) The diagram 3 shows all elements of height 0 and

1. Clearly p, q, s, and t are pseudo-indecomposable. Moreover, O(p) = O(q) = 1 p

s t

o

Figure 3. The elements of height one II.

as we noted in 2.2.4. (Incidentally, the element r of that example has order

2.) On the other hand O(s) = O(t) = 00. For example, r(s,s) = {s} ~ D2,

so that ,(s2) = {s}, and K(s2) = 2 arguing inductively, if ,(sn) = {sn-1},

and K(sn) = 2, then f(sn,s) = rt{sn,sn-1s} = {sn} and ,(sn+1) = {s"},

K(sn+l) = 2. Thus O(s) = 00, as claimed. The proof that O(t) = 00 is similar.

(2) It is not difficult to show that in the semigroup D, satisfying the full

system of axioms, there are only five elements that are simultaneously idempo­

tent and pseudo-indecomposable. They are represented by the following diagram

31

p

Figure 4. Idempotent and pseudo-indecomposable elements.

(Figure 4). To prove this assertion, let x and y be non-zero elements in ,(v)

where v2 = v and v is pseudo-indecomposable. Then xy ::; v2 = v and xy =1= o. Since v is pseudo-indecomposable, x ::; xy < v, y ::; xy < v, hence x = xy = y.

In particular, x 2 = x. Using 2.3.8 this argument shows that either ,(v) = 1;

(hence v = 0 or v = 1), ,(v) = {I} (hence, v = p), ,(v) = {O, I} (hence v = q),

or ,(v) = {O,x}, where x 2 = :r and K(x) = 1. In the last case, x t. 0 so that

x t. q and x t. t; also, x t. 8 because O(x) = 1 and 0(8) = 00 (see (1)). Thus,

x = q and v = n.

(3) Consider the diagram (Figure 5) representing a convex subset of D.

It is assumed that K(pd = 1 or 2 for n ~ 2, that ,(P2) = {pd or {PI, ed, and

that for 2 < k ::; n, ,(pd = {Pk-I, el(kd if K(pk-d = 1, and ,(pd = {Pk-d

or {Pk-l, el(kd if K(Pk-J) = 2. Note that l(l.~) is not necessarily defined if

K(pk-d = 2. In compliance with axiom B 2 , l(k) < k where l(k) is defined, and

l(j) < l(k) if l(j) and l(k) are defined, and j < k. Axiom C implies:

(i) P5 = Po, q5 = q,

(ii) e2 = Po qo ,

(iii) ek+l = PkPO = Pkqo for 1 ::; k ::; n,

32

(iv) PkPI = PHI for 0 ~ k ~ I ~ n.

LEMMA 2.3.10. For n 2: 1, T contains more than 2n - I pseudo-indecomposable

elements P such tilat O(p) = 2 and h(p) = n + 1.

PROOF: The values J((P2),'" ,J((Pn) can be assigned arbitrarily in 2.3.9 (3).

Each assignment gives rise to a distinct pn with required properties.

en+l

Pn en

• 0 0 8 0

e2

1 o

Figure 5. Elements of order 2.

(4) The following diagram (Figure 6) of a convex subset of D includes

an element a such O(a) = 3. It can be verified by axioms C and 2.1.6, 2.1.7

that this figure is correctly labeled, and riPo = riqo = riPi = riPoqo = TiP,

qo = 1'; = (riQo)2 for i = 0,1, aqo = apo = apI, aro = aro, and a2qo = a2po = a2p1 = .. , = a3 =a4 = .... Thus O(a) =3.

33

1

Figure 6. An element of order 3.

34

The topological interpretation of this example is that there is a Boolean

space X with the curious property that the Cartesian product of four copies of X

is homeomorphic to the Cartesian product of three copies of X, but X, X x X,

and X x X x X are topologically distinct.

Remark 2.3.11. Pierce, in his paper [P2], showed that for each natural

number m there exists an element d E D such that d, d2 , ••• ,dm - I , dm are

distinct and dm = dm +I . The construction of d was based on the following

convex subset of D represented by Figure 7, with n = 2m.

It is assumed that J{(P2) = ... = J{(Pn+J) = 2, ,(pd = {Pk-d for

k > 1, J{(Si) = J{(/d = 1, ,(Sj) = {Pn,PiqO} and ,(Ii) = {Si,Pn+IqO}. By the

technique used in example 3.3.9 (3), the diagram is correctly labeled. Moreover,

using axiom C, it is easy to verify that

Thus {/o, 11,··· ,In-d ~ E. Clearly Ii A /j = e in the distributibe lattice

E, so that {/o,iI,··· ,In-I} generates a 2n element Boolean algebra B, a sub­

lattice of E, Since every finite distributive lattice can be embedded in a finite

Boolean algebra, it follows that every finite distributive lattice is isomorphic to

a sublattice of E.

2.4. Torsion Free Quotient

'rVe retain the hypotheses and the notations of the previous section.

Explicitly, (D,·,:::;) is a partially ordered semi group that satisfies the axioms of

group A; D*={XED:x=/:O};

T = {x E D* : xn = xn+I for some n};

E = {e E T : e2 = e}.

f n-l

Figure 7. Anm element of giw>n ordf'[ m.

35

36

Definition 2.4.1. For x, y E D*, define x == y if there exists e E E such

that xe = yeo

LEMMA 2.4.2. x = y if and only if there exist u, vET such that xu = yv.

PROOF: Assume that xu = yv, with u, vET. There is n < w such that un = u n+1 E E, and v n = v n+1 E E. Then xunvn = xu . unvn = yvunvn = yunu n

and unvn E E. the converse is clear because E ~ T.

LEMMA 2.4.3. = is a semigroup congruence on D*.

PROOF: Clearly, = is symmetric and reflexive. Let x = y, y = z, then xe = ye,

yJ = zJ, for some e, J E E. Plainly, eJ E E, and xeJ = zeJ so that x = Z. If

x == y and z E D* then xe = ye for some e E E; and xze = yze, so that xz = yz.

Thus, = is a congruence on D*.

Notation 2.4.4. Q = D* /=, 7r : D* ~ Q the natural semi group homo­

morphism associated with the congruence ==. That is; 7rX = [xl = {y: x = V}.

Define 7rX ::; try in Q if there exist Xl ::; YI in D*, such that 7rX = 7rXI,

7ry = 7rYI.

TUEOREM2.4.5. (Q,' ::;) is a partially ordered commutative scmigroup that

satisfies the axioms of group A. Moreover, 7r1 is the only torsion element of Q.

PROOF: (1) 7rX ::; try if and only if there exists, e E E, such that xe ::; ye;

clearly x = xe, y _ ye, so that xe ::; ye implies 7rX ::; 7ry. Conversely assume

7rX ::; 7rY, then there exist Xl ::; YI in D* and e, J E E such that xe = Xl e and

YIJ = yJ. Then xeJ = Xl eJ ::; YI eJ = yeJ.

(2) ::; is a partially ordering of Q; reflexivity is clear. The transitivity

follows from (1): let 7rX ::; try and try ::; 7rZ, so that by (1) xe ::; ye, yJ ::; zJ,

for some c, J E E; since ef E E and xeJ ::; zeJ it follows that 7rX ::; 7rZ. If

37

7rX ~ 7rY and 7rY ~ 7rX, then there exist e, fEE such that xe ~ ye, yf ~ xf.

Then xef = yef implies x = y, so that 7rX = 7ry.

(3) 7r is isotone (i.e., x ~ y implies 7rX ~ 7rY). This is clear from the

definition of the order on Q.

(4) (Q,. ~) is a commutative, partially ordered semigroup; if 7rX ~ 7rY,

say xe ~ ye for suitable e E E, then (7rx)(7rz) = 7r(xe)(7rz) = 7r(xez) ~ 7r(yez) =

7r(y)7r(z) by (3). The statement (4) then follows from (2).

(5) b(7rx) = 7r(b(x)); that is, {7rY E Q: try ~ 7rx} = 7r{y E D* : y ~ x}.

Thus, b( 7rx) is finite, that is, A2 holds in Q. The inclusion 2 follows from

(3). If try ~ 7rX, then for some e E E, y ~ ye ~ xe. By A4 Y = zt, where

z ~ x and t ~ e. Hence ye = zte = ze, because e ~ et ~ e2 = e. Thus,

7ry = 7rze E 7r(b(x)).

(6) 7r1 is the identity of Q, and 7r1 ~ 7rX for all x E D*; by (3).

(7) Let 7rX ~ 7rYI7rY2 = 7r(YI Y2). Then x ~ xe ~ YI YI e for some e E E.

Hence, by A4, x = ZIZ2t, Zl ~ YI, Z2 ~ Y2, t ~ e. Then xe = ZlZ2te = ZIZ2e, so

that 7rX = 7rZI7rZ2 and 7rZI ~ 7rYI, 7rZ2 ~ 7rY2. Thus A4 is satisfied in Q.

(8) If t E T, then 7rt = 7r1, since te(t) = e(t) = 1· e(t).

(9) (7rx)n+1 = (7rx)n implies 7rX = 7r1. If (7rx)n+1 = (7rx)n, then

x n+l ~ xn+le = xne for some e E E. Hence, xn+2 ~ xn+le ~ x ne2 = xne, and

so on: xm ~ xne for all m, so that x E T, and 7rX = 7r1. Thus, 7r1 is the only

element of finite order in Q.

LEMMA 2.4.6. Eacll equivalence class [xl

Moreover, if x = min [x] then [x] = xT.

7rX has unique smallest element.

PROOF: Let x E D*. By the descending chain condition, there is a minimal

Xl ~ x with Xl = x. Suppose that Y == x. Then there exist u, vET such

that Xl ~ xu = yv. By A4 Xl = Z1O, where z ~ y, 10 ~ v. Hence Z = Xl == X,

38

Z ::; Xl ::; X, and the minimality of Xl implies Xl = Z ::; y. Thus, Xl is the

minimum element in 7rX. Note that if X = min[x], then xT ~ [x], by 2.4.2. Also,

y E [x] implies y ::; ye = xe, e E Ei hence y = xlt I , where Xl ::; X, tl E T. By

minimality of x, Xl = Xi and y = xt ExT.

THEOREM 2.4.7. If X is pseudo-indecomposable in D*\T, then 7rX is pseudo­

indecomposable in Q. Conversely, every pseudo-indecomposable element in Q is

of the form 7rX, where X is pseudo-indecomposable element of D*\T. If X and y

are distinct pseudo-indecomposable elements of D*\T then 7rX =1= 7rV.

PROOF: (1) Let X E D*\T be pseudo-indecomposable and assume that 7rX =

7ry7rZ. Then xe = yze for some e E E. Since X is pseudo-indecomposable and

X ~ T, it follows that X ::; Y or X ::; z. Hence 7rX ::; try ::; 7rX or 7rX ::; 7rZ ::; 7rX.

(2) Let 7rZ be a pseudo-indecomposable element of Q. Let x be the

minimum element of [z]. Assume that x = YI Y2 then YI ::; x, Y2 ::; X and

7r Z = 7rX = 7rYI7rY2 implies YI or Y2 is in the class [zl, since [z] is pseudo­

indecomposable. Since X is the minimum of [z] it follows that x = VI or x = Y2'

That is, x is pseudo-indecomposable. Moreover, since 7rX = 7rZ =1= 7r1, it follows

that x E D*\T.

(3) Let x =1= Y be pseudo-indecomposable elements of D*\T. If xe = ve,

for some e E E, then x ::; ye, y ::; xei and x, Y ~ T implies x ::; y and y ::; x,

contrary to the hypothesis that x =1= y. Thus 7rX =1= 7ry.

Since each congruence class mod := has unique minimum element, then

the following definition makes sense.

Definition 2.4.8. Define

p: Q ~ D*

39

by

p[X] = min[x].

Note that trp = 1Q, and p7rX S; x for all x E T.

PROPOSITION 2.4.9. 7rX S; try if and only if p7rX S; p7ry. Thus P is an order

isomorphism of Q to a subset of D* .

PROOF: 7rX S; try implies that there exist Xl S; YI such that 7rX = 7rX, 7rYI = 7rY.

Let X2 = p(7rx), Y2 = p(7rY). Then Xl = X2 i I, YI = Y2i2, iI, i2 E T by 2.4.6,

and X2 S; xzi = Xl S; YI = Y2 i 2 implies X2 = Y3 i3, with Y3 S; Y2 and i3 S; i 2·

Therefore, Y3 = X2 ? Y3, which implies X2 = Y3 S; Y2 because X2 is minimal

in 7rX. This means that p(7rx) S; p(7rY). Conversely, p(7rx) S; p(7rY) implies

7rX = 7rp(7rx) S; 7rp(7rY) = 7rY.

COROLLARY 2.4.10 .. X E p( Q), X S; yi, t E T implies X S; y.

PROOF: Since x E p(Q), we have x = p(7rx) S; p(7ryt) = p(7rY) S; y.

Remark 2.4.11. The exact sequence

~ p T~D~Q.

shows that as an ordered set, D* is a split extension of Q by T. Unfor­

tunately p is not a semigroup homomorphism, since it is clear from the diagram

(Figure 8) that

COROLLARY 2.4.12. x E p(Q), z < x implies 7rZ < 7rX.

PROOF: Since Z < x it follows that 7rZ < 7rX. If 7rZ = 7rX, then x = p( 7rx) =

p( 7r z) S; z. Contrary to the hypothesis.

2 r

r

p

1

Figure 8. p is not a homomorphism.

PROPOSITION 2.4.13. Ifx E p(Q), then i(IT:r) = Ji.(IT(i(;c))).

st

2 ps

2 s

40

PROOF: If ITY E i(ITx), then by 2.4.9, p(ITY) < p(ITx) = x. Thus there exists

Z E i(X) such that p(ITY) ::; Z < :/:. By 2.4.12,

Therefore, ITY = IT::: E IT(-y(x)), because ITY E i(IT:r). If IT:: is not maximal in

41

7r(-y(x)), then there exists W E ,(x) such that 7rZ < 7rW. By 2.4.12, 7rW < 7rX,

contrary to 7r Z E ,( 7rx). Therefore, ,( 7rx) ~ p( 7r(-y( x))). Assume that y E ,(x)

is such that 7ry E p(7r(-y(x))). By 2.4.12 7ry < 7rX. If7rY rt ,(7rx), then by the

first part of the proof, there exists Z E ,(x), satisfying 7ry < 7rZ < 7rX. Since this

contradicts 7ry E p(7r(-y(x))), it follows that p(7r(-y(x))) = '(7rx).

It would be interesting to have a characterization of the image of p. The

following alternative statement of the definition 2.4.8 is somewhat enlightening.

LEMMA 2.3.14. If x E D*\T, then p7rX = x if and only if x = yi, i E T implies

x = y.

In particular, every pseudo-indecomposable element of infinite order is

in the image of p.

Here is another sufficient condition for an element x to be in the image

of p.

LEMMA 2.4.15. If I,(x) n p(Q)1 > 1, then x E p(Q).

PROOF: If x rt p(Q), then x = Xli, where Xl < x and i E T. Let y E ,(x)np(Q).

Then y < Xli, so that y ~ Xl < x by 2.4.10. Therefore, y = Xl. This observation

gives the contradiction hex) n p(Q)1 ~ 1.

Another consequence of 2.4.13 and 2.2.3 is the observation that every el­

ement of p( Q) can be expressed as a product of pseudo-indecomposable elements

of infinite order.

PROPOSITION 2.4.16. Suppose that D* satisfies:

(*) S ~ D*, S is an antichain, 1 < lSI < 00 implies the existence of x E D*

such that S = ,(x).

Then Q also satisfies (*).

42

PROOF: Let T ~ Q, T an antichain, 1 < ITI < 00. Define S = p(T). Then

S is an antichain by 2.4.14, and lSI = ITI. By assumption S = ,(x) for some

x E D*. By 2.4.13 T = 7r(S)) = 7r,(x) = '(7rx).

Note that (*) holds when D satisfies the full system of axioms.

Definition 2.4.17. A partially ordered commutative semi group Q that

satisfies the axioms of group A is called torsion free if 1 is the only torsion

element of Q.

LEMMA 2.4.18. IfQ is torsion free then xyk = xyk+1 implies y = 1, for x, y E Q.

PROOF: Since xyk = xyk+1yn-(k+l) = xyn ;::: yn for all n > k and 6( xyk) is

finite, it follows that yn = yn+1 for some n. Thus y = 1.

PROPOSITION 2.4.19. If Q is torsion free, then every pseudo-indecomposable

element of Q is indecomposable.

PROOF: If Q is torsion free and x is pseudo-indecomposable then x = yz implies

x = y or x = z, say x = z. Thus x = yx. It follows from 2.4.18 that y = 1.

That is, x is indecomposable.

L n 4 20 A h Q. . r. ·f kl k2 k EMMAL.. . ssume t at 1S torslOn lree, 1 Y1 Y2 ... yrr

with ki ::; Ii, 1 =f. Yi E Q for all i, then ki = Ii for all i.

PROOF: Let

then k· /. k k k k· xy.' < xy.' = Y ly 2 ••• Y r < xy.' 1- 112 r- I'

and

43

If ki < Ii then xyfi = xyfi+ 1 so that by Lemma 2.4.7 Yi = 1. Thus ki = Ii.

THEOREM 2.4.21. If Q is torsion free, x E Q, tben tbe number of representa­

tions of x as a product of pseudo-indecomposable elements is finite.

PROOF: If x = p~l p~2 ... p~r with the Pi distinct pseudo-indecomposable ele­

ments of Q and ki ;:::: 1, then Pi :::; pfi :::; x. By 2.4.18, mx(p) = I{n < w : pn :::;

x} I is finite for all p =1= 1. It follows that there are at most n m x (p) distinct pE'l1(x)

representations of x as product of pseudo-indecomposable elements. (Using

2.4.20 this numerical estimate can be considerably improved.) Of course the

representation is not unique, as in the following diagram (Figure 9):

"

Figure 9. Nonuniqueness II.

44

CHAPTER 3

CRITERIA FOR FINITE ORDER

3.1. Convex Subsemigroups

To get deeper results on the structure of D, it is necessary to use more

than the axioms of group A. In this section, it is assumed that the full system

of axioms is satisfied by D.

Definition 3.1.1. A subset V of D is called a convex subsemigroup of D

if

(1) 0 E V and 1 E V,

(2) x E V and y E V implies xy E V,

(3) x ::; y E V implies x E V.

Clearly, any convex subsemigroup of D satisfies the axioms of group A axiom

C, and Theorem 2.1.6 (the uniqueness theorem). The Torsion subsemigroup

T U {OJ of D is an example of a convex subsemigroup.

Standing Hypothesis and Notation 3.1.2.

(1) V is a convex subsemigroup of Dj

(2) s E D\V and ,(s) = {V},V2,··· ,vn} ~ Vj

(3) V-I = {OJ, and for n < w, Vn = {xs k : x E V, k ::; n}j TV = U Vn. n<w

LEMMA 3.1.3. For -1 ::; n < w, Vn is convex: u ::; v E Vn implies 1l E Vn.

Moreover, W is a convex subsemigroup of D.

PROOF: For n = -1 and n = 0 the lemma is clear from the definition of the

hypothesis of 3.1.2. Assume that n > 0 and z ::; XS n, X E V. Then by axiom

A4 , Z = ZI Z2, where Zl ::; xs n- 1 and Z2 ::; s. By the induction hypothesis,

Zl E Vn- I . If Z2 = s, then Z E s Vn- 1 ~ Vn. If Z2 < 8, then Z2 E V by 3.1.2 (2),

45

and Z = Zl Z2 E VVn- 1 = Vn- I ~ Vn. Thus, every Vn is convex. Consequently,

W is convex. Obviously, W is closed under multiplication.

It follows form 3.1.2 (1) and (2) that s is pseudo-indecomposable, clearly,

the pseudo-indecomposable elements in Ware exactly the pseudo-indecomposable

elements of V, together with s.

The problem to which this section is addressed can now be described.

Roughly speaking, it is to relate the arithmetic in W to the arithmetic in V. By

an argument that is similar to the proof of 2.1.10, W is uniquely determined (as

a partially ordered semigroup) by V, the set ,( s) = {VI,'" ,vm }, and by the

specification of K( s). Therefore the arithmetic in V, together with the specifica­

tion of K(s) and VI, V2,'" ,Vm , determines the arithmetic in IV. However, the

relation between these arithmetics does not seem to be simple. The more modest

program of this section is to reduce questions such as "is xsn :::; ysm(x, y E V)?"

to sets of inequalities or equalities involving smaller exponents than m and n,

or coefficients Xl and YI with h(xt) < h(x), h(yJ) < h(y). These results will

be used in the next section to prove a theorem that has some practical use in

dealing with the struct ure of D.

LEMMA 3.1.4. Recall the notation f(x, y) = l1{x,(y) U y,(x)}.

(1) If x, Y E D, tilen ,(XV) ~ 8f(x, V).

(2) For all k :::: 1, ,(sk) ~ 8(sk-I,(s)).

(3) For all k:::: 1, f(s,sk-I) = f.l(sk-I,(s)).

PROOF: By 2.1.7, either ,(XV) = f(x,y) ~ 8(f(x,y)) or xy E f(x,y) ~

8(f( x, y)). In the latter case, ,( xy) ~ 8(f( x, y)), since 8(f( x, y)) is convex

subset of D. For k = 1, (2) and (3) are clear. If (2) holds for k, then by the

definition of f(s,sk), (3) holds for k + 1, and if (3) holds for k + 1, then so does

(2) (by (1)). Thus (2) and (3) follow by induction on k.

46

Second Hypothesis 3.1.5. n 2:: 1 and sn C/:. Vn- 1 •

LEMMA 3.1.6. xsk C/:. Vk- 1 for all x E D* and k ::; n.

P~OOF: If x E D* then x 2:: 1 and xsk ~ sk. If xsk E Vk- 1 , then sk E Vk- l by

3.1.3. Consequently, sn E Sn-kVk_1 ~ Vn- l contrary to 3.1.5.

PROOF: By 3.1.4, r(s, sk-l) = f..l(sk-l,(s )). By axiom C, either ,(sk)

J1(sk-I,(s)) and 1{(sk) = min{1{(s),1{(sk-l)} or sk E sk-l,(s) ~ Vk- 1 . The

latter option is ruled out by 3.1.6 whenever k ::; n. The lemma follows by

induction on k.

LEMMA 3.1.8. If xsn ::; ysn where x, y E V and x =I 0, then there exists z ::; y

such that xsn = zsn.

PROOF: By axiom A 4 , there is z ::; y and w ::; sn such that xsn = zw. If

w < sn, then w E Vn- l by 3.1.7. This would imply xsn E Vn- l . Contrary to

3.1.6. Thus, w = sn.

LEMMA 3.1.9. If xsn- l < ysn, where x, y E V and x =I 0, tl1en there exists

z E V and Vi E ,(s) such that z ::; yVi and xsn- l = zsn-l.

PROOF: Let U E V be minimal with the properties u ::; y, xsn- l < usn. Then

xs n- l E 8(,(us n)) ~ 8(r(u,sn)) = 8(usn- I,(s) U sn,(u)). By the minimality

of u and 3.1.6, xsn- l C/:. 8(sn,(u)). Hence, xs n- l ::; UViSn-1 for some Vi E ,(s).

By 3.1.8, there exists z ::; UVi < YVi such that xs n- l = zsn-l.

LEMMA 3.1.10. For x and y in V - {OJ, r(x, sn) = r(y, sn) if and only if

(1) J1(sn,(x)) = J1(sn,(y)), and

(2) f..l(xsn-l,(s)) = Il(ysn-l,(s)).

47

PROOF: By 3.1.7, r(x,sn) = J1(sn,(x)Uxsn- I,(s)) = J1(J1(sn,(x))UJ1(xs n- I,(s))),

and r(y,sn) = J1(J1(sn,(y)) U J1(x(sn-I,(s))). Thus, (1) and (2) are sufficient

conditions for r(x,sn) = r(y,sn). Moreover, r(x,sn) = r(y,sn) implies (1) by

3.1.6 and 3.1.3. Supose that XSn-IVj E xsn-l,(s). Then XSn-IVj ~ z for some

z E r(x,sn) = r(y,sn). Thus, either

(a) XSn-IVj ~ ysn-I Vj for some Vj E ,(s), or

(b) XSn-IVj < usn for some u E ,(y).

In case (b), it follows from 3.1.9 that XSn-IVj ~ UVjSn-1 ~ ysn-I Vj for some Vj E

,(s). This proves that xsn-l,(s) ~ 8(ysn-I,(s)). By symmetry, ysn-l,(s) ~

8(sxn-I,(s»). Thus, J1(xs n- I,(s) = J1( ysn-I,(s)); that is (2) holds.

Notation 3.1.11. lvh = {x E V - {OJ : if y < x then ysk < xs k}.

LEMMA 3.1.12. V - {OJ = 11'[0 ;2 All ;2 M2 ;2 .. , .

PROOF: x E lvh+l and y < x implies ysk+l < xsk+l. Consequently, ysk < xs k.

LEMMA 3.1.13. Let x E V - {OJ. Then x E 1I1n ifand only if,(xsn) = rex, sn).

In this case I\(xsn) = min{K(x),I\(s)}.

PROOF: Let x E 1I1n. Suppose that ,(xsn) i= r(x,sn). By 2.1.7 r(x,sn) = {xs n}. That is xsn E J1(sn,(x) U xsn-l,(s». By 3.1.6, xsn E sn,(x), so that

xsn = ysn for some y < x. Hence, x tf. Mn. Therefore, x E 1I1n implies

,(xsn) = r(x,sn) and I\(xsn) = min{K(x),I\(sn)} = min{I\(x),I\(s)} by

axiom C and 3.1.7. Conversely, if x tf. 1I1n , then xsn = ysn for some y E ,(x).

Then xsn E r( x, sn), so that ,( xsn) i= r( x, sn).

LEMMA 3.1.14. Let x, y ElvIn. Then xsn = ysn if and only if

(1) J{ ( s) = 1 or K ( x) = J{ (y ),

(2) J1b(x)sn) = J1b(y)sn), and

48

PROOF: By 2.1.6 and 3.1.13, xsn = ysn if and only if rex, sn) = r(y, sn) and

min{K(x),K(s)} = min{K(y),K(s)} which is equivalent to (1). By 3.1.10

r(x,sn) = r(y,sn) is equivalent to conjunction of (2) and (3).

It is obvious that if x E V - {OJ, then there exists y E Mn with y :::; x

and ysn = xsn. (Any y that is minimal in the set {z E V : z :::; x, zsn = xsn}

will do.)

The next result is a criterion for an equality ysn = xsn where y < x.

LEMMA 3.1.15. Let y < x in V. Then ysn = xsn if and only if

(1) K(ysn) = 1

(2) pCr(x)sn) = {ysn}

(3) p(xsn-l,(s)) = p(ysn-l,(s)).

PROOF: Suppose that ysn = xsn. Since y < x, there exists w E ,(x) such that

y :::; w. Consequently, xsn = ysn :::; wsn :::; xsn. Thus, ,(x, sn) = {xsn} = {ysn}, and by axiom C, K(ysn) = 1. Also {ysn} = I1Cr(X)Sn U xsn-l,(s))

implies (by 3.1.6) that (2) holds, and XSn-1Vj < ysn for all 'Vj E ,( s). Conse­

quently, by 3.1.9, XSn-1Vj :::; ysn-I Vj :::; XSn-1Vj for some Vj E ,(s). Hence (3)

holds. Conversely, if (1), (2) and (3) hold, then r( x, sn) = {ysn} ~ D1, so that

xsn = ysn by 2.1.7 and 2.1.8.

LEMMA 3.1.16. For x E M n, sn+l = xsn if and only if either

(1) I1(Sn,(S)) = {xsn} ~ Dl7 or

(2) (a) K(s) :::; K(x),

(b) I1(Sn,(S)) = I1(Sn,(X)), and

(c) I1(XS n- I,(s)) = p(sn-l,(s)r(x)).

49

PROOF: By3.1.4(3)f(s,sn) = Il(Sn,(S)). Assumefirstthat,(sn+l) =I r(s,sn).

By 2.1.7 and 2.1.8, Il(sn,(s)) = f(s,sn) = {sn+l} = {snx} ~ D I, that is, (1)

holds. Conversely, if (1) is satisfied than sn+1 = xsn by axiom C. Suppose that

,(sn+l) = f(s,sn). Then K(sn+l) = min{K(s),K(sn)} = K(s) by axiom C

and 3.1.13. Also, since x E M n, 3.1.13 yields ,(xsn) = f(x,sn) and K(xsn) =

min{K(x),K(s)}. By 2.1.6, sn+1 = xsn if and only if r(s,sn) = f(x,sn) and

K(s) = min{K(x),K(s)}, that is, K(s) ::; K(x). By an argument similar to

the proof of 3.1.10, it is easy to show that f(s,sn) = f(x,sn) if and only if the

conditions 2(b) and 2( c) are satisfied:

Hence Il(sn,(s)) = Il(sn,(x)) and for all Vi E ,(s) there exists w E ,(x)

such that XViS n- 1 < wsn. Thus for all Vi E ,(s), there is some Vj E ,(s)

such that XVjSn-1 ::; WVjSn-l. It follows that x,(s)sn-l ~ b(sn-l,(sh(x)) ~

b(x,(s)sn-l), and ll(xsn- 1 ,(s)) = Il(sn-l,(s hex )). Conversely, (b) and (c) im­

plyf(x,sn) = ll(sn,(x)Uxsn-I,(s)) = ll(sn,(x)Us n- 1,(sh(x)) = Il(sn,(x)) = Il(sn,(s)) = f(s,sn).

3.2. Elements of Infinite Order

The hypotheses 3.1.2, 3.1.5 and the notation that was introduced in Section

3.1 will be in effect in this section.

Definition 3.2.1. The order of s relative to V is Ov(s) = sup{m ~ 1 : sm ~

Vm-d with the supremum equal to 00 if sm ~ l'm-l for all m.

Since sm E Vm, it is clear that Ov(s) ::; O(s).

By 3.1.6, xs k =I ysl for all x, yin l' - {a} and 1 < k ::; n. Thus for 1 ::; k ::; n

multiplication by s maps l'k-l - Vk-2 onto Vk - Vk-l. It is convenient to use s

to designate this mapping as well as the distinguished element of D.

50

The question to be considered in this section is: When is Ov (s) = oo? The

problem is interesting because if Ov(s) = n < 00, then lV = Vn .

The following provisional terminology will be used.

Terminology 3.2.2. Let t E Vk - Vk-l, where 1 ~ k < n. Then t IS

monogenic if Is-I (t)1 = 1.

In the following lemmas, x, y, z and w denote elements of V - {OJ.

LEMMA 3.2.3. If x E M n- l and xsn is monogenic, then x E !YIn.

PROOF: If y < x and ysn = xsn , then ysn-I = xsn- l because xsn is monogenic.

This contradicts x E lIIn - l •

LEMMA 3.2.4. Suppose that xsn ~ ysn, where xsn is monogenic. Then xsn- l ~

ysn-I.

PROOF: By 3.1.8, xsn = zsn, where z ~ y. Then xs n- l = zsn-I ~ ysn-l, since

xsn is monogenic.

COROLLARY 3.2.5. Suppose that Xl,' .. ,Xb, y,' .. ,Yc are elements of V - {OJ

such that XjSn and YjSn are monogenic for all i and j. Denote S = {Xl, ... ,Xb},

T = {Yb ... ,Yc} then

(1) /1(sns) = S/1(sn-IS), and

(2) /1(sns) = /1(snT) implies Il(Sn-IS) = Il(sn- IT).

PROOF: By 3.2.4, s is an order isomorphism of sn-I S onto sn S, and sn-I T

onto snT, so that the result is clear.

LEMMA 3.2.6. Assume that n ~ 2, and every element of the form XVjSn-l(x E

V - {OJ, Vj E ,(s)) is monogenic. Then every element ofVn - Vn- I is monogenic,

that is s maps Vn- I - Vn- 2 bijectively to Vn - Vn- I .

51

PROOF: Suppose otherwise, and choose x to be a minimal element in {z E V­

{OJ : zsn is not monogenic }. The minimality of x implies that x E Mn ~ M n- 1

by 3.1.12. Since xsn is not monogenic, there is minimal y E {z E V - {OJ :

xsn = zsn, xs n- 1 =/:. zsn-I}. In this case, the minimality of y yields y E M n- l .

First assume that y rt. Mn. Then by the definition of M n, there exists z < y

such that zsn = ysn = xsn. It can be assumed without loss of generality that

z E Mn. Note that zsn-l = xs n- 1 by minimality of y. It follows from 3.1.15

and 3.1.13 that

(1) f{(zsn-l) = min{K(z),J{(s)} = f{(zsn) = 1

(2) p(-y(y)sn) = {zsn}, and

(3) p(ysn-l,(s)) = p(zsn-l,(s)).

By assumption, the elements in ysn-l,(s) and zsn-l,(s) are monogenic, so that

by 3.2.5

(4) /-l( ysn-2,(s)) = /-l( zs n-2,(s)).

If w E ,(y) and wsn = zsn = xsn, then also wsn- l = xs n- l by the minimality

of y. On the other hand, if wsn < zsn = xs n , then by 3.1.8, wsn = usn for

some u < x. It follows from the minimality of x that usn = wsn is monogenic.

Hence, ws n- l :::; xs n- l = zsn-l by 3.2.4. These remarks prove

(5) p(-y(y)sn-l) = {zsn-l}.

It follows from (1), (4), and (5), using 3.1.15 that ysn-l = zsn-l = xs n- 1. This

contradiction shows that y E 1\11n. Therefore, by 3.1.14

(6) f{ ( s) = 1 or K ( x) = K (y )

(7) /-l(-y(x)sn) = p(')'(y)sn), and

(8) p(xsn-1,(s)) = p(ysn-l,(s)).

By 3.2.5, (8) implies that

(9) p(xsn- 2,(s)) = p(ysn-2,(s)).

It follows from the minimality of x that the elements of ,(x )sn are monogenic.

52

By (7), the elements of ,",/(y)sn are also monogenic (using 3.1.8). Hence by 3.2.5

(10) fl('"'t( x )sn-l ) = fl('"'t(y )sn-l).

Combining (6), (9) and (10), it follows by 3.1.14 that xs n- 1 = ysn-l. This

contradiction proves the lemma.

The proof of 3.2.6 shows that '"'/( xsn) = s'"'/( xsn- 1 ) and K( xsn) = min{ K( x),

K(s)}. Thus the lemma provides an algorithm that constructs Vn - Vn- 1 from

Vn - 1 - Vn -2.

LEMMA 3.2.7. Let the hypotheses of Lemma 3.2.6 be satisfied. Then sn+l rf- Vn.

PROOF: Since sn+I 2: sn rf- Vn- 1 , it follows from 3.1.3 that sn+l rf- Vn-I'

Assume that s n+ 1 E Vn. Then there exists x E M n such that s n+ 1 = X S n. Since

x E M n , it follows from 3.1.13 that

(1) J((xsn) = min{J((x),K(s)} = K(xsn- 1 ).

Assume that 3.1.16 (1) holds, that is fl(Sn,",/(S)) = {xsn} ~ D 1 • Then by

3.2.5,3.2.6 and (1) fl(sn-I'"'/(s)) = {xsn-l} ~ D I . Therefore, by 3.1.16 sn = xs n - I E Vn - I , which is contrary to hypothesis. Consequently, alternative (2) of

3.1.16 must hold: K(s) :::; K(x), fl(Sn,",/(S)) = fl(Sn,",/(X)) , and fl(XSn-I'"'/(s)) = fl(sn-I'"'/(sh(x)). Again using 3.2.5, 3.2.6, and the monogeneity hypothesis, it

follows that fl(sn-I'"'/(s)) = fl(sn-I'"'/(x)) and fl(XS n- 2'"'/(s)) = fl(sn-2'"'/(sh(x)).

Therefore, by 3.1.16, sn = xsn- I E lin-I' This final contradiction proves the

lemma.

THEOREM 3.2.8. Let the hypotheses of 3.1.2 be satisfied. Assume that for some

n2:2

(1) sn rf- Vn- I and

(2) Is-I(xVjsn-I)1 = 1 for all x E V - {OJ and Vj E ,",/(s).

Then Ov (s) = 00. In particular, O( s) = 00.

53

PROOF: It follows by induction from 3.2.6 and 3.2.7 that sm+l r/:. Vm for all

m 2:: n.

LEMMA 3.2.9. In addition to the hypotheses 3.1.2 assume that V is finite. Then

O(s) = 00 if and only ifOv(s) = 00.

PROOF: By definition, O(s) = n < 00 means sn+l = sn E Vn so that Ov(s) :::; n.

If Ol/(s) = n < 00, then sn+l = xsn for some x E V. Consequently, by

induction on k, sn+k = xksn. Since V is finite and x E V, O(x) = k < 00.

Hence, sn+k+l = xk+lsn = xksn = sn+k so that O(s) < 00.

For an improved result, let 5 = {XVi: X E V- {O},Vi E ,(s)}, and

assume that 151 = m < 00. Since 151 2:: 15s1 2:: 15s2 1 2:: "', it follows that

15sn - 2 1 = 15sn - 11 for some n :::; m + 2, that is, the elements of 5sn - 1 are

monogenic. If also sn r/:. Vn- 1 , then by 3.2.8, O( s) = 00. Otherwise, sn = xsn- 1 ,

where X E V. As in the proof of 3.2.9, O(s) :::; n + k - 1, where k = O(x). This

argument proves the following:

COROLLARY 3.2.10. Let the hypotheses of 3.1.2 be satisfied with V finite. Let

m = I{xvi : x E V - {O},Vi E ,(s)}1 and k = max{O(x): x E V}. Then either

O(s) :::; m + k + 1, or O(s) = 00.

Remark 3.2.11. This corollary shows that the order function O( x) is recur­

sively computable relative to a suitable Godel numbering of the elements of D.

For example, let PI, P2, ... ,Pn,'" be the list of odd primes in increasing order,

and define inductively for xED,

where ,(x) = {VI,V2,'" ,vm } with #(VI) > #(V2) > ... > #(vn ), ei(x) =

#( Vi). An induction argument shows that # is an injective, isotone mapping of

54

D to N. Then K(x) is a recursive function of #(x), and the mapping #(x) ~

(el (.L), e2(x), ... ,em (x)) is recursive. It is easy to prove that #( xy) is a recursive

function of (#(x),#(y)). By induction, the numbers m and k in 3.2.10 are

recursive functions of #( s); and by the corollary, to compute O( s), it is only

necessary to evaluate #(s), #(S2), ... ,#(sn) for n ::; m + k + 2. Thus, O(s) is

a recursive function of #( s).

Example 3.2.12. (1) The following diagram (Figure 10) represents a convex

subsemigroup of D.

Figure 10. Elements of infinite order I.

To show that the labeling is correct:

" ."

" " "

r(p,p) = p{V)'(p)} = p{p} = {p} ~ D2 ,

so that ,(p2) = {p} and K(p2) = K(p) = 2.

55

implies ,(8p) = {8,p2} and K(8p) = 1.

so that 82 = 8p. Hence, if V = {0,I,p,p2,p3, ... }, then OV(8) = 1. Of course,

0(8) = 00. Note that 8 : x -+ 8X is injective on V, so that the elements of 8X are

monogenic. However, 8 2 E VI.

r pq

p q

Figure 11. Convex subsemigroup.

(2) Let V be the convex subsemigroup of D that is represented by the

diagram (Figure 11).

Suppose that 8 E D is such that h( 8) = 3, and ,(8) ~ V. Then by axioms B

and C, 8 is one of the following three elements (Figure 12).

Using axiom C, it is possible to construct VI - l'o and 8; (Figure 13).

It is clear from these diagrams that the elements of X8j with x E V are

monogenic, and 8; ~ Vi. Thus 0(81) = 0(82) = 0(83) = 00 by 3.2.8.

56

s

1 o 1 o

Figure 12. An element of height 3.

3.3. Application

In this section we consider the order of elements 8 such that 1'( 8) = {r}.

If O( r) = 00, then O( s) = 00, so it can be assumed that r has finite order. In

this case, the elements of 1'(1') have finite order, and therefore generate a finite,

convex subsemigroup of D. Moreover, by axiom C, K(r) = 2. These remarks

motivate the standing hyyotheses of this section.

Hypotheses and Notation (3.3.1). (a) V is finite, convex sub semi group of

D, where D is a partially ordered semigroup that satisfies the axioms A, Band

C.

(b) r E D\ V satisfies 1'(1') ~ V and K(7') = 2.

(c) With n 2: 1, VT! defined as in 3.1.2, and 1\11n defined as in 3.1.11, the

element r satisfies

(d) 8 E D\V, (V = U lid satisfies ,(8) = {r}. k<w

57

1 o

Figure 13. Elements of infinite order I I.

- k -(e) V m = {S z: z E V, k ::; m}.

(f) Wm = {rasb x : x E V, a + b:::; m}.

(g) Mab = {y E V : x < y implies rasb x < rasb y }.

(h) Mb = {z E V: w < z implies sbw < sb z }.

Note that Mao = Ma in the notation of 3.1.11.

PROPOSITION 3.3.2. If K(s) = 1, then O(s) ::; O(r) + 1.

58

PROOF: f(s,s) = p(s,(s)) = {rs} ~ D 1 , so that s2 = rs by 2.1.8. By

induction, sm+l = rm s for all m < w. In particular, if O(r) = n, then

sn+2 = r n+1 = rns = sn+l, so that O(s) ::; n + 1.

Henceforth, it is assumed that J« s) = 2.

The next lemma shows that the results of 3.1 are applicable to (s, V) as

well as (r, V).

LEMMA 3.3.3. ifm < n, then sm+l rt. V m'

PROOF: Let m be the minimal with the property sm+l E V m' If no such m

exists, or m ~ n, then the lemma is clearly true. Therefore assume that m < n.

The hypothesis s ~ V implies m ~ 1. Let sm+l = sm z with z E Mm. Write

z = rau, u E V. We can suppose that u E Va. By 3.1.16, either

(1) smr = sm z E D 1 , or

(2) (a) I«z) = 2,

(b) p(sm,(z)) = {sm r }, and

(c) p(sm-lr,(z)) = {sm-l rz }.

Suppose first that 1 holds, that is J« smr) = 1. The minimality of m im­

plies sm rt. V m- 1 , so that I«sm) = 2 by 3.1.7; and I«r) = 2 by hypothe­

sis. Therefore, by axiom C, {smr} = f(sm,r) = v(sm-l r2 U sm,(r)). Thus,

smr = smv for some v E ,(r) by the minimality of m. In this case r m +1 ::; s11lv,

59

and therefore r m +I = ZIZ2'" ZmVI, where Zi ~ s for all i and VI ~ v. The

equality Zi = S cannot hold for any i otherwise s E V. Hence Zi ~ r for

all i, and r m +I ~ rmvI E Vm , contrary to m < n. Thus, 1 cannot occur.

Assume that K ( s m r ) 2, so that the conditions of 2 are satisfied. Write

Z = rau with u E Ma. By 3.1.13, ,(z) = r(ra,u) = p(ra,(u) U ra-Iu'y(r) and

2 = K(z) = min{K(ra),K(u)} ~ K(u). Thus (b) becomes

(b') p(smra,(u) U smra-Iu,(r» = {sm r }.

Since r has finite order and K(r) = 2, there exists W E ,(r) such that W > 1.

Moreover, u 2: 1. Hence, if a > 1, then by (b'), smrp ~ smr ~ sm rp, where

p is an element of finite order that covers 1. Since K(smrp) ~ K(p) = 1

and K(smr) = 2, the assumption that a > 1 produces a contradiction. If

a = 0, then u > 1, because K(smr) = 2 implies sm+I > sm. In this case,

it follows from (b') that smr = smv for some v E v". As in the discussion of

Case 1, this equation leads to the contradiction r m +I E Vm . Hence a = 1 and

p(smr,(u) U smu,(r» = {sm r }. In particular, smrv ~ smr for all v E ,(u).

If some v E ,( u) satisfies v > 1, then the previous argument leads to the

contradiction smrp = smr for some p such that K(p) = 1. Thus, since u

has finite order and K(u) = 2, it follows that u = 1. Thus, (b') reduces to

p(sm,(r» = {smr}, so that smr = smw for some w E V. As in Case 1, this

equation leads to contradiction r m +I E Vm . All options being excluded, the

proof that m 2: n is complete.

I b' b LEMMA 3.3.4. If u', u E V* are such that r a s 1[' = r a s u, where b', b ~ n,

then b' = b. If also a' + b ~ a + b ~ n, then a' = a.

PROOF: If b > b', then sb ~ rasb u = ralsblu ' E Vbl ~ Vb-I contrary to 3.3.3.

Similarly, b' > b is excluded, so that b' = b. Assume that a' + b < a + b = m ~ n.

Then rm ~ rasb u = ral sbu'. Thus, rm = Zl'" Za'WI ... WbU" with Zi ~ r,

60

Wj :::; s, and u" :::; u'. If some Wj = s, then s E V, contrary to the hypothesis

3.3.1 (d). Consequently, Wj :::; , for all j, and ,m :::; ,a'+bu' E Vm - 1 . Since

m :::; n, this conclusion contradicts 3.3.1 (c). Therefore, a' = a.

, b' b LEMMA 3.3.5. If u', u E V* such that ,a s u' < ,a S u, where b < n, then

b' :::; b. If also a + b < n, then a' + b' :::; a + b.

PROOF: If b' > b, then SHI :::; ,a' sb' u' :::; ,asbu E Vb, contrary to 3.3.3 because

+1 ' b' b b < n. Thus, b' :::; b. If a' + b' > m = a + b, then,m :::;,a S U' :::; ,a S u.

This inequality implies ,m+1 E Vm , as in proof of 3.3.3. However, by 3.1.6,

,m+l ~ Vm because m < n. Thus, a' + b' :::; a + b.

LEMMA 3.3.6. If u', u E V are such that ,a sb u' :::; ,a sb u, Wit11 a + b :::; n, then

there exists v :::; u such that ,a 8 bU' = ,a sbv.

PROOF: Write ,asbu' = Zl··· ZaWI ... WbV, where Zj :::; " Wj :::; s, and v < u.

By 3.3.3, Wj = s for all j and Zj = , for all i, which is the asseretion.

LEMMA 3.3.7. For v E V, and a+b:::; n the following conditions are equivalent:

(a) v E lvIab

(b) vEMa and ,av E Ah.

PROOF: Assume that v E lvIab. If u < v, then ,a sbu < ,u sbv. In particular,

,Uu < ,U v . Thus, vEMa. If ,av ~ .Af b, then there exists z E V such that

z < ,U v and sb z = ,asbv . By 3.1.9, z = ,a' u', where a' :::; a and u' E V. The

equality ,u'sbu' = ,u sb v yields a' = a, according to 3.3.4. Moreover, ,uu' =

z < ,U v implies ,u u' = ,av ' for some v' < v by 3.1.8. Consequently, ,U sbv' =

,a sbu ' = ,U sbv , which contradicts the assmmption that v E Mab. Thus, ,av E

M b. Conversely, if v E V and v ~ lvIab, then there exists u < v such that

,a sbu = ,a sbv . If ,au = ,av , then v ~ Mu. If ,au < ,av , then ,av ~ AI b.

61

The following lemma is the key result that we need for the proof of our

main theorem.

LEMMA 3.3.8. Let a + b = m :::; n, a 2: 1, and u, v E V. Then

1vloreover, Mab = .l\1a - I b+I.

PROOF: Note that the validity of (*) implies Mab = .l\1a - I b+I. Indeed, v E .l\1ab

if and only if ra sbu < r a sbv for all u < v, if and only if (by (*)) r a - I sb+I u <

ra-Isb+Iv for all u < v if and only if v E Ma-Ib+I.

The proof of (*) is by triple induction, first on m, then on b, and finally

by the well ordering -< of the sets {h( u), h( v)} defined by:

{m,n} -< {m',n'} if max{m,n} < max{m',n'}

or

maxim, n} = maxim', n'} and minim, n} < minim', n'}.

The observation that {m, n} -< {m, n'} if and only if n < n' will be used several

times.

If m = 0, then the hypothesis a 2': 1 cannot be satisfied, so this

case IS vacuous. Assume that m > 0, and the result (*) holds for smaller

values of m (an empty assumption if m = 1). The induction begins with

b = O. That is, we have to show rmu :::; rmv if and only if r m - I su :::; r m - I sv.

Our induction hypothesis is that this equivalence holds when u and v are re­

placed by u' and v' satisfying {h(u'),h(v' )} -< {h(u),h(v)}. If rmu :::; rmv,

then there exists v' :::; v such that rmu = rmv' (by 3.1.8). If v' < v, then

by the induction hypothesis r m-

I su :::; r m- I sv' :::; r m

-l sv. Therefore, sup­

pose that v' = v, that is rmu = rmv. Let UI :::; u and VI :::; v be such

62

that rmUI = rmu = rmv = rmVI and UI, VI E Mm. By 3.1.14, K( UI) = K(vd, p(rm/(UI)) = p(rm/(VI)), p(rm-IUI/(r)) = p(rm-IVI/(r)). The in­

duction hypothesis yields p(rm-ls/(uI)) = p(rm-ls/(vJ)) and (if m > 1)

p(rm- 2suI/(r)) = p(rm- 2svI/(r)). Thus

r(rm-ls,uJ) = pb(rm-IS)UI Urm-ls/(UI))

= p(rmul U rm- 2 sUI/(r) U r m- l SIC uI))

(with the middle set omitted if m = 1)

= p(rmvI U rm- 2svI/(r) U rm-ls/(VI))

= r(rm-ls,vI)'

Since K(uJ) = J{(vJ) it follows that rm-lsul = rm-lsvI. If UI = U and VI = V,

this is the desired result. Suppose that UI < u. Since rmUI = rmu and K(r) = 2,

it follows from 3.1.15 and 3.1.13 that K(uI) = 1, p(rm/(u)) = {rmud, and

p(rm-Iu/(r)) = p(rm-IUI/(r)). If w E leu), then hew) < h(u), so that

maxi h( w), h( uJ)} < h( u) ::; maxi h( u), h( v)}. The induction hypothesis there­

fore applies, giving p(rm-ls/(u)) = {rm-lsud. Also, if m > 1, the primary

induction hypothesis implies p(1,m-2 su/(r)) = Il(rm- 2suI/(r)). Thus,

r(rm - l s, u) = p(rmu U r m- l s/(u) U 1,m-2 sH,(r))

= p(rmul Ur m- 1sul Ur m- 2sul,(r)) = {r m- 1suIJ

with J{(rm- 1suJ) = 1. Therefore rm-lsu = r m- 1sul. Similarly, rm-lsv

rm- 1 SVI' Hence r m- l su = rm-lsv. Conversely, assume that r m- 1 su ::; 1,m-1 st'.

By 3.3.6, there exists v' ::; V such that rm- 1su = rm-lsv'. If v' < v, then

{h(u), h(v')} -< {h(u), h(v)}, so that the induction hypothesis yields rmu ::;

rmv' ::; rmv. Thus, assume that VI = V, that is, vm- l su = rm- 1 sv. Let HI,

63

VI E M m - l , be such that rm-lsul = rm-lsu = rm-lsv = rm-lsvI. By 3.3.7

rm-1UI, rm-IVI E MI. It follows from 3.3.3 and 3.1.14 (3) that rmUI = rmVI. If

rm-Iu < rm-Iu, then by 3.3.3 and 3.1.15 (3), the equality rm-lsul = rm-lsu

implies rmUI = rmu. The same result follows of course if rm-1UI = r m - l u.

Similarly, rmVl = rmv. Therefore rmu = rmv. This completes the basis step

b= O.

The proof of (*) at the induction step is similar to the case b = O. As­

sume that 0 < b < m and (*) is valid for every pair (a', b') such that a' + b' < m

or a' + b' = m and b' < b. It is also assumed that (*) holds for (a, b) when

(u,v) is replaced by any (u',v') such that {h(u'),h(v')} -< {h(u),h(v)}. These

assumptions are the induction hypotheses. We wish to show that r a sbu ::::; r a sbv

implies ra-lsb+l u ::::; ra-lsb+lv. By 3.3.7 there exists v' ::::; V such that rasb u =

r a sbv '. As in the case b = 0, if v' < v, then the induction hypothesis im­

plies ra-lsb+1u ::::; ra-lsb+l v ' ::::; ra-lsb+lv. Thus, it can be assumed that

r a sbu = r a sbv . Let uI, VI E .A1ab satisfy Ul ::::; u, VI ::::; v, r a sbu = r a sbU1 ,

and r a sb v = va sb V1 . The desired conclusion r a- 1 sb+l u = r a- l sb+ l v is obtained

by showing that ra-lsb+IUI = ra-lsb+ l u (hence ra-lsb+1.VI = ra-lsb+l v by

symmetry) and ra-lsb+lul = ra-lsb+lvl' Consider the first point. If UI = u,

then there is nothing to show, so assume that UI < u. Moreover, if h( u) ::::; h( v),

then {h(uJ),h(u)} -< {h(u),h(v)} so that the desired result is a consequence of

the induction hypothesis. Thus, suppose that h( v) < h( u). If raUl = rau, then

since b > 0, the induction hypothesis yields r a - l SUI = r a - l su, and therefore

r a- l sb+l Ul = r a- l Sb+l u. Therefore, assume that raUl < rau. Consequently

3.1.15 implies that K(rasbuJ) = 1 and J.l(sb,(ra u )) = {rasbud. Since UI E lvJab ,

it follows from 3.3.7 that UI E Ala and raUl E Mb. Thus, K(ra s 1;uJ) = 1 im­

plies that K(uJ) = 1 by 3.1.13. Let rasbUI = sb z with z E ,(rau). By 3.3.4,

z = rau' where u' < w E 'V(u). Consequently r a sbu = r a sbu ' < r a sbw < , _ I ,I - -

64

ra SbU = ra SbUI , so that ra sbUI E ra sb,( u). On the other hand, if w E ,( u), then

rasbw :::; rasbu = rasbul' Thus J.l(rasb,(u)) = {rasbutJ. since {h(w),h(UI)} -<

{h(u),h(v)}, the induction hypothesis yields p(,a-lsHI,(u)) = {ra-lsHlud.

If a = 1, then this equality and the hypothesis rsbu = rsbul (plus K( ud =

1) gives sHlu = sHlul' by 3.1.15, the desired result in this case. Assume

that a > 1. Clearly, r a - I sbwu :::; r a sbu = r a sbUI , for all WE.,( r), so

that ra-lsbwu = sb z , z :::; raUl. It follows from 3.3.4 that z = ra-Iw' for

some w' E V, and ra-Iw' < rau implies that r a - l w' < ra-Iw"u w" E _ I_I,

Since w is any element

in ,( r), it follows from this inequality and the previously established equal­

ity J.l(ra-lsHI,(u)) = {ra-lsHlud that J.l(sHlr(ra-l,u)) = {ra-lsHlutJ.

If ra-Iu E r(ra-l,u), then ra-lsHlu = ra-lsb+IUI, as required. Otherwise,

J.l(sHI,(ra-Iu)) = {ra-lsHlutJ, that is, 3.1.15 (2) holds. Since K(UI) = 1,

3.1.15 (1) is also satisfied. Moreover, 3.1.15 (3) is our hypothesis rasbu = rasbul'

Thus 3.1.15 yields the desired conclusion ra-Isb+l u = ra-lsb+lul in the case

that a > 1. Next, consider the implication: if UI, VI E Mab and ra sbUI = ra sbVI ,

then ra-lsb+IUI = ra-lsb+IVI. Since raUl, raVI E Ah and uI, VI E M a, it fol-

lows from 3.1.14 and 3.1.13 that

(a) K(ud = K(vd, and

(b) J.l(rasb,(ud U ra-lsbun(r)) = J.l(rasb,(vt) U ra-lsbvI,(r)).

The equality (b) separates into two parts by virtue of 3.3.4:

(bd J.l(rasb,(ud) = J.l(rasb,(vd), and

(b2) J.l(ra-lsbuI,(r)) = J.l(ra-lsbvI,(r)).

The induction hypothesis applies to both (b l ) and (b2 ) giving

(CI) J.l(ra-lsHI,(ud) = J.l(ra-lsHI,(vd), and if a> 1,

(C2) J.l(ra- 2sHl ul,(r)) = J.l(ra- 2sH1 vI,(r)).

Hence, using Ul, VI E Ma ~ M a- l and 3.1.13,

(c) J.l(sb+l,(ra-lud) = J.l(sb+1,(r a- l vd).

65

Since r a sbUl = r a sbVl by hypothesis, it follows from (a) and( c) that r a- l sb+l Ul = r a- l sb+l VI' This completes the proof that r a sb u :::; r a sbv implies r a- l sb+l U :::;

r a- l sb+l v. The converse implication is easier to establish. As before, the in­

duction hypothesis reduces the problem of showing that r a - l sHl U = r a - l sH1 V

implies ru' sbu = r a sbv . Let Ul :::; U, VI :::; V satisfy UI, VI E Ma-IHI and

ra-Isb+lUl = ra-Isb+lu = ra-Isb+lv = ra-Isb+IVI. By 3.3.7 ra-luI, ra-Ivl E

M HI. Thus, by 3.1.14 (3), we get r a sbUI = r a sbVI . If ra-lul < ra-lu, then a

similar application of 3.1.15 (3) gives the conclusion rasbul = rasb u . (Note that

here as in previous applications of 3.1.15 and 3.1.14 we have tacitly used 3.3.3.)

If r a- I Ul = ra-Iu, then the equality 1'(1 sbUl = r a sbu is obviously satisfied.

Hence, r a sbu = r a sbv .

LEMMA 3.3.9. Let a + b = m :::; n, a 2:: 1, and U E V. Then f{(ra-lsHlu) =

f{(rasbu).

PROOF: Let Ul :::; U, Ul E Mab with r a sbUl = r a sbu . By 3.3.8, 1£1 E Ma-lHl

and ra-lsHlul = ra-1sHlu. If raul < rau, then by 3.1.13, 3.1.15 and

3.3.7 f{(ud = f{(raud = f{(rasbud = 1. In this case K(ra-lsHlu) =

f{(ra-IsHluJ) = 1 = K(rasbuI) = K(rasbu). Assume raul = raU. If Ul < u,

then again by 3.1.15 and 3.3.3, f{(ud = 1. As before, f{(ra-IsHlu) = 1 =

f{(rasbu). Finally, assume that UI = u that is, u E l\lab. Since f{(r) = I\(s) = 2, it follows from 3.3.3, 3.3.7 and 3.3.8 that K(rau) = K(u) = K(ra-lu) and

f{(rasb u ) = K(u) = ]{(ra-lsHlu).

LEMMA 3.3.10. Let r n+I = rnx, as in 3.3.1 (c). ]fO :::; k :::; n, then

(lh rn-k+lsk = rn-kskx.

66

PROOF: Case 1. 3.1.16 (1) is satisfied, that is x < rj then K(x) = 1 by 3.3.9.

We proceed by induction on k to prove (1 h together with

The case k = 0 of (1)0 is our hypothesis, and (2)0 follows because x < r. Assume

that k ~ 1 and (lh-l holds. By 3.3.3 and 3.1.7, ,(rn-ksk) = l1(rn-k-l-lsk-l U

rn- k- 1sk,(r)). Hence, (l)k-l implies

r( rn- k sk, r) = 11( rn- k sk,( r) U r,( rn- k sk))

= 11( rn- k sk,(r) U rn- k +2 sk-l)

= 11( rn- 1 sk,(r) U rn- k +l sk-l x)

= l1(rn- ksk,(r))

because r < s and x :::; w for some w E ,(r). It follows from axiom C, that

(l)k is equivalent to (2)k. Moreover, by (2h-1 that is, l1(rn-k+lsk-l,(r)) =

{rn- H1 sk-1 x}, we obtain 11( rn- k sk,(r)) = {r n- k sk x} (that is, (2h) from 3.3.8.

This completes the induction for the first case.

Case 2. 3.1.16 (2) is satisfied, that is, I«x) = 2, l1(rn,(r)) = l1(rn,(x)),

and l1(r n- 1x,(r)) = l1(rn- 1,(r),(x)). This is the case in which f(rn,r) =

f( rn , x) either has cardinality ~ 2, or is a singleton of the second kind. By

3.3.8, if 0 < k :::; n,

(3h l1(r n - ks k ,(r)) = l1(r n - ksk ,(x)), and

(4h l1(r n-

k-

1sk x,(r)) = l1(rn- k- 1sk,(r)')'(x)) (k < n).

We can assume inductively that k ~ 1 and (1 h-1 is satisfied. If k < n, then

f( rn-k, 8 k) = 11( r n - k - 1 sk,( r) U rn- k+1 sk-1 ) has cardinality ~ 2 by 3.3.3, 3.3.5,

and the results of 3.1.16. Also ,(sn) = {rsn-1} by 3.3.3 and 3.1. 7. Thus, for

k < n,

f( 1'n- k sk, x) = J-L(r n- k sk,( x) U 1' n- k- 1 sk x,(1') U 1'n- k+1 sk-l x)

=J-L(1'n-ksk,(x)U1'n-k+lsk-lx) by (4h

= J-L(rn-ksk,(1') U 1'n-k+2sk-l) by (3)k and (l)k-l

= f(1'n-ksk,1').

67

Again, by 3.3.3, 3.3.5, and the results of 3.1.16 this set has cardinality 2:: 2, so

that ,(rn-kskx) = ,(1'n-k+1sk), and K(1'n-kskx) = 2 = f{(1'n-k+1s k ). There-

fore, (lh is satisfied. Assume finally that k = n. Using (3)n and (l)n-l as

above, we have f(sn, x) = J-L(sn,(x) U 1'sn- 1 x) = J-L(sn,(1') U 1'2 sn-l) = f(sn, 1'),

so that snx = sn1'.

In the case 1 of 3.1.16, this lemma determines the order of s.

PROPOSITION 3.3.11. Let 1'n +1 = 1'nx, where x < l' and K(x) = 1. H sED

satisfies ,(s) = {1'}, K(s) = 2, then O(s) = 0(1').

PROOF: By (l)n of 3.3.10, K(snr) = K(snx) = 1. Since f(sn,s) = {sn1'}, it

follows from axiom C that sn+l = sn1' = snx. By induction on k 2:: 1, we have

rn+k = rkxk and sn+k = snxk. Moreover, 0(1') = n + k if and only if 1'nx k- 1 <

1'nxk = 1'nxk+l; and O(s) = n + k if and only if snxk-l < snxk = snxk+l. By

3.3.3, the conditions 1'nx k- 1 < rnxk = rnxk+l and snxk-l < snxk = snxk+l

are equivalent. Therefore, 0(1') = O(s).

3.4 Sufficient Conditions for Infinite Order

It is obvious that if x :::; y and O( x) = 00, then O(y) = 00. Therefore, the

program of characterizing elements of infinite order in terms of smaller elements

of D can be restricted to the case in which the covered elements have finite

order.

68

Throughout this section, the hypotheses of 3.1.2 and 3.1.3 are retained.

It is also assumed that V ~ T U {O}; and that ,( s) = {VI," . , Va} generates V

as a convex subsemigroup of D; equivalently, f = e( VI) ... e( Va) is the largest

element of V. In particular, if s 11+ 1 E Vn , then O( s) < 00. The remark in the

previous paragraph justifies this restriction on V.

Notation 3.4.1. (1) n = {l,2, .. ·a}, the index set of ,(s). Note that

Vet =f. vp if Q' =f. ,B in n. (2) For ¢ =f. ~ ~ n, denote VE = n Vet and Vq, = 1.

etEE

(3) For x, y E V* and k < w, write x ~k y if sk x ::; sky. In particular,

x ~o Y if and only if x ::; y.

It is convenient to record three obvious properties of ~.

LEMMA 3.4.2. (1) For k < w, ~k is a pre-order of V*.

(2) Xl ~k YI and X2 ~k Y2 implies XI X2 ~k YIY2.

(3) If j ::; k and x ~j y, then x :::Sk y.

LEMMA 3.4.3. Let k ::; n. Ifx, Y E V* satisfy x ~k y, then there exist mappings

e : n -+ nand TJ : n -+ n such that for all Q' E n: (1) Vetx ~k-I ve(et)Y;

(2) Vet ~k-I v 71 (et)x and

(3) Vet ~k-I V71 Wet»x.

PROOF: By 3.1.8, there exists z ::; Y such that skx = sk z and it can be as­

sumed that z E Ah. Assume first that x E lvh. In this case, p(sk-I,(s)X) =

p(sk-I,(s)z) by 3.1.14. Consequently mapping e, TJ: n -+ n exist satisfying

() k-I < k-I < k-I d a s VaX _ s ve(a)z _ V ve(a)Y, an

(b) sk-I VetZ ::; Sk-I V71 (a)x.

69

The inequalities (a) and (b) yield (1) and (2) directly, and also (3) because

k-I < k-I < k-I < k-l If d M th b 3 1 15 s Va _ S VaX _ s Ve(a)Z _ s Vf/(e(a»X' X 'F k, en y ..

J-l(sk-I,(s)x U sk,(x)) = {skx} = {skz}. In particular, sk-I vax :::; sk z for all

a E n. Hence, by 3.1.9, there is a mapping e : n ~ n such that sk-I vax :::;

Sk-IVe(a)Z for all a E n; that is, (a) holds. If w :::; X, is an element of Mk that

satisfies skw = xkx = xkz, then by 3.1.14, J-l(sk-I,(S)W) = J-l(sk-I,(s)z). Thus,

there is a mapping 7] : n ~ n that satisfies sk-I vaz :::; sk-Ivf/(a)W :::; sk-Ivf/(a)X

for all a E n. Therefore (b) holds. As before, (1), (2) and (3) are satisfied.

LEMMA 3.4.4. If sn+1 E Vn, then there exists ~ ~ n with I~I :::; 2 such that

Va :::5n VE for all a E n.

PROOF: If r(sn,s) = {snv,B} for some f3 E n, then snva :::; snv,B for all a E n;

that is Va :::Sn V{,B} for all a E n. Assume that Ir(sn, s)1 ~ 2. Since sn+1 E Vn,

there exists X E V* such that s n+ I = S n x. It can be assumed that X E M n'

By 3.1.13, J-l(sn,(s)) = r(sn,s) = ,(sn+l) = ,(snx) = J-l(sn,(x) U sn-I,(s)x).

Thus, f3 E n exists with the property that sn-I X :::; snv,B. It follows from

3.1.9 that sn-I X :::; Sn-Iv,Bv')' for some, E n. Consequently, for all a E n,

snva :::; sn+1 = snx :::; snv,Bv')'. That is, V :::5n V{,B,,),} for all a E n. Recall that f = e( VI) ... e( Va) denotes the largest element of V, and

w(f) = w n b(f) is the set of pseudo-indecomposable elements in V. Since

,(s) generates V, there is a natural number n such that f = vrv2' ··v~. Con­

sequently, if p :::; f is pseudo-indecomposable element then p :::; Va for some

a E n.

Notation 3.4.5. Let <I> ~ \J!(J).

(1) For p E <I>, denote np = {a: En: p:::; va}; for II ~ <I>, denote nil = n np. pEn

(2) For a E n, denote <I>a = {p E <I> : p :::; va}; for ~ ~ n denote <I>E = n <I> a • aEE

(3) For r E <I>, denote clq, (r) = <I>!1 r •

70

A routine check shows that the pair of mappings II ---+ nIT and ~ ---+ <I>}:

is a Galois connection between the lattice of all subsets of <I> and the lattice of

subsets of n. In particular, II ---+ <I>nn = cl.p (II) is closure operator on P( <I».

LEMMA 3.4.6. Assume that sn+l E Vn and <I> ~ \I!(f) satisfies:

(A(<I») 1 r/:. <I> and for alIa, (3 E!1, <I>O' U <I>,B =J <I>.

Tllen there exist mappings e, 7] : n ---+ n and r E <I> such that

(1) e(n) ~ nn (2) for all 0' E n, <I>O' ~ <I>7](€(0'» U cl.p(r),

(3) for all 0' E n, <I>O' ~ <I>7](0') U cl.p(r),

PROOF: Denote N = {j < w : 3p E <I>(p ::5j vn-np}'

(a) 0 r/:. N.

Indeed, if p E <I> and np = n, then vn-np = 1, and p 1:. 1 because 1 r/:. <I>. If p E <I>

and 0' E n - np , then p 1:. Va. Therefore, p 1:. n va = vn-np' because p is O'En-np

pseudo-indecomposable. Thus, p io vn-np for all p E <I>.

(b) n E N

In fact, by 3.4.4 there exist (3, ,En such that va ::5n v,Bv-y for all 0' E n. The

condition A(<I» guarantees the existence of p E <I>\(<I>/3 U <I>-Y). Hence V{/3,-y} ~

vn-np' By the remarks preceding 3.4.5, there exists 0' E n such that p ~ va.

Thus p ~ Va ::5n V{/3,-y} ~ vn-np' In particular, n E N. Denote k = minN. By

(a) and (b), k is a well defined natural number that satisfies 1 ~ k ~ n. Since

kEN, there exists r E <I> such that r::5k vn-np' By 3.4.3, there exist mappings

e, 7] : n ---+ n with the property that for all 0' E n.

(c) vO'r ::5k-1 Ve(O')vn-np' and

(d) Va ::5k-1 v7](O')r, va ::5k-1 v7]WO'»r.

On the other hand the minimality of k implies that q ik-l Vn-nq for all q E <I>.

In particular,

71

(e) for all q E <1> and 2; ~ n, if q :5k-l V~, then 2; n nq f:. <p.

Since r ::; vO'r :5k-l ve(O')Vn-nr by (c), it follows from (e) that ({e(a)} U (n -

nr)) n nr f:. <p for all a E n. Thus, e(n) ~ nn as asserted in (1). The assertions

(2) and (3) are consequences of (d) and (e). Indeed, if a and j3 are such that

Va :5k-l v{3r, then for any q E <1>0' and, E nr , q ::; Va -<k-l v{3r ::; V{3V-y. By

(e) f3 E nq or , E nq. That is, q E <1>{3 U <1>1'. Since q E <1>0' and r E nr are

unspecified, it follows that <1>0' ~ n <1>{3 U <1>1' = q>{3 U n <1>1' = <1>{3 U cl4>(r). -yEn r -yEn r

By taking f3 = 17( a)j this observation shows that (d) implies (3). Similarly, (d)

implies (2) when j3 = e(17(a)).

The result 3.4.6 will be the principal tool used to establish sufficient

conditions for the order of s to be infinite. Note that A( q» and the conditions

(1), (2), (3) in 3.4.6 do not depend on n. Therefore, if there is q> ~ n(J)

satisfying A( q>)- and the condition that for all r E q> there are no maps e, 17 : n -t n for which (1), (2) and (3) hold, then the order of s is necessarily

infinite. It is convenient to develop conditions for non-existence of e and 17. This

is a purely combinatorial matter that makes no use of the results in Section 3.1.

however, the notion and hypothesis 3.1.2 remain in effect. To state this result

succinctly, more notation is needed.

Notation 3.4.7. Let <1> ~ \I!(J) and r E <1>. The condition B( <1>, r) is

satisfied if there are no mappings e, 17 : n -t n such that all of the conditions

(1), (2) and (3) of 3.4.6 are satisfied. The condition B(<1» is the statement that

B(<1>,r) holds for all r E q>.

PROPOSITION 3.4.S. If <1> ~ \I!(!) exists so that A( <1» and B( <1» are satisfied,

then O(s) = 00.

PROOF: Assume that O(s) < 00. By 3.2.9 O\,(s) = n < OOj that is, sn ¢ Vn- 1

and sn+l E Vn . Since A(<1» holds by assumption, it follows from 3.4.6 that

72

B( q" r) fails for some r E q,. Since this conclusion contradicts the hypothesis of

the lemma, it follows that O( s) = 00.

3.5. The Condition B( q" r)

In order to use 3.4.8, it is necessary to have an effective criterion for

the validity of B( q" r). The purpose of this section is to develop conditions

that make it possible to decide whether or not B( q" r) holds. Throughout this

paragraph, q, is a fixed but unspecified subset of \J!(J). The notation of 3.4.1

(1),3.4.5 and 3.4.7 is contrived.

Notation 3.5.1. Let a, (3 E 0" r E q,. Denote:

(1) a <lo (3 if q,Q ~ q,i3 and a <lr (3 if q,0' ~ q,i3 U clq,(r); and

(2) aOo(3 if a <10 (3 and (3 <lo a; aO r(3 if a <lr (3 and (3 <lr a.

Clearly, <10 and <lr are preorders of 0,; also 00 and Or are the equivalence

relations induced by these preorders. Note that aOo/3 if and only if q,0' = q,i3.

The conditions (2) and (3) of 3.4.6 are conveniently formulated in terms

of <lr as

(2') a <lr TJ~( a) and

(3') a <lr TJ( a).

Since 0,/00 and n/O r are finite posets, it makes sense to speak of their

maximal elements and minimal elements, that is, the maximal and minimal

classes in 0, relative to the preordering <lo and <lr. An element a E 0, is a 0-

maximal (r-maximal) if the class of a in fl/O o (in 0,/ Or) is maximal. Similarly,

a is O-minimal (r-minimal) if a/Oo (if a/Or) is minial in 0,/00 (n/or).

LEMMA 3.5.2. Let a, (3 E 0" r E q,.

(1) a <lo (3 implies a <lr (3, aOo(3 implies aOr(3.

(2) If (3 E nr, then a <lr (3 implies a <10 (3; if a, (3 E nr, then aO r(3 if and only if

aOo(3·

73

(3) If 0 is r-maximal and 0 <10 (3, then oOr(3; if (3 is r-minimal and 0 <10 (3, then

oOr(3·

(4) If 0 E nr and 0 is r-maximal, then 0 is O-maximal; if (3 E nr and (3 is

O-minimal, then (3 is r-minimal.

These results are clear from the definitions, noting that if 0 E nn then

cl<I>(r)~<I>°.

Notation 3.5.3. (1) IIo = {01, 02, ... ,0 b} is a complete set of represen­

tatives ( mod (0) of the O-maximal elements of n.

(2) For r E <I>, IIr is a complete set of representatives of the r-maximal

elements of n that belong to IIo.

(3) II~r = IIo n n r , II~ = IIr n n r , II~r = IIo \II~r' II~ = IIr \II~.

(4) For 0 E n, rr(o) = {(3 E nr : (3 <lr o}, ~r(o) = {(3 E IIr : 0 <lr (3}.

For~~n,rr(~)= U rr(o)and~r(~)= U ~r(o). oEE oEE

LEMMA 3.5.4. (1) IIr is a complete set of representatives ( mod Or) of the

r-maximal elements of n.

(2) If (3 E nr is r-maximal, then (30ro for a unique 0 E II~.

PROOF: If (3 E n is r-maximal, then there exists 0 E n such that 0 is 0-

maximal and (3 <10 o. By 3.5.2, (30ro. By 3.5.3 (1), there exists OJ E IIo such

that oOOOj, hence oOrOj and (30rOj by transitivity. Finally, by 3.5.3 (2), there

exists O:j E IIr such that OjOrOj and (30rO:j. Thus, IIr is a complete set of

representatives of the r-maximal elements. If also (3 E n r , then (30ro j implies

OJ <10 (3 by 3.5.2, hence OjOo(3 because OJ is O-maximal. That is, r E <I>i3 = <I>0j

and OJ E II~. This proves (2).

LEMMA 3.5.5. B( <I> , r) fails if and only if there is a set of distinct representatives

of the family of sets {r r( 0) : 0: E II r }.

74

PROOF: If B(q"r) is not satisfied, then there are mappings e, TJ : n ~ n such

that e(n) ~ nr and a<lrTJe(a), a<lrTJ(a) for all a E n. If a E ITT) then a <lr TJe(a)

implies aOrTJe(a). Thus, 101 = e(a)<lrTJe(a)Ora, and 10 E r rea). If also (3 E ITr

and 101 = 1{3, then aOrTJIOI = TJ'{30r(3. Hence, a = (3 because ITr is a set of

representatives of the Or classes. This argument shows that {'o : a E ITr } is

a set of distinct relpresentatives of {r r( a) : a E ITr}. Conversely, assume that

there is an injective mapping a ~ 10 of ITr to nr such that 101 E r r( a) for 'all

a. Define TJ: n ~ n by TJ(lOI) = a; and if (3 E n\{,o : a E ITr }. Let TJ((3) = aj,

where j = min{i : 1 :::; i :::; b,(3 <lr ai E ITr }. Since the 10 are distinct and every

(3 E n satisfies (3 <lr ai for some ai E ITr by 3.5.4 (1), this prescription defines

a mapping of n to ITr such that (3 <lr TJ((3) for all (3 E n. Define e : n ~ nr by

e((3) = '0' where TJ((3) = a E Ilr . If TJ((3) = a, then (3 <lr a = TJ( '0) = TJe((3)·

Thus, e and TJ satisfy the condition (1), (2) and (3) of 3.4.6, and B(q"r) fails.

PROPOSITION 3.5.6. For r E q" the following conditions are equivalent

(a) B(q"r) is satisfied;

(b) There exists P ~ Ilr such that Irr(P)1 < IPI;

(c) There exists ~ ~ nr such that Inr \~I < IIlr \~r(~)I·

PROOF: By 3.5.5, B( q" r) fails if and only if there is a set of distinct represen­

tatives of {r r( a) : a E ITr}. By the theorem of Philip Hall [H], such a system

of representatives exists if and only if Irr(P)1 2 IPI for all subsets P of ITr .

Thus, (a) and (b) are equivalent. The equivalence of (b) and (c) is a conse­

quence of the observation that the maps P ~ nr \r rep) and ~ ~ ITr \~r(~)

are the polarities induced by the relation nr x Ilr\<lr ([MMT], p. 51). Thus,

these maps define a Galois connection between P(Il r ) and p(n r ). Hence, if '" '"

P ~ Ilr satisfies Irr(P)1 < IPI, then ~ = nr\rr(P) satisfies Ilr\~r(~) 2 P and

IIlr\~r(~)12 IPI > Irr(P») = Inr\~I· Similarly, (c) implies (b).

75

The conditions (b) and (c) can be improved somewhat.

LEMMA 3.5.7. If P ~ IIr and IPI > Ir r(P)I, then IP n II~I > Ir rep n II~)I.

PROOF: P\P n II~ ~ II~ = IIr n nr. Thus, a E P\P n II~ implies a E r rep).

Moreover, if (3 E P and (3 =/:- a, then a /Jr(3j hence a rt r rep n II~). This

argument shows that P\pnII~ ~ rr(P)\rr(pnII~). Consequently, IpnII~1 =

IPI-IP\P n II~I > If r(P)1 + If r(P)\r reP n II~)I = Ir rep n II~)I·

COROLLARY 3.5.8. For r E <I> , denote n~ = nr \r r(II~) = {(3 E nr : (3 <Ira for

all a E II~}. The following conditions are equivalent:

(a) B(<<I>,r) is satisfied

(b) there exists P ~ II~ such that Ir r(P)1 < IPI;

(c) there exists ~ 2 n~ such that Inr\~1 < IIIr\~r(~)I.

PROOF: The equivalence of (a) and (b) follows from 3.5.6 and 3.5.7. By 3.5.6,

(c) implies (a). Finally (b) implies (c), by the last part of the proof of 3.5.6

because P ~ II~ implies nr \r r( P) 2 nr \r r(II~) = n~.

COROLLARY 3.5.9. Each of the following conditions implies B( «I>, r):

(a) l~r(nr)1 < IIIrl;

(b) Inri < IIIrl;

(c) Ir r(II~)1 < III~I·

The results follow by taking ~ = nil ~ = <p in 3.5.6, and P = II~ in

3.5.8. (Note that by 3.5.7, Irr(IIr)1 < IIIrl implies Irr(II~)1 < III~I·)

COROLLARY 3.5.10. For r E <P-, the following conditions are equivalent:

(a) IIo ~ nr ;

(b) IIr ~ nr

(c) II~ = <p.

76

If these conditions hold, then B( ip, r) fails.

PROOF: By construction, TIr ~ TIo, so that (a) implies (b). If (b) is satisfied,

then TI~ = TIr\(TIr n n r) = ¢. Finally, (c) implies (a) by the equivalence of (a)

and (b) in 3.5.S.

A fairly mild restriction on the family {ipO' : 0:' En} leads to a somewhat

simpler version of 3.5.6.

Terminology 3.5.11. ip satisfies the condition C( ip) if ipO' !f ipP whenever 0:'

and 13 are distinct.

For ip = w = w(f), the condition C( ip) can be related directly to order

structure of V. For simplicity write eO' for e(vO'). If VO' = PIP2'" Pd with Pi E W

(hence, Pi E wO'), then eO' = p~l p~2 ... p~d. By pseudo-indecomposability, for an

element q E W, then conditions q E wand q :::; eO' are equivalent. Moreover, since

eO' is idempotent, eO' = II e(q). These remarks prove the following result. qEIJI"

LEMMA 3.5.12. For 0:', 13 E n, eO' 1:. ep if and only if wO' !f wp. Thus, C(\lI) is

satisifed if and only if eO' 1:. ep for all 0:' f= 13 in n.

Notation 3.5.13. For r E ip, denote Ar = nr U TIr

LEMMA 3.5.14. IfC(ip) is satisfied, tllen Ar is a poset of lengtll at most two

that contains nr as an anti chain.

PROOF: Let 0:', 13 E Ar satisfy 0:' <lr 13. If 13 E nn then Q' = 13 by 3.5.2 (2) and

C( ip). If 0:' E TI r , then 0:' = 13 by the definition of TIr . In particular, Q' <lr 13 and

13 <lr 0:' implies 0:' = 13. Hence, Ar is a poset. If 0:', 13, , E Ar and 0:' <lr 13, 13 <lr "

then either 0:' = 13 or 13 = ,. Thus, Ar has length of at most two.

PROPOSITION 3.5.15. If C(ip) is satisfied, then the follwing conditions are

equivalent for eac11 r E ip:

77

(a) B( CP, r) is satisfied;

(b) There is an antichain T ~ Ar such that ITI > Inri; (c) Ar cannot be written as a disjoint union of Inri chains.

PROOF: If B( CP, r) is satisfied, then by 3.5.8 there exists P ~ IT~ such that

Ir r(P)1 < IPI· Let T = P U (nr \r r(P)). By 3.5.14 and the definitions of

ITr and r r(P), T is an antichain, and ITI = IPI + Inr \r r(P)1 > Inri because

P ~ IT~ implies P n nr = ¢. Thus, (b) holds. Conversely, if (b) is satisfied, then

~ = nr n T has the property that T - ~ ~ ITr \~r(~) because T is an antichain.

Moreover IITr - ~r(~)1 2:: IT - ~I = ITI- I~I > Inrl-I~I = Inr - ~I· By 3.5.6,

B( CP, r) holds. The equivalence of (b) and (c) is an immediate consequence of

Dilworth's theorem ([H], p. 62).

3.6. Reducing cP

If 0 is a subset of CP, then it is of interest to decide how the conditions

A(cp) and A(0), B(cp) and B(0), C(cp) and C(0) are related. These questions

are the topics of this section.

LEMMA 3.6.1. Let 0 ~ CPo Then fora: E n, eO' = cp O' n8 and cle(r) = cl<l>(r)ne

for all r E e. If the preorderings of a: <l~ (3 and a: <lr (3 are defined on n by

a: <J~ (3 if and only if 00' ~ 8f3 U cle(r)

a: <Jr (3 if and only if cpO' ~ cpf3 U cl<l> (r),

tllen a: <Jr (3 implies a: <J~ (3.

These assertions are clear from the definitions eO' = {p E 0 : p ~ vO' },

cpO'={pECP:p~vO'},cle(r)= n 00', andcl<l>(r) = n cpO'. O'En r O'En r

PROPOSITION 3.6.2. Let 8 ~ CP, r E e (1) A(e) implies A( cp).

(2) B(8,r) implies B(CP,r).

78

(3) G(8) implies G(~).

PROOF: (1) By 3.6.1, p E 8\(80' U 8~) implies p E <I>\(<I>O' U <I>~).

(2) If B(<I>, r) fails, then there exist mapping e, 7] : n ~ n with e(n) ~ nr , 0:' <lr 7]eO:', 0:' <lr 7]0:'. By 3.6.1, 0:' <l~ 7]eO:' and 0:' <l~ 7]0:'. Hence B(8, r) fails by

3.4.6.

(3) If p E 8\ 8~, then p E <I> 0'\ <I>~ by 3.6.1.

The condition B( 8) does not generally imply B( <I» since there can be

elem~nts r E <I> \ 8 for which B ( <I> , r) fails. The important question to ask is:

does <I> contain a subset 8 such that A(8) and B(8) are satisfied. By 3.6.2, it

is necessary that A( <I» holds. The next observation provides an algorithm for

determining the existence of 8 ~ <I> such that A( 8) and B( 8) are fulfilled.

PROPOSITION 3.6.3. Assume that A( <I» is satisfied. Denote <I>(I) = {r E <I> :

B( <I> , r) is satisfied}, and define inductively

(1) <I> = <I>(O) ;2 <I>(1) ;2 <I>(2) 2 <I>(3) 2 ...

(2) There is an n < w such that <I>(n) = <I>(n+l); in this case <I>(m) = <I>(n) for all

m~n.

(3) <I>(n+l) = <I>(n) Hand only if B(<I>(n» holds; in this case, if A(q>(n») is satisfied,

then O(s) = 00.

(4) If there exists 8 ~ <I> SUell that A(8) and B(8) are satisfied, then e ~ <I>(k)

f01 all k; thus, A(<I>(k») holds for all k.

PROOF: (1) and (3) are clear from the definition of the prime operation and

3.4.8; and (2) follows because <I> is finite. (4) follows by induction on k from

3.6.2.

79

Starting with <I> = iJ!(J), the proposition provides an algorithm by which

it can be decided whether the criterion of 3.4.8 can be used to show that O( s) = 00. Indeed, there exists e ~ iJ!(J) satisfying A(e) and B(e) (hence O(s) = 00)

if and only if A(<I>(n») is satisfied for the least n such that <I>(n) = <I>(n+l).

Plainly, n :::; I<I>I. Moreover, 3.5.6 and 3.5.8 determine finite algorithms for

constructing <I>(n). The results in the rest of this section and the next one

provide improvements in this algorithm.

It will be useful to have conditions that provide converses of the impli­

cations in 3.6.2. The preorderings <l~ and <lr corresponding to e and <I> (as in

3.6.1) wiII be used below in several places.

LEMMA 3.6.4. For p, q E <I>, define p f'V q if 0.p = 0.q. Then f'V is an equivalence

relation. Let e ~ <I> be a full set of representatives of the equivalence classes

modulo "". Then A(e) if and only if A(<I»; for r E e, B(e, r) if and only if

B( <I> , r); B(e) if and only if B( <I»; and G(e) if and only if G( <I».

PROOF: By 3.6.2, it suffices to prove A(<I» implies A(e), B(<I>, r) implies B(e, r),

and G( <I» implies G( e). If p E <I> \ <I>,8, then there exists q E e such that q f'V p,

that is, nq = 0.p • Hence, 0:' r/:. 0.q and f3 r/:. 0.q. Thus q E e \ eO' u e,8. This

argument shows that A(<I» implies A(e). A similar proof gives G(<I» implies

G(e). Let r E e. To prove that B(<I>,r) implies B(e,r), it suffices to show

that 0:' <l~ (3 implies 0:' <lr (3. Thus by 3.6.1, the preorderings <l~ and <lr coincide,

and therefore by 3.5.6 so do the conditions B(e, r) and B( <I> , r). If 0:' 1r(3, then

<I>O'\ (<I>,8 u clq, (r)) f. <p, say p E <pO' \ (<I>,8 U clq, (r)). Let q E e satisfy q "" p. Thus,

q E eO', and q r/:. e,8. Moreover, p r/:. clq,(r) implies p r/:. <I>1' for I E 0. r . Therefore,

q r/:. e1', so that q r/:. cle(r). Thus, 0:' 1r f3, which completes the proof.

LEMMA 3.6.5. Denote!:l = {p E <P : 0.p = 0.}, e = <I>\!:l. Then A(e) if and

only if A(<I»; for r E e, B(e, r) if and only if B(<I>, r); B(e) if and only if B(<I»;

80

and C(E» if and only ifC(cp).

PROOF: Note that p E b. if and only if p E cpa for all a E n. Hence, E>\(E>a U

8 P) = cp\(cpaUcpP), E>a\E>P = q>a\cpP, and E>a\(E>PUcle(r)) = cpa\(cpPUcl<l>(r))

(that is <l~ = <lp). The assertion of the lemma follow from these observations.

Terminology 3.6.6. cp is reduced if

(a) p # q in cp implies np # nq , and

(b) np 1= n for all p E CPo

The content of 3.6.4 and 3.6.5 is that there exists a reduced set E> ~ cp

such that A(E» iff A(cp), B(E» iff B(cp) and C(E» iff C(CP). Thus, for the

application of 3.4.8, it can be assumed that cp is reduced. This observation leads

to a new viewpoint: () = {np : p E cp} is a family of proper non-empty subsets of

n that uniquely determines the sets cpa, a E n. Indeed, for p E Cp, the relations

p E cpa and a E np are equivalent.

Notation 3.6.7. Let () = {np : p E cp} be a family of subsets of n. For

r E Cp, the r - kernel of a E n is

The kernel of a E n is

(The empty intersection is interpreted as n in these definitions.)

PROPOSITION 3.6.S. Let cp ~ 'iJ!(f).

(1) A( cp) is satisfied if and only if for all a, (J E n, tllere exists p E cp such that

{a,(J} En - np.

(2) p E cl<l>(r) if and only if np "2 nr .

81

(3) a <lr fJ if and only if fJ E kerr(a)j and fJ E kerr(a) if and only ifkerr(fJ) ~

kerr ( a).

(4) C«P) is satisfied if and only if ker( a) = {a} for all a E n.

PROOF: (1) is clear from the definitions. Since clq, is a closure operation, P E

clq, (r) if and only if clq, (p) ~ clq, (r). the fact that n ~ nil, L: ~ <p~ define a

Galois connection then implies that np = nc/of>(p) 2 nc/<l>(r) = nr is equivalent

to clq,(p) ~ clq,(r). For the proof of (3), note that a <lr fJ iff <p a ~ <p.B U clq,(r)

iff for all p (p ¢. <p a or p E <p.B or p E clq,(r» iff for all p (a ¢. np or fJ E np or

nr ~ np) iff for all p (a E np and nr !f np implies fJ E np) iff p E ker r( a); and

fJ E kerr(a) is equivalent to kerr(fJ) ~ kerr(a) because kerr is a closure operation

on n. Finally, C( <p) is satisfied iff, for every a i- fJ, there exists p E <p a\ <p.B, that

is, a E np and f3 ¢. np. This condition is plainly equivalent to ker(a) = {a}.

LEMMA 3.6.9. Assume that nq = np1 U··· U nPm where q i- PI, P2,··· ,Pm. If

e = <p\ {q}, then A( <p) implies A(e), B( <P, r) implies B(e, r) for r E e, and

C(<P) implies C(e).

PROOF: It is clear from 3.6.8 that A(<p) implies A(e), and C(<p) implies C(e).

Moreover, n{np : a E n p, nr !f np} = n{np : a E n p, nr !f np,p i- q}, so that

<l~ = <lr by 3.6.8 (3). Thus, B(<p,r) implies B(e,r).

3.7. Minimal Elements of Infinite Order

The results of Sections 3.4, 3.5 and 3.6 can now be used to give some

explicit sufficient conditions for the element s to have infinite order. The nota­

tion of Section 3.4 is retained, but it will be assumed that <p is reduced. By the

development in Section 3.6, it is possible to adopt a different approach to the

problem of infinite order. As before n = {I, 2, ... ,a} is the indexing set for the

elements of ,( s). It is assumed that a ~ 3. The hypothesis that <p is reduced

means that we can limit our attention to the family () = {nr : r E <p} of distinct,

82

non-empty, proper subsets of n. In fact, the discussion can be carried out at the

abstract level of an arbitrary family of proper, non-empty subsets of n. With

this revised emphasis, it is appropriate to refer to the properties: A( 0), B( 0),

and C( 0) instead of A( <I», B( <I» and C( <I».

We begin this section with some conditions for the validity of B(<I>, r).

LEMMA 3.7.1. The validity of A(O) implies IlIrl > 1 for all r E <I>.

PROOF: Assume that IIr = {a}. Since nr =I- f/J, there exists /3 E n r . By 3.6.7,

A( 0) implies that there exists q E <I> such that a, /3 ct nq. Let, E nq. Then

nr ~ nq because /3 E nr\nq. Thus, kerrh) ~ nq , and therefore a ct kerrh).

Hence, kerr(a) ~ kerr(,), ,<Ira, contrary to IIr = {a}.

LEMMA 3.7.2. If A(O) is satisEed, and Inri = 1, then B(<I>,r) is satisEed.

This result follows directly from 3.7.1 and 3.5.9(b).

PROPOSITION 3.7.3. The following conditions are equivalent for r E <I>

(a) for distinct a, /3 in n, there exists, E nr such that <I>O' ~ <I>(3 U <I>i.

(b) For distinct a, /3 in n, there exists p E <I> such that a E np , /3 ct np , and

nr ~ nl)'

(c) kerr(a) = {a} for all a E n.

(d) IIr = n.

If these conditions hold then B(<I>, r) is satisfied.

PROOF: If (a) holds and p E «1>0'\( <I> (3 u <I>i), where, E nT) then a E np , /3 ct np ,

and, E nr \np , that is, (b) is satisfied. It is clear from the definition of kerr( a),

that (b) implies (c). By 3.6.7 (3), (c) implies (d). Finally, if (d) is satisfied,

then a =I- /3 implies a 1r/3; that is <I>O' ~ <I>(3 U cl<I> (r). It follows that there exists

83

I E nr such that <pO' ~ <p i3 U <P'. The last statement of the lemma follows from

(d), the fact that nr is proper, and 3.5.9.

COROLLARY 3.7.4. Assume that 0 separates elements ofn from pairs in n; that

is, if G' =f. f3, I in n, then there exists p E <P such that G' E np and f3, I r/:. np.

Then A( 0) and B( 0) are satisfied; hence O( s) = <Xl.

PROOF: Since Inl ~ 3, it is clear that A(O) is satisfied. Let r E <P. If Inri = 1,

then B(<P,r) is satisfied by 3.7.2. Assume that Inri> 1. If G' =f. f3 in n, then

there exists I E nr such that I i= G'j and the hypothesis implies that for a

suitable p E <P, G' E np and f3, I r/:. np- Consequently, nr ~ np. By 3.7.3,

B( <P, r) is satisfied. Thus, B( 0) is satisfied, and O( s) = <Xl by 3.4.8.

LEMMA 3.7.5. If nr = n\{G'}, then B(<I>, r) holds if and only if

(a) for all f3 =f. G', there exists p E <I> such that G' E np and f3 r/:. np;

(b) for all f3 =f. G', there exists q E <I> such that f3 E nq , G' r/:. nq , and q =f. r.

PROOF: Since nr = n\{G'}, it follows from 3.5.3 (3) that II~ ~ {G'}. By 3.5.9

and 3.5.10, B( <I> , r) is satisfied if and only if II~ = {G'} and r r(II~) = cP. These

conditions translate to G' 'Af3 and f3 ,6ra for all f3 =f. G'. By 3.6.8 (3), these

properties are equivalent to kerr (f3) ~ kerr(a) and kerr(G') ~ kerr (f3) for f3 =f. a.

Finally, referring to the definition 3.6.7 of ker r leads to the equivalence of B( <I> , 7')

with the conditions (a) and (b).

Note that the condition nr = f2\ {a} is equivalent to 7' E <I>i3 for all

f3 =f. aj (a) is satisfied if and only if <I>O' ~ q,i3 for all f3 =I- G'j and (b) is equivalent

to <I>i3 ~ <I>O' U {7'} for all f3 =f. a.

The algorithm of 3.6.3 makes it possible to determine whether a given

family 0 = {np : p E <I>} of subsets of n contains a subfamily :J = {np : P E 8},

(8 ~ <I» such that A(:J) and B(:J) are satisfied. A more desirable goal is to

84

survey the collection of all () ~ p(n) such that A(..1) and B(..1) hold for some '"

..1 ~ (). For this purpose it suffices to find all ' 'minimally good" families () in

the following sense.

Definition 3.7.6 A family () ~ ~(n) is minimally good (abbreviated

M.G.) if A(()) and B(()) are satisfied, but either A(..1) or B(..1) fails if..1 c ().

The following trivial observation is useful.

PROPOSITION 3.7.7. Let a be apennutation ofn. If() ~ !:(n), then () is M.G.

if and only if a() is M.G.

(The notation a() means {a~ : ~ E ()}, where a~ = {a(a) : a E ~}.)

This result is immediate: the permutation a has the effect of relabeling

the elements of ~, and the condition A( ~) and B( ~) are clearly invariant under

such a change.

Example 3.7.8. If a, (3" are distinct elements of 0" then () = {{a}, {(3},

{I}} is M.G.

PROOF: Denote np = {a}, nq = {(3}, nr = {I}' Clearly, A(()) is satisfied; and

A(..1) fails if ..1 is properly contained in (). It follows from 3.7.2 that B( ()) is

satisfied.

COROLLARY 3.7.9. If 0, = {1,2,3}, then {{1},{2},{3}} is the unique M.G.

family in !:(n).

PROOF: If () ~ p(n) satisfies A(()) then clearly {I}, {2}, {3} E (). The result '"

therefore follows from 3.7.7.

COROI,LARY 3.7.10. If/,(s) n q,(f)/;::: 3, then O(s) = 00.

COROI,I,ARY 3.7.11. If /,(s)/ = 3 and A(q,(f)) is satisfied then O(s) = 00.

85

PROOF: The hypothesis that A(w(f)) is satisfied implies the existence of p, q,

r E w(f) such that np = {I}, nq = {2}, nr = {3}. Thus if <I> = {p, q, r}, then

A( <I» and B( <I» hold.

Example 3.7.12. If Inl ~ 4 and () = {n\{O',,B} : 0' =J,B in n}, then () is

M.G.

PROOF: Clearly, () satisfies A((})j and for any proper subset .:J of (), A(.:J) fails.

Using the assumption Inl ~ 4, it is easy to check that the hypotheses of 3.7.3

(b) are satisfied for all r. Thus B( (}) is valid.

Example 3.7.13. Let n = {I, 2, 3, 4}, and () = {nr : r E <I>}, a family of

proper, non-empty subsets of n, indexed without repetition. We will describe

those families of () that are minimally good. It is convenient to classify the M.G.

families of () according to t.he number

n = I{r E <I> : Inri = Ill·

(N ote that ° :::; n :::; 4.)

(a) If n = 3 or 4, then () is M.G. if and only if () is a permutation of

{{I}, {2}, {3}}.

This assertion is a consequence of 3.7.7 and 3.7.8.

(b) If n = 0, then () is M.G. if and only if () = {{I, 2}, {I, 3}, {I, 4}, {2, 3},

{2,4}, {3,4}}.

Since Inl = 4, the condition A( <I», together with the hypothesis that

n = 0, forces a M.G. set to include all two element subsets of Q. The statement

(b) then follows from 3.7.12.

(c) If n = 1, then () is M.G. if and only if () is a permutation of

{{1},{2,3},{2,4},{3,4}}.

86

Indeed, it can be assumed that {I} E () and {a} rt () for a > 1. The

hypothesis A(<I» then forces {2,3}, {2,4}, {3,4} to be members of (). Thus, it

suffices to prove that {I}, {2,3}, {2,4}, {3,4}} satisfies B(()).

Denote np = {2,3}, nq = {2,4}, nr = {3,4} and nu = {I}. By

3.7.2, B(<I>,u) is satisfied. By calculation, kerp(I) = {I}, kerp(2) = {2,4},

kerp(3) = {3,4}, and kerp(4) = {4}. Thus, IIp = {I,4} and rp(I) = cP. By

3.5.6, B(<I>,p) is satisfied. The symmetry between p, q and r guarantees that

B( <I» is satisfied.

(d) If n = 2 then () is M.G. if and only if () is a permutation of one of

the two families {{I}, {2}, {3, 4}, {I, 3}{2, 3}} or {{I}, {2}, {3, 4}, {I, 3}{I, 4}}.

For the proof of the assertion (d), it can be assumed tht () is M. G. with

{I}, {2} E () and {3}, {4} rt (). The condition A(()) forces {3,4} E (). Denote

nr = {I}, nu = {2}, n t = {3,4}. If <I> = {r, u, t}, then an easy calculation gives

kert(1) = {I}, kert(2) = {2}, kert(3) = kert(4) = n, lIt = {I, 2}, 30 t4<lt 1, 2. By

3.5.6, B( <I>, t) fails. Thus, () must contain additional subsets of n. By 3.6.9 and

the minimality of (), {I, 2}, {I,3,4}, and {2,3,4} are not members of (). Suppose

that {3,4} is the only two element subset of (). Then either np = {I, 2, 3}, nq =

{I, 2, 4}, or both np and nq belong to (); say np E (). However, it is clear that the

condition (b) of 3.7.5 cannot be satisfied if a = 4, f3 = 3 (and p replaces r). Thus,

B ( <I> , p) is not satisfied. This argument shows that there are at least two sets in ()

that have two elements. By symmetry, it can be assumed that np = {I, 3} E (). If

<I> = {r, u, t,p}, then kert(1) = {I}, kert(2) = {2}, kert(3) = {I, 3}, kert(4) = n; so by 3.6.8, lIt = {I,2} and 4 <It 3 <It 1,2. By 3.5.G, B( <I> , t) fails. Thus, () contains

additional sets. Again, by 3.6.9, {I, 2, 3} rt (). Another application of 3.7.5 rules

out the possibility that () = {{I}, {2}, {3,4}, {I, 3}, {I, 2, 4}}. Hence, () contains

another two element subset of n. The possibilities are {2,3}, {I,4}, and {2,4}.

If Qq = {2,3} E (), then kert(1) = {I}, kert(2) = {2} and kert(3) = {3}, so that

87

ITt 2 {I,2,3}, IITtl > Inti, and B(fP,t) is satisfied by 3.5.9. Similarly, B(fP,p)

and B(fP, q) are satisfied. Thus {{I}, {2}, {3,4}, {I, 3}, {2, 3}} is M.G., as is

any permutation of this set (by 3.7.7). Therefore, assume that {2,3} rt B, and

exactly one of {I,4}, {2,4} is a member of B. Assume first that nq = {1,4} E B,

that is, B = {{I}, {2}, {3,4}, {1,3}, {I,4}}, these sets are labeled n r , n u , nt, n p, nq , respectively. Then kert(1) = {I}, kert(2) = {2}, kert(3) = {I,3},

kert(4) = {I,4}, so that ITt = {I,2}. Since f t(2) = cp, it follows from 3.5.6

that B(fP,t) is satisfied. Also, kerp(I) = {I}, kerp(2) = {2} and kerp(4) = {4}

implies B(fP,p) holds (by 3.5.9). Similarly, B(fP,q) is satisfied. Thus, () is

M.G. in this case. Finally, assume that B = {{I}, {2}, {3, 4}, {I, 3}, {2, 4}},

labeled n r , n u , nt, n p, nq , respectively. Then kert(I) = {I}, kert(2) = {2},

kert(3) = {I,3}, kert(4) = {2,4}. Note that C(fP) is satisfied, and At is the

union {I,3} U {2,4} of two chains. By 3.5.15, B(fP, t) fails. Thus, B is not

M.G. Moreover, by 3.7.9, there is no extension of () by sets of three elements

that is M.G. Consequently, the permutations of {{I}, {2}, {3,4}, {I,3}, {l,4}}

and {{I}, {2}, {3, 4}, {I, 3}, {2, 3}} are the only M.G. subsets of pen) such that '"

n = 2.

88

REFERENCES

[B-M] Bonnett, R. and J. D. Monk, Handbook of Boolean Algebra, Elsevier, Amsterdam, 1988.

[H] Hall, M. Combinatorial Theory, Blaisdell, Waltham, 1967.

[MMT] McKenzie, R. N., G. F. McNulty, and W. F. Taylor, Algebras, Lattices, Varieties, Vol. I, Wadsworth and Brooks/Cole, Monterey, 1987.

[PI] Pierce, R. Arithmetical Properties of Certain Partially Ordered Semi­groups, Semigroup Forum, Vol. 11, (1975), 115-129.

[P2] Pierce, R., Tensor Products of Boolean Algebras, Univ. Alg. and Lattice Theory Proc. Fourth Int. Cong., Puebla, Mexico, Lecture Notes in Mathematics, Vol. 1004, 232-239.

[P3] Pierce, R., Introduction to the Theory of Abstract Algebras, Holt, Rinehart and Winston, New York, 1968.