Prime Ideals in Semigroups

University of Tennessee, Knoxville TRACE: Tennessee Research and Creative Exchange

Masters Theses Graduate School

3-1950

Helen Bradley Grimble University of Tennessee - Knoxville

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Recommended Citation Grimble, Helen Bradley, "Prime Ideals in Semigroups. " Master's Thesis, University of Tennessee, 1950. https://trace.tennessee.edu/utk_gradthes/1119

This Thesis is brought to you for free and open access by the Graduate School at TRACE: Tennessee Research and Creative Exchange. It has been accepted for inclusion in Masters Theses by an authorized administrator of TRACE: Tennessee Research and Creative Exchange. For more information, please contact [email protected]. To the Graduate Council:

I am submitting herewith a thesis written by Helen Bradley Grimble entitled "Prime Ideals in Semigroups." I have examined the final electronic copy of this thesis for form and content and recommend that it be accepted in partial fulfillment of the equirr ements for the degree of Master of Arts, with a major in Mathematics.

D. D. Wilson, Major Professor

We have read this thesis and recommend its acceptance:

Wallace Givens, O. G. Harrold Jr. Walter S. Snyder

Accepted for the Council: Carolyn R. Hodges

Vice Provost and Dean of the Graduate School

(Original signatures are on file with official studentecor r ds.) lIarch 9, 1950

To the Committee on Graduate StudYI

I am submItting to you a theaia wrItten by Helen Bradley Grimble entItled ·Prime Ideal. in 8 •• 1- groups." I reoommend that it be acoepted for nine quarter hours or ored1t 1n partial fulfIllment ot the requ1rements for the degree of Mas t er of Arts, WIth a major In Mathematios.

. D.j).� Major 'rotea80r

We have read thi8 the.ie and recommend its acoeptance:

b £1;Jra1&?,L4� /

Aooepted tor the CommIttee " PRIME IDEALS IN SEMIGROUPS

A THESIS

Submitted to The Commit tee on Graduate Study of The University of Tennessee in Partial Fulfillment of the Requirements for the degree of Master of Arts

He len Bradley Grimble

March 1950 ACKNOWLEDGMENT

The author wishes tc express her appreciation for the va luable assistance rendered her by Professor Do Do Miller, under whose direc tion this paper was written.

� �t"-� /i .f/" }£" r<., � . �!�JJf.... . ' ) o. Introductiono The concept of prime ideal, which arises in the theory of rings as a generalization of the concept of prime number in the ring of integers, plays a highly important role in that theory, as might be expected from the central position occupied by the primes in arithmetic. In the present paper, the concept is de fined for ideals in semigroups, the simplest of the algebraic systems of single composition, and some analogies and differences between the ring and semigroup theories are brought outo We make only occasional references to ring theory, however; a reader ac quainted with that theory will perceive its relation to our theorems without difficulty, and a reader unacquainted with it will find that the logical development of our results is entirely independent of it. Our first section is a collection of definitions of basic terms and well known theorems concerning semigroups. The second is devoted to a few theorems on divisors of identity ele ments, and the third to zero divisorso In the fourth, we intro duce prime ideals, and di scuss their unions, intersections, and products. The fifth section concerns prime ideals in semigroups with identity elements, and in the sixth we consider relations between prime ideals and maximal ideals. Throughout the paper we have occasion to consider

"one-sided" concepts. It will be clear that any theorem in volving such notions may be converted into another of the same truth-value by interchanging throughout the worda "left" and "right." 2

1. Preliminaries 'J

We she..ll ex va. il ourselves of the standard terminology

and notation of the theory of sets , with which we assume the reader to be acquaintedo In particular, we shall us e cap ital Roman letter s to denote sets, small Raman letters for the ir

elements, small Greek letters for mappings , E: for the rela-

tion of elementhood , � for inclusion, c= for proper inclu

v n sion , for un ion ( = class sum = set-theoretic sum), for intersection , and p for the null set; when al l sets under

discussion are subsets of some set S,we shall denote by

C(A) the complement of A with respect to S. The symbol

= will be used both for a relation between sets and for a re-

lation between elements: A = B will mean that both A c: B

and B = A , and a = b will ,mean that a and b are the same element ; it may be noted that equality (logical identity)

of elements ma.y often be replaced. by an equivalence relation, or at least by a regularl equivalenc e, but we shall not advert to this fact in the sequel. We also assume that the reader is familiar with su ch ordinary algebraic notions as that of arith- metical congruence, denoted by = .

lFor definition and properties of regular equivalence relatio::1S .. see Paul Dubreil, Algebre , v .. I (Paris: Gauthier Villars� 1946), pQ 80. For an axiomatic treatment of seml groups in wh ich an e�uivalence relation replaces equality , see H .. SQ Vandiver, The Elements of a theory of abstract discrete Semi-groups," Viertel ahrsschrift der Naturforschenden Gesellschaft in Zurich, 85 1940 , pp. 71-Se:- - 3

Defin1 tion 1., L. A semigroup is a system consisting of a non empty set S and a single-valued binary associative operation which associates with each ordered pair of elements of S an unique element of S " Ordinarily, we shall call this operation multiplication, and call the element as'sociated wi th an ordered pair of elements a, b their Eroduct, written a·b or abo In this notation, the associativity of the operation is de scr'i bed by the equality a (bc) = {ab)o for· all elements

a, b, c of S .. If ab = ba we say that the eiements a . 2 and b commute with each other; if ab = ba for all elements 3 a, b E S I we cal1 s a commutative semigroup. Occasionally we may find it convenient to write a ... b instead of ab, in which case we speak of an additive instead of a multiplica tive semigroup, but we shall do so only when referring to com mutative semigroupso The order of a semigroup is the cardinal number of its set of elements.

A few well-known examples of semigroups may be men tioned: (1) all integers, under addition or under multiplica tion; (2) the positive integers greater than some fixed inte

ger k I under addition or under multiplication; (3) all real functions on a given interval (finite or infinite) under

- multiplication, the product fg of two such functions being

2 It will be convenient to use the n9.tation "a, b E gft as an abbreviation for "a £ S and b ( Sit. 3No confusion ne�d arise from the use of S to denote the semigroup as well as the set of all its elements. 4 defined to be th e function whose val ue at any point a of the given interval is f(a')og(a) ; (4) ,the integers under mul tiplication modulo n ; (5} all m x n matrices (for fixed

" m and nJ over an arbitrary ring, under matrix addition;

{51 all square matrices of order n over an arbitrary ring, un'der matrix multiplication; (7) all single-valued mappings of a set E into itse1f, under iteration <.!.�., consecutive performance of mappings r. The first five of th ese are ex amples of commutative semigroups; the ftrst three are examples of infinite semigroups, and the fourth of a fi nite semigroup;

(5j and (6) are finite if the ring is finite, otherwise in finite; and (7) is' fini te if the set E is finite, otherwise infinite. Additional examples will be found in the Appendix, which is a list, compiled by K. 3. Carman, J. C. Harden, and

E� E. Posey, of all semigroups of orders less than five; we shall refer to these se migroups by number, semigroup m.n being th the n in the list of semigroups of order m •

If a subset 31 of a semigroup S is itself a semi group under the defining operation in S, $1 will be called a subsemi�roup of S 0 Since associativity holds throughout

S , it holds in SI I and a necessary and sufficient condition 81 that a subset of 8 be a subsemigroup of S is that Sl be closed -'llnder the multip11-Otrt1on defined in S. If 81 is a subsemigroup'of S and Sl c: S , we shall say that 51 is a proper sub semigroup of S. Clearly, if 82 is a subsemi group of 81 and 81 is a subsemigroup of S then 82 is a 5 subsemigroup of S. Thus I in Example (2) above, the even integers in the semigroup form a subsemigroup regardless of whether addition or multiplication is'ohosen as the defining operation, while the odd integers form a subsemigroup under multiplioation but not under addition; and the entire semi group (2) , in turn, und'er either operation is a subsemigroup of (1) under the same operation. The other examples listed above also contain. . noteworthy. subsemigroups: the continuous functions on th e interval are a subsemigroup of (:3); the in tegral multiples of d, where d is a divisor of n, are a subsemigroup of (4); the matrix all of whose elements are

. zero constitutes by itself a subsemigroup of (5); the non- singular matrioes in (6) form a subsemigroup, and the singu lar matrioes form another; in (7)' the one-to-one mappings oon stitute a subsemigroup .

It is obvious that the intersection of an arbitrary oolt�ction of subsemigroups of a semigroup 8 is empty or a subsemigroup of 8. But the union of subsemigroups need not be a subsemigroup; for example, in the additive semigroup of integers modulo 6, if 81 consists of the multiples of 2, and 82 of the multiples of :3, then 2 + :3 = 5 � 81 v 82 •

Definition 1.2. In a semigroup 8, an element e suoh that ex = x for every x £ S is oalled a lefb i�entitl element of 8 0 Dually, an element f suoh that xf = x for every x c 8 is oalled a right identity element of S. If e 6

is both a le ft identity element and a right identity element

of S # !.�o" if xe = ex = x for all xeS" then e is called a two-sided identity element of S , or simply an identity element of S oAsemig roup S may have any num ber of left (or of right) identity elements; but if it has at least one le ft identity element e and at least one right

identity element f then e = ef = f # whence S has an unique two-sided identity element and no other le ft or right identity element.

Definition 103. In a semigroup S # an element y such that 4 yx = y for every x ( S is called a le ft zero element of

S # and an element z such that xz = � for every x es

is called a right � element of S 0 If X1 = 'Ix = 2. for every x € S , so that z is both a left and a right zero element of S" then z is called a two-sided � element,

or simply a � element, of S 0 A se migroup S may have any number of le ft (or of right) zero elements, but if it has a le ft ze ro element y and a right zero element z then y = yz = 2 , whence S has an unique two-sided zero element and no other le ft or right zero element .

Definition 1040 An element a of a semigroup S is said 2 n to be idempotent if a = a 0 If a, b c S and a = b ,

' 4""Same ,"" writers pre fer" the term annihilator or �nnulator instead of "zero element, " presumably to avoid confusion with the identity element of an. additive s'6migroup . Cfo Vandiver,

.£E. cit., p .. 76. 7 5 where n is a positive integer, then a is oalled an nth root of b ; the roots of a two-sided zero element of

S are oalled nilpotent elements. An idempotent element a is obviously an nth root of a for every positive integer n. The square roots of a left or right identity element of

S are often oalle-d involutory elements. Clearly, al l left and right identity elements, and al l left and right zero elements, are idempotent, and the former are involutory.

Definition 1.5. If a and b are elements of a semigroup

E: :::: s , and. if there exists an eieme-nt x S suoh that ax b ,

a . :::: then is said to be a left divisor of b , if ya b for

. some y e S , them a is said to be a right iu visor of b , if a is both a left and. a right divisor of b , it is oalled a two-s1;ded divisor of b. If a is a left divisor of every element of S,we may oall it a universal left divisor of S; universal right di visors and universal two-sided. divisors of

S are defined similarly. If a � b � x and ax = b , we say that a is a proper left divisor of b ; proper right divisors and proper two-sided. divisors are defined similarly.

Every left (right} identity elemeht of S is of oourse a th universal left (right) divisor of S, and any n root

(n > 1) of an element b is a two-sid.ed divisor of b (in

5positive integral powers of an element of a semigroup n n l are defined in the usuat way (bl = b; b = gb - for n > 1) .and n n n obey the rules �b = tiM+ =.b tiM and (bM)n = bmn, eaoh_of wh ioh oan be derived from olosure and. assooiativity by.an easy induotive argument .. Non-integral.exponents and integral non positive exponents are not defined for elements of general semigroups. 8

particular, any idempotent element is a two-sided divisor of

itself). If S contains a right (left) identity element e

then every element a e S is a left (right ) divisor of i t-

�

self, for ae = a (ea = a)o But in the absence of the appro-

� priate identity elements, an element need not divide itself.

For example, in No. 4068 there are two right identity elements (1 and 2), which of course are universal right divisors of the

semigroup, and each is a two-sided divisor of itself since

each is idempotent; but neither is a left divisor of the other.

The element 3 in this semigroup is idempotent (in fact, it

is a two-sided zero element) and hence is a two-sided divisor

of' itself 0 But the semigroup has no 'left identity element,

and the element 4 " although a left divisor of itself, is not a right divisor of itself. It may be remarked that an ele ment (such as 4 in No. 4G44) may be a two-sided divisor of

" itself without being idempotent.

Definition 1.6. The usual definition of group may be formu

lated in our present terminology as follows: a group G is

a semigroup containing a left identity element e of which

every element of G is a right divisor. Under these condi

tions it follows readily6 that e is a (necessarily unique)

6For a neat proof, see Hans Zassenhaus, Lehrbuch der Gruppentheorie, v. I (Leipzig u. Berlin: Teubner, 1937}-,- po 20 rEnglish translation, The Theory of Groups (New York: Chelsea, 1949), same pageJ 9 two-sided identity element of G and that for each a ( G there is an unique inverse element a-l such that I l ? a- a = a a- = e • An equivalent definition states that a group G is a semigroup in which, for arbitrary a, b (G , the equations ax = b and ya = b are solvable; the unique.;. ness of the solutions is easily pr oved. In our present termi nology the . latter definition may be stated thus: a group is a semigroup in which each element is a two-sided divisor of every element. We note that a group cannot contain either a left or a right zero element, and that its only idempotent element is the identity. Commutative groups are usually re ferred to as abelian. If a group G is a subsemigroup of a semigroup S, we shall say that G is a subgroup of S •

In any group the cancellation laws hold: ax = ay implies x = y , and xa = ya implies x = y ; conversely, a finite semigroup in which both cancellation laws hold is a group,8 but this need not be true for infinite semigroups <"�.• .s., the multiplicative semigroup of all positive integers ) .

Definition l.? The product AB of non-empty subsets A . and B of a semigroup S is defined to be the set of all elements ab , where a , A and b E B 0 It is almost an

?The equivalence of these definitions was first noted by Eo Vo Huntington, ItSimplified definition of a group," Bull. Amer .. Math .. Soc .. 8 (1901-2), pp .. 296-300. .A. proof is given in Zassenhaus, £E.. cit., p. 3.

8For a proof, see Zassenhaus, loco cit. 10 immediate consequence9 of the associativity of the multiplica tion of elements defined in S' that the multiplication of subsets is associative: A(BC) = (AB)C • The following rules of calculation, which we shall Use constant1YI we shall list l without proof: O (1) if A �A and B � B , then 1 1 A B G; AB ; (2) if ?n is any c lass of subsets then 1 1

A • .}4",M = ,H".AM; (3) A' "� ...M C; ,.o"AM . The inc1usi on re- 11 1ation in (3) is n�t-in general reversib1e.

Definition 1.8. In terms of multiplication of subsets, sub semigroups of a semigroup S may be characterized as non 2 empty subsets A such that -A s:1 . If the stronger co n dition AS � A is satisfied, A is called a right ideal of s· ; if SA -= A then A is called a left ideal of S ; if

A is both a right and a left ideal of S it is called a two-sided ideal of S. If A is a two-sided ideal of S then SAS = (SA)S � AS sa A. Conversely, if SAS sa A and

S has a left identity element e then '.AS = eAS sa SAB SA, so that A is a right ideal of S ; dually, if S has a right

9 A detailed proof 1s given in ·W. A. Rutledge's Univer sity of Tennessee Masterts Thesis, Normality in Semigroups, 19481 p. 4. . . . 10 �. For proofs, see Dubrei1, .£E. cit., p. 31.

11! finite example in which the left-hand' side of the inclusion is empty while the right-hand side is not may be found in No. 4.49; it is discussed in detail in K. S. Carmants University of Tennessee Masterts Thesis, Semigroup Ideals, 1949, p. 11 . An example in which. both sides are non-empty is given by Dubrei1, �. cit., p. 32. 11 identity element and SAS � A then A is a left ideal of

S ; hence if S has a two-sided identity element and

SAS sa A then A is a tWo--sided ideal of S. But, in the absence of a.ppropriate identity elements, SAS Si A need not imply either SA s; A or AS sa A. For example, in the 12 commutative semigroup whose multiplication table is exhibited

below, let A be the subset [1 2 3 4J • Then 2 SAS :: S A :: r 1 2 4 5]· r 1 2 3 4J = (1 4J c· A ; but

AS == SA :: r 1 2 4 5 J ¢. A •

1 2 3 4 5 6

1 1 1 1 444

2 1 1 144 4

3 112 445

4 4 441 1 1

5 4 4 4 1 1 1

6 4451 1 2

However, it should be notioed that, for an arbitrary non empty subset A of S, the subsets SA , AS , and SAS

are respectively left, right, and two-sided ide�ls of S •

Any left (right) zero element of S is evidently oon tained in every left (right) ideal of S J and hence in every

12This semigroup was found in J. C. Harden's Univer sity of Tennessee Master's Thesis, Direct and Semidirect Products of SemigrOU�S, 1949, p. 14. It iathe direct product (a term we s all define later) of Nos. 2.1 and 3.2. 12 two-sided ideal; any two-sided zero element therefore lies in every idealo Indeed, a left (right, two-sided) zero ele ment is itself a right (left, two-sided) ideal; apd any left (right, two-sided) ideal consisting of a single element is a right (left, two-sided) zero element. . We cite without proof 13 the following well-known and fundamental theorems concerning ideals, of which we shall make constant use: ( 1) the produc t of two left (right, two- sided) ideals of S is a left (right, two-sided) ideal of S

. [this statement can obviously be extended to the product of any finite set of left (right, two-sided) ideals, taken in any order) ; (2) if L is a left ideal and R a right ideal

of S 1 then LR is a two-sided ideal of S , but RL need be neither a left nor a right ideal of S (cf. No. 4.64); (3) the intersection of any collection of left (right, two sided) ideals of S is either the null set or a left (right, two-sided) ideal of S, but the intersection of a left ideal and a right ideal need not be an ideal (cf. No. 4.111); (4) the intersection of any finite collection of two-sided ideals of S is a two-sided id.eal of S,and furthermore the intersection contains all the products obtained by multiplying all the ideals in the collection in any order; (5) the union

13These results are easily obtained. Perhaps the least obvious is the fact that (even in the absence of a zero element) the intersection of a finite collection of two-sided ideals is non-empty; a detailed proof may be found in Carman, 2E. cit., p. 13. 13

of any collection of left (right# two-sided) ideals of S

is a left (right# two-sided) ideal of S. That (4) fails

if the finiteness condition be dropped is shown by the fol

lowing example: let S be the additive semigroup of all

positive integers, and let Ak be the set of integers greater

than k (k = I, 21 •••); then each Ak is a two-sided

= ideal of S 1 but n Ak � 0 Furthermore1 (4) need not It�S hold for one-sided idealsl for left (right) ideals may be

� disjoint (cf. no. 4.66); howeverl it follows from (2) that

if S contains at least one left ideal L an d at le ast

one right ideal R then S contains at least one two-sided

ideall and si nce RL S; L" R it follows that every left

- ideal intersects every right ideal.

The principal left ideal generated El � element

b E S is the ideal Sb; if S contains a left identity

element e # then b = eb E. Sb , but in the absence of a left identity element we may have b � Sb (for example,

- the principal ideal generated by 2 in the multiplicative

semigroup of even integers). Dually, the principal right

ideal generated EI b is defined to be bS; and if S

contains a right identi ty element then b € bS . We also

define the principal two-sided ideal generated £I b to be

SbS , which contains b if S contains a two-sided identity

element.

Definition 1.9. A subset A of a set M is said to be

maximal in M with respect to � property P if A is 14

not a proper subset of any proper subset of M having prop

erty P..Thus" A is a maximal subsemigroup (subgroup"

ideal) of a semigroup S if A C B � S "where B is a

subsemigroup (subgroup" ideal) of S" implies B = S 0

Definition 1 .. 10.. If a is an element of a semigroup S"

and if" for all x" yES" ax = ay implies x = y " then

a is said to be le ft cancellable in S ; dually" if xa = ya necessarily implies 'x = y;;,�., a is right cancellable in S.

A cancellable element is one which is both left and right can

cellable. We have already remarked (Definition 1.6) that if every element of a finite semigroup S is cancellable then

S is a group.

2. Divisors of identity ..

We proceed now to a few theorems on left and right divisors of left and right identity elements.. In most cases we shall state only one of each pair of dual theorems ..

Theorem 2.1.. If e is either � left identi ty element or a right identitl element of � semigroup S" and if b is a root of e , then the left divisors of b form a subseroi- group of S •

n Proof. Let cx = dy = b. By hypotheSiS, b = e

. e for so me positive integer n � 2 , for if b = then

2 2 b = e = e .. Now

n l n l n l n (cd)( yb - x) = c(dy)b - x = cbb - x = cb x = cex .. But if 15

e is a left identi ty element then oex = o{ex} = ox = b ; and if e is a right identity element then oex = (oe}x = ox = b. Henoe, in either case, n l (cd)(yb - x) = b , whenoe od is a left divisor of b •

By left-right duality, the right divisors of a left or right identity element b form a semigroup; and si noe the set of two-sided divisors of b is just the interseo tion of the semigroups of left and of right divisors, the two-sided divisors also oonstitute a semigroup.

In partioular (and this is the oase in whioh we are most interested) the left (right, two-sided) divisors of a . left or right identity element oonstitute a subsemigroup of S 0

We remark that while in a oommutative semigroup the idempotent elements (and, more generally, the elements a n suoh that a = a , where n is a fi xed positive int eger) oonstitute a subsemigroup, this need not be the case in a non-oommutative semigroupo For example, in No. 4.113 the elements 1, 2, and 4 are idempotent, while 3 is not, but 4-2 = ;) 0

Theorem 2.2. If a semigrouE S oontains at ieast one left identity element, then all left identity elements of S have � oommon set D of left divisors, and D is exactly the set of all universal left divisors of S • Furthermore,

D is � semigro� and oo ntains as a subsemigroup the set E 16 of all left ident ity elements of S ; the set of le ft identity elements of D is jus t E , which g also the set of all idem potent elements in D. Proofo Let e be a left identity element of S, let dx = e , and let s £ S. Then d(xs) = (dx)s = es = s ,

. - whence d is a le ft divisor of the arbitrary element s, and hence is a universal left divisor of S. It is tri vial that, conversely, any universal le ft divisor of S is a left divisor of the arbitrary le ft identity element e. By Theorem 2.1, D is a semigroup ; and since an y left ident ity element is a left divisor of itself, E � D. But if e and f are left identity elements of S then ef = f E E , whence E is a subsemigroup of D " Let c be a left identi ty element

s . of D , let d E D , let e E E , and let E S , then dx = e for some x E S . Now cs = c(es) = ces = c(dx)s = (cd)( xs) = d(xs) = (dx}s = es = s ,

- whence c is a left identity e_lement of S , i.e., c E - - E ; the converse is trivial" Hence E is the set of all left identity elements of D 0 Finally, if d E D and s E S 2- then dx = s for some XES ; henoe if d = d we have 2 s = dx = d x = d(dx) = ds I whence dEE . Conversely, every element of E is idempotent, so we conclude that E is the set of all idempotent elements of D 0

Corollaryo A commutative semigroup S ha s � identity ele ment if and onll if S contains at least one universal 17 divisor. If S has an identity element e, the divisors of e � just the universal divisors of S, and they form � subsemigroup of S •

Proof. An identity element of S is itself a univer<- sal divisor of S •.. Conversely, if d is a universal divisor of S, then d divides itself, so that dc ::: d for some c E S ; and if s E S then xd::: s for some XES.

Now sc ::: (xd)c ::: x{dc) ::: xd ::: s , wh ence c is an identity element of the commutative semigroup S. The rest of the

Corollary follows immediately from Theorem 2020

Theorem 2.30 The left cancellable elements of a semigroup S form � semigroup, of which the right divisor s of any left identity element of S constitute � subsemigrouE·

Proof .. Let a and b be left cancellable elements of

S • Then, for any x, y E: S , ax ::: ay and bx = by each

::: .. = imply x y Hence (ab)x (ab)y implies a{bx) ::: a(by} .I

- which by hypothesis implies bx ::: by , and this implies x ::: y •

Hence the left cancellable elements of S form a semigroup

C..If e is a left identity element of S" the right divisors of e are left cancellable in S; for if sd = e and x" yES then if dx = dy we have x ::: ex = (sd)x ::: s(dx) = s(dy) = {sd)y ::: ey ::: y , whence d is left cancellable. But, by Theorem 2.1" the right divisors of e form a semigroup, and hence a subsemigroup of C • 18

No. 4065 assure s us that the right divisors of a left identity element may be a proper subsemigroup of the left can cellable elements" for here the on ly left identity element is 3 , and its onl y right divisors are 3 and 4" but every element of the semigroup is left cancellab leo We remark that if a semigroup S contains a left ident ity element and a right cancellable element then the left ident ity element is unique. For" if e and f are left ident ity elements and a is right cancellable in S"

then ea = a = fa " whence e = f 0

Corollary. In a finite commutative semigroup S I the �-

cellable elements form � grouE C ; and if S has an iden-

tity element e then the divisors of e form � subgroup of C 0 Proofo Since C is a finite semigroup (by Theorem

� 203) in which the cance llation laws hold, C is a groupo But" as is well known,,14 any subsemigroup of a finite group

is a subgroupo Hence» by The orem 203" the divis ors of e

constitute a subgroup of C 0 We sha ll improve upon the second half of this Corollary shortly" by showing that even in a non-commutative finite semigroup the right divisors of any left identity element

14See" for example" Garrett Birkhoff and Saunders MacLane" A Survey of Modern Algebra (New York: Macmillan, -- 1941)" po-143o 19 form a groupo But before proceeding to this and other theorems on divisors of identity in finite semigroups we present a theo 5 rem of Rees, 1 which we shall find useful and which may be of some interest on its own account. We have added slight embel lishments to the theorem, and we give a new and quite detailed proof. We first adopt

Definition 2. 1. A semigroup all of whose elements are posi tive integral powers of som e element b is called a cyclic semigroup. If all the elements in the infinite sequence b, 2 b , b3, • • • are distinct, then the correspondence bi � i 2 3 between the elements of th e semigroup b, b , b , • •• and the additive semigroup of positive integers is an isomorphism, i j i j for b b =b + �i + j ..

Theorem 2 .4. Every finite cyclic semigroup contains an uni que maximal subgroup, which is cyclic and is the intersection of all ideals of the semigroup .. 2 Proof. Let B be th e cyclic semigroup b, b , b3 ,

• • . generated by b • By the associative law, B is commu- i j tati ve. If i r j implies b r b for all positive integers i and j , then B is the (unique) infinite cyclic semigroup,

- and is isomorphic to the additive semigroup of positive integers. Hence if B is finite there exist positive integers i and j such that i � j and bi = bj.. Let n be the

15D• Rees, ffOn Semi-groups, " Proc. Carob. Phil. Soc., 36(1940), p. 3 88 . 20 n k least positive integer such that b = b for some k � n ; n i n i then if k i i � n we have b i b , for if b = b then

with botb i" n and k '" n , contrary to the choice 2 n-l 0 o of n Hence the elements b, b , • • , b are distinct.

Let m = n - k ; we show now that if j and q are positive j j integers and j > k then b bqm = b , and later that if j' k t then bjb i bj for every positive integer t. Assuming

j � k , we may write j = k + s , where s is a non-negative

integer. If s = 0 and q = 1 , qm k m k n bJb = b b = b +m = b = bk = bj 0 If s = 0 and q > 1 , ' k (q-l}m k we make the inductive assumption that b b = b • But k qm k (q-l}m m k m k b b = b b b = b b = b , whence the induction is k qm k complete and b b = b for every positive integer q. k t k Now if t is any positive integer and b b = b then if k s t k t s k s k s s > 0 we have b + b = b b b = b b = b + • Letting t = qm , we see that for any positive q and non-negative s,

k+s qm = k s b b b + 0 Hence ( 1)

Let p be a non-negative integer, and let p = qm � where and r are non-negative integers and + I q k+p k+r r " m 0 We show now that b = b • If p "- m then

= r p � and the result is trivial. If p _m and p =. 0

(mod m) then p = qm , q > 0 , and by (1) we have k p k+qm k qm k k o k+r b + = b = b b = b = b + = b 0 If p>m and

p =1= 0 (mod m) then q > 0 , r > 0 I and k+p k+qm+r' k r k r k r = = m = = b b b bq b b b b + Q Since r '" m I 21

k + r � k + m , and an immediate consequence is tpat the 2 n - 1 elements b .. b , • . .. constitut� the whole j i of B..Furthermore, if j '" k then b f. b fpr all

positi ve integers i, i:l j " It follows 1.mmedif3,tely.. as

forecast in the preceding paragraph .. that if j L k then j t j b b :I b for all positive integers' t •

Now we show that if u and v !!! �-negative k+u k+V integers then b = b if and only if u!!.v (mod m) •

tet u = q m + r {q � 0 , . a � r "'- m}' and l l l l

= k+u k+r v q m + r (q 0, a 4 r J:. m) . Then b = b l 2 2 2 � 2 k+v k+r and b = b 2 , whence if bk+u = bk+v then k+rl k+r2 b == b .. But a

r k + rl = k + r2 , rl = r2 .. Hence u - v = (ql - q2)m, so � that um, v (mod m) .. Conversely, if u,.v (mod m) then k+u k+rl k+r2 k+v rl = r2 , whence b = b = b = b "

We now prove a converse of (1) : if j and t are positive integers and bjbt = bj then t = qm for so me posi- t tive integer q. If bjb = bj then .. as we have seen,

k+{ -k+t) = +t t ( -k) � k ; hence b j bj = bjb = bj = bk+ j j ..

whence by the preceding paragraph j - k + t = j - k

(mod m) .. whence t= a (mod m) • But t '> a .. so that

= t qm for some positive integer q .. Combined wi th (1),

this result yields the following: if j and t !!! Eositive j t = integers then b b bj if and only if both j ). k and t = qm for � positive integer q. 22

Let G denote the set of m distinct elements. We have already seen that if i i .... k then b E: G :I whence G is closed under multiplication and so is a subsemigroup of B. We proceed to prove that G is a group; since B (and hence G) is commutative, we need only prove that for arbitrary • G the equa- i j tion b x = b is solvable in G. Let k = q' m + r, q' � 0, 0,r 4 m , and let q = 2+ qt . Then qm = {2 + qt)m = 2m + q'm = 2m + k - r, and, since r "" m , i j' qm � 0 • Since b , b C G , we may take k6i" k+m-l and k 6 j � k + m - 1 , whence j - i � 1 - m. Hence, since

- r � 1 - m , qm + j - i = 2m + k - r + j - i �. 2m + k + (1 - m) + (1 - m) = k + 2.

qM+j-i i +j-i qm+j qm j � whence b c G. But b bqm = b = b b j = b , qm+j-i so that x = b is the desired solution, and therefore G is a group and the solution is unique in G. The exponent pm+j-i qm+j-i qm+j-i is not unique, of course, for b = b for any integer p � q. It may be that qm + j - i > k + m - 1 , and it might seem desirable to give the solution in the form t b with k" t � k + m - 1 ; but this is not possible in general (l.e., for all i and j in the range from k to t qm+j-i k + m - I), as the following argument shows. If b = b

+ - = then t =. qm j - i - j - i (mod ml, whence t pm + j - i for some integer p 0 If P � q then pm � (q - l)m = (1 + q'lm = m + q'm = m + k - r , and

- pm + j i, m + k - r + j - i • But if r � 2 and 23 j - i = 1 - m then m ,. k - r ,. j - i � k - 1 I and t t = pm ,. -j i :!! k. - 1 I whenoe b ttl G • Henoe

where t L qm ,. j - i , is not a solution in G for all ohoioes of i and j unless r" 2. But if r "" 2 and

p = q - 1 == 1 ,. q' then pm = m ,. k - r � k ,. m - 1; and if j - i > 0 then t == pm ,. 'j - i > k ,. m 1 I whenoe t

lies outside the range from k to k,.m - 1. The remain-

ing possibility is the oase r < 2 I P < q - 1. But if p � q - 2 = q' I then r � 2 implies pm, q'm = k - r � k , whenoe t == pm ,. -j i. k ,. -j i I so that if j - i � 0 t then t � k ; henoe either t � 0 and b is not defined, t or 0 � t � k and b ¢ G 0 We oonolude that for no oyolio semigroup B is it possible to write the solution of i j t b x == b I for all i and j, in the form b where k�t�k "m -l ..

Defining q I as in the preoeding paragraph, to be 2,.(k - r) /m ,where r is the least non-negative residue of k modulo m' it is obvious that the identity qm element of G is b , where qm = 2m ,. k - r. Even when

2m ,. k - r is as small as possible (i.�", when r = m - 1 and 2m ,. k - r == k ,. m + I), qm > k + m - 1; henoe qm is

never in the range from k to k + m - 1 • However, i (q-l)m i (q - l) m = {I + q')m > 0 and b b = b for all (q-l}m i � k , so that b is the identity element of G ..

And sinoe 0 4: r t6. m - 1 and (q - l) m == k ,. m - r I we (q - l) have k,.1 � (q - l)m � k ,. m..He noe b m represents 24 the identity element of G in a form in wh ich the exponent lies in the ranee from k to k + m - 1 unless the identity k k+m element is b , in which case it is represented as b • Clearly" no smaller exponent than {q - l)m will suffice,

t = (q-1)m for if b b then t =: {q - l)m-=: a (mod m) , so -- that the greatest eligible t less than {q - l)m is

(q - 2)m ; but (q - 2)m = q'm , and if ql = a then (q-2) b m is not defined. i The inverse of an element b € G is given by (2q-1)m-i b ; for

i (2 q- 1)m-i = (2q- 1)m = (q-1)m +qm = (q-1)m qm = (q-1)m • b b b b b b b , and since i Ii. k + m and r � m we have

( 2 q - l)m - i '> {2 q - l)m - k - m = 2 qm - 2m - k = 2m of. k - 2 r > k" so that b(2q-1)m-i eGo No smaller exponent than i (2q - l)m - i will suffice for the inverses of all b ; for, as we have argued b e fare, if bibt = b(q-1)m then t c (q - l)m - i (mod m) , and the greatest such t � (2q - l)m - i is (2q - 2)m - i ; but if i = k of. m - 1 and r = m - 1 then

(2q-2)m-i = (2q-2)m-k-m+1 = 2qm-3m-k+1 = m+k-2r+1 = k-m+3 , whence if m > 3 then (2q -2)m - i 4 k and {2 q-2)m-i b • G • i We observe that, since bJb , G for all j > a and all bi E G, G is not only a group but an ideal

(necessarily two-sided since B is commutative) of B, 25

1 and hence 6 is the zeroid group (lo�ol the intersection of all two-sided ideals) of B 0 It follows thatl for every

j i ... G i b e- B and every b I the equation bjx::::: b has 1 a solution in G � It also follows 7 that any group con tained in B but not contained in G must lie entirely

G outside I and from th is we conclude that every group in

G • Gt B is a subgroup of , for if is a group lying out-

G h side of I and b i8 the element of Gt having the

( h 2- 2h greatest exponentl then b ) = b t- Gr since by hypoth-

G esis f. B and so h f. 1 0 Thus G is the unique maximal subgroup of B 9

FinallYI we prove that the group G is cyclic.

Among the integers of the form xm + 1 , select the least

+ one not less than k I and call it tm 1 0 Since m t l t = t l) tm ... 1 � k I b m+ EGo Now (b m+�_ bm( m+ is the

G identity element of since tm + 1 � k 0 On the other

tm+l s(tm l) hand, if (b ) S = b ... is the identity element, for

+ some positive integer s I then s(tm 1) -5 0 (mod m) ; andl since tm + 1 is prime to m, this implies 8 S 0

(mod m) I lo�ol s::::: hm for some positive integer h 0 s Hence the least positive integer s such that (btm+�

l 6For definition and properties of the zeroid group I see A. H. Clifford and Do D. Miller, "Semigroups having zeroid elementsl" Amero_Jour. Math.1 70(1948), pp 0 11 -250 7

l7Ibidol po 1230 26 is the identity element of G is m. Thus the order of the + + element btm l equals the order of th e group , whence btm l generates G, and G is cyclic .

Corollary 10 Every finite semigroup contains at least one idempotent element.

Corollary 2. Every subgroup of a cyclic semigroup is cyclic. Proof. The infinite cyclic semigroup has no idempotent element and he nce no subgroup , whenc e the corollary is true vacuously. In a finite cyclic semigroup, we have shown that every group is a subgroup of a cyclic group, and it is well known18 that every subgroup of a cyclic group is cyclic ..

Definition 2 ..2 . We shall say that an element b of a semigroup is of finite order if the cyclic semigroup generated by b is finite, and that the order of this cyclic semigroup is the order of b 0

Lemma 2.1.. If a univers al left divisor b of a semigroup

S is of finite order, then the cyc lic semigroup B gener ated � b is a grouE. Proof. By Theorem 2 ..4 , B contains an unique maximal subgroup G consisting of the powers

• • 0 I where n is the least positive integer

18 See, for example, Zassenhaus, �. cit., p. 15. 27 n such that b = bk for same k � n. Now if k = 1 then

G = B and our result is attained. If k > 1 � we show first that bk is a left divisor of b • Since b is a

universal left divisor of S � we may write b = bd for some d e S 1 and d = be for some c c. S • Now 2 2 b = bd ;:: b(bc) ;:: b c , whence b is a left divisor of b 0 -l We make the inductive assumption that bk is a left divi- -l sor 0 f b ; th en, f or some a. S , we h ave b = bk a •

But a = br for some rES 1 whence -l -l b = bk a = bk br = bkr . Thus by complete induction we have shown that bk is a left divisor of b. Now let bq (k � q � n) be the identity element of G. Then q bqb = b (bkr) � (bqbk) r = bkr = b. But we have seen in i i proving Theorem 2.4 that bqb = b if and only if i � k , i 1.eol if and only if b e G. Hence b E: G and G = B •

Theorem 2.5. A semigroup S has � left identitl element if and only if S contains a universal left divisor of finite order.

Proof. The necessity is immediate, for any left identity element is a universal left divisor of S and generates a group of order 1 . To prove th e sufficiency, let b be a universal left divisor of finite order. Now, by Lemma 2.1, b generates a group G; let e be the identity element of G, and let s 8 S 0 Then bx = s for some x f: S 1 and es ;:: e{bx) = (eb)x = bx = s , whence e is a left identity element of S • 28

Corollar�. A finite semigroup S has � left identity ele-

S has a universal le ft divisor.

Lemma 2 .2. In a finite semigroup S I an element b is left cancellable in S if and only if b is a un iversal le ft divisor of S • Proof of necessity. Let the distinct elements sll •••I sn be all the elements of S. By hypothesis, i -:J j implies (i, j = 1, . . . , nL whence the n elements bSl' • • • ,bsn are distinc t and there

s fore constitute all of S. Hence bx = is solvable for every s E S ,whence b is a universal left divisor of S. Proof of sufficiency. If b is a universal left divisor of S I i .�., if the equation bx = s is solvable for every s c S , then there are in S exactly n distinct elements bXl' . . . I bXnl where n is the order of

S , and i -:J j implies xi -:J Xj • Hence the elements xlI • • • , xn constitute all of S I whence bx = by im-

x == £ plies y for any x, y S I !.� ., b is le ft can cellable in S.

Theorem 206. The righ t divisors of a left identity element of � finite semigroup S form � subgroup of the semigroup of left canc ellable elements of S. Proof. Let e be any left identity element of S. By The orem 2.3, the right divisors of e form a subsemi group of the semigroup of le ft cancellable elements of S • 29

But in a finite semigroup, by Lemma 2.2, the left cancellable elements coincide with the universal left divisors; and, by

Lemma 201, each universal left divisor in a finite semigroup generates a cyclio group. Hence each right divisor of e generates a cyclic group. Since the right divi sors of e form a semigroup, any positive integral power of a right divisor of e is also a right divisor of e, whence all elements lying in the cyclic groups generated by right divi sors of e are right divisors of e. Hence if r is any righ t divisor of e and k is an y positive integer then

= k e yr for some Y ES. Now if i is the identity e18- ment of the cyclic group generated by r, we have

k k k e = yr = y(r i) = (yr )i = ei = i , whence e is a common identity element for al l the cyclic groups generated by right divisors of e. Now let Re be the set of all right divisors r ' •••,r of e I and let R I •••, � l m I be the respective cyclic groups generated by these right

= v .•. \wi '�'C: di visors 0 Then Re Rl Rm.Let a, b Re ; -1 then a € R (1 tIi j " m) and we may denote by a the j , inverse of a in the group R • Now j

l = l = l � a(a - b) (aa- )b = eb b , and a- b Re since Re is a semigroupo Hence ax = b is solvable in Re for arbitrary l l a, b ti' Re And (ba- ) a = b{a- a) = be = b , the last

E � Ia equality arising from the fact that b Rk (1 k m) and e is the identity element of Bk ; hence the equation 30

ya = b is solvable in Re 0 Therefore Re is a group, as was to be proved.

We note that an example may be given (the direct 19 product of Nos. 2.1 and 4057) to show that the right

divisors of a left identity element need not all generate

the same cyclic groupo We note also that in a finite

semigroup the left di visors of a left identity element

( = universal left divisors = left canc�llable elements )

need not form a group; example: No. 40&5. In fact, we have

Theorem 2.7. If the left divisors of � left identity ele

ment e of a finite semigroup S form � group, then e

� the only left identity element of S.

Proof. By Theorem 2.2, all left identity ele-

ments of S have a common set D of left divisors. But

every left identity element lies in D I and each such ele

ment is idempotent. Since a group can contain only one idem

potent element, there can be only one left identity element

e in S I and it is the identity element of th e group of

left divisors of e •

By left-right duality the foregoing theorems yield

Theorem 208. In � finite semigroup S with a two-sided iden-

tity element e I the left divisors of e, right divisors of

e , left cancellable elements, and right cancellable elements

19 The multiplication table of this semigroup may be found in Harden, ££. cito, po 200 31

all coinc ide and constitute a subgrou� of S 0

3. Divisors of zero.

Theorem 3.1. The left divisors of a right zero element of a semigroup S form a le ft ideal of S • Proof. Let z be a right zero element of S, let

£ dx = z , and let s S. Then (sd)x = s(dx) = sz = z , whenc e sd is a left divisor of z. Hence if D is the set of all left divisors of z we have SD ,; D , i.�., D is a left ideal of S. We note that the le ft divisors of a le ft zero element of S need not form an ideal, nor even a subsemigroup, of S..Fo r example, in No .. 4.63 the element 2 is a left zero element , and its le ft divisors are 2, 3, and 4; but

4·2 = 1 •

. ... Theorem 3.2. If !. semigroup S has a two-s ided zero element z form a left ideal of S , the right divisors of z form!. right ideal of S, and the

. . . . two-sided divisors of z form!. subsemigroup 2! S.

--- Proof.. f'he first two statements are immediate from The orem 3. 1 and its le ft-right dual . Let L be the left ideal of left divisors of z and R the right ideal of right divisors of z..Then the set of all two-sided divi sors of z is the intersection t� R. But any le ft or right ideal of S is a subsemigroup of S , and the intersection of subsemigroups, if non-empty, is a subs�migroup. 32

Now L 1\ R 1- ¢ since z c L "'" R I whence L..I'\R is a

subsemigroup of S •

We observe that the semigroup of two- sided divisors

of a tw o-sided zero element ne ed not be a two- sided ideal of

S. An example may be extracted from the transforma tion semi

group of degree four ( = semigroup of all single-valued mappings

of a set of four ob jects into itself), the multiplication table

of which has been given in full by P�sey;20 in Posey's notation ,

the exemplary subsemigroup consists of the mappings Al l '7 1 B2, B8 1 B271 B47 1 B551 Bl15 1 B1211 °11 041 0271 0331 0341

055 1 0611 and Dl • Among the divisors of a two-sided zero element are

the nilpotent element s. In a commutative semigroup S the se m n form a subsemigroup , for if a, b C S and a = a = z I where z is a two-sided zero element of S , then nm ron mn m n = a b = (a ) (b ) = z·z = z • In a non-commutative semigroup , however, this need not be the case. For example,

in the semigroup whose multiplication table is exhibited be-

low the element 1 is a two-sided zero element and 2 and

2 2 = 3 are nilpotent, but ( 2 03 } = 4 4 •

2O:E 0 E. Posey, .. Endomorphisms and Translations of Semigroups, Un iversity of Tenne ssee laster 's Thesis , 194� 1 2 345 33 1 1 1 2 1 III 1 4 1 2 3 1 1 3 1 4125 1 4 1

5 1 1 3 1 5

We may find an example in the multiplicative semigroup

of all second-order square matrices over the ring of int egers

to sh ow that the left divisors of a nilpotent element need

not form a semigroup .. Let

(- 10 -20, (-2 4 ) N = 5 10) , B = 1 -2 ..

Then N is nilpotent , and A and B are left divi sors of N since -18) (-1O -20) (-2 12) _ _ (� �)(-� 14 - 5 10 - 1 -t)(i 1 • 10) = But AB (-� ° ' whence AB cannot be a Ie ft divis or of

N , for if X is any matrix then the second row of ABX must

consist entirely of zeros .

4. Un ion, intersection, product, and direct product of prime ideals .

Definition 4.1. A left (right , two-sided ) ideal P of a semi-

group S is said to be Erime if a, b E S and ab E P

jointly imply that either a E P or b E P..Cl early, a prime ideal might be defined equivalently as an ideal whose 34 complement is either empty (in case the ideal is S itself, which is obviously prime ) -or a subsemigroup of S..It is also clear that a left (right, two-sided ) zero element of S is a prime right (left, tw o-sided) ideal of S if and only if it has no proper divisors o

Theorem 401. The union of an arbitrary collection of prime right ideals of a semigroup S is � prime risht ideal of S .. Proofo Let [R�J be a collection of prime right ideals of S , where L ranges over an index set M of arbitrary cardinality . Then if a, b E S and ab E �VH p�

E we must have ab PL, for some L € M • But P1. is prime,

e whence eithe r a E P� or b PI. .. Therefore either a E U PI. or b E U P� , whence U PI. is prime. Simi- larly for left and tw o-sided prime ideals . The uni on of two or more ideals may be prime, howe ver, even though none of the given ideals be prime ; for example, in No . 4.66 each of the elements 1 and 2 is a left zero element and hence a right ideal, and neither is prime sinc e · 3°2 = 1 and 401 = 2 ; but their union is prime . We now raise the question whether the product of two or more prime right ideals need be prime . Without further hypothe ses, the an swer is negative , for in No . 4.71 the right ideals [1, 3, 4] and [4] are prime, while their product [1, 4] is not . On the other hand, in No. 4.67 the product [3, 4J of the prime right ideals [2, 3, 4J 35 and [4J is also prime . We observe that in both these ex amples one of the fac tors is properly contained in the prod

uct, and we find that when this situation does not obtain we

can find necessary and suffi cient conditions that a produc t

of prime right ideals be prime .

Theorem 4.2. f � Eroduct of a finite set of prime right ideal s of a semigrouE S is Erime, then the Eroduct ideal

contains at least one of the given ideals .

Proof. Let Rl , •. 0 , Rn be prime right ideal s of K S , and let DR be an abbreviation for the product J i

:!ff. • Rn (1 .6 k n ) We wish to prove that if � n V Ri is prime then Rk .s; 17 Ri for some k (1 � k �n) . Suppose, to the contrary, that 'n Rk $. RTJ i for all k = 1, . , n. Then for each k the re is an element such that Now 11 'rl 11 a a = a E R J whence E since l • ".2 i 11 i TI i 11 al ;. V Ri and t Ri is prime . Now If we make the in- " duc tive assumption that a • rr a = k '11'+1 i '1'l 1\ " since a R we have TT a E 1JR 0 Hence by com- k � 1l i /1.. 1 i i " plete induction we conclude that an E 1l Ri ' contrary to

our choice of an ' and the theorem is proved. We have already cited an example (No. 4.71 ) which shows that the converse statement "if a produc t of prime right ideals - contains one of them then the product is prime" is false . How- ever, we may state Theorem 402 in a slightly different form, 36 and combine it with the tr ivial observation that if the product of prime ideals is one of those ideals then the product is prime, to obta in

Theorem 4 ,,3� If £ .E.E0duc! of .E�!!.l!£ Eight J-deals .� f a semi group does !2ot EEoperll ccp.tain any of them � then the Erodu./·,t is prime if and only if it is � of' the .a!ven ideals " The proof of Theorem 402 holds mutatis mutandis for two-sided idea1 s, and for these we may drop the hypothesis concerning proper inc lusion in Theorem 4.3. For, as we have

- already noted (in (4) of Definition 1.8), any product of two- sided ideals is contained in their intersec tion, so that the product cannot contain any of them properly and th� inclusion hypothesis in Theorem 4.0 is sat isfied automatically. Hence we have

Theorem 4.4. A product of prime tw o-sided ideals of � semi group 1! prime if and only if g 1! � of the given ideals .

Corollar1_ If a Eroduct of prime �H o-sided ideals of a semigro't!E. .!! primel then th� product of the ideals is just their intersection .

rrhe converse of thL J corollary fails s however , because the in�er8ect ion of the ideals may be a proper subset of each of them . For example, in the commutative semigroup No . 4.23 the ideal [1 1 4J is both the product and the inter'section of the prime ideals [1, 2, 4J and [1, 0, 4] , but is not prime . 37

The intersection of prime right ideals need not be

prime ; for example, in No . 4066 the prime right ideals

[I, 2J and [I , 'OJ intersect in the right ideal [ lJ I which is not prime . However, the que st ion whether the non

empty intersection of a set of prime right ideals of a semi

group S is prime in S may be reduc ed to the question whether the intersection is prime in the union of the given

ideals .

The orem 405.. The intersection of � set of prime right ideals

of � semigroup S l! a prime right ideal of S if and only

if it � a prime right ideal of the un ion of the given ideals.

Proofo The necessity being obvious , we proceed to prove the sufficiency. Let [RJ be a set of prime right ideals of S,whe re L ranges over an arbitrary index set M By hypothesis, n R1. is a prime right ideal of LfM

R(, and hence is non-empty. Let x, y E S I xy E (\ R 0 y,., L Ii M t Then if x n R the re is some R;.. ( A E M) such that � {€� l x R and if y f n R then there is some R E M) I A l eM l ft (}J. such that y l Rfl • Hence x, y ¢ R� A R,M • But C xy E n RL RfJ"\ fI'and both R" and R,M are prime , !.Hl R whence either x E RI- or y E R" , and either x E R fA or y €. H}4 Therefore , we have either x E R1- and y E Rft or else y E Rt. and x E R,M • In either case, x, y E R" v Rf'A S U R 0 But, by hypothe sis, n R LeM I. LIM l is prime in U R I whenc e either x E n HI.. or LEM !. �'M

Y E n Rio 0 Hence n HI. is a prime ideal of S .. I.EM LfM 38 Clearly the foregoing theorem is equally valid for tw o-s ided ideals » but for finite sets of these we have the following mo re specific resulto

Theorem 4.,60 The lntersec clon of � finite set Ef prime tw o� sided ls.eals of a semigrour. S if and E..:..g...1: 5_ f !!:e is I:r ime intersection is � of the �iv�� eala Q

ProofG The suffic iency is tr ivial. To prove the neces-

. ? • sity, let AI' , An be prime two-sided ideals of S , and recall that their intersection mus t be non-empty. If 'n k = I I A Ak for all 1, . n then for each k I(t i -f 'T1 there is an element ak e Ak such that ak ,. IJ Ai • Now , exactly as in the proof of Theorem 4.2 , we may prove induct ively that if "..a i E Q Ai then an E Q Ai ' contrary to the choice of an I and hence conclude that 71 'II 1\ a Ai . But "IT a i E A Ai • Hence 11 i ¢ Q i V i � Q 1'\ the supposition that k = • . n n Ai Ak for all I , . , 1 -f is contradicted, and our the orem is proved.

Definition 4.2. The direct product of two semigroups S and

T is defined to be a system whose elements are all the ordered

t pairs (s" t) I w1 s E: S and E T " and with multiplication defined by ( s11 tl ) (s21 t2 ) ;:;; ( s ls2, tlt2) " The definiti on may be extended readily to the direct product of any finite set of semigroups o It is we ll known (and easily ' 21 proved ) that a direct pr oduct of semigroups is a semlgroup ,, 39 and that a subset of such a direct produc t is a left (right, tw o-s ided ) ideal thereof if and only if the subset is the di rect produc t of le ft (right, two-sided ) ideals of the factors ..22 Our purpose in introduc ing di rect multiplication at this point is simp ly to point out that the direct product o.f prime ideals of two semigroups need not ·be prime in the direct product of the semigroups . For example, the commutative semi groups No. 2 . 2 and No . 3016 each have a zero element , and the ordered pair of zero elements is (necessarily ) a zero element of the direct product,23 and hence a two-sided ideal; but it haa proper divisors and therefore ·ia not prime .

5. Prime ideals in aemigroups with identity elements .

In this secti on and the next we shall be concerned with maximal ideals as well as with prime ideal s, and a few words concerning 'the property of maximality seem to be in order here .. We note first that while a two-sided ideal which is a maximal right ideal is necessarily a maximal tw o- sided ideal, the conve rse need not hold o For example , in No . 4.74 the ideal [3 " 4J is a maximal tw o-s ided ideal but is properly contained in the proper maximal right ideals [I, 3, 4J and

[ 2 , 3, 4] •

22Ibid .. , p. 7. 23Ibido, p. 15, the multiplication table is given . 40

A similar situat ion arises whenever we discus s maximality with respect to the conjunction of two or more properties" In particular, under Definition 109 a maximal prime right ideal of a semigroup S is a prime right ideal which is not properly contained in any proper prime right ideal of S. Such an ideal need not be a maximal right ideal of S for example D in No o 4025 the maximal prime two-sided ide al [4J is properly contained in the proper two-sided ideal

[lp 2, 4J 0 To avoid ambiguity: we might speak of a maximal prime right ideal as maximal am ong the prime right ideals of S .. and similarly for left and two-sided ideals o On the other hand .. a prime maximal righ t (left" two- sided ) ideal is a maximal right (left, two-sided ) ideal of S which is also a prime id eal of S ; this concept will recur frequently in the sequela We shall ha ve occasion to use repeatedly the obvi ous fact that if a left (right , tw o- sided ) ideal A of S

, contains every proper left (right , two-sided ) ideal of S, then A is an un ique maximal left (right, tw o-sided) ideal of S " We no te in pass ing that maximal ity of ideals (even of prime id eals ) is not pres erved under direct multiplicationo 24 For example , in the direct product of NosG 202 and 3016 the zero eleme nt is not a maximal ideal, al though the zero 41 elements of the re spective factors are both maximal and prime.

Definition 5,,1.. We extend De finition 1.5 by introduc ing a new kind of divisoro If a, b E S we shall say that a is an

internal divisor of b l. f there exist elements x, y f: S

such that xay = b..An eI.ement which is 'not a left (right ,

internal ) divisor of D may be called a left (right , internal ) non-divisor of b ..

Theorem 5 ..1. The set of all right �-divisors of !£ arbitrary element b of a semigroup S either is empty (in which � b is a left zeroid element of S and hence25 lies in the intersection of all le ft ideals of ideal of S ; and dually .. The set of all internal non-divisors of b either is empty or is � two-sided ideal of S .. Proof. Let a be a right non-divisor of b and let x E S .. Then if sa is a right divisor of b , 1"� ", if

= E x{ sa) b for some x S , we have {xs)a = b , contrary to the hypothe sis that a is a right non-d ivisor of b"

Hence sa is a right non-di visor of b for every s E S and every right non-divi sor a;1 ..� " , the right non-divis ors form a le ft ideal of S ..

an Now let a be internal non-divisor of b I and let s e S..The n if sa is an internal divisor of b we have

X b = x{ sa)y = {xs )ay for some I yES; and if as is an

25 Clifford and Miller, £Eo cit., p. 118. 42 internal divisor of b we have b = u(aa )v = ua (sv) for some u, v E S a In each case the hypothe sis that a is an internal non-divisor of b is contradic ted . There fore sa and as are int ernal non-divisors of b,wh ence the internal non-divisors of b constitute a tw o-sided ideal of

S 0

It may be noted that when an element b has at least one right (left, internal ) non-divisor, the left (right, tw o sided) ideal of all such non-divisors need not be prime ; for example, in No o 4069 the right divisors of 2 do not form a semigroup , whence the left ideal of right non-divisors of 2 is not prime o It may al so be remarked that the right non- divisors of b,when they exist, ne ed not form a right ideal of S ; for example , in No o 4084 the only right non-divi sor of 1 is 4,whi ch is a right zero element, and hence a left ideal, but is not a right ideal o

Lemma 5010 If a right ideal A of � aemigroup S contains a left divisor of a left identity el eme nt e of S, then

A = S ; and duallyo If a two-sided ideal M of S contains an internal divisor of e , then M = S 0

Proof 0 If a EA· and ax = e for some xES , then e = ax E AS c::: A , whenc e S = eS SAS S;; A , so that

A = S If m E M and xmy = e for some x, YES , then e = xmy E SMS 5 M , whence S = S( eS ) = SeS S 3MS S M , so that M = S 0 43 Throughout the rest of this section, we shal l be con cerned with non-divisors of one-sided and two-sided identity elementso To avoid tiresome repetition, we prescribe the following notation for the rema ining developments of this section: in a semigroup S with le ft , right , or two-sided identity element e Ii L wtll denote the set of all right �- divisors of e , R the set of �l l left non-divi sors of e , and T the set of all internal non-divisors of e 0 It should be noted that L� R, and T are proper subsets of

e S ; for eee = ee = e , whenc e has at least one int erna l divisor , at least one right divi sor, and at least one le ft divisor o We shall use this fact repeatedly wi th out further explicit mention of it in the proofs o

Theorem 5Q20 In a semigroup S co:r:taining a left identity element e,if R 1 ¢ then R is an un ique proper maximal right ideal of S I ,!lnd is prime Q

Proof" By Theorem 501", R is a right ideal of S 0 If A is any right ideal of S , and A¢R , then, since

R is the set of all left non-divisors of e I there is an

(; element E A such that ax = e for some x E: S 0 He nce I

0 0 by Lemma 5 1 , A = S Therefore R contains every proper right ideal of S and hence is an unique maxima l right ideal of S 0 By Theorem 201, C(R) is a semigroup, whence R is a prime ideal" 44

Corollaryo If a semigroup S contains a le ft identity ele ment e and contains a proper right ideal A, then S con tains an unique proper maximal righ t ideal, which is prime .

Froofo Since A is proper by hypothesis, it follows from Lemma 501 that A contains no left divisor of e 0

Hence A � R , whence R � ¢ and Theorem 502 yields the des ired conclus ion o

The orem 5030 In � semigroup S containing .! le ft identity element e I if L � ¢ then L 1s .! proper prime left ideal of S; if S is finite, L is a maximal left ideal of S • ------

Proofo By Theorem Sol, L is a left ideal of S I and by Th eorem 201 it is prime o If S is finite, suppose

L CB S S I where B is a left ideal of S •.Th en, since

L is the set of all right non-divisors of e I there is an element b E B such that yb = e for some y E. S , whence

= e yb E SB S; B 0 But , by Theorem 206, C ( L ) is a group 0 with e as its identity element, whence C ( L } e = C ( L ) •

Hence C C L ) = C C L ) 0 e s: C ( L ) 0 B � SB s= B , whence

LvC ( L ) S; B 0 But L uC( L ) = S , whence B = S 0 There- fore L is a proper maximal left ideal of S 0

We note that the proper maximal ideal L,un like R, need not be un ique ; see, for example, Noo 4.570

Theorem 5040 In a sem1group S containing � le ft identity element e, if T � ¢ then T � an unique proper maxima l two-sided ideal of S 0 45

Proof o Le t T � ¢ and let M be any proper two -

sided a1 of S..If M � T then M contains at least

one int e rnal divi sor of e;when cel by Lemma 501� M = S •

T conta ave�y pr oper two-sided ideal of S,whenc e

T Is an un ique proper maximal tw o'-sided ideal of S <>

identit:r e16-

un -"_.an -...... - proper ptBximal tw,?,", sided ideal "

Proof o By Lemma 5alt M contains no proper internal

e C divisor of � whence M T, T � ¢ I and Theorem 504

applies"

The orem 5050 In � samigrouE S containins � left identitl

5'leme n� a I L::¢ if and qnll if S is � group ; if S

R :: T = ¢ ; T S R I _a _n _d _i _f s is fini te

than T = R L Q

Proof o If S is a group � then S contains no proper left or right or two- side d ideals; henca L::R :: T == ¢ 0 If

L =.: ¢ then every element of S is a right divisor of e I

e i () eo eV6ry lament ha s a Ie ft inverse with respect to the

e left iden '� ity element J whence S is a group o If t € T and tx = e for some x E S then

p = :: == T etx �, at jx tx e 5 whence t ¢ , hence if t E T

tx for x 0 then f: e all E S " !"��I T £. R If S is c fini te then" by Lemma 2,,2 and Theorem 2,,6, R L � 1IIbence

- - - 0 :f if R ¢ then T -f(-¢ And if R i ¢ (still assuming 46 s to be finite ) let a E R ; then, if xay = e for some

eeL) C (R x, yES 3 we have ay E C } ,1.�., ay is a left divisor of e ,whence a is a left divi�or of e, contra.ry to the hypothesis that a E R. Therefore if S is fini te then R £; T , so that T=R S;; L.

We may now classify semigroups containing left identity eleme nts according to the emptiness or non-emptiness of the sets L, R, and T. Although ! priori eight cases could arise, it follows from the relations exhibited in The orem 5.5 ' that only four of these can arise: (1) L = R = T = ¢, S being a group in this case and only in this case; L R = T = 65 (2) 1 ¢ I ¢ , No o 4 .. being an example of this (3) 4.60 case; L 1 ¢ J R 1 ¢, T 1 ¢ , No. being an ex (4) R ample of thi s case; L 1 ¢, 1 ¢, T = ¢ , there being no finite example of this case according to The orem 5.5.

We have not found an example of Case (4), but we can give an example to show tha t if S is infinite we may have T ¥ R •

Let S be the semigroup of all single-valued mappings of the set N of non-negative integers into itsel f, let l denote the identity mapping n -7 n (a two-sided ident ity element of S ), and for any elements t:p,1JI E S define q:>)V to be the result of performing first VI J then cp.No w let cP be defined by cp (n) = n + 1 for even integers n

= Q and cp(n) n for odd n Since �(N) 1 N J rp cannot be a left divisor of L .. Let y be defined by �(n) = 2n for all n e: N , and let -x.. be defined by (n) = 0 for 47

even n and X(n) = (1/2)( n - l) for odd n. Then

for all n EN, wh ence Xcpi.f= l • Therefore cp e R while

The se find ings ; toge ther with Th eorem 2.2, may be

summarized in

Theorem 5�6o If � semigroup S con�aining � le ft identity

element is not � group ; then S contains at least � proper

prime le ft ideal, and, un less every eleme nt of S is a left

divisor of all elements of S , contains � unique proper

maximal �ight ideal . If S is finite , and is not � group,

it contains a proper maxima l le ft ideal , and its proper maxi ma l righ t ideal , if � exists , is al so an unique maximal two

sided ideal.

For semigroups ha ving two<�sided ident ity' e lements, we may comb ine the foregoing theorems with the ir left -right

dua ls to ob tain

Theorem 5Q7o In � semigroup S ha ving a tw o- sided ident ity

element , if S is not a group then S contains proper maximal

left anq Ei�h � ideals� each unique and prime, and an unique max imal tw o-sided ideal whi9h � prime if S is finite .

We can improve upon the last part of The orem 5 .. 7 by

proving

Theorem 5.8. If T is the maximal proper tw o- s ided ideal of

� S semigrou£ wit h two-sided ident it� element e I and if 48

C(T) is finites then T is a Frim� ideal and CCT) is a grou:p..

Proof. Let A be the set of all left divisors of e, B the set of all right divisors of e a and D the

e 0 set of all internal d sors of Then C(T) = D and

AvB D c. By Theorem 2�3 and its dual� A ana B are finite semigroups ; and each eleme nt of A is a left divisor of th e right identity element e; and each element of B a right divisor of the left identity element e 0 Hence� by

Theorem 2q6 and its dual " A and B are groupsQ Therefore", -1 for arbitrary a E: A .: there exists an element a E A

-1 such that a a = e :I so that a is a right divisor of e and hence A S- B .. SimilarlY:I B 5 A :I whence A = B ..

. Now let d E D J then xdy = e for some X s Y E:. S " and necessari ly x, y e CCT) since x is a left and y a right divisor of e" Hence xd e A = B s whence xd I

a and therefore d:lis right divisor of e 0 Therefore

D S; B , whence D = A = B, OCT) is a grouP I and T is prime ..

The proof just completed shows that the set D of all divisors of a tw o�s idej ident ity element of a semigroup is a g:r olJ.p if D is finite ., A familiar example is the group [± 1, ± iJ in the semigroup of Gaus sian integers o

6. Prime ideals and maximal idealsQ

In this concluding sectl on we investigate the conditions und er which a maximal ideal is prime� wi thout assuming 49

the existenc e of an identity e1ement Q

t c omnlement of M is commu t atl va aw.J. grouE S, and !! ,,"�,� --,,- --

is contained in S2 then M is a prime ideal of S 0

ProofQ The theo rem is trivial if M = S Q Le t M

be a proper maximal id,: al an d suppose M is not prime ;

then there exist elements x and y (not necessarily dis-

tinct ) in C(M) suc h that xy ¢. M" Let x be some fixed

eleme nt of C(M) such that; for at least on e element

y £ C(M) I xy E M I and let Y be the set of all suc h

elements y 0 We di st ingui sh tw o cases : (1) Y c: C (It)

and (2 ) Y = C (M ) ..

Case (1): In this case we shal l show that M v Y

is a two -sided ide al of S,!� �Q$ that (M U Y)S � M VY

�

an d SCM v Y) S 1M! u y" Sinc e S =:: M v C (M) , we may

- accomp lish thi s by show ing that (M v Y)M � M v Y and

M(M u Y) � 1'4 VI Y I whi ch is obvious since M is a two-

sided ide al ; and that (M v Y)oC(M) � M uY and

C (M ) .. (M U Y) C M V Y. Now if Y It Y and c £ C eM)

then x( yc ) ::-.: {xy)c £ 11 , for xy eM; hence either

yc €' M or yo .;; Y }i so the.t (M v Y) oC (M) � M U Y ..

And since we have as sumed CeM) to be commutat ive,

x ( cy) = x(yc ) = ( xy ) o EM"wh en ce either cy c M or

cy £ :( � and th erefore G(M)· (M v Y) � M v Y � Th is

comple tes the proof that M \.) Y is a tw o-sided ideal ..

Since Y is non-empty and M ....., Y ;:: ¢ I M eMU Y ; 50 and M v Y C S since in this case we have assumed

Y <:: C(M) � Therefore M v Y is a proper two-s ided

ideal of S 6 properly containing M I contrary to the hypothesis that M is maximal ..

Case ( 2 ): If Y = C{M) then, by virtue of the hypothesis that C(M) is commutative , yx = xy £ M for all y E. C (M ) I whence xS � M and Sx S;; M..In par 2 2 c ticular I x M J whenc e x � x .. But since 2 CeM) C S , there exist elements a and b (not necessarily distinct) in S such that x = ab ; and both a and b lie in CeM) since x t£ COlI) .. Now either a � x 2 = = ;::: . or b � x I for if a == b x then x ab x , say a � x 0 Hence M v x is a two-sided ideal of S I for

(M v x)S ;::: MS v xS � M v M � M � M v x and

SCM v x) = 8M v Sx '= M \wi M � M s;. M v x..But

M c. M u x since x E C(M) ; and M u xe S since a € C(M) and a f x..The refore M v x is a proper two- sided ideal of S I properly containing M I contrary to the hypothesis that M is maximal ..

The supposition that M is not prime having led to a contradiction in both cases, we conclude that M is prime .

We observe that in the above proof of Case (1) no use 2 was ma de of th e hypothesis that C(M) � S .. Hence by com

bining the maximality of M I the commutativity of C(M) , and the condition defining Case (1 ) , �e ob tain the 51

Corollary 10 If M is a maximal tw o-sided ideal of � semi

gr oup, if the comp lement O(M} is commutat ive " and if" for

E: g ) every element x C(M} , x C (M ¢: M , then M is a

prime idea10

The condition X"O( M) � M whi ch suffices in the

presenc e of the maxima1 ity of M and the commutativity of

the comp lement may be contrasted with the much stronger de

fining condition for prime ideals" viz "" [C (Mn 2 � OeM) "

An immediate cons equenc e of The orem 6 .. 1 is

I � S 82 = 8 Corollary 2 .. n semigroup such that " every

maximal two-sided ideal whose canp1ement 1! commu tative 1!

prime .. In par ticular" in � semigroup having either a �

E! � right identity element every maximal two-sided ideal

� In whose comp leme nt commutative is pr ime .. --a commutative � � semigroup such that S2 = S (and fortiori in commuta

tive semigroup with identi ty )" every max ima l ideal is pr ime .

We now prove a converse of Theorem 601" and are able

to drop the hypothe sis that the comp lement of the ideal be

commutative 0

6 M � The orem ..2 . If is maximal and prime two-sided ideal

of � sem�grouE S" then the comp lement of M is contained

in 82 ..

Proof .. Supp ose C(M ) � s2 1 and le t x be any ele

ment of C(M) such that x � S2 .. Then x 'I x2 " and

x2 E C (M) since M is prime Q Now the set 52

M u x2 v Sx2 v .xZS v Sx2S is a tw o-si de d ideal of S ; M for }IS " M I 8M S;; I S( S.x2 } =S2x2 G Sx2 , 2 {x2S )S = .x2g2 � x2S I ( Sx2 )S = ' S(x2S) = Sx S ,

S (Sx2S ) = S2x25 G 8x2S I and (8x2S)S-= Sx252 g S.x2S •

But M C (M \oJ x2 v Sx2 v .x2S vSx2S ) si nc e x2 � M ;

an d (M v :x2 u ·Sx2 v x2S v Sx2S ) cS si nce x f. M and

x f. 52. The refore M v x2 v - Sx2 v x2S V S.x2S is a. prope r

tw o-side d ideal of S properly containing M, contrary to

the hypothe si s that M is maxi mal. We note in passing th at if A is an y le ft or ri ght

ide al of a se migroup S I and if C(A) c S2 , then

D(A) C S2 ; for S2 is a two -sided ideal of S I whence

S2 "':' A f. ¢ while O(A) "..... A = ¢ I and so O(A) I- S2 •

- - He nc e the conc lusion of The orem 6.2 could be strengthe ne d

to re ad "the complem ent of M is contained prope rly in 82 ". 2 We note also that, in any sem i group S I the ide al S is

prime if and only if S2 =S ; and that 82 is maximal if

and on ly if 0(S2) contains at most a si ngle element, the

"only if" follo wing from the fac t that if x, y £ 0(S2)

- �nd x I- y the n S2 v x is a prope r ideal of S prope rly

containing S2 •

The orems 6.1 an d 602 toge the r yield

� The ore m 6Q30 In semigroup S I a maximal two-sided ideal whose complement is commutative 1! � prime ide al if 'and on ly

if the comp lement is contained in S2 0 53 It is a well-kn own theorem26 in the theory of rings that if M is an ideal in a commutative ring R then the re s idue class ring RIM is a field if and only if both

(1) M is a maximal ideal of R and ( 2 ) x2 £ M implies x e: M 0 Another familiar result is the theorem2'7 that an ideal in a commutative ring R is a prime ideal of � if and only if RIM is an integral domain. Since every field is an integral domain (but not conve rsely), it follows that if conditions (1) and ( 2 ) hold then M is a prime ideal ; the converse fails because, although every prime ideal sat isfies condition ( 2 ), a prime ideal need not be maximal. However, it follows that if M is a maximal ideal in a com mutative ring then M is prime if and only if ( 2 ) holds, and thi s result we shal l prove both for one-sided and for two-sided ideals in semigroups, dispensing with the hypothesis of commutativityo Before preceeding to the proofs , we remark that it is easy to show28 that in a commut ative ring with identity element condition ( 2 ) follows from condition (1), whence we conclude that in such a ring every maxima l ideal is prime . This result carries over to commutative semigroups with identity , as we have seen in Corollary 2 to Theorem 6 .li

26Neal H .. McCoy, Rings and Ideals (Buffalo: Math . As sn .. of Am er o, 1948), ppo 80-81: 2'7Ibid ., po 98. 28Ib id .. , p. 81. 54 ind eed, we ha ve seen ( loc. Cit. ) that in any semig roup with id entity a maximal id ea l whose comp lem ent is commutative must be prime. The com muta tivity ca nnot be dropped altog ether however, unless some such condition as finitenes s be Bubsti- tuted fo r it, as in Theorem 5.8. We sh all give an example later to show that in an infinite semigroup with tw o- sid ed identity element a maximal tw o-s ided id eal need not be prime.

We now st ate and prove the theorems announced in the foregoing paragraph, giving separate proofs for one- sided and tw o-sid ed id eals ; neither proof seem s readily adaptable to the other theorem.

Theorem 6.4. A maximal right id ea l M of a semigroup S is 2 a prime right id eal of S if and only if x E. M imp lies x € M • Proof. The neces sity of the condition follow s imm ed i- ately from the definition of prime id eal . To prove the sufficiency, suppos e that M is not prime. Then there exist elements x and y lying in the complement of M such 2 that xy C M • Since x £ M implies x 4£ M by hypothe- 2 2 Sis, x y and x , y c OeM) Now x yS . for if r .. f- , 2 E S ( = ( = (£ x = ys for some s th en xy ) s x ys) x o (M) , contrary to our supposition that xy lies in the right id ea l M" Hence we have x ¢ M, X:fY ,and x � yS ,

� (If v v v whence x Y yS ) and therefore (M y \wi yS) c::. S •

And M c:: (M v Y v yS) , for y ;, M..But th e set 55

M v Y vyS is a right ideal of S, for 2 (M v y 'owl y8 )8 = MS ....; y8 v y8 J:;. M u Y '" y3 •

Therefore M v y v y3 is a proper right ideal of 8 3 properly containing M, contrary to the hypothesis that M is a maximal right ideal of S

In th e case of two-sided ideals, the above proof breaks down because, although a two�sider1 ideal is a right ideal, a maximal tw o-sided ideal need not be a maximal right ideal, 1.�., a tw o-sided ideal may be contained properly in a proper right ideal even though it is not a proper subset of any proper tw o-sided ideal. For example, in No. 4 0 67 the proper two-sided ideal (3 , 4J is maximal, but is proper ly contained in the proper right ideals [I , 3, � and

[2 , 3, 4] •

Lemma 6.1. If M � � two �sided ideal of � semigroup S, 2 and if x E:. M implies X EM, then xy E:. M implies yx IE M ; if x € S , if Y is the set of all elements yES such that xy E: M , and if Z is the set of all elements z £ S such that zx EM, the� Y = Z .. 2 Proof. If xy € M then (yx) = {yx )(yx } = y{xy ) x E M I whence yx € M. Hence if y € Y then y c. Z , and vice- versA, so that Y = Z •

Theorem 6.5. A maximal tw o-sided ideal M of a semigrouE 2 S is � prime ideal of 8 if and only if x E M implies x E. M • 56 Proofo The necessity of the condition is immediate.

If M = S the sufficiency of th e condition is trivial, so

0 E: we assume M e. S Let x C (M ) I and le t the sets Y

4E and Z be defined as in Lemma 601. For any s S , Y E: Y I

E: X(ys) = (xy)s eM, whence ys Y ; and by Lemma 6.1 I

E (sy)x = s(yx) M I whence sYeZ = Y. The refore

YS � Y and SY S Y , wh ence Y is a two-sided ideal of

c 2 • S 0 Now x .. Y ; for x Y implies x = xx M , whence x € M 0 Therefore Y C! S , and since M � Y

E. and M is maximal we have M = Y 0 Hence xy M implies y € M ,whence M is prime 0

From Theorems 603 and 605 we have at once the

Corollary ., If M is a maximal two-sided ideal in a semi group S I and if the complement C(M) of M is oommuta 2 2 tive , th en C(M) 8 if and only if x £ M implies - s;. ------

We present in conclus ion the example, to which we have alluded earlier, of a (necessarily infinite and non commutative ) semigroup with two -sided identity element in which the maximal proper two-sided ideal is not prime . Let N be the set of al l non-negative integers , and let S be the semigroup of all single-valued mappings of N into it se lf, of which L : n -+ n is a two-sided identity element ., Let E be the class of all 9" € S such that

,(N) is a finite subset of N; � is clearly a proper two sided ideal of S I and we proc eed to prove that it is 57

maximal by proving th at every elem ent in its complement is

an inte rnal divisor of L "

Now C(E) consists of th e mappings � of N into

itself such that feN) is infinite . Since N is countably

infinite, every infinite subset of N is countably infinite.

Hence, given any two infinite subsets , and N of N, � 2

there exists at least one element IT E C (�) such that o-( 41" Nl) = N2 and is a one-to- one correspondence between

Nl and N2 " Let 'I? E;: C (� , , and let � (N) = M • For

each m E: M let Am be the complete inverse image

"a. (m) of m under p Since � is single-valued, the

sets Am are disjoint ; and since � is everywh ere defined

on N, the sets Am exhaust N Q Now, calling upon th e

Axiom of Choice, we may choose from each Am exactly one

element m ' ; let M' be the set of elements so chosen"

Then th e correspondence m � mT is a one- to-one corres-

pondence between M and M' " But, since M is countably

infinite, there is a one-to-one correspondence between N

and M. Hence there is a one-to-one correspondence between

N and M' 0 Let � be any su ch corresp ondence; regarded

as a mapping of N into itself, 41 Ii. C (!!) 0 Now

�'J'(N) = fJ[I.f(Nfl = �(M') = Itt , and fJ'I' is a one-to-one mapping

of N into itself o Let � be any element of C{�) such

that if n E: N and tl4Ie n ) = m then 'X. em) = n 0 Then, for

any n €. N , if t.p (n) = m' and �(m' ) = m and · "/.. em) = n - then ltlfJ(n) = n , whence XP'P = (. " Hence p is an internal 58 d '1vi s or of (. whence no proper tw o�slded ideal of S can contain � " 1'hol'� fore � is an unique maximal ideal

of S 0

Fi nally we sr: ·.:;w that £ i 3 not a prime ideal hy axhibiting 9. pa.i.r of elements of erE) whos e pro duct lies in E 0 Let � be defined by t:p(n) :: n + 1 fo r n even an d �(n) = n for n odd; then feN) 1s the set of all odd positi ve in tegers 0 Let 'P be de fin ed by y (n) = n ( for n even and � n ) = 0 for n odd; then ¥(N) is the set of al l evan non-negativa integers . Thus cP Ii "If E-. C(�) If

l: . But ;1 for any n E N $ 'fCP(n) == 0 ; hen ce yep €:

Th ere fore � is not a prime ideal of S • BIBLIOGRAPHY BIBLIOGRAPHY

Birkhoff, Garrett and MacLane , Saunders . A Survey of �odern Algebra. New York : Ma cmillan, 1941 .

Carman , Ke nneth Sc ott . "Semigroup Ideals." Master 's thesis, Knoxville, Un iversity of Tennessee, 1949 .

Clifford, Ao H. and Miller,1 Do D. "Semigroups Having Zeroid Elements," The American Journal of MathematiCS, 70:117-25, 1948 .

Dubreil, Paul . Alg�bre , v. I. Paris: Gauthier-Villars , 1946 .

Harden, John Charles . "Direct and Semidirect Produc ts of Semigroups ." Master 's thesis, Knoxvi lle, U�iversity of Tennessee, 1949 .

Huntington, Eo V. "Simplified Definition of a Group ." Bul letin of the Ame rican Math��a t��l Soc ietI, 8:296-300, 1901-2 0

McCoy" Neal H. Rings and Ideals . Buffalo: The Mathematical Assoc iation of America, 1948 .

Posey, Eldon Eu�ene o "Endomorphi sma and Translations of Semi groups . Master 's thesis, Knoxville, U�iversity of Tennessee, 1949 .

Rees, Do "On Semi-groups ." Proceedings of the Cambridge Philosophical Soc iety, 36 :387-400, 1940.

Rutledge, William Ao "Normality in Semigroups ." Master 's thesis, KnOXVille, University of Tennessee, . 1948 .

Vandiver, Ho S. "The Elements of a Theory of Ab stract Discrete Semi-groups ." Yi erteljahrsschrift der Naturforschenden Ge sellschaft in ZUrich, 85:71-86 , 1940.

Zassenhaus , Hans . Le hrbuch der Gruppentheorie , v. I. Le ipzig Uo Berlin : Teubner, 1937 . [Engl ish transla tion, The Theory of Groups . New York: Chelsea, 1949 J

APPENDIX APPENDIX

In this appendix, compiled jointly by Messrs . K. S. Carman1 Jo Co

Harden, and Eo Eo Posey, are listed what the authors bel ieve to be all

distinct semigroups of orders two, three, and four� and their proper su�

semigroups o Two semigroups are regarded as distinct if no isomorphism

(or anti-isomorphism) can be set up betwe en theme

Commutative semigroups of these orders (under the name of "finite

ovan ) were listed by A. Re Poole in his dissertation, l but were omitted 2 fro� the publ ished version . We have constructed independently all

semigroups of orders two and three (non-commutative as well as commuta-

tive), and our list of commutative semigroups of those orders agrees

perfectly with Poole !s.

In addition to constructing the non�c�tative semigroups of

order four, we have examined Pool.eis list of commutative semigroups of

that order, and have found wh at appear to be a few errors " It should be

pointed out that the copy of ¥oole 's thesis av ailable to us was a carbon

copy borrowed from the library of the California In stitute of Teclmology, with the subs cripts in the semigroup multiplication tables filled in by hand, so that it is entirely possible that tbJa. El,l"r0r8 noted do not occur

in the original dissertation; indeed, some of these errors are obviously me rely mistakes in copying . We did not construct the commutative

lA o Ro Poole, "Finite Ova," (Doctoral Dissertation 9 California Institute of Teclmology, 1935)0

2Ao Ro Poole , "Fini:t.e. Ova, " !!!: .American JOUl'!lal .2! Mathematics, ,9 :23-32, 19370 2a semigroups of orde F f011:!" in dependently, and cannot cl aim that our exam i- nation of Poole �s w-as suificiently exhaus tive to preclude the possibil ity that it c()n tnins errors ather than those noted below.

til u2 u3 112

u3 111 u3

l.l 'Ill u2 1 u2 which is non=commutativE' since 114u3 F u3u4 0 An examination of associativity shows that if uhuJ :=: then the sys tem is not a semigroup re gardl.ess to u3 :;J we have a COnl,-',u tative semigroup, distinct from all others in

Poo1e �s list.!' wh ich is given in our list as No . 4 .. 38 .

The mul ti.pl icatdon table of .t-'oole �s R 430 is identical with that of his R this semigroup appears in our list as No .. 4.90 Changing a 417 , in single subscript yields a new commutative semigroup, No o 4 ..5 our list:

it H 4 ,,5 , 430 \ U1 ul 11 11 u1 ul u1

11J ul 112 u1 u2 ul u2

u 11--. u u u u u l .I.. l 1 1 1 1

1 u u u u 111 u2 12 l 2 1 4

as Foo1e ls T and T are identical� and appear in our list 42 41." No. 4.220 Two changes produce a new coa�utative semigroup, our 3a

T T 42 , 43 4,, 23

U U U u ul tt 1 l 1 4 l 1lJ.. 114

Ul U2 Ul u4 UJ. u2 'llJ. 114 U U Ul l l u4 u1 UJ. u3 u4 u u4 nil, u4 til ul� u4 4 u4

Multi pl icatiJn x'oole t s list fails to satisf"J the

a,3sociati 'ri ty , but Gan be so modified as to become a commut,a-.

tive semi:;roup 9Ir0arently dis ti.llct from others in that list; we

include it as 403

ul 111 ul ul

u1 u1 'UI. u1 u u l ul 2 u2 u u l l u2 u2

Pool,; ' ;;;: st:mli.f,Yonps If r!.2 and I� h8 seem to be 'isomorphic ; this

our Po ole 's tables and an s!3migroup :Xl as No.1> 4.hJQ iao- mo rphism bet.wl)cm

,) ,,\, R 48 'th u u,.... U E ,. 11 u 11 tt 1 11]. .,t, l 1 1 UJ. 1 l u 11 u 11 ): uL. u u l 2 1 2 ( llJ. 2 2 llJ. .. u 11 ul u3 u3 11.'3 .( 2 1 -Ij'3 U". )10 u, u.., U u -E u - l'T l "- u4 't1�J 3 tt:l ,

In Table $.� '1re ! Le;ted four semigroupa of order two , of which

Table IIa contaLTJ.B the llon-Cl}!m;r;l·- only No . 2.3 non�commu.tat ive o six tative semi,�r(ups of o,,:,,jer three (Nos o 3 ..7, 3.9, a�2, 3,,13, 3014, 3.15'J 4a

and the twelve commutative semigroups of order three, the latter being

identical. (to within isomorphism) with PooleI s ova of order three " In

Table IlIa we list the SS commutative semigroups of order four (Nos � 4�1-

4.55) and the 66 non-commutative semigroups of order four (Nos . 4 ..56 - 4.121) .

In Table IVa, the columns are headed by the reference numbers of

the semigroups of order two , and the rows are labeled with the reference

numbers of the semigroups of order three. If No . 30z (71- 1, <>o�, 18)

. -

contains k subsemigroups isomorphic to Noo 20Y (y = 1, 00", 4) , the

letter k is placed at the intersection of the row 30z and the column

2 ..yo Table Va exhibits similarly the subsemigroups of orders two and .

three of the semigroups of order fouro Subsemigroups of order one are

simply the idempotent elements of a semigroup, and are not listed

separately since they are visible immediately upon inspection of the

multiplication table of the semigroup o

It would be futile to expect tables of this. sort to be entirely

free from errors 0 The authors have checked their. work repeatedly, and

have rectified several mistakes. but others hav e doubtless escaped their

notice 0 They will appreciate receiving eo�tians ,� users of the

tables Q

..- Sa

TABLE Ia

SE1ITGROUPS OF ORDER TWO

2.1 2.2 2.3 2 .. 4

1 2 1 1 11 11 2 1 1 2 2 2 11

TABLE IIa

SEMIGROUre OF ORDER THREE

3 .. 1 3.2 3.3 3 .4 3 ..5 306

1 2 2, III 123 113 113 III 211 III 2 3 1 11:3 123 III 211 112 312 331 3 3 1 113

3.7 308 3 .9 3.10 3.11 3Q12 ,1' 1 11 1 11 III 1 1 3 123 III 121 112 222 113 213 222 131 123 1 11 333 333 113

3.13 3.14 3.15 3 ..16 3.17 3.18

III III 113 III 1 11 III 222 222 223 122 121 11 1 123 333 333 123 113 1 11 6a

TABLE !IIa

SZ:lIGRuUl"S UF uRDER FOUR

401 402 4.3 4.4 405 4 .. 6 4.7

1 III III1 III1 III1 III1 1 1 1 1 1 1 1 1 1 2 2 2 1 2 1 1 1 1 1 1 1 214 121 2 1 211 III1 123 3 1 III 112 2 1 1 1 1 1 1 1 1 III 1 III 2 1 2 3 3 1 1 1 1 112 2 1 h. 1 1 1 214 III 3 112 2

4.8 409 4010 4011 4012 4013 4014

1 1 1 1 1 1 1 1 \/ 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 1 vI 2 1 1 111 3 121 2 214 3 2 1 h 3 2 314 231 2 212 2 1 III 11 1 1 3 4 2 1 3 h 1 2 3 1 2 4 312 3 1 2 3 4 1 3 1 2 1 2 1 2 4 3 1 2 4 3 2 1 4 4 4 4 1 2 3 4 1 2 4 3

4.15 4016 4017 4.18 4019 4,,20 4621

1212 1 2 1 1 ,1211 1 2 3 1 1 2 3 3 1234 1 114 2 1 2 1 212 2 212 2 2 1 3 2 2 1 3 3 2134 1 2 3 4 123 4 1 2 3 1 1 2 3 4 .3 333 333 3 3 3 3 4 1 314 2 1 4 3 1 2 1 h 1 244 1 2 3 h 3 3 3 4 444 4 4 441 4022 4.2.3 4,,24 402.5 4,,26 4,,27 4 ..28

III4 1 1 1 4 113 3 1 2 1 4 1 2 4 4 1 1 1 1 1111 1 2 1 4 1 2 1 4 1 2 3 4 2222 2 2 2 2 1 2 1 4 1 2 3 4 J 1 114 1 1 3 4. 3 .3 1 1 1 2 1 4 4 2 1 1 1 1 1 1 131 3 444 1 4 444 .3 411 h 2 it 1 4 211 141 2 143 2 4029 4030 4.31 4032 4033 4,,34 4.35

113 1 1 211 1 212 1 2 2 2 1 1 3 1 113 3 1233 1 2 3 2 212 2 2 1 2 1 2 III 1 131 113 3 231 1 3 3 1 3 1 2 1 1 1 212 2 111 3 31.3 3 3 1 1 312 2 1 2 3 2 121 1 2 1 2 1 2 1 1 1 113 2 3312 312 2

4 ..36 4 03'7 4038 4 ..39 4 ..40 4 .41 4,,42

, 1 III III1 III 1 III1 1 111 1111 1 111 1 2 3 3 1 2 3 U 12.32 1 2 3 4 1 2 2 2 1 2 2 2 '1 2 2 2 131 1 1 3 1 1 131 .3 1 3 1 1 1 2 2 2 1 2 2 2 1 2 3 3 1 311 14.11 1 2 .3 2 1 L. 1 .3 1 2 2 2 122 3 1 2 3 4

\ 7a

TABLE IIIa

,;:i unDER FOUR {Continued)

-- I/, 4.43 4 .44 4 .45 4 ,,)J6 4 ..47 4,,48 h.49

1111 1 111 1111 1 111 1 1 1 1 11 11 1 1 1. 1 1221 1 2 2 1. 1 2 2 4 1 2 2 2 1 2 2 2 1 2 1 1 121 2 1 2 J 1 1 2 34 1 2 3 4 1 2 3 2 1 2 3 1.). 1 1 3 1 1 1 3 3 III4 1 1 4 1 1 4 h 1 1 2 2 2 1 2 4 2 1 1 1 4 1 2 3 h

4 .50 4 .51. ).). .52 4 . 53 4.514 4.55 4056

1 III 1 1 1 1 1111 1 1 1 1 1111 1111. 1 1 1 h 1. 211 1 2 1 1 1 2 1 2 1 1 1 1 1 1 1 1 111 1 1 1 1 h 1 1. 3 1 J. 1 3 h 1 1 3 1 .1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1., 111 1 1 1 4 1 121 2 1 1 2 1 11 1 2 1111 1. 11.4

4.57 h .58 4.59 4 ..6 0 4 ..61 4,,62 4463

11.14 1 1. 3 4 1 1. 1 1 1 2 3 4 1 1 3 4 1 1. 1 1 1 1 1. 1 1. 2 3 4 1 1 3 h 222 2 2 1. 3 4 1 1. 3 4 2 2 2 2 2 2 2 2 1 3 2 4 3 334 111.2 3 3 3 4 3 3 3 4 3 3 3 3 1 2 3 4 1 1. 1 4 3 3 3 4 444 4 443 h 4 h 3 4 1. 2 2 h 214 3

4 064 4.65 4 ..66 4 .. 67 4.68 4069 h Q 70

113 3 1 2 3 4 1 1 1 1 113 3 11.33 1 1. 1 1 1 1 1 1 2 244 2 143 2 2 2 2 2 2 3 3 2 2 3 3 112 2 2 2 2 2 1. 1. 3 1. 2 3 4 1. 1. 3 3 3 3 3 3 3333 1 2 3 4 3 3 3 3 2 244 2 1 h 3 2 2 4 4 4 4 h 4 443 3 1 2 3 4 3 3 3 4 2 4 ..71 4 .7 4 ..73 4,,74 4 ..75 4 ..76 4 0 T?

111 1 1 1 3 h 1. 133 1 1 3 4 113 3 1 1 1 1 1 1. 1 1 2 222 2 2 3 4 2 2 33 2 2 34 2 2 33 1 1 1 3 222 ? 1 1 1 1 3 334 3 3 3 3 3 3 3 3 3 333 1. 1. 1 3 1 1 1. 1 4 444 3 3 3 4 3 3 3 4 3333 3 333 1. 2 3 4 1 1 1 1.

4.78 4.79 4 . 80 4 .. 81 4.82 4 .. 83 4.8h

1 1 1 1 1111 1 1 1 1 1111 1 111 1111 1 1. 1 4 2 2 2 2 2 2 2 2 2222 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 4 1 1 1. 1 1 1 1 1 1 1. 1 1 1. 1 1 3 1 2 3 1 111 3 1 314 2 222 1. 1 1 4 2 2 2 4 2 2 2 4 1. 1 1 1 1 214 1 1 1 4 8a

TABLE IITa

Sm.,.qGROUPS OF ORDER FOUR (Continued)

4.85 4686 4 ..87 4.88 4 089 4 .. 9.0 4�91

1111 III 4 1111 1111 1111 1114 III 1 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 III 3 III 4 1111 III 3 III 3 1 1 3 4 1131 1 1 3 4 444 4 1 2 3 4 1 234 III 4 444 4 1114

4.92 4693 4 ..94 4095 4.96 4.97 4,,98

1111 III1 III 1 1111 1 11 1 1111 1 1 3 4 2 2 2 2 2222 2 2 2 2 2 2 2 2 2 2 2 2 222 2 2 234 1 1 3 1 1 2 3 1 1 1 3 3 1133 3 3 3 3 1 2 3 3 333 4 2 2 2 4 1114 1134 1 2 3 4 1 2 3 4 1 2 3 4 443 4

4099 4 .. 100 4 ..101 40102 40103 40104 4 .. 105

III 4 1134 1111 1 L1 4 1134 1 1 3 4 113 4 , 2 2 2 4 2 2 3 4 2 222 2 2 2 4 2 234 2 2 3 4 1 1 3 4 1 2 3 4 333 3 3 3 3 3 333 4 3 334 3 3 3 4 333 3 4 444 444 4 4444 4444 4 4 4 3 444 4 333 3

4.106 4 .. 107 4,,108 4 ..109 4 ..110 40111 4 .. 112

1 234 111 4 1111 1111 1111 1111 III 1 2 1 3 4 1 214 1211 1211 1111 1 214 111 2 333 3 131 4 1311 1 3 1 1 1111 1311 1 1 1 1 333 3 444 1 131 1 1111 1 2 3 4 1111 1 2 3 4

4.113 4.114 4.n5 40116 4.117 4 ..118 40119

1111 1111 1111 1114 1111 1111 1111 121 1 1 2 1 1 1212 1 2 1 4 1211 1 2 1 2 111 2 1 3 1 1 1311 131 3 1 314 1311 131 3 1133 1 3 3 4 1134 1 214 4444 III 4 1 2 3 4 1 1 3 4

4.120 4 ..121

1 2 2 1 1 111 211 2 11 11 211 2 III 2 1 2 3 4 1 111 9a

TABLE IVa

J:'ROlJER SUBSEMIGROlrl>S uF SEllIGROUJ:'S of ORDER THREE

2.1 202 2.3 2 ..4

3.1 1

3 .2 1

3.3

3ol.j. 1 1

30> 1 1

3 06 1 1 1 3.7 1

3.8 1 1

3 .9 1 1

3 .10 1 1

3 oil 1 1 1 3.12 1

3.13 2 1

3.14 .3

3.15 2 1

3.16 3

3 0 17 2

3.18 2 lOa

TABLE Va

PROPER SUBSEMIGROUPS OF SEMIGROUPS OF ORDER FOUR

20 J . 2 b 4. 1 2 3 4 1 3 4 �. 7 " 9 I�� III [.I." �J 1 3 1 1l41-'02.r1 ll' 2 1 2 � _2 3 1 2 4 1 2 � � 1 5 3 1 _t£ b 1 1 1 1 7 2 1 � B 2 1 9 1 2 1 =r�t 10 1 II 2 f$ 12 ....:1: l- �3 � 1 14 2 � � 15 2 � 16 � 2 2 J. 17 1 .l. 2 .L IB ,1. .L J. 19 1 3 J. 1 20 1 3 2 � 21 1 1 1 1 � 1 22 � 1 1 1 1 1 23 .5 � � 24 � � 1 1 2� 1 1 1 � J. 1 20 1 � � J. 27 1 1 1 � 28 1 1 J. .L 29 1 � J. � 30 J. 2 2 1 31 � 1 1 � 32 � 2 33 1 � � 1 34 � 1 � 3.5 � 3b � 2 J. 37 1 2 � 3B � 2 J. -� m .L � � � 39 - 40 � 2 2_ � 41 1 � � J. l1a

TABLE Va

PROPER SUBSEMIGROUPS OF SEMIGROUPS OF ORDER FOUR (Continued)

20 3 0 40 1 1 2 6 8 �O [11 12 15 16 17 18 2 .3 4 3 4 .5 7 9 113 1'14 42 () 4 43 4 1 2 1 1 1 Wl .3 2 1 45 :3 46 .3 � 1 1 1 1. 1, 1. �7 j !ll:l :3 3 �9 4 2 1 �O ,c; 1 2 1. 5l 2: 1, 1 1. 1 52 2 1 1 53 J 2 54 :2 � 1. 55 :3 1 1 3 5_Q 1 1 - 51 1 1 1 1 1 >tl 1. J. 1 1. .l. >9 J 1. 1, 1, 60 1 1 1 Z 1. 2 61 2 1.� 1 62 2 J 1 oj 1. G 1 -hti Q4 J 0> 2: 1 66 1. 2 2 0" Z 'd, G. 1. _Qt 2 1 1. 2: 1 b� 2: 1, 1 2 1. 0 1. J 2 l i '1. 0" i 2 1. '2 2 1 '3 2 �#2 . )'- 2 !4 <::. 75 1. 1 2 1 :z ]. 1 r6 2 J., 2 gm� 1. '1 'tl 1. c: 1 1 1 1 1 79 1+ 1 1:10 1 til 1 m 1 1. tl2 2: J., l. IT 1 1. 12a

TABLE Va

PR01'ER SUBSEJ.'::IGROUPS OF SEMIGROUPS OF ORDER FOUR (Continued)

"Lo 30 40 1 2 3 4 ! '2 3 4 5 6 7 ts 9],0 'll r.l.G: I.l.J i.l.4 [.1.2 [1.� 1 7llr 83 2 1 1 1 1 1 84 1 1 1 1 1 65 1 1 1 1 1 1 tib 2 1 1 1 1 1 ti7 2 1 1 ._..,... 1 1 1 tltl :c: .J. .1 1 1 1 B9 1 , 1 1 1 1 90 4_ 1 i 1 J. 1 91 2 1 2 1 92 2 1 2 93 3 1 1 1 1 ;z4 :3 1 2 1 9 1. 1 1 1 5 14 . � 9b 1 } . 3 1 97 t 2 2 9ti L 2 2 99 I) 1 J. 2 tff 2 2 100 4 101 () 4 102 � } 1 3 2 1 � 1 103 1 104 .�. 1 2 2 105 .1 1 1 lOb 1 1 1 1 107 1 * 1 1 1 1 � _1_0 ti 1 1 ]. 109 2 ±fr- 1 1 1 liO 1 c, 2 1 .l.ll .L 2' 2 1 liZ' 1 -� 1 1. ]. 113_ 2 1 2. 114 2 1 2 1 1 2 115 3 1 -- lIb J 1. I 1 1 117 2 1 i._ I 1. 1 l1B - 1 1 1 ' = li9 3 1 1 ]. 120 1 1 1 Jg$ 2 121 1 --