ON FINITE COMMU?AT§VE SEMIGROUPS HAVING A GROUP-LEKE PROPERTY Thesis {or “no Down-c of pit. D. MICHIGAN STATE UNIVERSITY Richard L. Cantos 1965 — THESIS This is to certify that the thesis entitled "Cn Finite Commutative Semi-groups having a group-like property" presented by Richard L. Cantos has been accepted towards fulfillment of the requirements for Mdeqme inflathematics ‘ %Mwém E. Deskins Major professor anew ‘ 0-169 LIBRARY L" Michigan State University MSU LIBRARIES ”- RETURNING MATERIALS: Place in book drop to remove this checkout from your record. FINES will be charged if book is returned after the date stamped below. ABSTRACT ON FINITE COMMUTATIVE SEMIGROUPS HAVING A GROUP-LIKE PROPERTY by Richard Lo Gantos In the study of finite commutative groups, it is a fact that every homomorphic image of a finite commuta= tive group G can be imbedded in G, Moreover, if (3(G) is a homomorphic image of G, it itself has this same property. This thesis investigates when a finite commu- tative semigroup S will have these same imbedding prOp- erties. That is, we try to determine the structure of S so that every homomorphic image of S can be imbedded in S -- from now on referred to as property P —- and all homomorphic images of S also satisfy preperty P. Chapter 1 deals with finite commutative one—idem- potent (unipotent) semigroups S which belong to the following class of semigroupsg £3 is the collection of all finite commutative unipotent semigroups such that (1) if S 8 £2 , then S has property P and (2) if S 8 £3, then 9 (S) € E , where 8 is a homomorphism on S. Let S be a finite commutative unipotent semigroup, e its idempotent, G its unique maximal subgroup, (S 23G), Richard L. Gantos 2 and R = (S~G)~;e. The following necessary conditions are determined in order that SS ii: (1) If 88 t2, then R is a nilpotent semigroup of S‘ of class q > 1 with gr = g for every gtEG and re:R. (2) If 38 fig, then qu—ml = m+i and [qum' = [G[+m, where m = 1, 2, ..., qm2 and q is the nilpotent class of R. (3) If Se :3, then the nilpotent class q of R must satisfy the inequality 1< qg 4. From (10, (2), and (3) we are led to the complete structure of SE {:8 Se {Z if and only if either S is a group, 35 £2, SE $3, or SE 640‘ The classes (:3, C3, 5. consist of finite commutative semigroups S such that (h) S = G~'R, GrsR = e, where G is the maximal group in S, e is the idempotent in S, and R = (S-G)v e, (2) gr: g for every gEG and reR, and (3) R is nilpotent of class @ = 2, 3 or 4, reapectively. In the case of the classes=i23 and.‘¢¢, further condi- tions are imposed on the nilpotent subsemigroup R, Chapter 2 deals with finite semilattices and finite semilattices of groups” In a finite semilattice E ( a finite commutative semigroup every element of which is Richard L. Gantos 3 idempotent) we have a partial ordering 3 defined by: egr if and only if ef .—.- f‘e = e. ' Let 5% be the collection of finite commutative semigroups such that (1) if Se 3+, then S has preperty P and (2) if Se 3¥, then 9 (8)5 5¥, where 9 is a homomorphism on Sr Some of the theorems in Chapter 2 pertaining to this Class 3% are the followings If’ E28 3+ is a finite semilattice which contains elements a, b, e such that ai= be, h) a, cfl>a then (1) e<‘b or e< c for every eezE where the dimension of e is less than or equal to the dimension of a, (2) the number of incomparable elements ee E, with e incomparable to a< and having dimension less than or equal to a, is at most one. A finite semilattice of groups is a finite commuta- tive semigroup n? such that (1) S 29:4 G1, where Gi is the maximal subgroup of S containing the idempotent e1, (2) GinGj = o for 1 g j, and (3) GiGj = Gk where eiej = eko Some theorems on finite semilattices of groups: If SE fiH’ is a finite semilattice of groups, then gGi = {g}? for all g 2G1 (where e13 e for every idemw potent e in S) and i = 2, 3, 0009 no Richard L. Gantos 4 If 88 3+ is a finite chain semilattice of groups, then gig:l = gi, for all gie G1, 3;] 5G3, where ei< 93’ i =1, 29 0009 n and 3:: 29 39 0009 no ON FINITE COMMUTATIVE SEMIGROUPS HAVING A GROUPQLIKE PROPERTY By \“ '\ Richard L; Gantos A THESIS Submitted to Michigan State Universit in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1965 ACKNOWLEDGEMENTS The author is deeply indebted to his major profesa sor, Professor W, Eugene Deskins, for suggesting this thesis problem and for his helpful guidance during itSt preparation, Appreciation is due him especially for his kind consideration and encouragement, The author also extends his appreciation to his parents, Mr. and Mrs, Theodore E, Cantos, for providing for the typing and printing of the manuscript, ii Dedicated to Ted and Haseebie Gantos iii TABLE OF CONTENTS Page Notation Used in Chapters 1 and 2 ,..,,,.,.,.,,,,., 1 IntrOduCtionoooooooooooooocooooooooooooooooooooooo. 2 Chapter 1. Finite Commutative Unipotent Semigroups,,.,,,,. 7 1.1 Basic PrOperties of Unipotent Semigroups,. 7 1.2 Semigroups of Class § .,...,,,,,,,,.,... 11 2, Finite Commutative Semigroups of Class :N' .... 48 2.1 Certain Properties of Finite Semilattices. 48 2.2 Finite Semilattices in the Class 3+ ...., 59 2.3 Finite Semilattices of Groups in 3+ 79 Bibliography OOOOOOOOOOOOOOOOOOOOOO0000000000000... 85 iv Notation Used in Chapters 1 and 2 Square brackets are used for reference to the bibliography. Let A and B be sets. ACZB (or BZDA) means A is properly contained in B, AQB (or BQA) means ACB or A = B, A-B means the set of elements of A which are not in B. IAI means the cardinal number of the set A. C denotes the empty set, If A and B are subsets of a semigroup S, then AB means {abla€:A, be B}, If o is an equivalence relation on a set X, and if (a,b) e p, then we write apib and say that a and b are pbequivalent, and that they belong to the same pmclass. If p is a congruence relation on a semigroup S, then S/o denotes the factor semigroup of S modulo p. S/I denotes the Rees factor semigroup of S modulo an ideal I. 3 means "isomorphic". If 63 is a homomorphism on S and RCIS, then s R shall mean a restricted to R, Introduction The aim of this dissertation is to study finite commutative semigroups which have a property inherited by finite commutative groups, It is evident that every finite commutative group G has the preperty, we shall call it prOperty P, that all its homomorphic images can be imbedded in G, That is, for each homomorphic image H of G there is a subgroup of G isomorphic with H. Moreover H has prOperty P. We shall investigate the structure of finite commutative semigroups S for which (1) S has property P and (2) 8 (S) has prOperty P, where 9 is a homomorphism of S. As in group theory, every homomorphic image of a semigroup: S can be realized (differing only by an iso- morphism) as a factor semigroup. We therefore need to introduce the concept of congruences, A relation 0 on a semigroup S is said to be compatible if ap b (a,b in S) implies acp be and cap ob for every 0 in S. By a congruence on SI we mean an equivalence relation on S which is compatible. Denote by S/b the set of all equivalence classes of S mod p . Let K1 and K2 be members of S/p . Let a1, a25 K1 and let b1,b22 K2. From a1p a2 we have a1b1p a2b1. From b1f)b2 we have 32b1;)82b2o The transitivity of 9 gives us that 81b1p a2b2. 2 Therefore the set product K1K2 of K1 and K2 is con» tained in a unique equivalence class K3 mod p , ZBecause of this prOperty one may define in a natural way an op— eration in 8%) . Since this Operation does not coincide with the Operation of multiplication Of subsets, we shall use the sign 0 for denoting the result Of this Operation. Suppose that K1, K2, K3 are three equivalence classes mod p such that K1K2CIK3, In Sfla we put (1)1(10K2 = K3. Since, for any K1, K2, K35 SA) (K1<>K2)<3K3:D(K1K2)K3 = K1K2K3 K1 0(K2 OK3)ZDK1(K2K3) = K1K2K3 the Operation 0 is associative in s/b . Definition, For a congruence p on the semigroup S, the set S/b Of all equivalence classes mod p, con_ sidered relative to (1), is a semigroup, called the factor semigroup Of the semigroup S modulo 9 . Assign to each element aezs the equivalence class K of S/p which contains it, We Obtain a mapping of S onto the factor semigroup S/p», This mapping is a homomorphism; it is called the natural homomorphism Of S Onto S/p . Therefore every factor semigroup of a semi» group S is a homomorphic image of S, The following theorem (Clifford and Preston [1]) shows conversely that every homomorphic image of S is isomorphic with a factor semigroup of 8. Therefore, if we do not distinguish between isomorphic semigroups, the study of homomorphic images can be replaced by the study of congruences on S. (Main Homomorphism Theorem). Let 6) be a-homo~ morphism Of a semigroup S onto T, Let ag>b (a,b in S) if and only if 6 (a) = (3(b). Then p is a con- gruence on S and if we denote the natural homomorphism Of 8' onto S/p by p*, there is an isomorphimn 5 of s/o onto T such that Bp* = e . An important example which will be Of constant application throughout this dissertation is the following. Let I be an ideal Of a semigroup S. Define ap b (a,b in S) if and only if either a = b or else both a and b belong to I. p is called the figgg congruence modulo I. The equivalence classes Of S modulo 9* are the ideal I itself and all the one ele» ment sets {x)' with x in Sal. The ideal I, as an equivalence class Of S modulo p , is of course the zero element of the semigroup S/p , One may represent S/p as a semigroup Obtained from S by identifying with one another all the elements Of the ideal I. We shall write S/I instead of S/p , and we call S/I the Eggg factor semigroup of S modulo I. Now suppose S is a finite commutative semigroup. It has an idempotent element e. (i.e. e2 = e). In fact some power of every element of a finite semigroup is idempotent. This was first shown by Frobenius (fiber endliche Gruppen, Sitzungsber. Preuss. Akad. Berlin, 1895). Let E be the set of idempotents of S. For each e in E, let Se be the set of all x in S such that xn = e for some positive integer n. Then Se" Sf = 2‘ if e .4 f in E, and s = eLgse. Each 88 is a sub- semigroup of S containing 2 but no other idempotent and SeSf§;Sef for all e, f in E. Se is called a maximal one-idempotent (or unipotent) subsemigroup of S. All of the above facts can be found in either Clifford and Preston [1] or Lyapin [2]. Because of this decomposition of a finite commuta- tive semigroup we first studied finite unipotent commuta— tive semigroups which have prOperty P and whose homomorphic images also have prOperty P. We are able to completely describe the products in such semigroups for this case. That is, we are able to determine the fine structure of finite unipotent commutative semigroups having prOperty P and whose homomorphic images have prOperty P. This result is stated in Theorem 1.16. We next considered the case when the maximal uni~ potent semigroups 86 have only the idempotent e as an element. In this case, every element of S is idempotent. Such semigroups are called bands. Finally we close with 1 I I the possibility that the maximal unipotent semigroups are groups. Chapter 1 Finite Commutative Unipotent Semigroups In this chapter we shall characterize those finite commutative unipotent semigroups S which have the prop- erty that every homomorphic image of S can be imbedded in S and moreover all its homomorphic images have the same prOperty. 1.1 Basic Properties of Unipotent Semigroups We develop in this section some basic concepts of finite unipotent commutative semigroups. Some of these concepts can be found in either Clifford and Preston [1 ] or Lyapin [2’]. Definition. A two~sided ideal M of a semigroup S is said to be universally minimal in S if it is contained in every ideal of .8. Lemma 1.1 . If a semigroup S possesses a two- sided ideal G that is a group9 then G is the univer- sally minimal ideal of S. M. Let L be an arbitrary left ideal of 8. Since G is a two-sided ideal of S we have GL§;GS§;G 7 and since L is a left ideal it follows GLgSLgL. Moreover G(GL) = GZLGQGL which implies GB is a left ideal of G. However G is a group and a group has no ideals other than itself so we must have GL = G. There- fore G : GLSQL and thus G is contained in every left ideal of S. A similar argument shows G is contained in every right ideal of S. Lemma 1.2 . A semigroup S cannot have more than one two-sided ideal that is a group. 2522;. Suppose G1 and G2 are two—sided ideals of S which are also groups. By Lemma 1, G1 and G2 are both universally minimal ideals of S. Hence we would have G1§G2 and G2§G1 so that G1 = G20 Theorem 1.: . Let S be a finite unipotent coma mutative semigroup9 let 6 denote its unique idempotent element and let G = {as Slea = a and xa = e for some xe S}. Then 1) G = as = Se = eSe. 2) G is the unique maximal non-empty commutative subgroup of S. 3) G is the twomsided ideal which is universally .minimal in S. 4) Sn = G for some positive integer n. M. Since eeG then G is nonempty. Let xe:S. The cyclic subsemigroup (x) (i.e. the set of all positive powers of x) is finite so there exists positive integers m and r such that xm+r = xr and (x) = {x, x2. ..., xm+r-1}. Moreover the set xx = {xr’ xr+1, ..., xm+r-1} is a cyclic subgroup of 3.. Its identity element xk, r_<_ kgr+m-1, is an idem- potent element in S and hence xk = e, for S is unipotent. Therefore for every xezs there exists a positive integer k such that xk = e. 1) We show G = eS. First of all G = eG§;eS. Let x:eeS. Therefore x = ey where yeS9 ex = x and there exists a positive integer k such that xk = e. If k>1, then e=xk=xk'1x so that 2:36. If k = 1, then x = e and again xe G. Therefore eS§;G and equality follows. Moreover G = eS = Se = eSe since S is com~ mutative. 2) We show first that G is a group. Let g19 gzeaG. Then there exist elements u1 and u2 in S with e = g1u1 = g u g1e = g1 and g2e = g2. Therefore 2 2’ (g1g2)(u2u1) = g1(g2u2)u1 = g1eu1 = (g1e)u1 = g1‘11 = 8' (g1g2)e = 81(g26) = 3182 and we have g g‘2 5G. The element e 2G is the identity 1 element of G. Let gs G. There exists us S such that e = gu and g = ge. Now 10 ll e = e2 = (gu)e e1 such that Tn = {o}. It is said to have class n, (n:>1), if n is the least positive integer such that Tn = {o}. If the class of T is n = 2 we will call T a null semigroup. 13 Lemma 1.5 . Every finite unipotent commutative semigroup T with zero element 0 is nilpotent. 2523:. To show T is nilpotent we need to show Tm = {O} for some positive integer m. Since T is a finite unipotent commutative semigroup, then it follows from Theorem 1.3 that it has a unique maximal subgroup K which is an ideal universally minimal in T and more— over Tn = K for some positive integer n. But {0} is an ideal of T contained in every ideal of T and since K is universally minimal it follows K = {0}. Therefore Tn = {O }o Theorem 1.6 . If S has property P, then 1) R is a finite unipotent commutative subsemi- group of S with e as its zero element. 2) gr = g for every geG and rsR. 3) R is nilpotent of class q>'1. 2:33;. 1) Since G is a prOper ideal of S we can form S/G the Rees factor semigroup of S modulo G. By Lemma 1.4 we have S/G is a finite unipotent commuta- tive semigroup. Its elements, the equivalence classes modulo G, are every one-element set {a} where aeS—G and G itself. The element G in S/G is a zero ele- ment for s/G and IS/GI s lSl-lGl+1 = [R]. Now, by hypothesis, S/G can be imbedded in S so there exists subsemigroup T of S with zero element 14 z and [T] = [SI -[G|+1. The element 2 is an idem- potent in S so it follows 2 = e. Also Gr\T = {e} for e is the identity element of G and the zero element of T. But TQS= GvR and GnT= {e} implies T§;R. However [TI = [RI so that T = R. Therefore we have shown part (1), namely, R is a finite unipotent commutative subsemigroup of S with e its zero element. We have er = e for every reiR since e is the zero element of B. Let ge G, gyée and let reR. There exists a positive integer m such that gm = e and therefore g is the unique inverse element of gm‘1. On the other hand, gr'eG- fer Gt is an ideal of S and gm'1(gr) = gmr = or = e. Hence gr is also an inverse of gm"1 in G so we must have gr = g. By Lemma 1.5, since R is a finite unipotent commutative semigroup with zero element, we have that R is nilpotent of class q) 1. Lemma 1.7 . If S has property P, then (1) 31m = Gva and Gan = e for all positive integers m. (2) SmZDG for m = 1, 2, ..., (q-1) and Sm = G for qu‘, where q is the nilpotent class of R. (3) S/Sm is a nilpotent semigroup of class m, m = 29 39 090-9 q. 15 (4) [(s/sm)n| = [snl-lsm|+1 where n and m are positive integers such that 1gngmgq. Iggggf. To verify Sm = G..Rm we apply induction on m. The decomposition of our finite unipotent commu- tative semigroup 8, namely S = G~JR and Gr~R = e, verifies Sm = G~1Rm for m = 1. By Theorem 1.6, GR = G and consequently GRm = G for all positive integers m. Assume Sm = GsaRm. Then Sm+1 = sms = (GuRm)(GuR) = GvaGuGRva+1 =GURmH° Also Gr~Rm = e since e is the identity element of G and zero element of R. Hflso by Theorem 1.6, R is nilpotent. Let q be its class. We have Sm = G..RT for any positive in— teger m so that G§;Sm for any positive integer m. If G = Sn for some natural number n, then it follows, from Gan = e, that Rn = e. Hence, n_>_q. We now proceed to show that S/Sm is nilpotent m of class ni>1. 8/8 is a finite unipotent commutative semigroup with zero element; its zero element being the ideal Sm itself. Lemma 1.5 gives us that s/sm is nilpotent. Let 3* = S/Sm. Denote its zero element by 0* and the remaining elements by x* = {x}, where x eS-Sm .-. R-Rm.. Indeed (S*)m = 0* for (s-sm)mgsm. We next assert that (S*)m—1 £ 0*. Since RZDRZZD... 1 DRm' DRm 3... BBQ = e we can select x e (Rm-1—Rm). 16 2 Then x = 11x2...xm_1 where xie‘R-R (i = 1, 2, ..., m-1). However x*, xi, ..., x;_1 are non-zero elements of S* and x* = 1:101:50 ... 01514.. Consequently, x* e (8*)!!!-1 and x*,£ 0*w But (8*)m = 0* and (3*)m-1 £ 0* imply 3* is nilpotent of class m. n It remains to show that [(S/Sm) I = ISn|—|8m|+1. By part (3), this equation obviously holds for n = m. Using an argument similar to the one above, one can easily note that there is a one-to-one correspondence . n between the non-zero elements x* in (S/Sm) and the n elements x in Sn- Sm. Hence [(S/Sm) I = [Snl-ISm|+1. It has been shown, thus far, that if S is to have property P, it is necessary that R be a nilpotent semigroup of class q1>1. We now would like to determine the order of the ideals Rm where m = 2, 3, ..., q-1. To do this we need to introduce the following concept. Definition. An element u in S is said to have a factorization in Sm, (m = 2, 3, ..., q) if and only if u eSm. Two factorizations of u in Sm, u = x1x2...xm = y1y2...ym, are said to be distinct if and only if the sets {x1, x2, ..., xm} and {y1, ..., ym} are unequal. The number of distinct factorizations of u in Sm will be denoted by B(u;m). 17 Lemma 1.8 . If S has prOperty P, then (i) IRq"1 I: 2 and (2) [sq'1l = |G|+1 where q is the nilpotent class of R, q) 2. 23292. By Theorem 1.6, R is nilpotent of class q~ so that Rq-1 DRq = {e} and consequently [qu1 [22. Suppose [Rq-1I>’2. We can find at least two distinct elements in Rq-1 both different from the zero element e in R. Choose us (Rq-1-{e}) such that B (u;q-1) is maximal. Let v be any element in (Rq-1-{e}) with u ,1. v. Clearly S(ugq-1) 2. B(v;q-1)21. Let us consider that partition of the semigroup S determined by the disjoint subsets KO = Sq = G, K = {u,v} and all one element subsets Kx = {x} where xe:S-(Sq~I{u,v}). Let p be the equivalence relation induced by this partition. Moreover we can show that p is a congruence on S. Suppose a;>b, where a and b are elements in S such that a £ b. Then we must have both a and b belong to K0 = Sq or both a and b belong to K = {u,v}. If the former occurs then, since Sq = G is an ideal of S, it follows xagaxb for all xe.S. If the latter occurs, we would have a = u and b = v or a = v and b = u. However {u,v}§Rq"1§Sqm1 and hence xu and xv belong to Sq for every xe S. Therefore xug>xv which implies xaw>xb for every xe S. We have verified that the equiv- alence relation p induced by the disjoint subsets KO, K 18 and the one-element sets Kx = {x} is indeed a congruence. We form the factor semigroup S/p . Let us de- note its elements (equivalence classes mod p ) by K0, K and all Kx = {x} where xeS-(qu {u,v}). S/p is a finite unipotent commutative semigroup and, since Sq is an ideal of S, the idempotent KO of S/p is a zero element of S/p . We first show that Ks (S/p )q-1. Now ueRq-1-{e} which implies u = u1u2...u where each “i e R-R2. q-1 Indeed uie S-(Sq~J{u,v]) so that each of the one-ele- ment subsets Kfii = {hi} (i = 1, 2, ..., q-1) of s are non-zero elements of S/p different from the ele- ment K. However Ku1Ku2°NKuq_1 = {u1u2...uq_1}= {u}C{u,v} which implies Ku1‘3Ku2 o°°°<3KPq_1 = K . Hence Ks (S/p )q’1. Next we show that (8/9 )q KO. This result together with Ke:(S/p )q-1 tells us that S/p is nilpotent of class q. Let At:(S/p )q. Then A = A10A20...0A Where A18 8/9 (1 =1, 2, 0009 Q). q If any A1 = KO we would have A = KO so we may assume A1 s’xo for all i a 1, 2, ..., q. This implies that each A1, as a subset of S, is either a one-element set {x1}, where x1e: S-(qu {u,v}) or the set {u,v}. In any case, the set product A1A2...Aq must be a subset of Sqe Consequently, by the definition of the Operation 0 19 in S/p~, it follows A = A1 0A2 o... 0Aq = KC and we have (S/p )q= KO. Since Ke:(S/p=)q_1, then K has a factoriza- tion in (S/p )q-1. Let f3(K;q-1) be the number of distinct factorizations of K in (S/p )q-1. We now claim that B (u;q-1)+ B(v;q:—1)-<_ B(K;q-1). To show this we first shall show that distinct factorizations of both u and v in Rq-1 give rise to distinct factorizations of K in (S/p )q-1° Suppose u = x1x2...xq_1 = y1y2...yq_1 are two distinct factori- zations of u in Rq-1; i.e. the set {x1, x2, ..., xq_1} and the set {y1, y2, ..., yq_1)' are unequal. New xie(R-R2) and yie (R-RZ) for i = 1, 2, ..., q-1, since uezRq-1-{e}. Consequently each of the one-element subsets of S, Kx1 = {x1}, sz = {x2}, ..., qu_1= {xq_1}3 Ky1 = {y1}, Ky2 = {y2}, ..., qu_1 = {yq_1} is a non- zero element of 8/9 . Moreover the sets {Kx , ..., Kx } 1 q~1 and {K , ..., K } are unequal. But, from Y1 yq_1 Kx Kx ...Kx = {x1x2...an1} = {u}<::{u,v} 1 2 q-1 and Ky1Ky2'°0qu-1 = {y1y2ooqu_1}={u}C{u,v} it follows K=K 0K 0...0K =K o 0...0K . x1 x2 xq-1 Y1 Kyz Yq—1 This implies K has two distinct factorizations in (S/p )q-1. Using the same argument, one can obtain dis- tinct factorizations for K in (S/p~)q'1 from distinct factorizations of v in Rq-1. 0n the other hand, since 20 u £ v, a factorization of u in HQ”1 and a factori- zation of v in Rq-1 will give rise to two distinct factorizations of K in (s/. )q-1. This together with the above clearly imply that E3(u;q-1)+ 5(vsq-1)S.B(K;q-1). Up to this point we have shown that S/p is a nilpotent semigroup of class q and moreover there exists a non-zero element K of S/p which has a(K;q-1) Z B(u;q-1)+ B(v;q-1) factorizations in (S/p )q-1' Since S has prOperty P, then 3/, can be imbedded in S. Hence there exists a nilpotent subsemi- group: T of S of class q and in T there is an element t which has B(K;q-1) factorizations in Tq-1. But Tr~G = e for e is the zero element of T and the identity element of G. Hence Tc:R and t £ e. Therefore t eRq-1-{e}. However t has at least B (K;q-1) factorizations in HQ"1 and since B(K;q-1) ZB(u;q-1)+ S(v;q-—1)> B(u;q-1) we have a contradiction with the choice of u. Hence we must have IRq-1 I: 2. This proves part (1) of our lemma. By Lemma 1.7, 3‘1""1 = Gvuq‘1 and en Rq‘"1 = {e}. Therefore [Sq-1| = |G|+[Rq-1|-1 = [G[+1. This completes the proof of the lemma. The condition that S has prOperty P has given us the conclusion that the subsemigroup R is nilpotent of class q_>1 and moreover the ideal Rq"1 contains 21 exactly one non-zero element. By further imposing on S the condition that all homomorphic images of S also have property P we shall show that [Rq-2| = 3, qu-fi = 4, ..., [R2[ = q-1. This result will be used repeatedly in determining those semigroups S which have property P and whose homomorphic images have prOperty P. Definition. Let g: be the collection of finite commutative unipotent semigroups such that (1) if S a g, then S has prOperty P and (2) if S e g, then 0(8) 8 2: for all homomor- phisms 9 on S. Lemma1.9. If sec, then R8 {Z and 922: where G is the unique maximal subgroup of S and R is the nilpotent subsemigroup of S such that S = G~JR and Gr\R = {e}. 2339;. Let Se g and let G be its maximal subgroup with identity element e. Then, by Theorem 1.6, S = G~aR, Gr~R = e, R1 is the nilpotent subsemigroup of class q>1 and gr=g for all geG and reR. Since G is a finite commutative group it follows G and 6(G), 6 a homomorphism of G, both have prop- erty 1?. Hence Gags. Let 6:S—>R be the transformation of S into R defined by 22 x if xe:R 6(X) = e if xe G e is indeed single-valued and onto R. Since gr = g for every ge G and reiR it follows 8 is a homo- morphism of S onto R. Since Se CZ we have Re(: . Theorem 1.10 . If Sec , then 1) ,Rq-m, =m+1 for m: 1, 2, ..., q-2 and 2) I sq'ml= lG|+m for m = 1, 2, ..., q-2, where q is the nilpotent class of R. 3:23;. We have [sq‘ml = [Gqu_m| = |G[+[Rq-m|-1. Hence to show (2) we need only show [Rg-ml = m+1. By Lemma 1.9, Se C implies Re C . Therefore it suffices to prove this result for finite nilpotent unipotent com- mutative semigroups Rezc . Suppose the statement of the theorem is false. Let R e C be a minimal ecunterex- . ample. Let the nilpotent class of R be q) 2. By Lemma 1.8, [Rq-JI = 2 since R has property P. There- fore there exists a positive integer m, where 1< m_<_ q—2, such that [Rq—n, = n+1 for n = 1, 2, ..., m-t but [Rq-ml £ n+1. Since Rq-mZDRq-(m-1), then qu-m1> qu-(m-1)l = m. Consequently, 'Rq-n' = n+1 for n = 1, 2, ..., m—1 but qu-mI> m+1. Consider the Rees factor semigroup R/Rq-1. By Lemma 1.7 we know that it is nilpotent of class q-1 and moreover by (4) of this same lemma we have 23 [R/Rq'1l = [RI-2+1 = [RI-1, 1(R/Rq-1)q-nl = [Rq-nl-2+1 = n for n = 2, ..., m—1 and r(R/Bq'1)q"m[ = [Rq-ml—2+1 > m. However R/Rq"1e C since it is a homomorphic image of R. Therefore R/Rq'1 is also a counterex- ample to our desired result which has order less than the order of R. This contradicts the minimality of R so our assumption is not valid. Corollary 1.11 . If Se $3 and q is the nil- potent class of the subsemigroup R of S, then [s/sq'1l = [RI-1 and [(s/sq‘1)q-m| = m where m = 1, 2, ..., q—2. 2322;. By Theorem 1.10 we have lSq“1[ = |Gl+1 and I sq-q‘ = IGl+m for m = 2, 3, ..., q-2. Then I s/sq‘41 = Isl-lsq‘1|+1 = ([G|+|R[-1)-(|G[+1)+1 = [RI-1. Also by Lemma 1.7 we have m 1(s/sq-‘>“P" l = [Sqml- lSq"1l+11 -- m. where m = 1, 2, ..., qu2. Theorem 1.12 . If s e C’ and if q is the nilpotent class of the subsemigroup R, then 1< qu4. 23931;. Suppose qZ 5. R being nilpotent of class qz 5 gives the following chain of ideals 1DRq = {e}. 2, IRq'2[ = 3 and R DRq-3 BBQ-23 Rq’ By Theorem 1.10 it follows |Rq-1[ 24 [Rq-31 = 4. Therefore we can choose non-zero distinct elements v, u, and w from R—{e} such that 1-Rq = {v}, unz-Rq-1 % {u} and Rq‘3—Bq‘2 = {w}. That is, Rq = {e}, Rq-1 = {e,v}, qu2 = {e,u,v}, and RQ-3 Rq‘ = {e,w,u,v}. Vg‘Rq_1-Rq: implies v = r1r2...r where q-1 ritzR-R2 for i = 1, 2, ..., q-1. Consider the (q-l) -94 _ elements am = II ri where m = 1, 2, ..., q-1. First ‘In of all. each a;" is an element of Rq-z. Since v = r a and since ve‘Rq-1-Rq it follows;that aIn (m = 1, 2, ..., q-1) is not an element of Rq-1; for otherwise v = rmamezR°Rq-1 = Rq: which cannot happen. Therefore ame:Rq"'2-Rq-1 for all m = 1, 2, ..., q-1. Hence am = u for m = 1, 2, ..., q-1 so that v = rmu, $4 m = 1, 2, .099 q-19 and u a ll r1 for eaCh m = 192,000, i=1 q-l. However 1‘” 3-1 3-1 u =i.|1,|i.£i = rn an r1 = rn'an,m (n y! m) Clearly an9meRq-3 for all“?! and m, with n ,4 m, n = 19 29 .909 q*1 and m = 19 2, one, Q‘1o Since u = rnan m and since us qu2-Rq'”1 it follows 9 an meRq"3—qu2 for the above possible values of n and 9 m. Therefore an9m = w for‘ n.£ m, n = 1, 2, ..., q-1 and m = 1, 2, ..., q-1. We have thus shown that 1-1 1 1 3-. v=_]'T-ri = rmu , u='T[_ri = rnw and w=._[Tri '1... 1:1, in. 1“: mm} where m = 1, 2, ..., q~1 and n = 1, 2, ..., q-1. 25 We show further that q-1 q-1rq-1 V = r1 = r2 = .00 = rq~1 q-2 q-2 rq-2 (1201) u: r1 = r2 = 0.. = I‘q_1 w = ri" = .3-3 = -.- = .33: First of all, u = r1w = r1CTT'ri) = r1 II r1. By finite 181 1'2. induction we can derive 11.1.- (12.2) u = rk(17:ri ) for k: 1, 2, ..., q-3. We have already shown the truth of (12. 2) for k = 2. — 1-141 Suppose u = rk-ITJ'ri. Now rk1'TJ'r181R: 3 and since (rk‘II:I ri )rq k = rk‘ 1II r1: a-f'ki it follows rk-ITT‘r e R‘1"3-R‘-1'2 . This implies 1:3 w = rk"1 r and r1 i=1 114:4 1*‘1 u = r w = r1(rk"T1T-ri)=rkT17r . 1 1.2, 1 1:2. 1 Formula (12.2) when k: q-3 yields u = r? r2. Therefore rg-3ezRq-3—Rq—2 which implies w = rg-j. This in turn implies u = r1w = r%-2 and v = r1u = rg-1. Since R is commutative then by a mere relabeling we have certainly verified (12.1). We next show that r1 = r2 = ... = r Suppose q-1' two of the ri are distinct; say r1 £ r2. Consider the Rees factor semigroup S/Sqf1. It is a nilpotent semi- group of class q-1, I S/Sq‘1l = [RI-1 and moreover, by Corollary 1.11 we have |(s/sq‘5q"1| = 1, |(s/sq‘1)q'2| = 2 and I (s/sq'1)q-3l = 3. 26 Since s/sq"1 can be imbedded in 8, there exists a subsemigroup T of S such that (12.3) T is nilpotent of class q-1, (12.4) ITI = [RI-1, and (12.5) [Tq-1I = 1, ITq-ZI = 2 and [Tq"3l = 3. Actually T is a subsemigroup of R; for e is the zero element of T and hence Tr~G = e. Condition (12.5) implies we can choose two distinct elements '3 and t? in T such that Tq'2.-.Tq"'1 = {t} and Tq"3-.Tq‘2 = {t*}. Moreover t3 = e for all ts T.. Also .3: Tq-2C2Rq_2 = {e, u, v}; hence we have either .3 = u or .3 = v. Since we are assuming r1 # r2 and since ITI = [RI-1 we have either r12 T or rzezT. Therefore either r$-2e Tq"2 or rg-zequ_2.' But u = rqfl'2 = rg-Z, so we will always have us Tq-2. However uI: Rq"1 which implies ut Tq-1. Hence ue Tq"'2-Tq"'1 which yields u =.E° 0n the other hand, since either r16 T or r28 T we should have either e = rm; = r1u or e = r23 = r2u. But we know v = r1u = r2u and hence we have our con- tradiction. Consequently we have r1 = r2 = ... = rq_1 so that (12.6) v = r1 = r u, 11"- Our next step is to show that u = r%‘2 is the only factorization of u in Rq'z. That is, if 2 u = S182...8q_2, where sis RuR , is a factorization 27 of u in Rq-2’ then we shall show that 81 = 82 = ... = s = r1. Assume at least one of the above si £ r1; q-2 say 81 £ r1. 2 We have u — 8182...8q_ 2, where 81 ER-R , and 81 £ r1. The elements =II si, (m = 1, 2, ..., q-2), 3:1 are elements of Rq'3. Moreover u = s b for m m’ 1 it follows m = 1, 2, o o e 9 (1-2, and Since 11 8 Rq-2~Rq- meRq'3-Rq'2 for m = 1, 2, ..., q-2. Therefore w = bm for m = 1, 2, ..., q-2. However 51W = 81b = S 1:I::I18 = S182...8q_2 = u and 1‘3 11 = 81W = 81 I I 8 = S I I S o 11.1 11:2 1 Using finite induction in the same manner as employed in formula (12.2), we obtain 14-1 (1297) u = 811' I 51, k =1; 29 one, (1'30 22— For k = q-3 in (12.7), we find that u = s3‘3s2. Hence s?"3 is an element of R‘1"3-Rq'2 so that w = s%'3. This implies u s w = sq'2 and v s r1u = r1(s1w) 1 1 _ q-1 111—81 0 Again we consider the Rees factor semigroup = 81(r1V) = s S/Sq'1. As before we have the existence of a subsemi- group; T of R which has the prOperties (12.3), (12.4) and (12.5). Note that r1 is not in T; for otherwise v = rg'1e Tq"1 = {e} which is impossible. Since 31 £ r1, r1¢ T and I TI = IRI-1, it follows 818 T. But this is certainly impossible for 28 V 2 8511-18 Tq_1 = (61%» Therefore s1 = 52 = ... = sq_2 = r1 which gives u = r%-2 is the only factorization of u in Rq-Z. The above result gives us that w = rq"3 is the only factorization of w in Rq-B; for suppose w = 8182...Sq_3, where sis R—Rz. Then 2 1 and r1(s1s2...sq_3)e:Rq- -Rq_ . It follows from the unique factorization of u in Fig"2 that 81 = s2 = ... = sq_3 = r1. Hence w = rq”3 is the only fac- torization of w in Rq_3. Up to this point our arguments have shown that if Se 2; and C125. then we must have v = rg-I = r1u, where Rq-1-Rq = {v}, (12.8) u rg-2 = r1w, where Rq-2_Rq-1= {u} , w = r?'3, where Rq“3-Rq“2= {w}, u = rqm2 is the only factorization of u in HQ"2 and w = r3-3 is the only factorization of w in Rq_3. We will proceed to show that this situation cannot occur so that our assumption that S e I; and q2 5 is false. We again consider the Rees factor semigroup s/sq‘1. As before, it is nilpotent of class q-1,‘ IS/sq“l = [RI-1. [(8/sq‘1)q’zl = 2 and [(s/sq“)q"3I = 3. We therefore know that the sets (s/s‘l")‘1'2-(8/8‘1"‘>q'1 and (s/sq‘1)q‘3- (s/sq‘1)q"2 29 are one-element sets. It will help us to know exactly what these elements are and also the manner in.which they are factored in S/Sq'1. The elements of S/Sq"1 consist of the zero element KO = Sq”1 and all one- element sets Kr = {r} where rezR-Rq‘1. Therefore Ku = {u}, Kw = {w} and Kr1 = {r1} are all non-zero and distinct elements of S/Sq-I. However u = rg-Z and w = rg-B imply -2 -2 -1 q Ku=Kr10Kr1oooooKr1=(KI-1)q E(S/Sq ) and q-3 ee}. K1, K2, ..., Km be the equiv- alence classes of R3 mod 1). K8 is the zero element of RB/p . Case 1. Suppose aeKe. Then ap e implies u1apu1e and hence upe. Therefore us: Ke. Since a does not belong to Ki (1 = 1, 2, ..., m) it follows each class Ki (1 = 1, 2, ..., m), considered as a sub- set of R3, is actually a subset of the subsemigroup N. But N2 = {s} so that KinC;NN a {e} for i = 1, ..., m and j = 1, ..m, m. Thus KiOKJ = Ke for i = 1, ..., m and j = 1, ..., m. From each class Ki’ 1 = 1, ..., m, choose an element bi“ Let T = {9, b1, ..., bm}. Every b is an element of N so it follows bibj= e for i i = 1, ..., m and j = 1, ..., m. Consequently T is a subsemigroup of R3. Define the transformation a : R3/p—-> T as follows: a(Ke) = e and 0(Ki) = bi for i = 1, 2, ..., m. c is an isomorphism of RB/p onto T so we have Ra/p 37 imbedded in R3. Hence RB/p has prOperty P. Now T is nilpotent of class 2 and therefore R3/p is nilpotent of class 2. But we already know such finite unipotent nilpotent commutative semigroups have prOperty P. Hence 33/p has prOperty P. Since R3 and RB/p have prOperty P, it follows R3 5 If. Case 2. Suppose a does not belong to Ke but uezKe. Let ae.K . Then each equivalence class K 1 (i = 2, 3. ..., m), considered as a subset of R 19 3, is actually a subset of the subsemigroup N. Therefore KiOKJ = K6 for i = 29 3, 0099 m and J: 29 39 0.0 m.- Moreover K1<>K1 = Ke for i = 1, 2, ..., m; for let r1r2¢.;K1Ki where r1.c;K1 and r2eKi. Since 2 _ r1r22R3 — {e, u} and since ueKe, it follows r1r2p e. Therefore K K QK so that K OK = K , for 1 i e 1 i e i = 1, 2, ..., m. The elements of R3/p have the fol- lowing products: Ke 0Ki = K6 and KiOKj = Ke for all i and 3. From each equivalence class K i = 2, 3, .... m, 19 choose an element a1. Let T = {e, u, a2, ..., am}. Now a a = e for i = 2, ..., m and j = 2, ..., m since i 3 they all are elements of N. Moreover uai = e for all i = 2, ..., m. Therefore T is a subsemigroup of R3. Define the transformation a : R3,6-—> T as follows: a(Ke) = e, a(K1) = u and a(Ki) = a1 for i = 2, 00., mo 38 a‘ is an isomorphism of R3/p onto T so that R3 has prOperty P. Moreover R3/p has property P for it also is nilpotent of class 2. Thus we have R3 3 ?. Case 5. Suppose both a and u are not elements of Ke’ If apu, then 11131311111 which implies up 6. This cannot happen so that a and u belong to distinct equivalence classes. Let agK1 and us K2. Now each Ki’ 1 = 2, 3, ..., m, are actually subsets of N. Therefore Ki‘DKj = Ke for i = 2, ..., m and j = 2, ..., m. A‘slso K1oK2 = Ke for u£K2 and m: e for every rs R3. However either K1 oKj = Ke or K1on= K2 for 3 £ 2% This follows from the fact that either ar = e or ar = u for rsRB. The products in R3/p are KeoKi = Ke for i =1, ..., m, KiOKj 2 K6 for 2_<_ igm e e K2 for 3.42. K10 K2 = K and either K1 oKj = K or K1 oKJ = From each class Kj’ j = 3, 4, ..., m, choose one ele- .ment aj. Let T = {e, a, u, a3, a4, ..., am}. Since T2913; = {e, u}, it follows T is a subsemigroup of R3. Define meB/p*—) T as follows: a(Ke) = e, a(K1) = a, a(K2) = u, c(Ki) = a1 for i = 3. 4, ..., m. One can check that a is an isomorphism of R3/pl onto T. Hence R3 has property P. By studying the multiplication in R3/p one can 39 determine that R3/p will be either nilpotent of class 2 or will be a semigroup of the same type as R3. Hence RB/p has property P. These cases all yield the same result; namely R3 51;. By Lemma 1.15, it follows 33g; and hence (Saga: Theorem 1.15 . C.; {2. M. Let R42 (:4 and let p be a congruence on R4 and consider R4/p. Since R4 6 (F. it has prOp- erties (15.1) thru (15.5) introduced earlier. Let e, v, u, r1 and r0 be the elements of R4 as defined in properties (15.1) thru (15.5). The equivalence classes mod p) will be denoted by Ke’ K1, K2, ..., Km, where Ke = {x 8114pr e}. We again need to consider several cases. Case 1. Suppose r18 Ke' Then rip e and r?¢>e so that u and v both belong to Ke. Conse- quently, the equivalence class Ke contains 2 4 = {e, v, u}. Hence Kitin = Ke for all i = 1, ..., m and j = 1, ..., m. That is, R4/p is nilpotent of R class 2. Now IKe|_>_4 so that mg |R4|-4. But [MI = [R4I-2 where M = R4-{ro,r1}. Hence for each equivalence class Ki (1 = 1, ...,m) we can select an element a1, set M. Let T = {e, a1, ..., am}. PrOperty 15.5 gives where a1 £ aj if 1 £ 3, from the 4O aiaj = e for all i and 3. Therefore T is a sub- semigroup of R4. Define a :R4/p-9 T by a(Ke) = e and 0(K1) = a1 for i = 1, ..., m. a is an isomor- phism of H4/p= onto T. Therefore R4 has property P. Since R4/p is nilpotent of class 2, it follows R4/p has property P. Thus R4 s I:. Case 2. Suppose r1 does not belong to Ke but ueKe. ueKe implies r1upr1e so that vpe. Hence vs Ke. Therefore the set Ke contains the ideal R = {e, v, u}, and, as in the above case, we must have K “#N 0KJ = K3, where i = 1, ..., m and j = 1, ..., m. But [KeI_>_5 so that m$IR4I-5. Hence m<|MI and consequently for each equivalence class Ki (1 = 1, 2, ..., m) we can select an element a1, with ai # aJ when 1 £ 3, from the set M. Let T = {e, a1, a2, ...,am}. T is a subsemigroup of R4, by property (15.5). Define a. : R4/p—) T by 0-(Ke) z: e and a(Ki) = a1 for i = 1, 2, ..., m. a is an isomorphism of R4/p onto T and we conclude that R4 has property P. As in the previous case, R4/p is nilpotent of class 2 so that R4/p has prOperty P. and u do not belong to Ke but veKe. Suppose r on. Then 1‘2 1 1 implies uf>v. But this is impossible so that r1 and u' belong to distinct equivalence classes. Let r1ezK1 and ue:K2. Since r? = u it follows K10K1 = K2. Case 5. Suppose r1 p r1u which 41 On the other hand KioKj = Ke except when i = j = 1. This follows from the fact that KiKJg;RZ§;Ke. Again mgIMI so for each equivalence class K1, 1 = 5, ..., m, we can select an element a1 (a1 £ aJ if 1 £ 3) from the set M. Let T = {e, r0, v, a3, a4, ..., am}. T is a subsemigroup. Define a : R4/p —> T by (1(Ke) = e, 0(K1) = r0, 0(K2) = v and C(Ki) e ai for i = 5,..., m. One can check to see that a is an isomorphism of Rm/p onto T. Therefore R4 has property P. By studying the multiplication in R¢/p, one can determine that it is a nilpotent semigroup of class 5 and moreover of the same type as R3. That is, it sat- isfies properties (14.1) and (14.2). Hence R4/p .has property P. Case 4. Suppose r1, u and v do not belong to Ke‘ Then we must have r1, u and v belong to three distinct p—classes; say r18 K1, uEIK2 and veKB. Moreover r0 belongs to some p-class distinct from K1, K2 and K3. For each equivalence class Ki’ where i = 5, 6, ..., m, we select an element a1 (a1 £ 33 if 1 fi 3) from the set (M—{u,v}). Let T = {e, r0, r1, u, v, a5, a6, ..., am]. T is a sub- semigroup of R4 since T2gRi QT. Define a 2R4/p -->T as follows: cure) (1(K4) = r0 and a(Ki) = a1 for i = 5, 6, ..., m. e! “(K1) = r19 “(K2) = u9 0(K3) = V, 42 a can be checked to show it is an isomorphism of R4/p onto T. Therefore R4 has property P. The multiplication in T shows that R4/p is a semigroup of the same type as R4. That is, it has properties (15.1) thru (15.5). Hence R4/p has prOp- erty P. In each of the above cases, we have shown that if 842;“ then 1245;. Hence S46 C and we have CGQC. We are now ready to prove the main result of this section. That is, we shall now characterize those finite unipotent commutative semigroups S where Se;¢ . Theorem 1.16. S; C if and only if either 1) S is a finite commutative group, 2) Se:(fg, 3) SE $3, or 4) 88 (PM Iggggg. If S is a finite commutative group then SE C.. Theorem 1.14 and Theorem 1.15 show that when S is amember of the class C2, C3, or ('24., then S e(;. Suppose S s(: and S is not a group. Let s be its idempotent, let G be its maximal group and let R = (S-G)~a{e}. By Theorem 1.6, R is a finite unipotent 43 nilpotent commutative subsemigroup of S of class q) 1 and gr = g for every gs G and reR. Moreover, from Theorem 1.12, it follows 1< qg 4. We now determine the structure of S when q = 2, q = 5 and q = 4. Case 1. If q = 2, then R is nilpotent of class 2 and clearly S a $3. Case 2. Suppose q = 5. By Theorem 1.10, it fol- lows ISZI = IGI+1 and IR2I = 2. Consider the Rees factor semigroup S/S2. It is nilpotent of class 2 and, from Corollary 1.11, it follows IS/S2I = [3}1. But S a g implies 8/82 can be imbedded in S. Hence there exists a subsemigroup N of S such that N is nil- potent of class 2, e is the zero element of N and INI = [RI-1. But NrsG = e so that NC;R. Therefore R is a nilpotent unipotent commutative semigroup of class 5 such that |R2I = 2 and there is a nilpotent subsemigroup N of class 2 with INI = [RI-1. There- fore R has the preperties (14.1) and (14.2) so that 3.;CT.. Case 5. Suppose q = 4. By Theorem 1.10, it fol- lows IRZI = 5 and IR3I = 2. We proceed to show that R satisfies (15.2), (15.3). (15.4) and (15.5). Let 32- R3 = {u} and 33—34 {v}: that is, 32 = {e, v, u} and R3 = {e, v}. veR3 implies v = r1r2r3 where rieR—R2. Then 2 5 r r2, r1r3 and r2r3 are non-zero elements of R - R 1 44 so it follows u = r1r2 = r1r3 = r2r3. Consequently, v = r1u = r1r$ = r?. Likewise v = r3 = r; and we have _ 3 _ 3 _ 5 _ 2 _ 2 _ 2 v — r1 _ r2 _ 3 and u _ r1 _ r2 — r3. We now show that r1 = r2 = r3. Suppose two of these elements r1 are distinct; say r1 # r2. Con— sider the Rees factor semigroup S/SB. It is nilpotent of class 3 and by Corollary 1.11, IS/S3I = [RI-1 and |'(s/s3)2| = 2. Since 3. 2; it follows that 5/33 can be imbedded in S. Thus there is a nilpotent sub- semigroup T of class 5 in S with (16.1) ITI = [RI-1, (16.2) IT2I = 2. Actually T is a subsemigroup of R. By (16.2), there is an element t*e T with T2-T3 = {t*}. From T3 = {e} 2 it follows tvt* = e for all ta T. More- over t*s T2§;R2 = {e, v, u} implies t* = u or t* = v. and t*s T We are assuming r1 # r2 so it follows, from IT] = [RI-1, that either r18 T or r28 T. Hence either rfie T2 or 2 2 2 r2: T . Since u = r1 = rg then we must always have u 8T2. But 'u¢ T3 which gives us us T2-T3. Hence u = t*. On the other hand, since either r15 T or r25:T we have either e = r1t* = r1u = v or e = r2t* = r2u = v. This indeed is impossible so that our assumption is false. Hence r1 = r2 = r3 which implies v = r? and u = r?. Next we show that u = r2 is the only factorization 1 45 of u in R2. That is, if u = s1s2, where sis R-R2, is a factorization of u in R2, then we show that 81 = 32 = r1. Assume that at least one of the 31 £ r1; say 81 £ r1. Now v = r1u = r1s1s2 implies r1s1, r1s2 1s2 are elements of R2-R3. Hence u=r1s1=r and s _ _ _ 2 1 and v _ s1u — s1(r1s1) — r181. But v = r1s§ feRZ-R3 so that u = s? v = 8?. Again we consider the Rees factor semigroup 8/83. As before we have the existence of a semigroup 82 = 8182 implies s and T of R which is nilpotent of class 5 and satisfies properties (16.1) and (16.2). Note that r14 T; for otherwise v'= r?e T3: {e}. Therefore 815 T since ITI = [RI-1 and s1 # r1. But again we have a contra- diction since vI= s?e:T3 = {e}. Thus u = r? is the only factorization of u in T2. We have just shown that R has property (15.2). Since rr1 £ u unless r = r1, it follows 3 for every rgR with r;ér1. This r~r1g {e, v} = R shows that R has property (15.5) We proceed to show that R has property (15.4). Again we consider 8/8}. We know [(8/83)2I = 2 so that the set (8/83)2- (8/83)3 is a one-element set. It will help us to know exactly the element in this set and how it is factored in 8/83. The elements of 8/83 are the zero element Ko 2 S3 46 and all one-element sets Kr =-{r} where re:R-R3. Hence Kt = {u} and K1". = {r1} are elements of 8/83. Since u — r2 it foll - K <>K (S/SB)2 - 1 ows Ku — r1 1W e . 2 r in (S/S3) since u = r Moreover Ku = K is the only factorization of Ku 2 1 is the only factorization of u in R2. Therefore 2- 3 = {Kn}. Since s/s3 can be imbedded in s, it follows there exists a nilpotent subsemigroup T of class 5 in R which satisfies conditions (16.1) and (16.2). Condition (16.2) gives the existence of an element t*e with T2-T3 = {t*}. Moreover, since t* is the isomor- phic image of Ku’ it follows there exists tO.sT with 2 2 i = * = t to. Also t tO t* in T2. is the only factorization of From t*e T2g;R2 = {e, u, v}, it follows t* = u or t* = v. If t* = u, then t0 = r1 for r? is the 3 only factorization of u in R2. Then v = r12 T3 = {e} which clearly is not possible. Therefore t* = v so that v = t8 and this is the only factorization of v 2 in T . But r14 T so that T = R-{r1}. Hence rtO = e for all re:R-{t0, r1}. We just have proven property (15.4) holds in R by taking r0 = to. Let M = R-{r1, r0}. Let a and b be elements of M. The fact that r? is the only factorization of u in R2 shows that ab = e or ab = v. Since a # r1 T 47 and b # r1, it follows both a and b belong to T. But then v # ab for r3 is the only factorization of v in T2. Hence ab = e for every a and b in M. Therefore M is nilpotent of class 2 and we have shown R has property (15.5). Since R satisfies properties (15.1) thru (15.5) it follows S a C4,. Theorem 1.16 completely characterizes those finite unipotent commutative semigroups which have prOperty P and whose homomorphic images also have prOperty P. In our above results we did not need that all homomorphic images of S also have prOperty P, but only those homo- morphic images of the form S/Sk have prOperty P. It is still an Open question as to whether this condition is also necessary. Chapter 2 Finite Commutative Semigroups of Class 3V The semigroups which.will be discussed in this chapter are finite commutative bands and finite commu- tative semigroups which are unions of groups. For such semigroups S we will again impose the prOperty P (recall that S has property P if and only if S/p can be imbedded in S for all congruences p in S). We will try to determine the fine structure for the above mentioned semigroups S which have prOperty P and whose homomorphic images also have prOperty P. 2.1 Certain Preperties of Finite Semilattices In this section we will define a semilattice (introduced by Klein-Barman [6]) and deve10p several basic prOperties. Let E be a finite commutative band. Recall that a band is a semigroup every element of which is idempotent. Consider the relation. g on the band E defined by e$,f (e, f, in E) if and only if ef = fe = e. If egf we say that e is under f and that f is over e. The relation .3 on E is a 48 49 partial ordering of E. That is, g is a reflexive, antisymmetric and transitive relation on E. To see that g is a partial ordering on E, let e, f, g e E. (1) 62 = e and hence as e. (2) If eg f and fg e, then of = fe = e and fa = ef = f, consequently e = f. (5) If egf and fgg, then ef=fe==e and fg = gf = f, and hence ge = eg = (ef)g = e(fg) = of = e. Therefore as g. We shall call < the natural partial ordering of E. We define a meetasemilattice as follows. Let X be a partially ordered set. An element b of X is called a lower bound of a subset Y of X if 3'2 b for every y in Y. A lower bound b of Y is a greatest lower bound or meet of Y if b2,c for every lower bound 0 of Y. If Y has a meet in X, it is clearly unique. A partially ordered set X is called a meet-semilattice if every two-element subset {a,b} of X has a meet in X; consequently every finite subset of X has a meet. The meet of (a, b} is denoted by ar~b. A commutative band E is a meet-semilattice with respect to the natural partial ordering of E. The meet, a..b, of two elements a and b of E is just their product ab. From (ba)a = ba2 = be and (ab)b = ab2 = ab, we see that abfia and abgb. 50 Suppose cga and cgb. Then (ab)c = a(bc) = ac = c, and similarly e(ab) = c, whence cgab. This shows that the meet of (a, b} is precisely ab. It is evident that the converse is true. That is, every meet-semilattice is a commutative band with respect to the meet operation. In this thesis, the term.§gmi- lattice will mean meet-semilattice and consequently we will use the term semilattice as synonymous with commum tative band. We know that every finite subset of a semilattice has a meet. Therefore, if E is a finite semilattice it follows E itself has a greatest lower bound or meet; we denote it by 2. Then 25 e for every 6 2E so that ze = ez = 2. Clearly z is unique and hence z is the zero element of E. Lemma 2.1 . If E is a finite semilattice and a, b, c and d are elements of E, then (1) aZb and cad imply ac_>_bd, and (2) azb implies xa be for every xs E. M: (1?) azb and c_>_d gives ab = ba = b and cd 2 do = d, reSpectively. Therefore (ac)(bd) = (ab)(cd) (2) (xa)(xb) every x e E. bd so that ac 2_bd. x2(ab) = xb so that xa_>_xb for 51 In this section and the following one, E will always denote a finite semilattice, 2 its zero element and "z" the natural partial ordering of E. We now define several concepts which are needed in the following arguments. Definition. Let a and b be two arbitrary elements of E. If azb or as b, a and b are said to be comparable; in the opposite case, a and b are said to be incomparable elements, which is exe pressed by aIIb. Further symbols used are < and >, signifying aa, if agb but a£b. Definition. If for a pair of elements a and b of E, a<:b holds and there is no element x in E such that a< x< b, then it is said the element a is covered by b (or b covers a). This situation is expressed by the symbol a<>a). Accordingly, aggb will symbolize that "b either covers or equals a"; in short, "b at most covers a". Definition. A subset T of E is said to be a chain (or simply ordered set) if and only if for every pair a, b in T, either azb or b_>_a. By the length of a chain T consisting of m+1 elements (that is, being of the form x0< x1<:...< xm) we shall mean 52 the non-negative integer m. A chain T from a to b (a, b in T), of length m, is of the form a = x0) x1 > ...>'xm = b. It is said to be a maximal chain from a to b, if x1 covers xi+1 (i = O, 1, ..., m—1). (That is, being of the form a = x0>>...>>xm = b.) Definition. An element a of E is said to have dimension d (written d(a)), where d is the length of the longest maximal chain from a to z (the zero ele- ment of E). we Will make the convention d(z) = 0. Lemma 2.2 . (1) For every a and b in E, a>’b, there exists a maximal chain from a to b. (2) If a>’b, then, for the non-negative inte- gers d(a) and d(b). d(a)) d(b). PM. If a» b, we then have a maximal chain from a to b. If a does not cover b, then there exists an element x in E such that a) x) b. Con- sider the set T(a, b) of all comparable elements y in E with a>»y>>b. T(a, b) is a non-empty and finite chain. Choose x1e T(a, b) with x12 y for all ye:T(a, b). Then a>>x1> b. If x1 covers b, we would have a maximal chain from a to b. If not, cons sider the set T(x1, b) of all comparable elements y in E with x1) y>tb and choose x2 in T(x1, b) such that x22 y for all y e T(x1, b). We then have 53 a>>x1>>x2> b. By finite induction, we get a chain of elements a, x1, x2, ..., In, ... such that a>>x1>>x2>>...>>xn>>...>‘b. Since E is finite, it follows we get an element xm which covers b and hence a maximal chain a>>x1>>...>>xm>>b from a to b. (2) By part (1), there exists a maximal chain from a to b, say a = aO>>a1>>...>>am = b. If b = 2 we clearly have d(a)) d(b). For a>ib and b £ 2, choose a maximal chain from b to 2 having length d(b) = n; say b = b0>>b1>>...>>bn = z. Then a = a0>>a1>>...>>b>>b1>>...>>bn= z is a chain from a to 2 having length m+n. Since m21, it follows d(a)} m+n>n = d(b). This completes the proof of the lemma. A homomorphic image of a finite semilattice (commutative band) is clearly a finite semilattice (commutative band). If 9 is a homomorphism of a finite semilattice E onto E*, we shall use the same symbol "3" to denote the natural partial ordering in the semi- lattice E*. We end this section by deve10ping some inequalities involving the dimension function of ele- ments a in E and 6(a) in E*. 54 Lemma 2.5 . Let E be a finite semilattice and 6 a homomorphism of E. Then d(a); d(6(a)) for every ae:E. Iggggf. Let 6 be a homomorphism of E onto E*, as E and a* = 6(a) sE*. Let 2* be the zero element in E* (z* = 6(2)) and denote d(a*) = m. If m = 0, then d(a)3 d(a*) = 0 since d(a) is a non-negative integer. Suppose m>'O. Then we can find a maximal chain in E* from a* to z* having length m; a* = a5>>a:>>...>>a; = z* . Consider the following subsets of E: 6-1(a:) = {xe:E|6(x) = a:}, where i = O,1,...,m. Since 6 is a homomorphism of E onto E*, it follows 9 :(a*) is1a non-empty subsemigroup of E and 81(a*)n 61(a3) = H for i #1 j. Moreover e 1(a*)e (3* )c:e (a; 1) (i = o, 1, ..., m-1). Indeed if 12 e “1(a;) and yea“: "1(a:+1), then e(x) = and e(y)= reapectively. But ai+ 1’ =*§=* e(zy)= 6(1)6(y) aiai 1 a1+1 so that xyse 1(8115“). Now age 1(as). Choose y1ee-1(a*1*). Then 31 Choose yzs 6‘1(a§). Then a2 = a1y2 s 6'1(a§) and = ay1e;671(85)e'1(a:)§;e'1(a:). Consequently a> a1. a>ia1> a2. Continuing in this manner we obtain a chain a>a1>a2>...>am_1, where 8189-1(a‘i’)- Now zse'1(a;) 55 and z = am_1z. Thus we have constructed a chain a>a1)82)ooo>am_1>am= Z from a to z of length m. Since d(a) is the length of the longest maximal chain, it follows m_<_ d(a) and we have proven the lemma. Definition. An element aeIE is called an atom if and only if a covers the zero element of E. (That is, a>>Zc) Note. An element aezE is an atom if and only if d(a) = 1. Below we introduce some very important subsets of E. Definition. Let a be an arbitrary element of E. We denote the set of all elements x in E satis- fying the inequality xga by (a] and the set of all x in E satisfying xz,a by [a). Lemma 2.4 . Let a be an element of E. Then (1) [a) is a subsemigroup of E and (2) (a] is an ideal of E. 2522;. (1) Let x, ya [a). Then x_>_a and y_>_a so that xyz a2: a. Hence xye [a) which shows [a) is a subsemigroup of S. (2) Let eeE and xe(a]. Then x_<_a and 56 xegxga so it follows xee(a]. Thus (a] isan ideal of E. Lemma 2.5 . Let A be the set of all atoms in the semilattice E. Then B = A~Iz is an ideal of E. Moreover if’ x is the natural homomorphism of E onto E/B, then d().(a)) = d(a)—1 for every asE—B. M. Let peA. That is, let p be an atom in E. Then (p1 = (z, p} and, by Lemma 2.4, it is an ideal of E. However B = H‘fp] so it is also an ideal of E. Consequently we can consider the Rees factor semigroup of E modulo B. The ideal B is the zero element of E/B which we will denote by K2. The remaining elements of E/B are one-element sets {1:} where xEE-B. We denote these elements by Kx = {x}. Let A be the natural homomorphism of E onto E/B. Then K a Kz if aiiB . Let aeE—B. Then }.(a) {a} if a s E-B A(a) = Kae E/B. By Lemma 2.5, d(x(a))g d(a). Since a has d(a)) 1, then there is a maximal chain from a to 2 having length d(a) = m; a = aO>>a1>>...>>am_2>>am_1>>am = 2. Now am_1 isanatom of E so that am_1sB. But 81¢ B for i = O, 1, ..., m=2. Hence Ka = {a}, 57 Ka1 = {afl3 ..., Ham 2 = {am_2} are non-zero elements in E/B and moreover Ka = Kao>>Ka1>>o oo>>K > Kzo am—2 Indeed if there exists Ks E B such that K K ’ / 31-)- 2 Kai+1 (i = O, 1, coo, 111-3), then K = {X} Where X E EmB and KaiOK = K and KoK 8i+1 = Kai+1° Consequently KaiK = {aix}g;fx} and KKai+1 = {xai+1} §;{a1+1}. This implies aix = x and xai+1= a1+1 which yields aiz x2 ai+1. Since ai>>ai+1 we must have a1 = x or ai+1 = x. Therefore K = Kai or K = K so it follows Ka >>Ka for i = O, 1, ..., ai+1 1 1+1 m-5. Therefore we have a chain from x(a) = Ka to A(z) = KZ in E/B of length m-1 = d(a) —1. Hence d(x(a)) zd(a)-1 and more precisely d(a)-1g d(x(a)) S d(a) = m. ' Suppose d(x(a)) = d(a) = m. Then we have a maximal chain in E/B from 1(a) = Ka to K29 (5.1) Ma) = Ka = KO>>K1>>...>>Km_1>>Kz of length m. Since Ki # Hz for i = O, 1, ..., m-1 it follows Ki = {a1} where aie E—B, i = O, 1,..., ma1. But (5.1) implies a = a0>>a1>>...>>am_1. Now am_1¢ B and there exists an atom ame B such that am_1> am. Hence we have (5.2) a = a0>>a1>>...>>am_1>a.m> z which is a chain in E from a to z having length m+1. 58 But this is impossible since the longest maximal chain from a to z has length d(a) = m. Therefore d(x(a)) # m so that d(x(a)) = d(a)-1 and we have proven our lemma. Lemma 2.6 . Let M be a subsemigroup of E which contains the zero element z of E. Then the number of atoms in M is less than or equal to the number of atoms in E. 2593;. Since zeMgE, it follows z is the zero element of M. Thus if p is an atom of M, then p>>z. Let n denote the number of atoms in M and let a1, a2,..., an be the distinct atoms in M. For each a1, form the set T(ai, z), of all comparable elements esE such that a1}, a) z. T(ai, z) is a non-empty chain. From each chain T(ai, 2) choose a minimal ele— ment e1; i.e. eige for every eeT(ai, z). Then we have alas”. z for i = 1, 2,..., 11. However the e1 (1 = 1,..., n) are all distinct; for if e1 = e3, then aiz,ei and 832 e1 which in turn implies z = aiajz of = e1, a contradiction. Thus each atom a1 in M (i = 1,..., n) gives rise to an atom e1 in E (i = 1,..., n) so that n is less than or equal to the number of atoms in E. This completes the proof of Lemma 2.6. 59 ‘Egig. It might help the reader, when reading through a proof involving finite semilattices, to actually represent the semilattice by a diagram. To obtain a dia- gram of a semilattice E (Every non-void finite partly- ordered set can be represented by a diagram, [Birkoff [9], Theorem 4]), represent each element a of E by a small circle C in the plane of the drawing (denoted by the a letter a), such that if a< b (a, beE), the circle Cb is above Qfi now consider each pair of elements x, y in E for which x<a). If some a has no decomposition at all of the form (1), it is said to be irreducible. 61 If the semilattice E has at least two atoms a1 and a2, then clearly its zero element 2 is re~ ducible. Every non-zero element of a finite chain semi- lattice is irreducible. We now develop several neces- sary conditions for a semilattice E, which contains a reducible element a, to be in the class 3+. If a is reducible in E, then there exist elements 81, a2 in E such that a = a1a2 with a1) a and a2) a. Since E is finite we can find elements b and c in E such that a1z,bX>a and 322 c>>a. However bc$a1a2 = a and bcz a-a = a so that be = a. Conse- quently we shall study finite semilattices E 83+ which contain an element a with a = bc (b>>a, c>>a). Lemma 2.7 . Let E be a finite semilattice with EtiflW. Let a, b, and c be elements of E such that a is an atom in E and a = bc (b>>a, c>>a). Then p< b or p< c for every atom p in E. M. To show p>a and c>>a, but pb 14 p and 130 24 p . Clearly these elements are mutually distinct. Consider 62 the maximal chains b>>a>>z and c>>a>>z and the ideals (a] = {2, a}, (b] = {xsEngb} and (0] = {XEEIXS°}° The element p is not an element of either (b] or (0], since bp # p and cp # p. Moreover z, a, b belong to (b] and z, a, c belong to (c]. We need to study two separate cases. Case 1. Suppose (b] = (z, a, b} and (a] = (z, a, 0}. Then (b]r~(c] = (a]. Let q be the number of atoms in E. Then ZSQSJEI-Zv since a and p are atoms in E but b and c are not atoms in E. Let us form the Rees factor semigroup E/(a]. We denote its zero element (namely the ideal (a]) by K2 and the remaining elements by Kx = {x}, where x eE-(a]. The Rees factor semigroup E/(a] is a finite semilattice with I I E/(aJI = IEIa1. Since p, b and c are distinct elements of E not in (a], it follows Kp_= {p}, Kb = {b} and Ko = {c} are non-zero elements of E/(a]. We first will verify that Kb and K0 are atoms in E/(a]; that is, Kb covers K and Ko covers K Suppose we 2 2‘ have Kb?_K> Kz where K; E/(a]. Then K is a non-zero element of E/(a] so that K = {h}, ‘where he E-(aJ 63 and moreover Kb<>K = K and KIJKZ = K Consequently, z° KbK = {bh}c_:.{h} which implies bh = h. Therefore he:(b] = {z, a, b} and since h i a and h f z (for h is not in (a]) we must have h = b. Indeed Kb = K which proves Kb is an atom in E/(aJ. The same type of argument can be applied on Kc' Our next step is to show that each atom e in E (e # a) gives rise to an atom K8 in E/(aJ. Now e is not an element of (a] so that Ke = {e} is a non- zero element of E/(aJ. Suppose KeZK> K29 where Kle E/(aj. Then K = {x} where xe E-(aJ, and Ke OK = K, KOKZ = K Consequently ex = x. Since 2' xyéz we have e_>_x>z. But ecovers x so that e = 1. Therefore Ke = K and we have Ké is an atom in E/(a]. Our above arguments show that to each atom eeaE (e £ a), there corresponds an atom K6 in E/(aJ. Certainly, distinct atoms e in E correspond to dis- tinct atoms K8 in E/(a]. Moreover, Kb and K0 are also distinct atoms in E/(a] which are different from the above atoms Ke. If we let q* denote the number of atoms in E/(a], then our above remarks show that q*2_q+1, where q is the number of atoms in E. Since E 83%, there is a subsemigroup N of E isomorphic to E/(a). Therefore N is a finite subs semilattice of E, with zero element 'g, INI = IEI=1 64 and the number of atoms in N is greater than or equal to q+1. Indeed z = z; for we know zghg and if zi N it follows a and p belong to N (for [NI = IEI—1) so that z = apeaN, a contradiction. There- fore N is a subsemigroup of E, having z as its zero element, and the number of atoms of N is greater than or equal to q+1. From Lemma 2.6, this situation cannot occur and hence the situation in Case 1 is impossible. Case 2. We consider the remaining possibility; namely (b]:D{z, a, b} and (c]:3{z, a, c}. Let Q(b) = {x2 EIb>>x and x £ a} and Q(c) = {xe:E| c>>x, x £ a}. From pb £ p and pc £ p it follows p is not an element of either Q(b) or Q(c). Let R(b) .9qung and R(c) -.= Uch] . Both R(b) and R(c) are unions of ideals so that each is itself an ideal. Let w = R(b)v R(c). w is precisely the ideal of elements of E which are under the elements b and c with the exception of the ele- ment a. That is, if y<:b, or y< c and y % a, then yiaw. Also a, b, c and p are not members of W. We form the Rees factor semigroup E/W. As be— fore, KZ will denote its zero element and Ky = {y}, where ye EaW, its remaining elements. We have Ka = {a}, Kb = {b}, Kc = {c} and K = {p} are non- p zero elements of E/W. Moreover 65 K >>Ka9 Kc>>Ka and Ka, K are atoms in E/W. P Form the following ideals in E/W; (Kb] = {Ks E/wlxgxb} and (K01 = {KeE/WIKg KC}. b pose Ks (KM and K ,é Kz. Then Kz>a and c>>a. Then e>a, c>>a, and 1gd(a)>a, c>>a), d(e)gd(a) = n, eb )4 e and ec # e. We again need to consider two separate cases. Case 1. Suppose d(e))»1. Let A be the set of all atoms in E, let B = A~.z and form the Rees 67 factor semigroup E/B. (This is valid for, by Lemma 2.5, B is an ideal of E.) Denote the zero element of E/B by KZ and its remaining elements by Ky = {y} where ytaE-B. Ka9 Kb, Kc and K6 are non-zero elements of E/B such that Ka = KbOKc (Kb>>Ka9 Kc>>Ka)9 Kb 0Ke ;4 Ke and KO oKe ,4 Ke° Since a, e eEaB, then, by Lemma 2.5, it follows d(Ke) = d(e)-1 and d(Ka) = d(a)—1. Therefore d(Ke) = d(e)-1gd(a)-1 = d(Ka) and d(Ka) d(a)—1< d(a) = n . From E e 3.! we have E/B 55%. Moreover E/B has three elements K Kb, Kc which satisfy precisely the condi- a9 tions of our induction hypothesis. Since d(Ka)< n and d(Ke) gd(1g),it follows from our induction hypothesis that Kec’Kb = K9 and KeoKC = Ke. This contradicts what we have above so that our assumption is false in the case d(e) > 1. Case 2. Suppose d(e) = 1. Consequently we must have d(a)) 1; for otherwise we would contradict Lemma 2.7. Let A be the set of all atoms in E, B = Av z and R = Bmfe}. R is an ideal in E for l R = LJ(x] . 'Form the Rees factor semigroup E/R. Since a, b, c and e are not members of the ideal (R, it fol- lows Ka = {a}, Kb = {b}, Kc = {c} and K6 = {e} are 68 non—zero elements of E/R. Moreover Ka = Kb<3KC, Kb>>Ka, Kc>>Ka, Ke oxb .4 K6 and Ke oxc ,4 KB. We first assert that d(Ka) d(a)-1 = n-1. Recall we always have d(Ka)g d(a) n. Since d(a)>’1, there exists a maximal chain from a to z a = ao>>a1>>...>>an_2>>an__1>>an = z of length n. From be £ at it follows an_1 # 6. But an_1 is an atom in B; consequently an_1e:R. On the other hand, 31¢ R for i = O, 1,..., n-2. Therefore Ka = {a}, Ka1 = {a1}...., Kan_2 = {5111.2} are (rt-1) distinct non-zero elements of E/R and moreover Ka>>Ka1>>...>>Kan_2 >Kz. This implies we will have a maximal chain from Ka to K2 in E/R of length at least (n51). Hence d(Ka)2;n=1. Suppose d(Ka) = n. Then there is a maximal chain from K, to K2 in E/R. Ka = KCO>>KC1>> ,.,>>Kcn_1>> Ken = Kg. , of length n. From Kci fi Kz for i = O, 1,..., n-1, it follows KCi = {c1} (1 = o, 1,..., n-1) where ciezEmR. The above chain induces a chain in E from a to z, namely a = co>>c1>>...>>cn,1 >z. However there exists an atom ye:R such that cn_1> y) 2. Consequently a = co>>c1>>...>>cn_1> y) 2. Thus we will have a maximal chain from a to z of 69 length at least (n+1). This is not the case, for d(a) = n. Thus our assumption is false and d(Ka) = n-1. Now E/Rej‘I’ and E/R has four elements Ka, Kb, Kc and K8 of the same nature as those in Case 1. Moreover 1 = d(e)g_d(Ka) = d(a)—1< n. Our induction hypothesis applies and we obtain the same type of con- tradiction that occurred in Case 1. The arguments used in Case 1 and Case 2 complete the proof of this theorem. The previous theorem gives us a rather strong necessary condition for a finite semilattice E, ad- mitting a non-zero reducible element a, to be in the class 5%. Let a be a reducible element of E. We know that every element e a E, d(e) gd(a), is under one of the covers of a. We shall show that the number of incomparable elements e 8E, with d(e) gd(a) and eflIa, can be at most one. Before doing this,-we need some further preperties. Lemma 2.9 . Let E e.§( be a finite semilattice which has elements a, b, 0 such that a is an atom in E, a=bc (b>>a, c>>a). Let A be the set of atoms of E. 1) If eeA-a, and e<_b, then for every x3e we have either xzb or xgb. 7O 2) If esA-a and ego, then for every x2e we have either x2e or xgc. M. Suppose e eA—a, egb, x2e but x is neither above b nor under b. First of all, x is not under c; for otherwise we have e5 xgc which contra?- dicts that a is the atom in E under both b and c. Also x is not above c. Suppose x20. Then xszzeoe = e and b2xb2e‘. But xb ;4 z, xb .4 b, and xb )4 x. Therefore b>bx_>_bc = 3. Since b>>a, it follows bx = a and consequently a = bxz e> z. This is not possible since a is an atom. Moreover x is neither above a nor under a. Indeed, suppose agx. We have a = bagbx< b which implies a = bxz 92 = e. This again is not possible. Let J = {ye (x]Iy<>Ka9 Kc>>Ka and K8 is an atom in E/L. Moreover K is an atom in E/L since all elements I under x belong to L. Since E/L a 3+ and since there exist elements 71 K Kb, KceE/L with Ka=Kbch, Kb>>Ka, Kc>>Ka a’ and Ka an atom in E/L, it follows Lemma 2.7 applies; namely every atom in E/L is under Kb or Kc' However Kx is an atom of E/L which does not have this prOperty. We have therefore shown if eg b, e_<_ x then x_>_ b or xg b. A similar argument can be applied to show part (2) of this lemma. Lemma 2.10 . Let E5 54' be a finite semilattice which has elements a, b, c where a is an atom in E, a = bc (b>>a, c>>a). Then a is the only atom in E which admits such a decomposition. Proof. Suppose a1e E is an atom different from a which also has such a decomposition. That is, there exist elements b1. 0. in E with a1 = b1c1 (bf»a >>a1). We can apply Lemma 2.7 on both elements a 19 °1 and a1. Consequently, either a1g b or a1g c and either as hi or afgc We have four separate cases 1. to consider. (1) agb1 and a1gb, (2) agb1 and a1gc, (5) ago1 and a1gb, and (4)a.<_c1 and a1gc. In case (1) we have b_>_ bb gaoaa- a and b 2bb1za1a = a . 1 1 1 1 But b>>a and b1>>a1 so in order that both the above inequalities hold simultaneously we must have Using the same arguments for the remaining three cases, one can show that these cases yield the results 72 cb1 = c = b1, bc1 = b = c1 and cc1 = c = 01, respectively. In any event, we must have the two sets (b, c} and {b1, 01} have an element in common. It suffices to assume that b = b1. We have the situation b>>a, c>>a, b>>a1, c1>>a1, a and a1 atoms in E such that a = be and a1 = bc.. By Lemma 2.9, since a1 is an atom, a sA-a, a < b and c 1 1" a1gc1, it follows 0 2 b or 1 013 b. However both situations are not possible since 01 and b cover a1. (Of course, unless = b which in turn is not possible from c C b=a1)o 1 1 This ends the proof of our lemma. Lemma 2.11 . Let E EJW— be a finite semilattice which has elements a, b, c such that a is an atom in E, a = bc (b>>a, c>>a). Let A be the set of atoms of E and N = Ama. Then either N is an empty set or I! INI 1. M. Suppose INI 2.2. Note that from N £ 9 we can apply Lemma 2.7 to conclude that b and c are the only elements of E which cover a. Also, by Lemma 2.7, we know that each element of N is under either b or c. We therefore have two separate cases to consider. Case 1. Suppose there exist elements e1, e22 N '73 such that 913 b and e23 0. Form the Rees factor semigroup E/(a]. It has Kb = {b}, Kc = {c} as non- zero elements. We first show that there is a one-tonone correspondence between the atoms in N and the atoms of E/(a]. Suppose e is an atom of E with eezN. Then Ke = {e} is a nonmzero element of E/(a]. Suppose there is an element Kc E/(a] for which Ke.>_ K> Kz. Then K = {x} where x5 E-(a] and Ke<3K = K. This implies ex = x so that e_>_x> z. Since e is an atom, it follows e = x which in turn implies Ke = K. Thus K is an atom in E~(a]. Note that this implies e K b and K0 are nonmzero elements of E/(a] which are not atoms. Suppose Ke = {e} is an atom in E/(aJ, where ea Em(a]. Let x be an element of E such that e) x) z. If x = a, then e> ai’z. Consequently, e > b or e3 c since b and c are the only elements which cover a. Therefore e2,b2_a or ez,cz_a and this, in turn, implies Keg Kb or KeZKc. We know Kb and Kc are not atoms in E/(a] so we contradict the fact that K8 is an atom in E/(a). Hence x ¥ a. But xyé a and e_>_x>z implies x4 (a]. Therefore Kx = {x} is a nonmzero element of E/(a] and KeaKx> Kz. Since K8 is an atom, it follows e = x. Thus e is an atom of E. In the above two paragraphs we have shown there is a one-touone correspondence between the atoms e of E, ee N, and the atoms Ke of E/(a]. Therefore the number of atoms in E/(a] is INI. We next assert that E/(a] has no atom Ka* = {a*} for which there exist elements Kb* = {b*}, Kc* = {c*} such that Ka* = Kb*()Kc otherwise a* would be an atom in E, b*>>a*, c*>>a* * (Kb*»Ka*’ Kc*>>Ka*); for and a* a b*c*. From Lemma 2.10 we must have a = a* and hence b = b*, c = 0*. However a = a* is impos_ sible, since a*¢ (a]. This proves our desired assertion. From E e 31" we must have E/(a] can be imbedded in E. Therefore there exists a subusemilattice E1 of E which satisfies (11.1) 119,! = IEI_1 . (11.2) number of atoms in E1 is INI, and (11.5) there exists no atom. f’eE1 and elements g, h in B, such that f = gh, h>>f and g>>f. Suppose there is an atom peE and p4E1. From IE1I = IEIa1, b g p, c £ p, it follows b, c 8E1. But E1 is a subsemigroup of E so that a = bceIE1. This cannot occur for we would contradict property (11.5). Thus every atom of E is an element (and, hence, an atom) in E1. However the number of atoms in E is INI+1 (since N = Ana) which contradicts prOperty (11.2). This completes the proof for this case. Case 2. Suppose all atoms es N are under the element b. Then the only atom in E under c is the b atom a and every atom of E c \\//f is under b. a Form the Rees factor semigroup I E/(al. Again Kb = {b}, Kc = {c} are non—zero elements of E/(a). It is necessary to show that K0 is an atom in E/(aJ. Suppose K011i; Kz, where xx: {x} with xe:E~(a]. But KCZ K: implies c22x> 2. Since c covers a and no other element (x:£ a) it follows c = x. Therefore Ki = Kc and we have shown K0 is an atom in E/(a]. Using arguments exactly like the one employed in Case 1, one can show that there is a one—to-one corre- spondence between the atoms in N and the atoms in E/(a]. (Clearly Kb is not an atom in E/(aJ). Hence the number of atcms of E/(a] is equal to INI+1 which is the number of atoms in E. As in Case 1, we can apply Lemma 2.10 to assure us that no atom K3, in E/(a] .has a reducible decomposition. Let Ke be an atom in) E/(a] such that Ke # Kc’ Then e is an atom of E, e £ a. Hence eg;b which in turn implies Ke53Kb° Hence every atom of E/(aJ, with the exception of K0, is under the element Kb° It is clear that d(Kb) = d(b)3 2. Since F}efl¥, then E/(a] can be imbedded in E. Therefore there exists a submsemilattice E of E such 1 that (11.4) IE1I = IEIm1, (11.5) number of atoms in E1 is precisely the number of atoms in E, (11.6) there exists an atom eOe E1 and an element e1ezE., where d(e1) = d(b), such that every atom ptaE1, p £ e0, satisfies p> p, x2>>p). Let p be an atom of -E with p4 E1. Then, since IE,I = IEIm1. b £ p, 1c £ p, it follows b, c are elements in E1. Therefore a = bcsaE1 which is not possible by property (11.7). Hence every atom of E is an atom of E19 and since they are equal in number it follows all the atoms of E, are precisely the atoms of E. From INI32, there is an atom pEE (and thus in E1) such that p ,1 a and p .4 so. From (11.6), pge1. Also, by hypothesis, we have p:§b. Lemma 2.9 applies to give us e13 b or elg b. But d(e1) = d(b) and so 77 b = e1. However b has the property that every atom of E (and hence E1) is under b. This contradicts that there is an atom eoe E1 such that eOXe1 = b. This completes the proof of the lemma. The previous lemma essentially states if we have a finite semilattice E in the class 5% which has a reducible atom a, then E will have at most one other atom p, p £ a. The next theorem will give a similar conclusion removing the hypothesis that a is an atom. We will show that if a is reducible, then the number of incomparable elements eeE with d(e)gd(a), eIIa, is at most one. Theorem.2.32 . Let Fla 5¥ 'be a finite semilattice which has elements a, b, c such that a = bc, b>>a and c>>a. Let M be the set of incomparable elements etaE such that e” a and d(e)g,d(a). Then either M is an empty set or [MI 2 1. 2:32;. We prove this theorem by induction on the dimension of the reducible element a. Lemma 2.11 shows that the theorem is true for d(a) = 1. Assume the truth of the statement of the theorem for all finite semilattices in 39 which contain elements a, b, and c having the above properties with d(a)< n. Let E be a finite semilattice in 54 , let a, b, c be 78 elements in E such that a = bc (b>>a, c>>a, d(a) = n) and let M be the set of incomparable elements eeZE (eIIa, d(e)_<_d(a) = n) such that IMIZZ. Consequently, M contains at least two elements s1 and e2 such that e1, e2, a are mutually incomparable and d(e1) sd(a), d gas). We need to consider three separate cases. Case 1. Suppose d(e1)> 1 and d(e2)> 1. Let A be the set of atoms in E, let B = A v z and form the Rees factor semigroup E/B. We apply exactly the same arguments used in the proof of Theorem 2.8 (Case 1) to show that this case is impossible. Case 2. Suppose d(e.) = 1 and d(e2) = 1. Then d(a)) 1; for otherwise we would contradict Lemma 2.11. Let A be the set of atoms in E, B=:At.z and R = B~{e1, e2}. R is an ideal in E for R = fight] . Again we can apply the same arguments that were used in the proof of Theorem 2.8 (Case 2) to show that this situation cannot occur. Case 5. Suppose d(e1) = 1 and d(ez)>’1. Since d(a)2;d(62), it follows d(a)) 1. Let A be the set of all atoms in E, B = A~az and R = B - e1. We can form the Rees factor semigroup E/R and apply arguments similar to the one in Theorem 2.8 (Case 2) to show that this case also is not possible. 79 In this section we have deve10ped two rather strong necessary conditions (namely, the statements of Theorem 2.8 and Theorem 2.12) when a finite semi- lattice E, having a reducible element, will be a mem- ber of the class JW'. However we have not been able to completely classify the finite semilattices in the class 3*. Below is a diagram of a semilattice which satisfies our necessary conditions but does not belong to 3%. 1 I a c /,\\\/// o\ a \z 2.5 Finite semilattices of Groups in 3¥ This seaflon will deal briefly with the study of finite commutatits semigroups Se:j¥ whose maximal unim <+ . ent subsemigroups are groups. That is, if we let E T} C) etc its set of idempotents (E is a finite semilat- O (D :3 tice), Ge the maximal unipotent subsemigroup of S containing the idempotent eezE but no other, then git, Seonrzfl if e75f in E, Ger§;Gef for all e, f in E, S: and Ge is a group for all e5 E. For such semigroups S, 80 we shall use the abbreviated expression, "S is a semi— lattice of groups". For the case when E is a chain, we will say that S is a chain semilattice of groups. The purpose of this section is to determine nec» essary conditions when a finite semigroup S, which is a semilattice of groups, will be in the class gt. Throughout this section we will use the following notations S will denote a finite semilattice of groups, E = {e1, e2,..., en} its set of idempotents and Gm, (m = 1, 2,..., n), the group having identity element . n I Then S = I5 G19 GIG3€;Gk for eiej = 9k and e 1 m. Lemma 2.15 . Let SEZJW’ be a finite semilattice of groups, and e, the zero element of E. Then gG, = {g} for all geG1 and i = 2, 5,..., n. Egggf. Flrst of all, for e1< e3 we must have GiGj = G1. Let s, be the zero element of the finite semilattice E = {e1, e2...., en}. From . n n G1S = G1(1L.J1Gi) =12). 919199“ = G19 it follows C, is an ideal of S. Therefore, we are able to form the Rees factor semigroup S/G1. Denote its zero element (the ideal G1) by K91 and its re_ maining elements by Kx = {x} ‘where xe:S-G1. Note that K82 = {€2}9 K63 = {93% ..., Ken == {en} 81 are idempotents in S/G1. Moreover, these and the idem- potent Ke1 are precisely all the idempotents of S/G1. Consequently, the number of idempotents in S/G, is [E[ = n. - Since 5 a 9% it follows s/G1 can be imbedded in S. Hence there exists a subsemigroup T of S such that (1) [T] = [S/G1l = [SI—[G1[+1, (2) T has a zero element and (3) T contains [E[ = n idempotents. From (3), we must have E§;T and in particular e1ezT. Since e1 is the zero element of E and since T has a zero element, it follows 61 is the zero ele- ment for T. However G1r~T = e1; for e1 is the identity element of G1. But T<;S, Tr\G1 = e1 and G r\G 1 J . T€;(S-G1)~Je1. On the other hand, ITI = lSI-lG k1 = fl for j = 2,..., n, all imply that = I(S-G1)v e1l, which implies T = (S—G1)v e1. That 7‘ is, T: gait. e,. Since 6, is the zero element of T we have e1gm= e1 for all gmsGm, m: 2,..., 11. Let g1;v§e11 be an arbitrary element of G1, and gIn an arbitrary element of Gm (m = 2,..., n). Then g1sm = (g1e1)gm = g1(e1sm) = 8191 = $1 - This proves the lemma. Theorem 2.14 . Let Se 3%‘ be a finite chain semilattice of groups. Then 82 8183 = 81 for all gie G1, gje Gj where ei<'ej, i = 1, 2,..., n and j = 2, 3gooog no Proof. It suffices to show that for 91g; = 81 all gje:Gj with e, < 833 for if this is true, then 8183 = (Siei)gj = 81(9183) = giei = $10 Because S is a chain semilattice of groups, we have E = {e1<<...<b, a i b, and let xeS. Then xeGk, for some integer k; (a) If ek G; (1 ,4 k) defined by t(x) = x* is clearly an isomorphism of G1 onto G1. Therefore GI, (i = 1, 2,..., n), are the maximal groups of 8* containing the idempotents eI. * = a» . l- i- = (Gm {em} ) Also Girth G (1 fi 3) since Gin G3 =¢ (1£J)0 Evidently the semigroup S* has the decomposi- n tion, 8* = UGI, GfoG* == ¢ and G* oG*C_'_'_G* h1 * J 1 3 k where e; 083 = e11. Let g3 a G3 and let e: be the identity element of G; where ei< e3. For i = 1, 2,..., m-1, we have e: = {31}; consequently, by our assumption, ez<>g3 = e;. Since G g §;G , it follows e*<>g* = e* for all m j m m j m g* eG* where eg< e15. The above shows that J 3 e*0 * = e* for all *8 e*