a homomorphism theorem for finite semigroups

A Homomorphism Theorem for Finite Semigroups*

by

JOHN RHODES'["

1. Introduction

Let S and T be finite semigroups and let 0: S ) ) T be an epimorphism. 1 • It will be proved in Section 2 that 0 can be written in the form 0 = 0, , • • • 0 1 ,

where each Ok is an epimorphism such that each of 01, 03, 05, • • • , when restricted to any subgroup of its domain, will be one-to-one (in fact, each will be one-to-one on each ~g%class o f its domain), and 0~, 04, 0n, • • • will be such that 02k(al) -~- 02k(a2) implies that a I and a 2 both generate the same principal left ideal and the same principal right ideal (that is, 0 2 k ( a O =

0 2 k ( a 2 ) implies a~ X a2 in the domain o f 02k). We call epimorphisms of the first class y(,,~)-epimorphisms and

epimorphisms of the second class ,,~¢'-epimorphisms. Thus every epimorphism between two finite semigroups can be decomposed into y(~")- and Yf-epimorphisms.

In Section 5 we shall prove that the collection of all aq~'-epimorphisms on S forms a modular lattice (in fact, every two such congruences com- mute). Each ~,a-epimorphism 0 satisfies an "honest-classical" homomorphism theorem in the sense that a finite sequence of normal subgroups, Ker 0, can be assigned to 0 in such a way that Ker 0 determines 0 in a natural manner . No such theorem is valid for y(,,qV)-epimorphisms. However, the following classification theorem is very useful in s tudying y (~) - epimorphisms.

We say that the epimorphism 0: S • > T is a maximal proper epimorphism (MPE) if, whenever 0 = 0201, where 01, 02 are epimorphisms, then

• This research was sponsored in part by the Office of Naval Research, Information System Branch, Contract Number: Nonr-4705(00), the United States Air Force, Office of Scientific Research, Grant Number: AF-AFOSR-848-65 and AF 49(638)-1550 and the United States Air Force, Air Force System Command, System Engineering Group, Wright-Patterson AFB, Ohio, under Contract Number: AF 33(615)-3893.

? University of California, Berkeley, the Institute of Advanced Study, Princeton, New Jersey, and the Krohn-Rhodes Research Institute, 2118 Milvia Street, Berkeley, California. This paper was presented at the Conference on the Algebraic Theory of Machines, Lan- guages and Semigroups, August 29-September 8, 1966, at Asilomar, California, organized by the Krohn-Rhodes Research Institute.

In this paper an epimorphism means an onto homomorphisra.

289

MATHEMATICAL SYSTEMS THEORY, VOI. | , NO. 4. Published by Springer-Verlag New York Inc.

290 JOHN RHODES

exactly one of 02 and 01 is one-to-one. Clearly every epimorphism 0 which is not one-to-one can be written as 0 = 0,, • • • 01 with each Ok an MPE. The above results imply that each MPE is a T(a~)-epimorphism or an Yd'- epimorphism. Further, we prove that every MPE 0 has a J - c l a s s J in its domain such that 0 is one-to-one on the complement of J. Thus an anal- ysis of 0 via the Green-Rees theorems applied t o J is available. This allows us to prove in Section 3 that each MPE lies in exactly one of four classes I, II, III , or IV. Class I is the class of the a~-epimorphisms, and classes II, III and IV subdivide the MPE's which are "y(~)-epimorphisms.

The above classification of MPE's is extremely useful in proving (and discovering) statements true for all finite semigroups by the popular method of considering the "minimal counter-example" (i.e., induction) and then applying the classification theorem. Of course the statement is valid by induction for each MPE image, so that it is necessary only to "pull the statement through the MPE" via the classification theorem for MPE (thus contradicting the existence of a counter-example). The recent results in the theory of finite-state machines and finite semigroups an- nounced by the author in [ 1 ] are proved by the aid of this technique. For another general technique, see also the Examples and Remarks 4(k).

The author would like to thank Mr. Price Stiffler for his generous help in prepar ing this manuscript.

2. The Homomorphism Theorem

N o t a t i o n 2 .1 . In the following all semigroups are of finite order. The undef ined notation follows [2].

Let S be a semigroup. Following Green (see [3] or [2]) we write sl W s2 for sl, s2 E S if S l s l = Sis2 and s i S i = s2S I.

Definition 2.1. Let 0: S • • T be an epimorphism. We shall call 0 a T(~')- e p i m o r p h i s m if 0 is one-to-one on each W-equivalence class of S. In particular, every y(gP)-epimorphism is one-to-one when restricted to each s u b g r o u p of its domain.

0 is an ~ - ep imorph i sm if sl, s2 ~ S and O(sl) = O(s2) imply sx aZP s2.

T H E O R E M 1. L e t O: S • ) T be a n e p i m o r p h i s m . T h e n 0 c a n be w r i t t e n

as 0 = O, • " • 01, w h e r e 01, 03, 0~, • • • are y (agP)-ep imorphisms a n d 02, 04, 06, " ° "

are agP-epimorphisms.

N o t e 2 .1 . In [5, See{. 6] we proved a weaker form of this theorem but with a proof which can possibly be extended beyond the finite case. Here we shall use the assumption of finite order strongly and prove more. We shall also require the results proved in Section 2 for the proofs of Section 3.

To prove Theorem 1 we introduce the following definition.

Definition 2.2. Let 0: S ~ T be an epimorphism. Then 0 will be called a maximal proper epimorphism (MPE) if, whenever 0 = 0201, 01, 02 are epimorphisms, exactly one of 01 and 02 is one-to-one.

A Homomorphism Theorem for Finite Semigroups 291

Clear ly any e p i m o r p h i s m 0 which is not o n e - t o - o n e can be wr i t t en as 0 = 0,, • • • 01, w h e r e each Ok is an MPE. T h u s it fol lows easily tha t T h e o r e m 1 is equ iva l en t with the fol lowing.

T H E O R E M 2. Let 0: S ~ > T be an M P E . Then 0 is either a y(o~)-epimor-

phism or an a~-epimorphism. T o p r o v e T h e o r e m s 1 a n d 2 we r e q u i r e the fo l lowing def in i t ion a n d

l e m m a .

De f in i t i on 2.3. Le t 0: S ~ ~ T be an e p i m o r p h i s m . T h e n a J - c l a s s J o f S is O-singular i f a n d on ly if 0 is o n e - t o - o n e on S - J .

L E M M A 2.1. Let O: S • ) T be an M P E . Then there exists a O-singular J -

class o f S. P r o o f Let 11 be a m e m b e r which is m a x i m a l ( u n d e r set inclusion) o f

the set o f all I such tha t I is an ideal o f S a n d such tha t 0 res t r ic ted to I is one - to -one ; 11 # S since 0 is not o n e - t o - o n e on S, bu t I m a y be e m p t y ( for c o n v e n i e n c e we shall cons ide r the e m p t y set 0 to be an ideal). Le t J l be a ~<-minimal e l e m e n t o f the set { J : J is a ,,C-class o f S a n d J f3 11 = ~} (see [2] o r [3]) . Le t 12 = 11 U J1. T h e n 12 is an ideal o f S p r o p e r l y c o n t a i n i n g 11. T h u s f r o m the def in i t ion o f 11, 0 res t r i c ted to 12 is not one - to -one .

Le t m o d 0 be the c o n g r u e n c e on S g iven by sl -- s2 ( m o d 0) if a n d on ly if O(sO = O(s2). Let - be the equ iva l ence re la t ion on S g iven by sl -= s2 f o r sl, s2 ~ S if e i t he r

(2.1) sl = s2

or

(2.2) s, # s2, Sl, s2 E 12, O(sl) = 0(s2).

Clear ly --- is an equ iva lence re la t ion a n d

(2.3) sl --- s., impl ies sl - s2 ( m o d 0)

a n d

(2.4) sl, s2 e 12 impl ies tha t sl --- s2 if a n d on ly if Sl ---- s2 ( m o d 0).

Consequen t l y ,

f o r s o m e sl, s2 ~ 12, sl = s2 a n d Sl # s2. (2.5)

F u r t h e r ,

(2.6) s, -= s2 a n d sl # s2 imp ly sl, s2 e 12 a n d at least o n e o f sz, s2 lies in J1.

W e nex t ver i fy tha t - is a c o n g r u e n c e on S. Le t x e S, sl --- s2 a n d sl # s2. T h e n , by (2.6), sl, s2 ~ 12 a n d by (2.4), O(sl) = 0(s2). T h u s six, s2x, XSm, xs2 ~ 12, so tha t 0(XSl) = 0(xs2) a n d O(slx) = O(s2x). H e n c e , by (2.4), xsl = xs2 a n d s lx =- s2x. T h u s ~ is a c o n g r u e n c e .

Le t 01: S • ~ S~ ==- be the n a t u r a l e p i m o r p h i s m (see [2] ) , a n d let 02:

292 JOHN RHODES

S/-- ~ ~ S /mod 0 be the natural ep imorph i sm which is well def ined by (2.3). T h e n 0 = 0201, and 01 is not one- to-one on /2 by (2.5). T h u s 02 is one- to-one since 0 is an MPE. T h e r e f o r e = is equivalent to mod 0. Now it follows f rom (2.6) that J1 is a 0-singular J -c lass . This proves L e m m a 2.1.

We state two corollaries to Lemma 2.1.

C O R O L L A R Y 1. Let O: S ~- ~ T be an M P E . Let I be an ideal o f S such that 0 restricted to I is not one-to-one. Then 0 is one-to-one on S -- I and O(S - I) r) O(I) is empty.

C O R O L L A R Y 2. Let O: S > ~ T be an M P E . Then there exists a non-empty

proper ideal Is o f S, an ideal 11, perhaps empty, of S, and a J-c lass J o f S such that

(a) I 1 ~ /2, I1 U J = / 2 ; (b) 0 is one-to-one on I1 and 0 is not one-to-one on 12; (c) 0 is one-to-one on S - 12 and O(S - I 0 N 0(12) is empty.

In particular, these properties imply that 0 is one-to-one on S - J . We now prove T h e o r e m 2. Let 0: S > > T be an MPE wi thJ a 0-singular

J-class . I f 0 is a T(o%~')-epimorphism we are done, so that we may assume that 0 is not a y(fft0-epimorphism. Thus

(2.7) there exis t j l , j2 ~ J wi th j l # j 2 such that j l oZdj2 and O(j l )= O(j2).

Let ~ be the equivalence relation on S def ined by sl --- s2 for si, s2 ~ S if

(2.8) sl ~ s2 and O(sO = O(s2).

Clearly -- is an equivalence relation and f rom (2.7) we have

(2.9) there exist jl, j2 ~ J , j l # j2, such that j~ -~ j2-

We next verify that - is a congruence on S. Let x ~ S, sl ---- s2, Sl # s2. T h e n f rom (2.8) and f rom the fact t h a t J is 0-singular, we have that sl, s2 E J , Sl 9g ~ s2 and O(sl) = O(s2). Now si ~ s2 implies six ~ s2x so that e i ther

(2.10) six, s2x ~ S - J and O(slx) = O(s2x)

or

(2.1 1) six, s2x ~ J and O(slx) = O(s2x).

I f (2.10) holds, then six = s2x s inceJ is 0-singular, and thus six - s2x. I f (2.1 1) holds, then s~x ~d s2x since six J si implies six ~ si, which in tu rn implies six ~ s2x since sl ~ s2. Already sl ~¢ s2 implies six ~ s2x, hence six ~ s2x (see [3]). T h e n (2.11) implies six ~ s2x. By applying dual a rguments we may complete the p r o o f that - is a congruence .

Let 0~: S ~ >S/~- and 02: S / - ) ) S /mod 0 be the natural epimorphisms. T h e n 02 is well def ined by (2.8) and 01 is not one- to-one by (2.9). T h u s 02 is one-to-one, so that --- is equivalent to rood 0. But then 0 is an o¢t ~- ep imorph ism by (2.8).

This proves T h e o r e m 2 and, afortiori, T h e o r e m 1.


$. Classification of Maximal Proper Epimorphisms (MPE)

Notation 3.1. Let a(S) be an equivalence relation o f S (e.g., any of the Green relations). Let 0: S > ; T be an epimorphism. T h e n 0 will be called a y(a)-epimorphism i f 0 is one-to-one on each a(S) equivalence class. 0 will be called an a-epimorphism i f O(sl) = 0(s2) implies sla(S)s2 for all s~, s~ ¢ S.

Let X be a subset o f the semigroup S. T h e n

X ~ =(.J { J : J is a J -c lass o f S a n d J N X ~ 0 } .

Let 0: S > -" T be an epimorphism and l e t J be a J -c lass of S. T h e n o(J) is contained in a J - c l a s s J ' o f T, so that O(J) ~ = J ' . Clearly O-l(O(J) ~) is the union of disjoint J - c l a s s e s J =J~, • • • , Jk of S, i.e., o-l(o(J) ~) =J~ u • • • u

Jk. In this case we say t h a t J has O-order k. Notice that (]1, • • • ,Jk} = (]* : J~ is a j - c l a s s o f S and o(J #) _c o ( ] ) : = J ' } so that 0-1(0(J0 t ) - 0-1(0(,]1 ---J)>') for 1 ~ j ~ k and thus the 0-0rder of J~ is k for 1 ~ j ~ k. Also observe that for two J-classes J~, J2, o-l(o(Jl) 0~) and 0-1(0(J2) t ) are either disjoint or equal.

Notice that 0 is a J - e p i m o r p h i s m if and only if each J -c lass of the domain of 0 has 0-0taler 1.

T H E O R E M 3. Let O: S ) ) T be an MPE. Then there exists either one or two distinct O-singular J-classes of S. 0 has exactly two distinct O-singular J - classes i f and only i f 0 is a y(~)-epimorphism. I f 0 is a J-epimorphism, then 0 has exactly one O-singular J-class, although the converse is in general false.

Proof Lemma 2.1 implies the existence of at least one 0-singular J - class. Suppose J1, J2, J3 are three distinct 0-singular J-classes. Let sl, s2 S, sl # s~. Then there exists an integer k, 1 ~< k ~< 3, such that sl, s2 E S --Jk sinceJ1,J2,Ja are pairwise disjoint. Thus O(sa) ~ O(sz), whence 0 is one-to- one, which is a contradiction since 0 is an MPE. Thus there are at most two 0-singular J-classes. This proves the first s tatement of Theo rem 3.

I f 0 has two distinct 0-singular classesJ~,J2, then each J - c l a s s J o f S is contained in S - J 1 or S - J 2 so that 0 is one-to-one o n J . The converse follows f rom Theorem 4, which will be proved later. For suppose 0 is y ( J ) . Then 0 is not J , or else 0 would be one-to-one, so that there exist at least two J-classes J1,J2 which are mapped by 0 into the same J-class . From Theorem 4 we have then that the 0-0rders o f both J1 and J~ are 2, and that the 0-order of J is 1 f o r J #J~,J2 . From this and f rom the fact that 0 is y ( J ) it follows tha t J i , J~ are both 0-singular.

I f 0 is a J - e p i m o r p h i s m with two 0-singular classes, then, by the argu- ment of the preceding paragraph, 0 is also y ( J ) , so that 0 is one-to-one. This is a contradiction, so that 0 has exactly one 0-singular J-class . This, together with Example 4(e), proves the last s tatement o f Theorem 3.

T H E O R E M 4. Let O: S ) ) T be an MPE. Then exactly one of the following occurs.

(a) Every J-class of S has O-order ].

294 JOaN RHODES

(b) Exactly two distinct if-classes o f S have O-order 2 and the remaining have O-order 1.

Furthermore, case (a) occurs if and only if 0 is a if-epimorphism (and case (b) occurs if and only if 0 is not a if-epimorphism). In case (b) each O-singular if-class has O-order 2, although the converse is not in general true.

Proof L e t J be a if-class orS of 0-order k, so that 0-1(0(J) j) =J1 u • • • u Jk,J =J1, and letJ * denote a ~<-minimal member of {J~, • • • ,Jk}. Then , of course, o(Jj) c j ' for 1 ~< j ~ k, but fur thermore 0(J*) = J ' (see [5, Sect. 3]). Thus, since 0(Jj) _ c j ' for 1 ~<j ~< k, it follows that for eachji eJ i there exists j* E J* so that O(j*)= O(ji).

We next show that i f J is aif-class of S, then the 0-order of J is <~2. Sup- pose not. Let 0-1(J) ~ = J 1 u • • • u Jk, J~ = J , k i> 3. Let J* be a ~-minimal member of {J1," " " ,J~} and let J** be a ~<-minimal member of {J1, " " " ,Jk} - {J*}. Let I be the union of a l l J such t h a t J is a if-class o f S a n d J ~<J* o r J ~ J * * . Then I is an ideal o r S with J* u J** c I and

J* ~ J**. Thus, by the preceding paragraph, for any j** e J** there exists j* e J* such that ~0(1"*) = ¢p(j'**). Since j* # j**, 0 is not one-to-one on I. Then, by Corollary 1 to Lemma 2.1, 0 is one-to-one on S - - I and O(S-I) N 0(I) = o. Now letJ j e { J , , . • • , J k } - { J * , J * * } ; such aJ j exists since k >~ 3 by assumption. But then by the definition o f I , Jj f3 1=o. ThusJ j C_ S - - I and O(Jj) 0 o(J*) = o. But this is a contradiction since as before O(J*) =J'., a if-class, and o(Jj) c J' for all 1 ~<j ~< k. Hence we must have k <~ 2. Tha t is, i f J is any if-class of s, then the 0-order of J cannot exceed 2.

Now choose I~, 12 a n d J as in Corollary 2 to Lemma 2.1. LetJ~ be a if-class of S contained in S - Is. Then from what has preceded, either the 0-order of j1 is 1 or else o-l(o(J1) j) = J l u J2 for some if-class J2. We shall show that this last case cannot occur. I f j2 c s - Is, then by renaming if necessary we may assume without loss of generality thatJ~ is a ~<-minimal member of {J1, J.,}. But then there exist j2 aJ.~,j* ~J~, so that O(j*) = O(j2),

j,, ~ j * . But this is impossible since 0 is one-to-one on S --I~. ThusJ.2 C 12. Consequently,J2 must be a ~<-minimal member of {J1,J2}, so that 0(Jl) c o(J2). On the other hand, from the choice of Is and from the fact that

J1 c_ S - I~ , J2 c_ Is, it follows that 0(J2) f'l o(J~) c 0(12) n o ( s - I2 ) - - -o . This is a contradiction, so that if J1 is a if-class of s contained in S - Is, then the 0-order of J1 is 1.

LetJ~ c_ Is, and suppose that 0-1(0(J~)£) =J1 u J2,J1 #J2. We may also assume J2 c_C_ I2, for otherwise we can apply the above arguments. As before, we have that 0(Ja) f-I o(J2) ~ o. But then J1 and J2 cannot both lie in I1 since 0 is one-to-one on 11. Therefore , since Is = 11 U J , one of the classesJ1,J2 must be the class J , and of course the other must be the remaining if-class of 0-~(0(J) t) =J1 u J2. Thus we have shown that if any if-class J1 o r s has 0-order 2, then J also has 0-order 2, and ei therJ~ = J or J1 = J ' , where 0-1(0(J) j) = J u J ' . Tha t is ,J a n d J ' are the only if-classes with 0-order 2. Thus either (a) or (b) occurs.


Clearly any epimorphism ~0: S > > T is a J - e p i m o r p h i s m if the ,p-order of J is 1 for each J-c lass J of S. Conversely, suppose there exist J-classes Jl,J2,J1 #J~, with ~o(J1)/= ~ ( J 2 ) J = J ' c_ T. By [5, Sect. 3] at least one o f J1 and J2 must map onto J ' . Hence ,p(J~) N ~0(J2) # ~ cannot be a J - epimorphism. This proves all but the last s tatement of T h e o r e m 4.

I f J is a g-singular class of g-order 1, then 0(J) A O(S - J ) = ~, and, since 0 is one-to-one on S - J , it follows that 0 is a J - e p i m o r p h i s m . Thus in this case (a), and not (b), occurs. Finally, Example 4(e) furnishes an example of a J -c lass with g-order 2 which is not g-singular. This proves Theo rem 4.

T H E O R E M 5. Let 8: S , ~ T be an MPE. Then 0 lies in exactly one of the following classes I, H, III or IV:

~: S ' ~ T is of Class I / f ~o is an o~-epimorphism (which is not one-to-one) with a unique ~o-singular J-class J.

~0: S ) ~ T is of Class I I / f~o is a y(•)-epimorphism and a J-epimorphism (which is not one-to-one) with a unique ~o-singular J-class J.

~0: S ~ > T is of Class III i f ~o is a y(f'C)-epimorphism which is not a J - epimorphism (and so is not one-to-one), but there exists a ~o-singular J-class J2 of S and the ~p-order of each J-class of S is 1 except for exactly two J-classes, J, and J.,, whose ~p-orders are 2, and, moreover, Jm < J2.

~0: s ~ ; T is of Class IV/f~o is a y(J)-epimorphism which is not a J-epimorphism (and so is not one-to-one) but there exist exactly two ~o-singular J-classes of S, J1 and J2, and the ~o-order of each J-class J of S is 1, exceptJbrJ1 and J2 whose ~o-orders are 2. Moreover, Jm and J2 are <~-incomparable (i.e., both J1 ~ J~ and J2 <~ J1 are false).

Before giving the proof o f T h e o r e m 5 we make the following remark. Remark 3.1. (i) I f 0 lies in Class I, then ,p is a J - e p i m o r p h i s m , ~0(J) 71

~p(S - J ) = ~, ~, is one-to-one on s - J , and ~0 restricted toJ is not one-to-one but j , , j2 ~ S, ~o(jl) = ~o(j2),j~ #j2, implies thatj~,j2 ~J and j, g'~j., in S. (ii) I f 0 lies in Class II, then ~(J) N ~o(S - J ) = ~, ~0 restricted to S - J is one-to- one, and ~0 restricted t o J is not one-to-one. However, ifj~,j~ ~ S,j~ # j~ , ~o(j~) = ~0(j2), thenj~,j2 ~ J andj~ not ~-equiva len t to j2 in S. (iii) I f 0 lies in Class III, then

~-'(~o(J,) ~) = ¢-'(,p(/2) ~) = J , u J2,

while ~0-a(~p(J) ) = J f o r J # J~ , J , . T h e n it also follows that i f J is a J -c lass of S a n d J < J , , t h e n J <~J1 < J2,J~ ~ (s~J2 to J2S ~) ~ ~, ¢ is a y ( J ) - ep imor - phism if and only ifJa is a ~o-singular j -c lass , every ~o-singular J -c lass of S is ei therJ~ orJ~, and lastly ~o(J~) = J ' is a J -c lass of T with ~p(J.,) _c ~o(J0 = J ' . (iv) I f 0 lies in Class IV, then

¢-'(~o(J,)~) = ¢-'(~o(/~) ~) = j , to j~

and ~o-'(~o(J) ~) - J , w h e r e J #Jm, J=. T h e n it follows that ~o(1~) = ~o(J2) = J '

296 JOHN RHODES

is a J-c lass of T, J1 andJ~ are both non-regular J-classes of S, a n d J <J~ if and only i f J < J 2 for all J - c l a s ses J orS.

We now prove the assertions of both Theorem 5 and the above remark together.

First observe that Classes I, II, III, and IV are pairwise disjoint, since any epimorphism which is both an oT'-epimorphism and a , / (~)-epimorphism is one-to-one. By Theorem 2 of Section 2, 0 is either an ~,"-epimorphism or a y(•)-epimorphism.

First assume that 0 is an ~ ' -epimorphism. Then, in particular, 0 is a J - e p i m o r p h i s m so that Theorem 3 implies that there exists a unique 0- singular J -c la s s J . T h e n clearly 0 lies in Class I.

Now assume that 0 is a y(gg)-epimorphism and a J - ep imorph i sm. Again Theorem 3 implies that there exists a unique 0-singular g-c lassJ . Then clearly 0 lies in Class II.

Thus we have shown that when 0 is a J - ep imorph i sm, then 0 lies in Class I or II depending on whether 0 is an ~"-epimorphism or a y(a~")- epimorphism. Clearly, Remark 3.1 (i) and (ii) follow.

Now assume that 0 is not a J - ep imorph i sm. Since it is therefore not an aT'-epimorphism it must be a T(gP)-epimorphism by Theo rem 2. Also, in Theorem 4 above, 0 must be of type (b). LetJ1,J2 be the two f -c lasses of S with P-order 2, so that

(3.1) 0-'(0(J1) ~) = 0--1(0(J2) y) = J l to J2

(3.2) o-t(o(J) "f) =J, J # J1,J~

wheneverJ is a f - c l a s s of S. Of course, we may assume that either Jr <J2 or else that J1, J2 are ~<-incomparable. By Theorem 4(b) every 0-singular f - c lass lies among Jr, J2, and at least one of J1,J2 is P-singular.

Suppose that J l < J2. By what we have just stated above, either J , or J2 is O-singular. Suppose that J2 is not 0-singular. T h e n J , is 0-singular and by (3.1) 0 restricted to Jr is not one-to-one. We now consider the equivalence relation st --- s2 if Sl = s2 or sl # s2, sl, s2 C Jr , 0(S1) = 0('~2)" Then --- is a congruence and Or: S ~ ~ S / - , 02: S/-- ) ) S/rood 0 are well defined with 0 = 0201. 0t is not one-to-one so that 82 is one-to-one, whence --- is equivalent m mod 0. Hence 0 is a J - ep imorph i sm. This is a contradiction, so that

J2 must be 0-singular and 0 must be of Class III. From Theorem 3, ~0 is a y ( J ) - ep imorph i sm if and only if Jr is also a

~0-singular f -c lass , and it follows f rom Theo rem 4(b), (3.1) and (3.2) that every ~.0-singular J-clasPs of S is eitherJ~ orJ~. The last statement of Remark 3.1(iii) follows f rom Proposition 3.2(b) o f [5]. See below.

Now again, under the assumption that 0 is not a J - e p i m o r p h i s m , let us suppose that the two f-c lasses Jr,J2 are ~<-incomparable, that is, both

J t <~J2 and J2 ~<j~ are false. From [5, Prop. 3.2] we have that 0(Jr) = o(J2) = J ' , where J ' is a f - c l a s s of T. Now for k = 1, 2 let mk be the equivalence relation on S given by st mk S2 if St = S2 or st, s2 ~Jk and O(&) = 0(s2). We claim that -=k is a congruence. To verify this, suppose that x, s~, s2 ~ S, st # s~ and


s~ =~ s2. T h e n we must show that s~x =-k s~x and xs~ -=k xs2. Clearly s~ ---k s2 implies that O(s~) = O(s2) and thus O(s~x) = O(s2x). From (3.1) and (3.2) it follows that either (1) sax and SsX lie in J l u J2, or (2)six and s~ both lie in S - (J~ u J2). In case (2) s,x = s~x since 0 is one-to-one on S -- (J~ u J2). Now ifs~ ~ s2 and if s, -=k s~, then sl, s2 EJk. Thus in case (1) six, s2x EJ1 to J2 but s~x, s2x do not lie inJk,, where k' = 1, 2 and k' # k, sinceJk, < J r is false. Thus s~x, s2x E Jk and O(s~x) = O(s2x), so that s~x =-k S2X. There fore ----k, k = 1, 2, is a congruence on S.

Since s~ ---k s2 implies O(sl) = 0(s2) we may write 0 = 0201, where 01: S > S/=--k and 02: S/~k > ' T, where 0~ is the natural epimorphism, and 02Is]k= O(s), where [S]k is the ----k-equivalence class o f s. There fore exactly one of 0~ and 02 is one-to-one. Suppose that 02 is one-to-one. T h e n mod 0 is the same as ----k- But then 0 is a J - ep imorph i s m , and this is a contradiction. Thus 01 is one-to-one, which is to says1 =ks2 if and only if sl =s2 fork = 1,2. But this is equivalent with 0 being one-to-one when restricted toJk for k = 1, 2. Thus 0 is a y ( J ) -ep imorph ism, and J! and J2 are 0-singular J-classes.

Now suppose that ~0: S ~ ) T is an epimorphism of Class IV. We must verify the assertions o f Remark 3.1(iv). Proposition 3.2(b) of [5] implies that ~0(Jl) = ~0(J2) = J ' , w h e r e J ' is a J -c lass of T, and that bothJ~ andJ~ are non-regular.

Now suppose t h a t J <J~, so tha t J #Jx,Ju. Then there exist j ~J,j~ eJ1, s~, s~ e s I such t h a t j = sj,s~. Choose j2 eJ~ so that ~0(j~) = ~0(jx). Then ~0(j) = ~o(s~j,sz) = ~o(s,)~o(j,)~o(s~) = ~o(s,~(js)~o(s2) = ~o(s,j~s~). I f we use (3.1), (3.2) and the fact that ~0 is y ( j ) , then ~0(j) = ~0(sljzs2) wi th j # s,j~s~ implies that

j , s,j~s~ e J , t.J Jz. However , J #J , , J~ so t h a t j ~J, to Ju. Therefore we must h a v e j = slj~s~, which implies t h a t J < J z . This proves Theorem 5 and Re- mark 3.1.

The next theorem is an application of Theorem 5 under additional re- strictions on S.

THEOREM 6. (a) I f S is a union of groups, then every MPE is either of Class I, II, or IlL Furthermore, i f 0 is an MPE of S of Class III with respect to J1 and J2 then 0 restricted to J~ is also one-to-one. Thus 0 is a y(J)-ep/morphism and J1 and J2 are both O-singular, so that whenever S is a union of groups, every MPE is either an ogia-epimorphism, a J-epimorphism and a y(a°F)-epimorphism, or else is a y(J)-epimorphism. In particular, every MPE is a J-epimorphism or a y( J)-epimorphism.

(b) I f S is a regular semigroup, i f d~: S > ) T is an arbitrary epiraorphism, and i f 1~ and I~ are ideals of S such that qJ restricted to each of I~ and 12 is one-to- one, then ~ restricted to the ideal I1 to 12 is also one-to-one. In particular, there is a unique largest ideal of S (under inclusion) I , such that tO restricted to I , is one- to-one, where I , can be the empty set.

Proof The first assertion is trivial. It follows f rom [2] that s ---> S~sS ~ is a homomorph ism of S into (2 s, Iq), the set o f all subsets o f S under intersection, if and only i fS is a union o f groups. Thus ~: S ---> ({0, 1}, ")O(s) = 1

298 Joan RHODES

if and only ifJ2s _c J2 and qJ(s) = 0 if and only ifJ2s rl J2 = ~ is a homomorphism. Fur thermore , qJ(J2) = { 1 }, and rood qJ and rood 0 are incomparable. Thus, since 0 is an MPE, sl = s~ e S if and only if O(sl) = O(s2) and qJ(sl) = qJ(s2). Now Theorem 6(a) follows easily f rom Theorem 5 since qJQ'~)= qJ(/'2) forj2,j~ eJ .

We now prove part (b). It suffices to prove the following. (,) Let S be regular and let ~0: S ~ ~ T be an epimorphism. Le t l l , Is be two ideals of S such that ~p is one-to-one when restricted to lk for k = 1, 2. T h e n ~o is one- to-one when restricted to the ideal I1 U 12.

We may assume, with no loss of generality, that S = 11 k) 12. Suppose ~p is not one-to-one. Now ~o is one-to-one on each J - c l a s s J orS. Thus there exists a J - c l a s s J of T such that ~0-~(J) consists of more than one J -c lass and ~0-1(J) has a unique minimal J - c l a s s J ' [5, Prop. 3.2(d)]. Thus there exist J-classes of S, JI a n d J ' , such thatJ~ > J ' and ~0(Jt) c ~0(J'). Now Jl c l k for some k, so t ha t J ' C_ Iks inceJ ' <J1; thus J l LJJ ' c 11<. But this is impossible since ~0 is one-to-one on Ik. This proves Theo rem 6.

In the next section we shall give numerous examples of MPE's.

4. Examples and Remarks

All the following epimorphisms are MPE's. (a) Let Z2 be the cyclic group of order 2. Let {a, b} r be the semigroup

of order 2 with xy = y for all x, y ~ {a, b}. We consider the direct product S = Z2 × {a, b} r, where (zl, xl)(zz, x2) = (zl + z2, x2) for (zl, x0, (z2, x2) e S. T h e n

P2:Z2 × {a, b}" ~ {a, b} r

is an MPE of Class I, where P2(z, x) = x, and

PI: Z2 X {a, b} r ~ Z2

is an MPE of Class II, where Pl(z, x) = z. (b) Let S = FR(Xn) denote the semigroup of all mappings of X, =

{ 1, • • • , n} into itself under ( f . g)(x) = g(f(x)). Then K(S) is the collection o f all constant mappings {Ck: k e X,}, where Ck(x) = k for all x e Xn.

Then 0: S ~ S/K(S), where S/K(S) is the Rees quotient and lq is the natural epimorphism, is an MPE of Class II. It is the only MPE of S.

(c) Let S/K(S) act on the right of the set X consisting of all subsets of Xn having exactly two ,elements together with 0, where

0 . f = 0 for a l l f e S

{a,b} "f= {~f(a),f(b)} if f (a) ~ f (b) otherwise.

Then each member o f K(S) acts like zero so that S/K(S) acts on S. Let 8: S/K(S) ~ FR(X), where 8(f)(x) = x -f , i.e., 8(fl) = 6(f2) if and only if f l and f2 act the same on X. Then~i is an MPE of Class I. It is a special case of the following class of MPE's.


(d) Let I be a 0 - m i n i m a l ideal o f S such tha t each n o n - z e r o ~g"-class o f S in I has exac t ly two e lement s . Le t - be the equ iva l ence re la t ion on $ g iven b y s l - s2 fo r sl, s2 E S if sl = s 2 ors1 ~ $2, $1, $2 E / , $1 ~g's2. T h e n - - " i s a c o n g r u e n c e a n d 7: S ~ > S / - is an MPE o f Class I.

(e) Le t C be the Rees r e g u l a r m a t r i x s e m i g r o u p ~ t ° ({ l , . . . , n}, {1, • • • , n}, {1}, I,,), w h e r e I,, is the n x n iden t i ty m a t r i x with I,,(i,l ) = 8~j. W e a s s u m e tha t n t> 2. T h e n

"C ' > { 1 }

is an MPE o f Class I I I . T h i s is a ,special case o f the f o l l o w i n g c l a s s o f MPE's . ( f ) Le t S ~ { 1 } be a finite s e m i g r o u p . T h e n S 7: ~ {' | }.:is~an~ M'PE if a n d

on ly if o n e o f (1), (2) a n d (3) holds: (1 )S~is~a i f in i t~s im.p le~grayp # {1}; (2) S = { 0, 1 } w h e r e aft = 0 fo r a , fl e { 0, 1}, L e.~S i s ! t h e : ~ i m , n u l l semi- g r o u p ; (3) S is a 0 - s imple s e m i g r o u p i s o m o r p h i c t o ~ ( { 1; . . . . , n}, { 1, • • • , m}, { 1}, C), a r e g u l a r Rees ma t r i x s e m i g r o u p with n, m 1> 1 a n d with all rows a n d c o l u m n s o f C dist inct , i.e., C(b, a) = C(b', a) fo r all .a {1, • • • , n} if a n d only if b = b', a n d C ( b , a ) = C ( b , a ' ) f o r a l l b ~ {1, • • • , m} i f a n d on ly if a = a ' . F u r t h e r m o r e , each row a n d c o l u m n o f C con ta in s at least o n e n o n - z e r o en t ry . T h u s n = m = 1 a n d ]SI = 2, o r n, m I> 2. I n case (1) S ,> {1} is o f C l a s s I . I n cases (2) a n d (3) S ~ , { 1 } i s o f C l a s s I I I .

(Proof Clear ly S ) ) { 1 } , S ~ {1} , i s an MPE if a n d ordy if the on ly c o n g r u e n c e s o n S a r e the two trivial c o n g r u e n c e s . U s i n g [ 2 ] , w e see : tha t the s e m i g r o u p s sa t is fying (1), (2) o r (3) satisfy this condi t ion . Conve r se ly , let S sat isfy this condi t ion . Le t I be a p r o p e r i d e a l o f S . T h e n S ;~;~Sli,~is an e p i m o r p h i s m , so tha t e i t he r S = I o r I = {0}; w h e r e 0 is ,the~zero, of'~S. T h u s e i t he r S is s imple o r else its on ly ideals a r e S a n d {0}, so~ith.at e i d e r S is s imple , S is 2 -po in t null , o r S is 0-s imple . N o w b y a p p l y i n g the Rees T h e o r e m , w h e r e S is s imple o r 0-s imple , t o g e t h e r wi th t h e classif ication o f e p i m o r p h i s m s on r e g u l a r Rees m a t r i x s e m i g r o u p s (see [2] ) , we see tha t ( f ) follows easily.)

(g) Le t S be any s e m i g r o u p sa t i s fy ing (f)(3). Le t G be a g r o u p , G # { 1}. Le t P2: S × G > > G. T h e n P2 is an MPE o f Class I I I . K(S × G ) , the ke rne l o r m i n i m a l 2-s ided ideal o f S x G, is {(0, g): g ~ G}. T h e n S × G/K(S × G) ~- ..it'°({ 1, • • • , n}, { 1, • • • , m}, G,C) , w h e r e S =. .~°({ 1, • • • , n}, { 1, • • . , m}, {1}, C). F u r t h e r m o r e , q: S x G ~ > S × G/K(S × G) is an MPE if a n d on ly if G is a s imp le g r o u p ; in this case, ~ is o f Class I.

(h) Le t G be a g r o u p G ~ {1}. Le t H be a s u b g r o u p o f G. T h e n S = H × {1} t3 G × {0} is a s u b s e m i g r o u p o f G × {0, 1}, w h e r e {0, 1} is a semi- g r o u p with two e l emen t s , 0 the ze ro a n d 1 the ident i ty . T h e n O: S ~, G, d e f i n e d by O(g, x) = g, is an MPE o f Class I I I .

(i) Le t N , = {0, 1, 2, • • • , n -- 1} be the n - p o i n t n u l l s emigToup w h e r e a/3 = 0 fo r a , fl ~ Nn. Le t 0n: Nn ' ~ N , - I , n t> 2, w h e r e O , ( j ) = ' m i n (n -- 1 ,j). T h e n 0, is an MPE o f Class IV.

(j) Le t S be a finite s e m i g r o u p , S # { 1 }. T h e n t h e r e exist e p i m o r p h i s m s

S = S o ~>$1 ' ' $ 2 ' > ' ' " ~ ' S . = { 1 } ,

300 JOHN RHODES

w h e r e each e p i m o r p h i s m is o f Class I, Class I I o r Class I I I with j l = {0}, so t h a t f 2 is a O-minimal ideal.

(Proo f Let us a s s u m e tha t (j) is false, a n d let S be a s e m i g r o u p o f small- est o r d e r fo r which (j) is false. Le t I = K(S) if K ( S ) ~ {0}; o the rwi se let I be any 0 - m i n i m a l ideal o f S, so tha t I # {0}. T h e n e i the r (1): the non- zero X-c lasses o f S c o n t a i n e d in I consist o f exact ly o n e poin t , o r (2): (1) is false. In case (2) cons ide r the ep imorph i sm. lq : S ~ S / - , w h e r e -~ is the c o n g r u e n c e g iven by sl ---- s2 if sl = Sz or sl ~ s2, sl, Sz E I , sl ~,D s2. By (2) ~ is not o n e - t o - o n e , h e n c e S~ =- is o f strictly smal le r o r d e r t h a n S a n d so by in- duc t ion S~ =- ~ ~ { 1 } can be d e c o m p o s e d into e p i m o r p h i s m s o f the p r o p e r type. But n o w wri te ~ = ~k " " " "ql, w h e r e ~q~ is an MPE f o r j = 1, • • • , k. By the def in i t ion o f -=, ~ is an ~ - e p i m o r p h i s m , so tha t it is easy to see tha t "ql, " " " , 7/k m u s t also be ~ - e p i m o r p h i s m s a n d thus all o f Class I. H e n c e (j) ho lds fo r S. T h e r e f o r e (2) m u s t be false. A s s u m e tha t (1) holds. Con- s ider S ~ S / I , which is a y ( ~ ) - e p i m o r p h i s m by (1) a n d is no t one - to -one . W r i t e S ~- ~ S / l as Om " . . Ol, w h e r e O t i s a n M P E f o r j - - 1 , - . , , m . N o w i t is easy to ver i fy tha t each Oj m u s t be o f Class I I o r Class I I I with J1 = {0}. T h u s it follows as b e f o r e tha t aga in (j) ho lds fo r S. T h u s (1) m u s t also be false. H e n c e S does not exist a n d (j) is t r u e . )

(k) Clear ly fo r each S t h e r e exist s e m i g r o u p s S1, • " • ,Sm a n d e p i m o r -

p h i s m s Oj: S ' ) St such tha t 0: S ~ Sx x • • • x Sin, w h e r e O(s)= (01(s), • • • ,

Ore(s)), is an i s o m o r p h i s m o f S in to S1 x • • • x S,, with Pt(O(S)) = Sj f o r j = 1, • • • , m, w h e r e Pt is the j t h p ro jec t ion m a p , a n d w h e r e each St has (up to c o n g r u e n c e ) exactly one M P E . In o t h e r words , S can be wr i t t en as a sub- d i rec t p r o d u c t o f h o m o m o r p h i c images , each c o m p o n e n t o f which has exact ly o n e (up to c o n g r u e n c e ) MPE.

Not ice tha t S has a u n i q u e MPE 0 if a n d on ly i f0 : S ~ T is an M P E a n d tha t ql: S ~ R, w h e r e qJ is any e p i m o r p h i s m which is no t o n e - t o - o n e , implies tha t qJ = 010 fo r s o m e e p i m o r p h i s m 0.

FR(X, ) ' ) FR(Xn) /K(FR(X, ) )

(see (b) above) is the u n i q u e MPE o f FR(X, ) fo r n >/ 2.

5. Elementary Theory of ~-Epimorphisms

W e n o w wish to p r o v e an "hones t " h o m o m o r p h i s m t h e o r e m fo r ~g'- e p i m o r p h i s m s a n d to show tha t the col lect ion o f ~ " - e p i m o r p h i s m s f o r m s a m o d u l a r lattice with a m a x i m u m a n d m i n i m u m e lemen t . T h i s is a sl ight genera l i za t ion o f the m e t h o d s o f [5, Sect. 6, P rop . 6.9] which t h e n yield a slight genera l i za t ion o f [4] .

Le t S be a finite s e m i g r o u p . Let HI , H2, • • • , Hn be a choice o f ~ ' - classes o f S, w h e r e we have chosen exact ly o n e f r o m each J - c l a s s o f S. L e t J k be tha t J - c l a s s o f S c o n t a i n i n g H k. I f J e is r egu l a r , we choose He to be a m a x i m a l s u b g r o u p o f S, which is always possible (see [ 2]). Le t Ge = ~ L ( H e )

be the so-cal led Sch i i t z enbe rge r g r o u p o f H e c o n s i d e r e d as a t rans i t ive


group o f regular permuta t ions acting faithfully on the left o f Hk. I f J ~ is regular so that Hk is a maximal subgroup o f S, then Gk is the left regular representa t ion o f Hk. We write G <J S if G <~ Gk (G is a normal subgroup o f Gk) for some k. Fur ther , we choose a point hk ~ I lk with h~ = hk if this is possible (i.e., if Hk is a subgroup), and otherwise arbitrary. In what follows H1, H2, • • • , H , , hi, h2, • • • , h , are held fixed for S.

Let 0: S > > T be an ~"-epimorphism, and let 05 deno te the restriction o f 0 to Hs. T h e n we define

K e r 0 = { K I , K2," " ' , K n }

where each Kj is Ker 0j in a natural manner . More precisely, Kj <1 Gj, and Ks is def ined by

(5.1) Kj = {~r ~ Gs: O(~(h~)) = 0(hj)},

i.e., Ks is the kernel o f the homomorph i sm 0~: Gj,--* 5Pt.(Rj) def ined by

(5.2) OOrhs) = OjOr)O(hj),

where Hs is the 6T'-class o f T containing O(H~) = Os(Hs). Thus when Hj is a group , Ks = Ker ~ ~ Ker 0j. We note at this point that not only does 0 s de te rmine Ks (via (5.1)) but conversely, Ks de termines 0~, since for s~, s2 E Hj with, say, sl = ~r~hj, s2 = ~r2hs, we have, by (5.2), Oj(~rlhs) = OjOr2hj) if and only if ~0r l ) = ~(~r2) if and only if 7rl~r~ 1 ~ Ker ~ = Ks.

T H E O R E M 7. Two oqg'-epimorphisms 01, 02 induce the same congruence

on S i f and only i f Ker 01 = Ker 02. Proof By the remarks above, Ker 01 = Ker 02 if and only if 01 = 02 on

each Hs. Thus it suffices to show that rood 01 = mod 02 on each Hj if and only if mod 01 ---- mod 0., on S. Now mod 01 = rood 0., on S implies at once that mod 0~ = mod 02 on each Hi. It remains to show the converse. Let Hj be ano the r )Y-class o f S with H~ J H s. T h e n there exist xl, x2, zl, z . , , S 1 such that a --> XaaX2, fl --'> zlflz2 are one-to-one, mutually inverse mappings H~ H~, Hj ~ H~, respectively. Thus for sl, s2 ~ H~ we have that Ok(s1) = O~(s2) implies that Ok(XlSlX2)= Ok(XlS2X2), which implies that Ok(z,xlslx2z2) = Ok(ZlX~S2X2Z2), which in turn implies that O(sO = O(s~) since zlxls~x.zz2 = si. T h a t is, O,(sl) = O~(s2) if and only if O~(xlslx2) = O~(xls.,x2), where xlslx.,, xls2x2 ~ H~. Thus rood Ok on H~ de termines rood 0~ on any o ther )g-class H~ in J~, and the re fore determines mod Ok on Js since 0k is an ~g'-epimorphism. Th e re - fore rood 01 = mod 02 on each H~ means that mod 01 = rood 0., on eachJ~, so that mod 0~ = mod 0., on S.

T h e following construct ion is o f interest.

Definition 5.1. Let Ks <~ S. T h e n T/(K~): S > >" S / K s is the ep imorph ism ~ffi_: S > • S / - - , where - is given as follows. First, for s~, s2 eJ~ (the J -c l a s s containing Hi) we say that sl -- ' s2 if sl g~'s2 and there exist xl, x2 ~ S 1 and • r ~ Ks such that xls~x2, xls2x2 ~ Hs and ~r(xlslx2) = xls2x2 (i.e., xls2x2 ~ Kj(xls~x2),

3 0 2 J O H N RHODES

or XlSlX2, x1.$2x2 are in the same "coset" d e t e r m i n e d by KS). T h e n letj~ be the m a p on S given by

j3(s) = {~s]~, i f s ~ J j if s ~Jj ,

where [s] _~, is the - ' - e q u i v a l e n c e class ors. We then define sl --- s2 i f j ] (as j3)= J](as2/3) for all a , / 3 ~ S 1.

R e m a r k 5.1 In the above construct ion - is the same as - ' onJ j . Clearly - is f iner than - - ' onJs . Now suppose that s, ---' sz for some sl, s2 eJ j . T h e n sl ~ s2 and, accord ing to [2] , e i ther axJ3, axz/3 ~J j or axl/3 Yg' axe/3 e J j for all a , / 3 e S 1. In the first casej](ax,/3) ----J](ax2/3) = 0. In the second case let &,/~ ~ s 1 be such that &as~/3/3 = si, i = 1, 2. Let Xl, x2 e S 1 and rr e Ks be such that x~slx2, xls2x2 ~ H~ with 1r(xls,x2) = xlssX~. T h e n xl&, ~x2 E S 1, ~r e Ks, and

^ '~ A ^' A A (xla)(asafl)([3x2) e Hs, (x1cl~)(~2/3)(~x2) ~ H j and 7r(xlaasl/3/3x2) = 7r(xlslx2) = xls2x2 = xl&as2/3/~x2, so that as j3 -~' as2/3, nencef j (as l /3) =j~(as2fl), so that

$1 -~" $2"

T h u s ---' is the natura l equivalence relat ion induced by Kj o n J j , and =- is the coarsest extension o f - ' to a congruence on S which separates e lements of Js f r o m those not i n J j a n d which agrees with - ' onJ j .

Remark 5.2 Suppose that s,, s2 EJ j and that s~ aegs2. I f there existxl, x2 ~ S 1 and 7r E Kj such that x,s~x2, XlSlX2 ~ Hs with ~r(X,SlX2) = xls2x2, then for any x;, x~ ~ S 1 for which x~s,x~, x~s2x~ ~ H j there exists a r r ' ~ K~ such that ~r'(x~slX;~) = x~s2x~. One way to p rove this is to in t roduce Rees coordinates on Js and view xl, x~, x2, x' 2 as r ight and left translations, f r o m which it follows (see [2]) that there exist ~rl ~ 5~L(H~) = G~ and zr2 ~ 5~R(Hs) (the r ight Schi i tzenberger g roup) such that x~slx~ = ~q(XlSlX2)¢r2 and x~s2x~ = ,//'l(X1$2x2)7/'2. (Note that ~rl((S)~r2) = 0h(s))rr2 for all s ~ H~, ~rl ~ o'°~L(Hj), 77"2 ,~PR(H~).) T h e r e f o r e the desired 7r' is 7T17]'71"1 1 since 7rlTr~r-il(x~slx~)= ~r,frcr-ilrrl(xlsax2)~r2 = ~rl(xls2x~)~r2 = x~s2x~ and rr' ~ ~r,KsTr] -1 = K s.

T h u s in the definit ion o f - - ' in Definition 5.1 the phrase " there exist x,, x2 e S ~ and ~r ~ Ks" is equivalent to the phrase "for xl, x~ ~ S 1 there exists a 7r ~ K s " .

T H E O R E M 8. Let O: S > > T be an ~-ep imorphism, and let Ker 0 =

{K1, K2, • • • , Kn}. Then

~A --~ ( ' 0 ( / < 1 ) X T / ( K 2 ) X • • • X ' r / ( K n ) ) A : S > ' " r / A ( S )

is such that m o d ~A ---- m o d 0, where A is the n-component diagonal func t ion s ~ ( s , . . . , s ) .

P r o o f First, m o d "o(Ks) is f iner than rood 0 o n J j . For let s~, s2 eJ~ with sl -- s2. T h e n s~ - ' s2 by R e m a r k 5.1 so that there exist Xl, xz ~ S 1 and 7r e Ks with xls~x2, xls~x2 e H~ and ¢r(x,slxz) = xas~2. Since ~r e Ks it follows f r o m (5.1) and ( 5 . 2 ) t h a t O(xaszx2) = O(~r(x~s~x2)) = O(XlSlXz). Choos ing ~71, ks e S 1 so that klxlsix~i¢~, = s~ we get O(sO = O(~q(x~s,xz)ic2) = O(sz), whence m o d "o(Kj) is f iner , than m o d 0 o n J j . T h u s rood ~A <~ m o d 0, i.e., S > • S/rood r~A * •


S/mod 0. We must show that mod 0 <~ mod "0A for which it suffices to show that mod 0 <~ rood 7/(K~) for e ach j . Let sl, sz ~ S with O(sl) = 0(s2). Th en , for any a , fl ~ S ~, O(as~fl) = 0(aszfl), so that aslfl ~g' as2fl since 0 is an oT'- epimorphism. I f aslfl, as2fl ~J~, thenj~(aslfl) =J](as2fl) = 0. I f otslfl, as2fl e J j there exist xl, x2 e S ~ such that xlas~flx2, x~as2flx2 e Hj , and hence there exists or ~ Gj such that or(xlaslflx2) = xzots2flx2. But O(or(xlaslflx~)) = O(xlas2flx2) = 0(xlaslflx2) since O(sO = 0(s2). Hence or ~ Ks so that aslfl - ' as2fl and thus

J)(aslf l) = J)(as2fl) in this case. Thus O(sl) = O(s2) implies that "o(Kj)(st) =

TI(K j)(s2). We proceed next to show that the collection o f ~,~-epimorphisms, or

ra ther their associated congruences u n d e r the operat ions o f intersection and transitive closure, form a modula r lattice with minimal and maximal element.

T h e n , as is well known, there exists a minimal aT'-epimorphism Sar o f S, as shown, for example, in [5, Remark 6.1 (b)]. T h e identity map is trivially the maximal ,g~-epimorphism. T h e lattice a rguments follow f rom the following.

T H E O R E M 9. Let Ore: S ~ • Ore(S) be ~f~"-epimorphisms f o r m = 1, 2, and

let g e r Om= {Kin1, Km~, " " " , Kin,}. Then (a) (01 X 02)A: S ;-~ (0a × 02)A(S) is an ~gD-epimorphism with kernel

{K~, f3 K.,~," • " ,K I , CI Ks,}. (b) 0': S ) > S / ( m o d 0 , V mod 02), i.e.

:/ S

\

S/mod 01

\ O' , , S/(mod Ol V rood 02)

/ S / m o d 0.2

is an acia-epimorphism with kernel {KlIK12 = K1~Ku, • • • , K1 ,K2 , = K2 ,K1 , } . P r o o f (a) Clearly (01 × 02)A is an oZg-epimorphism since S ' ) (01 ×

0z)A(S) ~-~- O~(S) and 01 is an vZ~'-epimorphism. Let Ker (01 × 0~)A = {K~, K~, • • • , K,'~}. Now using (5.1) we have or ~ KIj f3 Koj if and only if both O~(orhj) = O~h~ and 02(orhj) = 02hj, which hold if and only if (01 × 02)A(orhj) = (0~ × 02)A(hj), or, what is the same, or ~ K~. HenceK~ =-KIj f3 K2~.

(b) Let Sa 'be the minimal aOF-epimorphism o f S. T h e n S • • S /mod Om - ~ S ~ e f o r m = 1, 2 implies that S • • S / ( m o d 0 1 V mod02) • • S ~ , a n d

thus 0' is an ~ - e p i m o r p h i s m since S ~ • S ~r is an aT'-epimorphism. Next we note that for any j , K~jK2~ = K2jKaj since KIj, K ~ <3 Gj. This is impor tan t because it means that the two congruences =-m ---- rood Ore, m = 1, 2, com- mute. T h a t is, let o deno te the usual p roduc t of relations, i.e. h~(--- o -~')h2

304 JOHN RHODES

if t he re exists an h such that hi --- h a nd h ---' h2. T h e n h1(--1 o -=2)h~ implies tha t hi ---1 h -=2 h2 fo r some h. H e n c e hi ~g'h g /h2 and hi g'° h2. T h u s if ht, h2 J j t hen fo r suitable h ~Jj , 7rl, ~r~ ~ KI~, 7r2, ~r2 ~ K2j, hl(-=~ ° --2)h2 ~oh~ -=~ h -=3 h2 •) (lrihl h a nd ~r2h = h2) ffi) 1r27rahl = h2 •) lrlcr2h1 = h2 •) (~r~(hl) = 1r2hl a n d ~r~(Ir~h~) = h2) •oha -=3 lr~h~ -=~ h2 =ohm(-=2 ° -=a)h2. C o n s e q u e n t l y -=1 ° -=3 is the same as -=2 ° -=1. N o w assume tha t Ker 0' = {K~, • • • , K~}. T h e n it follows at once f r o m (5.1) tha t Ir~ ~ K~j, and ~r2 ~ K2j imply tha t lr~, 7r2 Kj, so tha t 7rlTr2 ~ Kj a nd hence KljKzj C_ Kj. Converse ly , m o d 0' =

m o d 0~ V m o d 03 = 6 (-=1 ° -=2) n is the same as -=~ o -=3 since ---~ c o m m u t e s )t= I

with -=2. T h u s ~" ~ Kj implies by (5.1) tha t "ffhj ~ hj ( m o d 0') =o ~rh~(-=l ° ~-2)hj ~ Irhj -=1 h -=3 h~ fo r some h, which implies tha t ~'2¢qTrhj = hi, w h e r e lrllrhj = h, 7r2h = hj a nd 7r,, ~ Kmj. T h e r e f o r e ¢r2~rllr = 1, w h e n c e 1r = ~r~-~r~ -1 K1jK2j. H e n c e Kj = K~jK2j.

Now by T h e o r e m 9 a n d its p r o o f we see tha t the set o f c o n g r u e n c e s o f ~F-ep imorph i sms is closed u n d e r in tersect ion a n d transit ive c losure , and f u r t h e r m o r e the c o n g r u e n c e s c o m m u t e . Hence , as is well known, this implies tha t they f o r m a m o d u l a r lattice. Aga in the c o n g r u e n c e s cor re - s p o n d i n g to the ident i ty a nd to S ) > S ~r a re the min imal and max imal e lements . This general izes the results o f M u n n [4] .

T h e r e a d e r shou ld r e f e r to [5, Sect. 6 ] , especially Defini t ion 6.3, P ropos i t ion 6.9 a nd R e m a r k 6.4.

R E F E R E N C E S

[1] M. A. ARBIB (ed.), Machine, Languages and Semigroups, Academic Press, New York (to appear).

[2] A. H. CLIFFORD and G. B. PRESTON, The Algebraic Theory ofSemigroups, Vol. I, Math. Sur- veys, No. 7, Amer. Math. Soc., Providence, R. I., 1961.

[3] J. A. GREEN, On the structure of semigroups, Ann. of Math. (2) 54 ('1951), 163-172. [4] W. D. MUNN, A certain sublattice of the lattice of congruences on a regular semigroup,

Proc. Cambridge Philos. Soc. 60 (1964), 385-394. [5] JOHN RHODES, Some results on finite semigroups,J. Algebra 4 (1966), 471-504. [6] R. J. WARNE, Extensions of completely 0-simple semigroups by completely 0-simple

semigroups, Proc. Amer. Math. Soc. 17 (1966), 524-526.

(Received 15 June 1967)

a homomorphism theorem for finite semigroups

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