Birkh~iuserVerlag, Basel

Aequationes Mathematicae 20 (1980) 198-223 University of Waterloo

S o m e theorems on hemigroups K. P. C'vlmOA

This work is from the author's dissertation [4], and was directly motivated by that of Furstenberg in [7], where he studied non associative binary systems with the property that ( x y ) ( z y ) = xz, which he called half groups. W e use the word " h e m i g r o u p " instead of "half-group". W e introduce topological hemigroups and extend certain results in [7]. A m o n g other things, Furstenberg defined principal hemigroup and two important natural relations on a hemigroup M which we called 1: and J; he showed that M/J is a principal hemigroup and that principal hemigroups are just groups in disguise and conversely. W e topologise all his results for M compact, and go on to study the natural projections of M into M/J and M/F; the subsets eM, Me and e M N M e when e is an idempotent; compact totally disconnected hemigroups; and finite ones.

1. Introduction Throughout we will write x for {x}. A B will denote the collection of all elements of the form ab with a ~ A and b ~ B. Thus if A = {a} we write aB and B a instead of {a}B and B{a}. Also, sometimes we will omit parentheses or use as a separator in products when the meaning is clear; for example, x 2- y2 means (xx)(yy), and x y . z w means (xy)(zw). We begin by making the following definitions, which are topological analogues of Furstenberg's. (1.1) D E F I N I T I O N S . A hemigroup is a Hausdortt space M with a continuous multiplication, denoted by juxtaposition, such that (xy)(zy) = xz for all x, y, z e M. A principal hemigroup is a hemigroup M such that x M = M x = M for some x e M. A n idempotent is an element x such that x 2= x, and E will denote the set of idempotents of M.

AMS (1970) subject classification: lh'imary 22A15, 22D30. Received August 11, 1977 and, in final form, March 15, 1978. 198

Vol. 20, 1980

Some theorems on hemigroups

199

Actually, Furstenberg defined M to be principal if x M = M x = M for each x e M, b u t (v) of the following l e m m a shows that this is so if it is true for o n e element. (1.2) L E M M A . L e t M be a hemigroup with idempotents E. (i) x 2 e E for each x e M . (ii) The square o f each element o f x M is x 2, and x M c x2M. (iii) I f M = M 2 then x M = x2M. (iv) In a principal hemigroup M, there exists a unique idempotent e and x = e (ex) = xe for each x ~ M. (v) I f M is principal, then x M = M x = M for each x ~ M. Proof. (i) x2x 2 = ( x x ) ( x x ) = xx = x 2, by the h e m i g r o u p property. (ii) Similarly, (xy) 2 = (xy)(xy) = xx = x2; and x M = x x . M x c x2M. (iii) L e t z ~ x2M, so that z = x 2 m for s o m e m e M. Since M = M 2, x = yt for s o m e y, t e M. H e n c e x = y 2 - t y and, x 2 = y2; thus x = x 2- ty. So z = x 2 m = ( x 2- ty)(m • t y ) = x ( m • t y ) ~ x M . This shows that x 2 M c x M which with (ii) implies x M = x2M. (iv) L e t x ~ M be such that x M = M x = M, and let e = x 2. T h e n y2 = e for each y e x M b y (ii); e M = xx • M x = x M = M ; and M e = M x • xx = M x = M . T h u s if y ~ M , t h e n y = e m for s o m e m ~ M ; h e n c e y = ( e m ) ( m m ) = y m ~ = ye; and t h e n y = ye = (yy)(ey) = e(ey) also. (v) L e t x e M such that x M = M x = M and let e = x 2. T h e n e is the unique i d e m p o t e n t by (iv). Fix y ~ M and let us show that M c My, which implies M = M y ; by (iv), if z ~ M, z = ze = (z, ey)(e • ey) = (z • ey)y ~ My. Using this we see that y M = y y . M y = xx • M = xx • M x = x M = M

also.

(1.3) D E F I N I T I O N S . A morphism is a c o n t i n u o u s function 4 , : M - - - ~ M ' w h e r e M and M ' are h e m i g r o u p s and 4~(mm') = 4~(m)~b(m') for all m, m ' ~ M. A n isomorphism is a m o r p h i s m which is also a h o m e o m o r p h i s m . A congruence o n a h e m i g r o u p M is an equivalence relation R o n M such that (x, y ) ~ R implies (xa, ya) and (ax, a y ) ~ R for all a e M . (1.4) L E M M A . L e t M be a compact or discrete hemigroup. I f R is a closed congruence on M, if ¢b : M - - ~ M / R is the natural m a p a n d if M / R has the quotient topology, then M / R is a hemigroup a n d ~ is a surmorphism. Conversely if 6 : M ~ M ' is a surmorphism, then R = {(x, y ) : 6 ( x ) = ~b(y)} is a closed congruence on M and the induced function from M / R to M ' is an isomorphism. T h e p r o o f is categorical; the hypothesis of c o m p a c t o r discrete is used to show that M f R is H a u s d o r l t , and t h e n to s h o w that ~ is continuous. Hereafter all spaces

200

K. P, CHINDA

AEQ. MATH.

will be Hausdorff, all functions will be continuous, all quotient spaces will have the usual quotient topology and all product spaces the usual product topology. (The theorems will hold if one assumes all spaces discrete instead.) Notice that any continuous bijection wilt be a homeomorphism if the spaces are compact. (1.5) E X A M P L E . (a) Let M be any space and define xy = x. This happens to be associative and is known as the left zero semigroup on M. Note that M = E here. Conversely whenever M is a hemigroup consisting only of idempotents then its multiplication is this left zero product: for x x = xy • xy since M is a hemigroup, and since M = E this says x = xy. (b) Let (M, *) be a group and define xy = x*y -1. This is a principal hemigroup. Conversely, every principal hemigroup M has a unique idempotant e and admits a group structure defined by x*y =x(ey). Furthermore, these constructions are inverse to each other. These algebraic assertions are in Theorem 1 in [7]; that the induced functions are continuous is dear. Thus a principal hemigroup is nothing but a group in disguise and conversely. Notice that assertions like (1.2) (111) and (iv) are obvious if one assumes this group structure known; for example, e(ex) = e*x = x because e is the identity for the group (M, *). (c) Let E be a space. The simple hemigroup on E x E is defined by (e, f)(e', f') = (e, e') (p. 994[7]). Clearly this is continuous. Note that the set of idempotents of E x E is its diagonal zi, and this is not a subhemigroup---in fact za2= E x E. (d) Any direct product of hemigroups is again one (p. 994 [7]) and continuity is clear. In particular, we will use, as Furstenberg did, one of the form E x E x M, where this is the direct product of a simple hemigroup E x E and another given hemigroup M. (e) Equivalent extensions (p. 995 [7]): Let M be a hemigroup, M' a set containing M, and 4': M' ~ M a retraction of M ' onto M. Then a hemigroup product can be defined on M' by xoy = 4~(x)~b(y). Such M' is called an equivalent extention of M, and 4~ will be a morphism. Furstenberg showed that for any hemigroup M, M ~ is a subhemigroup such that (M2) 2= M 2 and that M is an equivalent extention of M 2 (Theorem 8 [7]). Because of this and because equivalent extentions can be so arbitrary, it therefore seemed practical to him to xestriet attention to hemigroups M such that M = M 2. We, however, will do this bnly in particular instances.

2. The Relation F This is the same as Furstenberg's relation ~ . In this section, assume all spaces compact.

Vol. 20, 1980

Some theorems on hemigroups

(2.1) D E F I N I T I O N (p. 993 [7]). Let M be a hemigroup. Let (x,

201

y)eF iff

xa = ya for some a e M.

(2.2) T H E O R E M (Theorem 5 [7]). Let M be a hemigroup. T h e n (x, y ) e F implies xa = ya a n d ax = ay for all a e M. In Theorem 4 [7] Furstenberg proved that algebraically M/F is a hemigroup; F is closed by continuity; and these with (1.4) above yield the following. (2.3) T H E O R E M . The relation F is a closed congruence, so M / F is a h e m i group and the natural m a p 4~ : M---~ M [ F is a surraorphism. (2.4) T H E O R E M . Let e be an arbitrary idempotent in a hemigroup M. Then 4~ maps M e bijectively onto M / F and 4~(x) = ¢b(xe) for each x e M , i.e. (x, x e ) ~ F . M e m a y not be a subhemigroup of M , but the bijection ~b l M e induces a hemigroup on M e which can also be defined directly by x oy = (xy)e. Proof. d p l M e is one to one because 6 ( x e ) = 6(ye) implies x e . e = ye. e by (2.2); hence xe • ee = ye • ee and hence xe = ye. For any x e M, xe = xe • ee = xe • e, so (x, xe) e F by (2.1). So oh(Me) = M / F and so 4~ I M e is a bijection as asserted. For ease of notation, let ~b denote 4~ I M e in this paragraph, and let ~ denote ~b(x). Clearly, 4~ induces a hemigroup multiplication (*) on Me, by x*y = 4~-1(:~), and for this multiplication, 4~ is an isomorphism. Also clear is that x o y = (xy)e is well defined on M e . Finally, these operations are the same: for ~ b ( x * Y ) = Y ~ = x y = c b ( x y , e) ~b(xoy), and hence x * y = x o y because 4~ is one to one. Example (1.5) (c) shows that M e need not be a subhemigroup.

(2.5) C O R O L L A R Y . Let M be a hemigroup. Let e e E a n d let M e have the multiplication x oy = (xy)e. Define g : M ~ M e by g ( x ) = xe. T h e n g is a morphism a n d 4)' = rk I M e is the unique isomorphism such that ck = ck' g. M ....................

j

M/F -

f

/

/ (Me,o)

Proof. Since

/

/

g(xy)=xy.e=(xe- ye)e=(g(x)g(y))e=g(x)og(y), g is a morphism. Because ~b'g(x)= ~ ' ( x e ) = cb(xe) and cb'(xe)= 4~(x), by (2.4), we see that ~ ' g = 4~. Also by (2.4), ~' is an isomorphism, and dearly, 4¢ is unique.

202

K . P . CHINDA

AEQ. MATH.

(2.6) T H E O R E M . Let M be a hemigroup and let a, b ~ M be arbitrary. Then card M a = card Mb. Proof. Define h : M A ~ Mb by h(xa) = xb for all x e M. Then h is one-to-one because xb = yb implies (x, y ) ~ F (2.1) and hence xa = ya (2.2). Also h is onto because for any zb ~ Mb, za ~ M a and h(za) = zb. Hence card ma = card Mb. (2.7) C O R O L L A R Y . I[ e, f e E and multiplication on M e and M f is defined by x o y = (xy)e and z o w = ( z w ) f respectively for all x, y ~ M e and z, w ~ M[ and if M is compact or discrete then M e and M f are isomorphic. Proof. Replacing a by e and b by f in the above theorem we get

h(xe) = xf It is obvious that h is continuous. We now show that h preserves multiplication. For xe, ye ~ Me, h(xe oye)= h((xe . ye)e)= h(xy " e ) = xy " f

= ( x f . y f ) [ = x f o y f = h(xe)oh(ye). Hence h is a morphism.

3. The Relation J This is the same as Furstenberg's relation --. In this section, assume all spaces compact. (3.1) DEFINITIONS. Let M be a hemigroup. On M define a relation J as follows: (x, y ) ~ J iff xy ~ E 2. Let j : M -~, M/J be the natural m. Furstenburg proved on p. 993 of [7] that M/J is algebraically a hemigroup. J is closed by continuity, and these facts together with (1.4) yield the following. (3.2) L E M M A . I is closed congruence, so M/J is a hemigroup and the natural map ]: M ~ M/J is a morphism. (3.3) L E M M A (i) F c J ,

(ii) (x, y ) ~ l igxy =x2y ~, (iii) ] ( E ) = j ( E 2) and this is a point.

Vol. 20, 1980

Some theorems on hemigroups

Proo[. (i) (x, y ) e F implies xa = ya for all a e M take a = y then we have x y = y y = y Z = ( y Z ) ( y 2 ) e E 2,

hence

203

by (2.2). In particular, if we

(x,y)el.

Thus F c $. (ii) If xy = x 2 y 2 then d e a r l y (x, y ) e $ . Conversely, assume (x, y ) e ] , so that xy = e[ for some e, f e E ; then x2=(xy)(xy) = (ef)(eD = e. H e n c e xy = e f = x 2. f. Also xy = e l = el" flf = ex" f so that xy = x 2" f = x y - f. Therefore (x 2, cy) e F so that xy • a = x 2. a for all a e M by (2.2). Taking a = y2 we get xy • y2 = x 2 . yZ, and this says xy = x 2. y2 as asserted. (iii) By definition of $, j(E) is a point. If e, f e E then (e, el) e F by (2.4). Hence (e, eD e l by (i) above. So j(e[)= j ( e ) - i . e , j ( E 2) = ](E).

T h e following theorem is immediate from (3.2) and Furstenberg's T h e o r e m 4, in the case M = M 2. However, we provide a brief proof using (3.2) and (3.3) and also show it without assuming M = M 2. (3.4) T H E O R E M . M/$ is a Principal Hemigroup.

Proof. First ](M 2) = ](M) since j(xe) = j(x) for any x e M and e e E by 3.3 (iii). Next, M "zc E M 2 because ab = a 2" ba, hence j ( M ) = j(EM~). Hence j(E)(M/$)= M/I. Similarly ab = a b . b 2, so M 2 c M2E and (M/J)j(E)= M/$. Since j(E) is a point, we have shown that M/J is Principal. (3.5) C O R O L L A R Y . Assume M is a hemigroup satisfying M 2 = M; then (i) ( e ( f x ) , x ) e ] for all e, f e E and x e M , (ii) ( a , b ) e J iff x ( y a ) = x ( y b ) for all x, y e M .

Proof. ( i ) B y the notation of (3.4), j(e(fx))=j(e)j(f-x)=j(e)(j(f)j(x))= 1(1. i(x)) where 1 is the unique idempotent of M/l, and (1(. j(x)) equals j(x) by (1.2) (iii). (ii) If x(ya) = x(yb) for all x, y e M then it holds for x, y e E, so (a, b) e J by (i). Conversely, suppose (a, b ) e l . T h e n ab = a2b z by (3.3) (ii), and also ab = a b . b 2, hence a b . b2=a2b 2. T h e r e f o r e (ab, a 2 ) e F so by (2.2) we know that y b . ab= yb • a 2. But yb • ab = ya = ya • a 2. Hence ya • a 2= yb • a 2. Thus (ya, yb) e F ; then using (2.2) again we get x(ya)=x(yb)

for aU

xeM.

204

K.P. CHINDA

AEQ. MATH,

(3.6) DEFINITION. (p 994, [7]). M ' is said to be an equivalent retraction of M if there exists a surmorphism ¢b:M---~M' such that ~b(x)=4~(y) implies (x, y ) e F. Such a morphism is also called an equivalent retraction. (3.7) T H E O R E M . Let M satisfy M 2 = M and let E be its set of idempotents. Let E x E × M be the direct product of the simple hemigroup E × E with M, and define

h:E×E×M---~M

by

h(e,f,x)=e(fx).

Then h is an onto morphism. Proof. Consider (e, f, x) and (e', f', x') in E × E × M. We have h((e, f, x)(e', f', x')) = h(e, e', xx') = e(e'. xx'), and we wish to prove that this equals h(e, f, x)h(e', f', x') = e(fx), e'(f'x'). Let p = e(fx) and q = e'(f'x') and note that p2 = e 2 = e and q2 = e 2 = e'. Thus pq = p2. qp = p2(q2, pq) = e(e'. pq). Also ](pq) = ](p)j(q)= j ( x ) j ( x ' ) = j(xx') by two applications of (3.5) (i), so

e(e'. pq) = e(e'. xx') by (3.5) (ii), i.e. pq = e(e' . xx'). Thus h is a morphism. Now for any x e M = M 2, x = ab = a 2. ba = a2(b 2. ab) = a2(b2x). Hence x = h(a 2, b 2, x), so that h is onto. The following is essentially Furstenberg's Theorem 10, however, we have defined our retraction differently then he did in order to obtain continuity easily. (3.8) T H E O R E M . If M = M 2 then M is an equivalent retraction of E x E × (M/l) via the [unction g(e, f, ](x)) = e(fx).

Proof. Consider the following diagram, in which i is the identity map on E and j is the natural map defined earlier. By (3.5) (ii) h(e,f, x ) = h(e, f, y) if j ( x ) = j ( y ) . Thus there is a well defined function g as indicated, and dearly g is unique with respect to making the diagram commute. Because h and i × i x ] are morphisms, g is one too. Because all spaces are compact Hausdorff, g is continuous. (So far this is just Sierpinski's Lemma [11] at work.) E×Ex M ....... .h_........ ,..~M f ixi×i

jJ~ I"

ExE~M/]

7

J

Vol. 20, 1980

Some theorems on hemigroups

205

Now we will show that g is an equivalent retraction, so suppose g(e, f, j(x)) = g ( e ' , f ' , j(x')). We need to show that (e,f, j(x)) and (e', f', j(x')) are in the F relation on E x E x M/3. First e(fx) = e'(['x') is true from the commuting diagram. Then squaring yields e = e'; and j(x) = j ( e . fx) and j(x') = j ( e ' - f'x') by (3.5) (i), so j(x) =j(x'). Then if we let a = (e, f, j(y)) we see that (e, f, j ( x ) ) , a = (e', f, j ( x ' ) ) , a and the latter product equals (e',f', j ( x ' ) ) . a because second coordinates are irrelevant in a simple product. W e will now show that for each idempotent e ~ M, e M n M e is a maximal principal subhemigroup of M and isomorphic with M/I. (3.9) L E M M A . Let e ~ E be arbitrary. ( e M ) e , a n d also (eM)2~ e(eM) 2.

Then

(eM)2=eMOMe=e(eM)

=

Proof. Let x e ( e M ) 2 ; so x = ( e m ) ( e m ' ) = e m . e m ) ( e m ' . e m ) = e ( e m ' . e m ) e e ( e M ) 2. This shows the inclusion asserted and also shows x ~ eM. Similarly x = ( e r a . e m ' ) ( e m ' , e m ' ) = ( e m , e m ' ) e e M e . So we have ( e M ) 2 c e M O M e . Continuing, if y ~ e M O M e , then y = e m * m ' e , m, m ' e M , so y 2 = e = m ' m ' , and then y = m ' e = m ' m ' . e m ' = e . e m ' e e( eM). Hence e M n M e ~ e( eM). If z e e( e M ) then z = e ( e m ) = ( e . e m ) ( e m . era) = (e . e m ) e ~ (eM)e, so e(eM) c (eM)e. Obvi-

ously, (eM)e c e(eM) so we have proved the equalities. (3.10) T H E O R E M . I f M is a hemigroup a n d e ~ E then e M O M e is a m a x i m a l principal subhemigroup o f M. Proof. L e t S = e M O Me. W e will show that S is a subhemigroup, that x S = Sx = S for some x ~ S and that if T is any other subset of M satisfying these

properties and containing S, then T = S. First note that S c e M so S 2 c (eM) 2 = S by (3.9); hence S is a subhemigroup. Clearly e e S, and S c eS by (3.9), so S = eS. x ~ M e implies x = m e = ( m e ) ( e e ) = xe, so S = Se. Therefore, S is principal. If T is a principal subhemigroup of M containing S, then e e S ~ T implies e T = Te = T by (1.2) (v), so certainly T c e T A T e c e M n M e = S. Therefore T = S and hence S is maximal as asserted. (3.11) T H E O R E M . L e t e e E be arbitrary. T h e n the m a p J l ( e M O M e ) is an isomorphism o f e M n M e onto M / I . T h u s for a n y e, f ~ E, e M O M e a n d f M N M f are isomorphic. Proof. Recall e M n M e = e ( e M )

by (3.9). Since j ( e ( e M ) ) = j ( M ) by 2.5(0, the restriction map is onto. T o see that it is one-to-one, let x, y e e ( e M ) . Then x = e ( e x ) and y = e(ey) by 1.2(iv), because e(eM) is a principal hemigroup with

206

K . P . CHINDA

AEQ. MATH.

idempotent e by (3.10). If j ( x ) = j(y), then e(ex)= e(ey) by (3.5)(ii), so x = y and thus the restriction map is one-to-one. (3.12) C O R O L L A R Y . / / e M n M e for each f ~ E.

is trivial for one e e E , then

[Mf'IMf

=f

4. The Relation Q Although E need not be a subhemigroup of M, we will see that M/Q is a hemigroup and is bijective with E in a natural way. (4.1) D E F I N I T I O N S . L e t M be a hemigroup and define Q = { ( x , y): x 2= y2, x, y ~ M}, q : M --> M / Q as the natural map, and tr : M---> E as the squaring map, viz. tr(m) = m z, m e M. (4.2) T H E O R E M . I f M is a compact or discrete hemigroup then Q is a closed congruence and q is a surmorphism. I f M = M 2, q [ E is a bijection and the unique map which makes the following diagram commute. q M ................. M / Q

o'

!

/

/

f

"

/ / /

I f M = M 2 then cr is an open mapping. I f in addition M is connected, then q is monotone.

Proof. Clearly Q is an equivalence relation. Since ( a x ) E = ( a x ) ( a x ) = a a = (ay)(ay) = (ay) 2 and (xa) 2 = (xa)(xa) = xx -~x 2 = y2 = yy = (ya)(ya) for each a e M and each (x, y ) e Q , it follows that Q is a cofigruence. Moreover, since Q = ( ~ x ~ ) - l ( A ) , where Zl represents the diagonal of E, Q is closed. H e n c e the canonical map q :M---~ M/Q is a surmorphism. Obviously q tE is a bijection. Since Q = ker ~, E and M I Q are isomorphic as sets. Existence and uniqueness of the map commuting the following diagram follows from Sierpinski's lemma [11], since tr(x) = tr(y) implies q(x) = q(y). q M . . . . . . . . . . . . . . . . . -**--M / O

a

t

E/

//" //

/

/

/

"

Vol. 20, 1980

Now

Some theorems on hemigroups

207

q(x) = x M = x2M(1.2iii) = or(x)M = (q l Eoor)(x).

H e n c e q = q I Eoor, implying that q I E is the unique map commuting the above diagram. Since or-l(x) = x2M, which is connected if M is connected, or is monotone. We now show or is open if M = M 2. Let W be an open set in M. We have to show that or(W) is open in E. W e first show that

E\o-(W) ={x ~ M I x M c tV~W}NE. Let y e E \ o r ( W ) . T h e n y r t r ( W ) and y 2 = y so o ' - l ( y ) O W = t - 1 . H o w e v e r or-l(y) = y2M = yM. so y M n W = [], which means y M c E \ W and y e E. Hence E\or( W) c {x ~ M l x M c M~ W} n E. Conversely if z e {X e M I x M D M~ W} A E, then z 2 = z and z M c M ~ W so that z M A W = I Z ] and hence z 2 M N W = [ Z ] because z = z 2. So o - - l ( z ) n W=1-1 implying z~o-(W) and hence z eE\or(W). Hence the equality as desired. { x ~ M [ x M c M ~ W } is closed by (1.2)(iv), [6], and it follows that or(W) is open.

5. Cancellable Hemigroups We now study hemigroups for which cancellation laws hold. It is shown that every such hemigroup is topologically homeomorphic to a cross product of a trivial left multiplication hemigroup and a principal hemigroup; and that if the multiplication is defined appropriately on the cross product, then the homeomorphism is indeed an isomorphism also. It then follows that if a hemigroup has condimension one and is right cancellable, then it is either principal or is equal to the set of all its idempotents. Following Aczrl [1] we show that every hemigroup that is left cancellable is principal. (5.1) D E F I N I T I O N . A hemigroup M is said to be right cancellable if x = y whenever xa = ya holds for all x, y, a e M and is said to be le[t cancellable if x = y whenever ax = ay for all x, y, a e M. (5.2) T H E O R E M . Let M be a right cancellable hemigroup. Then M is

homeomorphic to E × eM where E is the set of all idempotents o[ M and e ~ E is arbitrary.

208

K . P . CHINDA

AEQ. MATH.

Proof. Define qJ :M---~ E x e M by qJ(m) = ( m 2, era). Since multiplication on M is continuous, the map q~ is continuous. qJ is one-to-one. Assume qffm')= 4,(m'9, that is, (m '2, era')= ( m 'r2, emO. This means m '2 = tn 'a and era' = era". Hence m '2 • era' = m 'a . era", from which it follows that m'e = m"e. Since M is right cancellable, ra' = ra". Hence the map q~ is one-to-one. rk is onto. L e t ( e ' , e m ' ) ~ E x eM, and take ra = e'(era'). Then ra2= e ' ( e m ' ) , e'(em') = e " e ' = e' and era = e(e'(era')) = e2(e ' ' era'). Now since e 2 = era'. era', from the last equality we have era = (era'. era')(e' . era') = era" e'. Then era" e' = (era'. e')e' = (era'. e')(e', e') = e m " e'. Hence era = era' from right canceUability. Hence 6(ra) = (m 2, em) = (e', era'). This proves that 4, is onto. q,-~ is continuous. Clearly 4,-1(e ', era')= e'(era'), as shown above. Hence qJ-~ is nothing but multiplication restricted to E x eM, and hence is continuous. Thus qJ is indeed a homeomorphism.

5.3) C O R O L L A R Y . I[ multiplication on E x e M is defined by (el, emO e2, era2)= (el, e((el" eml)(e2" era2))), then E x e M is a hemigroup and q, is an isomorphism. Hence M is isomorphic to E x eM. Proo[. We show that q, preserves multiplication, from which it will follow that E x e M is a hemigroup. For m t , m 2 ~ M we have qJ(mlm2) = ((re,m2) 2,

e ( m l m 2 ) ) = (m~, e(mim2))

(1)

and *(ml)ql(m2) = (m~, e m O ( m 2, era2) = (m~, e((m~, eml)(m~, em2))) = (m~, e((m~, eml)(m~, era2)))

= (m~, e((mle)(m2e))) = (m21, e(mlm2)).

(2)

From (1) and (2) it follows that $(ralm2)= t~(ml)~b(m2). Since ~ is one-to-one and o n t o it follows that $ is an isomorphism. For each n w r i t e / ~ ( X ) for the n 'h Alexander-Wallace-Spanier Cohomology groups of X with arbitrary but fixed coefficient group.

Vol. 20, 1980

Some theorems on hemigroups

209

(5.4) D E F I N I T I O N . Let X be a non-empty compact Hausdorff space. W e define the cohomological dimension, or more briefly the codimension of X to be

cd(X) = inf {n I n is a non-negative integer and H'+I(X, A ) = 0 for every closed subset A of X}. W e define cdOS])=-1. For an arbitrary Hausdortt space X, we then define the codimensic'~x of X to be cd(X)=sup{cd(K)l K

is a compact subset of

X}.

The following result is known [10]. If X and Y are Hausdorff spaces with c d ( X ) = m and c d ( Y ) = 1, then c d ( X x Y ) = r e + l , X, Y being continua. (5.5) C O R O L L A R Y . Let M be a compact and connected right cancellable Hemigroup having codimension one. Then either M is principal or M is an idempotent hemigroup.

Proof. Since M is right cancellable, by (5.3) above M is homeomorphic to E x eM. Since cd(M) = 1, at least one of E, e M must have codimension less than one. In case cd(eM) is less than one is since e ~ e M , e M # O . Hence cd(eM)=O. But this implies, since e M is compact and connected, that e M = e. Thus M is homeomorphic to E x e, from which it follows that M = E, the set of idempotents. On the other hand, if c d ( E ) is less than one, then since E#[S] cd(E)=O. As before, since E is compact and connected, E is a singleton. In that case M has exactly o n e idempotent, say, e. Hence M = eM. Also M = Me since for any x e M we have xe = (xe)(ee)= x e . e and M is right cancellable. Thus M = e M N M e , which was shown to be principal in (3.10) above. (5.6) T H E O R E M . If M is a right cancellable hemigroup and if (eM)= e or e (eM) = e for some e ~ E then M = E. Proof. Since by (3.9) above ( e M ) e = e ( e M ) , it will be enough to prove the result for the case when (eM)e = e for some e e E . Now by (3.12) this implies (gM)g = g for all g ~ E. Now let x e M. Then (x2M)x 2 = x 2 from above since x2~ E. Therefore, (x2x)x 2= x 2= x 2" x 2. Hence, as M is right cancellable, we have x 2 • x = x 2,

from which it follows that

x2(x 2 • x) = x2x 2.

210

K.P. CH1NDA

AEQ. MATH.

But the first of these terms is x - x 2. T h e r e f o r e we now have X " X 2 = X 2 " X 2.

Using right cancellability one more time we have X -~. X 2

showing that M = E. (5.7) R e m a r k . That the above result is not true without right cancellability can be seen from the following: (5.8) E X A M P L E . L e t M = X × X . On M define multiplication by (a, b ) x (c, d) = a, c), a, b, c, d, ~ M. T h e n M is a hemigroup with E = {(a, a) J a ~ X}, and clearly ((a, a ) M ) ( a , a ) = (a, a). However, M # E and M is not right cancellable because (a, b)(c, cl) = (a, c) = (a, e)(c, d) and yet (a, b) # (a, e) if X has more than one point. (5.9) T H E O R E M . I f a hemigroup M is left cancellable then it is principal. Proof. Let e be any idempotent in M and x any arbitrary d e m e n t of M. Consider ex = ( e e ) ( x e ) = e ( x e ) . This gives x = xe by left cancellability. Hence x ~ Me. Since x was arbitrary, M = Me. Also x 2= x2e by the same argument as above. Then x2e = x 2= x2x 2 and on cancelling on the left we get x 2= e. Hence x = xe = x2(ex) = e(ex) ~ eM. T h e r e f o r e M = e M also. So M = e M n M e

and so by (3.10) M is principal.

6. Connected Hemigroups The relation ( x y ) ( z y ) = x z - ( i ) has also been considered by others, especially b y Acz61, Hosszu and Kuczma, references to which may be found on pages 73, 76 and 274 of [1]. T h e case that interested them most was when M is the set of all real numbers. Acz61 in [1] calls (i) the transitivity equation and writes it as F(F(x, y), F(z, y)) = F(x, z), which is the same as (i) if we suppress F and write xy for F(x, y).

Vol. 20, 1980

Some theorems on hemigroups

211

Chinda in [4] has obtained all possible algebraic structures on the real line or on the unit circle (in the complex plane) that satisfy (i) above using only the hypothesis that M ~ = M which Furstenberg in [7] showed was none too strict. We now extend that result to spaces of cd one. In doing so we use a theorem of Day and Hofmann in [5]. W e begin by defining an act. (6.1) D E F I N I T I O N . Let S be a sernigroupmthat is, a Hausdorff space with continuous associative multiplication S x S --~ S, denoted by juxtaposition. L e t X be any other Hausdortt space. Then S is said to act on the right on X if there exists a continuous function Ix : X x S ~ X satisfying

~(x, st)= ~(~(x, s), t). The map ~ is called an act. Let M be a hemigroup, and fix e ~ E. Define a map t h : M x M---~ M by f f t ( x , y ) = x ( e y ) . For ease of notation, we write r h ( x , y ) = x o y . Note that rh is continuous, and that, restricted to the principal hemigroup e(eM), this is a group multiplication by (3.9), (3.10) and (1.5)(b). The following lemmas can be proved without much difficulty. (6.2) L E M M A . I f z e e M , then x o ( y o z ) = ( x o y ) o z for all x, y e M . In other words o is associative on eM. Also e M o e M c eM, so (eM, o) is a semigroup. Proof. z e e M means z = e m for some m e M and hence z 2 = (em)(em) = e 2 = e. H e n c e ez

= (ez(zz)

= ez

. z 2 = ez

. e.

Now

x o(y o z ) = x o(y

ez) = x ( e ( y • ez))

x(ey) ( e ( y - ez). (ey)) x(ey) ( e ( y " ez))(ey) x(ey) ( e ( y - ez))((e, e z ) ( y • ez) x(ey) (e(e. ez)) x(ey) ((ez "ez)(e . ez)) x(ey) • (ez" e) x(ey) • (ez) x(ey)oz

=(xoy)oz.

212

K.P. C H I N D A

Also, if m, m' e M

AEO, M A T H .

then

(em o(em') = (em)(e . era') = (em)2[(e • em')(em)]

= e[(e. e m ' ) ( e m ] e eM, since M ~ c M. (6.3) L E M M A . Let M be a hemigroup. Fix e ~ E with Fa defined as in the paragraph preceeding (6.2). W e have ~h I M x e M : M × e M - - ~ M is an act. Moreover, rh ( M x e M ) ~ Me. Proof. Let x ~ M and y, z ~ eM. Then

#z(y, y oz) = xo(y oz) = ( x o y ) o z = #~(xoy, z ) ,

by (6.2), since z e e M . Hence rh is an act. Also since x o ( y o z ) = x ( e ( y ( e z ) ) ) = x ( e ( y , e z ) ) . e, th(x, y o z ) e M e . (6.4) Remark. Let T = (eM)e. Then (T, o) is a group with identity e, as noted above, and a subgroup of (era, o). It then follows from (6.3) that the map rh I X x T: X x T ~ X

is an act where

X = M e c M.

The following result is an immediate consequence of (2.5), [5]. (6.5) T H E O R E M . I f X is a continuum with c d ( X ) = n, and if G is a compact group acting on X on the right then either X G = x G for each x ~ X or else c d ( x G ) < n for each x e X . (6.6) C O R O L L A R Y . I[ M = M ~ is a compact and connected hemigroup with c d M = 1, and if y(eM) is not a singleton for at least one y ~ M and e ~ E then M is principal. Proof. Let ~a M e x T--~ Me, where T = (eM)e, be as in (6.4). Then as asserted there, (T, o) is a group. Since M is compact and connected, M e is compact and

Vol. 20, 1980

Some theorems on hemigroups

213

connected also and so is T. Moreover, rh is an act. In (6.5) take X = Me and G = T. T h e n either M e = x T for all x ~ M e o r else c d ( x T ) < 1 for each x ~ Me. Now ye e M e and ye • T = ( y e ) ( e M . e) = y • eM, which is not a singleton by hypothesis. H e n c e we must have M e = x T for all x e Me. W e show that this implies M is principal. W e have ( m e ) ( e M . e) = M e for all m ~ M. Thus m ( e M ) = M e for all m e M and for any e ~ E. In particular, if m = e, then e ( e M ) = Me. which means e = e 2 = e(eb)" e(eb) = ae • ae = a 2. Hence, for all a e M, a 2 = e. Now let y e M. T h e n since y = z w for some z, w e M, y = z:. wz =e(wz)eeM. Thus M = eM. But then, by definition, M is principal. (6.7) T H E O R E M . If M is a locally connected continuum wtth M 2 = M and if C is a component of M ~ E , then o'(C*) = Bdr C, where C* is the closure of C and tr is the map defined in 4.1 above. Proof. Since ~r is a retraction, E is closed in M and hence M \ E is o p e n and, C being a c o m p o n e n t of M \ E , it follows that C is also open [2]. Since C* is the closure of C and C is connected, C* is also connected and Bdr C = C * \ C . Since C c C* and C is a c o m p o n e n t of M ~ E it follows that C * ¢ MkE. In other words C * A E # V q . In fact, for any p e C * \ C . C U p is connected and therefore p ~ E. H e n c e C * \ C c E, and therefore t r ( C * \ C ) = C * \ C T h e r e f o r e or(C*) = or(C) o c r ( C * \ C ) = or(C) o

C*\C

(1)

Now let x e C. T h e n since M = M 2, we have x ~ x 2 M and t r ( x 2 M ) = t r ( x ) = x 2. Also x 2 M N C ~ D , since x belongs to both of them. Since x 2 belongs to both E and x2M, it follows that x 2 M N E ~ V q . Since M is connected, x Z M is connected; and since x 2 M meets both C and its complement, it meets the boundary of C. In other words x2 M f~ ( C * \ C) ¢ I-1. Let a ~ x ~ M n ( C * \ C ) . T h e n or(a) = x 2 on the one hand and or(a) = a on the other hand. Hence, ~r(x~M) = ~ ( a ) = ~ ( x ~) = ~ ( x ) = a.

214

K, P. C H I N D A

AEQ. M A T H .

Thus cr(x)e C*\C. But x~ C was arbitrary. Therefore tr(C)= C*\C. Then by equation (i) we have =

r(C) U o ' ( C * ) \ C )

= o'(C) U C*\C = C * \ C = Bdr C (6.8) DEFINITION. A point p belonging to a space M is said to be of countable order if every neighbourhood of p contains a neighbourhood V of p that contains only countably many points of M. (6.9) PROPOSITION. Let M be a hemigroup. If M is a continuum with M ~ = M and E nontrivial, and i[ p ~ M is o[ countable order, then p ~ E. Proof. Assume on the contrary that p~ E. Then there exists an open set V containing p such that V N E = I-q. Since p is of countable order there exists a neighbourhood of p, say W, such that W c V and such that W contains at most countably many points of M. Since M 2 = M implies that x ~ x2M,

w c U {x2MI x Since M is connected, therefore each x2M is connected. Since M is compact, each x 2 M is compact, and hence closed. Hence W, an open subset of a compact space M, is the countable union of closed sets. Since the sets x2M are pairwise disjoint, it follows by Bake's theorem [2] that at least one x 2 M must contain an open set Vx2. But then tr(Vx2)c cr(x2M)= X2. This is impossible since the map cr is open and x 2 is closed in the nontrivial continuum E. Hence p e E. (6.9) PROPOSITION. Let M be a hemigroup with E nontrivial. If M is connected and if a point p in M cuts M, then p ~ E. Proof. Assume that p ~ E . Then since p cuts M, there exists two mutually disjoint open sets W and V such that M {p} = W U V.

Since M is connected and E is a continuous image of M, E is also connected. Since E c M~{p}, either E c W or E c V. Without loss of generality it may be assumed that E c W. Since x 2 M is connected for each x ~ M, for each y e V, y2M must contain p. Since the sets y2M are pairwise disjoint, it follows that for each

Vol. 20, 1980

Some theorems on hemigroups

215

V E V, v E p 2 M . However, this means tr(V) = p2. But that is impossible since the m a p (r is o p e n and points are not o p e n in a nontrivial connected Hausd0rff space. H e n c e we must have p ~ E.

7. Compact Totally Disconnected Hemigroups It will be shown in this section that, just as in topological groups and semigroups, a compact totally disconnected hemigroup with ~ = M is the inverse limit of finite hemigroups. W e start with some definitions. (7.1) D E F I N I T I O N . A directed set is a non-void set A and a relation -< such that a --- a whatever a c A, a -----b, and b --- c imply a --- c whenever a, b, c ~ A, and for each a, b e A there is some c ~ A such that a --- c, b - c. (7.2) D E F I N I T I O N . A n inverse system is a collection of spaces Xa, indexed by a directed set A, and m a p s (called bonding maps)

~:Xb--~ X,~,

a<_b

satisfying f~ = identity, whatever a e A,

f~ffb=f~, whatever

a-
in

A.

T h e inverse limit of such a system is the set of all those x ={xo}elI{Xala c A } such that ~ ( x b ) = x ~ for a, b e A with a<_b. If X is the inverse limit then fa : X ~ Xa will denote the restriction to X of the projection m a p of the product space onto X. T h e following t h e o r e m in [2] is well known. (7.3) T H E O R E M . Any compact totally disconnected topological group is the inverse limit of finite groups with surjective bonding maps. T h e corresponding t h e o r e m on semigroups is in [9].

Any compact totally disconnected semigroup is the inverse limit of finite semigroups with surjective bonding maps. (7.4) T H E O R E M .

T h e following two L e m m a s a p p e a r in [9]. (7.5) L E M M A . Let S be a compact Hausdorff space and R an open equivalence on S. Then SIR is a finite discrete space.

216

g . P . CHINDA

AEQ. MATH.

(7.6) L E M M A . Let S be compact, totally disconnected Hausdorff space and D an open subset of S x S containing the diagonal. Then there exists, in D, an open subset R of S x S which is an equivalence.

Note. By an equivalence we mean a relation which is reflexive, symmetric and transitive. (7.7) L E M M A . Let M be a compact hemigroup satisfying M = M 2 and let U be an open congruence. Then M/U is a finite discrete hemigroup satisfying (M/U) 2 = M/U.

Proof. By (7.5), M / U is finiteand discrete. Also since U is a congruence the natural map f : M-'~ M / U is a morphism. H e n c e for any a, b, c ~ M,

(Ka)f(b))(f(c)f(b)) = [(ab)f(cb) = f((ab )(cb)) = f(ac) = f(a)f(c). H e n c e M / U is a hemigroup. Also since M = M 2,

f(a) = / ( x y ) = f(x)f(y) because a e M implies a = xy for some x, y ~ M. Hence (M/U) 2 = M/U. (7.8) L E M M A . Let M be a compact totally disconnected hemigroup and D an open subset of M × M containing the diagonal A. Then D contains an open congruence. Proof. L e t R be an open equivalence contained in D, which exist by (7.6). Let U be the union of all congruences contained in R. U # [ ] because A c U. Then U is the largest congruence in R and by the proof of L e m m a 1 of [8] it follows that R is open. This proves the lemma. The following well known result appears in [11]. (7.9) P R O P O S I T I O N . The inverse limit of an inverse system of compact spaces is non-void and compact. We are now ready to prove a theorem. (7.10) T H E O R E M . Let M be a compact totally disconnected hemigroup with M 2 = M. Then M is the inverse limit of finite discrete hemigroups M~ each satisfying

Vol. 20, 1980

S o m e t h e o r e m s on h e m i g r o u p s

217

Proof. By (7.8) we can find a family {J~ : h e A} of open congruences with the property that i'lx~AJx =A, the diagonal of M × M . If we define a---/3iff Ja ~ J , then A becomes a directed set. Let a e A and define M~ = M/J,,. Then by (7.5) each of M~ is finite and discrete. H e n c e {M~:a e A} is a family of finite and discrete hemigroups each satisfying M2,=M~ (7.7). Define bonding maps as follows: for a - / 3 define f a: MB __~M,~ as in the diagram where rl,~ and "qa are the natural maps. ~ is unique by Sierpinski's L e m m a [11]. If a <--/3<- 3', then ~ = faf~ because Mt3 f~ M~ rl,~ M rl~ ~%=n~,

~%=~,

~=n~

and

M ........

-~-". . . . . . . . .

-~'~ M.

Hence ~ = f ~ because of the uniqueness. Hence we have an inverse system {M~ : ~ , A}. Each of M~ is compact hence the inverse limit is non-empty, by proposition 7.9. Let M~ denote this limit. We claim M~ is isomorphic to M. For define ~ : M ~ M~ by ~(m) = (~1,,(m))a~A. It is an easy exercise to check that the map g' is continuous, one to one, onto and preserves multiplication.

8. Finite Hemigroups We know that to every group there corresponds a principal hemigroup. In this section we intend to discuss the existence of an operation on a finite set M so that M becomes a hemigroup. W e will be mainly concerned with those operations which are different from left trivial (xy = x). Also we will not consider multiplications that produce only one idempotent or all idempotents. For, as already shown, M 2 = M and a unique idempotent implies M is principal, and M = E implies left trivial multiplication. W e will show how the various entries in the multiplication table can be chosen and how to cross check if any entry is correct or not. It will also be shown that if the finite set M has n. + 2 d e m e n t s , then there are precisely n non-isomorphic multiplications satisfying (xy)(zy) = xz and M 2 = M and such that E contains n + 1 elements.

218

K. P. CHINDA

AEQ. MATH.

Finally w e s h o w that a h e m i g r o u p M consisting o f ( 2 p - 1 ) elements, p a prime, a n d satisfying M 2 = M c a n n o t have a principal s u b - h e m i g r o u p of o r d e r p. W e begin with the following L e m m a . (8.1) L E M M A . Let M be a finite h e m i g m u p consisting of n elements so that the multiplication tabulation on M looks like an n × n matrix. Let a~j be the entries in this matrix, 1 <- i, j <- n, c~ being an idempotent for all i. T h e n a i i = auaii and a~i = ajjaij.

Proof. A s s u m e w i t h o u t loss of generality that a~i = xy and therefore ai~ = yx. T h e n obviously a~i = x 2 a n d aii = y2. T h e r e f o r e ai~a~ = x 2- yx = xy = a~i. Similarly ai~a~ = y 2 . xy = yx = ai~. This proves the lemma. (8.2) T w o - e l e m e n t Hemigroups. It is claimed that these m u s t be left trivial. F o r assume M = { a , b}. T h e n either there is exactly o n e i d e m p o t e n t , say a, in which case a 2 = a and b 2 = a or there are two idempotents, in which case a 2 = a and b 2 = b. In the first case the only possible multiplication is the g r o u p multiplication and in the s e c o n d case it is left trivial. T h u s there is n o nontrivial h e m i g r o u p multiplication o n a set of two elements. (8.3) Three-element hemigroups. W e assume M = { e , f, a} with e 2 = e, f 2 = f and a 2 = e (or f w i t h o u t a n y prejudice). Since x e I M implies that x 2 = f and since there is exactly o n e e l e m e n t in M w h o s e square is f, ~ f = { t ~ . Also e M = { e , a}. H e n c e ea = a o r ea = e. W e discuss each case individually. T h e multiplication so far looks thus:

f a e

e

e

ffff (1) A s s u m e e a = a . T h e n e ( e a ) = e a . Thus, a e = a 2" e a = e ( e a ) = e a ; that is ae = ea = a. N o w e f = a o r e f = e. W e claim this is impossible because if e f = a

Vol. 20, 1980

Some theorems on hemigroups

219

then a f = e for otherwise by (2.2) all c o r r e s p o n d i n g elements in the first two rows are equal. N o w a f = e implies a z . f a = e which in turn implies e ( f a ) = e. B u t f a = f, h e n c e e f = e, a contradiction. H e n c e e f = a is n o t possible. Similarly, if e l = e then a [ = a (as a r g u e d above). B u t then a 2- f a = a f = a implies that e • f a = a, that is e f = a, because f a = f. B u t this contradicts the fact that e f = e. This shows that situation (1) is impossible. W e n o w turn to the s e c o n d case. (2) A s s u m e e a = e. In that case e • e a = ee = e. B u t we can write e = a 2. H e n c e a 2 • e a = a e = e. H e n c e a e = e a = e. B u t ee = e so that we n o w have a e = e e and hence, by (2.2), a x = e x and x a = x e for all x ~ M. T h u s e f = af. N o w e f = a [ = a or e l = a f = e. But e l = e is n o t possible since then e M = e and M 2 ~ M. H e n c e e [ = a [ = a and we arrive at the following multiplication for M.

f t2

e

e

cl

e

e

e

a

f f f f (8.4) F o u r - e l e m e n t H e m i g r o u p s . W e consider the case w h e n M has either two o r three idempotents, the cases of o n e and four i d e m p o t e n t s being trivial. By using an a r g u m e n t similar to that described in (8.3), we arrive at the following multiplication table w h e n M has exactly two i d e m p o t e n t s e and f:

f a

e

a

e

a

e

a

e

a

e

b

f

b

f

b

f

b

f

b

f

Finally, we p r o v e below two t h e o r e m s o n finite hemigroups, o n e giving the precise n u m b e r of n o n i s o m o r p h i c h e m i g r o u p multiplication u n d e r certain restraints a n d the o t h e r concerning the existence of s u b - h e m i g r o u p s of a hemigroup.

220

K, P, CHINDA

AEQ. MATH.

(8.5) T H E O R E M . Let M be a set consisting o f (n + 2) elements. There exist exactly n nonisomorphic multiplications that can be defined on M such that M becomes a Hemigroup having exactly (n + 1) idempotents and satisfying M 2 = M. Proof. L e t M = { e , a, w l , w 2 , . . . , w,}. O n M d e f i n e m u l t i p l i c a t i o n such t h a t w 2 = w , e 2 = e a n d a 2 = e. T h e n w e h a v e

E = {e, wt, w 2 , . . •, w,}. N o w f o r a n y x ~ M, (wlx) 2 = (w~x)(w~x) = w~wi = w~. H e n c e f o r a n y x ~ M , w~x = w~ for i = 1, 2, 2, 3 . . . . . n, b e c a u s e t h e o n l y e l e m e n t satisfying t h e c o n d i t i o n t h a t its s q u a r e is w~ is w~ itself. W e t h u s h a v e t h e f o l l o w i n g m u l t i p l i c a t i o n Wn

a

e

e

e

W1

Wt

W1

V¢1

W1

W2

W2

W2

W2

W2 .....

Wn

Wn

Wn

Wn

Wn

°*°

w

~,~

T h e o n l y rows r e m a i n i n g t o b e filled a r e t h e first t w o rows. W e first s h o w t h a t an e n t r y in t h e s e r o w s c a n o n l y b e filled w i t h an a o r e. F o r if x ~ M a n d if y = a o r e w e h a v e (yx) 2 = ( y x ) ( y x ) = yy = y2 = e. H e n c e yx = a o r yx = e. N o w let x # a. T h e n x 2 = x. H e n c e (ex)x = (ex)x 2 = ( e x ) ( x x ) = ex. So if ex = a f r o m t h e a b o v e , it w o u l d f o l l o w t h a t ex = ex. B u t t h e n b y (2.1) a n d (2.2) it w o u l d f o l l o w t h a t e m = a m f o r all m e M. T h a t c o n t r a d i c t s t h e fact t h a t ea = a a n d a a = e. H e n c e n o c o l u m n , e x c e p t p o s s i b l y t h e first, in t h e m u l t i p l i c a t i o n t a b l e b e g i n s e, a , . . N e x t a s s u m e a w i = a a n d ew~ = e. T h e n a = aw~ = a2(w~a) = e(w~a) = ewi, b e c a u s e a 2 = e a n d w~a = w~. H e n c e a = ewe, so ew~ c a n n o t b e e. H e n c e n o c o l u m n , e x c e p t p o s s i b l y t h e s e c o n d , b e g i n s a, e . . . . .

Vol. 20, 1980

Some theorems on hemigroups

221

N o w consider the p r o d u c t ( a w ~ ) ( e w ~ ) . Clearly aw~ = ew~ because w~a = w ~ so that a 2 ( w ~ a ) = a2w~ = ew~ since a 2 = e. H e n c e aw~ = ( a a ) ( w ~ a ) = ewe. S o if aw~ = a = ew~ o r if aw~ = e = ew~ then ( a w i ) ( e w i ) = a a o r ee. In either case ( a w ~ ) ( e w ~ ) = e. Also since M is a h e m i g r o u p , ( a w l ) ( e w ~ ) = a e . H e n c e a e = e. T h e n e ( a e ) = ee f r o m which it follows that e a = a e = e. T h u s the first two columns begin (e, e . . . ) and e v e r y o t h e r c o l u m n begins (e, e . . . ) o r (a, a . . . ) , except n o t all c o l u m n s m a y begin (e, e . . . ) in view of the supposition that M 2 = M. W i t h o u t loss of generality let aw~ = e if i < - S , a w i = a if i > S , w h e r e S is an integer varying f r o m 0 to n, assuming possibly the value 0 but n e v e r the value n. T h u s S m a y take o n n distinct values, which d e t e r m i n e the n distinct h e m i g r o u p multiplication. W e claim that every such multiplication satisfies the condition (xy)(zy) = x z f o r all x, y, z e M. F o r if x = wi, then it is obvious because (xy)(zy) = (w~y)(zy) = wi(zy) = w~ = w~z = x z . If x = a or x = e, then we have (xy)(zy) = ( a y ) ( z y ) or

(xy)(zy) = (ey)(zy). B u t since a m = e m for all m ~ M, a y = ey, a n d h e n c e we have (xy)(zy)

=

(ay)(zy).

Since a y = a o r e, it follows that ( a y ) ( z y ) = a ( z y ) o r e(zy). But, again, since a m = era, a ( z y ) = e(zy). H e n c e we have always (xy)(zy)

=

a(zy).

Also x z = a z o r e z ; b u t since the two are equal, x z = a z , always. W e now show that a z = a ( z y ) . F o r that will p r o v e that (xy)(zy) = a ( z y ) = a z = x z , which will p r o v e o u r claim. If z = w~ we have n o t h i n g to prove, because in that case zy = z. If z = a then a(zy)=a(ay) and if z = e then a(zy)=a(ey). B u t a y = e y ; and therefore,

222

K . P . CHINDA

AEQ. MATH.

whether z = a o r z = e , we have a ( z y ) = a ( a y ) . Also a y = a or ay = e ; therefore a ( z y ) = a a or a ( z y ) = a e . But from a m = e r a , it follows, if we take m = e that ae = ee = aa. H e n c e a z = a(zy). Hence there exist precisely n hemigroups. Since groups of prime order have unique group structures, it follows that every principal hemigroup of prime order has a unique structure too. This follows immediately from (1.5)(b). (8.6) T H E O R E M . L e t M be a hemigroup with M 2 = M and a s s u m e the cardinality o f M is 2 p - 1, where p is a prime number. T h e n M cannot have a principal subhemigroup o f cardinality p. Proof. Assume M has a principal subhemigroup N of order p and let e be the unique idempotent of N. Then by the remarks preceding the theorem, N has a unique structure which is derived from the corresponding group structure. Let N = e, al, a2 . . . . . . ap-1 and let the remaining members of M be wl, w2 . . . . . wp-1. Since the group structttre derived from the hemigroup structure is cyclic, we have the following multiplication diagram:

e

e

%-1

%-2

"'"

al

al

al

e

a~_ 1

...

a2

a2

a2

al

e

...

a3

ap-1

av_ ~

a ~

a2

""

e

Wl

W2

Vol. 20, 1980

Some theorems on hemigroups

223

Here the group structure for N is assumed to be such that ao_,a, = e. W e have to fill the remaining squares. As in 8.5 the first p rows can only be filled with members of N. Also no two rows among the first p rows can have the same element in any one column. For otherwise the two rows would have to be the same. (Thm.2.2). The last p - 1 rows can be filled with only those elements of M that are not in N. Thus the last ( p - l ) rows have to be filled with only ( p - l ) elements wl, w2 . . . . . wp_l. H e n c e at least two of the first p columns must have the same entry in one of the last ( p - 1) rows. Assume the ith row has the same entry in the ]th and kth column, p + 1 -< i --- 2p - 1, i <- j, k <- p. Let the entry be X, where X = w , for some r such that l < - r < - p - 1 . Then by L e m m a 8.1, a i w ~ = e w , and akw~=ew,, H e n c e a j w i = a k w ~ . . B u t then ajy = a k y for all y e M by (2.2). This is impossible. Hence M cannot have a Principal Sub Hemigroup of order p. The author wishes to express his sincere gratitude to Professor Alexander D. Wallace not only for the very many ideas, guidance, suggestions and encouragement he continuously offered but also for the personal interest he took in developing the professional ability of the author and for making the author's stay at Gainesville an experience to be cherished.

REFERENCES [1] ACZI~L, J., Lectures on functional equations and their applications. Academic Press, New YorkLondon, 1966. [2] BOURBAm,N., General Topology. Addison-Wesley, Reading, Mass., 1966. [3] CHINA, K. P., The equation F(F(x, y), F(z, y))=F(x, z). Aequationes Math. 11 (1974), 196198. [4] CHn~DA, K. P., Structure theory of finite, compact and connected hemigmups. Dissertation, University of Florida, 1973. [5] DAY, JANEM. and H o ~ , K. H., Clan acts and codimension. Semigroup Forum 4 (1972), 206-214. [6] DAY,J. M. and WALLACE,A. D., Semigmups acting on continua. J. Austral. Math. Soc. 7 (1967), 327-340. [7] FUaSTEr~ERG, HARRY, The inverse operation in groups. Proc. Amer. Math. Soc. 6 (1955), 991-997. [8] KOCH, R. J. and WALLACE,A. D., Maximal ideals in compact semigmups. Duke Math. J. 21 (1954), 681-685. [9] NUMAKUaA,KAXSUNI,Theorems on compact totally disconnected semigroups and lattices. Proc. Amer. Math. Soc. 8 (1957), 623-626. [t0] StOMON,K E I ~ , Algebraic topology notes. University of Florida, 1969. [11] WALLACE,A. D., General topology notes. University of Florida, 1970. Hindu College, University of Delhi, Delhi 110007, India.