ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS. Jaroslav Ježek

Transcription

1 ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS Jaroslav Ježek Abstract. Groupods satsfyng the equaton x(yz) = (xy)(xz) are called left dstrbutve, or LD-groupods. We gve an algorthm for ther enumeraton and prove several results on the collecton of LD-groupods extendng a gven monounary algebra. 0. Introducton Suppose that we want to enumerate all groupods (.e., algebras wth one multplcatvely denoted bnary operaton) on a gven fnte set A of n elements that satsfy a gven fnte collecton E of equatons. The groupods can be dentfed wth ther multplcaton tables, and we can suppose that A = {0,1,...,n 1}. A smple algorthm can do the task: generate the tables of all groupods on A accordng to ther lexcographc order, and for each groupod check f t satsfes the equatons from E. Of course, ths can work n a reasonable tme for very small numbers n only, as the number of all tables s n n2. Wth n = 5, we would have to check 5 25 tables, whch s already too much. If the collecton E s suffcently strong, the number of groupods satsfyng E on a set of n elements can be essentally smaller than the number of all groupods. So, one can ask for a faster algorthm, avodng large ntervals n the lexcographc orderng for whch t s clear from a smple reason that they cannot contan a table satsfyng E. One such algorthm s gven n Secton 3 of ths paper. The algorthm s wrtten n the language C, but does not use any hard to understand features of the language. It s formulated for the specal case of E consstng of the left dstrbutve law. Ths algorthm makes t possble to fnd the number of somorphsm types of n-element left dstrbutve groupods for both n = 5 and n = 6. The numbers are gven n Secton 4. Let P beapartal groupod. By an LD-extenson of P we mean aleft dstrbutve groupod G wth the underlyng set P, such that xy = z n P mples xy = z n G. If xy s defned n P for the pars x,y wth x = y and no other ones, then P can be dentfed wth a monounary algebra. In Secton 1 we nvestgate the collecton of LD-extensons of a gven monounary algebra. We determne under what condtons there exsts precsely one LD-extenson, and for some monounary algebras we are able to gve a smple descrpton of all the LD-extensons. The conjectures to some oftheresultswereobtanedbyrunnngaversonofthealgorthm gvennsecton3. In Secton 2 we are concerned wth fnte zeropotent left dstrbutve groupods. Weprovethatallofthemsatsfytheequatonx(yz) = 0, andfndrecursveformulas for ther enumeraton. 1

2 2 JAROSLAV JEŽEK In two related papers [6] and [7], fnte left dstrbutve groupods wth one generator are completely descrbed. Related are also the papers [1], [2], [3], [4], [5], [8], [9], and [10]. 1. Fnte left dstrbutve groupods extendng a gven monounary algebra Let (A,x ) be a monounary algebra (.e., an algebra wth one unary operaton). By an LD-extenson of (A,x ) we shall mean an LD-groupod (A,xy) such that xx = x for all x A. Let (A,x ) be gven. If a A, then a () s defned for any nonnegatve nteger recursvely n the followng way: a (0) = a and a (+1) = a (). The set {a () : 0} s called the orbt of a. Two elements of A are called connected f ther orbts are not dsjont. Ths relaton on A s an equvalence; ts blocks wll be called components of (A,x ). Let C be a component of (A,x ). The ntersecton of all orbts of elements of C s called the cycle of C. The cycle s nonempty and concdes wth the orbt of any of ts elements. An element a A s called rreducble f there s no element b wth a = b. Of course, every LD-groupod s an LD-extenson of precsely one monounary algebra. Gven an LD-groupod G, we put x = xx for every x G, ntroduce the notaton x () and speak about orbts and components wth respect to ths monounary algebra Lemma. Let (A,xy) be an LD-extenson of (A,x ). Then: (1) (ab) = ab for any a,b A; (2) f b s n the orbt of a, then ab = b. Proof. (1) (ab) = ab ab = a bb = ab. (2) Let us prove aa () = a (+1) by nducton on. For = 0 t s clear. If > 0, then aa () = a a ( 1) a ( 1) = aa ( 1) aa ( 1) = a () a () = a (+1) Theorem. Every monounary algebra (A,x ) has at least one LD-extenson (A,xy), e.g., xy = y. A monounary algebra (A,x ) has a unque LD-extenson f and only f t has only one component and, for any rreducble element a A, a = b mples b = a. Proof. If xy = y for all x,y A, then x yz = z = xy xz and (A,xy) s an LD-extenson of (A,x ). Let (A,x ) have more than one component. We are gong to show that then (A,x ) has at least two dfferent LD-extensons. If every component conssts of a sngle element (so that x = x for all x), then xy = x and xy = y are two dfferent LD multplcatons on A, both satsfyng xx = x. So, we can assume that there s a component D wth at least two elements. Take another component C D. We can defne multplcaton on A by xy = y f x C and y D, and xy = y n all other cases. Clearly, xx = x for all x, and there are pars x,y wth xy y. It remans to prove x yz = xy xz for all x,y,z. If z / D, then x yz = z = xy xz. Let z D. If x / C and y / C, then x yz = z = xy xz. If x C and y C, then x yz = z = xy xz. If precsely one of the elements x and y belongs to C, then x yz = z = xy xz.

3 ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS 3 Let a A be an rreducble element and suppose that there s an element b a such that a = b. Defne multplcaton on A by a a = b and xy = y whenever (x,y) (a,a). If b a, then t s easy to see that x yz = z for all x,y,z; but then also xy xz = z and we get x yz = xy xz. If b cannot be taken dfferent from a, then A has only two elements and t s easy to verfy that the multplcaton s also left dstrbutve n ths case. Clearly, xx = x for all x and xy = y fals for (x,y) = (a,a). Hence (A,x ) has at least two dfferent LD-extensons. Now let (A,x ) have only one component and let a = b mply b = a for any rreducble element a. Let (A,xy) be an LD-extenson of (A,x ). It remans to prove that xy = y for all x,y A. By Lemma 1.1(2), f a belongs to the cycle, then xa = a for any x A. Let b be an rreducble element. Put b = b () for all. By Lemma 1.1(2), b b j = b j for j. Let us prove, by nducton on, that b b j = b j for all j. For = 0 ths has been proved, so let > 0. If j > 0, then b b j = b 1 b 1 b 1 b j 1 = b 1 b 1 b j 1 = b 1 b j = b j, so t remans to prove b b 0 = b 0. We have (b b 0 ) = b b 0 = b b 1 = b 2 = b ; snce b s rreducble, ths mples b b 0 = b = b 0. In partcular, we get ay = y for any element a of the cycle and any element y. By the depth of an element x A we shall mean the least nonnegatve nteger such that x () belongs to the cycle. Let us prove xy = y by nducton on the depth of x. If x s of depth 0, then x belongs to the cycle and we are through. Let x be of a postve depth. The depth of x s less then the depth of x, so x y = y for all y by nducton. Let b be an rreducble element. As before, t s suffcent to prove xb = b +1 for all, where b = b (). Let us proceed by nducton on. Take any element a n the cycle. Snce (xb) = a xb = ax ab = x b = b and b s rreducble, we have xb = b,.e., xb 0 = b 1. For > 0, xb = x b 1 b 1 = xb 1 xb 1 = b b = b +1. Ths proves xy = y for all x,y A Lemma. Let G be a fnte LD-groupod and let a,b be two elements of G such that b belongs to a cycle and ab = b (k) for some k 0. Then a () b (j) = b (k+j) for all and j. Proof. By Lemma 1.1, ab (j) = b (k+j) for all j. Now t s suffcent to prove aa b = b (k). We have aa b (k) = aa ab = a ab = ab (k) = b (k+k), so that aa b (k+j) = b (k+k+j) for all j 0. Snce b belongs to a cycle, for a sutable j ths means that aa b = b (k). Let n 1,...,n d (where d 1) be postve ntegers; for every par,j of elements of {1,...,d} wth j let m,j be a number from {0,...,n j 1}. We denote by H n,m,j thegroupodwththeunderlyngset{(,k) : {1,...,d},k {0,...,n 1}} and multplcaton gven by (,k)(j,l) = (j,l+ j m,j ) where m, = 1 and + j denotes addton modulo n j.

4 4 JAROSLAV JEŽEK 1.4. Theorem. H n,m,j s an LD-groupod of cardnalty n 1 + +n d and wth cycles of cardnaltes n 1,...,n d. Let n 1,...,n d be postve ntegers such that n s not a multple of n j for any j. Then every LD-groupod of cardnalty n 1 + +n d wth cycles of cardnaltes n 1,...,n d s somorphc to H n,m,j for some collecton of numbers m,j ( j) as above. There are precsely (n 1 n 2...n d ) d 1 somorphsm types of such groupods. Proof. The frst asserton s straghtforward. Let G be an LD-groupod whch s the dsjontunonofcyclesofmutuallyprmecardnaltesn 1,...,n d. Defneabjecton f of {(,k) : {1,...,d},k {0,...,n 1}} onto G n the followng way. For every {1,...,d} let f(,0) be an arbtrarly chosen element of the cycle of cardnalty n, and put f(,k) = f(,0) (k) for all k. By Lemma 1.1, f(,k)f(,l) = f(,l+ 1). Let j. By Lemma 1.1, (f(,0)f(j,0)) (nj) = f(,0)f(j,0); snce n j s not a multple of any other number from {n 1,...,n d }, t follows that f(,0)f(j,0) = (j,m,j ) for precsely one number m,j {0,...,n j 1}. By Lemma 1.3 we get f(,k)f(j,l) = (j,l+ j m,j ) for all k and l. The last asserton follows easly Example. The partton of an LD-groupod nto components s not necessarly a congruence. The followng fve-element LD-groupod s a counterexample: a b c d e a b c d e b a d d d b a e d d a b c d e a b c d e a b c d e Let us take n+2 elements a,b,c 1,...,c n for some postve nteger n. For every par M,N of subsets of {c 1,...,c n } and every reflexve relaton R on {c 1,...,c n } denote by E M,N,R the groupod wth the underlyng set {a,b,c 1,...,c n } and multplcaton gven by xa = a xb = a { b for c M ac = a for c / M { b for c N bc = a for c / N { b for (,j) R c c j = a for (,j) / R 1.6. Theorem. Let (A,x ) be the monounary algebra wth the underlyng set A = {a,b,c 1,...,c n } and the operaton a = b = a, c = b. The collecton of groupods E M,N,R s just the collecton of all LD-extensons of (A,x ). Proof. The groupods E M,N,R are left dstrbutve, snce x yz = a for all x,y,z; clearly, they are LD-extensons of (A,x ). Conversely, let G be an LD-extenson of (A,x ).

5 ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS 5 By Lemma 1.1 we have xa = a for all x, and c b = a. Suppose ab = b. If also bc = b for some, then a = bb = b bc = bb bc = ab = b, a contradcton. Snce b c c = a, we cannot have bc = c j. Hence bc = a for all. Now a = c a = c bc = c b c c = ab = b, a contradcton. We have proved ab b. If ab = c, then a bb = c c,.e., aa = b, a contradcton. Hence ab = a. In total, xb = a for all x. Snce xc xc = x c c = xb = a, we get xc {a,b} for all x. Now t s clear that G = E M,N,R for some M,N,R Corollary. There are precsely 2 n(n+1) LD-extensons of the monounary algebra (A,x ) from 1.6. The number of somorphsm types of the LD-extensons s the same as the number of somorphsm types of n-element relaton systems wth two unary relatons and one bnary reflexve relaton. 2. Fnte zeropotent left dstrbutve groupods A groupod s called zeropotent f t satsfes xx y = y xx = xx. Equvalently, a groupod G s zeropotent f t contans a zero element 0 and xx = 0 for all x G Theorem. A fnte zeropotent groupod (wth zero 0) s left dstrbutve f and only f t satsfes x yz = 0. Proof. Let G be a fnte zeropotent groupod wth zero 0. If G satsfes x yz = 0, then x yz = 0 = xy xz and G s left dstrbutve. Conversely, let G be left dstrbutve. By a bad trple we shall mean a trple a,b,c of elements of G wth a bc 0; we need to prove that there s no bad trple. Snce ab ac = a bc, we get: f a,b,c s a bad trple, then also ab,a,c s a bad trple. For any par a,b of elements of G, defne an nfnte sequence P a,b ( = 0,1, ) of elements of G as follows: It s easy to see that P a,b 0 = a, P a,b 1 = ab, P a,b = P a,b 1 Pa,b 2 for 2. P ab,a = P a,b +1 for any a,b G and any 0. Let us prove by nducton on that ap a,b ap a,b 1 = a ab = aa ab = 0 ab = 0. For 2, ap a,b 00 = 0. Let us prove by nducton on 1 that P a,b = 0. We have ap a,b 0 = aa = 0 and = a P a,b 1 Pa,b 2 = apa,b 1 apa,b 2 = P a,b 1c = a bc for any elements a,b,c G. For = 1 ths s just the left dstrbutve law. If 2, then we can use the nducton hypothess: P a,b P a,b 1 c = Pa,b 1 Pa,b 2 Pa,b 1 c = Pa,b 1 Pa,b 2 c = a bc. Let us prove by nducton on n 0 that f a,b,c s a bad trple, then the elements P a,b 0, Pn a,b are parwse dfferent. For n = 0 there s nothng to prove. Let n 1. By the nducton hypothess appled to a,b,c, the elements P a,b 0,,P a,b n 1 are parwse dfferent. By the nducton hypothess appled to the trple ab, a, c (whch s also bad), the elements P ab,a 0,,P ab,a n 1 are also parwse dfferent. Snce

6 6 JAROSLAV JEŽEK are parwse dfferent. So, t remans to prove P a,b 0 Pn a,b. Suppose, on the contrary, that a = Pn a,b. Then a bc = Pn a,b P a,b n 1 c = a Pa,b n 1 c = apa,b n 1 ac = 0 ac = 0, a contradcton. So, we have proved that f there s a bad trple a,b,c, then all members of the nfnte sequence P a,b ( = 0,1, ) are parwse dfferent, a contradcton wth the fnteness of G. P ab,a = P a,b +1, the elements Pa,b 1,,Pn a,b 2.2. Corollary. The number of zeropotent left dstrbutve groupods wth a gven underlyng set of n elements and a gven zero element 0 s gven by n 2 ( ) n 1 C, =0 where C k,0 = 1, C k,m = (m+1) (n k 1)(n 2) m 1 =0 ( ) m C k, for 0 < m k. Proof. It s easy. Let G be a set of n elements wth a fxed element 0 G. For a gven k-element subset K of G {0} and a gven m-element subset M of K, C k,m s the number of the groupods such that K = {xy : x,y G} {0} and xa = 0 for all x G, a K Example. For n = 2,3,4,5,6 the numbers are, respectvely, 1, 3, 52, 5681, and Although t seems probable that the assumpton of fnteness n Theorem 2.1 cannot be elmnated, the author has not been able to fnd the correspondng example. The queston remans open: 2.4. Conjecture. There are nfnte zeropotent left dstrbutve groupods not satsfyng x yz = 0. Ths s equvalent to the followng: 2.5. Conjecture. The term x yz s not equvalent to any term contanng a subterm tt, for any term t, wth respect to the equatonal theory of left dstrbutve groupods. It s easy to see that the two conjectures are equvalent. The negaton of 2.4, together wth Theorem 2.1, would be equvalent to sayng that x yz = uu s a consequence of the left dstrbutve law together wth xx y = y xx = xx. If x yz = uu s a consequence, then there exsts a formal proof of the equaton, a fnte sequence of terms w 0,,w n such that w 0 = x yz, w n = uu and, for any > 0, the equaton w 1 = w s an mmedate consequence of ether x yz = xy xz or xx y = xx or y xx = xx; f n s the last ndex such that w 1 = w s an mmedate consequence of the dstrbutve law, then clearly w n contans a subterm tt for some term t, and x yz = w n s a consequence of the left dstrbutve law. The converse mplcaton s clear.

7 ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS 7 3. The enumeraton algorthm The followng fragment of a C program can be used to fnd the number of LDextensons of a gven monounary algebra on n elements. We need two arrays A and B of the sze n 2. A holds the multplcaton table, whch s generated n the lexcographc order. The array B holds nformaton on the current state of A; f B[]= j <, then the value of A[] was forced by A[j] and should be kept unchanged untl A[] s changed; ths makes t possble to skp over ntervals n the lexcographc order. In partcular, B[]= 1 means that A[] should never be changed. ThefunctonVerfy()dealswththetableAwhchmayalsocontanthenumber 1, meanng that the correspondng place s undefned andas the table of a partal groupod. The functon returns -1 f the partal groupod was found contradctory, 0 f no completon was done, 1 f at least one completon was done. Here s the fragment (for n 26): #nclude <stdo.h> nt n,n1,nn,nn1,p; long nt N=0L,NI=0L; nt A[676]; nt B[676]; nt Verfy(I){ nt,j,k,a,b,c,d,e,p,q,z=0; for(=0;<n;++)for(j=0;j<n;j++){ c=a[n*+j];f(c>=0)for(k=0;k<n;k++){ d=a[n*+k];a=a[n*j+k];f(a>=0&&d>=0){ p=n*+a;q=n*c+d;b=a[p];e=a[q]; f(b>=0&&e<0){a[q]=b;b[q]=i;z=1;} else f(b<0&&e>=0){a[p]=e;b[p]=i;z=1;} else f(b>=0&&b!=e)return -1;}}} return Z;} nt FndNext(I){ whle(i>=0&&(a[i]==n1 B[I]<I))I--; f(i>=0)a[i]++; return I;} vod MakePartal(I){ nt ; for(=i+1;<nn;++)f(b[]>=i){a[]=-1;b[]=nn;p++;}} vod MakeComplete(I){ nt ; P=0;for(=I+1;<nn;++)f(A[]==-1)A[]=0;} vod Process(){ N++;} vod man(){ nt,i,v,a,b,c; char s[80];] prntf("\ninput cardnalty of groupod: "); scanf("%d",&n); n1=n-1;nn=n*n;nn1=nn-1; for(=0;<nn;++){a[]=0;b[]=nn;} do{scanf("%s",s); f(strlen(s)==3){a=s[0]- a ;b=s[1]- a ;c=s[2]- a ;

8 8 JAROSLAV JEŽEK A[n*a+b]=c;B[n*a+b]=-1;}} whle(strlen(s)==3); I=nn1;P=0; whle(i>=0){ do V=Verfy(I); whle(v==1&&p); f(!v&&p){makecomplete(i);i=nn1;v=verfy(i);} f(!v)process(); I=FndNext(I);MakePartal(I);} prntf("n=%ld\n",n);} Of course, one may ask for more than smply countng the number of multplcaton tables. The changes should be done n the functon Process(), whch s called each tme when a new vald table was found. For example, one may be nterested n fndng the number of somorphsm types. It s not necessary to store all the precedng tables n order to check f the current table s somorphc to one of them. Snce the tables are found n the lexcographc order, t s suffcent to ask f the new table can be somorphc to one whch came n the lexcographc order earler, and ths can be done by checkng all permutatons of the underlyng aset. There are no requrements on ether space or memory n the program. To make the program more user frendly, changes n the functon man() should be done. One s often nterested n enumeratng not all tables, but only those that are extensons of a gven partal groupod. For that purpose one can append A[n*+j]=k; B[n*+j]=-1; to the ffth lne of the functon man(), for each tem j = k of the gven partal groupod. 4. Sx-element groupods The numbers of all left dstrbutve groupods and of all somorphsm types of left dstrbutve groupods on a gven set of two, three, four, fve and sx elements are gven n the followng table: Elements: Groupods: Iso types: For sx elements, we dd not count the number of groupods. Instead, the somorphsm types were dvded nto 130 groups accordng to the somorphsm types of ther dagonals. There are 130 somorphsm types of monounary algebras on sx elements. For each of them, the number of somorphsm types of LD-extensons was computed usng the algorthm descrbed n Secton 3. Let us denote the sx elements by a,b,c,d,e,f. For a 6-tuple a 1,...,a 6, denote by N(a 1,...,a 6 ) the ordered par (n,m) where n s the number of LD-extensons of the monounary algebra on a,b,c,d,e,f wth a = a 1,b = a 2,...,f = a 6, and m s the number of ther somorphsm types. Among the 130 cases, there were nneteen wth n 10 5 : N(baaabb) = (122263, 15426) N(aaaaaa) = ( , ) N(acaaaa) = ( , ) N(addaaa) = ( , )

9 ENUMERATING LEFT DISTRIBUTIVE GROUPOIDS 9 N(acaeaa) = ( , ) N(aeeeaa) = ( , ) N(addafa) = ( , ) N(affffa) = ( ,45960) N(abbbbb) = ( , 61166) N(abdbbb) = (384558, ) N(abeebb) = (226518, ) N(abdbf b) = (138240, 69489) N(aacccc) = (178839, 31039) N(abcccc) = (250078, 24965) N(abcddd) = (294766, 31215) N(abccee) = (110569, 29150) N(abcdf e) = (121736, 5891) N(abcdee) = ( , 51254) N(abcdef) = ( , 32541) References 1. R. Bashr, A. Drápal, Quastrval left dstrbutve groupods, Commentatones Math. Unv. Carolnae 35 (1994), P. Dehornoy, The adjont representatons of left dstrbutve structures, Comm. Algebra 20 (1992), P. Dehornoy, A normal form for the free left dstrbutve law, preprnt. 4. P. Dehornoy, On the syntactc algorthm for the word problem of left dstrbutvty, preprnt. 5. P. Dehornoy, Constructon of left dstrbutve operatons, preprnt. 6. A. Drápal, Fnte left dstrbutve algebras wth one generator, preprnt. 7. A. Drápal, Fnte left dstrbutve groupods wth one generator, preprnt. 8. A. Drápal, T. Kepka, M. Musílek, Group conjugaton has non-trval LD-denttes, Commentatones Math. Unv. Carolnae 35 (1994), D. M. Larue, Left-dstrbutve dempotent algebras, preprnt. 10. R. Laver, The left dstrbutve law and the freeness of an algebra of elementary embeddngs, Advances n Math. 91 (1992), MFF UK, Sokolovská 83, Praha 8