Adhikary, D. N.

Kishore, Nawal

English

Kishore Nawal

Adhikary D. N.

University of Szeged

J Independence of the conditions of associativity in ternary operations By N. KISHORE and D. N. ADHIKARY in Berhampur (Orissa, India) 1. Let S be a set of elements {a, b,c, ...}. A ternary operation / i s a mapping of SX >SX S to S. The o p e r a t i o n / i s said to be associative, if for every a, b,c,d,e£S we have (*) [a, b, (c, d, 6 ^ ] / = [a, (b, c, d ) f , e]f = [(a, b, c)f, d, e]f, where (x, y, z)f denotes the image by / of the ordered triplet (x, y, z) in S. The equalities in (1. 1) are called associativity conditions for the elements a, b, c, d, e. Associativity conditions in a set are said to be independent if any of them is not implied by the rest [1]. Thus, for ternary operations in S, the associativity conditions are independent if for every sequence {a, b, c, d, ej of five elements in S it is possible to define a ternary o p e r a t i o n / i n S in such a way that the equalities ( * ) hold for every sequence of elements {p,q,r,s,t} of S different f rom {a, b, c, d, e} whereas for this latter sequence ( * ) does not hold. 2. ( j . SZASZ [2] has investigated the independence of associativity conditions for binary operations and has established the following theorem: T h e o r e m . If the number of elements in a set S is greater than three, then the associativity conditions in S are independent. In what follows we study the same problem for ternary operations and prove: T h e o r e m . If the number of elements in a set S is greater than five, then the associativity conditions for ternary operations in S are independent. 3. P r o o f . For simplicity we drop the mapping l e t t e r / for the ternary operation. Let {a, b, c, d, e} be an arbitrary sequence of five elements of S. We shall prove the theorem by defining ternary operations over S such that the associativity con-ditions hold in all other cases except for this sequence. In what follows we consider separately the various alternatives for the elements of the sequence {a, b, c, d, e} to prove the theorem. 8» 276 N. Kishore—D. N. Adhikary 3. 1. Let a = b = c=d=e. Since S contains more than five elements, we can choose three more distinct elements u, v, w different f rom a, and define the following operation in 5 : (a,a,a)=u, (a,a,u) = v, and (x, y, z) = w in all other cases. Clearly, thé associativity conditions do not hold for the given sequence of elements, as [{a, a, a), a, a]=(u, a, a) = w, [a, (a, a, a), d\=(a, u,a) = w, but [a, a, (a, a, a))=(a, a,u) — v. We show that for every sequence {p, q, r, s, t} different f rom {a, a, a, a, a), the associativity conditions do hold. Since (p, q, r) and (q,r,s) are never a, we have [(p, q,r), s, t] = w and [p,(q,r,s),t] = w. Now, if (r,s,t)^u, then as (r, s, O^a, [p, q, {r, s, t)] = w. And if (r, s, t) = u but p^a, even then •(/?, q,u) = w. And even if (r, s, t)=u, p = a, then (a, q, u) = w, except when q = a, in which case {p, q, r, s, t} is the same as {a, a, a, a, a}, the given sequence. This proves the contention. 3. 2. Let a^b and b = c = d=e (all but one element being the same). Since 5 contains more than five elements, we can choose an element w dif-ferent f r o m a and b, and define the following operation in S: (a,b,b)=a and (x,y,z) = w in all other cases. We have for the given sequence {a, b, b, b, b): [a,b,(b,b,b)] = w and [a, (b, b, b), b] = w, but [(a, b, b), b, b]=(a, b, b)=a, so that the associativity conditions do not hold in this case. For any sequence {p, q, r, s, t} different f rom {a, b, b, b, b} we show that the associativity conditions do hold. Since (iq,r,s)9ib and ( r , s , t ) ^ b , [p(q, r, s), t] = w and [p, q, (r, s, f)] = w. Now, if (p,q,r)^a, then [(p, q, r), s, t] = w. And if ( p , q , r ) = a, but s^b, then [(p, q, r), s, t]=(a, s, t) = w. And even if (p, q, r)—'a, s = b, then [(/7, q, r), s, t] = = (a, b, t) = w except when t=b, in which case {p,q,r,s,t} is the . same as {a, b, b, b, b}, the given sequence. 3.2.1. For another permutation of the same sequence of 3 . 2 we prove the theorem as follows: Independence of the condition of associativity 277 When the sequence is {b, a, b, b, b}, we can choose three more distinct elements u, v and w different f rom a and b (as S contains more than five elements) and then define the following operation in S : (b,b,b)=u, (b,a,u) = v, and (x, y, z)=w in all other cases. As [{b, a, b), b, b] = w, [b, (a, b, b),b] = w, and [b, a, (b, b, b)]=(b, a, u)=v, we see that the associativity conditions do not hold for the given sequence {b, a, b, b, b}. For any sequence {p, q, r, s, t ) different f rom {b, a, b, b, b} we show that the conditions do hold. Since (p, q, r) and (q, r, J ) can never be equal to a or b, [{p, q, r), s, t] = w and [p, (q, r,s),t] = w. ' Now, if (r, s, then [p, q, (r, s, t)] = w, as (r, s, t^^b. And, if (r, s, t)=u, but p^b, then [p, q, (/*, s, i)] = w. And even if (r, s, t) = u, p = b, [p, q, (r, s, t)] = = (b, q, li) = w, except when q=a, in which case {p, q, r, s, t} is the same as {b, a, b, b, b}, the given sequence. By defining a similar operation in S, we can demonstrate the truth of the theorem in the same way for- every other sequence of five elements in which four elements are equal but the fifth is different*). 3. 3. Let a=b, c—d=e and a^c. Since S contains more than five elements we can choose three more distinct elements u, v, and w different from a and c and define the operation as follows: (c, c, c)=w, (a, a, u)=v, and (x, y, z)=w in all other cases. We have for the given sequence [{a, a, c), c, c] = w. and [a, (a,c, c), c] = w, but [a, a, (c, c, c)]=(a, a, u)=v, so that the associativity conditions do not hold, whereas we show that for any sequence {p, q, r,s,t} different f rom {a, a, c, c, c} the conditions do hold. Since (p, q, r) and (q, r, s) can never be equal to a or c, we get [(p, q, r), s, t] = w and [p, (q, r, s), t] = w. Now if (r,s,t)^u, then [p, q, (r, s, t)] = w as (r, s, Also, if (r , s , t ) = u, but p^a, [p, q, (r,t)] = (p, q, u) = w, and even if (r, s, t) = u and p=a, [p, q, (r, s, = *) The authors have investigated each of these permutations also, but for brevity all these have not been incorporated here. 278 N. Kishore—D. N. Adhikary = (a, q, u) = w except when q=a, in which case {p, q, r, s, t} is the same as the given sequence {a, a, c, c, c}. We can similarly prove the statement of the theorem for other arrangements of the elements in the sequence a, c, c, c} such as {¿z, c, o, c, c}, {a, c, c, a, c} etc. by defining similar operations. 3. 4. Let a=b, c=d and let e be different f rom a and c. Since, under the hypothesis, S has more than five elements, we can choose three more distinct elements 11, v, and w different from a, c, and e and define the following operation: (c,c,e) = u, (a, a,u) = v, and (x,y,z) = w in all other cases. We have for the given sequence [(a, a, c), c, e]=w, [a, (a, c, c,), e] = w, but [a, a, (c, c, e)]=(a, a, u)=v, so that the associativity conditions do not hold, whereas we show that for any sequence {p, q, r, s, t} different f rom {¿, a, c, c, e} the conditions do hold. Since (p, q, r) and (q, r, s) can never be equal to a or c, we have [(p, q, r), s, t] — w and [p, (q, r, s), t] = w. Now if (r, s, then [p, q, (r, s, t)] = w as (r,s,t)7±e. Also, if (r, s, t) — u, but p^a, then [p, q, (r, s, t)] = w. And even if (r, s, t) = u, p = a, then [p, q, (r, s, /)] = w except when q = a, in which case {p, q ,r,s, t} is the same as the given sequence {a, a, c, c, e}. We can have a similar proof for ..other arrangements of the elements in the sequence {a, a, c, c, e}. 3. 5; Let a = b = c, d^e, and let d, e be different f rom a. Since S contains more than five elements, we can choose three distinct elements u, v, w different f rom a, d, e and then define the following operation : (a,d,e)=u, (a, a, u) = v, and (x, y, z) — w in all other cases. Here we see that the associativity conditions do not hold for the given sequence. For, [(a, a, a), d, e] = H>, [a, {a, a, d), e]=w, but [a, a, (a, d, e)]=(a, a, u) = v. For any sequence {p, q, r, s, t} different f rom {a, a, a, d, e), we show that the conditions do hold. Since (p, q, r) can never be equal to a and (q, r, s) can never be equal to a or d, we have [(p,q,r)s,t] = w and [p, (q, r, s), t] = w. Now if (r, s, then [p, q, (r, s, t)] = w as (r, s, tj^e. Also if (r, 5, t)=u, but Independence of the condition of associativity 28II p^a, then [p, q, {r, s, i)] = w. And even if (r, s, t) = u, p=a, then [p, q, (r, s, i)] = — (a, q,u) = w, except when q = a, in which case {p, q, r, s, t} is the same as the given sequence {a, a, a, d, e}. We can have a similar proof for other arrangements of the elements in the sequence {a, a, a, d, e}. 3. 6. Let a=c, and let b, d, e be all distinct and different f rom a. Since S contains more than five elements we can choose two distinct elements u and w in S, different f rom a, b, d, and e, and define the operation as follows: (a,b,a)=a, (b,a,b)=b, (a,d,e)=u, and (x, y, z) = w in all other cases. We see that the associativity conditions do not hold for the given sequence, for [a, b, {a, d,e)] = (a, b, u) — w, [a, (b, a, d), e]=(a, w, e) = w, but [(ia, b, a), d, e]=(a, d, e) = u. For any sequence {p, q, r, s, t} different f rom {a, b, a, d, e} and containing an element different f rom a, b, d, e, the conditions do hold, as in all such cases [(p, q, r), s, t] = w, [p, {q, r, s), t] = w, [p, q, (r, s, ?)] = w. Hence taking p, q, r, s, t all f rom a, b, d, e only, we show that the associativity conditions do hold in each case. We distinguish five cases, denoted by the letters (A)—(E). (A) When p^a, we have [{p, q, r), .s, = i, t), say, where x=b or x = w. (i) Let x=b i.e. (p, q, r)=(b, a, b). Then [(b, a, b), s, t]=(b, s, t)=b when s=a, t=b in which case [b, (ia, b, a), b]=(b, a,b)=b and [b, a, (b, a, b)] = (6, a, b)=b. And when s ^ a or t ^ b , [(6, a, b)s, t] = [b, (a, b, s), i] = [b, a, (b, s, *)] = w. (ii) Let x=w. Then [(p,q,r),s,t] = (w,s,t) = w; and [p, (q, r, s), t] = (p, a, t) or ( p , b , t ) , or ( p , u , t ) , or (p, w, t). :280 N. Kishore—D. N. Adhikary If (q, r, s)=a i.e. (q, r, s) = (a, b, a),p cannot be equal to b, for otherwise (p, q, r) = b. Then (p, a, t)=(p, b, t)=(p, u, t)=(p, w,t) = w and [p, q, (r, i, f)] = 0 , q, a) or (p, q, b) or ( p , q , u ) or (p,q,w). Now, if (r, s, t)—b,p and q can never have values b and a, respectively, for (/?, q, r ) = = w. Therefore (P, q, a) = (p, q, b) - (p, q, u) = (p,q,w) = w, Hence [(/», q, r),s, t] = [p, (q, r, s), /] = [p, q, (r, s, i)] = w. (B) When p = a but q'^b, we have [(a, q, r), s, t] = [a, (q, r, s), t] = [a, q, (r, s,t)] = w, for ( a , q , r ) ^ a or b, (q,r,s)?^b or d, (r,s,t)^e. (C) When p = a, q — b but r^a, we have [(a, b, r) , 5, t]=[a, (b, r, s), t] = [a, b, (r, s, t)]=w. (D) When p=a, q=b, r=a but s^d, we have [(p, q, r), s, t] = [(a, b, a), s, t] = {a, s, t) = a (when s = b, t = a) in which case [a, (b, a, b), a]=(a, b, a)=a and [a, b, (a, b, a)]=(a, b, a)=a, and (when s^b) in all other cases [(a, b, a), s, t]=[a, (b, a, s), t] = [a, b, (a, s, = (E) When p = a, q = b, r = a, s = d but t^e, we have [(a, b, a), d,t] = [a, (b, a, d),t]=[a, b, {a, d,t]) = w. This proves the contention. 3. 7. Let a, b, c, d, e be all different f rom one another. As S contains more than five elements, we can choose an element w different f rom the given five elements and then define the operation in S as follows: (c, d,c) = c, (a, b, e)=a, (d,e,b) = d, (d, c, d)~d, (b, e, b)=b, (e, b, c) = c, (c,d,e) = e, (e,b,e) = e, (b, c, d)=b, and (x, y, z) = w in all other cases. Independence of the condition of associativity 28II For the given sequence [a, b, (c, d, e)]—{a, b, e)—a, [a, (b, c, d), e]=(a, b, e)=a, but [(a, b, c), d, e] — (vv, d,e) = w so that the associativity conditions do not hold. For any sequence {p, q, r, s, t \ different f rom {a, b, c, d, e} and containing an element different f rom a, b, c, d and e, we have [(/>, q,r),s,t] = [p, (q, r, s), t] = [p, q, (r, s, t)] = w. Hence taking p, q, r, s, t all f rom a, b, c, d, e only, we demonstrate that the asso-ciativity conditions do hold in each case as follows: (A) When p^a, [(p, q, r), s, t] = (x, s, t), say, then x^a. (i) Let x = b i.e. (p, q, r) = (b, e, b) or (p, q, r) = (b, c, d). Then in the first: case [(p, q, r), s, t] — [(b, e, b), s, f\ — (b, s, t) = b (when s=e, t = b or s = c, t = d), and hence [b, (e, b, e), b] — (b, e,b)—b and [b, e, (b, e, b)] = (b, e, b)=b, [b, (e, b, c), d] = (b, c, d) = b and [b, e, (b, c, d)] = (b, e, b) = b, while in the second case [(p, q, r), s, = [{b, c, d), s, t] = (b, s, t)=b (when s = e, t = b, or s = c, t = d)> and hence [b, (c, d, e), b] = (b, e, b) = b and [b, c, (d, e, b)] = (b, c, d)=b, lb, (c, d, c), d] = (b, c, d) = b and [b, c, (d, c, d)] = (b, c, d)=b, and [(p, q, r), s, t] =[/>, (q, r, s), t]=[p, q, (r, s, ?)] = w in other cases. (ii) Let x=c, i.e. (p, q, r) = (c, d, c) or (p, q\ r) = (e, b, c). Then [(/>, q, r), s, /] = [(c, d, c), s, t] = (c, s, ' ) = {*' ^ Z j :282 N. Kishore—D. N. Adhikary -.in these cases [c, (d, c, d), c]=(c, d, c) = c and [c, d, (c, d, c)] = (c, d, c) = c, and [c, (d, c, d), e] = (c, d, e) = e and [c, d, (c, d, e)] — (c, d, e)~e, •and [(p, q, r), s, t] = [(e, b, c), s, t] = (c, s, t) = c (when s = d and t = c) ox=e (when a=d and t=e), in which case [e, (b, c, d), c] = (e, b, c) = c and [e, b, (c, d, c)] = (e, b, c) = c, . [e, (b, c, d),e] = (e, b, e) = e and [e, b, (c, d, e)] = (e, b, e) = e, ;and [{p, q, r), s, t] = [p, (q, r, s), t] = [p, q, (r, s, t)] = w in other cases. (iii) Let x=d, i.e. (p, q, r) = (d, c, d) or (p, q, r) = (d, e, b). Then [{p, q, r),s, t] = = (d, s, t) = d (when s=c, t = d, or, when s=e, t = b), :in which case [d, (c, d, c), d] = (d, c, d)=d and [d, c, (d, c, d)] = (d, c, d)~d, [d, (c, d, e), b] = (d, e, b) = d and [d, c, (d, e, b)] — (d, c, d) = d, [d, (e, b, c),d]= (d, c,d) = d and [d, e, (b, c, d)] = {d, e, b) = d, [d, (e, b, e), b] = (d, e, b) = d and [d, e, (b, e, b)] = (d, e, b) = d, ;and [(/?, q, r), s, i] = [/>, (q, r, 5), t] = [p, q, (r, s, 0 ] = w in all the other cases. (iv) Let x=e, i.e. (p , q, r) = (c, d, e) or (p, q, r) = (e, b, e).. Then ,, . . , . i c, if s = b, t — c, . [ ( p , * , r ) , M ] = fcM) = | e f i f s = b t t = e t .and in each case [c, {d, e, b), c] = (c, d, c) = c and [c, d, (e, b, c)] = (c, d, c) = c, [c, (d, e, b), e] =(c , d, e) = e and [c, d, (e, b, e)]=(c, d, e) = e, [e, (b, e, b), c] = (e, b, c) = c and [e, b, (e, b, c)] = (e, b, c) = c, [e, (b, e, b), e]=(e, b, e) = e and .[e, b, (e, b, e)] = (e, b, e) = e, ;and [(/>, q, r, )s, Z] = [p, {q, r, 5), t] = [p, q, (r, s, Z)] = vv in all other cases. Independence of the condition of associativity 28II (v) Let x=w, i.e. (p,q,r) = w. Then [O, q, r), s, t] = (w, s, t) = w. Further-more, [/?, (q, r, s), t]=w, for (p, a, t) = w for all values of p and t, (p, b, t) = w as / ¿¿a and p^e, (p, c, t) = w as p^b and pj^d, (p, d, t) = w as p^c, (/>, e, t) = w as p^b and p^d, (p, w, t) = w for all values of p and t. Again, in each case [p, q, (r, s, t)] = w, because (p, q, a) = w for all values of p and q, (p,q,b) = w as (p^b, qj^e) and (p^d, q^e), (p, q, c) = w as (p^c, q^d) and (p^e, q^b), (p, q, w) = w for all values of p and q, (p,q,d) = w as (p^d^^c) and (p^b, q^c), (p, q, e) = w as (p^e, q^b) and (p^a, q^b), (p^c, q^d). (B) When p = a, but q^b, then [(a, q, r), s, t] = (w, s,t) = w and [a, (q, r, s), t] = w, [a, q, (r, s, ?)] = w. (C) When p=a, q=b, but r^c, then [(p,q,r),s,t] = [(a,b,r),s,t] = (a,s,t) = a if r = e, s = b, t = e, in which case . [a, (b, e, b), e]=(a, b,e)=a and [a, b, (e, b, e)]=(a, b, e)=a, and [{a, b, r), s, t] = [a, (b, r, s), ;] = [a, b, (r, s, /)] = w in all other cases. (D) When p = a, q = b, r = c, but s^d, then [(a, b, c), s, t] = [a, 0b, c, s), t] = [a, b, (c, s, t)] = w. 284 N. Kishore—D. N. Adhikary: Independence of the condition of associativity (E) W h e n p=a, q=b, r=c, s~d, but tj^e, t h e n [{a, b, c), d, t] = [a, (b, c, d), t] = [a, b, (c, d, /)] = w. C o m b i n i n g 3. 1 a n d 3. 7 toge ther c o m p l e t e s the p r o o f o f the t h e o r e m . References [1] E. S. LJAPIN, Semigroups (Providence, Rhode Island, 1963). [2] G. SzÄsz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta Sei. Math., 15 (1953), 2 0 — 2 8 . KHALLIKOTE COLLEGE, BERHAMPUR UNIVERSITY, INDIA (Received March 12, 1970) 

Independece of the conditions of associativity in ternary operations

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Independece of the conditions of associativity in ternary operations

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