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3URAL MATHEMATICAL JOURNAL, Vol. 9, No. 1, 2023, pp. 64-77
DOI: 10.15826/umj.2023.1.005
TERNARY *-BANDS ARE GLOBALLY DETERMINED
Indrani Dutta^, Sukhendu Kar^
Jadavpur University, 188, Raja S. C. Mallick Road, Kolkata - 700032, India
[email protected] ^[email protected]
Abstract: A non-empty set S together with the ternary operation denoted by juxtaposition is said to be ternary semigroup if it satisfies the associativity property ab(cde) = a(bcd)e = (abc)de for all a, b, c,d,e € S. The global set of a ternary semigroup S is the set of all non empty subsets of S and it is denoted by P(S). If S is a ternary semigroup then P(S) is also a ternary semigroup with a naturally defined ternary multiplication. A natural question arises: "Do all properties of S remain the same in P(S)?" The global determinism problem is a part of this question. A class K of ternary semigroups is said to be globally determined if for any two ternary semigroups Si and S2 of K, P(Si) = P(S2) implies that Si = S2. So it is interesting to find the class of ternary semigroups which are globally determined. Here we will study the global determinism of ternary *-band.
Keywords: Rectangular ternary band, Involution ternary semigroup, Involution ternary band, Ternary *-band, Ternary projection.
1. Introduction
In our previous paper [7] we have discussed the global determinism of ternary groups and finite left zero ternary semigroups. Here we will discuss some properties of a rectangular ternary band and of a proper rectangular ternary band and also discuss the global determinism problem of ternary *-band.
Let us briefly present the literature on the problem of global determinism. In 1960 B.M. Shane formulated the importance of studying the problem of global determinism. In 1967, T. Tamura and J. Shafer [11] proved that groups are globally determined. In 1984, T. Tamura [10] proved that rectangular groups are globally determined. In 1984, M. Gould and J.A. Iskra [4] also studied some globally determined classes of semigroups. M. Gould, J.A. Iskra, C. Tsinakis [5, 6] also studied the global determinism problem of semigroup theory. In 1984, Y. Kobayashi [9] proved that semilattices are globally determined. At present, the problem of global determinism is a well-known research problem. M. Vincic [13] established in 2001, that *-bands are globally determined. In 2014, A. Gan, X. Zhao and Y. Shao [1] proved that clifford semigroups are globally determined. In 2015, A. Gan, X. Zhao and M. Ren [3] studied the global determinism of semigroups having regular globals. A. Gan, X. Zhao and Y. Shao [2] also discussed the globals of idempotent semigroups in 2016 and in 2017, B. Yu, X. Zhao, A. Gan [12] proved that idempotent semigroups are globally determined.
So the problem of global determinism is important and relevant in the ternary theory of semigroups. Here we will prove that ternary *-bands are globally determined.
2. Preliminaries
First we provide the basic definitions and results which are used in the rest of the paper.
Definition 1. A ternary semigroup S is said to be left (resp. right) zero ternary semigroup if for a,b,c € S, abc = a (resp. abc = c).
Definition 2. A ternary semigroup S is said to be a ternary band if every element of S is idempotent, i.e. a3 = a for all a € S.
Definition 3. A ternary semigroup S is said to be rectangular ternary band if aba = a for all a,b € S.
Although the definition of rectangular ternary band and rectangular band in binary are similar, but all the rectangular ternary bands are not rectangular bands in binary. The following example illustrates this fact.
Example 1. Let M2 (r) is the set of all 2 x 2 matrices over r. This is a ternary semigroup w.r.t. the natural ternary matrix multiplication.
(i) | ( 01 o) ' ( 01 01) | € M2(r). This is a rectangular ternary band w.r.t. natural ternary matrix; multiplication.
(«){( ^ , ^ 01 01) | € M2(r). This is a rectangular ternary band w.r.t. natural ternary
matrix multiplication.
Lemma 1. A ternary semigroup S is rectangular ternary band if and only if ababa = a and abcde = ace for all a,b,c,d,e € S.
Proof. Let S be a rectangular ternary band. Then aba = a for all a, b € S. Therefore,
ababa = (aba)ba = aba = a for all a,b € S.
Now
abcde = a(b(adc)b)(cde) = (aba)(d(cbc)d)e = a(dcd)e = (adc)(d(ace)d)e = (a(dcd)a)c(ede) = (ada)ce = ace.
Conversely, suppose that ababa = a and abcde = ace. Then
aba = (ababa)ba = a(bab)aba = aaa = a3 = a.
Therefore, S is the rectangular ternary band. □
Lemma 2. A ternary semigroup S is a rectangular ternary band if and only if it can be expressed as a cartesian product of left zero and right zero ternary semigroups.
Proof. Let S be a rectangular ternary band and u be a fixed element of S. Define two sets L, R such that
L = {xuu : x € S}, R = {uux : x € S}.
Since
(xuu)(yuu)(zuu) = x(uuy)u(uzu)u = xuu
for all xuu, yuu, zuu € L, we have, L is left zero ternary semigroup. Similarly,
(uux)(uuy)(uuz) = u(uxu)u(yuu)z = uuz
for all uux, uuy, uuz € R implies that R is right zero ternary semigroup.
Define a mapping 0 : S —> L x R such that 0(x) = (xuu, uux) for all x € S. Here the ternary operation on L x R is as follows:
(a, b)(c, d)(e, f) = (ace, bdf) = (a, f) for all (a, b), (c, d), (e, f) € L x R.
Let 0(x) = 0(y). This implies that xuu = yuu, uux = uuy. Now
x = xux = xuuux = (xuu)ux = (yuu)ux = yu(uux) = yu(uuy) = yuuuy = yuy = y.
Therefore, 0 is one-to-one mapping.
0(xuz) = (xuzuu,uuxuz) = ((xuu)(uuu)(zuu), (uux)(uuu)(uuz)) = (xuu, uuz).
Therefore, 0 is an onto mapping.
0(x)0(y)0(z) = (xuu,uux)(yuu,uuy)(zuu,uuz) = (xuuyuuzuu,uuxuuyuuz) = (xuu, uuz) = 0(xyz).
Thus 0 is a ternary homomorphism. Hence 0 is an isomorphism and S = L x R.
Conversely, suppose that S is isomorphic to L x R, where L, R are left zero, right zero ternary semigroups respectively. Now (a,c)(e,g)(a,c) = (aea,cgc) = (a, c). Therefore, L x R is a rectangular ternary band. Since S is isomorphic to a rectangular ternary band, S is also rectangular ternary band. □
Let S = L x R where L be the left zero, R be the right zero ternary semigroup and / be an isomorphism from S to L x R. The ternary operation on L x R is defined as
(a, b)(c, d)(e, f) = (ace, bdf) = (a, f) for all (a, b), (c, d), (e, f) € L x R.
There are some notions defined as follows: Let A € P(S), where P(S) is the global of S.
nL(A) = {i € L : 3 k € R such that (i, k) € /(A)}. ■kr(A) = {k € R : 3 i € L such that (i, k) € /(A)}.
If S = L x R then for any A € P (S), we have
nL(A) = {i € L : 3 k € R such that (i, k) € (A)}. nR(A) = {k € R : 3 i € L such that (i, k) € (a)}.
Definition 4. A rectangular ternary band S is said to be a proper rectangular ternary band if it is not left zero, right zero and lateral zero ternary semigroup. By the notation TRB2 we mean the proper rectangular ternary band.
Definition 5. (i) A ternary semigroup S is said to be an involution ternary semigroup if it is equipped with a unary operation * such that (xyz)* = z*y*x* and (x*)* = x.
(ii) An idempotent involution ternary semigroup is said to be an involution ternary band.
Definition 6. (i) Let S be an involution ternary semigroup. If for each x € S, xx x — x, x*xx* — x* and x2y — xy2 for all x,y € S, then S is said to be a ternary *-semigroup.
(ii) A ternary semigroup S is said to be ternary *-band if S is an idempotent ternary *-semigroup.
Definition 7. Let S be an involution ternary semigroup. An element x € S is said to be a projection of S if it is idempotent and is a fixed point of involution, i.e. x3 — x and x* — x.
Definition 8. Let S be an involution ternary semigroup. Then X c S is said to be an involution ternary subsemigroup if X is a subsemigroup of S and X* c X, where X* — {x* : x € X}.
Two important notations of this paper are as follows: The set of all subsemigroups of a ternary *-band S is denoted by S(S) and
Ch(S) — {X € S(S) : X — Y3 —^ X — Y for all Y € P(S)}.
Remark 1. Every bijection between two left (resp. right) zero ternary semigroups is the isomorphism between them.
We have already discussed in [7], that finite left (resp. right) zero ternary semigroups are globally determined. In this paper, we generalize this result for arbitrary left (resp. right) zero ternary semigroups.
Here we assume the generalized continuum hypothesis which states that if cardinality of an infinite set lies between that of an infinite set A and that of the power set P(A) of A, then it has the same cardinality as either A or P(A).
Lemma 3. If P(S1) = P(S2) then |S1| — |S2| where |S1| and |S2| denote cardinality of S1, S2 respectively.
Proof. To prove the result we consider the following three cases.
Case 1. Suppose that S1, S2 both are finite sets. Let |S1| — m and |S21 — n.
Since P(S1) ^ P(S2), so |P(S1 )| — |P(S2)|. Again |P(s^| — 2m and |P(S2)| — 2n. Therefore, 2m — 2n. This implies that m — n.
Case 2. Suppose that S1 is a finite set and S2 is an infinite set and |S1| — m. Then |P(S1 )| — 2m, i.e. a finite number but |P(S2)| is not a finite number. Therefore, |P(S1 )| — |P(S2)|. Hence Case 2 is not true.
Case 3. Let us assume that both S1, S2 are infinite sets. Then the following three situations may arise.
(i) If S1 and S2 both are countable then |S11 — |S2| — ^01. So there is nothing to prove.
(ii) Suppose S1 is a countable set and S2 is an uncountable set. Then |S1| — —^ |P(S1 )| — 2Ho and |S21 > 2Ho. Therefore, |P(S2)| > 2H° — |P(S1)|. But this is not possible.
(iii) Suppose S1 and S2 both are uncountable. If possible, let |S1| — |S2|. Then |S1| — c and iS2i — ci where c,ci > 2H°. Therefore, |P(S1)| — 2c and |P(S2)| — 2C1. Since c — ci thus 2C — 2C1. This contradicts our assumption. Therefore, |S1| — |S2|. □
Theorem 1. Left (resp. right) zero ternary semigroups are globally determined.
Proof. Let S1 and S2 be two left zero ternary semigroups and 0 : P(S1) —> P(S2) is an isomorphism, i.e. P(S1) = P(S2). This implies that |P(S1)| — |P(S2)|. Hence by Lemma 3, |S1| — |S2|. Thus there is a bijection from S1 to S2. Since S1 and S2 are left zero ternary semigroups, by Remark 1, it follows that the bijection is an isomorphism. So it is clear that S1 = S2. Hence the class of all left zero ternary semigroups is globally determined.
Similarly, we can show that right zero ternary semigroups are globally determined. □
1K is the cardinality of the set of all natural number.
3. Main result
A rectangular ternary band is said to be a proper rectangular ternary band if it is not a left zero, right zero or lateral zero ternary semigroup. In this section, we provide some results of proper rectangular ternary bands and show that the proper rectangular ternary band satisfies the strong isomorphism property. By strong isomorphism property we mean that any isomorphism 0 from P(S) to P(Si) is also an isomorphism from S to Si. Here we also discuss that ternary *-bands are globally determined. Unless otherwise stated, in this section, we assume that S is a proper rectangular ternary band.
Lemma 4. Let S and Si be two ternary semigroups such that P(S) = P(Si) and 0 is an isomorphism from P(S) to P(Si). Then the restriction 0 \ch(s) is a bijection from Ch(S) to Ch(Si).
Proof. Let A € Ch(S). This implies that A € S(S). Therefore, 0(A) € S(Si).
Let 0(A) = Ai. If possible, there exists Bi € P(Si) such that Bf = Ai. Since 0 is an isomorphism there exists B € P(S) such that 0(B) = Bi. Therefore, Ai = Bf = (0(B))3 = 0(B3).
Hence -0(A) = 0(B3). This implies that A = B3. Since A € Ch(S), we have A = B. Therefore, ■0(A) = 0(B). This implies that Ai = Bi. Hence Ai € Ch(Si). Therefore, 0 \Ch(S) is a bijection from Ch(S) to Ch(Si). □
Lemma 5. Let S be a proper rectangular ternary band such that S = L x R. Then A € Ch(S) if and only if |nL(A)| = 1 or |nR(A)| = 1.
Proof. The proof is similar to the binary result of [2]. □
Theorem 2. Rectangular ternary bands are globally determined.
Proof. Proof of the theorem immediately follows from the binary result of [10]. □
Theorem 3. Proper rectangular ternary band satisfies the strong isomorphism property.
Proof. The proof is similar to the binary result of [2]. □
A restricted class of a involution ternary semigroup is ternary *-band. Unless otherwise stated, in the rest of this section, B denotes an involution ternary band and S(B) denotes the set of all involution ternary subsemigroups of B.
Lemma 6. For any involution ternary band B, S(B) coincides with the set of all projections of P(B). Therefore, if Bi and B2 be two involution ternary bands, then every isomorphism 0 : P(Bi) —> P(B2) induces a bijection from S(Bi) to S(B2).
Proof. Since B is a involution ternary band, for any subset X of B, we have X c X3 and (X * )* = X.
Now let X € s(B). This implies X3 c X, X* c X and X c B. Thus X3 = X and (X*)* c X*. Hence X3 = X and X c X *.
Therefore, X3 — X and X* — X. So X is a projection of P(B).
Conversely, if X is a projection of P(B) then X3 — X and X — X*. Therefore, X3 c X and X* c X. Thus X € s(B).
Hence s(B) is the set of all projections of P(B).
Let ^ : P(B1) —^ P(B2) be an isomorphism, where B1, B2 be two involution ternary bands. We have to show that X € s(B1) implies that ^(X) € s(B2). Let us extend ^ on (P(B1))* as ^ : (P(B1))* (P(B2))* such that ^(X*) — (^(X))*.
If X — X* then ^(X) — ^(X*) — (0(X))*. Hence X € s(B1) implies that ^(X) € s(B2). Thus ^(s(B1)) c s(B2). Similarly, ^-1(s(B2)) c s(B1). This implies that
^(r1 (s(B2))) c ^(s(B1)) s(B2) c ^(s(B1)).
Therefore, s(B2) — ^(s(B1)). Thus ^ induces a bijection between s(B1) and s(B2). □
Lemma 7. Let B1, B2 be two involution ternary bands. Any isomorphism from P(B1) to P(B2) induces a bijection from Ch(B1) to Ch(B2).
Proof. If we are able to show that for any isomorphism ^ : P(B1) —► P(B2), ■0(Ch(B1)) — Ch(B2) then ^ : Ch(B1) —► Ch(B2) becomes onto mapping. Again since ^ is an isomorphism from P(B1) to P(B2) and Ch(B1) c P(B1) so ^ : Ch(B1) —► Ch(B2) is one-to-one mapping hence a bijection.
Let X € Ch(B1). If possible there exists Y' € P(B2) such that Y'3 — ^(X). Then there exists Y € P(B1) such that ^(Y) — Y'. Therefore,
^(X) — Y'3 — (^(Y ))3 — ^(Y3).
This implies that X — Y3 because X € Ch(B1). Hence X — Y. Thus ^(X) — ^(Y). Therefore, ■0(X) — Y'. So X € Ch(B1) implies that ^(X) € Ch(B2). Hence ^(Ch(B^) c Ch(B2).
Since ^ is an isomorphism, is also an isomorphism. Hence Y € Ch(B2) implies that (Y) € Ch(B1). Thus
^-1(Ch(B2)) c Ch(B1) —^ (Ch(B2))) c ^(Ch(B1)) —^ Ch(B2) c ^(Ch(B1)).
Hence ^(Ch(B1)) — Ch(B2).
Therefore, ^ is a bijection from Ch(B1) to Ch(B2). □
Let us a define partial ordering and a chain on a ternary band as follows.
Definition 9. Let B be a ternary band. A partial order < on a ternary band B can be defined as a < b if and only if
a — a2b — ab2 — b2a — ba2.
Definition 10. Let A be a non empty subset of a ternary band B. Then A is said to be a chain of B if for all a, b € A either a < b or b < a.
Lemma 8. Let B be a ternary *-band. Then X € Ch(B) if and only if X is a chain of projections.
Proof. Let B be a ternary *-band. Then
x3 — x (x*)* — x x*xx* — x* xx*x — x and
Jb - Jb ^ \ 'JLj I - Jb ^ Jb Jb Jb - <Ju , Jb Jb Jb - <Ju Clilvi
x2y — xy2 for all x,y £ B.
Suppose X € Ch(B) and x,y € X. If possible let x2y € {x,y}. Construct Y = X \ {x2y}. Now x2y € Y3 and Y C X. Since B is ternary *-band and Y C X C B, we have Y C Y3. Thus it follows that Y3 = X. Again
Y C X —^ Y3 C X3 — X.
Now
Y = X \ {x2y} C Y3 c X.
Since x2y € Y3, it follows that Y3 = Y. Hence Y3 = X. By definition of Ch(B) we find that Y = X. This contradicts our assumption that x2y € {x,y}. Therefore, x2y € {x,y}. Similarly, xy2,yx2,y2x,xyx,yxy all are in {x,y}. Hence X is a chain.
Now our aim is to show that x € X implies that x is a projection. Since X € Ch(B), X3 = X. Therefore, x* € X for all x € X. Now x2x* € {x,x*}. Suppose This implies that
(x2x*)* — x*
* *
xx X Jb Jb Jb
Therefore, x — x*. Hence x is the projection. Again if
x2x*
* * xx X
Jb Jb Jb
x2x*
x2x*
x.
Hence x* = x. Therefore, x is the projection. This implies that X is a chain of projections.
Conversely, suppose that X € P(B\) is a chain of projections. Suppose there exists Y € P(B\) such that Y3 = X. It is clear that Y C Y3. Therefore, Y C X. Subset of a chain must be a chain. Hence Y3 c Y. This implies X c Y C X. Thus X = Y. Therefore, X € Ch(B). □
■i
■i
x
x
*
x
x
Let B be a ternary *-band. Let define a partial ordering " < " on P(B) as follows: X < Y if and only if X = X2Y = YX2 for all X,Y € P(B).
X : Y if and only if X < Y in S(B) and there does not exist any Z € S(B) such that X < Z <Y.
Again X —^ Y if and only if X < Y and there does not exist any Z € Ch(B) such that X <Z <Y.
It is clear that
X : Y =^ X —► Y.
Remark 2. If B is a ternary band then B is also a ternary semigroup. So ideal of a ternary band is the same as the ideal of a ternary semigroup.
Lemma 9. Let B be a ternary band. Thus there exists .some ternary semilattice S such that there is a homomorphism a : B —> S such that a(B) — S.
Proof. Let B be a ternary band. Define
Ia — {xay : x,y £ B} for any a £ B. Then Ia is an ideal of B, generated by a.
Let us define a relation p on B such that ap6 if and only if Ia = Ib. There is no doubt that p is an equivalence relation on B. Now let Iai = Ibl, Ia2 = Ib2, Ia3 = Ib3. Therefore,
6i = xiaiyi, 62 = X2a2y2, 63 =
Hence
616263 = xiaiyix2a2y2x3a3y3 = xi(aiyix 2a2 2)a2y2X3a3y3 = Xi(aiyiX2a22)(aiyiX2a22)(aiyiX2a22)a2y2X3a3y3 = xiX (a2aiyiX2a2y2X3)a3y3 = xiX (a2aiYa32)a3y3 = xiX (a2aiYa32)(a2aiYa32)(a2 aiYa32)a3y3 = (xiXa2aiYa32a2aiYa3)(a3 a2ai)(Ya3y3) = Xi(a3a2ai)Yi,
where
X = aiyix2a22aiyix2a2, Y = yix2a2y2x3, Xi = xiXa2aiYa32a2aiYa3, Yi = Ya3y3.
Therefore, 6i6263 € Ia3a2ai. Similarly, we can show that a3a2ai € Iblb2b3. Thus it is clear that Ia3a2ai = Iblb2b3 • Now
(i) a6c = a6ca6ca6c = a(6ca)6ca6c € Ibca.
Similarly, we can show that 6ca € Iabc. This implies that Iabc = Ibca. Again
(ii) a6c = a6ca6ca6c = a(6ca6c)(6ca6c)(6ca6c)a6c = (a6ca6)(c6ca6c6ca)(6ca6c) = (a6ca6)(c6cxa)(6ca6c) = (a6ca6)(c6cxa)(c6cxa)(c6cxa)(6ca6c) = (a6ca6c6cx)ac6(cxac6cxa6ca6c),
where x = a6c6c. Therefore, a6c € Iacb. Hence Iabc = Iacb. Thus
Iabc — Xacb — Ibac — Ibca — Xcab — Icba-
This shows that Iaia2a3 = Iblb2b3. Therefore, p is a ternary congruence relation on B.
Now B/p be the set of all equivalence classes of the congruence relation and the elements are denoted by a for a G £>. Define a ternary operation on £>/p by abc = abc.
Now we show that B/p is a ternary semilattice w.r.t. above defined ternary operation. This is clear from the above discussion that B/p is a commutative ternary semigroup. Again since B is ternary band,
aaa = aaa = a3 = a.
Thus B/p is also a ternary band. Now
O O Q / 0\ O Q o o
a 6 = a 6 = a(a6 )6 € Iab2, a6 = a 6 = a(a 6)6 € Ia2b.
Therefore, /a2fc = /afc2. This implies that a26 = ab2. Hence a26 = ab2. Thus B/p is a ternary semilattice.
Now we define a mapping a : B —► B/p such that a(a) = a. Then a is an epimorphism. If we consider S = B/p then there exists a ternary semilattice which is homomorphic image of B. □
Lemma 10. Let B be a ternary *-band and S be a ternary semilattice image of B. If X, Y € Ch(B) are such that X < Y and a(X) : a(Y) [resp. a(X) a(Y)] holds in P(S) then X : Y [resp. X —> Y], where S is the semilattice image of B and a is the corresponding epimorphism from B to S.
Proof. Let a : B —> S be an epimorphism. Then
a(X) : a(Y) a(X) < a(Y).
Suppose Z € S(B) be such that X < Z < Y. Now Z € S(B) implies that a(Z) € S(S), since (a(Z))3 = a(Z3) = a(Z). Therefore,
X = X 2Z = XZ2 = ZX2 = Z2X.
This implies that
a(X) = a(X )2a(Z) = a(X )a(Z)2 = a(Z )a(X )2 = a(Z)2 a(X).
Hence a(X) < a(Z). Thus it follows that a(X) < a(Z) < a(Y). This contradicts a(X) : a(Y). Hence a(X) : a(Y) =^ X : Y.
Again let X,Y € Ch(B) such that a(X) —> a(Y). If possible, there exists Z € Ch(B) such that X < Z <Y. This implies a(X) < a(Z) < a(Y). Since Z € Ch(B) implies that a(Z) € Ch(S), this contradicts that a(X) —> a(Y). Hence X —> Y. □
Lemma 11. Let B be a ternary *-band and X € Ch(B). If x € X is not a maximal element of X then X : X \{x}.
Proof. Let X € Ch(B). This implies that X3 = X. Now
X2(X \ {x}) = X(X \ {x})2 c X3 = X.
Let h € X and h = x. Then h = h3 € X2(X \ {x}). Again if h = x then there exists some y € X such that x = x2y = xy2. Therefore, h = x = x2y € X2(X \ {x}). This implies that X c X2(X \ {x}). Hence X2(X \ {x}) = X. Thus X < X \ {x}. Since X € Ch(B), X \ {x} € Ch(B). Since {x} is not a maximal element of X, a({x}) is also not a maximal element of a(X). Then from [8], we can write
a(X) : a(X) \ a({x})= a(X \{x}).
Hence by Lemma 10, it follows that X : X \ {x}. □
Lemma 12. Let B be a ternary *-band and X € Ch(B). If X has a greatest element xi and there exists a projection y € B such that xi —> y, then X —> X u {y}.
Proof. Let B be a ternary *-band. Then y € B implies that
/ N< \ H< * N< N< N<
(y ) = y, yy y = y, y yy = y .
Let X € Ch(B). This implies that X is a chain of projections. If y be a projection of B such that xi —> y then it is clear that X u {y} is also a chain of projections. Hence X u {y} € Ch(B). Now
X2(X U{y}) = X3 u X2{y} = X. Similarly, (X U {y})X2 = X. Again
X(X u {y})2 = X3 u X2{y} u X{y}2 u X{y}X = X, (X u {y})2X = X.
Therefore, X < (X U {y}).
If possible, there exists Y € Ch(B) such that X < Y < (X u {y}). Thus
X2Y = YX2 = X and Y2(X u {y}) = (X u {y})Y2 = Y.
Therefore, Y2X u Y2{y} = Y. This implies that
X u Y2{y} = Y X c Y.
If X = Y then there exists z € Y such that z / X. Again let z € Y \ X. If z < xi then
z = zxi = z2xi = xiz2 = xfz € YX2 = X.
So z € X. This contradicts our assumption that X = Y. Hence xi < z. Since
X u Y2{y} = X u {y}Y2 = Y, z € Y2{y} = {y}Y2.
Therefore, z = yiy2y = yy3y4 for some yi,y2,y3,y4 € Y. This implies that
zy2 = yiy2yy2 = yiy2y = z, y2z = y2(yy3y4) = y3y3y4 = yy3y4 = z.
Hence z = zy2 = y2z. Therefore, z < y.
Thus we get xi < z < y. This contradicts the relation xi —> y. Hence our assumption is not true and so X —> X u {y}. □
Lemma 13. Let B be a ternary *-band and let x € B be a projection. If {x} —> Y for some Y € Ch(B) then Y = {x, y} with x —> y.
Proof. Since {x} —> Y, we get {x}2Y=Y{x}2={x}. Hence for any y € Y, x2y = yx2 = x. This implies that x < y for all y € Y. Let Yi = {x} u Y. Now
Y2Yi = Y2{x} u Y3 = {x} u Y = Yi = YYi2, YiY2 = {x}Y2 u Y3 = {x} u Y = Yi = Yi2Y.
This implies that Yi < Y. Again
Yi{x}2 = {x}3 u Y{x}2 = {x} = Yi2{x}, {x}2Yi = {x}3 u {x}2Y = {x} = {x}Yi2.
This implies that {x} < Yi. Therefore, {x} < Yi < Y. This contradicts the relation {x} —> Y. Hence Yi = Y. This implies x € Y.
Next we assume that z € Y \ {x} is an arbitrary element. Consider the set
Z = {y € Y : y < z}.
Since Y € Ch(B), Z is also a chain of projections. Now z € Z implies that Z is nonempty. Hence
x2Z = {x2y : y < z} = {x}.
Similarly, xZ2 = {x}. Therefore, {x} < Z.
Again let u € Z2Y = ZY2. Therefore, u = ziz2yi for some yi € Y and zi, z2 € Z. This implies either yi < z or yi > z.
Case 1. Let yi < z then yiz2 = yi2z = z2yi = zyi2 = yi. Hence
2 2 2 2 z u = z ziz2yi = ziz2yi = u, Uz = ziz2yiz = ziz2yi = U.
This implies u < z.
Case 2. Let yi > z then yiz2 = yi2z = z2yi = zyi2 = z. Hence
2 2 2 2 2 z u = z ziz2yi = ziz2yi = u, uz = z^yiz = ziz2z yi = z^yi = u.
Therefore, u < z. Thus u Z. Hence Z2Y = ZY2 c Z.
Similarly, v € YZ2 = Y2Z v = yiziz2 = y2y3z3. Then either yi < z or yi > z. Case 1. Let yi < z. Then
2 2 2 2 yi = yiz = yi z = z yi = zyi ,
2 2 2 2 2 2 z V = z yiziz2 = yiziz2 = V = zv , Vz = yiz^z = yiziz2 = V = V z.
Hence v < z.
Case 2. Let yi > z. Therefore,
2 2 2 2 z = yiz = yi z = z yi = zyi ,
2 2 2 2 2 Vz 2 = yiziz2z 2 = yiziz2 = V, z 2V = z 2yiziz2 = yiz 2ziz2 = yiziz2 = V.
Hence v < z. This implies that v € Z. Thus YZ2 = Y2Z c Z.
Conversely, Z = Z3 c Y2Z = YZ2 and Z = Z3 c Z2Y = ZY2. Hence
Z = Y 2Z = YZ2 = ZY2 = Z 2Y.
This implies that Z < Y. Therefore, {x} < Z < Y. This contradicts the relation {x} —> Y. Hence Y = Z. Since z is an arbitrary element, Y has only two elements say {x,y}. It is clear that x < y. If possible x ^ y. Then there exists z € Ch(B) such that x < z < y. Then
x2{x, z} = x{x, z}2 = {x, z}2x = {x, y}x2 = {x}, {x, z}Y2 = xY2 u zY2 = {x} u {z} = {x, z} = {x, z}2Y = Y2{x, z} = Y{x, z}2.
Therefore, {x} < {x, z} < Y. This is a contradiction. Hence Y = {x, y} and x —> y. □
Proposition 1. Let B be a ternary *-band and X € Ch(B) such that |X| > 3. Then X has a topknot.
Proof. Let X € Ch(B) and |X| > 3. Then there exist x,y,z € X such that x < y < z. Since {x} and {y} are not maximal elements of X, X : X\{x} and X\{y}, by Lemma 11. Again X \ {x} : X \ {x, y} and X \ {y} : X \ {x, y}. Therefore, we have the following topknot:
X \{x}
X _ X \ {x,y}
X \{y}
□
Proposition 2. Let B be a ternary *-band. If X € Ch(B) and \X\ = 2 then X has either a maximal hair of length 1 or a topknot.
Proof. Let X = {x,y}, with x < y. By Lemma 11, it follows that X : X \ {x} = {y}. If X has no maximal hair of length 1 then there is an element z € Ch(B) such that y —> z. Again by Lemma 13, {y} —> {y, z}. Also by Lemma 12, X —> X u {z} = {x, y, z}. Then by Lemma 11, {x,y,z} : {y,z}. Hence we can construct the following topknot:
□
Proposition 3. Let B is a ternary *-band and X € Ch(B). Then \X\ =1 if and only if X has neither maximal hair of length 1 nor topknots.
Proof. Suppose that X = {x} and if possible, X has a maximal hair of length 1. Then there exists Y € Ch(B) such that X : Y. Thus by Lemma 13, we get Y = {x, y} for some y with x —> y. Since x is not maximal in Y, by Lemma 11, it is clear that {x} : Y : Y \ {x}. This contradicts our assumption that X has a maximal hair of length 1. Now suppose that X has a topknot as follows:
Again by Lemma 13, we have Y = {x, y} with x —> y and Z = {x, z} with x —> z and y = z so that y2z = yz2 = zy2 = z2y.
Now T € Ch(B). Consider W = {x, y, z}. Since y2z = yz2 = zy2 = z2y = x, W / Ch(B). So xyz = x implies that W3 = W. Hence W € S (S). Now
{x,y}2W = {x3,x2y,x2z,xyx,xy2,xyz,yx2,yxy,yxz,y2x, y3,y2z} = {x,y}.
Similarly,
{x,y}W2 = W{x,y}2 = W2{x, y} = {x,y}.
This shows that {x,y} < W. Again W 2T = WT2 = W. Thus {x,y} <W <T. This contradicts the existence of topknot. Hence X has no topknot.
Conversely, suppose that X has neither maximal hair of length 1 nor topknot. If possible, IX| > 1. Then Proposition 1 and Proposition 2 contradicts our assumption. Thus the result holds. □
Theorem 4. Ternary *-band is globally determined.
Proof. Let Bi, B2 be two ternary *-bands such that ^ : P(Bi) —> P(B2) be an isomorphism. Let Bb B2 be the set of all singleton subsets of B1 and B2 respectively. Let us define
: B1 —> B such that ^1(x) = {x} and : B2 —> B2 such that ^2({y}) = y. Now from the construction of it follows that and be two isomorphisms from B1 to B and from B2
to B2 respectively.
If we are able to show that ^ B is a bijection from B1 to B2 then it follows that ^ B: B —> B2 is an isomorphism.
Let X € Ch(B1) such that |X| = 1. Then X € B1. If possible, ^(X) = X' / B2, i.e., |X'| > 2. Then by Proposition 1 and Proposition 2, it follows that X' has either a maximal hair of length 1 or a topknot.
Case 1. Suppose X' has a maximal hair of length 1. Then X has also a maximal hair of length 1. This contradicts that X € B1.
Case 2. Suppose X' has a topknot as follows:
where Y',Z',W' € Ch(B2) and Y' = Z'. Then there exists Y,Z,W € Ch(B1) and Y = Z such that Y' = ^(Y), Z' = ^(Z) and W' = ^(W). Then the above topknot can be written as follows:
■0(X)
^(Y)
■0(Z)
^(W)
Hence we have the following topknot:
From the Proposition 3, it follows that X / B. This contradicts our assumption. Therefore, |^(X)| = |X'| = 1 and X' € Ch(B2). Thus ^ is a bijection from the singleton subset of projection of B1 to the singleton subset of projection of B2.
Now let x € B1 such that x is not projection. Then (x2x*)* *. Therefore, x2x* is
a projection.
Similarly, x*x2 is also a projection and (x2x*)(x2x*)(x*x2) = x. Therefore, any element of B can be written as a product of three projections, say x = Imn, where l, m, n are projections. So
^({x}) = ^({Imn}) = ^({¿}{m}{n}) = ^({1})^({m})^({n}) = 11m1n1 = x1.
Therefore, {x} € B implies that ^({x}) € B2, i.e., ^(i?1) c B2. Similarly, ^-1(B2) c B^ This implies that ^(^-1(i52)) c ^(i?1) and B2 c ^(B1). Hence ^(i51) = B2. Therefore, ^ is an onto
mapping from Bi to B2 and since 0 is an isomorphism from P(Bi) to P(B2) and B1 c P(Bi), it follows that 0 \b1 : B1 —> B2 is an isomorphism.
Therefore, 020 \b>1 01 : B1 —> B2 is an isomorphism. Hence B1 = B2. Thus we conclude that ternary *-bands are globally determined. □
4. Conclusion
Throughout this paper we investigated the on global determinism of ternary *-bands and successfully proved that ternary *-bands are globally determined. This research enriches the study of global determinism problem on different classes of ternary semigroup. In future we will be able to study the global determinism problem of another class of ternary semigroup with the help of those results that we have proved in this paper. We hope this work will flourish the field of ternary semigroup, specially the global determinism problem on various classes of ternary semigroupes.
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