Siberian Mathematical Journal, Vol. 56, No. 5, pp. 827–840, 2015 c 2015 Zhuchok A.V. Original Russian Text Copyright
DIMONOIDS AND BAR-UNITS c A. V. Zhuchok
UDC 512.57; 512.579
Abstract: A. P. Pozhidaev proved that each dialgebra may be embedded into a dialgebra with a barunit. As is known, a dialgebra is a vector space with two binary operations satisfying the axioms of a dimonoid. It is natural in this situation to pose the problem about the possibility of adjoining bar-units to dimonoids in a given class and the problem of embedding dimonoids into dimonoids with bar-units. In the present article these problems are solved for some classes of dimonoids. In particular, we show that it is impossible to adjoin a set of bar-units to a free dimonoid. Also, we solve the problem of embedding an arbitrary dimonoid into a dimonoid with bar-units. DOI: 10.1134/S0037446615050055 Keywords: dimonoid, bar-unit, adjoining a set of bar-units, free dimonoid, free rectangular dimonoid, free commutative dimonoid, free n-(di)nilpotent dimonoid, semigroup, automorphism group
1. Introduction. One of the classical algebraic problems is the embedding problem of one algebraic system into the other. For example, it is known that each commutative semigroup may be embedded into a group if and only if it is a semigroup with cancelations, and an integral domain may be embedded into a field. Malcev found necessary and sufficient conditions of embeddability of an arbitrary semigroup into a group. Ore proved that every right reversive ring without zero divisors is embeddable into a skew-field. Dimonoids were first introduced by Loday [1] and used to construct free dialgebras and cohomologies of dialgebras. Recently there have appeared some articles on the properties of dimonoids and dialgebras which are linear analogs of dimonoids (for example, see [2–6]). In particular, it was shown in [5] that each dialgebra may be embedded into a dialgebra with a bar-unit. The cohomologies of dialgebras with a bar-unit were studied in [6]. The relatively free dimonoids play an important role in the theory of dimonoids, since every dimonoid is a homomorphic image of some relatively free dimonoid. Therefore, we may acquire thorough knowledge of properties of every concrete dimonoid studying the properties of relatively free dimonoids (for example, see [7–11]). This article studies the questions of adjoining bar-units to the dimonoids of a given class and embedding dimonoids into dimonoids with bar-units. We solve the problems of adjoining a set of bar-units to the left zero and right zero dimonoid [8], to the dimonoid Sf [12], and also we solve the problem of embedding an arbitrary dimonoid into a dimonoid with bar-units. We prove that it is impossible to adjoin a set of bar-units to the free dimonoid [1], to the free commutative dimonoid [9], to the free n-nilpotent dimonoid (n > 1) [10], and to the free n-dinilpotent dimonoid (n > 1) [11]. Moreover, we establish that the semigroups of the free rectangular dimonoid, the free dimonoid, the free n-nilpotent dimonoid, and the free n-dinilpotent dimonoid are anti-isomorphic; the semigroups of the free commutative dimonoid are isomorphic, and the automorphism groups of the free rectangular dimonoid, the free dimonoid, the free commutative dimonoid, the free n-nilpotent dimonoid, and the free n-dinilpotent dimonoid are isomorphic to the symmetric group. These results can be applied to the dialgebras, and some of them may be useful in consideration of the embedding problem of a dimonoid into a digroup [13]. The main results of Subsections 3–6 were announced in [14]. 2. Preliminaries. Recall that a dimonoid is a nonempty set D with the two binary operations Starobelsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 56, No. 5, pp. 1037–1053, September–October, 2015; DOI: 10.17377/smzh.2015.56.505. Original article submitted August 25, 2014. Revision submitted May 25, 2015. 0037-4466/15/5605–0827
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and satisfying the axioms
(x y) z = x (y z), (x y) z = x (y z), (x y) z = x (y z), (x y) z = x (y z), (x y) z = x (y z)
for all x, y, z ∈ D. In the case when the operations coincide, the dimonoid is a semigroup; i.e., the dimonoids are a generalization of semigroups. There exist numerous examples of dimonoids with various operations (for example, see [1, 2, 7–12]). An element e of a dimonoid (D, , ) is a bar-unit if x e = x = e x for every x ∈ D [1]. Note that in contrast to monoids a dimonoid may possess not only one bar-unit; and the set of all its bar-units (if the latter is nonempty) forms a subdimonoid. If a dimonoid has a unit then its operations coincide, which follows from the axioms of a dimonoid. The presence of a bar-unit in a dialgebra plays an important role in studying the dialgebra homologies [6]. The following two lemmas give new examples of dimonoids with bar-units, and we will use these examples in the sequel. Let S be an arbitrary semigroup, and let ε ∈ / S be an arbitrary symbol. Denote by U = UεS the set of all words of the shape wεv in the alphabet S ∪ {ε}, where w, v ∈ S ∪ {ω} (ω is the empty word, ω∈ / S ∪ {ε}). Consider the monoid (S ∪ {ω}, ∗) obtained from S by adjoining ω as a unit. Define the operations and on U by putting w1 εv1 w2 εv2 = w1 ε(v1 ∗ w2 ∗ v2 ),
w1 εv1 w2 εv2 = (w1 ∗ v1 ∗ w2 )εv2
for all w1 εv1 , w2 εv2 ∈ U. Lemma 1. The algebra (U, , ) is a dimonoid with the bar-unit ε. Proof. Take w1 εv1 , w2 εv2 , w3 εv3 ∈ U . Then (w1 εv1 w2 εv2 ) w3 εv3 = w1 ε(v1 ∗ w2 ∗ v2 ) w3 εv3 = w1 ε(v1 ∗ w2 ∗ v2 ∗ w3 ∗ v3 ) = w1 εv1 w2 ε(v2 ∗ w3 ∗ v3 ) = w1 εv1 (w2 εv2 w3 εv3 ), w1 εv1 (w2 εv2 w3 εv3 ) = w1 εv1 (w2 ∗ v2 ∗ w3 )εv3 = w1 ε(v1 ∗ w2 ∗ v2 ∗ w3 ∗ v3 ), (w1 εv1 w2 εv2 ) w3 εv3 = (w1 ∗ v1 ∗ w2 )εv2 w3 εv3 = (w1 ∗ v1 ∗ w2 )ε(v2 ∗ w3 ∗ v3 ) = w1 εv1 w2 ε(v2 ∗ w3 ∗ v3 ) = w1 εv1 (w2 εv2 w3 εv3 ), (w1 εv1 w2 εv2 ) w3 εv3 = (w1 ∗ v1 ∗ w2 )εv2 w3 εv3 = (w1 ∗ v1 ∗ w2 ∗ v2 ∗ w3 )εv3 = w1 εv1 (w2 ∗ v2 ∗ w3 )εv3 = w1 εv1 (w2 εv2 w3 εv3 ), (w1 εv1 w2 εv2 ) w3 εv3 = w1 ε(v1 ∗ w2 ∗ v2 ) w3 εv3 = (w1 ∗ v1 ∗ w2 ∗ v2 ∗ w3 )εv3 . Thus, (U, , ) satisfies the axioms of a dimonoid. From the definitions of operations it follows that ε is a bar-unit. The lemma is proved. Denote the above-obtained dimonoid by S[ε]. Let {S[εi ]}i∈I be a set of dimonoids S[εi ] with bar-units εi , i ∈ I. Define the operations and on i∈I S[εi ] by putting w1 εi v1 w2 εj v2 = w1 εi (v1 ∗ w2 ∗ v2 ), for all w1 εi v1 , w2 εj v2 ∈ i∈I S[εi ]. 828
w1 εi v1 w2 εj v2 = (w1 ∗ v1 ∗ w2 )εj v2
Lemma 2. The algebra i∈I S[εi ], , is a dimonoid with bar-units εi , i ∈ I. Proof. By analogy with the proof of Lemma 1 we can show that this algebra is a dimonoid. From the definitions of operations it follows that each element εi , i ∈ I, is a bar-unit. The lemma is proved. Denote the above-obtained dimonoid by S[εi ]i∈I . Corollary 1. Let S be a finite semigroup, and let I be a nonempty finite set. Then |S[εi ]i∈I | = + |I|(2|S| + 1). Let (D, , ) be an arbitrary dimonoid, and let Y be an arbitrary nonempty set such that D ∩Y = ∅. If we can extend the binary operations and from D to D ∪ Y in such way that x e = x = e x for all x ∈ D ∪ Y , e ∈ Y , and (D ∪ Y, , ) is a dimonoid then we call the transition from (D, , ) to (D ∪ Y, , ) as adjoining the set of bar-units Y to (D, , ). In the opposite case we say that it is impossible to adjoin the set of bar-units Y to (D, , ). It is well known that a unit can be adjoined to an arbitrary semigroup, and every semigroup may be embedded into a semigroup with a unit. Recall that a dialgebra is a vector space with two binary operations satisfying the axioms of a dimonoid [1]. For a dialgebra and a dimonoid the definitions of a bar-unit coincide. It was proved in [5] that each dialgebra may be embedded into a dialgebra with a bar-unit. In this situation it is natural to put the question about the possibility of adjoining a set of bar-units to the dimonoids of a given class. Denote by [X] the symmetric group on a set X; and by Aut(D, , ), the automorphism group of a dimonoid (D, , ). Denote the set of all positive integers by N. |S|2
3. A left zero and right zero dimonoid. In this subsection we solve the problem of adjoining a set of bar-units to the left zero and right zero dimonoid. A semigroup is a left (right) zero semigroup provided that it satisfies the identity xy = x (xy = y). A dimonoid (D, , ) is a left zero and right zero dimonoid provided that (D, ) is a left zero semigroup, and (D, ) is a right zero semigroup [8]. A left zero and right zero dimonoid (D, , ) is denoted by Dz,rz . It is known that every left zero and right zero dimonoid is free [8]. Clearly, each element of the dimonoid Dz,rz is a bar-unit, its semigroups are anti-isomorphic; and the endomorphism semigroup (automorphism group) of Dz,rz is isomorphic to the symmetric semigroup (symmetric group) on D (see [8]). Let D and Y be arbitrary nonempty sets such that D ∩ Y = ∅. Lemma 3. It is possible to adjoin the set of bar-units Y to the left zero and right zero dimonoid Dz,rz . Proof. Obviously, we can adjoin the set of bar-units Y to Dz,rz by putting x y = x,
x y = y,
z ε = z = ε z,
z ε = ε = ε z
for all x, y ∈ D, z ∈ D ∪ Y , and ε ∈ Y . The lemma is proved. Thus, the problem of adjoining a set of bar-units Y to an arbitrary dimonoid (D, , ) (see Subsection 2) is reduced to the problem of defining such a dimonoid on D ∪ Y that contains the dimonoids (D, , ) and Yz,rz as subdimonoids. 4. The free rectangular dimonoid. In this subsection we solve the problem of adjoining a set of bar-units to the free rectangular dimonoid. We show that the semigroups of the free rectangular dimonoid are anti-isomorphic, and its automorphism group is isomorphic to the symmetric group. Consider the free rectangular dimonoid [8]. Let X be an arbitrary nonempty set, and put X 3 = X × X × X. Define the operations and on X 3 by the rules (x1 , x2 , x3 ) (y1 , y2 , y3 ) = (x1 , x2 , y3 ), for all (x1 , x2 , x3 ), (y1 , y2 , y3 ) ∈
X 3.
The algebra
(x1 , x2 , x3 ) (y1 , y2 , y3 ) = (x1 , y2 , y3 ) 3 (X , , ) is denoted by FRct(X). 829
Theorem 1 [8, Theorem 1]. FRct(X) is the free rectangular dimonoid. The rectangular dimonoids were used in solving the problem of decomposition of dimonoids into dibands of subdimonoids [8]. The following lemma establishes a connection between the semigroups of the free rectangular dimonoid. Lemma 4. (X 3 , ) and (X 3 , ) are anti-isomorphic semigroups. Proof. Define the mapping μ : (X 3 , ) → (X 3 , ) by putting (a, b, c)μ = (c, b, a) for all (a, b, c) ∈ X 3 . It is immediate that μ is an anti-isomorphism. The lemma is proved. Clearly, the set {(a, a, a) | a ∈ X} generates FRct(X). Hence, we get the following description of the automorphism group of the free rectangular dimonoid: Lemma 5. Aut FRct(X) ∼ = [X]. Let B be an arbitrary nonempty set such that FRct(X) ∩ B = ∅. Theorem 2. It is possible to adjoin the set of bar-units B to the free rectangular dimonoid FRct(X). Proof. Fix d ∈ X. Define the operations and on FRct(X) ∪ B by putting w1 w2 = w1 w 2 , w ε = w = ε w,
w1 w2 = w1 w2 ,
ε (a, b, c) = (a, d, c) = (a, b, c) ε
for all w1 , w2 , (a, b, c) ∈ FRct(X), w ∈ FRct(X) ∪ B, and ε ∈ B. Prove that (FRct(X) ∪ B, , ) is a dimonoid containing the set of bar-units B. For this, it suffices to show that the axioms of a dimonoid hold in the case when every side of axioms contains one or two elements from B. The remaining cases are obvious. Let (a, b, c) and (x, y, z) be arbitrary elements of FRct(X), and take ε, e ∈ B. Using the definitions of and we get (ε (a, b, c)) (x, y, z) = ε ((a, b, c) (x, y, z)) = ε ((a, b, c) (x, y, z)) = (a, d, z), (ε (a, b, c)) (x, y, z) = ε ((a, b, c) (x, y, z)) = (a, b, z), (ε (a, b, c)) (x, y, z) = ε ((a, b, c) (x, y, z)) = (ε (a, b, c)) (x, y, z) = (a, y, z), ((a, b, c) ε) (x, y, z) = (a, b, c) (ε (x, y, z)) = (a, b, c) (ε (x, y, z)) = (a, b, z), ((a, b, c) ε) (x, y, z) = (a, b, c) (ε (x, y, z)) = (a, d, z), ((a, b, c) ε) (x, y, z) = (a, b, c) (ε (x, y, z)) = ((a, b, c) ε) (x, y, z) = (a, y, z), ((a, b, c) (x, y, z)) ε = (a, b, c) ((x, y, z) ε) = (a, b, c) ((x, y, z) ε) = (a, b, z), ((a, b, c) (x, y, z)) ε = (a, b, c) ((x, y, z) ε) = (a, y, z), 830
((a, b, c) (x, y, z)) ε = (a, b, c) ((x, y, z) ε) = ((a, b, c) (x, y, z)) ε = (a, d, z), (ε e) (x, y, z) = ε (e (x, y, z)) = ε (e (x, y, z)) = (x, d, z), (ε e) (x, y, z) = ε (e (x, y, z)) = (x, d, z), (ε e) (x, y, z) = ε (e (x, y, z)) = (ε e) (x, y, z) = (x, y, z), ((x, y, z) ε) e = (x, y, z) (ε e) = (x, y, z) (ε e) = (x, y, z), ((x, y, z) ε) e = (x, y, z) (ε e) = (x, d, z), ((x, y, z) ε) e = (x, y, z) (ε e) = ((x, y, z) ε) e = (x, d, z), (ε (x, y, z)) e = ε ((x, y, z) e) = ε ((x, y, z) e) = (x, d, z), (ε (x, y, z)) e = ε ((x, y, z) e) = (x, y, z), (ε (x, y, z)) e = ε ((x, y, z) e) = (ε (x, y, z)) e = (x, d, z). Thus, (FRct(X) ∪ B, , ) is a dimonoid containing the set of bar-units B. The theorem is proved. 5. The dimonoid Sf . Here we solve the problem of adjoining a set of bar-units to the dimonoid Sf constructed in [12] and study endomorphisms of Sf . Recall the definition of Sf . Let S be a semigroup, and let f be its idempotent endomorphism. Define the two multiplications on S by the rules x y = x(yf ),
x y = (xf )y
for all x, y ∈ S. Proposition 1 [12, Proposition 1]. (S, , ) is a dimonoid. An analogous situation takes place in construction of dialgebras and appears naturally on a differential associative algebra, a graded superalgebra, and an A-bimodule (see [1]). The above-constructed dimonoid is denoted by Sf . The following lemma characterizes the endomorphisms and automorphisms of Sf in the case when S is a semigroup with left (or right, or two-sided) cancelation. Lemma 6. Let S be a semigroup with left (or right, or two-sided) cancelation. Then an endomorphism (automorphism) τ of S is an endomorphism (automorphism) of Sf if and only if τ f = f τ . Proof. Assume that an endomorphism τ of a semigroup S with left cancelation is an endomorphism of Sf . For all x, y ∈ S we have (xy)τ = (x(yf ))τ = xτ (yf τ ),
xτ yτ = xτ (yτ f ),
whence xτ (yf τ ) = xτ (yτ f ). Since S is left cancelable, the last equality gives yf τ = yτ f for all y ∈ S, i.e., f τ = τ f . Conversely, let τ be an endomorphism of S such that τ f = f τ . For all x, y ∈ S we obtain (xy)τ = (x(yf ))τ = xτ (yf τ ) = xτ (yτ f ) = xτ yτ, (xy)τ = ((xf )y)τ = (xf τ )yτ = (xτ f )yτ = xτ yτ. Analogously, we prove the cases when S is a semigroup with right or two-sided cancelation, and also the case when τ is an automorphism of S. The lemma is proved. Let S be an arbitrary semigroup, let f be a nonidentity idempotent endomorphism of S, and let B be an arbitrary nonempty set such that S ∩ B = ∅. 831
Theorem 3. It is possible to adjoin the set of Proof. Let and be binary operations on x y = x y, z ε = z = ε z,
bar-units B to Sf . S ∪ B given by the rules x y = x y, ε x = xf = x ε
for all x, y ∈ S, z ∈ S ∪ B, and ε ∈ B. In order to prove that (S ∪ B, , ) is a dimonoid containing the set of bar-units B, it suffices to verify the axioms of a dimonoid in the case when every side of axioms contains one or two elements from B. The remaining two cases are obvious. Let x and y be arbitrary elements of S, and let ε, e ∈ B. Using the definitions of and , and also the properties of f we obtain (ε x) y = ε (x y) = ε (x y) = (xf )(yf ), (ε x) y = ε (x y) = x(yf ), (ε x) y = ε (x y) = (ε x) y = (xf )y, (x ε) y = x (ε y) = x (ε y) = x(yf ), (x ε) y = x (ε y) = (xf )(yf ), (x ε) y = x (ε y) = (x ε) y = (xf )y, (x y) ε = x (y ε) = x (y ε) = x(yf ), (x y) ε = x (y ε) = (xf )y, (x y) ε = x (y ε) = (x y) ε = (xf )(yf ), (ε e) x = ε (e x) = ε (e x) = xf, (ε e) x = ε (e x) = xf, (ε e) x = ε (e x) = (ε e) x = x, (x ε) e = x (ε e) = x (ε e) = x, (x ε) e = x (ε e) = xf, (x ε) e = x (ε e) = (x ε) e = xf, (ε x) e = ε (x e) = ε (x e) = xf, (ε x) e = ε (x e) = x, (ε x) e = ε (x e) = (ε x) e = xf. Thus, (S ∪ B, , ) is a dimonoid containing the set of bar-units B. The theorem is proved. Note that if f is the identity automorphism then Sf is a semigroup. In this case, applying the method of adjoining bar-units from the proof of Theorem 3, we infer that B = {ε} and (S ∪ B, , ) is a semigroup with the unit ε. 6. The free dimonoid. Here we prove that it is impossible to adjoin a set of bar-units to the free dimonoid. We show that the semigroups of the free dimonoid are anti-isomorphic, and its automorphism group is isomorphic to the symmetric group. Consider the free dimonoid constructed in [1]. Let X bean arbitrary nonempty set, n ∈ N. Denote by Yn the union of n different copies of X n and ˘ i . . . xn the element in the ith component of Yn , we define the put D(X) = n≥1 Yn . Denoting by x1 . . . x operations on D(X): ˘ i . . . xk ) (xk+1 . . . x ˘ j . . . xl ) = x1 . . . x ˘ i . . . xl , (x1 . . . x ˘ i . . . xk ) (xk+1 . . . x ˘ j . . . xl ) = x1 . . . x ˘ j . . . xl (x1 . . . x for all x1 . . . x ˘ i . . . xk , xk+1 . . . x ˘ j . . . xl ∈ D(X). Then (D(X), , ) is the free dimonoid on X (see [1, p. 15]). The following lemma establishes a connection between the semigroups of the free dimonoid. 832
Lemma 7. (D(X), ) and (D(X), ) are anti-isomorphic semigroups. Proof. Define the mapping π : (D(X), ) → (D(X), ) by putting ˘ i . . . xk )π = xk . . . x ˘ i . . . x1 (x1 . . . x for all x1 . . . x ˘ i . . . xk ∈ D(X). An immediate verification shows that π is an anti-isomorphism. The lemma is proved. It is easy to notice that the free dimonoid is uniquely defined (up to isomorphism) by the cardinality of X, whence we get the description of the automorphism group of the free dimonoid: Lemma 8. Aut(D(X), , ) ∼ = [X]. Let Y be an arbitrary nonempty set such that D(X) ∩ Y = ∅. Theorem 4. It is impossible to adjoin the set of bar-units Y to the free dimonoid (D(X), , ). Proof. Given w = x1 . . . x ˘ i . . . xk ∈ D(X), put w = x1 . . . xi . . . xk . Now, let and be binary operations on D(X) ∪ Y satisfying the conditions w 1 w 2 = w1 w 2 ,
w1 w2 = w1 w2 ,
w ε = w = ε w
for all w1 , w2 ∈ D(X), w ∈ D(X) ∪ Y , and ε ∈ Y . Prove that (D(X) ∪ Y, , ) is not a dimonoid. We can proceed by way of contradiction. Assume that (D(X)∪Y, , ) is a dimonoid. Let ε w = t and w ε = t for some ε ∈ Y , w ∈ D(X). We have (w ε) w = w w = w (ε w) = w t.
(1)
The three cases for t are possible: 1) t = e ∈ Y ; 2) t ∈ D(X), t = w; 3) t = w. Consider Case 1. By (1), w w = w w = w t = w e = w, whence w is an idempotent of (D(X), ). This contradicts the fact that (D(X), ) has no idempotents. Hence, ε w = e for all ε ∈ Y and w ∈ D(X). Consider Case 2. By (1), = w˜t = w t = w t, w w = w w = w w whence Further,
w = ˜t.
(2)
t, w (ε w) = w t = w t = w
(3)
(w ε) w = t w.
(4)
The three cases for t are possible: (a) t = e ∈ Y ; (b) t ∈ D(X), t = w; and (c) t = w. In Case (a) we have (w ε) w = e w = w,
w (ε w) = w w = w w,
whence w w = w, which is impossible since (D(X), ) has no idempotents. Therefore, t = e. . Then, using (3), we have w t = t w . Consider Case (b). By (4), (w ε) w = t w = t w = t w The latter equality is false since w = t by (2). Thus, Case (b) is impossible. . Then by (3) we deduce Let us turn to Case (c). By (4), (w ε) w = w w = w w = ww w t = ww , which is false since w = t by (2). So, Case (c) is impossible. 833
Thus, Condition 2 is false. By previous arguments, Condition 3 holds, i.e., ε w = w for all ε ∈ Y and w ∈ D(X). Since (D(X) ∪ Y, , ) is a dimonoid by hypothesis, ε (w1 w2 ) = ε (w1 w2 ) for all w1 , w2 ∈ D(X), whence w 1 w 2 = w1 w 2 = w1 w 2 = w 1 w 2 . It means that the operations of the free dimonoid coincide, which contradicts its definition. Hence, Condition 3 is false either. Summing up, we arrive at the conclusion that our assumption for the algebra (D(X) ∪ Y, , ) is false. Thus, (D(X) ∪ Y, , ) is not a dimonoid. So, we cannot adjoin the set of bar-units Y to the free dimonoid (D(X), , ). The theorem is proved. 7. The free commutative dimonoid. Here we prove that it is impossible to adjoin a set of barunits to the free commutative dimonoid. We show that the semigroups of the free commutative dimonoid are isomorphic, and its automorphism group is isomorphic to the symmetric group. Consider the free commutative dimonoid constructed in [9]. Let F ∗ [X] be the free commutative semigroup on a set X, and let G be the set of all unordered pairs (p, q), p, q ∈ X. Define the operations and on F ∗ [X] ∪ G by the rules a1 . . . am b1 . . . bn = a1 . . . am b1 . . . bn , a1 . . . am b1 . . . bn =
a1 . . . am b1 . . . bn , (a1 , b1 ),
mn > 1, m = n = 1,
a1 . . . am (p, q) = a1 . . . am (p, q) = a1 . . . am pq, (p, q) a1 . . . am = (p, q) a1 . . . am = pqa1 . . . am , (p, q) (r, s) = (p, q) (r, s) = pqrs for all a1 . . . am , b1 . . . bn ∈ F ∗ [X] and (p, q), (r, s) ∈ G. Theorem 5 [9, Theorem 3]. (F ∗ [X] ∪ G, , ) is the free commutative dimonoid. The following lemma establishes a connection between the semigroups of the free commutative dimonoid. Lemma 9. (F ∗ [X] ∪ G, ) and (F ∗ [X] ∪ G, ) are isomorphic semigroups. Proof. Define the mapping λ : (F ∗ [X] ∪ G, ) → (F ∗ [X] ∪ G, ) by putting ⎧ ⎨ (a, b), if w = ab, a, b ∈ X, if w = (a, b) ∈ G, wλ = ab, ⎩ w otherwise. An immediate verification shows that λ is an isomorphism. The lemma is proved. Note that the free commutative dimonoid is uniquely defined (up to isomorphism) by the cardinality of X, whence we get the following description of the automorphism group of the free commutative dimonoid: Lemma 10. Aut(F ∗ [X] ∪ G, , ) ∼ = [X]. Let Y be an arbitrary nonempty set such that (F ∗ [X] ∪ G) ∩ Y = ∅. 834
Theorem 6. It is impossible to adjoin the set of bar-units Y to the free commutative dimonoid ∪ G, , ). Proof. Let and be binary operations on F ∗ [X] ∪ G ∪ Y satisfying the conditions
(F ∗ [X]
w 1 w 2 = w1 w 2 ,
w1 w2 = w1 w2 ,
w ε = w = ε w
for all w1 , w2 ∈ F ∗ [X] ∪ G, w ∈ F ∗ [X] ∪ G ∪ Y , and ε ∈ Y . By way of contradiction we will show that (F ∗ [X] ∪ G ∪ Y, , ) is not a dimonoid. Assume that (F ∗ [X] ∪ G ∪ Y, , ) is a dimonoid. Let ε x = t and x ε = t for some ε ∈ Y and x ∈ X. The four cases for t are possible: 1) t = e ∈ Y ; 2) t = (a, b) ∈ G; 3) t = a1 . . . am ∈ F ∗ [X], m > 1; and 4) t = c ∈ X. In Case 1 we get (x ε) x = x x = xx,
x (ε x) = x e = x,
whence xx = x, which is false. Hence, ε x = e for all ε ∈ Y and x ∈ X. Consider Case 2: ε (x x) = x x = (x, x),
(ε x) x = (a, b) x = abx,
whence (x, x) = abx, which is impossible. Therefore, ε x = (a, b) for all ε ∈ Y and x ∈ X. Turn to Case 3. We have ε (x x) = (x, x),
(ε x) x = a1 . . . am x = a1 . . . am x,
whence (x, x) = a1 . . . am x, which is false. Consequently, ε x = a1 . . . am for all ε ∈ Y , x ∈ X, and m > 1. By arguments above we infer that Case 4 holds. Analogously, we can show that t = d ∈ X. Further, (x ε) x = d x = dx, x (ε x) = x c = (x, c), whence dx = (x, c), which is impossible again. It means that Case 4 is impossible. Thus, it is a false assumption that (F ∗ [X]∪G∪Y, , ) is a dimonoid. Therefore, (F ∗ [X]∪G∪Y, , ) is not a dimonoid. It follows from here that we cannot adjoin the set of bar-units Y to the free commutative dimonoid (F ∗ [X] ∪ G, , ). The theorem is proved. 8. The free n-nilpotent dimonoid. Here we prove that it is impossible to adjoin a set of bar-units to the free n-nilpotent dimonoid (n > 1). We show that the semigroups of the free n-nilpotent dimonoid are anti-isomorphic, and its automorphism group is isomorphic to the symmetric group. Consider the free n-nilpotent dimonoid of an arbitrary rank [10]. Let X be an alphabet, and let F [X] be the free semigroup on X. Denote the length of a word w ∈ F [X] by lw . Take n ∈ N and put F Nn = {(w, m) ∈ F [X] × N | m ≤ lw ≤ n} ∪ {0}. Define the operations and on F Nn by (w1 , m1 )(w2 , m2 ) = (w1 , m1 )(w2 , m2 ) =
(w1 w2 , m1 ), 0,
lw1 w2 ≤ n, lw1 w2 > n,
(w1 w2 , lw1 + m2 ), 0,
lw1 w2 ≤ n, lw1 w2 > n,
(w1 , m1 ) ∗ 0 = 0 ∗ (w1 , m1 ) = 0 ∗ 0 = 0 for all (w1 , m1 ), (w2 , m2 ) ∈ F Nn \{0} and ∗ ∈ {, }. The algebra (F Nn , , ) is denoted by F Nn (X). 835
Theorem 7 [10, Theorem 1]. F Nn (X) is the free n-nilpotent dimonoid. The following lemma establishes a connection between the semigroups of the free n-nilpotent dimonoid. Lemma 11. (F Nn , ) and (F Nn , ) are anti-isomorphic semigroups. Proof. Define the mapping χ : (F Nn , ) → (F Nn , ) by putting (xm xm−1 . . . x1 , m − s + 1), if w = (x1 x2 . . . xm , s) ∈ F Nn \{0}, xi ∈ X, 1 ≤ i ≤ m, wχ = 0, if w = 0. An immediate verification shows that χ is an anti-isomorphism. The lemma is proved. It is easy to notice that the free n-nilpotent dimonoid is uniquely defined (up to isomorphism) by the cardinality of X, whence we get the description of the automorphism group of the free n-nilpotent dimonoid: Lemma 12. Aut F Nn (X) ∼ = [X]. Let Y be an arbitrary nonempty set such that F Nn (X) ∩ Y = ∅. Theorem 8. It is impossible to adjoin the set of bar-units Y to the free n-nilpotent dimonoid F Nn (X) (n > 1). Proof. Let and be binary operations on F Nn (X) ∪ Y satisfying the conditions w 1 w 2 = w1 w 2 ,
w1 w2 = w1 w2 ,
w ε = w = ε w
for all w1 , w2 ∈ F Nn (X), w ∈ F Nn (X) ∪ Y , and ε ∈ Y . Prove by contradiction that (F Nn (X) ∪ Y, , ) is not a dimonoid. Assume that (F Nn (X) ∪ Y, , ) is a dimonoid. Let ε (x, 1) = t and (x, 1) ε = t for some ε ∈ Y and (x, 1) ∈ F Nn \{0}, where x ∈ X. The three cases for t are possible: 1) t = 0; 2) t = e ∈ Y ; and 3) t ∈ F Nn \{0}. Consider Case 1. Let (y, 1) ∈ F Nn \{0}, where y ∈ X. We have ε ((x, 1) (y, 1)) = (x, 1) (y, 1) = (xy, 2), (ε (x, 1)) (y, 1) = 0 (y, 1) = 0, whence (xy, 2) = 0, which is false. Hence, ε (x, 1) = 0 for all ε ∈ Y and (x, 1) ∈ F Nn \{0}. In Case 2 we obtain ((x, 1) ε) (x, 1) = (x, 1) (x, 1) = (xx, 1), (x, 1) (ε (x, 1)) = (x, 1) e = (x, 1), whence (xx, 1) = (x, 1), which is impossible. Therefore, ε (x, 1) = e for all ε ∈ Y and (x, 1) ∈ F Nn \{0}. By analogy with Case 2 we have in Case 3 ((x, 1) ε) (x, 1) = (xx, 1),
(x, 1) (ε (x, 1)) = (x, 1) t,
whence (xx, 1) = (x, 1) t. It follows from the last equality that t = (x, 1). Thus, ε (x, 1) = (x, 1). Analogously, we can show that t = (x, 1). Further, ((x, 1) ε) (x, 1) = (x, 1) (x, 1) = (xx, 1), (x, 1) (ε (x, 1)) = (x, 1) (x, 1) = (xx, 2), whence (xx, 1) = (xx, 2), which is false again. It means that the case t ∈ F Nn \{0} is impossible. 836
Thus, it is a false assumption that (F Nn (X) ∪ Y, , ) is a dimonoid. Hence, (F Nn (X) ∪ Y, , ) is not a dimonoid. It follows from here that we cannot adjoin the set of bar-units Y to the free n-nilpotent dimonoid F Nn (X) with n > 1. The theorem is proved. 9. The free n-dinilpotent dimonoid. Here we establish that it is impossible to adjoin a set of barunits to the free n-dinilpotent dimonoid (n > 1). We show that the semigroups of the free n-dinilpotent dimonoid are anti-isomorphic, and its automorphism group is isomorphic to the symmetric group. Consider the free n-dinilpotent dimonoid of an arbitrary rank [11]. We use the notations of Subsection 8. Fix n ∈ N and define the operations and on F Dn = {(w, m) ∈ F [X] × N | m ≤ lw , m ≤ n, lw − m + 1 ≤ n} ∪ {0} by the rules
(w1 , m1 )(w2 , m2 ) = (w1 , m1 )(w2 , m2 ) =
(w1 w2 , m1 ), 0,
lw1 w2 − m1 + 1 ≤ n, lw1 w2 − m1 + 1 > n,
(w1 w2 , lw1 + m2 ), 0,
lw1 + m2 ≤ n, lw1 + m2 > n,
(w1 , m1 ) ∗ 0 = 0 ∗ (w1 , m1 ) = 0 ∗ 0 = 0 for all (w1 , m1 ), (w2 , m2 ) ∈ F Dn \{0} and ∗ ∈ {, }. The algebra (F Dn , , ) is denoted by F Dn (X). Theorem 9 [11, Theorem 1.1]. F Dn (X) is the free n-dinilpotent dimonoid. The following two lemmas can be proved by analogy with Lemmas 11 and 12. Lemma 13. (F Dn , ) and (F Dn , ) are anti-isomorphic semigroups. Lemma 14. Aut F Dn (X) ∼ = [X]. Let Y be an arbitrary nonempty set such that F Dn (X) ∩ Y = ∅. Theorem 10. It is impossible to adjoin the set of bar-units Y to the free n-dinilpotent dimonoid F Dn (X) (n > 1). The proof is the same as the proof of Theorem 8. 10. Embedding an arbitrary dimonoid into a dimonoid with bar-units. Theorems 4, 6, 8, and 10 show that there exist such dimonoids that we cannot adjoin a set of bar-units to them. In connection with this the question whether it is always possible to embed an arbitrary dimonoid into a dimonoid with bar-units is of a great interest. The following theorem gives the positive answer to this question showing also that Theorem 3 of [5] has an analog in the class of dimonoids. Given a relation ρ on a dimonoid (D, , ), the congruence generated by ρ is the least congruence on (D, , ) containing ρ. This congruence is denoted by ρ∗ and may be characterized as the intersection of all congruences on (D, , ) containing ρ. If α is a congruence on a dimonoid (D, , ) such that the operations of the quotient dimonoid (D, , )/α coincide and this dimonoid is a semigroup then we say that α is a semigroup congruence. Let (D, 1 , 1 ) be an arbitrary dimonoid and ρ = {(x1 y, x1 y) | x, y ∈ D}. Obviously, ρ∗ is the ¯ if it contains a ∈ D; least semigroup congruence on (D, 1 , 1 ). A congruence class of ρ∗ is denoted by a and the quotient dimonoid (D, 1 , 1 )/ρ∗ is denoted by D. Now, consider the monoids (D ∪ {θ}, 2 ) and (D ∪ {θ}, 2 ) obtained from the semigroups (D, 1 ) and (D, 1 ), respectively, by adjoining θ as a unit. Passing from the operations 2 and 2 to the operations 1 and 1 and using the axioms of a dimonoid, we can prove 837
Lemma 15. The following hold: (i) (b 2 c2 d)1 x = b1 (c2 d2 x), b, x ∈ D, c, d ∈ D ∪ {θ}; (ii) (b2 c2 d)1 x = b1 (c2 d2 x), b, x ∈ D, c, d ∈ D ∪ {θ}; (iii) (b1 c)2 d2 x = b1 (c2 d2 x), b, c ∈ D, d, x ∈ D ∪ {θ}; (iv) (b2 c2 d)1 x = b2 c2 (d1 x), d, x ∈ D, b, c ∈ D ∪ {θ}; (v) (b2 c2 d)2 x2 y = b2 c2 (d2 x2 y), d ∈ D, b, c, x, y ∈ D ∪ {θ}; (vi) b2 (c2 d2 x)2 y = b2 c2 d2 x2 y, y ∈ D, b, c, d, x ∈ D ∪ {θ}; (vii) b2 c2 d2 x2 y = b2 (c2 d2 x)2 y, b ∈ D, c, d, x, y ∈ D ∪ {θ}. Let ε be an arbitrary symbol, ε ∈ / D. Consider the set of all words in the alphabet D ∪ {ε} of the / D ∪ {ε}), and the monoid (D ∪ {θ}, ∗) obtained shape a ¯εb, where a, b ∈ D ∪ {θ} (θ is the empty word, θ ∈ from D by adjoining θ as a unit. Clearly, a ¯ ∗ b = a 2 b = a 2 b
(5)
for all a, b ∈ D ∪ {θ}, and if a1 , a2 , b1 , b2 ∈ D then a1 εb1 = a2 εb2 ⇐⇒ (a1 , a2 ) ∈ ρ∗ ∧ (b1 , b2 ) ∈ ρ∗ . Furthermore, D[ε] is a dimonoid with the bar-unit ε by Lemma 1. Using (5) and the associativity of 2 and 2 , it is easy to prove Lemma 16. The following hold: (i) (b1 c)2 d = b2 (c2 d), b, c ∈ D, d ∈ D ∪ {θ}; (ii) (b2 c)2 d = b2 (c1 d), c, d ∈ D, b ∈ D ∪ {θ}; (iii) b2 c2 d2 x = b2 (c2 d)2 x = b2 (c2 d2 x), b, c, d, x ∈ D ∪ {θ}; (iv) b2 c2 d2 x = b2 (c2 d)2 x = (b2 c2 d)2 x, b, c, d, x ∈ D ∪ {θ}. Theorem 11. For every dimonoid (D, 1 , 1 ) there exists a dimonoid VεD,I with the set of bar-units i {εi }i∈I containing (D, 1 , 1 ) as a subdimonoid. Proof. Consider the set {D[εi ]}i∈I of dimonoids D[εi ] with bar-units εi , i ∈ I. Define the operations 3 and 3 on (D, 1 , 1 ) ∪ D[εi ]i∈I (see Subsection 2) by putting t3 g = t1 g,
t3 g = t1 g,
¯ = t2 c2 d, t3¯cεj d
¯ = t2 cεj d, ¯ t3¯cεj d
¯ 3 t = ¯cεj d2 t, ¯cεj d
¯ 3 t = c2 d2 t, ¯cεj d
¯=a ¯εi b2 c2 d, a ¯εi¯b3¯cεj d
¯ = a2 b2 cεj d ¯ a ¯εi¯b3¯cεj d
¯ ∈ D[εi ]i∈I . Using the definition of ρ∗ on (D, 1 , 1 ) and the axioms of for all t, g ∈ D, a ¯εi¯b, ¯cεj d a dimonoid, we can show that t1 g = t1 z and g1 t = z1 t for all t, g ∈ D and z ∈ g¯. It follows from . here that the operations 3 and 3 are correctly defined. Denote the so-constructed algebra by VεD,I i D,I Show that Vεi is a dimonoid. By Lemma 2 D[εi ]i∈I is a dimonoid. Therefore, it suffices to verify the axioms of a dimonoid only in the case when the elements from D appear on all sides of the axioms once or twice. We have ¯ 3 g) = t3¯cεj d2 g = t2 c2 d2 g t3 (¯cεj d ¯ 3 g, = (t2 c2 d)1 g = (t2 c2 d)3 g = (t3¯cεj d) 15(i) ¯ 3 g) = t3 (c2 d2 g) = t1 (c2 d2 g) Lemma = (t2 c2 d)1 g, t3 (¯cεj d
¯ = t 3 (g 2 c 2 d) = t 1 (g 2 c 2 d) = t2 (g2 c2 d) t 3 (g 3 ¯cεj d) ¯ = (t3 g)3¯cεj d, ¯ = (t1 g)2 c2 d = (t1 g)3¯cεj d 838
¯ = t 3 g 2 cεj d ¯ = t 2 (g 2 c) 2 d Lemma=15(vii) t 2 g 2 c 2 d, t 3 (g 3 ¯cεj d) ¯ 3 (t3 g) = ¯cεj d ¯ 3 (t1 g) = ¯cεj d2 (t1 g) = ¯cεj d2 (t2 g) ¯cεj d ¯ 3 t)3 g, = ¯cεj (d2 t)2 g = ¯cεj d2 t3 g = (¯cεj d 16(ii) ¯ 3 (t3 g) = ¯cεj d ¯ 3 (t1 g) = ¯cεj d2 (t1 g) Lemma = ¯cεj (d2 t)2 g, ¯cεj d ¯ =a ¯εi b2 (t2 c2 d) ¯εi b3 (t2 c2 d) = a a ¯εi b3 (t3¯cεj d)
¯ = (¯ ¯ =a ¯εi (b2 t)2 c2 d = a ¯εi b2 t3¯cεj d aεi b3 t)3¯cεj d, Lemma 16(iii) ¯ =a ¯=a a ¯εi b3 (t3¯cεj d) = a ¯εi b2 t2 c2 d, ¯εi b3 t2 cεj d ¯εi b2 (t2 c)2 d ¯ 3 t) = a ¯εi b3¯cεj d2 t = a ¯εi b2 c2 (d2 t) a ¯εi b3 (¯cεj d
¯ 3 t, =a ¯εi (b2 c2 d)2 t = a ¯εi b2 c2 d3 t = (¯ aεi b3¯cεj d) Lemma 16(iii) ¯ 3 t) = a a ¯εi b3 (¯cεj d ¯εi b3 (c2 d2 t) = a ¯εi b2 (c2 d2 t) = a ¯εi b2 c2 d2 t, ¯ = t3 a t3 (¯ aεi b3¯cεj d) ¯εi b2 c2 d = t2 a2 (b2 c2 d)
¯ = (t3 a ¯ = (t2 a2 b)2 c2 d = (t2 a2 b)3¯cεj d ¯εi b)3¯cεj d, ¯ = t3 a2 b2 cεj d ¯ = t2 (a2 b2 c)2 d Lemma=15(vii) t2 a2 b2 c2 d. aεi b3¯cεj d) t3 (¯ Verify, for example, the axiom (x y) z = x (y z): ¯ 3 g = t2 cεj d ¯ 3 g = t2 cεj d2 g = t3¯cεj d2 g = t3 (¯cεj d ¯ 3 g), (t3¯cεj d) ¯ = (t1 g)3¯cεj d ¯ = (t1 g)2 c2 d (t3 g)3¯cεj d Lemma 15(iii)
=
¯ t1 (g2 c2 d) = t3 (g2 c2 d) = t3 (g3¯cεj d),
¯ 3 t)3 g = (c2 d2 t)3 g = (c2 d2 t)1 g (¯cεj d Lemma 15(iv)
=
¯ 3 (t1 g) = ¯cεj d ¯ 3 (t3 g), c2 d2 (t1 g) = ¯cεj d
¯ = (a2 b2 t)3¯cεj d ¯ = (a2 b2 t)2 c2 d (¯ aεi b3 t)3¯cεj d Lemma 15(v)
=
¯ a2 b2 (t2 c2 d) = a ¯εi b3 (t2 c2 d) = a ¯εi b3 (t3¯cεj d),
¯ 3 t = a2 b2 cεj d ¯ 3 t = a2 b2 cεj d2 t (¯ aεi b3¯cεj d) ¯ 3 t), =a ¯εi b3¯cεj d2 t = a ¯εi b3 (¯cεj d ¯ = t2 aεi b3¯cεj d ¯ = t2 aεi b2 c2 d (t3 a ¯εi b)3¯cεj d ¯ = t3 a ¯εi b2 c2 d = t3 (¯ aεi b3¯cεj d). The remaining two axioms may be verified analogously, using Lemmas 15(ii),(vi) and 16(i),(iv). Thus, VεD,I is a dimonoid. From the definitions of operations it follows that each element εi , i ∈ I, is a bar-unit. i with the set of bar-units {εi }i∈I . Hence, the dimonoid (D, 1 , 1 ) is a subdimonoid of the dimonoid VεD,I i The theorem is proved. Corollary 2. Let A be a dialgebra. Then there exists a dialgebra with a set of bar-units containing A as a subdialgebra. From the results of the article it follows that there exist these dimonoids that we can adjoin a set of bar-units to them (Lemma 3 and Theorems 2 and 3) and it is impossible either to adjoin bar-units (Theorems 4, 6, 8, and 10). In our opinion, in connection with these facts the question naturally appears of finding necessary and sufficient conditions for the possibility of adjoining a set of bar-units to an arbitrary dimonoid. 839
References 1. Loday J.-L., “Dialgebras,” in: Dialgebras and Related Operads, Springer-Verlag, Berlin, 2001, pp. 7–66 (Lecture Notes in Math.; 1763). 2. Zhuchok A. V., “Dimonoids,” Algebra and Logic, 50, No. 4, 323–340 (2011). 3. Bokut L. A., Chen Y.-Q., and Liu C.-H., “Gr¨ obner–Shirshov bases for dialgebras,” Int. J. Algebra Comput., 20, No. 3, 391–415 (2010). 4. Kolesnikov P. S., “Varieties of dialgebras and conformal algebras,” Siberian Math. J., 49, No. 2, 257–272 (2008). 5. Pozhidaev A. P., “0-Dialgebras with bar-unity and nonassociative Rota–Baxter algebras,” Siberian Math. J., 50, No. 6, 1070–1080 (2009). 6. Frabetti A., “Dialgebra (co)homology with coefficients,” in: Dialgebras and Related Operads, Springer-Verlag, Berlin, 2001, pp. 67–103 (Lect. Notes Math.; V. 1763). 7. Zhuchok A. V., “Free dimonoids,” Ukrainian Math. J., 63, No. 2, 196–208 (2011). 8. Zhuchok A. V., “Free rectangular dibands and free dimonoids,” Algebra Discrete Math., 11, No. 2, 92–111 (2011). 9. Zhuchok A. V., “Free commutative dimonoids,” Algebra Discrete Math., 9, No. 1, 109–119 (2010). 10. Zhuchok A. V., “Free n-nilpotent dimonoids,” Algebra Discrete Math., 16, No. 2, 299–310 (2013). 11. Zhuchok A. V., “Free n-dinilpotent dimonoids,” Probl. Fiz. Mat. Tekhn., 17, No. 4, 43–46 (2013). 12. Zhuchok A. V., “Commutative dimonoids,” Algebra Discrete Math., No. 3, 116–127 (2009). 13. Phillips J. D., “A short basis for the variety of digroups,” Semigroup Forum, 70, 466–470 (2005). 14. Zhuchok A. V., “Adjoining bar-units to dimonoids,” in: Abstracts: Intern. Algebraic Conf. Dedicated to the 100th Anniversary of L. A. Kaluzhnin, Kyiv, 2014, p. 100. A. V. Zhuchok Luhansk Taras Shevchenko National University, Starobelsk, Ukraine E-mail address: zhuchok
[email protected]
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