In his paper "On a construction of semigroups", M. Kolibiar gives a
construction for a semigroup $T$ (beginning from a semigroup $S$) which is said
to be derived from the semigroup $S$ by a $\theta$-construction. He asserted
that every semigroup $T$ can be derived from the factor semigroup $T/\theta
(T)$ by a $\theta$-construction, where $\theta (T)$ is the congruence on $T$
defined by: $(a, b)\in \theta (T)$ if and only if $xa=xb$ for all $x\in T$.
Unfortunately, the paper contains some incorrect part. In our present paper we
give a revision of the paper.Comment: Version v5 differs from the published and the earlier versions in the
  following: some part of the proof of Theorem 2 are clarified, and some typing
  errors are correcte

Nagy, Attila

Periodica Mathematica Hungarica

English

arXiv

Attila Nagy

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Remarks on the paper “M. Kolibiar, on a construction of semigroups”

In his paper ”On a construction of semigroups”, M. Kolibiar gives a construction for a semigroup T (beginning from a semigroup S) which is said to be derived from the semigroup S by a θ-construction. He asserted that every semigroup T can be derived from the factor semigroup T /θ(T) by a θ-construction, where θ(T) is the congruence on T defined by: (a,b) ∈ θ(T) if and only if xa = xb for all x ∈ T . Unfortunately, the paper contains some incorrect part. In our present paper we give a revision of the paper

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arXiv:1504.07181v5  [math.GR]  29 Jan 2016REMARKS ON THE PAPER1”M. KOLIBIAR, ON A CONSTRUCTION OF SEMIGROUPS”Attila NagyDepartment of AlgebraBudapest University of Technology and EconomicsP.O. Box 911521 BudapestHungaryAbstractIn his paper ”On a construction of semigroups”, M. Kolibiar gives aconstruction for a semigroup T (beginning from a semigroup S) which issaid to be derived from the semigroup S by a θ-construction. He assertedthat every semigroup T can be derived from the factor semigroup T/θ(T )by a θ-construction, where θ(T ) is the congruence on T defined by: (a, b) ∈θ(T ) if and only if xa = xb for all x ∈ T . Unfortunately, the paper containssome incorrect part. In our present paper we give a revision of the paper.A semigroup S is called a left reductive semigroup ([1], [4]) if, for everya, b ∈ S, whenever xa = xb holds for all x ∈ S then a = b. It is known that, foran arbitrary semigroup S, the relation θ(S) defined by (a, b) ∈ θ(S) for somea, b ∈ S if and only if xa = xb for all x ∈ S is a congruence on S. In [3], theauthor examined this congruence. He defined a sequence θn (n = 1, 2, . . . ) ofcongruences on an arbitrary semigroup S as follows: θ1 = θ(S), and if θn isgiven, θn+1 is the congruence relation on S, induced by the congruence relationθ(S)/θn ([2]). In Lemma 2 of [3] it is shown that (a, b) ∈ θn for some a, b ∈ S ifand only if xa = xb for all x ∈ Sn.In Theorem 1 of [3], it is asserted that θ∗ = ∪∞n=1θn is the least elementin the set of all congruence relations θ on S such that S/θ is a left reductivesemigroup. The proof of this theorem is not correct. The author asserts that,from the result (xa, xb) ∈ θ∗ for all x ∈ S, it follows that txa = txb for all x ∈ Sand all t ∈ Sk for some k ∈ N (that is (xa, xb) ∈ θk for all x ∈ S). This is notcorrect, because (xa, xb) ∈ θ∗ for all x ∈ S means that, for every x ∈ S, there isa positive nx such that (xa, xb) ∈ θnx . From this it does not follow that thereis a positive integer k such that (xa, xb) ∈ θk for all x ∈ S.Theorem 2 of [3] is also incorrect. In part (a) of Theorem 2, the followingconstruction is given: For a semigroup S and each x ∈ S, let Tx 6= ∅ be a setsuch that Tx ∩ Ty = ∅ if x 6= y. Let a mapping fyx : Tx 7→ Ty be given for allcouples x, y ∈ S such that y = xu (in the paper y = ux, but it is a missprint)for some u ∈ S. Suppose further fzy ◦ fyx = fzx . Given a, b ∈ ∪x∈STx = T , seta◦b = fxyx (a), where a ∈ Tx, b ∈ Ty. Then (T ; ◦) is a semigroup (this semigroup1This version differs from the published one that we clarified some part of the proof of The-orem 2, and corrected the typing errors. Key words: semigroups, left reductive semigroups,congruences; AMS Classification: 20M10; e-mail: nagyat@math.bme.hu1is said to be derived from S by a θ-construction), and each set Tx is containedin a θ-class of T . If S is left reductive then the θ-classes of T are exactly the setsTx (x ∈ S). The assertions of this part of the theorem is correct. In part (b) ofTheorem 2, it is asserted that, for an arbitrary semigroup T , if S denotes thefactor semigroup T/θ(T ) and T[x]θ denotes the θ(T )-class [x]θ of T containingx, then f[xy]θ[x]θ : a 7→ ab (a ∈ [x]θ, b ∈ [y]θ) is a mapping on [x]θ to [xy]θ and,for all c, d ∈ T , cd = f[cd]θ[c]θ (c). This last equation means that cd = c ◦ d forall c, d ∈ S. The next example shows that the assertion cd = f[cd]θ[c]θ (c) for allc, d ∈ T is not correct.Example 1 Let T = {e, a, u, v, 0} be a semigroup defined by Table 1.· e a u v 0e e a 0 0 0a a e 0 0 0u u v 0 0 0v v u 0 0 00 0 0 0 0 0Table 1:Let θ denote the congruence θ(T ). It is easy to see that the factor semigroupS = T/θ is a left reductive semigroup. We note that S is a semigroup which canbe obtained from a two-element group {[e]θ, [a]θ} by adjunction the zero [0]θ.Consider the semigroup (T ; ◦) which is derived from T/θ by a θ-construction(see (b) of Theorem 2 of [3]). As [0]θ = [0]θ[0]θ, [0]θ = [0]θ[e]θ, [0]θ = [0]θ[a]θ,we have three possibility to define a mapping f[0]θ[0]θ from [0]θ into [0]θ.In case f[0]θ[0]θ x 7→ x · 0 = 0 for all x ∈ [0]θ, we haveu ◦ e = f[u·e]θ[u]θ (u) = f[0]θ[0]θ (u) = 0 6= u = u · e.In case f[0]θ[0]θ x 7→ x · e for all x ∈ [0]θ, we getu ◦ a = f[u·a]θ[u]θ (u) = f[0]θ[0]θ (u) = u · e = u 6= v = u · a.In case f[0]θ[0]θ x 7→ x · a for all x ∈ [0]θ, we haveu ◦ e = f[u·e]θ[u]θ (u) = f[0]θ[0]θ (u) = u · a = v 6= u = u · e.Under any possible choosing of the mapping f[0]θ[0]θ we have a contradiction.Thus T can not be derived from T/θ by a θ-construction.In the next we formulate a new and correct version of Theorem 2 of [3].2Theorem 2 (a) Let S be a semigroup. For each x ∈ S, associate a set Tx 6=∅. Assume that Tx ∩ Ty = ∅ for all x 6= y. Assume that, for each triple(x, y, xy) ∈ S × S × S is associated a mapping f(x,y,xy) of Tx into Txy actingon the left. Suppose further f(xy,z,xyz) ◦ f(x,y,xy) = f(x,yz,xyz) for all triples(x, y, z) ∈ S × S × S. Given a, b ∈ ∪x∈STx = T , set a ◦ b = f(x,y,xy)(a), wherea ∈ Tx and b ∈ Ty. Then (T ; ◦) is a semigroup, and each set Tx is contained ina θ(T )-class of T . If S is left reductive then the θ(T )-classes of the semigroupT are exactly the sets Tx (x ∈ S).(b) Let (T ; ·) be a semigroup. Let θ denote the congruence θ(T ), and letS = T/θ. For an element [x]θ ∈ S, let T[x]θ be the θ-class [x]θ of T . Forarbitrary triple ([x]θ, [y]θ, [x · y]θ) ∈ S × S × S, let f([x]θ,[y]θ,[x·y]θ) : a 7→ a · b(a ∈ [x]θ), where b is an arbitrary element of [y]θ. (We note that a · b =a · [b]θ for all a, b ∈ T . We also note that we consider all of the mappingsf([x]θ,[z]θ,[x·z]θ), where [z]θ ∈ S satisfies [x]θ[y]θ = [x]θ[z]θ.) For all c, d ∈ T ,let c ◦ d = f([c]θ,[d]θ,[c·d]θ)(c). Then (T ; ◦) is a semigroup which is isomorphic tothe semigroup (T ; ·). Consequently, every semigroup (T ; ·) is isomorphic to asemigroup derived from the semigroup S = T/θ using the construction in part(a) of the theorem.Proof. (a): Let (x, y, z) ∈ S × S × S be an arbitrary triple and let a ∈ Tx,b ∈ Ty, c ∈ Tz be arbitrary elements. Thena ◦ (b ◦ c) == a◦f(y,z,yz)(b) = f(x,yz,xyz)(a) = (f(xy,z,xyz)◦f(x,y,xy))(a) = (f(x,y,xy)(a))◦c == (a ◦ b) ◦ c.Thus (T ; ◦) is a semigroup. Let x ∈ S be arbitrary. We show that Tx is in aθ(T )-class of T . If a, b ∈ Tx are arbitrary elements then, for every y ∈ S and forall t ∈ Ty,t ◦ a = f(y,x,yx)(t) = t ◦ bwhich implies that (a, b) ∈ θ(T ).Assume that S is left reductive. Let Tx and Ty be sets such that they are inthe same θ(T )-class of T . Let a ∈ Tx and b ∈ Ty be arbitrary elements. Then,for every z ∈ S and t ∈ Tz, t ◦ a = t ◦ b ∈ Tzx ∩ Tzy. Thus zx = zy for everyz ∈ S. As S is left reductive, we get x = y. Hence the θ(T )-classes of T areexactly the sets Tx (x ∈ S).(b): It is clear that, for every x, y ∈ T , f([x]θ,[y]θ,[x·y]θ) maps [x]θ = T[x]θ intoT[x·y]θ = [x · y]θ. For every (x, y, z) ∈ T × T × T and a ∈ [x]θ,(f([x·y]θ,[z]θ,[x·y·z]θ) ◦ f([x]θ,[y]θ,[x·y]θ))(a) == (a · y) · z = a · (y · z) = f([x]θ,[y·z]θ,[x·y·z]θ)(a).Thus (T ; ◦) is a semigroup by part (a) of the theorem.For every x, y ∈ T and every a ∈ [x]θ, b ∈ [y]θ,a ◦ b = f([x]θ,[y]θ,[x·y]θ)(a) = a · b.Thus (T ; ·) is isomophic to the semigroup (T ; ◦). ⊓3Example 3 Let T be the semigroup defined in Example 1. Let θ denote thecongruence θ(T ). Then S = T/θ is a semigroup which can be obtained froma two-element group by adjunction of a zero. Let e, a, 0 denote the θ-classes[e]θ = {a}, [a]θ = {a}, [0]θ = {0, u, v} of T , respectively. Then the Cayley-tableof S ise a 0e e a 0a a e 00 0 0 0Table 2:Apply our construction when the beginning semigroup is S. The sets Tx(x ∈ S) are Te = {e}, Ta = {a}, T0 = {0, u, v}. The mappings (see (b)of Theorem 2) are: f(e,e,e) : e 7→ e, f(e,a,a) : e 7→ a, f(e,0,0) : e 7→ 0,f(a,e,a) : a 7→ a, f(a,a,e) : a 7→ e, f(a,0,0) : a 7→ 0, f(0,e,0) : x 7→ x · e forevery x ∈ [0]θ, f(0,a,0) : x 7→ x · a for every x ∈ [0]θ, f(0,0,0) : x 7→ 0 for everyx ∈ [0]θ. It is easy to see that the semigroup T is isomorphic to the semigroup(T ; ◦) derived from S by applying our construction in Theorem 2.We formulate the new version of Theorem 3 of [3].Theorem 4 Let S be a semigroup whose congruence relations satisfy the as-cending chain condition. Then there are semigroups S0 = S, S1, . . . Sn such thatSn is left reductive, and each Si−1 can be derived from Si (i = 1, . . . , n) by theconstruction defined in (a) of Theorem 2.Proof. Let S be a semigroup. For a congruence ̺ on S, consider the congruence̺∗ defined by (a, b) ∈ ̺∗ for some a, b ∈ S if and only if (xa, xb) ∈ ̺ for all x ∈ S.(see [5]). Let̺(0) ⊆ ̺(1) ⊆ · · · ⊆ ̺(n) ⊆ . . .be a sequence of congruences on S defined by ̺(0) = ̺ and, for every nonnegativeinteger i, let ̺(i+1) = (̺(i))∗ (see [5]). By Lemma 2 of [3], ι(i)S = θi for everypositive integer i, where ιS denotes the identity relation on S and θi are thecongruences examined above. As the congruence relations on S satisfy theascending chain condition, there is a (least) nonegative integer n such thatι(n)S = ι(n+1)S = · · · . Let Si = S/ι(i)S for every i = 0, 1, . . . , n. Thus S = S0. ByTheorem 1 of [5], ι(n)S = θn is a left reductive congruence which means that thefactor semigroup Sn = S/θn is left reductive. For every i = 1, . . . , n,Si = S/ι(i)S∼= (S/ιi−1S )/(ι(i)S /ι(i−1)S )∼= Si−1/(ι(i)S /ιi−1)S )∼= Si−1/((ι(i−1)S )∗/ι(i−1)S )using also Theorem 5.6 of [2]. By Lemma 7 of [5],(ι(i−1)S )∗/ι(i−1)S = (ιSi−1)∗ = θ(Si−1).4ThusSi ∼= Si−1/θ(Si−1)for every i = 1, . . . , n. By (b) of Theorem 2, the semigroup Si−1 (i = 1, . . . , n)can be derived from Si by the construction in (a) of Theorem 2. ⊓An addendum: It may be that the semigroup Sn = S/θn in Theorem 4has only one element, that is, θn is the universal relation ωS on S. The nexttheorem is about this case.Theorem 5 For a semigroup S, θn = ωS for some positive integer n if andonly if S is an ideal extension of a left zero semigroup by a nilpotent semigroup.Proof. Let ιS denote the identity relation on a semigroup S. By Theorem 5of [5], ι(n)S = ωS for some non-negative integer n if and only if S is an idealextension of a left zero semigroup by a nilpotent semigroup. As θ(S) = ι(1)S , ourassertion is a consequence of Theorem 5 of [5]. ⊓References[1] A.H. Clifford and G.B. Preston, The Algebraic Theory of Semigroups I,American Mathematical Society, Providence, R. I., 1961[2] J.M. Howie, An Introduction to Semigroup Theory, Academic Press, London,1976[3] M. Kolibiar, On a construction of semigroups, Scripta Fac. Sci. Nat. UjepBrunensis, Arch. Math., 22:99-100, 1971[4] A. Nagy, Special Classes of Semigroups, Kluwer Academic Publishers, Dor-drecht, Boston, London, 2001[5] A. Nagy, Left reductive congruences on semigroups, Semigroup Forum,87:129-148, 20135

Remarks on the paper "M. Kolibiar, On a construction of semigroups"

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