A Steiner loop, or a sloop, is a grupoid (L; · ,1), where · is a binary operation and 1 is a constant, satisfying the identities 1 · x = x, x · y = y · x, x · (x · y) = y. There is a one-to-one correspondence between Steiner triple systems and finite sloops. Two constructions of free objects in the variety of sloops are presented in this paper. They both allow recursive construction of a free sloop with a free base X, provided that X is recursively defined set. The main results besides the constructions are: Each subsloop of a free sloop is free two. A free sloop S with a free finite bases X, |X| ≥ 3, has a free subsloop with a free base of any finite cardinality and a free subsloop with a free base of cardinality ω as well; also S has a (non free) base of any finite cardinality k ≥ |X|. We also show that the word problem for the variety of sloops is solvable, due to embedding property

Ana Sokolova

Smile Markovski

English

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GLASNIK MATEMATIČKIVol. 36(56)(2001), 85 – 93FREE STEINER LOOPSSmile Markovski, Ana SokolovaFaculty of Sciences and Mathematics, Republic of MacedoniaAbstract. A Steiner loop, or a sloop, is a groupoid (L; ·, 1), where ·is a binary operation and 1 is a constant, satisfying the identities 1 · x =x, x ·y = y ·x, x ·(x ·y) = y. There is a one-to-one correspondence betweenSteiner triple systems and finite sloops.Two constructions of free objects in the variety of sloops are presentedin this paper. They both allow recursive construction of a free sloop witha free base X, provided that X is recursively defined set. The main resultsbesides the constructions, are: Each subsloop of a free sloop is free too. Afree sloop S with a free finite base X, |X| ≥ 3, has a free subsloop with afree base of any finite cardinality and a free subsloop with a free base ofcardinality ω as well; also S has a (non free) base of any finite cardinalityk ≥ |X|. We also show that the word problem for the variety of sloops issolvable, due to embedding property.1. PreliminariesA Steiner loop, or a sloop, is an algebra (L; ·, 1), where · is a binary op-eration and 1 is a constant, that satisfies the following identities(S1) 1 · x = x(S2) x · y = y · x(S3) x · (x · y) = yA Steiner triple system (STS) is a pair (L,M) where L is a finite set,M is a set containing three-element subsets of L with the property that forany a, b ∈ L (a 6= b) there is a unique c ∈ L such that {a, b, c} ∈ M . It isevident that any STS on a set L enables a construction of a sloop on the setL∪{1} where 1 /∈ L, and vice versa. So, there is a one-to-one correspondencebetween Steiner triple systems and finite sloops (see [4], [7]).2000 Mathematics Subject Classification. 08B20,05B07,08A50.Key words and phrases. free algebra, sloop, Steiner loop, variety, word problem.8586 SMILE MARKOVSKI, ANA SOKOLOVAThe class of involutory commutative loops is defined by the laws:1x = x, xx = 1, xy = yx, ∀x∀y∃!z∃!u(xz = y ∧ ux = y).Proposition 1. The variety of sloops is a proper subvariety of the classof involutory commutative loops.Proof. If (L; ·, 1) is a sloop then the equation ax = b for any a, b ∈ Lhas a unique solution x = ab. What follows is an example of an involutorycommutative loop which is not a sloop:· 1 a b c d e1 1 a b c d ea a 1 c d e bb b c 1 e a dc c d e 1 b ad d e a b 1 ce e b d a c 1Further on we use the term base for a minimal generating set of an algebra,and free base for a base of an algebra in a given variety which has the universalmapping property. So, a set X is a free base of a sloop S = (S; ·, 1) iff X isits base and each mapping from X to L, where L = (L; ·, 1) is a sloop, can beextended to a homomorphism from S into L.2. Free sloops - construction 1Let X be a given set. We define a chain of sets Xi and a set FX by:X1 := X, Xi+1 := Xi ∪ {{u, v} ⊆ Xi| u 6= v, u /∈ v, v /∈ u},FX := (∪ (Xi| i ≥ 1)) ∪ {1} where 1 /∈ ∪ (Xi| i ≥ 1).Proposition 2. An element x ∈ Xi+1 \ Xi iff x = {u, v} for someuniquely determined u and v such that u ∈ Xi \Xi−1 or v ∈ Xi \Xi−1.Define an operation ∗ on FX as follows. If u, v ∈ FX \ {1} thenu ∗ v :={u, v} u 6= v, u /∈ v, v /∈ u1 u = vt v = {u, t} or u = {v, t}and 1 ∗ u := u, u ∗ 1 := u, 1 ∗ 1 := 1.Theorem 2.1. FX = (FX ; ∗, 1) is a free object in the variety of sloopswith free base X.Proof. The commutativity is obvious. We check the identity u∗(u∗v) =v in the following cases.FREE STEINER LOOPS 871) u 6= v, u /∈ v, v /∈ u : u ∗ (u ∗ v) = u ∗ {u, v} = v,2) v = {u, t} : u ∗ (u ∗ v) = u ∗ t = {u, t} = v,3) u = {v, t} : u ∗ (u ∗ v) = u ∗ t = v.In every other case, the statement is straightforward. So, FX is a sloop.It is clear that X is a base of FX and it is a free one too. Namely, let(L; ·, 1) be a sloop and φ : X −→ L a mapping. Define inductively a chain ofmappings (φi : Xi −→ L | i ≥ 1) as follows. φ1 = φ and if φi is defined, thenfor x ∈ Xi+1,φi+1(x) :={φi(x) x ∈ Xiφi(u) · φi(v) x = {u, v} ∈ Xi+1 \XiBy Proposition 2, φi is well defined for each i ≥ 1.Let φ∗ := ∪(φi | i ≥ 1) ∪ {(1, 1)}. In order to prove that φ∗ is a homo-morphism we consider the following cases.1) u 6= v, u /∈ v, v /∈ u (u, v ∈ Xi for some i ≥ 1) : φ∗(u ∗ v) =φ∗({u, v}) = φi+1({u, v}) = φi(u) · φi(v) = φ∗(u) · φ∗(v).2) u = {v, t} ∈ Xi for some i > 1 : φ∗(u ∗ v) = φ∗(t) = φi−1(t) =(φi−1(v) · φi−1(t)) · φi−1(v) = φi({v, t}) · φi−1(v) = φ∗(u) · φ∗(v), since t =v ∗ (t ∗ v) by (S2) and (S3).The remaining cases are trivial.Assuming that the set X is already well ordered (i.e. we work with setsof ZFC set theory (see [8])), we define an order on FX extending the order ofX , by induction on the number of pairs of braces, in the following way.The element 1 is the smallest in FX . If α, β ∈ FX and α has smallernumber of (pairs of) braces than β, then α < β. If {α, β} 6= {γ, δ} ∈ FX ,{α, β}, {γ, δ} have the same number of pairs of braces and α < β, γ < δ,then we set{α, β} < {γ, δ} if either α < γ or α = γ, β < δ and{γ, δ} < {α, β} if either γ < α or α = γ, δ < β.Proposition 3. (FX ,≤) is a well ordered set.Proof. Let A ⊆ FX . If A contains an element without braces, then thesmallest element in (X ∪ {1})∩A is the smallest in A. Else, let k > 0 be thesmallest number of braces of an element of A and A′ = {a ∈ A | the number ofbraces in a is k}. Consider the set A′′ = {u ∈ FX | {u, v} ∈ A′, u < v}. By theinductive hypothesis A′′ has the least element α and A′′′ = {v ∈ FX | {α, v} ∈A′} has the least element β. Then {α, β} is the least element in A′ i.e. A.Note that if X is a recursive set, then FX is recursive too.3. Free sloops - construction 2Here we will present another description of the free sloops by using thefree term algebra TermX = (Term; ·, 1) (i.e. the absolutely free algebra) overa set of free generators X , in the signature ·, 1. Any free sloop with free base88 SMILE MARKOVSKI, ANA SOKOLOVAX can be obtained as a quotient algebra of TermX ([7, 2]). Instead of that,our new construction will use a subset of Term as a universe of a free sloop.Define inductively a mapping d : Term −→ N, where N is the set ofnon-negative integers, by:d(1) := 0, d(x) := 0 for x ∈ X , d(t1 · t2) := d(t1) + d(t2) + 1.We shall refer to d(t) as weight of the term t ∈ Term.By induction on weight, define a mapping C : Term −→ FX in thefollowing way:C(t) :=1 t = 1 or t = t1 · t2, C(t1) = C(t2)t t ∈ XC(t1) t = t1 · t2, C(t2) = 1C(t2) t = t1 · t2, C(t1) = 1C(t3) t = t1 · t2, C(t2) = {C(t1), C(t3)} ort = t1 · t2, C(t1) = {C(t2), C(t3)}{C(t1), C(t2)} t = t1 · t2 and none of the previous holdsProposition 4. The mapping C is an epimorphism from TermX ontoFX .Now, by the homomorphism theorem we have TermX/kerC ∼= FX . Fur-ther on we will determine a canonical representative for each congruence classas follows.Assuming that X is a well ordered set we define a mapping T : FX −→Term using the well ordering of FX , by:T (1) := 1, T (x) := x for x ∈ X, T ({u, v}) := T (u) ·T (v) where u < v.Proposition 5. T is injective.Proposition 6. TCT = T, CTC = C.Proof. Let α ∈ FX . If α = 1 or α ∈ X , the statement holds trivially.Let α = {u, v}, u, v ∈ FX , u < v. Assume that the statement holds forany element of FX smaller than α. So, TCT (u) = T (u), TCT (v) = T (v).Then CT (u) = u, CT (v) = v by Proposition 5, and since α ∈ FX we haveCT (u) 6= CT (v), CT (u) /∈ CT (v), CT (v) /∈ CT (u). Hence, TCT (α) =TC(T (u) · T (v)) = T ({CT (u), CT (v)}) = T ({u, v}) = T (α).Now, CTC = C follows by Proposition 5 and TCT = T .For an element t ∈ Term we say that it is reduced if TC(t) = t. Themapping R = TC will be called reduction. Note that (R(t), t) ∈ kerC andin each congruence class there is only one reduced element which will be thecanonical representative of the class.The mapping R has the following properties.Proposition 7. Rn = R, for each n ≥ 2, and for all t, s ∈ Term wehave:(i) R(1 · t) = R(t);FREE STEINER LOOPS 89(ii) R(t · s) = R(s · t);(iii) R(t · (t · s)) = R(s);(iv) R(t · s) = t · s =⇒ R(t) = t, R(s) = s;(v) R(R(t) · s) = R(t · s);(vi) R(R(t) ·R(s)) = R(t · s).Proof. Rn = R for n ≥ 2 follows from Proposition 6 . (i), (ii) and (iii)are straightforward since C is a homomorphism and FX is a sloop, and (vi)is a consequence of (ii) and (v).(iv) R(t · s) = TC(t · s) = t · s implies that C(t · s) = {α, β} whereα < β, T (α) = t, T (β) = s. Now, TC(t) = TCT (α) = T (α) = t byProposition 6 , and in the same way TC(s) = s.(v) Since CR(t) = CTC(t) = C(t) we haveC(R(t) · s) =1 C(t) = C(s)C(s) C(t) = 1C(t) C(s) = 1C(l) C(t) = {C(s), C(l)} or C(s) = {C(t), C(l)}{C(t), C(s)} otherwisei.e. C(R(t) · s) = C(t · s) and hence TC(R(t) · s) = TC(t · s).Let GX be the set of reduced terms i.e. GX = R(Term) = T (FX). Definean operation ◦ on GX byt ◦ s := R(t · s) for all t, s ∈ GX .Theorem 3.1. GX = (GX ; ◦, 1) is a free sloop with free base X.Proof. We will prove that the bijective mapping T is an isomorphismbetween (FX ; ∗, 1) and (GX ; ◦, 1). For each t, s ∈ TermX by Proposition 4and 5 we have that t/kerC · s/kerC = (t · s)/kerC = (R(t · s))/kerC =R(R(t) · R(s))/kerC = (R(t) ◦ R(s))/kerC. Since TermX/kerC ∼= FX weobtain C(t) ∗ C(s) = C(R(t) ◦ R(s)) and if u = C(t), v = C(s), then T (u ∗v) = T (C(t) ∗ C(s)) = T (C(R(t) ◦ R(s))) = R(R(t) ◦ R(s)) = R(t) ◦ R(s) =TC(t) ◦ TC(s) = T (u) ◦ T (v).Note that if X is a recursive set, then since C and T are recursivelydefined, we have that GX is a recursive set too.4. Some properties of free sloopsProposition 8. If X is a free base of a free sloop S, then S is finite ifand only if |X | ≤ 2.Proof. S = {1} for X = ∅, S = {1, a} for X = {a} and S = {1, a, b, ab}for X = {a, b}. If X = {a, b, c, . . .} where |{a, b, c}| = 3, consider the setM = {xi | i ≥ 1} where x1 = a, x2 = b, x2n+1 = ax2n, x2n+2 = cx2n+1 forn ≥ 1. We have M ⊆ S, and M is infinite since xi 6= xj for i 6= j.90 SMILE MARKOVSKI, ANA SOKOLOVATheorem 4.1. Every subsloop of a free sloop is free too.Proof. Let GX = (GX ; ◦, 1) be a free sloop as in Construction 2, andlet G′ be a subsloop of GX . Recall that R(t) = t for each t ∈ GX .If x, y ∈ G′ \{1}, then we say that x is a divisor of y if and only if there isa t ∈ Term \ {1} such that y = t · x or y = x · t. Then also t ∈ G′ \ {1}, sinceby Proposition 7, (iv), the definition of ◦ and (S2), (S3) we have t ∈ GX andt = x ◦ y. Note that if x is a divisor of y then d(x) < d(y), which implies thatany sequence t1, t2, . . . , tn, . . . such that ti+1 is a divisor of ti, i ≥ 1, is finite.We shall prove that B = {t ∈ G′ \ {1}| t has no divisors} is a free basefor G′.At first, by an induction on weight we show that B is a generating set ofG′. Let z ∈ G′ \ {1}. If z /∈ B, then z has divisors, i.e. z = x · y for somex, y ∈ G′ and z = R(z) = R(x · y) = x ◦ y. By the inductive hypothesis x andy are generated by B and so is z.Next we show that B is a base of G′. Namely, let b ∈ B and let G′′ bethe subsloop of G′ generated by B \ {b}. Then G′′ = ∪(G′′i | i ≥ 1) whereG′′1 = B \ {b}, G′′i+1 = {t ◦ s | t, s ∈ G′′i }. Now, b /∈ G′′2 since if b = t ◦ s forsome t, s ∈ G′′1 = B \ {b}, then b = t · s. If b ∈ G′′i+1 \G′′i for some i ≥ 2, thenb = t ◦ s for some t, s ∈ G′′i such that t ∈ G′′i \G′′i−1 (or s ∈ G′′i \G′′i−1). Wehave to consider several cases. The case t = s is not possible, since t ◦ t = 1and b 6= 1. If t = s ◦ u (or s = t ◦ u) for some u ∈ G′′i−1, then b = u ∈ G′′i−1.The only case left is b = t · s, contradicting b ∈ B.Let (L; ∗, 1) be an arbitrary sloop and f : B → L a mapping. We extendf to homomorphism f ′ : G′ → L by an induction on weight in the followingway: f ′(1) := 1, f ′(b) := b for each b ∈ B, f ′(t) := f ′(x) ∗ f ′(y) whent = x · y ∈ G′ \B.Then for any t, s ∈ G′ we have:t ◦ s = R(t · s) =1 t = ss t = 1t s = 1l s = t · l or s = l · t ort = s · l or t = l · st · s if none of the previousholds and C(t) < C(s)s · t otherwiseIn all of the cases listed, from the definition of f ′ and the fact that (L; ∗, 1) isa sloop, it follows that f ′(t ◦ s) = f ′(t) ∗ f ′(s).Corollary 1. Every free sloop with at least 3 element free base has afree subsloop with infinite free base, and a free subsloop with free base of anyfinite cardinality.FREE STEINER LOOPS 91Proof. Let GX be the free sloop with free base X obtained by theconstruction 2, and let a, b, c ∈ X . Let M = {xi | i ≥ 1} ⊆ Term, wherex1 = ab, x2 = ac, x2n+1 = (x2n−1c)(x2nb), x2n+2 = (x2nb)(x2n+1c). Let G′be the subsloop of GX generated by M . Since M is the set of elements of G′that have no divisors, by Theorem 3 we have that G′ is a free subsloop of GXwith infinite free base M . Out of the same reason, if K = {x1, x2, . . . , xk} ⊂M , then the subsloop of GX generated by K is a free one with k-element freebase K.Proposition 9. A free sloop with free base X, |X | ≥ 3, has infinitelymany free bases.Proof. Let X = {a, b, c} be a free base of a free sloop S. Denote asequence of elements of S by b0 = b, b2k+1 = ab2k, b2k+2 = cb2k+1, k ≥ 0.Then Xi = {a, bi, c} is a free base of S as well.The variety of sloops has nontrivial finite algebras, so there are no twoisomorphic free sloops with finite free bases of different cardinality [7]. Nev-ertheless, we will show that any free sloop with finite base X, |X | ≥ 3, has abase of any finite cardinality greater than |X |. Namely, it is a consequence ofthe following property, where GX denotes the free sloop of construction 2.Proposition 10. If X = {b1, b2, b3, . . . , bk}, k ≥ 3, is a base of GX ,then GX has also a base {b1, b2, b′3, b′′3 , b4, . . . , bk}, whereb′3 = (b1 · (b2 · b3)) · (b2 · (b1 · b3)), b′′3 = b′3 · b3.Proof. Let S be the subsloop of GX generated by {b1, b2, b3}, and letS′ be the subsloop of GX generated by {b1, b2, b′3, b′′3}. Since b3 = b′3 · b′′3 , it isclear that S′ = S. We shall prove that {b1, b2, b′3, b′′3} is a base for S′.Let S′′ be the subsloop of GX generated by {b1, b2, b′3}. We shall provethat b3 /∈ S′′.For this purpose, first note that S ′′ = ∪(S′′i | i ≥ 1) where S′′1 = {b1, b2, b′3}and S′′i+1 = S′′i ∪ {x ◦ y | x, y ∈ S′′i }.It is clear that b3, t1 = b1·(b2 ·b3), t2 = b2·(b1 ·b3) /∈ S′′2 . Let b3, t1, t2 /∈ S′′i .Then b3 /∈ S′′i+1 since in order to extract b3, b′3 must be multiplied by t1 or t2.Also, since b3 /∈ S′′i we have t1, t2 /∈ S′′i+1.In a similar manner, it follows that the subsloops of S ′ generated by eachof the sets {b1, b2, b′′3}, {b2, b′3, b′′3}, {b1, b′3, b′′3} are proper subsets of S ′.5. The word problem for sloopsWe show that the word problem for the variety of sloops is solvable.Namely, we use the following T. Evans’ result ([3]):If V is a variety with the property that any incomplete V -algebra can beembedded in a V -algebra, then the word problem is solvable for V .According to Evans’ definition of incomplete algebras, an incomplete sloopwith universe G is a quadruple (G, ·, 1, D), where D ⊆ G2, 1 ∈ G, · : D → G92 SMILE MARKOVSKI, ANA SOKOLOVAis a mapping (called an incomplete operation on G), satisfying the followingconditions:(IS1) (x, x) ∈ D =⇒ x · x = 1(IS2) (x, y) ∈ D =⇒ (y, x) ∈ D, x · y = y · x(IS3) (x, 1) ∈ D =⇒ x · 1 = x(IS4) (x, y) ∈ D =⇒ (x, x · y) ∈ D, x · (x · y) = yProposition 11. Any incomplete sloop can be embedded into a sloop.Proof. Let (G, ·, 1, D) be an incomplete sloop. Denote G0 = G, D0 =D ∪ {(x, x)|x ∈ G} ∪ {(1, x), (x, 1)|x ∈ G} and let ·0 : D0 → G be defined byx ·0 y := x · y, for (x, y) ∈ D, x ·0 x := 1, x ·0 1 := x, 1 ·0 x := x for x ∈ G. Then(G0, ·0, 1, D0) is an incomplete sloop such that D ⊆ D0 ⊆ G20.If (Gi, ·i, 1, Di) is defined incomplete sloop, we form a new one as follows.Denote Ci = {{x, y}|x, y ∈ Gi, (x, y) /∈ Di} and put Gi+1 = Gi ∪ Ci(assuming that Ci ∩Gi = ∅). Define an incomplete operation ·i+1 by:(x, y) ∈ Di =⇒ x ·i+1 y := x ·i y,(x, y) ∈ G2i \Di =⇒ x ·i+1 y := {x, y},x ∈ Gi+1 =⇒ x ·i+1 x := 1, x ·i+1 1 := 1 ·i+1 x := x,x ∈ Gi, {x, y} ∈ Ci =⇒ x ·i+1 {x, y} := y, {x, y} ·i+1 x := y.Let Di+1 be the set of all (x, y) ∈ Gi+1 for which x ·i+1 y is defined.It is clear that (IS1) - (IS3) hold for (Gi+1, ·i+1, 1, Di+1). Several caseshave to be considered in order to check (IS4) and the nontrivial ones are:(x, y) ∈ G2i \Di =⇒ x·i+1y = {x, y} =⇒ x·i+1 (x·i+1y) = x·i+1{x, y} =y;x ∈ Gi, y = {x, z} ∈ Ci =⇒ x ·i+1 y = x ·i+1 {x, z} = z =⇒x ·i+1 (x ·i+1 y) = x ·i+1 z = {x, z} = y.That way we obtained chains of sets (Gi| i ≥ 0), (Di| i ≥ 0), (·i| i ≥ 0),with the properties:Gi ⊆ Gi+1, Di ⊆ G2i ⊆ Di+1, ·i ⊆ ·i+1.LetG∗ =⋃i≥0Gi, D∗ =⋃i≥0Di, ·∗ =⋃i≥0·i.Now for x, y ∈ G∗, there exists i ≥ 0 such that x, y ∈ Gi, so (x, y) ∈ Di+1, i.e.(x, y) ∈ D∗. Hence, D∗ = (G∗)2 i.e. (G∗, ·∗, 1) is a sloop in which (G, ·, 1, D)is embedded.As a corrolary of Proposition 11 and [3] we get the following result.Theorem 5.1. The word problem for the variety of sloops is solvable.FREE STEINER LOOPS 93References[1] R. U. Bruck: A survey of Binary Systems, Berlin - Götingen - Heidelberg, 1958[2] Ǵ. Čupona, S. Markovski: Primitive varieties of algebras, Algebra universalis 38(1997), 226-234[3] Trevor Evans: The word problem for abstract algebras , The Journal of The LondonMathematical Society, vol. XXVI, 1951, 64-71[4] P. M. Hall: Combinatorial Theory, Blaisdell publishing company, Walthand Mas-sachusetts, Toronto, London, 1967[5] S. Markovski, A. Sokolova: Free Basic Process Algebra, Contributions to GeneralAlgebra, vol. 11, 1998, 157-162[6] S. Markovski, A. Sokolova: Term rewriting system for solving the word problem forsloops, Matematički bilten 24(L), 2000, 7–18[7] R. N. McKenzie, W. F. Taylor, G. F. McNulty: Algebras, Lattices, Varieties,Wadsworth & Brooks, Monterey, California, 1987[8] J. R. Shoenfield: Mathematical Logic, Addison-Wesley Publ. Comp., 1967Faculty of Sciences and Mathematics,Institute of Informatics, p.f.162 Skopje,Republic of MacedoniaE-mail : smile,anas@pmf.ukim.edu.mkReceived : 07.07.1999.Revised : 29.02.2000.

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