aa r X i v : . [ m a t h . G R ] A p r Lee monoid L is non-finitely based Inna A. Mikhailova ∗ and Olga B. Sapir † Abstract
We establish a new sufficient condition under which a monoid is non-finitely based and apply this condition to show that the 9-element monoid L is non-finitely based. The monoid L was the only unsolved case in the finitebasis problem for Lee monoids L ℓ , obtained by adjoining an identity elementto the semigroup L ℓ generated by two idempotents a and b subjected to therelation 0 = abab · · · (length ℓ ). We also prove a syntactic sufficient conditionwhich is equivalent to the sufficient condition of Lee under which a semigroupis non-finitely based. This gives a new proof to the results of Zhang-Luo andLee that the semigroup L ℓ is non-finitely based for each ℓ ≥ : 20M07, 08B05 Keywords and phrases : Lee monoids, identity, finite basis problem, non-finitely based, variety, isoterm
An algebra is said to be finitely based (FB) if there is a finite subset of its identitiesfrom which all of its identities may be deduced. Otherwise, an algebra is said tobe non-finitely based (NFB). By the celebrated McKenzie’s result [6] the classes ofFB and NFB finite algebras are recursively inseparable. It is still unknown whetherthe set of FB finite semigroups is recursive although a very large volume of work isdevoted to this problem (see the survey [10]).Recently, Lee suggested to investigate the finite basis property of the semigroups L ℓ = h a, b | aa = a, bb = b, ababab · · · | {z } length ℓ = 0 i , ℓ ≥ L ℓ obtained by adjoining an identity element to L ℓ . ∗ email: [email protected], Ural Federal University, Ekaterinburg, Russia The firstauthor was supported by the Russian Foundation for Basic Research, project no. 17-01-00551, theMinistry of Education and Science of the Russian Federation, project no. 1.3253.2017, and theCompetitiveness Program of Ural Federal University. † email: [email protected] L = A is long known to be finitely based [2]. Zhangand Luo proved [11] that the 6-element semigroup L is NFB and Lee generalized thisresult into a sufficient condition [4] which implies that for all ℓ ≥
3, the semigroup L ℓ is NFB [5].As for the monoids L ℓ , the 5-element monoid L was also proved to be FB byEdmunds [1], while the 7-element monoid L is recently shown to be NFB by Zhang[12]. It is proved in [9] that for each ℓ ≥ L ℓ is NFB. The goal of thisarticle is to prove that L is NFB.To this aim we establish a new sufficient condition under which a monoid isNFB. Throughout this article, elements of a countably infinite alphabet A are called variables and elements of the free monoid A ∗ and free semigroup A + are called words .We say that a word u has the same type as v if u can be obtained from v by changingthe individual exponents of variables. For example, the words x yxzx y xzx and xy x zxyx zx are of the same type.An island formed by a variable x in a word u is a maximal subword of u whichis a power of x . For example, the word xyyx yx has three islands formed by x andtwo islands formed by y . We use x + to denote x n when n is a positive integer andits exact value is unimportant. If u is a word over a two-letter alphabet then the height of u is the number of islands in u . For example, the word x + has height 1, x + y + has height 2, x + y + x + has height 3, and so on. For each ℓ ≥ S . • (C ℓ ) If the height of u ∈ { x, y } + is at most ℓ , then u can form an identity of S only with a word of the same type.We use var S to refer to the variety of semigroups generated by S . The followingresult from [9] gives us a connection between Lee semigroups, Lee monoids andProperties (C ℓ ). Fact 1.1. [9, Corollary 7.2] Let ℓ ≥ and S be a semigroup (resp. monoid). Then S satisfies Property (C ℓ ) if and only if var S contains L ℓ (resp. L ℓ ). In view of Fact 1.1, the sufficient condition of Lee [4] (see Fact 6.1 below) isequivalent to the following sufficient condition.
Sufficient Condition 1. (cf. Theorem 6.2) Let S be a semigroup that satisfiesProperty (C ) and k ≥ . If for each n ≥ , S satisfies the identity x k y k y k . . . y kn x k ≈ x k y kn y kn − . . . y k x k (1) then S is NFB. Note that every monoid that satisfies (1) violates Property (C ). Therefore,Sufficient Condition 1 cannot be used to establish the non-finite basis property ofany monoid. Theorem 2.7 in [8] implies the result of Zhang [12] that L is NFB andcan be reformulated as follows. 2 ufficient Condition 2. [8] Let M be a monoid that satisfies Property (C ). If foreach n > , M satisfies the identity ( x x . . . x n − x n )( y y . . . y n − y n )( x n x n − . . . x x )( y n y n − . . . y y ) ≈ (2)( y y . . . y n − y n )( x x . . . x n − x n )( y n y n − . . . y y )( x n x n − . . . x x ) , then M is NFB. Note that for n = 1 the identity (2) fails on L and consequently on L ℓ for each ℓ ≥
4. Let π denote the special permutation on { , , . . . , n } used by Jackson toprove Lemma 5.4 in [3]. The next theorem implies that for each ℓ ≥ L ℓ is NFB [9]. Sufficient Condition 3. [9, Theorem 2.1] Let M be a monoid that satisfies Prop-erty (C ). If for each n > , M satisfies the identity ( x x . . . x n − x n ) ( x kπ x kπ . . . x kπn ) ( x n x n − . . . x x ) ≈ (3)( x x . . . x n − x n ) ( x kπn . . . x kπ x kπ ) ( x n x n − . . . x x ) for some k ≥ , then M is NFB. Since L satisfies xyxyyx ≈ xyxyxy , it does not satisfy Property (C ). Therefore,Sufficient Condition 3 cannot be used to establish the non-finite basis property of L . The following theorem gives us a new sufficient condition under which a monoidis NFB and will be proved in Section 5. Theorem 1.2.
Let M be a monoid that satisfies Property (C ). If for each n > , M satisfies the identity U n = ( x x . . . x n )( x n x n − . . . x )( x x . . . x n ) ≈ ( x x . . . x n )( x x . . . x n ) = V n , then M is NFB. If τ is an equivalence relation on the free semigroup A + then we say that a word u is a τ -term for a semigroup S if u τ v whenever S satisfies u ≈ v . Recall [7] that u is an isoterm for S if u = v whenever S satisfies u ≈ v . If u is an isoterm for S thenevidently, u is a τ -term for S for every equivalence relation τ on A + . It is shownin [9] that for ℓ ≤ L ℓ carry no information about the non-finitebasis property of L ℓ . However, the non-finite basis property of Lee semigroups L ℓ and Lee monoids L ℓ for ℓ ≥ τ -terms, where τ isthe equivalence relation on A + defined by u τ v if u and v are of the same type.In particular, Sufficient Condition 3 is proved in [9] by analyzing τ -terms formonoids for which all words in { x, y } + of height at most 5 are τ -terms. Likewise,we prove Theorem 1.2 by analyzing τ -terms for monoids for which all words in { x, y } + of height at most 4 are τ -terms. In Section 6, we prove Sufficient Condition1 by analyzing τ -terms for semigroups for which all words in { x, y } + of height atmost 3 are τ -terms. 3 Lee monoids L ℓ are NFB for all ℓ > We use Con( u ) to denote the set of all variables contained in a word u . Theorem1.2 implies the following. Corollary 2.1.
The monoid L = h a, b, | aa = a, bb = b, abab = 0 i is NFB.Proof. In view of Fact 1.1, it is enough to verify that L satisfies the identity U n ≈ V n for each n > U n and V n . Fixsome substitution Θ : A → L . If for some 1 ≤ i ≤ n , the set Con(Θ( x i )) containsboth a and b then both Θ( U n ) and Θ( V n ) contain ( ab ) or ( ba ) as a subwordand consequently, both are equal to zero. Therefore, we may assume that for each1 ≤ i ≤ n we have Θ( x i ) ∈ { a, b, } . To avoid some trivial cases we may also assumethat Θ( x x . . . x n − x n ) contains both letters a and b . Consider two cases. Case 1 : Θ( x x . . . x n − x n ) contains ab as a subword.In this case, both Θ( U n ) and Θ( V n ) contain abab as a subword and consequently,both are equal to zero. Case 2 : Θ( x x . . . x n − x n ) = ba .In this case, Θ( U n ) = ( ba )( ab )( ba ) = baba = Θ( V n ).Notice that U ≈ V fails on L . Indeed, substitute x → b , x → a , x → b . Theorem 2.2.
Lee monoid L ℓ is FB if and only if ℓ = 2 .Proof. The 5-element monoid L was proved to be FB by C. Edmunds [1]. The7-element monoid L is NFB by the result of W. Zhang [12]. The 9-element monoid L is NFB by Corollary 2.1. For each ℓ ≥ L ℓ is NFB by the result ofthe second-named author [9]. ) If a variable t occurs exactly once in a word u then we say that t is linear in u .If a variable x occurs more than once in u then we say that x is non-linear in u .Evidently, Con( u ) = Lin( u ) ∪ Non( u ) where Lin( u ) is the set of all linear variablesin u and Non( u ) is the set of all non-linear variables in u . A block of u is a maximalsubword of u that does not contain any linear variables of u . Fact 3.1. [9, Lemma 3.4] Let M be a monoid that satisfies Property (C ). If M | = u ≈ v then(i) Lin( u ) = Lin( v ) , Non( u ) = Non( v ) and the order of occurrences of linearletters in v is the same as in u .(ii) The corresponding blocks in u and v have the same content. In other words,if u = a t a t . . . t m − a m − t m a m , here Non( u ) = Con( a a . . . a m − a m ) and Lin( u ) = { t , . . . , t m } , then v = b t b t . . . t m − b m − t m b m such that Con( a q ) = Con( b q ) for each ≤ q ≤ m . If Con( u ) ⊇ { x , . . . , x n } we write u ( x , . . . , x n ) to refer to the word obtainedfrom u by deleting all occurrences of all variables that are not in { x , . . . , x n } andsay that u deletes to u ( x , . . . , x n ). Lemma 3.2. [9, Lemma 3.6] Let M be a monoid that satisfies Property (C ). Let u be a word with Non( u ) = { x, y } such that(i) u ( x, y ) has height at most ;(ii) every block of u has height at most .Then u can form an identity of M only with a word of the same type. We use D x ( u ) to denote the result of deleting all occurrences of variable x in aword u . The next lemma is a special case of Lemma 3.5. Lemma 3.3.
Let M be a monoid that satisfies Property (C ). Let u be a word with3 non-linear variables such that(I) for each { x, y } ⊆ Con( u ) , the height of u ( x, y ) is at most ;(II) no block of u deletes to x + y + x + y + ;(III) no block of u deletes to x + y + x + z + x + ;(IV) If some block B of u deletes to x + y + z + x + then u satisfies each of thefollowing:(a) if there is an occurrence of y to the left of B then there is no occurrence of z to the right of B ;(b) if there is an occurrence of z to the left of B then there is no occurrence of y to the right of B .Then u can form an identity of M only with a word of the same type.Proof. Suppose that Non( u ) = { x, y, z } and that M satisfies u ≈ v .In view of Fact 3.1, in order to prove that u and v are of the same type, it isenough to show that every block B of u is of the same type as the correspondingblock B ′ of v . In view of Condition (I)–(III), modulo duality and renaming variablesthere are only five possibilities for B . Case 1: B involves only y and z .In this case, the words D x ( u ) and D x ( v ) are of the same type by Lemma 3.2.Therefore, the corresponding blocks of D x ( u ) and D x ( v ) are also of the same type.In particular, B is of the same type as the corresponding block B ′ of v . Case 2: B = x + y + z + x + .In this case, in view of Lemma 3.2, we have B ′ ( y, z ) = y + z + , B ′ ( y, x ) = x + y + x + and B ′ ( x, z ) = x + z + x + . So, if B ′ and B are not of the same type then B ′ = x + y + x + z + x + .Modulo duality, there are four possibilities for the word u . Subcase 2.1:
Neither y nor z occurs in u to the left of B .In this case, Condition (I) implies the following.5 laim 1. (a) the occurrences of y can form at most one island (denoted by y + ifany) to the right of B ;(b) the occurrences of z can form at most one island (denoted by z + if any) tothe right of B ;(c) the last occurrence of x in u precedes y + and z + . Let Θ : A → A ∗ be a substitution such that Θ( y ) = Θ( z ) = y , Θ( x ) = x andΘ( p ) = 1 for each p
6∈ { x, y } . Then Θ( v ) has height at least 5, but in view of Claim1, Θ( u ) is either x + y + x + or x + y + x + y + . This contradicts Property (C ). Subcase 2.2:
Both y and z occur in u to the left of B .In this case, Condition (IV) implies that neither y nor z occurs in u to the rightof B . Consequently, this case is dual to Subcase 2.1. Subcase 2.3: y occurs in u to the left of B but z does not occur in u to theleft of B .In this case, there is no y to the right of B by Condition (I), and no z to theright of B by Condition (IV). Thus, this case is also dual to Subcase 2.1. Subcase 2.4: z occurs in u to the left of B but y does not occur in u to theleft of B .In this case, there is no z to the right of B by Condition (I), and no y to theright of B by Condition (IV). Thus, this case is also dual to Subcase 2.1. Case 3: B = x + y + z + y + x + .In this case, in view of Lemma 3.2, we have B ′ ( y, z ) = y + z + y + and B ′ ( y, x ) = x + y + x + . Therefore, B ′ must be of the same type as B . Case 4: B = x + y + x + z + In this case, in view of Lemma 3.2, we have B ′ ( x, y ) = x + y + x + and B ′ ( x, z ) = x + z + . Therefore, B ′ must be of the same type as B . Case 5: B = x + y + z + In this case, in view of Lemma 3.2, we have B ′ ( x, y ) = x + y + and B ′ ( y, z ) = y + z + .Therefore, B ′ must be of the same type as B . Lemma 3.4.
Two words u and v are of the same type if and only if Con( u ) =Con( v ) and for each set of three variables { x, y, z } ⊆ Con( u ) , the words u ( x, y, z ) and v ( x, y, z ) are of the same type.Proof. If u and v are of the same type then evidently, Con( u ) = Con( v ) and foreach X ⊆ Con( u ) the words u ( X ) and v ( X ) are also of the same type.Now suppose that for each set of three variables { x, y, z } ⊆ Con( u ), the words u ( x, y, z ) and v ( x, y, z ) are of the same type. Then u and v begin with the sameletter. If u and v are not of the same type then u = a xy b and u = a ′ xz b ′ for somepossibly empty words a , a ′ , b and b ′ such that a x and a ′ x have the same type and { x, y, z } are pairwise distinct variables. Then the words u ( x, y, z ) and v ( x, y, z ) arealso not of the same type. To avoid a contradiction, we must assume that u and v are of the same type. Lemma 3.5.
Let M be a monoid that satisfies Property (C ). Let u be a word suchthat I) for each { x, y } ⊆ Con( u ) , the height of u ( x, y ) is at most ;(II) no block of u deletes to x + y + x + y + ;(III) no block of u deletes to x + y + x + z + x + ;(IV) If some block B of u deletes to x + y + z + x + then u satisfies each of thefollowing:(a) if there is an occurrence of y to the left of B then there is no occurrence of z to the right of B ;(b) if there is an occurrence of z to the left of B then there is no occurrence of y to the right of B .Then u can form an identity of M only with a word of the same type.Proof. Suppose that M satisfies u ≈ v . Let B and B ′ be the corresponding blocksin u and v . Then for each { x, y, z } ⊆ Con( u ) the words B ( x, y, z ) and B ′ ( x, y, z )are of the same type by Lemma 3.3. Therefore, the words B and B ′ are also of thesame type by Lemma 3.4. n Fact 4.1. [9, Fact 4.2] Given a word u and a substitution Θ : A → A + , onecan rename some variables in u so that the resulting word E ( u ) has the followingproperties:(i) Con( E ( u )) ⊆ Con( u ) ;(ii) Θ( E ( u )) is of the same type as Θ( u ) ;(iii) for every x, y ∈ Con( E ( u )) , if the words Θ( x ) and Θ( y ) are powers of thesame variable then x = y . Lemma 4.2. [9, Lemma 4.3] Let U be a word such that for each { x, y } ⊆ Con( U ) ,the height of U ( x, y ) is at most . Let Θ : A → A + be a substitution which satisfiesProperty (ii) in Fact 4.1. If Θ( u ) = U then u satisfies Condition (I) in Lemma 3.5,that is, for each { x, y } ⊆ Con( u ) , the height of u ( x, y ) is at most . A word u is called a scattered subword of a word v whenever there exist words u , . . . , u k , v , v , . . . , , v k − , v k ∈ A ∗ such that u = u . . . u k and v = v u v . . . v k − u k v k ;in other terms, this means that one can extract u treated as a sequence of lettersfrom the sequence v . For example, x x is a scattered subword of x x x x x .For the rest of this section, for each n >
2, we use U n to denote a word of thesame type as ( x x . . . x n )( x n x n − . . . x )( x x . . . x n ). The following properties of U n can be easily verified: • (P1) U n ( x i , x j ) = x + i x + j x + i x + j for each 1 ≤ i < j ≤ n ; • (P2) U n ( x i , x j , x k ) = x + i x + j x + k x + j x + i x + j x + k for each 1 ≤ i < j < k ≤ n ; • (P3) x i x j appears exactly twice in U n as a scattered subword and x j x i appearsexactly once in U n as a scattered subword for each 1 ≤ i < j ≤ n ; • (P4) U n contains x + n x + n − . . . x +2 x +1 as a subword between the two scatteredsubwords x i x j for each 1 ≤ i < j ≤ n ; 7 (P5) the occurrences of x and x n form exactly two islands in U n and for each1 < i < n , the occurrences of x i form exactly three islands in U n . We refer to theseislands as x +1 , x +1 , x + n , x + n , x + i , x + i and x + i ; • (P6) for each 1 < i < n , U n contains x + n x + n − . . . x + i +2 x + i +1 as a subword between x + i and x + i and, contains x + i − . . . x +1 as a subword between x + i and x + i . Also, U n contains x + n x + n − . . . x +2 as a subword between x +1 and x +1 and, contains x + n − . . . x +1 as a subword between x + n and x + n ; • (P7) for each 1 < i = j < n , if x j occurs in U n to the left of x + i or between x + i and x + i then j < i ; if x j occurs in U n between x + i and x + i or to the right of x + i then j > i . Lemma 4.3.
Let u be a word in less than n − variables for some n > . Let Θ : A → A + be a substitution which satisfies Property (ii) in Fact 4.1.If Θ( u ) = U n , then u satisfies Condition (II) in Lemma 3.5, that is, no block of u deletes to x + y + x + y + .Proof. Suppose that some block B of u deletes to x + y + x + y + , where { x, y } ⊆ Con( u ). In view of Fact 4.1, there are 1 ≤ i = j ≤ n such that x i ∈ Con(Θ( x )) and x j ∈ Con(Θ( y )). Since Θ( B ) contains x i x j twice as a scattered subword, Property(P3) implies that i < j . Then Θ( B ) contains x + n x + n − . . . x +2 x +1 between the twoscattered subwords x i x j by Property (P4).Since Con( u ) involves less than n − t ∈ Con( B ) such that Θ( t ) contains x k +1 x k for some 1 < k < n . In view of Property(P3), t is linear in u . Therefore, there is a linear letter between the two scatteredsubwords xy in B . A contradiction. Lemma 4.4.
Let u be a word in less than n − variables for some n > . Let Θ : A → A + be a substitution which satisfies Property (ii) in Fact 4.1.If Θ( u ) = U n , then u satisfies Condition (III) in Lemma 3.5, that is, no blockof u deletes to x + y + x + z + x + .Proof. Suppose that some block B of u deletes to x + y + x + z + x + , where x, y, z arethree distinct non-linear variables that belong to Con( u ).Due to Fact 4.1, there are pairwise distinct 1 ≤ i, j, k ≤ n such that x i ∈ Con(Θ( x )), x j ∈ Con(Θ( y )) and x k ∈ Con(Θ( z )). Therefore, the occurrences of x i form at least three islands in Θ( B ). By Property (P5), 1 < i < n and theoccurrences of x i form exactly three islands in Θ( B ). Property (P6) implies thatΘ( B ) contains x + n x + n − . . . x + i +2 x + i +1 between the first two islands formed by x i and x + i − . . . x +1 between the last two islands formed by x i .Since Con( u ) involves less than n − t ∈ Con( B ) such that Θ( t ) contains x r +1 x r for some 1 ≤ r < i − i + 1 ≤ r < n . Inview of Property (P3), t is linear in u . Therefore, there is a linear letter in B eitherbetween the first two islands formed by x or between the last two islands formed by x . A contradiction. 8 emma 4.5. Let u be a word in less than n − variables for some n > such thatsome block B of u deletes to x + y + z + x + . Let Θ : A → A + be a substitution whichsatisfies Property (ii) in Fact 4.1.If Θ( u ) = U n , then u satisfies Condition (IV) in Lemma 3.5, that is, each ofthe following holds:(a) if there is an occurrence of y to the left of B then there is no occurrence of z to the right of B ;(b) if there is an occurrence of z to the left of B then there is no occurrence of y to the right of B .Proof. If Θ( x ) is not a power of a variable, then Θ( x ) contains x k x k +1 for some1 ≤ k < n and x appears twice in u by Property (P3). Since u involves less than n − x in u due toProperty (P4). This contradicts the fact that B is a block of u . So, we can assumethat Θ( x ) = x + i for some 1 ≤ i ≤ n .Due to Property (P5), the occurrences of x i form at most three islands in U n .We refer to the two islands formed by x in B as x + and x + . Since Θ satisfiesProperty (ii) in Fact 4.1, four cases are possible. Case 1: i = 1 or i = n , Θ( x + ) is a subword of x + i and Θ( x + ) is a subword of x + i .Since u involves less than n − x in u due to Property (P6). This contradicts the fact that B is ablock of u . Case 2: < i < n , Θ( x + ) is a subword of x + i and Θ( x + ) is a subword of x + i .Use the same arguments as for Case 1. Case 3: < i < n , Θ( x + ) is a subword of x + i and Θ( x + ) is a subword of x + i .In this case, in view of Property (P7), neither y nor z occurs to the left of B . Case 4: < i < n , Θ( x + ) is a subword of x + i and Θ( x + ) is a subword of x + i .In this case, in view of Property (P7), neither y nor z occurs to the right of B . Lemma 5.1. [9, Lemma 5.1] Let τ be an equivalence relation on the free semigroup A + and S be a semigroup. Suppose that for infinitely many n , S satisfies an identity U n ≈ V n in at least n variables such that U n and V n are not τ -related.Suppose also that for every identity u ≈ v of S in less than n − variables, everyword U such that U τ U n and every substitution Θ : A → A + such that Θ( u ) = U we have U τ Θ( v ) . Then S is NFB. We use Con ( u ) to denote the set of all variables which occur twice in u andCon > ( u ) to denote the set of all variables which occur at least 3 times in u . Thenext lemma is similar to Lemma 4.1 in [9] (Lemma 6.6 below).9 emma 5.2. Let u and v be two words of the same type such that Lin( u ) = Lin( v ) , Con ( u ) = Con ( v ) and Con > ( u ) = Con > ( v ) .Let Θ : A → A + be a substitution that has the following property:(*) If Θ( x ) contains more than one variable then x ∈ Lin( u ) ∪ Con ( u ) .Then Θ( u ) and Θ( v ) are also of the same type.Proof. Since u and v are of the same type, the following is true. Claim 2.
Suppose that y ∈ Con ( u ) . If there is an occurrence of x between the twooccurrences of y in u then there is an occurrence of x between the two occurrencesof y in v . Since u and v are of the same type, for some r ≥ u , . . . , u r , v , . . . , v r > u = c u c u . . . c u r r and v = c v c v . . . c v r r , where c , . . . , c r ∈ A are such that c i = c i +1 .First, let us prove that for each 1 ≤ i ≤ r the words Θ( c u i i ) and Θ( c v i i ) areof the same type. Indeed, If c i is linear in u (and in v ) then u i = v i = 1 andΘ( c u i i ) = Θ( c v i i ). If c i occurs twice in u (and in v ), then in view of Claim 2, either u i = v i = 1 or u i = v i = 2. In either case, we have Θ( c u i i ) = Θ( c v i i ). If c i occurs atleast 3 times in u (and in v ) then Θ( c u i ) = x + for some variable x and Θ( c v i i ) is apower of the same variable.Since Θ( u ) = Θ( c u c u . . . c u r r ) = Θ( c u )Θ( c u ) . . . Θ( c u r r )and Θ( v ) = Θ( c v c v . . . c v r r ) = Θ( c v )Θ( c v ) . . . Θ( c v r r ) , we conclude that Θ( u ) and Θ( v ) are of the same type. Proof of Theorem 1.2.
Let τ be the equivalence relation on A + defined by u τ v if u and v are of the same type. First, notice that the words U n and V n are not of thesame type. Indeed, U n contains x n x n − as a subword but V n does not have thissubword.Now let U be of the same type as U n . Let u ≈ v be an identity of M in lessthan n − A → A + be a substitution such that Θ( u ) = U .Notice that E ( u ) also involves less than n − E ( u ) ≈ E ( v ) is also anidentity of M .Due to Property (P1), for each 1 ≤ i < j ≤ n the height of U ( x i , x j ) is atmost 4. So, by Lemma 4.2, E ( u ) satisfies Condition (I) in Lemma 3.5, that is, foreach { x, y } ⊆ Con( E ( u )) the height of E ( u ( x, y )) is at most 4. Also, E ( u ) satisfiesConditions (II)–(IV) in Lemma 3.5 by Lemmas 4.3–4.5.Therefore, Lemma 3.5 implies that the word E ( v ) is of the same type as E ( u ).Due to Property (P3), for each 1 ≤ i = j ≤ n the word x i x j appears at most twicein U as a scattered subword. Consequently, the word E ( u ) and the substitutionΘ satisfy Condition (*) in Lemma 5.2. According to Fact 3.1 in [9], Property (C )implies that the word x is an isoterm for M . Thus all other conditions of Lemma10.2 are also met. Therefore, the word Θ( E ( v )) has the same type as Θ( E ( u )) byLemma 5.2. Thus we have U = Θ( u ) F act . τ Θ( E ( u )) Lemma . τ Θ( E ( v )) F act . τ Θ( v ) . Since Θ( v ) is of the same type as U , M is NFB by Lemma 5.1. The following sufficient condition implies that for each ℓ ≥ L ℓ isNFB [5]. Fact 6.1. [4, Theorem 11] Fix k ≥ . Let S be a semigroup such that var S contains L . If for each n ≥ , S satisfies the identity x k y k y k . . . y kn x k ≈ x k y kn y kn − . . . y k x k then S is NFB. In view of Fact 1.1, the following sufficient condition is a slight generalization ofFact 6.1.
Theorem 6.2.
Let S be a semigroup that satisfies Property (C ). If for infinitelymany n ≥ , S satisfies the identity U n = x k y k y k . . . y kn x k ≈ x k y kn y kn − . . . y k x k = V n for some k ≥ , then S is NFB. The goal of this section is to prove Theorem 6.2 directly using Lemma 5.1. Tothis aim, we establish some consequences of Property (C ) for semigroups. Lemma 6.3.
Let S be a semigroup that satisfies Property (C ). If S | = u ≈ v then Con( u ) = Con( v ) and Lin( u ) = Lin( v ) .Proof. If x ∈ Con( u ) but x Con( v ) then for some y ∈ Con( v ) and r > u ≈ v implies y r ≈ w such that the height of w ∈ { x, y } + is at least 2. Toavoid a contradiction to Property (C ) we conclude that Con( u ) = Con( v ).If x is linear in u but appears at least twice in v then substitute xy for x and y forall other variables. Then for some c + d > u ≈ v implies y c xy d ≈ w such that the height of w ∈ { x, y } + is at least 4. To avoid a contradiction toProperty (C ) we conclude that Lin( u ) = Lin( v ). Lemma 6.4.
Let S be a semigroup that satisfies Property (C ). If each variableforms only one island in a word u then u can form an identity of S only with aword of the same type. roof. Since each variable forms only one island in the word u we may assume that u = x +1 x +2 . . . x + r for some distinct variables x , x , . . . , x r . Suppose that u forms anidentity of S with a word v . By Lemma 6.3, Con( u ) = Con( v ). Note that if wereplace x by x and any other letter by y , then the word u turns into x + y + . Since S satisfies Property (C ) the word v also starts with x .First, let us prove that each variable forms only one island in the word v . Supposethe contrary, there is a letter a ∈ Con( u ) = Con( v ) that forms at least two islandsin v . We replace a by y and other letters by x . Thus, the identity u ≈ v implies u ′ ≈ v ′ . Notice that the height of u ′ ∈ { x, y } + is at most 3 and that y forms onlyone island in u ′ . On the other hand, y forms at least two islands in v ′ . Since S satisfies Property (C ), this is impossible.Second, let us take two consecutive letters x i , x i +1 ∈ Con( u ) and replace themby y , while other letters by x . Note that the word u transforms into a word u ′ ∈{ x, y } + of height at most 3 with one island of y . Therefore, by Property (C ) thecorresponding word v ′ also has one island of y implying the word v has either x i x i +1 or x i +1 x i as a subword. Since u and v start with the same letter x we concludethat u and v have the same type. Lemma 6.5.
Let S be a semigroup that satisfies Property (C ). Let u be a wordthat begins and ends with x such that x forms exactly two islands in u . Suppose alsothat u contains a linear letter and that each variable other than x forms only oneisland in u . Then u can form an identity of S only with a word of the same type.Proof. Suppose that S | = u ≈ v . If we substitute y for all variables in Con( u ) =Con( v ) other than x then u turns into x + y + x + . Since S satisfies Property (C ), thisimplies that v starts and ends with x and x forms exactly two islands in v . Thuswe have u = x + a t b x + and v = x + a ′ t b ′ x + where Con( ab ) = Con( a ′ b ′ ) ∩ { x, t } = ∅ .We have b = y +1 y +2 . . . y + r for some { y , . . . , y r } ⊆ Con( ab ). If y is not the firstletter in b ′ then substitute xy for t , y for y and x for all other variables. Then u turns into x + y + x + while v becomes a word that contains at least two islands of y . To avoid a contradiction to Property (C ) we conclude that b ′ starts with y .If b ′ does not have y y as a subword, then substitute xy for t , y for y and for y and x for all other variables. Then u turns into x + y + x + while v becomes a wordthat contains at least two islands of y . To avoid a contradiction to Property (C )we conclude that b ′ starts with y +1 y . And so on. Eventually we conclude that thewords b and b ′ are of the same type.In a similar way, one can show that a and a ′ are also of the same type. Conse-quently, u and v are of the same type.Finally, in order to prove Theorem 6.2 we need the following statement similarto Lemma 5.2. Lemma 6.6. [9, Lemma 4.1] Let u and v be two words of the same type such that Lin( u ) = Lin( v ) and Non( u ) = Non( v ) .Let Θ : A → A + be a substitution that has the following property: *) If Θ( x ) contains more than one variable then x is linear in u .Then Θ( u ) and Θ( v ) are also of the same type.Proof of Theorem 6.2. Let τ be the equivalence relation on A + defined by u τ v if u and v are of the same type. First, notice that the words U n and V n are not ofthe same type. Indeed, U n contains y y as a subword but V n does not have thissubword.Now let U be of the same type as U n . Let u ≈ v be an identity of S in less than n variables and let Θ : A → A + be a substitution such that Θ( u ) = U . Notice that E ( u ) also involves less than n variables and E ( u ) ≈ E ( v ) is also an identity of S .If every variable forms only one island in E ( u ) then by Lemma 6.4 the word E ( v ) is of the same type as E ( u ). If some variable x forms more than one island in E ( u ) then in view of Fact 4.1, x forms exactly two islands in E ( u ) and E ( u ) beginsand ends with x . Also, E ( u ) contains a linear letter because it involves less than n variables. So, in this case, the word E ( v ) is of the same type as E ( u ) by Lemma6.5.Therefore, the word Θ( E ( v )) has the same type as Θ( E ( u )) by Lemma 6.6. Thuswe have U = Θ( u ) F act . τ Θ( E ( u )) Lemma . τ Θ( E ( v )) F act . τ Θ( v ) . Since Θ( v ) is of the same type as U , S is NFB by Lemma 5.1. Acknowledgement
The authors thank an anonymous referee for helpful comments.
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