Epimorphisms, dominions and seminormal identities II

Semigroup Forum (2013) 87:97–119DOI 10.1007/s00233-012-9461-8

R E S E A R C H A RT I C L E

Epimorphisms, dominions and seminormal identities II

Noor Mohammad Khan · Aftab Hussain Shah

Received: 15 February 2012 / Accepted: 16 October 2012 / Published online: 3 January 2013© Springer Science+Business Media New York 2013

Abstract We associate two natural numbers with a seminormal identity dependingon the first and the last occurrences of two two-letter subwords of the left hand sidewhich are not subwords of its right hand side. Using these natural numbers, we findsome classes of heterotypical identities where both sides contain repeated variablesthat are preserved under epis in conjunction with a seminormal identity.

Keywords Epimorphism · Epimorphically closed · Preserved under epis ·Heterotypical identity · Seminormal identity

1 Introduction and preliminaries

A morphism α : S → T in the category of all semigroups is called an epimorphism(epi for short) if for all morphisms β,γ , αβ = αγ implies β = γ . Let U and S beany semigroups with U a subsemigroup of S. Following Isbell [4], we say that U

dominates an element d of S if for every semigroup T and for all homomorphismsα,β : S → T , uα = uβ for all u ∈ U implies dα = dβ . The set of all elements of S

dominated by U is called the dominion of U in S, and we denote it by Dom(U,S).It may easily be seen that Dom(U,S) is a subsemigroup of S containing U . A semi-group U is said to be saturated if Dom(U,S) �= S for every properly containingsemigroup S, and epimorphically embedded or dense in S if Dom(U,S) = S. It may

Communicated by Benjamin Steinberg.

N.M. Khan (�)Department of Mathematics, Aligarh Muslim University, Aligarh 202002, Indiae-mail: [email protected]

A.H. ShahDepartment of Mathematics, Central University of Kashmir, Srinagar 190004, Indiae-mail: [email protected]

mailto:[email protected]

mailto:[email protected]

98 N.M. Khan, A.H. Shah

be easily checked that α : S → T is epi if and only if the inclusion map i : Sα → T

is epi, and the inclusion map i : U → S is epi if and only if Dom(U,S) = S. Everyonto morphism is epi but the converse is not true in general. A variety V of semi-groups is said to be epimorphically closed or closed under epis if whenever U ∈ Vand Dom(U,S) = S, it follows that S ∈ V .

An identity of the form

x1x2 · · ·xn = xi1xi2 · · ·xin (n ≥ 2), (1)

is called a permutation identity, where i is any permutation of the set {1,2,3, . . . , n}and ik , for each k (1 ≤ k ≤ n), is the image of k under the permutation i. A permu-tation identity of the form (1) is said to be nontrivial if the permutation i is differentfrom the identity permutation. Further, a nontrivial permutation identity of the form(1) is said to be left semicommutative if i1 �= 1, right semicommutative if in �= n,and seminormal if i1 = 1 and in = n. It is easy see that every nontrivial permuta-tion identity is either left semicommutative, right semicommutative, or seminormal.A semigroup S satisfying a nontrivial permutation identity is said to be permuta-tive, and a variety V of semigroups is said to be permutative if it admits a nontrivialpermutation identity.

The study of epimorphisms and dominions in the theory of semigroups was firstconsidered by Isbell [4]. It was revived in the 1980’s, when it was studied extensivelyby Hall, Higgins, Khan, and others, resulting in the appearance of various interestingresearch articles (see for example, Howie [3], for all these references). It was provenby Khan [7], jointly with Higgins [1], that any semigroup variety which satisfies apermutation identity x1x2 · · ·xn = xi1xi2 · · ·xin , where i1 �= 1 or in �= n, is epimor-phically closed, while Higgins [2] found an example of an identity where both sidescontain repeated variables and which is not preserved under epis in conjunction withthe identity xyzt = xzyt (a seminormal identity). Thus the problem of finding thosesemigroup identities where both sides contain repeated variables and are preservedunder epis in conjunction with a seminormal identity appears to be worthy of men-tion. This problem went through a dormancy period of almost three decades. Recently[8], the authors revisited this problem and found some sufficient conditions for cer-tain classes of semigroup identities to be preserved under epis in conjunction witha seminormal identity. There, the authors dealt almost exclusively with homotypicalidentities. The objective of the present paper is to extend the results to certain classesof heterotypical identities where both sides contain repeated variables.

Towards this end, we obtain some partial results by establishing some sufficientconditions for such identities to lie in this class (Theorems 2.2, 2.5 and 2.14). How-ever, a full determination of all such semigroup identities that are preserved underepis in conjunction with a seminormal identity remains an open problem.

In the remainder of this section, we quote some results that will be used in restof the paper. Our notation will be standard and we refer the reader to Howie [3] forany unexplained symbols and terminology. Further, in what follows we will oftenspeak of a semigroup “satisfying (1)” to mean that the semigroup in question satisfiesan identity of that type. We will say that an identity u = v is preserved under epis tomean that if U satisfies u = v and Dom(U,S) = S, then S also satisfies u = v. For anyword u, the content of u is the set (necessarily finite) of all variables appearing in u

Epimorphisms, dominions and seminormal identities II 99

and is denoted by C(u). An identity u = v is said to be heterotypical if C(u) �= C(v);and homotypical otherwise.

Result 1.1 ([7, Theorem 3.1]) All permutation identities are preserved under epis.

A most useful characterization of semigroup dominions is provided by Isbell’sZigzag theorem.

Result 1.2 ([4, Theorem 2.3] or [3, Theorem VII.2.13]) Let U be a subsemigroup ofa semigroup S and let d ∈ S. Then d ∈ Dom(U,S) if and only if d ∈ U or there existsa series of factorizations of d as follows:

d = a0t1 = y1a1t1 = y1a2t2 = y2a3t2 = · · · = yma2m−1tm = yma2m, (2)

where m ≥ 1, ai ∈ U (i = 0,1, . . . ,2m), yi, ti ∈ S (i = 1,2, . . . ,m), and

a0 = y1a1, a2m−1tm = a2m,

a2i−1ti = a2i ti+1, yia2i = yi+1a2i+1 (1 ≤ i ≤ m − 1).

Such a series of factorization is called a zigzag in S over U with value d , length m

and spine a0, a1, . . . , a2m.

In whatever follows, we refer to the equations in Result 1.2 as the zigzag equations.

Result 1.3 ([6, Result 3]) Let U be any subsemigroup of a semigroup S and letd ∈ Dom(U,S) \U . If (2) is a zigzag of minimal length m over U with value d , thenyj , tj ∈ S \ U for all j = 1,2, . . . ,m.

In the following results, let U and S be any semigroups with U dense in S.

Result 1.4 ([6, Result 4]) For any d ∈ S \ U and k any positive integer, if (2) is azigzag of minimal length over U with value d , then there exist b1, b2, . . . , bk ∈ U

and dk ∈ S \ U such that d = b1b2 · · ·bkdk .

Result 1.5 ([6, Corollary 4.2]) If U be permutative, then

sx1x2 · · ·xkt = sxj1xj2 · · ·xjkt

for all x1, x2, . . . , xk ∈ S, s, t ∈ S \ U and any permutation j of the set {1,2, . . . , k}.

Result 1.6 ([8, Proposition 2.2]) Let U be any permutative semigroup with n ≥ 3.Then for each g ∈ {2,3, . . . , n} such that xg−1xg is not a subword of xi1xi2 · · ·xin , forall m ≥ g − 1 and for all u ∈ S(m), v ∈ S \ U , we have

ux1x2 · · ·x�v = uxλ1xλ2 · · ·xλ�v


for all x1, x2, . . . , x� ∈ S (� ≥ 2), where λ is any permutation of the set {1,2, . . . , �}.Symmetrically, for all p ≥ h − 1 such that xn−hxn−(h−1) is not a subword ofxi1xi2 · · ·xin and for all v ∈ S(p), u ∈ S \ U , we have

ux1x2 · · ·x�v = uxλ1xλ2 · · ·xλ�v

for all x1, x2, . . . , x� ∈ S (� ≥ 2), where λ is any permutation of the set {1,2, . . . , �}.

Result 1.7 ([8, Lemma 2.3]) Let U be any permutative subsemigroup which is densein S. If U satisfies

xru(x1, x2, . . . , x�)ys = xrv(x1, x2, . . . , x�)y

s, (3)

where u and v are any words in x1, x2, . . . , x� and r , s any positive integers, then theidentity (3) is also satisfied for all x, y ∈ S and x1, x2, . . . , x� ∈ U .

Remark Result 1.7 does not require |xi |u > 0 nor |xj |v > 0, so it applies to bothhomotypical and heterotypical identities.

We shall freely use the following results in the text without explicit mention.

Result 1.8 ([8, Corollary 1.8]) For any d ∈ S and positive integer k, if d =b1b2 · · ·bkdk for some b1, b2, . . . , bk ∈ U and dk ∈ S \ U such that b1 = y1

′c1 forsome y1

′ ∈ S \ U , c1 ∈ U , then dp = b1pb2

p · · ·bkpdk

p for any positive integer p.

The symmetrical statement in the following result does not appear in the original,but is immediate.

Result 1.9 ([7, Proposition 4.6]) Assume that U is permutative. If d ∈ S \ U and (2)is a zigzag of length m over U with value d such that y1 ∈ S \ U , then dk = ak

0 tk1 foreach positive integer k; in particular, the conclusion holds if (2) is of minimal length.Symmetrically, if d ∈ S \ U and (2) is a zigzag of length m over U with value d

such that tm ∈ S \ U , then dk = ykmak

2m for each positive integer k; in particular, theconclusion holds if (2) is of minimal length.

2 Main results

Let (1) be any seminormal permutation identity. Then there exists g (2 ≤ g ≤ n − 2)

such that xg−1xg is not a subword of xi1xi2 · · ·xin . Thus the set P of all positiveintegers g (2 ≤ g ≤ n − 2) such that xg−1xg is not a subword of xi1xi2 · · ·xin isnon-empty and, so, will have a minimum element; let g0 = minP , the minimumof P . Similarly, the set Q of all positive integers h (1 ≤ h ≤ n − g0 − 1) such thatxn−hxn−(h−1) is not a subword of xi1xi2 · · ·xin , is non-empty; let h0 = minQ.

In all that follows, g0 and h0 will be as defined above. Also, for any word w andany variable x of w, |x|w will denote the number of occurrences of x in w.


To avoid introduction of new symbols, we shall treat, wherever is appropriate,x, y, x1, x2, . . . , x� etc. both as variables and as the members of a semigroup withoutexplicit mention of the distinction.

The following lemma easily follows from Result 1.6, where the bracketed state-ment is dual to the previous statement (and likewise where other bracketed statementsoccur elsewhere in the text).

Lemma 2.1 Let S be a semigroup satisfying a seminormal permutation identity andlet u be any word in the variables x1, x2, . . . , x�. Let r , s be any positive integers suchthat r ≥ g0 − 1 and s ≥ h0 − 1. If xj = s1a [xj = bs2] for some a, b, s1, s2 ∈ S andfor some j ∈ {1,2, . . . , �}, then

xru(x1, . . . , xj , . . . , x�)ys = xr(s1)

|xj |uu(x1, . . . , a, . . . , x�)ys

[xru(x1, . . . , xj , . . . , x�)y

s = xru(x1, . . . , , b, . . . , x�)(s2)|xj |uys

]

for all x, y, x1, x2, . . . , x� ∈ S.Further if xj = s1cs2 for some c, s1, s2 ∈ S and for some j ∈ {1,2, . . . , �}, then

xru(x1, . . . , xj , . . . , x�)ys = xr(sk1)

|xj |u(sk2)|xj |uu(x1, . . . , c, . . . , x�)y

s

and

xru(x1, . . . , xj , . . . , x�)ys = xru(x1, . . . , c, . . . , x�)(sk1)

|xj |u(sk2)|xj |uys

for all x, y, x1, x2, . . . , x� ∈ S, where k is any permutation of the set {1,2}.

Theorem 2.2 Let (1) be any seminormal identity and let u, v be any words inx1, x2, . . . , x� and z1, . . . , zp respectively. Let r , s be any positive integers such thatr ≥ g0 − 1, s ≥ h0 − 1 with min{|xi |u, |zj |v} ≥ min{r, s} for all i ∈ {1,2, . . . , �} andfor all j ∈ {1,2, . . . , p}. Then all heterotypical identities of the form

xru(x1, x2, . . . , x�)ys = xrv(z1, z2, . . . , zp)ys (4)

are preserved under epis in conjunction with (1).

Proof We shall prove the theorem for case when min{r, s} = r , the proof in theother case follows along similar lines. So, assume that U , and hence S by Re-sult 1.1, satisfy (1). We shall show that if U satisfies equality (4), then so does S.Let x, y, x1, x2, . . . , x�, z1, z2, . . . , zp ∈ S. If all of x1, x2, . . . , x�, z1, z2, . . . , zp ∈ U ,then the result holds by Result 1.7. So, assume that not all of x1, . . . , x�, z1, z2, . . . , zp

are in U . Now, to show that equality (4) is satisfied by S, we shall first prove that

xru(x1, x2, . . . , x�)ys = xrv(v1, v2, . . . , vp)ys (5)

for all x, y, x1, x2, . . . , x� ∈ S and v1, v2, . . . , vp ∈ U . We prove equality (5) by in-duction on k assuming that arguments x1, x2, . . . , xk of the word u are from S andthe remaining arguments xk+1, . . . , x� are from U . When k = 0, equality (5) holds byResult 1.7.


Next assume inductively that equality (5) holds for all x, y, x1, x2, . . . , xk−1 in S

and xk, xk+1, . . . , x� ∈ U . From this we shall prove that equality (5) also holds forall x1, x2, . . . , xk−1, xk, x, y ∈ S and xk+1, . . . , x� ∈ U . If xk ∈ U , then equality (5)follows by the inductive hypothesis. So, assume that xk ∈ S \ U . Let (2) be a zigzagof minimal length m over U with value xk . Now, for any v1, v2, . . . , vp ∈ U , we have

xru(x1, x2, . . . , xk−1, xk, xk+1, . . . , x�)ys

= xru(x1, x2, . . . , xk−1, yma2m,xk+1, . . . , . . . , x�)ys (by the zigzag equations)

= xr(ym)|xk |uu(x1, . . . , xk−1, a2m,xk+1, . . . , x�)ys (by Lemma 2.1)

= xr(ym)|xk |uv(v1, . . . , vp)ys

(by inductive hypothesis as ym ∈ S \ U, |xk|u ≥ r)

= xr(ym)|xk |uu(x1, . . . , xk−1, a2m−1, xk+1, . . . , x�)ys

(by the inductive hypothesis)

= xru(x1, . . . , xk−1, yma2m−1, xk+1, . . . , x�)ys (by Lemma 2.1)

= xru(x1, . . . , xk−1, ym−1a2m−2, xk+1, . . . , x�)ys (by the zigzag equations)

= xr(ym−1)|xk |uu(x1, . . . , xk−1, a2m−2, xk+1, . . . , x�)y

s (by Lemma 2.1)

...

= xry|xk |u1 u(x1x2, . . . , xk−1, a2, xk+1, . . . , x�)y

s

= xry|xk |u1 v(v1, . . . , vp)ys (by inductive hypothesis as y1 ∈ S \ U, |xk|u ≥ r)

= xry|xk |u1 u(x1, . . . , xk−1, a1, xk+1, . . . , x�)y

s (by the inductive hypothesis)

= xru(x1, x2, . . . , xk−1, y1a1, xk+1, . . . , x�)ys (by Lemma 2.1)

= xru(x1, x2, . . . , xk−1, a0, xk+1, . . . , x�)ys (by the zigzag equations)

= xrv(v1, v2, . . . , vp)ys (by the inductive hypothesis)

as required.Similarly, we may prove that

xru(u1, u2, . . . , u�)ys = xrv(z1, z2, . . . , zp)ys (6)

when x, y, z1, z2, . . . , zp ∈ S and u1, u2, . . . , u� ∈ U .Now, using Result 1.7 and (5) and (6), we have

xru(x1, . . . , x�)ys = xrv(v1, . . . , vp)ys = xru(u1, . . . , u�)y

s = xrv(z1, . . . , zp)ys.

This completes the proof of Theorem 2.2. �

Proposition 2.3 Let U be a semigroup satisfying a seminormal identity and dense inS. Let u, v be any words in x1, x2, . . . , x�, z1, z2, . . . , zp and x1, x2. . . . , x� (�,p ≥ 1)


respectively and let r, s be any positive integers with min{|xi |u, |zj |v} ≥ min{r, s} forall i ∈ {1,2, . . . , �} and j ∈ {1,2, . . . , p}. If U satisfies

xru(x1, x2, . . . , x�, z1, z2, . . . , zp)ys = xrv(x1, x2, . . . , x�)ys, (7)

then (7) is also satisfied for all x, y, x1, x2, . . . , x� ∈ S and for all z1, z2, . . . , zp ∈ U .

Proof We shall prove the theorem for case when min{r, s} = r , the proof in othercase follows along similar lines. As U satisfies a seminormal identity, so doesS by Result 1.1. We shall show that if U satisfies (7), then (7) is also satis-fied for all x, y, x1, x2, . . . , x� in S and z1, z2, . . . , zp ∈ U . If x, y ∈ S and all ofx1, x2, . . . , x�, z1, z2, . . . , zp ∈ U , then (7) holds by Result 1.7. So, assume first thatnot all of x1, x2, . . . , x� ∈ U . We prove equality (7) by induction on k assuming thatarguments x1, x2, . . . , xk of the words u and v are from S while remaining argumentsxk+1, . . . , x� lie in U . For k = 0, equality (7) holds by Result 1.7.

Now assume inductively that the result is true for all x, y, x1, x2, . . . , xk−1 ∈ S andxk, xk+1, . . . , x�, z1, . . . , zp ∈ U . From this, we shall prove that the result is also truefor all x, y, x1, x2, . . . , xk−1, xk ∈ S and xk+1, . . . , x�, z1, . . . , zp ∈ U . If xk ∈ U , thenequality (7) follows by inductive hypothesis. So, assume that xk ∈ S \ U . Let (2) be azigzag of minimum length m over U with value xk . Now

xru(x1, . . . , xk, . . . , x�, z1, z, . . . , zp)ys

= xru(x1, . . . , yma2m, . . . , x�, z1, z2, . . . , zp)ys (by the zigzag equations)

= xr(ym)|xk |uu(x1, . . . , a2m, . . . , x�, z1, z2, . . . , zp)ys (by Lemma 2.1)

= xr(ym)|xk |uv(x1, . . . , a2m, . . . , x�)ys

(by the inductive hypothesis as |xk|u ≥ r)

= xr(ym)|xk |uv(xk, . . . , a2m−1tm, . . . , x�)ys (by the zigzag equations)

= xr(ym)|xk |u(tm)|xk |v v(x1, . . . , a2m−1, . . . , x�)ys (by Lemma 2.1)

= xr(ym)|xk |u(tm)|xk |vu(x1, . . . , a2m−1, . . . , x�, z1, . . . , zp)ys


= xr(tm)|xk |vu(x1, . . . , yma2m−1, . . . , x�, z1, . . . , zp)ys (by Lemma 2.1)

= xr(tm)|xk |vu(x1, . . . , ym−1a2m−2, . . . , x�, z1, . . . , zp)ys

(by the zigzag equations)

...

= xr(t2)|xk |vu(x1, . . . , y1a2, . . . , x�, z1, . . . , zp)ys

= xr(y1)|xk |u(t2)|xk |vu(x1, . . . , a2, . . . , x�, z1, z2, . . . , zp)ys (by Lemma 2.1)

= xr(y1)|xk |u(t2)|xk |v v(x1, . . . , a2, . . . , x�)y

s (by the inductive hypothesis)

= xr(y1)|xk |uv(x1, . . . , a2t2, . . . , x�)y

s (by Lemma 2.1)


= xr(y1)|xk |uv(x1, . . . , a1t1, . . . , x�)y

s (by the zigzag equations)

= xr(y1)|xk |u(t1)|x1|v v(x1, . . . , a1, . . . , x�)y

s (by Lemma 2.1)

= xr(y1)|xk |u(t1)|x1|vu(x1, . . . , a1, . . . , x�, z1, z2, . . . , zp)ys


= xr(t1)|x1|vu(x1, . . . , y1a1, . . . , x�, z1, z2, . . . , zp)ys (by Lemma 2.1)

= xr(t1)|xk |vu(x1, . . . , a0, . . . , x�, z1, . . . , zp)ys (by the zigzag equations)

= xr(t1)|xk |v v(x1, . . . , a0, . . . , x�)y

s (by the inductive hypothesis as |xk|v ≥ r)

= xrv(x1, x2, . . . , a0t1, . . . , x�)ys (by Lemma 2.1)

= xrv(x1, x2, . . . , xk, . . . , x�)ys (by the zigzag equations)

as required. This completes the proof of the proposition. �

Proposition 2.4 Assume that U satisfy a seminormal identity and is dense in S. Letu, v be any words in x1, x2, . . . , x�, z1, z2, . . . , zp and x1, x2, . . . , x� respectively andlet r, s be any positive integers such that min{|xi |u, |xi |v, |xj |u} ≥ min{r, s} and r ≥g0 − 1, s ≥ h0 − 1 for each i ∈ {1,2, . . . , �} and j ∈ {1,2, . . . , p}. If U satisfies

xru(x1, x2, . . . , x�, z1, z2, . . . , zp)ys = xrv(x1, x2, . . . , x�)ys, (8)

then so does S.

Proof We shall prove the theorem for case when min{r, s} = r , as the proof in othercase follows along similar lines. Since U satisfies a seminormal identity, so does S

by Result 1.1. We shall show that if U satisfies (8), then so does S. If x, y ∈ S andall of x, y, x1, x2, . . . , x�, z1, z2, . . . , zp ∈ U , then (8) holds by Result 1.7; and if allof x, y, x1, x2, . . . , x� ∈ S and z1, z2, . . . , zp ∈ U , then (8) holds by Proposition 2.3.So assume that not all of z1, z2, . . . , zp ∈ U . We prove equality (8), by induction onk assuming that arguments z1, z2, . . . , zk of the word u lie in S and the remainingarguments zk+1, . . . , zp of the word u are from U . For k = 0, equality (8) holds byProposition 2.3.

Next, assume inductively that the result is true if zk, zk+1, . . . , zp ∈ U andx, y, x1, . . . , x�, z1, . . . , zk−1 ∈ S. From this, we shall prove that the result is alsotrue if x, y, x1, x2, . . . , x�, z1, z2, . . . , zk−1, zk ∈ S and zk+1, . . . , zp ∈ U . If zk ∈ U ,then equality (8) follows by inductive hypothesis. So, assume that zk ∈ S \ U . Let (2)be a zigzag of minimal length m over U with value zk . Now

xru(x1, x2, . . . , x�, z1, z2, . . . , zp)ys

= xru(x1, . . . , x�, z1, . . . , yma2m, . . . , zp)ys (by the zigzag equations)

= xr(ym)|zk |uu(x1, . . . , x�, z1, . . . , a2m, . . . , zp)ys (by Lemma 2.1)

= xr(ym)|zk |uv(x1, x2, . . . , x�)ys (by the inductive hypothesis as |zk|u ≥ r)

= xr(ym)|zk |uu(x1, . . . , x�, x1, . . . , a2m−1, . . . , zp)ys


(by the inductive hypothesis as |zk|u ≥ r)

= xru(x1, . . . , x�, z1, . . . , yma2m−1, . . . , zp)ys (by Lemma 2.1)

= xru(x1, . . . , x�, z1, . . . , ym−1a2m−2, . . . , zp)ys (by the zigzag equations)

= xr(ym−1)|zk |uu(x1, . . . , x�, z1, . . . , a2m−2, . . . , zp)ys (by Lemma 2.1)

= xr(ym−1)|zk |uv(x1, x2, . . . , x�)y

s (by the inductive hypothesis as |zk|u ≥ r)

= xr(ym−1)|zk |uu(x1, . . . , x�, z1, . . . , a2m−3, . . . zp)ys

(by the inductive hypothesis as |zk|u ≥ r)

...

= xr(y1)|zk |uu(x1, . . . , x�, z1, . . . , a2, . . . , zp)ys

= xr(y1)|zk |uv(x1, x2, . . . , x�)y

s (by the inductive hypothesis as |zk|u ≥ r)

= xr(y1)|zk |uu(x1, . . . , x�, z1, . . . , a1, . . . , zp)ys (by the induction as |zk|u ≥ r)

= xru(x1, . . . , x�, z1, . . . , y1a1, . . . , zp)ys (by Lemma 2.1)

= xru(x1, . . . , x�, z1, . . . , a0, . . . , zp)ys (by the zigzag equations)

= xrv(x1, x2, . . . , x�)ys (by the inductive hypothesis as |zk|u ≥ r)

as required. Thus the proof of Proposition 2.4 is complete. �

Now, by combining Propositions 2.4 and 2.5, we get the following:

Theorem 2.5 Let u and v be any words in x1, . . . , x�, z1, z2, . . . , zp and x1, . . . , x�

respectively; r, s be any positive integers such that min{|xi |u, |xi |v, |xj |u} ≥ min{r, s}for each i ∈ {1,2, . . . , �} and j ∈ {1,2, . . . , p} and let (1) be any seminormal identity.If r ≥ g0 − 1 and s ≥ h0 − 1, then all heterotypical identities of the form

xru(x1, x2, . . . , x�, z1, z2, . . . , zp)ys = xrv(x1, x2, . . . , x�)ys,

are preserved under epis in conjunction with (1).

Proposition 2.6 Let U , S be semigroups such that U satisfy a seminormal identityand is dense in S and let u, v be any words in p1,p2, . . . , p�. Let r , s be positiveintegers such that r ≥ g0 − 1 and s ≥ h0 − 1. If U satisfies

x1x2 · · ·xru(p1,p2, . . . , p�)z1z2 · · · zs = xi1 · · ·xir v(p1,p2, . . . , p�)zλ1 · · · zλs , (9)

where i and λ are any permutations of the sets {1,2, . . . , r} and {1,2, . . . , s} re-spectively, then (9) is also satisfied for all x1, . . . , xr ∈ S and p1,p2, . . . , p�, z1, . . . ,

zs ∈ U .

Proof For k = 1,2, . . . , r ; consider the word xi1xi2 · · ·xik of length k. We shall provethe proposition by induction on the length of these words, assuming that the remain-ing elements xik+1, . . . , xir ∈ U .


First for k = 1, i.e., when xi1 ∈ S and xi2, . . . , xir ∈ U , we wish to show thatidentity (9) holds. When xi1 ∈ U , (9) holds trivially. So, assume that xi1 ∈ S \ U . Let(2) be a zigzag of minimal length m over U with value xi1 .

For the ease of writing, we introduce some notation:

w1(xi1, xi2, . . . , xir ) = xi1xi2 · · ·xir = u1(x1, x2, . . . , xr );w2(xi1, xi2, . . . , xir ) = x1x2 · · ·xr = u2(x1, x2, . . . , xr );

(p1,p2, . . . , p�) = p.

Using this notation, we need to prove that

w2(xi1 , xi2, . . . , xir )u(p)z1z2 · · · zs = w1(xi1, xi2, . . . , xir )v(p)zλ1 · · · zλs

or, equivalently, that

u2(x1, x2, . . . , xr )u(p)z1z2 · · · zs = u1(x1, x2, . . . , xr )v(p)zλ1 · · · zλs .

�

Case (i) i1 = 1. Now

xi1xi2 · · ·xir v(p)zλ1zλ2 · · · zλs

= yma2mxi2 · · ·xir v(p)zλ1zλ2 · · · zλs (by the zigzag equations)

= ymw1(a2m,xi2, . . . , xir )v(p)zλ1zλ2 · · ·yλs

= ymw2(a2m,xi2, . . . , xir )u(p)z1z2 · · · zs (as U satisfies (9))

= yma2mx2 · · ·xru(p)z1z2 · · · zs

= x1x2 · · ·xru(p)z1z2 · · · zs (by the zigzag equations)

as required.

Case (ii) 1 < i1 ≤ r . put j = i1. The word xj+1 · · ·xr is the empty word whenj = r . Now


= yma2mxi2 · · ·xir v(p)zλ1zλ2 · · · zλs (by the zigzag equations)

= ymw1(a2m,xi2, . . . , xir )v(p)zλ1zλ2 · · · zλs

= ymw2(a2m,xi2, . . . , xir )u(p)z1z2 · · · zs (as U satisfies (9))

= ymx1x2 · · · (xj−1a2m)xj+1 · · ·xru(p)z1z2 · · · zs

= ymx1x2 · · · (xj−1a2m−1tm)xj+1 · · ·xru(p)z1z2 · · · zs


= yma2m−1x1 · · · (xj−1tm) · · ·xru(p)z1 · · · zs (10)


= ym−1a2m−2x1 · · · (xj−1tm) · · ·xru(p)z1 · · · zs (by the zigzag equations)

= ym−1x1 · · · (xj−1a2m−2tm) · · ·xru(p)z1 · · · zs (11)

= ym−1x1 · · · (xj−1a2m−3tm−1) · · ·xru(p)z1 · · · zs (by the zigzag equations),

where equalities (10) and (11) above follow by Result 1.6 as z1z2 · · · zs ∈ S(s) and asym and ym−1 ∈ S \ U , respectively.

Continuing in this way, where equality (12) follows by Result 1.4 for somec1, c2, . . . , cs in U and t ′1 ∈ S \ U , as t1 ∈ S \ U , and where z = z1z2 · · · zs , we obtain


= ym−1x1 · · ·xj−1a2m−3tm−1xj+1 · · ·xru(p)z1z2 · · · zs

...

= y1x1x2 · · ·xj−1a1t1xj+1 · · ·xru(p)z1z2 · · · zs

= y1x1 · · ·a1 · · ·xru(p)t1z1 · · · zs

(by Result 1.6 as y1 ∈ S \ U and z1 · · · zs ∈ S(s))

= y1x1 · · ·xj−1a1xj+1 · · ·xru(p)c1c2 · · · cst′1z (12)

= y1u2(x1, . . . x2, . . . , xj−1, a1, xj+1, . . . , xr )u(p)c1c2 · · · cst′1z

= y1u1(x1, . . . , xj−1, a1, xj+1, . . . , xr )v(p)cλ1 · · · cλs t′1z (as U satisfies (8)).

Since the word u1(ξ1, ξ2, . . . , ξr ) begins with ξi1 = ξj , the above product in S con-tains y1a1 as a subproduct. Thus, using the fact that y1a1 = a0 from the zigzag equa-tions, we have


= y1u1(x1, . . . , xj−1, a1, xj+1, . . . , xr )v(p)cλ1 · · · cλs t′1z

= u1(x1, . . . , xj−1, a0, xj+1, . . . , xr )v(p)cλ1cλ2 · · · cλs t′1z

= u2(x1, . . . , xj−1, a0, xj+1, . . . , xr )u(p)c1c2 · · · cst′1z (as U satisfies (9))

= x1 · · ·xj−1a0xj+1 · · ·xru(p)c1c2 · · · cst′1z1 · · · zs (by the definition of z)

= x1 · · ·xj−1a0xj+1 · · ·xru(p)t1z1z2 · · · zs (as c1c2 · · · cst′1 = t1)

= x1 · · ·xj−1a0t1xj+1 · · ·xru(p)z1 · · · zs (13)

= x1x2 · · ·xj−1xjxj+1 · · ·xru(p)z1z2 · · · zs (by the zigzag equations)

where equality (13) above follows by Result 1.6 because a0 = y1a1, y1 ∈ S \ U andz1z2 · · · zs ∈ S(s), as required.

This is the end of the proof in Case (ii) and, thus, of the base of the induction.Assume inductively that the result holds for all xi1, xi2, . . . , xiq−1 ∈ S and for all

xiq , xiq+1 , . . . , xir ∈ U . From this assumption, we shall prove that the result also holdsfor all xi1, xi2, . . . , xiq ∈ S and xiq+1 , . . . , xir ∈ U . Again, we need not consider the


case where xiq ∈ U . So, we assume that xiq ∈ S \ U . Let (2) be a zigzag of minimallength m over U with value xiq . Put j = iq and � = iq−1.

Case (1) � = j − 1. Now


= xi1xi2 · · ·xiq−1yma2mxiq+1 · · ·xir v(p)zλ1zλ2 · · · zλs (by the zigzag equations)

= xi1xi2 · · · (xiq−1ym)a2mxiq+1 · · ·xir v(p)zλ1zλ2 · · · zλs

= x1x2 · · ·xj−2(xiq−1ym)a2mxj+1 · · ·xru(p)z1z2 · · · zs

(by inductive hypothesis)

= x1x2 · · ·xj−2xiq−1(yma2m)xj+1 · · ·xru(p)z1z2 · · · zs

= x1x2 · · ·xj−2xj−1xjxj+1 · · ·xru(p)z1z2 · · · zs (by the zigzag equations)

as required.

Case (2) � < j − 1 and j ≤ r . When j = r , xj+1 · · ·xr denotes the empty word.Now


= xi1 · · ·xiq−1yma2mxiq+1 · · ·xir v(p)zλ1 · · · zλs (by the zigzag equations)

= xi1 · · · (xiq−1ym)a2mxiq+1 · · ·xir v(p)zλ1 · · · zλs

= w1(xi1, . . . , xiq−1ym,a2m,xiq+1 , . . . , xir )v(p)zλ1 · · · zλs

= w2(xi1, . . . , xiq−1ym, a2m,xiq+1 , . . . , xir )u(p)z1 · · · zs


= x1x2 · · ·x�−1(xiq−1ym)x�+1 · · ·xj−1a2mxj+1 · · ·xru(p)z1z2 · · · zs

= x1 · · ·x�−1(xiq−1ym) · · ·a2m−1tm · · ·xru(p)z1 · · · zs (by the zigzag equations)

= x1 · · ·xiq−1(yma2m−1) · · · tm · · ·xru(p)z1 · · · zs

(by Result 1.5 as ym, tm ∈ S \ U)

= x1 · · ·xiq−1(ym−1a2m−2) · · · tm · · ·xru(p)z1 · · · zs (by the zigzag equations)

= x1 · · ·xiq−1(ym−1) · · ·a2m−2tm · · ·xru(p)z1 · · · zs

(by Result 1.5 as ym−1, tm ∈ S \ U)

= x1 · · ·xiq−1(ym−1) · · ·a2m−3tm−1 · · ·xru(p)z1 · · · zs (by the zigzag equations).

Continuing in this way, we obtain


= x1 · · ·x�−1xiq−1ym−1 · · ·xj−1a2m−3tm−1xj+1 · · ·xru(p)z1 · · · zs


...

= x1 · · ·x�−1xiq−1y1 · · ·xj−1a1t1xj+1 · · ·xru(p)z (where z = z1 · · · zs)

= x1 · · ·x�−1xiq−1y1 · · ·xj−1a1 · · ·xru(p)t1z (by Result 1.6 as y1 ∈ S \ U).

From here, we obtain the following, where equality (14) below follows by Result 1.4for some c1, c2, . . . , cs ∈ U , t ′1 ∈ S \ U , as t1 ∈ S \ U .


= x1 · · ·x�−1xiq−1y1x�+1 · · ·xj−1a1xj+1 · · ·xru(p)c1 · · · cst′1z (14)

= u2(x1, . . . , x�−1, (xiq−1y1), x�+1, . . . , xj−1, a1, . . . , xr

)u(p)c1 · · · cst

′1z

= u1(x1, . . . , x�−1, (xiq−1y1), x�+1, . . . , xj−1, a1, . . . , xr

)v(p)cλ1 · · · cλs t

′1z,

where last equality follows by inductive hypothesis. Now the word u1(ξ1, ξ2, . . . , ξr )

contains ξiq−1ξiq as a subword, so the above product in S contains (xiq−1y1)a1 asa subproduct. Thus, using the fact that y1a1 = a0 from the zigzag equations, andas equalities (15) and (16) below follow by Result 1.4 as c1c2 · · · cst

′1 = t1, and by

Result 1.6 as a0 = y1a1 with y1 ∈ S \ U respectively, we have


= u1(x1, . . . , x�−1, (xiq−1y1), x�+1, . . . , xj−1, a1, . . . , xr

)v(p)cλ1 · · · cλs t

′1z

= u1(x1, . . . , x�−1, xiq−1 , x�+1, . . . , xj−1, a0, . . . , xr )v(p)cλ1 · · · cλs t′1z

= u2(x1, . . . , x�−1, . . . , xj−1, a0, . . . , xr )u(p)cs · · · cst′1z


= x1x2 · · ·x�−1xiq−1x�+1 · · ·xj−1a0xj+1 · · ·xru(p)csc2 · · · cst′1z

= x1x2 · · ·x�−1xiq−1x�+1 · · ·xj−1a0xj+1 · · ·xru(p)t1z1z2 · · · zs (15)

= x1x2 · · ·x�−1xiq−1x�+1 · · ·xj−1a0t1xj+1 · · ·xru(p)z1z2 · · · zs (16)

= x1 · · ·x� · · ·xj−1xj · · ·xru(p)z1 · · · zs

(by the zigzag equations and xiq−1 = x�)

as required.

Case (3) j + 1 = �. Now

xi1xi2 · · ·xir v(p)zλ1zλ2 · · ·yλs

= xi1xi2 · · ·xiq−1yma2mxiq+1 · · ·xir v(p)zλ1 · · · zλs (by the zigzag equations)

= xi1xi2 · · · (xiq−1ym)a2mxiq+1 · · ·xir v(p)zλ1 · · · zλs

= x1 · · ·xj−1a2m(xiq−1ym)x�+1 · · ·xru(p)z1 · · · zs (by the inductive hypothesis)

= x1 · · ·xj−1a2m−1tm(xiq−1ym) · · ·xru(p)z1 · · · zs (by the zigzag equations)


= x1x2 · · ·xj−1a2m−1(tmxiq−1ym)x�+1 · · ·xru(p)z1z2 · · · zs

= xi1 · · ·xiq−2(tmxiq−1ym)a2m−1 · · ·xir v(p)zλ1 · · · zλs


= xi1 · · ·xiq−2(tmxiq−1ym−1)a2m−2 · · ·xir v(p)zλ1 · · · zλs


= x1 · · ·xj−1a2m−2(tmxiq−1ym−1)x�+1 · · ·xru(p)z1 · · · zs


= x1 · · ·xj−1a2m−3(tm−1xiq−1ym−1) · · ·xru(p)z1 · · · zs

(by the zigzag equations).


xi1xi2 · · ·xir v(p)zλ1zλ2 · · ·yλs

= x1 · · ·xj−1a2m−3(tm−1xiq−1ym−1) · · ·xru(p)z1 · · · zs

...

= x1x2 · · ·xj−1a1(t1xiq−1y1)x�+1 · · ·xru(p)z1z2 · · · zs

= xi1 · · ·xiq−2(t1xiq−1y1)a1xiq+1 · · ·xir v(p)zλ1 · · · zλs


= xi1 · · ·xiq−2(t1xiq−1)a0xiq+1 · · ·xir v(p)zλ1 · · · zλs


= x1x2 · · ·xj−1a0(t1xiq−1)x�+1 · · ·xru(p)z1z2 · zs


= x1x2 · · ·xj−1xjxj+1 · · ·xru(p)z1z2 · · · zs (by the zigzag equations)

as required.

Case (4) j + 1 < � and � ≤ r . When j = r , xj+1 · · ·xr denotes the empty word.Now


= xi1xi2 · · ·xiq−1yma2mxiq+1 · · ·xir v(p)zλ1 · · · zλs (by the zigzag equations)

= xi1xi2 · · · (xiq−1ym)a2mxiq+1 · · ·xir v(p)zλ1zλ2 · · · zλs

= w1(xi1, xi2, . . . , xiq−1ym,a2m,xiq+1 , . . . , xir )v(p)zλ1zλ2 · · · zλs

= w2(xi1, . . . , xiq−1ym, a2m, . . . , xir )u(p)z1 · · · zs (by the inductive hypothesis)

= x1x2 · · ·xj−1a2mxj+1 · · ·x�−1xiq−1ymx�+1 · · ·xru(p)z1z2 · · · zs


= x1 · · ·a2m−1tmxj+1 · · ·x�−1xiq−1ym · · ·xru(p)z1 · · · zs


As equality (17) below follows by Result 1.4 for some b(m)j+1, . . . , b

(m)j+(�−j−1) ∈ U and

t ′m ∈ S \ U , as tm ∈ U , we have


= x1 · · ·a2m−1tmxj+1 · · ·x�−1xiq−1ym · · ·xru(p)z1 · · · zs


= x1 · · ·a2m−1b(m)j+1 · · ·b(m)

j+(�−j−1)t ′mxj+1 · · ·x�−1xiq−1ym · · ·xru(p)z1 · · · zs

(17)

= u2(x1, . . . , a2m−1, b

(m)j+1, . . . , b

(m)j+(�−j−1)

, t ′mxj+1 · · ·xiq−1ym, . . . , xr

)

× u(p)z1 · · · zs

= u1(x1, . . . , a2m−1, b

(m)j+1, . . . , b

(m)j+(�−j−1), t

′mxj+1 · · ·xiq−1ym, . . . , xr

)

× v(p)zλ1 · · · zλs ,

where last equality follows by inductive hypothesis. Since u1(ξ1, ξ2, . . . , ξr ) containsξiq−1ξiq as a subword, the above product in S contains (xiq−1ym)a2m−1 as a subprod-uct, which by zigzag equations, equals to (xiq−1ym−1)a2m−2. Therefore


= u1(x1, . . . , a2m−1, b

(m)j+1, . . . , b

(m)j+(�−j−1), t

′m · · ·xiq−1ym, . . . , xr

)v(p)zλ1 · · · zλs

= u1(x1, . . . , a2m−2, b

(m)j+1, . . . , b

(m)j+(�−j−1), t

′m · · ·xiq−1ym−1, . . . , xr

)

× v(p)zλ1 · · · zλs .

As equalities (18) and (19) follow by inductive hypothesis and by Result 1.4 respec-tively as b

(m)j+1 · · ·b(m)

j+(�−j−1)t′m = tm, we have


= u1(x1, . . . , a2m−2, b

(m)j+1, . . . , b

(m)j+(�−j−1), t

′m · · ·xiq−1ym−1, . . . , xr

)

× v(p)zλ1 · · · zλs

= u2(x1, . . . , a2m−2, b

(m)j+1, . . . , b

(m)j+(�−j−1), t

′m · · ·xiq−1ym−1, . . . , xr

)

× u(p)z1 · · · zs (18)

= x1x2 · · ·xj−1a2m−2b(m)j+1 · · ·b(m)

j+(�−j−1)t′m · · ·x�−1xiq−1ym−1 · · ·xru(p)z1 · · · zs

= x1x2 · · ·xj−1a2m−2tmxj+1 · · ·x�−1xiq−1ym−1x�+1 · · ·xru(p)z1z2 · · · zs (19)


= x1 · · ·a2m−3tm−1 · · ·x�−1xiq−1ym−1 · · ·xru(p)z1 · · · zs


Continuing in this way, as equality (20) below follows by Result 1.4 for someb

(1)j+1, . . . , b

(1)j+(�−j−1)

∈ U and t ′1 ∈ S \ U , as t1 ∈ S \ U , we obtain


= x1 · · ·xj−1a2m−3tm−1 · · ·x�−1xiq−1ym−1 · · ·xru(p)z1 · · · zs

...

= x1x2 · · ·xj−1a1t1xj+1 · · ·x�−1xiq−1y1x�+1 · · ·xru(p)z1z2 · · · zs

= x1 · · ·a1b(1)j+1 · · ·b(1)

j+(�−j−1)t′1xj+1 · · ·x�−1xiq−1y1 · · ·xru(p)z1 · · · zs (20)

= u2(x1, . . . , a1, b

(1)j+1, . . . , b

(1)j+(�−j−1), t

′1xj+1 · · ·x�−1xiq−1y1, . . . , xr

)

× u(p)z1 · · · zs

= u1(x1, . . . , a1, b

(1)j+1, . . . , b

(1)j+(�−j−1)

, t ′1xj+1 · · ·x�−1xiq−1y1, . . . , xr

)

× v(p)zλ1 · · · zλs ,

where last equality follows by the inductive hypothesis. As before, u1(ξ1, ξ2, . . . , ξr )

contains ξiq−1ξiq as a subword, the above product in S contains (xiq−1y1)a1 as a sub-product which, by the zigzag equations, equals to xiq−1a0. Thus


= u1(x1, . . . , a1, b

(1)j+1, . . . , b

(1)j+(�−j−1), t

′1xj+1 · · ·x�−1xiq−1y1, . . . , xr

)

× v(p)zλ1 · · · zλs

= u1(x1, . . . , a0, b

(1)j+1, . . . , b

(1)j+(�−j−1)

, t ′1xj+1 · · ·x�−1xiq−1 , . . . , xr

)

× v(p)zλ1 · · · zλs .

Now, as equalities (21) and (22) below follow by the inductive hypothesis and asb

(1)j+1 · · ·b(1)

j+(�−j−1)t ′1 = t1, we have


= u1(x1, . . . , a0, b

(1)j+1, . . . , b

(1)j+(�−j−1), t

′1xj+1 · · ·x�−1xiq−1 , . . . , xr

)

× v(p)zλ1 · · · zλs

= u2(x1, . . . , a0, b

(1)j+1, . . . , b

(1)j+(�−j−1)

, t ′1xj+1 · · ·x�−1xiq−1 , . . . , xr

)

× u(p)z1 · · · zs (21)

= x1x2 · · ·xj−1a0b(1)j+1 · · ·b(1)

j+(�−j−1)t′1xj+1 · · ·x�−1xiq−1 · · ·xru(p)z1z2 · · · zs


= x1x2 · · ·xj−1a0t1xj+1 · · ·x�−1xiq−1x�+1 · · ·xru(p)z1z2 · · · zs (22)

= x1x2 · · ·xj−1xjxj+1 · · ·x�−1x�x�+1 · · ·xru(p)z1z2 · · · zs,

where last equality follows by the zigzag equations as xiq = a0t1 = xj and xiq−1 = x�,as required.

By arguments similar to the proof of Proposition 2.6, we have

Proposition 2.7 Let U,S be semigroups such that U satisfy a seminormal identityand is dense in S and let u,v be any words in p1,p2, . . . , p�. Let r, s be any positiveintegers such that r ≥ g0 − 1 and s ≥ h0 − 1. If U satisfies

x1x2 · · ·xru(p1,p2, . . . , p�)z1z2 · · · zs = xi1 · · ·xir v(p1,p2, . . . , p�)zλ1 · · · zλs ,

where i and λ are any permutations of {1,2, . . . , r} and {1,2, . . . , s} respectively,then the above identity is also satisfied for all xi1, xi2, . . . , xir , zλ1 , zλ2 , . . . , zλs in S

and p1,p2, . . . , p� ∈ U .In the following, by length of a word u, denoted by �(u), we mean the sum of the

number of occurrences of all variables appearing in u.

Lemma 2.8 ([8, Lemma 2.7.1]) Let (1) be any seminormal identity, and let u,v andw be any words in the variables x1, x2, . . . , xk (k ≥ 2) such that �(u) ≥ g0 − 1 and�(v) ≥ h0 − 1. Take any a1, a2, . . . , ak ∈ U and t1, t2, . . . , tk ∈ S1 (the semigroupobtained by adjoining 1 to S, if necessary). If for each i such that ti ∈ S, ai = yibi

[ai = biyi] for some yi ∈ S \ U and bi ∈ S (i = 1,2, . . . , k), then for any choiced1, d2, . . . , dk for the variables x1, x2, . . . , xk in S respectively

u(d)w(a1t1, a2t2, . . . , aktk)v(d) = u(d)w(a1, a2, . . . , ak)w(t1, t2, . . . , tk)v(d)[u(d)w(t1a1, t2a2, . . . , tkak)v(d) = u(d)w(t1, t2, . . . , tk)w(a1, a2, . . . , ak)v(d)

],

where d = (d1, d2, . . . , dk).

The following lemma is an easy consequence of Result 1.6.

Lemma 2.9 Let S be any permutative semigroup and let u1, u2, u3,w and w′ be anywords in the variables x1, x2, . . . , xk such that �(w) ≥ g0 − 1, �(w′) ≥ h0 − 1. Then

w(x)u1(x)u2(x)u3(x)w′(x) = w(x)uj1(x)uj2(x)uj3(x)w′(x),

where x = (x1, x2, . . . , xk) ∈ S(k) and j is any permutation of the set {1,2,3}.

Lemma 2.10 Let U,S be semigroups such that U satisfy a seminormal identity andis dense in S. Let u, v and w be any words in the variables x1, x2, . . . , xk (k ≥ 2)

such that �(u) ≥ g0 − 1 and �(w) ≥ h0 − 1. If xi ∈ C(v), for some (1 ≤ i ≤ k), besuch that xi ∈ S \ U , then


u(x)v(x)w(x) = u(x)c1c2 · · · cq

(x′i

)|xi |v v(x(i)

)w(x)

[u(x)v(x)w(x) = u(x)v

(x(i)

)(x′i

)|xi |v c1c2 · · · cqw(x)]

in S1, for some c1, c2, . . . , cq ∈ U and any q ≥ 1 (in fact the two products are equalin S), where x = (x1, x2, . . . , xk) and for any 1 ≤ i ≤ k

x(i) = (x1, x2, . . . , xi−1,1, xi+1, . . . , xk),

∀x1, x2, . . . , xk ∈ S (thus the product v(x(i)) is obtained from the product v(x) byomitting all the occurrences of the element xi ).

Remark We shall make use of the above vector notation x and x(i) here and in therest of the text without further mention.

Proof Suppose that xi ∈ S \ U for some i ∈ {1,2, . . . , k}. Then xi = x′b = x′b′y′,for some b, b′ ∈ U and x′, y′ ∈ S \ U . Equalities (23), (24) and (25) below follow byResult 1.6 as y′ ∈ S \U , �(u) ≥ g0 −1; x′ ∈ S \U , �(w) ≥ h0 −1; and by Results 1.4and 1.6 for some b1, b2, . . . , bq ∈ U and x′

i ∈ S \ U , as xi ∈ S \ U and �(u) ≥ g0 − 1respectively, we have

u(x)v(x)w(x)

= u(x)v(x1, x2, . . . , xi−1, xi, xi+1, . . . , xk)w(x)

= u(x)v(x1, x2, . . . , xi−1, x

′b′y′, xi+1, . . . , xk

)w(x)

= u(x)(x′)|xi |v v

(x1, x2, . . . , xi−1, b

′y′, xi+1, . . . , xk

)w(x) (23)

= u(x)(x′b′y′)|xi |v v(x1, x2, . . . , xi−1,1, xi+1, . . . , xk)w(x) (24)

= u(x)(xi)|xi |v v(x1, x2, . . . , xi−1,1, xi+1, . . . , xk)w(x)

= u(x)(b1)|xi |v (b2)

|xi |v · · · (bq)|xi |v (x′i

)|xi |uv(x(i)

)w(x) (25)

= u(x)c1c2 . . . cq

(x′i

)|xi |v v(x(i)

)w(x)

where c1 = (b1)|xi |v , . . . , cq = (bq)|xi |v , as required.

The dual statement may be proven analogously. �

Let

d = (d1, d2, . . . , d�).

If T [γ ], for any semigroup T and for any integer γ ≥ 2, denotes the Cartesianproduct of the γ -copies of T , then, as established in [5, Lemma 4.3], d has a zigzag

over U [�] in (S1)[�]

of length m as follows:

d = a0 t1, a0 = y1a1,

yka2k = yk+1a2k+1, a2k−1 tk = a2k tk+1 (1 ≤ k ≤ m − 1, 1 ≤ i ≤ m − 1);


a2m−1 tm = a2m, yma2m = d;(26)

where at ∈ U [�] (t = 0,1, . . . ,2m) and yq , tq ∈ (S1)[�]

(q = 1,2, . . . ,m).

Proposition 2.11 Let U be a semigroup satisfying a seminormal identity anddense in S. Let u, v be any words in p1,p2, . . . , p� such that C(u) ∪ C(v) ={p1,p2, . . . , p�} and let r , s be any positive integers such that r ≥ g0 − 1 ands ≥ h0 − 1. If pj ∈ C(v) is such that pj ∈ S \ U for some j ∈ {1,2, . . . , �}, then

xi1xi2 · · ·xir v(d)zλ1zλ2 · · · zλs = x1x2 · · ·xru(yma2m−1)v(tm)zλ1zλ2 · · · zλs ,

where λ is a permutation of the set {1,2, . . . , s}.

Proof As U is dense in S, by Result 1.1, S satisfy a seminormal identity. Let pj ∈C(v) is such that pj ∈ S \ U for some j ∈ {1,2, . . . , �}. Then

xi1xi2 · · ·xir v(d)zλ1zλ2 · · · zλs

= xi1xi2 · · ·xir v(a0 t1)zλ1zλ2 · · · zλs (by (26))

= xi1xi2 · · ·xir v(a0)v(t1)zλ1zλ2 · · · zλs (by Lemma 2.8)

= xi1xi2 · · ·xir v(a0)c1c2 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z,

where last equality follows by Lemma 2.10 for some c1, c2, . . . , cs ∈ U , as t(j)

1 is inS \ U , and where z = zλ1zλ2 · · · zλs . Therefore


= xi1xi2 · · ·xir v(a0)c1c2 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z

= xi1 · · ·xir v(a0)cλ1 · · · cλs

(t(j)′1

)|pj |v v(t1(j)

)z

(by Result 1.6, as t(j)′1 in S \ U )

= x1 · · ·xru(a0)c1c2 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z (by Proposition 2.7)

= x1 · · ·xru(y1a1)c1c2 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z (by (26))

= x1 · · ·xru(y1)u(a1)c1c2 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z (by the dual of Lemma 2.8)

= x1 · · ·xru(a1)c1c2 · · · csu(y1)(t(j)′1

)|pj |v v(t1(j)

)z

(by Result 1.6 as t(j)′1 ∈ S \ U )

= xi1xi2 · · ·xir v(a1)cλ1cλ2 · · · cλs u(y1)(t(j)′1

)|pj |v v(t1(j)


= xi1 · · ·xir u(y1)v(a1)c1 · · · cs

(t(j)′1

)|pj |v v(t1(j)

)z

(by Result 1.6 as t(j)′1 ∈ S \ U )


= xi1xi2 · · ·xir u(y1)v(a1)v(t1)z (by Lemma 2.10)

= xi1xi2 · · ·xir u(y1)v(a1 t1)z (by Lemma 2.8)

= xi1xi2 · · ·xir u(y1)v(a2 t2)z (by (26)).



= xi1xi2 · · ·xir u(y1)v(a2 t2)z

...

= xi1xi2 · · ·xir u(ym−1)v(a2m−2 tm)z

= xi1xi2 · · ·xir u(ym−1)v(a2m−2)v(tm)z (by Lemma 2.8)

= xi1xi2 · · ·xir v(a2m−2)v(tm)u(ym−1)z (by Lemma 2.9)

= xi1xi2 · · ·xir v(a2m−2)u1u2 · · ·us

(t(j)′m

)|pj |v v( ˜tm(j)

)u(ym−1)z,

where last equality follows by Lemma 2.10 for some u1, u2, . . . , us ∈ U , as t(j)m is in

S \ U . Therefore


= xi1xi2 · · ·xir v(a2m−2)u1u2 · · ·us

(t(j)′m

)|pj |v v( ˜tm(j)

)u(ym−1)z

= xi1 · · ·xir v(a2m−2)uλ1 · · ·uλs

(t(j)′m

)|pj |v v( ˜tm(j)

)u(ym−1)z (by Result 1.6)

= x1 · · ·xru(a2m−2)u1u2 · · ·us

(t(j)′m

)|pj |v v( ˜tm(j)

)u(ym−1)z

(by Proposition 2.7)

= x1x2 · · ·xru(a2m−2)v(tm)u(ym−1)z (by Lemma 2.10)

= x1x2 · · ·xru(ym−1)u(a2m−2)v(tm)z (by Lemma 2.9)

= x1x2 · · ·xru(ym−1a2m−2)v(tm)z (by the dual of Lemma 2.8)

= x1x2 · · ·xru(yma2m−1)v(tm)z (by (26))

as required. �

Proposition 2.12 Let U be a semigroup satisfying a seminormal identity anddense in S. Let u, v be any words in p1,p2, . . . , p� such that C(u) ∪ C(v) ={p1,p2, . . . , p�}, pj ∈ C(v), for some j (1 ≤ j ≤ �), be such that pj ∈ S \ U andlet r , s be any positive integers with r ≥ g0 − 1 and s ≥ h0 − 1. If pj ∈ C(u), then

x1x2 · · ·xru(yma2m−1)v(tm)zλ1 · · · zλs = x1x2 · · ·xru(d)z1z2 · · · zs

for any permutation λ of the set {1,2, . . . , s}.


Proof As U is dense in S, by Result 1.1, S satisfy a seminormal identity. Lettingz = zλ1 · · · zλs and z′ = x1x2 · · ·xr , we have

x1x2 · · ·xru(yma2m−1)v(tm)zλ1 · · · zλs

= z′u(ym)u(a2m−1)v(tm)z (by the dual of Lemma 2.8)

= z′u(ym(j)

)(y

(j)′m

)|pj |ub1b2 · · ·bru(a2m−1)c1c2 · · · cs

(t(j)′m

)|pj |v v( ˜tm(j)

)z,

where last equality follows by Lemma 2.10 and its dual for some b1, . . . , br andc1, c2, . . . cs ∈ U . As equalities (27) and (28) below follow by Proposition 2.7, andby Lemma 2.10 for some q1, q2, . . . , qr ∈ U respectively, we have

x1x2 · · ·xru(yma2m−1)v(tm)zλ1 · · · zλs

= z′u(ym(j)

)(y

(j)′m

)|pj |ub1b2 · · ·bru(a2m−1)c1c2 · · · cs

(t(j)′m

)|pj |v v( ˜tm(j)

)z

= z′u(ym(j)

)(y

(j)′m

)|pj |ubi1 · · ·bir v(a2m−1)cλ1 · · · cλs

(t(j)′m

)|pj |v v( ˜tm(j)

)z (27)

= z′u(ym(j)

)(y

(j)′m

)|pj |ub1 · · ·brv(a2m−1)c1 · · · cs

(t(j)′m

)|pj |v v( ˜tm(j)

)z

(by Result 1.5)

= z′u(ym)v(a2m−1)v(tm)zλ1 · · ·yλs (by Lemma 2.10 and by the definition of z)

= z′u(ym)v(a2m−1 tm)zλ1zλ2 · · · zλs (by Lemma 2.8)

= z′u(ym)v(a2m)zλ1zλ2 · · · zλs (by (26))

= z′u(ym(j)

)(y

(j)′m

)|pj |uq1 · · ·qrv(a2m)zλ1 · · · zλs (28)

= z′u(ym(j)

)(y

(j)′m

)|pj |uqi1 · · ·qir v(a2m)zλ1zλ2 · · · zλs (by Result 1.6)

= z′u(ym(j)

)(y

(j)′m

)|pj |uq1q2 · · ·qru(a2m)z1z2 . . . zs (by Proposition 2.7)

= z′u(ym)u(a2m)z1z2 · · · zs (by the dual of Lemma 2.10)

= x1 · · ·xru(yma2m)z1 . . . zs (by Lemma 2.8 and by the definition of z′)

= x1x2 · · ·xru(d)z1z2 . . . zs (by (26))

as required. �

Proposition 2.13 Let U be a semigroup satisfying a seminormal identity anddense in S. Let u, v be any words in p1,p2, . . . , p� such that C(u) ∪ C(v) ={p1,p2, . . . , p�} and let r , s be any positive integers with r ≥ g0 − 1 and s ≥ h0 − 1.If C(u) �= C(v) and there exists pj ∈ C(v) such that pj ∈ S \ U , then

x1x2 · · ·xru(yma2m−1)v(tm)zλ1 · · · zλs = x1x2 · · ·xru(d)z1z2 · · · zs

for any permutation λ of the set {1,2, . . . , s}.Proof As U is dense in S, by Result 1.1, S satisfy a seminormal identity. If pk ∈C(u), for some k ∈ {1,2, . . . , �}, is such that pk ∈ S \ U , then y

(k)m ∈ S \ U , and the


proof follows on the lines similar to the proof of Proposition 2.12. So, assume thatpk /∈ C(u) ∩ (S \ U) for all k ∈ {1,2, . . . , �}. Letting z = zλ1zλ2 · · · zλs , we have

x1x2 · · ·xru(yma2m−1)v(tm)zλ1zλ2 · · · zλs

= x1x2 · · ·xru(a2m−1)v(tm)z (since all variables in u are replaced from U)

= x1x2 · · ·xru(a2m−1)b′1b

′2 . . . b′

s

(t(j)′m

)|pj |v v( ˜tm(j)

)z,

where last equality follows by Lemma 2.10 as b′1, b

′2, . . . , b

′s ∈ U . As equality (29)

below follows by Result 1.6, as t(j)′m ∈ S \ U , we have

x1x2 · · ·xru(yma2m−1)v(tm)z

= x1x2 · · ·xru(a2m−1)b′1b

′2 · · ·b′

s

(t(j)′m

)|pj |v v( ˜tm(j)

)z

= xi1 · · ·xir v(a2m−1)b′λ1

· · ·b′λs

(t(j)′m

)|pj |v v( ˜tm(j)


= xi1 · · ·xir v(a2m−1)b′1b

′2 · · ·b′

s

(t(j)′m

)|pj |v v( ˜tm(j)

)z (29)

= xi1xi2 · · ·xir v(a2m−1)v(tm)z (by the dual of Lemma 2.10)

= xi1xi2 · · ·xir v(a2m−1 tm)z (by the dual of Lemma 2.8)

= xi1 · · ·xir v(a2m)zλ1 · · · zλs (by (26) and as z = zλ1 · · · zλs )

= x1x2 · · ·xru(a2m)z1z2 · · · zs (by Proposition 2.7)

= x1 · · ·xru(yma2m)z1 · · · zs (since all the variables in u are replaced from U)

= x1x2 · · ·xru(d)z1z2 · · · zs (by (26))

as required. This completes the proof of Proposition 2.13. �

Theorem 2.14 Let u, v are any words in p1,p2, . . . , p� such that C(u) ∪ C(v) ={p1,p2, . . . , p�} and let (1) be any seminormal identity. If r , s be any positive integerssuch that r ≥ g0 − 1 and s ≥ h0 − 1, then all heterotypical identities of the form

x1x2 · · ·xru(p1,p2, . . . , p�)z1z2 · · · zs = xi1 · · ·xir v(p1,p2, . . . , p�)zλ1 · · · zλs ,

(30)where i and λ are any permutations of the sets {1,2, . . . , r} and {1,2, . . . , s} respec-tively, are preserved under epis in conjunction with (1).

Proof Let U and S be semigroups with U dense in S and let U , and hence S byResult 1.1, satisfy a seminormal identity. Now we shall show that the identity (30)is also satisfied by S. Let p = (p1,p2, . . . , p�). With this notation, the identity (30)becomes

x1x2 · · ·xru(p)z1z2 · · · zs = xi1xi2 · · ·xir v(p)zλ1zλ2 · · · zλs .

Now we wish to show that

x1x2 · · ·xru(d)z1z2 · · · zs = xi1xi2 · · ·xir v(d)zλ1zλ2 · · · zλs , (31)


holds in S.So assume that xi1, xi2, . . . , xir , zλ1 , zλ2 , . . . , zλs and p1,p2, . . . , p� ∈ S. If all of

xi1, xi2, . . . , xir ∈ S and zλ1 , zλ2, . . . , zλs ,p1,p2, . . . , p� lie in U , then (31) holds byProposition 2.6. On the other hand, if all of xi1, xi2, . . . , xir , zλ1 , zλ2 , . . . , zλs are in S

and p1,p2, . . . , p� ∈ U , then (31) holds by Proposition 2.7.So assume that not all of p1,p2, . . . , p� ∈ U . Then, there exists at least one pj ,

1 ≤ j ≤ � such that pj ∈ S \ U . Therefore, y(j)i , t

(j)i ∈ S \ U for all i such that 1 ≤

i ≤ m. Without loss of generality, we may assume that pj ∈ C(v). The proof of thetheorem now follows by Propositions 2.11, 2.12 and 2.13. �

Acknowledgement We sincerely thank the referee for his hard work, constructive criticism and advicesall along which has led, significantly, to improve clarity of notions and results as well as presentation ofthe paper.

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Epimorphisms, dominions and seminormal identities II