Identities over the two-element alphabet

Semigroup Forum Vol. 52 (1996), 365–379c© 1996 Springer-Verlag New York Inc.

RESEARCH ARTICLE

Identities Over the Two-Element Alphabet

Miroslav Ciric and Stojan Bogdanovic*

Communicated by Boris M. Schein

AbstractIn [4] and [5] the authors considered semigroup identities that induce various typesof decompositions. In this paper we continue this work for identities over the two-element alphabet. In particular, we solve some problems that are not solved in thegeneral case and we solve several types of well-known Tamura’s problems.

1. Introduction and preliminaries

In this section we introduce basic notions, notations and results.By Z+ we denote the set of all positive integers. By Reg(S)(Gr(S), E(S)) wedenote the set of all regular (completely regular, idempotent) elements of asemigroup S . If S is a semigroup and a, b ∈ S , then a | b iff b = xay forsome x, y ∈ S1 , a |

rb iff ax = b for some x ∈ S1 , a |

lb iff xa = b for some

x ∈ S1 . By “ -” we denote the complement of the relation “ |”. A semigroup Sis Archimedean (left Archimedean, right Archimedean, t-Archimedean) if for alla, b ∈ S there exists n ∈ Z+ such that b | an (b |

lan, b |

ran, b |

lan and b |

ran).

By S = S0 we denote that S is a semigroup with the zero 0. An elementa ∈ S0 is a nilpotent if there exists n ∈ Z+ such that an = 0. A semigroup S0

is a nil-semigroup if all of its elements are nilpotents. An ideal extension S of asemigroup K is a nil-extension if S/K is a nil-semigroup. An ideal extension Sof a semigroup K is a retract extension (or retractive extension) of K if thereexists a homomorphism ϕ of S onto K such that ϕ(a) = a for all a ∈ K . Sucha homomorphism we call a retraction. A retractive extension S of a semigroupK is an inflation of K if S2 ⊆ K .

A regular semigroup is orthodox if its idempotents form a subsemigroup.An orthodox union of groups we call an orthogroup, a band of groups we calla cryptogroup and an orthodox band of groups we call an orthocryptogroup. IfS is a union of groups, then Green’s J -classes of S we call completely simplecomponents of S . A band S is right (left) regular if xa = axa (ax = axa) forall a, x ∈ S . A semigroup S is π -regular if for any a ∈ S there exists n ∈ Z+

such that an is regular.By B2 we denote the semigroup

B2 =⟨

a, b | a2 = b2 = 0, aba = a, bab = b⟩

.

It is not a semilattice of Archimedean semigroups. By L2 (R2 ) we denote thetwoelement left (right) zero semigroup.

Let X1 and X2 be classes of semigroups. By X1 ◦X2 we denote the Mal’cev’sproduct of the classes X1 and X2 , i.e. the class of all semigroups S on whichthere exists a congruence ρ such that S/ρ is in X2 and every ρ -class that is a

* Supported by Grant 0401A of RFNS through Math. Inst. SANU

366 Ciric and Bogdanovic

subsemigroup is in X1 . The related decomposition is an X1 ◦ X2 -decomposition.We introduce the following notations:

Notation Class of semigroups Notation Class of semigroupsA Archimedean M×G rectangular groupsLA left Archimedean πR π -regularT A t-Archimedean N nil-semigroupsUG unions of groups S semilatticesCS completely simple

Clearly X ◦ S is the class of all semilattices of semigroups from the class X . IfX2 is a subclass of the class N , then X1 ◦ X2 is a class of all semigroups thatare ideal extensions of semigroups from X1 by semigroups from X2 . Also, in such acase, by X1 ~X2 we denote the class of all semigroups that are retract extensionsof semigroups from X1 by semigroups from X2 .

By A+ we denote the free semigroup over an alphabet A and by A∗ wedenote the free monoid over an alphabet A . By |u| we denote the length of aword u ∈ A+ and by |x|u we denote the number of appearances of the letterx in the word u . A word v ∈ A+ is a subword (left cut, right cut) of a wordu ∈ A+ if v | u (v |

ru, v |

lu). If u ∈ A+, |u| ≥ 2, then by h(2)(u) (t(2)(u))

we denote the left (right) cut of u of length 2. By h(u) (t(u)) we denote thehead ( tail), by l(u) (r(u)) we denote the left (right) part and by c(u) we denotethe content of the word u [7]. Since in this paper we consider free semigroupsover finite alphabets, we introduce the following notations: For n ∈ Z+, n ≥ 3,An = {x1, x2, . . . , xn} , and A2 = {x, y} .

By [u = v] we denote the variety determined by the identity u = v .Identities u = v and u′ = v′ over an alphabet A+

n are p-equivalent if u′ = v′ canbe obtained from u = v by some permutation of letters. Clearly, p-equivalentidentities determine the same variety. If X is a class of semigroups, then (1) isan X -identity if [u = v] ⊆ X . If X1 and X2 are classes of semigroups, then (1)is a X1 . X2 -identity if [u = v] ∩ X1 ⊆ X2 . If u = v is an identity over thealphabet An, n ≥ 3, then we introduce notations pi = ||xi|u − |xi|v|, 1 ≤ i ≤ n ,and the number p = gcd(p1. . . . , pn) will be called the period of u = v . If u = vis an identity over A2 , then px = ||x|u − |x|v|, py = ||y|u − |y|v| , and the periodof u = v is p = gcd(px, py).

Let ϑ be an equivalence on a semigroup A+2 determined by the partition

Ca = {(xy)nx | n ∈ Z+ ∪ {0} }, Cb = {(yx)ny | n ∈ Z+ ∪ {0} },Cab = {(xy)n | n ∈ Z+}, Cba = {(yx)n | n ∈ Z+},

C0 = A+2 − (Ca ∪ Cb ∪ Cab ∪ Cba).

Then ϑ is a congruence on A+2 and the factor semigroup A+

2 /ϑ is isomorphic toB2 [4].

For undefined notions and notations we refer to [1] and [7].In the next considerations the following result will be very useful:

Shevrin-Veronesi’s theorem. [9], [13] A semigroup S is a semilattice of nil-extensions of completely simple semigroups if and only if S is π -regular andReg(S) = Gr(S) .

Ciric and Bogdanovic 367

Identities which induce various types of semigroup decompositions have beenstudied by the authors in [4] and [5]. In [4] the authors described all identities thatinduce decompositions into semilattices of Archimedean semigroups. Here wegive a finer description of such identities over the two-element alphabet (Theo-rem 1). Also, in [4] the authors described identities that induce decompositionsof π -regular semigroups into semilattices of nil-extensions of left groups. But inthe general case, the problem of recognition of identities that induce decompo-sitions into semilattices of left (right, twosided) Archimedean semigroups is notsolved. By Theorem 2 and by Corollary 1 of this paper we solve this problem foridentities over the two-element alphabet. Furthermore, we describe identitiesover A2 that induce orthodoxity in completely simple semigroups.

In Section 3 we first consider the problem of ”transmission of identitiesthrough several semigroup compositions”. Namely, we describe orthogroups andcryptogroups that satisfy some semigroup identity, by Theorems 4 and 5. Usingthis and the results from Section 2 we find solutions for several types of the well-known Tamura’s problems, stated in [12]. Some special types of these problems areconsidered by M. S. Putcha and J. Weissglass [8], J. Clarke [6] and S. Bogdanovic[3], and about an importance of these problems for varieties of semigroups thereader can be informed in articles by L. N. Shevrin and M. V. Volkov [10] andL. N. Shevrin and E. V. Suhanov [11].

2. Decompositions induced by identities over A2

In this section we shall consider the identity

(1) u = v

over the alphabet A2 for which c(u) = c(v) = A2 .

The following theorem gives a characterization of identities over A2 thatinduce a decomposition into a semilattice of Archimedean semigroups.

Theorem 1. The identity (1) is an A ◦ S -identity if and only if it isp-equivalent to one of the following identities:

(A1) xy = w , where w ∈ A+2 − {xy} ;

(A2) (xy)k = w , where k ∈ Z+, k ≥ 2 and w ∈ A+2 − {(xy)m | m ∈ Z+} ;

(A3) (xy)kx = w , where k ∈ Z+ and w ∈ A+2 − {(xy)mx | m ∈ Z+} ;

(A4) xyk = w , where k ∈ Z+, k ≥ 2 and w ∈ A+2 − {xym | m ∈ Z+} ;

(A5) xky = w , where k ∈ Z+, k ≥ 2 and w ∈ A+2 − {xmy | m ∈ Z+} .

Proof. Let (1) be a A ◦ S -identity. Then by Theorem 2 [4] we obtain thatthere exists a homomorphism T : A+

2 → A+2 such that

(2) (T (u), T (v)) /∈ ϑ.

Then it is clear that T (u) /∈ C0 or T (v) /∈ C0 . Without loss of generality wecan assume that

(3) T (u) /∈ C0.

Also, without loss of generality we can assume that h(u) = x . Then clearlyxy | u .

Let x2, y2 | u . By (3) it follows that

T (x), T (y) ∈ Cab or T (x), T (y) ∈ Cba.


But now we obtain that

T (u), T (v) ∈ Cab or T (u), T (v) ∈ Cba,

that is in contradiction with (2). Therefore, x2 - u or y2 - u . Assume that

x2 - u and y2 | u.

By (3) it follows that T (y) ∈ Cab ∪ Cba . Let T (y) ∈ Cab . Then by xy | uand by (3) it follows that T (x) ∈ Cab ∪ Cb . If yx | u , then by (3) we obtainthat T (x) ∈ Cab ∪ Ca , so T (x) ∈ Cab , whence it follows that (2) does not hold.Therefore, yx - u , so

(4) u = xyk , k ∈ Z+, k ≥ 2 .

In a similar way we consider the case with T (y) ∈ Cba . Let

x2 | u and y2 - u.

Then in a similar way we prove that

(5) u = xky , k ∈ Z+, k ≥ 2 .

Finally, letx2 - u and y2 - u.

Then it is clear that

(6) u = (xy)k , k ∈ Z+,

or

(7) u = (xy)kx , k ∈ Z+.

Assume that (4) hold. If v = xym for some m ∈ Z+, m ≥ 2, then it is easyto verify that the semigroup B2 satisfies (1), that is not possible. Therefore, itremains that v = xy , so (A1) holds, or that v /∈ {xym | m ∈ Z+} , so (A4) holds.If (5) holds, then similarly we obtain that (A1) or (A5) holds.

Let (6) hold. If k = 1, then v 6= xy (otherwise B2 satisfies (1) ), whence itfollows that (A1) holds. Let k ≥ 2. Since B2 does not satisfy (1), then it followsthat (A1) or (A2) holds.

Finally, let (7) hold. If v = (xy)mx for some m ∈ Z+ , then it is easy toshow that B2 satisfies (1), which is not possible. Therefore, (A3) holds.

Conversely, let the identity (1) is p-equivalent to the identity of one of theforms (A1)–(A5). Without loss of generality we can assume that (1) has one ofthe forms (A1)–(A5).

If (1) is of the form (A2) or (A3), then by Theorem 2 [4] it follows that (1)is a A ◦ S -identity.

Let (1) be of the form (A1), i.e. let u = xy , v = w . Then v 6= xy , whence

x2 | v ory2 | v or yx | v.


If x2 | v or y2 | v , then by Theorem 2 [4] we obtain that (1) is a A ◦ S -identity.Assume that yx | v . Let T : A+

2 → A+2 be the homomorphism determined by

the mapping

(8)(

x yx yx

)

of A2 into A+2 . Then T (u) = xyx and x2 | T (v) , so by Theorem 2 [4] it follows

that (1) is an A ◦ S -identity.Let (1) be of the form (A4), i.e. let u = xyk , k ∈ Z+ , k ≥ 2 and v = w .

If v = xy , then we obtain the case (A1). Let v 6= xy . Then by (A4) we obtainthat

x2 | v or yx | v.

If T is the homomorphism determined as in (9), then T (u) = x(yx)k = (xy)kxand x2 | T (v), so by Theorem 2 [4] it follows that (1) is an A ◦ S -identity.Similarly we consider the case when (1) is of the form (A5).

By the following theorem we describe all identities over A2 that inducedecompositions into a semilattice of left Archimedean semigroups. Such identitiesare not characterized in the general case for identities with more than two letters.

Theorem 2. The following conditions for the identity (1) are equivalent:(i) (1) is an LA ◦ S -identity;(ii) (1) is not satisfied in the semigroups B2 and R2 ;(iii) (1) is an A ◦ S -identity and t(u) 6= t(v) .

Proof. (i)⇒(ii) and (ii)⇒(iii). This follows immediately.(iii)⇒(i). Let (iii) hold, let S be a semigroup that satisfies (1) and let

a, b ∈ S . By Theorem 1 we obtain that (1) is p -equivalent to some identity ofthe form (A1), (A2), (A3), (A4) or (A5).

Let (1) be p-equivalent to some identity of the form (A1) or (A2). Then (1)is p -equivalent to the identity

(xy)k = w′x,

w′ ∈ A+2 . Let F : A+

2 → S be the homomorphism determined by the mapping(

x ya b

)

.

Then by F (u) = F (v) it follows that

(9) (ab)m ∈ Sa for some m ∈ Z+.

Assume that (1) is p -equivalent to some identity of the form (A3). Then it isclear that (1) is p-equivalent to the identity

(xy)kx = w′y,

w′ ∈ A+2 . Let F : A+


x yb a

)

.


Then by F (u) = F (v) it follows that (ba)kb ∈ Sa , whence we obtain that (9)holds.

Let (1) be p-equivalent to some identity of the form (A5). Then (1) isp-equivalent to the identity

xky = w′x,

w′ ∈ A+2 . Let F : A+


x yba a

)

.

Then by F (u) = F (v) it follows that (ba)kb ∈ Sba ⊆ Sa , whence we obtainthat (9) holds.

Finally, let (1) be p-equivalent to some identity of the form (A4). Then it isclear that (1) is p-equivalent to the identity

(10) xyk = w′x,

w′ ∈ A+2 . Suppose that (10) is an aperiodic identity. Then we obtain that

w′ = yk , so by the permutation of letters we obtain the previous case. Let (1) be aperiodic identity. Then S is periodic semigroup, so by Theorem 4 [4] and byTheorem 1.1 [2] it follows that (9) holds. Therefore, (9) holds in any of the cases,so by Theorem 1.1 [2] it follows that S is a semilattice of right Archimedeansemigroups, i.e. (i) holds.

Corollary 1. The following conditions for the identity (1) are equivalent:(i) (1) is a T A ◦ S -identity;(ii) (1) is not satisfied in semigroups B2, R2 and L2 ;(iii) (1) is a A ◦ S -identity, t(u) 6= t(v) and h(u) 6= h(v) .

By the following theorem we describe identities that induce orthodoxity incompletely simple semigroups.

Theorem 3. The identity (1) is a CS .M× G -identity if and only if one ofthe following conditions holds:

(B1) h(u) 6= h(v) or t(u) 6= t(v) ;(B2) (1) is p-equivalent to some identity of the form

xm1yn1xm2yn2 · · ·xmhynh = xk1yl1xk2yl2 · · ·xksyls

mi , ni , kj , lj ∈ Z+ , with

gcd(px, py, h− s) = 1 ,

where px =∑ h

i=1 mi −∑ s

j=1 kj and py =∑ h

i=1 ni −∑ s

j=1 lj .(B3) (1) is p-equivalent to some identity of the form

xm1yn1xm2yn2 · · ·xmhynhxmh+1 = xk1yl1xk2yl2 · · ·xksylsxks+1

mi , ni , kj , lj ∈ Z+ , with

gcd(px, py, h− s) = 1 ,


where px =∑h+1

i=1 mi −∑s+1

j=1 kj and py =∑ h

i=1 ni −∑ s

j=1 lj .

Proof. Let (1) be a CS .M×G -identity. If h(u) 6= h(v) or t(u) 6= t(v), thenwe obtain (B1). Assume that h(u) = h(v) and t(u) = t(v).

First we will consider the case with h(u) 6= t(u). Then it is clear that theidentity (1) is p -equivalent to some identity of the form (B2). Let we prove thatgcd(px, py, h− s) = 1. Assume that

gcd(px, py, h− s) = r ≥ 2.

Let G be a cyclic group of order r and let I,Λ be sets such that |I|, |Λ| ≥ 2.Since G is commutative, we then have that for all a, b ∈ G it is

am1bn1 . . . amhbnh = ambn

= akbl (since m ≡ k (mod r), n ≡ l (mod r) ),

= ak1bl1 . . . aksbls ,

where m =h

∑

i=1

mi, n =h

∑

i=1

ni, k =s

∑

j=1

kj and l =s

∑

j=1

lj . Fix i ∈ I, λ ∈ Λ and

define a Λ× I matrix P over G by

pµj ={

c if µ = λ, j = i

e otherwise,

where e is the identity element of G and c ∈ G such that c 6= e . LetS = M(G; I,Λ; P ) be the Rees matrix semigroup over G with the sandwichmatrix P . Let (a; j, µ), (b; t, ν) ∈ S . Then by the definition of the multiplicationon S and by the commutativity in G we obtain that

(a; j, µ)m1(b; t, ν)n1 . . . (a; j, µ)mh(b; t, ν)nh = (ambnpm−hµj ph−1

νt ph−1νj ; j, ν),

and

(a; j, µ)k1(b; t, ν)l1 . . . (a; j, µ)ks(b; t, ν)ls = (akblpk−sµj ps−1

νt ps−1νj ; j, ν).

Since r is a common divisor for px = m− k, py = n− l and h− s , then

m ≡ k (mod r), n ≡ l (mod r) and h ≡ s (mod r),

so

m− h ≡ k − s (mod r), n− h ≡ l − s (mod r) and h− 1 ≡ s− 1 (mod r).

Thus,

am = ak, bn = bl, pm−hµj = pk−s

µj , pn−hνt = pl−s

νt , ph−1µt = ps−1

µt , ph−1νj = ps−1

νj ,

whence it follows that S satisfies (1).Let j ∈ I, j 6= i and let µ ∈ Λ, µ 6= λ . Then (e; j, λ) and (e; i, µ) are

idempotents and (e; j, λ)(e; i, µ) = (pλi; j, µ) = (c; j, µ) is not an idempotent.Therefore, S is not orthodox, so S is not a rectangular group, which is incontradiction with our hypothesis. The similar contradiction we obtain in thecase with px = py = 0, h = s . Hence, (B2) holds.


If h(u) = t(u), then similarly we show that (B3) holds.Conversely, let one of the conditions (B1)–(B3) holds, let S be a completely

simple semigroup that satisfies (1) and let e, f ∈ E(S).Let (B1) hold. Assume that h(u) 6= h(v) (the similar proof we have for

t(u) 6= t(v) ). Without loss of generality we can assume that h(u) = x, h(v) = y .If F : A+


x yf ef

)

,

then F (u) = f(ef)m and F (u) = (ef)n , where m = |y|u, n = |y|v , whence

f(ef)m = (ef)n.

Now we obtain that

(ef)n = e(ef)n = ef(ef)m = eff(ef)m = ef(ef)n = (ef)n+1.

Since ef is a group-element, then we obtain that ef = (ef)2 , so S is a rectan-gular group, i.e. (1) is a CS .M×G -identity.

Let (B2) hold. Since ef is a group-element, then by r we will denote theorder of ef in the subgroup of S . Since S satisfies the identity from (B2), wethen obtain that

(ef)h = (ef)s, (ef)m = (ef)k, (ef)n = (ef)l ,

and by this it follows that r divides the integers m−k, n− l and h− s . Thus, by(B2) we obtain that r = 1, i.e. ef = (ef)2 . Therefore, S is orthodox, so it isa rectangular group.

Similarly we prove the case (B3).

Corollary 2. The identity (1) is a πR . (M×G ◦ N ) ◦ S -identity if and onlyif (1) is a A ◦ S -identity and a CS .M×G -identity.

Proof. This follows by Theorem 2 [4] and by Theorem 3.

The following two propositions we obtain immediately by the results from[5].

Proposition 1. The identity (1) is a UG ◦ N -identity if and only if it isp-equivalent to an identity of one of the following forms:

(C1) xy = w , where w ∈ A+2 and w /∈ {xym | m ∈ Z+}∪

{xmy | m ∈ Z+} ∪ {yx} ;(C2) xym = xny , where m,n ∈ Z+, m, n ≥ 2 .

Proof. This follows by Theorem 1 [5].

Proposition 2. The identity (1) is a UG ~ N -identity if and only if it isp-equivalent to an identity of one of the following forms:

(D1) xy = w , where w ∈ A+2 , |w| ≥ 3 and h(2)(w) 6= xy 6= t(2)(w) ;

(D2) xym = xny , where m,n ∈ Z+, m, n ≥ 2 .

Proof. This follows by Theorem 2 [5].


3. Some types of Tamura’s problems

In this section we solve several types of known Tamura’s problems.If w ∈ A+

n , S is a semigroup, a = (a1, . . . , an) ∈ Sn and F : A+n → S is

a homomorphism determined by F (xi) = ai, i ∈ {1, . . . , n} , then we will usenotation w(a) = F (w).

Theorem 4. Let S be an orthogroup. Then S satisfies the identity u = vover the alphabet An if and only if all of its subgroups satisfy u = v and

(11) l(u)(a) L l(v)(a) ∧ r(u)(a) R r(v)(a)

for all a ∈ Sn .

Proof. Let S satisfies u = v . Then it is clear that all of its subgroups satisfyu = v . Let a1, . . . an ∈ S , let a = (a1, . . . , an) and let A be the J -class of theelement a1 · · · an . Then

u(a), v(a), l(u)(a), l(v)(a), r(u)(a), r(v)(a) ∈ A.

Since A is a rectangular group, then A is a rectangular band I × Λ of groupsHiλ, i ∈ I, λ ∈ Λ. Then Hiλ are Green’s H -classes of A , Li = ∪{Hiµ |µ ∈ Λ} ,i ∈ I are Green’s L -classes of A and Rλ = ∪{Hjλ | j ∈ I} , λ ∈ Λ are Green’sR-classes of A . Let E(Hiλ) = {eiλ} , i ∈ I , λ ∈ Λ. Assume that l(u)(a) ∈ Hiλ ,l(v)(a) ∈ Hjµ for some i, j ∈ I , λ, µ ∈ Λ. Then

u(a) = l(u)(a)c for some c ∈ A,

∈ HiλA ⊆ Li

v(a) = l(v)(a)d for some d ∈ A,

∈ HjµA ⊆ Lj .

Since u(a) = v(a), then we have that i = j , i.e. l(u)(a) L l(v)(a). In a similarway we prove that r(u)(a) R r(v)(a). Therefore (11) holds.

Conversely, let G satisfies u = v for every subgroup G of S and let (11)hold. Let a1, . . . , an ∈ S , let a = (a1, . . . , an) and let A be the J -class of theelement a1 · · · an . Let we use previous notations for A . Then

u(a), v(a), l(u)(a), l(v)(a), r(u)(a), r(v)(a) ∈ A.

By (11) it follows that l(u)(a), l(v)(a) ∈ Li , r(u)(a), r(v)(a) ∈ Rλ for somei ∈ I, λ ∈ Λ. Then it is clear that u(a), v(a) ∈ LiA ⊆ Li and u(a), v(a) ∈ARλ ⊆ Rλ , so

u(a), v(a) ∈ Li ∩Rλ = Hiλ.

Assume that u(a) = bc for some b, c ∈ 〈a1, . . . , an〉 . Then bc, eiλb, ceiλ ∈ A and

eiλb = eiλeiλb ∈ HiλA ⊆ Li,

ceiλ = ceiλeiλ ∈ AHiλ ⊆ Rλ,

so eiλb ∈ Hiµ, ceiλ ∈ Hjλ for some j ∈ I, µ ∈ Λ. Now we have that

bc = eiλbceiλ = eiλbeiµejλceiλ = eiλbeiλceiλ,


since A is orthodox. By induction it follows that

(12) u(a) = eiλu(a1, . . . , an)eiλ = eiλu(a1eiλ, a2eiλ, . . . , aneiλ)eiλ.

In a similar way we prove

(13) v(a) = eiλv(a1, . . . , an)eiλ = eiλv(a1eiλ, a2eiλ, . . . , aneiλ)eiλ.

Since u(b1, . . . , bn) H v(b1, . . . , bn) for all b1, . . . , bn ∈ A , where H is the Green’sH -relation on A , then we have that I × Λ ∼= A/H satisfies u = v . SinceA is isomorphic to a direct product of I × Λ and of some of its subgroups,then by hypothesis, A satisfies u = v . Since A satisfies u = v and a1eiλ ,a2eiλ, . . . , aneiλ ∈ A , then by (12) and (13) we have that u(a) = v(a). Therefore,S satisfies u = v .

Corollary 3. Let S be an orthocryptogroup. Then S satisfies the identityu = v if and only if all of its subgroups satisfy u = v and S/H satisfies u = v .

Proof. Since S/H satisfies u = v , then

ϕ(u(a)) = u(ϕ(a1), . . . , ϕ(an)) = v(ϕ(a1), . . . , ϕ(an)) = ϕ(v(a)) ,

where a = (a1, . . . , an) ∈ Sn and ϕ is the natural homomorphism of S ontoS/H . Therefore, u(a) H v(a), whence it easy to verify that (11) holds, so byTheorem 4 it follows that S satisfies u = v .

The converse follows immediately.

Theorem 5. Let S be an cryptogroup and let u = v be an identity over thealphabet A2 such that h(u) = h(v) and t(u) = t(v) . Then S satisfies u = v ifand only if all of its completely simple components satisfy u = v .

Proof. Let S be a band I of groups Gi, i ∈ I and let by ei we denote theidentity element of Gi, i ∈ I .

Let all completely simple components of S satisfy u = v and let a, b ∈ S .Then a ∈ Gi , b ∈ Gj for some i, j ∈ I and ab ∈ Gij . Since aeiji ∈ GiGiji , then(aeiji)r = areiji , for all r ∈ Z+ . In a similar way we prove that (ejijb)t = ejijbt ,for all t ∈ Z+ . Now we have that

arbt = eijarbteij (since arbt ∈ Gij),

= eijareijiejijbteij (since eijar ∈ Giji, bteij ∈ Gjij),

= eij(aeiji)r(ejijb)teij

= (aeiji)r(ejijb)t (since (aeiji)r(ejijb)t ∈ Gij).

for all r, t ∈ Z+ .Assume that u = v is p -equivalent to the identity

xm1yn1xm2yn2 · · ·xmhynh = xk1yl1xk2yl2 · · ·xksyls ,

for some m1, . . . , mh, n1, . . . nh, k1, . . . , ks, l1, . . . , ls ∈ Z+ . Then

am1bn1am2bn2 · · · amhbnh =

= (aeiji)m1(ejijb)n1(aeiji)m2(ejijb)n2 · · · (aeiji)mh(ejijb)nh

= (aeiji)k1(ejijb)l1(aeiji)k2(ejijb)l2 · · · (aeiji)ks(ejijb)ls

= ak1bl1ak2bl2 · · · aksbls


since aeiji and ejijb are in the same completely simple component of S (i.e.aeijiJ ejijb). Therefore, S satisfy u = v .

Let u = v be p -equivalent to the identity

xm1yn1xm2yn2 · · ·xmhynhxmh+1 = xk1yl1xk2yl2 · · ·xksylsxks+1 ,

for some m1, . . . , mh+1, n1, . . . nh, k1, . . . , ks+1, l1, . . . , ls ∈ Z+ . Then

arbtaq = (aeiji)r(ejijb)teijaq

= (aeiji)r(ejijb)teijaqeiji (since eijaq ∈ Giji)

= (aeiji)r(ejijb)teij(aeiji)q

= (aeiji)r(ejijb)t(aeiji)q

for all r, t, q ∈ Z+ . Thus

am1bn1am2bn2 · · · amhbnhamh+1 =

= (aeiji)m1(ejijb)n1(aeiji)m2(ejijb)n2 · · · (aeiji)mh(ejijb)nhaeijimh+1

= (aeiji)k1(ejijb)l1(aeiji)k2(ejijb)l2 · · · (aeiji)ks(ejijb)lsaeijiks+1

= ak1bl1ak2bl2 · · · aksblsaks+1

since aeiji and ejijb are in the same completely simple component of S (i.e.aeijiJ ejijb). Therefore, S satisfy u = v .

The converse follows immediately.

In the next considerations we will study the identity

(14) xy = xm1yn1xm2yn2 · · ·xmhynh ,

with h, mi, ni ∈ Z+, i ∈ {1, 2, . . . , h} and h = 1 ⇒ m1, n1 ≥ 2. Also, we willuse the following notations:

px =h

∑

i=1

mi − 1 , py =h

∑

i=1

ni − 1 , p = gcd(px, py).

(i.e. p is the period of the identity (14) ).

Theorem 6. A semigroup S satisfies the identity (14) with

(15) gcd(px, py, h− 1) = 1

if and only if S2 is an orthogroup whose subgroups satisfy (14) and

(16) ab L am1b ∧ ab R abnh ,

for all a, b ∈ S .

Proof. Let S satisfies (14) and let (15) hold. By Proposition 1 and by Theorem3 [5] we obtain that S2 is a union of groups, by Theorem 3 we obtain that S2

is an orthogroup and it is clear that all of its subgroups satisfy (14). Also, bythis and by Shevrin-Veronesi’s theorem it follows that S is a semilattice Y ofsemigroups Sα and for all α ∈ Y , S2

α = Kα = S2 ∩ Sα is a rectangular group.


Let a, b ∈ S . Then ab , am1b , abnh ∈ Sα for some α ∈ Y , i.e. ab, am1b,abnh ∈ Kα . Assume that Kα is a rectangular band I × Λ of groups Hiλ , i ∈ I ,λ ∈ Λ. Then ab ∈ Hiλ, am1b ∈ Hjλ, abnh ∈ Hiµ for some i, j ∈ I, λ, µ ∈ Λ, soby (14) it follows that i = j and λ = µ , whence (16) holds.

Conversely, let S2 be an orthogroup whose subgroups satisfy (14) and let (16)hold. Then by Shevrin-Veronesi’s theorem we have that S is a semilattice Y ofsemigroups Sα , α ∈ Y , and for all α ∈ Y , S2

α = Kα is a rectangular group.Let a, b ∈ S , let ab ∈ Sα for some α ∈ Y and let Kα be a rectangular bandI × Λ of groups Hiλ , i ∈ I , λ ∈ Λ. Then ab , am1b , abnh ∈ Kα , so ab ∈ Hiλ ,am1b ∈ Hjλ , abnh ∈ Hiµ for some i, j ∈ I , λ, µ ∈ Λ. By (16) we obtain thati = j and λ = µ , so ab, am1b, abnh ∈ Hiλ . Let e be the identity element ofHiλ . As in the proof of Theorem 4 we obtain that

ab = eabe = e(ae)(be)

= e(ae)m1(be)n1 · · · (ae)mh(be)nh

= e(am1bn1 · · · amhbnh)e

= am1bn1 · · · amhbnh ,

since ae, be ∈ Kα and Kα satisfies (14). Therefore, S satisfies (14).

Corollary 4. A semigroup S satisfies the identity (14) with m1, nh = 1 ifand only if S2 is an orthogroup whose subgroups satisfy (14) .

Theorem 7. A semigroup S satisfies the identity (14) with m1, nh ≥ 2 and

(17) gcd(px, py, m1 − 1) = gcd(px, py, nh − 1) = 1,

if and only if S is an inflation of a cryptogroup whose completely simple compo-nents satisfy (14) .

Proof. By Proposition 2 it follows that S is an inflation of a union of groupsK and by Theorem 6 [4] it follows that K is a cryptogroup. It is clear thatcompletely simple components of K satisfy (14).

Conversely, let S be an inflation of a cryptogroup K whose completelysimple components satisfy (14). Then by Theorem 5 it follows that K satisfies(14) and since there exists a retraction of S onto K = S2 , then we easy provethat S satisfy (14).

Corollary 5. (i) A semigroup S satisfies the identity (14) with p = m1 =nh = 1 if and only if S2 is a band.(ii) A semigroup S satisfies the identity (14) with p = 1 and m1, nh ≥ 2 if and

only if S is an inflation of a band if and only if S satisfies the system ofidentities xy = x2y = xy2 if and only if S satisfies the identity xy = x2y2 .

(iii) A semigroup S satisfies the identity (14) with p = 1 and m1 ≥ 2, nh = 1(m1 = 1, nh ≥ 2) if and only if S2 is a band and S satisfies the identityxy = x2y (xy = xy2) .

Corollary 6. A semigroup S satisfies the identity (14) with m1, nh ≥ 2 and

gcd(px, py,m1 − 1) = gcd(px, py, nh − 1) = gcd(px, py, h− 1) = 1

if and only if S is an inflation of an orthocryptogroup whose subgroups satisfy(14) .


Let us consider the identity

(18) xy = yn0xm1yn1xm2yn2 · · ·xmhynh ,

with h, n0, mi, ni ∈ Z+, i ∈ {1, 2, . . . , h} . Also, we will use the followingnotations:

px =h

∑

i=1

mi − 1 , py =h

∑

i=0

ni − 1 , p = gcd(px, py).

(i.e. p is the period of the identity (18) ).

Theorem 8. A semigroup S satisfies the identity (18) if and only if S2 is asemilattice of right groups whose subgroups satisfy (18) and

(19) ab R abnh

for all a, b ∈ S .

Proof. Let S satisfies (18). Then by Proposition 1 and by Theorem 3 [5] weobtain that S2 is a union of groups and it is clear that S is periodic, so by thedual of Theorem 2 we obtain that S2 is a semilattice of right groups. It is clearthat all of its subgroups satisfy (18). As in the proof of Theorem 6 we provethat (19) holds.

Conversely, let S2 be a semilattice of right groups and let (19) hold. Bythis and by Shevrin-Veronesi’s theorem it follows that S is a semilattice Y ofsemigroups Sα, α ∈ Y , and for all α ∈ Y , Kα = S2

α = S2 ∩ Sα is a right group.Let a, b ∈ S . Then ab, abnh ∈ Kα for some α ∈ Y . Assume that Kα is a rightzero band I of groups Hi, i ∈ I . Then by (19) it follows that ab, abnh ∈ Hi

for some i ∈ I . Let e be the identity element of Hi . Then we obtain thatae = aee ∈ KαHi ⊆ Hi and be = bee ∈ KαHi ⊆ Hi , so ae = eae and be = ebe .Since ab, abnh , ae, be ∈ Hi and since Hi satisfies (18), then we obtain that

ab = abe = (ae)(be) = (be)n0(ae)m1(be)n1 . . . (ae)mh(be)nh

= bn0am1bn1 . . . amhbnhe

= bn0am1bn1 . . . amhbnh .

Therefore, S satisfies (18).

Corollary 7. [3] A semigroup S satisfies the identity (18) with nh = 1 ifand only if S2 is a semilattice of right groups whose subgroups satisfy (18) .

Corollary 8. (i) A semigroup S satisfies the identity (18) with p = nh = 1if and only if S2 is a right regular band.(ii) A semigroup S satisfies the identity (18) with p = 1 and nh ≥ 2 if and only

if S is an inflation of a right regular band.

Theorem 9. A semigroup S satisfy the identity (18) with nh ≥ 2 and

(20) gcd(px, py, nh − 1) = 1

if and only if S is an inflation of a right regular band of groups whose subgroupssatisfy (18) .


Proof. Let S satisfies (18) and let (20) hold. Then by Proposition 2 it followsthat S is an inflation of a union of groups K . Also, by Theorem 6 [4] we obtainthat K is a band I of groups and by Proposition II 3.10 [7] it follows that I isa right regular band. It is clear that all of subgroups of K satisfy (18).

Conversely, let S be an inflation of a semigroup K and let K be a rightregular band I of groups Gi , i ∈ I that satisfy (18). Let a, b ∈ K , i.e., let a ∈ Gi ,b ∈ Gj for some i, j ∈ I , and let e be the identity element of Gij . Then ab ∈ Gijand since I is a right regular band, then ij = jij , so ae, be ∈ Gij . Therefore,ae = eae and be = ebe , and since Gij satisfies (18) and since ab, abnh ∈ Gij ,then we obtain that

ab = abe = (ae)(be) = (be)n0(ae)m1(be)n1 . . . (ae)mh(be)nh

= bn0am1bn1 . . . amhbnhe

= bn0am1bn1 . . . amhbnh .

Therefore, K satisfies (18). Since there exists a retraction of S onto K = S2 ,we then easy prove that S satisfies (18).

Finally, we will consider the identity

(21) xym = xny,

with m,n ∈ Z+, m, n ≥ 2. By p = gcd(m − 1, n − 1) we will denote the periodof the identity (21).

Lemma 1. Let S be a nil-semigroup. Then(i) S satisfies the identity xyn+k = xny , with k, n ∈ Z+ , n ≥ 2 , if and only if

anb = 0 , for all a, b ∈ S .(ii) S satisfies the identity xyn = xn+ky , with k, n ∈ Z+ , n ≥ 2 , if and only if

abn = 0 , for all a, b ∈ S .

Proof. (i). Let S satisfies xyn+k = xny and let a, b ∈ S . Then bn+k+1 = bn+1

and since S is a nil-semigroup, we then easy prove that bn+1 = 0, so bn+k = 0.Now we obtain that anb = abn+k = 0.

Conversely, let anb = 0 for all a, b ∈ S . Then it is clear that bn+1 = 0for all b ∈ S , whence anb = 0 = abn+k for all a, b ∈ S . Thus, S satisfies theidentity xyn+k = xny .

(ii). The proof is symmetric to the proof for (i) .

Theorem 10. A semigroup S satisfies the identity (21) if and only if S isa retractive extension of a semigroup that satisfies x = xp+1 by a nil-semigroupthat satisfies (21) .

Proof. Let S satisfies (21). Then by Proposition 2 we obtain that S is aretractive nil-extension of a union of groups K . It is clear that S/K is a nil-semigroup that satisfies (21). Let a ∈ K . Then a ∈ Ge for some e ∈ E(S),whence a = ae = aem = ane = an and a = ea = ena = eam = am . Therefore,am−1 = an−1 = e , so by the definition of the period of the identity we obtainthat ap = e , i.e. a = ap+1 . Hence, K satisfies the identity x = xp+1 .

Conversely, let S be a retractive nil-extension of a semigroup K that satisfiesx = xp+1 by a nil-semigroup that satisfies (21). It is clear that K is a union ofgroups. Let ϕ be the retraction of S onto K .

Let a, b ∈ K and let a ∈ Ge , b ∈ Gf for some e, f ∈ E(S). Then ap+1 = a ,bp+1 = b , so ap = e , bp = f , whence an−1 = e , bm−1 = e , since p divides m− 1

379

and n − 1. Thus, an = a and bm = b , whence abm = ab = anb . Therefore, Ksatisfies (21).

Let a ∈ K, b ∈ S −K . Since K satisfies (21), we then obtain that

abm = aϕ(bm) = aϕ(b)m = anϕ(b) = anb.

Similarly we consider the case a ∈ S −K, b ∈ K .Finally, let a, b ∈ S −K . If abm, anb ∈ K , then

abm = ϕ(a)ϕ(bm) = ϕ(a)ϕ(b)m = ϕ(a)nϕ(b) = ϕ(an)ϕ(b) = anb.

Assume that abm /∈ K . Then abm 6= 0 in S/K and abm = anb , since S/Ksatisfies (21). Thus, abm, anb /∈ K in S and abm = anb in S . Similarly weconsider the case anb /∈ K . Hence, S satisfies (21).

References

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[3] Bogdanovic, S., Semigroups of Galbiati-Veronesi II, Facta Universitatis(Nis), Ser. Math. Inform. 2 (1987), 61–66.

[4] Ciric, M. and S. Bogdanovic, Decompositions of semigroups induced byidentities, Semigroup Forum 46 (1993), 329–346.

[5] Ciric, M. and S. Bogdanovic, Nil-extensions of unions of groups inducedby identities , Semigroup Forum 48 (1994), 303–311.

[6] Clarke, G., Semigroup varieties of inflations of union of groups , SemigroupForum, 23 (1981), 311–319..

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[8] Putcha, M. S. and J. Weissglass, Semigroups satisfying variable identities ,Semigroup Forum 3 (1971), 64–67.

[9] Shevrin, L. N., On decompositions of quasi-periodic semigroups into a bandof Archimedean semigroups , XIV Vsesoyuzn. algebr. konf. Tezisy dokl,Novosibirsk, 1977, Vol. 1, 104–105 (in Russian).

[10] Shevrin, L. N. and M. V. Volkov, Identities on semigroups , Izv. vuzov.Matematika 11 (1985), 3–47 (in Russian).

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Faculty of Economics18000Nis, Yugoslavia

Received October 6, 1994and in final form March 1, 1995

Identities over the two-element alphabet

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