/compile/output.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 3, 2016, 277-291

ISSN 1307-5543 – www.ejpam.com

On Generalized Ideals of Left Almost Semigroups

Waqar Khan1,2,3, Faisal Yousafzai2,∗, and Madad Khan3

1 School of Mathematics and Statistics, Southwest University, Beibei, Chongqing, China
2 School of Mathematical Sciences, University of Science and Technology of China, Hefei, China
3 Department of Mathematics, COMSATS Institute of Information Technology, Abbottabad, Pakistan

Abstract. In this paper, we study (m, n)-ideals of an LA -semigroup in detail. We characterize (0,2)-

ideals of anLA -semigroup S and prove that A is a (0,2)-ideal of S if and only if A is a left ideal of some

left ideal of S. We also show that an LA -semigroup S is 0 − (0,2)-bisimple if and only if S is right

0-simple. Furthermore we study 0-minimal (m, n)-ideals in an LA -semigroup S and prove that if R,

(L) is a 0-minimal right (le f t) ideal of S, then either Rm Ln = {0} or Rm Ln is a 0-minimal (m, n)-ideal

of S for m, n≥ 3. Finally we discuss (m, n)-ideals in an (m, n)-regularLA -semigroup S and show that

S is (0,1)-regular if and only if L = SL where L is a (0,1)-ideal of S.

2010 Mathematics Subject Classifications: 20M10, 20N99

Key Words and Phrases: LA -semigroups, left invertive law, left identity and (m, n)-ideals.

1. Introduction

A left almost semigroup (LA -semigroup) is a groupoid Ssatisfying the left invertive law

(ab)c = (cb)a for all a, b, c ∈ S. This left invertive law has been obtained by introducing

braces on the left of ternary commutative law abc = cba. The concept of an LA -semigroup

was first given by Kazim and Naseeruddin in 1972 [3]. AnLA -semigroup satisfies the medial

law (ab)(cd) = (ac)(bd) for all a, b, c, d ∈ S. Since LA -semigroups satisfy medial law, they

belong to the class of entropic groupoids which are also called abelian quasigroups [12]. If

an LA -semigroup S contains a left identity (unitary LA -semigroup), then it satisfies the

paramedial law (ab)(cd) = (dc)(ba) and the identity a(bc) = b(ac) for all a, b, c, d ∈ S [7].

An LA -semigroup is a useful algebraic structure, midway between a groupoid and a

commutative semigroup. AnLA -semigroup is non-associative and non-commutative in gen-

eral, however, there is a close relationship with semigroup as well as with commutative struc-

tures. It has been investigated in [7] that if an LA -semigroup contains a right identity, then

it becomes a commutative semigroup. The connection of a commutative inverse semigroup

∗Corresponding author.

Email addresses: wkyousafzai@163.com (W. Khan), yousafzaimath@gmail.com (F. Yousafzai),

madadmath@yahoo.com (M. Khan)

http://www.ejpam.com 277 c© 2016 EJPAM All rights reserved.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 278

with an LA -semigroup has been given by Yousafzai et al. in [16] as, a commutative inverse

semigroup (S, ·) becomes an LA -semigroup (S,∗) under a ∗ b = ba−1r−1, ∀a, b, r ∈ S. An

LA -semigroup S with left identity becomes a semigroup under the binary operation "◦e",
defined as x ◦e y = (xe)y for all x , y ∈ S [17]. An LA -semigroup is the generalization of

a semigroup theory [7] and has vast applications in collaboration with semigroups like other

branches of mathematics. Khan et al. studied an intra-regular class of an LA -semigroup

in [4] and proved some interesting problems by using different ideals. They proved that the

set of all two-sided ideals of intra-regular LA -semigroup forms a semilattice structure. They

characterized an intra-regularLA -semigroup by using left, right, two-sided and bi-ideals. An

LA -semigroup is the generalization of a semigroup theory [7]. Many interesting results on

LA -semigroups have been investigated in [5, 9–11, 15].

Yaqoob, Corsini and Yousafzai [13] extended the concept of LA-semigroups and introduced

a new structure called left almost semihypergroup. Further Yaqoob and Gulistan [14] de-

fined partial ordering on left almost semihypergroups. Gulistan et al. [2] defined Hv-LA -

semigroups which is a new generalization of LA -semigroups and LA -semihypergroups.

2. Preliminaries and Examples

If S is an LA -semigroup with product · : S × S −→ S, then ab · c and (ab)c both denote

the product (a · b) · c.

If there is an element 0 of an LA -semigroup (S, ·) such that x · 0 = 0 · x = x ∀x ∈ S, we

call 0 a zero element of S.

Example 1. Let S = {a, b, c, d, e} with a left identity d. Then the following multiplication table

shows that (S, ·) is a unitary LA -semigroup with a zero element a.

· a b c d e

a a a a a a

b a e e c e

c a e e b e

d a b c d e

e a e e e e

Example 2. Let S = {a, b, c, d}. Then the following multiplication table shows that (S, ·) is an

LA -semigroup with a zero element a.

· a b c d

a a a a a

b a d d c

c a c c c

d a c c c

The above LA -semigroup S has commutative powers, that is aa · a = a · aa for all a ∈ S

which is called a locally associativeLA -semigroup [8]. Note that S has no associative powers

for all a ∈ S because (bb · b)b 6= b(bb · b) for b ∈ S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 279

Assume that S is an LA -semigroup. Let us define a1 = a, am+1 = ama and

am = ((((aa)a)a) . . . a)a = am−1a for all a ∈ S where m≥ 1. It is easy to see that

am = am−1a = aam−1 for all a ∈ S and m ≥ 3 if S has a left identity. Also, we can show by

induction, (ab)m = am bm and aman = am+n hold for all a, b ∈ S and m, n≥ 3.

A subset A of an LA -semigroup S is called a right (left) ideal of S if AS ⊆ A (SA⊆ A), and

is called an ideal of S if it is both left and right ideal of S.

A subset A of an LA -semigroup S is called an LA -subsemigroup of S if A2 ⊆ A.

The concept of (m, n)-ideals of a semigroup and an LA -semigroup was given in [6] and

[1] respectively.

An LA -subsemigroup A of an LA -semigroup S is said to be an (m, n)-ideal of S if AmS ·
An ⊆ A where m, n are non-negative integers such that m= n 6= 0. Here Am or An are suppressed

if m= 0 or n= 0, that is A0S = S or SA0 = S. Note that if m= n= 1, then an (m, n)-ideal A of

an LA -semigroup S is called a bi-ideal of S. If we take m = 0 or n = 0, then an (m, n)-ideal

A of an LA -semigroup S becomes a left or a right ideal of S.

An (m, n)-ideal A of an LA -semigroup S with zero is said to be 0-minimal if A 6= {0} and

{0} is the only (m, n)-ideal of S properly contained in A.

An LA -semigroup S with zero is said to be 0-(0,2)-bisimple if S2 6= {0} and {0} is the

only proper (0,2)-bi-ideal of S.

An LA -semigroup S with zero is said to be nilpotent if S l = {0} for some positive integer

l.

Let m, n be non-negative integers and S be an LA -semigroup. We say that S is (m, n)-

regular if for every element a ∈ S there exists some x ∈ S such that a = (am x)an. Note that

a0 is defined as an operator element such that a0 y = y and za0 = z for any y, z ∈ S.

3. 0-Minimal (0, 2)-Bi-Ideals in Unitary LA -Semigroups

If S is a unitary LA -semigroup, then it is easy to see that S2 = S, SA2 = A2S and A⊆ SA

∀A ⊆ S. Note that every right ideal of a unitary LA -semigroup S is a left ideal of S but the

converse is not true in general. Example 1 shows that there exists a subset {a, b, e} of S which

is a left ideal of S but not a right ideal of S. It is easy to see that SA and SA2 are the left and

right ideals of a unitary LA -semigroup S. Thus SA2 is an ideal of a unitary LA -semigroup

S.

Lemma 1. Let S be a unitary LA -semigroup. Then A is a (0,2)-ideal of S if and only if A is an

ideal of some left ideal of S.

Proof. Let A be a (0,2)-ideal of S, then SA · A= AA · S = SA2 ⊆ A and

A · SA= S · AA= SS · AA= SA2 ⊆ A. Hence A is an ideal of a left ideal SA of S.

Conversely, assume that A is a left ideal of a left ideal L of S, then

SA2 = AA · S = SA · A⊆ SL · A⊆ LA⊆ A,

and clearly A is an LA -subsemigroup of S, therefore A is a (0,2)-ideal of S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 280

Corollary 1. Let S be a unitary LA -semigroup. Then A is a (0,2)-ideal of S if and only if A is

a left ideal of some left ideal of S.

Lemma 2. Let S be a unitary LA -semigroup. Then A is a (0,2)-bi-ideal of S if and only if A is

an ideal of some right ideal of S.

Proof. Let A be a (0,2)-bi-ideal of S, then SA2 · A = A2S · A = AS · A2 ⊆ SA2 ⊆ A and

A · SA2 = SS · AA2 = A2A · SS = SA · A2 ⊆ SA2 ⊆ A. Hence A is an ideal of some right ideal SA2

of S.

Conversely, assume that A is an ideal of a right ideal R of S, then

SA2 = A · SA= A · (SS)A= A · (AS)S ⊆ A · (RS)R ⊆ AR ⊆ A,

and (AS)A⊆ (RS)A⊆ RA⊆ A, which shows that A is a (0,2)-ideal of S.

Theorem 1. Let S be a unitary LA -semigroup. Then the following statements are equivalent.

(i) A is a (1,2)-ideal of S;

(ii) A is a left ideal of some bi-ideal of S;

(iii) A is a bi-ideal of some ideal of S;

(iv) A is a (0,2)-ideal of some right ideal of S;

(v) A is a left ideal of some (0,2)-ideal of S.

Proof. (i) =⇒ (ii). It is easy to see that SA2 · S is a bi-ideal of S. Let A be a (1,2)-ideal of

S, then

(SA2 · S)A=(SA2 · SS)A= (SS · A2S)A= (S · A2S)A= A2S · A

=AS · A2 ⊆ A,

which shows that A is a left ideal of a bi-ideal SA2 · S of S.

(ii) =⇒ (iii). Let A be a left ideal of a bi-ideal B of S, then

(A · SA2)A=(S · AA2)A⊆ [S(SA · AA)]A= [S(AA · AS)]A

=[AA · S(AS)]A= [{S(AS) · A}A]A= [(AS · A)A]A

⊆[(BS · B)A]A⊆ BA · A⊆ A,

which shows that A is a bi-ideal of an ideal SA2 of S.

(iii) =⇒ (iv). Let A be a bi-ideal of an ideal I of S, then

SA2 · A2 =(A2 · AA)S = (A · A2A)S ⊆ [A · (AI)A]S = AA · S

=SA · A⊆ SI · S ⊆ I ,

which shows that A is a (0,2)-ideal of a right ideal SA2 of S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 281

(iv) =⇒ (v). It is easy to see that SA3 is a (0,2)-ideal of S. Let A be a (0,2)-ideal of a right

ideal R of S, then

A · SA3 =A(SS · A2A) = A(AA2 · S) ⊆ A[(SA · AA)S] = A[(AA · AS)S]

=(AA)[(A · AS)S] = [S · A(AS)]A2 = [A · S(AS)]A2

⊆RS · A2 ⊆ RA2 ⊆ A,

which shows that A is a left ideal of a (0,2)-ideal SA3 of S.

(v) =⇒ (i). Let A be a left ideal of a (0,2)-ideal O of S, then

AS · A2 = (AA · SS)A= SA2 · A⊆ SO2 · A⊆ OA⊆ A,

which shows that A is a (1,2)-ideal of S.

Lemma 3. Let S be a unitary LA -semigroup and A be an idempotent subset of S. Then A

is a (1,2)-ideal of S if and only if there exist a left ideal L and a right ideal R of S such that

RL ⊆ A⊆ R∩ L.

Proof. Assume that A is a (1,2)-ideal of S such that A is idempotent. Setting L = SA and

R= SA2, then

RL =SA2 · SA= A2S · SA= (SA · SS)A2 = (SS · AS)A2

=[S(AA · SS)]A2 = [S(SS · AA)]A2 = [S{A(SS · A)}]A2

=[A(S · SA)]A2 ⊆ AS · A2 ⊆ A.

It is clear that A⊆ R∩ L.

Conversely, let R be a right ideal and L be a left ideal of S such that RL ⊆ A⊆ R∩ L, then

AS · A2 = AS · AA⊆ RS · SL ⊆ RL ⊆ A.

Assume that S is a unitary LA -semigroup with zero. Then it is easy to see that every left

(right) ideal of S is a (0,2)-ideal of S. Hence if O is a 0-minimal (0,2)-ideal of S and A is a left

(right) ideal of S contained in O, then either A= {0} or A= O.

Lemma 4. Let S be a unitary LA -semigroup with zero. Assume that A is a 0-minimal ideal of

S and O is an LA -subsemigroup of A. Then O is a (0,2)-ideal of S contained in A if and only if

O2 = {0} or O = A.

Proof. Let O be a (0,2)-ideal of S contained in a 0-minimal ideal A of S. Then SO2 ⊆ O ⊆ A.

Since SO2 is an ideal of S, therefore by minimality of A, SO2 = {0} or SO2 = A. If SO2 = A, then

A= SO2 ⊆ O and therefore O = A. Let SO2 = {0}, then O2S = SO2 = {0} ⊆ O2, which shows

that O2 is a right ideal of S, and hence an ideal of S contained in A, therefore by minimality of

A, we have O2 = {0} or O2 = A. Now if O2 = A, then O = A.

Conversely, let O2 = {0}, then SO2 = O2S = {0}S = {0} = O2. Now if O = A, then

SO2 = SS ·OO = SA ·SA⊆ A= O, which shows that O is a (0,2)-ideal of S contained in A.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 282

Corollary 2. Let S be a unitary LA -semigroup with zero. Assume that A is a 0-minimal left

ideal of S and O is an LA -subsemigroup of A. Then O is a (0,2)-ideal of S contained in A if and

only if O2 = {0} or O = A.

Lemma 5. Let S be a unitary LA -semigroup with zero and O be a 0-minimal (0,2)-ideal of S.

Then O2 = {0} or O is a 0-minimal right (le f t) ideal of S.

Proof. Let O be a 0-minimal (0,2)-ideal of S, then

S(O2)2 = SS ·O2O2 = O2O2 · S = SO2 ·O2 ⊆ OO2 ⊆ O2,

which shows that O2 is a (0,2)-ideal of S contained in O, therefore by minimality of O, O2 = {0}
or O2 = O. Suppose that O2 = O, then OS = OO ·SS = SO2 ⊆ O, which shows that O is a right

ideal of S. Let R be a right ideal of S contained in O, then R2S = RR · S ⊆ RS · S ⊆ R. Thus R is

a (0,2)-ideal of S contained in O, and again by minimality of O, R= {0} or R= O.

The following Corollary follows from Lemma 4 and Corollary 2.

Corollary 3. Let S be a unitary LA -semigroup. Then O is a minimal (0,2)-ideal of S if and

only if O is a minimal left ideal of S.

Theorem 2. Let S be a unitaryLA -semigroup. Then A is a minimal (2,1)-ideal of S if and only

if A is a minimal bi-ideal of S.

Proof. Let A be a minimal (2,1)-ideal of S. Then

[(A2S · A)2S](A2S · A) =[{(A2S · A)(A2S · A)}S](A2S · A)

⊆[{(AS · A)(AS · A)}S](AS · A)

=[{(AS · AS)(AA)}S](AS · A)

=[(A2S · AA)S](AS · A)

⊆[(AS · AS)S](AS · A)

=(A2S · S)(AS · A)

⊆(AS · S)(AS · A) = (AS · AS)(SA)

=A2S · SA= AS · SA2 = (SA2 · S)A

=(A2S · S)A= (SS · AA)A= A2S · A,

and similarly we can show that (A2S ·A)2 ⊆ A2S ·A. Thus A2S ·A is a (2,1)-ideal of S contained

in A, therefore by minimality of A, A2S · A= A. Now

AS · A=(AS)(A2S · A) = [(A2S · A)S]A= (SA · A2S)A

=[A2(SA · S)]A⊆ A2S · A= A,

It follows that A is a bi-ideal of S. Suppose that there exists a bi-ideal B of S contained in A,

then B2S · B ⊆ BS · B ⊆ B, so B is a (2,1)-ideal of S contained in A, therefore B = A.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 283

Conversely, assume that A is a minimal bi-ideal of S, then it is easy to see that A is a (2,1)-

ideal of S. Let C be a (2,1)-ideal of S contained in A, then

[(C2S · C)S](C2S · C) =(SC · C2S)(C2S · C) = (SC2 · CS)(C2S · C)

=[C(SC2 · S)](C2S · C) = [(C2S · C)(SC2 · SS)]C

=[(C2S · C)(S · C2S)]C = [(C2S · C)(C2S)]C

=[C2{(C2S · C)S}]C ⊆ C2S · C .

This shows that C2S · C is a bi-ideal of S, and by minimality of A, C2S · C = A. Thus A =

C2S · C ⊆ C , and therefore A is a minimal (2,1)-ideal of S.

Theorem 3. Let A be a 0-minimal (0,2)-bi-ideal of a unitary LA -semigroup S with zero. Then

exactly one of the following cases occurs:

(i) A= {0, a}, a2 = 0;

(ii) ∀a ∈ A\{0}, Sa2 = A.

Proof. Assume that A is a 0-minimal (0,2)-bi-ideal of S. Let a ∈ A\{0}, then Sa2 ⊆ A. Also

Sa2 is a (0,2)-bi-ideal of S, therefore Sa2 = {0} or Sa2 = A.

Let Sa2 = {0}. Since a2 ∈ A, we have either a2 = a or a2 = 0 or a2 ∈ A\{0, a}. If a2 = a,

then a3 = a2a = a, which is impossible because a3 ∈ a2S = Sa2 = {0}. Let a2 ∈ A\{0, a}, we

have

S · {0, a2}{0, a2}= SS · a2a2 = Sa2 · Sa2 = {0} ⊆ {0, a2},

and

[{0, a2}S]{0, a2}= {0, a2S}{0, a2}= a2S · a2 ⊆ Sa2 = {0} ⊆ {0, a2}.

Therefore {0, a2} is a (0,2)-bi-ideal of S contained in A. We observe that {0, a2} 6= {0}
and {0, a2} 6= A. This is a contradiction to the fact that A is a 0-minimal (0,2)-bi-ideal of S.

Therefore a2 = 0 and A= {0, a}.
If Sa2 6= {0}, then Sa2 = A.

Corollary 4. Let A be a 0-minimal (0,2)-bi-ideal of a unitary LA -semigroup S with zero such

that A2 6= 0. Then A= Sa2 for every a ∈ A\{0}.

Lemma 6. Let S be a unitary LA -semigroup. Then every right ideal of S is a (0,2)-bi-ideal of

S.

Proof. Assume that A is a right ideal of S, then

SA2 = AA · SS = AS · AS ⊆ AA⊆ AS ⊆ A, AS · A⊆ A,

and clearly A2 ⊆ A, therefore A is a (0,2)-bi-ideal of S.

The converse of Lemma 6 is not true in general. Example 1 showed that there exists a

(0,2)-bi-ideal A= {a, c, e} of S which is not a right ideal of S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 284

Theorem 4. Let S be a unitaryLA -semigroup with zero. Then Sa2 = S ∀a ∈ S\{0} if and only

if S is 0-(0,2)-bisimple if and only if S is right 0-simple.

Proof. Assume that Sa2 = S for every a ∈ S\{0}. Let A be a (0,2)-bi-ideal of S such that

A 6= {0}. Let a ∈ A\{0}, then S = Sa2 ⊆ SA2 ⊆ A. Therefore S = A. Since S = Sa2 ⊆ SS = S2,

we have S2 = S 6= {0}. Thus S is 0 − (0,2)-bisimple. The converse statement follows from

Corollary 4.

Let R be a right ideal of 0-(0,2)-bisimple S. Then by Lemma 6, R is a (0,2)-bi-ideal of S

and so R= {0} or R= S.

Conversely, assume that S is right 0-simple. Let a ∈ S\{0}, then Sa2 = S. Hence S is

0-(0,2)-bisimple.

Theorem 5. Let A be a 0-minimal (0,2)-bi-ideal of a unitary LA -semigroup S with zero. Then

either A2 = {0} or A is right 0-simple.

Proof. Assume that A is 0-minimal (0,2)-bi-ideal of S such that A2 6= {0}. Then by using

Corollary 4, Sa2 = A for every a ∈ A\{0}. Since a2 ∈ A\{0} for every a ∈ A\{0}, we have

a4 = (a2)2 ∈ A\{0} for every a ∈ A\{0}. Let a ∈ A\{0}, then

(Aa2)S · Aa2 =a2A · S(Aa2) = [(S · Aa2)A]a2 ⊆ [(S · A)A]a2

=(AA · SS)a2 = SA2 · a2 ⊆ Aa2,

and

S(Aa2)2 =S(Aa2 · Aa2) = S(a2A · a2A) = S[a2(a2A · A)]

=(aa)[S(a2A · A)] = [(a2A · A)S]a2

⊆(AA · SS)a2 = SA2 · a2 ⊆ Aa2,

which shows that Aa2 is a (0,2)-bi-ideal of S contained in A. Hence Aa2 = {0} or Aa2 = A. Since

a4 ∈ Aa2 and a4 ∈ A\{0}, we get Aa2 = A. Thus by using Theorem 4, A is right 0-simple.

4. (m, n)-Ideals in Unitary LA -Semigroups

In this section, we characterize a unitaryLA -semigroup in terms of (m, n)-ideals with the

assumption that m, n≥ 3. If we take m, n≥ 2, then all the results of this section can be trivially

followed for a locally associative unitary LA -semigroup. If S is a unitary LA -semigroup,

then it is easy to see that SAm = AmS and AmAn = AnAm for m, n≥ 3 such that A0 = e if occurs,

where e is a left identity of S.

Lemma 7. Let S be a unitary LA -semigroup. If R and L are the right and left ideals of S

respectively, then RL is an (m, n)-ideal of S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 285

Proof. Let R and L be the right and left ideals of S respectively, then

(RL)mS · (RL)n =(Rm Lm · S)(Rn Ln) = (Rm Lm · Rn)(SLn)

=(LmRm · Rn)(SLn) = (RnRm · Lm)(SLn)

=(RmRn · Lm)(SLn) = (Rm+n Lm)(SLn)

=S(Rm+n Lm · Ln) = S(Ln Lm · Rm+n)

=SS · Lm+nRm+n = SLm+n · SRm+n

=Rm+nS · Lm+nS = SRm+n · SLm+n,

and

SRm+n · SLm+n =(S · Rm+n−1R)(S · Lm+n−1 L)

=[S(Rm+n−2R · R)][S(Lm+n−2 L · L)]

=[S(RR · Rm+n−2)][S(LL · Lm+n−2)]

⊆(SS · RRm+n−2)(SS · LLm+n−2)

⊆(SR · SRm+n−2)(SL · SLm+n−2)

⊆(Rm+n−2S · RS)(L · SLm+n−2)

⊆(Rm+n−2S · R)(S · LLm+n−2)

=(RS · Rm+n−2)(SLm+n−1)

⊆RRm+n−2 · SLm+n−1

⊆SRm+n−1 · SLm+n−1,

therefore

(RL)mS · (RL)n ⊆SRm+n · SLm+n ⊆ SRm+n−1 · SLm+n−1 ⊆ . . . ⊆ SR · SL

⊆(SS · R)L = (RS · S)L ⊆ RL,

and also

RL · RL = LR · LR= (LR · R)L = (RR · L)L ⊆ (RS · S)L ⊆ RL.

This shows that RL is an (m, n)-ideal of S.

Theorem 6. Let S be a unitary LA -semigroup with zero. If S has the property that it contains

no non-zero nilpotent (m, n)-ideals and R (L) is a 0-minimal right (le f t) ideal of S, then either

RL = {0} or RL is a 0-minimal (m, n)-ideal of S.

Proof. Assume that R(L) is a 0-minimal right (le f t) ideal of S such that RL 6= {0}, then by

Lemma 7, RL is an (m, n)-ideal of S. Now we show that RL is a 0-minimal (m, n)-ideal of S.

Let {0} 6= M ⊆ RL be an (m, n)-ideal of S. Note that since RL ⊆ R ∩ L, we have M ⊆ R ∩ L.

Hence M ⊆ R and M ⊆ L. By hypothesis, M m 6= {0} and M n 6= {0}. Since {0} 6= SM m = M mS,

therefore

{0} 6=M mS ⊆ RmS = Rm−1R · S = SR · Rm−1 = SR · Rm−2R



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 286

=RRm−2 · RS ⊆ RRm−2 · R= Rm,

and

Rm ⊆SRm = SS · RRm−1 = Rm−1R · S = (Rm−2R · R)S

=(RR · Rm−2)S = SRm−2 · RR ⊆ SRm−2 · R

=(SS · Rm−3R)R= (RRm−3 · SS)R= (RS · Rm−3S)R

⊆(R · Rm−3S)R= (Rm−3 · RS)R ⊆ Rm−3R · R= Rm−1,

therefore {0} 6= M mS ⊆ Rm ⊆ Rm−1 ⊆ . . . ⊆ R. It is easy to see that M mS is a right ideal of S.

Thus M mS = R since R is 0-minimal. Also

{0} 6= SM n ⊆ {0} 6= SLn = S · Ln−1 L = Ln−1 · SL ⊆ Ln−1 L = Ln,

and

Ln ⊆SLn = SS · LLn−1 = Ln−1 L · S = (Ln−2 L · L)S = SL · Ln−2 L

⊆L · Ln−2 L = Ln−2 · LL ⊆ Ln−2 L = Ln−1 ⊆ . . . ⊆ L,

therefore {0} 6= SM n ⊆ Ln ⊆ Ln−1 ⊆ . . . ⊆ L. It is easy to see that SM n is a left ideal of S. Thus

SM n = L since L is 0-minimal. Therefore

M ⊆RL = M mS · SM n = M nS · SM m = (SM m · S)M n

=(SM m · SS)M n = (S ·M mS)M n = (M m · SS)M n

=M mS ·M n ⊆ M .

Thus M = RL, which means that RL is a 0-minimal (m, n)-ideal of S.

Theorem 7. Let S be a unitary LA -semigroup. If R (L) is a 0-minimal right (le f t) ideal of S,

then either Rm Ln = {0} or Rm Ln is a 0-minimal (m, n)-ideal of S.

Proof. Assume that R(L) is a 0-minimal right (le f t) ideal of S such that Rm Ln 6= {0},
then Rm 6= {0} and Ln 6= {0}. Hence {0} 6= Rm ⊆ R and {0} 6= Ln ⊆ L, which shows that

Rm = R and Ln = L since R (L) is a 0-minimal right (le f t) ideal of S. Thus by Lemma 7,

Rm Ln = RL is an (m, n)-ideal of S. Now we show that Rm Ln is a 0-minimal (m, n)-ideal of S.

Let {0} 6= M ⊆ Rm Ln = RL ⊆ R∩ L be an (m, n)-ideal of S. Hence

{0} 6= SM2 = M M · SS = MS ·MS ⊆ RS · RS ⊆ R

and {0} 6= SM ⊆ SL ⊆ L. Thus R = SM2 = M M · SS = SM ·M ⊆ SM and SM = L since R (L)

is a 0-minimal right (le f t) ideal of S. Therefore

M ⊆Rm Ln ⊆ (SM)m(SM)n = SmM m · SnM n = SS ·M mM n

=M nM m · S = SM m ·M n = M mS ·M n ⊆ M ,

Thus M = Rm Ln, which shows that Rm Ln is a 0-minimal (m, n)-ideal of S.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 287

Theorem 8. Let S be a unitary LA -semigroup with zero. Assume that A is an (m, n)-ideal of S

and B is an (m, n)-ideal of A such that B is idempotent. Then B is an (m, n)-ideal of S.

Proof. It is trivial that B is an LA -subsemigroup S. Secondly, since AmS · An ⊆ A and

BmA · Bn ⊆ B, then

BmS · Bn =(BmBm · S)(BnBn) = (BnBn)(S · BmBm)

=[(S · BmBm)Bn]Bn = [(Bn · BmBm)(SS)]Bn

=[(Bm · BnBm)(SS)]Bn = [S(BnBm · Bm)]Bn

=[S(BnBm · Bm−1B)]Bn = [S(BBm−1 · BmBn)]Bn

=[S(Bm · BmBn)]Bn = [Bm(SS · BmBn)]Bn

=[Bm(BnBm · SS)]Bn = [Bm(SBm · Bn)]Bn

=[Bm{(SS · Bm−1B)Bn}]Bn = [Bm(BmS · Bn)]Bn

⊆[Bm(AmS · An)]Bn ⊆ BmA · Bn ⊆ B,

which shows that B is an (m, n)-ideal of S.

Lemma 8. Let 〈a〉(m,n) = amS · an, then 〈a〉(m,n) is an (m, n)-ideal of a unitary LA -semigroup

S.

Proof. Assume that S is a unitaryLA -semigroup and m, n are non-negative integers, then

�

{〈a〉(m,n)}
mH
�

{〈a〉(m,n)}
n =[{((amH)an)}mH] {(amH)an}n

=[{(ammHm)amn}H] {(amnHn)ann}

=[ann(amnHn)][H{(ammHm)amn}]

=[[H{(ammHm)amn}](amnHn)] ann

=[amn[[H{(ammHm)amn}]Hn]] ann

⊆(amnH)ann = (amnHn)ann

={(amH)an}n ⊆
�

〈a〉(m,n)

�n
⊆ 〈a〉(m,n) ,

and similarly we can show that
�

〈a〉(m,n)

�2
⊆ 〈a〉(m,n).

Theorem 9. Let S be a unitary LA -semigroup and 〈a〉(m,n) be an (m, n)-ideal of S. Then the

following statements hold:

(i)
�

〈a〉(1,0)

�m
S = amS;

(ii) S
�

〈a〉(0,1)

�n
= San;

(iii)
�

〈a〉(1,0)

�m
S ·
�

〈a〉(0,1)

�n
= (amS)an.



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 288

Proof. (i). As 〈a〉(1,0) = aS, we have

�

〈a〉(1,0)

�m
S =(aS)mS = (aS)m−1(aS) · S = S(aS) · (aS)m−1

=(aS)(aS)m−1 = (aS)[(aS)m−2(aS)]

=(aS)m−2(aS · aS) = (aS)m−2(a2S)

= . . .=

¨

(aS)m−(m−1)(am−1S) if m is odd

(am−1S)(aS)m−(m−1) if m is even

=amS.

Analogously, we can prove (ii) and (iii) is simple.

Corollary 5. Let S be a unitary LA -semigroup and let 〈a〉(m,n) be an (m, n)-ideal of S. Then

the following statements hold:

(i)
�

〈a〉(1,0)

�m
S = Sam;

(ii) S
�

〈a〉(0,1)

�n
= anS;

(iii)
�

〈a〉(1,0)

�m
S ·
�

〈a〉(0,1)

�n
= (Sam)(anS).

Let L(0,n), R(m,0) and A(m,n) denote the sets of (0, n)-ideals, (m, 0)-ideals and (m, n)-ideals of an

LA -semigroup S respectively.

Theorem 10. If S is a unitary LA -semigroup, then the following statements hold:

(i) S is (0,1)-regular if and only if ∀L ∈ L(0,1), L = SL;

(ii) S is (2,0)-regular if and only if ∀R ∈R(2,0), R= R2S such that every R is semiprime;

(iii) S is (0,2)-regular if and only if ∀U ∈ A(0,2), U = U2S such that every U is semiprime.

Proof. (i). Let S be (0,1)-regular, then for a ∈ S there exists x ∈ S such that a = xa. Since

L is (0,1)-ideal, therefore SL ⊆ L. Let a ∈ L, then a = xa ∈ SL ⊆ L. Hence L = SL. Converse

is simple.

(ii). Let S be (2,0)-regular and R be (2,0)-ideal of S, then it is easy to see that R = R2S.

Now for a ∈ S there exists x ∈ S such that a = a2 x . Let a2 ∈ R, then

a = a2 x ∈ RS = R2S · S = SS · R2 = R2S = R,

which shows that every (2,0)-ideal is semiprime.

Conversely, let R = R2S for every R ∈ R(2,0). Since Sa2 is a (2,0)-ideal of S such that

a2 ∈ Sa2, therefore a ∈ Sa2. Thus

a ∈Sa2 = (Sa2)2S = (Sa2 · Sa2)S = (a2S · a2S)S = [a2(a2S · S)]S

=(a2 · Sa2)S = (S · Sa2)a2 ⊆ Sa2 = a2S,

which implies that S is (2,0)-regular.

Analogously, we can prove (iii).



W. Khan, F. Yousafzai, and M. Khan / Eur. J. Pure Appl. Math, 9 (2016), 277-291 289

Lemma 9. If S is a unitary LA -semigroup, then the following statements hold:

(i) If S is (0, n)-regular, then ∀L ∈ L(0,n), L = SLn;

(ii) If S is (m, 0)-regular, then ∀R ∈R(m,0), R= RmS;

(iii) If S is (m, n)-regular, then ∀U ∈ A(m,n), U = (UmS)Un.

Proof. It is simple.

Corollary 6. If S is a unitary LA -semigroup, then the following statements hold:

(i) If S is (0, n)-regular, then ∀L ∈ L(0,n), L = LnS;

(ii) If S is (m, 0)-regular, then ∀R ∈R(m,0), R= SRm;

(iii) If S is (m, n)-regular, then ∀U ∈ A(m,n), U = Um+nS = SUm+n.

Theorem 11. Let S be a unitary (m, n)-regularLA -semigroup such that m= n. Then for every

R ∈R(m,0) and L ∈ L(0,n), R∩ L = Rm L ∩ RLn.

Proof. It is simple.

Theorem 12. Let S be a unitary (m, n)-regularLA -semigroup. If M (N) is a 0-minimal (m, 0)-

ideal ((0, n)-ideal) of S such that MN ⊆ M ∩ N, then either MN = {0} or MN is a 0-minimal

(m, n)-ideal of S.

Proof. Let M (N) be a 0-minimal (m, 0)-ideal ((0, n)-ideal) of S. Let O = MN , then clearly

O2 ⊆ O. Moreover

OmS ·On =(MN)mS · (MN)n = (M mN m)S ·M nN n ⊆ (M mS)S · SN n

=SM m · SN n = M mS · SN n ⊆ MN = O,

which shows that O is an (m, n)-ideal of S. Let {0} 6= P ⊆ O be a non-zero (m, n)-ideal of S.

Since S is (m, n)-regular, therefore by using Lemma 9, we have

{0} 6=P = PmS · Pn = (Pm · SS)Pn = (S · PmS)Pn = (Pn · PmS)(SS)

=(PnS)(PmS · S) = PnS · SPm = PmS · SPn.

Hence PmS 6= {0} and PmS 6= {0}. Further P ⊆ O = MN ⊆ M ∩ N implies that P ⊆ M and

P ⊆ N . Therefore {0} 6= PmS ⊆ M mS ⊆ M which shows that PmS = M since M is 0-minimal.

Likewise, we can show that SPn = N . Thus we have

P ⊆O = MN = PmS · SPn = PnS · SPm = (SPm · SS)Pn

=(S · PmS)Pn = PmS · Pn ⊆ P.

This means that P = MN and hence MN is 0-minimal.



REFERENCES 290

Theorem 13. Let S be a unitary (m, n)-regular LA -semigroup. If M (N) is a 0-minimal

(m, 0)-ideal ((0, n)-ideal) of S, then either M ∩N = {0} or M ∩N is a 0-minimal (m, n)-ideal of

S.

Proof. Once we prove that M ∩N is an (m, n)-ideal of S, the rest of the proof is same as in

Theorem 11. Let O = M ∩ N , then it is easy to see that O2 ⊆ O. Moreover

OmS ·On ⊆ M mS · N n ⊆ MN n ⊆ SN n ⊆ N . But, we also have

OmS ·On ⊆M mS · N n = (M m · SS)N n = (S ·M mS)N n = (N n ·M mS)S

=(M m · N nS)(SS) = (M mS)(N nS · S) = M mS · SN n

=M mS · N nS = N n(M mS · S) = N n · SM m = N n ·M mS

=M m · N nS = M m · SN n ⊆ M mN ⊆ M mS ⊆ M .

Thus OmS ·On ⊆ M ∩ N = O and therefore O is an (m, n)-ideal of S.

ACKNOWLEDGEMENTS The work of first author is supported by the NNSF (Grant No.

11371335) grant of China. The second author is highly thankful to CAS-TWAS President’s

Fellowship.

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