J. Semigroup Theory and Appl. 2013, 2013:10
ISSN: 2051-2937
COMPLETELY SIMPLE SEMIGROUP WITH BASIS PROPERTY
ALJOUIEE A., AL KHALAF A.∗
Department of Mathematics, Faculty of Science, Al Imam Muhammad Ibn Suad Islamic University, KSA
P.O. Box 9004, Riyadh 11623, Saudi Arabia
Copyright c2013 Aljouiee A. and Al Khalaf A. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract.An inverse semigroup (group)Sis called an inverse semigroup (group) with basis property, if each two
minimal (irreducible) generating sets (with respect to inclusion) of an arbitrary subsemigroup (group)H ofSis
equivalent (i.e. they have the same cardinality).
It is proved that every completely simple semigroup with basis property is either group with basis property or its
sandwich matrix has at most two rows or two column and its maximal subgroup is either trivial group or a primary
cyclic group.
Keywords: completely simple semigroup, basis property.
2000 AMS Subject Classification:20M10.
1. Introduction
LetGbe a finite group. we defined(G)to be the least number of generators required to generate the group.
This leads us to a simple definition of what it means for a generating set of a group to be minimal.
A generating setXis said to be minimal if it has no proper subset which is also form a generating set.
The subsetXof a semigroupSis called independent, if for allx∈S, x∈ h/ X\ {x}i. An independent setXis called a basis of the subsemigrouphXi.
∗Corresponding author
Received July 5, 2013
In 1978 Jones [5] introduced and initiated the study of semigroups with the basis property. Jones [6] states that
ifSis an inverse semigroup andU≤V≤Sthen aU-basis forV is a subsetX ofV which is minimal such that
hU∪Xi=S. So a minimal generating set forV is a /0-basis.
Definition1. A basis property of a universal algebraAmeans that every two minimal (with respect to inclusion)
generating sets (bases) of an arbitrary subalgebra ofAhave the same cardinality[2].
Definition 2. A semigroup (group)S is called a semigroup (group) with basis property if there exists a basis
minimal (irreducible) generating sets (with respect to inclusion) for every subsemigroup (group)HofSand every
two bases are equivalent (i.e. they have the same cardinality)[6].
LetGbe a p-group, then the minimal generating sets (that is, generating sets for which no proper subset also
generatesG) all have the same cardinality.
Notice that finitely generated vector spaces have the property that all minimal generating sets have the same
cardinality.
Jones[6]introduced another concept which was stated for inverse semigroups.
Definition 3. An inverse semigroupShas the strong basis property if for any inverse subsemigroupV ofSand
inverse subsemigroupUofV, any twoU-bases forV have the same cardinality.
Shiryaev[10]showed that the completely simple semigroupS is filtering if and only if eitherS signal band,
group with two elements, or combinatorial Brandt semigroup with five elements.
Also Jones[6]showed that a free inverse semigroup, semilattices and finite combinatorial inverse semigroups
posses the basis property.
Let(Z, +)be an additive abelian group, we see thatZ=h1i=h2,3ieven though 2∈ h/ 3iand 3∈ h/ 2i. Thus Zdoes not have the basis property, so complete semisimplicity alone is not sufficient to ensure the basis property. Hence even free groups do not have the basis property.
The first results on the basis property of groups was in[6]he showed that a group with basis property is solvable
and all elements of such a group have prime power order. A classification of a group with the basis property was
announced by Dickson to Jones in[6], but yet to be published. However a classification of finite groups with the
basis property was announced by Al Khalaf[2]exploiting Higman’s result, this classification requires a technical
condition on thep-group and he proved the following theorem:
Theorem 1[2]Let a finite groupGbe a semidirect product of ap-groupP=Fit(G)(Fitting subgroup) ofGby
a cyclicq-grouphyi, of orderqb, wherep6=q(pandqare primes),b∈N. Then the groupGhas basis property
if and only if for every element inhyi,y6=eand for any invariant subgroupHofPthe automorphismϕumust
define an isotopic representation on every quotient Frattini subgroup ofH.
In[4]some results of Maschke, Clifford and Krull-Schmidt common to both group and module theory, are used
Finally Jones[7,8] studied basis property from the point of view exchange properties, then semigroups are determined for regular and periodic semigroup. The main method of proof is to eliminate undesirable types of
semigroups by showing that each contains a certain special subsemigroup, for instance the bicyclic semigroup.
Definition 4. A semigroupSis a periodic if all cyclic subsemigrouphai,a∈Sare finite.
Proposition 1. LetS be a group, which is a semigroup with basis property. ThenS is a periodic group with
basis property.
Proof. LetSbe a group, which is a semigroup with basis property and letHbe a cyclic subgroup ofSsuch
thatH=hai,a∈S. We prove thatHis finite. Suppose thatHis infinite. We see that a seta,a−1 is a basis of H. Consider the setX=
a−6,a10,a15 . We havea= (a−6)4a10a15∈ hXianda−1=a5a−6∈ hXi, sohXi=H. Therefore
D
a−6,a10E⊆
a2,a−26=H,
D
a−6,a15E⊆
a−3,a3
6=H,
D
a10,a15E⊆Da5E6=H.
ThusXis a basis ofHwith three elements, soHhas two bases with different cardinality. But this is contradicting
with basis property forS. HenceSis periodic, we note that for everyY⊆Sthe subsemigrouphYiis a subgroup,
thereforeSis a group with basis property.
Theorem 2.[3]A semigroupSis completely simple if and only if it is isomorphic to a Rees matrix semigroup
M(G;I,Λ;P)for some groupG, nonempty setsI, Λ, and regular sandwich matrixP.
Thus in the following propositions we suppose thatSis a Rees matrix semigroupM(G,I,Λ;P)with sandwich
matrixP= (pλi)λ∈Λ,i∈Iover groupG.
Proposition 2. LetSbe a semigroup of typeM(G;I,Λ;P). IfSis a semigroup with basis property and satisfies the following condition:
∃i∈I,∃λ∈Λ(λ 6=µ∧pλi6=0∧pµi6=0). (2−1)
Then either|G|=1 orGis a primary cyclic group.
Proof Let S be a semigroup with basis property satisfying the condition of proposition, then consider that
S=I×G×Λ. PuttingHλ ={i} ×G× {λ},Hµ={i} ×G× {µ}andH=Hλ∪Hµ. Sincepλi6=0 andpµi6=0,
soHis a subsemigroup ofSisomorphic to right subgroup, which are isomorphic toG. Hence we will study the set
G× {1,2}denoting it byK, then forg1,g2∈Gandi,j∈ {1,2}the product inKgiven by
(g1,i) = (g2,j) = (g1g2,j),
So we have the homomorphismϕ:K→Gis given byϕ(g1,i) =g1.SinceSis semigroup with basis property, andGis a subsemigroup ofS, thenGis a group, which is a semigroup with basis property, hence by proposition 1
Suppose that|G|>1, i.e. Gdoes not primary cyclic group, thenGhas independent setX={x1,x2}. Putting G1=hXi,K1=G1× {1,2}. Assume thate∈Gis an identity ofG. SinceGis a periodic group, then for alli,j∈
{1,2}there existss1,s2∈Nsuch thatxs11=eandx s2
2 =e. Let us take the subsemigroupK1=h(x1,1),(x2,2)iof semigroupK. Notice that the setX1={(x1,1), (x2,2)}is a basis ofK1. Now letY={(e,1),(x1,2), (x2,2)}be a set ofK1. So we will prove that theYis a basis ofK1. Since(x1,1) = (x1,2) (e,1)and(x2,2)∈Ythen we have
hYi=K1on the other hand we will show thatY is an independent set . In fact we have(e,1)∈ h(/ x1,2),(x2,2)i. Suppose that(x1,2)∈ hY\ {(x1,2)}i. Then there exists a wordusuch that
(x1,2) =u((e,1),(x2,2)). (2−2)
Using the homomorphismϕfor(2−2)we getϕ(x1,2) =x1 =u(xs11,x2)∈ hX\ {x1}i, contradicting withX, which is an independent set.
Similarly(x2,2)∈ hY\ {(x2,2)}iand contradicting . HenceY is a basis ofK1with three elements. Let nowV={(x1,1) (x2,2)}, since(e,1) = xs11,1
= (x1,1)s1∈ hViand
(x2,2) = (x1,1) (e,2) = (x1,1) xs22,2
= (x1,1) (x2,2)s2 ∈ hVi.
ThenY⊆V. And so
K1=hYi ⊆ hVi ⊆K1.
HenceK1=hViandVis a generating set of semigroupK1, with two elements. SoK1has a basis with cardinality
2. ThusK1has two different bases and we get a contradiction with basis property forSandK. ThenGmust be
cyclic and periodic, so it is finite and using[9]we get thatGeither trivial i.e.|G|=1 or primary cyclic group.
Proposition 3. LetSbe a completely simple semigroup. IfSdoes not form a group, then its maximal subgroup
is either trivial or primary cyclic group.
Proof. If the matrix P has two non zero elements in some row or column, then the proposition holds by
proposition 2. If it is not, then sincePis regular andSis not a group, so in every row and every column there exists
only one non zero element and by cancelation we can gete. Thus consider thatI=ΛandPis an identity matrix,
i.e.Sis a Brandt semigroup. Suppose, thatGis not cyclic or primary group, then we have a contradiction. In fact
in this case the groupGhas an independent setX={x1,x2}with two elements.
TakingG1=hx1,x2i,K={i,j} ×G1× {i,j}, andi,j∈I, i6=j. Let|x1|=s1, |x2|=s2. Since the matrixPis an identity, and the product inSgiven by:
∀i1,i2,i3,i4 ∈I, g1,g2∈G1,
(i1,g1,i2)(i3,g2,i4) =
(i1,g1g2,i4), i f i2=i3 0, i f i26=i3
Consider the two sets;
X={(i,e,j),(j,x1,i),(j,x2,j)},
Y={(i,e,j),(j,e,i),(i,x1,i), (j,x2,j)}.
From(2−3)we have
(j,e,i) = (j,x1,i) (i,e,j)s1−1(j,x1,i)∈ hXi,
(i,x1,i) = (i,e,j) (j,x1,i)∈ hXi,
(j,x1,i) = (j,e,i) (i,x1,i) (i,e,j)∈ hYi.
Hence we gethXi=hYi. If we can prove thatY is an independent, then the semigroupShas a basis with four
elements and also has another basis with smaller elements and we have a contradiction with the basis property for
S. Hence the element
(i,e,j)∈ {(/ j,e,i),(i,x1,i),(j,x2,j)}.
Since the non zero product for these elements by the componentifrom the right may be only of the form(i,x1,i)m, i.e. the third component isi.
Similarly the element(j,e,i)∈ {(/ i,e,j),(i,x1,i),(j,x2,j)}and so(i,x1,i)∈ {(/ i,e,j),(j,e,i),(j,x2,j)}since the set{x1,x2}is an independent inG. Similarly the element(j,x2,j)∈ {(/ i,e,j),(j,e,i),(i,x1,i)}. Hence we get a contradiction.
Corollary 1. From the propositions 1,2,3, we see that the groupGeither trivial or primary cyclic group with a
single generating element, so we denote it byg. Then fori∈I,λ ∈Λif[H]is a classHiλ ={i} ×G× {λ}is a subgroup of semigroupS, so it has a unique generated elementhiλ . Hencehiλ = (i,giλ,λ), for somegiλ ∈G.
Proposition 4. For somei,j∈Iandλ,µ∈Λ, λ 6=µ. If the following conditionpλipµjpλj6=0 holds, then pµi6=0.
Proof. Suppose that
pλipµjpλj6=0, i6=j,λ6=µ, pµi=0. (2−4)
Then we get a contradiction. Multiplying the rows and the columns of matrixPby non zero elements, then we
consider, that
pλi=e, pµj=e, pλj=e. (2−5)
By proposition 3 the groupGis cyclic. Letgbe a generating element of the group. We study two sets:
X= {(i,g,λ),(j,g,µ),(j,g,λ)},
If we prove thathXi=hYiand the independent of the setX, then from proposition 3, we get a contradiction with
basis property forS. Assume that|G|=m. From(2−4)and(2−5)we have
(i,g,λ) = (i,e,µ) (j,g,λ),
(j,g,µ) = (i,g,λ) (i,e,µ),
(i,e,µ) = (i,g,λ)m−1(j,g,µ).
Then we gethXi ⊆ hYi ⊆ hXiandhXi=hYi. Sincepµi=0 then the element(j,g,λ)6= (i,g,λ) (j,g,µ). Thus
(i,g,λ)∈ h(/ j,g,µ) (j,g,λ)i,
(j,g,µ)∈ h(/ i,g,λ) (j,g,λ)i.
ThusXis an independent set.
Proposition 5. Leti,j,k∈I,i6=j6=kandλ,µ,γ∈Λ, λ6=µ6=γ. Then for somel∈ {i,j,k}andχ∈ {λ,µ,γ}
must be satisfies the equationpχl=0.
Proof. PuttingI1={i,j,k}andΛ1={λ,µ,γ}. Suppose that
∀l∈I1,∀χ∈Λ1,pχl6=0, (2−6)
Then every[H]classH={l} ×G× {χ}is a group isomorphic toGand lethlχ generating element ofG. From
lemma (3.2)[5]and using(2−6)we have
∀s,t∈I1,∀χ,ξ∈Λ1,HsχHtξ =Hsξ. (2−7)
Consider the sets
X=
hjλ,hkγ,hiµ,hjγ ,
Y =
hiλ,hjµ,hkγ .
We see thatX is an independence and from(2−7)we havehXi=hYi, so we get a contradiction with basis
property forS.
Proposition 6. Leti,j,k∈I,i6=j6=kandλ,µ,γ∈Λ, λ 6=µ6=γand pλi=pλk=pµj=pγj=0 and pµi6=0,pµk6=0, pγi6=0, pγk6=0. (2−8) Thenpλj=0.
Proof. PuttingI1={i,j,k} andΛ1={λ,µ,γ},H= I1×G×Λ1. Suppose that(2−8)holds andpλj6=0. Multiplying the rows and the columns of the matrixPby elements ofG, we get,
By proposition 3 the groupGis cyclic, so assume thatG=hgi,|G|=m, and also pγk=gl, l∈N. Consider the sets
X={(i,e,λ),(k,e,λ),(j,g,µ),(j,g,γ)},
Y={(k,e,λ),(j,g,µ),(i,g,γ)}.
Now using(2−9)we have
(i,e,λ) = (i,g,γ)m−1(k,e,λ)∈ hYi
(j,g,γ) = ((j,g,µ) (k,e,λ))m−1(j,g,µ) (i,g,γ)∈ hYi
(i,g,γ) = (i,e,λ) (j,g,γ)∈ hXi.
ThushXi=hYi. It is easy to see thatXis an independence, so we have a contradiction with basis property forS.
Hencepλj=0.
Proposition 7. Let{i,j,k} ⊆Iand{λ,µ,γ} ⊆Λ, and
pλj=pλk=pµi=pµk=pγi=pγj=0 andpλi6=0,pµj6=0. (2−9) Thenpγk=0.
Proof. Suppose thatI1={i,j,k},Λ1={λ,µ,γ}andpγk6=0. Thus we can assume thatpλi=pµj=pγk=e, then the subsemigroupH= I1× {e} ×Λ1∪ {0}is a combinatorial semigroup Brandt, so we can considerHas the
setI1×I1∪ {0}with operation:
(l,m) (n,q) =
(l,q), if m=n
0 , if m6=n,
and 0(l,m) = (l,m)0=0,∀l,m,q ∈I1, we want to prove thatHis a semigroup with basis property. Let the sets:
X={(i,j),(j,i),(k,j),(j,k)},
Y={(k,i),(i,j),(j,k)}.
Similarly, as in the proposition 6, we can prove thathXi=hYiandX is an independence. Thus we get a
contradiction with basis property forH.
Proposition 8. Let|I| ≥2 and|Λ| ≥2. Then min{|I|,|Λ|}=2 .
Proof. Suppose that|I| ≥3 , and|Λ| ≥3. SincePis a regular matrix, then there is a different indexes{i,j,k} ⊆I,
{λ,µ,γ} ⊆Λsuch that
pλi6=0,pµj6=0,pγk6=0. (2−10)
Denote byP1for a submatrixPwhich is the intersectionλ,µ,γ-rows andi,j-k-columns. Substitutionsλ,µ,γ and correspondencei,j,kby 1,2,3. Then from(2−10)we have that the leading diagonal of matrixP1consist from nonzero elements. From proposition 4, we have that the matrixP1is symmetric, if we substitute every element by
From proposition 5 we have in out the leading diagonal exists a 0. From proposition 7 we have that in out the
diagonal it is impossible to be all elements equal zero. Then we have a contradiction, since by proposition 6 and
proposition 4 there is no two elements equal zero in the same row or in the same column. Then in every row there
exists three non zeros , and in other there exists two nonzeros and two zeros are symmetric depend to the leading
diagonal. Hence we get a contradiction with proposition 4, because does not exists submatrix of rank 2 has only
one zero. Thus min{|I|,|Λ|}=2 .
Proposition 9. LetSbe a completely simple semigroup such that|I|=2≤ |Λ|andI={1,2}. Then for all
λ,µ∈Λ,λ6=µ, we have
D
pµip−1λipλjp−1µj
E
=G. (2−11)
Proof. It is sufficient to consider the subsemigroup
H∼=M(G;{i,j},{λ,µ};P1)
such that
P1=
pλi pλj
pµi pµj
∼
p0λi p0λj p0µi p0µj
,p0λi=p0λj=p0µi=e,p0µj=h, so
P1=
e e
e h
andhhi 6=G=hgi.
Consider the sets
X={(i,e,µ),(j,e,λ),(j,g,µ)},
Y ={(j,g,λ),(j,g,µ)}.
Now again it is sufficiency to see thathXi=hYi, and the setX is an independence, then we get a contradiction
with a basis property ofS. By proposition 2, we consider that the groupGis a cyclicp−group, where pis prime
and|G|=pα =m,
α≥1.Sincehhi 6=Gwe haveh=gk, wherep|k. Hence(2k+1, p) =1, then there exists u∈Nsuch thatu(2k+1)≡1(mod m). Now from
pλi=pλj=pµi=e, pµj=h,
we have
(i,e,µ) = (i,g,λ)m−1(j,g,µ)∈ hYi,
(j,e,λ) = (j,g,µ) (i,g,λ)m−1∈ hYi
(i,g,λ) =
i,g(2k+1)u,λ
=i,g2k+1,λ
u
= i,pµjgpµj,λu=
ThushXi=hYi. The independence ofX get from that the first and the second elements ofX under
multipli-cation can’t get the third fromX the another elements fromX have the same component in the first or third . A
contradiction give us that(2−11)holds for basis property ofS.
Corollary 2. IfS is a completely simple semigroup with a basis property and it is not a right, left group, or
rectangular band, thenSis finite and|S| ≤2pα+1such thatpα is the order of the maximal subgroup.
Proof. Without the loss of generality, consider that|Λ|=2≤ |I|and the groupGis an additive group of residue
modulopα. Then from(2−11)we find that the matrixPtransforms into the form
0 s1 ... sr ... 0 0 ... 0 ...
,
Such thatsi6≡0(mod p)andsj6≡0(mod p). Then the number of column smaller thanP, so|S| ≤2|G| |I| ≤2pα+1.
Proposition 10. Let|Λ|=2, a matrixPhas in every column 0. Then the groupGis trivial.
Proof. LetGbe a group of ordermand generating elementg. By the condition of the proposition we must
consider the case whenI=Λ={1,2}and the matrixPis an identity. We study the sets
X={(1,e,2),(2,e,1),(2,g,2)},
Y ={(1,e,2),(2,g,1)}.
Since
(2,e,1) = ((2,g,1) (1,e,2))m−1(2,g,1)∈ hYi,
(2,g,2) = (2,g,1) (1,e,2)∈ hYi,
(2,g,1) = (2,g,2) (2,e,1)∈ hXi.
We get thathXi=hYi. It is easy to see thatX is an independence and we have a contradiction.
The Main Theorem
Theorem 3. LetSbe a completely simple semigroup, thenSis semigroup with basis property if and only if it
is satisfying the following conditions:
(1) Sis group with basis property,
(2) Sis left or right group, which is the maximal subgroup either trivial or primary cyclic subgroup,
(3) S isomorphic to Rees matrix semigroupM(G;I,Λ;P)with sandwich matrixPover groupG, which is either trivial or primary cyclic, and then min{|I|, |Λ|}=2 . Also sandwich matrixP= (pλi)λ∈Λ,i∈I satisfying
∀i,j∈I∀λ,µ∈Λ
i6=j, λ6=µ⇒
D
pλip−1µipµjp−1λj
E
Proof. In fact, letSbe a semigroup with basis property. IfSa group thenSis a group with basis property by
proposition1. On the other hand if we assume thatSis not a group, thenSisomorphic to Rees matrix semigroup
M(G;I,Λ;P)with the sandwich matrix Pover groupG. SinceSis not a group, then|I| ≥1 or|Λ| ≥1 . If we
take|I|=1 or|Λ| ≥1, then by proposition 2 the groupGeither trivial or a primary cyclic group (also if|Λ|=1
,S is either rectangular band, or rectangular band primary cyclic group. Therefore let |I| ≥2 , |Λ| ≥2. Then
by proposition 8 we have min{|I|,|Λ|}=2. IfS is a completely simple semigroup, then by proposition 9 the
condition (2-11) holds, a groupGeither trivial or primary cyclic group by proposition 3.
The suffice condition proves in the same investigation .
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