Table Of ContentREFLEXIVE UNITARY SUBSEMIGROUPS OF
LEFT SIMPLE SEMIGROUPS
A. Nagy
5
1
0
2
Introduction
n
a
Ideal series of semigroups [1] play an important role in the examination of
J
semigroupswhichhavepropertwo-sidedideals. Butthecorrespondingtheorems
7
cannotbe usedwhenleft simple(or rightsimple orsimple)semigroupsarecon-
] sidered. Soitisanaturalideathatwetrytousethegroup-theoreticalmethods
R
(instead of the ring-theoretical ones) for the examination of these semigroups.
G
The purpose of this paper is to find a suitable type of subsemigroups of
. left simple semigroups which makes possible for us to generalize some notions
h
t (the notionofthe normalseriesandthe compositionseriesofgroups)andsome
a
results concerning the groups to the left simple semigroups. We note that the
m
subsemigroupswearelookingforarethe reflexiveunitarysubsemigroupsofleft
[
simple semigroups.
1 For notations and notions not defined here, we refer to [1] and [2].
v
1
3
Reflexive unitary subsemigroups
4
1
0
. As it is known [1], a subset H of a semigroup S is said to be reflexive in S
1
0 if ab∈H implies ba∈H for all a,b∈S.
5 We say that a subset U of a semigroupS is left [right] unitary in S (see [1])
1 if ab,a∈U implies b∈U [ab,b∈U implies a∈U] for all a,b∈S. A subset of
:
v S which is both left and right unitary in S is said to be unitary in S.
i WenotethatifA⊆B aresubsemigroupsofasemigroupS andB isunitary
X
in S, then A is unitary in B if and only if A is unitary in S. In this paper we
r
a shall use this fact without comment.
It can be easily verified that, in a group, the unitary subsemigroup are
exactlythe subgroups,andthe reflexiveunitarysubsemigroupsareexactly the
normal subgroups. So the notion of the reflexive unitary subsemigroup of a
semigroup can be considered as a generalization of the notion of the normal
subgroup of a group.
1
Reflexive unitary subsemigroups are important in the description of the
group congruences of left simple semigroups. To show this importance, we
need the following.
Let S be a semigroup and H a non-empty subset of S. Let
H...a={(s,t)∈S×S : sat∈H}, a∈S.
We can define a relation P (S) on S as follows:
H
P (S)={(a,b)∈S×S : H...a=H...b}.
H
It can be easily verified that P (S) is a congruence on S such that WH ={c∈
H
S : H...c=∅} is a P (S)-classand anidealof S. P (S) is called the principal
H H
congruence on S determined by H (see [1]).
Ifwe considerandfix asemigroupS then, forasubsetH ofasubsemigroup
N of S, the principal congruence on N determined by H will be denoted by
P (N). we shall use P instead of P (S). For short, we shall use N/P
H H H H
instead of N/P (H). P |N will denote the restriction of P to N, that is,
N H H
P |N =P (S)∩(N ×N).
H H
We note that if α is a congruence on a semigroup S then we shall not
distinguish the congruence classes of S modulo α from the elements of the
factor semigroup S/α.
Let S denote a left simple semigroup. Then, by Theorem 10.34, Corollary
10.35 and Exercise for §10.2 of [1], if H is a reflexive unitary subsemigroup of
S then the factor semigroup S/P is a group with identity element H and,
H
conversely,if P is a congruence on S such that S/P is a group with identity H
then H is a reflexive unitary subsemigroup of S and P =P .
H
The following theorem is important for the next.
Theorem 1 Every right unitary subsemigroup of a left simple semigroup is left
simple.
Proof. Let N be a right unitary subsemigroupof a left simple semigroupS.
Consider two arbitrary elements a an b in N. Then there is an element s in
S such that sa = b. As N is right unitary in S, it follows that s ∈ N which
implies that N is left simple. Thus the theorem is proved. ⊓
Thefollowingtheoremshowsthat,fixingareflexiveunitarysubsemigroupH
of a left simple semigroup S, there is a one-to-one correspondence between the
set of all unitary subsemigroups of S containing H and the set of all subgroups
of S/P . In this connection the reflexive unitary subsemigroups correspond to
H
the normal subgroups of S/P .
H
Theorem 2 Let H ⊆ N be subsemigroups of a left simple semigroup S such
that H is reflexive and unitary in S. Then N is unitary in S if and only if N
is saturated by P and N/P is a subgroup of S/P . In this case P (N) =
H H H H
P |N. If N is unitary in S, then it is also reflexive in S if and only if N/P
H H
2
is a normal subgroup of S/P . If this is the case, then (S/P )/(N/P ) is
H H H
isomorphic with S/P .
N
Proof. LetH be a reflexiveunitary subsemigroupofa left simple semigroup
S. Assume that N is an unitary subsemigroup of S with H ⊆ N. Let a ∈ N
be arbitrary. As N is left simple by Theorem 1, there are elements u,v ∈ N
such that uav ∈ H. As H is reflexive in S, vua ∈ H. Since N is unitary in
S, vu ∈ N. So, for every b ∈/ N, it follows that vub ∈/ N from which we get
vub ∈/ H. Thus ubv ∈/ H and so (a,b) ∈/ P which means that N is saturated
H
byP . Sincethe inclusionP |N ⊆P (N)followsfromthe definitionofP |N
H H H H
and P (N), it is sufficient to show that P (N) ⊆ P |N. To show this last
H H H
inclusion,letaandbbearbitraryelementsinN with(a,b)∈P (N). Weprove
H
(a,b) ∈ P |N. Assume, in an indirect way, that (a,b) ∈/ P |N. Then there
H H
are elements x,y ∈ S such that either xay ∈ H and xby ∈/ H or xay ∈/ H and
xby ∈H;weconsideronlythefirstcase. SinceH isreflexiveinS,itfollowsthat
yxa∈H and yxb∈/ H. As a∈N and N is unitary in S, we get yx∈N. Thus,
for anarbitraryelement h in H, it followsthat yxah∈H and yxbh∈/ H. Since
yx and h are in N, we have (a,b) ∈/ P (N) which contradicts the assumption
H
(a,b) ∈ P (N). Consequently, (a,b) ∈ P |N, that is, P (N) = P |N. It is
H H H H
evident that N/P is a subsemigroup of S/P . Let [b] ∈ N/P be arbitrary,
H H H
where [b] denotes the congruence class of S modulo P containing b. From
H
[b][b]−1 ⊆ H it follows that [b]−1 ⊆ N/P , because N is unitary in S and
H
[b]⊆N. Thus N/P is a subgroup of S/P .
H H
Conversely,ifasubsemigroupN withH ⊆N issaturatedbyP andN/P
H H
is a subgroup of S/P , then N is a unitary subsemigroup of S. Next, consider
H
two arbitrary elements a and b in S, and let ω denote the canonical homomor-
phism of S onto S/P . Then ab ∈ N if and only if aωbω ∈ Nω. So N is
H
reflexive in S if and only if Nω is a normal subgroup of S/P .
H
To prove the isomorphism between S/P and (S/P )/(N/P ) = G, let β
N H H
denote the canonical homomorphism of S/P onto G. Then
H
′
S ={s∈S : sωβ is the identity of G}
equals N and the kernel of ωβ is P (see Theorem 10.34 of [1]). Thus S/P is
N N
isomorphic with G. Thus the theorem is proved. ⊓
Theorem 3 If H and N are subsemigroups of an arbitrary semigroup S such
that H is reflexive and unitary in S and H ∩N 6=∅, then H ∩N is a reflexive
unitary subsemigroup of N and hH,Ni/PH is isomorphic with N/PH∩N.
Proof. Let H an N be subsemigroups of a semigroup S such that H is re-
flexive and unitary in S and H ∩N 6= ∅. It can be easily proved that H ∩N
is a reflexive unitary subsemigroup of N. We may assume S = hH,Ni. If [N]
denotes the union of the P -classes A of S with A∩N 6=∅, then S =[N]. Let
H
(a,b) ∈ P for some a,b ∈ N. Then, for every x,y ∈ N, xay ∈ H ∩N if and
H
onlyifxby ∈H∩N,because(a,b)∈P ,a,b∈N andN isasubsemigroup. So
H
3
(a,b) ∈ PH∩N, that is, PH|N ⊆ PH∩N(N). Next we show that PH∩N(N) ⊆
PH|N. Let a,b ∈ N with (a,b) ∈ PH∩H(N). Assume, in an indirect way,
that (a,b) ∈/ P |N. Then there are elements x,y ∈ S such that, for example,
H
xay ∈H and xby ∈/ H. Since H is the identity element S/P and S =hH,Ni,
H
there are elements u and v in N with (x,u) ∈ P and (y,v) ∈ P . Using
H H
(xay,uav) ∈ P and (xby,ubv) ∈ P , we have uav ∈ H and ubv ∈/ H. Since
H H
u,v ∈ N, the last result contradicts (a,b) ∈ PH∩N(N). Thus (a,b) ∈ PH|N.
Consequently PH|N = PH∩N(N). Nothing that, for any x ∈ S, there exists
u ∈ N such that (x,u) ∈ P , we can see that hH,Ni/P is isomorphic with
H H
N/PH∩N. ⊓
Theorem 4 IfH andN areunitarysubsemigroupsofaleftsimplesemigroupS
suchthat H is reflexive in S, then H∩N 6=∅, hH,Ni is a unitarysubsemigroup
of S and hH,Ni=HN. If N is also reflexive in S, then HN is reflexive in S.
In this case hH,Ni=HN =NH.
Proof. Let H and N be unitary subsemigroups of a left simple semigroup
S such that H is reflexive in S. Consider two elements b and b in N with
1 2
(b ,b ) ∈ P . By Theorem 1, N is left simple. Thus there is an element x in
1 2 H
N suchthat xb =b . So (xb ,ub )∈P for every u∈H. Since P is a group
1 2 1 2 H H
congruenceon S, we have (x,u)∈P which means that H∩N 6=∅. Evidently
H
H ∩N is a unitary and reflexive subsemigroup of N. Let
K ={s∈S : (s,n)∈P for some n∈N}.
H
Then K is a subsemigroup of S. We show that K is unitary in S. Let a, b
be arbitrary elements in S. Assume a,ab ∈ K. Then there is an element a
1
in N such that (a,a ) ∈ P . Since H ∩N is a reflexive unitary subsemigroup
1 H
of N and N is left simple, PH∩N(N) is a group congruence on N. Thus there
is an element r in N such that ra ∈ H ∩N. Then ra ∈ H which implies
1
(rab,b) ∈ P . Since rab = r(ab) ∈ NK ⊆ K, it follows that b ∈ K. Thus K
H
is left unitary in S. To show that K is right unitary in S, assume a,ba ∈ K.
Let k denote the product ba. Then there are elements a ,k ∈ N such that
2 1
(a ,a) ∈ P and (k ,k) ∈ P . Since N is left simple, there is an element q in
2 H 1 H
N such that qa =k . Then (ba,qa)∈P . Since P is a group congruence on
2 1 H H
S,itfollowsthat(b,q)∈P ,thatis,b∈K. ThusK isrightunitaryinS. SoK
H
isunitaryinS. The proofofthe firstassertionwillbe complete ifweshowthat
HN = K. Since H,N ⊆ K, it follows that HN ⊆ hH,Ni ⊆ K. Let k be an
arbitraryelementofK. Fixanelementv ofN suchthat(v,k)∈P . SinceS is
H
leftsimple,thereisanelementsinS suchthatsv =k whichmeansthats∈H
and so k ∈HN. Thus K ⊆HN and so HN =hH,Ni=K. Assume that N is
alsoreflexiveinS. ThenH∩N isareflexiveunitarysubsemigroupofS andboth
H and N are saturatedby PH∩N. Denote ω the canonicalhomomorphismof S
onto S/PH∩N. Then Hω and Nω are normal subgroups of the group S/PH∩N
and so NωHω =(NH)ω is a normal subgroup of S/PH∩N. Let L={s∈S :
4
sω ∈ (NH)ω}. Then, by Theorem 2, L = NH = hH,Ni = HN is a reflexive
unitary subsemigroup of S. ⊓
Theorem 5 (Zassenhauslemma)LetS bealeftsimplesemigroupandA,B,M,N
subsemigroups of S. Assume that A and B are unitary in S and A∩B 6= ∅.
If N and M are reflexive unitary subsemigroups in A and B, respectively, then
N(A∩M) and M(B ∩N) are reflexive unitary subsemigroups in N(A∩B)
and M(A∩B), respectively, such that N(A∩B)/PN(A∩M) is isomorphic with
M(A∩B)/PM(B∩N).
Proof. FirstweprovethatA∩M 6=∅. BytheassumptionforAandB,A∩B
is an unitary subsemigroup in B. Since M is a reflexive unitary subsemigroup
in B, we get M ∩ A = M ∩ (A ∩ B) 6= ∅ (see also Theorem 4). Similarly,
B∩N 6=∅. UsingalsoTheorem3, A∩M andB∩N areunitarysubsemigroups
ofA∩B. ItcanbeeasilyverifiedthattheyarealsoreflexiveinA∩B. Then,by
Theorem4, hA∩M,B∩Ni=(A∩M)(B∩N)=(B∩N)(A∩M)is areflexive
unitary subsemigroupinA∩B. Let C, D andH denote the semigroupsA∩B,
(A∩M)(B ∩N) and C/P . As C is left simple by Theorem 1, H is a group.
D
We give a homomorphism p of NC onto H. Let x be an arbitrary element
of NC. Then there exist elements a ∈ N and c ∈ C such that x = ac. Let
xp be the congruence class of C modulo P (C) containing the element c. We
D
prove that p is well defined. Let a ∈ N and c ∈ C be arbitrary elements
1 1
with a c = x. We prove that (c,c ) ∈ P (C). Since H is a group whose
1 1 1 D
identity element is D, there is an element c′ in C such that c c′ ∈ D. Thus
1
acc′ = xc′ = a c c′ ∈ ND. Since D is unitary in A ∩ B, it is unitary in
1 1
A. By Theorem 4, hH,Di = ND is a unitary subsemigroup of A and so of
S. Since acc′ ∈ ND and a ∈ hN,Di = ND then it follows that cc′ ∈ ND
′
and so cc ∈ ND∩C. As ND∩C = [((N ∩D)∪(N \D))D]∩C = [((N ∩
D)D)∩C]∪[((N \D)D)∩C] = [((N ∩D)D)∩C]∪∅ = [((N ∩D)D)∩C]=
[(N ∩B)(N ∩B)(M ∩A)]∩C = [(N ∩B)(M ∩A)]∩C = D ∩C = D, we
get cc′ ∈ D. This result together with c c′ ∈ D imply that the P (C)-classes
1 D
[c] and [c ] of C containing c and c , respectively, the inverses of the P (C)-
1 1 D
class [c′] of C containing c′. As the factor semigroup C/P is a group, we get
D
[c] = [c ] which implies that xp is well defined. Consequently, the mapping p
1
is well defined. To prove that p is a homomorphism, let x and y be arbitrary
element of NC. Then xy ∈ NC because NC = hN,Ci is a subsemigroup.
Thus there are elements a ,a ,a ∈ N ⊆ A and c ,c ,c ∈ C ⊆ A such that
1 2 3 1 2 3
x = a c , y = a c , xy = a c . We show that (c c ,c ) ∈ P (C). Since
1 1 2 2 3 3 1 2 3 D
a c a c = a c ∈ A then we have (a c a c ,a c ) ∈ P (A). Since N is the
1 1 2 2 3 3 1 1 2 2 3 3 N
identity element of the groupA/P then it follows that (c c ,c )∈P (A). To
N 1 2 3 N
prove(c c ,c )∈P (C), we have to that, for all u,v ∈C =A∩B, both of the
1 2 3 D
elements uc c v and uc v are either in D or in C\D. First we show that, for
1 2 3
5
every t,s∈C, the condition (t,s)∈P (A) implies that both t and s are either
N
in D or in C \D. Since N ⊆hNDi=ND and ND is a unitary subsemigroup
ofAbyTheorem4,thenwegetthatND issaturatedbyP (A)byTheorem2.
N
Since ND ∩C = D (see above) then (t,s) ∈ P (A) implies either t,s ∈ D
N
or t,s ∈/ D for every t,s ∈ C. Let u and v be arbitrary elements in C. As
u and v re in A, taking into consideration that (c c ,c ) ∈ P (A), we get
1 2 3 N
(uc c v,uc v) ∈ P (A). As uc c v,uc v ∈ C, both uc c v and uc v ar in
1 2 3 N 1 2 3 1 2 3
either D or in C\D (as above). Consequently (c c ,c )∈P (C) which means
1 2 3 D
thatpisahomomorphismofNC ontoH. RecallthatD isthe identityelement
of H. Let Y = {y ∈ NC : yp = D}. We prove that Y = N(A∩M). Since
A∩M ⊆D(andsoY ⊆ND),wehaveN(A∩M)⊆Y. ToproveY ⊆N(A∩M),
let y be an arbitrary element in Y. Then there are elements a ∈ N and d ∈D
with y =ad. Thus y ∈ND = N(B∩N)(A∩M)⊆N2(A∩M)=N(A∩M),
because N is an unitary subsemigroup of S and so (by Theorem 1) it is left
simple,fromwhichitfollowsthatN2 =N. ThusY ⊆N(A∩M). Consequently
Y =N(A∩M),thatis,N(A∩M)istheidentityelementofthefactorgroupH ∼=
N(A∩B)/ker . ThenN(A∩M)isareflexiveunitarysubsemigroupofN(A∩B)
p
and N(A∩B)/PN(A∩M) ∼=H. We can prove, in a similar way, that M(B∩N)
isareflexiveunitarysubsemigroupofM(B∩A)andM(B∩A)/PM(B∩N) ∼=H.
By the transitivity of the isomorphism, the theorem is proved. ⊓
Normal series and composition series
Definition 6 Let S be a left simple semigroup. By a normal series of S we
mean a finite sequence
S =S ⊇S ⊇···⊇S (1)
0 1 k
of subsemigroups S of S with the property that S is a reflexive unitary sub-
i i
semigroup of Si−1 for i = 1,...,k. The integer k is called the length of the
normal series (1). The factor of (1) are the groups Si−1/PSi, i=1,...,k.
We say that the normal series (1) and a normal series
S =H ⊇H ⊇···⊇H (2)
0 1 n
are isomorphic if n = k and there exists a permutation i 7→ i∗ of the integers
i=1,...,k suchthatSi−1/PSi andHi∗−1/PHi∗ areisomorphic fori=1,...,k.
Moreover, we say that (2) is a refinement of (1) if n ≥ k and every S equals
i
some H , i=1,...,k;j =1,...,n.
j
Theorem 7 In a left simple semigroup S, every two normal series have refine-
ments ending with the same subsemigroup of S.
6
Proof. Let
S =S ⊇S ⊇···⊇S (3)
0 1 k
and
S =H ⊇H ⊇···⊇H (4)
0 1 n
betwonormalseriesofaleftsimplesemigroupS. UsingTheorem4,wegetthat
S ∩H 6= ∅ for every i = 0,1,...,k and every j = 0,1,...,n. It can be easily
i j
verifiedthat,foreveryi=0,1,...k, S ∩H isareflexiveunitarysubsemigroup
i j
of Si ∩Hj−1, j = 1,...,n. Similarly, for every j = 0,1,...,n, Si ∩Hj is a
reflexive unitary subsemigroup of Si−1∩Hj, i=1,...,k. Thus
S =S ⊇S ⊇···⊇S ⊇S ∩H ⊇S ∩H ⊇···⊇S ∩H (5)
0 1 k k 1 k 2 k n
and
S =H ⊇H ⊇···⊇H ⊇H ∩S ⊇H ∩S ⊇···⊇H ∩S (6)
0 1 n n 1 n 2 n k
are two normal series of S such that (5) and (6) are refinements of (3) and (4),
respectively. ⊓
Theorem 8 (Schreier refinement theorem) In a left simple semigroup, every
two normal series have isomorphic refinements.
Proof. Let
S =S ⊇S ⊇···⊇S (7)
0 1 k
and
S =H ⊇H ⊇···⊇H (8)
0 1 n
be two normal series of a left simple semigroup S. By Theorem 7, we may
assume S =H . For every i=1,...,k and j =0,1,...,n, let
k n
Sj,i =Si(Si−1∩Hj)=hSi,(Si−1∩Hj)i
and, for every i=0,1,...,k and j =1,...,n, let
Hj,i =Hj(Hj−1∩Si)=hHj,(Hj−1∩Si)i.
By Theorem 4, S0,i = Si(Si−1∩H0) = SiSi−1 = hSi,Si−1i = Si−1 and Sn,i =
Si(Si−1 ∩Hn) = Si(Si−1 ∩Sk) = SiSk = hSi,Ski = Si for every i = 1,...,k.
Similarly, Hj,0 = Hj−1 and Hj,k = Hj for every j = 1,...,n. Theorem 5
implies that,foreveryi=1,...,k andj =1,...,n,the subsemigroupsS and
j,i
Hj,i are reflexive unitary subsemigroups of subsemigroups Sj−1,i and Hj,i−1,
respectively, such that Sj−1,i/PSj,i is isomorphic with Hj,i−1/PHji. Then
S =S ⊇S ⊇···⊇S =S =S ⊇S ⊇···⊇S =S (9)
0,1 1,1 n,1 1 0,2 1,2 n,k k
7
and
S =H ⊇H ⊇···⊇H =H =H ⊇H ⊇···⊇H =H (10)
1,0 1,1 1,k 1 2,0 2,1 n,k n
are isomorphic normal series of S such that (9) and (10) are refinements of (7)
and (8), respectively. ⊓
Definition 9 A normal series S = S ⊇ S ⊇ ··· ⊇ S of a left simple semi-
0 1 k
group S will be called a composition series of S if Si−1 6= Si and Si−1 has no
reflexive unitary subsemigroup T with Si−1 ⊃ T ⊃ Si, i = 1,...,k +1 (here
S denotes the empty set).
k+1
Remark. It is clear that a normal series S = S ⊇ S ⊇ ··· ⊇ S of a left
0 1 k
simple semigroup S is a composition series of S if and only if Si−1 6=Si, every
factor Si−1/PSi is a simple group (i=1,...,k) and Sk has no proper reflexive
unitary subsemigroups.
Theorem 10 (Jordan-H¨older theorem) If a left simple semigroup S has a com-
position series, then every two composition series of S are isomorphic with each
other.
Proof. By Theorem 8, it is obvious. ⊓
Example. Let(S ;◦),(S ;∗)be leftsimplesemigroupssuchthatS ∩S =
1 2 1 2
∅ and there is an isomorphism α of the semigroup S onto S . On the set
1 2
F =S ∪S , define an operation as follows: for every e,f ∈F, let
1 2
e◦f, if e,f ∈S
1
ef =e∗fα if e∈S2,f ∈S1
eα∗f if e∈S ,f ∈S
1 2
aα−1◦fα−1 if e,f ∈S2.
It can be shown that F is a left simple semigroup such that S is a reflexive
1
unitary subsemigroup of F.
Remarks. If
S =S ⊇S ⊇···⊇S (11)
0 1 k
is a normal series of a left simple semigroup S, then every S , i = 0,1,...,k
i
is a unitary subsemigroup of S and so they are left simple semigroups (see
Theorem 1). If (11) is a composition series, then S has no proper reflexive
k
unitary subsemigroups. So it is a natural problem to describe the structure of
left simple semigroups which have no proper reflexive unitary subsemigroups.
ByTheorem1.27of[1],asemigroupisleftsimpleandcontainsanidempotent
elementifandonlyifitisadirectproductofaleftzerosemigroupandagroup.
Sowecanprovethataleftsimplesemigroupwithanidempotentelementhasno
proper reflexive unitary subsemigroups if and only if it is a left zero semigroup.
A problem. Find all left simple semigroups without idempotent elements
which have no proper reflexive unitary subsemigroups.
8
References
[1] Clifford,A.H.andG.B.Preston,TheAlgebraicTheoryofSemigroups,Amer.
Math. Soc. Providence R.I. I(1961), II(1967)
[2] Kurosh, A.G., Theory of Groups, Moscow, 1953 (Russian)
Attila Nagy
Department of Algebra
Mathematical Institute
Budapest University of Technology and Economics
e-mail: [email protected]
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