1-Primitive Near-rings

0.12 MB·English

by  Gerhard Wendt

1-PRIMITIVE NEAR-RINGS
GERHARDWENDT
5 ABSTRACT. Wecombinetheconceptofsandwichnear-ringswiththatofcen-
1 tralizernear-ringstogetaclassificationofzerosymmetric1-primitivenear-rings
0 as dense subnear-rings of centralizer near-rings with sandwich multiplication.
2 Thisresult generalizes the well known density theorem for zero symmetric 2-
primitive near-rings with identity to the much bigger class of zero symmetric
n
a 1-primitivenear-ringsnotnecessarilyhavinganidentity.
J

8 1. INTRODUCTION

] We consider right near-rings, this means the right distributive law holds, but not
A
necessarilytheleftdistributivelaw. Thenotationisthatof[2]. Primitivenear-rings
R
play the same role in the structure theory of near-rings as primitive rings do in
.
h ring theory. Whenaprimitive near-ring happens tobearing, then itisaprimitive
t
a ring in the usual sense. However, in near-ring theory there exist several types of
m primitivity depending on the type of simplicity of a near-ring group. A complete
[ descriptionofsocalled2-primitivenear-ringswithidentityisavailable. Suchnear-

rings are dense subnear-rings of special types of centralizer near-rings (see [2]
1
v for athorough discussion). If one studies primitive near-rings without necessarily
2 havinganidentity, thenstillresultsareavailable butmuchmoretechnicalindetail
4
in comparison to the 2-primitive case with identity. Combination of the concepts
8
1 of centralizer near-rings and sandwich near-rings were used in [4] and in [5] to
0 describe 1-primitive near-rings which do not necessarily have an identity. In this
.
1 paper we will extend and simplify the results obtained in [4] and [5] and we will
0 alsoconsider 2-primitivenear-rings asaspecialcaseof1-primitivenear-rings.
5
Wewillnowbrieflygivethedefinitionsforzerosymmetricprimitivenear-rings to
1
settleournotation whichwewillkeepthroughout thepaper.
:
v Let N be a zero symmetric near-ring. Let G be an N-group of the near-ring N.
i
X AnN-ideal I ofG is anormal subgroup ofthe group (G ,+) such that ∀n∈N∀g ∈
r G ∀i∈ I : n(g +i)−ng ∈ I. When N is itself considered as an N-group, then an
a
N-ideal of N is just a left ideal of the near-ring. An N-group G of the near-ring N
is of type 0 if G 6={0}, if there are no non-trivial N-ideals in G , so G is a simple
N-group,andifthereisanelementg ∈G suchthatNg =G . Suchanelementg will
be called a generator of the N-group G . The N-group G is of type 1if it isof type
0and moreover wehave Ng =G orNg ={0} for any g ∈G . LetK be asubgroup
of the N-group G . K is called an N-subgroup of G if NK ⊆K. The N-group G is
called N-group of type 2ifNG 6={0} and there are no non-trivial N-subgroups in
G . It is easy to see that an N-group of type 2 is also of type 1. In case N has an
2000MathematicsSubjectClassification. 16Y30.
Keywordsandphrases. sandwichnear-rings,centralizernear-rings,primitivity.
This work has been supported by grant P23689-N18 of the Austrian National Science Fund
(FWF).
1
2 GERHARDWENDT
identity element, an N-group of type 1 is also of type 2 (see [2], Proposition 3.4
andProposition 3.7).
Givenan N-group G and asubset △⊆G then (0:△)={n∈N|∀g ∈△:ng =0}
willbecalledtheannihilator of△. G willbecalledfaithful if(0:G )={0}.
A near-ring N is called 1-primitive if it acts on a faithful N-group G of type 1. In
suchasituation wewillsaythatthenear-ringacts1-primitively ontheN-groupG .
It is a well known fact that any zero symmetric and 1-primitive near-ring with
identitywhichisnotaringisdense(i.e. equalto,inthefinitecase)inacentralizer
near-ring M (G ):={f :G →G |∀g ∈G ∀s∈S:s(f(g ))= f(s(g )) and f(0)=0},
S
where(G ,+)isagroup,0denotingitsneutralelementw.r.t. +,Sisafixedpointfree
automorphism group of G and the near-ring operations are the pointwise addition
of functions and function composition (see [2], Theorem 4.52 for a detailed dis-
cussion).
In case a 1-primitive near-ring contains no identity element, the situation gets
more complicated and one can use sandwich near-rings to get a classification of
1-primitive near-rings as certain kind of centralizer near-rings. We introduce the
concept of a sandwich near-ring in the next definition which we will need in the
nextsections. Theoperation symbol◦standsforcomposition offunctions.
Definition1.1. Let(G ,+)beagroup,X ⊆G asubsetofG containingthezero0of
(G ,+)andf :G −→X amapsuchthatf (0)=0. Definethefollowingoperation◦′
onG X (G X denotingthesetofallfunctionsmappingfromX toG ): f◦′g:= f◦f ◦g
for f,g∈G X. Then (G X,+,◦′) is a zero symmetric near-ring. Let M (X,G ,f ):=
0
{f :X →G |f(0)=0}. With respect to the operations + and ◦′, M (X,G ,f ) is a
0
zero symmetric subnear-ring of (G X,+,◦′). We call M (X,G ,f ) a sandwich near-
0
ringandf iscalledthesandwichfunction.
We give two examples of such sandwich near-rings in the following. They will
also serve as examples explaining the concepts of primitivity by doing concrete
calculations. LetN :={f ∈M (Z )|f(2)= f(3)=0}. Withrespect topointwise
0 4
addition of functions and function composition N is a zero symmetric near-ring
which acts faithfully on the N-group Z . For g ∈{0,2,3} we have Ng ={0} and
4
if g =1 we have Ng =Z . {0,2} is not an N-ideal of the N-group. To see this,
4
let f ∈N be such that f(1)=3. Then, f(1+2)− f(1) =16∈{0,2}. So, N acts
1-primitively on Z . Clearly, {0,2} is an N-subgroup of the N-group Z . Thus,
4 4
N does not act 2-primitively on Z . Note that N has 4 elements and cannot be
4
isomorphic to a centralizer near-ring since it is missing an identity element. Now
let X := {0,1} and f : Z → X such that f (0) = f (2) = f (3) = 0 and f (1) =
4
1. Then, M (X,Z ,f ) is a sandwich near-ring. For f ∈ M (X,Z ,f ) we have
0 4 0 4
f(0)=0 and f(1)∈{0,1,2,3}. Thus, M (X,Z ,f ) has 4 elements. Let g ,g ∈
0 4 1 2
M (X,Z ,f )such thatg (1)=1andg (1)=2. Thenmultiplication ◦′ isdone as
0 4 1 2
g ◦′g =g ◦f ◦g . So,g ◦′g (1)=g (f (g (1)))=g (f (2))=g (0)=0. Thus
1 2 1 2 1 2 1 2 1 1
g ◦′g isthezerofunction. Let f ∈N andlety :X →Z ,x7→ f(x). Thefunction
1 2 f 4
h:N →M (X,Z ,f ),f 7→y iseasilyseentobeanear-ring isomorphism.
0 4 f
Similary,ifweletN :={f ∈M (Z )|f(3)=0}thenoneseesthatN is2-primitive
1 0 4 1
on Z because {0,2}, the only non-trivial subgroup of Z , is not an N -subgroup.
4 4 1
N has 16 elements. Let X ={0,1,2} and f :Z →X such that f (0)=f (3)=0,
1 4
f (1) = 1 and f (2) = 2. Then, M (X,Z ,f ) is a sandwich near-ring. For f ∈
0 4
M (X,Z ,f ) we have f(0) =0, f(1) ∈{0,1,2,3} and f(2)∈{0,1,2,3}. Thus,
0 4
M (X,Z ,f ) has 16 elements. Let f ∈N and let y :X →Z ,x7→ f(x). As in
0 4 1 f 4
1-PRIMITIVENEAR-RINGS 3
theexamplebefore thefunction h:N →M (X,Z ,f ),f 7→y iseasilyseen tobe
0 4 f
anear-ring isomorphism.
Combinations of the concepts of centralizer near-rings and sandwich near-rings
were used in [4] and in [5] to describe 1-primitive near-rings which do not nec-
essarily have an identity. In [4], 1-primitive near-rings are described as dense
subnear-rings of near-rings of the type of M (X,G ,f ,S):={f :X −→G | f(0)=
0
0and∀s∈S∀x∈X : f(s(x))=s(f(x))},seeDefinition2.1,whereSisanautomor-
phism group of G acting without fixed points on the set X\{0} and the near-ring
operations are pointwise + of functions and sandwich multiplication, but the re-
sult andproof in [4]requires that theprimitive near-ring hasamultiplicative right
identity. We will see in this paper that this restriction is not needed. Thus we
cangeneralizetheresultof[4]tonear-ringsnotnecessarilyhavingamultiplicative
rightidentity andobtain adescription ofallzerosymmetric1-primitive near-rings
as well as 2-primitive near-rings as dense subnear-rings of sandwich centralizer
near-rings. Thisconstruction simplifiestheconstruction of[5].
In[5]alsonomultiplicative rightidentity isrequired andthe1-primitivenear-ring
is described as dense subnear-ring of a sandwich near-ring M(X,N,f ,y ,B,C):=
{f :X −→N |∀s∈S∀x∈X : f(s(x))=y (s)(f(x))}, whereX isanon-emptyset,
(N,+)agroup, f thesandwich function, Basubgroup ofAut(N,+),Sagroup of
permutations on X and y ∈ Hom(S,B). This construction is more technical than
that in [4] and that we will use in our approach in this paper and the functions
in M(X,N,f ,y ,B,C) are not centralized by elements of S. For the details of the
construction werefertheinterested readerto[5].
The idea of combining the concepts of centralizer near-rings and sandwich near-
ringsusedinthispaperallowsustoexplicitelydescribeandconstructthesandwich
functionf whichdeterminesthemultiplicationintheprimitivenear-ring. Thiswill
be done in the last section of this paper and will give us the possibility to con-
structzerosymmetric1-primitivenear-ringswithoutnecessarilyhavinganidentity
systematically. Examplestodemonstrate thisconstruction areincluded.
2. SANDWICH CENTRALIZER NEAR-RINGS
The following definition introduces certain types of sandwich near-rings which
were used by the author in [3] and [4] to describe near-rings with amultiplicative
rightidentity.
Definition2.1. Let(G ,+)beagroup,X ⊆G asubsetofG containingthezero0of
(G ,+)andf :G −→X amapsuchthatf (0)=0. LetS⊆End(G ,+),Snotempty,
be such that ∀s ∈ S,∀g ∈ G : f (s(g )) = s(f (g )) and such that S(X) ⊆ X. Then
M (X,G ,f ,S):={f :X −→G | f(0)=0and∀s∈S,x∈X : f(s(x))=s(f(x))}is
0
a zero symmetric subnear-ring of M (X,G ,f ) as defined in Definition 1.1, which
0
wecallasandwichcentralizer near-ring.
Itisstraightforward tosee thatM (X,G ,f ,S) isindeed azero symmetric subnear-
0
ring of M (X,G ,f ) where the zero of M (X,G ,f ,S) is the zero function 0 on X.
0 0
SinceS(X)⊆X,M (X,G ,f ,S)isnotempty,since0iscontained inM (X,G ,f ,S).
0 0
Notethatthefunctionid:X →G ,x7→xiscontainedinM (X,G ,f ,S)andservesas
0
amultiplicativerightidentityofthenear-ring. Asshownin[3],anyzerosymmetric
near-ringwithamultiplicativerightidentityisisomorphictoasandwichcentralizer
near-ring M (X,G ,f ,S)withsuitable X,G ,f ,S.
0
4 GERHARDWENDT
3. THE EQUIVALENCE RELATION ∼
Given a 1-primitive near-ring and an N-group G of type 1 we now introduce an
equivalence relation ∼ on G . What we need for the proof of our theorems in the
next section isaspecial type of system ofrepresentatives w.r.t. ∼, being invariant
under the N-automorphisms of G . The existence of such a representative system
willbeguaranteed inLemma3.3.
Definition3.1. LetNbeanear-ringandletG beanN-group. Letg ,g ∈G . Define
1 2
g ∼g iff∀n∈N :ng =ng .
1 2 1 2
It is easy to see that ∼ is an equivalence relation. To introduce another notation,
we mention that when we have a function f with domain D and M ⊆D, then f
|M
meanstherestriction ofthefunction tothesetM.
In the proof of Lemma 3.3 we will use that given an N-automorphism s of an N-
group G , then also the inverse function s−1 is an N-automorphism of the N-group
G . Thisisstraightforward toseeasisshowninthenextproposition.
Proposition 3.2. Let N be a zero symmetric near-ring and G be an N-group. Let
s∈Aut (G ,+). Thenalsotheinverse functions−1∈Aut (G ).
N N
Proof. Let s∈Aut (G ). Clearly, s−1 ∈Aut(G ,+). Let n∈N and g ∈G . Then,
N
ng =n(s(s−1(g )))=s(n(s−1(g ))). Thus, s−1(ng )=n(s−1(g )). Soweseethatalso
s−1 isanN-automorphism. (cid:3)
Lemma3.3. LetN beazero symmetricnear-ring andG beanN-group oftype 1.
Let S:=Aut (G ,+). Then there is a set of representatives X of the equivalence
N
relation∼suchthatS(X)⊆X and0∈X.
Proof. Let q :={g ∈G |Ng =G } be the set of generators and q :={g ∈G |Ng =
1 0
{0}}bethesetofnon-generatorsofG . SinceG isanN-groupoftype1,G =q ∪q .
0 1
Let D be a set of representatives w.r.t. ∼. We let 0 be the representative for the
equivalence class of 0 and therefore, for any d ∈ q , d ∼ 0. Hence, D = X ∪
0 1
{0}, with X ⊆q . Let f be the function which maps every element in q to its
1 1 1
representative in X w.r.t. ∼. Let g ∈q and let S(g ):={s(g )|s∈S} be the orbit
1 1
of g under the action of S on G . It is easy to see that S(g )⊆q . Let s ,s ∈S and
1 1 2
supposethats (g )∼s (g ). Then,foralln∈N,s (ng )=ns (g )=ns (g )=s (ng ).
1 2 1 1 2 2
Since g ∈q we see that for all d ∈G , s (d )=s (d ) and therefore, s =s . This
1 1 2 1 2
implies that the restriction f|S(g ) of f to the orbit S(g ) is an injective map. Let
K := {∪g ∈JS(g )|J ⊆ q 1 and f|∪g∈JS(g ) isinjective}. As we have just shown, f is
injective onanysingle orbit S(g ), g ∈q . Consequently, K isnotthe emptyset. K
1
is ordered w.r.t. set inclusion ⊆. Let I be an index set such that for i∈I,C ∈K
i
and (Ci)i∈I forms achain in K. LetM :=∪i∈ICi. So, M =∪i∈I(∪g ∈MiS(g )) where
fori∈I,M ⊆q aresuitablesetssuchthat(C) formsachain. So,thesetMisa
i 1 i i∈I
unionofunionsoforbitsandconsequently, Misaunionoforbitsofelementsfrom
q . Ifwecanshowthat f isinjectiveonM,thenM∈K. Suppose f isnotinjective
1
on M. So there are x,y∈M, x6=y such that f(x)= f(y). Thus, there are j,l ∈I
such that x∈C and y∈C. Since (C) forms a chain, we either haveC ⊆C
j l i i∈I j l
orC ⊆C . So, either x∈C and y∈C or x∈C and y∈C . f is injective onC
l j l l j j j
aswellasonC andconsequently wehave f(x)6= f(y)whichisacontradiction to
l
theassumptionthat f isnotinjectiveonM. Thus,M∈K andM isanupperbound
forthechain(C) . ByZorn’sLemma,K containsamaximalelementR,say. We
i i∈I
1-PRIMITIVENEAR-RINGS 5
claim that R∪{0} is a set of representatives w.r.t. ∼ which is invariant under the
actionofN-automorphisms ofG .
Note that as an element of K, R is a union of orbits of S, so S(R)⊆R. Since f is
injectiveonR,anytwoelementsofRareindifferentequivalenceclasses. Suppose
there is an element a ∈q such that a 6∼r for any r ∈R. Let s∈S and suppose
1
there is an element r∈R such that s(a )∼r. Since s is an N-automorphism, also
theinversefunctions−1 isanN-automorphismandcontainedinS(seeProposition
3.2). Thus, for all n ∈ N, s(na ) = ns(a ) = nr and therefore na = ns−1(r), so
a ∼s−1(r)∈R. Thisisacontradiction totheassumption a 6∼r foranyr∈Rand
so we see that for any s∈S and any r∈R, s(a )6∼r. But then, f is injective on
R∪S(a ), so R∪S(a )∈K. Clearly, S(a )6⊆R and so R is properly contained in
R∪S(a ). This contradicts the maximality of R. Since wealso have d ∼0 for any
element d ∈q we see that X :=R∪{0} is a set of representatives w.r.t ∼. Since
0
S(0)=0andS(R)⊆R,S(X)⊆X. (cid:3)
Wekeep the notation ofLemma3.3 togive somecomments. Wehave seen in the
proofofthelemma,thatgivenanelementg ∈q ,thenS(g )isasetof∼inequivalent
1
elements. ItiseasytoseethatS(q )⊆q . Consequently, theresult ofLemma3.3
1 1
isimmediate andwewouldnothavetoapply Zorn’sLemmaiftheN-groupG has
finitelymanyorbitsw.r.t. theactionofSonG ,inparticular thisisthecasewhenG
isfinite.
4. DENSITY THEOREMS
Wewillnowprovetwotheorems whichareourmaintheorems ofthispaper. First
we show that up to isomorphism zero symmetric and 1-primitive near-rings show
up as dense subnear-rings of sandwich centralizer near-rings with special condi-
tions on X,G ,f and S. Following this theorem we then can easily prove a similar
result for 2-primitive near-rings. We should make clear what density means (see
also[2],Proposition 4.26)andfixsomemorenotation.
Definition 4.1. F is a dense subnear-ring of M (X,G ,f ,S) if and only if ∀s∈N
0
∀x ,...,x ∈X ∀g∈M (X,G ,f ,S)∃f ∈F: f(x)=g(x)foralli∈{1,...,s}.
1 s 0 i i
Also,weneedtheconceptoffixedpointfreeness.
Definition4.2. LetSbeagroup ofautomorphisms ofagroupG . LetM⊆G \{0}
such that S(M) ⊆ M. S is called fixedpointfree on M if for s ∈ S and m ∈ M,
s(m)=mimpliess=id,id beingtheidentityfunction. Siscalledafixedpointfree
automorphism groupofG ifitactsfixedpointfree onG \{0}.
Let (G ,+)be a group. If I is anormal subgroup of G we willdenote this as I⊳G .
Ford ∈G ,d +I isthecosetofd . 0/ standsfortheemptyset.
As already pointed out in the introduction, 1-primitive near-rings which are rings
are primitive rings in the ring theoretical sense (see [2], Proposition 4.8). So, we
restrict our discussion to non-rings. Weare now ready to formulate our first theo-
rem.
Theorem 4.3. LetN be a zero symmetric near-ring which is not a ring. Then the
followingareequivalent:
(1) N is1-primitive.
(2) Thereexist
6 GERHARDWENDT
(a) agroup(G ,+),
(b) asetX ={0}∪X ⊆G ,X 6=0/,06∈X and0beingthezeroofG ,
1 1 1
(c) S ≤Aut(G ,+), with S(X)⊆X and S acting without fixed points on
X ,
1
(d) afunctionf :G →X withf =id,f (0)=0andsuchthat∀g ∈G ∀s∈
|X
S:f (s(g ))=s(f (g )),
such that N isisomorphic to adense subnear-ring M of M (X,G ,f ,S)
S 0
whereX,G ,f ,Sadditionally satisfythefollowingproperty (P):
Let G 0 :={g ∈G |f (g )=0} andC:={I⊳G |I ⊆G 0 and G 0 =∪d ∈G 0d +
I and ∀g ∈ G \G ∀i ∈ I : S(f (g +i)) = S(f (g ))}. Then I ∈C ⇒ (I =
0
{0}or∃i∈I∃g ∈G \G ∃s∈S∃g ∈G :f (g +i)=s(f (g ))ands(g )−g 6∈
1 0 1 1
I).
Proof. (1)⇒(2): LetNact1-primitivelyontheN-groupG . Letq :={g ∈G |Ng =
1
G }bethesetofgeneratorsandq :={g ∈G |Ng ={0}}bethesetofnon-generators
0
ofG . By1-primitivity ofN,G =q ∪q . LetS=Aut (G ,+). OnG wedefinethe
1 0 N
equivalence relation ∼ as in Definition 3.1 by g ∼g iff for all n∈N, ng =ng .
1 2 1 2
According to Lemma 3.3 we choose a set of representatives X of the equivalence
relation ∼ in a way that S(X)⊆X and 0 is the representative of the equivalence
class of 0. Note that any element in q is equivalent to 0 w.r.t. ∼. Thus, X =
0
X ∪{0},whereX ⊆q . NotethatX 6=0/ because asanN-groupoftype1,G has
1 1 1 1
agenerator.
Letf :G →X,g 7→x where g ∼x. Then f =id and f (0)=0 because the repre-
|X
sentativeofthezeroequivalence classwastakentobezero.
Next we show that for all g ∈ G and s∈ Aut (G ,+) we have s(f (g )) = f (s(g )).
N
To show this, we first prove that s(f (g )) ∼s(g ) for any g ∈G . Let n∈N. Then,
ns(f (g ))=s(nf (g )). Now,f (g )∼g , sos(nf (g ))=s(ng )=ns(g ). Thisshowsthat
s(f (g ))∼s(g ). Consequently, by the definition of f , f (s(f (g ))) =f (s(g )). Since
S(X)⊆X andf =id wehavethatf (s(f (g )))=s(f (g ))whichprovesthedesired
|X
property.
NextweprovefixedpointfreenessofSonX . NotethatS(X )⊆X becauseS(X)⊆
1 1 1
X and S is a group of automorphisms, so only the zero 0 in X is mapped to zero.
Letg ∈X ⊆q . ThenNg =G . Suppose thats(g )=g . Therefore, foralln∈N we
1 1
getns(g )=s(ng )=ng . Thus,foralld ∈G wehaves(d )=d ands=id.
Now weshow how to embed N into M (X,G ,f ,S). For every n∈N,let f be the
0 n
function f :X −→G ,x7→nx. We now prove that the mapping h:n7→ f is an
n n
embedding ofN intoM (X,G ,f ,S).
0
Firstweshow that forn∈N, f ∈M (X,G ,f ,S), sofor alls∈S,for allx∈X we
n 0
must have s(f (x))= f (s(x)). Since s is an N-automorphism we get f (s(x))=
n n n
n(s(x))=s(nx)=s(f (x))andconsequently, f ∈M (X,G ,f ,S)sincealso f (0)=
n n 0 n
0. So,hmapsN intoM (X,G ,f ,S).
0
h is a near-ring homomorphism: Let j and k be arbitrary elements of N. Then
h(j+k)= f = f + f byrightdistributivity ofN. Letx∈X. Thenh(jk)(x)=
(j+k) j k
f (x)=(jk)x. Ontheotherhand,h(j)◦′h(k)= f ◦f ◦f . So,foreveryx∈X, f ◦
jk j k j
f ◦f (x)= j(f (kx)). Bydefinitionoff weknowthatf (kx)∼kxandconsequently,
k
j(f (kx))= j(kx)=(jk)x. Thisshowsthath(j)◦′h(k)= f =h(jk).
jk
h is injective: Since h is a near-ring homomorphism, it suffices to show that the
kernel of h is zero. Suppose there exists an element j∈N such that f is the zero
j
function. Thismeansthat f (x)=0forallx∈X. SinceX isasetofrepresentatives
j
1-PRIMITIVENEAR-RINGS 7
w.r.t. ∼, this implies jG ={0}. By faithfulness of G we get j =0. Hence, h is
injectiveandthisfinallyprovesthathisanembedding.
SowecanembedN intothenear-ring M (X,G ,f ,S)andweleth(N)=:M . Con-
0 S
sequently, N ∼= M and it remains to show that M is a dense subnear-ring of
S S
M (X,G ,f ,S)whereX,G ,f ,Sadditionally satisfy theproperty (P).
0
Supposex ∈X andx ∈X arefromdifferentorbitsofSactingonX andsuppose
1 1 2 1 1
x and x havethesameannihilator. Then, Nx =Nx =G ands:G −→G ,nx 7→
1 2 1 2 1
nx isawelldefinedN-automorphism ofG ,whichisstraightforward tosee. Since
2
x ∈q , there is an element k∈N such that kx =x . Consequently, s(x )=kx .
1 1 1 1 1 2
For any n∈N we have nx =nkx and therefore n−nk ∈(0:x )=(0 :x ). It
1 1 1 2
follows that nx =nkx for all n∈N which means that s(x )=kx ∼x . Since
2 2 1 2 2
s(x )∈ X , s(x )=f (s(x )) =x which contradicts the assumption that x ∈X
1 1 1 1 2 1 1
andx ∈X arefromdifferentorbitsofSactingonX . So,elementsfromdifferent
2 1 1
orbitsofSactingonX musthavedifferentannihilators.
1
Take x ,...,x ∈X , l ∈N, from finitely many different orbits of S acting on X .
1 l 1 1
Thenallelementsin{x ,...,x }havedifferentannihilatorsaswehavejustshown.
1 l
SinceN isnotaringwecanapplyTheorem4.30of[2]togetthatforallg ,...,g ∈
1 l
G there exists some n∈N such that nx =g for all i∈{1,...,l}. Hence, for all
i i
i∈{1,...,l},h(n)(x)= f (x)=nx =g .
i n i i i
We now have to show that ∀l ∈ N ∀x ,...,x ∈ X ∀f ∈ M (X,G ,f ,S)∃m ∈ M :
1 l 0 S
m(x)= f(x)foralli∈{1,...,l}.
i i
For any function f ∈ M (X,G ,f ,S) and any function m ∈ M we have f(0) =
0 S
m(0)= 0. So it suffices to consider the case that for l ∈N, x ,...,x ∈ X . Let
1 l 1
f ∈ M (X,G ,f ,S). Let v ∈ N and let z ,...,z be a set of orbit representatives
0 1 v
for the elements x ,...,x ∈X under the action of S on X . Thus, z ,...,z have
1 l 1 1 1 v
different annihilators and so there is an element m ∈ N such that f (z) = f(z)
m i i
for i∈{1,...,v}. For k∈{1,...,l} there exists a unique j∈{1,...,v} such that
x ∈S(z ). Then x =s(z ) for some, by fixedpointfreeness of S on X , unique s
k j k j 1
andso, f (x )= f (s(z ))=s(f (z ))=s(f(z ))= f(s(z ))= f(x ). So, f and
m k m j m j j j k m
f areequalfunctionswhenrestricted totheset{x ,...,x }andalso f ∈M . This
1 l m S
provesdensity ofM inM (X,G ,f ,S).
S 0
Finally wehave toshowthatX,G ,f ,S satisfy theproperty (P).LetI ∈C. Assume
that I 6= {0}. I 6= G since G 6= G , so I is a non-trivial normal subgroup of G .
0
Supposeproperty(P)doesnothold,soassumethatforalli∈I,forallg ∈G \G ,
1 0
for all s ∈ S and for all g ∈ G either f (g +i) 6= s(f (g )) holds or s(g )−g ∈ I
1 1
holds. Since I ∈C, for all i∈I and for all g ∈G \G there is an s∈S such that
1 0
f (g +i)=s(f (g ))andtherefore, forthiss∈S,s(g )−g ∈I forallg ∈G .
1 1
Letn∈N and g ∈G . Thus, g +I ⊆G because bydefinition of theelements in
0 0 0 0
C, G is a union of cosets of I. Consequently, for all j ∈I, g + j∈g +I ⊆G .
0 0 0 0
Thus,n(g + j)−ng =nf (g + j)−nf (g )=0−0∈I. Letg ∈G \G andi∈I.
0 0 0 0 1 0
Consequently, f (g +i) = s(f (g )) for some s ∈ S and so, since property (P) is
1 1
assumed not tohold, s(g )−g ∈I forallg ∈G . So,foralln∈N,n(g +i)−ng =
1 1
nf (g +i)−nf (g )=ns(f (g ))−nf (g )=s(nf (g ))−nf (g )∈I. Thisshowsthat
1 1 1 1 1 1
I isanon-trivial andproper N-idealofG ,contradicting thatN is1-primitive onG .
Hence,property(P)musthold.
(2)⇒(1): We have to show that N ∼=M is a 1-primitive near-ring. G is an M -
S S
groupinanaturalwaybydefiningtheaction⊙ofM onG asm⊙g :=m(f (g ))for
S
m∈M andg ∈G . Sincef =id wehaveX =f (G ),soG isafaithful M -group.
S |X S
8 GERHARDWENDT
Let g ∈G and suppose f (g ) =0, so g ∈G . Then clearly M ⊙g ={0}. On the
0 S
otherhandthereexistelementsg ∈G ,suchthat06=f (g )∈X becausef =idand
1 |X
X 6=0/. Let g be such that f (g )∈X , so g ∈G \G . Let d ∈G . We now define a
1 1 0
function f :X −→G in the following way: f(f (g )):=d . Let s∈S. Since S acts
withoutfixedpointsonX ,wecanwelldefine f(s(f (g ))):=s(f(f (g )))=s(d )and
1
f(X\S(f (g ))):={0}. From the definition of f wesee that f ∈M (X,G ,f ,S), so
0
by density of M , there is a function m∈M with m(f (g ))= f(f (g ))=d . Since
S S
d ∈G was chosen arbitrary, this shows that M ⊙g =G . Consequently, for d ∈G
S 0
wehaveM ⊙d ={0}andforg ∈G \G wehaveM ⊙g =G .
S 0 S
Wenowshowthattherearenonon-trivial M -idealsinG . SupposethatI isanon-
S
trivialM -idealofG . ThenI isanon-trivial normalsubgroup of(G ,+),andforall
S
m∈M ,g ∈G andi∈Iwehavethatm⊙(g +i)−m⊙g =m(f (g +i))−m(f (g ))∈I.
S
Sincem(f (0))=0forallm∈M ,I beinganM -idealimpliesthatm(f (i))∈I for
S S
all i∈I and all m∈M . This implies that I ⊆G . Let d ∈G , i∈I. Then, for all
S 0 0
m∈M ,m(f (d +i))−m(f (d ))=m(f (d +i))∈I. Thisshowsthat d +i∈G and
S 0
so,G isaunionofcosetsofI.
0
AssumethatI isnotcontainedinthesetC. Thus,thereexistsanelementg ∈G \G
0
and an element i ∈ I such that S(f (g +i)) 6= S(f (g )). Since S is a group, this
implies that foralls∈S, f (g +i)6=s(f (g )). Thismeans that f (g )and f (g +i) are
in different orbits of S acting on X. Suppose that g +i∈ G . Since I is an M -
0 S
ideal, this implies that m(f (g ))∈I for all m∈M , hence we must have f (g )=0,
S
acontradiction tog ∈G \G . Sowehave thatf (g +i)6=0aswellasf (g )6=0and
0
wenowdefinetwofunctions f :X −→G and f :X −→G .
1 2
Let f (f (g )):=d ∈I,fors∈Slet f (s(f (g ))):=s(f (f (g )))and f (X\S(f (g ))):=
1 1 1 1 1
{0}. Let f (f (g +i)):=d 6∈I, fors∈S let f (s(f (g +i))):=s(f (f (g +i))) and
2 2 2 2
f (X\S(f (g +i))):={0}. f and f arewelldefinedbecauseoffixedpointfreeness
2 1 2
ofSonX . Itisaroutinechecktoseethat f and f areelementsinM (X,G ,f ,S).
1 1 2 0
Since for all s ∈ S, f (g +i) 6= s(f (g )) we have that f (f (g +i)) = 0 as well as
1
f (f (g ))=0.
2
Wenowhavethat(f +f )(f (g +i))−(f +f )(f (g ))=f (f (g +i))+f (f (g +i))−
1 2 1 2 1 2
f (f (g ))− f (f (g ))=0+d −0−d 6∈I. By density of M in M (X,G ,f ,S), there
2 1 2 1 S 0
is an element m∈M such that m(f (g +i))=(f + f )(f (g +i)) and m(f (g ))=
S 1 2
(f + f )(f (g )). Consequently, I isnotanM -ideal.
1 2 S
AssumingthatIisnotcontainedinthesetCcontradictsourassumptionthatIisan
M -ideal. So, wenow assume that I iscontained inC. Consequently, by property
S
(P),thereexistsi∈I,thereexistsg ∈G \G ,thereexistss∈Sandthereexistsg ∈G
1 0
such that f (g +i)=s(f (g )) and s(g )−g 6∈I. Since g ∈G \G , M ⊙g =G , so
1 1 1 0 S 1
there exists an element m∈M such that m(f (g ))=g . Consequently, m(f (g +
S 1 1
i))−m(f (g ))=m(s(f (g )))−m(f (g ))=s(m(f (g )))−m(f (g ))=s(g )−g 6∈I.
1 1 1 1 1
Thisisagainacontradiction totheassumption thatI isanM -ideal.
S
Thisshowsthatthereexistnonon-trivialM -idealsinG ,soM is1-primitiveonG .
S S
(cid:3)
Property (P)ofTheorem 4.3isatechnical condition whichexcludes subgroups of
G to be M -ideals, in the language of Theorem 4.3. Since any 2-primitive near-
S
ring is also 1-primitive, Theorem 4.3 applies to zero symmetric and 2-primitive
near-rings also. When considering 2-primitive near-rings the technical condition
of property (P) can be much more simplified. This leads to an especially simple
versionofthetheorem.
1-PRIMITIVENEAR-RINGS 9
Theorem 4.4. LetN be a zero symmetric near-ring which is not a ring. Then the
followingareequivalent:
(1) N is2-primitive.
(2) Thereexist
(a) agroup(G ,+),
(b) asetX ={0}∪X ⊆G ,X 6=0/,06∈X and0beingthezeroofG ,
1 1 1
(c) S ≤Aut(G ,+), with S(X)⊆X and S acting without fixed points on
X ,
1
(d) afunctionf :G →X withf =id,f (0)=0andsuchthat∀g ∈G ∀s∈
|X
S:f (s(g ))=s(f (g )),
such that N isisomorphic to adense subnear-ring M of M (X,G ,f ,S)
S 0
whereG :={g ∈G |f (g )=0} does not contain any non-trivial subgroups
0
ofG .
Proof. (1)⇒(2):SinceN is2-primitiveitisalso1-primitiveandhenceTheorem
4.3 and its proof of (1) ⇒ (2) applies. So, let G be the N-group of type 2 the
near-ring acts on 2-primitively. Let f be as in the proof of (1)⇒(2) of Theorem
4.3. It only remains to show that G does not contain any non-trivial subgroups
0
of G . Suppose K ⊆G is a subgroup of G . Since f (K)={0} we know from the
0
definitionoff thatK⊆q ={g ∈G |Ng ={0}}. Thus,NK={0}⊆K andK isan
0
N-subgroupofG . Itfollowsfrom2-primitivity ofN thatK ={0}.
(2)⇒(1): As in the proof of (2)⇒(1) of Theorem 4.3 one shows that M acts
S
faithfully on G with the action ⊙ and such that for d ∈G we have M ⊙d ={0}
0 S
and for g ∈G \G we have M ⊙g =G . Also, G \G 6=0/. Suppose that K is an
0 S 0
M -subgroup of G and K 6=G . It follows that K ⊆ G . By assumption, G does
S 0 0
not contain anynon-trivial subgroups ofG . Hence, K ={0}. Thus, G contains no
non-trivial M -subgroups andM is2-primitiveonG .
S S
(cid:3)
5. CONSTRUCTION OFf AND EXAMPLES
We keep the notation of the proof of Theorem 4.3 throughout the whole section.
Inordertoconstruct1-primitivenear-ringsassandwichcentralizernear-ringswith
thehelpofTheorem 4.3onemustbeassured thatX,G ,f ,Ssatisfyproperty (P).In
caseX =G andf =id wehavethatG ={0}. Inthiscaseproperty (P)istrivially
0
fulfilled because only the trivial group {0} is contained inC. In fact, in this case
M (X,G ,f ,S)=M (G ):={f :G →G |∀g ∈G ∀s∈S:s(f(g ))= f(s(g ))and f(0)=
0 S
0}. So, M (X,G ,f ,S) = M (G ) is a 1-primitive near-ring with identity element
0 S
(andthus2-primitive). Thefactthatallzerosymmetric1-primitivenear-ringswith
identity element which arenot rings show up asdense subnear-rings ofnear-rings
of the type M (G ) with S a group of fixedpointfree automorphisms acting on G is
S
certainly the most well known density theorem for primitive near-rings (see [2],
Theorem 4.52). This result is also covered by Theorem 4.3, Theorem 4.4 respec-
tively, because in case of a near-ring with identity, f as constructed in the proof
of (1)⇒ (2) of Theorem 4.3 is just the identity function. So, X =G , G ={0}
0
andM (X,G ,f ,S)=M (G ). SisactingwithoutfixedpointsonG \{0}because of
0 S
condition (2c)inTheorem4.3.
Weneednotonlyconsidernear-rings withidentitytoobtainsituations whenprop-
erty(P)iseasilyfulfilled. Property(P)isobviously fulfilledwhenC onlycontains
thetrivialgroup {0}. C willonlycontain thetrivial group forexample whenG is
0
10 GERHARDWENDT
not a union of cosets of some non-trivial normal subgroup of G . Hence, probably
theeasiestwaytoobtain1-primitivenear-ringsistotakeS={id}andanyfunction
f mapping from G to a subset X of G containing the zero 0 such that G is not a
0
union ofcosets ofsomenon-trivial normalsubgroup ofG . Notethat theexamples
afterDefinition1.1areofthattype.
To be in a position to construct 1-primitive near-rings or 2-primitive near-rings
with the help of our main theorems and S6={id} one has to find a suitable sand-
wich function f which commutes with the automorphisms in S and S has to act
without fixed points on the set X =X\{0}. If one has found such a function f
1
and the group S, then primitivity of the near-ring M (X,G ,f ,S) only depends on
0
thesubgroups contained inG . Inthefollowing twopropositions weshow howto
0
construct the sandwich function f with the desired properties and show that any
suchf canbeconstructed thisway.
Proposition 5.1. Let(G ,+)be agroup and S≤Aut(G ,+). LetG⊆G \{0} such
that S(G)⊆G and S acts without fixed points on G. Let {e|i∈I}, I a suitable
i
index set, be a complete set of orbit representatives of the orbits of S acting on
G. So, G=∪ S(e). Let 0/ 6=J, J ⊆I. Let X :=∪ S(e ) and X :={0}∪X .
i∈I i 1 j∈J j 1
Let K :=I\J. If 0/ 6=K, let f :{e |k ∈K}−→∪ S(e ) be a function. Define
k j∈J j
f :G −→X as
0 ifg ∈G \G
f (g ):= g ifg ∈∪ S(e )
 j∈J j
s(f(e )) ifK isnotemptyandg =s(e )∈∪ S(e )
k k k∈K k

Then,f isawelldefinedfunction suchthatf =id and∀g ∈G ∀s∈S:f (s(g ))=
|X
s(f (g )). Furthermore, SactswithoutfixedpointsonX andS(X )⊆X .
1 1 1
Proof. Since X ⊆G, S acts without fixed points on X . Also, since X is a union
1 1 1
oforbits ofS,S(X )⊆X . SupposeK :=I\J isnottheemptyset. Let f :{e |k∈
1 1 k
K}−→∪ S(e )beafunction. Letg ∈∪ S(e ). Byfixedpointfreeness ofSon
j∈J j k∈K k
Gthereisauniques∈Sandauniquek∈Ksuchthatg =s(e ). Thenthedefinition
k
f (g ):=s(f(e ))makesf awelldefinedfunction. Since0∈G \∪ S(e)wehave
k i∈I i
f (0) = 0 and so, f = id. It remains to show that for all g ∈ G and all s ∈ S,
|X
s(f (g ))=f (s(g )). Let g ∈G \∪ S(e) and s∈S. Then, s(f (g ))=s(0)=0 and
i∈I i
since s(g )∈G \∪ S(e) we also have f (s(g ))=0. Let g ∈∪ S(e ) and s∈S.
i∈I i j∈J j
Then, s(f (g )) = s(g ). On the other hand, s(g ) ∈ ∪ S(e ) and so we also have
j∈J j
f (s(g ))=s(g ) bydefinition off . Suppose K isnot empty. Letg ∈∪ S(e ). So,
k∈K k
g = s (e ) for a unique s ∈ S. Let s ∈ S. Then, s(s (e )) ∈ ∪ S(e ). Then,
1 k 1 1 k k∈K k
f (s(g ))=f (s(s (e )))=s(s (f(e ))). Onthe other hand, wealso have s(f (g ))=
1 k 1 k
s(f (s (e )))=s(s (f(e ))). This finally shows that f (s(g ))=s(f (g )) for all g ∈
1 k 1 k
G . (cid:3)
Wegiveaconcreteexampletodemonstratetheconstructionprocessofthefunction
f inProposition 5.1. WeusethenotationofProposition5.1. Let(G ,+):=(Z ,+)
7
and S:={id,−id} where −id :Z →Z ,x7→−x. S is a fixedpointfree automor-
7 7
phism group of (Z ,+). Let G:={1,6,2,5} be the union of the orbits of 6 and
7
5. Let e := 6 and e := 5, so I := {1,2}. Let J := {1} and K := {2}. Thus,
1 2
X :={1,6},theorbitof6,andX :={0,1,6}. Let f :{5}→{1,6},f(5)=1(here
1
we could also define f(5)=6 resulting in a different f ). Then f :Z →X is de-
7
finedasfollows: 0=f (0)=f (3)=f (4),f (1)=1,f (6)=6. Since2=−id(e )=
2