QUASI-INJECTIVE NEAR-RING GROUPS

We extend the concepts of quasi-injective modules and their endomorphism rings to near-ring groups. We attempt to derive the near-ring character of the set of endomorphism of quasi-injective N -groups under certain conditions and this leads us to a near-ring group structure which motivates us to study various characteristics of the structure. If E is a quasi-injective N-group and S = End(injective hull of E) then we study the structure ES and various properties of ES. It is proved that ES is a minimal quasi-injective extension of E and any two minimal quasi-injective extensions are equivalent. This structure motivates to study the Jacobson radical of endomorphism near-ring of quasi-injective N -group E. It is established that the near-ring modulo the Jacobson radical is a regular near-ring. Some properties of quasi-injective N groups relating essentially closed N -subgroups and complement N -subgroups are established. AMS Subject Classification: 16Y30


Prerequisites
All basic concepts used in this paper are available in Pilz [4].In this section we define the basic terms and results that are needed for the sequel.Definition 1.1.For a right near-ring (N, +, .)and a corresponding Ngroup E, suppose there is an x ∈ E such that {nx|n ∈ N } = E. Then E is a monogenic N -group and x is a generator.Since every left ideal is a left N -subgroup, a strictly small left ideal of N is also a small left ideal of N .
Definition 1.4.The intersection of all maximal ideals maximal as Nsubgroups of N -group E is called radical of E and is denoted by J(E).
Lemma 1.1.[3]: If the radical ideal J(N) is strictly small in N then the following conditions are equivalent-(i) Y ∈ J(N ) (ii) 1-xy is left invertble for all x ∈ N (iii) yM = 0 for any irreducible left N-group M. Definition 1.5.An N -subgroup (ideal) I of E is said to be a essentially closed N-subgroup (ideal) of E if I has no proper essential extension in E. Definition 1.6.An N -subgroup (ideal) I of E is said to be a essentially closed N -subgroup (ideal) of E if I has no proper essential extension in E.
Theorem 1.3.[Clay]: For a near-ring (N, +, .)with identity 1, suppose E is a monogenic unitary N-group with generator x and suppose that T = {m ∈ N/Ann(x)m ∈ Ann(x)} is a subgroup of (N , +, .).Then the N -endomorphisms E of N -group E forms a right near-ring where

Endomorphism Near Ring of Quasi-Injective N-Groups
In this section we investigate various characteristics of endomorphism near-ring of quasi-injective N -groups.We also study Jacobson radical of endomorphism near-ring of quasi-injective N -groups.Throughout this section unless and otherwise mention we assume E satisfies the condition of theorem 1.3 and N is a dgnr.
If Ê is injective hull of E and For this S-group we get the following: Proof.(a) Let M be an N -subgroup of ES and f : M → ES.We take the inclusion map i : ES → Ê.Then the composite map h = if : M → Ê.Since Ê is injective , so h can be extended by some λ : Ê → Ê such that for Therefore (ES)λ ⊆ ES. λ induces λ : ES → ES. i.e. λ can be restricted by some λ : (b) Let P be any quasi-injective N -subgroup of Ê containing E. We wish to show ES = ∩P .Since by (a) ES is quasi-injective.So ∩P ⊆ ES.Now to show ES ⊆ ∩P .We will show ES ⊆ P , so it is sufficient to show that P α ⊆ P ∀α ∈ S. Since if ∀α ∈ S, P α ⊆ P then P S ⊆ P .But And E ⊆ ES is obvious by inclusion map: ES = E Definition 2.1.(P, E, f ) denotes a N -monomorphism f : E → P and is called an extension of E. An extension (P, E, f ) of an N -group E is a minimal quasi-injective extension in case P is quasi-injective and the following condition is satisfied: If (A, E, g) is any quasi-injective extension of E, then there exists a monomorphism φ : P → A such that: Then (g(ES))Ω ⊆ g(ES) and we conclude that (B)Ω ⊆ B where Since ES is the smallest quasi-injective extension of E contained in Ê, we conclude that g −1 (B) = (ES).So B = g(ES) ⊆ A. This establishes that ES is a minimal quasi-injective extension.
Next if (A, E, g) is also a minimal quasi-injective extension of E, then (A, E, g) is also equivalent to ES. ES minimal quasi-injective extension of Definition 2.2.A near-ring N is said to be a regular near-ring if for every element x ∈ N , there exists an element y ∈ N such that xyx = x.Theorem 2.1.Let E be quasi-injective N-group let Λ = Hom(E, E) and let J = J(Λ) denote the Jacobson radical of Λ and is strictly small in Λ.Then J = {λ ∈ Λ/E essential extension of Kerλ}.If for γ ∈ J, λ ∈ Λ, γλ ∈ J then Λ/J is a regular near-ring.Where addition of two N -subgroups is again N -subgroup of E and N need not be dgnr.
However if λ ∈ I, Ker(1+ µλ) = 0 for Kerλ∩Ker(1+ µλ) = 0.For if, λ ∈ I we have E essential extension of Kerλ.x ∈ Kerλ∩Ker(1 This establishes that I ⊆ J. Next let λ be arbitrary element of Λ, let L be a complement N-subgroup of E corresponding to K = Ker(λ) and consider the correspondence λx → x ∀ x ∈ L. If λx = λy with x, y ∈ L, then λ(x − y) = 0 and then (x − y) Thus J = I is asserted.Also I is an ideal by given condition.Thus Λ is a regular modulo I.

Some Properties of Quasi-Injective N-Groups
This section contains some properties of quasi-injective N -groups related to essentially closed N -subgroups and complement N -subgroups.Let M be an N -subgroup of E. We consider F = {P/P N -subgroup ofE, If sum of two N -subgroups is again an N -subgroup of an N -group we get the following: So by Zorn's Lemma Q ∈ F , maximal element exists.Thus Q in the first sentence exists.
Now to prove the second part.
Let T be any non-zero N -subgroup of Q and assume that T ∩ M = 0. Since ] This contradicts the definition of K.This proves that Q is an essential extension of M .If P is an N -subgroup of E properly containing Q , then P ∩ K = 0 and (P ∩ K) ∩ M = P ∩ (K ∩ M ) = P ∩ 0 = 0. Thus P is not essential extension of M , completing the proof.Proof.Let M be a essentially closed N -subgroup and K is any complement of M .Then by lemma 3.1 there exists a complement Next let M be complement of an N -subgroup P .Then ∃ a complement K of M which contains P .
If possible let M / ≤ E such that M ⊆ M / & K∩M / = 0. Then P ∩M / = 0. ∵ P ⊂ K, which contradicts (1).Therefore M is also maximal such that K ∩ M = 0. Therefore M is complement of K. Then M is essentially closed by lemma3.1.This also proves the last statement.Theorem 3.1.Let E be quasi-injective and let M be a essentially closed N -subgroup, then for each N -subgroup K of E, N -homomorphism w : K ⇒ M can be extended to N -homomorphism u : E → M Proof.Let F = {L/w is extended to a map of T into M for N -subgroup T of E containing L} By Zorn's lemma we can assume that K is such that w cannot be extended to a map of T into M for any N -subgroup T of E which properly contains K. Since, E is quasi-injective, w is induced by a map u : (1) If M is essentially closed N -subgroup of E, then M is a direct summand of E and M is quasi-injective.Also M has a complement in E.
(2) If P is any N -subgroup of E, then there exists a quasi-injective essential extension of P contained in E.
(3) Each minimal quasi-injective extension of an N -group K is an essential extension of K.

Definition 1 . 2 .
An N -subgroup B of E is called fully invariant if for each N -homomorphism f : E → E, f (B) ⊂ B. Definition 1.3.A left ideal A of N is called small (strictly small) if N = B for each left ideal (N -subgroup) B such that N = A + B.
E. (A, E, g) is also quasi-injective extension of E. By definition for E e. g = φf Again (A, E, g) is minimal quasi-injective extension of E. ES is also quasiinjective extension of E. By definition for E e. f = ωg.Now f = ωg ⇒ f = ωφf .So I = ωφ.Again g = φf ⇒ g = φωg.So I = φω.Thus ω and φ both are invertible which implies both ω and φ are isomorphic.Hence ES ∼ = A.

Lemma 3 . 2 .
The essentially closed N -subgroups of an N -group E coincide with the complement N -subgroup of E. If M and K are complement N -subgroups and if K is a complement of M in E then M is a complement of K in E.

Corollary 3 . 1 .
a contradiction.Therefore a / ∈ M and a = x − b ∈ L + M .Now T = y ∈ E/u(y) ∈ L + M is an N -subgroup of E containing K. ∵ x ∈ K ⇒ w(k) ∈ M ⇒ u(k) ∈ M ∀k ∈ K. Therefore T contains K.If y ∈ E is such that u(y) = a then y ∈ T , but y / ∈ K since a / ∈ M .[∵ y ∈ T ⇒ u(y) = a ∈ L&y ∈ K ⇒ w(y) ∈ M ⇒ u(y) ∈ M ∀y ∈ K, contradiction to a / ∈ M ].Let π denote the projection of L+M on M .Then πu is a map of T in M and πu(y) = u(y) = w(y) ∀y ∈ K. [∵ y ∈ K ⇒ w(y) ∈ M ⇒ u(y) ∈ M ∀y ∈ K].Thus πu is a proper extension of w, a contradiction.Therefore u(E) ⊆ M, so u is the desired extension.For quasi-injective N -group E.

Proof. ( 1 )
If e : E → M is the extension given by theorem 3.1 of the injection map M → M then E = M Ker(e) where e(m) =m, m ∈ M 0, m / ∈ M So that M is a direct summand of E. Therefore M is quasi-injective by theorem 1.2 Moreover Ker(e) is complement of M .Since M ∩ Ker(e) = (0)(2)M essentially closed ⇒ M complement of some N -subgroup K ⇒ K is complement of M , i.e. max M max K = (0)