EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 1, 2018, 238-243
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Beta G-Star Relation on Modules

Celil Nebiyev1,∗, Nurhan Sökmez1

1 Department of Mathematics, Ondokuz Mayıs University, Kurupelit-Atakum, Samsun, Turkey

Abstract. In this work, we say submodules X and Y of M are β∗
g equivalence, Xβ∗

gY , if and only
if Y + K = M for every K E M such that X + K = M and X + T = M for every T E M such
that Y +T = M . It is proved that the β∗

g relation is an equivalent relation and has good behaviour
with respect to addition of submodules and homomorphisms.

2010 Mathematics Subject Classifications: 16D10, 16D70

Key Words and Phrases: Small Submodules, Generalized Small Submodules, Supplemented
Modules, G-Supplemented Modules

1. Introduction

Throughout this paper all rings will be associative with identity and all modules will
be unital left modules.

Let R be a ring and M be an R−module. We will denote a submodule N of M by
N ≤ M . Let M be an R−module and N ≤ M . If L = M for every submodule L of M
such that M = N +L, then N is called a small submodule of M and denoted by N �M .
Let M be an R−module and N ≤M . N is called essential submodule of M and denoted
by N EM in case K∩N 6= 0 for every submodule K 6= 0. Let M be an R−module and K
be a submodule of M . K is called a generalized small (briefly, g-small) submodule of M if
for every essential submodule T of M with the property M = K +T implies that T = M ,
then we write K �g M . (in [11], it is called an e-small submodule of M and denoted by
K �e M). It is clear that every small submodule is a generalized small submodule but the
converse is not true generally. M is called a (generalized) hollow module if every proper
submodule of M is (generalized) small in M . Here it is clear that every hollow module is
generalized hollow module. Let M be an R−module and U, V ≤ M . If M = U + V and
V is minimal with respect to this property, or equivalently, M = U + V and U ∩ V � V ,
then V is called a supplement of U in M . M is called a supplemented module if every
submodule of M has a supplement in M . Let M be an R−module and U, V ≤ M . If
M = U +V and M = U +T with T E V implies that T = V , or equivalently, M = U +V
and U ∩V �g V , then V is called a g-supplement of U in M . M is called g-supplemented

∗Corresponding author.

Email addresses: cnebiyev@omu.edu.tr (Celil Nebiyev), nozkan@omu.edu.tr (Nurhan Sökmez)

http://www.ejpam.com 238 c© 2018 EJPAM All rights reserved.



C. Nebiyev, N. Sökmez / Eur. J. Pure Appl. Math, 11 (1) (2018), 238-243 239

if every submodule of M has a g-supplement in M . Let M be an R−module and U ≤M .
If for every V ≤M such that M = U + V , U has a g-supplement V

′
with V

′ ≤ V , we say
U has ample g-supplements in M . If every submodule of M has ample g-supplements in
M , then M is called an amply g-supplemented module. SocM indicates the socle of M
(the sum of all simple submodules of M).

Lemma 1. Let M = U + V and M = U ∩ V + T . Then M = U + V ∩ T = V + U ∩ T .

Proof. See [4, Lemma 1.24].

2. The β∗g Relation

Definition 1. We define the relation ′β∗g
′ on the set of submodules of an R−module M by

Xβ∗gY if and only if Y +K = M for every K EM such that X+K = M and X+T = M
for every T EM such that Y + T = M .

Proposition 1. Let M be an R−module and X,Y ≤M . If Xβ∗Y , then Xβ∗gY .

Proof. Clear from definitions. (See [2]).

Lemma 2. The β∗g relation is an equivalence relation.

Proof. The reflective and symmetric properties are clear. For transitive property,
assume Xβ∗gY and Y β∗gZ. Let K EM and X +K = M . Since Xβ∗gY , then Y +K = M ,
and since Y β∗gZ, then Z + K = M . Let T E M and Z + T = M . Since Y β∗gZ, then
Y + T = M , and since Xβ∗gY , then X + T = M . Hence Xβ∗gZ.

Lemma 3. Let X,Y ≤M . The following statements are equivalent.
(i) Xβ∗gY .
(ii) For every T EM such that X + Y + T = M , X + T = M and Y + T = M .

Proof. (i) =⇒ (ii) Let T E M and X + Y + T = M . Since T E M , then Y + T E M
and X + T E M . Then by Xβ∗gY , M = X + Y + T = X + X + T = X + T and
M = X + Y + T = Y + Y + T = Y + T .

(ii) =⇒ (i) Let K EM and X +K = M . Then X + Y +K = M and by hypothesis,
Y +K = M . Similarly we prove that for every T EM such that Y +T = M , X+T = M .

Proposition 2. Let X,Y ≤M . If Xβ∗gY , then X+Y
X �g

M
X and X+Y

Y �g
M
Y .

Proof. Let X+Y
X + T

X = M
X for T

X E M
X . Clearly, we can see that T E M . Since

X+Y
X + T

X = M
X , then M

X = X+Y
X + T

X = Y+T
X and Y + T = M . Then by Xβ∗gY ,

X + T = M , and since X ≤ T, T = M . Hence X+Y
X �g

M
X . Similarly, we can prove that

X+Y
Y �g

M
Y .



C. Nebiyev, N. Sökmez / Eur. J. Pure Appl. Math, 11 (1) (2018), 238-243 240

Remark 1. The converse of the Proposition 2 is not true in general. For example, con-
sider the Z-module ZZ and let p and q be primes with p 6= q. Since Z

Zp and Z
Zq are simple,

Zp+Zq
Zp = Z

Zp �g
Z
Zp and Zp+Zq

Zq = Z
Zq �g

Z
Zq . But Zpβ∗gZq is not true.

Theorem 1. Let X,Y ≤ M such that X ≤ Y + A and Y ≤ X + B, where A,B �g M .
Then Xβ∗gY .

Proof. Let T EM and X+Y +T = M . Then (Y +A)+Y +T = M and A+Y +T = M.
Since T E M , then Y + T E M . Then, by A �g M , Y + T = M . Similarly, we can see
that X + T = M .

Lemma 4. Let X ≤M . X �g M if and only if Xβ∗g0.

Proof. (=⇒) Let X �g M and let X + 0 + T = X + T = M for T E M . Since
X �g M and X + T = M , then 0 + T = T = M . Then, by Lemma 3 Xβ∗g0.

(⇐=) Let Xβ∗g0. Let X + T = M for T E M . Since Xβ∗g0, then T = 0 + T = M .
Hence X �g M .

Corollary 1. Let X,Y ≤M and Xβ∗gY . If X �g M , then Y �g M .

Proof. Since X �g M , then by Lemma 4, Xβ∗g0, and since Xβ∗gY , then by Lemma 2,
Y β∗g0. Then, by Lemma 4, Y �g M .

Corollary 2. Let M be an R−module. Then M is generalized hollow if and only if Xβ∗g0
for every proper submodule X of M .

Proof. Clear from Lemma 4.

Corollary 3. Let M be an R−module. Then M is generalized hollow if and only if Xβ∗gY
for every proper submodules X, Y of M .

Proof. Clear from Lemma 4.

Remark 2. Let M be a nonzero semisimple R−module. Since M have no proper essential
submodules, M �g M and by Lemma 4, Mβ∗g0. But Mβ∗0 is not true.

Corollary 4. Let M be an R−module. Then SocMβ∗g0.

Lemma 5. Let X1, X2, Y1, Y2 ≤M such that X1β
∗
gY1 and X2β

∗
gY2. Then (X1 +X2)β

∗
g (Y1 + Y2).

Proof. Let X1 +X2 +K = M for K E M . Since K E M , then X2 +K E M . Then,
by X1β

∗
gY1, Y1 + X2 + K = M . Since K E M , then Y1 + K E M . Then, by X2β

∗
gY2,

Y1 + Y2 + K = M . Similarly, we can see that X1 + X2 + T = M for every T E M such
that Y1 + Y2 + T = M .

Corollary 5. Let X1, X2, ..., Xn, Y1, Y2, ..., Yn ≤ M and Xiβ
∗
gYi for every i = 1, 2, ..., n.

Then X1 +X2 + ...+Xnβ
∗
gY1 + Y2 + ...+ Yn.



C. Nebiyev, N. Sökmez / Eur. J. Pure Appl. Math, 11 (1) (2018), 238-243 241

Proof. Clear from Lemma 5.

Corollary 6. Let X1, X2, ..., Xn, Y ≤ M and Xiβ
∗
gY for every i = 1, 2, ..., n. Then

X1 +X2 + ...+Xnβ
∗
gY .

Proof. Clear from Lemma 5.

Lemma 6. Let f : M −→ N be an R−module epimorphism and X,Y ≤ M . If Xβ∗gY ,
then f (X)β∗gf (Y ).

Proof. Let f (X) + f (Y ) + T = N for T E N . Then X + Y + f−1 (T ) = M . Since
T E N , then we can see that f−1 (T ) E M . Then, by Lemma 3, X + f−1 (T ) = M and
Y + f−1 (T ) = M . Since X + f−1 (T ) = M and Y + f−1 (T ) = M , then f (X) + T = N
and f (Y ) + T = N . Hence, by Lemma 3, f (X)β∗gf (Y ).

Corollary 7. Let X,Y, Z ≤M . If Xβ∗gY, then X+Z
Z β∗g

Y+Z
Z .

Proof. Clear from Lemma 6.

Corollary 8. Let M be an R−module, A be a direct summand of M and X,Y ≤ A. If
Xβ∗gY in M , then Xβ∗gY in A also holds.

Proof. Clear from Lemma 6.

Proposition 3. Let X,Y ≤ M . If Xβ∗gY and Y is an essential maximal submodule of
M , then X ≤ Y .

Proof. Assume X � Y . Then, because Y is an essential maximal submodule of M ,
X + Y = M and since Xβ∗gY , Y = Y + Y = M . This contradicts maximality of Y .

Definition 2. Let M be an R−module and U, V ≤M . If U + V = M and U ∩ V �g M ,
then V is called a weak g-supplement of U in M . If every submodule of M has a weak
g-supplement in M , then M is called a weakly g-supplemented module. (See [8])

Proposition 4. Let Xβ∗gY in M .
(i) If X has an essential g-supplement V in M , then V is also a g-supplement of Y in

M .
(ii) If X has an essential weak g-supplement V in M , then V is also a weak g-

supplement of Y in M .

Proof. (i) Since M = X+V and V EM , then by Xβ∗gY , Y +V = M . Let M = Y +T
with T E V . Since T E V and V E M , then we can see that T E M . Then by Xβ∗gY ,
X + T = M . Since X + T = M and T E V , then T = V . Hence V is a g-supplement of
Y in M .

(ii) Since M = X + V and V EM , then by Xβ∗gY , Y + V = M . Let Y ∩ V + T = M
with T EM . Since M = Y +V and M = Y ∩V +T , then by Lemma 1, M = Y +V ∩T .



C. Nebiyev, N. Sökmez / Eur. J. Pure Appl. Math, 11 (1) (2018), 238-243 242

Since V E M and T E M , then V ∩ T E M . Then by Xβ∗gY , X + V ∩ T = M .
Since M = V + T and M = X + V ∩ T , then by Lemma 1, X ∩ V + T = M. Because
X ∩ V + T = M and T EM and X ∩ V �g M , then T = M . Hence Y ∩ V �g M and V
is a weak g-supplement of Y in M .

Proposition 5. Let M be an amply g-supplemented module and X,Y ≤ M . If g-
supplements of X and Y in M is the same, then Xβ∗gY .

Proof. Let X +K = M with K EM . Since M is amply g-supplemented, there exists
a g-supplement K

′
of X with K

′ ≤ K. By hypothesis, K
′

is a g-supplement of Y in M .
Then Y +K

′
= M and since K

′ ≤ K, Y +K = M . Similarly, we can see that X+T = M
for every T EM such that Y + T = M .

Proposition 6. Let M be weakly g-supplemented module and X,Y ≤ M . If weak g-
supplements of X and Y in M is the same, then Xβ∗gY .

Proof. Let X + K = M with K E M . Since M is weakly g-supplemented, by [8,
Proposition 1] there exists a weak g-supplement K

′
of X with K

′ ≤ K. By hypothesis,
K
′

is a weak g-supplement of Y in M . Then Y +K
′

= M and since K
′ ≤ K, Y +K = M .

Similarly, we can see that X + T = M for every T EM such that Y + T = M .

Proposition 7. Let M be an R−module, X ≤ Y ≤ M and C be an essential weak
g-supplement of X in M . If Xβ∗gY , then Y ∩ C �g M .

Proof. Since Xβ∗gY and C is an essential weak g-supplement of X in M , then by
Proposition 4, C is also a weak g-supplement of Y in M . Hence Y ∩ C �g M .

Lemma 7. Let M be an R−module, X ≤ Y ≤M and C be a weak g-supplement of X in
M . If Y ∩ C �g M , then Xβ∗gY .

Proof. Let Y + T = M with T E M . Since C is a weak g-supplement of X in M ,
C +X = M . Since X ≤ Y , by Modular Law, Y = Y ∩M = Y ∩ (C +X) = Y ∩ C +X.
Then M = Y +T = Y ∩C+X +T and since Y ∩C �g M and X +T EM , X +T = M .
If X +K = M with K EM , Y +K = M also holds since X ≤ Y . Hence Xβ∗gY .

Proposition 8. Let M = M1 ⊕M2 and M1 ≤ X ≤M . If X ∩M2 �g M , then Xβ∗gM1.

Proof. Clear from Lemma 7.

Proposition 9. Let M be an R−module. If every submodule of M equivalent to an
essential weak g-supplement in M by β∗g relation, then M is weakly g-supplemented.

Proof. Let X ≤ M . By hypothesis, there exists an essential weak g-supplement V in
M such that Xβ∗gV . Let V be a weak g-supplement of U in M . By hypothesis, there
exists an essential weak g-supplement Y in M such that Uβ∗gY . Since V is an essential
weak g-supplement of U in M , by Proposition 4, V is a weak g-supplement of Y in M .
Then Y is an essential weak g-supplement of V in M and since Xβ∗gV , by Proposition 4,
Y is a weak g-supplement of X in M . Hence M is weakly g-supplemented.



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