Journal of Indian Acad. Math.  
Vol. 47, No. 1 (2025) pp. 1-16  
ISSN: 0970-5120  
P. Tamilarasi1  
and  
TOPOLOGICAL SIMPLE ROUGH  
GROUPS  
R. Selvi2  
Abstract: In this paper, we introduce the concept of topological simple  
rough group using simple rough group. Also, we explore the filter of  
identity neighborhood in topological simple rough groups and we discuss  
some results related to these concepts.  
Keywords: Topological Rough Group, Topological Rough Group  
Homomorphism, Topological Rough Normal Subgroup,  
Simple Rough Group, Topological Simple Rough Group,  
Filter of Identity Neighbourhood.  
Mathematics subject Classification (2020) No.: 22A05, 54A05, 54D70,  
20D05.  
1. Introduction  
The rough set theory, initially proposed by Pawlak (1982) [19], has been  
utilized as an effective mathematical tool for modeling and processing incomplete  
information. In recent years, rough sets have been integrated with various  
mathematical theories such as algebra and topology.  
Algebraic structures of rough sets have been studied by several authors,  
including Bonikowaski. Z, Kuroki. N, Wang. PP and Li. Z et al. [5, 14, 15, 16].  
In 1994, Biswas and Nanda [4] introduced the concept of rough group and rough  
subgroups, which are based on upper approximation and are independent of lower  
approximation. Miao et al. [7] have enhanced the definitions of rough group and  
2
P. TAMILARASI AND R. SELVI  
rough subgroup, and have demonstrated their new properties. Conversely, Kuroki and  
Wang [15] outlined certain properties of lower and upper approximations in relation  
to the normal subgroups in 1996. Bagirmaz et al. [17] proposed the concept of  
topological rough groups, expanding the idea of a topological group to encompass the  
algebraicstructuresofroughgroups.  
In group theory, A group is simple if its only normal subgroups are the identity  
subgroup and the group itself [12]. The notion of a simple group was introduced by  
Galois about 180 years ago. Simple groups are the building blocks of all groups. In  
the concept of topological group, filters provide a powerful tool for understanding the  
topological properties like convergence, continuity, compactness, etc..  
In this paper, we investigate the key principles of topological simple rough  
groups,whichmergethestructuresofsimplegroupandtopologicalroughgroup. We give  
some examples to illustrate this concept and discuss the basis of topological simple  
rough group, which forms the foundation for studying their local properties, Also we  
explore the filter of identity neighborhoods, underscoring their role in analyzing the  
structureoftopologicalsimpleroughgroups.  
2. Preliminaries  
Definition 2.1 ([7]): Let U be a universe, C be a family of subsets of U ,  
C {X1,X2,...,Xn}. C is called a classification of U if the following properties are  
satisfied:  
1. X1 X2 ...Xn U  
;
2. Xi Xj ,(i j)  
.
Definition 2.2 ([7]): Let K (U,R) be an approximation space and X be a  
subset of U. The sets  
1. X {x |[x]R X }  
;
2. X {x |[x]R X}  
;
3. BN(X) X X  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
3
are called upper approximation, lower approximation and boundary region of X in K,  
respectively.  
Definition 2.3 ([7]): Let K (U,R) be an approximation space and  
be a  
binary operation defined on U. A subset G of universe U is called a rough group if  
the following properties are satisfied:  
1. x, y G, x y G  
;
2. Association property holds in  
G
;
3. e G such that x G,x e e x x ; e is called the rough identity  
element of rough group G;  
4. x G, y G such that x y y x e ; y is called the rough  
inverse element of x in G;  
Definition 2.4 ([7]): A non-empty subset H of rough group G is called its  
rough subgroup, if it is a rough group itself with respect to operation  
.
There is only one guaranteed trivial rough subgroup of rough group G, i.e.,  
G itself. A necessary and sufficient condition for {e} to be a trivial rough subgroup  
of rough group G is e G  
.
Definition 2.5 ([7]): A rough group is called a commutative rough group if  
for every x,y G , we have x y y x  
.
Definition 2.6 ([7]): A rough subgroup N of rough group G is called a rough  
invariant subgroup, if a G,a N N a  
.
Definition 2.7 ([9]): Let G be a rough group andA G . We say that A is  
symmetric if A A1  
.
4
P. TAMILARASI AND R. SELVI  
Definition 2.8 ([9]): Let G1 U1 and G2 U2 be rough groups. We say  
that G1 and G2 be rough homomorphism if there exists a surjection mapping  
:G1 G2 such that the following conditions (1)-(3) hold:  
1.|G1 is a surjection mapping from G1 to G2  
;
2. For any x, y G1 {e} , we have (x 1 y) (x)2 (y)  
;
3. For any subset H of G1,H 1((H))  
.
If a rough homomorphism is a bijection, then we say that G1 and G2 are  
rough isomorphism.  
Definition 2.9 ([3]): Let G be a topological group. A filter on G is a family  
of non-empty subsets of G satisfying the two conditions:  
1. If U and V are in  
then U V is also in  
;
2. If U and U W G , then W   
.
Definition 2.10 ([3]): Let G be a topological group. A family  
is called an  
open filter on G if there exists a filter  
in G such that  
is the intersection of  
with the family of all open subsets of G.  
Of course, this definition is equivalent to the following one:  
is an open  
filter on G if  
is a family of non-empty open subsets of G such that the intersection  
of any finite number of elements of  
is also in  
, and for each U and for  
every open subset W of G such that U W , W also belongs to ξ.  
Definition 2.11 ([17]): A topological rough group is a rough group (G,)  
together with a topology T on  
G
satisfying the following two properties:  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
5
1. the mapping f :G G G defined by f(x,y) xy is continuous with  
respect to product topology on G G and the topology TG on  
G
induced by T ,  
2. the inverse mapping g :G G defined by g(x) x1 is continuous with  
respect to the topology TG on G induced by T .  
Definition 2.12 ([17]): let G be a topological rough group and let H be a  
subgroup of G. Then, H is called a topological rough subgroup of G if  
1. the mapping fH : H H H defined by fH (x,y) xy is continuous  
where  
H
carries the topology induced by  
G
,
2. the inverse mapping gH : H H defined bygH (x) x1 is continuous.  
Definition 2.13 ([1]): A mapping :G1 G2 is called a topological rough  
group homomorphism, if  
to the topology 2 on G2 inducing G2 on G2 and the topology 1 on G1 inducing  
G1 on G1  
is a rough homomorphism and continuous with respect  
.
Definition 2.14 ([17]): Let G be a topological rough group and let N be a  
normal subgroup of G. Then, N is called a topological rough normal subgroup of G  
if a G,aN Na  
.
Throughout this paper, we consider X be the universal set, GR be a rough  
group with identity e and Gꢀ be the upper rough approximation of Gꢀ  
.
3. Topological Simple Rough Group  
Definition 3.1: A rough group Gꢀ  
contains no proper non-trivial rough normal subgroups. That is, Gꢀ  
rough normal subgroups are {e} and Gꢀ  
is called a simple rough group if it  
has only the  
.
6
P. TAMILARASI AND R. SELVI  
Example 3.2: Let X {1,2,3,4,5,6} be the set all integers with respect to  
the multiplication modulo 7. A classification of X isX/R {{1,6},{2,3},{4,5}}  
.
Let Gꢀ {1,2,4}. Then Gꢀ X and Gꢀ . Clearly, Gꢀ is a rough group and  
it has no proper rough normal subgroups, hence Gꢀ is a simple rough group.  
Example 3.3: Let X S4 be the set of all permutations of {1, 2, 3, 4} with  
the multiplication operation of permutations. Consider a classification of X is  
X/R {C1,C2,C3,C4}, where  
C1 = {(1), (12), (13), (14), (23), (24), (34)}  
C2 = {(123), (132), (124), (142), (134), (143), (234), (243)}  
C3 = {(1234), (1243), (1324), (1342), (1423), (1432)}  
C4 = {(12)(34), (13)(24), (14)(23)}  
Let A4 be the set all even permutations of S4 that is,A4 {(1),C2,C4}  
.
Then upper approximation of A4 , A4 C1 C2 C4 and lower approximation of  
A4  
,
A4 C2 C4 . Hence, A4 is a rough group.  
Also we get some proper rough normal subgroups of A4 , like  
{(1)},{(1),C2} and {(1),C4}. Therefore, A4 is not a simple rough group.  
Definition 3.4: A topological simple rough group is a simple rough group  
(Gꢀ ,) together with a topology on Gꢀ satisfying the following two properties:  
(i) The mapping f : Gꢀ Gꢀ Gꢀ defined by f(x,y) xy, x,y Gꢀ  
is continuous with respect to the product topology on Gꢀ Gꢀ and the  
topology  
on Gꢀ induced by   
(ii) The inverse mapping g :Gꢀ Gꢀ defined by g(x) x1  
,
x Gꢀ is  
continuous with respect to the topology τ on Gꢀ induced by .  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
7
Example 3.5: Let X {[0],[1],[2],[3],[4]} be the set of residue classes of  
modulo 5 and be the binary operation of residue addition. A classification of X is  
X/R {{[0],[2]},{[3],[4]}}. Let Gꢀ {[0],[1],[4]}, then Gꢀ X and Gꢀ   
Obviously, Gꢀ is a rough group and also Gꢀ has no proper rough normal  
subgroups. Therefore, Gꢀ is a simple rough group.  
.
Let {,Gꢀ ,{[0]},{[1],[2],[4]},{[0],[1],[2],[4]}. Then we get the  
induced topology on Gꢀ is {,Gꢀ , {[0]},{[1],[4]}}. Hence, Gꢀ is a topological  
simple rough group.  
Proposition 3.6: Let Gꢀ be a topological simple rough group and fix  
x Gꢀ  
.
Then  
(i) The map Lx :Gꢀ Gꢀ  
defined by Lx(y) xy is one-to-one and  
continuous, for all y Gꢀ  
;
(ii) The map Rx :Gꢀ Gꢀ  
defined by Rx(y) yx is one-to-one and  
continuous, for all y Gꢀ  
;
(iii) The map f :Gꢀ Gꢀ defined by f(a) a1 is homeomorphism, for  
all a Gꢀ  
.
Proof: (i) Let y1, y2 Gꢀ . Then Lx(y1) Lx(y2) implies xy1 xy2  
.
Since, Gꢀ is a topological simple rough group andx Gꢀ , x1 Gꢀ Gꢀ . Thus,  
x1(xy1) x1(xy2) which implies y1 y2 . Hence Lx is one-to-one. Now let us  
prove Lx is continuous. Let U be an open set of xy in Gꢀ . Then there exists open  
sets V1,V2 of x, y in Gꢀ  
such that V1V2 U . Since, xV2 V1V2 U  
,
Lx(V2) xV2 U . Hence, Lx is continuous on Gꢀ  
.
(ii) The proof of Rx is similar to Lx  
.
8
P. TAMILARASI AND R. SELVI  
(iii) Since Gꢀ  
is a topological simple rough group, the inverse mapping  
f :Gꢀ Gꢀ is continuous. Therefore, f 1 is also continuous. Hence,  
the map  
f
is homeomorphism of Gꢀ into Gꢀ  
.
Proposition 3.7: Let Gꢀ  
is an open set with e U , then there exists a symmetric open set V of e in Gꢀ such  
that VV U  
be a topological simple rough group. If U Gꢀ  
.
Proof: Since Gꢀ  
is a topological simple rough group, the mapping  
f :Gꢀ Gꢀ Gꢀ  
is continuous. Then f 1(U)  
is open in Gꢀ Gꢀ  
and  
(e,e) f 1(U). Therefore, there exists open sets V1,V2 in Gꢀ with e V1,e V2  
such that V1V2 U . Also the inverse mapping g :Gꢀ Gꢀ is continuous, so V11  
and V21 are open. Let V3 V1 V2 . Then V3 is open in Gꢀ and also V3V3 U  
.
Now we consider V V3 V31 be an open set in Gꢀ and e V . Hence, V V1  
and VV V3V3 U  
.
Proposition 3.8: Let Gꢀ  
be a topological simple rough group. Then for  
every open set W of e in Gꢀ , there exists a symmetric open set V of e in Gꢀ such  
that VV Gꢀ W  
.
Proof: Let W be an open set of e in Gꢀ . Then there exists an open set U of  
e in Gꢀ such that W U Gꢀ . Since the mapping f :Gꢀ Gꢀ Gꢀ is  
continuous and the inverse mapping g :Gꢀ Gꢀ is homeomorphism, there  
exists an open set V of e in Gꢀ  
and V V 1 such that VV U . Hence,  
VV Gꢀ W.  
Proposition 3.9: Let Gꢀ be a topological simple rough group. If Gꢀ  
are open sets of Gꢀ , then {e}is open in Gꢀ and Gꢀ is a discrete space.  
,
{e}  
Proof: Since Gꢀ  
is a topological simple rough group, the mapping  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
9
f :Gꢀ Gꢀ Gꢀ is continuous and e Gꢀ . Also, {e} is open in Gꢀ . Thus, we  
get f 1({e}) is open in Gꢀ Gꢀ and ee e {e} which implies UV {e}  
,
where U,V are open sets in Gꢀ  
and e U, e V . Suppose U {e}, we get  
UU {e}  
.
Hence,  
U V {e} Let x Gꢀ Since the mapping  
.
.
f :Gꢀ Gꢀ Gꢀ  
is continuous at (x,x1), there exists a neighborhood U of  
x Gꢀ  
such that UU1 e. So UU1 {e}. Hence, U {x}. Hence, Gꢀ  
is  
discrete.  
Proposition 3.10: Let Gꢀ be a topological simple rough group. If Gꢀ is  
open in Gꢀ , then H {U : U (e)} is a topological group.  
Proof: Let x, y H . Then x, y U and given U (e). Since Gꢀ  
is a  
topological simple rough group and Gꢀ  
is open in Gꢀ , there exists an open set  
V (e) such that VV U . Thus, xy VV U . Therefore, xy H . Since the  
inverse mapping f :Gꢀ Gꢀ is homeomorphism, there exists an open set V of e  
in Gꢀ such that V V 1  
.
Theorem 3.11: Let X be a topological group and Gꢀ  
simple rough group. If Hꢀ  
topological closure of Hꢀ , cl(Hꢀ ), in Gꢀ  
be a topological  
is a topological rough subgroup of Gꢀ , then the  
is a topological rough group of Gꢀ  
.
Proof: Let x, y cl(Hꢀ ) and U be an open set of xy . Then there exists  
open sets V1 and V2 of x and y such that V1V2 U . Since Hꢀ is a topological  
rough subgroup of Gꢀ , there exists an elements a, b Hꢀ such that a V1 Hꢀ  
and b V2 Hꢀ . Thus, we get ab V1V2 and ab Hꢀ , that is ab V1V2 Hꢀ  
.
So, V1V2 Hꢀ and also U cl(Hꢀ ) . Hence, xy cl(Hꢀ ). Let W be an  
open set of x1 in cl(Hꢀ ). Then there exists an open set V of x such that  
V
1 W . Since, x cl(Hꢀ ), there exists an element a of Hꢀ such that a V  
.
10  
P. TAMILARASI AND R. SELVI  
Then a V Hꢀ  
which implies a1 V 1 Hꢀ . So, W Hꢀ and hence,  
x1 cl(Hꢀ ).  
Remark 3.12: The topological closure ofHꢀ , cl(Hꢀ ), in Gꢀ  
topological rough subgroup in Gꢀ  
is also a  
.
Theorem 3.13: Let Gꢀ be a topological simple rough group and Gꢀ be an  
open set in Gꢀ . If S is a subset of Gꢀ and U is an open set in Gꢀ , then the sets  
SU Gꢀ and US Gꢀ are the open sets of S in Gꢀ  
.
Proof: Let x S . Then there exists an open set V U of x in Gꢀ such that  
xV xU Gꢀ . Therefore,  
xV SU Gꢀ . Hence, we get SU Gꢀ  
xS  
is  
an open set of S in Gꢀ . Likewise, SU Gꢀ  
is an open set of S in Gꢀ  
.
Definition 3.14: Let Gꢀ be a topological simple rough group and   
be a base for . For x Gꢀ , the family x {U Gꢀ :U , x U}   
is called a base at x in  
.
Theorem 3.15: Let Gꢀ be a topological simple rough group and Gꢀ be an  
open set in Gꢀ . Let ße be the family of base at e in Gꢀ . Then, for every x Gꢀ  
,
{(xU)Gꢀ : U }  
,
x {(Ux)Gꢀ : U }  
x
e
e
are two families of bases at x in Gꢀ  
.
Proof: Let U e . Since Gꢀ is a topological simple rough group and Gꢀ  
is an open set in Gꢀ , f :Gꢀ Gꢀ Gꢀ is continuous at (x,e). Then there exists  
an open set V such that V U and xV Gꢀ . It is enough to prove that  
e
xU Gꢀ and Ux Gꢀ are open sets in Gꢀ  
.
Since the map Lx  
:Gꢀ Gꢀ is  
1  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
11  
1  
one-to-one and continuous, L (V) is open in Gꢀ  
and xV Gꢀ . Then we  
1  
x
1  
get L (V) xV is open in Gꢀ . Hence, xV xU and xV G which implies  
1  
x
xU Gꢀ is an open set in Gꢀ . Similarly, Ux Gꢀ is an open set in Gꢀ .  
4. Filter of Identity Neighborhoods  
In this section, the filter  
be the set of all identity neighborhoods of Gꢀ  
.
e
Proposition 4.1: Let Gꢀ and Hꢀ be topological simple rough groups. Then  
rough homomorphism f :Gꢀ Hꢀ is topological rough group  
homomorphism if and only if it is continuous at the identity element.  
a
a
Proof: Let e,e' be the identity elements in Gꢀ  
Suppose f is a topological rough group homomorphism. That is, f is rough  
homomorphism and continuous. Since f is continuous, it is continuous at e in Gꢀ  
and Hꢀ  
respectively.  
.
Conversely, suppose f is continuous at e. Let a Gꢀ and V be a neighborhood of  
f(a) in Hꢀ . Let us prove for any neighborhood U of a in Gꢀ , f(U)V . Since f  
is a rough homomorphism, f(ax) f(a).f(x), for all x Gꢀ . Since f is continuous  
at e, there exists a neighborhood W of e such that f(W)V . Then U aW is an  
open neighborhood of a and f(U) V  
.
Proposition 4.2: Let Gꢀ and Hꢀ be topological simple rough groups and  
f : Gꢀ Hꢀ be a topological rough group homomorphism. Then the following  
conditions are equivalent:  
(i)  
f is open  
(ii)  
For each W (Gꢀ ), the image f(W) has a nonempty interior  
e
(iii) There is a basis  
interior  
of neighborhood W such that f(W) has a nonempty  
e
12  
P. TAMILARASI AND R. SELVI  
(iv) There is a basis  
of neighborhood W in Gꢀ such that f(W) is an  
e
identity neighborhood in Hꢀ  
(v)  
For all W (Gꢀ ) , we have f(W)(Hꢀ )  
e
e
Proof: (i) (ii): Let W be an identity neighborhood in Gꢀ . Since f is  
open, f(W) is open. Also, W (Gꢀ ). Hence, int(f(W)   
.
e
(ii) (iii): Suppose f(W) has a nonempty interior, for each neighborhood W of e in  
Gꢀ . Let be a basis in Gꢀ . Let W and W (Gꢀ ). Then  
e
e
e
int(f(W)) . Hence, the image f(W) has a nonempty interior.  
(iii) (iv): Let U,V be two identity neighborhoods in Gꢀ  
such that  
V int(U) . Then int(f(V)   
.
Consider x V and f(x)int(f(V))intf(int(U))  
.
Let W int(U)x1 and e,e' be identity elements in Gꢀ  
,
Hꢀ  
.
Then e xx1int(U)x1. So, W is an open neighborhood of identity  
element in Gꢀ and  
e' f(x)f(x)1 int(f(V))f(x)1 int(f(int(U)))f(x)1  
int(f(int(U))f(x)1)  
int(f(int(U)x1))  
.
int(f(W))  
.
Hence, f(W) is an identity neighbourhood of Hꢀ  
.
(iv) (v): From the above proof, f(W)(Hꢀ )  
e'  
.
TOPOLOGICAL SIMPLE ROUGH GROUPS  
13  
(v) (i): Obviously, this result follows from(v)  
.
Proposition 4.3: Let Gꢀ  
and Hꢀ  
be topological simple rough groups.  
Then the topological rough group homomorphism f :Gꢀ Hꢀ is both continuous  
and open if and only if f() e' , where  
and  
are the filter of identity  
e
e
e'  
neighborhoods in Gꢀ and Hꢀ respectively.  
Proof: Suppose the homomorphism f is continuous and open. Let prove  
f() e' . Since f is continuous, f() e' . Since f is open, f(e)  
.
e
e
e'  
then f is both open and  
continuous. Let U and V e' . Then f(U) V which implies U f(V)  
Therefore f() e' . Conversely, if f()   
e'  
e
e
.
e
Since U is an open neighborhood of e in Gꢀ , f 1(V) is an open set in Gꢀ . Also  
f(U)which implies f(U) is an open set containing the identity element e' in  
e'  
Hꢀ . Hence f is both open and continuous.  
Lemma 4.4: If U is an open neighborhood of the identity element  
in topological simple rough group Gꢀ , then U cl(U) UU,cl(U) means closure  
of U .  
Proof: We know that U cl(U). It is enough to prove that cl(U)UU  
.
Let a cl(U). Then there exists a symmetric neighborhood W of e in Gꢀ such that  
WW U that is, W U . Also a W and aW is an open neighborhood of a. So  
aW U and a aW cl(U) . Let b aW U . Then b aw , for some  
w W  
which implies a bw 1. Since W is symmetric and b U,  
w1 W1 W U and a UU  
.
Theorem 4.5: (First closure lemma) Let Gꢀ  
be a topological simple  
rough group such that Gꢀ is open in Gꢀ . If S is a subset of Gꢀ , then  
cl(S)   
SU   
cl(SU),  
U  
U  
e
e
14  
P. TAMILARASI AND R. SELVI  
where  
be the filter of identity neighborhood in Gꢀ  
.
e
Proof: Let a cl(S) and U e . Then aU Gꢀ is a neighborhood of a in  
Gꢀ and aU S . Since U U1  
,
aU1 S is a neighborhood of a. So, let  
b aU1 S that is b aU1 and b S . Then b au1, for some u U  
.
.
Therefore, a bu SU cl(SU). Let a   
cl(SU). Let us prove a   
(SU)  
Then there exists an identity neighborhoodV such that VV U . Therefore,  
e
a cl(SV) SVV SU . Proceeding this process we get,  
(SU)  
.
cl(SU)   
Suppose a   
(SU). Let W be an identity neighborhood of a in  
and  
e
W1a U . Since, a SU  
,
a hu  
,
for some  
h S,u U . Thus,  
e
h au 1 aU1 aa1W W which implies h S W . Hence, a cl(S)  
.
Theorem 4.6: Let Gꢀ be a finite topological rough group with identity e  
be the filter of identity neighborhood in Gꢀ . Then there exists a  
rough normal subgroup Nꢀ in Gꢀ such that  
{U Gꢀ /Nꢀ U} and the elements are symmetric.  
and let  
e
topological  
e
Proof: ConsiderNꢀ  
U . Since Gꢀ  
is finite, N is non empty in  
Ue  
and e Nꢀ . Leta,b Nꢀ . Then there exists W such that WW 1 Nꢀ .  
e
e
Since Nꢀ W  
,
a,b W . Thus, ab1 UU1 Nꢀ . Hence, ab1 Nꢀ . Let  
g Gꢀ  
.
Since Nꢀ , gNꢀ g1   
and  
g1Nꢀ g   
.
That implies  
e
e
e
Nꢀ gNꢀ g1 and Nꢀ gNꢀ g1 . Hence, Nꢀ  
is a topological rough normal  
subgroup.  
5. Conclusion  
In this paper, we studied topological simple rough group from the simple  
rough group structure and given some examples. Further we investigated the basis of  
topological simple rough groups and discussed the concept of filters in topological  
simple rough groups, which are essential for a deeper understanding of their  
topological properties.  
TOPOLOGICAL SIMPLE ROUGH GROUPS  
15  
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1. Research Scholar (Reg. No: 22211202092003),  
Department of Mathematics,  
(Received, September 21, 2024)  
(Revised, November 18, 2024)  
Sri Parasakthi College for Women,  
Courtallam - 627802,  
Affiliated by Manonmaniam Sundaranar University, Tirunelveli - 627012,  
Tamilnadu, India.  
E-mail: tamilarasiparamasivan@gmail.com  
2. Associate Professor,  
Department of Mathematics,  
Sri Parasakthi College for Women,  
Courtallam - 627802,  
Affiliated by Manonmaniam Sundaranar University, Tirunelveli - 627012,  
Tamilnadu, India.  
E-mail: r.selvimuthu@gmail.com