On the Line Graph of the Complement Graph for the Ring
of Gaussian Integers Modulo
n
Manal Ghanem1, Khalida Nazzal2
1Department of Mathematics, Irbid National University, Irbid, Jordan
2Department of Mathematics, Palestine Technical University (Kadoorie), Tulkarm, Palestine Email: [email protected], [email protected]
Received October 13,2011; revised November 2, 2011; accepted November 22, 2011
ABSTRACT
The line graph for the complement of the zero divisor graph for the ring of Gaussian integers modulo n is studied. The diameter, the radius and degree of each vertex are determined. Complete characterization of Hamiltonian, Eulerian, planer, regular, locally H and locally connected
[ ]
n
L i is given. The chromatic number when n is a power
of a prime is computed. Further properties for
[ ]
n
L i and
n[ ]i
are also discussed.Keywords: Complement of a Graph; Chromatic Index; Diameter; Domination Number; Eulerian Graph; Gaussian
Integers Modulo n; Hamiltonian Graph; Line Graph; Radius; Zero Divisor Graph
1. Introduction
The line graph L G
of a graph G is defined to bethe graph whose vertex set constitutes of the edges of G,
Where two vertices are adjacent if the corresponding edges have a common vertex in G. The importance of
line graphs stems from the fact that the line graph transforms the adjacency relations on edges to adjacency relations on vertices. For example, the chromatic index of a graph leads to the chromatic number of its line graph. The zero divisor graph of a commutative ring R ,
denoted by
R , is defined as the graph whose vertexset is the set of all non-zero zero divisors of R and
edge set E
R
=
xy x y: , R
0 and xy= 0
. This type of graphs provides an example showing that algebraic methods could be applied to problems about graphs. The set of Gaussian integers, denoted by [ ]i , is defined asthe set of complex numbers a bi , where a b, . If x is a prime Gaussian integer, then x is either
1)
1i
or
1i
, or2) q where q is a prime integer and q3 mod 4
, or3) a bi , a bi where a2b2 = p, p is a prime
integer and p1 mod 4
.Throughout this paper, p and pi denote prime
integers which are congruent to 1 modulo 4, while q
and and qi denote prime integers which are congruent
to 3 modulo 4. All rings in this paper are assumed to be commutative with unity. The zero divisor graph for the ring of Gaussian integers modulo n is studied in [1]
and [2], the complement of this graph is discussed in [3]. While the line graph of the zero divisor graph for the ring
of Gaussian integers modulo n is investigated in [4]. In
this paper it should be kept in mind that
2[ ] = {1 }
V i i , and hence, its line graph is K0,
[ ]
q i
is an integral domain, so
q[ ] =i
K0. Further,
q2[ ]i is a complete graph whose complement is totally disconnected and thus its line graph is K0. While
p[ ] =i
Kp1,p1 , so its complement is disconnected with two components each of which is isomorphic to
1
p
K . Finally, note that the graph
2q[ ]i
is bipartite,[1] and
1 2
2 2 1 1,2 1 [ ] =q q i Kq q
.
In this paper, we investigate properties of the graph
n[ ]
L i . We find the diameter, the radius of
n[ ]
L i . We determine which L
n[ ]i
is Eu-lerian, Hamiltonian, regular, locally H , locally connec-
ted or planer. Furthermore, the chromatic index and the edge domination number of
n[ ]i
where n is apower of a prime are computed. While the domination number of
n[ ]i
is given. On the other hand, a for- mula which gives the degree of each vertex in
n[ ]i
is derived, thus the degree of its complement as well as its line graph could easily be found.
2. When Is
L
n[ ]
i
Eulerian or
Planner
only if every vertex of G has even degree. For a finite
ring R, the line graph L
R of a connected graph
R is Eulerian if and only if all vertices of
Rhave the same parity ( see the proof of Lemma 3.10, [5]). On the other hand, if G has both even and odd vertices,
then so is its complement. So, for a connected graph
n[ ]i
, the graph L
n[ ]i
is Eulerian if andonly if all vertices in
n[ ]i
are either even or allvertices in
n[ ]i
are all odd. But
n[ ]i
is con-nected if , 2 , , 1 2
m m
n p q q q [3] and
n[ ]i
isEulerian if n= 2,p or n is a product of distinct odd
primes [1]. It is easy to show that all vertices of
n[ ]i
are odd if and only if = 2
n q . This proves the
following theorem.
Theorem 2.1 L
n[ ]i
is Eulerian if and only ifn is a product of distinct odd primes.
A planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way
that its edges intersect only at their endpoints.
Next we determine when the graph L
n[ ]i
isplanar.
In a graph G the maximum vertex degree and the
minimum vertex degree will be denoted by
G and
G , respectively.
The following theorem characterizes graphs G whose
line graph L G
is planer.Theorem 2.2 [6]
A nonempty graph G has a planer line graph L G
if and only if 1) G is planer.
2)
G
4, and3) if degG
v = 4, then v is a cut vertex.The graph
n[ ]i
is planer if and only if n= 2,5or 2
q [3]. For n= 2, q2, L
n[ ] =i
K0. Whilefor n= 5,
n[ ] =i
K4K4, this graph is regular ofdegree 3.
Thus we obtain the following.
Theorem 2.3 The graph L
n[ ]i
is planer ifand only is n= 5.
3. The Diameter of
L
n[ ]
i
For a connected graph G, the distance, d u v
, ,between two vertices u and v is the minimum of the
lengths of all u v paths of G. The eccentricity of a
vertex v in G is the maximum distance from
v
to any vertex in G. The diameter of G, diam G
, is themaximum eccentricity among the vertices of G. Since
n[ ]i
is connected if n p, 2 ,m qm,q q1 2 and each
of
p[ ]i
and
1 2[ ] q q i
is the union of two complete graphs, while
2m[ ]i
and
qm[ ] ,i
m 3 are the union of a nullgraph and a connected graph [3], we have the following.
Theorem 3.1 L
n[ ]i
is connected if and only if2 1 2
2, , ,
n p q q q .
Theorem 3.2 If n= 2 ,m m2 or n=qm,m3, then
n[ ]
= 2diam L i .
Proof. 1) Assume that n= 2 ,m m2 and
x= x1x i y2, = y1y i2
, z=z1z i w2, =w1w i2
are two nonadjacent vertices in V L
n[ ]i
. Sincefor every
2m[ ]
a bi i , a and b are both even or
odd [1], we have three cases:
Case I: for = 1, 2i , x y zi, ,i i and wi are odd. Then
we have the path [ , ]x y [ , ]x z [ , ]z w .
Case II: for = 1, 2i , xi or yi is odd(even) and zi
or wi is even (odd). Assume that x x1, 2 are even and 1, 2
z z are odd. Then we have the path
[ , ]x y [ , ]x z [ , ]z w .
Case III: for = 1, 2i , x y zi, ,i i and wi are even.
Then 1 1 2 2
1 1 2 2
[ , ] = 2t 2 ,s 2t 2s
x y i i and
3 3 4 4
3 3 4 4
[ , ] = 2t 2 ,s 2t 2s
z w i i where i, i
are odd and 1t si, im for 1 i 4. If t s t1, , ,1 2 or 2< 2
m
s , say t1, then t s t3, ,3 4 or s4<m t 1, say t3.
So, we have the path [ , ]x y [ , ]x z [ , ]z w . Now
suppose that m is odd. Then
a) If = = 1, 2
i i i i
m
t s , for i= 1 or 2, say for
= 1
i , then t s t3, ,3 4 or s4<m t 1, say t3. Hence, we
have the path [ , ]x y [ , ]x z [ , ]z w .
b) If ti or
1 =
2
i
m
s and ti si, for i= 1 or 2, say
for i= 1, then we have a path [ , ]x y [ , ]x z [ , ]z w or
[ , ]x y [ , ]x w [ , ]z w .
c) If = = 1, = 2
i i i i
m
t s , for i= 1 or 2, say for
= 1
i , then 2= 2= 1
2
m
t s implies that 22. Other-
wise t2 or 2
1 2
m
s . Then we have a path
[ , ]x y [ , ]y z [ , ]z w or
[ , ]x y [ , ]y w [ , ]z w .
2) Assume that n=qm,m3 and
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
, ,
, [ ]
t s t s
t s t s
n
q q i q q i
q q i q q i V L i
Then t s t1, ,1 2 or 2 <
2
m s
, say t1. Hence t s t3, ,3 4 or 4< 1
s m t , say t3. Then we have the path
1 1 2 2 1 1 2 2
3 3 1 1
1 1 3 3 3 3 4 4 3 3 4 4
,
[ , ]
,
t s t s
t s
t s
t s t s
q q i q q i
q q i q q i
q q i q q i
.
Theorem 3.3 Let R be a ring that is a product of two rings R1 and R2 with at least one of them is not ID with more than one regular element and the other has more than two regular elements. Then
= 3diam L R .
Proof. Suppose that R=R1R2 and R1 is not ID,
1 2reg R and reg R
2 3 . Let z1V
R1
and u2reg R
2 1 . Clearly,
1,0 , 1,0 , 0,1 , 0, 2
= 3d z u inL
R . So,
3diam L R . Now, let
a a1, 2
, b b1, 2
, c c1, 2
, d d1, 2
V L
R
,
then a b1 10 or a b2 2 0 and c d1 10 or c d2 20.
So, we have three cases:
Case I: a b1 10 and c d1 1 0. Then
1, 1 1
a c reg R implies that
z1,0 ,
a a1, 2
z1,0 ,
c c1, 2
E L
R
.
And a1 or c1Z R
1 , say a1 implies that
u1,0 ,
a a1, 2
u1,0 ,
c c1, 2
E L
R
where u1reg R
1 c1 .Case II: a b2 2 0 and c d2 2 0. Then there exists
2 2 2, 2
v reg R a c and hence
a a1, 2
, 0,v2
c c1, 2
, 0,v2
E L
R
.
Case III: a b1 10 and c d2 2 0 or a b2 20 and 1 1 0
c d . Let a b1 10 and c d2 2 0. Then
1 1
a reg R implies that
a a1, 2
, z c1, 2
V L
R
and
z c1, 2
= d d1, 2
or
d d1, 2
, z c1, 2
V L
R
.And if a1Z R
1 , then
a c1, 2
= b b1, 2
or
a c1, 2
, b b1, 2
V L
R
and
a c1, 2
= d d1, 2
or
a c1, 2
, d d1, 2
V L
R
.
For = m
n p , n[ ] m m
p p
i
[7] and for
1 2
=
n n n with g c d n n. .
1, 2
= 1,1 2
[ ] [ ] [ ]
n i n i n i
.
Moreover reg
2[ ] = 2i
and reg
m[ ]i
3 for2
m . An immediate consequence of Theorem 3.3 is the
following.
Theorem 3.4 Let n= pm,m2
or n is a composite such that nq q1 2. Then
n[ ]
= 3diam L i .
4. The Radius and the Girth of the Graph
n[ ]
L
i
For a connected graph G, the radius of G, rad G
, isthe minimum eccentricity among the vertices of G. So,
rad G diam G . Since for any
[ , ]a b V L n[ ]i , [ , ]a b and [ , ]ai bi are non ad-
jacent, rad L
n[ ]i
> 1. Using Theorem 3.2 gives for = 2 ,m 2n m or n=qm,m3,
n[ ]
= 2rad L i .
Theorem 4.1 If n= pm,m2
or = m
n t s where
1
m , t is prime integer, g c d t s. .
, = 1 and nq q1 2, then rad L
n[ ]i
= 2.Proof. Since rad L
n[ ]i
> 1 to show that
n[ ]
= 2rad L i it is enough to find a vertex
n[ ]
vV L i with eccentricity 2. If
2 2
= m, = , 2
n p p a b m , then
[ , ],[ , ]
2d abi a bi x y for every
[ , ]x y V L n[ ]i . So rad L
pm[ ]i
= 2
.
Now, assume that = m , 1
n t s m and
( , ),( , )x y w z
V L
n[ ]i
.Then we have four cases: Case I: = 2t . Then
1 ,1 , 1,0 , , , ,
2d i x y w z .
Case II: =t p. Then
,1 , ,1 , , , ,
2d abi a bi x y w z .
Case III: t=q1 and m= 1. Then sq2 and hence
there exists aV
s[ ]i
. So,
0,1 , 1, , , , ,
2d a x y w z .
Case IV: = ,t q m2. Then
,1 , 1,0 , , , ,
2d q x y w z .
Theorem 4.2 rad L
n[ ]i
= 2 if and only if 22, , ,
Vising [8], proved that for a connected simple graph
G with n-vertices and radius 2, the upper bound of the
number of edges of G is
2
2
n n
. Then Golberg [9] proved that the lower bound of numbers of edges of a simple connected graph G with radius 2 is 3
1
4
n
. So we can conclude the following.
Theorem 4.3 For n2, , ,p q q2 or q q1 2,
n[ ]
=L i t implies that
3 1 2
[ ]
4 n 2
t t t
E L i
.
The girth of a graph G, g G
is the length of ashortest cycle contained in the graph. If the graph does not contain any cycles (i.e.. it’s an acyclic graph), its
girth is defined to be infinity. If a b c a, , , is a cycle of
length three in G. Then [ , ],[ , ],[ , ],[ , ]a b b c c a a b is a
cycle of length 3 in L G
. So, g L G
= 3 when-ever g G
= 3. In [3] it is proved that the girth of
n[ ]i
equals 3 for n2, ,q q2. So, we have the fol-
lowing.
Theorem 4.4 For n2, ,q q2, g L
n[ ]i
= 3.5. The Locally Connected Property of the
Graphs
n[ ]
i
and
L
n[ ]
i
We say that a vertex v is locally connected if the
neighborhood of v, N v
, is connected; and G islocally connected if every vertex of G is locally con-
nected.
Theorem 5.1 If R=R1R2, reg R
i 2 for= 1, 2
i and either R1 or R2 is not ID, then
R is locally connected.Proof. Suppose that R1 is not ID and
x y, V
R . Then we have two cases:Case I: = 0x or = 0y . If = 0x , then there exists
1 1
z V R . So
z1,1 a b, E
R
for all
a b, N
x y,
. And if y= 0, then there exists
1 1 { }
u reg R x such that
u1,0
N
x y,
. There-fore,
u1,0
a b, E
R for every
a b, N
x y,
. So N
x y,
is connected.Case II: x0 and y0. Then there exist
1 1 { }
v reg R x , v2reg R
2 { }y and
1 1
z V R such that
v1,0 ,
z v1, 2
and
0,v2
N
x y,
. Moreover,
v1,0
z v1, 2
, 0,v2
z v1, 2
E
R . And for every
t s, Z R
,
v1,0
t s, or
0,v2
t s, E
R .So N
x y,
is connected.
Theorem 5.2 If R=R1R2, reg R
i 2 for= 1, 2
i and either R1 or R2 is not ID, then
L R is locally connected.
Proof. Suppose that R1 is not ID, z1V
R1
and
x y, , ,z w
V L
R
, then we have threecases:
Case I: x= = 0z . Then
z1,1 , ,x y z1,1 , ,z w
E L
R
.
Case II: y=w= 0. If x z, reg R
1 , then
z1,1 , ,x y z1,1 , ,z w
E L
R
.
Otherwise there exists u1reg R
1 { , }x z . So,
u1,0 , ,
x y u1,0 , ,
z w
E L
R
.
Case III: ,x y0 or ,z w0. Assume that xy0,
then z0 implies that there exists u1reg R
1 { }xsatisfies
u1,0 , ,
x y u1,0 , ,
z w
E L
R
.
While w0 implies that that there exists
2 2 { }
v reg R w satisfies
0,v2
, ,x y 0,v2
, ,z w
E L
R
.
From Theorem 5.1 and Theorem 5.2 we conclude the following.
Theorem 5.3 If n= pm,m1 or n is a composite integer such that nq q1 2, then both
n[ ]i
and
n[ ]
L i are locally connected.
6. When Is
L
n[ ]
i
Hamiltonian?
A Hamiltonian cycle is a cycle that visits each vertex exactly once (except the vertex which is both the start and end, and so is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph. The line graph of a graph G with more than 4 vertices and
diameter 2 is Hamiltonian [10]. But
2m[ ] ,i m 2
is disconnected with one isolated vertex
2m1 2m1
i
and the other component, call this component H, with
diameter 2 [3]. So,
2m[ ]
L i L H . Similarly,
qm[ ]i
has a connected subgraph H with diameter
2 and
m[ ]
q
L i L H
. Hence, the following
result is obtained.
Theorem 6.1 If n= 2 ,m m2 or n=qm,m3, then
n[ ]
L i is Hamiltonian.
a complete bipartite graph K1,3) is hamiltonian. Since
the line graph is claw free, using Theorem 5.3, we get the following.
Theorem 6.2 If n= pm, m2 or n is a composite integer such that nq q1 2, then L
n[ ]i
is hamil-tonian.
7. The Chromatic Number of the Graph
n[ ]
L
i
The edge coloring of a graph G is an assignment of
colors to the edges of the graph so that no two adjacent edges have the same color. The minimum required num- ber of colors for the edges of a given graph is called the chromatic index of the graph denoted by
G .Lemma 7.1 [12]
If G has order 2s and
G = 2s1, then
G =
G .
Theorem 7.2 If n= 2 ,m m2, then
[ ] = 2
2m1 3 n i
. Proof. Note that in
2m[ ]i
, the induced subgraph,
H , with
1 1
2
= [ ] 2m 2m
m
V H V i i
is con-
nected,
= 22m1 2V H , [1] and
H =
2m[ ]i
. Since the vertex 1i is adja-
cent to all other vertices in H , we have
=
1
= 22m1 3H deg i
. Using Lemma 6.1,
2 12 [ ] = 2 3
m m i
.
Since
q[ ]i
is empty graph and
2
2[ ] = 1 1
q i q K
is edgeless with 2 1
q verti-
ces, we consider the case
qm[ ] ,i
q3.Theorem 7.3 If = m, 3
n q m , then
[ ] =
2m 2 2 1n i q q
Proof. Let A=
qm1 qm1i: , q
0 .
Then A is the set of all isolated vertices in
m[ ]
q i
.
So the induced subgraph, H, with the vertices
=
qm[ ]
V H V i A
is a connected graph,
= 2m 2 2V H q q . Clearly the vertex q is adjacent to
all other vertices in H and hence,
= 2m 2 2 1deg q q q . Using Lemma 7.1,
[ ] =
2m 2 2 1.m
q i q q
Finally we find the chromatic index of
m[ ] ,
2p i m
.
A subset D of the vertex set V G
is said to beindependent if no two vertices in this set are adjacent. A clique of a graph is a maximal complete subgraph. A graph G is said to be split if it’s vertex set can be
partitioned into two subsets A and B such that A
induces a clique and B is independent in G.
Lemma 7.4 [13] Let G be a split graph. If
G is odd, then
G =
G .Theorem 7.5 If n= p2, then
[ ] = 2
3 2 1n i p p p
.
Proof. Since 2[ ] 2 2
p i p p
, it is enough to find
p2 p2
. First, we’ll show that
2 2
p p
is a split graph. Let
2
2
= , : and }
, : and ,
p p
p p
A u p u U p v v U
= , : and p 0,0
B p p .
Clearly, V
p2p2
=AB, A induces a cliqueand B is independent. Therefore,
2 2
p p
is a split graph. Moreover,
2 2
2 3 2
= 1,
= [ ] 0, : 0 1,
= 2 1
p p
p p
deg p
i p p
p p p
is odd. From Lemma 7.4,
[ ] = 2
3 2 1m
p i p p p
.
A graph G is said to be critical if G is connected
and
G =
G 1 and for every edge e of G,we have
G\{ } <e
G . The well-known Vizing’stheorem states that for a simple graph G,
G =
G or
G 1.Lemma 7.6 [14]
If G is a critical graph, then G has at least
G G 2 of vertices of maximum degree. Therefore, if G is a simple graph such that for every
vertex v of maximum degree there exists an edge vu
of vertices with maximum degree in G, we have
G =
G [13].
Theorem 7.7 If n= pm,m3, then
[ ] = 2
2m 1 2m 2 1n i p p p
.
Proof. Let , V
p
and u v, U
pm .Then the vertices of
pmpm
with maximum de-gree have the form
p v,
or
u,p
where 0and 0 and
1
, =
,0 : 0 ,
m m
p p
m
N p v V
p p v
and
1
, =
0, : 0 , .
m m
p p
m
N u p V
p u p
So,
= 2 2m1 2m 2 1m m
p p p p p
. And
the vertices of
pmpm
with minimum degreehave the form
m1,0
p or
0, m1
p where
m1,0 =
, i
: 1
N p up i and
0, m1
=
i,
: 1
N p p v i . So
[ ] =
m m1
2
mp i p p mp m p
.
Therefore,
1 1 2 1
1
[ ] 2
= 2 1
> 2 1
m m m
p p p
m m m m
m m
i
p p p mp m p p p
p p p
.
But the graph
pmpm
has only
1
2 m m 1
p p p vertices of maximum degree. So,
= 2 2m1 2m 2 1m m
p p p p p
.
Since pm[ ]i pmpm, the result holds.
Since the edge coloring of any graph leads to a vertex coloring of its line graph, we obtain the following.
Corollary 7.8 1) If n= 2 ,m m2, then
[ ]
= 22m 1 3 nL i
.
2) If = m, 3
n q m , then
[ ]
= 2m 2 2 1 nL i q q
.
3) If = m, 2
n p m , then
[ ]
= 2 2m 1 2m 2 1 nL i p p p
.
8. The Domination Number of
n[ ]
i
A subset D of the vertex set V G
of a graph G isa dominating set in G if each vertex of G, not in D,
is adjacent to at least one vertex of D. The minimum
cardinality of all dominating sets in G,
G , is calledthe domination number of G.
In
2m[ ] ,i m 2
, the vertex 2m1 2m1 i
is an
isolated vertex while the vertex 1i dominates all
vertices in the second component. Therefore,
2m[ ] = 2i
. The graph
2 2[ ] = 1 1
q i q K
,
thus
22[ ] = 1
q i q
. In
qm[ ] ,i
m3 thevertices m 1 m1
q q i
are isolated while the vertex
q
is adjacent to all other vertices in
[ ]
m1 m1 : ,
m q
q
V i q q i
,
so
[ ] =
2m
q i q
. Since
1 2
2 1 2 1 1 2[ ] =
q q i Kq Kq
and
p[ ] =i
Kp1Kp1,
q q1 2[ ] =i
p[ ] = 2i
.
The set D= 1,0 , 0,1
is a minimum dominating set for
pmpm
. And if n=n n1 2, where
1 2
. . , = 1
g c d n n , then
1 2
[ ] [ ] [ ]
n i n i n i
. This
graph is connected and the set D= 1,0 , 0,1
is a minimum dominating set for
1[ ] 2[ ]
n i n i
.
Theorem 8.1 1) If n2,qm, then
n[ ] = 2i
.
2)
22[ ] = 1
q i q
and
[ ] =
2, 3.m
q i q m
9. The Domination Number of
L
n[ ]
i
The independence number of G,
G , is the maxi-mum cardinality of all independent sets in G. A subset D of the edge set V G
of a graph G is an edgedominating set in G if each edge of G, not in D, is
adjacent to at least one edge of D . The minimum
cardinality of all edge dominating sets in G,
G , iscalled the edge domination number of G. The minimum
cardinality of all independent edge dominating sets,
i G
, is called the independence edge domination number of G. The study of the domination number of
the line graph of G leads to the study of edge or line
domination number of G, i.e.
L G
=
G . Onthe other hand, for any graph G, i
G =
G [15].If S is an independent set in G, then S induces a
complete graph in G. While if S induces a complete
graph in G, then it is independent in G. Recall that 2
2m[ ]i 2m
[2]. Then the sets,
22
= 2 :j
j m j
A U , j= 1, 2, , 2 m1 form a partition for the set V
22m
. Clearly, the set2 1 =
= m j j m
T
A is the maximum independent set in
22m , while the set = m=11 j j
S
A induces a maxi-mum complete subgraph in
22 m
. There are some edges joining S to T , no other adjacency exists in
22m . Any edge dominating set for
22m mustcontain at least S 2 element in order to dominate
S
. On the other hand, this dominating set for S do-
minates all other edges in
22m
. Since
2 1
= 2 m j j
A , then S and T , could easily be com-
puted to get the following theorem.
Theorem 9.1 For = 2 ,m 2
n m .
1)
[ ] = 2 2
m
m1 1
n i
.
2)
[ ] = 2
m 1n i
.
3)
1
1
[ ] = [ ]
= [ ] = 2 2 1 .
n i n
m m
n
L i i
i
To study the graph
[ ] ,
3m
q i m
, consider the
partition of
qm[ ]i
given by
= : and ,
1 , .
k j
kj qm k qm j
A q q i U U
k j m
and not both j k, =m. The set
= =
2 2
= m m
kj mm
m m
k j
T A A
is the maximum inde-
pendent set, while 2 1 2 1 =1 =1 = m m kj j k S A
induces amaximum complete subgraph in
qm[ ]i
. There aresome edges joining S to T , and
m[ ]
q i
has no other adjacency. Easy calculations give
2 2 2= 1 m k j
kj
A q q when 1k j, m 1, 1
= m j m j
mj
A q q and Akm =qm k qm k 1
when
,
k jm. While
2 2 = 1 m T q
and
2 2
2 2
= 1
m m
S q q
.
Thus we obtain the following theorem.
Theorem 9.2 If n=qm,m3, then
1)
2 2
2 2
[ ] = 1
m m
n i q q
.
2)
[ ] =
m 1ni q
if m is even and m 1
q if m is odd.
3)
2 22 2
[ ] = [ ]
1
= [ ] = 1 .
2
n i n
m m
n
L i i
i q q
Now, we move to the case = m
n p . Let
= k, j : and
kj pm k pm j
A p p U U .
Clearly, the sets Akj where 0k j, m and not
both k j, =m or 0, partition the vertices of
pm pm
and = 2m k j 2
1
2 kjA p p . Let
1
1 =1 0 =1 0 1 1
2 2
2 3
=1 =1 = = 2 2 1 1 2 2 4 5 =1 =1 = = 2 2 = = , = ,
= and = .
m m k k k k m m m m kj kj m m k j k j m m m m kj kj m m j k k j
S A A
S A S A
S A S A
pm pm
. Vertices in =11 0 m
k k A
are adjacent to all vertices except some vertices in m=11km k A
. Similarly, vertices in m=11 0k k A
are adjacent to all vertices except some vertices in m=11mk k A
, and vertices in Am0 areadjacent to all vertices except vertices in A0m. On the
other hand A0m induces a complete subgraph and
vertices in this set are adjacent to all other vertices except those of Am0. Clearly S2 induces a complete subgraph.
Vertices in S3 form an independent set, and are
adjacent to some vertices in S1S2S4S5A0m.
Each of S4 and S5 induces a complete subgraph and
are adjacent to some vertices in S1S2S3A0m.
Besides, there are some edges between S4 and S5. On
the other hand,
3
= = 2 2
= m m kj mm.
m m
k j
S A A
The above argument shows that
32
2 1 2 2 2
[ ] = [ ]
1
= [ ] = [ ]
2 1
= 2 2 .
2
m i m
p p
m m
p p
m m
m m
L i i
i i S
P p p
10. The Degree of the Vertices in
n[ ]
i
and
L
n[ ]
i
Now, we determine the cardinality of the annihilator of the element a bi , ann a
bi
in n[ ]i . This helpsfind the degree of each vertex in
n[ ]i
, its complement, as well as the degree of each vertex in their corresponding line graphs.Theorem 10.1 If a bi n[ ]i , then
= 2 2ann abi c d where g c d a. .
bi n,
=cdi. Proof. Let a bi n[ ]i and
. . , =
g c d abi n cdi. Then
=
[ ] :
0 mod
ann abi xi x abi n .
So, .x a bi 0 modn x.a bi 0 mod n
c di c di
.
But n [ ]
c di
a bi
U i
c di
. So, 0 mod
n x
c di
and hence there exists m[ ]i such that =x n m cdi .
Since m=t c
di
r where t r, [ ]i and the normof r is less than the norm of cdi,
= : c di[ ] =
c di[ ]ann a bi r r i i . By Theorem
2 of [7], 2 2
2 2
[ ] = =
c di i c d c d
, so the result
holds.
Theorem 10.2 Let vV
n[ ]i
and
. . , =
g c d v n cdi. Then
= 22 22 1, if 22 02, if = 0
c d v
deg v
c d v
.
The order of
n[ ]i
can be easily computed usingformulas given in [1]. Thus we can find the degree of each vertex in the complement of
n[ ]i
, here wegive the degree of each vertex in the line graph of
n[ ]i
, an analogous formula for the degree of vertices in L
n[ ]i
could be obtained.Corollary 10.3 Let
u v, V L
n[ ]i
,
. . , =
g c d u n abi and g c d v n. .
, =cdi. Then
2 2 2 2 2 2
2 2 2 2 2 2
2 2 2 2 2 2
[ , ]
4, if 0 and 0
= 5, if = 0 and 0
6, if = 0 and = 0
deg u v
a b c d u v
a b c d u v
a b c d u v
.
Proof. Note that, for any graph G and uvE G
,
( ) [ , ] = 2
L G G G
deg u v deg u deg v .
In the following we determine the degree of every vertex in the graphs
n[ ]i
when= 2 ,m 2, = m, 3
n m n q m and n= pm,m1. Theorem 10.4 Let = 2 ,m 3
n m and , are odd. Then in
n[ ]i
,1)
2
2
2 1
2 1
2 1, if 1 < < and < 1 = < and
2 2
2 2, if < < or = < and
2 2
2 2 =
2 2, if = < and = 2
2 1, if 1 = < and = 2
k
k
k s
k
k
m m
k s m k or k s
m m
k s m k s m
deg i
m
k s m m k s