Power graph of finite abelian groups
Let G be a group. The power graph ΓP(G) of G is a graph with vertex set V(ΓP(G)) = G and two distinct vertices x and y are adjacent in ΓP(G) if and only if either xˡ = y or yʲ = x, where 2 ≤ i, j ≤ n. In this paper, we obtain some fundamental characterizations of the power graph. Also, we characteri...
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Інститут прикладної математики і механіки НАН України
2013
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| Цитувати: | Power graph of finite abelian groups / T. Tamizh Chelvam, M. Sattanathan // Algebra and Discrete Mathematics. — 2013. — Vol. 16, № 1. — С. 33–41. — Бібліогр.: 8 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860221008502325248 |
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| author | Tamizh Chelvam, T. Sattanathan, M. |
| author_facet | Tamizh Chelvam, T. Sattanathan, M. |
| citation_txt | Power graph of finite abelian groups / T. Tamizh Chelvam, M. Sattanathan // Algebra and Discrete Mathematics. — 2013. — Vol. 16, № 1. — С. 33–41. — Бібліогр.: 8 назв. — англ. |
| collection | DSpace DC |
| container_title | Algebra and Discrete Mathematics |
| description | Let G be a group. The power graph ΓP(G) of G is a graph with vertex set V(ΓP(G)) = G and two distinct vertices x and y are adjacent in ΓP(G) if and only if either xˡ = y or yʲ = x, where 2 ≤ i, j ≤ n. In this paper, we obtain some fundamental characterizations of the power graph. Also, we characterize certain classes of power graphs of finite abelian groups.
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| first_indexed | 2025-12-07T18:17:50Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 16 (2013). Number 1. pp. 33 – 41
© Journal “Algebra and Discrete Mathematics”
Power graph of finite abelian groups1
T. Tamizh Chelvam and M. Sattanathan
Communicated by L. A. Shemetkov
Abstract. Let G be a group. The power graph ΓP (G) of G
is a graph with vertex set V (ΓP (G)) = G and two distinct vertices
x and y are adjacent in ΓP (G) if and only if either xi = y or yj = x,
where 2 ≤ i, j ≤ n. In this paper, we obtain some fundamental
characterizations of the power graph. Also, we characterize certain
classes of power graphs of finite abelian groups.
Introduction
The study of algebraic structures using the properties of graphs be-
comes an exciting research topic in the last twenty four years, leading
to many fascinating results and questions. There are many papers on
assigning a graph to a group or a ring, for instance, see [1, 2, 5, 8]. Also
investigation of algebraic properties of groups or rings using the associated
graph becomes an exciting topic. In 2002, The directed power graph of a
semi group S was defined by Kelarev and Quinn [7] as the digraph ~G(S)
with vertex set S, in which there is an arc from x to y if and only if x 6= y
and y = xm for some positive integer m. Motivated by this, Chakrabarty
et al.[5] defined the undirected power graph ΓP (G) of a group G. Actually
1The work reported here is supported by the Special Assistance Programme (No.
F510/ 9/ DRS/ 2007 (SAP-I)) of University Grants Commission, India awarded to
the Department of Mathematics, Manonmaniam Sundaranar University, India for the
period 2007-2012.
2010 MSC: 05C25.
Key words and phrases: power graph, planar graph, Eulerian graph, finite
group.
34 Power graph of finite abelian groups
the power graph ΓP (G) of G is the graph with vertex set V (ΓP (G)) = G
and two distinct vertices x, y ∈ G are adjacent in ΓP (G) if and only if
either xi = y or yj = x, where i and j are integers and 2 ≤ i, j ≤ n. In
this paper, we use the following notations and definitions.
By a graph Γ = (V, E), we mean an undirected graph Γ with vertex
set V , edge set E and has no loops or multiple edges. The degree degΓ(v)
of a vertex v in Γ is the number of edges incident to v and if the graph
is understood, then we denote degΓ(v) simply by deg(v). The order of
Γ is defined as |V (Γ)| and its maximum and its minimum degrees will
be denoted, respectively, by ∆(Γ) and δ(Γ). A graph Γ is regular if the
degrees of all vertices of Γ are the same. A subset X of the vertices of Γ
is called an independent set if the induced subgraph < X > on X has no
edges. The maximum size of an independent set in a graph Γ is called
the independence number of Γ and denoted by β0(Γ). The length of a
smallest cycle in a graph Γ is referred to as its girth, which is denoted
by gr(Γ). If a graph Γ contains no cycles, then gr(Γ) is taken as ∞. A
planar graph is a graph that can be embedded in the plane so that no two
edges intersect geometrically except at a vertex which both are incident.
An Eulerian graph has an Eulerian trail, a closed trail containing all
vertices and edges. Unicyclic graphs are graphs which are connected and
have just one cycle. The union Γ1 ∪ Γ2 of two graphs Γ1 = (V1, E1) and
Γ2 = (V2, E2) is the graph with vertex set V = V1 ∪ V2 and edge set
E = E1 ∪ E2. The join Γ1 + Γ2 of Γ1 = (V1, E1) and Γ2 = (V2, E2) is the
graph with V = V1 ∪ V2 and E = E1 ∪ E2 together with edges joining all
vertices in V1 with vertices in V2.
Let G be a group with identity e. The number of elements of a group
is called its order and it is denoted by o(G). The order of an element g
in a group is the smallest positive integer n such that gn = e. If no such
integer exists, we say g has infinite order. The order of an element g is
denoted o(g).
We state the following theorems for use in the subsequent discussions.
Theorem 1 ([3]). K5 and K3,3 are non-planar.
Theorem 2 ([5]). Let G be a finite group. Then ΓP (G) is complete if
and only if G is a cyclic group of order 1 or pm, for some prime number
p and for some m ∈ N.
T. Tamizh Chelvam, M. Sattanathan 35
1. Properties of power graph
In this section, we derive some properties of power graphs, which will
be used later for further study.
Proposition 1. Let G be a group with at least one non-self inverse
element. Then gr(ΓP (G)) = 3.
Proof. Let G be a group with identity e. Let x be a non-self inverse
element of G. Note that < x >=< x−1 >, e ∈< x > and thus the graph
induced by the set {e, x, x−1} is K3 in ΓP (G). Hence gr(ΓP (G)) = 3.
Remark 1. Let x ∈ G. Clearly x is adjacent to x2, x3, . . . , xo(x) in ΓP (G)
and so the number of edges in ΓP (G) is given by
|E(ΓP (G))| ≥
∑
x∈G,x 6=e
o(x)
2
.
Now we characterize groups G for which power graph ΓP (G) contains
exactly
∑
x∈G,x6=e
o(x)
2 edges.
Theorem 3. Let G be a finite group with n elements. Then ΓP (G) is a
graph with
∑
x6=e
o(x)
2 edges if and only if every element other than identity
of G is of prime order.
Proof. Assume that ΓP (G) is a graph with
∑
x6=e
o(x)
2 edges. It means that
deg(x) = o(x) − 1 for all e 6= x ∈ G in ΓP (G). Let x 6= e be any element
of G. Suppose order of x is not a prime. Without loss of generality, we
assume that o(x) = pq, where p, q are distinct primes. Consider the
subgroup H =< x >. Since p|o(H), H has an element say y such that
o(y) = p. From this deg(y) = o(y) − 1 = p − 1. Since x /∈< y > and
y ∈< x >, y is adjacent to at least x, y2, . . . , yp = e. This implies that
deg(y) > p − 1 = o(y) − 1, which is a contradiction. Hence every element
other than identity in the group G is of prime order.
Conversely, assume that every element other than identity of G is of prime
order. To conclude that ΓP (G) contains
∑
x6=e
o(x)
2 edges, it is enough to prove
that deg(x) = o(x) − 1 for all e 6= x ∈ G in ΓP (G). If deg(x) > o(x) − 1
for some x ∈ G − e, then there exists y /∈< x > and y is adjacent to x.
This implies that x ∈< y > and so < x >⊆< y >. Since o(x) and o(y)
36 Power graph of finite abelian groups
bb
b b
b
b
0
3
4
5 1
2
Figure 1. ΓP (Z6)
are prime, we get that < x >=< y >, a contradiction to y /∈< x >. Hence
deg(x) = o(x) − 1 for all e 6= x ∈ G in ΓP (G).
Proposition 2. Let G be a finite group with n elements and Z(G) be its
center. If deg(x) = n − 1 in ΓP (G), then x ∈ Z(G).
Proof. Let x ∈ G be a vertex with deg(x) = n−1 in ΓP (G) and H =< x >.
Since deg(x) = n − 1, x ∈< y > for all y ∈ G − H. Hence x commutes
with all elements in G and so x ∈ Z(G).
Remark 2. The converse of Proposition 2 is not true. For example,
consider the group (Z6, +6) and the graph ΓP (Z6) given below. Here 3
∈ Z(Z6), whereas deg(3) = 3 6= 5 = 6 − 1.
Theorem 4. Let G be a finite group with n elements. Then the following
are equivalent:
(i) ΓP (G) ∼= K1,n−1
(ii) ΓP (G) is a tree
(iii) Every element of G is its own inverse.
Proof. (i) ⇒ (ii) It is trivially true.
(ii) ⇒ (iii) Assume that ΓP (G) is a tree. Suppose that there exists an
element a ∈ G such that a 6= a−1. Then the graph induced by {e, a, a−1}
is a triangle in ΓP (G), which is a contradiction.
(iii) ⇒ (i) Since every element of G is its own inverse, < x >= {e, x} for
all x ∈ G − e. From this ΓP (G) ∼= K1,n−1.
T. Tamizh Chelvam, M. Sattanathan 37
Theorem 5. Let G be a finite group of order pq, where p < q, p and q
are two distinct primes, and φ is the Euler function. Then
(i) G is cyclic if and only if ΓP (G) ∼= (Kp−1 ∪ Kq−1) + Kφ(pq)+1
(ii) G is non-cyclic if and only if ΓP (G) ∼= K1 + (qKp−1 ∪ Kq−1).
Proof. (i): Let G be a cyclic group of order pq. Then G has a unique
p-Sylow subgroup namely H1 and a unique q-Sylow subgroup namely H2.
By Theorem 2, ΓP (H1) ∼= Kp and ΓP (H2) ∼= Kq. Note that all elements
in G − (H1 ∪ H2) are generators of G and so |G − (Hp ∪ Hq)| = φ(pq).
Since the generators and the identity element e of G have full degree in
ΓP (G) and every non identity element in H1 is not adjacent to every non
identity element in H2, ΓP (G) ∼= (Kp−1 ∪ Kq−1) + Kφ(pq)+1.
Conversely, assume that ΓP (G) ∼= (Kp−1 ∪ Kq−1) + Kφ(pq)+1. If G is
non-cyclic, then every non identity element of G has order either p or q.
From this, identity is the only vertex of full degree in ΓP (G), which is a
contradiction. Hence G is cyclic.
(ii): Let G be a non-cyclic group. Then the number of p-Sylow subgroups
of G is q and G has a unique q-Sylow subgroup. Also the identity element
of G has full degree in ΓP (G). Hence ΓP (G) ∼= K1 + (qKp−1 ∪ Kq−1).
Conversely, assume that ΓP (G) ∼= K1 + (qKp−1 ∪ Kq−1). If G is cyclic,
then G has φ(pq) generators and so ΓP (G) has φ(pq) + 1 full degree
vertices, which is a contradiction. Hence G is non cyclic.
Theorem 6. Let G be a finite group. Then ΓP (G) is Eulerian if and
only if o(G) is odd.
Proof. Assume that o(G) is odd. Clearly deg(e) is even in ΓP (G). For
any e 6= x ∈ G, clearly o(x) is odd and so o(x) − 1 is even. If deg(x) =
o(x) − 1 in ΓP (G), then deg(x) is even . If deg(x) > o(x) − 1, then
there exists an element y ∈ G such that y /∈< x > and x ∈< y >.
Since < y >=< y−1 >, x ∈< y−1 >. From this x is adjacent to
{e, x2, . . . , xo(x)−1, y1, y−1
1 , . . . , yk, y−1
k } for some k ≥ 1. Since o(G) is odd,
G has no self inverse element and so deg(x) is even in ΓP (G). Hence
ΓP (G) is Eulerian.
Conversely, assume that ΓP (G) is Eulerian. Since deg(e) = o(G) − 1
is even, o(G) is odd.
Theorem 7. Let G be a group and o(G) = pα1
1 pα2
2 . . . pαn
n , where
p1, p2, . . . pn are distinct primes. Then the independence number
β0(ΓP (G) ≥ n.
38 Power graph of finite abelian groups
Proof. By Cauchy’s Theorem, G has an element ai of order pi for all
i = 1, 2, . . . n. Note that < ai > ∩ < aj >= {e} for all i 6= j. From this
{a1, a2, . . . , an} is an independent set in ΓP (G) and hence the indepen-
dence number β0(ΓP (G) ≥ n.
2. Power graph of finite abelian groups
In this section, we study about power graph of finite abelian groups.
Theorem 8. Let G be an elementary abelian group of order pn for some
prime number p and positive integer n. Then ΓP (G) ∼= K1 + ∪ℓ
i=1Kp−1,
where ℓ = pn−1
p−1 and β0(ΓP (G)) = ℓ.
Proof. Note that there are pn − 1 elements in G each with order p. Since
a group of order p has exactly p − 1 elements of order p, G has exactly
pn−1
p−1 distinct subgroups of order p. Clearly if any two elements a and b
are adjacent in ΓP (G), then they are in the same subgroup of G. Since
G has ℓ distinct subgroups of order p, by Theorem 2, ∪ℓ
i=1Kp−1, where
ℓ = pn−1
p−1 , is an induced subgraph of ΓP (G). Since the identity element of
G is adjacent to all other elements of G, ΓP (G) ∼= K1 + ∪ℓ
i=1Kp−1, where
ℓ = pn−1
p−1 . From this β0(ΓP (G)) = ℓ.
Theorem 9. Let G be a finite abelian group. Then ΓP (G) is planar if and
only if either G is isomorphic to Z3 ×Z3 × . . . ×Z3 or Z2 ×Z2 × . . . ×Z2
or Z4 × Z4 × . . . × Z4 or Z2 × Z2 × . . . × Z2 × Z4 × Z4 × . . . × Z4.
Proof. If part:
Case (i): Let G ∼= Z3 ×Z3 × . . .×Z3. By Theorem 8, ΓP (G) ∼= ∪ℓK2 +K1,
where ℓ = 1 + 3 + 32 + . . . + 3n−1. Hence G is planar.
Case (ii): Suppose G ∼= Z2 × Z2 × . . . × Z2. Then o(G) = 2n for some
n ∈ Z
+. By Theorem 8, ΓP (G) ∼= K1,2n−1 and so ΓP (G) is planar.
Case(iii): Suppose G ∼= Z4 ×Z4 × . . . ×Z4 or Z2 ×Z2 × . . . ×Z2 ×Z4 ×
Z4 × . . . ×Z4. Then G can be partitioned into three sets, namely A,B and
C, where A = {e}, B = {x ∈ G/o(x) = 4} and C = {x ∈ G/o(x) = 2}.
Note that no two elements of order 2 are adjacent to each other so that
the set C is an independent set in ΓP (G). Let a, b ∈ B such that a 6= b.
Then a and b are adjacent if and only if a ∈< b > or b ∈< a >. Since
a 6= b and o(a) = o(b) = 4, a = b−1 or b = a−1 i.e., a and b are adjacent
if and only if they are inverse of each other. Thus the subgraph induced
by the set B in ΓP (G) is union of K2.
Claim: No two elements of C are adjacent to the same element of B.
T. Tamizh Chelvam, M. Sattanathan 39
Suppose not, let a ∈ B, b, c ∈ C and b 6= c such that b, c are adjacent to
a. Since o(b) = o(c) = 2 and o(a) = 4, a /∈< b > and a /∈< c >. Therefore
b, c ∈< a >. Since < a > has a unique element of order 2, b = c, which
is a contradiction. Now, we arrange the elements of B such that first
collection of elements which are adjacent to the first element of C, second
collection of elements which are adjacent to the second element of C and
so on. That the graph induced by the set B ∪ C is planar in ΓP (G). Since
ΓP (G) ∼=< A > + < B ∪ C >, ΓP (G) is planar.
Conversely assume that ΓP (G) is planar. Suppose p|o(G), for some
prime p ≥ 5. Then G has an element x of order p. By Theorem 2, the
subgraph induced by the subgroup < x > is Kp. Since p ≥ 5, by Theorem 1,
ΓP (G) is non-planar, a contradiction. Therefore o(G) = 2n13n2 , where
n1 ≥ 0, n2 ≥ 0 are two integers.
Suppose o(G) = 2n13n2 , for two integers n1 ≥ 1 and n2 ≥ 1. Since
2|o(G) and 3|o(G), G has two elements a and b such that o(a) = 2 and
o(b) = 3. Since G is abelian, we have o(ab) = o(a)o(b)
gcd(o(a),o(b)) , for all a, b ∈ G.
Therefore o(ab) = 6. Now the subgraph induced by < ab > must have
K3,3 as a subgraph. By Theorem 1, ΓP (G) is non-planar, a contradiction.
Therefore o(G) is either 2n or 3m, for some n, m ∈ Z
+.
Suppose o(G) = 3m and G 6∼= Z3 × Z3 × . . . × Z3. Since 9|o(G) and G
is abelian, Z9 must be a subgroup of G. Hence by Theorem 2, ΓP (G)
must have K9 as a subgraph, again ΓP (G) is non-planar, a contradiction.
Hence if o(G) = 3m, for some m ∈ Z
+ then G ∼= Z3 × Z3 × . . . × Z3.
Suppose o(G) = 2n, for some n ∈ Z
+ and G is not isomorphic to one of
the groups Z2×Z2×. . .×Z2, Z4×Z4×. . .×Z4 and Z2×Z2×. . .×Z2×Z4×
Z4 × . . .×Z4. In such a case 8|o(G), G must have Z8 as a subgroup and so
by Theorem 2, K8 must be a subgraph of ΓP (G), a contradiction. Hence
if o(G) = 2n for some n ∈ Z
+, then G is isomorphic to Z2 ×Z2 × . . . ×Z2
or Z4 × Z4 × . . . × Z4 or Z2 × Z2 × . . . × Z2 × Z4 × Z4 × . . . × Z4.
Theorem 10. Let G be a finite abelian group. Then β0(ΓP (G)) = 2 if
and only if G is a cyclic group of order pnq, where p and q are distinct
primes and n is a positive integer.
Proof. Assume that G is a cyclic group of order pnq, where p and q are
distinct primes and n is a positive integer. By Cauchy’s Theorem, there
exist two non-identity elements a and b in G such that o(a) = p and
o(b) = q. Since < a > ∩ < b >= {e}, a and b are non-adjacent in ΓP (G).
Hence β0(ΓP (G)) ≥ 2. Since G is cyclic, G has a unique subgroup H
of order pn. By Theorem 2, ΓP (H) ∼= Kpn . Let x, y ∈ G − H. Since the
40 Power graph of finite abelian groups
subgroup H is unique, o(x) = piq and o(y) = pjq for some 0 ≤ i, j ≤ n.
From this, we have either o(x)|o(y) or o(y)|o(x) and so either x ∈< y >
or y ∈< x >. Therefore ΓP (G − H) ∼= Kpn(q−1). Since identity element is
adjacent to all other elements of G, K1 +(Kpn−1 ∪Kpn(q−1)) is a subgraph
of ΓP (G) and so β0(ΓP (G)) ≤ 2. Hence β0(ΓP (G)) = 2.
Conversely, assume that β0(ΓP (G)) = 2. First we claim that G is
cyclic. Suppose G is not cyclic. Since G is a non-cyclic abelian group,
there exists a prime number p such that Zp × Zp is a subgroup of G
and so G has at least p2−1
p−1 distinct subgroups of order p and p2−1
p−1 > 2.
Hence G has at least three distinct elements a, b and c from three distinct
subgroups of order p. This implies that {a, b, c} is an independent set of
ΓP (G), which is a contradiction. Hence G is cyclic. Since G is cyclic and
β0(ΓP (G)) = 2, there are exactly two distinct prime divisors p and q of
o(G) and o(G) = pn1qn2 for some positive integers n1, n2 and n1 ≥ n2.
Suppose n2 > 1. Since G is cyclic, G has three elements a, b and c such
that o(a) = pn1 , o(b) = pqn2−1 and o(c) = qn2 . From this order of either
of a or b or c divide orders of others and so {a, b, c} is an independent set
of ΓP (G), which is a contradiction. Hence G is a cyclic group of order
pnq, where p and q are distinct primes and n is a positive integer.
Theorem 11. Let G be a finite abelian group. Then ΓP (G) is a unicyclic
graph if and only if G ∼= Z3.
Proof. Assume that G ∼= Z3. Clearly ΓP (G) ∼= K3 and hence ΓP (G) is a
unicyclic graph.
Conversely, assume that ΓP (G) is a unicyclic graph. Suppose there
exists a prime p > 3 such that p|o(G). Then Kp is a subgraph of ΓP (G), a
contradiction to ΓP (G) is unicyclic. Hence o(G) = 2n3m for some integers
n, m ≥ 0.
Suppose o(G) = 2n3m for some integers n, m ≥ 1. Since 2|o(G) and
3|o(G), G has two elements a and b such that o(a) = 2 and o(b) = 3 and
hence o(ab) = 6. Now the subgraph induced by < ab > must have K4 as
a subgraph, which is a contradiction. Therefore o(G) is either a power of
2 or a power of 3.
If G is an elementary abelian group of order 2n, then ΓP (G) ∼= K1,2n−1,
which is a contradiction. On the other hand, there exists a m > 1 such that
Z2m is a subgroup of G. By Theorem 2, K2m is a subgraph of ΓP (G), which
is a contradiction. If o(G) is a power of 3 and G is an elementary abelian
group of order 3n for some n > 1, then by Theorem 8, ΓP (G) ∼= K1 +∪
ℓ
K2
where ℓ = 3n−1
2 . Thus ΓP (G) is not unicyclic, which is a contradiction. On
T. Tamizh Chelvam, M. Sattanathan 41
the other hand, for some n > 1, Z3n is a subgroup of G. By Theorem 8,
K3n is a subgraph of ΓP (G), which is a contradiction. Hence o(G) = 3
and so G ∼= Z3.
References
[1] A. Abdollahi, S. Akbari and H.R. Maimani, Non-Commuting graph of a group, J.
Algebra, 298(2006), 468-492.
[2] S. Akbari and A. Mohammadian, On the zero-divisor graph of commutative ring, J.
Algebra, 274 (2)(2004), 847-855.
[3] J.A. Bondy, and U.S.R. Murty, Graph theory with applications, Elsevier, 1977.
[4] G. Chartrand and P. Zhang, Introduction to Graph Theory, Tata McGraw-Hill,
2006.
[5] I. Chakrabarty, S. Ghosh and M.K. Sen, Undirected power graphs of semi groups,
Semi group Forum,78(2009), 410-426.
[6] J.A. Gallian, Contemporary Abstract Algebra, Narosa Publishing House, 1999.
[7] A.V. Kelarev and S.J. Quinn, Directed graph and Combinatorial properties of semi
groups, J. Algebra, 251(2002), 16-26.
[8] S. Lakshmivarahan, Jung-Sing Jwo and S.K Dhall, Symmetry in interconnection
networks based on Cayley graphs of permutation groups: A survey, Parallel Comput.,
19(1993), 361-407.
Contact information
T. Tamizh Chelvam,
M. Sattanathan
Department of Mathematics,
Manonmaniam Sundaranar University,
Tirunelveli 627 012, Tamil Nadu, India
E-Mail: tamche59@gmail.com,
satta_nathan@gmail.com
Received by the editors: 27.10.2011
and in final form 15.07.2012.
|
| id | nasplib_isofts_kiev_ua-123456789-152306 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T18:17:50Z |
| publishDate | 2013 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Tamizh Chelvam, T. Sattanathan, M. 2019-06-09T17:12:46Z 2019-06-09T17:12:46Z 2013 Power graph of finite abelian groups / T. Tamizh Chelvam, M. Sattanathan // Algebra and Discrete Mathematics. — 2013. — Vol. 16, № 1. — С. 33–41. — Бібліогр.: 8 назв. — англ. 1726-3255 2010 MSC:05C25. https://nasplib.isofts.kiev.ua/handle/123456789/152306 Let G be a group. The power graph ΓP(G) of G is a graph with vertex set V(ΓP(G)) = G and two distinct vertices x and y are adjacent in ΓP(G) if and only if either xˡ = y or yʲ = x, where 2 ≤ i, j ≤ n. In this paper, we obtain some fundamental characterizations of the power graph. Also, we characterize certain classes of power graphs of finite abelian groups. The work reported here is supported by the Special Assistance Programme (No.F510/ 9/ DRS/ 2007 (SAP-I)) of University Grants Commission, India awarded tot he Department of Mathematics, Manonmaniam Sundaranar University, India for the period 2007-2012. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Power graph of finite abelian groups Article published earlier |
| spellingShingle | Power graph of finite abelian groups Tamizh Chelvam, T. Sattanathan, M. |
| title | Power graph of finite abelian groups |
| title_full | Power graph of finite abelian groups |
| title_fullStr | Power graph of finite abelian groups |
| title_full_unstemmed | Power graph of finite abelian groups |
| title_short | Power graph of finite abelian groups |
| title_sort | power graph of finite abelian groups |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/152306 |
| work_keys_str_mv | AT tamizhchelvamt powergraphoffiniteabeliangroups AT sattanathanm powergraphoffiniteabeliangroups |