Filomat 26:6 (2012), 1201–1208
DOI 10.2298/FIL1206201M
Published by Faculty of Sciences and Mathematics,
University of Ni
ˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
On the Power Graph of a Finite Group
M. Mirzargar, A. R. Ashrafi, M. J. Nadjafi-Arani
Department of Mathematics, Faculty of Mathematics, Statistics and Computer Science, University of Kashan, Kashan 87317-51167,
I. R. Iran
Abstract. The power graph P(G) of a group G is the graph whose vertex set is the group elements and
two elements are adjacent if one is a power of the other. In this paper, we consider some graph theoretical
properties of a power graph P(G) that can be related to its group theoretical properties. As consequences
of our results, simple proofs for some earlier results are presented.
1. Introduction
All groups and graphs in this paper are finite. Throughout the paper, we follow the terminology and
notation of [11, 12] for groups and [18] for graphs.
Groups are the main mathematical tools for studying symmetries of an object and symmetries are usually
related to graph automorphisms, when a graph is related to our object. Groups linked with graphs have
been arguably the most famous and productive area of algebraic graph theory, see [1, 11] for details. The
power graphs is a new representation of groups by graphs. These graphs were first used by Chakrabarty et
al. [4] by using semigroups. It must be mentioned that the authors of [4] were motivated by some papers of
Kelarev and Quinn [8–10] regarding digraphs constructed from semigroups. We also encourage interested
readers to consult papers by Cameron and his co-workers on power graphs constructed from finite groups
[2, 3].
Suppose G is a finite group. The power graph P(G) is a graph in which V(P(G)) = G and two distinct
elements x and y are adjacent if and only if one of them is a power of the other. If G is a finite group then it
can be easily seen that the power graph P(G) is a connected graph of diameter 2. In [4], it is proved that for
a finite group G, P(G) is complete if and only if G is a cyclic group of order 1 or p
m
, for some prime number
p and positive integer m.
Following [12, 13], two finite groups G and H are said to be conformal if and only if they have the same
number of elements of each order. In [13], the following question was investigated:
Question: For which natural numbers n are any two conformal groups of order n isomorphic?
Let G be a group and x ∈ G. We denote by o(x) the order of x and G is said to be EPO−group, if all
non-trivial element orders of G are prime. An EPPO−group is that its element orders are prime power.
2010 Mathematics Subject Classification. Primary 05C25.
Keywords. Power graph; Clique number; EPPO-group; Maximal cyclic subgroup.
Received: October 22, 2011; Accepted: November 27, 2011
Communicated by Dragan Stevanovi´c
Email address: ashrafi@kashanu.ac.ir (A. R. Ashrafi)
M. Mirzargar, A. R. Ashrafi, M. J. Nadjafi-Arani / Filomat 26:6 (2012), 1201–1208 1202
The set of all elements order of G is called its spectrum, denoted by π
e
(G), A maximal subgroup H of G is
denoted by H < · G and the set of all elements of G of order k is denoted by Ω
k
(G).
Suppose Γ is a graph. A subset X of the vertices of Γ is called a clique if the induced subgraph on X is a
complete graph. The maximum size of a clique in Γ is called the clique number of Γ and denoted by ω(Γ).
A subset Y of V(Γ) is an independent set if the induced subgraph on X has no edges. The maximum size of
an independent set is called the independence number of G and denoted by α(G). The chromatic number of Γ
is the smallest number of colors needed to color the vertices of Γ so that no two adjacent vertices share the
same color. This number is denoted by χ(Γ).
Throughout this paper our notation is standard and they are taken from the standard books on graph
theory and group theory such as [12, 18].
2. Main Results
Suppose G is a finite group of order n. Chakrabarty, Ghosh and Sen [4] proved that the number of edges
of P(G) can be computed by the following formula:
e =
1
2
a∈G
{2o(a) − ϕ(o(a)) − 1},
where ϕ is the Euler’s totient function. In the case that G is cyclic, we have:
e =
1
2
d|n
{2d − ϕ(d) − 1}ϕ(d).
Moreover, P(Z
n
) is nonplanar when ϕ(n) > 7 or n = 2
m
, m ≥ 3. Finally, if n ≥ 3 then P(Z
n
) is Hamiltonian.
Suppose D(n) denotes the set of all positive divisors of n. It is well-known that (D(n), |) is a distributive
lattice. D(n) is a Boolean algebra if and only if n is square-free. In the following theorem we apply the
structure of this lattice to compute the clique and chromatic number of P(Z
n
).
Lemma 1 Suppose G is a group and A ⊆ G. The vertices of A constitute a complete subgraph in P(G) if and
only if {⟨x⟩ | x ∈ A} is a chain.
Proof Suppose C is a clique in P(G). To prove that {⟨x⟩ | x ∈ C} is a chain, we proceed by induction on |V(C)|.
If |C| = 2 the result is obvious. If V(C) = {x
1
, x
2
, · · · , x
n
} then by induction hypothesis, {⟨x
i
⟩ | 1 ≤ i ≤ n − 1}
is a chain in P(G). Without loss of generality we can assume that 1 ⊆ ⟨x
1
⟩ ⊆ ⟨x
2
⟩ ⊆ · · · ⊆ ⟨x
n−1
⟩. Consider
t = max{i | ⟨x
i
⟩ ⊆ ⟨x
n
⟩}. If t = n − 1 then the result is proved. Otherwise, ⟨x
t
⟩ ⊆ ⟨x
n
⟩ ⊆ ⟨ x
t+1
⟩, as desired.
Conversely, by definition of power graph, every chain of cyclic subgroups is a clique.
Theorem 2 Suppose n = p
α
1
1
p
α
2
2
· · · p
α
r
r
, where p
1
< p
2
< ... < p
r
are prime numbers. Then
ω(P(Z
n
)) = χ(P(Z
n
)) = p
α
r
r
+
r−2
j=0
(p
α
r−j−1
r−j−1
− 1)
j
i=0
ϕ(p
α
r−i
r−i
)
.
Proof Define the relation ∼ on Z
n
by a ∼ b if and only if they have the same order. Then it can easily seen
that ∼ is an equivalence relation on Z
n
and
Z
n
∼
can be equipped with an order such that
Z
n
∼
D(n). Here
a
∼
≤
b
∼
if and only if o(a)|o(b). Choose an element a ∈ Z
n
. By our definition, the elements of
a
∼
are adjacent in
P(Z
n
). Moreover, for each chain
v
1
∼
,
v
2
∼
, · · · ,
v
t
∼
of elements in
Z
n
∼
,
t
i=1
v
i
∼
is a complete subgraph of P(Z
n
). For
an arbitrary element
u
∼
∈
Z
n
∼
, define d(
o
∼
,
u
∼
) to be the same as distance between corresponding elements of
D(n).
M. Mirzargar, A. R. Ashrafi, M. J. Nadjafi-Arani / Filomat 26:6 (2012), 1201–1208 1203
To find a maximal complete subgraph of P(Z
n
), by Lemma 1 it is enough to obtain a maximal chain
Q :
a
0
∼
=
o
∼
,
a
1
∼
,
a
2
∼
, · · · ,
a
l
∼
,
n
∼
=
a
l+1
∼
(1)
such that Q has the maximum length,
a
1
∼
∪
a
2
∼
∪ · · · ∪
a
l
∼
has the maximum possible size and l + 1 = α
1
+ · · · + α
r
.
To do this, it is enough to choose a
1
to be an element of order p
r
, a
2
to be an element of order p
2
r
, ...., a
α
r
to be
an element of order p
α
r
r
, a
α
r+1
to be an element of order p
α
r
r
p
r−1
and so on. Therefore,
ω(P(Z
n
)) = |
a
0
∼
| + |
a
1
∼
| + · · · + |
a
l+1
∼
|
= (ϕ(p
r
) + ϕ(p
2
r
) + · · · + ϕ(p
α
r
r
))
+ ϕ(p
α
r
r
)(ϕ(p
r−1
) + · · · + ϕ(p
α
r−1
r−1
))
+ · · ·
+ ϕ(p
α
r
r
) · · · ϕ(p
α
2
2
)(ϕ(p
1
) + · · · + ϕ(p
α
1
1
)) + 1
= p
α
r
r
+
r−2
j=0
(p
α
r−j−1
r−j−1
− 1)
j
i=0
ϕ(p
α
r−i
r−i
)
.
To complete the proof we have to prove that ω(P(Z
n
)) = χ(P(Z
n
)) and this is a direct consequence of the
strong perfect graph theorem [5].
The exponent of a finite group G is defined as the least common multiple of all elements of G, denoted
by Exp(G). It is easy to see that if G is nilpotent then there exists an element a ∈ G such that o(a) = Exp(G).
Such groups are said to be full exponent.
Theorem 3 Suppose that G is a full exponent group and n = Exp(G) = p
β
1
1
p
β
2
2
· · · p
β
r
r
, where p
1
< p
2
< ... < p
r
are prime numbers. If x is an element of order n then
ω(P(G)) = χ(P(G)) = p
β
r
r
+
r−2
j=0
(p
β
r−j−1
r−j−1
− 1)
j
i=0
ϕ(p
β
r−i
r−i
)
.
Proof By Lemma 1, a subset A of G constitutes a clique in P(G) if and only if {⟨x⟩ | x ∈ A} is a chain. To
obtain a maximal clique in P(G), we have to choose a chain 1 ⊆ ⟨x
1
⟩ ⊆ ⟨x
2
⟩ ⊆ · · · ⊆ ⟨x
t
⟩ such that o(x
t
) = o(x)
and 1 +
t
i=1
φ(o(x
i
)) has maximum value among all possible chains of subgroups of ⟨x⟩. Now a similar
argument as given in the proof of Theorem 2, completes the proof.
Our calculations on the small group library of GAP [15] suggest the following conjecture:
Conjecture 1: The Theorem 3 is correct in general.
Corollary 4 Let G be a finite group. Then the power graph P(G) is planar if and only if π
e
(G) ⊆ {1, 2, 3, 4}.
Proof Suppose P(G) is planar. Then P(G) does not have the complete graph K
5
as its induced subgraph and
the Theorem 3 concludes the result. Conversely, if π
e
(G) ⊆ { 1, 2, 3, 4} then it can easily seen that P(G) can be
embedded into sphere, as desired.
In [4, Lemma 4.7], the authors proved that if G is a cyclic group of order n, n ≥ 3 and ϕ(n) > n then P(G)
is not planar. Also, in [4, Lemma 4.8] it is proved that a cyclic group of order 2
n
, n ≥ 3, is not planar. In the
following corollary we apply Corollary 4 to find a simple classification for planarity of the power graph of
cyclic groups.
Corollary 5 The power graph of a cyclic group of order n is planar if and only if n = 2, 3, 4.
M. Mirzargar, A. R. Ashrafi, M. J. Nadjafi-Arani / Filomat 26:6 (2012), 1201–1208 1204
In what follows, U
n
denotes the groups of units in the ring Z
n
. In the following corollary a new simple
proof for [4, Lemma 4.10] is presented.
Corollary 6 The power graph of U
n
is planar if and only if n|240.
Proof Suppose n = p
e
1
1
p
e
2
2
· · · p
e
k
k
, where p
1
, p
2
, · · · , p
k
are distinct primes. Then by [7, Theorems 6.11, 6.13 and
Corollary 6.14] , U
p
e
is cyclic for odd p, U
2
1, U
4
Z
2
, U
2
n
Z
2
× Z
2
n−2
and U
n
U
p
e
1
1
× · · · × U
p
e
k
k
. Therefore,
by Corollary 4, n|240.
Consider the dihedral group D
2n
presented by
D
2n
= ⟨x, y | x
n
= y
2
= e & y
−1
xy = x
−1
⟩.
From [4, Corollary 4.3], we can deduce that the number of edges of P(D
2n
) is given by e =
1
2
d|n
{2dϕ(d) −
ϕ(d)
2
} + n. This graph is neither Eulerian nor hamitonian, since the group has elements of order 2.
By corollary 5, it is easy to prove the power graph of a dihedral group of order 2n is planar if and only
if n = 2, 3, 4.
Corollary 7 χ(P(D
2n
)) = ω(P(D
2n
)) = χ(P(Z
n
)).
Proof Notice that the power graph P(D
2n
) is a union of P(Z
n
) and n copy of K
2
that share in the identity
element of D
2n
.
The semi−dihedral group SD
2
n
is presented by
SD
2
n
= ⟨x, y | x
2
n−1
= y
2
= 1, yxy = r
2
n−2
−1
⟩.
Corollary 8 The power graph P(SD
2
n
) is a union of a complete graph of order 2
n
and 2
n
copies of K
2
that share in the identity vertex. This graph is non-Eulerian, non-hamiltonian and nonplanar, for n ≥ 3.
Moreover, χ(P(SD
2
n
)) = ω(P(SD
2
n
)) = α(P(SD
2
n
)) = 2
n
.
Following [6] we assume that P is a finite partially ordered set (poset for short) which possesses a rank
function r : P −→ N with the property that r(p) = 0, for some minimal element p of P and r(q) = r(p) + 1
whenever q covers p. Let N
k
:= {p ∈ P : r(p) = k} be its k
th
level and let r(P) := max{r(p) : p ∈ P} be
the rank of P. An antichain or Sperner family in P is a subset of pairwise incomparable elements of P. It is
clear that each level is an antichain. The width (Dilworth or Sperner number) of P is the maximum size d(P)
of an antichain of P. The poset P is said to have the Sperner property if d(P) = max
k
|N
k
|. A k−family in P,
1 ≤ k ≤ r(P), is a subset of P containing no (k + 1)−chain in P, and P has the strong Sperner property if for each
k the largest size of a k−family in P equals the largest size of a union of k levels.
Theorem 9 Suppose that n = p
β
1
1
· · · p
β
r
r
is the prime decomposition of n and m = β
1
+ · · · + β
r
. Then α(P(Z
n
))
is the coefficient of the middle or the two middle term of Π
m
j=1
(1 + x + · · · + x
β
j
).
Proof It is well-known that the lattice of divisors of a natural number, ordered by divisibility, has strong
Sperner property and so its largest antichain is its largest rank level.
Let Γ be a graph. The minimum number of vertices of Γ which need to be removed to disconnect the
remaining vertices of Γ from each other is called the connectivity of Γ, denoted by κ(Γ). If G is finite group
then we define:
M(G) = {x ∈ G ; ⟨x⟩ < · G}.
Theorem 10 Suppose G is a non-cyclic group and x ∈ G such that ⟨x⟩ < · G. Define r(x) = ∪
y∈M(G)−⟨x⟩
⟨x⟩ ∩ ⟨y⟩
.
Then,
κ(P(G)) ≤ Min{|r(x)| ; ⟨x⟩ < · G}.
M. Mirzargar, A. R. Ashrafi, M. J. Nadjafi-Arani / Filomat 26:6 (2012), 1201–1208 1205
Proof Suppose ⟨x⟩ is a maximal cyclic subgroup of G. We claim that r(x) is a cut set of P(G). Since G is
noncyclic, there exists another maximal cyclic subgroup ⟨y⟩ different from ⟨x⟩. If r(x) is not a cut set of P(G)
then there exists a shortest path Q : x = x
0
, x
1
, x
2
, ..., x
n−1
, x
n
= y in P(G) connecting x and y. Without loss
of generality we can assume that x
2k
, 0 ≤ k ≤ ⌈
n
2
⌉, are generators of maximal cyclic subgroups of G. Thus,
x
1
∈ ⟨x⟩ ∩ ⟨x
2
⟩ ⊆ r(x) contradict by our assumption. This completes the proof.
For a finite group G, the set of all maximal cyclic subgroups of G is denoted by MaxCyc(G).
Lemma 11 Suppose G is a non-cyclic finite group, S ⊆ G − M(G), MaxCyc(G) = {⟨x
1
⟩, ..., ⟨x
r
⟩} and A =
{x
1
, ..., x
r
}. S is a minimal cut set with this property that each component of P(G) − S has exactly one element
of A if and only if S = ∪
x∈M(G)
r(x).
Proof If S = ∪
x∈M(G)
r(x) then by an argument similar to the proof of Theorem 10, one can see that if x, y ∈ M(G)
and ⟨x⟩ , ⟨y⟩ then {x
1
, x
3
, · · · } ⊆ S, where x = x
0
, x
1
, x
2
, ..., x
n−1
, x
n
= y is a shortest path in P(G) connecting x
and y. Therefore, if x, y ∈ M(G), ⟨x⟩ , ⟨y⟩ then x and y are not in the same component of P(G) − S.
Conversely, we assume that S is a cut set with this property that each component of P(G) − S has exactly
one element of A and x, y ∈ A. Suppose ł ∈ ⟨x⟩ ∩ ⟨y⟩ and ł < S. Then ł is adjacent to x and y and so there
exists a component of P(G) − S containing both of x and y, a contradiction. Therefore, ∪
x∈M(G)
r(x) ⊆ S. On
the other hand, we assume that z ∈ S and ⟨t⟩ is a maximal cyclic subgroup of G containing z. By minimality
of S, there are at least two components X
1
and X
2
of P(G) − S such that z is adjacent to a vertex v
1
∈ X
1
and a
vertex v
2
∈ X
2
. Without loss of generality, we can assume that X
1
is the component containing t and v
1
= t.
Obviously, ⟨v
2
⟩ * ⟨t⟩ and so there exists a vertex t
′
∈ A ∩ X
2
such that ⟨v
2
⟩ ⊆ ⟨t
′
⟩. Since z is adjacent to v
2
,
⟨z⟩ ⊆ ⟨v
2
⟩ or ⟨v
2
⟩ ⊆ ⟨z⟩. If ⟨z⟩ ⊆ ⟨v
2
⟩ then z ∈ ⟨t⟩ ∩ ⟨t
′
⟩, as desired. If ⟨v
2
⟩ ⊆ ⟨z⟩ then v
2
is adjacent to t which
is impossible. This completes our argument.
It is easily seen that the power graph of a p−group Q is a union of some complete graphs of order p
which share in identity vertex if and only if Q has exponent p. In the following theorem we investigate the
same problem for an arbitrary group.
Theorem 12 P(G) is a union of complete graphs which share the identity element of G if and only if G is an
EPPO-group and for every maximal cyclic subgroup A and B with A , B, A ∩ B = {e}.
Proof Suppose there exist x ∈ G and prime numbers p
1
and p
2
such that p
1
, p
2
|o(x). Then the cyclic subgroup
⟨x⟩ is containing non-adjacent elements x
1
of order p
1
and x
2
of order p
2
. Since x
1
and x
2
are adjacent to x,
they are in the same block of P(G), a contradiction. If A = ⟨a⟩ and B = ⟨b⟩ are maximal cyclic subgroup of G
such that e , x ∈ A ∩ B then x, a and b are mutually adjacent and so A ⊆ B or B ⊆ A, which is impossible.
Conversely, we assume that maximal cyclic subgroups of G have prime power order and for every maximal
cyclic subgroup A and B with A , B, A ∩ B = {e}. By Lemma 11, S = ∪
x∈M(G)
r(x) = {e}. On the other hand,
if MaxCyc(G) = {⟨x
1
⟩, ..., ⟨x
r
⟩} and A = {x
1
, ..., x
r
} then by Lemma 11, each component of P(G) − {e} is of form
⟨x
i
⟩ − {e}, for some i, 1 ≤ i ≤ r, which is a complete subgraph of P(G). This completes the proof.
Corollary 13 If G is an EPO−group then P(G) is a union of some complete graphs which share in the identity
element of G.
Lemma 14 A finite group G is EPPO if and only if the vertices of every maximal clique of P(G) is a maximal
cyclic subgroup of G.
Proof (⇐=) Suppose H is a maximal clique in P(G) and x ∈ H. If o(x) has at least two prime divisors p and q
then there are elements of these orders in H which is impossible.
(=⇒) By Lemma 1, we map the maximal clique H in P (G) to the chain 1 ⊆ ⟨x
1
⟩ ⊆ ⟨x
2
⟩ ⊆ · · · ⊆ ⟨x
t
⟩. Then
x
t
has prime power order p
α
and since G is EPPO group, p
α
= 1 + φ(p) + · · · + φ(p
α
). This implies that
H = ⟨x
t
⟩.
A Chinese group theorist Wujie Shi [14] conjectured that a finite group and a finite simple group are