Automorphism group of certain power graphs of finite groups
Ali Reza Ashrafi
a, Ahmad Gholami
b, Zeinab Mehranian
baDepartment of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan 87317-53153, I. R. Iran
bDepartment of Mathematics, University of Qom, Qom, I. R. Iran
[email protected], [email protected], [email protected]
Abstract
The power graph P(G) of a group G is the graph with group elements as vertex set and two elements are adjacent if one is a power of the other. The aim of this paper is to compute the automorphism group of the power graph of several well-known and important classes of finite groups.
Keywords: power graph, automorphism group Mathematics Subject Classification: 05C25 DOI:10.5614/ejgta.2017.5.1.8
1. Introduction
The investigation of graphs associated to algebraic structures is very important, because such graphs are closely related to automata theory [8] and have valuable applications [9]. In this paper we describe the automorphism groups of the undirected power graphs of the dicylic groupsT4n, the semidihedral groupsSD8n, and the groupsU6nandV8nappearing in the textbook of James and
Received: 15 January 2016, Revised: 19 February 2017, Accepted: 22 February 2017.
Liebeck [7]. For evenn,V8nis described in [6]. These groups have the following presentations:
T4n = ha, b|a2n= 1, an=b2, b−1ab=a−1i, SD8n = ha, b|a4n=b2 = 1, bab=a2n−1i,
U6n = ha, b|a2n=b3 = 1, b−1ab=a−1i,
V8n = ha, b|a2n=b4 = 1, aba=b−1, ab−1a=bi.
Adirected graphΓ = (V, E)consists of a nonempty setV of vertices and a setE of ordered pairs of distinct vertices called edges. Anundirected graph(or graph for short) is given by a set V of vertices and setEof edges which are unordered pairs of distinct vertices.
The directed power graphof a semigroup G is a graph in which V(P(G)) = G and two distinct elements x andy are adjacent if and only if y is a power of x. This directed graph was firstly introduced by Kelarev and Quinn in their seminal paper [10]. In the mentioned paper, they gave a very technical description of the structure of the power graphs of all finite abelian groups.
Kelarev and his co-workers [11, 12, 13], studied some other classes of semigroups by directed power graph.
The undirected power graphP(G) (or power graph for short) of a semigroupGwas intro- duced by Chakrabarty et al. in [5]. It is proved that if Gis finite group then P(G) is connected if and only if Gis periodical and ifGis a finite group then P(G)is complete if and only if Gis cyclic of order1orpm, wherepis prime andm≥1is a natural number. They also obtained exact formula for the number of edges in a finite power graph. Cameron and Ghosh [3] proved that the only finite group whose automorphism group is the same as its power graph is the Klein group of order4. They also conjectured that two finite groups with isomorphic power graphs have the same number of elements of each order that proved by Cameron in [4]. Mirzargar et al. [15], consid- ered some graph theoretical properties of a power graph that can be related to its group theoretical properties and in [17], the authors considered the problem of2−connectivity of the power graphs into account and in [2] this problem is solved in some classes of finite simple groups.
The proper power graphP(G)of a groupGis obtained from the power graph ofGby deleting the identity element as a vertex. The power graph of a group is always connected, since identity element is adjacent to every other vertex. Moghaddamfar et al. [16] proved thatP?(G)is connected if and only ifG has a unique minimal subgroup. Further, P?(G)is shown to be bipartite if and only if no element of Ghas order greater than3, and to be planar if and only if no element ofG has order greater than6. We refer to [1] for a complete survey of recent results on this topic.
Throughout this paper our graph theory notations are standard and can be taken from the stan- dard books on graph theory or [8]. To simplify our figures, the power graph of the cyclic group of ordern is denoted byZn. Other notations are standard and can be taken from the book of James and Liebeck [7] or [8].
In this paper we prove the following results:
Theorem 1.1.
Aut(P(T4n)) ∼=
S2n−2×S2×(S2oSn), if nis a power of 2 Q
d|2nSφ(d)×(S2oSn), otherwise
, where n≥3,
Aut(P(SD8n)) ∼=
S4n−2×S2n×(S2oSn), if n is a power of 2 Q
d|4nSφ(d)×S2n×(S2oSn), otherwise
, where n≥2,
Aut(P(U6n)) ∼=
Q
d|3nSφ(d)×Q
d|2n,d-nSφ(d)oS3 k= 0 Q
d|2n,d-nSφ(d)oS3×Q
d|nSφ(d)×Q
d|n,d-tSφ(d)oS3 k= 1 Q
d|2n,d-nSφ(d)oS3×Q
d|nSφ(d)×Q
d|3t,d-tSφ(d)oS3
×Q
d|n,d-3tSφ(d)oS2 k≥2
,
where k is nonnegative integer and satisfies n = 3kt and some positive integer t such that 3-t,
Aut(P(V8n)) ∼=
S2n×S2oSn×Q
d|2n,d-nSφ(d)oS2×Q
d|2nSφ(d) k= 0, S2n+1×S2oSn×Qk−1
l=1 S22l×S2koS2 t= 1, k≥1 S2n×S2oSn×Q
d|tS4φ(d)×Qk s=2
Q
d|2st , d-2s−1tSφ(d)2
×Q
d|2k+1t , d-2ktSφ(d)oS2 t >1, k≥1 .
where n = 2kt for a nonnegative k and some positive odd integer t.
We also describe the automorphism groups of the undirected power graphs of the first Mathieu groupM11and first Janko groupJ1.
Theorem 1.2.
Aut(P(M11)) ∼= (S10oS144)×(S4oS396)×(S2oS55)× (S6oS3)×(S2oS4)×S2 oS165, Aut(P(J1)) ∼= (S10oS1596)×(S6oS4180)×(S18oS1540)
× S2×S8×S4×(S4 oS3)×(S2oS5) oS2
oS1463.
2. Power Graph Descriptions
Our proof of Theorem 1.1 relies upon the authors’ earlier description of the power graphs of these groups [14].
Theorem 2.1.
Aut(P(Zn))∼=
Sn n is a prime power
Sφ(n)+1×Q
d|n,d6=1,nSφ(d) otherwise .
Suppose Γ1 = (V1, E1) and Γ2 = (V2, E2) are graphs with disjoint vertex sets. The union Γ1∪Γ2 is the graph withV(Γ1∪Γ2) =V1∪V2andE(Γ1∪Γ2) =E1∪E2. For positive integers n, we writenΓto denote the union ofndisjoint copies ofΓ.
Theorem 2.2. SupposeΓ = n1Γ1∪n2Γ2 ∪ · · · ∪ntΓtwithΓi 6∼= Γj fori 6= j. Then, Aut(Γ) = Aut(Γ1)oSn1 ×Aut(Γ2)oSn2 × · · · ×Aut(Γt)oSnt.
LetΓ = (V, E)be a graph, and for eachv ∈ V , let∆v be a graph. Following Sabidussi [18, p. 396], theΓ−join of{∆v}v∈V is the graphΓ[{∆v}v∈V]with
V(Γ[{∆v}v∈V]) = {(x, y)|x∈V &y ∈V(∆x)},
E(Γ[{∆v}v∈V]) = {(x, y)(x0, y0)|xx0 ∈E or x=x0 &yy0 ∈E(∆x)}.
We represent theΓ−join pictorially by a graphΓ whose vertices are labeled by other graphs.
In graph join diagrams, we abbreviateZn =P(Zn). Note thatP(Z2)is an edge andP(Z4)is the complete graph on four vertices.
Theorem 2.3. The following are hold:
(1) InP(T4n), there is an elementaof order2adjacent to every other element.P(T4n)\{e, a} ∼= T ∪nZ2, whereT is the power graph ofZ2nwith the identity an element to order2removed.
(2) The power graph ofSD8nis depicted in 1. In this figure,Ddenotes the power graph ofZ4n
with the identity and elementato order2removed.
(3) The power graph ofU6nis depicted in Figure 2.
(4) For odd integersn,P(V8n)is isomorphic to the nested graph join of Figure 3.
(5) Forn = 2k,P(V8n)is isomorphic to the nested graph join of Figure 4.
(6) Forn = 2kt, where k is positive andt is positive odd integer P(V8n)is isomorphic to the nested graph join of Figure 5.
The identity of a group is adjacent to every vertex in its power graph. If there is no other such vertex, then the identity is fixed by every automorphism andP(G) = P?(G). More generally, ifA is the set of all vertices adjacent to every vertex in a graphΓ, then Aut(Γ)is isomorphic toS|A|× Aut(Γ−A). By [5], the power graph of a group is complete if and only if the group is isomorphic to a cyclic group of prime power order.
Observe that if two elements of a group generate the same cyclic subgroup, then they have the same neighbors in the power graph of the group. In particular, there is an automorphism of the power graph which swaps two such elements while fixing every other vertex of the power graph.
For the reason we often see direct summands of the formSφ(d). Cameron and Ghosh [3] proved that the only finite group whose automorphism group is the same as its power graph is the Klein group of order4.
Suppose Gis a finite group andx ∈ G. Then the degree of xin P(G) can be calculated by deg(x) = |{g ∈G| hxi ≤ hgiorhgi ≤ hxi}|.
Ifn= 1thenP(T4)∼=K4and soAut(K4)∼=S4, and ifn = 2thenAut(P(T8))∼=C2×C2× S4. The other cases are considered in the following theorem:
Theorem 2.4. Supposen≥3is a natural number. Then
Aut(P(T4n))∼=
S2n−2×S2×(S2oSn), nis a power of 2 Y
d|2n
Sφ(d)×(S2oSn), otherwise .
Proof. By Theorem 2.3, the power graph ofT4ncan be constructed from a copy ofP(Z2n)andn copies ofK4 that all of them have a common edgeeasuch thato(a) = 2. Note that verticeseand ahave maximum degrees between vertices ofP(T4n). On the other hand,
P(T4n)− {e, a} ∼= ˜P(Z2n)∪K2∪ · · · ∪K2
| {z }
n
,
whereP˜(Z2n)∼=P(Z2n)− {e, a}. We now assume thatnis a power of2. Thendeg(e) =deg(a) andP˜(Z2n)is complete graph of order2n−2. Therefore,
Aut(P(T4n)) = Aut( ˜P(Z2n))×Aut(K2)×Aut(K2)oSn∼=S2n−2×S2×S2oSn. Otherwise,deg(e)6=deg(a)and by Theorem 2.1,
Aut( ˜P(Z2n))∼=Sφ(2n)× Y
1,2,2n6=d|2n
Sφ(d) ∼=Y
d|2n
Sφ(d).
Therefore,
Aut(P(T4n)) =Aut( ˜P(Z2n))×Aut(K2)oSn ∼=Y
d|2n
Sφ(d)×(S2oSn),
which completes the proof.
It is easy to see that ifn= 1thenAut(P(SD8))∼=C2×(S2oS2). The other cases, are studied in the following theorem:
Theorem 2.5. Supposen≥2is a natural number. Then,
Aut(P(SD8n))∼=
S4n−2×S2n×(S2oSn), n is a power of 2
Y
d|4n
Sφ(d)×S2n×(S2 oSn), otherwise .
Proof. By Theorem 2.3, the power graph ofSD8n is a union of P(Z4n), n copies of P(Z4)that share an edge and 2n copies of P(Z2), all of them are connected to each other in the identity element ofSD8n, Figure 1. Supposen≥2andP˜(Z4n)∼=P(Z4n)− {e, a}, whereo(a) = 2. It is clear that the identity has maximum degree inP˜(Z4n)and soAut(P∗(SD8n)) = Aut(P(SD8n)).
On the other hand,
P∗(SD8n)− {a}= ˜P(Z4n)∪K2∪ · · · ∪K2
| {z }
n
∪K1 ∪ · · · ∪K1
| {z }
2n
.
We now assume thatnis a power of 2. ThenP˜(Z4n)∼=K4n−2 andAut( ˜P(Z4n))∼=S4n−2, So Aut(P(SD8n))∼=S4n−2×S2n×(S2oSn)
Otherwise, by Theorem 2.1,
Aut( ˜P(Z4n))∼=Sφ(4n)× Y
1,2,4n6=d|4n
Sφ(d) ∼=Y
d|4n
Sφ(d).
Therefore,
Aut(P(SD8n)) =Aut( ˜P(Z4n))×S2n×Aut(K2)oSn ∼=Y
d|4n
Sφ(d)×S2×(S2 oSn),
proving our result.
The power graph of the groupU6nis depicted in Figure 2 and in three cases that3-n,3|nand 9-n, and9|n, respectively.
Theorem 2.6. Supposen= 3kt, wherek is a non-negative integer andtis a positive integer such that3-tandk ≥0. Then
Aut(P(U6n))∼=
Y
d|3n
Sφ(d)× Y
d|2n,d-n
Sφ(d)oS3, k= 0
Y
d|2n,d-n
Sφ(d)oS3×Y
d|n
Sφ(d)× Y
d|n,d-t
Sφ(d)oS3, k = 1
Y
d|2n,d-n
Sφ(d)oS3×Y
d|n
Sφ(d)× Y
d|3t,d-t
Sφ(d)oS3× Y
d|n,d-3t
Sφ(d)oS2, k ≥2 .
Proof. Our main proof consider three cases as follows:
1. k = 0. In this case, the power graph can be constructed from a copy ofP(Z3n) and three copies of P(Z2n) all of them share a subgraph isomorphic toP(Zn), Figure 2. Since the identity has maximum degree,Aut(P∗(U6n))∼=Aut(P(U6n)). DefineP(˜ Z3n) =P(Z3n)− P(Zn)andP˜(Z2n) = P(Z2n)− P(Zn). Thus,
Aut(P∗(U6n)) =Aut( ˜P(Z3n))×Aut(P(Zn))×Aut( ˜P(Z2n))oS3,
∼= Y
d|3n,d-n
Sφ(d)×Y
d|n
Sφ(d)× Y
d|2n,d-n
Sφ(d)oS3,
∼=Y
d|3n
Sφ(d)× Y
d|2n,d-n
Sφ(d)oS3.
2. k = 1. By Theorem 2.3, the power graph can by constructed from three copies ofP(Z2n)and four copies ofP(Zn)in such a way that all copies ofP(Z2n)share a subgraph isomorphic to P(Zn)and three of four copies ofP(Zn)share in a subgraph isomorphic toP(Zt), Figure 2.
DefineP˜(Zn) =P(Zn)− P(Zt)andP˜(Z2n) =P(Z2n)− P(Zn). SinceAut(P∗(U6n))∼= Aut(P(U6n)),
Aut(P∗(U6n)) =Aut( ˜P(Z2n))oS3×Aut( ˜P(Zn))×Aut(P(Zt))×Aut( ˜P(Zn))oS3,
∼= Y
d|2n,d-n
Sφ(d)oS3× Y
d|n,d-t
Sφ(d)×Y
d|t
Sφ(d)× Y
d|n,d-t
Sφ(d)oS3,
∼= Y
d|2n,d-n
Sφ(d)oS3×Y
d|n
Sφ(d)× Y
d|n,d-t
Sφ(d)oS3.
3. k ≥ 2. By Theorem 2.3, the power graph can be constructed from three copies ofP(Z2n), three copies of P(Zn), four copies of P(Z3t) and a copy of P(Zt), see Figure 3. Define P˜(Z3t) = P(Z3t)− P(Zt), P˜(Zn) = P(Zn)− P(Z3t) andP˜(Z2n) = P(Z2n)− P(Zn).
Therefore,
Aut(P∗(U6n)) = Aut( ˜P(Z2n))oS3×Aut( ˜P(Zn))×Aut( ˜P(Z3t))
×Aut(P(Zt))×Aut( ˜P(Z3t))oS3×Aut( ˜P(Zn))oS2,
∼= Y
d|2n,d-n
Sφ(d)oS3 × Y
d|n,d-3t
Sφ(d)× Y
d|3t,d-t
Sφ(d)×Y
d|t
Sφ(d)
× Y
d|3t,d-t
Sφ(d)oS3× Y
d|n,d-3t
Sφ(d)oS2,
∼= Y
d|2n,d-n
Sφ(d)oS3 ×Y
d|n
Sφ(d)× Y
d|3t,d-t
Sφ(d)oS3× Y
d|n,d-3t
Sφ(d)oS2.
This completes the proof.
Supposen = 2kt, wherek is a non-negative andtis an odd positive integer. The power graph of the group V8n is depicted in Figures 3, 4 and 5, in three cases thatk = 0, (t = 1, k ≥ 1) and t > 1, k ≥ 1, respectively. It can easily be seen that if n = 1then Aut(P(V8)) ∼= S4 ×S3. The other cases will be studies in the following theorem:
Theorem 2.7. Supposen= 2kt, wherekis a non-negative andtis an odd positive integer. Then
Aut(P(V8n))∼=
S2n×S2oSn× Y
d|2n,d-n
Sφ(d)oS2×Y
d|2n
Sφ(d) k= 0
S2n+1×S2oSn×
k−1
Y
l=1
S22l×S2k oS2 t= 1, k≥1
S2n×S2oSn×Y
d|t
Sφ(d)4 ×
k
Y
s=2
Y
d|2st d-2s−1t
Sφ(d)2 × Y
d|2k+1t d-2kt
Sφ(d)oS2 t >1, k≥1 .
Proof. Supposen ≥ 3is odd. From Figure 3, one can see that the identity has maximum degree.
This shows thatAut(P∗(V8n)) =Aut(P(V8n)). On the other hand,P∗(V8n) =F∗∪K1∪ · · · ∪K1
| {z }
2n
and so Aut(P∗(V8n)) ∼= S2n × Aut(F∗). Consider the element a of order 2 in F and F∗∗ = F∗ − {a}. Since a has maximum degree, Aut(F∗∗) ∼= Aut(F∗). Define P˜(Z2n) = P(Z2n)− P(Zn)− P(Z2). We have:
Aut(F∗∗)∼=S2oSn×Aut( ˜P(Z2n))oS2×Aut( ˜P(Z2n))×Aut(P(Zn))
∼=S2oSn× Y
d|2n,d-n
Sφ(d)oS2× Y
d|2n,d-n
Sφ(d)×Y
d|n
Sφ(d)
∼=S2oSn× Y
d|2n,d-n
Sφ(d)oS2×Y
d|2n
Sφ(d).
This completes the proof of part(1).
For part(2), we first assume thatk = 1. ThenAut(P(V16))∼=S5×(S2oS2)oS2, as desired. If k ≥2then by Figure 4,Aut(P∗(V8n))∼=Aut(P(V8n))andP∗(V8n)) = K1∪ · · · ∪K1
| {z }
2n+1
∪R∗∪S∗ which implies thatP∗(V8n))∼=S2n+1×Aut(R∗)×Aut(S∗). On the other hand,Aut(S∗)∼=S2oSn. So, it is enough to calculateAut(R∗). To do this, we defineP(˜Z2l) =P(Z2l)− P(Z2l−1),2≤l ≤ k+ 1. Then
Aut(R∗) =
k
Y
l=2
(Aut(P(˜Z2l)))2 ×Aut(P(Z˜2k+1))oS2,
∼=
k
Y
l=2
S22l−1 ×S2k oS2,
∼=
k−1
Y
l=1
S22l×S2k oS2.
This proved the part(2).
To prove (3), we note that by Figure 5, one can see that Aut(P∗(V8n)) = Aut(P(V8n))and P∗(V8n)∼=K1∪ · · · ∪K1
| {z }
2n
∪J∗. HenceAut(P∗(V8n))∼=S2n×Aut(J∗). Again we assume thata is an element of order2inJ. It is clear that ahas maximum degree inJ. IfJ∗∗= J∗− {a}then Aut(J∗∗)∼=Aut(J∗). DefineP˜(Z2st) = P(Z2st)− P(Z2s−1t),1≤s≤k+ 1. Then
Aut(J∗∗)∼=S2oSn×Aut(P(Zt))×Aut(P(˜Z2t))3
×
k
Y
s=2
Aut(P(Z˜2st))2×Aut(P(Z˜2k+1t))oS2.
Sincetis odd,P(Zt) =P(˜Z2t)and soAut(P(Zt))∼=Aut(P(˜Z2t)). Therefore,
Aut(J∗∗)∼=S2oSn×Aut(P(Zt))4×
k
Y
s=2
Aut(P(Z˜2st))2×Aut(P(Z˜2k+1t))oS2
∼=S2oSn×Y
d|t
Sφ(d)4 ×
k
Y
s=2
Y
d|2st d-2s−1t
Sφ(d)2 × Y
d|2k+1t d-2kt
Sφ(d)oS2.
This completes the proof.
Figure 1. The power graph of the semi−dihedral groupSD8n.
Figure 2. The power graph ofU6n.
3. Concluding Remarks
In this paper the automorphism group of the power graphs of some families of finite groups are computed. The present authors [14] asked the following question:
Figure 3. The graphP(V8n), wherenis odd.
Figure 4. The graphP(V8n), wheren= 2k.
Figure 5. The graphP(V8n), wheren= 2kt,kis positive andtis positive odd integer.
Figure 6. The graphP(M11).
Figure 7. The graphP(J1).
Question. What is the automorphism group ofP(G), whereGis a sporadic group?
In this section, the automorphism group of P(M11) and P(J1) are computed. We start this section by computing the automorphism group of the first Mathieu group.
Proposition 3.1. The automorphism group of the power graph of the first Matheiu groupM11 is isomorphic to
(S10oS144)×(S4oS396)×(S2oS55)× (S6oS3)×(S2oS4)×S2 oS165.
Proof. In [14] it is proved that the power graph ofP(M11)has exactly 7920vertices. This graph can be constructed from 165 copies of a graph L, Figure 6, 55copies of P(Z3), 396 copies of P(Z5)and144copies ofP(Z11), all connected to each other in the identity group ofM11.
On the other hand, it is clear thatAut(P∗(M11))∼=Aut(P(M11))and so P∗(M11)∼= 165L∗∪144P∗(Z11)∪396P∗(Z5)∪55P∗(Z3).
Therefore,
Aut(P∗(M11))∼=Aut(L∗)oS165×S10oS144×S4oS396×S2oS55.
To complete the proof, we have to compute Aut(L∗). Suppose a is an element of order 2in L∗ andL∗∗ =L∗− {a}. Since this element has maximum degree inL∗,Aut(L∗)∼=Aut(L∗∗).Thus, Aut(L∗∗)∼=S6oS3×S2oS4×S2, which completes the proof.
Proposition 3.2. The automorphism group of the power graph of the first Janko group J1 is iso- morphic to
(S10oS1596)×(S6oS4180)×(S18oS1540)× S2×S8×S4×(S4oS3)×(S2oS5) oS2
oS1463.
Proof. In [14] it is proved that the Janko groupJ1has exactly175560elements and its power graph is a union of1463 copies of a graphK, Figure 7, 1540copies ofP(Z19), 1596copies ofP(Z11) and 4180 copies of P(Z7), all connected to each other in the identity group of J1. Obviously, Aut(P∗(J1))∼=Aut(P(J1))and
P∗(J1)∼= 1463K∗∪1596P∗(Z11)∪4180P∗(Z7)∪1540P∗(Z19).
Therefore,
Aut(P∗(J1))∼=Aut(K∗)oS1463×S10oS1596×S6oS4180×S18oS1540.
Supposeais an element of order2inK∗ andK∗∗ =K∗ − {a}. Sinceahas maximum degree in K∗,Aut(K∗)∼=Aut(K∗∗).Thus,
Aut(K∗∗)∼= (S2×S8 ×S4×(S4oS3)×(S2oS5) oS2. This completes the proof.
The following general problem is most important and worth considering.
Problem 1. Investigate properties of the diameters of the power graphs.
SupposeF is the set of all groups which can be constructed from symmetric group by direct or wreath product. Our calculations with small group theory of GAP [19] suggests the following conjecture.
Conjecture 1. The automorphism group of the power graph of each group is a member ofF.
Acknowledgement
We are very thankful from the referee for his/her suggestions and helpful remarks. The research of the authors is partially supported by the University of Kashan under grant number 159020/26.
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