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Classifying pentavalent symmetric graphs of order 12pq

  • Xiaorui Qian , Bo Ling EMAIL logo , Jinlong Yang and Yun Zhao
Published/Copyright: December 14, 2024

Abstract

A graph is said to be symmetric if its automorphism group is transitive on its arcs. Guo et al. (Pentavalent symmetric graphs of order 12p, Electron. J. Combin. 18 (2011), no. 1, #P233, DOI: https://doi.org/10.37236/720) and Ling (Classifying pentavalent symmetric graphs of order 24p, Bull. Iranian Math. Soc. 43 (2017), no. 6, 1855–1866) Li and Ling (Symmetric graphs and interconnection networks, Future Gener. Comput. Syst. 83 (2018), no. 1, 461–467, DOI: https://doi.org/10.1016/j.future.2017.05.016) determined all pentavalent symmetric graphs of order 12 p , 24 p , and 36 p . In this article, we shall generalize these results by determining all connected pentavalent symmetric graphs of order 12 p q , where p > q are distinct primes.

MSC 2010: 05C25; 05E18

1 Introduction

Throughout this article, all graphs are assumed to be finite, simple, connected, and undirected.

Let Γ be a graph. We denote by V Γ , E Γ , A Γ , and Aut Γ its vertex set, edge set, arc set, and full automorphism group, respectively. Note that an arc is an ordered edge, that is an ordered pair of adjacent vertices. For u , v V Γ , { u , v } denotes the edge incident to u and v in Γ . The valent of a vertex v in a graph Γ , denoted by d ( v ) , is the number of edges of Γ incident with v , each loop counting as two edges. In particular, if Γ is a simple graph, d ( v ) is the number of neighbours of v in Γ . We denote by δ ( Γ ) and Δ ( Γ ) the minimum and maximum valences of the vertices of Γ . A graph Γ is k -regular if δ ( Γ ) = Δ ( Γ ) = k for all v V . In particular, 5-regular graphs, also referred to as pentavalent graphs, play a prominent role in graph theory. We say Γ is vertex-transitive graph if Aut Γ is transitive on V Γ and Γ is arc-transitive graph or symmetric graph if Aut Γ is transitive on A Γ . Let s be a positive integer. An s-arc in a graph Γ is an ( s + 1 ) -tuple ( v 0 , v 1 , , v s ) of s + 1 vertices such that ( v i 1 , v i ) A Γ for 1 i s and v i 1 v i + 1 for 1 i s 1 . Let G be a subgroup of Aut Γ . We say Γ is ( G , s ) -arc-transitive if G is transitive on the s -arcs of Γ and Γ is ( G , s ) -transitive if it is ( G , s ) -arc-transitive but not ( G , s + 1 ) -arc-transitive. In the case where G = Aut Γ , we say an ( G , s ) -arc-transitive or ( G , s ) -transitive graph is an s-arc-transitive or s-transitive graph. In particular, 0-arc-transitive means vertex-transitive, and 1-arc-transitive means arc-transitive or symmetric.

Classifying symmetric graphs has a long history, beginning with a seminal work by Tutte [1,2] on the cubic case. Later on, lots of work was done on the classification of symmetric graphs and a series of results have been obtained. For example, all the possibilities of vertex stabilizers of pentavalent symmetric graphs are determined in [3,4]. Following this, by using deep group theory, the symmetric graphs of order p , 2 p , 3 p , and p q were classified in [58], where p and q are two prime numbers. Furthermore, classifications of arc-transitive pentavalent graphs of order 18 p , 12 p , 18 p , 24 p , 36 p , 2 p q , 4 p q are presented in [915]. Following these, Li gave a characterization of symmetric graphs of odd order or square-free order in [16,17]. Ling et al. gave a classification of pentavalent symmetric graphs of order four times an odd square-free integer in [18]. In this article, we classify pentavalent symmetric graphs of order 12 p q with p , q being distinct primes. The main result of this article is the following theorem.

Theorem 1.1

Let Γ be a pentavalent symmetric graph of order 12 p q , where p > q and p , q are distinct primes. Then one of the following statements holds.

  1. Aut Γ PSL ( 2 , p ) , PGL ( 2 , p ) , PSL ( 2 , p ) × Z 2 , or PGL ( 2 , p ) × Z 2 with p = 59 or 61.

  2. Γ is isomorphic to one of the graphs in Table 1.

The information in Table 1 is determined with the help of the Magma system [19].

Table 1

Pentavalent symmetric graphs of order 12 p q

Γ Aut Γ ( q , p )
C 72 1 PGL ( 2 , 9 ) ( 2 , 3 )
C 72 2 Aut ( A 6 ) × Z 2 ( 2 , 3 )
C 120 1 A 5 × D 10 × Z 2 ( 2 , 5 )
C 120 2 S 5 × D 10 ( 2 , 5 )
C 252 1 A 9 × Z 2 ( 3 , 7 )
C 252 2 S 9 × Z 2 ( 3 , 7 )
C 264 i PGL ( 2 , 11 ) × Z 2 ( 2 , 11 ) , i = 1 , 2 , 4
C 264 3 PSL ( 2 , 11 ) : D 8 ( 2 , 11 )
C 396 M 11 ( 3 , 11 )
C 408 1 PSO ( 4 , 4 ) ( 2 , 17 )
C 408 2 PSL ( 2 , 16 ) ( 2 , 17 )
C 684 i PSL ( 2 , 19 ) × Z 2 ( 3 , 19 ) , i = 1 , 2 , 3
C 684 4 PGL ( 2 , 19 ) × Z 2 ( 3 , 19 )
C 684 i PGL ( 2 , 19 ) ( 3 , 19 ) , 5 i 10
C 1428 PSL ( 2 , 71 ) × Z 2 ( 7 , 71 )
C 2436 i PSL ( 2 , 29 ) × Z 2 ( 7 , 29 ) , 1 i 7
C 2436 8 PGL ( 2 , 29 ) × Z 2 ( 7 , 29 )
C 2436 i PGL ( 2 , 29 ) ( 7 , 29 ) , 9 i 25
C 3444 i PSL ( 2 , 41 ) × Z 2 ( 7 , 41 ) , 1 i 3
C 3444 i PSL ( 2 , 41 ) ( 7 , 41 ) , 4 i 6
C 11748 PSL ( 2 , 89 ) × Z 2 ( 11 , 89 )
C 12324 PSL ( 2 , 79 ) ( 13 , 79 )

Remark 1.2

(Remarks on Theorem 1.1)

  1. The graph C 72 1 , C 72 2 , C 252 1 , C 252 2 , C 396 , and C 684 i with 1 i 10 are introduced in [13]; the graph C 120 1 , C 120 2 , C 264 i with i = 1 , 2 , 3 , 4 , C 408 1 or C 408 2 are introduced in [12].

  2. For convenience, other graphs in Table 1 are introduced in Example 2.11.

2 Preliminary results

We give some necessary preliminary results in this section.

Let Γ be a G -vertex-transitive graph with G Aut Γ . Let N be an intransitive normal subgroup of G on V Γ . Denote by V N the set of N -orbits in V Γ . The normal quotient graph Γ N is the graph with vertex set V N and two N -orbits B , C V N are adjacent in Γ N if and only if some vertex of B is adjacent in Γ to some vertex of C . The following Lemma ([20, Lemma 2.5]) provides a basic reduction method for studying our pentavalent symmetric graphs.

Lemma 2.1

Let Γ be an G-arc-transitive graph of prime valency p > 2 , where G Aut Γ , and let N G have at least three orbits on V Γ . Then, the following statements hold.

  1. N is semiregular on V Γ , G N Aut Γ N , and Γ N is an G N -arc-transitive graph of valency p;

  2. Γ is ( G , s ) -transitive if and only if Γ N is ( G N , s ) -transitive, where 1 s 5 or s = 7 .

The following lemma determines the vertex stabilizers of connected pentavalent symmetric graphs, refer to [3,14].

Lemma 2.2

Let Γ be a pentavalent ( G , s ) -transitive graph for some G Aut Γ and s 1 . Let α V Γ . Then, one of the following holds, where D 10 , D 20 , and F 20 denotes the dihedral groups of order 10, 20, and the Frobenius group of order 20, respectively.

(1) If G α is soluble, then G α 80 and s 3 . Moreover, ( G α , s ) lies in the following table:

s 1 2 3
G α Z 5 , D 10 , D 20 F 20 , F 20 × Z 2 F 20 × Z 4
G α 5,10,20 20,40 80

(2) If G α is insoluble, then G α 2 9 3 2 5 and 2 s 5 . Furthermore, ( G α , s ) lies in the following table:

s 2 3 4 5
G α A 5 , S 5 A 4 × A 5 , ( A 4 × A 5 ) : Z 2 , ASL ( 2 , 4 ) , AGL ( 2 , 4 ) , Z 2 6 : Γ L ( 2 , 4 )
S 4 × S 5 A Σ L ( 2 , 4 ) , A Γ L ( 2 , 4 )
G α 60, 120 720, 1440, 2880 960, 1920, 2880, 5760 23040

For reduction, we need some information about the connected pentavalent symmetric graphs of order 12 p , 18 p , 24 p , 36 p , n , and 4 n are given in the following lemmas, where p is a prime and n is an odd square-free integer.

By [9], we can describe pentavalent symmetric graphs of order 12 p in the following lemma. From [9], it is easy to see that there is a unique connected pentavalent symmetric graph of order 24, that is, the standard double cover I 12 ( 2 ) of the Icosahedron graph I 12 .

Lemma 2.3

Let Γ be a pentavalent symmetric graph. Let p be a prime. If V Γ = 12 p , then Γ is isomorphic to one of the graphs in Table 2.

Table 2

Pentavalent symmetric graphs of order 12 p

Γ Aut Γ Remark
I 12 ( 2 ) A 5 × Z 2 2 p = 2
C 36 Aut ( A 6 ) p = 3
C 60 A 5 D 10 p = 5
C 66 ( 2 ) PGL ( 2 , 11 ) × Z 2 p = 11
C 132 1 PSL ( 2 , 11 ) × Z 2 p = 11
C 132 i PGL ( 2 , 11 ) p = 11 , 2 i 5

From [13], we give some information about pentavalent symmetric graphs of order 18 p , in the following lemma.

Lemma 2.4

Let Γ be a pentavalent symmetric graph. Let p be a prime. If V Γ = 18 p , then Γ is isomorphic to one of the graphs in Table 3.

Table 3

Pentavalent symmetric graphs of order 18 p

Γ Aut Γ Remark
O 4 S 9 p = 7
C 342 1 PSL ( 2 , 19 ) p = 19
C 342 2 PGL ( 2 , 19 ) p = 19

From [12], pentavalent symmetric graphs of order 24 p are determined in the following lemma.

Lemma 2.5

Let Γ be a pentavalent symmetric graph of order 24 p , where p is a prime. Then, p = 2 , 3, 5, 11, or 17. Furthermore, Aut Γ , ( Aut Γ ) α , and Γ are described in Table 4, where α V Γ .

Table 4

Pentavalent symmetric graphs of order 24 p

Γ Aut Γ ( Aut Γ ) α Girth Diameter Bipartite? Cayley?
C 48 SL ( 2 , 5 ) : D 8 F 20 6 4 Yes Yes
C 72 1 PGL ( 2 , 9 ) D 10 4 4 No Yes
C 72 2 Aut ( A 6 ) × Z 2 F 20 × Z 2 6 5 Yes No
C 120 1 A 5 × D 10 × Z 2 D 10 6 6 Yes Yes
C 120 2 S 5 × D 10 D 10 4 6 Yes Yes
C 264 1 PGL ( 2 , 11 ) × Z 2 D 10 4 7 Yes No
C 264 2 PGL ( 2 , 11 ) × Z 2 D 10 6 6 Yes No
C 264 3 PSL ( 2 , 11 ) : D 8 D 20 6 6 Yes No
C 264 4 PGL ( 2 , 11 ) × Z 2 D 10 4 7 Yes No
C 408 1 PSO ( 4 , 4 ) D 20 6 6 No No
C 408 2 PSL ( 2 , 16 ) D 10 8 5 No No

From [13], we give some information about pentavalent symmetric graphs of order 36 p , in the following lemma.

Lemma 2.6

Let Γ be a pentavalent symmetric graph. Let p be a prime. If V Γ = 36 p , then p = 2 , 3, 7, 11, or 19. Then, Γ is isomorphic to one of the graphs in Table 5.

Table 5

Pentavalent symmetric graphs of order 36 p

Γ Aut Γ Remark
C 72 1 PGL ( 2 , 9 ) p = 2
C 72 2 Aut ( A 6 ) × Z 2 p = 2
C 108 ( ( Z 3 . A 6 ) : Z 2 ) : Z 2 p = 3
C 252 1 A 9 × Z 2 p = 7
C 252 2 S 9 × Z 2 p = 7
C 396 M 11 p = 11
C 684 i PSL ( 2 , 19 ) × Z 2 ( 3 , 19 ) , i = 1 , 2 , 3
C 684 4 PGL ( 2 , 19 ) × Z 2 (3, 19)
C 684 i PGL ( 2 , 19 ) (3, 19), 5 i 10

From [21], pentavalent symmetric graphs of order n are determined in the following lemma, where n is a square-free integer and has at least four prime factors.

Lemma 2.7

Let Γ be a connected arc-transitive pentavalent graph of order n, where n is a square-free integer and has at least four prime factors. Then, one of the following statements holds.

  1. Γ C D n , k (as in Example 2.11), n = 2 5 t p 1 p 2 p s and Aut Γ D n : Z 5 , where 0 t 1 , s 2 , and p 1 , p 2 , , p s are distinct primes such that 5 p i 1 for i = 1 , 2 , , s . Up to isomorphism, there are exactly 4 s 1 such graphs of order n .

  2. Aut Γ PSL ( 2 , r ) or PGL ( 2 , r ) with r 29 being a prime.

  3. The triple ( Γ , n , Aut Γ ) lies in Table 6.

Table 6

Pentavalent symmetric graphs of order n

Γ Aut Γ Remark ( Aut Γ ) α
C 390 PSL ( 2 , 25 ) n = 390 F 20
C 2926 J 1 n = 2926 A 5

From [18], pentavalent symmetric graphs of order 4 n are determined in the following lemma, where n is an odd square-free integer.

Lemma 2.8

Let Γ be a connected arc-transitive pentavalent graph of order 4 n , where n is an odd square-free integer. If n has at least three prime factors. Then, one of the following statements holds:

  1. A PSL ( 2 , r ) × Z 2 , PGL ( 2 , r ) × Z 2 , PSL ( 2 , r ) , or PGL ( 2 , r ) , with r 29 being a prime.

  2. The triple ( Γ , n , Aut Γ ) lies in Table 7.

Table 7

Pentavalent symmetric graphs of order four times an odd square-free integer

Γ Aut Γ n ( Aut Γ ) α
C 17556 i , 1 i 5 J 1 3 7 11 19 D 10
C 5852 J 1 × Z 2 7 11 19 A 5
C 780 i , 1 i 3 PSL ( 2 , 25 ) × Z 2 3 5 13 F 20

In order to construct pentavalent symmetric graphs, we need to introduce the so-called coset graph. Let G be a finite group and H G . Suppose D is a union of some double cosets of H in G . The coset graph Cos ( G , H , D ) of G with respect to H and D is defined to have vertex set [ G : H ] , the set of right cosets of H in G , and the edge set { { H g , H d g } g G , d D } . The graph Cos ( G , H , D ) has valency D H and is connected if and only if D generates group G . Clearly, the graph Cos ( G , H , D ) is undirected if and only if D = D 1 = { d 1 d D } and Cos ( G , H , D ) Cos ( G , H α , D α ) for each α Aut ( G ) . Let S be a generator subset of G with 1 S and S = S 1 . Clearly, the coset graph Cos ( G , 1 , S ) is a connected undirected simple graph, which is called the Cayley graph on G with respect to S and denoted by Cay ( G , S ) . The following lemma is well known, see [22].

Lemma 2.9

Let Γ be a graph and let G be a vertex-transitive subgroup of Aut Γ . Then, Γ is isomorphic to a coset graph Cos ( G , H , D ) , where H = G α is the stabilizer of α V Γ in G and D consists of all elements of G , which map V to one of its neighbors. Further,

  1. Γ is connected if and only if D generates group G;

  2. Γ is G -arc-transitive if and only if D is a single double coset. In particular, if g G interchanges V and one of its neighbors, then g 2 H and D = H g H ;

  3. the valency of Γ equal to D H = H : H H g .

The maximal subgroups of PSL ( 2 , q ) are known, see [23].

Lemma 2.10

Let N = PSL ( 2 , q ) , where q = p n 5 with p being a prime. Then, a maximal subgroup of N is isomorphic to one of the following groups, where d = ( 2 , q 1 ) .

  1. D 2 ( q 1 ) d , where q 5 , 7 , 9 , 11 ;

  2. D 2 ( q + 1 ) d , where q 7 , 9 ;

  3. Z q : Z ( q 1 ) d ;

  4. A 4 , where q = p = 5 or q = p 3 , 13 , 27 , 37 (mod 40);

  5. S 4 , where q = p ± 1 (mod 8);

  6. A 5 , where q = p ± 1 (mod 5), or q = p 2 1 (mod 5) with p an odd prime;

  7. PSL ( 2 , p m ) with n m being an odd integer;

  8. PGL ( 2 , p n 2 ) with n being an even integer.

By using the Magma program [19], we have the following example.

Example 2.11

  1. There is a unique connected pentavalent graph of order 1,428, which admits PSL ( 2 , 71 ) × Z 2 as an arc-transitive automorphism group. This graph is denoted by C 1428 . Similarly, we can define the graphs C 11748 and C 12324 , which satisfies the conditions in Table 1.

  2. The graphs C 2436 i ( 1 i 25 ) are pairwise non-isomorphic connected pentavalent symmetric graphs of order 2436 with Aut ( C 2436 i ) PSL ( 2 , 29 ) × Z 2 for 1 i 7 , Aut ( C 2436 8 ) PGL ( 2 , 29 ) × Z 2 and Aut ( C 2436 i ) PGL ( 2 , 29 ) for 9 i 25 , which satisfies the conditions in Table 1.

  3. The graphs C 3444 i ( 1 i 6 ) are pairwise non-isomorphic connected pentavalent symmetric graphs of order 3444 with Aut ( C 3444 i ) PSL ( 2 , 41 ) × Z 2 for 1 i 3 and Aut ( C 3444 i ) PSL ( 2 , 41 ) for 4 i 6 , which satisfies the conditions in Table 1.

  4. Let G = a , b a m = b 2 = 1 , a b = a 1 be a dihedral group, and let k be a solution of the equation

    x 4 + x 3 + x 2 + x + 1 0 ( mod m ) .

    Set

    S = { b , a b , a k + 1 b , a k 2 + k + 1 b , a k 3 + k 2 + k + 1 b } , C D 2 m , k = Cay ( G , S ) .

    The first two letters “ C D ” of the name of the graph C D 2 m , k stand for “Cayley graph of a dihedral group.”

  5. The graphs C 1218 i ( 1 i 3 ) are pairwise non-isomorphic connected pentavalent symmetric graphs of order 1218 with Aut ( C 1218 1 ) PGL ( 2 , 29 ) and Aut ( C 1218 i ) PSL ( 2 , 29 ) for i = 2 , 3 , which satisfies the conditions in Table 8.

  6. There is a unique connected pentavalent graph of order 1722 which admits PSL ( 2 , 41 ) as an arc-transitive automorphism group. This graph is denoted by C 1722 . Similarly, we can define the graphs C 2982 and C 5874 , which satisfies the conditions in Table 8.

  7. The graphs C 10266 i ( 1 i 7 ) are pairwise non-isomorphic connected pentavalent symmetric graphs of order 10266 with Aut ( C 10266 1 ) PGL ( 2 , 59 ) and Aut ( C 10266 i ) PSL ( 2 , 59 ) for 2 i 7 , which satisfies the conditions in Table 8.

  8. The graphs C 11346 i ( 1 i 7 ) are pairwise non-isomorphic connected pentavalent symmetric graphs of order 11346 with Aut ( C 11346 1 ) PGL ( 2 , 61 ) and Aut ( C 11346 i ) PSL ( 2 , 61 ) for 2 i 7 , which satisfies the conditions in Table 8.

Table 8

Pentavalent symmetric graphs of order 6 p q

Γ Aut Γ ( q , p )
C 36 Aut ( A 6 ) (2,3)
C 60 A 5 D 10 (2,5)
C 66 ( 2 ) PGL ( 2 , 11 ) × Z 2 (2,11)
C 132 1 PSL ( 2 , 11 ) × Z 2 (2,11)
C 132 i PGL ( 2 , 11 ) (2,11), 2 i 5
O 4 S 9 (3,7)
C 342 1 PSL ( 2 , 19 ) (3,19)
C 342 2 PGL ( 2 , 19 ) (3,19)
C 390 PSL ( 2 , 25 ) (5,13)
CD 6 p q , k D 6 p q : Z 5 ( 5 p 1 , 5 q 1 )
C 1218 1 PGL ( 2 , 29 ) (7,29)
C 1218 i PSL ( 2 , 29 ) (7,29), i = 2 , 3
C 1722 PSL ( 2 , 41 ) (7,41)
C 2982 PSL ( 2 , 71 ) (7,71)
C 5874 PSL ( 2 , 89 ) (11,89)
C 10266 1 PGL ( 2 , 59 ) (29,59)
C 10266 i PSL ( 2 , 59 ) (29,59), 2 i 7
C 11346 1 PGL ( 2 , 61 ) (31,61)
C 11346 i PSL ( 2 , 61 ) (31,61), 2 i 7

3 The proof of Theorem 1.1

In this section, we will prove Theorem 1.1 through a series of lemmas. To prove Theorem 1.1, we need information of pentavalent symmetric graphs of order 6 p q . Therefore, we first prove the following lemma.

Lemma 3.1

Let Γ be a pentavalent symmetric graph of order 6 p q , where p > q are distinct primes. Then, Γ is isomorphic to one of the graphs in Table 8.

Remark 3.2

(Remarks on Lemma 3.1)

  1. The graph C 36 , C 60 , C 66 ( 2 ) , C 132 1 , C 132 i with 2 i 5 are introduced in [9]; the graph O 4 or C 342 i with i = 1 , 2 are introduced in [13]; the graph C 390 is introduced in [24].

  2. For convenience, other graphs in Table 8 are introduced in Example 2.11.

Proof

Let Γ be a pentavalent symmetric graph of order 6 p q , where p > q are distinct primes. Let A = Aut Γ . Take α V Γ . By Lemma 2.2, A α 2 9 3 2 5 . Hence, A 2 10 3 3 5 p q . If q = 2 , then Γ has order 12 p and by Lemma 2.3, we have p { 3 , 5 , 11 } and Γ is isomorphic to C 36 , C 60 , or C 132 i with 1 i 5 . If q = 3 , then Γ has order 18 p and by Lemma 2.4, we have p { 7 , 19 } and Γ is isomorphic to O 4 or C 342 i with i = 1 , 2. If q = 5 , then Γ is isomorphic to C 390 by [24]. Thus, to finish the proof, we next consider the case where p > q > 5 .

We first suppose that A is soluble. Then, by Lemma 2.7, we have A D 6 p q : Z 5 and Γ is isomorphic to CD 6 p q , k , where 5 p 1 , 5 q 1 , as in row 10 of Table 8.

Thus, in what follows, we assume that A is insoluble. By Lemma 2.7, we have A PSL ( 2 , r ) , PGL ( 2 , r ) , or the triple ( Γ , n , Aut Γ ) lies in Table 6. For the latter case, there exists no pentavalent symmetric graph of order 6 p q for p > q > 5 in Table 6. For the former case, let N = PSL ( 2 , r ) be a normal subgroup of A . Assume that N has t orbits on V Γ , t 3 . Then, by Lemma 2.1, N α = 1 and so N 6 p q , which is a contradiction as N PSL ( 2 , r ) . Hence, N α 1 , N has at most two orbits on V Γ and 3 p q divides N : N α . Since Γ is connected, N A , and N α 1 , we have 1 N α Γ ( α ) A α Γ ( α ) , it follows that 5 divides N α , we thus have that 15 p q divides N . Since A : N 2 , we have A α : N α 2 . If A α is insoluble, then N α is also insoluble as A α : N α 2 . It implies that N α is an insoluble subgroup of PSL ( 2 , r ) . As Aut ( PSL ( 2 , r ) ) = PGL ( 2 , r ) has no subgroup isomorphic to A 5 Z 2 , by Lemma 2.10, N α = A α = A 5 . It implies that the order of N divides 360 p q . If A α is soluble, and by Lemma 2.2, A α divides 80, and so N α 80 . It implies that the order of N divides 480 p q .

We claim that r = p . Since V Γ = 6 p q = A A α , and A α = 60 or A α 80 , where p > q > 5 . We have the equation 6 p q = r ( r 1 ) ( r + 1 ) 2 A α or r ( r 1 ) ( r + 1 ) A α . Since r 29 is a prime and A α 2 9 3 2 5 , we have r = p or q . Assume that r = q . Then, 6 p = ( r 1 ) ( r + 1 ) 2 A α or ( r 1 ) ( r + 1 ) A α , it implies r + 1 = p , which is a contradiction as r + 1 is not a prime. Thus, r = p and N = p ( p 1 ) ( p + 1 ) 2 . Note that ( p + 1 2 , p 1 2 ) = 1 , if q p 1 2 , then p + 1 360 or 480, where p 29 and p is a prime. It follows that p = 29 , 31, 47, 59, 71, 79, 89, 179, 239, 359, or 479. If q p + 1 2 , then p 1 360 or 480. It follows that p = 31 , 37, 41, 61, 73, 97, 181, or 241. Note that 15 p q N and N 2 10 3 3 5 p q , where p > q > 5 . Then, p 31 , 37, 47, 73, 97, 181, or 239. Therefore, N is one of the following groups (Table 9):

Assume that p = 29 . Then, N = PSL ( 2 , 29 ) and ( q , p ) = ( 7 , 29 ) . Note that N : N α = 3 p q or 6 p q . By [25], N has no subgroup of index 3 p q , we have N is transitive on V Γ , thus N α = 10 . And by Lemma 2.2, we have N α = D 10 . Thus, by Lemma 2.9, Γ = Cos ( N , N α , N α g N α ) , where g is a two-element in N such that g 2 N α and N α , g = N . Computations performed by the Magma computational software show that there exist three graphs, which we denote as C 1218 i , where 1 i 3 . Furthermore, the information of these graphs is listed in Table 8. Similarly, assume that p = 71 or 89. By computation in Magma, Γ C 2982 or C 5874 in Table 8.

Assume that p = 41 . Then, N = PSL ( 2 , 41 ) and ( q , p ) = ( 7 , 41 ) . If N has two orbits on V Γ , then A = PGL ( 2 , 41 ) , thus A α = 40 and A α = F 20 × Z 2 by Lemma 2.2. This is impossible as PGL ( 2 , 41 ) has no subgroups isomorphic to F 20 × Z 2 . Therefore, N is transitive on V Γ , and by Lemma 2.2, N α = D 20 . Thus, by Lemma 2.9, Γ = Cos ( N , N α , N α g N α ) , where g is a two-element in N such that g 2 N α and N α , g = N . Computations performed by the Magma computational software show that there exists a graph, which we denote as C 1722 in Table 8. Similarly, assume that p = 179 , 359 or 479. Computation in Magma shows that there exists no graph, which is a contradiction.

Assume that p = 59 . Then, N = PSL ( 2 , 59 ) and ( q , p ) = ( 29 , 59 ) . If N has two orbits on V Γ , then A = PGL ( 2 , 59 ) . Hence, A α = 20 and A α = D 20 by Lemma 2.2. Computations performed by the Magma computational software show that there exists a graph, which we denote as C 10266 1 in Table 8. If N is transitive on V Γ and by Lemma 2.2, N α = D 10 . Computation in Magma shows that there exist six graphs, which we denote as C 10266 i in Table 8, with 2 i 7 , i is an integer. Similarly, assume that p = 61 . By computation in Magma, Γ C 11346 i in Table 8, where 1 i 7 .

Assume that p = 79 . Then, N = PSL ( 2 , 79 ) and ( q , p ) = ( 13 , 79 ) . If N has two orbits on V Γ , then A = PGL ( 2 , 79 ) , thus A α = 80 and A α = F 20 × Z 4 by Lemma 2.2. This is impossible as PGL ( 2 , 79 ) has no subgroups isomorphic to F 20 × Z 4 . If N is transitive on V Γ and by Lemma 2.2, N α = 40 . Hence, N α = F 20 × Z 2 by Lemma 2.2. This is impossible as PSL ( 2 , 79 ) has no subgroups isomorphic to F 20 × Z 2 , which is a contradiction.

Assume that p = 179 . Then, N = PSL ( 2 , 179 ) and ( q , p ) = ( 89 , 179 ) . If N is transitive on V Γ , and by Lemma 2.2, N α = 30 , which is a contradiction. If N has two orbits on V Γ , then A = PGL ( 2 , 179 ) , thus A α = 60 and A α = A 5 by Lemma 2.2. Thus, by Lemma 2.9, Γ = Cos ( A , A α , A α g A α ) , where g is a two-element in A such that g 2 A α and A α , g = A . By computation in Magma, there is no such g A , which is a contradiction. Similarly, p 359 or 479.

Thus, we complete the proof of Lemma 3.1.□

Table 9

Some possible simple groups N

N Order N Order
PSL ( 2 , 29 ) 2 2 3 5 7 29 PSL ( 2 , 79 ) 2 4 3 5 13 79
PSL ( 2 , 41 ) 2 3 3 5 7 41 PSL ( 2 , 89 ) 2 3 3 2 5 11 89
PSL ( 2 , 59 ) 2 2 3 5 29 59 PSL ( 2 , 179 ) 2 2 3 2 5 89 179
PSL ( 2 , 61 ) 2 2 3 5 31 61 PSL ( 2 , 359 ) 2 3 3 2 5 179 359
PSL ( 2 , 71 ) 2 3 3 2 5 7 71 PSL ( 2 , 479 ) 2 5 3 5 239 479

Now, let Γ be a pentavalent symmetric graph of order 12 p q , where p > q and p , q are distinct primes. Let A Aut Γ . Take α V Γ . By Lemma 2.2, A α 2 9 3 2 5 , and hence, A 2 11 3 3 5 p q . If q = 2 , then Γ has order 24 p and by Lemma 2.5, we have p { 3 , 5 , 11 , 17 } and Γ is isomorphic to C 72 1 , C 72 2 , C 120 1 , C 120 2 , C 264 i with 1 i 4 , C 408 1 , or C 408 2 in Table 4. If q = 3 , then Γ has order 36 p and by Lemma 2.6, we have p { 7 , 11 , 19 } and Γ is isomorphic to C 252 1 , C 252 2 , C 396 , or C 684 i with 1 i 10 in Table 5. Thus, to finish the proof, we consider the case p > q 5 .

By Lemma 2.8, we have A PSL ( 2 , r ) , PGL ( 2 , r ) , PSL ( 2 , r ) × Z 2 , PGL ( 2 , r ) × Z 2 , or the triple ( Γ , n , Aut Γ ) lies in Table 7, where r 29 and n is an odd square-free integer. We first consider the cases listed in Table 7, we have Γ is isomorphic to C 780 i with 1 i 3 . Now, we consider the case that A PSL ( 2 , r ) , PGL ( 2 , r ) , PSL ( 2 , r ) × Z 2 , PGL ( 2 , r ) × Z 2 , we claim that r = p . Since V Γ = 12 p q = A A α , then we have the equation 12 p q = r ( r 1 ) ( r + 1 ) 2 A α or r ( r 1 ) ( r + 1 ) A α . Since r 29 is a prime and A α 2 9 3 2 5 , we have r = p or q . Assume that r = q . Then, 12 p = ( r 1 ) ( r + 1 ) 2 A α or ( r 1 ) ( r + 1 ) A α , it implies r + 1 = p , which is a contradiction as r + 1 is not a prime. Thus, r = p .

The following lemma deals with the case of A PSL ( 2 , p ) × Z 2 or PGL ( 2 , p ) × Z 2 .

Lemma 3.3

Assume that A PSL ( 2 , p ) × Z 2 or PGL ( 2 , p ) × Z 2 . Then, Γ is isomorphic to one of the graphs in Table 1 or Aut Γ PSL ( 2 , 59 ) × Z 2 , PGL ( 2 , 59 ) × Z 2 , PSL ( 2 , 61 ) × Z 2 , or PGL ( 2 , 61 ) × Z 2 .

Proof

Let N Z 2 is a normal subgroup of A . Then, Γ N is a pentavalent symmetric graph of order 6 p q , and A N Aut Γ N . By Lemma 3.1, we have p {29, 41, 71, 89, 59, or 61}.

Assume that p = 29 . Then, A PSL ( 2 , 29 ) × Z 2 or PGL ( 2 , 29 ) × Z 2 and ( q , p ) = ( 7 , 29 ) . Since A α = A 12 p q , we have A α = 10 or 20. By Magma [19], Γ C 2436 i in Table 1, with 1 i 8 , i is an integer. Similarly, assume that p = 41 , 71 or 89. By computation in Magma, Γ C 3444 i with 1 i 3 , C 1428 , or C 11748 in Table 1.

Assume that p = 59 or 61. The automorphism group of the graph Γ might be isomorphic to one of the following groups: PSL ( 2 , 59 ) × Z 2 , PGL ( 2 , 59 ) × Z 2 , PSL ( 2 , 61 ) × Z 2 , or PGL ( 2 , 61 ) × Z 2 , where ( q , p ) = ( 29 , 59 ) or ( 31 , 61 ) .

Thus, we complete the proof of Lemma 3.3.□

The following lemma deals with the case of A PSL ( 2 , p ) or PGL ( 2 , p ) .

Lemma 3.4

Assume that A PSL ( 2 , p ) or PGL ( 2 , p ) . Then, Γ is isomorphic to one of the graphs in Table 1, or Aut Γ PSL ( 2 , 59 ) , PGL ( 2 , 59 ) , PSL ( 2 , 61 ) , or PGL ( 2 , 61 ) .

Proof

Let N = PSL ( 2 , p ) be a normal subgroup of A . Assume that N has t orbits on V Γ , t 3 . Then, by Lemma 2.1, N α = 1 , and thus, N 12 p q . We conclude that N = 4 p q , 6 p q , or 12 p q because N is insoluble. Noting that p > q > 5 , it is impossible for N = 4 p q by [26]. And thus, N has two orbits or a orbit on V Γ , which is a contradiction. Hence, N has at most two orbits on V Γ , then 6 p q N . Since V Γ is connected, N A and N α 1 , we have 1 N α Γ ( α ) A α Γ ( α ) , it follows that 5 divides N α , we thus have that 30 p q N . And, N 2 11 3 3 5 p q as N A . Since A : N 2 , we have A α : N α 2 . If A α is insoluble, then N α is also insoluble as A α : N α 2 . By Lemma 2.10, N α = A 5 . It implies that the order of N divides 720 p q . If A α is soluble, and by Lemma 2.2, A α divides 80, and so N α 80 . It implies that the order of N divides 960 p q .

Note that N = p ( p 1 ) ( p + 1 ) 2 and ( p + 1 2 , p 1 2 ) = 1 . If q p 1 2 , then p + 1 720 or 960, where p 29 and p is a prime. It follows that p = 29 , 31, 47, 59, 71, 79, 89, 179, 191, 239, 359, 479, or 719. If q p + 1 2 , then p 1 720 or 960. It follows that p = 31 , 37, 41, 61, 73, 97, 181, 193, or 241. Note that 30 p q N and N 2 11 3 3 5 p q , where p > q > 5 . Then, p 31 , 37, 47, 73, 97, 181, 193, 239, or 241. Therefore, N is one of the following groups (Table 10).

Assume that p = 29 . Then, N = PSL ( 2 , 29 ) and ( q , p ) = ( 7 , 29 ) . If N has two orbits on V Γ , then A = PGL ( 2 , 29 ) and A α = D 10 by Lemma 2.2. Thus, by Lemma 2.9, Γ = Cos ( A , A α , A α g A α ) , where g is a two-element in A such that g 2 A α and A α , g = A . Computation in Magma shows that there exist ten graphs, which we denote as C 2436 i , with 9 i 18 . If N is transitive on V Γ and by Lemma 2.2, N α = Z 5 . Computation in Magma shows that there exist six graphs, there exist seven graphs, which we denote as C 2436 i , with 19 i 25 , in Table 1. Similarly, assume that p = 41 or 79. By computation in Magma, Γ C 3444 i or C 12324 in Table 1, where 4 i 6 .

Assume that p = 71 . Then, N = PSL ( 2 , 71 ) and ( q , p ) = ( 7 , 71 ) . In [25], N has no subgroup of index 12 p q , whereas in this study N has two orbits on V Γ . Thus, A = PGL ( 2 , 41 ) and A α = 60 . Computations performed by the Magma computational software show that no such graph exists. Similarly, q 89 , 179, 191, 359, 479, or 719.

Assume that p = 59 or 61. Then, N = PSL ( 2 , 59 ) or PSL ( 2 , 61 ) and ( q , p ) = ( 29 , 59 ) or ( 31 , 61 ) . The automorphism group of the graph Γ might be isomorphic to one of the following groups: PSL ( 2 , 59 ) , PGL ( 2 , 59 ) , PSL ( 2 , 61 ) , or PGL ( 2 , 61 ) .

Thus, we complete the proof of Lemma 3.4.□

Table 10

Some possible simple groups N

N Order N Order
PSL ( 2 , 29 ) 2 2 3 5 7 29 PSL ( 2 , 89 ) 2 3 3 2 5 11 89
PSL ( 2 , 41 ) 2 3 3 5 7 41 PSL ( 2 , 179 ) 2 2 3 2 5 89 179
PSL ( 2 , 59 ) 2 2 3 5 29 59 PSL ( 2 , 191 ) 2 6 3 5 19 191
PSL ( 2 , 61 ) 2 2 3 5 31 61 PSL ( 2 , 359 ) 2 3 3 2 5 179 359
PSL ( 2 , 71 ) 2 3 3 2 5 7 71 PSL ( 2 , 479 ) 2 5 3 5 239 479
PSL ( 2 , 79 ) 2 4 3 5 13 79 PSL ( 2 , 719 ) 2 4 3 2 5 359 719

By combining Lemmas 3.1, 3.3, and 3.4, we have completed the proof of Theorem 1.1.

Acknowledgement

The authors are grateful to the constructive comments of references.

  1. Funding information: This work was partially supported by the National Natural Science Foundation of China (12061089, 11861076, 11701503, and 11761079) and the Natural Science Foundation of Yunnan Province (202201AT070022 and 2019FB139).

  2. Author contributions: Formal analysis: QX and YJ; Writing original draft: QX and YJ; Writing review and editing: BL. All authors have read and agreed to the published version of the manuscript.

  3. Conflict of interest: The authors state no conflicts of interest.

  4. Data availability statement: The data will be made available by the authors on request.

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Received: 2024-01-24
Revised: 2024-09-16
Accepted: 2024-11-06
Published Online: 2024-12-14

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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