A Survey on Symmetry Group of Polyhedral Graphs

Article A Survey on Symmetry Group of Polyhedral Graphs

Modjtaba Ghorbani 1,∗ , Matthias Dehmer 2,3,4, Shaghayegh Rahmani 1 and Mina Rajabi-Parsa1

1 Department of Mathematics, Faculty of Science, Shahid Rajaee Teacher Training University, 15614 Tehran, Iran; [email protected] (S.R.); [email protected] (M.R.-P.) 2 Steyr School of Management, University of Applied Sciences Upper Austria, 4040 Steyr, Austria; [email protected] 3 Department of Mechatronics and Biomedical Computer Science, UMIT, A-6060 Hall in Tyrol, Austria 4 College of Artﬁcial Intelligence, Nankai University, Tianjin 300071, China * Corresponding: [email protected]; Tel.: +98-21-22970029

Received: 13 February 2020; Accepted: 21 February 2020; Published: 2 March 2020

Abstract: Every three-connected simple planar graph is a polyhedral graph and a cubic polyhedral graph with pentagonal and hexagonal faces is called as a classical fullerene. The aim of this paper is to survey some results about the symmetry group of cubic polyhedral graphs. We show that the order of symmetry group of such graphs divides 240.

Keywords: polyhedral graph; fullerene; automorphism group; group action

1. Introduction In the present work, all graphs are connected without loops and parallel edges, which we call simple graphs. An automorphism of a graph is a permutation on the set of vertices which preserves the edge set. We denote the image of automorphism β of graph Γ on the vertex u by β(u). The set Aut(Γ) = {α(u) : α is an automorphism } with the operation of composition is a permutation group on V(Γ) and we call it as automorphism group of Γ. The ﬁrst mathematician who considered the graph automorphism was Frucht. In addition, the numerical measures based on automorphisms of a graph have been investigated in reference [1]. A fullerene is a cubic three-connected graph whose faces are pentagons and hexagons. All three connected cubic planar graphs with hexagons and pentagons are called as fullerenes and we donote them by PH-fullerenes, see [2,3]. On the other hand, a three-connected cubic planar graph whose faces are triangles and hexagons is denoted by a TH-fullerene and a SH-fullerene is a three connected cubic planar graph with quadrangles and hexagons. See the references [4–6] as well as [7–11], for studying problems concerning with fullerene graphs. In mathematical aspects, fullerenes are member of a big family of larger graphs, namely the polyhedral graphs. In general, a polyhedral graph is three connected planar but in this paper, we just consider cubic polyhedral graphs, see [12–15]. We denote a cubic polyhedral graph with t triangles, s quadrangles, p pentagonal and h hexagonal faces and no other faces by a (t, s, p, h)−polyhedral or brieﬂy a (t, s, p)-polyhedral graph. This yields that in a SPH-polyhedral graph all faces are tetragons, pentagons and hexagons. Here, we enumerate the number of edges of a SPH-polyhedral graph F. It is clear that each vertex lies in three faces, because F is 3-regular. In addition, we enumerate each edge two times. Hence, |V| = (4s + 5p + 6h)/3, |E| = (4s + 5p + 6h)/2 = 3n/2 and the number of faces is f = s + p + h. The Euler’s formula yields that n − m + f = 2 and so (4s + 5p + 6h)/3 − (4s + 5p + 6h)/2 + s + h + p = 2. This means that p = 12 − 2s and h = n/2 + s − 10. Since p ≥ 0 we get s ≤ 6 and the following cases hold:

Case 1. If s = 0, then F is a fullerene and |Aut(F)| divides 120.

Symmetry 2020, 12, 370; doi:10.3390/sym12030370 www.mdpi.com/journal/symmetry Symmetry 2020, 12, 370 2 of 30

Cases 2–6. If s is either the number 1 or 2 or 3 or 4 or 5, then p is 10 or 8 or 6 or 4 or 2, respectively.

Case 7. If s = 6, then one can deduce that p = 0 and thus F is a SPH-fullerene and |Aut(F)| divides 24.

By above notation, we denote a polyhedral graph with triangles, pentagons and hexagons by a TPH-polyhedral graph. Suppose t is the number of triangles in F, then above discussion yields that p = 12 − 3s, h = n/2 + s − 10 and s ≤ 4. If s = 0, then F is a fullerene. If s is one the integers 1 or 2 or 3, then p is 9 or 6 or 3, respectively. If s = 4, then p = 0 and thus F is a TH-fullerene. One can easily see that in a TSH-polyhedral graph with t triangles, s squares and h hexagons, we yield that n = (3t + 4s + 6h)/3, m = (3t + 4s + 6h)/2 and f = t + s + h. Consequently, we obtain either s = 0 or s = 3 or s = 6. If s = 0, then F is TH-fullerene. If s = 6, then t = 0 and F is SH-fullerene. If s = 3, then t = 2 and F is a polyhedral graph with exactly two triangles, three squares and h hexagons. The cube graph Q3 is the smallest SPH-polyhedral graph which has no hexagonal face, see Figure1a. The symmetry group of this graph is isomorphic to Z2 × S4, see [16]. The pyramid is the smallest TPH-polyhedral graph, see Figure1b. The smallest TSH-polyhedral graph has no hexagon, see Figure1c.

Figure 1. The smallest SPH, TPH and TSH polyhedral graphs.

Much research about fullerene was started after producing fullerenes in bulk quantities in 1990, see [17]. Fullerene chemistry is nowadays a well-established field of both theoretical and experimental investigations. The initial enchanting appeal of fullerenes goes back to the high symmetry of these carbon nanostructures, see [18,19] but nowadays the fullerene era is a field of both theoretical and experimental research. The most important problem, with many applications in a large number of area, is finding the symmetry of molecules. There are several ways to determine the symmetry of a molecule. For example, Randić[20,21] and then Balasubramanian [22–29] studied the Euclidean matrix of a chemical graph to find the symmetry group. However, there is a classic theorem in algebraic graph theory which yields the automorphism group of each graph. Suppose σ ∈ Sn is an arbitrary permutation. Then, the permutation matrix Pσ is defined as Pσ = [xij], where xij = 1 if i = σ(j) and 0, otherwise. It is easy to see that the set of all n × n permutation matrices is a group isomorphic to the symmetric group Sn on n symbols. Moreover, a permutation σ ∈ Sn is a graph automorphism, if it satisfies PσA = APσ, where A is the adjacency matrix of G. In general, the symmetry group and the point-group symmetry of a graph are not isomorphic but about the regular polehedral graphs they are the same. By usung this fact, the authors of [30] computed the symmetry of all fullerenes with up to 70 vertices. In the present paper, we deal with the symmetry properties of fullerene graphs and we compute the symmetry group of several infinite classes of fullerene graphs. One of the aim of computing the symmetry of fullerenes is to enumerate the number of isomers whose number increases very quickly with n. A number of papers deal with the symmetry of fullerene isomers and related species [31–33]. The first author computed the automorphism group of some infinite families of fullerene graphs by using GAP programs [34]. Here, we improve the mentioned algorithm to compute the automorphism group of these fullerene graphs. Further, this paper bears some novel results significant for algorithmization of the fullerene graph machine analysis. Symmetry 2020, 12, 370 3 of 30

2. Main Results A polyhedral graph with either isolated squares or isolated pentagons is called respectively as ISR or IPR polyhedral graph. Here, we consider only the ISR and IPR polyhedral graphs. In [7] Fowler et al. showed that there are only 28 point groups that a fullerene can be realized. In [35] the author investigated that for a PH-fullerene graph F, |Aut(F)| divides 120. Similar results obtained by Ghorbani et al. in [36,37] who showed that if F is a TH-polyhedral graph or a SH-polyhedral graph, then |Aut(F)| divides 24 or 48, respectively. They also determined the order of automorphism groups of all TSH, TPH and SPH-polyhedral graphs, see [38]. In other words, they proved the following results.

Proposition 1. [38] If F is both ISR and IPR, and it is either an SPH or TSH or TPH-polyhedral graph, then for the vertex u ∈ V(F), we yield that the stabilizer Aut(F)u is trivial or it is isomorphic to one of three groups: the cyclic group Z2, the cyclic group Z3 and the symmetric group S3.

Theorem 1. [38] Let F be both ISR and IPR, SPH-polyhedral graph, G be a TSH-polyhedral graph and L be a TPH-polyhedral graph. Then Aut(F) is a {2, 3, 5}-group and both Aut(G) and Aut(L) are {2, 3}-group. Moreover, |Aut(F)| divides 24 × 3 × 5, |Aut(G)| divides 24 × 3 and |Aut(L)| divides 23 × 3.

In [13] Deza et al. investigated the list of allowed symmetry groups of smallest polyhedral graphs, see [7,14,15].

Theorem 2. Fora cubic (t, s, p)-polyhedral graphs, the possible point groups and the number of vertices of the smallest one are

i. (t, s, p) = (4, 0, 0): Z2 × Z2, Z2 × Z2 × Z2, D8, A4, S4. ii. (t, s, p) = (0, 6, 0) : C1, Z2, Z2 × Z2, S3, D8, Z2 × Z2 × Z2, Z2 × S3, D12, D6, Z2 × D12, Z2 × D6, Z2 × Z2 × D6. iii. (t, s, p) = (0, 0, 12) : C1, Z2, A3,, Z4, Z2 × Z2, S3, S6, S3, Z2 × Z3, Z2 × Z2 × Z2, D8, Z2 × Z5, D12, Z2 × S3, A4, D20, Z2D12, D24, S4, A4 × Z2, A5, Z2 × A5.

Theorem 3. Suppose F is a cubic polyhedral graph with at least two face sizes chosen from {3, 4, 5} and no faces of size greater than 6. Then the possible point groups and vertex counts of minimal examples are

i. (t, s, p) = (3, 1, 1): C1, Z2. ii. (t, s, p) = (3, 0, 3) : C1, Z2, A3, S3, Z2 × Z3. iii. (t, s, p) = (2, 3, 0) : C1, Z2, Z2 × Z2, S3, D12. iv. (t, s, p) = (2, 2, 2) : C1, Z2, Z2 × Z2. v. (t, s, p) = (2, 1, 4) : C1, Z2, Z2 × Z2. vi. (t, s, p) = (2, 0, 6) : C1, Z2, Z2 × Z2, S3, Z2 × S3, D12. vii. (t, s, p) = (1, 4, 1) : C1, Z2. viii. (t, s, p) = (1, 3, 3) : C1, Z2, A3, S3. Symmetry 2020, 12, 370 4 of 30

ix. (t, s, p) = (1, 2, 5) : C1, Z2. x. (t, s, p) = (1, 1, 7) : C1, Z2. xi. (t, s, p) = (1, 0, 9) : C1, Z2, A3, S3. xii. (t, s, p) = (0, 5, 2) : C1, Z2, Z2 × Z2, Z2 × Z5, D20. xiii. (t, s, p) = (0, 4, 4) : C1, Z2, Z2 × Z2, Z2 × Z2 × Z2, D8, Z4. xiv. (t, s, p) = (0, 3, 6) : C1, Z2, Z2 × Z2, A3, Z2 × Z3, S3, D12. xv. (t, s, p) = (0, 2, 8) : C1, Z2, Z2 × Z2, D8, Z2 × D8, D16, Z2 × Z2 × Z2, Z4. xvi. (t, s, p) = (0, 1, 10) : C1, Z2, Z2 × Z2.

3. Symmetry Group and Group Action Symmetry has a significant role in the analysis of the structure, bonding and spectroscopy of molecules. A classification of molecules can be done by their symmetry. As we mentioned in the last section, about the fullerene graphs, the symmetry group and the point group are the same. The name point group is also given, because the symmetry elements such as points, lines and planes intersect at a single point. Symmetry elements include mirror planes, axes of rotation, centers of inversion and improper axes of rotation, and all of them are properties related to the structure of the molecule. All molecules have an operation which leaves the molecule where it is and we call it as identity operation. Certain physical properties of molecules are clearly linked to molecular symmetry. For example, molecules with a center of inversion or a mirror plane cannot be chiral. Let F be a fullerene with symmetry group G = Aut(F). The orbit of a vertex u ∈ V(F) is uG = {ug, g ∈ G}. g The stabilizer of the vertex u ∈ V(G) is defined as Gu = {g ∈ G : u = u}. Let H = Gu, then G for u, v ∈ V(F), Hv is denoted by Gu,v. The orbit-stabilizer theorem implies that |u |.|Gu| = |G|. For every g ∈ G, let f ix(g) = {u ∈ V(F), ug = u}, then we yield the following result.

Lemma 1. (Cauchy–Frobenius Lemma) Let G acts on set V(F), then the number of orbits of G is

1 O = | f ix(g)|. (1) | | ∑ G g∈G

For every edge e = uv and every automorphism α ∈ Aut(G), define α¯ (e) = {α(u), α(v)}. Thus, Aut(G) acts on the set of edges by above definition and the Cauchy–Frobenius Lemma for the set of edges can be rewritten as follows: 1 O¯ = | f ix(g¯)|. (2) |G¯| ∑ g¯∈G¯ The novelty of this work is that we apply Cauchy–Frobenius Lemma to compute the whole symmetry group of given infinite family of graphs. In the references [37,39], the symmetry group of several infinite families of fullerene graphs have been determined by aid of GAP [34] program, but in this paper we use from the method given in [36,38]. Here is the first attempt to compute the symmetry group by thus way. To explain our method, at first, we investigate the symmetry group of molecular graph C24 shown in Figure2. In [ 39] the symmetry group of C24 is computed and it is isomorphic with the group Z2 × S4. Symmetry 2020, 12, 370 5 of 30

Figure 2. 3-D graph of fullerene C24.

Example 1. Consider now the molecular graph of fullerene C80, Figure3. Here, we show that the symmetry group of fullerene graph C80 is isomorphic with a dihedral group of order 20, namely D20. To do this, by using concept of symmetry, one can see that the generators of fullerene graph C80 are as follows:

X = (2, 16)(4, 14)(5, 18)(6, 17)(7, 20)(8, 19)(9, 36)(10, 35)(11, 34)(12, 33)(13, 49) (15, 51)(21, 24)(22, 23)(25, 37)(26, 52)(27, 39)(28, 50)(29, 54)(30, 53)(31, 56) (32, 55)(38, 40)(41, 42)(43, 44)(45, 60)(46, 59)(47, 58)(48, 57)(62, 67)(63, 66) (65, 77)(68, 80)(69, 73)(70, 79)(71, 78)(72, 76)(74, 75), Y = (1, 65)(2, 66)(3, 68)(4, 67)(5, 48)(6, 45)(7, 46)(8, 47)(9, 42)(10, 43)(11, 44) (12, 41)(13, 69)(14, 70)(15, 72)(16, 71)(17, 36)(18, 33)(19, 34)(20, 35)(21, 30) (22, 31)(23, 32)(24, 29)(25, 73)(26, 74)(27, 76)(28, 75)(37, 77)(38, 78)(39, 80) (40, 79)(49, 61)(50, 62)(51, 64)(52, 63)(53, 60)(54, 57)(55, 58)(56, 59).

By a GAP program, one can see that X2 = Y2 = (XY)10 = 1 and X−1(XY)X =(XY)−1. Symmetry 2020, 12, 370 6 of 30

47 46 48 45 26 49 25 67 27 24 65 66 28 44 50 64 8 68 9 10 23 7 29 78 11 51 43 22 69 63 30 77 79 12 1 42 62 6 52 5 2 70 31 21 20 80 76 13 40 3 71 41 4 32 53 61 75 72 19 60 39 18 15 14 38 73 33 74 54 16 34 59 17 35 37 36 55 58 56 57

Figure 3. 3-D graph of fullerene C80.

In continuing, consider the molecular graph of fullerene C84, Figure4. We prove that the symmetry group of the C84 fullerene is isomorphic to the group S4. To do this, suppose G is the symmetry group of this fullerene. Then G = hX, Yi, where X and Y are the following permutations:

X = (1, 2)(3, 4)(5, 8)(6, 80)(7, 81)(9, 18)(10, 19)(11, 20)(12, 78)(14, 83)(15, 82) (17, 84)(21, 54)(22, 77)(23, 55)(24, 79)(25, 76)(26, 27)(28, 59)(29, 60)(30, 57) (31, 58)(32, 66)(33, 70)(34, 72)(35, 67)(36, 64)(37, 65)(38, 74)(39, 73)(40, 75) (41, 56)(42, 51)(43, 53)(44, 52)(45, 48)(46, 49)(47, 50)(61, 71)(62, 63)(68, 69), Y = (1, 76, 31, 69)(2, 59, 30, 40)(3, 79, 28, 68)(4, 58, 29, 39)(5, 51, 35, 17) (6, 84, 49, 66)(7, 83, 48, 65)(8, 80, 41, 71)(9, 77, 42, 61)(10, 78, 43, 62) (11, 81, 44, 63)(12, 82, 45, 64)(13, 55, 27, 33)(14, 20, 53, 36)(15, 19, 52, 37) (16, 54, 26, 34)(18, 56, 32, 38)(21, 72, 23, 70)(22, 74, 46, 67)(24, 73, 50, 57) (25, 75, 47, 60). Symmetry 2020, 12, 370 7 of 30

1 2 6 9 3 12 7 5 4 11 13 14 8 10 22 32 17 27 15 20 25 30 31 21 19 16 26 29 28 23 18 33 24 36 34 42 35 53 54 41 47 48 38 49 50 56 43 55 44 39 37 51 52 40 46 58 45 60 57 65 70 75 66 61 71 76 69 59 74 64 67 68 78 77 62 63 79 72 73 80

Figure 4. 3-D graph of fullerene C84.

Now consider the molecular graph of fullerene C150 as given in Figure5. In [ 39] it is proved that the symmetry group of C150 is isomorphic with dihedral group D20.

Figure 5. 3-D graph of fullerene C150. Symmetry 2020, 12, 370 8 of 30

Example 2. Consider the fullerene graph C10n (n is even) as depicted in Figure6. The vertices of central 1 1 1 1 1 pentagon is labeled by {1 , 2 , 3 , 4 , 5 }. These vertices compose the first layer of C10n. The vertices of the second layer are the boundary vertices of five pentagons adjacent with the central pentagon and so on. In [40], it is shown that the following elements are in the automorphism group of fullerene graph C10n. The automorphism α is a symmetry element in which f ix(α) = {11, 102, 103, ··· , 10n, 52, 53, ··· , 5n, 3n+1} and σ = (11, 1n+1, 21, 2n+1, 31, 3n+1, 41, 4n+1, 51, 5n+1)(12, 2n, 32, 4n, 52, 6n, 72, 8n, 92, 10n)(22, 3n, 42, 5n, 62, 7n, 82, 9n, 102, 1n)(13, 2n−1, 33, 4n−1, 53, 6n−1, 73, 8n−1, 93, 10n−1)(23, 3n−1, 43, 5n−1, 63, 7n−1, 83, 9n−1, 103, 1n−1) ··· (1n/2, 2(n+4)/2, 3n/2, 4(n+4)/2, 5n/2, 6(n+4)/2, 7n/2, 8(n+4)/2, 9n/2, 10(n+4)/2)(2n/2, 3(n+4)/2, 4n/2, 5(n+4)/2, 6n/2, 7(n+4)/2, 8n/2, 9(n+4)/2, 10n/2, 1(n+4)/2)(1(n+2)/2, 2(n+2)/2, 3(n+2)/2, 4(n+2)/2, 5(n+2)/2, 6(n+2)/2, 7(n+2)/2, 8(n+2)/2, 9(n+2)/2, 10(n+2)/2). It is clear that α2 = σ10 = 1, ασα = σ−1 and 1 G = hα, σi ≤ A = Aut(C10n). On the other hand, every symmetry element which fixes 1 , must also fix {102, 103, ··· , 10n, 52, 53, ··· , 5n, 3n+1}. The identity element and the symmetry element α do this. 1 A Hence, the orbit-stabilizer property ensures that |A| = | 1 |.|A11 | and thus |A| = 10 × 2 = 20 which ∼ implies that A = D20. The orbits of the automorphism group are given in Table1.

Figure 6. C10n, n is even.

Example 3. Consider the fullerene graph C10n (n is odd) as depicted in Figure7. Assume that α is a symmetry element with ﬁxpoints {11, 102, 103, ··· , 10n, 1n+1, 52, 53, ··· , 5n} and σ is a symmetry element by the following permutation presentation: σ = (11, 4n+1, 21, 5n+1, 31, 1n+1, 41, 2n+1, 51, 3n+1)(12, 7n, 32, 9n, 52, 1n, 72, 3n, 92, 5n)(22, 8n, 42, 10n, 62, 2n, 82, 4n, 102, 6n)(13, 7n−1, 33, 9n−1, 53, 1n−1, 73, 3n−1, 93, 5n−1)(23, 8n−1, 43, 10n−1, 63, 2n−1, 83, 4n−1, 103, 6n−1) ··· (1(n+1)/2, 7(n+3)/2, 3(n+1)/2, 9(n+3)/2, 5(n+1)/2, 1(n+3)/2, 7(n+1)/2, 3(n+3)/2, 9(n+1)/2, 5(n+3)/2)(2(n+1)/2, 8(n+3)/2, 4(n+1)/2, 10(n+3)/2, 6(n+1)/2, 2(n+3)/2, 8(n+1)/2, 4(n+3)/2, 10(n+1)/2, 6(n+3)/2). Similar to the last case, one can see G = hα, σi = Aut(C10n) is isomorphic with dihedral group D20. The orbits of the automorphism group are given in Table2. Symmetry 2020, 12, 370 9 of 30

Figure 7. C10n, n is odd.

In continuing, we count all orbits of Ω = {1,. . . ,10n} under the action of automorphism group. Again consider the fullerene graph C10n, where n is even, as depicted in Figure8. We find the presentation of elements Aut(C10n). It is not difficult to see that there are five symmetry elements of order 2 in Aut(C10n) denoted by αi, 1 ≤ i ≤ 5. Clearly, we have o(αi) = 2. 1 2 3 n 2 3 n n+1 In other words, one can easily check that f ix(α1) = {1 , 10 , 10 , ··· , 10 , 5 , 5 , ··· , 5 , 3 }, 1 2 3 n 2 3 n n+1 1 2 3 n 2 3 n n+1 f ix(α2) = {2 , 2 , 2 , ··· , 2 , 7 , 7 , ··· , 7 , 4 }, f ix(α3) = {3 , 4 , 4 , ··· , 4 , 9 , 9 , ··· , 9 , 5 }, 1 2 3 n 2 3 n n+1 1 2 3 n 2 3 n n+1 f ix(α4) = {4 , 6 , 6 , ··· , 6 , 1 , 1 , ··· , 1 , 1 }, f ix(α5) = {5 , 8 , 8 , ··· , 8 , 3 , 3 , ··· , 3 , 2 }. 1 n+1 1 This means that f ix(αi) = 2n, (1 ≤ i ≤ 5). Suppose β1 is an involution that maps 1 to 2 , 2 to 1n+1, 21 to 2n+1, 31 to 5n+1, 41 to 4n+1, 51 to 3n+1, 12 to 2n, 22 to 1n, 32 to 10n, 42 to 9n, 52 to 8n, 62 to 7n, 2 n 2 n 2 n 2 n 7 to 6 , 8 to 5 , 9 to 5 , 10 to 3 and so on. It is clear that fix β1 = φ. By continuing this method, all permutation presentations of βi’sare as follows:

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β1 = (1 , 2 )(2 , 1 )(3 , 5 )(4 , 4 )(5 , 3 )(1 , 2 )(2 , 1 )(3 , 10 )(4 , 9n)(52, 8n)(62, 7n)(72, 6n)(82, 5n)(92, 4n)(102, 3n)(13, 2n−1)(23, 1n−1)(33, 10n−1) (43, 9n−1)(53, 8n−1)(63, 7n−1)(73, 6n−1)(83, 5n−1)(93, 4n−1)(103, 3n−1) ··· (1n/2, 2(n+4)/2)(2n/2, 1(n+4)/2)(3n/2, 10(n+4)/2)(4n/2, 9(n+4)/2)(5n/2, 8(n+4)/2)(6n/2, 7(n+4)/2)(7n/2, 6(n+4)/2)(8n/2, 5(n+4)/2)(9n/2, 4(n+4)/2)(10n/2, 3(n+4)/2)(1(n+2)/2, 2(n+2)/2)(3(n+2)/2, 10(n+2)/2)(4(n+2)/2, 9(n+2)/2)(5(n+2)/2, 8(n+2)/2)(6(n+2)/2, 7(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β2 = (1 , 3 )(2 , 2 )(3 , 1 )(4 , 5 )(5 , 4 )(1 , 4 )(2 , 3 )(3 , 2 )(4 , 1n)(52, 10n)(62, 9n)(72, 8n)(82, 7n)(92, 6n)(102, 5n)(13, 4n−1)(23, 3n−1)(33, 2n−1) (43, 1n−1)(53, 10n−1)(63, 9n−1)(73, 8n−1)(83, 7n−1)(93, 6n−1)(103, 5n−1) ··· (1n/2, 4(n+4)/2)(2n/2, 3(n+4)/2)(3n/2, 2(n+4)/2)(4n/2, 1(n+4)/2)(5n/2, 10(n+4)/2)(6n/2, 9(n+4)/2)(7n/2, 8(n+4)/2)(8n/2, 7(n+4)/2)(9n/2, 6(n+4)/2)(10n/2, 5(n+4)/2)(1(n+2)/2, t4(n+2)/2)(2(n+2)/2, 3(n+2)/2)(5(n+2)/2, 10(n+2)/2)(6(n+2)/2, 9(n+2)/2)(7(n+2)/2, 8(n+2)/2), Symmetry 2020, 12, 370 10 of 30

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β3 = (1 , 4 )(2 , 3 )(3 , 2 )(4 , 1 )(5 , 5 )(1 , 6 )(2 , 5 )(3 , 4 )(4 , 3n)(52, 2n)(62, 1n)(72, 10n)(82, 9n)(92, 8n)(102, 7n)(13, 6n−1)(23, 5n−1)(33, 4n−1) (43, 3n−1)(53, 2n−1)(63, 1n−1)(73, 10n−1)(83, 9n−1)(93, 8n−1)(103, 7n−1) ··· (1n/2, 6(n+4)/2)(2n/2, 5(n+4)/2)(3n/2, 4(n+4)/2)(4n/2, 3(n+4)/2)(5n/2, 2(n+4)/2)(6n/2, 1(n+4)/2)(7n/2, 10(n+4)/2)(8n/2, 9(n+4)/2)(9n/2, 8(n+4)/2)(10n/2, 7(n+4)/2)(1(n+2)/2, 6(n+2)/2)(2(n+2)/2, 5(n+2)/2)(3(n+2)/2, 4(n+2)/2)(7(n+2)/2, 10(n+2)/2)(8(n+2)/2, 9(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β4 = (1 , 5 )(2 , 4 )(3 , 3 )(4 , 2 )(5 , 1 )(1 , 8 )(2 , 7 )(3 , 6 )(4 , 5n)(52, 4n)(62, 3n)(72, 2n)(82, 1n)(92, 10n)(102, 9n)(13, 8n−1)(23, 7n−1)(33, 6n−1) (43, 5n−1)(53, 4n−1)(63, 3n−1)(73, 2n−1)(83, 1n−1)(93, 10n−1)(103, 9n−1) ··· (1n/2, 8(n+4)/2)(2n/2, 7(n+4)/2)(3n/2, 6(n+4)/2)(4n/2, 5(n+4)/2)(5n/2, 4(n+4)/2)(6n/2, 3(n+4)/2)(7n/2, 2(n+4)/2)(8n/2, 1(n+4)/2)(9n/2, 10(n+4)/2)(10n/2, 9(n+4)/2)(1(n+2)/2, 8(n+2)/2)(2(n+2)/2, 7(n+2)/2)(3(n+2)/2, 6(n+2)/2)(4(n+2)/2, 5(n+2)/2)(9(n+2)/2, 10(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β5 = (1 , 1 )(2 , 5 )(3 , 4 )(4 , 3 )(5 , 2 )(1 , 10 )(2 , 9 )(3 , 8 )(4 , 7n)(52, 6n)(62, 5n)(72, 4n)(82, 3n)(92, 2n)(102, 1n)(13, 10n−1)(23, 9n−1)(33, 8n−1) (43, 7n−1)(53, 6n−1)(63, 5n−1)(73, 4n−1)(83, 3n−1)(93, 2n−1)(103, 1n−1 ··· (1n/2, 10(n+4)/2(2n/2, 9(n+4)/2)(3n/2, 8(n+4)/2)(4n/2, 7(n+4)/2)(5n/2, 6(n+4)/2)(6n/2, 5(n+4)/2)(7n/2, 4(n+4)/2)(8n/2, 3(n+4)/2)(9n/2, 2(n+4)/2)(10n/2, 1(n+4)/2)(1(n+2)/2, 10(n+2)/2)(2(n+2)/2, 9(n+2)/2)(3(n+2)/2, 8(n+2)/2)(4(n+2)/2, 7(n+2)/2)(5(n+2)/2, 6(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β6 = (1 , 3 )(2 , 4 )(3 , 5 )(4 , 1 )(5 , 2 )(1 , 6 )(2 , 7 )(3 , 8 )(4 , 9n)(52, 10n)(62, 1n)(72, 2n)(82, 3n)(92, 4n)(102, 5n)(13, 6n−1)(23, 7n−1)(33, 8n−1) (43, 9n−1)(53, 10n−1)(63, 1n−1)(73, 2n−1)(83, 3n−1)(93, 4n−1)(103, 5n−1 ··· (1n/2, 6(n+4)/2)(2n/2, 7(n+4)/2)(3n/2, 8(n+4)/2)(4n/2, 9(n+4)/2)(5n/2, 10(n+4)/2)(6n/2, 1(n+4)/2)(7n/2, 2(n+4)/2)(8n/2, 3(n+4)/2)(9n/2, 4(n+4)/2)(10n/2, 5(n+4)/2)(1(n+2)/2, 6(n+2)/2)(2(n+2)/2, 7(n+2)/2)(3(n+2)/2, 8(n+2)/2)(4(n+2)/2, 9(n+2)/2)(5(n+2)/2, 10(n+2)/2).

Aut(C10n) includes four rotational elements γi (1 ≤ i ≤ 4) and four permutations σi (1 ≤ i ≤ 4) of order 10 with the following permutation presentations:

1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 γ1 = (1 , 2 , 3 , 4 , 5 )(1 , 3 , 5 , 7 , 9 )(2 , 4 , 6 , 8 , 10 )(1 , 3 , 5 , 7 , 9 )(2 , 4 , 63, 83, 103) ··· (1n−1, 3n−1, 5n−1, 7n−1, 9n−1)(2n−1, 4n−1, 6n−1, 8n−1, 10n−1)(1n, 3n, 5n, 7n, 9n)(2n, 4n, 6n, 8n, 10n)(1n+1, 2n+1, 3n+1, 4n+1, 5n+1),

1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 γ2 = (1 , 3 , 5 , 2 , 4 )(1 , 5 , 9 , 3 , 7 )(2 , 6 , 10 , 4 , 8 )(1 , 5 , 9 , 3 , 7 )(2 , 6 , 103, 43, 83) ··· (1n−1, 5n−1, 9n−1, 3n−1, 7n−1)(2n−1, 6n−1, 10n−1, 4n−1, 8n−1)(1n, 5n, 9n, 3n, 7n)(2n, 6n, 10n, 4n, 8n)(1n+1, 3n+1, 5n+1, 2n+1, 4n+1),

1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 γ3 = (1 , 5 , 4 , 3 , 2 )(1 , 9 , 7 , 5 , 3 )(2 , 10 , 8 , 6 , 4 )(1 , 9 , 7 , 5 , 3 )(2 , 10 , 83, 63, 43) ··· (1n−1, 9n−1, 7n−1, 5n−1, 3n−1)(2n−1, 10n−1, 8n−1, 6n−1, 4n−1)(1n, 9n, 7n, 5n, 3n)(2n, 10n, 8n, 6n, 4n)(1n+1, 5n+1, 4n+1, 3n+1, 2n+1),

1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 γ4 = (1 , 4 , 2 , 5 , 3 )(1 , 7 , 3 , 9 , 5 )(2 , 8 , 4 , 10 , 6 )(1 , 7 , 3 , 9 , 5 )(2 , 8 , 43, 103, 63) ··· (1n−1, 7n−1, 3n−1, 9n−1, 5n−1)(2n−1, 8n−1, 4n−1, 10n−1, 6n−1)(1n, 7n, 3n, 9n, 5n)(2n, 8n, 4n, 10n, 6n)(1n+1, 4n+1, 2n+1, 5n+1, 3n+1).

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ1 = (1 , 5 , 5 , 4 , 4 , 3 , 3 , 2 , 2 , 1 )(1 , 10 , 9 , 8 , 7 , 6 , 5 , 4 , 3 , 2n)(22, 1n, 102, 9n, 82, 7n, 62, 5n, 42, 3n)(13, 10n−1, 93, 8n−1, 73, 6n−1, 53, 4n−1, 33, 2n−1)(23, 1n−1, 103, 9n−1, 83, 7n−1, 63, 5n−1, 43, 3n−1) ··· (1n/2, 10(n+4)/2, 9n/2, 8(n+4)/2, 7n/2, 6(n+4)/2, 5n/2, 4(n+4)/2, 3n/2, 2(n+4)/2)(2n/2, 1(n+4)/2, 10n/2, 9(n+4)/2, 8n/2, 7(n+4)/2, 6n/2, 5(n+4)/2, 4n/2, 3(n+4)/2)(1(n+2)/2, 10(n+2)/2, 9(n+2)/2, 8(n+2)/2, 7(n+2)/2, 6(n+2)/2, 5(n+2)/2, 4(n+2)/2, 3(n+2)/2, 2(n+2)/2), Symmetry 2020, 12, 370 11 of 30

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ2 = (1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 )(1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10n)(22, 3n, 42, 5n, 62, 7n, 82, 9n, 102, 1n)(13, 2n−1, 33, 4n−1, 53, 6n−1, 73, 8n−1, 93, 10n−1)(23, 3n−1, 43, 5n−1, 63, 7n−1, 83, 9n−1, 103, 1n−1 ··· (1n/2, 2(n+4)/2, 3n/2, 4(n+4)/2, 5n/2, 6(n+4)/2, 7n/2, 8(n+4)/2, 9n/2, 10(n+4)/2)(2n/2, 3(n+4)/2, 4n/2, 5(n+4)/2, 6n/2, 7(n+4)/2, 8n/2, 9(n+4)/2, 10n/2, 1(n+4)/2)(1(n+2)/2, 2(n+2)/2, 3(n+2)/2, 4(n+2)/2, 5(n+2)/2, 6(n+2)/2, 7(n+2)/2, 8(n+2)/2, 9(n+2)/2, 10(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n σ3 = (1 , 2 , 4 , 5 , 2 , 3 , 5 , 1 , 3 , 4 )(1 , 4 , 7 , 10 , 3 , 6 , 9 , 2 , 52, 8n)(22, 5n, 82, 1n, 42, 7n, 102, 3n, 62, 9n)(13, 4n−1, 73, 10n−1, 33, 6n−1, 93, 2n−1, 53, 8n−1)(23, 5n−1, 83, 1n−1, 43, 7n−1, 103, 3n−1, 63, 9n−1) ··· (1n/2, 4(n+4)/2, 7n/2, 10(n+4)/2, 3n/2, 6(n+4)/2, 9n/2, 2(n+4)/2, 5n/2, 8(n+4)/2)(2n/2, 5(n+4)/2, 8n/2, 1(n+4)/2, 4n/2, 7(n+4)/2, 10n/2, 3(n+4)/2, 6n/2, 9(n+4)/2)(1(n+2)/2, 4(n+2)/2, 7(n+2)/2, 10(n+2)/2, 3(n+2)/2, 6(n+2)/2, 9(n+2)/2, 2(n+2)/2, 5(n+2)/2, 8(n+2)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ4 = (1 , 4 , 3 , 1 , 5 , 3 , 2 , 5 , 4 , 2 )(1 , 8 , 5 , 2 , 9 , 6 , 3 , 10 , 7 , 4n)(22, 9n, 62, 3n, 102, 7n, 42, 1n, 82, 5n)(13, 8n−1, 53, 2n−1, 93, 6n−1, 33, 10n−1, 73, 4n−1)(23, 9n−1, 63, 3n−1, 103, 7n−1, 43, 1n−1, 83, 5n−1) ··· (1n/2, 8(n+4)/2, 5n/2, 2(n+4)/2, 9n/2, 6(n+4)/2, 3n/2, 10(n+4)/2, 7n/2, 4(n+4)/2)(2n/2, 9(n+4)/2, 6n/2, 3(n+4)/2, 10n/2, 7(n+4)/2, 4n/2, 1(n+4)/2, 8n/2, 5(n+4)/2)(1(n+2)/2, 8(n+2)/2, 5(n+2)/2, 2(n+2)/2, 9(n+2)/2, 6(n+2)/2, 3(n+2)/2, 10(n+2)/2, 7(n+2)/2, 4(n+2)/2).

n−2 Hence, C10n (n is even) has 2 × 2 + 2 = n orbits and each of them has 10 vertices, see Table1.

Table 1. The structure of orbit elements of C10n, n is even.

Vertex The Structure of Orbit Elements 11 11, 21, 31, 41, 51, 1n+1, 2n+1, 3n+1, 4n+1, 5n+1 12 12, 32, 52, 72, 92, 2n, 4n, 6n, 8n, 10n 22 22, 42, 62, 82, 102, 1n, 3n, 5n, 7n, 9n . . . . 1n/2 1n/2, 3n/2, 5n/2, 7n/2, 9n/2, 2(n+4)/2, 4(n+4)/2, 6(n+4)/2, 8(n+4)/2, 10(n+4)/2 2n/2 2n/2, 4n/2, 6n/2, 8n/2, 10n/2, 1(n+4)/2, 3(n+4)/2, 5(n+4)/2, 7(n+4)/2, 9(n+4)/2 1(n+2)/2 1(n+2)/2, 2(n+2)/2, 3(n+2)/2, 4(n+2)/2, 5(n+2)/2, 6(n+2)/2, 7(n+2)/2, 8(n+2)/2, 9(n+2)/2, 10(n+2)/2 Symmetry 2020, 12, 370 12 of 30

Figure 8. C10n, n is even.

It is not difficult to see that | f ix(αi)| = n + 2 (1 ≤ i ≤ 5), | f ix(βj)| = 2 (1 ≤ j ≤ 5) and | f ix(β6)| = | f ix(γk)| = | f ix(σl)| = 0 (1 ≤ k ≤ 4), (1 ≤ l ≤ 4). By considering the action of Aut(C10n) on the set of edges and using Equation (3.1), one can prove that the number of orbits is n + 1 for which n 1 1 2 2 (n−2)/2 (n−2)/2 n/2 n/2 (n+2)/2 is even. These are O(e1), O(e ), O(e1), O(e ), ··· , O(e1 ), O(e ), O(e1 , O(e ), O(e1 ). Hence, we proved the following theorem.

Theorem 4. Consider the fullerene graph C10n, then there are n+1 orbits for which n is even under the action of automorphism group on the set of edges.

Now consider the fullerene graph C10n, where n is odd, as depicted in Figure9. There are ﬁve symmetry elements of order 2 in Aut(C10n) denoted by αi, 1 ≤ i ≤ 5. Clearly, we have o(αi) = 2. In 1 2 3 n n+1 2 3 n other words, one can easily check that f ix(α1) = {1 , 10 , 10 , ··· , 10 , 1 , 5 , 5 , ··· , 5 }, f ix(α2) = 1 2 3 n n+1 2 3 n 1 2 3 n n+1 2 3 n {2 , 2 , 2 , ··· , 2 , 2 , 7 , 7 , ··· , 7 }, f ix(α3) = {3 , 4 , 4 , ··· , 4 , 3 , 9 , 9 , ··· , 9 }, f ix(α4) = 1 2 3 n n+1 2 3 n 1 2 3 n n+1 2 3 n n+1 {4 , 6 , 6 , ··· , 6 , 4 , 1 , 1 , ··· , 1 }, f ix(α5) = {5 , 8 , 8 , ··· , 8 , 5 , 3 , 3 , ··· , 3 , 2 }. This means that f ix(αi) = 2n, (1 ≤ i ≤ 5). Similar to the last one, the presentation of other elements of Aut(C10n) are as follows:

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n β1 = (1 , 1 )(2 , 2 )(3 , 3 )(4 , 4 )(5 , 5 )(1 , 1 )(2 , 2 )(3 , 3 )(4 , 4 ) (52, 5n)(62, 6n)(72, 7n)(82, 8n)(92, 9n)(102, 10n)(13, 1n−1)(23, 2n−1)(33, 3n−1)(43, 4n−1)(53, 5n−1)(63, 6n−1)(73, 7n−1)(83, 8n−1)(93, 9n−1)(103, 10n−1) ··· (1(n+1)/2, 1(n+3)/2)(2(n+1)/2, 2(n+3)/2)(3(n+1)/2, 3(n+3)/2)(4(n+1)/2, 4(n+3)/2)(5(n+1)/2, 5(n+3)/2) (6(n+1)/2, 6(n+3)/2)(7(n+1)/2, 7(n+3)/2)(8(n+1)/2, 8(n+3)/2)(9(n+1)/2, 9(n+3)/2) (10(n+1)/2, 10(n+3)/2), Symmetry 2020, 12, 370 13 of 30

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β2 = (1 , 3 )(2 , 2 )(3 , 1 )(4 , 5 )(5 , 4 )(1 , 3 )(2 , 2 )(3 , 1 )(4 , 10n)(52, 9n)(62, 8n)(72, 7n)(82, 6n)(92, 5n)(102, 4n)(13, 3n−1)(23, 2n−1)(33, 1n−1) (43, 10n−1)(53, 9n−1)(63, 8n−1)(73, 7n−1)(83, 6n−1)(93, 5n−1)(103, 4n−1 ··· (1(n+1)/2, 3(n+3)/2)(2(n+1)/2, 2(n+3)/2)(3(n+1)/2, 1(n+3)/2)(4(n+1)/2, 10(n+3)/2) (5(n+1)/2, 9(n+3)/2)(6(n+1)/2, 8(n+3)/2)(7(n+1)/2, 7(n+3)/2)(8(n+1)/2, 6(n+3)/2) (9(n+1)/2, 5(n+3)/2)(10(n+1)/2, 4(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β3 = (1 , 4 )(2 , 3 )(3 , 2 )(4 , 1 )(5 , 5 (1 , 5 )(2 , 4 )(3 , 3 )(4 , 2n)(52, 1n)(62, 10n)(72, 9n)(82, 8n)(92, 7n)(102, 6n)(13, 5n−1)(23, 4n−1)(33, 3n−1) (43, 2n−1)(53, 1n−1)(63, 10n−1)(73, 9n−1)(83, 8n−1)(93, 7n−1)(103, 6n−1) ··· (1(n+1)/2, 5(n+3)/2)(2(n+1)/2, 4(n+3)/2)(3(n+1)/2, 3(n+3)/2)(4(n+1)/2, 2(n+3)/2) (5(n+1)/2, 1(n+3)/2)(6(n+1)/2, 10(n+3)/2)(7(n+1)/2, 9(n+3)/2)(8(n+1)/2, 8(n+3)/2) (9(n+1)/2, 7(n+3)/2)(10(n+1)/2, 6(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β4 = (1 , 5 )(2 , 4 )(3 , 3 )(4 , 2 )(5 , 1 )(1 , 7 )(2 , 6 )(3 , 5 )(4 , 4n)(52, 3n)(62, 2n)(72, 1n)(82, 10n)(92, 9n)(102, 8n)(13, 7n−1)(23, 6n−1)(33, 5n−1) (43, 4n−1)(53, 3n−1)(63, 2n−1)(73, 1n−1)(83, 10n−1)(93, 9n−1)(103, 8n−1) ··· (1(n+1)/2, 7(n+3)/2)(2(n+1)/2, 6(n+3)/2)(3(n+1)/2, 5(n+3)/2)(4(n+1)/2, 4(n+3)/2) (5(n+1)/2, 3(n+3)/2)(6(n+1)/2, 2(n+3)/2)(7(n+1)/2, 1(n+3)/2)(8(n+1)/2, 10(n+3)/2) (9(n+1)/2, 9(n+3)/2)(10(n+1)/2, 8(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β5 = (1 , 1 )(2 , 5 )(3 , 4 )(4 , 3 )(5 , 2 )(1 , 9 )(2 , 8 )(3 , 7 )(4 , 6n)(52, 5n)(62, 4n)(72, 3n(82, 2n)(92, 1n)(102, 10n)(13, 9n−1)(23, 8n−1)(33, 7n−1) (43, 6n−1)(53, 5n−1)(63, 4n−1)(73, 3n−1)(83, 2n−1)(93, 1n−1)(103, 10n−1) ··· (1(n+1)/2, 9(n+3)/2)(2(n+1)/2, 8(n+3)/2)(3(n+1)/2, 7(n+3)/2)(4(n+1)/2, 6(n+3)/2) (5(n+1)/2, 5(n+3)/2)(6(n+1)/2, 4(n+3)/2)(7(n+1)/2, 3(n+3)/2)(8(n+1)/2, 2(n+3)/2) (9(n+1)/2, 1(n+3)/2)(10(n+1)/2, 10(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 β6 = (1 , 2 )(2 , 1 )(3 , 5 )(4 , 4 )(5 , 3 )(1 , 1 )(2 , 10 )(3 , 9 )(4 , 8n)(52, 7n)(62, 6n)(72, 5n)(82, 4n)(92, 3n)(102, 2n)(13, 1n−1)(23, 10n−1)(33, 9n−1) (43, 8n−1)(53, 7n−1)(63, 6n−1)(73, 5n−1)(83, 4n−1)(93, 3n−1)(103, 2n−1) ··· (1(n+1)/2, 1(n+3)/2)(2(n+1)/2, 10(n+3)/2)(3(n+1)/2, 9(n+3)/2)(4(n+1)/2, 8(n+3)/2) (5(n+1)/2, 7(n+3)/2)(6(n+1)/2, 6(n+3)/2)(7(n+1)/2, 5(n+3)/2)(8(n+1)/2, 4(n+3)/2) (9(n+1)/2, 3(n+3)/2)(10(n+1)/2, 2(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ1 = (1 , 2 , 3 , 4 , 5 , 1 , 2 , 3 , 4 , 5 )(1 , 3 , 5 , 7 , 9 , 1 , 3 , 5 , 7 , 9n)(22, 4n, 62, 8n, 102, 2n, 42, 6n, 82, 10n)(13, 3n−1, 53, 7n−1, 93, 1n−1, 33, 5n−1, 73, 9n−1)(23, 4n−1, 63, 8n−1, 103, 2n−1, 43, 6n−1, 83, 10n−1) ··· (1(n+1)/2, 3(n+3)/2, 5(n+1)/2, 7(n+3)/2, 9(n+1)/2, 1(n+3)/2, 3(n+1)/2, 5(n+3)/2, 7(n+1)/2, 9(n+3)/2)(2(n+1)/2, 4(n+3)/2, 6(n+1)/2, 8(n+3)/2, 10(n+1)/2, 2(n+3)/2, 4(n+1)/2, 6(n+3)/2, 8(n+1)/2, 10(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ2 = (1 , 3 , 5 , 2 , 4 , 1 , 3 , 5 , 2 , 4 )(1 , 5 , 9 , 3 , 7 , 1 , 5 , 9 , 3 , 7n)(22, 6n, 102, 4n, 82, 2n, 62, 10n, 42, 8n)(13, 5n−1, 93, 3n−1, 73, 1n−1, 53, 9n−1, 33, 7n−1)(23, 6n−1, 103, 4n−1, 83, 2n−1, 63, 10n−1, 43, 8n−1) ··· (1(n+1)/2, 5(n+3)/2, 9(n+1)/2, 3(n+3)/2, 7(n+1)/2, 1(n+3)/2, 5(n+1)/2, 9(n+3)/2, 3(n+1)/2, 7(n+3)/2)(2(n+1)/2, 6(n+3)/2, 10(n+1)/2, 4(n+3)/2, 8(n+1)/2, 2(n+3)/2, 6(n+1)/2, 10(n+3)/2, 4(n+1)/2, 8(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ3 = (1 , 4 , 2 , 5 , 3 , 1 , 4 , 2 , 5 , 3 )(1 , 7 , 3 , 9 , 5 , 1 , 7 , 3 , 9 , 5n)(22, 8n, 42, 10n, 62, 2n, 82, 4n, 102, 6n)(13, 7n−1, 33, 9n−1, 53, 1n−1, 73, 3n−1, 93, 5n−1)(23, 8n−1, 43, 10n−1, 63, 2n−1, 83, 4n−1, 103, 6n−1) ··· (1(n+1)/2, 7(n+3)/2, 3(n+1)/2, 9(n+3)/2, 5(n+1)/2, 1(n+3)/2, 7(n+1)/2, 3(n+3)/2, 9(n+1)/2, 5(n+3)/2)(2(n+1)/2, 8(n+3)/2, 4(n+1)/2, 10(n+3)/2, 6(n+1)/2, 2(n+3)/2, 8(n+1)/2, 4(n+3)/2, 10(n+1)/2, 6(n+3)/2),

1 n+1 1 n+1 1 n+1 1 n+1 1 n+1 2 n 2 n 2 n 2 n 2 σ4 = (1 , 5 , 4 , 3 , 2 , 1 , 5 , 4 , 3 , 2 )(1 , 9 , 7 , 5 , 3 , 1 , 9 , 7 , 5 , 3n)(22, 10n, 82, 6n, 42, 2n, 102, 8n, 62, 4n)(13, 9n−1, 73, 5n−1, 33, 1n−1, 93, 7n−1, 53, 3n−1)(23, 10n−1, 83, 6n−1, 43, 2n−1, 103, 8n−1, 63, 4n−1) ··· (1(n+1)/2, 9(n+3)/2, 7(n+1)/2, 5(n+3)/2, 3(n+1)/2, 1(n+3)/2, 9(n+1)/2, 7(n+3)/2, 5(n+1)/2, 3(n+3)/2)(2(n+1)/2, 10(n+3)/2, 8(n+1)/2, 6(n+3)/2, 4(n+1)/2, 2(n+3)/2, 10(n+1)/2, 8(n+3)/2, 6(n+1)/2, 4(n+3)/2).

n−1 So, C10n (n is odd) has 2 × 2 + 1 = n orbits which each have 10 vertices, see Table2.

Table 2. The structure of orbit elements of C10n, n is odd.

Vertex The Structure of Orbit Elements 11 11, 21, 31, 41, 51, 1n+1, 2n+1, 3n+1, 4n+1, 5n+1 12 12, 32, 52, 72, 92, 1n, 3n, 5n, 7n, 9n 22 22, 42, 62, 82, 102, 2n, 4n, 6n, 8n, 10n . . . . 1(n+1)/2 1(n+1)/2, 3(n+1)/2, 5(n+1)/2, 7(n+1)/2, 9(n+1)/2, 1(n+3)/2, 3(n+3)/2, 5(n+3)/2, 7(n+3)/2, 9(n+3)/2 2(n+1)/2 2(n+1)/2, 4(n+1)/2, 6(n+1)/2, 8(n+1)/2, 10(n+1)/2, 2(n+3)/2, 4(n+3)/2, 6(n+3)/2, 8(n+3)/2, 10(n+3)/2 Symmetry 2020, 12, 370 15 of 30

Figure 9. C10n, n is odd.

By similar argument, we have the following theorem.

Theorem 5. Consider the fullerene graph C10n (n is odd), then there are n+1 orbits under the action of automorphism 1 1 2 2 (n−1)/2 (n+1)/2 (n+1)/2 group on the set of edges. These are O(e1), O(e ), O(e1), O(e ), ··· , O(e ), O(e1 ), O(e ).

Now consider the fullerene graph C12n (n is odd) as depicted in Figure 10.

Figure 10. The fullerene graph C12n (n is odd).

Theorem 6. Consider the fullerene graph C12n (n is odd), then there are n + 1 orbits under the action of automorphism group on the set of edges. Symmetry 2020, 12, 370 16 of 30

Proof. Suppose n is an odd number. The graph C12n has n + 1 levels such that the first and the last levels have six edges and the other levels have 12 edges. There are six edges between i-th and i + 1-th levels. We call them as i-th set of vertical edges. By rotation ρ = 60◦ around the central hexagon, one can see that the edges of the first level and the last level are in the same orbit. In addition, suppose α is a reflection which maps the central hexagon to the last one. One can prove that i-th and n + 1-th (1 ≤ i ≤ n − 1/2) set of edges of vertical edges are in the same orbit under the action of <ρ, α>. Note that in the case that i = (n + 1)/2, we have an orbit of size 6 and the other orbits are of size 12. In addition, by a similar argument, one can see that the edges of the j-th and n + 2-th (1 ≤ j ≤ n + 1/2) levels are in the same orbit. In other words, C12n has 1 + (n − 1)/2 orbits of size 12, n − 1/2 orbits of size 24 and one orbit of size 6. This means that the number of orbits of C12n on the set of edges is n + 1.

Similarly, if n is even, see Figure 11, the following theorem can be obtained.

Figure 11. The fullerene graph C12n (n is even).

Theorem 7. Consider the fullerene graph C12n (n is even), then there are n + 1 orbits under the action of automorphism group on the set of edges.

Now consider the fullerene graph C24n as given in Figure 12. Symmetry 2020, 12, 370 17 of 30

Figure 12. C24n, n is even.

Theorem 8. For the fullerene graph C24n (n ≥ 2), we have ∼ Aut(C24n) = D24 .

Proof. First, we compute the full automorphism group G = Aut(C24n) of fullerene C24n, where n is even as depicted in Figure 12. Suppose a is a symmetry element in which f ix(a) = {11, 12, 41, 102} and b denotes the element of order 12 such that permutation presentation of b is as follows: b = (11, 1n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2) (12, 2n+1, 42, 5n+1, 72, 8n+1, 102, 11n+1,..., 162, 17n+1) (22, 3n+1, 52, 6n+1, 82, 9n+1, 112, 12n+1, . . . , 172, 18n+1) (32, 4n+1, 62, 7n+1, 92, 10n+1, 122, 13n+1, . . . , 182, 1n+1) (1i, 3n−i+3, 5i, 7n−i+3, 9i, 11n−i+3, 13i, 15n−i+3, . . . , 21i, 23n−i+3) (2i, 4n−i+3, 6i, 8n−i+3, 10i, 12n−i+3, 14i,1 6n−i+3, . . . , 22i, 24n−i+3) (3i, 5n−i+3, 7i, 9n−i+3, 11i, 13n−i+3, 15i, 17n−i+3,..., 23i, 1n−i+3) (4i, 6n−i+3, 8i, 10n−i+3, 12i, 14n−i+3, 16i, 18n−i+3, . . . , 24i, 2n−i+3), where (3≤i≤(n+2)/2). 1 G Clearly, G ≥ ha, bi and the orbit-stabilizer property shows that |G| = |(1 ) | × |G11 |. By considering the action of subgroup G1 on the set of vertices, it can be found that any symmetry of fullerene 1 2 1 2 C24n which ﬁxes two vertices 1 and 1 must also ﬁxes the opposite vertices 4 and 10 . Applying 2 G 1 again the orbit-stabilizer property yields |G 1 | = |(1 ) 1 | × |G 1 |. It is easy to prove that |G 1 | = 1 1 12 1 12 G G 2, |(12) 11 | = 1 and |(11) | = 12. Hence |G| = 24. On the other hand, | ha, bi | = 24 and this leads ∼ 1 2 1 2 us to conclude that G = ha, bi = D24. Now suppose n is odd (see Figure 13), f ix(a) = {1 , 1 , 4 , 10 } and b = (11, 1n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2, 21, 2n+2) (12, 2n+1, 42, 5n+1, 72, 8n+1, 102, 11n+1, . . . , 162, 17n+1) (22, 3n+1, 52, 6n+1, 82, 9n+1, 112, 12n+1, . . . , 172, 18n+1) (32, 4n+1, 62, 7n+1, 92, 10n+1, 122, 13n+1, . . . , 182, 1n+1) (1i, 3n−i+3, 5i, 7n−i+3, 9i, 11n−i+3, 13i, 15n−i+3, . . . ., 21i, 23n−i+3) (2i, 4n−i+3, 6i, 8n−i+3, 10i, 12n−i+3, 14i, 16n−i+3, . . . , 22i, 24n−i+3) (3i, 5n−i+3, 7i, 9n−i+3, 11i, 13n−i+3, 15i, 17n−i+3, . . . , 23i, 1n−i+3) (4i, 6n−i+3, 8i, 10n−i+3, 12i, 14n−i+3, 16i, 18n−i+3, . . . , 24i, 2n−i+3) (1(n+3)/2, 3(n+3)/2, 5(n+3)/2, 7(n+3)/2, . . . , 21(n+3)/2, 22(n+3)/2) (2(n+3)/2, 4(n+3)/2, 6(n+3)/2, 8(n+3)/2,..., 22(n+3)/2, (n+3)/2 ∼ 24 ), where (3≤i≤(n-3)/2). By similar argument G = ha, bi = D24. In general, we can conclude that the symmetry elements C24n have the cycle types as given in Table3. Symmetry 2020, 12, 370 18 of 30

Table 3. The cycle type of automorphism group of fullerene C24n.

Permutations Cycle Type 1 124n 6 14212n−2 7 212n 2 38n 2 46n 2 64n 4 122n

Figure 13. C24n, n is odd.

By applying Equation (1) the fullerene graph C24n has (24n + 6 × 4)/24 = n + 1 orbits under the action of automorphism group on the set of vertices.

Theorem 9. [41] The automorphism group of fullerene graph A12n+4(n ≥ 4) is isomorphic with symmetric group S3.

Proof. Consider the labeling of the fullerene graph A12n+4 as shown by Figure 14. Suppose α = ∼ (2, 5, 8)(3, 6, 9) ... (12n + 2, 12n + 4, 12n), and β = (2, 8)(3, 7) ... (12n + 2, 12n). Clearly, S3 = hα, βi ≤ A A = Aut(A12n+4). On the other hand, |A| = |1 ||A1|. Clearly, each automorphism which ﬁxes points A A 1 and 2 must ﬁxes {7, 11, 17, ··· 12n − 1, 12n + 2}, |A1| = |A1,2||2 1 | = 2 × 3. Hence 1 = {1} and, ∼ thus, |A| = 6 which implies that A = S3. Symmetry 2020, 12, 370 19 of 30

12n+4 12n-1

12n-3 12n-2 12n-13 . 12n-4 . 69 70 59 12n-12 . 58 . 57 47 . . 46 . 68 35 . 45 48 60 34 56 33 23 36 22 24 44 21 11 10 61 12n 12n+3 32 2 12 20 37 12n-11 25 49 12n-5 67 9 3 55 1 13 43 31 8 19 4 14 26 5 50 7 18 6 38 15 62 30 42 16 . 54 17 . . 66 27 . .. 29 28 39 41 40 51 12n-10 53 63 12n-6 52 65 .64 . . 12n-9 12n-7 12n-8

12n+2 12n+1

Figure 14. Labeling the vertices of fullerene graph A12n+4.

Theorem 10. [41] The automorphism group of fullerene graph B12n+6(n ≥ 6) is isomorphic with group Z2 × Z2.

Proof. Consider the graph B12n+6 shown in Figure 15. Clearly, α, β are automorphisms of fullerene graph B12n+6:

α = (1, 5)(2, 4)(6, 8) ... (12n + 1, 12n + 4)(12n + 2, 12n + 3)(12n + 5, 12n + 6), β = (2, 8)(3, 7)(4, 6) ... (12n + 3, 12n + 5)(12n + 2, 12n + 6).

Then G = hα, βi ≤ A = Aut(B12n+6). Since every automorphism that ﬁxes point 3 also ﬁxes the points A {7, 26, 27, 33, ... , 12n − 8, 12n − 2}, the orbit-stabilizer property implies that |A| = |3 ||A3| = 2 × 2. ∼ Therefore, A = Z2 × Z2. Symmetry 2020, 12, 370 20 of 30

12n+1

12n-11

12n-10 12n 12n-23 . 12n-22 12n+2 12n+6 . 12n-12 . 12n-9 73 12n-1 84 74 61 12n-21 12n-13 62 72 49 83 60 50 37 75 71 48 38 23 63 59 51 12n-20 47 36 24 12n-8 39 12n-2 35 9 25 82 21 22 10 11 12n-14 70 58 1 40 64 20 2 12 76 46 34 8 26 52 33 7 3 27 6 4 19 5 13 18 14 45 32 17 15 28 53 16 41 ...... 69 57 65 81 29 77 12n-19 12n-15 31 30 42 44 43 12n-7 56 54 68 55 66 67 12n-3 80 78

12n+5 79 12n-18 12n+3 12n-16

12n-6 12n-4 12n-17

12n-5

12n+4

Figure 15. Labeling the vertices of the fullerene graph B12n+6.

Further, this group is isomorphic to the Abelian group Z2 × Z2 of order 4 and the cycle types of elements of S are as given in Table4.

Table 4. Cycle type of automorphism group of fullerene C12n+6.

Permutations Cycle Type 1 112n+6 1 1426n+1 1 1626n 1 26n+3

Theorem 11. [42] The automorphism group of the fullerene graph A12n(n ≥ 4) is isomorphic with dihedral group D24.

Proof. Let n be an even number and consider the fullerene A12n in Figure 16. Consider two permutations α, β with the following permutation representations:

α = (2, 6)(3, 5) ··· (12n − 5, 12n)(12n − 4, 12n − 1)(12n − 3, 12n − 2), β = (1, 12n − 2, 2, 12n − 1, 3, 12n, 4, 12n − 5, 5, 12n − 4, 6, 12n − 3) ... (6n − 5, 6n + 2, 6n − 3, 6n + 4, 6n − 1, 6n + 6, 6n + 1, 6n − 4, 6n + 3, 6n − 2, 6n + 5, 6n).

2 12 −1 ∼ One can prove that α = β = 1, αβα = β and thus A = Aut(A12n) ≥ hα, βi = D24. On the other hand, the identity element and the symmetry element α ﬁxes 1. Hence, the orbit-stabilizer property A ∼ implies that |A| = |1 |.|A1| and thus |A| = 12 × 2 and thus A = D24. By a similar method, if n is odd (as depicted in Figure 17), the automorphism group is isomorphic with dihedral group D24. Symmetry 2020, 12, 370 21 of 30

12n

12n-6 12n-18 12n-7 12n-17

12n-1 12n-5 12n-29 12n-30 12n-19 12n-16 12n-41

12n-31 12n-42 12n-8 . 12n-28 12n-53 12n-20 . 12n-43 . 12n-40 12n-32 6n+6 6n-5 12n-52 12n-44 6n+5 . . 54 43 6n+4 53 6n-4 12n-9 41 42 31 12n 12n-21 52 44 12n-33 30 -15 29 12n-27 12n-45 40 19 32 12n-51 6n+3 17 28 18 20 6n-3 51 7 45 12n-39 16 8 39 6 1 27 33 15 5 2 9 21 4 3 14 10

38 26 13 12 22 34 6n+2 .50 11 .. 46 25 . ... 23 .. 6n-2 12n-46 24 . 12n-50 37 . 35 . 12n-34 49 36 47 12n-38 12n-10 6n+1 48 12n-22 6n-1 12n-26 12n-47 12n 6n 12n-49 -14

12n-48 12n-35 12n-37 12n-2 12n-4 12n-25 12n-23 12n-13 12n-36 12n-11

12n-24

12n-12

12n-3

Figure 16. Labeling the vertices of fullerene graph A12n, n is even.

12n

12n-6

12n-18 12n-7 12n-17

12n-1 12n-19 12n-5 12n-30 12n-29 12n-8 12n-41 12n-16 12n-31 12n-42 12n-20 . 12n-28 12n-43 . 12n-53 6n+11 . 6n+12 12n-32 6n+10 12n-40 12n-44 . 12n-52 6n-1 . 6n 6n-2 . 6n+1 6n+9 54 53 43 12n-9 12n-21 6n-3 6n-11 41 42 31 52 44 12n-33 30 12n-15 29 19 6n+8 40 32 12n-27 12n-45 6n-4 17 12n-51 28 18 20 6n+2 51 6n-10 7 45 16 8 12n-39 39 6 27 1 15 5 2 9 21 33 4 3 14 10 38 26 12 6n-5 13 22 34 ...50 11 . 6n-9 . 25 46 . . 23 . . 6n+3 6n+7 24 12n-34 12n-46 37 . 35 36 . 49 . 12n-50 47 6n-6 6n-8 12n-38 12n-10 48 12n-22 6n+6 6n-7 12n-26 6n+4 12n-14 12n-49 12n-2 12n-47 6n+5 12n-37 12n-4

12n-35 12n-48 12n-25

12n-23 12n-13 12n-36 12n-11

12n-24

12n-12

12n-3

Figure 17. Labeling the vertices of fullerene graph A12n, n is odd.

Theorem 12. [42] The automorphism group of the fullerene graph A10n(n ≥ 7) is isomorphic with dihedral group D20. Symmetry 2020, 12, 370 22 of 30

Proof. Consider the fullerene graph A10n, n is even, as depicted in Figure 18. Two permutations α and β by the following presentation are in the automorphism group:

α = (1, 2)(3, 5) ... (10n − 3, 10n)(10n − 2, 10n − 1), β = (1, 10n)(2, 10n − 1) ··· (3, 10n − 2)(4, 10n − 3)(5, 10n − 4), γ = (1, 2, 3, 4, 5)(6, 7, 8, 9, 10) ... (10n − 4, 10n − 3, 10n − 2, 10n − 1, 10n).

2 10 −1 Suppose βγ = ρ, it is clear that α = ρ = 1, αρα = ρ and thus Γ = hα, ρi ≤ A = Aut(A10n) is isomorphic with dihedral group D20. A symmetry element which ﬁxes 4 (such as identity and α) must A also ﬁxes {9, 10n-9, 10n-4}. Apply the orbit-stabilizer property to obtain |A| = |4 |.|A4| and thus ∼ |A| = 10 × 2 = 20. This yields A = D20. Now, consider the fullerene graph A10n, where n is odd, as ∼ depicted in Figure 19. A similar argument shows that Aut(A10n) = D20.

10n-4

10n-9

10n-19 10n-18

10n-29 10n-28

10n-39 10n-38

5n+1 5n+2 5n-9 5n-8 . .. .. 10n- 10n . 3 10n-17 10n-10 31 10n-27 10n-20 32 10n-30 10n-37 10n-5 21 10n-8 5n 22 5n-7 5n+3 10n-11 5n+10 11 12 10n-16 23 10n-26 10n-21 30 40 33 10n-36 10n-31 5n+9 6 7 5n+4 20 13 5n-6 5n-1 39 34 19 1 2 14 24 29 5 3 8 4 10

28 18 9 15 25 17 16 35 38 27 26 37 36

5n-2 5n-5

5n+8 5n+5 5n-3 5n-4 5n+7 . 5n+6 10n-32 . 10n-35 10n-25 10n-22 10n-33 10n-34

10n-12 10n-23 10n-24 10n-15

10n-6 10n-13 10n-14 10n-7

10n-1 10n-2

Figure 18. Labeling the vertices of fullerene graph A10n, n is even. Symmetry 2020, 12, 370 23 of 30

10n-4

10n-9 10n-19 10n-10 10n-29 10n-20 10n-18 10n-11 10n-28 10n-21 . .. . . 10n-3 . 5n+5 5n-4

10n 5n-3 5n+4 10n-8

10n-5 31 40 30 21 32 39 20 6 11 29 19 12 22

7 10 1 23 28 5 2 18 34 13 33 38 17 14 5n-2 5n+3 9 8 34 24 37 27 10n-27 1615 . . 10n-22 . 25 . 10n-17 .. 26 5n-1 10n-12 5n+2 36 35 10n-26 10n-16 10n-23 10n-13

5n+1 5n

. 10n-6 . 10n-7

10n-24 10n-25

10n-14 10n-15

10n-1 10n-2

Figure 19. Labeling the vertices of fullerene graph A10n, n is odd.

4. Leapfrog Operation

Suppose Cn is a fullerene on n vertices. By means an operation which is called the Leapfrog principle, we can construct bigger fullerenes with the same symmetry group. To do this, put an extra vertex into the centre of each face of Cn. Then connect these new vertices with all the vertices surrounding the corresponding face. Then the dual polyhedron is again a fullerene with 3n vertices, 12 pentagonal and (3n/2)-10 hexagonal faces, see [43,44]. For example, in Figure 20 it is shown that Le(C20) = C60.

Figure 20. The Fullerene C20 and Le(C20).

Here, we compute the symmetry group of fullerenes constructed by Leapfrog. Consider the molecular graph of the fullerene C3n×20 as given in Figure 21. From the Leapfrog principle, one can see that the symmetry group G of these fullerenes is isomorphic to the group Ih = Z2 × A5 of order 120 and the cycle types of elements of G are as given in Table5. Symmetry 2020, 12, 370 24 of 30

Table 5. Cycle type of automorphism elements of fullerene C3n×20.

Permutations Cycle Type n 1 13 ×20 n 16 23 ×10

n−1 n−1 15 13 ×423 ×28

n−1 20 33 ×20 n 24 53 ×4

n−1 20 63 ×10

Figure 21. 3-D graphs of fullerene C3n×20, n = 2.

Now consider the molecular graph of the fullerene F3n×34, Figure 22. In [40] it is proved that the symmetry group G of these fullerenes is isomorphic to the group S3 of order 6 and the cycle types of elements of G are as given in Table6.

Figure 22. 3-D graph of fullerene C3n×34, n = 1. Symmetry 2020, 12, 370 25 of 30

Table 6. Cycle type of automorphism group of fullerene F3n×34.

Permutations Cycle Type n 1 13 ×34 n 3 16n217×3 −3n

n−1 2 33 ×34

Finally, consider the molecular graph of the fullerene C3n×34, Figures 23 and 24. In [40] it is shown that the symmetry group of this fullerene is isomorphic with a cyclic group of order 2, namely Z2 and the cycle types of elements are given in Tables7 and8.

Table 7. Cycle type of automorphism group of fullerene C3n×34 (n is even).

Permutations Cycle Type n 1 13 ×34

n/2 n n/2 1 16×3 234×3 −6×3

Table 8. Cycle type of automorphism group of fullerene C3n ×34 (n is odd).

Permutations Cycle Type n 1 13 ×34

(n−1)/2 n (n−1)/2 1 14×3 234×3 −4×3

Figure 23. 2-D and 3-D graphs of fullerene C34.

Figure 24. 2-D and 3-D graphs of fullerene C3n×34, n = 2. Symmetry 2020, 12, 370 26 of 30

5. The Symmetry Group of Non-Classical Fullerenes Here, we introduce an inﬁnite class of cubic polyhedral graph composed of quadrilateral, hexagonal and octagonal faces, see Figure 25. This class of cubic polyhedral graphs has exactly 16n vertices where n ≥ 3. We denote this new family of cubic polyhedral graph by F16n. Let s, h, o, n and m be the number of squares, hexagons and octagons, carbon atoms and bonds between them, in a given (4, 6, 8) cubic polyhedral graph F. Since each atom lies in exactly 3 faces and each edge lies in 2 faces, the number of atoms is n = (4s + 6h + 8o)/3, the number of edges is m = (4s + 6h + 8o)/2 = 3/2n and the number of faces is f = s + h + o. By Euler’s formula n − m + f = 2 and then s + h + o = 2 + n/2. This leads us to conclude that s = 12, h = 8n − 16 and o = 6, for n ≥ 3.

Example 4. Here, we compute the order of automorphism group of polyhedral graph F48 depicted in Figure 25. We compute the order of G = Aut(F16n) of symmetries of the polyhedral graph F16n, for n = 3 depicted in Figure 26. Clearly, the automorphism group of F16n for n ≥ 3 can be computed similarly. If α denotes the rotation of the F16n through an angle of 90◦ around an axis through the midpoints of the front and back faces, then the corresponding permutation is α = (1, 3, 5, 7)(2, 4, 6, 8)(9, 15, 26, 21)(10, 16, 27, 32)(11, 17, 28, 22)(12, 18, 29, 23)(13, 19, 30, 24)(14, 20, 31, 25)(33, 45, 41, 37)(34, 46, 42, 38)(35, 47, 43, 39)(36, 48, 44, 40). Thus the orbit of the subgroup hαi is 1hαi = {1, 3, 5, 7}. Now, consider the axis symmetry element which fixes no vertices, it is as follows: β = (1, 2)(3, 8)(4, 7)(5, 6)(9, 20)(10, 19)(11, 18)(12, 17)(13, 16)(14, 15)(21, 31)(22, 29)(23, 28)(24, 27)(25, 26) (30, 32)(33, 40)(34, 39)(35, 38)(36, 37)(41, 48)(42, 47)(43, 46)(44, 45). Clearly, G ≥ hα, βi. According to G the orbit-stabilizer property |G| = |2 | × |G2| while no element fixes 2. This means that |G2| = 1 and so |G| = |2G|. It is clear that 2hα,βi = {1, 2, 3, 4, 5, 6, 7, 8} and thus, |2hα,βi| = 8. On the other hand, every automorphism preserves the adjacency and thus there is no vertex u other than {1, 2, 3, 4, 5, 6, 7, 8} G and a permutation σ ∈ Aut(F16n) with σ(u) = 2. This implies that 2 = {1, 2, 3, 4, 5, 6, 7, 8} and hence |G| = 8. It is not difficult to see that |hα, βi| = 8 where α4 = β2 = 1 and βαβ = α−1. One can verify then ∼ G = hα, βi = D8 and the proof is complete.

Figure 25. The polyhedral graph F16n, where n = 3. Symmetry 2020, 12, 370 27 of 30

Figure 26. Labeling of fullerene F96.

In general, we have the following theorem.

Theorem 13. The automorphism group Aut(F16n) is isomorphic with dihedral group D8.

Now, consider an inﬁnite class of polyhedral graphs with exactly 12n + 72 vertices, where n is an integer greater than or equal with 2. That’s why we name this new family of polyhedral graphs by F12n+72.

Example 5. Consider the fullerene graph F96 depicted in Figure 26. If α denotes the rotation of F96 through an angle of 60◦ around an axis through the midpoints of the front and back faces, then the corresponding permutation is

α = (1, 2, 3, 4, 5, 6)(7, 10, 14, 17, 20, 24)(8, 11, 15, 18, 21, 25)(9, 12, 16, 19, 23, 26) (13, 50, 58, 74, 66, 42)(22, 71, 47, 39, 55, 63)(27, 28, 29, 30, 31, 32) (33, 48, 56, 72, 64, 40)(34, 49, 57, 73, 65, 41 )(35, 51, 59, 75, 67, 43) (36, 52, 60, 76, 68, 44)(37, 53, 61, 77, 69, 45)(38, 54, 62, 78, 70, 46) (79, 80, 81, 82, 83, 84)(85, 86, 87, 88, 89, 90)(91, 96, 95, 94, 93, 92). Symmetry 2020, 12, 370 28 of 30

Thus, an orbit of hαi containing the vertex 1 is 1hαi = {1, 2, 3, 4, 5, 6}. Now, consider the axis symmetry element which ﬁxes vertices {1, 4, 8, 18, 43, 44, 59, 60, 85, 88, 92, 95}, the corresponding permutation is

β = (2, 6)(3, 5)(7, 9)(10, 26)(11, 25)(12, 24)(13, 71)(14, 23)(15, 21)(16, 20) (17, 19)(22, 50)(27, 28)(29, 32)(30, 31)(33, 70)(34, 69)(35, 67)(36, 68) (37, 65)(38, 64)(39, 66)(40, 46)(41, 45)(42, 47)(48, 78)(49, 77)(51, 75) (52, 76)(53, 73)(54, 72)(55, 74)(56, 62)(57, 61)(58, 63)(79, 80)(81, 84) (82, 83)(86, 90)(87, 89)(91, 93)(94, 96).

G Let G = Aut(F96), clearly G ≥ hα, βi and the orbit-stabilizer property implies that |G| = |1 | × |G1|. Any symmetry of the polyhedral graph F96 which ﬁxes vertex 1 must also ﬁxes the opposite vertex 4. By again G applying the orbit-stabilizer property, we found that |G1| = |4 1 | × |G1,4|. It is easy to prove that |G1,4| = 2 and hence |G| = 12. On the other hand, |hα, βi| = 12, where α4 = β2 = 1, βαβ = α−1. This leads us to ∼ conclude that G = hα, βi = D12.

In general, we have the following theorem.

Theorem 14. The automorphism group Aut(F12n+72) is isomorphic with dihedral group D12.

Author Contributions: Methodology and writing-original draft preparation and editing, M.G.; investigation, review and funding acquisition, M.D.; writing-original draft, S.R.; software, M.R.-P. All authors have read and agreed to the published version of the manuscript. Funding: Matthias Dehmer thanks the Austrian Science Funds for supporting this work (project P30031). Conﬂicts of Interest: The authors declare no conﬂict of interest.

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