Power graphs of (non)orientable genus two
PPower graphs of (non)orientable genus two
Xuanlong Ma , Gary L. Walls , Kaishun Wang Sch. Math. Sci. & Lab. Math. Com. Sys., Beijing Normal University, Beijing, 100875, China Department of Mathematics, Southeastern Louisiana University, Hammond, LA 70402, USA
Abstract
The power graph Γ G of a finite group G is the graph whose vertex setis the group, two distinct elements being adjacent if one is a power of theother. In this paper, we classify the finite groups whose power graphs have(non)orientable genus two. Keywords:
Power graph, finite group, genus.
MSC 2010:
Throughout this paper, every graph is finite, simple and connected. A graph Γ iscalled a planar graph if Γ can be drawn in the plane so that no two of its edges crosseach other. In addition, we say that Γ can be embedded in the plane. A non-planargraph can be embedded in some surface obtained from the sphere by attaching somehandles or crosscaps. We denote by S k a sphere with k handles and by N k a spherewith k crosscaps. Note that both S and N are the sphere itself, and S and N are a torus and a projective plan, respectively. The smallest non-negative integer k such that a graph Γ can be embedded on S k is called the orientable genus or genus of Γ, and is denoted by γ (Γ). The nonorientable genus of Γ, denoted by γ (Γ), is thesmallest integer k such that Γ can be embedded on N k .The problem of finding the graph genus is NP-hard [23]. Many research articleshave appeared on the genus of graphs constructed from some algebraic structures.For example, Wang [24] found all rings of two specific forms that have genus at mostone. Chiang-Hsieh et al. [7] characterized the commutative rings of genus one. Very E-mail addresses: [email protected] (X. Ma), [email protected] (G.L. Walls)[email protected] (K. Wang). a r X i v : . [ m a t h . G R ] D ec ecently, Rajkumar and Devi [20] classified the finite groups whose intersection graphsof subgroups have (non)orientable genus one. Afkhami et al. [2] classified planar,toroidal, and projective commuting and noncommuting graphs of finite groups.Here we study the genus of the power graph of a finite group. The undirected powergraph Γ G of a group G has the vertex set G and two distinct elements are adjacentif one is a power of the other. The concept of a power graph was first introducedand considered by Kelarev and Quinn [13]. Note that since the paper deals onlywith undirected graphs, for convenience, throughout we use the term “power graph”to refer to an undirected power graph defined as above. Recently, many interestingresults on power graphs have been obtained, see [4–6, 9, 10, 14–18]. Furthermore, [1]is a survey that is a detailed list of results and open questions on power graphs.Doostabadi and Farrokhi D.G. [8] classified the finite groups whose power graphshave (non)orientable genus one. The goal of the paper is to find all power graphs of(non)orientable genus two. Our main results are the following theorems. Theorem 1.1.
Let G be a finite group. Then Γ G has orientable genus two if andonly if G is isomorphic to one group in Table 1. Table 1:
All finite groups G with γ (Γ G ) = 2 GAP ID Group[8 , Z [12 , Z × Z [16 , D [16 , QD [16 , Q [18 , Z × S [24 , Z × ( Z (cid:111) ϕ Z )[24 ,
8] ( Z × Z ) (cid:111) ϕ Z [24 , Z × Z × S [36 , Z × A [72 ,
43] ( Z × A ) (cid:111) ϕ Z Theorem 1.2.
There is no finite group G such that Γ G has nonorientable genus two. In this section we briefly recall some notation, terminology, and basic results andprove a lemma which we need in the sequel.Let Γ be a graph. Denote by V (Γ) and E (Γ) the vertex set and the edge set of Γ,respectively. An edge of Γ is denoted simply by ab where a, b ∈ V (Γ). If V (cid:48) ⊆ V (Γ),2e define Γ − V (cid:48) to be the subgraph of Γ obtained by deleting the vertices in V (cid:48) andall edges incident with them. Similarly, if E (cid:48) ⊆ E (Γ), then Γ − E (cid:48) is the subgraphof Γ obtained by deleting the edges in E (cid:48) . For two vertex-disjoint graphs Γ and ∆,Γ ∪ ∆ denotes the graph with vertex set V (Γ) ∪ V (∆) and edge set E (Γ) ∪ E (∆), andΓ + ∆ consists of Γ ∪ ∆ and all edges joining a vertex of Γ and a vertex of ∆. A unionof k isomorphic graphs Γ is denoted by k Γ. We use the natation (cid:100) x (cid:101) to denote theleast integer that is greater than or equal to x . Denote by K n and K m,n the completegraph of order n and the complete bipartite graph, respectively.For any subgraph ∆ of a graph Γ, one easy observation is that γ (∆) ≤ γ (Γ) and γ (∆) ≤ γ (Γ). The following result gives the (non)orientable genus of a completegraph and a complete bipartite graph. Theorem 2.1. ([25, p. 58, p. 152])
Let n be an integer at least . Then ( a ) γ ( K n ) = (cid:100) ( n − n − (cid:101) . ( b ) γ ( K n ) = (cid:100) ( n − n − (cid:101) if n (cid:54) = 7 ; γ ( K ) = 3 . ( c ) γ ( K m,n ) = (cid:100) ( m − n − (cid:101) . ( d ) γ ( K m,n ) = (cid:100) ( m − n − (cid:101) . A block of a graph Γ is a maximal connected subgraph B of Γ with respect to theproperty that removing any vertex of B does not disconnect B . The following resulttells us how to compute the (non)orientable genus of a graph by its blocks. Theorem 2.2. ([3, Theorem 1], [21, Corollary 3])
Let Γ be a connected graph with n blocks B , · · · , B n . Then (1) γ (Γ) = n (cid:88) i =1 γ ( B i ) . (2) If γ ( B i ) = 2 γ ( B i ) + 1 for each i , then γ (Γ) = 1 − n + n (cid:88) i =1 γ ( B i ) . Otherwise, γ (Γ) = 2 n − n (cid:88) i =1 µ ( B i ) , where µ ( B i ) = max { − γ ( B i ) , − γ ( B i ) } . Now we state Euler’s formula for S k and N k .3 heorem 2.3. Suppose that Γ is a connected graph that is embedded on a surface S ,resulting in f faces. Then ( a ) If S = S γ (Γ) , then | V (Γ) | − | E (Γ) | + f = 2 − γ (Γ) . ( b ) If S = N γ (Γ) , then | V (Γ) | − | E (Γ) | + f = 2 − γ (Γ) . All groups considered in this paper are finite. An element of order 2 in a groupis called an involution . Let G be a group and g be an element of G . Denote by | G | and | g | the orders of G and g , respectively. Let S be a subset of G . The set of primedivisors of | S | is denoted by π ( S ) and the set of natural numbers consisting of ordersof elements of S is denoted by π e ( S ). Also Z mn is used for the m -fold direct productof the cyclic group Z n with itself. Denote by Z n and D n the cyclic group of order n and the dihedral group of order 2 n , respectively.Planar power graphs were characterized in [19]. Theorem 2.4. ([19, Corollary 4])
Let G be a group. Then Γ G is planar if and onlyif π e ( G ) = { , , , } . For a subset S of a group G , Γ G [ S ] denotes the induced subgraph of Γ G by S andif the situation is unambiguous, then we denote Γ G [ S ] simply by Γ S . Lemma 2.5.
Let G be a group. Suppose that S is a union of some subgroups of G such that π e ( G \ S ) ⊆ { , , } . If γ (Γ S ) = r ( resp. γ (Γ S ) = r ) , then γ (Γ G ) = r ( resp. γ (Γ G ) = r ) .Proof. We first prove that if γ (Γ S ) = r , then γ (Γ G ) = r . Since γ (Γ S ) = r , Γ S canbe embedded on S r . Now we fix an embedding E of Γ S on S r . Clearly, if G \ S = ∅ ,then γ (Γ G ) = r . Thus, we may assume that G \ S (cid:54) = ∅ . Case 1. Γ G has no edge ab such that a ∈ S \ { e } and b ∈ G \ S , where e is theidentity element of G .Take a face F containing e in E . By Theorem 2.4, we may insert all vertices in G \ S and all edges incident with them in F without crossings. This implies that Γ G also can embed on a surface of genus r and so γ (Γ G ) = r , as required. Case 2. Γ G has an edge ab such that a ∈ S \ { e } and b ∈ G \ S .It is easy to see that a and b are an involution and an element of order 4, respec-tively. Let b, b − , b , b − , · · · , b t , b − t be all elements that are adjacent to a in G \ S .Note that | b i | = 4 for each i = 1 , . . . , t . Since e and a are adjacent, we may take a face F containing edge ea in E . Note that for any vertex of { b, b − , b , b − , · · · , b t , b − t } ,there are exactly three vertices that are adjacent to it. Then we can insert the vertices b, b − , b , b − , · · · , b t , b − t and and all edges incident with them in F without crossings,4 . .ea b b − b b − b − t b t Figure 1: Insert some vertices and edges in face F .as shown in Figure 1. It follows that if G has other involutions in S \ { e } that areadjacent to some elements in G \ S , then we may insert them in some faces of E without crossings. For the remainder vertices and edges in Γ G \ S , by Theorem 2.4 wealso can insert them in a face containing e of E without crossings. This implies that γ (Γ G ) = r , as desired.Similarly, we have that if γ (Γ S ) = r , then γ (Γ G ) = r . In this section we prove some results on finite groups with some cyclic subgroups oforder 6.
Lemma 3.1.
There is no group that has precisely two cyclic subgroups of order .Proof. Suppose, towards a contradiction, that there exists a group G that has exactlytwo cyclic subgroups of order 6, say (cid:104) a (cid:105) and (cid:104) b (cid:105) . Note that G has precisely fourelements a, a − , b, b − of order 6. Then b a = a, a , b or b , where b a is the conjugate of b by a , that is, b a = a − ba . Since (cid:104) a (cid:105) (cid:54) = (cid:104) b (cid:105) , one has that b a = b or b − . It follows that (cid:104) a (cid:105)(cid:104) b (cid:105) is a subgroup of order 36, 18 or 12. We check all groups of order 36, 18 and12 (for example using the computer algebra system GAP [11]) and find that there isno such group that has only four elements of order 6, which is a contradiction. Lemma 3.2.
Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G hasprecisely three cyclic subgroups. If there exist two cyclic subgroups of order of G such that their intersection has order , then the intersection of any two cyclicsubgroups of order of G is of order . roof. Let (cid:104) a (cid:105) , (cid:104) b (cid:105) , (cid:104) c (cid:105) be the three cyclic subgroups of order 6 of G and assumewithout loss of generality that (cid:104) b (cid:105) ∩ (cid:104) c (cid:105) = (cid:104) b (cid:105) = (cid:104) c (cid:105) .If (cid:104) a (cid:105) is a normal subgroup of G , then [ G : C G ( a )] ≤ N/C
Theorem (cf.[22, Theorem 1.6.13], here [ G : C G ( a )] denotes the index of C G ( a ) in G ). It followsthat b ∈ C G ( a ). Hence, | ab | = 6. If ab ∈ (cid:104) a (cid:105) , then b ∈ (cid:104) a (cid:105) so b = a and wehave a = b = c , as required. If ab ∈ (cid:104) b (cid:105) , then a ∈ (cid:104) b (cid:105) , a contradiction. Similarly,if ab ∈ (cid:104) c (cid:105) , then again a ∈ (cid:104) c (cid:105) , since b = c and this is a contradiction. The resultfollows in this case.So assume that (cid:104) a (cid:105) is not normal in G . Then, without loss of generality there isa g ∈ G , so that (cid:104) a (cid:105) g = (cid:104) b (cid:105) . Now (cid:104) a (cid:105) g ∩ (cid:104) c (cid:105) = (cid:104) b (cid:105) . It follows that (cid:104) a (cid:105) ∩ (cid:104) c (cid:105) g − hasorder 3. Now (cid:104) c g − (cid:105) cannot be (cid:104) a (cid:105) , so it must be either (cid:104) b (cid:105) or (cid:104) c (cid:105) . It follows thateither (cid:104) a (cid:105) = (cid:104) b (cid:105) or (cid:104) a (cid:105) = (cid:104) c (cid:105) and result follows.In GAP [11], the GAP ID [ n, m ] which is a label that uniquely identifies the groupin GAP, the first number n in the square brackets is the order of the group, and thesecond number m simply enumerates different groups of the same order. Theorem 3.3.
Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G hasprecisely three cyclic subgroups (cid:104) a (cid:105) , (cid:104) b (cid:105) , (cid:104) c (cid:105) of order . Then there exist two cyclicsubgroups of order in G such that their intersection has order if and only if G isisomorphic to one group in Table 2. Table 2:
All finite groups satisfying the conditions
GAP ID Group π e ( G )[12 , Z × Z { , , , } [18 , Z × S { , , , } [24 , Z × ( Z (cid:111) ϕ Z ) { , , , , } [24 ,
8] ( Z × Z ) (cid:111) ϕ Z { , , , , } [24 , Z × Z × S { , , , } [36 , Z × A { , , , } [72 ,
43] ( Z × A ) (cid:111) ϕ Z { , , , , } Proof.
We check all groups of order at most 144 (for example using the computeralgebra system GAP [11]) and find that the groups G in Table 2 are precisely thegroups of order at most 144 satisfying the conditions:( a ) π e ( G ) ⊆ { , , , , } .( b ) G has precisely three cyclic subgroups of order 6.( c ) There exist two cyclic subgroups of order 6 of G such that their intersectionhas order 3. 6n order to complete the proof, next we prove if G is a group satisfying theconditions, then | G | ≤ (cid:104) b (cid:105) ∩ (cid:104) c (cid:105) = (cid:104) a (cid:105) ∩ (cid:104) b (cid:105) = (cid:104) a (cid:105) ∩ (cid:104) c (cid:105) = (cid:104) a (cid:105) = (cid:104) b (cid:105) = (cid:104) c (cid:105) .Thus, without loss of generality, we may assume that a = b = c . Case 1. a ∈ Z ( G ), the center of G .Note that for any prime divisor p of | G | , in general, the number of subgroups oforder p of G is congruent to 1 modulo p . Since a ∈ Z ( G ) and G has exactly threecyclic subgroups of order 6, G has a unique subgroup (cid:104) a (cid:105) of order 3. This impliesthat (cid:104) a (cid:105) is normal in G .Suppose that a ∈ Z ( G ). Since π e ( G ) ⊆ { , , , , } , G has no elements of order4. If G has a unique involution, then it is easy to see that G ∼ = Z , a contradiction.Therefore, G has exactly three involutions. It follows that G = Z × Z .Now suppose that a / ∈ Z ( G ). Then C G ( a ) (cid:54) = G . So, by N/C
Theorem we seethat | G | = 2 | C G ( a ) | . Since a = b = c , we have a, b, c ∈ C G ( a ). Thus, C G ( a )also satisfies the conditions ( a )–( c ). Moreover, note that a , a ∈ Z ( C G ( a )) and theproof above, it follows that C G ( a ) ∼ = Z × Z . Consequently, in this case we have | G | = 24, as desired. Case 2. a / ∈ Z ( G ).By Case 1 we may assume that b , c / ∈ Z ( G ). Let S = { a , b , c } . It followsthat S is a conjugacy class of G . Thus, we obtain that [ G : C G ( a )] = [ G : C G ( b )] =[ G : C G ( c )] = 3. Furthermore, it is clear that C G ( a ) ∩ C G ( b ) ⊆ C G ( c ). Hence,one has that C G ( a ) ∩ C G ( b ) ∩ C G ( c ) = C G ( a ) ∩ C G ( b ). Similarly, one can get C G ( a ) ∩ C G ( c ) = C G ( a ) ∩ C G ( b ) = C G ( b ) ∩ C G ( c ).Now let η be the group action of G on S , where each element of G acts on theconjugacy class S by conjugation. ThenKer( η ) = C G ( a ) ∩ C G ( b ) ∩ C G ( c ) = C G ( a ) ∩ C G ( b ) . Note that η is a homomorphism from G to S . So we have [ G : C G ( a ) ∩ C G ( b )] ≤ C G ( a ) : C G ( a ) ∩ C G ( b )] ≤ Subcase 2.1. [ C G ( a ) : C G ( a ) ∩ C G ( b )] = 1.Then [ C G ( b ) : C G ( c ) ∩ C G ( b )] = 1, [ C G ( c ) : C G ( c ) ∩ C G ( a )] = 1 and C G ( a )is normal in G . This implies that C G ( a ) = C G ( b ) = C G ( c ) and so b , c ∈ C G ( a ).Since a = b = c , one has that a, b, c ∈ C G ( a ). This implies that C G ( a ) is a groupsatisfying the conditions ( a )–( c ). Note that a ∈ Z ( C G ( a )). By Case 1 we have | C G ( a ) | ≤
24. It follows that | G | = 3 | C G ( a ) | ≤
72, as desired.
Subcase 2.2. [ C G ( a ) : C G ( a ) ∩ C G ( b )] = 2.7hen [ C G ( b ) : C G ( b ) ∩ C G ( c )] = [ C G ( c ) : C G ( c ) ∩ C G ( a )] = 2. Note that C G ( a ) ∩ C G ( b ) is clearly a normal subgroup of G .Suppose that a ∈ C G ( a ) ∩ C G ( b ). Then S ⊆ C G ( a ) ∩ C G ( b ) and a ∈ C G ( a ) ∩ C G ( b ). It follows that a, b, c ∈ C G ( a ) ∩ C G ( b ). So C G ( a ) ∩ C G ( b ) is a groupsatisfying the conditions ( a )–( c ) and a ∈ Z ( C G ( a ) ∩ C G ( b )). In view of Case 1, wehave that | C G ( a ) ∩ C G ( b ) | ≤
24. Thus, | G | = 6 | C G ( a ) ∩ C G ( b ) | ≤ a / ∈ C G ( a ) ∩ C G ( b ). Note that S is a conjugacy class. Then c , b / ∈ C G ( a ) ∩ C G ( b ). Note that C G ( b ) (cid:54) = C G ( a ). Thus, a b / ∈ C G ( a ). Nowit is easy to check that | a b | = 3. This implies that (cid:104) a , b (cid:105) ∼ = S . Note that (cid:104) a , b (cid:105) ∩ C G ( a ) ∩ C G ( b ) = 1 and C G ( a ) ∩ C G ( b ) is normal in G . We have G ∼ = (cid:104) a , b (cid:105) × ( C G ( a ) ∩ C G ( b )). Since G has precisely three cyclic subgroups of order6, C G ( a ) ∩ C G ( b ) has no involutions and has only two elements of order 3. So C G ( a ) ∩ C G ( b ) ∼ = Z . This implies that G ∼ = S × Z , as desired. Theorem 3.4.
Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G hasprecisely four cyclic subgroups of order . Then there is no finite group G suchthat the intersection of two cyclic subgroups of order of G has order , and theintersection of the remaining two cyclic subgroups of order also has order .Proof. Suppose, for a contradiction, that G is a such group of minimal order. Let (cid:104) a (cid:105) , (cid:104) b (cid:105) , (cid:104) c (cid:105) , (cid:104) d (cid:105) be the four cyclic subgroups of order 6 of G , and |(cid:104) a (cid:105) ∩ (cid:104) b (cid:105)| = 3 and |(cid:104) c (cid:105) ∩ (cid:104) d (cid:105)| = 3. Without loss of generality, we may assume that a = b and c = d .Then (cid:104) a , b (cid:105) ⊆ C G ( a ). Note that (cid:104) a , b (cid:105) has at least 3 involutions. Thus, (cid:104) a, b (cid:105) hasat least 3 cyclic subgroups of order 6. So, without loss of generality, we may assumethat c ∈ (cid:104) a, b (cid:105) .If d ∈ (cid:104) a, b (cid:105) , then it is clear that (cid:104) a, b, c, d (cid:105) ⊆ C G ( a ). Suppose that d / ∈ (cid:104) a, b (cid:105) .Then (cid:104) a, b (cid:105) has precisely three cyclic subgroups of order 6 and |(cid:104) a (cid:105) ∩ (cid:104) b (cid:105)| = 3. ByLemma 3.2, we may assume that a = b = c . Hence, we have a = d . It followsthat d ∈ C G ( a ) and so (cid:104) a, b, c, d (cid:105) ⊆ C G ( a ). This implies that we always may assumethat (cid:104) a, b, c, d (cid:105) ⊆ C G ( a ).Now note that G is of minimal order. We have G = C G ( a ). This implies that a ∈ Z ( G ). Since G has precisely four cyclic groups of order 6, G has at most fourinvolutions. Now it is easy to check that the number of involutions of G is 2 or 4.This contradicts the fact that the number of involutions of a finite group of evenorder is odd. Lemma 4.1. γ (Γ Z n ) = 2 if and only if n = 8 . In particular, γ (Γ Z n ) ≥ for n ≥ . roof. Note that for a group G , Γ G is complete if and only if G is a cyclic group ofprime power order by [6, Theorem 2.12]. Thus if n = 8, then γ (Γ Z ) = 2 by Theorem2.1.Now suppose that γ (Γ Z n ) = 2. Since γ ( K ) = 1, one has γ (Γ Z n ) ≤ n ≤ n ≥
8. If n ≥
9, by [19, Theorem 2] one easy calculationshows that the clique number of Γ Z n is greater than 8 , and so its genus is not twoby Theorem 2.1. This implies that n = 8.The generalized quaternion group Q of order 16 which is given by (cid:104) x, y : x = y , x = 1 , y − xy = x − (cid:105) . QD denotes the group of order 16 that is given by (cid:104) a, b : a = b = 1 , bab = a (cid:105) . Lemma 4.2.
Let G be a p -group, where p is a prime. Then γ (Γ G ) = 2 if and only if G is isomorphic to one of the following groups: Z , D , Q , QD . (1) Proof.
By verifying we know that every group G in (1) has a unique cyclic subgroup S of order 8, and π e ( G \ S ) ⊆ { , , } . Note that γ (Γ S ) = γ ( K ) = 2. By Lemma2.5, we have γ (Γ G ) = 2.Now we suppose that γ (Γ G ) = 2. By Lemma 4.1, every element of G has order atmost 8. Thus, we may assume that p ≤ p = 7. Note that γ (Γ Z ) = 1. Then | G | = 7 n for some n ≥
2. Thisimplies that G has a subgroup A isomorphic to Z × Z or Z . Since G has noelements of order 49, A ∼ = Z × Z . So Γ A is isomorphic to K + 8 K that has genus8 by Theorems 2.2, a contradiction. Similarly, we obtain p (cid:54) = 5.If p = 3, then π e ( G ) = { , } , a contradiction by Theorem 2.4. Thus, now wemay suppose that | G | = 2 n for some n ≥
3, and so π e ( G ) = { , , , } . If G hastwo distinct cyclic subgroups of order 8, then the subgraph of Γ G induced by the twocyclic subgroups contains a subgraph isomorphic to K + ( K ∪ K ) which has genus3 by Theorem 2.2, a contradiction. This implies that G has a unique cyclic subgroupof order 8, which is normal in G . Let g ∈ G with | g | = 8. If there exists an element x in G \ (cid:104) g (cid:105) such that x ∈ C G ( g ), then G has a subgroup isomorphic to Z × Z , whichis a contradiction since Z × Z has two cyclic subgroups of order 8. Therefore, wehave that C G ( g ) = (cid:104) g (cid:105) . This implies that the quotient group G/ (cid:104) g (cid:105) is isomorphic toa subgroup of the full automorphism group of (cid:104) g (cid:105) . Thus, we conclude that | G | = 8,16 or 32. If | G | = 8, then G ∼ = Z . If | G | = 16, then G ∼ = D , Q or QD . Sincethere is no such group G of order 32 such that π e ( G ) = { , , , } and G has a uniquecyclic subgroup of order 8, we get the desired result.9 emma 4.3. Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G has pre-cisely three cyclic subgroups of order . Then γ (Γ G ) = 2 if and only if the intersectionof any two cyclic subgroups of order is of order .Proof. Suppose that γ (Γ G ) = 2. If the intersection of any two cyclic subgroups oforder 6 is of order 1 or 2, then it is easy to see that Γ G has a subgraph isomorphic to K + 3 K that has genus three by Theorem 2.2, and so a contradiction. Therefore,there exist at least two cyclic subgroups of order 6 in G such that their intersectionhas order 3. Now by Lemma 3.2, we get the desired result.For the converse, let H , H and H be the three cyclic subgroups of G . We assumethat H = { e, f , f , x, g , g } , H = { e, f , f , y, g , g } and H = { e, f , f , z, g , g } ,where | f i | = 6 , | g j | = 3 and | x | = | y | = | z | = 2 for i = 1 , . . . , j = 1 , G induced by ∪ i =1 H i . Denote by Γ (cid:48) the subgraph ofΓ obtained by deleting the vertices x, y, z and the edges eg , eg , g g , f f , f f , f f in Γ. Then Γ (cid:48) is isomorphic to K , . Note that γ (Γ (cid:48) ) = 1, the boundary of each faceis a 4-cycle when drawing Γ (cid:48) without crossings on a torus, and any two faces of Γ (cid:48) have at most one boundary edge in common (see [7, Remark 1.4]). Now we proceedto prove γ (Γ) ≥ γ (Γ) = 1. Then Γ can be embedded in a toruswithout crossings. Fix an embedding E of Γ on a tours. By deleting some verticesand edges from the embedding of Γ, we can get an embedding E (cid:48) of Γ (cid:48) on a tours.This implies that all faces of E can be recovered by inserting some vertices and edgesinto the faces of E (cid:48) . Let F (cid:48) be the face of E (cid:48) into which x is inserted during therecovering process from E (cid:48) to E . Note that xf , xf ∈ E (Γ). We obtain the face F (cid:48) as shown in Figure 2, where { u, v } ⊆ { e, g , g } . Since { e, g , g } is a clique, one has u vf f x Figure 2: Insert a in F (cid:48) .that u is adjacent to v in Γ. Thus, in order to insert the edge uv without crossings, E (cid:48) must have a face different with F (cid:48) such that its boundary is a 4-cycle containing u and v . This implies that there are two faces of E (cid:48) such that they have two boundaryedges in common, a contradiction. Thus, γ (Γ) ≥ S as shown in Figure 3. This means10hat γ (Γ) = 2. Now note that π e ( G \ ∪ i =1 H i ) ⊆ { , , } . By Lemma 2.5 one hasthat γ (Γ G ) = 2. e f ef f f g ee e e eef g x f yyf f f g x g zf Figure 3: An embedding of Γ on S . Lemma 4.4.
Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G hasprecisely four cyclic subgroups H , . . . , H of order . If γ (Γ G ) = 2 , then | H i ∩ H j | = 3 and | H s ∩ H t | = 3 , where { i, j, s, t } = { , , , } .Proof. By considering all possible cases and Theorem 2.2, it is not difficult to get thedesired result.
Lemma 4.5.
Let G be a group with π e ( G ) ⊆ { , , , , } . If G has at least fivecyclic subgroups H , . . . , H of order , then γ (Γ G ) ≥ .Proof. If | ∩ i =1 H i | (cid:54) = 3, we check all possible cases and then by Theorem 2.2,it is easy to see that γ (Γ G ) ≥
3. Thus, we may suppose that | ∩ i =1 H i | = 3.Let H = { e, f , f , a, g , g } , H = { e, f , f , b, g , g } , H = { e, f , f , c, g , g } , H = { e, f , f , d, g , g } and H = { e, f , f , w, g , g } , where | a | = | b | = | c | = | d | = | w | = 2, | f i | = 6 and | g j | = 3 for i = 0 , . . . , j = 1 ,
2. Denote by ∆the subgraph of Γ G induced by the vertices { e, f i , g , g : i = 0 , . . . , } . Write ∆ (cid:48) =∆ − { f f , f f , f f , f f , eg , eg , g g } and ∆ (cid:48)(cid:48) = ∆ (cid:48) − { f , f } . Then ∆ (cid:48)(cid:48) ∼ = K , .Since γ ( K , ) = 2, one has γ (∆) ≥
2. Next we prove γ (∆) ≥ γ (∆) = 2. Then γ (∆ (cid:48) ) = 2. Fix an embedding E (cid:48) of ∆ (cid:48) on S . Thus, we may get an embedding E (cid:48)(cid:48) of ∆ (cid:48)(cid:48) on S such that E (cid:48) canbe recovered by inserting f , f and all edges incident with f , f into the embedding E (cid:48)(cid:48) . Since γ (∆ (cid:48)(cid:48) ) = 2, by Theorem 2.3 there are 11 faces when drawing ∆ (cid:48)(cid:48) without11rossings on S and thereby, each of the faces is a 4-cycle or a 6-cycle. Let F (cid:48)(cid:48) t denotethe face of ∆ (cid:48)(cid:48) into which f is inserted during the recovering process from E (cid:48)(cid:48) to E (cid:48) . Since f is adjacent to f in ∆ (cid:48) , f should be also inserted into F (cid:48)(cid:48) t to avoid anycrossing. Moreover, since ef i , g f i , g f i ∈ E (∆ (cid:48) ) for i = 0 ,
9, one has that e, g , g liein the boundary of F (cid:48)(cid:48) t . This implies that F (cid:48)(cid:48) t is a 6-cycle. Then after inserting f , f and f f , ef , ef , g f , g f , f g into F (cid:48)(cid:48) t we get Figure 4 as below. But, it is easy to eg g f f Figure 4: The face F (cid:48)(cid:48) t .see from Figure 4 that we can not insert the edge f g into F (cid:48)(cid:48) t without crossings, acontradiction. Thus, we conclude that γ (∆) (cid:54) = 2 and so γ (∆) ≥
3. It follows that γ (Γ G ) ≥ Proof of Theorem 1.1.
Suppose that γ (Γ G ) = 2. Then by Lemma 4.1 we see that π e ( G ) ⊆ { , , , , , , , } . Suppose that P (cid:54) = 1 is a Sylow 5-subgroup of G . Then γ (Γ P ) ≤ γ (Γ G ) = 2.If | P | ≥
25, then G has a subgroup K isomorphic to Z × Z and so γ (Γ K ) = 6by Theorem 2.2, a contradiction. It follows that P ∼ = Z . Furthermore, by Sylow’sTheorem, the number of Sylow 5-subgroups of G is 5 k + 1 for some integer k . Notethat by Theorem 2.2, the subgraph of Γ G induced by all Sylow 5-subgroups of G has genus 5 k + 1. Therefore, we have that k = 0 and so P is normal in G . Since γ (Γ Z ) = 2, G has no elements of order 8 by Theorem 2.2. Note that every groupof order 30 has an element of order 15, and every group of order 35 is isomorphicto Z . So G has no elements of order 7 and 6. Consequently, we conclude that π e ( G ) ⊆ { , , , , } . Since G has precisely one subgroup of order 5 and γ (Γ P ) = 1,one has γ (Γ G ) = 1 by Lemma 2.5, a contradiction. This implies that 5 / ∈ π ( G ).Similarly, we have that 7 / ∈ π ( G ).Suppose that G has an element g of order 8. Then it follows from Theorem 2.2 that6 / ∈ π e ( G ) and G has only one cyclic subgroup of order 8. If there exists an element12 in G such that | a | = 3, then (cid:104) a, g (cid:105) is of order 24 and π e ( (cid:104) a, g (cid:105) ) = { , , , , } , acontradiction since such a group does not exist. Therefore, in this case G is a 2-group.By Lemma 4.2 we get the required result.Now we may assume that π e ( G ) ⊆ { , , , , } . By Theorem 2.4 G has someelements of order 6. If G has exactly one cyclic subgroup of order 6, then by Lemma2.5 it is easy to see that Γ G has genus one. Therefore, we conclude that G has atleast two cyclic subgroups of order 6. Now the necessity follows from Theorems 3.3and 3.4, and Lemmas 3.1, 4.3, 4.4 and 4.5.For the converse, it follows from Theorems 3.3, Lemmas 4.2 and 4.3. In this section we show Theorem 1.2. We begin with the following lemma.
Lemma 5.1. γ (Γ Z n ) (cid:54) = 2 for each n . In particular, γ (Γ Z n ) ≥ for n ≥ .Proof. Note that if n ≥
7, then the clique number of Γ Z n is greater than or equal to7. Now the result follows from Theorem 2.1. Lemma 5.2.
Let G be a group with π e ( G ) ⊆ { , , , , } . Suppose that G has atleast three cyclic subgroups of order . Then γ (Γ G ) > .Proof. Let H , H , H be three cyclic subgroups of order 6 of G . If | H s ∩ H t | (cid:54) = 3 foreach two s, t , then Γ G has a subgraph isomorphic to K + ( K ∪ K ∪ K ) that hasnonorientable genus three by Theorem 2.2 and hence γ (Γ G ) >
2, as desired. Thus,we now may assume that there exist two cyclic subgroups of order 6 in G such thattheir intersection is of order 3. Without loss of generality, let | H ∩ H | = 3. Denoteby ∆ the subgraph induced by H ∪ H . Next we prove that γ (∆) = 2.Let H = { e, g , g , f , f , a } and H = { e, g , g , f , f , b } , where | g | = | g | = 3, | a | = | b | = 2 and | f i | = 6 for i = 1 , . . . ,
4. Then ∆ − { a, b } is the graph as shown inFigure 5, which is isomorphic to the graph B in [12]. By the main result of [12], onehas γ (∆) ≥
2. On the other hand, we can embed ∆ on N as shown in Figure 6, so γ (∆) = 2.Suppose that | H ∩ H | (cid:54) = 3. Then Γ G contains a subgraph Γ that has two blocks ∆and a graph isomorphic to K . Clearly, γ (∆) ≥
1. Since we can embed ∆ on a torusas shown in Figure 7, one has γ (∆) = 1. Now by Theorem 2.2 we have that γ (Γ) = 3and thereby γ (Γ G ) ≥
3, as required. Therefore, we may suppose that | H ∩ H | = 3.Then | H ∩ H ∩ H | = 3. Write H = { e, g , g , f , f , c } where | f | = | f | = 6 and | c | = 2. Let Λ denote the subgraph induced by ∪ i =1 H i . Then Λ − { f , f , c } = ∆. Itis clear that γ (Λ) ≥ γ (∆) = 2. In order to end the proof, next we show γ (Λ) ≥ g g f f f f Figure 5: An obstruction for the projective plane. g g g g eee ef f f f f b a Figure 6: An embedding of ∆ on N e eee g g g g a f f bf f Figure 7: An embedding of ∆ on a torus14uppose, towards a contradiction, that γ (Λ) = 2. By Theorem 2.3 there are 15faces when drawing Λ without crossings on N . Fix an embedding of Λ on N and let { F , . . . , F } be the set of all faces of Λ corresponding to this embedding. So the facesobtained by deleting f , f , c and all edges incident with them from { F , . . . , F } forman embedding of ∆ on N . In other words, the faces { F , . . . , F } can be recoveredby inserting f , f , c and all edges incident with them into an embedding of ∆ on N . We first insert f into a face F . Since f and f are adjacent in Λ, f must beinserted into F . Note that { f e, f g , f g } ⊆ E (Λ). So e, g , g lie in the boundaryof F . However, we can not insert the edges f e, f g , f g into F without crossings(similarly, cf. Figure 4), a contradiction. Proof of Theorem 1.2.
Suppose for a contradiction that γ (Γ G ) = 2 for some group G . Then by Lemma 5.1 we have π e ( G ) ⊆ { , , , , , } .Suppose that G has a Sylow 5-subgroup P . If | P | ≥
25, then Γ G has a subgraphisomorphic to Γ Z × Z that has 6 blocks and each of its blocks is isomorphic to K ,and so γ (Γ Z × Z ) = 6 by Theorem 2.2, a contradiction. Therefore, we have | P | = 5.Considering the subgraph induced by all Sylow 5-subgroups, similarly, we have that G has a unique Sylow 5-subgroup. This implies that P is normal in G . If G hasan element of order 6, then G has a subgroup of order 30, a contradiction sinceevery group of order 30 has an element of order 15. Thus, in this case one has π e ( G ) ⊆ { , , , , } . Note that π e ( G \ P ) ⊆ { , , } . In view of Lemma 2.5, onehas γ (Γ G ) = 1, a contradiction.Thus, now we may assume that π e ( G ) ⊆ { , , , , } . If G has precisely onecyclic subgroup of order 6, then by Lemma 2.5 it is easy to see that γ (Γ G ) = 1, acontradiction. So G has at least two cyclic subgroups of order 6. Now we can get thefinal contradiction by Lemmas 3.1 and 5.2. Acknowledgement
K. Wang’s research is supported by National Natural Science Foundation of China(11271047, 11371204) and the Fundamental Research Funds for the Central Univer-sity of China.
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