aa r X i v : . [ m a t h . G R ] S e p K -free character graphs with sevenvertices Mahdi Ebrahimi ∗ School of Mathematics, Institute for Research in Fundamental Sciences (IPM) , P.O. Box: 19395–5746, Tehran, Iran
Abstract
For a finite group G , let ∆( G ) denote the character graph builton the set of degrees of the irreducible complex characters of G . Inthis paper, we determine the structure of all finite groups G with K -free character graph ∆( G ) having seven vertices. We also obtain aclassification of all K -free graphs with seven vertices which can occuras character graphs of some finite groups. Keywords:
Character graph, Character degree, Simple group.
AMS Subject Classification Number:
Let G be a finite group and R ( G ) be the solvable radical of G . Also let cd( G )be the set of all character degrees of G , that is, cd( G ) = { χ (1) | χ ∈ Irr( G ) } ,where Irr( G ) is the set of all complex irreducible characters of G . The set ofprime divisors of character degrees of G is denoted by ρ ( G ). It is well knownthat the character degree set cd( G ) may be used to provide information onthe structure of the group G . For example, Ito-Michler’s Theorem [13] statesthat if a prime p divides no character degree of a finite group G , then G hasa normal abelian Sylow p -subgroup. Another result due to Thompson [15]says that if a prime p divides every non-linear character degree of a group G ,then G has a normal p -complement.A useful way to study the character degree set of a finite group G is toassociate a graph to cd( G ). One of these graphs is the character graph ∆( G ) ∗ [email protected] G [12]. Its vertex set is ρ ( G ) and two vertices p and q are joined by an edgeif the product pq divides some character degree of G . We refer the readersto a survey by Lewis [8] for results concerning this graph and related topics.If we know the structure of ∆( G ), we can often say a lot about thestructure of the group G . For instance, Casolo et al. [3] has proved thatif for a finite solvable group G , ∆( G ) is connected with diameter 3, thenthere exists a prime p such that G = P H , with P a normal non-abelianSylow p -subgroup of G and H a p -complement. For another instance, allfinite solvable groups G whose character graph ∆( G ) is disconnected havebeen completely classified by Lewis [7]. In [2], it was shown that when thecharacter graph ∆( G ) of a finite group G is K -free, then ∆( G ) has at most 7vertices. In this paper, we wish to determine the structure of all finite groups G with K -free character graph ∆( G ) having seven vertices.The solvable group G is said to be disconnected if ∆( G ) is disconnected.Also G is called of disconnected Type n if G satisfies the hypotheses of Ex-ample 2 .n in [7]. Now let Γ be a finite graph. If Γ , Γ , ..., Γ n are connectedcomponents of Γ, we use the notation Γ = Γ ∪ Γ ∪ ... ∪ Γ n to determine theconnected components of Γ. Also note that for an integer n >
1, the set ofprime divisors of n is denoted by π ( n ). Now we are ready to state the mainresult of this paper. Main Theorem.
Let G be a finite group and ∆( G ) be a K -free graph withseven vertices. Then For some integer f > , G ∼ = PSL (2 f ) × R(G) , where | π (2 f ± | = 1 , or . Alsoi) If | π (2 f ± | = 1 , then for some disconnected groups A and B of discon-nected Types 1 or 4, R ( G ) ∼ = A × B and ∆( G ) ∼ = K c ⋆ C .ii) If | π (2 f ± | = 2 , then R ( G ) is a disconnected group of disconnected Type1 or 4 and ∆( G ) ∼ = ( K ∪ K ∪ K ) ⋆ K c .iii) If | π (2 f ± | = 3 , then R ( G ) is abelian and ∆( G ) ∼ = K ∪ K ∪ K . In this paper, all groups are assumed to be finite and all graphs are simpleand finite. For a finite group G , the set of prime divisors of | G | is denoted by π ( G ). If H G and θ ∈ Irr(H), we denote by Irr(G | θ ) the set of irreduciblecharacters of G lying over θ and define cd(G | θ ) := { χ (1) | χ ∈ Irr(G | θ ) } . Wefrequently use, Clifford’s Theorem which can be found as Theorem 6.11 of[6] and Gallagher’s Theorem which is corollary 6.17 of [6]. Also if N ⊳ G and θ ∈ Irr(N), the inertia subgroup of θ in G is denoted by I G ( θ ). We beginwith Corollary 11.29 of [6]. 2 emma 2.1. Let N ⊳ G and ϕ ∈ Irr(N) . Then for every χ ∈ Irr(G | ϕ ) , χ (1) /ϕ (1) divides [ G : N ] . Lemma 2.2. [17] Let N be a normal subgroup of a group G so that G/N ∼ = S ,where S is a non-abelian simple group. Let θ ∈ Irr(N) . Then either χ (1) /θ (1) is divisible by two distinct primes in π ( G/N ) for some χ ∈ Irr(G | θ ) or θ isextendible to θ ∈ Irr(G) and
G/N ∼ = A or PSL (8) . Lemma 2.3.
Suppose G is a finite group, N ⊳ G , θ ∈ Irr(N) is G -invariantand P/N is a normal Sylow p -subgroup of G/N . If all Sylow subgroups of
G/P are cyclic, then either θ extends to G or all character degrees in cd(G | θ ) are divisible by pθ (1) .Proof. If θ extends to P , then as all Sylow subgroups of G/P are cyclic, θ isextendible to G . Thus we assume that θ does not extend to P . Hence usingLemma 2.1, all character degrees in cd(P | θ ) are divisible by pθ (1). Nowlet χ ∈ Irr(G | θ ). There is ϕ ∈ Irr(P | θ ) such that χ ∈ Irr(G | ϕ ). Thus as ϕ (1) | χ (1) , χ (1) is divisible by pθ (1).Let Γ be a graph with vertex set V (Γ) and edge set E (Γ). The comple-ment of Γ and the induced subgraph of Γ on X ⊆ V (Γ) are denoted by Γ c and Γ[ X ], respectively. If E (Γ) = ∅ , Γ is called an empty graph. Now let ∆be a graph with vertex set V (∆) for which V (∆) ∩ V (Γ) = ∅ . The join Γ ∗ ∆of graphs Γ and ∆ is the graph Γ ∪ ∆ together with all edges joining V (Γ) and V (∆). We use the notations K n for a complete graph with n vertices and C n for a cycle of length n . If for some integer n >
2, Γ does not contain a copyof K n as an induced subgraph, then Γ is called a K n -free graph. We nowstate some relevant results on character graphs needed in the next sections. Lemma 2.4. (Palfy’s condition) [14] Let G be a group and let π ⊆ ρ ( G ) . If G is solvable and | π | > , then there exist two distinct primes u, v in π and χ ∈ Irr(G) such that uv | χ (1) . Lemma 2.5. [9] Let G be a solvable group with ∆( G ) ∼ = C . Then G ∼ = A × B , where A and B are disconnected groups. Also for some distinctprimes p, q, r and s , ρ ( A ) = { p, q } and ρ ( B ) = { r, s } . Lemma 2.6. [9] Let G be a solvable group. If ∆( G ) has at least vertices,then either ∆( G ) contains a triangle or ∆( G ) ∼ = C . The structure of the character graph of PSL (q) is determined as follows: Lemma 2.7. [18] Let G ∼ = PSL (q) , where q > is a power of a prime p . a) If q is even, then ∆( G ) has three connected components, { } , π ( q − nd π ( q + 1) , and each component is a complete graph. b) If q > is odd, then ∆( G ) has two connected components, { p } and π (( q − q + 1)) .i) The connected component π (( q − q + 1)) is a complete graph if and onlyif q − or q + 1 is a power of .ii) If neither of q − or q + 1 is a power of , then π (( q − q + 1)) can bepartitioned as { } ∪ M ∪ P , where M = π ( q − − { } and P = π ( q + 1) − { } are both non-empty sets. The subgraph of ∆( G ) corresponding to each of thesubsets M , P is complete, all primes are adjacent to , and no prime in M is adjacent to any prime in P . When ∆( G ) c is not a bipartite graph, then there exists a useful restrictionon the structure of G as follows: Lemma 2.8. [1] Let G be a finite group and π be a subset of the vertex setof ∆( G ) such that | π | > is an odd number. Then π is the set of verticesof a cycle in ∆( G ) c if and only if O π ′ ( G ) = S × A , where A is abelian, S ∼ = SL (u α ) or S ∼ = PSL (u α ) for a prime u ∈ π and a positive integer α ,and the primes in π − { u } are alternately odd divisors of u α + 1 and u α − . Lemma 2.9. [4] Let G be a simple group. If | π ( G ) | = 3 , then G is isomorphicto one of the groups A , A , PSL (7) , PSL (8) , PSL (17) , PSL (3) , PSU (3) and PSU (2) . Lemma 2.10.
Let G be a disconnected group. If | ρ ( G ) | = 2 and / ∈ ρ ( G ) ,then G is of disconnected Type or .Proof. Using descriptions of the six types of disconnected groups [7], we aredone.
PSL (q) In order to the proof of Main Theorem, we will need some properties ofPSL (q), where q is a prime power. We will make use of Dickson’s list of thesubgroups of PSL (q), which can be found as Hauptsatz II.8.27 of [5]. Wealso use the fact that the Schur multiplier of PSL (q) is trivial unless q = 4or q is odd, in which case it is of order 2 if q = 9 and of order 6 if q = 9.We now state some relevant results on ∆(PSL (q = 2 f )) needed in the nextsections. We will use Zsigmondy’s Theorem which can be found in [19]. Lemma 3.1. [11] If S = PSL (2 f ) and | ρ ( S ) | = 4 , then either: a) f = 4 , f + 1 = 17 and f − . or b) f > is prime, f − r is prime, and f + 1 = 3 .t β , with t an oddprime and β > odd. emma 3.2. If S = PSL (q = 2 f ) and | π ( q ± | = 2 , then either: a) f is a prime. b) f = 6 or .Proof. Let | π ( f ) | >
2. Then for some distinct primes p and p , p p | f . Thusone of the following cases occurs:a) p p = 6. If f = p p , then we are done. Thus we suppose f = p p . Hencethere is a prime p so that p p p | f . Therefore by Zsigmondy’s Theorem, thereis a prime s such that s | p − s ∤ −
1. Thus 3 . .s | f − p p = 6. Let p < p . By Zsigmondy’s Theorem, there are primes s , s and s so that s | p p − s ∤ p − s ∤ p − s | p − s ∤ p − s | p −
1. Hence s , s and s are distinct and s s s | f −
1. It is acontradiction.Therefore for some prime p and positive integer n , f = p n . If f = 9, we aredone. Now let n > f = 9. If p is odd, then by Zsigmondy’s Theorem, | π (2 f + 1) | > p = 2 and q = 2 n . If n = 2,then q + 1 = 17 which is impossible. Thus n >
3, then 255 = 2 − | f − | π ( q − | > f = p is a primeand the proof is completed.In the sequel of this section, we let p be a prime, f > q = p f and S ∼ = PSL (q). When q is an odd prime power and | π ( S ) | = 4,there is a useful result on possible values for q . Lemma 3.3. [11] If q is odd and | π ( q − | = 3 , then either: a) q ∈ { , , } , b) p = 3 and f is an odd prime, or c) p > and f = 1 . Lemma 3.4. [10] Let q > , f > and q = 9 . If S G Aut(S) ,then G has irreducible characters of degrees ( q + 1)[ G : G ∩ PGL (q)] and ( q − G : G ∩ PGL (q)] . Suppose G is a finite group, q >
11 and
G/R ( G ) = S . In the sequel,we consider a given character θ ∈ Irr(R(G)) and try to determine cd(G | θ )in some special cases. For this purpose, we assume that I := I G ( θ ) and N := I/R ( G ). Lemma 3.5.
Suppose N is a Frobenius group whose kernel is an elementaryabelian p -group. Then one of the following cases holds: a) θ is extendible to I and cd(G | θ ) = { θ (1)[G : I] , θ (1)b } , for some positiveinteger b divisible by ( q − / (2 , q − . ) θ is not extendible to I and all character degrees in cd(G | θ ) are divisibleby p ( q + 1) θ (1) .Proof. Let
Q/R ( G ) be the p -Sylow subgroup of N . If θ extends to Q , then θ extends to I and by Gallagher’s Theorem, cd(I | θ ) = { θ (1) , θ (1)[I : Q] } . Henceusing Clifford’s Theorem, cd(G | θ ) = { θ (1)[G : I] , θ (1)[G : Q] } . Therefore as( q − / (2 , q −
1) divides [ G : Q ], cd(G | θ ) = { θ (1)[G : I] , θ (1)b } , where b is apositive integer divisible by ( q − / (2 , q − θ is notextendible to Q . By Lemma 2.3, all character degrees in cd(I | θ ) are divisibleby pθ (1). Therefore by Clifford’s Theorem, all character degrees in cd(G | θ )are divisible by θ (1) p ( q + 1). Lemma 3.6.
Suppose N ∼ = PSL (p m ) , where m = f is a positive divisor of f , p m = 9 and p m > . Then θ (1)( p m − G : I ] , θ (1)( p m + 1)[ G : I ] ∈ cd(G | θ ) ,where [ G : I ] = p f − m ( p f − / ( p m − .Proof. Using Clifford’s Theorem and this fact that SL (p m ) is the Schurrepresentation group of N , we have nothing to prove. Lemma 3.7.
Let N ∼ = PGL (p m ) , where p is odd, m is a positive divisorof f and p m = 3 . Then some m ∈ cd(G | θ ) is divisible by θ (1) p f − m ( p f + 1) .Proof. Since N ∼ = PGL (p m ), there is a normal subgroup H of N such that H ∼ = PSL (p m ). Thus for some normal subgroup J of I , J/R ( G ) = H .It is easy to see that θ is J -invariant. Therefore looking at the charactertable of the Schur representation group Γ of H , for some λ ∈ Irr(Z(Γ)),cd(J | θ ) = { θ (1)m | m ∈ cd(Γ | λ ) } . It is clear that for every λ ∈ Irr(Z(Γ)),there exists m ( λ ) ∈ cd(Γ | λ ) such that m ( λ ) is even. Therefore for some λ ∈ Irr(Z(Γ)), n := m ( λ ) θ (1) ∈ cd(J | θ ) is even. There is ϕ ∈ Irr(J | θ ) sothat ϕ (1) = m ( λ ) θ (1). Now let χ ∈ Irr(I | θ ) be a constituent of ϕ I . Clearly, χ (1) is divisible by 2 θ (1). Hence m := χ G (1) ∈ cd(G | θ ) is divisible by2 θ (1) p f − m ( p f + 1)( p f − / p m − f is divisible by 2 m , we deducethat m is divisible by θ (1) p f − m ( p f + 1).Now we state a useful assertion which will be required in the next sections. Lemma 3.8.
Let G be a finite group, p be a prime larger than , q := 2 p , S ∼ = PSL (q) and G/R ( G ) = Aut(S) . Also let θ ∈ Irr(R(G)) , I := I G ( θ ) and N := I/R ( G ) . If S ⊆ N , then p ( q − θ (1) , p ( q + 1) θ (1) ∈ cd(G | θ ) .Proof. If N = S , then using Clifford’s Theorem and this fact that the Schurmultiplier of S is trivial, we have nothing to prove. Thus we can assumethat N = Aut(S). Using Fermat’s Lemma, we deduce p / ∈ π ( S ). Thus asthe Schur multiplier of N is trivial, θ is extendible to G and by Gallagher’s6heorem, cd(G | θ ) = { m θ (1) | m ∈ cd(Aut(S)) } . Therefore by lemma 3.4, p ( q − θ (1) , p ( q + 1) θ (1) ∈ cd(G | θ ).We end this section with an interesting result. For this purpose we requirethe following lemmas. Lemma 3.9.
Let G be a finite non-abelian group such that ∆( G ) c is notbipartite. Then there exists a normal subgroup N of G/R ( G ) in which forsome prime u and positive integer α > , N ∼ = PSL (u α ) . Also there exist u ′ ∈ π ( u α − − { } and u ′′ ∈ π ( u α + 1) − { } such that ∆( G ) c [ { u, u ′ , u ′′ } ] is a triangle.Proof. Since ∆( G ) c is not bipartite, there exists π ⊆ ρ ( G ) such that π is theset of vertices of an odd cycle with minimal length of ∆( G ) c . Using Lemma2.8, L := O π ′ ( G ) = S × A , where A is abelian, S ∼ = SL (u α ) or S ∼ = PSL (u α )for a prime u ∈ π and a positive integer α , and the primes in π − { u } arealternately odd divisors of u α + 1 and u α −
1. Thus N := LR(G) / R(G) ∼ =PSL (u α ) is a normal subgroup of G/R ( G ). We claim that in ∆( G ) c , u isadjacent to all vertices in π − { u } . On the contrary, we assume that thereexists x ∈ π − { u } such that x and u are adjacent vertices in ∆( G ). Then forsome χ ∈ Irr(G), xu | χ (1). Now let θ ∈ Irr(L) be a constituent of χ L . UsingLemma 2.1, χ (1) /θ (1) divides [ G : L ]. Thus as G/L is a π ′ -group, xu | θ (1). Itis a contradiction as x and u are non-adjacent vertices in ∆( L ). Thus | π | = 3and we are done. Lemma 3.10.
Let G be a finite group, R ( G ) < M G , S := M/R ( G ) beisomorphic to PSL (q) , where for some prime p and positive integer f > , q = p f , | π ( S ) | > and S G/R ( G ) Aut(S) . Also let θ ∈ Irr(R(G)) , I := I M ( θ ) and N := I/R ( G ) . If ∆( G ) c is not bipartite, then N = S .Proof. Since ∆( G ) c is not bipartite, by Lemma 3.9, there exists a normalsubgroup L of G/R ( G ) in which for some prime u and positive integer α , L ∼ = PSL (u α ). Also there exist u ′ ∈ π ( u α − − { } and u ′′ ∈ π ( u α + 1) − { } such that the induced subgraph of ∆( G ) c on π := { u, u ′ , u ′′ } is a triangle.Thus as L is a non-abelian minimal normal subgroup of G/R ( G ), q = u α .On the contrary, we assume that N = S . Then using Dickson’s list, one ofthe following cases holds. Note that d := (2 , q − N is an elementary abelian p -group. Then by Clifford’s Theorem,there is m ∈ cd(M | θ ) such that m is divisible by θ (1)( q − /d . It is acontradiction as ∆( G ) c [ π ] is a triangle.Case 2. N is contained in a dihedral group. Using Clifford’s Theorem, forsome m ∈ cd(M | θ ) and ǫ ∈ {± } , m is divisible by θ (1) p ( q + ǫ ) /d . It is acontradiction as ∆( G ) c [ π ] is a triangle.7ase 3. N is a Frobenius group whose kernel is an elementary abelian p -group. Then using Lemma 3.5, there is m ∈ cd(M | θ ) so that m is divisibleby θ (1)( q − /d or θ (1) p ( q + 1) and we again obtain a contradiction.Case 4. N ∼ = A . Then as SL (5) is the Schur representation group of A ,there exists m ∈ cd(M | θ ) such that m is divisible by q ( q − θ (1) / d whichis impossible as ∆( G ) c [ π ] is a triangle.Case 5. N ∼ = A . Then using Clifford’s Theorem, all character degrees incd(M | θ ) are divisible by θ (1) q ( q − / d . It is again a contradiction.Case 6. N ∼ = S and 16 | q −
1. Then using Clifford’s Theorem, all characterdegrees in cd(M | θ ) are divisible by θ (1) q ( q − / d which is impossible.Case 7. N ∼ = PSL (p m ), where m = f is a positive divisor of f , p m > p m = 9. By Lemma 3.6, b := θ (1) p f − m ( p m + 1)( p f + 1)( p f − / ( p m − ∈ cd(M | θ ). Thus as p ( q + 1) divides b , we again obtain a contradiction.Case 8. N ∼ = PSL (9), p = 3 and f > | θ ) are divisible by 3( q − /
80. It isa contradiction as ∆( G ) c [ π ] is a triangle.Case 9. N ∼ = PGL (p m ), where 2 m is a positive divisor of f and p m = 3.By Lemma 3.7, some m ∈ cd(M | θ ) is divisible by θ (1) p f − m ( p f + 1) which isagain a contradiction. Lemma 3.11.
Assume that f > is an integer, q = 2 f , S ∼ = PSL (q) and G is a finite group such that G/R ( G ) = S . If ∆( G )[ π ( S )] = ∆( S ) , then G ∼ = S × R ( G ) .Proof. Let H be the last term of the derived series of G . Since G is non-solvable, H is non-trivial. Let N := H ∩ R ( G ). As H/N ∼ = HR ( G ) /R ( G ) ⊳ G/R ( G ) ∼ = S , we deduce that H/N ∼ = S . We claim N = 1. On the contrary,we assume that N = 1. Thus there exists a non-principal linear character λ ∈ Irr(N). Now let f = 2 or 3. Hence S ∼ = A or PSL (8) and ∆( S ) is anempty graph. Using Lemma 2.2, for some χ ∈ Irr(H | λ ), χ (1) is divisible bytwo distinct primes in π ( S ). Therefore ∆( G )[ π ( S )] is a non-empty subgraphof ∆( G ). It is a contradiction. Therefore f >
4. Since ∆( G )[ π ( S )] = ∆( S ),∆( H ) ⊆ ∆( G ) is not a bipartite graph. Hence by Lemma 3.10 and this factthat the Schur multiplier of S is trivial, λ extends to H . It is a contradictionas H is perfect. Hence N = 1 and G ∼ = H × R ( G ) ∼ = S × R ( G ). In this section, we wish to prove our main result.
Lemma 4.1.
Suppose G is a finite group and ∆( G ) is a K -free graph witheseven vertices. Then there exists a normal subgroup R ( G ) < M G so that /R ( G ) is an almost simple group with socle S := M/R ( G ) ∼ = PSL (q) ,where q is a prime power and ∆( G ) c [ π ( S )] is not a bipartite graph. Also if π ⊆ ρ ( G ) is the set of vertices of an odd cycle in ∆( G ) c , then π ⊆ π ( S ) .Proof. As ∆( G ) is a K -free graph with seven vertices, ∆( G ) c is not bipartite.Now assume that π ⊆ ρ ( G ) is the set of vertices of an odd cycle in ∆( G ) c .By Lemma 2.8, N := O π ′ ( G ) = R × A , where A is abelian, R ∼ = SL (q) or R ∼ = PSL (q) for a prime power q and π ⊆ π (PSL (q)). Let M := N R ( G ).Then S := M/R ( G ) ∼ = N/R ( N ) ∼ = PSL (q) is a non-abelian minimal normalsubgroup of G/R ( G ). Note that π ⊆ π ( S ) and ∆( G ) c [ π ( S )] is not a bipartitegraph.Let C/R ( G ) = C G/R ( G ) ( M/R ( G )). We claim that C = R ( G ) and G/R ( G )is an almost simple group with socle S = M/R ( G ). Suppose on the contrarythat C = R ( G ) and let L/R ( G ) be a chief factor of G with L C . Then L/R ( G ) ∼ = T k , for some non-abelian simple group T and some integer k > L C , LM/R ( G ) ∼ = L/R ( G ) × M/R ( G ) ∼ = S × T k . Let F := π ( S ) ∩ π ( T ).Since ∆( G ) is K -free, | F |
3. Note that | ρ ( S ) | , | ρ ( T ) | > ∈ F . Nowone of the following cases occurs:Case 1. | F | = 1. Then F = { } . Hence 3 does not divide | T | . Therefore by[16], T is a Suzuki simple group. By Lemma 2.9, | π ( T ) | >
4. Thus ∆( S × T k )contains a copy of K and it is a contradiction.Case 2. | F | = 2. Then as | π ( S ) | , | π ( T ) | >
3, we again deduce that ∆( S × T k )contains a copy of K and it is a contradiction.Case 3. | F | = 3. If | π ( S ) | or | π ( T ) | is larger than 3, then ∆( S × T k )contains a copy of K which is a contradiction. Thus ρ ( S ) = ρ ( T ) = F .Hence ∆( S × T k ) is a triangle. It is a contradiction as ∆( G ) c [ π ( S )] is not abipartite graph. It completes the proof.Let G be a finite group and ∆( G ) be a K -free graph with seven vertices.By Lemma 4.1, there exists a normal subgroup R ( G ) < M G so that G/R ( G ) is an almost simple group with socle S := M/R ( G ) ∼ = PSL (q)where q is a prime power and ∆( G ) c [ π ( S )] is not a bipartite graph. Also if π ⊆ ρ ( G ) is the set of vertices of an odd cycle in ∆( G ) c , then π ⊆ π ( S ).Since ∆( G ) is K -free, ∆( S ) is too. The structure of ∆( S ) is determinedby Lemma 2.7. Note that as ∆( G ) c [ π ( S )] is not a bipartite graph, ∆( S ) ≇ K ∪ K , K ∪ K . In the sequel of this paper, we use the notation π R for theset ρ ( R ( G )) − π ( G/R ( G )). Lemma 4.2.
Let | π R | > and | π ( S ) | > . Then G does not exist.Proof. By Lemma 2.4, there are p, q ∈ π R so that p and q are adjacent verticesin ∆( R ( G )) ⊆ ∆( G ). There is θ ∈ Irr(R(G)) with pq | θ (1). Therefore using9emma 2.2, for some χ ∈ Irr(M | θ ), χ (1) /θ (1) is divisible by at least twodistinct prime divisors of | S | . Thus | π ( χ (1)) | > Lemma 4.3. If ∆( S ) has three vertices, then G ∼ = S × A × B , where a) S ∼ = A or PSL (8) , b) A and B are disconnected groups of disconnected Types 1 or 4. Also | ρ ( A ) | = | ρ ( B ) | = 2 , and c) ∆( G ) ∼ = C ⋆ K c .Proof. Since ∆( G ) c [ π ( S )] is not a bipartite graph, ∆( G )[ π ( S )] is an emptygraph. Thus using Lemma 2.9, we can see that G = M and S ∼ = A orPSL (8). Hence by Lemma 3.11, G ∼ = S × R ( G ). Thus as ∆( G )[ π ( S )] isan empty graph, π ( S ) ∩ ρ ( R ( G )) = ∅ and ρ ( R ( G )) = π R . Also as ∆( G ) is K -free, ∆( R ( G )) is triangle free. Hence as | ρ ( G ) | = 7, using Lemma 2.6,∆( R ( G )) ∼ = C . Thus using Lemma 2.5, R ( G ) ∼ = A × B , where A and B aredisconnected groups with | ρ ( A ) | = | ρ ( B ) | = 2. Therefor by Lemma 2.10, A and B are of disconnected Types 1 or 4. It is easy to see that ∆( G ) ∼ = C ⋆K c and the proof is completed. Lemma 4.4. ∆( S ) ≇ K ∪ K ∪ K .Proof. By Lemmas 2.7 and 3.1, either S ∼ = PSL (16) or PSL (2 p ), where p > t := 2 p − | π (2 p + 1) | = 2. Nowone of the following cases occurs:Case 1. S ∼ = PSL (16) or G = M . It is easy to see that π ( G/R ( G )) = π ( S ).Thus as | π ( S ) | = 4 and | ρ ( G ) | = 7, | π R | = 3. Hence by Lemma 4.2, we havenothing to prove.Case 2. S ∼ = PSL (2 P ) and G = M . Clearly, G/R ( G ) = Aut(S). Thusas | π (Aut(S)) | = 5 and | ρ ( G ) | = 7, π R = {∅} . Let b ∈ π R . There exists θ b ∈ Irr(R(G)) such that b | θ b (1). By Lemma 3.10, θ b is M -invariant. Thusby Lemma 3.8, m := θ b (1) p (2 p + 1) ∈ cd(G | θ b ) which is a contradiction as | π ( m ) | > Lemma 4.5. If ∆( S ) ∼ = K ∪ K ∪ K , then G ∼ = S × R ( G ) , where a S ∼ = PSL (2 ) , PSL (2 ) or PSL (2 p ) , where p is prime and | π (2 p ± | = 2 , b) R ( G ) is a disconnected group of disconnected Type 1 or 4, and c) ∆( G ) ∼ = ( K ∪ K ∪ K ) ⋆ K c .Proof. By Lemmas 2.7 and 3.2, S is isomorphic to PSL (2 f ), where f is either6 , p with | π (2 p − | = | π (2 p + 1) | = 2. Thus we can see that | π ( G/R ( G )) | = 5 or 6. Hence as | ρ ( G ) | = 7, π R is non-empty. Suppose q ∈ π R and θ ∈ Irr(R(G)) with q | θ (1). Using Lemma 3.10, θ is M -invariant.Then as the Schur multiplier of S is trivial, by Gallagher’s Theorem, q is10djacent to all vertices in π ( S ). Now let G = M . Thus [ G : M ] is divisibleby either 2 , p . By Lemma 3.4, [ G : M ](2 f − G : M ](2 f + 1) ∈ cd(G).Note that π R = ∅ . Let t ∈ π R . If f = 6 or 9, then the induced subgraphof ∆( G ) on π ( S ) ∪ { t } contains a copy of K which is impossible. Thuswe assume that f = p . Then using Lemma 3.8 and this fact that every θ ∈ Irr(R(G)) with t | θ (1), is M -invariant, the induced subgraph of ∆( G ) on X := { p, t }∪ π (2 p −
1) is a copy of K and we have a contradiction. Therefore G = M . Since π ( G/R ( G )) = π ( S ), | ρ ( G ) | = 7 and | π ( S ) | = 5, there existprimes q and q ′ such that π R = { q, q ′ } . Note that q and q ′ are adjacent to allvertices in π ( S ). Now we claim that ∆( G )[ π ( S )] = ∆( S ). On the contrary,we assume that there are x, y ∈ π ( S ) such that x and y are joined by anedge in E (∆( G )) − E (∆( S )). Then for some χ ∈ Irr(G), xy divides χ (1).Let ˜ θ ∈ Irr(R(G)) be a constituent of χ R ( G ) . Then using Lemma 3.10, ˜ θ is G -invariant. Then as the Schur multiplier of S is trivial, by Gallagher’sTheorem, cd(G | ˜ θ ) = { m˜ θ (1) | m ∈ cd(S) } . Hence as χ (1) ∈ cd(G | ˜ θ ), ˜ θ (1)is divisible by x or y . Without loss of generality, we assume that x divides˜ θ (1). There is ǫ ∈ {± } so that x / ∈ π (2 f + ǫ ). Thus the induced subgraphof ∆( G ) on X := π (2 f + ǫ ) ∪ { x, q } is a copy of K and it is a contradiction.Therefore ∆( G )[ π ( S )] = ∆( S ). Hence by Lemma 3.11, G ∼ = S × R ( G ). If ρ ( R ( G )) ∩ π ( S ) = ∅ , then the induced subgraph of ∆( G ) on π ( S ) ∪ { q } contains a copy of K and it is a contradiction. Thus as ∆( G ) is K -free, itis easy to see that ρ ( R ( G )) = π R = { q, q ′ } and ∆( R ( G )) ∼ = K c . Thereforeby Lemma 2.10, R ( G ) is a disconnected group of disconnected Type 1 or 4.Finally, it is easy to see that, ∆( G ) ∼ = ∆( S ) ⋆ K c ∼ = ( K ∪ K ∪ K ) ⋆ K c andthe proof is completed. Lemma 4.6. ∆( S ) ≇ K ∪ K ∪ K .Proof. By Lemma 2.7, for some ǫ ∈ {± } , there exists a prime t such that π ( q − ǫ ) = { t } and | π ( q + ǫ ) | = 3. Let x ∈ π ( G/R ( G )) − π ( S ). Then byLemma 3.4, for some m ∈ cd(G / R(G)) ⊆ cd(G), m is divisible by x ( q + ǫ ).It is a contradiction as | π ( m ) | >
4. Hence π ( G/R ( G )) = π ( S ). Thereforeas | π ( S ) | = 5 and | ρ ( G ) | = 7, there exist distinct primes x and y suchthat π R = { x, y } . Suppose x and y are adjacent vertices in ∆( G ). Thenthere exists θ ∈ Irr(R(G)) such that xy | θ (1). Thus using Lemma 2.2, forsome χ ∈ Irr(M | θ ), | π ( χ (1)) | > x and y are non-adjacent vertices in ∆( G ). We claim that no prime in π R isadjacent to any prime in π ( q + ǫ ). On the contrary, we assume that thereexist v ∈ π R and w ∈ π ( q + ǫ ) such that v and w are adjacent verticesin ∆( G ). Thus for some χ ∈ Irr(G), vw | χ (1). Now let ϕ ∈ Irr(M) and θ ∈ Irr(R(G)) be constituents of χ M and ϕ R ( G ) , respectively. By Lemma 2.1, v | θ (1). Using Lemma 3.10, θ is M -invariant. As the Schur multiplier of S is11rivial, by Gallaghers Theorem, ( q + ǫ ) θ (1) ∈ cd(M | θ ). Which is impossibleas | π ( θ (1)( q + ǫ )) | > π R is adjacent to any primein π ( q + ǫ ). Thus for every z ∈ π ( q + ǫ ), π := { x, y, z } is the set of verticesof an odd cycle in ∆( G ) c . Hence using Lemma 4.1, π ⊆ π ( S ) which is acontradiction as π R ∩ π ( S ) = ∅ . Lemma 4.7. ∆( S ) ≇ K ∪ K ∪ K .Proof. Using Lemma 2.7, there exists ǫ ∈ {± } such that | π ( q − ǫ ) | = 2 and | π ( q + ǫ ) | = 3. Let x ∈ π ( G/R ( G )) − π ( S ). Then by Lemma 3.4, for some m ∈ cd(G / R(G)) ⊆ cd(G), m is divisible by x ( q + ǫ ). It is a contradictionas | π ( m ) | >
4. Hence π ( G/R ( G )) = π ( S ). Therefore as | π ( S ) | = 6 and | ρ ( G ) | = 7, there exists a prime x such that π R = { x } . Thus for some θ ∈ Irr(R(G)), x | θ (1). Hence using Lemma 3.10, θ is M -invariant. Thusas the Schur multiplier of S is trivial, Gallagher’s Theorem implies that m := θ (1)( q + ǫ ) ∈ cd(M | θ ) which is a contradiction as | π ( m ) | > Lemma 4.8. If ∆( S ) ∼ = K ∪ K ∪ K , then G ∼ = PSL (q) × R(G) , where R ( G ) is abelian, for some integer f > , q = 2 f and | π ( q − | = | π ( q + 1) | = 3 .Also ∆( G ) ∼ = K ∪ K ∪ K .Proof. By Lemma 2.7, S ∼ = PSL (q), where for some integer f > q = 2 f and | π ( q − | = | π ( q + 1) | = 3. We first show that G = M . On thecontrary, we assume that G = M . Then there is a prime r so that r divides[ G : M ] = [ G/R ( G ) : S ]. Since q − q + 1 are co-prime, there is ǫ ∈ {± } with r does not divide q + ǫ . Using Lemma 3.4, for some m ∈ cd(G), m isdivisible by r ( q + ǫ ). Hence | π ( r ( q + ǫ )) | = 4 and ∆( G )[ π ( m )] has a copyof K which is impossible. Thus G = M and G/R ( G ) = S . We claimthat ∆( G )[ π ( S )] = ∆( S ). On the contrary, we assume that there exist non-adjacent vertices x and y of ∆( S ) such that x and y are adjacent verticesin ∆( G ). Thus for some χ ∈ Irr(G), χ (1) is divisible by xy . Let θ bea constituent of χ R ( G ) . Using Lemma 3.10, θ is G -invariant. Thus as theSchur multiplier of S is trivial, θ extends to G . By Gallagher’s Theorem,cd(G | θ ) = { m θ (1) | m ∈ cd(S) } . Thus as χ (1) ∈ cd(G | θ ) is divisible by xy , θ (1) is divisible by x or y . Hence one of the character degrees θ (1)( q −
1) or θ (1)( q + 1) of G is divisible by four distinct primes which is a contradiction.Therefore ∆( G )[ π ( S )] = ∆( S ). Thus using Lemma 3.11, G ∼ = S × R ( G ).Now let p ∈ ρ ( R ( G )). There is ǫ ∈ {± } such that p / ∈ π ( q + ǫ ). Thus∆( G )[ π ( p ( q + ǫ ))] is a copy of K which is a contradiction. Hence R ( G ) isabelian and we are done. Lemma 4.9. ∆( S ) ≇ (( K ∪ K ) ⋆ K ) ∪ K . roof. Using Lemma 3.3 and Fermat’s Lemma, π ( G/R ( G )) = π ( S ), unless S ∼ = PSL (3 f ), where f > f | [ G/R ( G ) : S ]. Wefirst assume that π ( G/R ( G )) = π ( S ). Then as | π ( S ) | = 4 and | ρ ( G ) | = 7, | π R | = 3. Hence by Lemma 4.2, we are done. Now let S ∼ = PSL (3 f ) where f > f | [ G/R ( G ) : S ]. Since [ G : M ] = f or 2 f , ρ ( G/R ( G )) = π ( S ) ∪ { f } . Hence as | ρ ( G ) | = 7, for some distinct primes p and q , π R = { p, q } . For every b ∈ π R , let θ b ∈ Irr(R(G)) be a character sothat b divides θ b (1). By Lemma 3.10, θ p and θ q are M -invariant. Thus usingLemma 3.4 and this fact that SL (3 f ) is the Schur representation group of S ,every prime in π := { p, q, f } is adjacent to all vertices in π (3 f − G ) is K -free, ∆( G ) c [ π ] is a triangle. Therefore by Lemma 4.1, π ⊆ π ( S )which is impossible as π R ∩ π ( S ) = ∅ . Lemma 4.10. ∆( S ) ≇ (( K ∪ K ) ⋆ K ) ∪ K or (( K ∪ K ) ⋆ K ) ∪ K .Proof. By Lemma 2.7, S ∼ = PSL (q), where q is an odd prime power. If forsome prime p , p ∈ π ( G/R ( G )) − π ( S ), then as ∆( S ) has a triangle, usingLemma 3.4, we can see that ∆( G/R ( G )) ⊆ ∆( G ) has a copy of K and itis a contradiction. Thus π ( G/R ( G )) = π ( S ) and π R = ∅ . Hence there isa prime p with p ∈ π R . There is θ ∈ Irr(R(G)) so that p divides θ (1). ByLemma 3.10, θ is M -invariant. Thus as SL (q) is the Schur representationgroup of S , θ (1)( q − θ (1)( q + 1) ∈ cd(M | θ ). There exists ǫ ∈ {± } with | π ( q + ǫ ) | = 3. Hence ∆( M )[ π ( θ (1)( q + ǫ ))] ⊆ ∆( G ) contains a copy of K and it is a contradiction. Acknowledgements
This research was supported in part by a grant from School of Mathemat-ics, Institute for Research in Fundamental Sciences (IPM). I would like toexpress my gratitude to the referee for valuable comments which improvedthe original version. I also would like to thank Sadegh Nazardonyavi for hishelp during the preparation of the manuscript.
References [1] Z. Akhlaghi, C. Casolo, S. Dolfi, E. pacifici, L. Sanus, On the characterdegree graph of finite groups, Annali di Mat. Pura Appl. (1923-), (2019)1-20.[2] Z. Akhlaghi, H.P. Tong-Viet, Finite groups with K -free prime graphs,Algebr. Represent. Theory, 18(1) (2015) 235-256.133] C. Casolo, S. Dolfi, E. Pacifici, L. Sanus, Groups whose character degreegraph has diameter three, Israel J. Math, 215 (2016) 523-558.[4] M. Herzog, On finite simple groups of order divisible by three primesonly, J. Algebra, 10 (1968) 383-388.[5] B. Huppert, Endliche Gruppen I, Die Grundlehren der MathematischenWissenschaften, Band 134. Springer-Verlag, Berlin-New York, 1967.[6] I.M. Isaacs, Character Theory of Finite Groups, AMS Chelsea Publish-ing, Providence, RI, 2006. Corrected reprint of the 1976 original [Aca-demic Press. New York; MR0460423].[7] M.L. Lewis, Solvable groups whose degree graphs have two connectedcomponents, J. Group Theory, 4(3) (2001) 255-275.[8] M.L. Lewis, An overview of graphs associated with character degreesand conjugacy class sizes in finite groups, Rocky Mountain, J. Math.,38 (1) (2008) 175-211.[9] M.L. Lewis, Q. Meng, Square character degree graphs yield direct prod-ucts, J. Algebra, 349 (2012) 185-200.[10] M.L. Lewis, D.L. White, Non-solvable groups with no prime dividingthree character degrees, J. Algebra, 336 (2011) 158-183.[11] M.L. Lewis, D.L. White, Four-vertex degree graphs of non-solvablegroups, J. Algebra, 378 (2013) 1-11.[12] O. Manz, R. Staszewski, W. Willems, On the number of components ofa graph related to character degrees, Proc. Amer. Math. Soc., 103 (1)(1988) 31-37.[13] O. Manz, T.R. Wolf, Representations of Solvable Groups, volume 185 ofLondon Mthematical Society Lecture Note Series. Cambridge UniversityPress, Cambridge, 1993.[14] P. Palfy, On the character degree graph of solvable groups. I. Threeprimes, Period. Math. Hungar., 36 (1) (1998) 61-65.[15] J.G. Thompson, Normal p-complements and irreducible characters, J.Algebra, 14 (1970) 129-134.[16] I. Toborg, R. Waldecker, Finite simple 3 ′′