aa r X i v : . [ m a t h . G R ] N ov ON THE ORDERS OF VANISHING ELEMENTS OF FINITEGROUPS
SESUAI Y. MADANHA
Abstract.
Let G be a finite group and p be a prime. Let Vo( G ) denote the set of theorders of vanishing elements, Vo p ( G ) be the subset of Vo( G ) consisting of those ordersof vanishing elements divisible by p and Vo p ′ ( G ) be the subset of Vo( G ) consisting ofthose orders of vanishing elements not divisible by p . Dolfi, Pacifi, Sanus and Spigaproved that if a is not a p -power for all a ∈ Vo( G ), then G has a normal Sylow p -subgroup. In another article, the same authors also show that if if Vo p ′ ( G ) = ∅ ,then G has a normal nilpotent p -complement. These results are variations of the wellknown Ito-Michler and Thompson theorems. In this article we study solvable groupssuch that | Vo p ( G ) | = 1 and show that P ′ is subnormal. This is analogous to thework of Isaacs, Mor´eto, Navarro and Tiep where they considered groups with justone character degree divisible by p . We also study certain finite groups G such that | Vo p ′ ( G ) | = 1 and we prove that G has a normal subgroup L such that G/L a normal p -complement and L has a normal p -complement. This is analogous to the recentwork of Giannelli, Rizo and Schaeffer Fry on character degrees with a few p ′ -characterdegrees. Bubboloni, Dolfi and Spiga studied finite groups such that every vanishingelement is of order p m for some integer m >
1. As a generalization, we investigategroups such that gcd( a, b ) = p m for some integer m >
0, for all a, b ∈ Vo( G ). We alsostudy finite solvable groups whose irreducible characters vanish only on elements ofprime power order. ‘ 1. Introduction
One of the interesting problems in character theory of finite groups is determine thestructure a group using information from its character table. Below are some resultsto this effect. Let G be a finite group and p be a prime. Let Irr( G ) denote the set ofcomplex irreducible characters of G , cd( G ) the set of character degrees of G , cd p ( G )the subset of cd( G ) consisting of character degrees divisible by p and cd p ′ ( G ) the subsetof cd( G ) consisting of those character degrees not divisible by p . A classical theoremof Thompson [17, Corollary 12.2] states that if | cd p ′ ( G ) | = 1, then O pp ′ ( G ) = 1, where O pp ′ ( G ) = O p ′ ( O p ( G )). Let O pp ′ pp ′ ( G ) = O pp ′ ( O pp ′ ( G )). Recently, Giannelli, Rizo andSchaeffer Fry [13] proved a variation Thompson’s theorem by showing the followingresult: Theorem 1.1. [13, Theorem A]
Let G be a finite group and let p > be a prime.Suppose that | cd p ′ ( G ) | = 2 . Then G is solvable and O pp ′ pp ′ ( G ) = 1 . We are interested in studying sets of vanishing elements with some restrictions cor-responding to these results. An element g ∈ G is a vanishing element of G if thereexists χ ∈ Irr( G ) such that χ ( g ) = 0. The set of all vanishing elements of G is denoted Date : November 26, 2020.2010
Mathematics Subject Classification.
Primary 20C15.
Key words and phrases. orders of vanishing elements, p ′ -elements, normal p -complements. by Van(G). A classical theorem of Burnside [17, Theorem 3.15] states that Van( G ) isnon-empty if there is a non-linear χ ∈ Irr( G ). Let Vo( G ) denote the set of the ordersof elements in Van( G ), Vo p ( G ) be the subset of Vo( G ) consisting of orders of vanishingelements divisible by p and Vo p ′ ( G ) be the subset of Vo( G ) consisting of those ordersof vanishing elements not divisible by p .Dolfi Pacifi, Sanus and Spiga proved the following result analogous to Thompson’sTheorem: Theorem 1.2. [8, Corollary B]
Let G be a finite group and let p be a prime. Supposethat Vo p ′ ( G ) = ∅ . Then O pp ′ ( G ) = 1 . As a variation of Theorem 1.2, we study certain finite groups G such that | Vo p ′ ( G ) | =1 with an extra condition and obtain our first result. Theorem A.
Let G be a finite group and let p and q be distinct primes. Suppose that | Vo p ′ ( G ) | = 1 . Then G is solvable. Moreover, suppose that one of the following holds: (a) Vo p ′ ( G ) = { b } such that b is divisible by at least two primes; (b) Vo p ′ ( G ) = { q n } for some positive integer n and Q ′ is subnormal, where Q is aSylow q -subgroup of G .Then O pp ′ pp ′ ( G ) = 1 . We remark that it is not clear if the hypothesis that Q ′ is subnormal in G is necessary.In our proof of Theorem A, we use a theorem of Franciosi, de Giovanni, Heineken andNewell (see Lemma 2.9) which states that if G = AB is a product of an abelian group A and a nilpotent group B , then A F ( G ) is normal in G and G has Fitting height at most3. In general, it is well known that for any positive integer n , there is group of order p a q b and Fitting height n . Hence a Fitting height of a product of two Sylow subgroupscan be arbitrarily large.The opposite of Thompson’s theorem is the famous Ito-Michler theorem. It statesthat if cd p ( G ) = ∅ , then G has a normal and abelian Sylow p -subgroup. As a gen-eralization of this, Isaacs, Mor´eto, Navarro and Tiep [18, Theorem A] proved thatif | cd p ( G ) | = 1, then P ′ is abelian and subnormal in G . In particular, P/ O p ( G ) iscyclic when G is solvable. Another generalization of the Ito-Michler theorem in termsof vanishing elements was proved by Dolfi, Pacifi, Sanus and Spiga [8]. They showedthat if all p -elements are non-vanishing elements of G , then G has a normal Sylow p -subgroup and that the Sylow p -subgroup need not be abelian. Their result impliesthat if Vo p ( G ) = ∅ , then G has a normal Sylow p -subgroup. As an analogue of thetheorem of Isaacs, Mor´eto, Navarro and Tiep [18, Theorem A(b)] for solvable groups,we consider the following for solvable groups. Theorem B.
Let G be a finite solvable group. If | Vo p ( G ) | = 1 , then P ′ is subnormalin G . In particular, P/ O p ( G ) is cyclic. We now discuss our next problem. Bubboloni, Dolfi and Spiga studied finite groupssuch that every vanishing element is of order p m for some integer m >
1. They provedthe following result:
Theorem 1.3. [5, Corollary B]
Let G be a finite group and p be a prime. If a = p m for some integer m > , for all a ∈ Vo( G ) , then one of the following holds: (a) G is a p -group. (b) G/ Z ( G ) is a Frobenius group with a Frobenius complement of p -power order and Z ( G ) = O p ( G ) . RDERS OF VANISHING ELEMENTS Observe that groups described in Theorem 1.3 are such that O pp ′ ( G ) = 1. Considerfinite groups G with the following extremal property for a fixed prime p :gcd( a, b ) = p m for some integer m > , for all a, b ∈ Vo( G ) . ( ⋆ )Property ( ⋆ ) is a generalization of the property in Theorem 1.3. For if a = p m forsome integer m >
1, for all a ∈ Vo( G ), then gcd( a, b ) = p k for some integer k >
1, forall a, b ∈ Vo( G ). Conversely, S and A are examples of groups that satisfy property( ⋆ ) for p = 2 but do not satisfy the property in Theorem 1.3. Property ( ⋆ ) is also ageneralization of a property studied in [19].Using the classification of finite simple groups, we show that if G satisfies property( ⋆ ) for an odd prime p >
7, then O pp ′ pp ′ ( G ) = 1: Theorem C.
Let G be a finite group and p > be a prime. If G satisfies property ( ⋆ ) for p , then G is solvable and O pp ′ pp ′ ( G ) = 1 . The condition that p > shows the theorem fails forany p p -power order or they are groups of order p a q b for some primes p and q . Theorem D.
Let G be a finite non-abelian solvable group and p, q be distinct primes.If every irreducible character of G vanishes on elements of prime power order only, thenone of the following holds. (i) G is a p -group; (ii) G/ Z ( G ) is a Frobenius group with a Frobenius complement of p -power order and Z ( G ) = O p ( G ) ; (iii) G is a Frobenius group with a Frobenius complement of p -power order and akernel of q -power order, that is, π ( G ) = { p, q } ; (iv) G is a nearly -Frobenius group and π ( G ) = { p, q } . Higman [16] described finite groups in which every element is of prime power order.In particular, he proved that the solvable groups with this property have orders thatare divisible by at most two distinct primes. Hence Theorem D generalizes Higman’sresult for solvable groups. Note that groups in Theorem D(ii) have orders that may bedivisible by more than two distinct primes (see [5, Proposition 5.4]).2.
Preliminaries
In this section we shall list some properties of vanishing elements.
Lemma 2.1. [22, Lemma 2]
Let G be a finite solvable group. Suppose M, N are normalsubgroups of G . (a) If M \ N is a conjugacy class of G and gcd( | M : N | , | N | ) = 1 , then M is aFrobenius group with kernel N and prime order complement. (b) If G \ N is a conjugacy class of G , then G is a Frobenius group with an abeliankernel and complement of order two. SESUAI Y. MADANHA
The existence of p -defect zero characters is guaranteed in finite simple groups G forall primes p > | G | as the following result shows: Lemma 2.2. [14, Corollary 2.2]
Let G be a non-abelian finite simple group and p be aprime. If G is a finite group of Lie type, or if p > , then there exists χ ∈ Irr( G ) of p -defect zero. Lemma 2.3. [3, Lemma 2.2]
Let G be a finite group, N a normal subgroup of G and p a prime. If N has an irreducible character of p -defect zero, then every element of N of order divisible by p is a vanishing element in G . Lemma 2.4. [2, Lemma 5]
Let G be a finite group, and N = S × · · · × S k a minimalnormal subgroup of G , where every S i is isomorphic to a non-abelian simple group S .If θ ∈ Irr( S ) extends to Aut( S ) , then ϕ = θ × · · · × θ ∈ Irr( N ) extends to G . Lemma 2.5. [20, Theorem 1.1]
Suppose that N is a minimal normal non-abelian sub-group of a finite group G . Then there exists an irreducible character θ of N such that θ is extendible to G with θ (1) > . Given a finite set of positive integers Y, the prime graph Π( Y ) is defined as theundirected graph whose vertices are the primes p such that there exists an element of Y divisible by p , and two distinct vertices p, q are adjacent if and only if there existsan element of Y divisible by pq . The vanishing prime graph of G , denoted by Γ( G ), isthe prime graph Π(Vo( G )). For a group G , let ω ( G ) be the set of orders of elements of G . We shall state a result on non-solvable groups with disconnected vanishing primegraphs. Let n ( G ) be the number of connected components of the graph G . Theorem 2.6. [9, Theorem B]
Let G be a finite non-solvable group. If Γ( G ) is dis-connected, then G has a unique non-abelian composition factor S , and n (Γ( G )) n (Π( ω ( S ))) unless G is isomorphic to A . We shall also state a result on solvable groups with disconnected vanishing primegraphs. We first recall two definitions:A group G is said to be a 2 -Frobenius group if there exists two normal subgroups F and L of G such that G/F is a Frobenius group with kernel
L/F and L is a Frobeniusgroup with kernel F .A group G is said to be a nearly -Frobenius group if there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group. Theorem 2.7. [10, Theorem A]
Let G be a finite solvable group. Then Γ( G ) has atmost two connected components. Moreover, if Γ( G ) is disconnected, then G is either aFrobenius group or a nearly -Frobenius group. The following is a classification of Frobenius complements.
Theorem 2.8. [4, Theorem 1.4]
Let G be a Frobenius group with Frobenius complement M . Then M has a normal subgroup N such that all Sylow subgroups of N are cyclicand one of the following holds: (a) M/N ∼ = 1 ; (b) M/N ∼ = V , the Sylow -subgroup of the alternating group A ; (c) M/N ∼ = A ; RDERS OF VANISHING ELEMENTS (d) M/N ∼ = S ; (e) M/N ∼ = A ; (f) M/N ∼ = S . Lemma 2.9. [12, Lemma 3]
Let the finite group G = AB be the product of an abeliansubgroup A and a nilpotent subgroup B . Then A F ( G ) is a normal subgroup of G . Inparticular, the Fitting height of G is at most . A reduction theorem
We shall prove a reduction theorem in this section. We shall first prove the propositionbelow. Let S = { A , A , A , PSL (7) , PSL (8) , PSL (4) , B (8) } . Proposition 3.1.
Let N be a non-abelian minimal normal subgroup of G such that N / ∈ S . Then one of the following holds (a) There are distinct primes p , p | | N | and vanishing elements x, y, z of G con-tained in N such that ord( x ) = p , ord( y ) = p and p p | ord( z ) . (b) There are distinct primes p , p | | N | and vanishing elements x , x , y , y of G such that ord( x i ) = ord( y i ) , p i | ord( x i ) and p i | ord( y i ) for i = 1 , . (c) There is a prime p = 3 such that p | | N | and vanishing elements x, y, z of G such that ord( x ) = 3 , ord( y ) = 3 p and ord( z ) = p . (d) There is an odd prime p | | N | and vanishing elements x, y, z of G such that ord( x ) = p , ord( y ) = 2 p and ord( z ) = 4 p .Proof. Let N = S × S × · · · × S k , where S i ∼ = S , S is a simple group. Suppose that k >
2. Using [17, Theorem 3.10], π ( | S | ) >
3. Suppose that π ( | S | ) >
4. Then there aredistinct primes p , p > p i | | N | and so N has an irreducible character of p i -defect zero for i = 1 , x i ∈ S i be a p i -element for i = 1 ,
2. Notethat x i is a vanishing element of G by Lemma 2.3. The element x x is a vanishingelement and ord( x x ) is divisible by p p . This is case (a).We may assume that | π ( G ) | = 3. Using a result of Herzog in [15], we have that S ∈ { PSL (5) , PSL (7) , PSL (8) , PSL (9) , PSL (17) , PSL (3) , PSU (3) , PSU (2) } . Inparticular, S is a simple group of Lie type. By Lemma 2.2, S and hence by extension N has an irreducible character of p -defect zero for all p | | N | . Let x ∈ S be a 2-elementand x ∈ S be a p -element for some odd prime p . Now x , x and x x are vanishingelements and the order of x x is divisible by 2 p . This is case (a).We may assume that N is a simple group. Let N be a sporadic simple group or F (2) ′ . Table 3.1 below contains an irreducible characters θ and θ of N of p -defectzero and q -defect zero for some primes p and q , respectively, and three elements ofdistinct orders divisible by p , q and pq , respectively. This is case (a) and it follows fromLemma 2.3. We shall use the character tables and notation in the Atlas [7].
SESUAI Y. MADANHA
Table 3.1
N θ (1) θ (1) ord( v C ) ord( v C ) ord( v C ) M χ (1) = 45 χ (1) = 16 3 A A AM χ (1) = 45 χ = 896 3 A A AM c L χ (1) = 1750 χ = 896 5 A A ACo χ (1) = 1771 χ = 1835008 7 A A AHe χ (1) = 1275 χ (1) = 21504 5 A A AF i χ (1) = 1001 χ (1) = 1441792 7 A A AF i χ (1) = 106743 χ (1) = 504627200 7 A A AF i ′ χ (1) = 249458 χ (1) = 197813862400 11 A A AHN χ (1) = 133 χ (1) = 3424256 7 A A AT h χ (1) = 30628 χ (1) = 4096000 13 A A AM χ (1) χ (1) 17 A A AJ χ (1) = 120 χ (1) = 56 3 A A AO ′ N χ (1) = 32395 χ (1) = 175616 5 A A AJ χ (1) = 85 χ (1) = 1920 5 A A ALy χ (1) = 48174 χ (1) = 120064 7 A A AJ χ (1) = 1187145 χ (1) = 1981808640 5 A A A F (2) ′ χ (1) = 325 χ = 2048 5 A A A For the sporadic simple groups in Table 3.2, we have two irreducible characters θ and θ of N of p -defect zero and p -defect zero, respectively. We also have two p i -elementsof distinct orders for each θ i on which θ i vanishes. This is case (b). For the case when N ∼ = Suz , we have that N has an irreducible character of 5-defect zero and elementsof orders 5 ,
10 and 20. In the case when N ∼ = HS , N has an irreducible character of3-defect zero and elements of orders 3 , N θ (1) ord( v C ) ord( v C ) θ (1) ord( v C ) ord( v C ) M χ (1) = 55 5 A A χ (1) = 54 3 A AM χ (1) = 45 5 A A χ (1) = 10395 3 A AJ χ (1) = 225 5 A A χ (1) = 189 3 A ACo χ (1) = 896 7 A A χ (1) = 31625 5 A ACo χ (1) = 1771 11 A A χ (1) = 299 13 A AB χ (1) = 3214743741 11 A A χ (1) = 347643114 13 A ARu χ (1) = 378 3 A A χ (1) = 378 7 A A We are left with the case when N ∼ = M . There exists θ = χ , θ = χ which areextendible to G by Lemma 2.4. Note that χ (3 A ) = χ (6 A ) = 0 and χ (4 A ) = 0 andhence this is case (c).Suppose that N is an alternating group A n , n >
5. Since
N / ∈ S , n >
8. Suppose N ∼ = A or A . Then N has two irreducible characters χ and χ of 3-defect zero and5-defect zero, respectively. The result follows because N has elements of orders 3, 5and 15. This is case (a).Suppose N ∼ = A . By Lemma 2.2, N has irreducible characters θ and θ , of 5-defectzero and 7-defect zero, respectively. Since N has elements of orders 5, 10, 7 and 21, theresult follows using Lemma 2.3. This is case (b).Suppose that N is an alternating group A n , n >
11. By Lemma 2.2, N has anirreducible character is of 5-defect zero. Note that N has three elements (12345), RDERS OF VANISHING ELEMENTS (12345)(6789)(10 11) and (12345)(67)(89) of orders 5, 10 and 20, respectively. UsingLemma 2.3, we have three elements are vanishing elements of G . This is case (d).We may assume that N is simple group of Lie type. Suppose that N has an elementof order 2 s , s an odd prime. Then N has irreducible characters (not necessarily distinct) θ and θ of 2-defect zero and of s -defect zero by Lemma 2.2. Since N has elements oforders 2, s and 2 s , the result follows since these two elements are vanishing elementsby Lemma 2.3. This is case (a). We may assume that N has no element of order 2 s , s an odd prime. Then the centralizer of each involution contained in N is a 2-group. Itfollows from [24, III, Theorem 5] that N is isomorphic to one of the following: PSL ( r ),where r is a Fermat or Mersenne prime; PSL (9); N ∼ = PSL (4); N ∼ = B (2 n +1 ), n > N / ∈ S , we may assume that N ∼ = PSL ( r ), where r is a Fermat or Mersenneprime and r >
17 or N ∼ = B (2 n +1 ), n >
2. Suppose that N ∼ = PSL ( r ), r > N ∼ = PSL (17). Then G has vanishing elements of orders 2 , , N ∼ = PSL (31), then G has vanishing elements of orders 3 , N ∼ = PSL ( r ), where r >
127 is a Fermat or Mersenne prime.The character tables of PSL ( q ) are well known (see for example [11, Chapter 38]). Inparticular N has vanishing elements orders which are factors of ( r + 1) / r − / r is a Fermat or Mersenne prime either r − n or r + 1 = 2 n for positiveinteger n . This means that N has elements of orders 2 and 4. Hence either r + 1 or r − r > r ± k , k > r ± N has vanishing elements of orders 2 , , , , q, q for some prime q = 3. In other words, this is case (b) or case (a).Suppose that N ∼ = B (2 n +1 ) with n >
2. In this case, N has cyclic Hall subgroups H , H and H with | H | = 2 n +1 − n +1 + 1 = r , | H | = 2 n +1 + 2 n +1 + 1 = r and | H | = 2 n +1 − r .Orders of elements of N consist of factors of r , r and r . Note that all the non-trivialelements of N are vanishing elements of G . If r i is not prime for some i = 1 , ,
3, thenthe result follows since N has elements of order 2 and 4. This is case (b). Hence we mayassume that r , r and r are prime. This means | π ( N ) | = 4. Using [6], we have that N ∼ = B (32). Now N has elements of orders 2, 4, 5 and 25 which are vanishing elementsof G and the result follows. This is case (b) and this concludes our argument. (cid:3) We are now ready to prove our reduction theorem below:
Theorem 3.2. ( Reduction Theorem ) Let G be a finite group and let p be a prime.If G satisfies property ( ⋆ ) for p , then there exists a solvable normal subgroup N of G such that one of the following holds: (a) G is solvable; (b) p = 2 , or , G/N ∼ = A and one of the following holds: (i) N/ O ( N ) is an elementary abelian group; (ii) N/ O π ′ ( G ) is a π -group with π = { , , } , that is, O π ′ ( G ) is a normal Hall π ′ -subgroup of G ; (c) p = 2 and G/N ∈ { A , A :2 } ; (d) p = 2 , , or and G ∼ = A . (e) p = 2 and G/N ∼ = PSL (7) ; (f) p = 3 and G/N ∼ = PSL (8) ; (g) p = 2 and G/N ∼ = PSL (4) ; (h) p = 2 and G/N ∼ = B (8) . SESUAI Y. MADANHA
Proof.
Suppose that G has no composition factor isomorphic to S ∈ S . We prove that G is solvable using induction on | G | . Let N be a non-trivial normal subgroup of G .Then G/N has no composition factor isomorphic to S and satisfies property ( ⋆ ) andhence G/N is a solvable group. If N and N are two minimal normal subgroups of G ,then G/N and G/N are solvable. Hence G is solvable since the class of finite solvablegroups is a formation. We may assume that G has a unique non-abelian minimal normalsubgroup N . Then by Proposition 3.1, G does not satisfy property ( ⋆ ) for any prime p .We may assume that G has a composition factor isomorphic to S ∈ S . Since G satisfies property ( ⋆ ), Γ( G ) is disconnected. By Theorem 2.6, G has a unique non-abelian composition factor. Letting N to be the solvable radical of G , we have that G/N is an almost simple group and there exists a normal subgroup M of G such that M/N is isomorphic to S ∈ S .Suppose that M = G and G/N ∼ = A . Then p = 2 , G has a 2-element, a 3-element and a 5-element which are vanishing elements. Let π = { , , } . Suppose that there exist an r element, r / ∈ π prime, which is a vanishingelement. Then r is an isolated vertex in the vanishing prime graph of G by property( ⋆ ). In other words, n (Γ( G )) > > n (Γ(A )) by Theorem 2.6, a contradiction.Hence if G has another vanishing element other than q -elements, q ∈ π , then it has tobe a q -singular element g such that some prime r / ∈ π divides ord( g ), using property ( ⋆ ).This means that G has a normal Sylow r -subgroup. Therefore either Van( G ) consists of2-elements, 3-elements and 5-elements or G has an additional vanishing element whoseorder is divisible by qr , q ∈ π and r / ∈ π . The first case is (b)(i) by [26, Theorem A].For the latter case we have that G has a normal Hall π ′ -subgroup. This is case (b)(ii)of the theorem.Suppose that M < G and
G/N ∼ = S . Then there exist elements of orders 2, 3 and6 which are vanishing elements of G/N . Hence
G/N does not satisfy property ( ⋆ ) andthe result follows.If M/N ∼ = A , then M/N has an irreducible character of 2-defect zero and so van-ishing elements of orders 2 and 4 by Lemma 2.3. This means that G does not satisfyproperty ( ⋆ ) for an odd prime p . Suppose that p = 2. Then checking the charactertables in the Atlas [7], for an almost simple group
G/N such that | G/M |
2, our resultfollows except for A and A :2 . For G such that | G/M | = 4, we have the charactertable from GAP [25] and can conclude that G does not satisfy property ( ⋆ ). Hence (c)follows.Suppose that M = G and G/N ∼ = A . Then G/N has vanishing elements of orders3, 4, 5 and 7. So Γ(
G/N ) has four components. Suppose N is a normal subgroupsuch that N/N is a chief factor of G . In particular, N/N is a r -group for some prime r . By [10, Lemma 5.1], Γ( G/N ) has at most two connected components. This meansthat G/N does not satisfy property ( ⋆ ). We have that G does not satisfy property ( ⋆ ).Hence p = 2 , , N = 1 and this is case (d) of the theorem.Suppose that M < G and
G/N ∼ = S . Then there exist elements of orders 2 , G/N and the result follows.Now if
M/N ∈ {
PSL (7) , PSL (4) , B (8) } , then M/N has an irreducible characterof 2-defect zero and elements of orders 2 and 4 and so
G/N does not satisfy property( ⋆ ) for an odd prime p . We may assume that p = 2.Suppose that M/N ∼ = PSL (7) or M/N ∼ = B (8). Then by consulting character tablesin the Atlas [7], our result follows. Suppose
M/N ∼ = PSL (4). Then for an almost simple G such that | G/M |
6, our result follows by checking the explicit character tables in
RDERS OF VANISHING ELEMENTS the Atlas [7]. For the case when | G/M | = 12, we obtain the character table from GAP [25] and our result follows.Suppose that
G/N ∼ = PSL (8). Then p = 3 and G satisfies property ( ⋆ ). If G/N ∼ =PSL (8):3, then G/N has vanishing elements of orders 2 , (cid:3) We prove the first part of Theorem A below. First we prove a lemma.
Lemma 3.3.
Let N be a non-abelian minimal normal subgroup of G . Suppose that p is a prime such that p | | N | . Then there exists a vanishing element of G that is a p -element.Proof. Since N is non-abelian, N = S × S × · · · × S k , where S i ∼ = S , and S isnon-abelian simple group. Let S be a finite group of Lie type or p >
5. By Lemma2.2, S has an irreducible character θ of p -defect zero. Let x be a p -element of S .Then ψ = θ × θ × · · · × θ is an irreducible character of N of p -defect zero. Hence g = x × x × · · · × x ∈ N is a p -element and by Lemma 2.3, g is a vanishing element of G .Suppose that p ∈ { , } and S is a sporadic simple group or an alternating groupA n , n >
7. By [8, Lemma 2.3] and [8, Proposition 2.4], there exists an irreduciblecharacter θ of S that extends to Aut( S ) and θ ( x ) = 0 for some p -element x ∈ S . Let g = x × x × · · · × x ∈ N and note that g is a p -element. By [8, Proposition 2.2], θ × θ × · · · × θ ∈ Irr( N ) extends to some χ ∈ Irr( G ). Hence χ ( g ) = ( θ ( x )) k = 0 andthis concludes our proof. (cid:3) Theorem 3.4.
Let G be a finite group and let p be a prime. Suppose that | Vo p ′ ( G ) | = 1 .Then G is solvable.Proof. We prove that G is solvable using induction on | G | . Let N be a non-trivialnormal subgroup of G . Then | Vo p ′ ( G/N ) | G/N is a solvable group. If N and N are two minimal normal subgroups of G , then both G/N and G/N aresolvable. Hence G is solvable. We may assume that G has a unique non-abelian minimalnormal subgroup N . Note that | π ( N ) | > | Vo p ′ ( G ) | >
2. Hence the result follows. (cid:3) Solvable groups
In this section we shall restate and prove our main theorems. We start with TheoremD.
Proof of Theorem D . Suppose that Γ( G ) has one component. Then all vanishingelements are p -elements for some p . By [5, Theorem A], G is either a p -group or G/ Z ( G )is a Frobenius group with a Frobenius complement of p -power order and Z ( G ) = O p ( G ).This is (a) and (b).Suppose Γ( G ) is disconnected. By Theorem 2.7, Γ( G ) has at exactly two connectedcomponents. This means that G has vanishing elements which are p -elements and q -elements for some distinct primes p and q . Note that G is a Frobenius group or a nearly2-Frobenius group.Suppose that G is a Frobenius group. Let the Frobenius complement be of p -powerorder for some prime p . Suppose that the order of the kernel K is divisible by atleast two distinct primes. There exists a q -element which is a vanishing element of G contained in K . Using [10, Proposition 2.1], there exists g which is a vanishing element such that π (ord( g )) = π ( | K | ), a contradiction. Therefore | K | is of q -power order asexpected in (c). If the Frobenius complement is divisible by two distinct primes, p and q , then there exists a prime r that divides | K | . Using [10, Proposition 3.2], there existsa vanishing element g such that ord( g ) is either divisible by pr or qr , a contradiction.Suppose that G is a nearly 2-Frobenius group. Then there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group. Since G/F is a 2-Frobenius group and G/F is a Frobenius group with kernel
L/F , it followsthat
L/F is a Frobenius group with kernel F/F . By the argument above, | G/L | isa p -power and | L/F | is a q -power. If | F | is divisible by three distinct primes, thenthere exists a prime r different from p and q . Using [10, Proposition 3.2], there existsa vanishing element g such that ord( g ) is either divisible by pr or qr , a contradiction.Hence | F | is divisible by at most two primes p and q . The result then follows. (cid:3) We note here that the converse of this theorem is not true (see [5, Example 2]). Thefollowing example shows a group which satisfies properties in part (iv) of the theorem.4.1.
Example.
Let G = ((C × C ) ⋊ C ) ⋊ C , L = (C × C ) ⋊ C , F = C × C , F = C and F = (C × C ) × C . Then G/F ∼ = S and G/F ∼ = D , a Frobenius groupof with kernel L/F ∼ = C . Also G/F ∼ = S , a 2-Frobenius group. In other words, G isa nearly 2-Frobenius group and Vo( G ) = { , , } using GAP [25].
Theorem 4.1.
Let G be a finite solvable group and p be a prime. If G satisfies property ( ⋆ ) for p , then O pp ′ pp ′ ( G ) = 1 .Proof. We first consider when Γ( G ) is connected. Suppose that Γ( G ) consists of a singlevertex. If the vertex is q , then by Theorem 1.3, G is a q -group or G/ Z ( G ) is a Frobeniusgroup with a Frobenius complement of q -power order and Z ( G ) = O q ( G ). If p = q , then O pp ′ ( G ) = 1. If p = q , then G has a normal Sylow p -subgroup. Hence O p ( O pp ′ ( G )) = 1.Suppose that Γ( G ) has at least two vertices. Then p | a for all a ∈ Vo( G ). UsingTheorem 1.2, O pp ′ ( G ) = 1.Suppose that Γ( G ) is disconnected. By Theorem 2.7, Γ( G ) has two connected com-ponents. Since G satisfies property ( ⋆ ) for p , there exists exactly one b ∈ Vo( G ) suchthat p ∤ b . If b is divisible by two distinct primes p , p , then G has a normal Sylow p i -subgroup, i = 1 , p i -elements as vanishing elements.Hence G has a normal Hall p ′ -subgroup H and H is a normal nilpotent p -complementof G .We may assume that b ∈ Vo( G ) and b is a q -power for the rest of the proof. If a = p m for all a ∈ Vo( G ) \ { b } , then by Theorem D, G is a Frobenius group with aFrobenius complement of r -power order and a kernel of s -power order or G is a nearly2-Frobenius group and π ( G ) = { r, s } . Assume that G is a Frobenius group with aFrobenius complement of r -power order and a kernel K of s -power order. If r = p , thenthe result follows. If s = p , then O p ( G ) = G and O p ′ ( G ) = K . Hence O p ( K ) = 1 andthe result follows. Assume that G is a nearly 2-Frobenius group and π ( G ) = { r, s } .Then there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group. In particular, | G/L | is an r -power, | L/F | is an s -power and | F | is an r -power. If r = p , then O p ( G ) L , O p ′ ( L ) F and RDERS OF VANISHING ELEMENTS O p ( F ) = 1. Assume that s = p . Then O p ( G ) = G , O p ′ ( G ) L , O p ( L ) F and O p ′ ( F ) = 1.Assume that there exist a, c ∈ Vo( G ) such that c = p m and c = p k n , where k , m and n are positive integers and n >
2. Let π = { p, q } . Then the Hall π ′ -subgroupof G is nilpotent and normal. By Theorem 2.7, G is a Frobenius group or a nearly2-Frobenius group. Suppose that G is a Frobenius group KM with Frobenius kernel K and Frobenius complement M . If n | | M | , let r be a prime such that r | n . Thenconsider r and q . If s is a prime such that s | | K | , then by [9, Proposition 3.2], thereexists an vanishing element g of G such that ord( g ) either divisible by rs or qs , acontradiction. Hence | M | is divisible by two primes p and q . Using Theorem 2.8, wehave that M has a unique normal subgroup N such that all the Sylow subgroups of N are cyclic and M/N ∈ { , V , A , S , A , S } . Since G is solvable we need not considerthe cases when M/N ∼ = A or S . Note that N is metacyclic and supersolvable by [23,p. 290].Suppose that M/N ∼ = 1. Let P be a Sylow p -subgroup of G and Q be a Sylow q -subgroup of G . If p > q , then by [1, Theorems 6.2.5 and 6.2.2], P is a normal subgroupof M and hence KP is normal in G . Hence O pp ′ pp ′ ( G ) = 1. If q > p , then by [1,Theorems 6.2.5 and 6.2.2], Q is a normal subgroup of M and so K Q is normal in G .Hence O pp ′ ( G ) = 1. Assume that M/N ∼ = V . So π ( M ) = { , r } , where r is an oddprime. Let R be a Sylow r -subgroup of G . Since N is supersolvable, R is normal in M by [1, Theorems 6.2.5 and 6.2.2]. It follows that KR is normal in G . If p = 2,then O pp ′ ( G ) = 1. If p = r , then O pp ′ pp ′ ( G ) = 1. Assume that M/N ∼ = A . Then π ( M ) = { , } . If p = 2, then N is a 2-group since the Sylow 3-subgroup of M iscyclic, and O pp ′ pp ′ ( G ) = 1. If p = 3, then N is a 3-group. This is because if a Sylow 2-subgroup T has order greater than 4, then T is a generalized quaternion and hence haselements of order 2 and 4 which are vanishing elements of G . Therefore O pp ′ pp ′ ( G ) = 1,as required. Assume that M/N ∼ = S . Then π ( M ) = { , } , p = 2 and N is a 2-group.Hence O pp ′ pp ′ ( G ) = 1.Suppose that G is a nearly 2-Frobenius group. Then there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group.Since G/F is a 2-Frobenius group and G/F is a Frobenius group with kernel
L/F , itfollows that
L/F is a Frobenius group with kernel F/F . By [9, Remark 1.2], G/L iscyclic and
L/F is cyclic with | L/F | odd. Since G/F is a Frobenius group with kernel L/F , we have gcd( | G/L | , | L/F | ) = 1. Since G \ F ⊆ Van( G ), we have that | G/L | isan r -power and | L/F | is s -power, where r and s are primes. If p = r , then O p ( G ) L , O p ′ ( L ) F . Hence O pp ′ pp ′ ( G ) = 1. If p = s , then O pp ′ ( G ) L . Note that the Hall p ′ -subgroup H of L is normal in L . Then O p ( L ) H . Hence O p ′ ( H ) = 1. Supposethat p ∤ a for all a ∈ Vo( G ). Then Vo( G ) = { a, b } . Either a or b is a prime power by[21, Theorem B]. Assume that a is an r -power. If b is not a prime power, then the Hall r ′ -subgroup is nilpotent and normal in G . It follows that O pp ′ pp ′ ( G ) = 1.Assume that there exists c ∈ Vo( G ) such that c = p k n , where k and n are positiveintegers and n > a ∈ Vo( G ) such that a = p m . Then the Hall q ′ -subgroup is normal and nilpotent, and the result follows.Suppose that p ∤ a for all a ∈ Vo( G ). Then Vo( G ) = { a, b } where gcd( a, b ) = 1.Either a or b is a prime power by [21, Theorem B]. We may assume that b is a q -power without loss of generality. Suppose further that a is an r -power for some prime r . TheHall { q, r } ′ -subgroup H is nilpotent and normal in G . Then O pp ′ ( G ) H and hence O pp ′ pp ′ ( G ) = 1. If a is not a prime power, then the Hall q ′ -subgroup is nilpotent andnormal in G . It follows that O pp ′ pp ′ ( G ) = 1. (cid:3) Proof of Theorem C . Since p > G is solvableusing Theorem 3.2. Now Theorem C follows from Theorem 4.1. (cid:3) Proof of Theorem A . By Theorem 3.4, G is solvable. Suppose that (a) holds. Thensince there are no q -elements that are vanishing elements for all q = p , the Hall p ′ -subgroup of G is normal and nilpotent by [8, Theorem A]. Hence O pp ′ ( G ) = 1Suppose that (b) holds. Suppose that Γ( G ) is connected. Then Vo( G ) = { q n } orVo( G ) = { q n } ∪ X , where X is a finite subset of { c ∈ N : pq | c } . Then the Hall q ′ -subgroup H is a normal and nilpotent subgroup of G using [8, Theorem A]. So O pp ′ ( G ) H . It follows that O pp ′ pp ′ ( G ) = 1.We may assume that Vo( G ) = X ∪ { q n } ∪ Y , where X and Y are finite subsets of { a ∈ N : a is a p -power } and { c ∈ N : pq | c } , respectively. Let π = { p, q } and supposethat P is a Sylow p -subgroup of G . Then the Hall π ′ -subgroup H is a normal andnilpotent subgroup of G . Then G = G/N is a π -group, where N = H O p ( G ) O q ( G ).Since Q ′ is subnormal, we have that Q ′ O q ( G ) and G is a product of an abelian q -group Q = QN/N and a p -group P = P N/N . By Lemma 2.9, Q F ( G ) is a normalsubgroup of G . Then G/Q F ( G ) is a p -group and Q F ( G ) / O p ( G ) is a q -group. Consider O p ( G ) = P N/N . Then P N/H O q ( G ) is a p -group and H O q ( G ) is a p ′ -group. Hence O pp ′ pp ′ ( G ) = 1.We may assume that Γ( G ) is disconnected. Suppose that for all a ∈ Vo( G ) \ { q n } , a is not a p -power. Then the Hall q ′ -subgroup H is a normal and nilpotent subgroup of G . Therefore O pp ′ pp ′ ( G ) = 1.Suppose that for all a ∈ Vo( G ) \ { q n } , a is a p -power. Then by Theorem D, G is a Frobenius group with a Frobenius complement of r -power order and a kernel of s -power order or G is a nearly 2-Frobenius group and π ( G ) = { r, s } . Assume that G is a Frobenius group with a Frobenius complement of r -power order and a kernel K of s -power order. If r = p , then O pp ′ ( G ) = 1. If s = p , then O p ( G ) = G and O p ′ ( G ) = K .Hence O p ( K ) = 1 and the result follows. Assume that G is a nearly 2-Frobenius groupand π ( G ) = { r, s } . Then there exist two normal subgroups F and L of G with thefollowing properties: F = F × F is nilpotent, where F and F are normal subgroupsof G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobeniusgroup with kernel L/F , and G/F is a 2-Frobenius group. In particular, | G/L | is an r -power, | L/F | is an s -power and | F | is an r -power. If r = p , then O p ( G ) L , O p ′ ( L ) F and O p ( F ) = 1. Assume that s = p . Then O p ( G ) = G , O p ′ ( G ) L , O p ( L ) F and O p ′ ( F ) = 1.Suppose that there exist a, b ∈ Vo( G ) \ { q n } such that a = p m and b = p k t , where k , m and t are positive integers and t >
2. Let π = { p, q } . Using [8, Theorem A],the Hall π ′ -subgroup of G is a nilpotent and normal subgroup of G . By Theorem 2.7, G is a Frobenius group or a nearly 2-Frobenius group. Suppose that G is a Frobeniusgroup KM with Frobenius kernel K and Frobenius complement M . If t | | M | , let r bea prime such that r | t . Since M \ { } ⊆ Van( G ), G has some r -elements which arevanishing elements. If r = q , then | Vo p ′ ( G ) | >
2, a contradiction. So r = q and Γ( G ) isconnected, another contradiction from our hypothesis. RDERS OF VANISHING ELEMENTS Suppose that G is a nearly 2-Frobenius group. Then there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group.Since G/F is a 2-Frobenius group and G/F is a Frobenius group with kernel
L/F , itfollows that
L/F is a Frobenius group with kernel F/F . By [9, Remark 1.2], G/L iscyclic and
L/F is cyclic with | L/F | odd. Since G/F is a Frobenius group with kernel L/F , we have gcd( | G/L | , | L/F | ) = 1. Since G \ F ⊆ Van( G ), we have that | G/L | isan r -power and | L/F | is s -power, where r and s are primes. If p = r , then O p ( G ) L , O p ′ ( L ) F . Hence O pp ′ pp ′ ( G ) = 1. If p = s , then O pp ′ ( G ) L . Note that the Hall p ′ -subgroup H of L is normal in L . Then O p ( L ) H . Hence O p ′ ( H ) = 1 and theresult follows. (cid:3) Proof of Theorem B . We consider the vanishing prime graph Γ( G ) of G . Supposethat Γ( G ) is connected. Assume that Γ( G ) consists of a single vertex. By Theorem1.3, G is a p -group or G/ Z ( G ) is a Frobenius group with a Frobenius complementof p -power order and Z ( G ) = O p ( G ). If G is a p -group then the result follows. Wemay assume that G/ Z ( G ) is a Frobenius group with a Frobenius complement of p -powerorder and Z ( G ) = O p ( G ). If p is odd, then Sylow p -subgroups are abelian and P/ O p ( G )is cyclic. If p = 2, then P/ O p ( G ) is either a generalized quaternion or cyclic since it isisomorphic to a Frobenius complement. If P/ O p ( G ) is a generalized quaternion, then G has vanishing elements of distinct orders divisible by 2 and 4, a contradiction. Henceour result follows.If Γ( G ) has more than one vertex, then it follows that the p -elements G are non-vanishing. Hence P is normal in G .We may assume that Γ( G ) is disconnected. By Theorem 2.7, G is either a Frobeniusor a nearly 2-Frobenius group. Also note that p is an isolated vertex.Suppose that G is a Frobenius group with a kernel K and a complement H . If p | | K | ,then P is normal in G and the result follows. So p | | H | . Then either P is a cyclicor a generalized quaternion. If P is generalized quaternion, then G contains vanishingelements of orders 2 and 4, a contradiction. Then P is cyclic as required.Suppose that G is a nearly 2-Frobenius group. Then there exist two normal subgroups F and L of G with the following properties: F = F × F is nilpotent, where F and F are normal subgroups of G . Furthermore, G/F is a Frobenius group with kernel
L/F , G/F is a Frobenius group with kernel L/F , and G/F is a 2-Frobenius group. Notethat G/L and
L/F are both cyclic and F is nilpotent. Hence the result follows. (cid:3) Acknowledgements
The author would like thank the referee for the careful reading of this article andtheir comments.
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