aa r X i v : . [ m a t h . G R ] N ov AVERAGE NUMBER OF ZEROS OF CHARACTERS OF FINITEGROUPS
SESUAI YASH MADANHA
Abstract.
There has been some interest on how the average character degree affectsthe structure of a finite group. We define, and denote by anz( G ), the average numberof zeros of characters of a finite group G as the number of zeros in the character table of G divided by the number of irreducible characters of G . We show that if anz( G ) < G is solvable and also that if anz( G ) < , then G is supersolvable. Wecharacterise abelian groups by showing that anz( G ) < if and only if G is abelian. Dedicated to the memory of Kay Magaard Introduction
Let G be a finite group and Irr( G ) be the set of complex irreducible characters of G . Let T ( G ) be the sum of degrees of complex irreducible characters of G , that is, T ( G ) = P χ ∈ Irr( G ) χ (1). Denote by k ( G ) the number of conjugacy classes of G . Then k ( G ) = | Irr( G ) | . Define the average character degree of G byacd( G ) := T ( G ) | Irr( G ) | .Recently, a lot of authors have investigated the average character degree of finite groupsand how it influences the structure of the groups (see [13, 15, 14]). In fact, Magaard andTong-Viet [13] proved that G is solvable whenever acd( G ) < G )
3. This conjecture was settled by Isaacs, Loukakiand Moret´o [10, Theorem A] and they obtained some sufficient conditions for a groupto be supersolvable and also to be nilpotent. In the same article, the authors of [10]conjectured that the best possible bound is when acd( G ) < for G to be solvable.Moret´o and Nguyen showed in [15, Theorem A], that indeed this was the best bound.We shall state the results on average character degrees with the best bounds below. Theorem 1.1. [15, Theorem A]
Let G be a finite group. If acd( G ) < , then G issolvable. Theorem 1.2. [10, Theorem B]
Let G be a finite group. If acd( G ) < , then G issupersolvable. Theorem 1.3. [10, Theorem C]
Let G be a finite group. If acd( G ) < , then G isnilpotent. More work on average character degrees of finite groups is found in [8, 6, 7]. Anotherinvariant that has been studied is the so-called average class size. We shall refer thereader to [5, 10, 18] for more bibliographic information.
Date : November 26, 2020.2010
Mathematics Subject Classification.
Primary 20C15, 20D10.
Key words and phrases. zeros of characters, solvable groups, supersolvable groups, nilpotent groups,abelian groups.
In this article, we investigate the corresponding problem for zeros of characters offinite groups. We have to define a new invariant first. Recall that if χ ( g ) = 0 for some g ∈ G and χ ∈ Irr( G ), we say χ vanishes on g . So χ vanishes on conjugacy classes. Inthe same spirit, we define nz( G ) to be the number of zeros in the character table of G and the average number of zeros of characters of G by:anz( G ) := nz( G ) | Irr( G ) | .Since linear characters do not vanish on any conjugacy class, anz( G ) = 0 for an abeliangroup G . A classical theorem of Burnside [9, Theorem 3.15] shows that χ ( g ) = 0for some g ∈ G and a non-linear irreducible character χ , that is, χ vanishes on someconjugacy class. This means anz( G ) > Theorem A. If N is a minimal non-abelian normal subgroup of G , then there exists anon-linear character χ ∈ Irr( G ) such that χ N is irreducible and χ vanishes on at leasttwo conjugacy classes of G . We need Theorem A to prove our second result below which corresponds to Theorem1.1:
Theorem B.
Let G be a finite group. If anz( G ) < , then G is solvable. Note that since anz(A ) = 1, this bound is sharp.We show that the converse of Theorem B does not necessarily hold. Let Q be a Sylow3-subgroup of S = PSL (7) and suppose G = N S ( Q ). We have that anz( G ) = 7 / > ) = and acd(S ) = .Since anz(A ) = and anz(S ) = , the following corresponding results hold: Theorem C.
Let G be a finite group. If anz( G ) < , then G is supersolvable. Theorem D.
Let G be a finite group. If anz( G ) < , then G is nilpotent. Note that for an abelian group G , acd( G ) = 1 and anz( G ) = 0. We show in Remark4.1 that there is no number c > G ) < c implies that G is abelian.Contrary to acd( G ), we show that there exists a bound c >
0, such that anz( G ) < c implies that G is abelian. Hence we obtain a non-trivial characterisation of abeliangroups in terms of zeros of characters: Theorem E.
Let G be a finite group. Then G is abelian if and only if anz( G ) < . When G is of odd order we obtain a bound that is better than that in Theorem C.The following result is analogous to [10, Theorem D(a)]. Theorem F.
Let G be a finite group of odd order. If anz( G ) < , then G is supersolv-able. However, the bound in Theorem F might not be optimal. Indeed, in [10, TheoremD(a)], the bound is best possible since there exists a group G of odd order which is notsupersolvable such that acd( G ) = and also anz( G ) = ( G is the unique non-abeliangroup of order 75). Hence we propose the following conjecture: Conjecture 1.
Let G be a finite group of odd order. If anz( G ) < , then G issupersolvable. VERAGE NUMBER OF ZEROS OF CHARACTERS 3
The article is organized as follows. In Section 2, we list some preliminary results. InSection 3, we prove Theorems A and B and in Section 4 we prove Theorems C to F.We also show that there exist an infinite family of non-abelian nilpotent groups G suchthat acd( G ) < in this last section.2. Preliminaries
Lemma 2.1.
Suppose that N is a minimal normal non-abelian subgroup of a group G .Then there exists an irreducible character θ of N such that θ is extendible to G with θ (1) > . In particular, if N is simple such that (a) N is isomorphic to A , then θ (1) = 5(b) N is isomorphic to A n , n > , then θ (1) = n − N is isomorphic to a finite group of Lie type defined over a finite field distinctfrom the Tits group F (2) ′ and PSL (5) , then θ is the Steinberg character.Proof. The first assertion is the statement of [13, Theorem 1.1] and the second assertionfollows from the proof of [13, Theorem 1.1]. (cid:3)
Lemma 2.2. [9, Corollary 6.17]
Let N be a normal subgroup of G and let χ ∈ Irr( G ) besuch that χ N = θ ∈ Irr( N ) . Then the characters βχ for β ∈ Irr(
G/N ) are irreducible,distinct for distinct β and are all of the irreducible constituents of θ G . Recall that Irr( G | K ) = { χ ∈ Irr( G ) : K * ker χ } . For χ ∈ Irr( G ) we write nυ ( χ )for the number of conjugacy classes of G on which χ vanishes. Then nz( G | K ) := P χ ∈ Irr( G | K ) nυ ( χ ). Lemma 2.3.
Let K be a normal subgroup of G such that K G ′ . If anz( G ) < , then anz( G/K ) anz( G ) .Proof. Let a = anz( G/K ). Note that k ( G ) = | Irr( G ) | = | Irr(
G/K ) | + | Irr( G | K ) | . If | Irr(
G/K ) | = c , | Irr( G | K ) | = d , nz( G/K ) = m and nz( G | K ) = n , then a = mc . Since K G ′ , Irr( G | K ) is a set of non-linear irreducible characters of G and also usingBurnside’s Theorem, | Irr( G | K ) | = d n = nz( G | K ). Since the character table of G/K is a sub-table of the character of G , we have that nz( G/K ) nz( G ). Hence1 > anz( G ) = m + nc + d > m + dc + d > mc = a as required. (cid:3) We show that Theorem B holds for perfect groups.
Lemma 2.4.
Let G be a finite group such that G = G ′ . Then anz( G ) > .Proof. Since G has one linear character, it is sufficient to show that there exists anon-linear irreducible character of G that vanishes on at least two conjugacy classes.Suppose the contrary, that is, every non-linear irreducible character of G vanishes onexactly one conjugacy class. Then by [2, Proposition 2.7], G is a Frobenius groupwith a complement of order 2 and an abelian odd-order kernel, that is, G is solvable,contradicting the hypothesis that G is perfect. Hence the result follows. (cid:3) Let G be a finite group and χ ∈ Irr( G ). Recall that υ ( χ ) := { x ∈ G | χ ( x ) = 0 } . Lemma 2.5.
Let G be a non-abelian finite group and χ ∈ Irr( G ) be non-linear. Supposethat N is a normal subgroup of G such that G ′ N < G . If χ N is not irreducible, thenthe following two statements hold: (a) There exists a normal subgroup K of G such that N K < G and G \ K ⊆ υ ( χ ) . (b) If ( G \ G ′ ) ∩ υ ( χ ) consists of n conjugacy classes of G , then SESUAI YASH MADANHA | G : G ′ | − | K : G ′ | n .Proof. For (a), note that since
G/N is abelian, it follows that χ is a relative M -characterwith respect to N by [9, Theorem 6.22]. This means that there exists K with N K G and ψ ∈ Irr( K ) such that χ = ψ G and ψ N ∈ Irr( N ). Hence G \ K ⊆ υ ( χ ), Since χ N is not irreducible we have that K < G and the result follows.For (b), let g , g , . . . , g m be a complete set of representatives of the cosets of G ′ in G , with g , g , · · · , g k a complete set of representatives of the cosets of G ′ in K so that m = | G : G ′ | and k = | K : G ′ | . Since χ vanishes on G \ K , we have that χ vanisheson g k +1 , g k +2 , . . . , g m . But these elements are not conjugate and so m − k ≤ n asrequired. (cid:3) In the following results we will consider groups with an irreducible character thatvanishes on one conjugacy class. We denote by k G ( N ), the number of conjugacy classesof G in N where N is a subset of G . We first prove an easy lemma: Lemma 2.6.
Let N be a normal subgroup of G . If k G ( G \ N ) = 1 , then | G : N | = 2 and G is a Frobenius group with an abelian kernel N of odd order.Proof. Since k G ( G \ N ) = 1, all the elements in G \ N are conjugate. It follows that if x ∈ G \ N , then the size of the conjugacy class containing x , say c , is | G | − | N | . Since N is a proper subgroup of G , c ≥ | G | − | G | / | G | /
2. But the size of any conjugacy classof a non-trivial group is a proper divisor of the group, so we deduce that | N | = | G | / C G ( x ) = h x i is of order 2, so G = h x i N is a Frobenius group with a complement h x i of order 2 and hence the Frobenius kernel N is abelian of odd order. (cid:3) We also need this result:
Proposition 2.7.
Let G be an almost simple group. If χ ∈ Irr( G ) vanishes on exactlyone conjugacy class, then one of the following holds: (a) G = PSL (5) , χ (1) = 3 or χ (1) = 4 ; (b) G ∈ { A :2 , A :2 } , χ (1) = 9 for all such χ ∈ Irr( G ) ; (c) G = PSL (7) , χ (1) = 3 ; (d) G = PSL (8):3 , χ (1) = 7 ; (e) G = PGL ( q ) , χ (1) = q , where q ≥ ; (f) G = B (8):3 , χ (1) = 14 .Proof. It was shown in [11, Theorem 5.2] and the proof of [12, Theorem 1.2] that allirreducible characters of G that vanishes on exactly one conjugacy class are primitive,the result follows. Hence the list in [12, Theorem 1.2] is a complete one. The resultthen follows. (cid:3) Lemma 2.8.
A finite group G has exactly one non-linear irreducible character if andonly if G is isomorphic to one of the following: (a) G is an extra-special -group. (b) G is a Frobenius group with an elementary abelian kernel of order p n and acyclic complement of order p n − , where p is prime and n a positive integer.Moreover, anz( G ) = m − m +1 for some integer m > .Proof. The statements (a) and (b) follows from [19, Theorem]. Let G be an extra-special 2-group of order 2 k +1 with | G/G ′ | = | G/Z ( G ) | = 2 k . Then G has 2 k linearcharacters and so 2 k + 1 irreducible characters. Note that the non-linear irreducible VERAGE NUMBER OF ZEROS OF CHARACTERS 5 character χ of G is fully ramified with respect to Z ( G ). In particular, χ vanishes on G \ Z ( G ). Hence χ vanishes on at least 2 k − G has theidentity and a non-trivial central element which are non-vanishing elements, χ vanisheson exactly 2 k − G ) = k − k +1 and the result follows.Suppose G is a Frobenius group with an elementary abelian kernel G ′ of order p n and a cyclic complement of order p n −
1. Note that | G/G ′ | = p n − G has p n − G has p n irreducible characters. Since the non-linearirreducible character χ of G is induced from irreducible character of G ′ , χ vanishes on G \ G ′ . It follows that χ vanishes on at least p n − G ′ has onenon-trivial conjugacy class of G , χ does not vanish on G ′ . Otherwise, χ vanishes on G \{ } , a contradiction. Since G has p n conjugacy classes, we have that χ vanishes onexactly p n − G ) = p n − p n as required. (cid:3) We shall need the following two results to prove Theorem F.
Lemma 2.9. [16]
A finite group G has exactly two non-linear irreducible characters ifand only if G is isomorphic to one of the following: (a) G is an extra-special -group. (b) G is a -group of order k +2 , | G ′ | = 2 , | Z ( G ) | = 4 , and G has two non-linearirreducible characters with equal degree k . (c) G is a Frobenius group with an elementary abelian kernel of order and Frobe-nius complement Q . (d) G is a Frobenius group with an elementary abelian kernel of order p k and acyclic Frobenius complement of order ( p k − / , where p is odd prime. (e) G/Z ( G ) is a Frobenius group with an elementary abelian kernel of order p k anda cyclic Frobenius complement of order p k − , where p is prime and | Z ( G ) | = 2 . Lemma 2.10. [17, Theorem 2.6]
Let G be a finite group of odd order. Then G has anirreducible character that vanishes on exactly two conjugacy classes if and only if G isone of the following groups: (a) G is a Frobenius group with a complement of order . (b) There are normal subgroups M and N of G such that: M is a Frobenius groupwith the kernel N ; G/N is a Frobenius group of order p ( p − / with the kernel M/N and a cyclic complement of order ( p − / for some odd prime p . In thiscase, χ M is irreducible. Lemma 2.11.
Let ℓ, m, n, b be positive integers with ℓ, n > and b > ℓ + 1 . Supposethat b = mn . Then b − m > ℓ + 1 .Proof. Suppose b is odd. If b is prime, then b − > ℓ + 1 − > ℓ + 1. If b is not prime,then the greatest m is when n = 3 and b − m > b − b > b > b = ℓ . If b is even, then b > ℓ + 2 and the greatest m is when n = 2. Hence b − m > ℓ + 2 − ( ℓ + 1) = ℓ + 1as required. (cid:3) Non-solvable and solvable groups
Proof of Theorem A . We begin the proof by showing that we may assume that N is simple. For if N = T × T · · · × T k , where T i ∼ = T , T is a non-abelian simple group,for i = 1 , , . . . , k and k >
2, then by Lemma 2.1, there exists θ i ∈ Irr( T i ) such that χ N = θ × θ × · · · × θ k , where χ ∈ Irr( G ). Since θ (1) >
1, there exists x ∈ T such that SESUAI YASH MADANHA θ ( x ) = 0 by Burnside’s Theorem. But χ vanishes on ( x, , . . . ,
1) and ( x, x, , . . . , G , χ vanishes on at least two conjugacyclasses of G .We may assume that N is simple. Let C = C G ( N ). We claim that C = 1. Otherwise, N × C is a normal subgroup of G . There exists a non-linear θ ∈ Irr( N ) such that θ N = χ by Lemma 2.1. There exists x ∈ N such that θ ( x ) = 0. Then χ ( xc ) = 0 for any c ∈ C .Let c ∈ C \{ } . Since x and xc are not conjugate in G , χ vanishes on at least twoconjugacy classes. Hence C = 1 and we have that G is almost simple.By Lemma 2.1, there exists χ such that χ N is irreducible and χ (1) >
5. If χ vanisheson two conjugacy classes, then the result follows. We may assume that χ vanishes on oneconjugacy class. By Proposition 2.7 and considering the choice of χ from Lemma 2.1, weare left with the following cases: G ∈ { PGL ( q ) , A :2 , A :2 } (note that PSL (9) ∼ = A ).Note that N = PSL ( q ). We will choose another irreducible character of N that extendsto G . Obviously that alternative character vanishes on at least two conjugacy classes.For G = S , let θ ∈ Irr(A ) be such that θ (1) = 4. Then θ is extendible to Aut(A ) = S .Using the Atlas [3], we have that χ G vanishes on two elements of distinct orders. Hencethe result follows.For G ∈ { A :2 , A :2 } , using the Atlas [3], we can choose an irreducible character θ of N of degree 10 that extends to Aut(A ). Lastly, χ vanishes on more than twoconjugacy classes of G . Hence the result follows.Suppose G = PGL ( q ), where q >
7. It is well known that PSL ( q ) has irreduciblecharacters of degree q − q +1. By [20, Theorem A], an irreducible character θ of de-gree q +1 is extendible to Aut(PSL ( q )) except when N = PSL (3 f ) and G = PGL (3 f ),with f an odd positive integer (case (iii) of [20, Theorem A]). If N = PSL (3 f ) and G = PGL (3 f ) with f an odd positive integer, then we choose an irreducible character θ of PSL (3 f ) of degree q − (3 f )). Thus the resultfollows. (cid:3) Proof of Theorem B . We shall prove our result by induction on | G | . We may assumethat G is non-solvable. If G = G ′ , then the result follows by Lemma 2.4, so G = G ′ .Let G ∞ be the solvable residual of G . Note that G ∞ is perfect. If N < G ∞ G ′ is aminimal normal subgroup of G , then G ∞ /N is perfect and so G/N is non-solvable. Butanz(
G/N ) < G/N is solvable by induction, a contradiction.We may assume that N = G ∞ is a non-abelian minimal normal subgroup of G . ByTheorem A, there exists χ ∈ Irr( G ) such that χ N is irreducible and χ vanishes on twoconjugacy classes. Suppose the two conjugacy classes are represented by elements g and g of G . Since χ G ′ is irreducible, we have that the character βχ is irreduciblefor every linear β of G by Lemma 2.2. The βχ ’s are distinct for distinct characters β . We show that every character of the form βχ also vanishes on g and g . Then βχ ( g i ) = β ( g i ) χ ( g i ) = β ( g i ) · i ∈ { , } .Hence for every linear character β of G , there is a corresponding non-linear irreduciblecharacter of G of the form βχ that vanishes on two conjugacy classes of G . Let a bethe number of linear characters of G and b be the number of non-linear irreduciblecharacters not of the form βχ (note that these irreducible characters may vanish onmore than one conjugacy class). Then | Irr( G ) | = 2 a + b and nz( G ) = 2 a + b + c , where c is a non-negative integer. Therefore anz( G ) = a + b + c a + b > a + b a + b > a a = 1, concludingour argument. (cid:3) VERAGE NUMBER OF ZEROS OF CHARACTERS 7 Supersolvable, nilpotent and abelian groups
Proof of Theorem C . Suppose G is non-abelian, that is, G ′ >
1. We shall useinduction on | G | to show that G is supersolvable. By Theorem B, G is solvable sinceanz( G ) < <
1. Let N G ′ be a minimal normal subgroup of G . By Lemma 2.3,anz( G/N ) anz( G ) < . Using the inductive hypothesis, G/N is supersolvable. If N is cyclic or N Φ( G ), the Frattini subgroup, then G is supersolvable and the resultfollows.Suppose that χ ∈ Irr( G ) be non-linear such that χ G ′ is irreducible. By Lemma 2.2,for every linear character β i ∈ Irr(
G/G ′ ), there exists a non-linear β i χ ∈ Irr( G ), that is,the number of linear characters of G is less than or equal to the number of non-linearirreducible characters of G . Every non-linear irreducible character of G vanishes onat least one conjugacy class by [9, Theorem 3.15]. Let | Irr( G ) | = a + b , where a isthe number of the non-linear characters and b is the number of linear characters andnz( G ) = a + c , where c is non-negative integer. Then anz( G ) = a + ca + b > a + c a > , since a > b . This contradicts our hypothesis.Hence every non-linear irreducible character χ of G is such that χ G ′ is not irreducible.Note that G is an M -group by [9, Theorems 6.22 and 6.23 ] since G/N is supersolvableand N is abelian. By Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( χ ). If υ ( χ ) is a conjugacy class, then G is a Frobeniusgroup with an abelian kernel and a complement of order two by Lemma 2.6. Thus by[2, Proposition 2.7], every irreducible character of G vanishes on at most one conjugacyclass. Note that G has two linear characters. Since anz( G ) < , G can only have onenon-linear character. It follows that G has three conjugacy classes. This implies that G ∼ = S . Now S is supersolvable group and we are done.Therefore every non-linear irreducible character of G vanishes on at least ℓ conjugacyclasses, where ℓ >
2. Suppose G has an irreducible character χ that vanishes on exactly ℓ conjugacy classes. Let b be the number of linear characters of G and m be the numberof non-linear irreducible characters of G . If b ℓ −
1, then anz( G ) > mℓb + m > ,contradicting our hypothesis. We may assume that b = 2 ℓ . Then G has only onenon-linear irreducible character χ . By Lemma 2.8, G is a Frobenius group with anelementary abelian kernel of order p n and cyclic complement of order p n − p . Also note that | Irr( G ) | = 2 ℓ + 1. By Lemma 2.8, anz( G ) = ℓ − ℓ +1 > since ℓ >
2, a contradiction.Suppose that b > ℓ + 1. Then by Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( ϕ ) for every non-linear irreducible character ϕ of G . We also have that | G : G ′ | − | K : G ′ | = b − | K : G ′ | > ℓ + 1 by Lemma 2.11,again contradicting our hypothesis that G has an irreducible character that vanisheson exactly ℓ conjugacy classes. This concludes our proof. (cid:3) Proof of Theorem D . Suppose anz( G ) < . By Theorem C, G is supersolvablegroup. In particular, G is an M -group. Suppose that χ ∈ Irr( G ) be non-linear suchthat χ G ′ is irreducible. Then by the argument in the second paragraph of the proofof Theorem C, anz( G ) > > , a contradiction. Hence every non-linear irreduciblecharacter χ of G is such that χ G ′ is not irreducible. There exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( χ ) by Lemma 2.5. If υ ( χ ) is a conjugacyclass, then G is a Frobenius group with an abelian kernel and a complement of order twousing Lemma 2.6. By [2, Proposition 2.7], we have that every irreducible character of G SESUAI YASH MADANHA vanishes on at most one conjugacy class. Since G has two linear characters, anz( G ) > ,contradicting our hypothesis.Hence every non-linear irreducible character of G vanishes on at least ℓ conjugacyclasses, where ℓ >
2. Suppose G has an irreducible character χ that vanishes on exactly ℓ conjugacy classes. Let b be the number of linear characters of G . If b ℓ − G ) > , a contradiction. We may assume that b = 3 ℓ . Then G has onlyone non-linear irreducible character χ . By Lemma 2.8, G is a Frobenius group with anelementary abelian kernel of order p n and a cyclic complement of order p n − p . By Lemma 2.8, anz( G ) = ℓ − ℓ +1 > since ℓ >
2, a contradiction.Suppose that b > ℓ + 1. Then by Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( ϕ ) for every non-linear irreducible character ϕ of G . We also have that | G : G ′ | −| K : G ′ | = b −| K : G ′ | > ℓ +1 by Lemma 2.11, contradictingour hypothesis that G has an irreducible character that vanishes on exactly ℓ conjugacyclasses. This concludes our proof. (cid:3) Remark 4.1.
Let L = { G | G is an extra-special 2-group of order 2 k +1 for somepositive integer k } and G ∈ L . Since | G/G ′ | = 2 k and G has only one non-linearirreducible character, we have acd( G ) = k +2 k k +1 . Note that L is an infinite family ofnilpotent groups G such that acd( G ) < . Hence acd( G ) → G → ∞ . In otherwords there does not exist c > G ) < c implies that G is abelian.Our last result shows that there does not exist a non-abelian nilpotent group G suchthat anz( G ) < . We shall restate Theorem E below. Theorem 4.2.
Let G be a finite group. Then G is abelian if and only if anz( G ) < .Proof. If G is abelian, then anz( G ) = 0 < . Suppose that anz( G ) < . Then G isnilpotent group by Theorem D. Using induction on | G | , we have that G/N is abelianfor some minimal normal subgroup N of G , that is, N = G ′ . If χ G ′ is irreducible,then χ is linear. Hence every non-linear irreducible character χ is such that χ G ′ is notirreducible.By Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( χ ). If υ ( χ ) is a conjugacy class, then G is a Frobenius group with anabelian kernel and a complement of order two by Lemma 2.6, a contradiction since G is nilpotent.Hence every non-linear irreducible character of G vanishes on at least ℓ conjugacyclasses, where ℓ >
2. Suppose G has an irreducible character χ that vanishes on ℓ conjugacy classes. Let b be the number of linear characters of G . If b ℓ −
1, thenanz( G ) > , a contradiction. We may assume that b = 3 ℓ . Then G has only onenon-linear irreducible character χ . By Lemma 2.8, G is an extra-special group 2-group.Using the second part of Lemma 2.8, we have that anz( G ) = ℓ − ℓ +1 > since ℓ >
2, acontradiction.Suppose that b > ℓ + 1. Then by Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( ϕ ) for every non-linear irreducible character ϕ of G . We have that | G : G ′ | − | K : G ′ | = b − | K : G ′ | > ℓ + 1 by Lemma 2.11, contradictingour hypothesis that G has an irreducible character that vanishes on exactly ℓ conjugacyclasses. This concludes our proof. (cid:3) Proof of Theorem F . Suppose that G is non-abelian. Note that G is solvable. Alsonote that every non-linear irreducible character of G vanishes on at least ℓ conjugacy VERAGE NUMBER OF ZEROS OF CHARACTERS 9 classes, ℓ >
2. Suppose that G has an irreducible character χ that vanishes on twoconjugacy classes. By Lemma 2.10, either G is a Frobenius group with a complement oforder 3 or there are normal subgroups M and N of G such that: M is a Frobenius groupwith the kernel N ; G/N is a Frobenius group of order p ( p − / M/N and a cyclic complement of order ( p − / p . If G is a Frobeniusgroup with a complement of order 3, then G can only have at most two non-linearirreducible characters. Otherwise anz( G ) > , a contradiction (note that a group ofodd order has an even number of non-linear irreducible characters). By Lemma 2.9, G is the group in the case (d) of Lemma 2.9 with ( p k − / p k = 7and | G | = 21. Hence G is supersolvable. We may assume that G is the group in thesecond case above. Then M = G ′ , χ M is irreducible and by Lemma 2.2, for every linearcharacter α of G , there exists a corresponding αχ ∈ Irr( G ) and hence anz( G ) > G vanishes on at least ℓ conjugacyclasses, ℓ >
3. Suppose that G has irreducible character vanishing on exactly ℓ conju-gacy classes. If there exists χ such that χ G ′ is irreducible, then for every linear character α of G , there exists a corresponding αχ ∈ Irr( G ) and hence anz( G ) >
1, contradictingour hypothesis. We may assume that for every irreducible character χ of G , χ G ′ is re-ducible. By Lemma 2.5, there exists a normal subgroup K of G such that G ′ K < G and G \ K ⊆ υ ( ϕ ) for every non-linear irreducible character ϕ of G . If b ℓ −
1, then G has at most two non-linear irreducible characters. Otherwise, anz( G ) > ℓ ℓ +2 >
1, acontradiction. Hence G has exactly two non-linear irreducible characters. By Lemma2.9, G is a Frobenius group with an elementary abelian kernel G ′ of order p k and acyclic Frobenius complement of order ( p k − /
2, where p is odd prime. Note that everynon-linear irreducible character of G vanishes on G \ G ′ . Thus b = ( p k − / | G : G ′ | = b , we have that b − ℓ using Lemma 2.5. Hence anz( G ) > b − b +2 >
1, acontradiction to our hypothesis.If b > ℓ + 1, we have that | G : G ′ | − | K : G ′ | = b − | K : G ′ | > ℓ + 1 by Lemma 2.11,a contradiction to the hypothesis that G has an irreducible character that vanishes onexactly ℓ conjugacy classes. (cid:3) Acknowledgements
The author would like to thank the reviewer for providing simpler arguments in theproofs of several lemmas and Theorem A.
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