Applied Mathematical Sciences, Vol. 9, 2015, no. 41, 2031-2035 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2015.4121008 A Class of Z4C-Groups Jinshan Zhang 1 School of Science Sichuan University of Science and Engineering Zigong, 643000, P. R. China Dandan Liu School of Mechanical Engineering Science Sichuan University of Science and Engineering Zigong, 643000, P. R. China Copyright c 2014 Jinshan Zhang and Dandan Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract The aim of this note is to classify a class of finite solvable groups whose every irreducible character vanish on at most four conjugacy classes in the character table. 1 Introduction Let G be a finite group and υ(χ) := {g G χ(g) = 0}, where χ is an irreducible complex character of G. A classical theorem of Burnside asserts that υ(χ) is non-empty for all χ Irr 1 (G), where Irr 1 (G) denotes the set of non-linear irreducible complex characters of G. We say that an element g G is a vanishing element of G if there exists χ in Irr(G) such that χ(g) = 0, and otherwise we call x a non-vanishing element. Let Van(G) denote the set of vanishing elements of G, that is, 1 Corresponding author Van(G) = {g G χ(g) = 0 for some χ Irr(G)}.
2032 Jinshan Zhang and Dandan Liu Clearly, Van(G) is a proper normal subset of G. We denote k G (N) the number of conjugacy classes of G contained in N, where N is a normal subset of G. Recently, in [4] we classify the finite groups whose every irreducible character of G vanishes on at most three conjugacy classes. Furthermore, the finite groups with k G (Van(G)) 4 have been studied (see [5]). Clearly, if G is a V nc-group, then in particular every irreducible character of G vanishes on at most n conjugacy classes of G. In this paper, we classify a class of finite solvable groups whose every irreducible character vanish on at most four conjugacy classes in the character table. Generally, we define Definition 1.1. A group G is called a ZnC-group if every irreducible character vanishes on at most n conjugacy classes of G. Remark 1.2. A V 4C-group is a Z4C-group. but the converse is not true. For example, G = a, b a 8 = b 2 = 1, b 1 ab = a 1. The Z3C-groups have been very well characterized (see [4]). result of this paper are as follows. The main Theorem A. Let N be a subgroup of G such that G : N = 2. Assume that G is non-nilpotent and that N is abelian. Then G is a Z4C-group but not a Z3C-group if and only if G = G P, where G is a normal abelian 2-complement of G, P Syl 2 (G), P = 8, Z(G) = 4, and G/Z(G) is a Frobenius group with kernel (G/Z(G)) = G and complement P/Z(G) of order 2. In this paper, G always denotes a finite group. Notation is standard and taken from [1]. In particular, denote k G (N) the number of conjugacy classes of G contained in N, where N is a normal subset of G. For N G, set Irr(G N) = Irr(G)\Irr(G/N). 2 Proof of Theorem A We will use frequently the following lemma (see [3, Theorem 2.1]). Lemma 2.1. Let G be non-abelian, and let χ Irr 1 (G). Assume that N is a normal subgroup of G such that G N < G. If χ N is not irreducible, then the following two statements hold: (1) There exists a normal subgroup H of G such that N H < G and G\H υ(χ). (2) If (G\G ) υ(χ) consists of n conjugacy classes of G, then [H : G ] ([G : H] 1) n.
A class of Z4C-groups 2033 The following Lemma characterizes the group G when its normal subgroup N contains all but two conjugacy classes. We only need and record the following weaker description of this classification. Lemma 2.2. ([2, Theorem 2.2]). Let N be a normal subgroup of a nonabelian solvable group G. Then k G (G\N) = 2 if and only if G is one of the following solvable groups. (1) N = 1 and G = S 3. (2) G/N = 3 and G is a Frobenius group with kernel N. (3) G/N = 2 and C G (x) = 4 for all x G N. Proposition 2.3. ([5, Theorem 2.3]). Suppose that G is a non-abelian nilpotent group. If G is a Z4C-group, then G is one of the following groups: (1) G = D 8 or Q 8. (2) G = a, b a 8 = b 2 = 1, b 1 ab = a 1. (3) G = a, b a 8 = 1, b 2 = a 4, b 1 ab = a 1. (4) G = a, b a 8 = b 2 = 1, b 1 ab = a 3. Lemma 2.4. ([4, Lemma 2.7]). Let G be a meta-abelian group. If [G : G ] = p, then G is a Frobenius group with kernel G and complement of order p. Proof of Theorem A. Since G : N = 2 and N is abelian, G = KP, where K is a normal abelian 2-complement of G and P Syl 2 (G). If G/K = P is non-abelian, then by Proposition 2.3, G/K satisfies types (1), (2), (3) or (4) in Proposition 2.3. Set N/K = (G/K). Then [G : N] = 4. Take an irreducible character ξ of G/K with υ(ξ) = G\N. Then it follows by Lemma 2.1 that N = G. Suppose that G := G/K is of order 16. Then k G (υ(ξ)) = 4, and thus the hypothesis yields that k G (υ(ξ)) = k G (υ(ξ)) = 4, and so χ vanishes only on υ(ξ) for every χ Irr(G K). Observe that C G (g) = 4 or 8 for every g G\G. Then we easily obtain a contradiction (note that [N : G ] = 2 and N is abelian). Suppose that G := G/K is of order 8. Then k G (υ(ξ)) = 3. If k G (υ(ξ)) = 3, then arguing as the above paragraph, we also obtain a contradiction. Hence k G (υ(ξ)) = 4. Observe that there exists an element g in G\G such that C G (g) = 6. On the other hand, we easily see that Z(G) = 2, thus it is easy to conclude that 4 divides C G (z), a contradiction. Hence we may suppose that P is abelian. Note that P is abelian, consequently, G K. Applying Lemma 2.1, we conclude that P 8 and that one of the following two cases occurs: (i) F (G) : G = 1, 2 or 4; (ii) F (G) : G = 3.
2034 Jinshan Zhang and Dandan Liu Case 1. N : G = 3. Recall now that G K N, then K = N and so P = 2. Observe that G\N = xg + yg + zg, where x, y, z G\N. Suppose that k G (G\N) = 3. We have G\N = G C G (x) + G C G (y) + G C G (z), and C G (x) = C G (y) = 6 = C G (y) = 6. On the other hand, by the second orthogonality relation we have 6 = C G (g) = G/G + { χ(g) 2 χ Irr 1 (G)}, for all g G\N. Hence χ(g) = 0 for all g G\N and all χ Irr 1 (G). Let P =< t >, where t is an involution. By Fitting Lemma, we have N = C N (P ) [N, P ]. Obviously, C N) (t) = C N (P ) = Z(G). Since C G (g) = 6 for every g G\N, we conclude that Z(G) = 3. So, G = B Z(G), where B = [N, P ]P. Observe that B is a Frobenius group with kernel B = G = [N, P ] and complement of order 2. Then G is a Z3C-group. If k G (G\N) = 4, we easily see that there exists an element g in G\G such that 4 divides C G (z), a contradiction. Case 2. N : G = 1, 2 or 4. First, assume that N = G. Since G is abelian, it follows by Lemma 3.2 that G is a Frobenius group with kernel G and complement of order 2. Thus G is a Z3C-group. Second, assume that N : G = 2. Then G : G = 4, and thus G = G P, where G is a normal 2-complement of G and P = 4. Observe that G/O 2 (G) is a Frobenius group with Frobenius kernel (G/O 2 (G)) = G and complement of order 2. Clearly, G is a Z3C-group. Finally, assume that N : G = 4. Thus G : G = 8 and so G = G P, where G is a normal 2-complement of G and P = 8. Clearly, O 2 (G) = 4, and thus G/O 2 (G) : (G/O 2 (G)) = 2. Furthermore, applying again Lemma 2.4, we see that G/O 2 (G) is a Frobenius group with Frobenius kernel (G/O 2 (G)) = G and complement of order 2, and we are done. Acknowledgment. This paper is supported by the NNSF of China (11201401, 11301532), the Opening Project of Sichuan Province University Key Laborary Bridge Non-destruction Detecting and Engineering Computing (2013QYJ02) and the Sichuan Provincial Education Department Foundation of China (12ZB291).
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