Products of conjugacy classes in simple groups more

PRODUCTS OF CONJUGACY CLASSES IN SIMPLE GROUPS JAMSHID MOORI AND HUNG P. TONG-VIET † Abstract. Let G be a finite group. For a ∈ G, let aG = {ag | g ∈ G} be the conjugacy class of a in G. In this paper, we study a conjecture due to Arad and Herzog which asserts that in a finite non-abelian simple group the product of two nontrivial conjugacy classes is never a single conjugacy class. In particular, we will verify this conjecture for several families of finite simple groups of Lie type. 1. Introduction and Notation All groups considered are finite and all characters are complex characters. Let G be a group. Denote by Irr(G) the set of all irreducible characters of G. For a ∈ G, we denote by aG the conjugacy class of a in G, that is, aG = {ag | g ∈ G}. Let A = aG for some a ∈ G. We call a or any conjugate of a in G, a representative of A, and A is said to be nontrivial if its representatives are nontrivial. For a, b ∈ G, aG bG = {xy | x ∈ aG , y ∈ bG } is the product of two conjugacy classes aG and bG in G. In [2], Arad and Herzog conjectured that in a non-abelian simple group, the product of two nontrivial conjugacy classes can never be a single conjugacy class. This conjecture arose when the authors studied the Szep’s Conjecture about the non-simplicity of groups which are the product of two centralizers, that is a group G satisfying G = CG (a)CG (b). Now the latter condition obviously implies that aG bG = (ab)G . Thus the truth of the conjecture mentioned earlier will settle Szep’s Conjecture. Although the Szep’s Conjecture was proved by Fisman and Arad in [7] using other methods, the Arad and Herzog conjecture still needs to be verified. This conjecture has been proved for the following simple groups: alternating groups, Suzuki groups, P SL2 (q), non-abelian simple groups of order less than one million and 15 out of 26 sporadic simple groups (see [2, 7]). In this note, using elementary character theory, we will show in Lemma 1 that for nontrivial elements a, b, c in G, aG bG = cG if and only if χ(a)χ(b) = χ(c)χ(1) for every irreducible character χ ∈ Irr(G). Using this equivalent condition, we can easily verify the Arad and Herzog Conjecture provided that the character table of the group is known. In fact, we will prove the conjecture for P SL3 (q), P SU3 (q), 2 G2 (q), P Sp4 (q), P Sp2n (3), n ≥ 2 and P SUn (2) with (n, 3) = 1. For non-simple groups, it was also proved in [7] that the product of two nontrivial conjugacy classes in the symmetric groups Sn , with n ≥ 5, is never a conjugacy class (see also [1]). Moreover Dade and Yadav [5] classified all groups such that the product of any two non-inverse conjugacy classes Date: September 7, 2011. 2000 Mathematics Subject Classification. Primary 20C15, 20D5, 20E45. Key words and phrases. conjugacy class, simple groups. Supported by research grants from NRF and North-West University. † Supported by the North-West University (Mafikeng) and the NRF (South Africa). 1 2 J. MOORI AND H.P. TONG-VIET is always a conjugacy class. For more results about products of conjugacy classes in finite groups, we refer the readers to [2]. 2. Preliminaries Let G be a group and let a, b, c ∈ G be nontrivial elements of G. We consider the following conditions. (A) a G bG = c G (B) χ(a)χ(b) = χ(c)χ(1) for all χ ∈ Irr(G). Denote by A, B and C the corresponding conjugacy classes in G with represen˜ ˜ ˜ tatives a, b and c, respectively. Let A, B and C be the corresponding class sums. For χ ∈ Irr(G), let ωχ be the algebra homomorphism from C[G] to C defined as in ˜ [8, p. 35]. Let g ∈ G and let K be a conjugacy class of G containing g and let K ˜ = χ(g)|K|/χ(1). the corresponding class sum. We have ωχ (K) Lemma 1. Conditions (A) and (B) are equivalent. Proof. Retain the notation above. Assume first that AB = C. It follows that ˜˜ ˜ ˜ ˜ AB = αC for some integer α. We obtain, for any χ ∈ Irr(G), ωχ (A)ωχ (B) = ˜ αωχ (C) and hence χ(a)|A|χ(b)|B|/χ(1)2 = αχ(c)|C|/χ(1). Choose χ to be the principal character of G, we obtain |A| · |B| = α|C|. Substituting this into the previous equation, we get χ(a)χ(b) = χ(c)χ(1), which is (B). Conversely, assume m ˜˜ ˜ that χ(a)χ(b) = χ(c)χ(1) for all χ ∈ Irr(G). Write AB = i=1 αi Ci , where Ci , i = 1, · · · , m, are conjugacy classes of G with ci ∈ Ci . By [8, p. 45] we have αi = |A||B| |G| χ(a)χ(b)χ(c−1 ) i . χ(1) χ∈Irr(G) As χ(a)χ(b) = χ(c)χ(1) for all χ ∈ Irr(G), we have αi = |A||B| |G| χ(c)χ(c−1 ) = i χ∈Irr(G) |A||B| |C| 0 if c is conjugate in G to ci otherwise ˜˜ ˜ Thus AB = αC, where α = |A||B|/|C|. Hence AB = C. The proof is complete. Let Irrrat (G) be a subset of Irr(G) consisting of rational valued irreducible characters of G. Define S(G) to be a subset of Irrrat (G) consisting of rational valued irreducible character χ such that there exists a prime p such that p | χ(1) but p χ(g) for any 1 = g ∈ G with χ(g) = 0. Finally if χ ∈ Irr(G) then define V (χ) to be the set of zeros of χ in G. We next describe a procedure to verify (B) using the character table of G provided that the set S(G) is non-empty. For χ ∈ S(G), if χ(c) = 0, then as χ(a)χ(b) = χ(c)χ(1), both χ(a) and χ(b) are non-zero, and since χ ∈ S(G), there exists a prime p such that p | χ(1) but p χ(g) for any 1 = g ∈ G with χ(g) = 0. Hence p divides χ(c)χ(1) but p χ(a)χ(b), which is a contradiction. Thus c ∈ V (χ). Let χ ∈ S(G) − {χ}. Arguing as above, c ∈ V (χ ) and so c ∈ V (χ) ∩ V (χ ). If V (χ) ∩ V (χ ) = ∅ then we are done. Otherwise, continuing the same procedure. As G has only a finite number of irreducible characters, this must stop at some step. If there is a subset T of PRODUCTS OF CONJUGACY CLASSES 3 S(G) such that the intersection of all V (χ), where χ ∈ T, is empty, then (B) cannot happen. If this is not the case, then we need to seek some other irreducible character to obtain a contradiction. We remark that using [4] and the method described here, it is easy to show that the product of any two nontrivial conjugacy classes of sporadic simple groups can never be a single conjugacy class. 3. Products of conjugacy classes in some simple groups of Lie type We next consider condition (B) for P SL3 (q), P SU3 (q), 2 G2 (q) and P Sp4 (q). In these cases, the character tables are available in [11], [12], [6, 10], respectively. Theorem 2. If G = P SL3 (q) or P SU3 (q), where q is a power of a prime p, then the product of any two nontrivial conjugacy classes of G cannot be a single conjugacy class. Proof. In view of Lemma 1, it suffices to show that G does not satisfy condition (B) for any triple (a, b, c) of nontrivial elements in G. We will follow the notation in [11]. Recall that a semi-simple element of G is an element whose order is coprime to the defining characteristic p. Assume that there exists a nontrivial triple (a, b, c) in G such that (B) holds, we will work to get a contradiction. As P SL3 (2) ∼ P SL2 (7) = and P SU3 (2) is not simple, we will assume that q ≥ 3. We first recall some notation from [11]. First of all, δ = 1 if G = P SL3 (q) and δ = −1 if G = P SU3 (q), r = q − δ, d = (3, r), s = q + δ, t = q 2 + δq + 1, r = r/d, s = s/d, t = t/d, d = (3 − d)/2, t r r r = r (1 + δ)/2, t = (t − 1)/6, r = 1, γ1 = , γ = γ1 and γ2 = γ1 . We first show that c is not a semi-simple element. By way of contradiction, assume that c is semi-simple. Take χ to be the Steinberg character of G. Then χ(1) = |G|p = q 3 and χ(a)χ(b) = χ(c)χ(1) = q 3 χ(c). As c is semi-simple, it follows from [3, Theorem 6.4.7] that χ(c) = 0 so that both χ(a) and χ(b) are nonzero. Thus by [3, Theorem 6.4.7] again, a and b are both semi-simple and hence χ(a) = ±|CG (a)|p and χ(b) = ±|CG (b)|p . However we can see that the p-part of the size of the centralizer of nontrivial semi-simple elements of G is at most q so that |χ(a)χ(b)| ≤ q 2 < q 3 |χ(c)|, a contradiction. Thus c is not semi-simple. It follows that χ(a)χ(b) = 0 and so a or b is not semisimple. Without loss of generality, assume that a is not semi-simple. Note that if 1 = χ ∈ Irr(G), and χ(a)χ(b)χ(c) = 0, then |χ(a)| = |χ(c)| and |χ(b)| = |χ(c)|, otherwise, |χ(b)| = χ(1) or |χ(a)| = χ(1), which is impossible as G is simple. By [11, Table 2], a and c belong to one of the following conjugacy classes C2 , C3 (0 ≤ l ≤ d − 1) or C5 (1 ≤ k ≤ r − 1). We note that the value of the Steinberg character χq3 at C2 must be 0 as C2 is a unipotent class. (i) Case c ∈ C2 . Let χ ∈ Irr(G) be an irreducible character labeled by the symbol (k) χqs . Then χ(1) = qs and χ(c) = δq. As 0 = |χ(a)| = |χ(c)|, we deduce that a ∈ C5 for some 1 ≤ k ≤ r − 1. Thus χ(a) = 1 and so χ(b) = χ(c)χ(1), which implies that |χ(b)| = qχ(1) > χ(1), a contradiction. (k) (ii) Case c ∈ C5 , where 1 ≤ k ≤ r − 1. It follows that r − 1 ≥ 1. Let χ be an (u) irreducible character of G labeled by χqt , where 1 ≤ u ≤ r − 1. Then χ(1) = qt and χ(c) = δ 3uk . As 0 = χ(a) = χ(c), we deduce that a ∈ C2 . Thus χ(a) = q and (l) (k) 4 J. MOORI AND H.P. TONG-VIET 3uk so χ(b) = δt = 0. Observe that if 1 = g ∈ G, such that χ(g) = 0, then 3uk χ(g) ∈ {q, s +q −6uk , ±δ 3uk , 3δ, δ( 3uk + 3ul + 3um )}. It follows that |χ(g)| ≤ q + s = 2q + δ, since q ≥ 3, r = 1 and δ = ±1. However |χ(b)| = t = q 2 + δq + 1 > 2q + δ, for any q ≥ 3. Hence this case cannot happen. (l) (iii) Case c ∈ C3 , where 0 ≤ l ≤ d − 1. Observe that 2t = (t − 1)/3 = 2 (t − d)/3d = (q + δq + 1 − d)/3d > 0 as q ≥ 3. Hence the character χ labeled by the (u) symbol χr2 s exists since its multiplicity is 2t > 0. We have χ(1) = r2 s and χ(c) = δ. It follows that a ∈ C2 so that χ(a) = −r and so χ(b) = −rs. By [11, Table 2], the (k) q2 δquk uk only possibility for b is that b ∈ C8 and hence χ(b) = Buk = δ(γ2 +γ2 +γ2 uk ), q2 δquk rt uk + γ2 uk )|, a contradiction. where γ2 = 1. But then rs = q 2 − 1 > 3 ≥ |δ(γ2 + γ2 Thus condition (B) cannot happen. Theorem 3. If G = 2 G2 (q), q = 32m+1 , then the product of any two nontrivial conjugacy classes of G cannot be a single conjugacy class. Proof. It suffices to show that G does not satisfy condition (B) for any nontrivial triple (a, b, c) in G. We will follow the notation in [12]. Assume that there exists a nontrivial triple (a, b, c) in G such that (B) holds, we will work to get a contradiction. As 2 G2 (3) is not simple, we assume m ≥ 1 and hence q ≥ 27. Let χ ∈ Irr(G) be an irreducible character of G labeled by the symbol ξ3 . Then χ is the Steinberg character of G with χ(1) = q 3 . We observe that if 1 = g ∈ G, such that χ(g) = 0, then χ(g) ∈ {±1, q}. Hence if c is semi-simple then both a and b must be nontrivial semi-simple so that |χ(a)χ(b)| ≤ q 2 < q 3 ≤ |χ(c)χ(1)|. Thus c is not semi-simple. As in the proof of Theorem 2, we can assume that neither a nor c is semi-simple. It follows from [12] that a, c ∈ {X, Y, T, T −1 , Y T, Y T −1 , JT, JT −1 }. Now let χ be an irreducible character of G labeled by ξ2 . Then χ (1) = q 2 −q+1 and χ (a), χ (c) ∈ {±1, 1−q}. As χ (a)χ (b) = χ (c)χ (1), and χ (c) = 0, we deduce that 0 = χ (b). Hence χ (a), χ (b) ∈ {±1, 1 − q, 3}. Obviously |χ (a)χ (b)| ≤ (1 − q)2 = q 2 −2q +1 < q 2 −q +1 ≤ |χ (c)χ (1)|, a contradiction. Thus (B) cannot happen. Theorem 4. If G = P Sp4 (q), where q = pf for some prime p, then the product of any two nontrivial conjugacy classes of G cannot be a single conjugacy class. Proof. It suffices to show that G does not satisfy condition (B) for any triple (a, b, c) of nontrivial elements in G. We will follow the notation in [6, 10]. Assume that there exists a nontrivial triple (a, b, c) in G such that (B) holds, we will work to get a contradiction. As P Sp4 (2) ∼ S6 is not simple, we assume that q ≥ 3. = We first show that c is not semi-simple. By way of contradiction, assume that c is semi-simple. Let χ be the Steinberg character of G. As c is semi-simple, χ(a)χ(b) = χ(c)χ(1) = 0. It follows that both χ(a) and χ(b) are nonzero so that a and b are also semi-simple. Assume at first q is even. By [6, Table IV -2], χ is labeled by θ4 . If 1 = g ∈ G such that χ(g) = 0, then χ(g) ∈ {±1, ±q}. As both a and b are semi-simple, χ(a), χ(b) ∈ {±1, ±q}. But then |χ(a)χ(b)| ≤ q 2 < q 4 ≤ |χ(c)χ(1)|. Therefore c is not semi-simple. Now assume q is odd. By [10, Table 3, 4], χ is exactly the character labeled by θ13 . If 1 = g ∈ G such that χ(g) = 0, then χ(g) ∈ {±1, ±q, q 2 }. Moreover the only conjugacy class with χ(g) = q 2 is D1 with representative diag(1, 1, −1, −1) (see [10, Table 1, 2]). Observe that if g ∈ D1 then g 2 = 1. As χ(a)χ(b) = χ(c)χ(1) = q 4 χ(c) = 0, the only possibility for (a, b, c) is that a, b ∈ D1 and χ(c) = 1. This forces b = ax for some x ∈ G and so aG bG = aG aG . It PRODUCTS OF CONJUGACY CLASSES 5 follows that 1 = a2 ∈ aG bG = cG . Thus c = 1, a contradiction. Therefore c is not semi-simple. (i) q is even. Let ψ, ψ ∈ Irr(G) be irreducible characters of G, labeled by the symbols θ1 , θ3 , respectively as in [6, Table IV -2]. We observe that both ψ and ψ are rational valued and since ψ(1) = ψ (1) = q(q 2 + 1)/2, there exists a primitive prime divisor r which divides ψ(1) and ψ (1) but it does not divides any other values of ψ and ψ , hence {ψ, ψ } ⊆ S(G), where S(G) is defined as in Section 2. We note that the values of character labeled by θ1 at B5 (i) and D3 (i) must be swapped. Applying the method described in Section 2, we deduce that c ∈ V (ψ) ∩ V (ψ ) and hence c ∈ B5 (i). By [6, Table IV -1], there are q 2 /4 such classes and the orders of the centralizers of these classes are q 2 + 1. Thus the order of c must divide q 2 + 1 which is coprime to p, and so c is semi-simple, a contradiction. (ii) q is odd. In this case, let ψ and ψ be irreducible characters of G labeled by the symbols θ11 and θ12 , respectively as in [10, Table 3, 4]. We also have {ψ, ψ } ⊆ ¯ ¯ ¯ S(G) and V (ψ) ∩ V (ψ ) = ∪{B1 (i) | i ∈ R1 , i even} ∪ A41 ∪ A42 . As in previous case, ¯ ¯ c ∈ V (ψ) ∩ V (ψ ) and since c is not semi-simple, we deduce that c ∈ A41 ∪ A42 . Next we choose χ and χ to be irreducible characters of G labeled by the symbols χ1 (j) and χ2 (l) as in [10, Table 3, 4], where j ∈ R1 , j even and l ∈ R2 , l even. Note that the values of χ are obtained by negating the values of the columns labeled by χ2 (l). Observe that if 1 = g ∈ G such that χ(g) = 0, then χ(g) ∈ ˜ ˜ ˜ ˜ {1 − q 2 , 1 ± q, 1, ζ ij + ζ −ij + ζ qij + ζ −qij } =: W, where ζ is a q 2 + 1-root of unity ¯41 ∪ A42 , we have χ(c) = 1 so that χ(a)χ(b) = χ(1) = (q 2 − 1)2 . It ¯ in C. As c ∈ A follows that χ(a), χ(b) ∈ W. As q ≥ 3, we see that |χ(a)χ(b)| < (q 2 − 1)2 = χ(1) ¯ ¯ unless a, b ∈ A21 ∪ A22 . But then |χ (a)χ (b)| = (q 2 − 1)2 < q 4 − 1 = |χ (c)χ (1)|. Thus condition (B) cannot happen. Theorem 5. If G = P Sp2n (3), n ≥ 2, or G = P SUn (2), (n, 3) = 1, n ≥ 4, then the products of any two nontrivial conjugacy classes of G cannot be a single conjugacy class. Proof. It suffices to show that G does not satisfy condition (B) for any triple (a, b, c) of nontrivial elements in G. By way of contradiction, assume that there exists a nontrivial triple (a, b, c) in G such that (B) holds. By [9], there exists a nontrivial ¯ character ψ ∈ Irr(G) such that χ = ψ ψ − 1 is irreducible in G, where ψ(1) = (3n + (−1)n )/2 and χ(1) = 3(3n − (−1)n )(3n−1 − (−1)n−1 )/4, when G = P Sp2n (3) and ψ(1) = (2n − (−1)n )/3 and χ(1) = 8(2n−2 − (−1)n )(2n−1 − (−1)n−1 )/9, ¯ ¯ ¯ ¯ when G = P SUn (2). We have ψ(a)ψ(b) = ψ(c)ψ(1) and ψ(a)ψ(b) = ψ(c)ψ(1) as ¯ ∈ Irr(G). Multiplying both sides of these two equations, we obtain ψ, ψ ¯ ¯ ¯ ¯ ψ(a)ψ(a)ψ(b)ψ(b) = ψ(c)ψ(c)ψ(1)ψ(1). ¯ Since ψ ψ = 1 + χ, we deduce that (1) (2) (3) (1 + χ(a))(1 + χ(b)) = (1 + χ(c))(1 + χ(1)). χ(a)χ(b) = χ(c)χ(1). χ(a) + χ(b) = χ(c) + χ(1). As χ ∈ Irr(G), we have It follows from Equations (1) and (2) that 6 J. MOORI AND H.P. TONG-VIET Squaring both sides of Equation (3) and subtracting 4 times Equation (2), we obtain (χ(a) − χ(b))2 = (χ(c) − χ(1))2 . Hence (4) or (5) χ(a) − χ(b) = χ(1) − χ(c). If case (4) holds then it follows from (3) that χ(b) = χ(1), which is impossible as b = 1 and G is simple. If case (5) holds then it follows from (3) that χ(a) = χ(1), which leads to a contradiction. Thus condition (B) cannot happen. Acknowledgment. The authors are grateful to the referee for the comments. References [1] E. Adan-Bante, H. Verrill, Symmetric groups and conjugacy classes, J. Group Theory 11 (2008), no. 3, 371–379. [2] Z. Arad and M. Herzog, Lecture Notes in Mathematics, 1112. Springer-Verlag, Berlin, 1985. [3] R. W. Carter, Finite Groups of Lie Type. Conjugacy classes and complex characters, A WileyInterscience Publication. John Wiley and Sons, New York, 1985. [4] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups, Oxford University Press, Eynsham, 1985. [5] E. C. Dade and M. K. Yadav, Finite groups with many product conjugacy classes, Israel J. Math. 154 (2006), 29–49. [6] H. Enomoto, The characters of the finite symplectic group Sp(4, q), q = 2f , Osaka J. Math. 9 (1972), 75–94. [7] E. Fisman and Z. Arad, A proof of Szep’s conjecture on nonsimplicity of certain finite groups, J. Algebra 108 (1987), no. 2, 340–354. [8] M. Isaacs, Character theory of finite groups, Corrected reprint of the 1976 original [Academic Press, New York]. AMS Chelsea Publishing, Providence, RI, 2006. [9] G. Malle, Almost irreducible tensor squares, Comm. Algebra 27 (1999), no. 3, 1033–1051. [10] M.A. Shahabi and H. Mohtadifar, The characters of finite projective symplectic group PSp(4, q), Groups St. Andrews 2001 in Oxford. Vol. II, 496–527, London Math. Soc. Lecture Note Ser., 305, Cambridge Univ. Press, Cambridge, 2003. [11] W.A. Simpson and J.S. Frame, The character tables for SL(3, q), SU (3, q 2 ), P SL(3, q), P SU (3, q 2 ), Canad. J. Math. 25 (1973), 486–494. [12] H.N. Ward, On Ree’s series of simple groups, Trans. Amer. Math. Soc. 121 (1966) 62–89. E-mail address, J. Moori: Jamshid.Moori@nwu.ac.za School of Mathematical Sciences, North-West University (Mafikeng), Mmabatho 2735, South Africa E-mail address, H.P. Tong-Viet: 23611294@nwu.ac.za School of Mathematical Sciences, North-West University (Mafikeng), Mmabatho 2735, South Africa χ(a) − χ(b) = χ(c) − χ(1)