Revista Colombiana de Matem´aticas Volumen 46(2012)2, p´aginas97-111 On Conjugacy Classes of SL(2; q) Sobre las clases conjugadas de SL(2; q) Edith Adan-Bante1;B, John M. Harris2 1University of Saint Thomas, Minnesota, USA 2University of Southern Mississippi, Mississippi, USA Abstract. Let SL(2; q) be the group of 2 × 2 matrices with determinant one over a finite field F of size q. We prove that if q is even, then the product of any two noncentral conjugacy classes of SL(2; q) is the union of at least q − 1 distinct conjugacy classes of SL(2; q). On the other hand, if q > 3 is odd, then the product of any two noncentral conjugacy classes of SL(2; q) is the union q+3 of at least 2 distinct conjugacy classes of SL(2; q). Key words and phrases. Conjugacy classes, Matrices over a finite field, Products of conjugacy classes, Special linear group. 2010 Mathematics Subject Classification. 15A33, 20E45, 20G40. Resumen. Sea SL(2; q) el grupo de las matrices 2 × 2 con determinante uno sobre un campo finito F de tama~no q. Se prueba que si q es par, entonces el producto de cualesquiera dos clases conjugadas no centrales de SL(2; q) es la uni´onde al menos q − 1 distintas clases conjugadas de SL(2; q). Por otro lado, si q > 3 es impar, entonces el producto de cualesquiera dos clases conjugadas q+3 no centrales de SL(2; q) es la uni´onde al menos 2 distintas clases conjugadas de SL(2; q). Palabras y frases clave. Clases conjugadas, matrices sobre un campo finito, producto de clases conjugadas, grupo especial lineal. 1. Introduction Let G be a finite group, A 2 G and AG = AB : B 2 G be the conjugacy class of A in G. Let X be a G-invariant subset of G, i.e. X A = BA : B 2 X = X for all A 2 G. Then X can be expressed as a union of n distinct conjugacy classes of G, for some integer n > 0. Set η(X ) = n. 97 98 EDITH ADAN-BANTE & JOHN M. HARRIS Given any conjugacy classes AG, BG in G, we can check that the product AGBG = XY : X 2 AG;Y 2 BG is a G-invariant subset and thus AGBG is the union of ηAGBG distinct conjugacy classes of G. It is proved in [2] that for any integer n > 5, given any nontrivial conjugacy Sn Sn classes α and β of the symmetric group Sn of n letters, that is α; β 2 Sn Sn Sn r feg, if n is a multiple of two or of three, the product α β is the union of at least two distinct conjugacy classes, i.e. ηαSn βSn ≥ 2, otherwise the product αSn βSn is the union of at least three distinct conjugacy classes, i.e. Sn Sn η α β ≥ 3. A similar result is proved for the alternating group An in [1]. Fix a prime p and an integer m > 0. Let F = F(q) be a field with q = pm elements and S = SL(2; q) = SL(2; F) be the special linear group, i.e. the group of 2 × 2 invertible matrices over F with determinant 1. Given any non-central conjugacy classes AS , BS in S, is there any relationship between ηAS BS and q? Arad and Herzog conjectured in [3] that the product of two nontrivial con- jugacy classes is never a conjugacy class in a finite nonabelian simple group. Thus, when q ≥ 4 is even we have that S = SL(2; q) = PSL(2; q) is simple and so we must have that ηAS BS > 1 unless A = I or B = I. In what follows, we expand and refine this statement. Theorem 1. Fix a positive integer m. Let A and B be matrices in S = SL(2; 2m). Then exactly one of the following holds: (i) AS BS = (AB)S and at least one of A, B is a scalar matrix. (ii) AS BS is the union of at least 2m − 1 distinct conjugacy classes, i.e. ηAS BS ≥ 2m − 1. Theorem 2. Fix an odd prime p and an integer m > 0 such that q = pm > 3. Let A and B be matrices in S = SL(2; q). Then exactly one of the following holds: (i) AS BS = (AB)S and at least one of A, B is a scalar matrix. S S q+3 (ii) A B is the union of at least 2 distinct conjugacy classes, i.e. S S q+3 η(A B ) ≥ 2 . Given any group G, denote by min(G) the smallest integer in the set G G η(a b ): a; b 2 G r Z(G)g. In Proposition 12, given any integer m > 0, we m m present matrices A, B in SL(2; 2m) such that ηASL(2;2 )BSL(2;2 ) = 2m − 1 and thus Theorem 1 is optimal. Also, given any q = pm > 3, where p is an odd prime and m is a positive integer, in Proposition 18 we prove that Theorem 1 is optimal by presenting matrices where min SL(2; q) is attained. Also, using GAP [5], we can check that min SL(2; 3) = 2 and thus Theorem 2 cannot apply when q = 3. Volumen 46, N´umero2, A~no2012 ON CONJUGACY CLASSES OF SL(2;Q) 99 When q is even, SL(2; q) = PSL(2; q) is a simple group of Lie type of char- acteristic two. Hence, if we require that both A and B are not involutions in Theorem 1, the conclusion of Theorem 1 follows from Theorem 2 of [4]. We thank Rod Gow for pointing this out to us. 2. Proofs Notation. We will denote with uppercase letters the matrices and with low- ercase letters the elements in F. Remark 3. We can describe matrix representatives of conjugacy classes in S = SL(2; F) by four families or types ([6]): r 0 (i) , where r 2 F and r2 = 1. 0 r r 0 (ii) , where r; s 2 F and rs = 1. 0 s s u (iii) , where s 2 F, s2 = 1 and u is either 1 or a non-square element of 0 s 2 F, i.e. u 2 F r x : x 2 F . 0 1 (iv) , where w = r + rq and 1 = r1+q for some r 2 E F, where E −1 w r is a quadratic extension of F. That is, any conjugacy class AS of S must contain one of the above matrices. Remark 4. By Lemma 3 of [2], we have that AS BS = BS AS . Thus if we want to prove that given any non-central conjugacy classes AS and BS of S, ηAS BS ≥ n for some integer n, it suffices to prove that the statement holds for each of the six combinations of conjugacy classes containing matrices of type (ii), (iii) and (iv). Remark 5. Two matrices in the same conjugacy class have the same trace. Thus, if the matrices do not have the same trace, then they belong to distinct conjugacy classes. a b e f Lemma 6. Let C = 2 S and A = 2 S. Then c d g h a(de − bg) + c(df − bh) b(de − bg) + d(df − bh) AC = C−1AC = ; a(−ce + ag) + c(−cf + ah) b(−ce + ag) + d(−cf + ah) and therefore Revista Colombiana de Matem´aticas 100 EDITH ADAN-BANTE & JOHN M. HARRIS (i) C r 0 adr − bcs bd(r − s) = ; 0 s −ac(r − s) ads − bcr (ii) C s u s + ucd ud2 = ; 0 s −uc2 s − ucd (iii) C 0 1 ab + c(d − bw) b2 + d2 − bdw = : −1 w −a2 − c2 + acw −ab + d(−c + aw) d −b Proof. Observe that C−1 = Hence −c a d −b e f a b C−1AC = −c a g h c d de − bg df − bh a b = −ce + ag −cf + ah c d a(de − bg) + c(df − bh) b(de − bg) + d(df − bh) = : X a(−ce + ag) + c(−cf + ah) b(−ce + ag) + d(−cf + ah) a b Lemma 7. Let C = 2 S. Then c d C ! r 0 u 0 (i) Trace = ad(r − s)(u − v) + (us + vr). 0 s 0 v C ! r 0 t u (ii) Trace = t(r + s) − ac(r − s)u. 0 s 0 t C ! r 0 0 1 (iii) Trace = (ac + bd)(s − r) + w(ads − bcr). 0 s −1 w C ! r u t w (iv) Trace = 2rt − uwc2. 0 r 0 t C ! r u 0 1 (v) Trace = −ud2 − uc2 + s(r − ucd): 0 r −1 s Volumen 46, N´umero2, A~no2012 ON CONJUGACY CLASSES OF SL(2;Q) 101 C ! 0 1 0 1 (vi) Trace −1 w −1 v = −a2 − b2 − c2 − d2 + bdw + acw + v − ab + d(−c + aw). Proof. The result follows by Lemma 6 and 6(i) C ! r 0 u 0 adr − bcs bd(r − s) u 0 (i) Trace = Trace 0 s 0 v −ac(r − s) ads − bcr 0 v = u(adr − bcs) + v(ads − bcr) = ad(ur + vs) − bc(us + vr) = ad(ur + vs) + (1 − ad)(us + vr) = ad(ur + vs − us − vr) + (us + vr) = adu(r − s) − v(r − s) + (us + vr) = ad(u − v)(r − s) + (us + vr): C ! r 0 t u adr − bcs bd(r − s) t u (ii) Trace = Trace 0 s 0 t −ac(r − s) ads − bcr 0 t = t(adr − bcs) − ac(r − s)u + t(ads − bcr) = ad(rt + st) − bc(st + rt) − ac(r − s)u = (ad − bc)t(r + s) − ac(r − s) = t(r − s) − ac(r − s)u: C ! r 0 0 1 (iii) Trace 0 s −1 w adr − bcs bd(r − s) 0 1 = Trace −ac(r − s) ads − bcr −1 w = −bd(r − s) − ac(r − s) + w(ads − bcr) = (ac + bd)(s − r) + w(ads − bcr): C ! r u t w r + ucd ud2 t w (iv) Trace = Trace 0 r 0 t −uc2 r − ucd 0 t = t(r + ucd) + w − uc2 + t(r − ucd) = 2rt − uwc2: C ! r u 0 1 r + ucd ud2 0 1 (v) Trace = Trace 0 r −1 s −uc2 r − ucd −1 s = −ud2 − uc2 + s(r − ucd): Revista Colombiana de Matem´aticas 102 EDITH ADAN-BANTE & JOHN M.