2010 Mathematics Subject Classification: 20F05, 20D06.We prove that the group PSL6(q) is (2,3)-generated for any q. In fact, we provide explicit generators x and y of orders 2 and 3, respectively, for the group SL6(q)

Tabakov, K.

Tchakerian, K.

Bulgarian Digital Mathematics Library at IMI-BAS

Serdica Math. J. 37 (2011), 365–370(2, 3)-GENERATION OF THE GROUPS PSL6(q)K. Tabakov, K. TchakerianCommunicated by V. DrenskyAbstract. We prove that the group PSL6(q) is (2, 3)-generated for any q.In fact, we provide explicit generators x and y of orders 2 and 3, respectively,for the group SL6(q).1. Introduction. A group G is called (2, 3)-generated if G = 〈x, y〉for some elements x and y of orders 2 and 3, respectively. Such groups attractattention mostly by the fact that a group is (2, 3)-generated if and only if it isa homomorphic image of the famous modular group PSL2(Z). This generationproperty is known to hold for a number of series of finite simple groups. The mostpowerful result in this direction is the theorem of Liebeck-Shalev and Lu¨beck-Malle which states that all finite simple groups, except the symplectic groupsPSp4(2m), PSp4(3m), the Suzuki groups Sz(2m) (m odd), and finitely manyother groups, are (2, 3)-generated (see [11]). Concerning the projective special2010 Mathematics Subject Classification: 20F05, 20D06.Key words: (2,3)-generated group.This work is partially supported by the Scientific Research Fund of the “St. Kl. Ohridski”University of Sofia under Contract No 027, 2012.366 K. Tabakov, K. Tchakerianlinear groups PSLn(q), (2, 3)-generation has been proved in the cases n = 2, q 6= 9[8], n = 3, q 6= 4 [4], [1], n = 4, q 6= 2 [12], [13], [9], n = 5, any q [14], n ≥ 5, oddq 6= 9 [2],[3], and n ≥ 13, any q [10]. The present paper is another contributionto the problem. We prove the following:Theorem. The group PSL6(q) is (2, 3)-generated for any q.We note that our approach is quite different from that of the authors of[2]. Their approach is based on the classification of finite irreducible linear groupsgenerated by root subgroups, while we make use of the known list of maximalsubgroups of PSL6(q).2. Proof of the Theorem. Let G = SL6(q) and G = G/Z(G) =PSL6(q), where q = pm and p is a prime. Set d = (6, q − 1), also Q = q5 − 1if q 6= 3, 7 and Q = (q5 − 1)/2 if q = 3 or 7. The group G acts naturally ona six-dimensional vector space V over the field F = GF (q) and G acts on thecorresponding projective space P (V ). Fix a basis e1, e2, e3, e4, e5, e6 of V .We shall need the following result.Lemma 1. Let M be a maximal subgroup of the group G. Then eitherM is reducible on the space P (V ) or M has no element of order Q/(d,Q).P r o o f. The maximal subgroups of PSL6(q) are determined (up to con-jugacy) in [5]. In particular, this implies that one of the following holds:(i) M belongs to the family C1 of reducible subgroups of G;(ii) M is a member of one of the remaining families C2, C3, C4, C5, C8 of (irre-ducible) geometric subgroups of G;(iii) M ∼= PSL3(q) if q is odd orM ∼= PSL2(11), A7, M12, PSL3(4)·Z2, PSU4(3),or PSU4(3) · Z2 for specific values of p and q.Zsigmondy’s well-known theorem provides a primitive prime divisor ofp5m − 1, i.e., a prime r which divides p5m − 1 but does not divide pi − 1 for0 < i < 5m. We have r ≥ 11 (as r − 1 is a multiple of 5m) and hence r dividesQ/(d,Q). Now it is easy to verify that in case (ii) the only subgroup of orderdivisible by r is M ∼= PSU6(q0) · Z(2,q0+1) if m is even and q = q20 (in fact, forq > 2 this is done in [7], Section 2.4). However, then|M | = q150 (q20 − 1)(q30 + 1)(q40 − 1)(q50 + 1)(q60 − 1)/(3, q0 + 1)(2, 3)-generation of the groups PSL6(q) 367and it is not difficult to see that |M | is not divisible by Q/(d,Q). As for thegroups in case (iii), r does not divide |PSL3(q)| = q3(q2 − 1)(q3 − 1)/(3, q − 1)and the remaining groups have elements of order at most 24 whereas Q/(d,Q) ≥25 − 1 = 31.The lemma is proved. 2.1. We first assume that q 6= 2, 4. Let ω ∈ GF (q5)∗ be of order Q andf(t) = (t− ω)(t− ωq)(t− ωq2)(t− ωq3)(t− ωq4) = t5 − αt4 + βt3 − γt2 + δt − ε.Then f(t) ∈ F [t] and the polynomial f(t) is irreducible over F . Note thatε = ωq5−1q−1 has order q − 1 if q 6= 3, 7, ε = 1 if q = 3, and ε3 = 1 6= ε ifq = 7.Now letx =−1 0 0 γε−1 0 γ0 −1 0 βε−1 0 β0 0 0 αε−1 −1 δ0 0 0 0 0 ε0 0 −1 δε−1 0 α0 0 0 ε−1 0 0, y =0 0 1 0 0 01 0 0 0 0 00 1 0 0 0 00 0 0 0 0 10 0 0 1 0 00 0 0 0 1 0.Then x and y are elements of G of orders 2 and 3, respectively. Denotez = xy =0 0 −1 0 γ γε−1−1 0 0 0 β βε−10 0 0 −1 δ αε−10 0 0 0 ε 00 −1 0 0 α δε−10 0 0 0 0 ε−1.Then the characteristic polynomial of z is fz(t) = (t−ε−1)f(t) and the character-istic roots ε−1, ω, ωq, ωq2, ωq3, ωq4of z are pairwise distinct. Then, in GL6(q5),z is conjugate to diag(ε−1, ω, ωq, ωq2, ωq3, ωq4) and hence z is an element ofG of order Q.Let H = 〈x, y〉, H ≤ G.Lemma 2. The group H acts irreducibly on the space V .P r o o f. Assume thatW is an H-invariant subspace of V and k = dimW ,1 ≤ k ≤ 5.368 K. Tabakov, K. TchakerianLet first k = 1 and 0 6= w ∈ W . Then y(w) = λw where λ ∈ F andλ3 = 1. This yieldsw = µ1(e1 + λ2e2 + λe3) + µ2(e4 + λ2e5 + λe6) (µ1, µ2 ∈ F ).Now x(w) = νw where ν = ±1. This yields consecutively µ2 6= 0, λ = νε−1,µ1 = µ2(α+ νδ − ε−1), and(1) (ν + 1)(α+ νδ − βε− ε−1) = 0,(2) (ν + 1)(α + νδ − γε−1 − ε−1) = 0.In particular, we have ε3 = ν and ε6 = 1. This is impossible if q = 5 orq > 7 since then ε has order q − 1. Thus q = 3 (and ε = 1) or q = 7 (andε3 = 1 6= ε). So ν = 1 and (1), (2) produce γ = βε2 and δ = −α+ βε+ ε2. Thenf(−1) = −(β + 1)(ε2 + ε + 1) = 0 both for q = 3 and q = 7, hence ω = −1, animpossibility.Now let 2 ≤ k ≤ 5. Then the characteristic polynomial of z|W has degreek and must divide fz(t) = (t− ε−1)f(t). The irreducibility of f(t) over F impliesthat this polynomial is f(t) and k = 5. Now the subspace U = 〈e1, e2, e3, e4, e5〉of V is 〈z〉-invariant. If W 6= U then U ∩W is 〈z〉-invariant and dim(U ∩W ) =4 which is impossible. Thus W = U but obviously U is not 〈y〉-invariant, acontradiction.The lemma is proved. (Note that the assertion is false for q = 2 or 4.) Now, in G, the elements x and y have orders 2 and 3, respectively, and (aseasily seen by the above-mentioned diagonal matrix) z = x ·y has order Q/(d,Q).So the group H = 〈x, y〉 has an element of order Q/(d,Q) and H is irreducibleon P (V ) as H is irreducible on V by Lemma 2. Then Lemma 1 implies that Hcannot be contained in any maximal subgroup of G. Thus H = G and G = 〈x, y〉is a (2, 3)-generated group.2.2. Let now q = 2 or 4. We keep the above element y ∈ G of order threebut this time choose the involution x ∈ G to bex =1 0 0 0 0 01 1 0 η 0 η20 0 0 η 1 η20 0 0 0 0 η0 0 1 η 0 η20 0 0 η2 0 0,(2, 3)-generation of the groups PSL6(q) 369where η is a generator of F ∗. Nowz = xy =0 0 1 0 0 01 0 1 0 η2 η0 0 0 1 η2 η0 0 0 0 η 00 1 0 0 η2 η0 0 0 0 0 η2.As in the proof of Lemma 2, one verifies that the group H = 〈x, y〉acts irreducibly on the space V . Indeed, assume first that W = 〈w〉 is a one-dimensional H-invariant subspace of V . Then y(w) = λw (λ ∈ F, λ3 = 1) yieldsagain w = µ1(e1+λ2e2+λe3)+µ2(e4+λ2e5+λe6) (µ1, µ2 ∈ F ). Now x(w) = wimplies consecutively µ2 6= 0, λ = η−1, µ1 = 0, and η = 0, an impossibility.And if W is an H-invariant subspace with 2 ≤ dimW ≤ 5 then one reaches acontradiction just as in the proof of Lemma 2.The characteristic polynomial of z is fz(t) = (t+η2)g(t) where g(t) = t5+η2t4+η2t3+η2t2+(η2+η)t+η. If q = 2 then the polynomial g(t) = t5+t4+t3+t2+1is irreducible over F and hence its roots have order 25−1 in GF (25)∗. Let q = 4,then g(t) = t5+η2t4+η2t3+η2t2+t+η, and set g(t) = t5+ηt4+ηt3+ηt2+t+η2.Now the polynomial h(t) = g(t)g(t) = t10 + t9 + t7 + t6 + t4 + t+ 1 is irreducibleover GF (2) and its roots have order 210 − 1 in GF (210)∗ (see [6], Table C). Itfollows that both for q = 2 and q = 4 the element z has order q5 − 1 = Q.Now, in G, the elements x, y, and z have orders 2, 3, and Q/d, respec-tively. So the group H = 〈x, y〉 has an element of order Q/d and H is irreducibleon P (V ). Again Lemma 1 implies thatH = G and G = 〈x, y〉 is a (2, 3)-generatedgroup.This completes the proof of the theorem. REFERENCES[1] J. Cohen. On non-Hurwitz groups and noncongruence of the modular group.Glasgow Math. J. 22 (1981), 1–7.[2] L. Di Martino, N. A. Vavilov. (2,3)-generation of SL(n, q). I. Casesn = 5, 6, 7. Comm. Alg. 22, 4 (1994), 1321–1347.[3] L. Di Martino, N. A. Vavilov. (2,3)-generation of SL(n, q). II. Casesn ≥ 8. Comm. Alg. 24, 2 (1996), 487–515.370 K. Tabakov, K. Tchakerian[4] D. Garbe. U¨ber eine Klasse von arithmetisch definierbaren Normalteilernder Modulgruppe. Math. Ann. 235, 3 (1978), 195–215.[5] P. Kleidman. The Low-Dimensional Finite Simple Classical Groups andTheir Subgroups. Ph.D. Thesis. Cambridge, 1987.[6] R. Lidl, H. Niederreiter. Finite Fields. Encyclopedia of Mathematicsand its Applications vol. 20. Reading, Massachusetts etc., Addison-WesleyCompany. Advanced Book Program, 1983.[7] M. W. Liebeck, C. E. Praeger, J. Saxl. The Maximal Factorizationsof the Finite Simple Groups and Their Automorphism Groups. Mem. Amer.Math. Soc. 86, 432 (1990).[8] A. M. Macbeath. Generators of the linear fractional group. Proc. Symp.Pure Math. 12 (1969), 14–32.[9] P. Manolov, K. Tchakerian. (2,3)-generation of the groups PSL4(2m).Ann. Univ. Sofia, Fac. Math. Inf. 96 (2004), 101–104.[10] P. Sanchini, M. C. Tamburini. Constructive (2,3)-generation: a permu-tational approach. Rend. Sem. Mat. Fis. Milano 64 (1994), 141–158.[11] A. Shalev. Asymptotic group theory. Notices Amer. Math. Soc. 48, 4(2001), 383–389.[12] M. C. Tamburini, S. Vassallo. (2,3)-generazione di SL4(q) in caratteris-tica dispari e problemi collegati. Boll. Un. Mat. Ital. B(7) 8 (1994), 121–134.[13] M. C. Tamburini, S. Vassallo. (2,3)-generazione di gruppi lineari. Scrittiin onore di Giovanni Melzi. Sci. Mat. 11 (1994), 391–399.[14] K. Tchakerian. (2,3)-generation of the groups PSL5(q). Ann. Univ. Sofia,Fac. Math. Inf. 97 (2005), 105–108.Faculty of Mathematics and Informatics“St. Kliment Ohridski” University of Sofia5, J. Bourchier Blvd1164 Sofia, Bulgariae-mail: ktabakov@fmi.uni-sofia.bgkerope@fmi.uni-sofia.bg Received April 25, 2012

(2,3)-generation of the groups PSL6(q)

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(2,3)-generation of the groups PSL6(q)

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