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Department of Mathematics and Applied Mathematics

arXiv:math/9306201v1 [math.GR] 10 Jun 1993(p, q, r)-GENERATIONS FOR THE JANKO GROUPS J1AND J2JAMSHID MOORIDepartment of Mathematics and Applied MathematicsUniversity of NatalP.O.Box 375, Pietermaritzburg3200 South Africa1. IntroductionSuppose that G is a group which is generated by two elements a andb such that o(a) = l and o(b) = m and their product ab has order n.We say that G is a (l, m, n)-generated group and it is obvious that inthis case G must be a quotient of the triangle group ∆(l, m, n) givenby the presentation∆(l, m, n) =< a, b, c | al = bm = cn = abc = 1 > .If G is a (l, m, n)-generated group, then it is also a (r, s, t)-generatedgroup whenever (r, s, t) is a rearrangement of (l, m, n).

Thus we mayassume that l ≤m ≤n. It is well-known that ∆(l, m, n) is finite if andonly if 1/l + 1/m + 1/n > 1.

(see [11] and [4]). Finite ∆(l, m, n) are∆(1, n, n) the cyclic group of order n, ∆(2, 2, n) the dihedral group oforder 2n, ∆(2, 3, 3) the alternating group A4, ∆(2, 3, 4) the symmetricgroup S4, and ∆(2, 3, 5) the alternating group A5.

If 1/l+1/m+1/n =1, namely in the cases of ∆(2, 3, 6), ∆(2, 4, 4) and ∆(3, 3, 3), then thetriangle group is infinite but soluble. In the case when 1/l+1/m+1/n <1, then the triangle group is infinite and insoluble.These trianglegroups have a remarkable wealth of interesting finite quotient groups.We encourage reader to consult [4], [14] and [15] for discussion andbackground material on the triangle groups.

A (2, 3, 7)-generated groupG which give rise to compact Riemann surfaces of genus greater than 2with automorphism groups of maximal order, are called Hurwitz groups([7] and [15]).Woldar in [16] determined that J1, J2, He, Ru, Co3, HN, Ly are Hur-witz, while M11, M12, M22, M23, HS, J3, M24, McL, Suz, O’N, Co1, Co21991 Mathematics Subject Classification. 20D08, 20F05.supported by a research grant from the University of Natanl.1

2JAMSHID MOORIare non-Hurwitz. In [8] Kleidman, Parker and Wilson proved that F23is not Hurwitz, while Linton [9] and Wilson [10] showed that Th andF ′24 are Hurwitz.

In [17] Woldar completes the problem, except for Band M where the question is unresolved, by proving that J4 andF22 areHurwitz.If G is a finite group, C1, C2, C3 conjugate classes of G, and c a fixedrepresentative of C3, then ξG(C1, C2, C3) denotes the structure constantof the group algebra Z(C[G]) which is equal to the number of distinctordered pairs (a, b) with a ∈C1, b ∈C2 and ab = c. Using the charactertable of G, the number ξG(C1, C2, C3) can be calculated by the formulaξG(C1, C2, C3) = |C1|.|C2||G|kXi=1χi(a)χi(b)χi(c)χi(1)where χ1, χ2, . .

. , χk are the irreducible complex characters of G. Letξ∗G(C1, C2, C3) denote the number of distinct ordered pairs (a, b) witha ∈C1, b ∈C2, ab = c and G =< a, b >.

Obviously G is a (l, m, n)-generated group if and only if there exist conjugacy classes C1, C2, C3with representatives a, b, c, respectively such that o(a) = l, o(b) = mand o(c) = n, for which ξ∗G(C1, C2, C3) > 0. In this case we say that Gis (C1, C2, C3)-generated.

If H is a subgroup of G containing c and Bis a conjugacy class of H such that c ∈B, then σH(C1, C2, B) denotesthe number of distinct ordered pairs (a, b) such that a ∈C1, b ∈C2,ab = c and < a, b >≤H.It is well-known that every finite non-abelian simple group can begenerated by two suitable elements. (For details see [5]).

If G is non-abelian finite simple group and l, m, n are divisors of |G| such that1/l+1/m+1/n < 1, then the following question arises: Is G a (l, m, n)-generated group? We are a great distance away from a complete answerto the question, however we are aiming at a partial answer to it byconsidering the case when l = p, m = q and n = r are primes and thegroup G is isomorphic to one of the sporadic simple groups.In the present paper we investigate (p, q, r)-generations for the Jankgroups J1 and J2 where p, q and r are distinct primes satisfying p < q < r.We prove the following results.Theorem 2.8 .

The group J1 is (p, q, r)-generated for p, q, r ∈{2, 3, 5, 7, 11, 19}with p < q < r, except when (p, q, r) = (2, 3, 5).Theorem 2.9 . The group J1 is generated by three involutions a, b, c ∈2A such that abc ∈11A.Theorem 3.5 .

(a) The group J2 is (2B, 3B, 7A)-, (2A, 5C, 7A)-, (2B, 5A, 7A)-, (2B, 5C, 7A)-,(3A, 5C, 7A)-, (3B, 5A, 7A)-, and (3B, 5C, 7A)-generated. (b) The group J2 is not (2A, 3A, 7A)-, (2A, 3B, 7A)-, (2B, 3A, 7A)-,(2A, 5A, 7A)-, or (3A, 5A, 7A)-generated.

GENERATIONS FOR J1 AND J23The remaining cases , namely when p = q = r, p ̸= q = r orp = q ̸= r will be treated separately in another paper by the author.For the description of the conjugacy classes, the character tables andinformation on the maximal subgroups readers are referred to ATLAS[3]. Computations were performed with the aid of CAYLEY and GAP(see [1] and [13]) running on a SUN GX2 computer.1.1.

Acknowledgments. I am very grateful to the referee for his valu-able and constructive remarks, and to Jonathan Hall for several of thearguments in this paper.

I also thank Andrew Woldar for his privatecommunication on the Lemma 2.1.2. (p, q, r)-generations for J1We list bellow the structure constants for the Janko group J1.ξJ1(C1, C2, C3)C1C2C3492A3A7A552A3A11A382A3A19A492A5A7A442A5A11A572A5A19A2092A7A11A2092A7A19A1332A11A19A1893A5A7A1983A5A11A1713A5A19A8583A7A11A8363A7A19A4943A11A19A8585A7A11A8365A7A19A5135A11A19A22997A11A19AWe would also like to mention here that if G is a (p, q, r)-generatedgroup where p, q and r are distinct primes, then G has no solublequotient.

This is an obvious consequence of the following elementaryresult:Lemma 2.1. Let G be a (l, m, n)-generated group with l, m, n pairwiseco-prime.

Then G has no soluble quotient.

4JAMSHID MOORIProof. Assume that G is generated by a, b such that o(a) = l, o(b) = mand o(ab) = n. If G has soluble quotient, then it also has a non-trivialabelian quotient A.

Let ¯a,¯b and ab be the respective images of a, b andab in A. Then o(ab) divides lm and n. Since lm and n are relativelyprime, ab = ¯e, where e is the identity of G. This implies that ¯a = (¯b)−1.So if ¯b ̸= ¯e then o(b) divides both l and m, a contradiction to co-primeness.But if ¯b = ¯e, then ¯a = ¯e.Hence A =< ¯a,¯b >= {¯e},contradicting the non-triviality of A.

This completes the proof of thelemma.We now deal with the (p, q, r)-generations of J1.Lemma 2.2. The group J1 is (2, 3, 7)-generated.Proof.

The fact that J1 is (2A, 3A, 7A)-generated is given by Woldarin [16].Lemma 2.3. The group J1 is (2, 3, 11)-generated.Proof.

The only non-soluble maximal subgroups of J1 with order di-visible by 11 are isomorphic to the group L2(11).If H ≤J1 withH ∼= L2(11), thenσH(2A, 3A, 11A) = ξH(2A, 3A, 11A) = 11.We also observe that NH(11A) = 11:5 and NJ1(11A) = (11:5):2. Let2 =< α >.

Then a fixed element in 11A is contained in H and Hg forg ∈J1 −H if and only if g = h−1βα for some h ∈H and β ∈NH(11A).Thus a fixed element of order 11 in J1 is contained in precisely twocopies of H, namely H and Hα. Hence H and its conjugates contribute11×2 = 22 to the ξJ1(2A, 3A, 11A).

Since ξJ1(2A, 3A, 11A) = 55 > 22,ξ∗J1(2A, 3A, 11A) ≥33. Therefore J1 is (2A, 3A, 11A)-generated.Lemma 2.4.

The group J1 is (2, 3, 19)-generated.Proof. The only maximal subgroup of J1 with order divisible by 19, upto isomorphism, is the group 19 : 6.

Since a (2, 3, 19)-generated groupcannot have a soluble quotient, J1 posses no proper (2, 3, 19)-generatedsubgroup. Now the result follows from the fact that ξJ1(2A, 3A, 19A) =ξ∗J1(2A, 3A, 19A) = 38.Lemma 2.5.

The group J1 is (2, 5, 11)-generated.Proof. Up to isomorphism, L2(11) is the only non-soluble maximal sub-group of J1 with order divisible by 11 × 10 = 110.We also haveσL2(11)(2A, 5A, 11A) = ξL2(11)(2A, 5A, 11A) = 11, and as in Lemma 2.3,L2(11) and it’s conjugates contribute 11×2 = 22 to the ξJ1(2A, 5A, 11A).Since ξJ1(2A, 5A, 11A) = 55, result follows.

GENERATIONS FOR J1 AND J25Lemma 2.6. The group J1 is (2, 5, 7)-, (2, 5, 19)-, (2, 7, 11)-, (2, 7, 19)-,(2, 11, 19)-, (5, 7, 11)-, (5, 7, 19)-, (5, 11, 19)-, (3, 5, 7)-, (3, 5, 19)-,(3, 7, 11)-, (3, 7, 19)-, (3, 11, 19)-, and (7, 11, 19)-generated.Proof.

The group J1 has no proper subgroups of order divisible by2 × 5 × 7, 2 × 5 × 19, 2 × 7 × 11, 2 × 7 × 19, 2 × 11 × 19, 5 × 7 × 11,5 × 7 × 19 , 5 × 11 × 19, 3 × 5 × 7, 3 × 5 × 19, 3 × 7 × 11, 3 × 7 × 19,3 × 11 × 19, and 7 × 11 × 19 respectively. Since ξJ1(C1, C2, C3) for thecorresponding conjugacy classes are positive, the result follows.Lemma 2.7.

The group J1 is (3, 5, 11)-generated.Proof. Up to isomorphism, L2(11) is the only maximal subgroup ofJ1 with order divisible by 3 × 5 × 11.Now σL2(11)(3A, 5A, 11A) =ξL2(11)(3A, 5A, 11A) = 22 implies that L2(11) and its conjugates con-tribute 22 × 2 = 44 to the ξJ1(3A, 5A, 11A).

Since ξJ1(3A, 5A, 11A) =198, the result follows.Theorem 2.8. The group J1 is (p, q, r)-generated for p, q, r ∈{2, 3, 5, 7, 11, 19}with p < q < r, except when (p, q, r) = (2, 3, 5).Proof.

This follows from the Lemmas 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8 andthe fact ∆(2, 3, 5) ∼= A5.If C1, C2, C3, C4 are conjugacy classes of a group G, and d a fixedrepresentative of C4 then ξG(C1, C2, C3, C4) denotes the number of dis-tinct triples (a, b, c) with a ∈C1, b ∈C2 and c ∈C3 such that abc = d.This number is computed using the formulaξG(C1, C2, C3, C4) = |C1||C2||C3||G|kXi=1χi(a)χi(b)χi(c)χi(d)(χi(1))2when χ1, χ2, . .

. , χk are the irreducible complex characters of G. Weuse this fact to prove the following theorem.Theorem 2.9.

The group J1 is generated by three involutions a, b, c ∈2A such that abc ∈11A.Proof. Using the character table of J1 we have ξJ1(2A, 2A, 2A, 11A) =17908.

In J1 we have two maximal subgroups, up to isomorphism, withorder divisible by 11, namely 11:10 and L2(11). Here NJ1(11A) = 11:10.

We also haveσL2(11)(2A, 2A, 2A, 11A) = ξL2(11)(2A, 2A, 2A, 11A) = 242.A fixed element of order 11 in J1 lies in two conjugates of L2(11). HenceL2(11) contributes 242 × 2 = 484 to the number ξJ1(2A, 2A, 2A, 11A).

6JAMSHID MOORIIt is routine to compute the character table of the group 11 : 10. Werepresent a part of this character table, giving the values of irreduciblecharacters on the classes 1A, 2A, and 11A in Table 1.Centralizer1101011Class1A2A11Aχ1111χ21-11χ3111χ41-11χ5111χ61-11χ7111χ81-11χ9111χ101-11χ111001Table 1.

Partial character table of 11:10Using Table 1 we haveσ11:10(2A, 2A, 2A, 11A) =|2A|311 × 10kXi=1(χi(2A))3χi(11A)(χi(1))2=|2A|311 × 1010Xi=1(−1)i+1= 0.Hence 11 : 10 does not contribute to the number ξJ1(2A, 2A, 2A, 11A).Since ξJ1(2A, 2A, 2A, 11A) = 17908 > 484, the group J1 is (2A, 2A, 2A, 11A)-generated. This completes the theorem.3.

(p, q, r)-generations for J2We list below the structure constants for the group J2.

GENERATIONS FOR J1 AND J27ξJ2(C1, C2, C3)C1C2C302A3A7A72A3B7A02B3A7A702B3B7A02A5A7A72A5C7A72B5A7A492B5C7A03A5A7A143A5C7A563B5A7A3433B5C7ALemma 3.1. The group J2 is not (2A, 3A, 7A)-, (2B, 3A, 7A)-, (2A, 5A, 7A)-, or (3A, 5A, 7A)-generated.Proof.

This follows trivially since ξJ2(C1, C2, C3) = 0 for the corre-sponding conjugacy classes of J2.Lemma 3.2. The group J2 is (2A, 5C, 7A)-, (2B, 5A, 7A)-, (2B, 5C, 7A)-,(3A, 5C, 7A)-, (3B, 5A, 7A)-, (3B, 5C, 7A)-generated.Proof.

The only maximal subgroups of J2 with order divisible by 7, upto isomorphism, are the groups U3(3) and L3(2) : 2. Since |U3(3)| =25.32.7 and |L3(2) : 2| = 24.

3. 7 and 5 does not divide neither |U3(3)|nor |L2(3) : 2|, we can say that no proper subgroup of J2 is (p, 5, 7)-generated when p ∈{2, 3}.

AsξJ2(2A, 5C, 7A) = ξJ2(2B, 5A, 7A) = 7,ξJ2(2B, 5C, 7A) = 49ξJ2(3A, 5C, 7A) = 14,ξJ2(3B, 5A, 7A) = 56,ξJ2(3B, 5C, 7A) = 343the result will immediately follow.Lemma 3.3. The group J2 is (2B, 3B, 7A)-generated.Proof.

This is given in [6].Lemma 3.4. The group J2 is not (2A, 3B, 7A)-generated.Proof.

Here we use a theorem of Ree (see [12] and [2]) which states: If Gis a transitive permutation group generated by permutations g1, g2, . .

. , gsacting on a set of n elements such that g1g2 · · · gs is the identity per-mutation, and if generator gi has exactly ci cycles for 1 ≤i ≤s, thenc1 + c2 + · · · + cs ≤(s −2)n + 2.

8JAMSHID MOORIThe group J2 acts as a transitive rank-3 group on a set X of 100elements. The point stabilizer is isomorphic to the group U3(3) withorbits of lengths 1, 36, and 63.If we denote the character of thepermutation representation of J2 on the set X by χ, then by referringto the character table of J2 we obtain χ = 1a + 36a + 63a where ma isthe first irreducible character of degree m in the ATLAS character tableof J2.

It is easy to verify that χ(2A) = 20, χ(3B) = 4, and χ(7A) = 2.This implies that, in the action of J2 on the set X, the elements in2A, 3B and 7A induce permutations with cycle types 120240, 14332 and12714 respectively. Nowc1 + c2 + c3 = (20 + 40) + (4 + 32) + (2 + 14) = 112and(s −2)n + 2 = (3 −2) × 100 + 2 = 102imply that c1 + c2 + c3 > (s −2)n + 2.

This contradicts Ree’s theoremstated above. Hence J2 is not (2A, 3B, 7A)-generated.Theorem 3.5.

For the Janko group J2 we have the following. (a) J2 is (2B, 3B, 7A)-, (2A, 5C, 7A)-, (2B, 5A, 7A)-, (2B, 5C, 7A)-,(3A, 5C, 7A)-, (3B, 5A, 7A)-, and (3B, 5C, 7A)-generated.

(b) J2 is not (2A, 3A, 7A)-, (2A, 3B, 7A)-, (2B, 3A, 7A)-, (2A, 5A, 7A)-,or (3A, 5A, 7A)-generated.Proof. This follows from the Lemmas 3.1, 3.2, 3.3 and 3.4.References[1] J. J. Cannon, An introduction to the group theory language CAYLEY, Compu-tational Group Theory (M. Atkinson, ed.

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