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rX-Complementary generations of the jankogroups j1 J2 And j3Shahiem Ganief a & Jamshid Moori ba University of the Western Cape , Private Bag XI7, Bellville, 7535, South Africa E-mail:b University of Natal , Private Bag X01, Scottsville, Pietermaritzburg, 3209, SouthAfrica E-mail:Published online: 27 Jun 2007.

To cite this article: Shahiem Ganief & Jamshid Moori (1996) rX-Complementary generations of the janko groups j1 J2

And j3 , Communications in Algebra, 24:3, 809-820, DOI: 10.1080/00927879608825602

To link to this article: http://dx.doi.org/10.1080/00927879608825602

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COMMUNICATIONS IN ALGEBRA, 24(3), 809-820 (1996)

r X - C O M P L E M E N T A R Y G E N E R A T I O N S O F THE J A N K O G R O U P S Jl, Jz A N D J3

SHAHIEM GANIEF University of the Western Cape, Prrvate Bag XI7, Bellville,

7535, South Africa, rnganief@math.uwc.ac.za

JAMSHID MOORI Unzversity of Natal, Pnvate Bag XOI, Scottsville, Pietermaritzburg,

9209, South Afrzca, moor~Ounpsunl.cc.unp.ac.za

ABSTRACT. A finite group G with conjugacy class r X is said to be r X - complementary generated if, given an arbitrary z E G - {I}, there is a y E r X such that G = (2, y). The rX-complementary generation of the simple groups was first introduced by Woldar in [17] to show that every sporadic simplegroup can be generated by an arbitrary element and another suitable element. It is conjectured in [5] that every finite simple group can be generated in this way. In this paper we investigate the rX-complementary generation of the first three Janko groups in an attemp to further develop the techniques of finding rX-complementary generation of the finite simple groups. As a consequence, we obtained all the (p, q , r)-generations of the Janko group J 3 , where p, q , r are distinct primes.

Let G be a group and r X a conjugacy class of G (notation as in [3]). The group G is said to be rX-complementary generated (so defined by Woldar in [17]) if, given an arbitrary non-identity element x E G, there exist a y E r X such that G = (x, y). The element y = y ( z ) for which G = (x, y) is called complementary. A verification of the rX-complementary generation of the finite simple groups in the affirmative will settle a weaker conjecture in [5]

- - - - - -

1991 Maihematvx Subject Classtficatzon. 20D08, 20F05. The first author was supported by a research grant from the FRD (SA) and DAAD (Germany). The second author was supported by a research grant from University of Natal.

Copyright 0 1996 by Marcel Dekker, Inc.

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810 GANIEF AND MOON

which states that every finite simple group is 3/2-generated, that is, given an arbitrary x E G - {I), complemenary y always exist. This conjecture is partially answered in [17], where it is proved that every sporadic simple group is PA-complementary generated, where p is the maximal prime divisor of the order of the group.

With this conjecture in mind, a natural question to ask is: Find all conju- gacy classes r X of a finite simple group G such that G is rX-complementary generated. Not much is known on rX-complementary generation and in this paper we will further develop the techniques of finding rX-complementary generation of the finite simple groups, initiated in [17]. We will investigate all the rX-complementary generation of the Janko groups J1, J z and J3. It turns out that we require the (p, q, r)-generations of these three Janko groups, where p, q, r are distinct primes. The (p, q, r)-generations of the groups J1 and Jz were studied in [ l l ] .

We shall use the ATLAS-notation ([3]) for conjugacy classes (eg., 2X de- notes a general class of involutions), character tables, maximal subgroups, etc. Computations were carried out with the aid of and GAP ([13]) running on a Sun GX2 computer.

2. PRELIMINARY RESULTS

In this section we discuss techniques that are useful in resolving generation- type questions for finite groups and set up the notation used in subsequent sections. We encourage the reader to consult [I], [8] and [17] for background material on triangular groups, rX-complementary generation and computa- tional techniques.

Lemma 2.1. [17] Let r X be a conjugacy class of a finite group G. Then G is rX-complementary generated if and only if for every element x of prime order there is a y E r X such that G = (x, y).

Consider the G-conjugacy classes r X and sY, with ( rX)n = sY, for some integer n. If G is not rX-complementary generated, then there exits an element x of prime order such that (x, y) < G, for all y E r X . Since r , yn E (2, y), it follows that (x, yn) < (x, y) < G, for all yn E sY. Thus we have proved the following result.

Lemma 2.2. If G is sY-complementary generated and ( rX)n = sY, then G is rX-complementary generated.

A group G is said to be (r,s,t)-generated if it can be generated by two elements x and y such that o(x) = r , o(y) = s and o(xy) = t . In this case we also say G is ( rX , sY, tZ)-generated, whenever x E r X , y E s Y and xy E t Z . From the above result, G is rX-cunlplementary generated if and oiily if G is (pY, r X , t,Z)-generated. for all conjugacy classes with elements of prime order

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rX-COMPLEMENTARY GENERATIONS 81 1

p and some conjugacy class tpZ (depending on pY). We therefore reduced the problem of pX-complementary generation of a finite group to one of triangular presentation.

For G-classes r X , s Y and tZ, with c a fixed representative of tZ, the struc- ture constant Ac(rX, sY, tZ) gives the the number of distinct ordered pairs (a, b) satisfying a E rX, b E sY and ab = c. As is well known, Ac(rX, sY, tZ) can readily be computed from the character table of G (see [9]). The num- ber of such pairs which additionally satisfy G = (a, b) will be denoted by Ab(rX, sY, tZ). Clearly a group G admits a (r, s, t)-generation if and only if there exist G-conjugacy classes rX, sY, t Z for which Az( rX, sY, t Z) > 0.

R e m a r k 2.3. (i) It is obvious that if a finite group G is (r, s, t)-generated, then it is

(a(.), ~ ( s ) , a(t))-generated, for any a E S3. (ii) For any positive integer n, the triangular group T(2,2, n) % D,, the

dihedral group of order 2n. Thus if G is a finite group not isomorphic to some dihedral group, then G is not (2X, 2X, nY)-generated, for all classes of involutions and any G-class nY. So by Lemma 2.1, G is not 2X-complementary generated.

In most instances it will be clear from the context to which conjugacy classes r X , sY and t Z of G we are referring. We shall often suppress the conjugacy classes, using A(G) and A*(G) as abbreviated notation for Ac(rX, sY,tZ) and Ac*(rX, sY, tZ), respectively. For any family {HI, . . . , H,) of subgroups of G, we denote C(H1 U . . . U H,) the number of pairs (2, y), x E r X and y 6 sY, which generates a subgroup of H,, for some 1 5 i 5 s. Then clearly A*(G) = A(G) - C(H1 U . . . U H,), where { H I , . . . , H,) is the family of all maximal subgroups of G containing a fixed c E tZ. We easily calculate C(H) from the character table of H stored in GAP.

Theorem 2.4. [9] Let G be a finite simple group and H a maximal subgroup of G containing a fixed element x. Then the number h of conjugates of H containing x is xH(x), where X H is the permutation character of G with action on the conjugates of H. In particular,

where 11,. . . ,x, are representatives of the H-conjugacy classes that fuse to the G-conjugacy class [XI.

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812 GANIEF AND MOOR1

The following result is, in certain situations, very effective a t establishing non-generations.

L e m m a 2.5. (21 Let G be a finite centerless group and suppose lX, m y , n Z are G-conjugacy classes for which A*(G) = A;Z(lX, my , n Z ) < ICc(x)l, x E n Z . Then A*(G) = 0 and therefore G is not (lX, m y , nZ)-generated.

A further useful result that we shall often use is;

L e m m a 2.6. [2] Let G be a (2X, sY, tZ)-generated simple group. Then G is (sY, sY, (tZ)')-generated.

In order to apply Lemma 2.1 we will first find the triangular presentations of J1 that will allow us to deduce its TX-complementary generation. The conju- gacy classes of J1 with elements of prime order are 2A, 3A, 5A, 5B, 7A, 11A, 19A, 19B and 19C.

L e m m a 3.1. [14] The group J1 is (2,3,7)-generated.

L e m m a 3.2. [ll] The group J1 is (2,5,7)-, (2,5,19)-, (2,7,1 I)-, (2,7,19)-, (2,117 lg)- , (5,7,11)-> (5, 7, lg)-, (5,11,19)-, (3,5,7)- , (3,5,19)-, (3,7,11)-, (3,7,19)-, (3,11,19)- and (7,11,19)-generated.

In [ I l l , the above result was proved for the conjugacy classes 5A and 19A (where applicable). If we replace 5A by 5 B = (5A)-', and 19A by 19B = (19A)' or 19C = (19A)4, then it turns out that the corresponding structure constants are unchanged. For example, AJ, (2A, SX, 19Y) = 57, for X E {A, B ) and Y E {A, B , C). Now by applying the result given for the proof of Lemma 2.6 in [ I l l , we deduce that the result mentioned above in Lemma 3.2 is true for all conjugacy classes with elements of appropriate order (eg., the group J1 is (2A,5X, 19Y)-generated, for X E {A, B ) and Y E {A, B, C)) .

Coro l l a ry 3.3. The group J, is (3A, 3.4,7A)-, (5A, 5A, 7A)-, (5R, 5B, 7A)-, (7A,7A, 1lA)-, ( l l A , 11A, 19A)-, (19A, 19A, 11A)-, (19B, 19B, 1 lA)- and (19C, 19C, 11A)-generated.

Proof. The proof is an immediate application of Lemma 2.6 and Remark 2.3(i) t o the results obtained in Lemmas 3.1 and 3.2.

L e m m a 3.4. The group Jl is (5A, 5B1 19A)- and (19A, 19B, 19C)-generated.

Proof. As argued in [ l l ] , the group J1 contains no maximal subgroup tha t is (p, q, 19)-generated, where p and q are primes distinct from 19. Thus for these cases A*(Jl) = A(J1). We easily calculate AJ, (5A, 5B, 19A) = 228 and AJ, (19A, 19B, 19C) = 573 and the result follows. 0

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rX-COMPLEMENTARY GENERATIONS

TABLE 1. Structure constants for J2

We are now ready to prove the main result of this section.

Theorem 3.5. The group J1 is rX-complementary generated if and only if r > 2.

Proof. It follows from Remark 2.3(ii) that J1 is not 2A-complementary gen- erated. We now show Jl is 5A-complementary generated. We proved above (Lemmas 3.2 and 3.4, Corollary 3.3) that J1 is (2A, 5A, 7A)-, (3A, 5A, 19B)-, (5A, 5A, 7A)-, (5A, 5B, 19A)-, (5A, 7A, 11A)-, (5A, 11A, 19C)-generated. Re- arranging these triangular presentation, using Remark 2.3(i), it follows that J1 is (pX, 5A1 q,Y)-generated for all conjugacy classes pX with elements of prime order. So by Lemma 2.1 the group J1 is 5A-complementary generated.

Using similar arguments we can show that J1 is 3A-, 5B-, 7A- , 11 A-, 19A-, 19B- and 19C-complementary generated. Also (6A)2 = 3A, (10A)2 = 5B, (10B)' = 5A1 (15A)5 = 3A = (15B)5 and the result follows from Lemma 2.2. 0

4. TX-COMPLEMENARY GENERATION OF THE GROUP J2

The conjugacy classes of J 2 with elements of prime order are 2A, 2B, 3A, 3B15A, 5 8 , 5C1 5D and 7A.

Lemma 4.1. The group J2 is not rX-complementary generated, where r X E {3A, 3B, 4A, 5A, 5B).

Proof. If the group Jz is rX-complementary generated, then it must be (2A, rX , tY)-generated for some t >_ 7. The appropriate structure constants of J 2 are listed in Table 1.

It is shown in [7] and [ll] that J2 is not (2A,3B,7A)-generated. Let H be a maximal subgroup of J 2 with H 2 22+4:(3 x 5'3). Then we calculate CH(2A, 3B, 8A) = 16 and hence

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814 GANIEF AND MOOR1

Thus J2 is not (2A, 3B,8A)-generated. Let Ii' be a maximal subgroup of J2 such that Ii E 5':D12. Then Cx(2A, 3B, 10X) = 10, for X E {C, D ) and consequently A*(2A, 3B, 10X) = 0, proving Jz is not (2A, 3B, 10X)- generated. Furthermore, if U r U3(3) is a maximal subgroup of J2, then A, (2A, 4A, 7A) = 7 = Cu(2A, 4A, 7A) and non-generation of Jz by this triple follows.

For the remaining triples in Table 1, we have AJ , (~A,TX, tY) < ICJ,(tX)[, and non-generation by these triples follows from Lemma 2.5. Thus for every r X E {3A, 3B, 4A,5A, 5B) , the group Jz is not (2A, r X , tY)-generated, for any t 2 7, and the result follows immediately from Lemma 2.1.

L e m m a 4.2. The group J2 is 6X-complementary generated, where X E {A, B) .

Proof. We will first consider the case X = A. We calculate the structure con- stant A J, (2A, 6A, 10C) = 10. The maximal subgroups of J 2 with non-empty intersection with the classes 2A, 6A and 10C are, up to isomorphisms, 2'+*:A5 and A5 x DIG However, we calculate x ~ ( 2 A , 6 A , 10C) = 0, where M 2 21+4:A5 or A5 x DI0. Thus A;2(2A,6A, 10C) = 10, and Jz is (2A,6A, 10C)- generated.

The only maximal subgroups with non-empty intersection with the classes 6A and 7A are isomorphic to U3(3). Furthermore, if pY E {2B, 5A, 5B, 5C, 5 0 1 , then pY f l U3(3) = 0 and therefore Aj2(pY,6A, 7A) = AJ2(pEr,6A,7A) > 0 (cf. Table 2), proving Jz is generated by these triples. Also a fixed element of order 7 is contained in 2 conjugates of a U3(3) subgroup of J2. Thus

Similarly, A;2 (3B, 6A, 7A) 2 700 - 2(56) = 588 and A;, (7A, 6A, 7A) 2 3756 - 2(168) > 0. The 6A-complementary generation of J2 now follows from Lemma 2.1.

Next we consider the case X = B. The (pX,6B, 7A)-generated proper sub- groups of J2 are contained in the maximal subgroups isomorphic to L3(2):2. Also a fixed element of order 7 is contained in a unique conjugate of a L3(2):2 subgroup. Thus

For any pY E (3A, 5A,5B,5C, 501 , we have pY n L3(2):2 = 0 and conse- quently A;2(pY, 6B, 7A) = Aj2(pY,6B, 7A) > 0 (cf. Table 2). For the re- maining triples we calculate Aj2(2A, 6B, 7A) = 28, A;, (2B, 6B, 7A) = 196, Aj2(2B,6B, 7A) = 1358 and Aj,(7A, 6B, 7A) = 7238, proving J2 is 6B- complementary generated.

L e m m a 4.3. The group J2 is 8A-complementary generated.

Proof. The only maximal subgroups of Jz that meet the classes 7A and 8A non-trivially are, up to isomorphisms, U3(3) and L3(2):2. Since the order of

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rX-COMPLEMENTARY GENERATIONS 815

TABLE 2. Structure constants for J2

both these groups does not divide 5 and A J2(5X, 8A, 7A) > 0 (cf. Table 2), the group Jz is (5X, 8A, 7A)-generated, for all X E {A, B, C, D) . Furthermore,

Similarly, A;,(2B, 8A, 7A) 2 294, Aj,(3A, 8A, 7A) 2 42, Aj2(3B, 8A, 7A) 2 1736 and A;,(7A, 8A, 7A) 2 10304. Thus we shown that Jz is (pX, 8A, 7A)- generated, for all classes pX with prime order representatives, proving the result.

Lemma 4.4. The group J2 is rX-complementary generated, where r X E {5C,5D, 10A, 10B, 15A, 15B).

Proof. The groups U3(3) and L3(2):2 do not contain elements of order 5, 10 or 15. Since these are the only maximal subgroups of J2, up to isomorphisms, with order divisible by 7, it follows that A;, (pY, rX , 7 A ) = A J, (pY, rX , 7A), for all r E {5,10,15). The result follows immediately from Lemma 2.1 and Table 2.

We now summarize the above results.

Theorem 4.5. The group Jz is rX-complementary generated if and only if r X E {5C, 5 0 ) or r > 6.

Proof, It is proved in [I71 that J2 is 7A-complementary generated. The result follows from Remark 2.3(ii) and Lemmas 4.1, 4.2, 4.3 and 4.4 and the fact that (10C)' = 5 0 , ( 1 0 0 ) ~ = 5C and (12A)' = 6A. 0

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GANIEF AND MOOR1

TABLE 3. Structure constants for J3

5. (p, q, T)-GENERATIONS OF THE GROUP J3

In this section we investigate the (p, q, r)-generations of the Janko group J3 and it consequences, where p, q and r are distinct primes satisfying p < q < r. In Section 6 we will consider some additional cases where p = q. Since we may assume l / p + l / q + l / r < 1 (for details, see [I]) it follows that r = 17 or 19. The maximal subgroups of J3 with order divisible by 17 are, up to isomorphisms, L2(16):2 and Lz(17). The group Lz(19) (two non-conjugate copies) is the only maximal subgroup of J3, up to isomorphisms, with order divisible by 19. Throughout this section H, K, L1 and Lz will denote subgroups of J3 with H L2(16):2, It' 2 L2(17), L1 Z Lz(19) L2 and L1 # L;, for all g E J3. Furthermore, for M a maximal subgroup of J3 we denote the permutation character of J3 with action on the conjugates of M by XM. The (2,3, t)-generations of J3 were discussed in [8].

We list in Table 3 the structure constants for the group J3 and in Table 4 a fusion maps of Lz(17) into J3 that we will use later. A general conjugacy class of elements of order n in a subgroup M of J3 will be denoted by nx.

L e m m a 5.1. [S] For the group J3 we have the following.

(i) J3 is (2A, 3B, 17X)-, (2A, 3A, 19X)-, (2A, 3B,l9X)-generated, where X E {A, B).

(ii) J3 i s not (2A, 3A, 17X)-generated, where X E { A , B).

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rX-COMPLEMENTARY GENERATIONS 817

Proof. This is proved in 181, Lemmas 3.10, 3.12 and 3.13.

TABLE 4. A fusion map of Lz(17) into J3

Corol lary 5.2. The group J3 is ( 3 8 , 3B, 17X)-, (3A, 3A, 19X)-, and (3B,3B, 19X)-generated, where X E {A, B).

- Lz(17)-class

CK(nz)

+ Jq

Proof. The result follows immediately from Lemma 2.6 and the fact that (17A)' = 17A, (17B)' = 17B, (19A)' = 19B and (19B)2 = 19A. 0

l a 2a 3a 4a 8a 8b 9a 9b 9c 17a 176

2448 16 9 8 8 8 9 9 9 17 17

1A 2A 3 8 4A 8A 8 B 9A 9 B 9 C 17A 1 7 8

Lemma 5.3. The group J3 is (2A,5X, 17Y)-generated, where X , Y E {A, B ) .

Proof. The only maximal subgroups of J3 with order divisible by 17 are, up to isomorphisms, H and I<. Since \I(\ is not divisible by 5, I( and its sub- groups are not (2A, 5X, 17Y)-generated. Now a fixed element x of order 17 is contained in xH(x) = 2 conjugate copies of H. We calculate C ( H ) = 17 and consequently A'(J3) 2 A(J3) - 2C(H) = 833 > 0 and therefore J3 is (2A, 5X, 17Y)-generated.

Corol lary 5.4. The group J3 is (5X, 5X, 17Y)-generated where X , YE{A, B ) .

Proof. This follows immediately from Lemmas 2.6 and 5.3. 0

Lemma 5.5. The group J3 is (2A,5X, 19Y)-generated, where X, Y E {A, B).

Proof. The permutation character of J3 on the conjugates of L1 is XL, = l a + 85ab+ 1140aa + 1215ab+ 1615a + 1920abc + 2432a. Using Theorem 2.4, a fixed element x of order 19 in L1 is contained in XL, (x) = 1 conjugate copy of L1. We easily calculate that C(L1) = 19 and A(J3) = 874. Since L1 and L2 are the only maximal subgroups of J3 that can contain x and since XL, = xbz ,

we have A'(J3) 2 A(J3) - C(L1) - C(L2) = 836 > 0 and the result follows.

Corollary 5.6. The group J3 is (5X,5X, 19Y)-generated, where X , YE{A, B).

Proof. This follows immediately from Lemmas 2.6 and 5.5.

Lemma 5.7. The group J3 i s ( 3 X , 5Y, 172)-generated, where X, Y, ZE{A, B).

Proof. The order of the subgroup I( is not divisible by 5 and therefore H is the only maximal subgroup of J3, UP to isomorphisms, that contains (3X, 5Y, 172)-

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818 GANIEF AND MOON

generated subgroups. We will first show that J3 is (3A, 5Y, 17Z)generated. In this case C(H) = 17 and a fixed element of order 17 in H is contained in exactly two conjugate copies of H . Thus A'(J3) 2 A(J3) - 2C(H) = 1496 - 34 > 0 and hence J3 is (3A, 5Y, 172)-generated.

Next we show that the group J3 is (3B, 5Y, 172)-generated. For this we easily calculate ~ ~ ( 3 8 ) = 0 and hence C(H) = 0. Therefore A*(J3) = A(J3) = 6902 and the result follows.

Corollary 5.8. If M is a poper(3A, 5Y, 17Z)-generated subgroup of J3, where Y, 2 E {A, B), then M Z Lz(16).

Proof. It is clear from the previous result that if M is (3A, 5Y, 172)-generated, then M 5 Hg, for some g E G. Since Hg Z H 2 L2(16):2, Hg has a soluble quotient isomorphic to the cyclic group 2. Thus Hg is not (3A, 5Y, 172)- generated (see [ll], Lemma 2.1). Further, C(H) = C(Lz(l6)) = 17 and L2(16) contains no maximal subgroup of order divisible by 3 x 5 x 17. Thus M E

Lz(l6). 0

L e m m a 5.9. The group J3 is (3X, 5Y, 192)-generated, where X, Y, ZE{A, B).

Proof. The subgroups L1 and Lz have trivial intersection with the class 3A since x ~ ~ ( , ~ ) ( ~ A ) = 0. Thus the (3A, 5Y, 19.7)-generation of J3 follows from the fact that A'(J3) = A(J3) = 1501.

We now show that J3 is (38,5Y, 19Z)generated. A fixed element of order 19 in L1 is contained in exactly one copy of L1. Hence each non-conjugate copy of Lz(19) contributes C(Ll) = 38 to A(J3) = 6840. Since L1 and Lz are the only maximal subgroups that contain elements of order 19, it follows that A*(J3) 2 6840 - 2 x 38 = 6764 > 0 and the result follows. 13

L e m m a 5.10. The group J3 is (2A, 17X, 19Y)-, (3X, 17Y, 192)-, (5X, 17Y, 192)-, (17X, 17Y, 192)- and (17X, 19Y, 192)-generated, X, Y, 2 E {A, B ) .

Proof. We calculate the structure constants Ah(17X, 17Y, 192) = 175769 and AJ,(17X, 19Y, 192) = 155629. Using these and the structure constants listed in Table 4, and the fact that the group J3 has no maximal subgroup of order divisible by 17 x 19, we deduce that A(J3) = A*(J3) > 0 in all cases mentioned in the statement of the Lemma. Now the result follows. [7

Corollary 5.11. The group J3 is (17X, 17X, 19Y)-generated, where X, Y E {A, B ) .

Proof. This follows immediately from Lemmas 2.6 and 5.10.

We now summarize the above lemmas in the following theorem.

T h e o r e m 5.12. The group J3 is (p, (I, 1)-generated for p, q, r E {2,3,5,17,19) with p < q < r , except when (p,q, r ) = (2,3,5).

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rX-COMPLEMENTARY GENERATIONS

TABLE 5. Structure constants for J3

Proof. This follows from Lemmas 5.1, 5.3, 5.5, 5.7, 5.9, 5.10 and the fact that the triangle group T(2,3,5) is isomorphic to As.

6. TX-COMPLEMENTARY GENERATION OF THE GROUP J3

Before we prove the main theorem of this section, we need the following result.

Lemma 6 .1 . The group J3 is (3A, 3B, 19X)- and (5A, 5B, 19X')-generated, where X E {A, B ) .

Proof. The proof is similar to that of Lemma 5.5 with C L , ( ~ A , 3B, 19X) = 0, EL, (5A, 5B, 19X) = 38, AJ , (~A, 3B,19X) = 228 and A J , ( ~ A , 5B, 19X) = 55 898. 0

Theorem 6.2. The group J3 is rX-complementary generated if and only if r > 2.

Proof. We proved in the above lemmas that the group J3 is (2A,38,19Y)-, (3X, 3X, 19Y)-, (5X, 5X, 19Y)-, (17X, 17X, 19Y)- and (17X, 19Y, 192)- generated, where X,Y,Z E {A, B) . Indeed, using Remark 2.3(i), we have shown that J3 is (pX, 19Y,qpZ)-generated, for all conjugacy classes pX with elements of prime order. It therefore follows from Lemma 2.1 that J3 is 19Y- complementary generated, where Y E (A, B). Similar arguments will show that J3 is 3X-, 5X- and 17X-complementary generated, for X E {A, B). Remark 2.3(ii) yields that J3 is not 2A-complementary generated.

We show next J3 is 4A-complementary generated. The group Lz(19) contains no elements of order 4 and therefore A;,(pX, 4A, 19A) = Aj, (pX, 4A, 19A) > 0 (cf. Table 5), for all classes p X with prime order representatives, proving J3 is 4A-complementary generated.

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820 GANIEF AND MOOR1

The result now follows from Lemma 2.2 since (6A)2 = 3A, (8A)2 = 4A, (9A)3 = (9B)3 = (9C)3 = 3A, (10A)2 = 5A, (10B)2 = 5A, (12A)' = 6A, ( I ~ A ) ~ = 5 8 and (15B)3 = 5A. 0

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Received: April 1995

Revised: September 1995

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