Applied Mathematics, 2013, 4, 1290-1295 http://dx.doi.org/10.4236/am.2013.49174 Published Online September 2013 (http://www.scirp.org/journal/am)

On (2, 3, t)-Generations for the Rudvalis Group Ru

Faryad Ali Department of Mathematics and Statistics, Faculty of Science, Al Imam Mohammad Ibn Saud Islamic University, Riyadh, Saudi Arabia Email: [email protected]

Received June 19, 2013; revised July 19, 2013; accepted July 26, 2013

Copyright © 2013 Faryad Ali. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

ABSTRACT A group G is said to be 2,3,t-generated if it can be generated by an involution x and an element y so that oy   3 and oxy t. In the present article, we determine all 2,3,t -generations for the Rudvalis sporadic Ru, where t is any divisor of Ru .

Keywords: Generating Triple; ; Simple Group; Rudvalis Group; 2,3,t -Generation

1. Introduction divisor of Fi22 . Further, Ganief and Moori determined the 2,3,t -generations for the Janko’s third sporadic A group G is said to be pqr,, -generated if it can be    simple group J (see [5]). Recently, the author with oth- generated by two of its elements x and y so that 3 ers computed 2,3,t -generations for the Held’s spo- ox p, oy q and oxy r. It is well known      radic simple group He, Tits simple group 22F  and that every finite simple group can be generated by just 4  Conway’s two sporadic simple groups Co and Co (see two of its elements. Since the classification of all finite 3 2 [6-8]). Darafsheh and Ashrafi [9] computed generating simple groups, more recent work in has pairs for the sporadic group Ru. In the present article, we involved the study of internal structure of these group compute all the 2,3,t -generations for the Rudvalis and generation type problems have played an important   simple group Ru, where t is any divisor of Ru . role in these studies. Recently, there has been consider- able amount of interest in such type of generations. A 2,3 -generated group is a homomorphic of the 2. Preliminaries projective PSL2, Z . It has been In this article, we use same notation as in [6]. In particu- known since 1901 (see [1]) that the alternating groups An lar, for C1, C2 and C3 conjugacy classes of elements the n  6,7,8 are 2,3-generated. Macbeath [2] proved that group Ru and g3 is a fixed representative of C3, we define projective special linear groups PS L2,q , q  9 are  G CCC12,,3 to be the number of distinct pairs 2,3 -generated. With the exception of Matheiu groups   g12, gC 1C 2 such that g12gg 3. We can com- M , M , M , and Maclarin’s group McL, all sporadic 11 22 23 pute the structure of G, G CCC123,, for the conju- simple groups are 2,3-generated (Woldar [3]). Gural- gacy classes C1, C2 and C3 from the character table of G nick showed that any non-abelian finite simple group can by the following formula be generated by an involution and a Sylow 2-. In addition, a large number of Lie groups and classical CC m g gg CC,,C 12 ii12 i3, linear groups are 2,3 -generated as well. Recently, G 123   G i1 iG1 Liebeck and Shalev proved that all finite classical groups

(with some exceptions) are 2,3 -generated. where 12,,, m are the irreducible complex char-  We say that a group G is 2,3,t -generated (or acters of the group G. Further let, G CCC123,, de-

2,3-generated) if it can be generated by just two of its notes the number of distinct tuples g12, gC 1C2 elements x and y such that x is an involution, y  3 and such that g12gg 3 and Gg 12,,,g gk 1. If  xy t. Moori in [4] computed all 2,3, p-generations G CCC12,,3 >0, then clearly G is CCC123,, -gen- for the smallest Fischer group Fi22 , where p is a prime erated. If H any subgroup of G containing the fixed ele-

Copyright © 2013 SciRes. AM F. ALI 1291

ment g33C , then H CCC123,, denotes the number classes of maximal (see Table 1) as also listed of distinct ordered pairs g12, gCC1 2 such that in the  of (see [12]). It has pre- g12g g 3 and g12, gH where H CC123,,C is cisely two classes of involution, namely 2A and 2B and a obtained by summing the structure constants unique class 3A of elements of 3 in Ru.

H cc12,,, ck  of H over all H-conjugacy classes It is a well known that if G is 2,3,t-generated finite cc12,,,ck1 satisfying cHii  C for 1ik1. simple group, then 12 13 1t <1. It follows that we A general conjugacy class of elements of order n in G need to consider the cases when is denoted by nX. For examples, 2A represents the first t 7,8,10,12,13,14,15,16,20,24,26,29 . conjugacy class of involutions in a group G. Most of the time, it will clear from the context to which conjugacy The cases when t is prime has already been discussed in classes lX, mY and nZ we are referring. In such case, we [9], so we need to investigate the cases when t  7,13, 29. suppress the conjugacy classes, using  G and G Next, we investigate each case separately starting with  as abbreviated notation for G lX,, mY nZ  and the conjugacy class 15A of Ru.

G lX,, mY nZ , respectively. Lemma 3.1 The sporadic simple Rudvalis group Ru is Lemma 2.1 ([10]) Let G be a finite centerless group 2,3,15X AA -generated for all X AB, . and suppose lX, mY, nZ are G-conjugacy classes for Proof: The maximal subgroups of the group Ru hav-  which GlGGXmYnZ,,Cz , znZ . Then ing elements of order 15, up to isomorphism, are H3, H8,  G 0 and therefore G is not lX,, mY nZ -gen- H11 and H15. erated. First we consider the case 2,3,15A AA. Using  Theorem 2.2 ([5]) Let G be a finite group and H a [13], we compute the structure constant subgroup of G containing a fixed element x such that Ru 2,3,15AA A  190. From the above list of maximal gcdox,  NG HH: 1 . Then the number h of con- subgroups, Ru class 2A does not meet the maximal sub- jugates of H containing x is H x , where H is the group H9. The fusion map of the maximal subgroups H3 permutation character of G with action on the conjugates into the group Ru yields of H. In particular, 2aAaAbAaAb 2 , 3 3 , 3 3 ,15  15 ,15  15A m Cx  h   G , where 2a, 3a, 3b, 15a, 15b and 2A, 3A, 15A are conju- i1 Cx gacy classes of elements in the groups H3 and Ru, re- NHG i spectively. With the help of this fusion map, we calculate where x ,, x are representatives of the NH- 1 m G   the structure H 2AA ,3 ,15 A  15 . Further, since a conjugacy classes that fuse to the G-class x . 3  G fixed element zA15 in Ru is contained in a two con- jugates of the maximal subgroup H3, the total contribu- 3. Main Results tion from the maximal subgroup H3 to the structure con- The Rudvalis group Ru is a sporadic simple group of stant Ru 2,A 3,15AA is 2 × 15. Similarly by consid- order ering the fusion maps from the maximal subgroups H8, H and H we compute that 2,3,15AA A 5, 14 3 3 11 15 H8  145926144000 2 3 5 7 13 29.  2AA ,3 ,15A 10 and 2,AA3,15 A 0. Since H11   H15  Wilson [11] completely determined the maximal sub- the fixed element z is contained in two conjugates of H8 groups of the group Ru. It has exactly 15 conjugacy and in a unique conjugate copy of H11, we obtain

Table 1. Maximal subgroups of rudvalis group Ru.

Group Order Group Order

 12 3 2 6 12 3 HF14222 23513 HU23 2: 3:2 237

2 8 38 14 HS3 28:z  3 235713 HL43 2: 2 237

523 46 14 HU5352 2357 H 6522 :S 235

2 52 62 HL7225 2 23513  H88 A 2357

2 2 2 HL92 29 235729 H10 5:4S 5 235729 

2 53 12 5 53 HA11 326  235 H12  5:2 25

3 2 52 HL13 2 13 2 23713 HA14 6  2 235

42 H15 5: 4 A5 235

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 order to investigate these triples, we construct the group Ru 2,3,15AA A Ru explicitly by using its standard generators given by Ru 2,3,15AA A  2 H 2,3,15 AA A 3 Wilson [14]. The Rudvalis group Ru has a 28 dimen- 22,3,15A AA  2,3,15 AAA sional irreducible representation over  2 . Using HH811   , we generate the group Ru ab , , where a and 190 2 15  2  5  1 10  0. b are 28 × 28 matrices over 2 with orders 2 and 4 Hence, the group Ru is 2,3,15A AA-generated. respectively such that ab has order 13. We see that Next, consider the case 2,3,15BA A. We compute aB2 , b4A and ab13A . We produce caba 2 , the algebra structure constant as 2ABA ,3 ,15 510. 8 Ru  2232 2 From the maximal subgroups of Ru, we see that the da  babbabab , maximal subgroups that may contain 2,ABA3,15 -gen- 3222 erated proper subgroups are isomorphic to H3, H9, H11 eab babab, and H15. By considering the fusion maps from the these 4 maximal subgroups into the group Ru and the values of h f  ab 22ba b22 ab  , which we compute using Theorem 1, we obtain le acb , me d and n fb such that cel,, 2 A, 2,3,15BA A 45, H3 df,3A , cd 12A , f 12B . Let Pmn , then 2BA ,3 ,15 A 30 , mA 2 , n3A and mn 8C such that P 0.    Hence by Lemma 2.1, we obtain Ru 2,3,8AAC 0, Therefore, 2,3,15BA A is generating triple of the proving that Ru is not generated by the triple group Ru and the proof is complete. □ 2,3A AC,8 . By using the standard generators a and b Lemma 3.2 The Rudavalis group Ru is 2,3,8X AZ- of Ru together with above produced elements c, d, e, f, l, generated if and and only if X ,,,,ZBBBC , m, n in a similar way we explicitly generate where X AB,  and Z ABC,, . 2,3X AX,8 -subgroups and observed that Ru is also Proof: Our main proof will consider a number of not generated by the triple 2,X 3,8AX. This com- cases. pletes the lemma. □ Case 2,3,8BAB: From the list of maximal sub- Lemma 3.3 The sporadic group Ru is 2,3,10X AZ- groups of Ru (Table 1) we observe that, up to isomor- generated for all X ,,ZA B . phism, H4 and H6 are the only maximal subgroups that Proof: We will investigate each triple separately. admit  2,3,8BAB-generated subgroups. From the struc- Case 2,3,10A AA : We compute the algebra struc- ture constant we calculate 2BAB ,3 ,8 576,  H   4  ture constant for this triple as Ru 2AA ,3 ,10 A 340 . 0 and H6 64 . Since a fixed element z8B in Ru Amongst the maximal subgroups of Ru having non- is contained in three conjugate copies of subgroup H6, we empty intersection with the classes 2A, 3A and 10A are have  Ru Ru3>0H , and therefore Ru   6 isomorphic to H1, H5, H6, H7, H10 and H15. Now, by con- is B2,BA3,8 -generated. sidering the fusion maps from these maximal subgroups Case 2,3,8BAC: For this triple we calculate the  into the group Ru we calculate H1 120 , H6   structure constant Ru 672 . Up to isomorphism, H4,  20, H7  15 and HH51 01H5 0 . H6 and H14 are the only maximal subgroups subgroups of Since a fixed element of order 10 is contained in two Ru that meet the conjugacy classes 2B, 3A and 8C. We conjugate copies of H1, four conjugate copies of H7 and a compute that H 8 and H 0 H . A 14  4 6 unique conjugate copy of H6. Therefore fixed element a fixed element of order 8 in Ru-class 8C  is contained in eight copies of the subgroup H14. We ob- Ru  Ru 2H16 HH4  7  tain Ru Ru8 H14 >0, showing that 340 2 120 20 415 > 0, 2,BA3,8C is a generating triple of the group Ru. Cases 2,3,8A AC, 2,3A AB,8 , 2,3,8BAA: In proving the generation of Ru by the triple 2,3,10A AA.

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Case 2,3,10A AB: The only maximal subgroups of not 2,3,12A AZ -generated by using the 28-dimen- Ru that may contain 2,3,10A AB-generated subgroups, sional irreducible representation of Ru over 2 as up to isomorphisms, are H3, H11, H14 and H15. However, we used in Lemma 3.2 above. We generate K  lf, we calculate that HH3111HH4 15 0. with lA 2 , f 3A such that f 12B . We have That is, no maximal subgroup of Ru is 2,3,10A AB- K  768 and NKRu   1536 . By investigating the  generated and we obtain RuRu 240 > 0 , group K, we see that K 2,3,AA12B  96 and conse- showing that 2,3,A AB10 is a generating triple for the quently group Ru.  Ru 2A,3AB ,12 288 4K 2AAB ,3 ,12  Case 2,3,10BA A: From the list of maximal sub- group (see Table 1) of Ru, observe that, up to isomor- <12.CBRu  phism, H6, H7 and H15 are only maximal subgroups that Hence by Lemma 2.1 Ru is not generated by the triple may admit 2,3,1BA0A-generated subgroups. We 2,A 3,12AB . Similar technique and arguments show  compute in that H6 40 , H7 0 and that 2,3A AB,12 is also not a generating triple for Ru. H15 5 . Thus by we have The lemma is complete. □  Lemma 3.5 The group Ru is 2,3,14X AZ-gener- Ru  RuHH 4  615 ated for all X AB,  and Z ABC,, . 5204045>0. Proof: We calculate the structure constants

Ru 2AA ,3 ,14Z 280, Ru 2,3,14BA Z 532. From This shows that Ru is 2,3,10BA A-generated. Table 1, the only maximal subgroups of Ru that meet the Case 2,3,10BA B: For this triple we compute  classes 2A, 3A and 14Z are isomorphic to H3, H4 and H13. B2BA ,3 ,10 540 and H3, H11, H14 and H15 are the  Further, H4 is the only maximal subgroup that contribute only maximal subgroups of Ru that meet the classes in to the structure constant Ru 2AA ,3 ,14 Z 280 as this triple. We calculate H11  30 , H14 10 , H 4   56 and HH310 3. Thus, we have  HH30 15  and a fixed element zB10 of Ru Ru 2H4  > 0 and the generation of order 10 in the group Ru is contained in exactly two Ru by this triple follows. copies of each of H11 and H14. Therefore, Next, we consider the triple 2ZBA,3,14 . For this tri-  ple, the maximal subgroups that meet the Ru classes 2B, Ru  Ru 2HH11  214 > 0. 3A and 14Z, up to isomorphism, are H3, H4, H9 and H13. Hence the group Ru is 2,3,10BA B-generated and the Our computation shows that H9 28 , H13  14 lemma is complete. □ and H34 0 H  . Thus, for this triple  Lemma 3.4 Let X ,,ZAB. The group Ru is Ru Ru 2HH912 3> 0 proving that 2,3,12X AZ-generated if and only if X  B . Z2,BA3,14 is a generating triple for the group Ru. Proof: We will consider each case separately. This completes the proof. □ Case 2,3,12BA A: The maximal subgroups of Ru Lemma 3.6 The Rudvalis group Ru is 2,3,16X AZ- with order divisible by 12 and non-empty intersection generated if and only if X  B , where X ,,ZAB  . with the classes 2B and 3A are isomorphic to H4, H6, H7 Proof: We calculate structure constant 2AA ,3 ,16 Z 288 and 2,3BA,16Z 576. and H11. We calculate that H 4  24 , H6 48 Ru   Ru  First we consi der the case when X  B . From the fusion while H70 H11. It follows maps of maximal subgroups in Table 1 into the Rudvalis  Ru  Ru 2 HH46 > 0 , group Ru, we observed that the 2,3,16BA Z-generated proper subgroups are contained in the maximal sub- proving the generation by this triple. groups isomorphic to H4 and H6. Further, since Case 2,3,12BA B : Up to isomorphism H3, H4, H6, Ru 2,3,16BA Z 0Ru  2,3,16 BA Z we obtain that H7, H11, H13 and H15 are maximal subgroups of Ru that H4 and H6 are not 2,BA3,16Z -generated. Hence we may contain 2,3,1BA2B-generated proper subgroups. have  2,3,16BA Z2,3B A,16 Z 576>0 and We compute that 2BA,3 ,12B 672 , H 72 , Ru Ru 3 generation of Ru by the triple 2,3,16BA Z follows. H 24 , H 48 ,  H 12 and 4 6  13  For the triple 2,3,16A AZ , we use random elemen t H71H1 0 H15. It follows that generation method as described in Conder [15] to show that Ru is not 2,3,16A AZ-generated. Since    Ru  Ru 4H346 2 HHH  13 2AA ,3 ,16Z 288 . This is, there are 288 pairs Ru   672 4 72  2  24  48 2  12 > 0  x, y with x  2A , y 3A and xyz16 Z. We apply a procedure (an analogous procedure given in Hence the group Ru is 2,3,12BA B -generated. Conder [15] for CAYLEY), in the computer algebra sys- Case 2,3,12A AZ : We show that the group Ru is tem  (see [16]). It turns out that all 288 pairs

Copyright © 2013 SciRes. AM 1294 F. ALI generate proper subgroup of Ru and so 2,3,16A AZ is   Ru 2,3,2BA0Y not a generating triple of Ru. □ 2,3,20BA A   2,3,20 BA Y Lemma 3.7 The group Ru is 2,3,20X AZ-gen- Ru H6 erated where X AB, and Z ABC,, . 600 40 > 0, Proof: The maximal su bgroups of the group R u which and the generation of Ru with the triples 2,3,20BA B contains element of order 20, up to isomorphism, are H1, and 2,3,20BA C follows. This completes the lemma. H5, H10 and H15 (see Table 1). We now consider each □ case separately. Lemma 3.8 The Rudvalis group Ru is 2,3,24X AZ- Case 2,3,A A20A: We observed that 20A class of generated, where X ,,ZAB  . Ru does not meet the maximal subgroup H15. So, the Proof: The maximal subgroup s of Ru having elements maximal subgroups of Ru having non-empty intersection of order 24 are isomorphic to H6, H7, H10 and H11 (see with the classes 2A, 3A and 20A are, up to isomorphism, Table 1). H1, H5, H6 and H10. We compute that First we consider the triple 2,A 3,24AZ. By looking 2,3,20AA A 0, 2AA ,3 , 20A25 , H1 H5  at the fusion maps from the above maximal subgroups 2AA,3 , 20 A 20 and 2AA ,3 ,20A 25. H6  H10 into the group Ru, in each case, we obtain Thus,  HHHH6  00 7  10 11 . Ru 2,3,20AA A Now, since Ru 240 we have  Ru Ru > 0. Ru 2,3,20A AA 2 H  2AAA,3,20 5 Thus, Ru is 2,3,2A A4A -, and 2,3A AB,24 -gener-  2A ,3AA ,20 2AAA,3 ,20 HH610 ated. 220 2 25  20 4  25 > 0, Next, for the triple Z2,3,2BA4 , we compute Ru 2BA ,3 , 24 Z 624 . From the above maximal sub- and therefore Ru is 2,3,20A AA-generated. groups H10 does the Ru-class 2B. For the maximal sub-

Case 2,3,20A AY  where YBC, : The only maxi- groups H6 and H11 we obtain HH610. 1 For mal subgroups of Ru that may contain 2,3,20A AY- the maximal subgroup H7 we calculate H7  24 .  generated proper subgroups are isomorphic to H6 and H15. Hence Ru  Ru 624 24 > 0, proving the Further since 2AA ,3 ,20 Y 20 and generation of Ru by the triples A2B,3,24A and H6   2,3,2AA0Y 0 we have 2,3,24BA B. This completes the proof. □ H15     Lemma 3.9 The Rudvalis simple group Ru is  2,3,2AA0Y Ru 2X ,3,26AZ -generated, where X AB, and  2,3,20A AA 2AAY,3,20 Ru H6  Z ABC,, . 220 20 > 0, Proof: Up to isomorphism, H3 and H7 are the only maximal subgroups of Ru which contain an element of and so Ru is 2,3,20A AB-, and 2,3,20A AC-gen- order 26 (see Table 1). erated. Case 2,3,26A AZ : In this case we compute Case 2,3,20BA A: We compute the algebra struc- Ru Z2A ,3A , 26  312 . However, in both cases we ture constant as Ru 2AAA,3,20 520. The only maxi- obtain H3 0H7. This implies that there is mal subgroup of Ru having non-empty intersection with no contribution from these maximal subgroups to the the classes 2B, 3A and 20A is isomorphic to H6. Since structure constant Ru . Hence generation of Ru by 2BA ,3 , 20 A 40 and a fixed element zRu is  H6  the triple follows as we hav e Ru  Ru > 0. contained in a unique conjugate of M6, we have Case 2,3,26A AZ : For this triple we have  Z2,B3,26A  57 2. By considering the fusion ma ps Ru 2,3,20BA A Ru   from the maximal subgroups H3 and H7 into Ru we cal- Ru 2BA,3 , 20BBH  2 ,3A , 20A 6 culate H3   0 and H7   26 . Thus 220 1 40 > 0,  Ru Ru  2 H7  57 2  2 26 > 0, proving that Ru is 2,3,20BA A -generated. and we conclude that Ru is 2,3,26BA Z-generated, Case 2,3,20BA Y where YBC, : First we cal- which completes the proof. culate that the values of structure constant 2BA ,3 ,20 Y 600 . Again, the only maximal sub- Ru  4. Acknowledgements groups of Ru which may contain 2,3,20BA Y-gener- ated proper subgroups are isomorphic to H6 and H15. We The author gratefully acknowledges partial financial compute 2,3,20BA Y 40 and support from the Deanship of Academic Research (Pro- H6  2,B3,20AY 0. Hence we obtain ject No. 301209) at Al Imam Mohammad Ibn Saud Is- H15 

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