﻿On (2, 3, <i>t</i>)-Generations for the Rudvalis Group Ru

Applied Mathematics
Vol. 4  No. 9 (2013) , Article ID: 36532 , 6 pages DOI:10.4236/am.2013.49174

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

Department of Mathematics and Statistics, Faculty of Science, Al Imam Mohammad Ibn Saud Islamic University, Riyadh, Saudi Arabia

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

Keywords: Generating Triple; Sporadic Group; Simple Group; Rudvalis Group; -Generation

ABSTRACT

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

1. Introduction

A group G is said to be -generated if it can be generated by two of its elements x and y so that, and. It is well known that every finite simple group can be generated by just two of its elements. Since the classification of all finite simple groups, more recent work in group theory has involved the study of internal structure of these group and generation type problems have played an important role in these studies. Recently, there has been considerable amount of interest in such type of generations. A -generated group is a homomorphic image of the projective special linear group. It has been known since 1901 (see [1]) that the alternating groups An are -generated. Macbeath [2] proved that projective special linear groups, are -generated. With the exception of Matheiu groups M11, M22, M23, and Maclarin’s group McL, all sporadic simple groups are -generated (Woldar [3]). Guralnick showed that any non-abelian finite simple group can be generated by an involution and a Sylow 2-subgroup. In addition, a large number of Lie groups and classical linear groups are -generated as well. Recently, Liebeck and Shalev proved that all finite classical groups (with some exceptions) are -generated.

We say that a group G is -generated (or -generated) if it can be generated by just two of its elements x and y such that x is an involution, and. Moori in [4] computed all -generations for the smallest Fischer group, where p is a prime divisor of. Further, Ganief and Moori determined the -generations for the Janko’s third sporadic simple group J3 (see [5]). Recently, the author with others computed -generations for the Held’s sporadic simple group He, Tits simple group and Conway’s two sporadic simple groups Co3 and Co2 (see [6-8]). Darafsheh and Ashrafi [9] computed generating pairs for the sporadic group Ru. In the present article, we compute all the -generations for the Rudvalis simple group Ru, where t is any divisor of.

2. Preliminaries

In this article, we use same notation as in [6]. In particular, for C1, C2 and C3 conjugacy classes of elements the group Ru and g3 is a fixed representative of C3, we define to be the number of distinct pairs such that. We can compute the structure of G, for the conjugacy classes C1, C2 and C3 from the character table of G by the following formula where are the irreducible complex characters of the group G. Further let, denotes the number of distinct tuples such that and. If , then clearly G is -generated. If H any subgroup of G containing the fixed element, then denotes the number of distinct ordered pairs such that and where is obtained by summing the structure constants of H over all H-conjugacy classes satisfying for.

A general conjugacy class of elements of order n in G is denoted by nX. For examples, 2A represents the first conjugacy class of involutions in a group G. Most of the time, it will clear from the context to which conjugacy classes lX, mY and nZ we are referring. In such case, we suppress the conjugacy classes, using and as abbreviated notation for and , respectively.

Lemma 2.1 ([10]) Let G be a finite centerless group and suppose lX, mY, nZ are G-conjugacy classes for which,. Then and therefore G is not -generated.

Theorem 2.2 ([5]) Let G be a finite group and H a subgroup of G containing a fixed element x such that. Then the number h of conjugates of H containing x is, where is the permutation character of G with action on the conjugates of H. In particular,

where are representatives of the - conjugacy classes that fuse to the G-class.

3. Main Results

The Rudvalis group Ru is a sporadic simple group of order

Wilson [11] completely determined the maximal subgroups of the group Ru. It has exactly 15 conjugacy classes of maximal subgroups (see Table 1) as also listed in the of Finite Group (see [12]). It has precisely two classes of involution, namely 2A and 2B and a unique class 3A of elements of order 3 in Ru.

It is a well known that if G is -generated finite simple group, then. It follows that we need to consider the cases when

.

The cases when t is prime has already been discussed in [9], so we need to investigate the cases when. Next, we investigate each case separately starting with the conjugacy class 15A of Ru.

Lemma 3.1 The sporadic simple Rudvalis group Ru is -generated for all.

Proof: The maximal subgroups of the group Ru having elements of order 15, up to isomorphism, are H3, H8, H11 and H15.

First we consider the case. Using [13], we compute the structure constant . From the above list of maximal subgroups, Ru class 2A does not meet the maximal subgroup H9. The fusion map of the maximal subgroups H3 into the group Ru yields

where 2a, 3a, 3b, 15a, 15b and 2A, 3A, 15A are conjugacy classes of elements in the groups H3 and Ru, respectively. With the help of this fusion map, we calculate the structure. Further, since a fixed element in Ru is contained in a two conjugates of the maximal subgroup H3, the total contribution from the maximal subgroup H3 to the structure constant is 2 × 15. Similarly by considering the fusion maps from the maximal subgroups H8, H11 and H15 we compute that, and. Since the fixed element z is contained in two conjugates of H8 and in a unique conjugate copy of H11, we obtain

Table 1. Maximal subgroups of rudvalis group Ru.

Hence, the group Ru is -generated.

Next, consider the case. We compute the algebra structure constant as. From the maximal subgroups of Ru, we see that the maximal subgroups that may contain -generated proper subgroups are isomorphic to H3, H9, H11 and H15. By considering the fusion maps from the these maximal subgroups into the group Ru and the values of h which we compute using Theorem 1, we obtain

,

,

.

Hence,

Therefore, is generating triple of the group Ru and the proof is complete.

Lemma 3.2 The Rudavalis group Ru is - generated if and and only if, where and.

Proof: Our main proof will consider a number of cases.

Case: From the list of maximal subgroups of Ru (Table 1) we observe that, up to isomorphism, H4 and H6 are the only maximal subgroups that admit -generated subgroups. From the structure constant we calculate, 0 and. Since a fixed element in Ru is contained in three conjugate copies of subgroup H6, we have, and therefore Ru is -generated.

Case: For this triple we calculate the structure constant. Up to isomorphism, H4, H6 and H14 are the only maximal subgroups subgroups of Ru that meet the conjugacy classes 2B, 3A and 8C. We compute that and. A fixed element a fixed element of order 8 in Ru-class 8C is contained in eight copies of the subgroup H14. We obtain, showing that is a generating triple of the group Ru.

Cases, ,: In order to investigate these triples, we construct the group Ru explicitly by using its standard generators given by Wilson [14]. The Rudvalis group Ru has a 28 dimensional irreducible representation over. Using, we generate the group, where a and b are 28 × 28 matrices over with orders 2 and 4 respectively such that ab has order 13. We see that, and. We produce,

,

,

,

, and such that, , ,. Let then, and such that with and. By investigating maximal subgroups of P and looking at the fusion maps of these maximal subgroups into the groups P and Ru we calculate. Consequently we obtain

Hence by Lemma 2.1, we obtain, proving that Ru is not generated by the triple. By using the standard generators a and b of Ru together with above produced elements c, d, e, f, l, m, n in a similar way we explicitly generate -subgroups and observed that Ru is also not generated by the triple. This completes the lemma.

Lemma 3.3 The sporadic group Ru is - generated for all.

Proof: We will investigate each triple separately.

Case: We compute the algebra structure constant for this triple as. Amongst the maximal subgroups of Ru having nonempty intersection with the classes 2A, 3A and 10A are isomorphic to H1, H5, H6, H7, H10 and H15. Now, by considering the fusion maps from these maximal subgroups into the group Ru we calculate, 20, and. Since a fixed element of order 10 is contained in two conjugate copies of H1, four conjugate copies of H7 and a unique conjugate copy of H6. Therefore

proving the generation of Ru by the triple.

Case: The only maximal subgroups of Ru that may contain -generated subgroups, up to isomorphisms, are H3, H11, H14 and H15. However, we calculate that. That is, no maximal subgroup of Ru is - generated and we obtain, showing that is a generating triple for the group Ru.

Case: From the list of maximal subgroup (see Table 1) of Ru, observe that, up to isomorphism, H6, H7 and H15 are only maximal subgroups that may admit -generated subgroups. We compute in that, and. Thus by we have

This shows that Ru is -generated.

Case: For this triple we compute and H3, H11, H14 and H15 are the only maximal subgroups of Ru that meet the classes in this triple. We calculate, , and a fixed element of order 10 in the group Ru is contained in exactly two copies of each of H11 and H14. Therefore,

Hence the group Ru is -generated and the lemma is complete.

Lemma 3.4 Let. The group Ru is -generated if and only if.

Proof: We will consider each case separately.

Case: The maximal subgroups of Ru with order divisible by 12 and non-empty intersection with the classes 2B and 3A are isomorphic to H4, H6, H7 and H11. We calculate that, while. It follows

proving the generation by this triple.

Case: Up to isomorphism H3, H4, H6, H7, H11, H13 and H15 are maximal subgroups of Ru that may contain -generated proper subgroups. We compute that, , , , and

. It follows that

Hence the group Ru is -generated.

Case: We show that the group Ru is not -generated by using the 28-dimensional irreducible representation of Ru over as we used in Lemma 3.2 above. We generate with, such that. We have and. By investigating the group K, we see that and consequently

Hence by Lemma 2.1 Ru is not generated by the triple. Similar technique and arguments show that is also not a generating triple for Ru. The lemma is complete.

Lemma 3.5 The group Ru is -generated for all and.

Proof: We calculate the structure constants ,. From Table 1, the only maximal subgroups of Ru that meet the classes 2A, 3A and 14Z are isomorphic to H3, H4 and H13. Further, H4 is the only maximal subgroup that contribute to the structure constant as and. Thus, we have and the generation of Ru by this triple follows.

Next, we consider the triple. For this triple, the maximal subgroups that meet the Ru classes 2B, 3A and 14Z, up to isomorphism, are H3, H4, H9 and H13. Our computation shows that, and. Thus, for this triple

proving that is a generating triple for the group Ru. This completes the proof.

Lemma 3.6 The Rudvalis group Ru is - generated if and only if, where.

Proof: We calculate structure constant and. First we consider the case when. From the fusion maps of maximal subgroups in Table 1 into the Rudvalis group Ru, we observed that the -generated proper subgroups are contained in the maximal subgroups isomorphic to H4 and H6. Further, since we obtain that H4 and H6 are not -generated. Hence we have and generation of Ru by the triple follows.

For the triple, we use random element generation method as described in Conder [15] to show that Ru is not -generated. Since . This is, there are 288 pairs with, and. We apply a procedure (an analogous procedure given in Conder [15] for CAYLEY), in the computer algebra system (see [16]). It turns out that all 288 pairs generate proper subgroup of Ru and so is not a generating triple of Ru.

Lemma 3.7 The group Ru is -generated where and.

Proof: The maximal subgroups of the group Ru which contains element of order 20, up to isomorphism, are H1, H5, H10 and H15 (see Table 1). We now consider each case separately.

Case: We observed that 20A class of Ru does not meet the maximal subgroup H15. So, the maximal subgroups of Ru having non-empty intersection with the classes 2A, 3A and 20A are, up to isomorphism, H1, H5, H6 and H10. We compute that , , and Thus,

and therefore Ru is -generated.

Case where: The only maximal subgroups of Ru that may contain - generated proper subgroups are isomorphic to H6 and H15. Further since and we have

and so Ru is -, and -generated.

Case: We compute the algebra structure constant as. The only maximal subgroup of Ru having non-empty intersection with the classes 2B, 3A and 20A is isomorphic to H6. Since and a fixed element is contained in a unique conjugate of M6, we have

proving that Ru is -generated.

Case where: First we calculate that the values of structure constant . Again, the only maximal subgroups of Ru which may contain -generated proper subgroups are isomorphic to H6 and H15. We compute and . Hence we obtain

and the generation of Ru with the triples and follows. This completes the lemma.

Lemma 3.8 The Rudvalis group Ru is - generated, where.

Proof: The maximal subgroups of Ru having elements of order 24 are isomorphic to H6, H7, H10 and H11 (see Table 1).

First we consider the triple. By looking at the fusion maps from the above maximal subgroups into the group Ru, in each case, we obtain

.

Now, since we have. Thus, Ru is -, and -generated.

Next, for the triple, we compute. From the above maximal subgroups H10 does the Ru-class 2B. For the maximal subgroups H6 and H11 we obtain For the maximal subgroup H7 we calculate. Hence proving the generation of Ru by the triples and. This completes the proof.

Lemma 3.9 The Rudvalis simple group Ru is -generated, where and .

Proof: Up to isomorphism, H3 and H7 are the only maximal subgroups of Ru which contain an element of order 26 (see Table 1).

Case: In this case we compute . However, in both cases we obtain This implies that there is no contribution from these maximal subgroups to the structure constant. Hence generation of Ru by the triple follows as we have

Case: For this triple we have . By considering the fusion maps from the maximal subgroups H3 and H7 into Ru we calculate and. Thus

and we conclude that Ru is -generated, which completes the proof.

4. Acknowledgements

The author gratefully acknowledges partial financial support from the Deanship of Academic Research (Project No. 301209) at Al Imam Mohammad Ibn Saud Islamic University (IMSIU), Riyadh, Saudi Arabia.

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