On the Covering Number of Small Symmetric Groups and Some Sporadic Simple Groups

Luise-Charlotte Kappe, Daniela Nikolova-Popova, and Eric Swartz

ABSTRACT. A set of proper subgroups is a covering for a group if its union is the whole group. The minimal number of subgroups needed to cover G is called the covering number of G, denoted by σ(G). Determining σ(G) is an open problem for many nonsolvable groups. For symmetric groups Sn, Maroti´ determined σ(Sn) for odd n with the exception of n = 9 and gave estimates for n even. In this paper we determine σ(Sn) for n = 8, 9, 10 and 12. In addition we ﬁnd the covering number for the Mathieu group M12 and improve an estimate given by Holmes for the Janko group J1.

1. Introduction

Let G be a group and A = {Ai | 1 6 i 6 n} a collection of proper subgroups of G. If n [ G = Ai, then A is called a cover of G. A cover is called irredundant if, after the removal of i=1 any subgroup, the remaining subgroups do not cover the group. A cover of size n is said to be minimal if no cover of G has fewer than n members. According to J.H.E. Cohn [7], the size of a minimal covering of G is called the covering number, denoted by σ(G). By a result of B.H. Neumann [22], a group is the union of finitely many proper subgroups if and only if it has a finite noncyclic homomorphic image. Thus it suffices to restrict our attention to finite groups when determining covering numbers of groups. Determining the invariant σ(G) of a group G and finding the positive integers which can be covering numbers is the topic of ongoing research. It even predates Cohn’s 1994 publication [7]. It is a simple exercise to show that no group is the union of two proper subgroups. Al- ready in 1926, Scorza [23] proved that σ(G) = 3 if and only if G has a homomorphic image isomorphic to the Klein-Four group, a result many times rediscovered over the years, see, e.g., [5, 16]. In [12], Greco characterizes groups with σ(G) = 4 and in [13] and [14] gives a partial characterization of groups with σ(G) = 5. For further details we refer to the survey article by Serena [24], and for recent applications of this research see for instance [3] and [4]. In [7], Cohn conjectured that the covering number of any solvable group has the form pα +1, where p is a prime and α a positive integer, and for every integer of the form pα+1 he determined a solvable group with this covering number. In [26], Tomkinson proves Cohn’s conjecture and suggests that it might be of interest to investigate minimal covers of non-solvable and in particular simple groups. Bryce, Fedri and Serena [6] started this investigation by determining the covering number for some linear groups such as PSL(2, q), PGL(2, q) or GL(2, q) after Cohn [7] had already shown that σ(A5) = 10 and σ(S5) = 16. In [20], Lucido investigates Suzuki groups and determines their covering numbers. For the sporadic groups, such as M11, 2010 Mathematics Subject Classification. Primary 20D06, 20D60, 20D08, 20-04, Secondary 20F99. Key words and phrases. symmetric groups, sporadic simple groups, finite union of proper subgroups, minimal number of subgroups. 1 2 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

0 M22, M23, Ly and O N, the covering numbers are established in [17] by Holmes and she gives c estimates for those of J1 and M L. Some of the results in [17] are obtained with the help of GAP [11], a first in this context. The covering numbers of symmetric and alternating groups were investigated by Maroti´ in n−2 [21]. For n 6= 7, 9, he shows that for the alternating group σ(An) > 2 with equality if and only if n is even but not divisible by 4. For n = 7 and 9, Maroti´ establishes σ(A7) 6 31 and σ(A9) > 80. For the symmetric groups he proves that, if n is odd and n 6= 9, then n−1 n−2 σ(Sn) = 2 , and, if n is even, then σ(Sn) 6 2 . It is a natural question to ask what are the exact covering numbers for alternating and symmetric groups for those values of n where Maroti´ only gives estimates. In [19] and [9] this was done for alternating groups in case of small values of n. As mentioned earlier, Cohn [7] already established σ(A5) = 10. In [19] it is shown that σ(A7) = 31 and σ(A8) = 71. Furthermore, Maroti’s´ bound for A9 is improved by establishing that 127 6 σ(A9) 6 157. Recently, it was shown in [9] that σ(A9) = 157 and σ(A11) = 2751. Moreover, the third author has determined the covering numbers for Sn when n is divisible by 6 and n > 18 [25]. The topic of this paper is to determine the covering numbers for symmetric groups of small degree and some sporadic simple groups. We determine the covering numbers for Sn in the cases when n = 8, 9, 10, and 12. In particular, we show σ(S9) = 256, establishing that Maroti’s´ n−1 result that σ(Sn) = 2 for odd n holds without exceptions. For n = 8, 10 and 12 we have σ(S8) = 64, σ(S10) = 221, and σ(S12) = 761, respectively. We observe that Maroti´ [21] gave already 761 as an upper bound for σ(S12). Since we can use the same methods, we establish in addition that the Mathieu group M12 has covering number 208 and improve the estimate given for the Janko group J1 in [17]. It should be noted that finding a minimal cover of M12 came down to finding a minimal cover of a particular conjugacy class of elements, and the minimal cover found for these elements contains subgroups from three different conjugacy classes of maximal subgroups. This seems to be a first in this context and explains why the covering number for the group M12 was not determined any earlier despite its relatively small order. We observe that σ(S4) = 4 and σ(S6) = 13 by [1]. Moreover, by [21], we know for large 1 n even values of n that σ(Sn) ∼ 2 n/2 ; indeed, Maroti´ proved that for any > 0 there exists N such that if n > N and n is even, then

bn/3c bn/3c 1 n 1 X n 1 n X n + − < σ(S ) + . 2 n/2 2 i n 6 2 n/2 i i=0 i=0 Our current methods rely on explicit tables for the symmetric groups in question and computer calculations to carry out certain linear optimizations. There are limits to the size of the group on how far these methods can carry us and statements for general values of n are extremely difficult and require entirely different methods than those used for small values of n. This will become clearer when we discuss our methods in the following. The methods employed here are an extension of those used in [19]. In determining a minimal covering of a group we can restrict ourselves to finding a minimal covering by maximal subgroups. The conjugacy classes of subgroups for the groups in question can be found in GAP [11]. To determine a minimal covering by maximal subgroups, it suffices to find a minimal covering of the conjugacy classes of maximal cyclic subgroups by such subgroups of the group. Already in [17] this method is used to determine the covering numbers of sporadic simple groups. Here this method is adapted to the case of symmetric groups where the generators of maximal cyclic subgroups can easily be identified by their cycle structure. At this stage we may want to raise the question whether further results on covering numbers can only be established one at a time, or if one can find methods to give general results for larger classes as it has been done in [21] and [25]. Some insight into the problem can be obtained by COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 3 taking a look at [25]. There some powerful theorems are used where the assumption holds only if n is sufficiently large, and the cases not covered by these theorems for small n have to be checked individually. It is the hope of the authors that the large values of n for which the covering number of Sn is unknown can be resolved using similar techniques, leaving only a small number of cases for small values of n to be resolved using computation or individual inspection. As of now, the smallest values of n for which the covering numbers of Sn and An are not known are n = 14 and n = 12, respectively. The following notation is used for the disjoint cycle decomposition of a nontrivial permutation. Let m1, m2, . . . , mt ∈ N with 1 < m1 < m2 < . . . < mt and k1, . . . , kt ∈ N. If α is a permutation with disjoint cycle decomposition of ki cycles of length mi, i = 1, . . . , t, then k1 kt 1 we denote the class of α by (m1 , . . . , mt ). If ki = 1, we just write mi instead of mi . As is customary, we suppress 1-cycles and the identity permutation is denoted by (1). For example, the permutation with disjoint cycle decomposition (12)(34)(5678) belongs to the class (22, 4). In the case of symmetric groups all elements of a given cycle structure are contained in the subgroups of a conjugacy class of maximal subgroups and the elements with the respective cycle structure are either partitioned into these subgroups or there exists an intersection between some of the subgroups of the conjugacy class. For the groups S8, S9, S10, S12, and M12, we provide two tables which are obtained with the help of GAP [11]. For the group J1, we refer to previous work in [17]. The first table gives the information on the conjugacy classes of maximal subgroups of the group, such as the isomorphism type and order of the class representative and the size of each class. The second table lists the order and cycle structure of each permutation generating a maximal cyclic subgroup as well as the total of such elements in the group together with the distribution of these elements over the various conjugacy classes. For each conjugacy class we list how many of these elements are contained in a class representative. If elements are partitioned over the representatives, we indicate this with P , and if each element is contained in k class representatives and each representative contains s such elements, we indicate this with sk. For some of the groups it suffices to give the second table in abbreviated form. For finding the covering number, the goal is to determine an irredundant covering and show that it is minimal. If the elements of a certain cycle structure are partitioned into the subgroups of a particular conjugacy class, it is not hard to find a minimal covering for such elements. The difficulty arises if the elements in question occur in several class representatives. In this case we interpret the subgroups and group elements as an incidence structure with the subgroup representatives as the sets and the group elements with the specific cycle structure as elements. This leads to a problem in linear optimization. Here are some of the details. Given a group G with a collection of maximal subgroups C, to each maximal subgroup Mi we associate a binary variable mi, which takes on the value of 1 if Mi is in C and 0 if Mi is not in C. Let g be an element of G that is contained in the maximal subgroups Mi1 , Mi2 , ... , Min . The collection C covers the element g if and only if the constraint

n X mij > 1 j=1 P is satisfied. Therefore, determining a minimal cover corresponds to minimizing mi subject to satisfying these constraints. Sometimes we are able to use a combinatorial argument to find a minimal cover, such as when the Erdos-Ko-Rado˝ Theorem [10] is applied in the case of S10. For instance, it is possible to transform the covering number problem into a question involving incidence matrices, and it should be noted that the covering number of S10 was originally determined using these methods. However, we prefer a more uniform approach, and so the proofs in Section4 avoid the notation 4 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ of incidence matrices, even though the fundamental ideas are the same. Otherwise, we have to resort to the help of computers to find optimal solutions, e.g., in the case of S9, M12 and J1. (The covering numbers for S8, S10, and S12 are obtained without the use of computers.) Roughly speaking, a system of linear inequalities with binary variables is prepared by GAP [11] and the optimal solution is found with the help of Gurobi [15]. Naturally, this approach puts a limit on how large our groups can be. In addition, the structure of the cover heavily depends on the arithmetic nature of n. The calculations for this paper were done on a variety of machines. Those for S9 and M12 were completed using a standard Dell desktop with a Core i-7 processor and 8 GB of RAM. Calculations for J1 were attempted on this machine, but, when this proved insufficient, we continued the calculations on Gordon Royle’s machine at the University of Western Australia, which has 256 GB of RAM. Attempts on this machine also proved insufficient for J1. Finally, calculations for J1 were carried out on KoKo, part of the High Performance Computing (HPC) cluster at Florida Atlantic University. The technical specifications of KoKo include four hundred Intel Xeon Cores and 128 GB of RAM per node. Our final bounds on σ(J1) are based on calculation carried out on KoKo. For more information about the computations, see Section8 and http://www.math.wm.edu/˜eswartz/coverings.

2. The Symmetric Group S8

The smallest symmetric group for which the covering number is not known is S8. Here we determine σ(S8) and show that it equals the upper bound given by Maroti´ in [21].

THEOREM 2.1. The covering number of S8 is 64.

PROOF. First we will show that there exists an irredundant covering of S8 by 64 subgroups. As can be seen from Table 2.2, all odd permutations of the group generating maximal cyclic subgroups are contained either in MS3 or MS6. Thus the union of MS3 and MS6 contains all odd permutations in question. We observe that this union does not contain all even permutations generating maximal cyclic subgroups, e.g., the permutation with cycle structure (3, 5) is only contained in MS1 and MS2. Thus MS1, MS3, and MS6 cover all of S8, and

σ(S8) 6 |MS1| + |MS3| + |MS6| = 64. Let C be the union of MS1, MS3, and MS6, and define Π to be the union of all elements with cycle structure (8), (3, 5), or (2, 32). The elements of Π are partitioned among the 64 groups of C, so C is an irredundant covering. We will now show that C is in fact a minimal cover. We first show that MS1, which only contains the subgroup A8, must be contained in any minimal cover. Suppose that we have a minimal cover B that does not contain MS1. By Table 2.2, since elements with cycle structure (3, 5) are only contained in MS1 and MS2, our minimal cover B must contain all 56 subgroups of MS2. However, none of the elements with cycle structure (8) for instance have been covered, and this requires at least 35 additional subgroups. This is a contradiction, since 35 + 56 > 64, the number of subgroups in C. Hence MS1 is included in any minimal cover. Suppose now that there is, in fact, a cover B of the elements of S8 with |B| < 64. Let B = (B ∩ C) ∪ B1 and let C = (B ∩ C) ∪ C1. This means that |B1| < |C1|, and, since MS1 must be present in any minimal cover, we may assume that C1 consists only of subgroups from MS3 and MS6. Let |C1| = c3 + c6, where c3 denotes the number of subgroups from MS3 in C1 and c6 denotes the number of subgroups from MS6 in C1. Since C consists of all subgroups from MS1, MS3, and MS6, we note that B1 consists only of subgroups from MS2, MS4, MS5, and MS7. Hence we let |B1| = c2 + c4 + c5 + c7, where ci denotes the number of subgroups from MSi in B1 for i = 2, 4, 5, 7. Since |B1| < |C1|, our first observation is that

(2.1) c2 + c4 + c5 + c7 + 1 6 c3 + c6. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 5

Label Isomorphism Type Group Order Class Size

MS1 A8 20160 1

MS2 S3 × S5 720 56

MS3 S2 × S6 1440 28

MS4 S1 × S7 5040 8

MS5 S2wr S4 384 105

MS6 S4wr S2 1152 35 MS7 PGL(2, 7) 336 120

TABLE 2.1. Conjugacy classes of maximal subgroups of S8.

MS1 MS2 MS3 MS4 MS5 MS6 MS7 Order C.S. Size (1) (56) (28) (8) (105) (35) (120) ODD 2 4 (2 , 4) 1260 0 0 902 0 363 1805 0

6 (2, 3) 1120 0 1005 1604 4203 0 963 0 2 6 (2, 3 ) 1120 0 402 40,P 0 323 0 0

6 (6) 3360 0 0 120,P 8402 32,P 0 562

8 (8) 5040 0 0 0 0 48,P 144,P 842 10 (2, 5) 4032 0 72,P 144,P 504,P 0 0 0 12 (3, 4) 3360 0 60,P 0 420,P 0 96,P 0 EVEN

4 (2, 4) 2520 P 902 1802 6302 24,P 72,P 0

6 (2, 6) 3360 P 0 120,P 0 32,P 1922 0 7 (7) 5760 P 0 0 720,P 0 0 48,P 15 (3, 5) 2688 P 48,P 0 0 0 0 0

TABLE 2.2. Inventory of elements generating maximal cyclic subgroups in S8 across conjugacy classes of maximal subgroups.

Now, since B is a cover, all elements with cycle structure (8) must be covered. By removing c6 subgroups of MS6 from C, this leaves 144c6 such elements that are not yet covered. By Table 2.2, since these 144c6 elements must be covered by the subgroups in B1, this implies that

144c6 6 48c5 + 84c7, which implies that 1 7 (2.2) c c + c . 6 6 3 5 12 7 6 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

Similarly, removing c3 subgroups of MS3 from C leaves 144c3 elements with cycle structure (2, 5) uncovered, and from Table 2.2 we see that

144c3 6 72c2 + 504c4, which implies that 1 7 (2.3) c c + c . 3 6 2 2 2 4 Putting Equations (2.1), (2.2), and (2.3) together, we see that 1 7 1 7 c + c + c + c + 1 c + c c + c + c + c , 2 4 5 7 6 3 6 6 2 2 2 4 3 5 12 7 which implies that 1 2 5 5 c + c + c + 1 c . 2 2 3 5 12 7 6 2 4 Eliminating the fractional coefﬁcients, we see that this implies that

(2.4) 6c2 + 8c5 + 5c7 + 12 6 30c4. On the other hand, examining the elements with cycle structure (2, 32) that are uncovered by removing c3 subgroups of MS3 from C, we see by Table 2.2 that

40c3 6 40c2 + 32c5, and so 4 (2.5) c c + c . 3 6 2 5 5 Combining Equations (2.1), (2.2), and (2.5), we see that 17 7 c + c + c + c + 1 c + c c + c + c , 2 4 5 7 6 3 6 6 2 15 5 12 7 which in turn implies that 5 2 c + 1 + c c . 4 12 7 6 15 5 Eliminating the fractional coefﬁcients, we see that

(2.6) 60c4 + 60 + 25c7 6 8c5.

However, since all the ci are nonnegative integers, Equations (2.4) and (2.6) imply that

60c4 + 60 6 8c5 < 6c2 + 8c5 + 5c7 + 12 6 30c4, a contradiction. Therefore, we conclude that no such cover B can exist, the collection of subgroups C is a minimal cover of the elements of S8, and σ(S8) = 64, as desired.

3. The Symmetric Group S9

In this section we will determine the exact covering number of S9, the case missing in [21], where the covering numbers for Sn with n odd were determined with the exception of n = 9.

THEOREM 3.1. The covering number of S9 is 256. This together with Theorem 1.1 in [21] yields the following corollary. n−1 COROLLARY 3.2. Let n > 3 be an odd integer. Then σ(Sn) = 2 . To prove the main result of this section, we need the following proposition.

PROPOSITION 3.3. The 84 subgroups of MS3 form a minimal cover of the elements with cycle structure (3, 6) in S9. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 7

Label Isomorphism Type Group Order Class Size

MS1 A9 181440 1

MS2 S4 × S5 2880 126

MS3 S3 × S6 4320 84

MS4 S2 × S7 10080 36

MS5 S1 × S8 40320 9

MS6 S3wr S3 1296 280 MS7 AGL(2, 3) 432 840

TABLE 3.1. Conjugacy classes of maximal subgroups of S9.

MS1 MS2 MS3 MS4 MS5 MS6 MS7 Order C.S. Size (1) (126) (84) (36) (9) (280) (840) ODD 2 4 (2 , 4) 11340 0 1802 2702 6302 1260,P 1624 0

6 (2, 3) 2520 0 22011 2709 4907 11204 364 0 2 6 (2, 3 ) 10080 0 1602 3603 280,P 1120,P 36,P 0

6 (6) 10080 0 0 120,P 8403 33603 36,P 0 3 6 (2 , 3) 2520 0 603 30,P 2103 0 364 0

6 (3, 6) 20160 0 0 240,P 0 0 2884 723

8 (8) 45360 0 0 0 0 5040,P 0 1082

10 (2, 5) 18144 0 144,P 4322 10082 40322 0 0

12 (3, 4) 15120 0 3603 180,P 420,P 33602 0 0 14 (2, 7) 25920 0 0 0 720,P 0 0 0 20 (4, 5) 18144 0 144,P 0 0 0 0 0 EVEN

6 (2, 6) 30240 P 0 360,P 840,P 3360,P 108,P 722 9 (9) 40320 P 0 0 0 0 144,P 0

12 (2, 3, 4) 15120 P 120,P 180,P 420,P 0 1082 0 15 (3, 5) 24192 P 192,P 288,P 0 2688,P 0 0

TABLE 3.2. Inventory of permutations generating maximal cyclic subgroups in S9 across conjugacy classes of maximal subgroups.

PROOF. We prove this computationally with the help of the software GAP [11] and Gurobi [15]. Using the GAP program as given in Function 8.1 for G = S9 and the conjugacy classes MS3, MS6, and MS7 of maximal subgroups, we are setting up the equations readable by Gurobi for the elements of type (3, 6). The Gurobi output shows that a minimal covering of these elements consists of 84 subgroups from MS3, MS6, and MS7. Since the elements with 8 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ cycle structure (3, 6) are partitioned into the subgroups of MS3, these 84 subgroups constitute a minimal covering of these elements. We note that the GAP output addressed in the above proposition as well as an abbreviated Gurobi output of these calculations are given at the end of Section8. These calculations were done on a Dell desktop machine with a Core i-7 processor and 8 GB of RAM; as can be read from the output, this calculation took 453.03 seconds. For further details we refer to http: //www.math.wm.edu/˜eswartz/coverings. Now we are ready to prove our theorem.

PROOFOF THEOREM 3.1. We will show ﬁrst that there exists a cover of S9 by 256 subgroups. We claim ﬁrst that the collection of all subgroups from one of MS1, MS2, MS3, MS4, or MS5 constitutes a cover of the elements of S9. To see this, we note that every even permutation is contained in A9, the unique subgroup in MS1, and, by Table 3.2, every maximal cyclic subgroup generated by an odd permutation is contained in a subgroup in MS2, MS3, MS4, or MS5. Hence this collection of subgroups is a cover, and

σ(S9) 6 |MS1| + |MS2| + |MS3| + |MS4| + |MS5| = 256. We must now show that this collection is in fact a minimal cover. As can be seen from Table 3.2, the only maximal subgroups containing the elements with cycle structure (2, 7) are in MS4, and the only maximal subgroups containing the elements with cycle structure (4, 5) are in MS2. Thus these two classes of maximal subgroups must be contained in any cover of the elements of S9 that uses only maximal subgroups. Examining Table 3.2, we see that the only classes of maximal subgroups of S9 containing 9-cycles are MS1 and MS6, and the 9-cycles are partitioned in each. This means that, if the subgroup in MS1 isomorphic to A9 is not contained in a cover of S9, all 280 subgroups isomorphic to S3wr S3 must be in the cover. However, any such cover would already contain more than 256 subgroups, and so a minimal cover of the elements of S9 must contain A9, the unique subgroup in MS1. We note that, since any minimal cover of the elements of S9 contains A9, we may restrict ourselves now to ﬁnding a cover of the odd permutations in S9. Examining Table 3.2, the only elements generating maximal cyclic subgroups that are not covered by subgroups in MS1, MS2, or MS4 have cycle structure (8) or (3, 6). Suppose that a minimal cover does not contain every subgroup from MS5. By Table 3.2, the 8-cycles are partitioned among the subgroups in class MS5, and each subgroup in MS5 contains 5040 different 8-cycles. Since the 8-cycles are contained only in subgroups in MS5 and MS7, and each subgroup in MS7 contains only 108 different 8-cycles, this means that at least d5040/108e = 47 different subgroups from MS7 in the cover. Assume that there are c7 subgroups from MS7 in the cover. By Table 3.2, there are still at least 20160 − 72c7 elements with cycle structure (3, 6) that need to be covered. Since there are at most 288 elements with cycle structure (3, 6) in a single maximal subgroup, if c is the number of subgroups in this cover from MS3 and MS6, then c > (20160 − 72c7)/288 = 70 − c7/4. However, this cover now contains at least 3c |MS1| + |MS2| + |MS4| + c + c = 163 + c + c 233 + 7 > 233 + 35 > 256, 7 7 > 4 a contradiction to this cover being minimal. Hence a minimal cover of the elements of S9 must contain all subgroups from classes MS1, MS2, MS4, and MS5. The only elements generating maximal cyclic subgroups in S9 that are not contained in subgroups in MS1, MS2, MS4, and MS5 have cycle structure (3, 6), and hence the problem of determining a minimal cover of S9 reduces to the problem of ﬁnding a minimal cover of the elements with cycle structure (3, 6). By Proposition 3.3, the 84 subgroups from MS3 form a minimal cover of these elements. Therefore, the subgroups from MS1, MS2, MS3, MS4, and MS5 form a minimal cover, and σ(S9) = 256. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 9

4. The Symmetric Group S10

In this section we determine the covering number of S10. It turns out to be less than the upper bound of 210−2 given by Maroti´ in [21].

THEOREM 4.1. The covering number of S10 is 221. Before we can prove Theorem 4.1, we have to establish some preparatory results involving combinatorics and incidence matrices leading to an application of a result due to Erdos,˝ Ko and Rado [10] (see also Theorem 5.1.2 in [2]). 1 THEOREM 4.2.[ 10] Let A1,...,Am be m k-subsets of an n-set S, k 6 2 n, which are n−1 pairwise nondisjoint. Then m 6 k−1 . The upper bound for m is best possible. It is attained when the Ai are precisely the k-subsets of S which contain a chosen ﬁxed element of S.

Label Isomorphism Type Group Order Class Size

MS1 A10 1814400 1

MS2 S4 × S6 17280 210

MS3 S3 × S7 30240 120

MS4 S2 × S8 80640 45

MS5 S1 × S9 362880 10

MS6 S2wr S5 3840 945

MS7 S5wr S2 28800 126 MS8 PΓL(2, 9) 1440 2520

TABLE 4.1. Conjugacy classes of maximal subgroups of S10.

In order to apply Theorem 4.2, we ﬁrst have to associate subgroups of S10 to subsets of {1, ..., 10}. In fact, there is a canonical bijection between subgroups M in MS3 and subsets of {1, ..., 10} of size three. Indeed, each subgroup M in MS3 is isomorphic to S3 × S7 and contains all elements of S10 that preserve a unique decomposition of the set {1, ..., 10} into a subset of size three and a subset of size seven. The subset of size three that is preserved by M is {1,...,10} the unique subset that we associate with M. Hence we deﬁne a function F : MS3 → 3 , {1,...,10} where 3 denotes the collection of all sets of size 3 of {1, ..., 10}, and, given M in MS3, we denote by F (M) the subset of size three that is preserved by every element of M. The following lemma shows how the number of elements with cycle structure (32, 4) in a subgroup M in MS3 depends on F (M). (The proof is straightforward and left to the reader.)

LEMMA 4.3. (i) Let g be an element with cycle structure (32, 4) that can be written as (k1, k2, k3)(k4, k5, k6)(k7, k8, k9, k10),

where the ki are distinct elements of {1, ..., 10}. The subgroup M in MS3 contains g if and only if F (M) equals {k1, k2, k3} or {k4, k5, k6}. In particular, each element g with cycle structure (32, 4) is contained in exactly two subgroups of MS3, namely, the subgroup M1 with F (M1) = {k1, k2, k3} and the subgroup M2 with F (M2) = {k4, k5, k6}. 10 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

(ii) Let M1 and M2 be subgroups in MS3. If F (M1) and F (M2) are disjoint, then there 2 are exactly 24 elements with cycle structure (3 , 4) contained in both M1 and M2. If F (M1) and F (M2) are not disjoint, then there are no elements with cycle structure 2 (3 , 4) contained in both M1 and M2.

MS1 MS2 MS3 MS4 Order C.S. Size (1) (210) (120) (45) ODD 2 4 (2 , 4) 56700 0 10804 18904 37803 2 4 (2, 4 ) 56700 0 5402 0 1260,P 3 6 (2 , 3) 25200 0 4804 8404 16803 2 6 (2, 3 ) 50400 0 12005 16804 22402 2 6 (2 , 6) 75600 0 360,P 0 33602 6 (3, 6) 201600 0 960,P 1680,P 0 8 (8) 226800 0 0 0 5040,P 10 (10) 362880 0 0 0 0 2 12 (3 , 4) 50400 0 240,P 8402 0 14 (2, 7) 259200 0 0 2160,P 5760,P 20 (4, 5) 181440 0 864,P 0 0 30 (2, 3, 5) 120960 0 0 1008,P 2688,P EVEN

6 (2, 6) 151200 P 720,P 25202 67202 8 (2, 8) 226800 P 0 0 5040,P 9 (9) 403200 P 0 0 0 12 (4, 6) 151200 P 720,P 0 0

12 (2, 3, 4) 151200 P 14402 25202 3360,P 21 (3, 7) 172800 P 0 1440,P 0 TABLE 4.2. Inventory of elements generating maximal cyclic subgroups in S10 across conjugacy classes of maximal subgroups.

The following result demonstrates the usefulness of Theorem 4.2 here.

PROPOSITION 4.4. A collection F of subgroups of MS3 is a cover of the elements with 2 cycle structure (3 , 4) if and only if F (M1) ∩ F (M2) 6= ∅ for any two subgroups M1 and M2 in the complement F. That is, the subsets in the collection {F (M): M ∈ F} are pairwise nondisjoint. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 11

MS5 MS6 MS7 MS8 Order C.S. Size (10) (945) (126) (2520) ODD 2 4 (2 , 4) 56700 113402 1803 9002 0 2 4 (2, 4 ) 56700 0 3005 18004 904 3 6 (2 , 3) 25200 2520,P 0 6003 0 2 6 (2, 3 ) 50400 100802 1603 8002 0 2 6 (2 , 6) 75600 0 2403 24004 0

6 (3, 6) 201600 20160,P 0 0 2403

8 (8) 226800 453602 240,P 0 1802 10 (10) 362880 0 384,P 2880,P 144,P 2 12 (3 , 4) 50400 0 1603 0 0 14 (2, 7) 259200 25920,P 0 0 0 20 (4, 5) 181440 18144,P 0 1440,P 0 30 (2, 3, 5) 120960 0 0 960,P 0 EVEN

6 (2, 6) 151200 302402 160,P 0 0

8 (2, 8) 226800 0 240,P 36002 1802 9 (9) 403200 40320,P 0 0 0

12 (4, 6) 151200 0 160,P 24002 0 12 (2, 3, 4) 151200 15120,P 0 1200,P 0 21 (3, 7) 172800 0 0 0 0

TABLE 4.3. Inventory of elements generating maximal cyclic subgroups in S10 across conjugacy classes of maximal subgroups.

PROOF. Assume ﬁrst that there exist two subgroups M1 and M2 in F such that F (M1) ∩ 2 F (M2) = ∅. By Lemma 4.3 (ii), there are 24 elements with cycle structure (3 , 4) in both M1 and M2. Now, Lemma 4.3 (i) implies that these 24 elements are in M1 and M2 but no other subgroups of MS3. Therefore, these 24 elements with cycle structure (32, 4) are not covered by the subgroups in F, and F is not a cover of the elements with cycle structure (32, 4). Conversely, assume that we have F (M1) ∩ F (M2) 6= ∅ for any M1,M2 ∈ F. Let g be an element with cycle structure (32, 4). By Lemma 4.3 (i), the element g is contained in 0 0 exactly two subgroups of MS3, and by Lemma 4.3 (ii), the two subgroups M1 and M2 in MS3 0 0 containing g have F (M1)∩F (M2) = ∅. Since for any two subgroups M1 and M2 of F we have F (M1) ∩ F (M2) 6= ∅, the collection F can contain at most one of the subgroups that contain g. Hence for every element g with cycle structure (32, 4), one of the subgroups containing it is 2 in F, and so F is a cover of the elements with cycle structure (3 , 4), as desired. We may now combine this last result with the Theorem 4.2 to produce a minimal cover of 2 the elements with cycle structure (3 , 4). Deﬁne the collection Di to be the set of all subgroups M of MS3 such that i ∈ F (M), where i ∈ {1, ..., 10}. 12 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

COROLLARY 4.5. For any i ∈ {1, ..., 10}, the collection Di is a minimal cover of the 2 elements with cycle structure (3 , 4) in S10. That is, each Di is a collection of 84 subgroups of MS3 that covers the elements with cycle structure (32, 4), and no collection of fewer than 84 subgroups covers the elements with cycle structure (32, 4).

PROOF. Assume that C is a collection of subgroups from MS3 that covers the elements with cycle structure (32, 4). By Proposition 4.4, the subsets of size three associated to the subgroups 9 in the collection C must be pairwise nondisjoint. By the Theorem 4.2, |C| 6 2 = 36. This implies that any cover of the elements with cycle structure (32, 4) by subgroups from MS3 has size at least 120 − 36 = 84. On the other hand, consider the collection Di, where i is ﬁxed. By deﬁnition, i ∈ F (M) for any subgroup M in Di, and hence the collection of subsets {F (M): M ∈ Di} are pairwise nondisjoint. By Proposition 4.4, this means that Di is a cover 2 9 of the elements with cycle structure (3 , 4). Note further that |Di| = 36, since there are 2 different subsets of size three of {1, ..., 10} that contain i. Therefore, each collection Di is a 2 minimal cover of the elements with cycle structure (3 , 4) in S10, as desired. Now we are ready to prove Theorem 4.1.

PROOFOF THEOREM 4.1. By Tables 4.2 and 4.3 and Corollary 4.5, it can be seen that MS1 ∪ MS5 ∪ MS7 ∪ D1 = S10, since all permutations generating maximal cyclic subgroups in S10 are contained in the union of these subgroups. Hence

σ(S10) 6 |MS1 ∪ MS5 ∪ MS7 ∪ D1| = 1 + 10 + 126 + 84 = 221. It remains to be shown that the covering obtained by the 221 subgroups is minimal, i.e., that σ(S10) > 221. We begin first by considering MS1, which contains the single subgroup A10. Assume that MS1 is not included in a minimal cover C. Referring to Tables 4.2 and 4.3 and examining the elements with cycle structure (3, 7), we conclude that, if MS1 is not part of C, then all 120 subgroups in MS3 are in C. We note that C must also cover all the 10-cycles, which are contained only in subgroups of the classes MS6, MS7, and MS8, and the 10-cycles are partitioned among the subgroups in each class. It requires at least 126 subgroups from these three classes (specifically, all the subgroups in MS7) to cover the 10-cycles, and hence |C| > 120 + 126 = 246. However, since we know that a minimal cover of the elements of S10 has at most 221 subgroups, this is a contradiction. Hence any minimal cover of S10 contains the unique subgroup A10 in MS1. Next, we will show that any minimal cover of S10 contains the entire class MS7 of subgroups isomorphic to S5wr S2. First, we know that MS1 is included in any minimal cover, so we need only consider how many odd permutations are covered by a particular subgroup. By Table 4.3, we see that every subgroup M in MS7 contains 2880 different 10-cycles. The 10- cycles must all be covered, and the only other subgroups that contain 10-cycles are in MS6 and MS8. However, the subgroups in MS8 only contain 1440 elements in total, and, using GAP [11], we see that any subgroup in MS6 contains only 1920 odd permutations, i.e., a subgroup in either MS6 or MS8 contains fewer odd permutations than a subgroup in MS7 does 10-cycles. Therefore, all of the subgroups from MS7 must be included in any minimal cover. Suppose now that there is a cover C of S10 of size at most 220. Since C contains the classes MS1 and MS7, this implies that there is a cover C1 of size at most 220 − 1 − 126 = 93 of the elements of S10 that are not in MS1 or MS7; namely, by Tables 4.2 and 4.3, there is a cover 2 C1 of size at most 93 of the elements with cycle structures (3, 6), (8), (3 , 4), and (2, 7). Define 2 Π to be the set of all permutations with cycle structure (8) or (3 , 4), and we note that C1 must cover all the elements of Π. (k) Define Bk to be the set of all the subgroups in MS5 except for the subgroup S9 that fixes the element k and permutes the elements {1, ..., 10}\{k}. For each i, k such that 1 6 i, k 6 10, COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 13 we have that |Di ∪ Bk| = 84 + 9 = 93 and Di ∪ Bk covers the elements of Π. Note first that Di constitutes a minimal cover of just the elements with cycle structure (32, 4) by Corollary 4.5. Moreover, Bk is a minimal cover of the 8-cycles for each 1 6 k 6 10. To see this, consider what happens when precisely m subgroups of MS5 are removed. By Table 4.3, every 8-cycle is contained in exactly two subgroups of MS5, so, if m = 1, then the collection of subgroups still covers the 8-cycles. Every 8-cycle fixes exactly two of the elements in {1,..., 10} and moves the other eight. Suppose that g is an 8-cycle, and g fixes the elements i, j ∈ {1,..., 10}; note that there are 7! = 5040 such elements g for a given pair {i, j}. This means that the two (i) (j) m subgroups in MS5 that contain g are S9 and S9 . Hence, if m > 2, there are 2 pairs {i, j} m such that the 8-cycles that fix i and j are not covered, i.e., there are exactly 5040 2 different 8-cycles that now are not covered. No maximal subgroup outside of MS5 contains more than (k) 5040 different 8-cycles, and hence the set of subgroups of MS5 without S9 is a minimal cover of the 8-cycles. Note, however, that the sets Bk are not the only minimal covers of the 8-cycles. If Hj,k is the subgroup of MS4 isomorphic to S2×S8 that stabilizes the breakdown of {1, ..., 10} into {k, j} and its complement, then the collection Bj,k = (Bj ∩ Bk) ∪ {Hj,k} also covers the 8-cycles. However, by the above, no more than two subgroups in MS5 can be removed, and, by Tables 4.2 and 4.3, only the subgroups of MS4 have enough elements to replace two subgroups of MS5 being removed. Therefore, these are the only minimal covers of the 8-cycles of S10. We now claim that C1 must consist of a minimal cover of the 8-cycles and a minimal cover of the elements with cycle structure (32, 4). Examining Tables 4.2 and 4.3, we see that 8-cycles are only contained in MS4, MS5, MS6, and MS8, while the elements with cycle structure 2 (3 , 4) are only contained in MS2, MS3, and MS6. If C1 is not just a minimal cover of the 8-cycles along with a minimal cover of the elements with cycle structure (32, 4), then there must be at least one subgroup H in C1 that contains both 8-cycles and elements with cycle structure (32, 4), and there must be at least one 8-cycle and one element with cycle structure (32, 4) in H that are not contained in any other subgroup in C1. In particular, this means that C1 contains at least one subgroup H from MS6. We now examine the manner in which C1 covers the 8-cycles. Since there is at least one 8-cycle in H that is not contained in any other subgroup of C1, this (i) (j) means that for some i, j with 1 6 i < j 6 10, we have that none of S9 , S9 , or Hi,j are in C1. This means that the 5040 different 8-cycles fixing i, j are covered only by subgroups from MS6 and MS8. However, since the subgroups in MS6 are isomorphic to S2wr S5, a simple counting argument shows that no subgroup of MS6 contains more than 48 different 8-cycles that fix i and j. Moreover, it can be verified that every subgroup of MS8 contains exactly four 8-cycles that fix i and j using GAP [11]. Hence, in order to cover all the 8-cycles, the collection C1 must contain at least 5040/48 = 105 > 93 subgroups, a contradiction. Hence C1 consists of a minimal cover of the 8-cycles and a minimal cover of the elements with cycle structure 2 (3 , 4). Moreover, by symmetry we may assume that C1 contains D1. In order for a cover of the elements of S10 to have size 220 or less, one of these possibilities for C1 must in fact cover all the odd permutations of S10. We now consider the different possibilities for C1. If C1 = D1 ∪ B1, then we note that the element (1, 2)(3, 4, 5, 6, 7, 8, 9) is not covered by C1. If C1 = D1 ∪ Bk, where k > 1, then without loss of generality we may assume that k = 10. In this case, the element (1, 2)(3, 4, 5, 6, 7, 8, 9) is not covered by C1. If C1 = D1 ∪ B1,j, then without loss of generality we may assume that j = 10. In this case, the element (1, 2)(3, 4, 5, 6, 7, 8, 9) is not covered by C1. Finally, if C1 = D1 ∪ Bk,j, where k, j > 1, then without loss of generality we may assume that k = 10 and j = 9. In this case, the element (1, 2)(3, 4, 5, 6, 7, 8, 9) is not covered by C1. Hence no such C1 can exist, and σ(S10) > 221. Therefore, σ(S10) = 221, as desired. 14 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

5. The Symmetric Group S12

In [21], Maroti´ gives an upper bound for the covering number of S12, which is lower than the general upper bound given there. We will show here that this bound is indeed the covering number of S12.

THEOREM 5.1. The covering number of S12 is 761.

PROOF. As noted by Maroti´ [21, p. 104], the covering number of S12 is at most 761, since S12 may be written as the union of all those subgroups conjugate to S6wr S2, S1 × S11, S2 × S10, S3 × S9, and A12, which correspond to the classes MS5, MS2, MS3, MS4, and MS1, respectively, of Table 5.1. Indeed, we will show that this is in fact a minimal cover of S12 by demonstrating that there is a particular class of maximal cyclic subgroups that is minimally covered by one of these five classes. Define Π1 to be the set of elements with cycle structure (5, 7), Π2 to be the set of elements 2 with cycle structure (4, 7), Π3 to be the set of elements with cycle structure (2, 5 ), Π4 to be the set of elements with cycle structure (3, 4, 5), and Π5 to be the set of elements with cycle S5 structure (12), i.e., the 12-cycles of S12. We define Π := i=1 Πi, and we will show that the collection C of subgroups from MS1, MS2, MS3, MS4, and MS5 is a minimal cover of the elements of S12 by showing that at least 761 subgroups are needed to cover the elements of Π. Note that the elements of Π are partitioned among the members of C. Let B be a minimal cover of Π. First, we examine the elements with cycle structure (12). It is not hard to see that the classes of maximal subgroups containing 12-cycles are all imprimitive subgroups in the classes MS5, MS8, MS9, and MS10 (a 12-cycle preserves such an imprimitive decomposition of twelve elements), and also the subgroups of class MS11. More- over, it is easy to see that the 12-cycles must be partitioned in each of the classes MS5, MS8, MS9, and MS10, respectively, since a 12-cycle stabilizes a unique imprimitive decomposition of twelve elements. Since the 12-cycles are partitioned among the subgroups in classes MS5, MS8, MS9, and MS10, respectively, and MS5 has the fewest number of subgroups, removing n subgroups from MS5 from the cover would require at least n + 1 replacements from the other classes. On the other hand, the 12-cycles are not partitioned in MS11. As can be seen from Table 5.2, each such subgroup contains 220 different 12-cycles. Removing even one subgroup from MS5, which contains 86400 different 12-cycles, would require at least d86400/220e = 393 different subgroups from MS11 to replace it. Since none of the subgroups in classes MS8, MS9, MS10, or MS11 meet Πi for 1 6 i 6 4, this means that any minimal cover B of the elements of Π contains all subgroups isomorphic to S6wr S2 from MS5. We next show that MS1, which consists of the subgroup A12, is contained in any minimal cover B of the elements of Π. If MS1 is not in a minimal cover B of Π, then B must contain every subgroup of class MS7, since these are the only two classes that meet Π1, and Π1 is partitioned in each class. However, there are 792 subgroups in MS7, and 792 > 761 = |C|. Hence MS1 is in B. Suppose that the minimal cover B of Π is not C. We may write C = (B ∩ C) ∪ C1 and B = (B ∩ C) ∪ B1, where C1 consists of all subgroups of C that are not in B, and B1 contains all subgroups of B that are not in C. We have seen above that both MS1 and MS5 are in B ∩ C and that no subgroup of MS8, MS9, MS10, or MS11 is in B. Hence C1 consists of subgroups from MS2, MS3, and MS4, while B1 consists of subgroups from MS6 and MS7. Let C1 consist of precisely c2 subgroups from MS2, c3 subgroups from MS3, and c4 subgroups from MS4, and let B1 consist of precisely c6 subgroups from MS6 and c7 subgroups from MS7. Since C is not a minimal cover of Π, we have

c6 + c7 = |B1| < |C1| = c2 + c3 + c4. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 15

Label Isomorphism Type Group Order Class Size

MS1 A12 239500800 1

MS2 S1 × S11 39916800 12

MS3 S2 × S10 7257600 66

MS4 S3 × S9 2177280 220

MS5 S6wr S2 1036800 462

MS6 S4 × S8 967680 495

MS7 S5 × S7 604800 792

MS8 S4wr S3 82944 5775

MS9 S2wr S6 46080 10395

MS10 S3wr S4 31104 15400 MS11 PGL(2, 11) 1320 362880

TABLE 5.1. Conjugacy classes of maximal subgroups of S12.

Order C.S. Size MS1 MS2 MS3 MS4 MS5 (1) (12) (66) (220) (462) 35 (5, 7) 13685760 P 0 0 0 0 28 (4, 7) 17107200 0 1425600,P 0 0 0 10 (2, 52) 4790016 0 0 72576,P 0 0 60 (3, 4, 5) 7983360 0 0 0 36288,P 0 12 (12) 39916800 0 0 0 0 86400,P

Order C.S. Size MS6 MS7 MS8 MS9 MS10 MS11 (495) (792) (5775) (10395) (15400) (362880) 35 (5, 7) 13685760 0 17280,P 0 0 0 0 28 (4, 7) 17107200 34560,P 21600,P 0 0 0 0 2 10 (2, 5 ) 4790016 0 120962 0 0 0 0 60 (3, 4, 5) 7983360 16128,P 10080,P 0 0 0 0

12 (12) 39916800 0 0 6912,P 3840,P 2592,P 2202

TABLE 5.2. Inventory of certain elements generating maximal cyclic subgroups in S12 across conjugacy classes of maximal subgroups. 16 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

On the other hand, we can compare elements from Π2, Π3, and Π4 covered by these groups. Since the elements of Π are partitioned among elements of C, using Table 5.2 we have:

1425600c2 6 34560c6 + 21600c7, 72576c3 6 12096c7, 36288c4 6 16128c6 + 10080c7.

Putting this all together, it follows that:

c6 + c7 < c2 + c3 + c4 34560 16128 21600 12096 10080 ( + )c + ( + + )c 6 1425600 36288 6 1425600 72576 36288 7 232 91 = c + c , 295 6 198 7 a contradiction. Therefore, C is the unique minimal cover of Π, and hence C is the unique minimal cover of the elements of S12 and σ(S12) = 761, as desired.

6. The Mathieu Group M12 Only as recently as 2010, it was shown by Holmes and Maroti´ in [18] that for the Mathieu group M12 we have 131 6 σ(M12) 6 222. Here we will determine the exact covering number of M12.

THEOREM 6.1. The covering number of M12 is 208.

1+4 The notation 2+ used in Table 6.1 to describe a normal subgroup of a subgroup in MS9 comes from [8] and refers to an extraspecial group of order 32 that is isomorphic to the central product of Q8 with Q8. We also note that M12 has two conjugacy classes of elements of order 11, which is why there are two identical rows in Table 6.2. Before we can prove this theorem, we need a proposition which gives a minimal covering for the elements with cycle structure (6, 6). (We note that M12 is represented here as a permutation group embedded into S12.) In fact, the minimal cover found contains subgroups from three different conjugacy classes of subgroups. This seems to be a ﬁrst in this context and explains why the covering number for the group M12 was not determined any earlier despite its relatively small order. The use of GAP and Gurobi led to this breakthrough. COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 17

Label Isomorphism Type Group Order Class Size

MS1 M11 7920 12

MS2 M11 7920 12 MS3 PΓL(2, 9) 1440 66 MS4 PΓL(2, 9) 1440 66 MS5 PSL(2, 11) 660 144

MS6 (C3 × C3):(C2 × S4) 432 220

MS7 (C3 × C3):(C2 × S4) 432 220

MS8 S5 × C2 240 396 1+4 MS9 2+ : S3 192 495

MS10 (C4 × C4): D12 192 495

MS11 A4 × S3 72 1320

TABLE 6.1. Conjugacy classes of maximal subgroups of M12.

PROPOSITION 6.2. There exists a covering of the elements with cycle structure (6, 6) in M12 by 130 subgroups, and this covering is minimal. This covering is made up of 120 subgroups isomorphic to PSL(2, 11) from MS5, eight subgroups isomorphic to C2 × S5 from MS8, and two subgroups that are isomorphic to (C4 × C4): D6 in MS10.

PROOF. Using the GAP [11] program listed in Function 8.1 for the elements with cycle structure (6, 6) in M12 and the appropriate maximal subgroups of M12, Gurobi [15] ﬁnds that there exists a covering of the elements with cycle structure (6, 6) by 130 subgroups in MS5, MS8, MS10, and that this covering is minimal.

A list of generators for the subgroups of M12 contained in this covering can be found on line at http://www.math.wm.edu/˜eswartz/coverings. It should be noted that it took Gurobi 815089.46 seconds to solve the system of equations created by GAP on a Dell Optiplex 9020 machine with a Core i-7 processor and 16 GB of RAM. Now we are ready to prove our theorem.

PROOFOF THEOREM 6.1. It can be easily seen from Table 6.2 that the subgroups in MS1 and MS4 cover all elements in M12 generating maximal cyclic subgroups with the exception of elements of cycle structure (6, 6). By Proposition 6.2 there exists a covering of the elements of cycle structure (6, 6) by 130 subgroups in MS5, MS8, and MS10. Thus

σ(M12) 6 |MS1| + |MS4| + 130 = 12 + 66 + 130 = 208.

It remains to be shown that any covering of M12 contains at least 208 subgroups. As can be seen from Table 6.2, a covering of the 11-cycles needs to contain at least 12 subgroups of MS1 or MS2. Similarly, a covering of the elements of cycle structure (2, 10) needs to contain the 66 subgroups of MS3 or MS4. Since the covering of the elements with cycle structure (6, 6) by the 130 subgroups from MS5, MS8, and MS10 is minimal by Proposition 6.2, it follows that σ(M12) > 12 + 66 + 130 = 208. We conclude σ(M12) = 208. 18 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

Order C.S. Size MS1 MS2 MS3 MS4 MS5 (12) (12) (66) (66) (144) 6 (2, 3, 6) 15840 1320,P 1320,P 240,P 240,P 0

6 (6, 6) 7920 0 0 0 0 1102

8 (2, 8) 11880 0 19802 0 3602 0

8 (4, 8) 11880 19802 0 3602 0 0 10 (2, 10) 9504 0 0 144,P 144,P 0 11 (11) 8640 720,P 720,P 0 0 60,P 11 (11) 8640 720,P 720,P 0 0 60,P

Order C.S. Size MS6 MS7 MS8 MS9 MS10 MS11 (220) (220) (396) (495) (495) (1320)

6 (2, 3, 6) 15840 1442 1442 0 32,P 0 242

6 (6, 6) 7920 0 0 603 0 322 6,P

8 (2, 8) 11880 0 1082 0 24,P 24,P 0

8 (4, 8) 11880 1082 0 0 24,P 24,P 0 10 (2, 10) 9504 0 0 24,P 0 0 0 11 (11) 8640 0 0 0 0 0 0 11 (11) 8640 0 0 0 0 0 0

TABLE 6.2. Inventory of elements generating maximal cyclic subgroups in M12 across conjugacy classes of maximal subgroups.

7. The Janko Group J1

In [17] it was shown by Holmes that 5165 6 σ(J1) 6 5415 for the Janko group J1. Using similar methods employed in this paper for S9 and M12, we were able to improve these bounds. It should be noted here that longer computation times on more powerful machines would likely improve these bounds; however, we have pushed the limits of these methods to the extreme, and resolving σ(J1) will likely require new techniques. To better utilize the results from [17], we will follow Holmes and use notation from the Atlas [8] rather than representing the groups as a permutation group as done in the previous cases. Recall that conjugacy classes of elements are named by the orders of their elements and a capital letter. They are written in descending order of centralizer size. Here is our improved estimate for σ(J1).

THEOREM 7.1. For the Janko group J1 we have 5316 6 σ(J1) 6 5413.

PROOF. In [17] it is determined that all 1540 maximal subgroups isomorphic to C19 : C6 and all 2926 maximal subgroups isomorphic to S3 × D10 are needed in a minimal covering. The only remaining elements generating maximal cyclic subgroups that need to be covered are those of type 11A and 7A. Holmes shows in [17] that only maximal subgroups isomorphic to PSL(2, 11) are needed to cover all elements of type 11A, and also only maximal subgroups 3 isomorphic to C2 : C7 : C3 are needed to cover elements of type 7A. Using the GAP program COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 19

[11] as given in Function 8.1 for G = J1 and the maximal subgroups isomorphic to PSL(2, 11), we are setting up the equations readable by Gurobi [15] for the elements of type 11A. The Gurobi output then tells us that a minimal covering of the elements of this type consists of exactly 196 subgroups isomorphic to PSL(2, 11). Similarly, preparing the linear equations for 3 Gurobi using Function 8.1 for G = J1 and the maximal subgroups isomorphic to C2 : C7 : C3 for the elements of type 7A, the Gurobi output shows that the number of subgroups of this type needed to cover the respective elements is between 654 and 752. Therefore, we find that the subgroup covering number of J1 is between 1540 + 2926 + 196 + 654 = 5316 and 1540 + 2926 + 196 + 751 = 5413. We have included the files produced by GAP which are read by Gurobi on http://www. math.wm.edu/˜eswartz/coverings. The bounds that we have now were obtained by running Gurobi on KoKo, part of the High Performance Computing (HPC) cluster at Florida Atlantic University. The technical specifications of KoKo include four-hundred Intel Xeon Cores and 128 GB of RAM per node. It took KoKo 8579.61 seconds to determine that 196 subgroups are needed to cover the elements of type 11A, and it took KoKo 2244640 seconds to reach the bounds for the number of subgroups required to cover the elements of type 7A. 8. GAP Code In this section, we start with the code used in GAP [11] to create the output files read by Gurobi [15]. Any solution to the system of equations encoded in the output corresponds to a subgroup cover of the elements. Whenever the “best objective” and the “best bound” found by Gurobi are identical, Gurobi has found a minimal subgroup cover. In short, GAP is used to find the optimal solution to a system of linear inequalities, which corresponds to a minimal cover. Gurobi then performs a linear optimization on this system of linear inequalities. For the case of S9, addressed in Proposition 3.3, we include the output of Function 8.1 as well as an abbreviated table of the Gurobi output. A complete table of this output can be found at http://www.math.wm.edu/˜eswartz/coverings. The corresponding output of Function 8.1, together with the generators for the subgroups in the minimal cover of elements with cycle structure (6, 6) in M12 and the linear programs produced for J1, can be found at the same website. FUNCTION 8.1. GAP function to create the output files to be read by Gurobi. #SubgroupCoveringNumber takes as input a group $G$, a list of #elements $L$, a list of maximal subgroups $M$, and the name of a #file of type .lp to which output is written.

SubgroupCoveringNumber:= function(G, ElementList, MaximalSubgroupList, filename) local maxs, maxconjs, x, y, temp, elts, eltconjs, output, NumberSubgroups, NumberElements, i, j, FilteredSubgroupIndices;

#Subgroup covering number first computes all conjugate subgroups #of those in the list MaximalSubgroupList. maxs:= []; for x in MaximalSubgroupList do maxconjs:= ConjugateSubgroups(G,x); for y in maxconjs do Add(maxs, y); od; od; 20 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

NumberSubgroups:= Length(maxs);

#All cyclic subgroups generated by the conjugates of the elements #in ElementList are stored in the irredundant list elts. elts:= []; for x in ElementList do eltconjs:= AsList(ConjugacyClass(G,x)); for y in eltconjs do if not Group(y) in elts then Add(elts, Group(y)); fi; od; od;

NumberElements:= Length(elts);

#SubgroupCoveringNumber now begins writing to the output file. #Each variable r1, r2,... represents a binary variable that takes #on the value 0 or 1. (A 1 represents the subgroup being included # in the covering; a 0 means it’s not included.)

#First, we write that we want to minimize the sum of all the #variables, i.e., we want to minimize the number of subgroups #included in the covering. output := OutputTextFile( filename, false );; SetPrintFormattingStatus(output, false); AppendTo(output,"Minimize\n"); for i in [1..NumberSubgroups] do AppendTo(output, Concatenation( " + r", String(i))); od; AppendTo(output,"\n Subject To\n");

#For each subgroup H in elts, we require that H is a subgroup #of at least one maximal subgroup in the covering. This #corresponds to the sum over all the variables representing #maximal subgroups containing H being at least 1. Note that #Gurobi interprets > as ‘‘less than or equal." for i in [1..NumberElements] do FilteredSubgroupIndices:= Filtered([1..NumberSubgroups], j -> (IsSubgroup(maxs[j],elts[i]))); for j in FilteredSubgroupIndices do AppendTo(output, " + r", String(j)); od; AppendTo(output, " > 1\n"); od;

#This last part specifies that each variable is ‘‘Binary," i.e., that #it can only take on the value 0 or the value 1.

AppendTo(output, "\\ Variables\n"); AppendTo(output,"Binary\n"); COVERING NUMBERS OF SMALL SYMMETRIC GROUPS 21

for i in [1..NumberSubgroups] do AppendTo(output, Concatenation( "r", String(i), "\n")); od;

AppendTo(output,"End\n"); CloseStream(output); return maxs;

#The function returns the list of maximal subgroups. end;; As a sample of the output of Function 8.1 we will show how the calculations proceed for the elements with cycle structure (3, 6) in the group S9. First, we use GAP to create a ﬁle that is readable by the optimization software Gurobi: OUTPUT 8.2. Output from GAP showing the use of Function 8.1. gap> G:= SymmetricGroup(9); Sym( [ 1 .. 9 ] ) gap> max:= MaximalSubgroupClassReps(G); [ Alt( [ 1 .. 9 ] ), Group([ (1,2,3,4,5), (1,2), (6,7,8,9), (6,7) ]), Group([ (1,2,3,4,5,6), (1,2), (7,8,9), (7,8) ]), Group([ (1,2,3,4,5,6,7), (1,2), (8,9) ]), Group([ (1,2,3,4,5,6,7,8), (1,2) ]), Group([ (1,2,3), (1,2), (4,5,6), (4,5), (7,8,9), (7,8), (1,4,7)(2,5,8)(3,6,9), (1,4)(2,5)(3,6) ]), Group([ (4,7)(5,8)(6,9), (2,7,6)(3,4,8), (1,2,3)(4,5,6)(7,8,9) ]) ] gap> M:= [max[3], max[6], max[7]]; [ Group([ (1,2,3,4,5,6), (1,2), (7,8,9), (7,8) ]), Group([ (1,2,3), (1,2), (4,5,6), (4,5), (7,8,9), (7,8), (1,4,7)(2,5,8)(3,6,9), (1,4)(2,5)(3,6) ]), Group([ (4,7)(5,8)(6,9), (2,7,6)(3,4,8), (1,2,3)(4,5,6)(7,8,9) ]) ] gap> g:= (1,2,3)(4,5,6,7,8,9); (1,2,3)(4,5,6,7,8,9) gap> L:= [g]; [ (1,2,3)(4,5,6,7,8,9) ] gap> Read("Programs/SubgroupCoveringNumber.g"); gap> l:= SubgroupCoveringNumber(G,L,M, "S9.lp");; gap> time; 218128 Note that only one element with cycle structure (3, 6) is needed in the list L since all elements with the same cycle structure are conjugate in a symmetric group. We next use Gurobi to optimize this system of linear equations. We have removed some lines of the output here for the sake of brevity, although the full output is available online at http://www.math.wm.edu/˜eswartz/coverings. OUTPUT 8.3. Output from Gurobi showing the solution to the ﬁle created by GAP. gurobi> m = read("S9.lp") Read LP format model from file S9.lp Reading time = 0.09 seconds (null): 10080 rows, 1204 columns, 80640 nonzeros gurobi> m.optimize() Optimize a model with 10080 rows, 1204 columns and 80640 nonzeros Found heuristic solution: objective 423 Presolve time: 0.10s Presolved: 10080 rows, 1204 columns, 80640 nonzeros Variable types: 0 continuous, 1204 integer (1204 binary) 22 LUISE-CHARLOTTE KAPPE, DANIELA NIKOLOVA-POPOVA, AND ERIC SWARTZ

Root relaxation: objective 7.000000e+01, 2182 iterations, 0.19 seconds

0 0 70.00000 0 280 423.00000 70.00000 83.5% - 0s H 0 0 180.0000000 70.00000 61.1% - 0s H 0 0 123.0000000 70.00000 43.1% - 0s H 0 0 84.0000000 70.00000 16.7% - 0s 0 0 70.65138 0 285 84.00000 70.65138 15.9% - 1s 0 0 70.78547 0 289 84.00000 70.78547 15.7% - 14s 0 0 70.98212 0 290 84.00000 70.98212 15.5% - 28s . . 104 36 79.66667 12 272 84.00000 77.51814 7.72% 1228 389s 345 56 82.42501 8 376 84.00000 78.08711 7.04% 839 412s 490 28 80.48307 7 349 84.00000 78.08715 7.04% 761 430s 606 3 cutoff 7 84.00000 78.08715 7.04% 724 441s 698 2 79.66022 9 315 84.00000 79.66022 5.17% 688 450s

Cutting planes: Zero half: 107

Explored 717 nodes (514185 simplex iterations) in 453.03 seconds Thread count was 8 (of 8 available processors)

Optimal solution found (tolerance 1.00e-04) Best objective 8.400000000000e+01, best bound 8.400000000000e+01, gap 0.0% The “Best objective” is the best actual solution that was found by Gurobi, and the size of this solution is 84. The “best bound” is the size of the best lower bound that Gurobi could determine for a solution to this system of equations, and this lower bound is also 84. Therefore, we conclude that the covering of the elements with cycle structure (3, 6) in S9 by the 84 subgroups in MS3 is minimal.

ACKNOWLEDGMENTS. The third author acknowledges the support of the Australian Research Council Discovery Grant DP120101336 during his time spent at The University of Western Australia. We are very thankful to Eric Borenstein, the administrator of the High Performance Computing Initiative at Florida Atlantic University, for gaining us access to KoKo and for help- ing us implement Gurobi on KoKo, which included ﬁnding the best parameters for optimal performance. Finally, we would like to thank Gordon Royle for giving us access to his machine at The University of Western Australia.

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DEPARTMENT OF MATHEMATICAL SCIENCES,STATE UNIVERSITYOF NEW YORK AT BINGHAMTON, BINGHAMTON, NY 13902-6000, USA. E-mail address: [email protected]

DEPARTMENT OF MATHEMATIAL SCIENCES,FLORIDA ATLANTIC UNIVERSITY,BOCA RATON, FL 33431, USA. E-mail address: [email protected]

DEPARTMENT OF MATHEMATICS,COLLEGEOF WILLIAMAND MARY,WILLIAMSBURG, VA 23185, USA. E-mail address: [email protected]