ERASMUS MUNDUS MASTER ALGANT UNIVERSITÀ DEGLI STUDI DI PADOVA HAMILTONIAN CYCLES


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1 ERASMUS MUNDUS MASTER ALGANT UNIVERSITÀ DEGLI STUDI DI PADOVA FACOLTÀ DI SCIENZE MM. FF. NN. CORSO DI LAUREA IN MATEMATICA ELABORATO FINALE HAMILTONIAN CYCLES IN THE GENERATING GRAPHS OF FINITE GROUPS RELATORE: PROF. A.LUCCHINI DIPARTIMENTO DI MATEMATICA PURA E APPLICATA LAUREANDO: AGLAIA MYROPOLSKA ANNO ACCADEMICO 2009/2010
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3 Contents 1 Graphs Solvable groups Some tools in finite group theory Diagonal subgroups Monolithic groups Imprimitive groups Classification of finite simple groups Probabilistic generation of finite simple groups Wreath product
4 4 CONTENTS Introduction In recent years there has been considerable interest in generation properties of finite simple groups and permutation groups, and many results were obtained using counting and probabilistic methods. The roots of the subject lie in a series of 7 papers by Erdos and Turan in which they study the properties of random permutations. For example they show that most permutations in the symmetric group S n have order about n 1 2 log n, and have about log n cycles. Dixon [11] used the ErdosTuran theory to settle an old conjecture of Netto proving that two randomly chosen elements of the alternating group A n generate A n with probability tending 1 as n and he proposed the following generalization: Conjecture 1 (Dixon, [11]) Two randomly chosen elements of a finite simple group G generate G with probability 1 as G. At the time the Classification of Finite Simple Groups was not yet available so the statement of conjecture was left without proof. Anyway a breakthrough was made in 1990 by Kantor and Lubotzky, who proved Dixon s conjecture for classical groups in [13] and the remainig exceptional groups of Lie type were handled by Liebeck and Shalev in 1995 in [14]. So we have: Theorem 1 (Dixon, Kantor, Lubotzky, Liebeck, Shalev) Dixon s conjecture holds. A related questions raised in [13] is as follows. Let G be a finite simple group, and let x G be a nonidentity element. Let P x (G) denote the probability that x and a randomly chosen element of G generate G. What can be said about P x (G)? Guralnick and Kantor in [9] show that for each finite simple group G there is a conjugacy class C G such that each nontrivial element of G generates G together with any of more than 1/10 of the members of C G, namely, every nonidentity element of a finite simple group sits in some generating pair. For a finite group G let Γ(G) denote the graph defined on the nonidentity elements of G in such a way that two distinct vertices are connected by an edge if and only if they generate G. In the literature many deep results about finite simple groups G can equivalently be stated as theorem about Γ(G). The results described above essentially say that the generating graph of a nonabelian simple group contains many edges with high probability, in particular there is no isolated vertex. Later Guralnick and Shalev [8] showed that for sufficiently large G the graph Γ(G) has diameter at most 2. Such results lead to conjecture that the generating graph of finite simple group contains also a Hamiltonian cycle. In [1] the following theorem was proved: Theorem 2 For every sufficiently large finite simple group G, the graph Γ(G) contains a Hamiltonian cycle. The previous theorem has been proved with asymptotical argument, so it works only when the order of G is large enough. However the conjecture is that Γ(G) contains a Hamiltonian cycle for each non abelian simple group. Breuer proved this conjecture ([16]) for non abelian simple groups of order at most 10 7, for sporadic simple groups and for P SL(2, q) for each prime power q. This has been obtained with the help of the GAP system. In several case that result on simple group have been extended to their automorphism group, for example, the following has been obtained.
5 CONTENTS 5 Theorem 3 [1] For every sufficiently large symmetric group S n the graph Γ(S n ) contains a Hamiltonian cycle. However how can one produce other examples of nonsimple finite group G which generating graph contains a Hamiltonian cycle? If Γ(G) contains a Hamiltonian cycle, then G/N must be cyclic for any normal subgroup N of G with N 1. Indeed, if n N, n 1, then n cannot be isolated in Γ(G), so G = n, g for some g G and consequently G/N = ng is cyclic. It has been conjectured that the previous condition is not only necessary, but also sufficient. Conjecture 2 Let G be a finite simple nonabelian group. Then the generating graph Γ(G) contains a Hamiltonian cycle if and only if for any normal subgroup N the factorgroup G/N is cyclic. This conjecture has been verified for the group of order at most 10 6 and in [1] it is proved that it is true in the solvable case. So the problem is to check whether a nonsolvable group has the property that Γ(G) contains a Hamiltonian cycle if for any nontrivial normal subgroup N G the factorgroup G/N is cyclic. We will show that a group with the latter property is monolithic (i.e. it has an unique minimal normal subgroup, say M). It turns out that M is isomorphic to a direct product of isomorphic non abelian simple group (M = S t ) and that S t G AutS C t with C t the subgroup of the symmetric group of degree t generated by the cycle (12...t). The simplest case to consider is therefore G = S C t. The study of this case has been started in [1], when the following is proved: Theorem 4 For every sufficiently large nonabelian finite simple group S the graph Γ(S C m ) contains a Hamiltonian cycle where m denotes a prime power. In the thesis this result was generalized: Theorem 5 For every sufficiently large nonabelian finite simple group S the graph Γ(S C m ) contains a Hamiltonian cycle where m = p α q β, p, q  odd primes or m = 2q, qprime. It is expected that using the similar ideas one can obtain the result for any composite number m. In the first chapter there are the basic definitions and criterions if the generating graph contains a Hamiltonian cycle. One of the easiest case to detect if the generating graph contains Hamiltonian cycle is for Solvable groups and the proof of this fact can be found in the chapter 2. In the third chapter there is some preliminary information which contains subsections about diagonal subgroup, monolithic group, imprimitive groups and the classification of finite simple groups. The history of the generation problem of finite simple groups is given in the fourth chapter. And the main result is proved in the last chapter.
6 6 CONTENTS 1 Graphs Definition 1 For a finite group G let Γ(G) denote the graph defined on the nonidentity elements of G in such a way that two distinct vertices are connected by an edge if and only if they generate G. Definition 2 A Hamiltonian cycle is a cycle in an undirected simple graph which visits each vertex exactly once. A graph is called Hamiltonian if it contains a Hamiltonian cycle. Definition 3 A simple graph with m vertices and list of vertex degrees d 1 d 2... d m satisfies Posa s criterion if d k k + 1 for all positive integers k with k < m/2 The following criterions are widely used in [1] to detect if the generating graph contains a Hamiltonian cycle. Also they were used in the proof of the main result of the thesis. Proposition 1 [6] A graph contains a Hamiltonian cycle if it satisfies Posa s criterion. Definition 4 For a simple graph Γ with m vertices let d(γ, v) denote the degree of the vertex v. The closure cl(γ) of Γ is the graph (on the same set of vertices) constructed from Γ by adding for all nonadjacent pairs of vertices u and v with d(γ, u) + d(γ, v) m the new edge uv. One of the best characterization of Hamiltonian graphs is Theorem 6 (Bondy, Chvatal, [2]) A graph is Hamiltonian if and only if its closure is Hamiltonian. For a simple graph Γ, let us set cl (1) (Γ) = cl(γ) and inductively set cl (i) (Γ) = cl(cl (i 1) (Γ)) for every positive integer i larger than 1. Remark 1 Let G be a finite group. Then if G is not generated by two elements then the graph Γ(G) is empty. Example 1 Consider the cyclic group of prime order C p. Then the generating graph Γ(C p ) is complete. In particular, it contains a Hamiltonian cycle g, g 2, g 3,..., g n 1, g. Example 2 Consider the alternating group A 5. The elements of A 5 \{1} are divided into three groups: 1. V 1 = {elements of order 5}, V 1 = V 2 = {elements of order 3}, V 1 = V 3 = {elements of order 2}, V 1 = 15. We can distinguish three maximal subgroups in A 5 ([3]): 1. Onepoint stabilizer A 4 ; 2. 2setstabilizer; 3. N A5 (C 5 ) = D 10. And in order to say that two elements x, y A 5, x, y = H generate the whole group A 5 is enough to show that H is not contained in any of these maximal subgroups. Fix any element v in V 1 then it can be connected to any element of order 5 which doesn t
7 2 Solvable groups 7 belong to the Sylow subgroup v and the number of such elements is 20; It is easy to see that v is connected to any element in V 2 because the group generated by elements of orders 3 and 5 can t be contained in any of maximal subgroups; At last we claim that v is connected to 10 elements in V 3. Consider any element y V 3 and y = 2. Define the subgroup H = x, y. It either is equal to A 5 or H doesn t contain any element of order 3 and H = 10. In the second case H = N A5 ( x ) and it is isomorphic to the dihedral group D 10 which contains exactly 5 elements of order 2. Now let v V 2. It is connected to any element in V 1 with the same argument as earlier. Assume that v = (i 1 i 2 i 3 ) and u is an arbitrary element in V 2. Then in order to generate the group u must shift i 4 and i 5 where {i 4, i 5 } = {1, 2, 3, 4, 5} \ {i 1, i 2, i 3 }. Notice that u, v can t be contained in any of maximal subgroups and the number of such elements u V 2 is 6. Now let u V 3. In order to generate A 5 u must not stabilize {i 4, i 5 }, besides i 4 and i 5 can t belong to the same cycle (otherwise H = u, v will coincide with the second maximal subgroup). Thus u has to be of the following form: u = (j 1 i 4 )(j 2 i 5 ) where j 1, j 2 are different and belong to {1, 2, 3, 4, 5} \ {i 4, i 5 } and the number of such u is 6. Fix an element v V 3. Then let count the number of edges between V 2 and V 1. On the one hand it is 15x where x is the number of edges between any fixed element of V 2 and V 1, on the other it is so x = 16. Assume that v = (i 1 i 2 )(i 3 i 4 ) and take an arbitrary element u V 2. Then in order to generate A 5 u must not fix i 5 = {1, 2, 3, 4, 5} \ {i 1, i 2, i 3, i 4 } so u has the following form (i 5 jk) and to avoid the coincidence with the second maximal subgroup j, k should belong to the different sets: {i 1, i 2 } or {i 3, i 4 }. Thus the number of such u is 8. At last we show that v can t be connected to any u V 3. Assume by contradiction that v, u = A 5. Notice that (vu) v = uv and (vu) u = uv then the subgroup vu is normal in A 5 which contradicts the fact that A 5 is simple. Remember that two vertices from V 1 and V 2 are always connected in Γ 0 (A 5 ). Now consider u, v V 1 then d(γ 0, u) + d(γ 0, v) = 2( ) = 60 = A 5. So two vertices in V 1 are connected in Γ 1 (A 5 ). Consider u V 1 and v V 3 then d(γ 1, u) + d(γ 1, v) = = 77 > 60 = A 5 which means that any vertex in V 3 is connected to any vertex in V 1 in Γ 2 (A 5 ). Consider u V 2 and v V 3 then d(γ 0, u) + d(γ 0, v) = = 60 = A 5 which means that any vertex in V 2 is connected to any vertex in V 3 in Γ 1 (A 5 ). Consider u, v V 2 then d(γ 0, u)+d(γ 0, v) = 2 36 > A 5 so any two vertices in V 2 are connected in Γ 1 (A 5 ). If u, v V 3 then d(γ 2, u) + d(γ 2, v) = 2( ) > A 5 so any two vertices in V 3 are connected in Γ 3 (A 5 ) which means that Γ 3 (A 5 ) is a complete graph and hence the generating graph Γ(A 5 ) is Hamiltonian. 2 Solvable groups Let G be a finite solvable group with at least 4 elements. To prove that Γ(G) contains a Hamiltonian cycle we may assume that G/N is cyclic for all nontrivial normal subgroups N of G. If G is cyclic, then any generator g of G is connected to every other vertex of Γ(G) and g 1, g 2,..., g n 1, g 1 determines a Hamiltonian cycle in Γ(G) where n = G. Hence we may
8 8 CONTENTS assume that G is noncyclic. If G has two distinct minimal normal subgroups, A and B, then G embeds in G/A G/B and so is Abelian. Since G is not cyclic, the Frattini subgroup of G must be trivial. Thus, G is a direct product of cyclic groups of prime order. It follows that G is elementary Abelian of order p 2 for some prime p. Then each vertex in Γ(G) has degree p 2 p and so there is Hamiltonian cycle in Γ(G) by Posa s criterion. So we may assume that G has a unique minimal normal subgroup M. It follows that M is an elementary Abelian pgroup for some prime p: indeed, since M is solvable then the commutator M char M G which leads us to the conclusion M = M; in addition, since M < then there exists pprime divisor of M and M must coincide with the subgroup of elements of order dividing p. Let H be a p Hall subgroup of G and let P/M be the Sylow psubgroup of G/M which is cyclic. It follows that P G and G = P H, moreover H is cyclic since H is an epimorphic image of G/P. Let F be the Frattini subgroup of P. Since F char P G then F G. And there are two possibilities for F : 1. F 1. In this case M F and by assumption G/F is cyclic hence H centralizes P/F thus by theorem below H centralizes P which means that H G and by uniqueness of minimal subgroup M it follows that H = 1 and G = P. Moreover P/F is cyclic then P = G is cyclic. Theorem 7 [7] Let A act via automorphisms on G, where A and G are finite groups, and assume that ( A, G ) = 1. If the induced action of A on the Frattini factor group G/Φ(G) is trivial, then the action of A on G is trivial. 2. F = 1 then P is elementary abelian pgroup. By Maschke Theorem (see below) P = M Y with Y G then P = M and G = M H. Theorem 8 (Maschke) Let K be a group of finite order m, and suppose that K acts via automorphisms on a group V with subgroups U and W such that V = U W, where U is abelian and Kinvariant. Assume that the map u u m is both injective and surjective on U. Then there exists a Kinvariant subgroup N V such that V = U N. Remark 2 If KM = G then either K = G or K M = 1: indeed K M is normal in K and it is normalized by M. In particular, H max G and H = N G (H). Remark 3 If m M and m 1 then H H m = 1 Proof. Assume that there exist h 1, h 2 H such that h 1 = h m 2. Since H is cyclic we must have h 1 = h 2, hence m normalizes h 1 and [m, h 1 ] M h 1 = 1 and m h 1 = m. Notice that C M (h 1 ) G: indeed if m C M (h 1 ) and h H then (m h ) h 1 = m hh 1 = (m h 1 ) h = m h 2. Hence C M (h 1 ) = M but then h 1 G and we came to a contradiction. Put m = M. Let H 1,..., H m be the distinct conjugates of H in G. Notice that m 3. For each i with 1 i m the cyclic group H i is maximal in G. Put n = H and let h be a generator of H m. For each k with 1 k m let v k be the unique element of M with v 1 k H mv k = H k. Let j be an arbitrary positive integer with 1 j m n. If j is a multiple of n, then set g j = v k where k is such that k j(mod m). Otherwise, if j is not a multiple of n, then set g j = v 1 k hi v k where i and k are so that i j (mod n) and k j
9 3 Some tools in finite group theory 9 (mod m). Then g mn is the identity element of G and g 1,..., g mn 1 are precisely the nonidentity elements of G. We claim that the vertices g 1,..., g mn 1, g 1 determine a Hamiltonian cycle in Γ(G). To show this claim, let L = g i, g i+1. By the way in which these elements are defined LM = G so by Remark 2 either L = G or L is a complement: the second possibility cannot occur since g i, g i+1 are not in the same conjugate of H. 3 Some tools in finite group theory 3.1 Diagonal subgroups Definition 5 Let G = n i=1 S i be a direct product of groups. A subgroup H of G is said to be diagonal if each projection π i : H S i, i = 1,..., n, is injective. If each projection π i : H S i is an isomorphism, then the subgroup H is said to be a full diagonal subgroup. If H is a full diagonal subgroup of G = n i=1 S i, then all the S i are isomorphic. Observe that if x = (x 1,..., x n ) H, then x i = x π i, for all i = 1,..., n, and then x = (x 1, x π 1 1 π 2 1,..., x π 1 1 πn 1 ). All ϕ i = π1 1 π i are isomorphisms of S 1 and then ϕ = (ϕ 1 = 1, ϕ 2,..., ϕ n ) Aut(S 1 ) n. Conversely, given a group S and ϕ = (ϕ 1, ϕ 2,..., ϕ n ) Aut(S) n, it is clear that {x ϕ = (x ϕ 1, x ϕ 2,..., x ϕn ) : x S} is a full diagonal subgroup of S n. More generally, given a direct product of groups G = n i=1 S i such that all S i are isomorphic copies of a group S, to each pair (, ϕ), where = {I 1,..., I l } is a partition of the set I = {1,..., n} and ϕ = (ϕ 1,..., ϕ n ) Aut(S) n, we associate a direct product D (,ϕ) = D 1... D l, where each D j is a full diagonal subgroup of the direct product i I l S i defined by the automorphisms {ϕ i : i I j }. Proposition 2 ([5]) Suppose that H is a subgroup of the direct product G = n i=1 S i, where the S i are nonabelian simple groups for all i I = {1,..., n}. Assume that all projections π i : H S i, i I, are surjective. Then there exists a partition of I such that the subgroup H is the direct product H = D H π D, where a) each H π D is a full diagonal subgroup of i D S i, b) the partition is uniquely determined by H in the sense that if H = D Hπ D = G Γ Hπ G for and Γ partitions of I, then = Γ 3.2 Monolithic groups Definition 6 A group G is said to be monolithic if there exists a nontrivial normal subgroup of G which is contained in every nontrivial normal subgroup of G. The minimal normal subgroup in this case is termed a monolith. We need the following result to proof the proposition 4:
10 10 CONTENTS Proposition 3 [5] Let S be a nonabelian simple group and write S n = S 1... S n for the direct product of n copies S 1,..., S n of S, for some positive integer n. Then Aut(S n ) = Aut(S) Sym(n), where Sym(n) is the symmetric group of degree n. Proposition 4 Assume that G is non solvable and G/N is cyclic for any nontrivial normal subgroup N. Then S t G AutS C t, where C t Sym(t) is a cyclic group generated by σ = (12...m). Proof. We may assume that any minimal normal subgroup N is non abelian (otherwise since G/N is cyclic then G would be solvable). Let show that G is monolithic: assume that N 1, N 2 are two different minimal normal subgroups of G. Since N 1 N 2 = 1 then N 2N 1 N 1 = N2 which is not abelian. We came to the contradiction with the fact that G/N 1 is cyclic. Now consider the unique minimal normal subgroup N of G. Then N is isomorphic to the direct product N = S t : indeed, assume that S is a minimal subnormal subgroup which is normal in N; then S is simple and moreover N = S t for some nonnegative integer t. Since G/C G (N) AutN then by uniqueness of minimal normal subgroup G AutN. And at last using the previous proposition we get S t G (AutS) Sym(t). Consider the projection π : G Sym(t). π(g) is a cyclic subgroup of Sym(t): indeed, π(g) is a subgroup of G/S t which is cyclic. 3.3 Imprimitive groups Suppose that G acts transitively on Ω, and let B 1,..., B r Ω be nonempty subsets. Definition 7 We say that {B 1,..., B r } is a partition of Ω if for any g G there exists 1 j r: B g i = B j Remark 4 In an arbitrary transitive action, the whole set Ω is a partition and so is every onepoint subset of Ω. These are referred to as trivial partition. Definition 8 We say that the given transitive action of G on Ω is imprimitive if there exists a nontrivial partition of Ω; otherwise, the action is primitive. {B 1,..., B r } is called a set of imprimitivity. Lemma 1 Let the cyclic group C n = (12...n) act on the set {1, 2,..., n}. Let A = {sets of imprimitivity of the action C n on {1, 2,..., n}} and B = {nontrivial divisors of n}. Then there exists a bijection ϕ : A B. Proof. Take a nontrivial divisor m of n and consider the action σ m on {1, 2,..., n}. Denote by Ω = {O 1,..., O m } the set of orbits of the action σ m on {1, 2,..., n}. Notice that Ω is a partition and for any 1 k n 1 there exists 1 j m: Oi σk = O j. On the other hand take any imprivity set B = {B 1,..., B r } for C n. Consider the stabilizer H = St G (B 1 ) = {x C n B1 x = B 1 } and H = σ m for some nontrivial divisor m of n. We can assume that 1 B 1 so then 1 H B1 H B 1 and it suffices to prove that 1 H = B 1. Notice that r = C/H then B 1 = n = n H = H. Since 1 H = H therefore 1 H = B r C 1.
11 4 Probabilistic generation of finite simple groups Classification of finite simple groups The classification theorem for finite simple groups Each finite simple group is isomorphic to one of the following groups: the cyclic groups of prime order the alternating groups of degree at least 5 the groups of Lie type one of 26 sporadic groups. In order to understand the classification, one must define what one means by group of Lie type and sporadic groups. The list of sporadic groups can be found in [3], but they may be roughly sorted as the Mathieu groups, the Leech lattice groups, Fischer s 3transposition groups, the further Moster centralizers, and the halfdozen oddments. In 1955 Chevalley discovered a uniform way to define bases for the complex simple Lie algebras in which all their structure constants were rational integers. It follows that analogues of these Lie algebras and the corresponding Lie groups can be defined over arbitrary fields. The resulting groups are now known as the adjoint Chevalley groups. Over finite fields, these groups are finite groups which are simple in almost all cases. But Chevalley s construction did not give all of the known classical groups. Steinberg found a modification of Chevalley s construction that gave these groups and a few new families. A further modification yields the infinite families of simple groups discovered by Suzuki and Ree. 4 Probabilistic generation of finite simple groups A nineteenthcentury conjecture of Netto asserts that almost all pairs of permutations in A n generate A n. This was confirmed by J.D.Dixon in While this beautiful result settles a vintage conjecture, there is a more modern aspect of it which is relevant in grouptheoretic algorithms: it is very easy to find generating pairs in A n ; after a few random choices you are very likely to end up with one. And there arises the question if this property shared by other important groups, for instance, by the finite simple groups. In Dixon [11] generalized Netto s conjecture as follows. Conjecture 3 (Dixon 1969) Two randomly chosen elements of a finite simple group G generate G with probability 1 as G. Here G and its Cartesian powers are regarded as probability spaces with respect to the uniform distribution, and the random elements are chosen independently (allowing repetitions). Denote by P (G) the respective generation probability, namely, the number of generating pairs (x, y) G G divided by G 2 (the number of all ordered pairs). It turns out that the maximal subgroups of the ambient group G provide a key to the solution of Dixon s conjecture. To understand their relevance, suppose x, y G are chosen at random. If these elements do not generate G, then they both lie in some maximal (proper) subgroup M of G. Given M, the probability that x, y lie in M is M 2 G 2. Summing up over all maximal
12 12 CONTENTS subgroups M of G, we obtain the following upper bound on the nongeneration probability: 1 P (G) G : M 2. M G,M maximal Define a zetafunction associated with G by ζ G (s) = M G,M maximal G : M s. Then, in order to confirm Dixon s conjecture, it suffices to show that, for finite simple groups G, ζ G (2) 0 as G. In other words, this would follow if we could show that simple groups do not have many maximal subgroups (so the sum in question has few summands) and that the indices of these subgroups are usually rather large (so that these summands are usually small). The study of the maximal subgroups of the finite simple groups is an ongoing project dating back to the nineteenth century. Two major results, by O NanScott and by Aschbacher respectively, describe the maximal subgroups of alternating groups and of classical groups [12]. Extensions for nonclassical simple groups of Lie type were provided by Seitz and others. The Dixon s conjecture in the case of alternating group is a result due to Dixon, however, his theorem is proved using nineteenthcentury results concerning permutation groups, the proof of the general theorem uses the classification of finite simple groups. Theorem 9 ([11], [13], [14]) Two randomly chosen elements of a finite simple group G generate G with probability 1 as G. In [13] Kantor and Lubotzky proposed a more subtle conjecture, namely, that a randomly chosen involution and a randomly chosen additional element of a finite simple group g generate G with probability 1 as G. At the time this conjecture was posed it was not even clear that every finite simple group can be generated by an involution and another element. The first result of the paper by Liebeck and Shalev [14] confirms the KantorLubotzky conjecture. Theorem 10 Let G be a finite simple group. Then the probability that a randomly chosen involution and a randomly chosen additional element of G generate G tends to 1 as G. But not all results in this field are of an asymptotic nature. Instead of studying the behavior of certain generation probabilities when the group order tends to infinity, some authors obtained lower bounds on such probabilities which hold for all finite simple groups. As an important example, consider the following result which was proved by Guralnick and Kantor a few years ago. Theorem 11 ([9]) In every finite simple group G there is a conjugacy class C G such that, for any nonidentity element g G, the probability that g and a random element of C G generate G is at least 1/10. This theorem has several interesting applications. One of them is following positive solution to a problem which has been open for many years regarding the socalled oneandahalf generation of finite simple groups: every nonidentity element of a finite simple group can be extended to
13 4 Probabilistic generation of finite simple groups 13 a generating pair. This result is yet another indication of the ease with which generators for finite simple groups can be found. One of the important contributions was made by Breuer in [16]. It was obtained that the generating graphs of nonabelian simple groups of order up to 10 6 satisfy Posa s criterion (which implies that they contain a Hamiltonian cycle), and that the same holds for those nonabelian simple groups of order between 10 6 and 10 7 that are not isomorphic with some P SL(2, q) while the generating graph of P SL(2, q) satisfies Posa s criterion for any prime power q. The probabilistic generation of finite group is developed in [8] introducing the spread of a finite group. Definition 9 The spread s(g) of a finite group G is the maximal integer k so that for any k nonidentity elements x 1,..., x k G there exists an element y G such that x i, y = G. By [9] s(g) 1 for any finite simple group. The spread of symmetric and alternating groups is studied in some papers by G.Binder, J.L.Brenner and J.Wiegold. It was shown that (with some small exceptions), s(s 2n ) = 3, s(s 2n+1 ) = 3 and s(a 2n ) = 4. It was suggested by J.L.Brenner and J.Wiegold that s(a 2n+1 ) might tend to infinity with n but in [8] it was shown that it s not the case. More specifically, it was shown that s(a n ) tends to infinity if and only if the smallest prime divisor p(n) of n tends to infinity. In general, the following theorem was proved: Theorem 12 ([8]) Let G i be a sequence of finite simple groups with G i tending to infinity. Then s(g i ) if and only if there does not exist an infinite subsequence of the G i consisting either of odd dimensional orthogonal groups over a field of fixed size or alternating groups of degree all divisible by a fixed prime. However there are some explicit estimates of s(g) for the various finite simple groups G. For example, for A n, the spread is largest when n is prime and otherwise can be estimated between two polynomials of small degree in the smallest prime divisor of n. One can study the structure of arbitrary finite groups of spread at least 1. If G is abelian, then G is either cyclic or is an elementary abelian pgroup of rank 2 for some prime p. If G is cyclic, then s(g) is infinite. If G is elementary abelian pgroup of rank 2, then s(g) = p. Let G be such a group that is not abelian. Clearly, if N is any minimal normal subgroup of G, then G/N is cyclic. It follows that N is unique and either is a nonabelian simple group or is an elementary abelian psubgroup for some prime p. This latter situation was studied by G.Szekeres in On a certain class of metabelian groups. In that case, s(g) 2. It is still open whether s(g) 1 for the case that N is nonabelian simple and G/N is cyclic (and there are no examples with s(g) 1). The results in [9] show that given any nontrivial element in G, there exists a mate so that the pair generates a subgroup containing N and in many but not all cases generates G. There is a deep theorem from [10] which is equivalent to saying that for a nonabelian finite simple group G the generating graph Γ(G) is a graph of diameter equal to 2. Theorem 13 (Breuer, Guralnick, Kantor, [10]) Let G be a nonabelian finite simple group. For every pair of nonidentity elements x 1 and x 2 in G there exists a y in G so that x 1, y = x 2, y = G
14 14 CONTENTS There are some results for specific finite simple groups. Suppose that G is a finite simple group of Lie type. By a random element of a nonempty finite set S we mean an element chosen uniformly from S. For a finite group G let P (G) be the probability that a random pair of elements of G generate G. For a finite group G and an element x G, define P x (G) to be the probability that x and a randomly chosen element y generate G. Note that for a nonidentity element x in a noncyclic finite group G the number P x (G) G is the degree of the vertex in Γ(G) corresponding to x in G. Theorem 14 (Fulman, Guralnick, [4]) There exists a universal positive constant c 2 so that c 2 < P x (G) for an arbitrary nonidentity element x in a finite simple group G of Lie type. Now let G be a subgroup of S n. Theorem 15 [15] For every ɛ > 0 there exists δ > 0 and a threshold n 0 such that for every n n 0, if G S n has fewer that [δn] fixed points then the probability that G and a random element σ S n generate A n or S n is at least 1 ɛ. The following direct consequence of Theorem 9 is also indicated in [15]. Let π be a permutation in A n. Corollary 1 For every ɛ > 0 there exists δ > 0 and a threshold n 0 such that for every n n 0, if π A n has fewer than [δn] fixed points then the probability that π and a random element σ A n generate A n is at least 1 ɛ. 5 Wreath product Let G and H be groups and Ω be a set, acted on by G. Let B be the set of all functions from Ω into H, and make B into a group by defining multiplication pointwise. Thus if f, g B, then fg is defined by the formula fg(α) = f(α)g(α) for α Ω. Then B is the external direct product of Ω copies of H. The action of G on Ω induces an action via automorphisms of G on B. If we view B as a direct product, the action of G can be described simply as a permutation of coordinates: given f B and x G, the function f x on Ω is defined so that f x (α) = f(α x 1 ). Definition 10 Now let W = B G with respect to this action. Then W is the wreath product of H with G, and B, viewed as a subgroup of W, is called the base group of the wreath product. Notation 1 H G We re interested in such a nonsimple finite group which generating graph contains a Hamiltonian cycle. It was shown earlier that the necessary condition for such a group G is the following fact that the factorgroup G/N is cyclic for any normal subgroup N G. And by Proposition 4 the simplest way to get an example of the nonsimple group which generating graph contains a Hamiltonian cycle may be by taking S C m where S is a simple group and C m is a cyclic group of order m.
15 5 Wreath product 15 Let S be a nonabelian finite simple group and let C m be the cyclic subroup of S m generated by cyclic permutation σ = (1, 2,..., m) with m = p α q β. Consider the wreath product G = S C m. Denote the base subgroup of G by N = S 1... S m and let π i : N S i be the projection on the ith factor. Moreover let A = Aut(S), r 1 = p α 1 q β, r 2 = p α q β 1, u = r 1 + 1, t = r and Λ 1, Λ 2 the sets {1 + r 1 i 0 i p 1}, {1 + r 2 i 0 i q 1} respectively. Lemma 2 A subgroup H of G coincides with G if the following properties are satisfied: (1) HN/N = C m ; (2) π i (H N) = S for some i; (3) there exists (y 1,..., y m ) H N and b Λ 1, c Λ 2 such that y 1, y b are not Aconjugated and y 1, y c are not Aconjugated. Proof. Assume that H satisfies these three properties and H G. Notice that H N H since N G and H N N. It is not restrictive to assume that π 1 (H N) = S. Consider the element (s 1,..., s m ) H N where s 1 π 1 (H N). Since H satisfies first condition then there exists element (x 1,..., x m )σ H: (s 1,..., s m ) (x 1,...,x m)σ = (s x 1 1,..., s xm m ) σ H N (because of normality). Notice that s x 1 1 π 2 (H N) and since this inclusion holds for any s 1 π 1 (H N) then π 1 (H N) = π 2 (H N). With the same arguments π i 1 (H N) = π i (H N) for any 2 i m thus π i (H N) = S for any 1 i m which means that H N is a subdirect product of N = S 1... S m. Let B 1,..., B t be a partition of {1, 2,..., m} and D j = {(s, s b 2,..., s bν ) i B j S i s S, b i A} such that for any i exists j: Di h = D j for any h H. Since H G then H N j D j is a diagonal subgroup of i Bj S i with B j 1 by Proposition 2. By first assumption of the lemma there exists h = (x 1,..., x m )σ H and since for any i there exists j such that: Di h = D j then Bi σ = B j which turns {B 1,..., B t } into a set of imprimitivity of action σ on {1, 2,..., m}. But using Lemma 1 notice that for any choice of B js, 1 and b or 1 and c belong to the same block. Lemma 3 Let i be an integer not divisible by p and q, ρ = σ i, τ = σ r 1, and g = (x 1,..., x m )τ G where (x 1,..., x m ) N. The probability that there is an edge in Γ(G) between g and a randomly chosen element in the coset ρn is at least η, where η is the probability that x 1, x 2, x 3 S satisfy the conditions: x 1, x 2 generate S, x 1 and x 2 are not Aconjugated, x 1 and x 3 are not Aconjugated. Proof. For illustrativeness take the example G = S C 6 : C 6 = (123456), τ = σ r 1 = (123456) 2 = (135)(246) and consider the element (x 1, x 2, x 3, x 4, x 5, x 6 )τ = (x 1, x 2, x 3, x 4, x 5, x 6 )(135)(246) S C 6. Then take any element y = (y 1, y 2, y 3, y 4, y 5, y 6 ) S 6 and conjugate x by y: x y = (y 1, y 2, y 3, y 4, y 5, y 6 ) 1 (x 1, x 2, x 3, x 4, x 5, x 6 )(135)(246)(y 1, y 2, y 3, y 4, y 5, y 6 ) = (y 1, y 2, y 3, y 4, y 5, y 6 ) 1 (x 1, x 2, x 3, x 4, x 5, x 6 )(y 5, y 6, y 1, y 2, y 3, y 4 )τ = (y 1 1 x 1 y 5, y 1 2 x 2 y 6, y 1 3 x 3 y 1, y 1 4 x 4 y 2, y 1 5 x 5 y 3, y 1 6 x 6 y 4 )τ.
16 16 CONTENTS Notice that we can choose y 5 = 1 and y 3 S such that x 5 y 3 = 1. Then we can choose uniquely y 1 such that y3 1 x 3 y 1 = 1. By analogy we can choose y 6 = 1 and y 4 S such that x 6 y 4 = 1 and then uniquely we can choose y 2 such that y4 1 x 4 y 2 = 1. So it is not restrictive to assume that x i = 1 for each i > r 1. Now come back to the lemma. Consider the element g = (x 1,..., x m )τ G then it is not restrictive to assume that x i = 1 for each i > r 1 : (just substitute g with a conjugate g n for a suitable choice of n N): g n = (n 1,..., n m ) 1 (x 1,..., x m )τ(n 1,..., n m ) = (n 1 1 x 1 n r1 +1,..., n 1 m x m n r1 )τ = (x 1,..., x r1, 1,..., 1)τ. Consider h = ρ(y 1,..., y m ). Since ρ = σ i and i is not divisible by p and q then there exist k < m and (h 1,..., h m ) N such that h k = (ρ(y 1,..., y m )) k = τ 1 (h 1,..., h m ). Let H be the subgroup generated by g = (x 1,..., x m )τ and h = ρ(y 1,..., y m ). H satisfies the first condition of Lemma 2. Notice that H N contains the element w = h k g = (x 1 h 1,..., x r1 h r1, h r1 +1,..., h m ) and w g. Notice also that π u (w g ) = h 1 x 1. In particular the second condition of Lemma 2 is satisfied if π u (w) = h u and π u (w g ) = h 1 x 1 generate S. The third condition is satisfied if h 1 x 1, h u and h t x t are not Aconjugate. It s sufficient to show that h 1, h u and h t are independent in order to state that there are η S 3 possible choices for h 1, h u, h t. Consider the case when r 1 > r 2 i.e. p < q and we may assume that i = 1 and ρ = σ. Then k = m r 1 and h 1 = y 1 y 2...y m r1, h t = y t y t+1...y m+r2 r 1 +1 and h u = y u y u+1...y m y 1. Thus we can choose y 1, y m r1 +1 and y m relatively to h 1, h t and h u such that all conditions of the previous lemma hold for them. Consider the case when r 2 > r 1 i.e. p > q, and we may assume that i = 1 and ρ = σ 1. Then we can choose y 1, y u and y r1 +r 2 +1 relatively to h 1, h u and h t such that all conditions of the previous lemma hold for them. Besides there are S m 3 possible choices for all other elements h i where 1 i m, i 1, i u, i t. So we find that the probability that there is an edge in Γ(G) between g and a randomly chosen element in the coset ρn is η N. Corollary 2 If g G \ N, then the degree of g as a vertex of Γ(G) is at least φ(m) N η = p α 1 q β 1 (p 1)(q 1) N η Lemma 4 [1] Let µ be the probability that two randomly chosen elements from S generate S and are not Aconjugate. Then lim S µ = 1. Lemma 5 Let η be as above. Then lim S η = 1 Proof. Let S be a nonabelian finite simple group and let A be the automorphism group of S. Notice that η 1 p 2q where p is the probability that a random pair of elements of S does not generate S and q is the probability that a random pair of elements of S is Aconjugate. By Lemma 4 p and q tend to 0 as S tends to infinity hence lim S η = 1.
17 5 Wreath product 17 Lemma 6 Let g = (x 1,..., x m )σ G. The probability that there is an edge in Γ(G) between g and a randomly chosen element of G is at least η. Proof. It is not restrictive (by substituting g with a suitable conjugate) to assume that x 1 =... = x m 1 = 1. Take an arbitrary element x = (y 1,..., y m )σ i G; there exist k N and (z 1,..., z m ) N with g k = σ i (z 1,..., z m ). Clearly g, x = G if and only if g, g k x = g, (y 1 z 1,..., y m z m ) = G. Assume H = g, (y 1 z 1,..., y m z m ). Notice that g r 1 g k x H. Then the first condition of Lemma 2 holds because g H. The second one holds if π 1 ((y 1 z 1,..., y m z m )) = y 1 z 1 and π 1 (g r 1 g k x) = y u z u. The third condition holds if π 1 (g k x) = y 1 z 1, π u (g k x) = y u z u and π t (g k z) = y t z t are not A conjugate. In particular g, x = G if we choose y 1, y u, y t such that: 1. y 1 z 1 and y u z u are not Aconjugate and generate S 2. y 1 z 1 and y t z t are not Aconjugate. Definition 11 C A (x) = {ξ A : x ξ = x} Lemma 7 [1] Assume that n = (x 1,..., x m ) N, with n 1. Choose i {1,..., m} with the property that P xi (S) P xj (S) for each 1 j m and let x = x i. Given a generator τ of C m = σ, the number of edges connecting n with elements of the coset Nτ is at least N µ with µ = max(p x (S) C A(x) S, Px(S) C S(x) S ρ x ), where ρ x = 1 if C A (x)s = A and ρ x = 0 otherwise. Lemma 8 Assume that n = (x 1,..., x m ) N with n 1. Choose i {1,..., m} with property that P xi (S) P xj (S) for each 1 j m and let x = x i. Given a generator τ of C m = σ, the number of edges connecting n with elements of the coset Nτ is at least N µ with µ = max(p x (S) 2 C A(x), P x (S) C S(x) 2 ρ S S 2 x ), where ρ x = 1 if C A (x)s = A and ρ x = 0 otherwise. Proof. It is not restrictive to assume that i = 1 and τ = σ. First we claim that the number of edges connecting n with elements of the coset Nσ is at least N µ 1 with µ 1 = P x (S) 2 C A(x). S It suffices to prove that, for any y 2,..., y m S m 1, there exist at least µ 1 S choices for y 1 such that if g = (y 1,..., y m )σ then n, g =G. We have g m = (h 1,..., h m ) with h 1 = y 1 y 2...y m. In particular the second condition of Lemma 2 is satisfied if x and h 1 generate S and there are at least S P x (S) choices for y 1 for which this is ensured. If there exist λ 1 Λ 1, λ 2 Λ 2 with x 1 and x λ1 are not Aconjugate, x 1 and x λ2 are not A conjugate then the third condition is automatically satisfied and we are done. If there exists λ 1 Λ 1 with x and x λ1 not Aconjugate but λ 2 1, λ 2 Λ 2 : x 1 and x λ2 are Aconjugate, then we can use Lemma 7 and the number of edges connecting n with elements
18 18 CONTENTS of the coset Nτ is at least N (P x (S) C A(x) ). S The same number of edges is if there exists λ 2 Λ 2 with x and x λ2 not Aconjugate but λ 1 1, λ 1 Λ 1 : x and x λ1 are Aconjugate. Otherwise for each 1 i p 1 : x ir1 +1 and x are Aconjugate so there exist α i A with x ir1 +1 = x α i and for each 1 j q 1 there exist γ i A with x ir2 +1 = x γ i. In this case we need an extra condition on y 1 to avoid that π Λ1 = (s, s β 1,..., s β p 1 ) and π Λ2 = (s, s ξ 1,..., s ξ q 1 ) with β i, ξ i A. Assume that this is the case. Since (x, x r1 +1,..., x r1 (p 1)+1) = π Λ1 (n) (x, x r2 +1,..., x r2 (q 1)+1) = π Λ2 (n) we must have β i C A (x)α i and ξ i C A (x)γ i. Consider g r 1 = (k 1,..., k m )σ r 1 and g r 2 = (l 1,..., l m )σ r 2 and let ɛ 1 = (1, r 1 + 1,..., r 1 (p 1)), ɛ 2 = (1, r 2 + 1,..., r 2 (q 1)) be two relatively p and qcycles. Since g r 1 H so H gr 1 H, besides N gr 1 N so g r 1 normalizes H N and we have that (k 1, k r1 +1,..., k r1 (p 1)+1)ɛ 1 normalizes π Λ1 (H N) = (s, s β 1,..., s βp 1 ). In particular, setting z 1 = β p 1 k r1 (p 1)+1, we have that z 1 β 1 = k 1 and z 1 β i = β i 1 k (i 1)r1 +1 for each 2 i p 1 and consequently z p 1 = k 1 k r k r1 (p 1)+1 = h 1. Since k r1 (p 1)+1 depends only on y 2,..., y m, the set is independent from y 1. Same arguments are correct for g r 2 1 = {(tα p 1 k r1 (p 1)+1) p t C A (x)} so setting z 2 = ξ q 1 l r2 (q 1)+1 we get z q 2 = l 1 l r l r2 (q 1)+1 = h 1 and the set 2 = {(tγ 1 1 k r2 (q 1)+1) q t C A (x)} is independent from y 1. If we choice y 1 such that x, h 1 = S and h 1 doesn t belong to 1 and 2, then g, n = G. Clearly the number of y 1 for which h 1 satisfies the two previous conditions is at most S (P x (S) 2 C A(x) ) S This concludes the proof of the first claim. Now we want to show that the number of edges connecting n with elements of the coset Nσ is at least N µ 2 with µ 2 = P x (S) C A(x) 2 ρ S 2 x Note that there are at least µ 2 N choices of y 1,..., y m so that h 1, x = S and k r1 (p 1)+1 αp 1C A (x), l r2 (q 1)+1 γq 1C A (x). We claim that for any of these choices, g = (y 1,..., y m )σ
19 5 Wreath product 19 generates G together with n. By the argument that we have used above, and under the same notations, it suffices to prove that z p 1 h 1 and z q 2 h 1. Notice that z 1 = β p 1 k r1 (p 1)+1 β p 1 αp 1C A (x) C A (x), hence z p 1 = h 1 would apply that [h 1, x] = 1 against h 1, x = S. Same arguments work for z 2 : z 2 = ξ q 1 l r2 (q 1)+1 ξ q 1 γq 1C A (x) C A (x) hence z q 2 = h 1 would apply that [h 1, x] = 1 against h 1, x = S. Lemma 9 Let m = 2p where p is prime. Let x = (x 1,..., x m )σ 2. Then the probability that there is an edge in Γ(G) between x and a randomly chosen element y = (y 1,..., y m )σ p in Nσ p is at least η. Proof. It is not restrictive (by substituting x with a suitable conjugate) to assume that x 1 = x 5 =... = x 2p 1 = 1. Consider the p+1 power of x, then there exists (w 2 1,..., w m ) N such that x p+1 2 = (w 1,..., w m )σ p+1. Take an arbitrary element y = (y 1,..., y m )σ p then yx p+1 2 = (y 1 w i1,..., y m w im )σ. Then there exists (z 1,..., z m ) N such that x p 1 = (z 1,..., z m )σ 2p 2 and (yx p+1 2 ) 2 x p 1 = (y 1 w i1 y 2 w i2, y 2 w i2 y 3 w i3,..., y m w im y 1 w i1 )σ 2 (z 1,..., z m )σ 2p 2 = (y 1 w i1 y 2 w i2 z j1, y 2 w i2 y 3 w i3 z j2,..., y m w im y 1 w i1 z jm ) N. Recall that here u = p+1 and t = 3. Then π 1 ((yx p+1 2 ) 2 x p 1 ) = y 1 w i1 y 2 w i2 z j1, π u ((yx p+1 2 ) 2 x p 1 ) = y u w iu y u+1 w iu+1 z ju, π t ((yx p+1 2 ) 2 x p 1 ) = y t w it y t+1 w it+1 z jt while π 1 (((yx p+1 2 ) 2 x p 1 ) x 1 ) = y t w it y t+1 w it+1 z jt. Then x, y = G if and only if yx p+1 2, (y1 w i1 y 2 w i2 z j1, y 2 w i2 y 3 w i3 z j2,..., y m w im y 1 w i1 z jm ) = G : in particular x, y = G if we choose y 1, y u, y t (in case p = 3 we choose y 1, y u+1, y t ) such that: 1. y 1 w i1 y 2 w i2 z j1 and y t w it y t+1 w it+1 z jt are not Aconjugate and generate S 2. y 1 w i1 y 2 w i2 z j1 and y u w iu y u+1 w iu+1 z ju are not Aconjugate. Lemma 10 Let m = 2p where p is prime. Let x = (x 1,..., x m )σ p. Then the probability that there is an edge in Γ(G) between x and a randomly chosen element y = (y 1,..., y m )σ 2 in Nσ 2 is at least η. Proof. It is not restrictive (by substituting x with a suitable conjugate) to assume that x 1 = 1. Take the element yx = (y 1 w 1,..., y m w m )σ 2+p where w i s don t depend on y 1,.., y m and σ 2+p = (1 p p p p 1 p p p 3 p 1). Consider the element n = x(yx) p N and the projections of n N: π 1 (n) = y p+1 w k1 y 3 w k2...y p 1 w kp, π 3 (n) = x 3 y p+3 w j1 y 5 w j2...y p+1 w jp, while π p+1 (n) = x p+1 y 1 w i1 y p+3 w i2...y 2p 1 w ip π 1 (n x 1 ) = x p+1 y 1 w i1 y p+3 w i2...y 2p 1 w ip.
20 20 CONTENTS Denote by I πi the set of indexes of y s in π i (n). Then I π1 = {p + 1, 3, p + 5, 7,..., 2p 3, p 1}, I π3 = {p + 3, 5, p + 7, 9,..., 2p 1, p + 1}, I πp+1 = {1, p + 3, 5, p + 7, 9,..., 2p 1}. Notice that 1 I πp+1 \ I π3, p + 1 I π3 \ I πp+1 and 3 I π1 \ (I π3 I πp+1 ). Choose s 1, s 3, s p+1 such that s 1 and s p+1 are generators, s 1 and s 3 are not Aconjugate, s 1 and s p+1 are not A conjugate. Choose as you like {y 1,..., y m } \ {y 1, y 3, y p+1 }. Then there exists a unique y p+1 such that π 3 (n) = s 3, there exists a unique y 1 such that π p+1 (n) = s p+1. Finally there exists a unique y 3 with π 1 (n) = s 1. Theorem 16 For every sufficiently large nonabelian finite simple group S the graph Γ(S C m ) contains a Hamiltonian cycle, where m = p α q β, p, q  odd primes or m = 2q, qprime. Proof. We divide the vertices of Γ(G) into three disjoint subsets: 1) V 1 is the set of vertices corresponding to elements (y 1,..., y m )τ with τ = m; 2) V 2 is the set of vertices corresponding to elements (y 1,..., y m )τ with 1 < τ < m; 3) V 3 is the set of vertices corresponding to the non trivial elements of the vase group N of the wreath product. Let Γ 0 = Γ(G) and Γ i = cl (i) (Γ(G)) for i 1. By Lemma 6 and Corollary 2, if u V 1 and v V 1 V 2 then d(γ 0, u) + d(γ 0, v) η G + η G (1 1 p )(1 1 q ) > 4 3 η G Since η tends to 1 as S tends to infinity (by Lemma 5), we deduce that if S is large enough then any vertex in V 1 is connected to any other vertex in V 1 V 2 in the first closure Γ Assume p, q > 2. But then, if v 1, v 2 V 2, then d(γ 1, v 1 ) + d(γ 1, v 2 ) 2 V 1 = 2(1 1 p )(1 1 ) G G q (in the worst case when p = 3, q = 5 we have d(γ 1, v 1 ) + d(γ 1, v 2 ) 2(1 1 3 )( ) G = 5 15 G ) which means that Γ 2 induces a complete subgraph on V 1 V Assume that p = 2 and α = β = 1. Then divide the set V 2 into two subsets: a) W 1 is the set of vertices corresponding to elements (y 1,..., y m )τ with τ = 2; b) W 2 is the set of vertices corresponding to elements (y 1,..., y m )τ with τ = q. By Lemmas 6, 9, 10 and Corollary 2 if v W 1 and u W 2 then d(γ 1, v) + d(γ 1, u) (2φ(2q) + ηφ(2) + ηφ(q)) N = G ( η q )
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Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation
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