ERASMUS MUNDUS MASTER ALGANT UNIVERSITÀ DEGLI STUDI DI PADOVA HAMILTONIAN CYCLES


 Stephany Ramsey
 3 years ago
 Views:
Transcription
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
2 2
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 )
Group Theory. Contents
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
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
More informationON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationG = G 0 > G 1 > > G k = {e}
Proposition 49. 1. A group G is nilpotent if and only if G appears as an element of its upper central series. 2. If G is nilpotent, then the upper central series and the lower central series have the same
More informationON GENERALIZED RELATIVE COMMUTATIVITY DEGREE OF A FINITE GROUP. A. K. Das and R. K. Nath
International Electronic Journal of Algebra Volume 7 (2010) 140151 ON GENERALIZED RELATIVE COMMUTATIVITY DEGREE OF A FINITE GROUP A. K. Das and R. K. Nath Received: 12 October 2009; Revised: 15 December
More informationTHE AVERAGE DEGREE OF AN IRREDUCIBLE CHARACTER OF A FINITE GROUP
THE AVERAGE DEGREE OF AN IRREDUCIBLE CHARACTER OF A FINITE GROUP by I. M. Isaacs Mathematics Department University of Wisconsin 480 Lincoln Dr. Madison, WI 53706 USA EMail: isaacs@math.wisc.edu Maria
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationON FINITE GROUPS WITH THE SAME PRIME GRAPH AS THE PROJECTIVE GENERAL LINEAR GROUP PGL(2, 81)
Transactions on Algebra and its Applications 2 (2016), 4349. ISSN: 67562423 ON FINITE GROUPS WITH THE SAME PRIME GRAPH AS THE PROJECTIVE GENERAL LINEAR GROUP PGL(2, 81) ALI MAHMOUDIFAR Abstract. Let
More informationCOMBINATORIAL PROPERTIES OF THE HIGMANSIMS GRAPH. 1. Introduction
COMBINATORIAL PROPERTIES OF THE HIGMANSIMS GRAPH ZACHARY ABEL 1. Introduction In this survey we discuss properties of the HigmanSims graph, which has 100 vertices, 1100 edges, and is 22 regular. In fact
More informationFinite groups of uniform logarithmic diameter
arxiv:math/0506007v1 [math.gr] 1 Jun 005 Finite groups of uniform logarithmic diameter Miklós Abért and László Babai May 0, 005 Abstract We give an example of an infinite family of finite groups G n such
More informationCONTRIBUTIONS TO ZERO SUM PROBLEMS
CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg
More informationBest Monotone Degree Bounds for Various Graph Parameters
Best Monotone Degree Bounds for Various Graph Parameters D. Bauer Department of Mathematical Sciences Stevens Institute of Technology Hoboken, NJ 07030 S. L. Hakimi Department of Electrical and Computer
More informationChapter 7. Permutation Groups
Chapter 7 Permutation Groups () We started the study of groups by considering planar isometries In the previous chapter, we learnt that finite groups of planar isometries can only be cyclic or dihedral
More informationSHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH
31 Kragujevac J. Math. 25 (2003) 31 49. SHARP BOUNDS FOR THE SUM OF THE SQUARES OF THE DEGREES OF A GRAPH Kinkar Ch. Das Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, W.B.,
More informationLarge induced subgraphs with all degrees odd
Large induced subgraphs with all degrees odd A.D. Scott Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, England Abstract: We prove that every connected graph of order
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationDMATH Algebra I HS 2013 Prof. Brent Doran. Solution 5
DMATH Algebra I HS 2013 Prof. Brent Doran Solution 5 Dihedral groups, permutation groups, discrete subgroups of M 2, group actions 1. Write an explicit embedding of the dihedral group D n into the symmetric
More informationON INDUCED SUBGRAPHS WITH ALL DEGREES ODD. 1. Introduction
ON INDUCED SUBGRAPHS WITH ALL DEGREES ODD A.D. SCOTT Abstract. Gallai proved that the vertex set of any graph can be partitioned into two sets, each inducing a subgraph with all degrees even. We prove
More informationSUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by
SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples
More informationCARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE
CARDINALITY, COUNTABLE AND UNCOUNTABLE SETS PART ONE With the notion of bijection at hand, it is easy to formalize the idea that two finite sets have the same number of elements: we just need to verify
More informationU.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory
More informationA REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE. 1. Introduction and Preliminaries
Acta Math. Univ. Comenianae Vol. LXVI, 2(1997), pp. 285 291 285 A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE E. T. BASKORO, M. MILLER and J. ŠIRÁŇ Abstract. It is well known that Moore digraphs do
More informationGroups with the same orders and large character degrees as PGL(2, 9) 1. Introduction and preliminary results
Quasigroups and Related Systems 21 (2013), 239 243 Groups with the same orders and large character degrees as PGL(2, 9) Behrooz Khosravi Abstract. An interesting class of problems in character theory arises
More informationGROUPS WITH TWO EXTREME CHARACTER DEGREES AND THEIR NORMAL SUBGROUPS
GROUPS WITH TWO EXTREME CHARACTER DEGREES AND THEIR NORMAL SUBGROUPS GUSTAVO A. FERNÁNDEZALCOBER AND ALEXANDER MORETÓ Abstract. We study the finite groups G for which the set cd(g) of irreducible complex
More informationMore Mathematical Induction. October 27, 2016
More Mathematical Induction October 7, 016 In these slides... Review of ordinary induction. Remark about exponential and polynomial growth. Example a second proof that P(A) = A. Strong induction. Least
More informationMA651 Topology. Lecture 6. Separation Axioms.
MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples
More informationFINITE NONSOLVABLE GROUPS WITH MANY DISTINCT CHARACTER DEGREES
FINITE NONSOLVABLE GROUPS WITH MANY DISTINCT CHARACTER DEGREES HUNG P. TONGVIET Abstract. Let G be a finite group and let Irr(G) denote the set of all complex irreducible characters of G. Let cd(g) be
More informationAlexander Hulpke Department of Mathematics Colorado State University Fort Collins, CO, 80523, USA http://www.math.colostate.
Finding Subgroups Alexander Hulpke Department of Mathematics Colorado State University Fort Collins, CO, 80523, USA http://www.math.colostate.edu/~hulpke I A Popular Problem Given a finite (permutation,
More informationSEQUENCES OF MAXIMAL DEGREE VERTICES IN GRAPHS. Nickolay Khadzhiivanov, Nedyalko Nenov
Serdica Math. J. 30 (2004), 95 102 SEQUENCES OF MAXIMAL DEGREE VERTICES IN GRAPHS Nickolay Khadzhiivanov, Nedyalko Nenov Communicated by V. Drensky Abstract. Let Γ(M) where M V (G) be the set of all vertices
More informationThe determinant of a skewsymmetric matrix is a square. This can be seen in small cases by direct calculation: 0 a. 12 a. a 13 a 24 a 14 a 23 a 14
4 Symplectic groups In this and the next two sections, we begin the study of the groups preserving reflexive sesquilinear forms or quadratic forms. We begin with the symplectic groups, associated with
More informationSets and Cardinality Notes for C. F. Miller
Sets and Cardinality Notes for 620111 C. F. Miller Semester 1, 2000 Abstract These lecture notes were compiled in the Department of Mathematics and Statistics in the University of Melbourne for the use
More informationCOMMUTATIVITY DEGREE, ITS GENERALIZATIONS, AND CLASSIFICATION OF FINITE GROUPS
COMMUTATIVITY DEGREE, ITS GENERALIZATIONS, AND CLASSIFICATION OF FINITE GROUPS ABSTRACT RAJAT KANTI NATH DEPARTMENT OF MATHEMATICS NORTHEASTERN HILL UNIVERSITY SHILLONG 793022, INDIA COMMUTATIVITY DEGREE,
More information13 Solutions for Section 6
13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationGENERATING SETS KEITH CONRAD
GENERATING SETS KEITH CONRAD 1 Introduction In R n, every vector can be written as a unique linear combination of the standard basis e 1,, e n A notion weaker than a basis is a spanning set: a set of vectors
More informationMA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES
MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions
More informationEXERCISES FOR THE COURSE MATH 570, FALL 2010
EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime
More informationGraphs without proper subgraphs of minimum degree 3 and short cycles
Graphs without proper subgraphs of minimum degree 3 and short cycles Lothar Narins, Alexey Pokrovskiy, Tibor Szabó Department of Mathematics, Freie Universität, Berlin, Germany. August 22, 2014 Abstract
More informationGROUPS ACTING ON A SET
GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for
More informationA 2factor in which each cycle has long length in clawfree graphs
A factor in which each cycle has long length in clawfree graphs Roman Čada Shuya Chiba Kiyoshi Yoshimoto 3 Department of Mathematics University of West Bohemia and Institute of Theoretical Computer Science
More informationDegree Hypergroupoids Associated with Hypergraphs
Filomat 8:1 (014), 119 19 DOI 10.98/FIL1401119F Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat Degree Hypergroupoids Associated
More informationGalois Theory III. 3.1. Splitting fields.
Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where
More informationGroup Fundamentals. Chapter 1. 1.1 Groups and Subgroups. 1.1.1 Definition
Chapter 1 Group Fundamentals 1.1 Groups and Subgroups 1.1.1 Definition A group is a nonempty set G on which there is defined a binary operation (a, b) ab satisfying the following properties. Closure: If
More informationSolutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory
Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013
More informationCourse 421: Algebraic Topology Section 1: Topological Spaces
Course 421: Algebraic Topology Section 1: Topological Spaces David R. Wilkins Copyright c David R. Wilkins 1988 2008 Contents 1 Topological Spaces 1 1.1 Continuity and Topological Spaces...............
More informationOn Integer Additive SetIndexers of Graphs
On Integer Additive SetIndexers of Graphs arxiv:1312.7672v4 [math.co] 2 Mar 2014 N K Sudev and K A Germina Abstract A setindexer of a graph G is an injective setvalued function f : V (G) 2 X such that
More informationCycles in a Graph Whose Lengths Differ by One or Two
Cycles in a Graph Whose Lengths Differ by One or Two J. A. Bondy 1 and A. Vince 2 1 LABORATOIRE DE MATHÉMATIQUES DISCRÉTES UNIVERSITÉ CLAUDEBERNARD LYON 1 69622 VILLEURBANNE, FRANCE 2 DEPARTMENT OF MATHEMATICS
More informationSets and functions. {x R : x > 0}.
Sets and functions 1 Sets The language of sets and functions pervades mathematics, and most of the important operations in mathematics turn out to be functions or to be expressible in terms of functions.
More informationIdeal Class Group and Units
Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals
More informationADDITIVE GROUPS OF RINGS WITH IDENTITY
ADDITIVE GROUPS OF RINGS WITH IDENTITY SIMION BREAZ AND GRIGORE CĂLUGĂREANU Abstract. A ring with identity exists on a torsion Abelian group exactly when the group is bounded. The additive groups of torsionfree
More informationIntroduction to Finite Fields (cont.)
Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number
More informationMathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson
Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement
More informationChapter 7: Products and quotients
Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products
More informationMIXDECOMPOSITON OF THE COMPLETE GRAPH INTO DIRECTED FACTORS OF DIAMETER 2 AND UNDIRECTED FACTORS OF DIAMETER 3. University of Split, Croatia
GLASNIK MATEMATIČKI Vol. 38(58)(2003), 2 232 MIXDECOMPOSITON OF THE COMPLETE GRAPH INTO DIRECTED FACTORS OF DIAMETER 2 AND UNDIRECTED FACTORS OF DIAMETER 3 Damir Vukičević University of Split, Croatia
More information3. QUADRATIC CONGRUENCES
3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where
More informationOn cycle decompositions with a sharply vertex transitive automorphism group
On cycle decompositions with a sharply vertex transitive automorphism group Marco Buratti Abstract In some recent papers the method of partial differences introduced by the author in [4] was very helpful
More informationA Turán Type Problem Concerning the Powers of the Degrees of a Graph
A Turán Type Problem Concerning the Powers of the Degrees of a Graph Yair Caro and Raphael Yuster Department of Mathematics University of HaifaORANIM, Tivon 36006, Israel. AMS Subject Classification:
More informationMean RamseyTurán numbers
Mean RamseyTurán numbers Raphael Yuster Department of Mathematics University of Haifa at Oranim Tivon 36006, Israel Abstract A ρmean coloring of a graph is a coloring of the edges such that the average
More informationPythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers
Pythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers Amnon Yekutieli Department of Mathematics Ben Gurion University email: amyekut@math.bgu.ac.il Notes available at http://www.math.bgu.ac.il/~amyekut/lectures
More informationOn three zerosum Ramseytype problems
On three zerosum Ramseytype problems Noga Alon Department of Mathematics Raymond and Beverly Sackler Faculty of Exact Sciences Tel Aviv University, Tel Aviv, Israel and Yair Caro Department of Mathematics
More informationOn an antiramsey type result
On an antiramsey type result Noga Alon, Hanno Lefmann and Vojtĕch Rödl Abstract We consider antiramsey type results. For a given coloring of the kelement subsets of an nelement set X, where two kelement
More information1. R In this and the next section we are going to study the properties of sequences of real numbers.
+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real
More informationCourse 221: Analysis Academic year , First Semester
Course 221: Analysis Academic year 200708, First Semester David R. Wilkins Copyright c David R. Wilkins 1989 2007 Contents 1 Basic Theorems of Real Analysis 1 1.1 The Least Upper Bound Principle................
More informationMath 231b Lecture 35. G. Quick
Math 231b Lecture 35 G. Quick 35. Lecture 35: Sphere bundles and the Adams conjecture 35.1. Sphere bundles. Let X be a connected finite cell complex. We saw that the Jhomomorphism could be defined by
More informationRANDOM INVOLUTIONS AND THE NUMBER OF PRIME FACTORS OF AN INTEGER 1. INTRODUCTION
RANDOM INVOLUTIONS AND THE NUMBER OF PRIME FACTORS OF AN INTEGER KIRSTEN WICKELGREN. INTRODUCTION Any positive integer n factors uniquely as n = p e pe pe 3 3 pe d d where p, p,..., p d are distinct prime
More information11 Ideals. 11.1 Revisiting Z
11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(
More informationCOMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS
COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V. EROVENKO AND B. SURY ABSTRACT. We compute commutativity degrees of wreath products A B of finite abelian groups A and B. When B
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationAn example of a computable
An example of a computable absolutely normal number Verónica Becher Santiago Figueira Abstract The first example of an absolutely normal number was given by Sierpinski in 96, twenty years before the concept
More informationOdd induced subgraphs in graphs of maximum degree three
Odd induced subgraphs in graphs of maximum degree three David M. Berman, Hong Wang, and Larry Wargo Department of Mathematics University of New Orleans New Orleans, Louisiana, USA 70148 Abstract A longstanding
More informationGeneralizing the Ramsey Problem through Diameter
Generalizing the Ramsey Problem through Diameter Dhruv Mubayi Submitted: January 8, 001; Accepted: November 13, 001. MR Subject Classifications: 05C1, 05C15, 05C35, 05C55 Abstract Given a graph G and positive
More informationFACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS
International Electronic Journal of Algebra Volume 6 (2009) 95106 FACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS Sándor Szabó Received: 11 November 2008; Revised: 13 March 2009
More informationGROUPS SUBGROUPS. Definition 1: An operation on a set G is a function : G G G.
Definition 1: GROUPS An operation on a set G is a function : G G G. Definition 2: A group is a set G which is equipped with an operation and a special element e G, called the identity, such that (i) the
More informationPartitioning edgecoloured complete graphs into monochromatic cycles and paths
arxiv:1205.5492v1 [math.co] 24 May 2012 Partitioning edgecoloured complete graphs into monochromatic cycles and paths Alexey Pokrovskiy Departement of Mathematics, London School of Economics and Political
More informationClassification of Cartan matrices
Chapter 7 Classification of Cartan matrices In this chapter we describe a classification of generalised Cartan matrices This classification can be compared as the rough classification of varieties in terms
More informationPrime Numbers and Irreducible Polynomials
Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.
More informationTHE 0/1BORSUK CONJECTURE IS GENERICALLY TRUE FOR EACH FIXED DIAMETER
THE 0/1BORSUK CONJECTURE IS GENERICALLY TRUE FOR EACH FIXED DIAMETER JONATHAN P. MCCAMMOND AND GÜNTER ZIEGLER Abstract. In 1933 Karol Borsuk asked whether every compact subset of R d can be decomposed
More informationMAT2400 Analysis I. A brief introduction to proofs, sets, and functions
MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take
More information4. CLASSES OF RINGS 4.1. Classes of Rings class operator Aclosed Example 1: product Example 2:
4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets
More informationCONTINUED FRACTIONS AND FACTORING. Niels Lauritzen
CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents
More informationGROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS
GROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS NOTES OF LECTURES GIVEN AT THE UNIVERSITY OF MYSORE ON 29 JULY, 01 AUG, 02 AUG, 2012 K. N. RAGHAVAN Abstract. The notion of the action of a group
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationCOMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS
Bull Austral Math Soc 77 (2008), 31 36 doi: 101017/S0004972708000038 COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V EROVENKO and B SURY (Received 12 April 2007) Abstract We compute
More informationSets of Fibre Homotopy Classes and Twisted Order Parameter Spaces
Communications in Mathematical Physics ManuscriptNr. (will be inserted by hand later) Sets of Fibre Homotopy Classes and Twisted Order Parameter Spaces Stefan BechtluftSachs, Marco Hien Naturwissenschaftliche
More informationDivisor graphs have arbitrary order and size
Divisor graphs have arbitrary order and size arxiv:math/0606483v1 [math.co] 20 Jun 2006 Le Anh Vinh School of Mathematics University of New South Wales Sydney 2052 Australia Abstract A divisor graph G
More informationThis chapter is all about cardinality of sets. At first this looks like a
CHAPTER Cardinality of Sets This chapter is all about cardinality of sets At first this looks like a very simple concept To find the cardinality of a set, just count its elements If A = { a, b, c, d },
More informationPrimality  Factorization
Primality  Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.
More informationComplete graphs whose topological symmetry groups are polyhedral
Algebraic & Geometric Topology 11 (2011) 1405 1433 1405 Complete graphs whose topological symmetry groups are polyhedral ERICA FLAPAN BLAKE MELLOR RAMIN NAIMI We determine for which m the complete graph
More informationBasic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011
Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely
More informationPOLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS
POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).
More informationLabeling outerplanar graphs with maximum degree three
Labeling outerplanar graphs with maximum degree three Xiangwen Li 1 and Sanming Zhou 2 1 Department of Mathematics Huazhong Normal University, Wuhan 430079, China 2 Department of Mathematics and Statistics
More informationTreerepresentation of set families and applications to combinatorial decompositions
Treerepresentation of set families and applications to combinatorial decompositions BinhMinh BuiXuan a, Michel Habib b Michaël Rao c a Department of Informatics, University of Bergen, Norway. buixuan@ii.uib.no
More informationOn the Union of Arithmetic Progressions
On the Union of Arithmetic Progressions Shoni Gilboa Rom Pinchasi August, 04 Abstract We show that for any integer n and real ɛ > 0, the union of n arithmetic progressions with pairwise distinct differences,
More informationModule MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013
Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of
More informationSOLUTIONS TO ASSIGNMENT 1 MATH 576
SOLUTIONS TO ASSIGNMENT 1 MATH 576 SOLUTIONS BY OLIVIER MARTIN 13 #5. Let T be the topology generated by A on X. We want to show T = J B J where B is the set of all topologies J on X with A J. This amounts
More informationElements of Abstract Group Theory
Chapter 2 Elements of Abstract Group Theory Mathematics is a game played according to certain simple rules with meaningless marks on paper. David Hilbert The importance of symmetry in physics, and for
More informationPrime Numbers. Chapter Primes and Composites
Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are
More informationLecture 13  Basic Number Theory.
Lecture 13  Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are nonnegative integers. We say that A divides B, denoted
More informationA NOTE ON TRIVIAL FIBRATIONS
A NOTE ON TRIVIAL FIBRATIONS Petar Pavešić Fakulteta za Matematiko in Fiziko, Univerza v Ljubljani, Jadranska 19, 1111 Ljubljana, Slovenija petar.pavesic@fmf.unilj.si Abstract We study the conditions
More information