DOCSLIB.ORG

S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 1. Let Sn be the group of permutations on n elements f1; 2; 3; : : : ; ng. Let An be the subgroup of even permutations. Prove that An is normal subgroup of Sn. 2. Let S15 be the group of permutations on 15 elements f1; 2; 3;:::; 15g. Prove that S15 has elements of order 56 but does not have any elements of orders 49 or 50. Answer: Use the following facts: (a) Order of a k-cycle is k, i.e. jαj = k for α a k-cycle. (b) jαβj = lcm(jαj; jβj) if α and β are disjoint permutations. • S15 has elements of order 56: Let σ = αβ, where α is an 8-cycle and β is a 7-cycle. Then jσj = jαβj = lcm(jαj; jβj) = lcm(8; 7) = 56. • S15 does not have any elements of orders 49: In order for lcm(m; n) = 49, either m = 49 or n = 49. Therefore we would need a cycle of order 49, hence need a 49-cycle in S15, which is impossible. • S15 does not have any elements of orders 50: In order for lcm(m; n) = 50, either m or n 2 must be a multiple of 5 = 25. Therefore we would need a 25-cycle or 50-cycle in S15, which is impossible. 3. Prove that a factor group of a cyclic group is cyclic. Answer: Recall: A group G is cyclic if it can be generated by one element, i.e. if there exists an element a 2 G such that G =< a > (this means that all elements of G are of the form ai for some integer i.) Recall: Elements of a factor group G=H are left cosets fgH j g 2 G. Proof: Suppose G =< a >. Let G=H be any factor group of G. WTS: G=H is cyclic. Any element of G=H is of the form gH for some g 2 G. Since G is cyclic, there is an integer i such that g = ai. So gH = aiH. aiH = (aH)i by definition of multiplication in factor groups. Therefore gH = (aH)i for any coset gH. So G=H is cyclic, by definition of cyclic groups. 4. Prove that a factor group of an Abelian group is Abelian. Answer: Recall: A group G is Abelian if ab = ba for all a; b 2 G. Proof: Suppose G is Abelian. Let G=H be any factor group of G. WTS: G=H is Abelian. Let aH; bH be any elements of G=H. Then (aH)(bH) = (ab)H by definition of multiplication in factor groups. Then (ab)H = (ba)H since G is abelian. (ba)H = (bH)(aH) again definition of multiplication in factor group. Therefore (aH)(bH) = (bH)(aH). Therefore G=H is Abelian. 1 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 5. Prove that any subgroup of an Abelian group is normal subgroup. Answer: Recall: A subgroup H of a group G is called normal if gH = Hg for every g 2 G. Proof: Suppose G is Abelian. Let H be a subgroup of G. WTS: gH = Hg for all g 2 G. Let g 2 G. Then gH = fgh j h 2 Hg by definition of left coset. gh = hg for all h since G is Abelian. Therefore fgh j h 2 Hg = fhg j h 2 Hg = Hg by definition of right coset Hg. Therefore gH = Hg for all g 2 G. Therefore H is normal subgroup of G. 6. Let H =< 5 > be the subgroup of G = Z generated by 5. (a) List the elements of H = h5i. Answer: Notice: Operation in G = Z is addition. H = h5i = f0; ±5; ±10;::: g = f5k j k 2 Zg (b) Prove that H is normal subgroup of G. Answer: Addition in Z is commutative. So G = (Z; +) is Abelian group and by previous problem every subgroup of an Abelian group is normal. (c) List the elements of the factor group G=H = Z= < 5 >. Answer: Elements of G=H are left cosets of H in G: 0 + H = 0 + h5i = f0 + 0; 0 ± 5; 0 ± 10;::: g = f0; ±5; ±10;::: g = H 1 + H = 1 + h5i = f1 + 0; 1 ± 5; 1 ± 10;::: g = f:::; −9; −4; 1; 6; 11;::: g 2 + H = 2 + h5i = f2 + 0; 2 ± 5; 2 ± 10;::: g = f:::; −8; −3; 2; 7; 12;::: g 3 + H = 3 + h5i = f3 + 0; 3 ± 5; 3 ± 10;::: g = f:::; −7; −2; 3; 8; 13;::: g 4 + H = 4 + h5i = f4 + 0; 4 ± 5; 4 ± 10;::: g = f:::; −6; −1; 4; 9; 14;::: g (d) Write the Cayley table of G=H = Z= h5i. G=H = Z= h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 1 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 2 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 3 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 4 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i (e) What is the order of 2 + h5i in Z= h5i? Answer: j2 + h5i j = 5 as 2 + h5i 2 Z= h5i. 2 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 7. Let K = h15i be the subgroup of G = Z generated by 15. (a) List the elements of K = h15i. Answer: K = h15i = f15k j k 2 Zg (b) Prove that K is normal subgroup of G. Proof: (Z +) is Abelian group and any subgroup of an Abelian group is normal (from 5). (c) List the elements of the factor group G=K = Z= h15i. Answer: G=K = Z= h15i = fi + h15i j 0 ≤ i ≤ 14g. (There are 15 elements.) !!! This is just one way of expressing these cosets. Notice that there are many ways of expressing the same coset, eg. (2+h15i) = (17+h15i) = (32+h15i) = (−13+h15i) = ::: . (d) Write the Cayley table of G=K = Z= h15i. Answer: Don't have to actually write it, but make sure that you know what it would look like. (e) What is the order of 3 + K = 3 + h15i in Z= h15i? Answer: j3 + h15i j = 5 in Z= h15i, since (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) = (15 + h15i) = (0 + h15i). (f) What is the order of 4 + K = 4 + h15i in Z= h15i? Answer: j4 + h15i j = 15 in Z= h15i. (g) What is the order of 5 + K = 5 + h15i in Z= h15i? Answer: j5 + h15i j = 3 in Z= h15i. (h) What is the order of 6 + K = 6 + h15i in Z= h15i? Answer: j6 + h15i j = 5 in Z= h15i, (i) Prove that G=K is cyclic. Answer: Z= h15i is generated by 1 + h15i, hence it is cyclic. underlineAnswer 2: Z= h15i is cyclic since it is factor of the cyclic group (Z; +) (this group is generated by 1). 3 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: (j) Prove that G=K = Z= h15i is isomorphic to Z15. Answer: • Elements of Z= h15i are left cosets, and operation is addition of cosets (as defined for factor groups!) • Elements of Z15: f0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14g and addition is modulo 15. • Define the map: f : Z= h15i ! Z15 by f(i + h15i) = ri, where ri is the remainder after dividing i by 15. { Notice that f, as defined, is a mapping from Z= h15i to Z15 since the values of f are the remainders after dividing by 15, hence they are non-negative integers between 0 and 14. { Prove that f is well-defined: Suppose the same coset (i + h15i) is given by another integer j, i.e. (i + h15i) = (j + h15i). Then f(i + h15i) = ri 2 Z15 and f(j + h15i) = rj 2 Z15. To show that f is well defined we must show that f(i + h15i) = f(j + h15i), i.e. WTS ri = rj 2 Z15. From (i+h15i) = (j +h15i) it follows that j 2 (i+h15i) and therefore j = i+k15 for some integer k 2 Z. By definition of remainders: i = n·15+ri and j = m·15+rj, with 0 ≤ ri; rj < 15. Therefore from j = i + k15, we have m · 15 + rj = n · 15 + ri + k15. Hence rj − ri = (k + n − m) ·15 which is an integer multiple of 15. Since the remainders are 0 ≤ ri; rj < 15, it follows that rj = ri. Therefore f is a well defined function. { Let f : Z= h15i ! Z15 defined by f(i+h15i) = ri, where ri is the remainder after dividing i by 15. Then f is "one-to-one" function. Proof: Suppose there are a; b 2 Z= h15i such that f(a) = f(b). WTS a = b. From a; b 2 Z= h15i, it follows that a = i + h15i and b = j + h15i. From f(a) = f(b) it follows f(i + h15i) = f(j + h15i), and by definition of f it follows that ri = rj. So i = n · 15 + ri and j = m · 15 + rj. Therefore i − j is a multiple of 15, hence and element of the subgroup h15i. Therefore a = i + h15i = j + h15i = b. Therefore f is a one-to-one function. { f : Z= h15i ! Z15 is onto. Proof: Let y 2 Z15. WTS: there is an x 2 Z= h15i such that f(x) = y.
Load more

Recommended publications
  • Classification of Finite Abelian Groups
    Math 317 C1 John Sullivan Spring 2003 Classification of Finite Abelian Groups (Notes based on an article by Navarro in the Amer. Math. Monthly, February 2003.) The fundamental theorem of finite abelian groups expresses any such group as a product of cyclic groups: Theorem. Suppose G is a finite abelian group. Then G is (in a unique way) a direct product of cyclic groups of order pk with p prime. Our first step will be a special case of Cauchy’s Theorem, which we will prove later for arbitrary groups: whenever p |G| then G has an element of order p. Theorem (Cauchy). If G is a finite group, and p |G| is a prime, then G has an element of order p (or, equivalently, a subgroup of order p). ∼ Proof when G is abelian. First note that if |G| is prime, then G = Zp and we are done. In general, we work by induction. If G has no nontrivial proper subgroups, it must be a prime cyclic group, the case we’ve already handled. So we can suppose there is a nontrivial subgroup H smaller than G. Either p |H| or p |G/H|. In the first case, by induction, H has an element of order p which is also order p in G so we’re done. In the second case, if ∼ g + H has order p in G/H then |g + H| |g|, so hgi = Zkp for some k, and then kg ∈ G has order p. Note that we write our abelian groups additively. Definition. Given a prime p, a p-group is a group in which every element has order pk for some k.
    [Show full text]
  • Mathematics 310 Examination 1 Answers 1. (10 Points) Let G Be A
    Mathematics 310 Examination 1 Answers 1. (10 points) Let G be a group, and let x be an element of G. Finish the following definition: The order of x is ... Answer: . the smallest positive integer n so that xn = e. 2. (10 points) State Lagrange’s Theorem. Answer: If G is a finite group, and H is a subgroup of G, then o(H)|o(G). 3. (10 points) Let ( a 0! ) H = : a, b ∈ Z, ab 6= 0 . 0 b Is H a group with the binary operation of matrix multiplication? Be sure to explain your answer fully. 2 0! 1/2 0 ! Answer: This is not a group. The inverse of the matrix is , which is not 0 2 0 1/2 in H. 4. (20 points) Suppose that G1 and G2 are groups, and φ : G1 → G2 is a homomorphism. (a) Recall that we defined φ(G1) = {φ(g1): g1 ∈ G1}. Show that φ(G1) is a subgroup of G2. −1 (b) Suppose that H2 is a subgroup of G2. Recall that we defined φ (H2) = {g1 ∈ G1 : −1 φ(g1) ∈ H2}. Prove that φ (H2) is a subgroup of G1. Answer:(a) Pick x, y ∈ φ(G1). Then we can write x = φ(a) and y = φ(b), with a, b ∈ G1. Because G1 is closed under the group operation, we know that ab ∈ G1. Because φ is a homomorphism, we know that xy = φ(a)φ(b) = φ(ab), and therefore xy ∈ φ(G1). That shows that φ(G1) is closed under the group operation.
    [Show full text]
  • The General Linear Group
    18.704 Gabe Cunningham 2/18/05 [email protected] The General Linear Group Definition: Let F be a field. Then the general linear group GLn(F ) is the group of invert- ible n × n matrices with entries in F under matrix multiplication. It is easy to see that GLn(F ) is, in fact, a group: matrix multiplication is associative; the identity element is In, the n × n matrix with 1’s along the main diagonal and 0’s everywhere else; and the matrices are invertible by choice. It’s not immediately clear whether GLn(F ) has infinitely many elements when F does. However, such is the case. Let a ∈ F , a 6= 0. −1 Then a · In is an invertible n × n matrix with inverse a · In. In fact, the set of all such × matrices forms a subgroup of GLn(F ) that is isomorphic to F = F \{0}. It is clear that if F is a finite field, then GLn(F ) has only finitely many elements. An interesting question to ask is how many elements it has. Before addressing that question fully, let’s look at some examples. ∼ × Example 1: Let n = 1. Then GLn(Fq) = Fq , which has q − 1 elements. a b Example 2: Let n = 2; let M = ( c d ). Then for M to be invertible, it is necessary and sufficient that ad 6= bc. If a, b, c, and d are all nonzero, then we can fix a, b, and c arbitrarily, and d can be anything but a−1bc. This gives us (q − 1)3(q − 2) matrices.
    [Show full text]
  • On Some Generation Methods of Finite Simple Groups
    Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography On Some Generation Methods of Finite Simple Groups Ayoub B. M. Basheer Department of Mathematical Sciences, North-West University (Mafikeng), P Bag X2046, Mmabatho 2735, South Africa Groups St Andrews 2017 in Birmingham, School of Mathematics, University of Birmingham, United Kingdom 11th of August 2017 Ayoub Basheer, North-West University, South Africa Groups St Andrews 2017 Talk in Birmingham Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography Abstract In this talk we consider some methods of generating finite simple groups with the focus on ranks of classes, (p; q; r)-generation and spread (exact) of finite simple groups. We show some examples of results that were established by the author and his supervisor, Professor J. Moori on generations of some finite simple groups. Ayoub Basheer, North-West University, South Africa Groups St Andrews 2017 Talk in Birmingham Introduction Preliminaries Special Kind of Generation of Finite Simple Groups The Bibliography Introduction Generation of finite groups by suitable subsets is of great interest and has many applications to groups and their representations. For example, Di Martino and et al. [39] established a useful connection between generation of groups by conjugate elements and the existence of elements representable by almost cyclic matrices. Their motivation was to study irreducible projective representations of the sporadic simple groups. In view of applications, it is often important to exhibit generating pairs of some special kind, such as generators carrying a geometric meaning, generators of some prescribed order, generators that offer an economical presentation of the group.
    [Show full text]
  • Orders on Computable Torsion-Free Abelian Groups
    Orders on Computable Torsion-Free Abelian Groups Asher M. Kach (Joint Work with Karen Lange and Reed Solomon) University of Chicago 12th Asian Logic Conference Victoria University of Wellington December 2011 Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 1 / 24 Outline 1 Classical Algebra Background 2 Computing a Basis 3 Computing an Order With A Basis Without A Basis 4 Open Questions Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 2 / 24 Torsion-Free Abelian Groups Remark Disclaimer: Hereout, the word group will always refer to a countable torsion-free abelian group. The words computable group will always refer to a (fixed) computable presentation. Definition A group G = (G : +; 0) is torsion-free if non-zero multiples of non-zero elements are non-zero, i.e., if (8x 2 G)(8n 2 !)[x 6= 0 ^ n 6= 0 =) nx 6= 0] : Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 3 / 24 Rank Theorem A countable abelian group is torsion-free if and only if it is a subgroup ! of Q . Definition The rank of a countable torsion-free abelian group G is the least κ cardinal κ such that G is a subgroup of Q . Asher M. Kach (U of C) Orders on Computable TFAGs ALC 2011 4 / 24 Example The subgroup H of Q ⊕ Q (viewed as having generators b1 and b2) b1+b2 generated by b1, b2, and 2 b1+b2 So elements of H look like β1b1 + β2b2 + α 2 for β1; β2; α 2 Z.
    [Show full text]
  • GROUP ACTIONS 1. Introduction the Groups Sn, An, and (For N ≥ 3)
    GROUP ACTIONS KEITH CONRAD 1. Introduction The groups Sn, An, and (for n ≥ 3) Dn behave, by their definitions, as permutations on certain sets. The groups Sn and An both permute the set f1; 2; : : : ; ng and Dn can be considered as a group of permutations of a regular n-gon, or even just of its n vertices, since rigid motions of the vertices determine where the rest of the n-gon goes. If we label the vertices of the n-gon in a definite manner by the numbers from 1 to n then we can view Dn as a subgroup of Sn. For instance, the labeling of the square below lets us regard the 90 degree counterclockwise rotation r in D4 as (1234) and the reflection s across the horizontal line bisecting the square as (24). The rest of the elements of D4, as permutations of the vertices, are in the table below the square. 2 3 1 4 1 r r2 r3 s rs r2s r3s (1) (1234) (13)(24) (1432) (24) (12)(34) (13) (14)(23) If we label the vertices in a different way (e.g., swap the labels 1 and 2), we turn the elements of D4 into a different subgroup of S4. More abstractly, if we are given a set X (not necessarily the set of vertices of a square), then the set Sym(X) of all permutations of X is a group under composition, and the subgroup Alt(X) of even permutations of X is a group under composition. If we list the elements of X in a definite order, say as X = fx1; : : : ; xng, then we can think about Sym(X) as Sn and Alt(X) as An, but a listing in a different order leads to different identifications 1 of Sym(X) with Sn and Alt(X) with An.
    [Show full text]
  • Mathematics of the Rubik's Cube
    Mathematics of the Rubik's cube Associate Professor W. D. Joyner Spring Semester, 1996{7 2 \By and large it is uniformly true that in mathematics that there is a time lapse between a mathematical discovery and the moment it becomes useful; and that this lapse can be anything from 30 to 100 years, in some cases even more; and that the whole system seems to function without any direction, without any reference to usefulness, and without any desire to do things which are useful." John von Neumann COLLECTED WORKS, VI, p. 489 For more mathematical quotes, see the first page of each chapter below, [M], [S] or the www page at http://math.furman.edu/~mwoodard/mquot. html 3 \There are some things which cannot be learned quickly, and time, which is all we have, must be paid heavily for their acquiring. They are the very simplest things, and because it takes a man's life to know them the little new that each man gets from life is very costly and the only heritage he has to leave." Ernest Hemingway (From A. E. Hotchner, PAPA HEMMINGWAY, Random House, NY, 1966) 4 Contents 0 Introduction 13 1 Logic and sets 15 1.1 Logic................................ 15 1.1.1 Expressing an everyday sentence symbolically..... 18 1.2 Sets................................ 19 2 Functions, matrices, relations and counting 23 2.1 Functions............................. 23 2.2 Functions on vectors....................... 28 2.2.1 History........................... 28 2.2.2 3 × 3 matrices....................... 29 2.2.3 Matrix multiplication, inverses.............. 30 2.2.4 Muliplication and inverses...............
    [Show full text]
  • The Classical Groups and Domains 1. the Disk, Upper Half-Plane, SL 2(R
    (June 8, 2018) The Classical Groups and Domains Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ The complex unit disk D = fz 2 C : jzj < 1g has four families of generalizations to bounded open subsets in Cn with groups acting transitively upon them. Such domains, defined more precisely below, are bounded symmetric domains. First, we recall some standard facts about the unit disk, the upper half-plane, the ambient complex projective line, and corresponding groups acting by linear fractional (M¨obius)transformations. Happily, many of the higher- dimensional bounded symmetric domains behave in a manner that is a simple extension of this simplest case. 1. The disk, upper half-plane, SL2(R), and U(1; 1) 2. Classical groups over C and over R 3. The four families of self-adjoint cones 4. The four families of classical domains 5. Harish-Chandra's and Borel's realization of domains 1. The disk, upper half-plane, SL2(R), and U(1; 1) The group a b GL ( ) = f : a; b; c; d 2 ; ad − bc 6= 0g 2 C c d C acts on the extended complex plane C [ 1 by linear fractional transformations a b az + b (z) = c d cz + d with the traditional natural convention about arithmetic with 1. But we can be more precise, in a form helpful for higher-dimensional cases: introduce homogeneous coordinates for the complex projective line P1, by defining P1 to be a set of cosets u 1 = f : not both u; v are 0g= × = 2 − f0g = × P v C C C where C× acts by scalar multiplication.
    [Show full text]
  • Semigroups and Monoids 
    S Luis Alonso-Ovalle // Contents Subgroups Semigroups and monoids Subgroups Groups. A group G is an algebra consisting of a set G and a single binary operation ◦ satisfying the following axioms: . ◦ is completely defined and G is closed under ◦. ◦ is associative. G contains an identity element. Each element in G has an inverse element. Subgroups. We define a subgroup G0 as a subalgebra of G which is itself a group. Examples: . The group of even integers with addition is a proper subgroup of the group of all integers with addition. The group of all rotations of the square h{I, R, R0, R00}, ◦i, where ◦ is the composition of the operations is a subgroup of the group of all symmetries of the square. Some non-subgroups: SEMIGROUPS AND MONOIDS . The system h{I, R, R0}, ◦i is not a subgroup (and not even a subalgebra) of the original group. Why? (Hint: ◦ closure). The set of all non-negative integers with addition is a subalgebra of the group of all integers with addition, because the non-negative integers are closed under addition. But it is not a subgroup because it is not itself a group: it is associative and has a zero, but . does any member (except for ) have an inverse? Order. The order of any group G is the number of members in the set G. The order of any subgroup exactly divides the order of the parental group. E.g.: only subgroups of order , , and are possible for a -member group. (The theorem does not guarantee that every subset having the proper number of members will give rise to a subgroup.
    [Show full text]
  • 1.6 Cyclic Subgroups
    1.6. CYCLIC SUBGROUPS 19 1.6 Cyclic Subgroups Recall: cyclic subgroup, cyclic group, generator. Def 1.68. Let G be a group and a ∈ G. If the cyclic subgroup hai is finite, then the order of a is |hai|. Otherwise, a is of infinite order. 1.6.1 Elementary Properties Thm 1.69. Every cyclic group is abelian. Thm 1.70. If m ∈ Z+ and n ∈ Z, then there exist unique q, r ∈ Z such that n = mq + r and 0 ≤ r ≤ m. n In fact, q = b m c and r = n − mq. Here bxc denotes the maximal integer no more than x. Ex 1.71 (Ex 6.4, Ex 6.5, p60). 1. Find the quotient q and the remainder r when n = 38 is divided by m = 7. 2. Find the quotient q and the remainder r when n = −38 is divided by m = 7. Thm 1.72 (Important). A subgroup of a cyclic group is cyclic. Proof. (refer to the book) Ex 1.73. The subgroups of hZ, +i are precisely hnZ, +i for n ∈ Z. Def 1.74. Let r, s ∈ Z. The greatest common divisor (gcd) of r and s is the largest positive integer d that divides both r and s. Written as d = gcd(r, s). In fact, d is the positive generator of the following cyclic subgroup of Z: hdi = {nr + ms | n, m ∈ Z}. So d is the smallest positive integer that can be written as nr + ms for some n, m ∈ Z. Ex 1.75. gcd(36, 63) = 9, gcd(36, 49) = 1.
    [Show full text]
  • 12.6 Further Topics on Simple Groups 387 12.6 Further Topics on Simple Groups
    12.6 Further Topics on Simple groups 387 12.6 Further Topics on Simple Groups This Web Section has three parts (a), (b) and (c). Part (a) gives a brief descriptions of the 56 (isomorphism classes of) simple groups of order less than 106, part (b) provides a second proof of the simplicity of the linear groups Ln(q), and part (c) discusses an ingenious method for constructing a version of the Steiner system S(5, 6, 12) from which several versions of S(4, 5, 11), the system for M11, can be computed. 12.6(a) Simple Groups of Order less than 106 The table below and the notes on the following five pages lists the basic facts concerning the non-Abelian simple groups of order less than 106. Further details are given in the Atlas (1985), note that some of the most interesting and important groups, for example the Mathieu group M24, have orders in excess of 108 and in many cases considerably more. Simple Order Prime Schur Outer Min Simple Order Prime Schur Outer Min group factor multi. auto. simple or group factor multi. auto. simple or count group group N-group count group group N-group ? A5 60 4 C2 C2 m-s L2(73) 194472 7 C2 C2 m-s ? 2 A6 360 6 C6 C2 N-g L2(79) 246480 8 C2 C2 N-g A7 2520 7 C6 C2 N-g L2(64) 262080 11 hei C6 N-g ? A8 20160 10 C2 C2 - L2(81) 265680 10 C2 C2 × C4 N-g A9 181440 12 C2 C2 - L2(83) 285852 6 C2 C2 m-s ? L2(4) 60 4 C2 C2 m-s L2(89) 352440 8 C2 C2 N-g ? L2(5) 60 4 C2 C2 m-s L2(97) 456288 9 C2 C2 m-s ? L2(7) 168 5 C2 C2 m-s L2(101) 515100 7 C2 C2 N-g ? 2 L2(9) 360 6 C6 C2 N-g L2(103) 546312 7 C2 C2 m-s L2(8) 504 6 C2 C3 m-s
    [Show full text]
  • And Free Cyclic Group Actions on Homotopy Spheres
    TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 220, 1976 DECOMPOSABILITYOF HOMOTOPYLENS SPACES ANDFREE CYCLICGROUP ACTIONS ON HOMOTOPYSPHERES BY KAI WANG ABSTRACT. Let p be a linear Zn action on C and let p also denote the induced Z„ action on S2p~l x D2q, D2p x S2q~l and S2p~l x S2q~l " 1m_1 where p = [m/2] and q = m —p. A free differentiable Zn action (£ , ju) on a homotopy sphere is p-decomposable if there is an equivariant diffeomor- phism <t>of (S2p~l x S2q~l, p) such that (S2m_1, ju) is equivalent to (£(*), ¿(*)) where S(*) = S2p_1 x D2q U^, D2p x S2q~l and A(<P) is a uniquely determined action on S(*) such that i4(*)IS p~l XD q = p and A(Q)\D p X S = p. A homotopy lens space is p-decomposable if it is the orbit space of a p-decomposable free Zn action on a homotopy sphere. In this paper, we will study the decomposabilities of homotopy lens spaces. We will also prove that for each lens space L , there exist infinitely many inequivalent free Zn actions on S m such that the orbit spaces are simple homotopy equiva- lent to L 0. Introduction. Let A be the antipodal map and let $ be an equivariant diffeomorphism of (Sp x Sp, A) where A(x, y) = (-x, -y). Then there is a uniquely determined free involution A($) on 2(4>) where 2(4») = Sp x Dp+1 U<¡,Dp+l x Sp such that the inclusions S" x Dp+l —+ 2(d>), Dp+1 x Sp —*■2(4>) are equi- variant.
    [Show full text]