DOCSLIB.ORG

The Linear Algebra of Free Abelian Groups

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
The Linear Algebra of Free Abelian Groups The Linear Algebra of Free Abelian Groups Recall that a non-trivial free abelian group with basis X = {x1, x2, ··· , xd} is iso- morphic to Z × Z × · · · × Z, where we have d factors of Z. To simplify notation, we will consider the free abelian groups Z × Z × Z × Z and Z × Z × Z with bases X = {x1, x2, x3, x4} and Y = {y1, y2, y3}, respectively. However, the results in this note are valid in general. A function φ : Z × Z × Z × Z → Z × Z × Z will be a homomorphism exactly if φ(α1x1 + α2x2 + α3x3 + α4x4) = α1φ(x1) + α2φ(x2) + α3φ(x3) + α4φ(x4) for all αi ∈ Z. That is, φ will be a homomorphism exactly if it is linear with respect to integer coefficients. Since φ is determined by its values on the generators of its domain, that is, by the values φ(xi), we only have to express each such φ(xi) in terms of the basis Y , in order to fully describe φ. Since Y = {y1, y2, y3} is a basis for Z × Z × Z, there are unique integers aji such that φ(x1) = a11y1 + a21y2 + a31y3 φ(x2) = a12y1 + a22y2 + a32y3 φ(x3) = a13y1 + a23y2 + a33y3 (1) φ(x4) = a14y1 + a24y2 + a34y3 We assemble the transpose of these entries into a matrix a11 a12 a13 a14 A = a21 a22 a23 a24 . a31 a32 a33 a34 Proposition 1. Let g ∈ Z × Z × Z × Z. Then φ(g) can be computed as follows. Express g = α1x1 + α2x2 + α3x3 + α4x4 in the basis X and compute α1 a11 a12 a13 a14 β1 α2 a21 a22 a23 a24 = β2 . α3 a31 a32 a33 a34 β3 α4 Then φ(g) = β1y1 + β2y2 + β3y3 in the basis Y . Proof. 4 ! 4 4 3 X X X X φ(g) = φ αixi = αiφ(xi) = αi ajiyj i=1 i=1 i=1 j=1 3 4 ! 3 X X X = ajiαi yj = βjyj j=1 i=1 j=1 1 Proposition 2. Changing either of the bases X = {x1, x2, x3, x4} or Y = {y1, y2, y3} has the following effect on the matrix A. For simplicity, we choose specific locations for the changes to occur. However, the results clearly remain valid in general. • Replacing X by X = {x1, x2, −x3, x4} changes the entries of the third column of A to their additive inverse. • Replacing Y by Y = {y1, −y2, y3} changes the entries of the second row of A to their additive inverse. • Replacing X by X = {x3, x2, x1, x4} interchanges column one and column three. • Replacing Y by Y = {y3, y2, y1} interchanges row one and three. • Replacing X by X = {x1, x2, x3 + 5x1, x4} adds 5 times column one to column three. • Caution: replacing Y by Y = {y1 − 5y3, y2, y3} adds 5 times row one to row three. Proof. We only prove the last two statements. The remaining four statements are left as an exercise. Suppose we replace X by X = {x1, x2, x3 + 5x1, x4}. Then Equation (1) turns into φ(x1) = a11y1 + a21y2 + a31y3 φ(x2) = a12y1 + a22y2 + a32y3 φ(x3 + 5x1) = (a13 + 5a11)y1 + (a23 + 5a21)y2 + (a33 + 5a31)y3 φ(x4) = a14y1 + a24y2 + a34y3 So, the entries of A change as predicted, since A is the transpose of these entries. However, if replace Y by Y = {y1 − 5y3, y2, y3}, we get φ(x1) = a11(y1 − 5y3) + a21y2 + (a31 + 5a11)y3 φ(x2) = a12(y1 − 5y3) + a22y2 + (a32 + 5a12)y3 φ(x3) = a13(y1 − 5y3) + a23y2 + (a33 + 5a13)y3 φ(x4) = a14(y1 − 5y3) + a24y2 + (a34 + 5a14)y3, which amounts to adding five times row one to row three in the matrix A. Remark. There is no change of bases that would result in a row or a column to be multiplied by an integer other than ±1. 2 Proposition 3. Using the above changes of bases, the matrix A can be brought into the following form: n1 0 0 0 0 n2 0 0 , 0 0 n3 0 with ni > 0 and n1|n2|n3. Proof. We may assume that there is an entry different from zero. Let a be a non- zero entry of the matrix. By interchanging rows and columns we can move a into the upper left corner of the matrix. Make a positive by changing the entire first row to their additive inverse, if necessary. Now, if b is any entry in the first column, below a, we can divide b = qa + r with 0 6 r < a. By adding (−q) times the first row to the row containing b, we can therefore make r appear in place of b. If r is not zero, then it is smaller than a. If that happens, we switch rows, so as to place r into the upper left corner instead of a. Every time we do this, the entry in the upper left corner decreases. So, this cannot go on forever. We must reach a point where the remainder r is zero for all the entries in the first row below the upper left corner. So, we can now assume that we have zeroes in the entire first column below a. Arguing in the other direction, we can arrange for the first row to consist of zeroes, to the right of a. We now arrange for a to divide all the remaining entries in the matrix. Let c be any entry of the matrix, other than a. We divide with remainder: c = qa + r where 0 6 r < a. Then we apply some clever row and column operations to replace c by r. (Which ones?) If r is not zero, we can move it into the upper left corner of the matrix where it will replace a. Again, we clear all entries below and to the right. This procedure keeps reducing the size of our upper left entry. So, sooner or later, we can only get remainders of zero. We have now found n1, namely n1 = a, because we have reached the point, where the matrix looks like this n1 0 0 0 0 0 0 0 a22 a23 a24 0 0 0 0 a32 a33 a34 and n1 divides all other entries. To finish the proof, simply apply the same logic to the smaller matrix 0 0 0 a22 a23 a24 0 0 0 . a32 a33 a34 3 Proposition 4. Suppose there are bases with respect to which the homomorphism φ : Z × Z × Z × Z → Z × Z × Z has a matrix of the form n1 0 0 0 0 n2 0 0 , 0 0 n3 0 with ni > 0 (and n1|n2|n3). Then Z × Z × Z/Im φ is isomorphic to Zn1 × Zn2 × Zn3 . Proof. Let X = {x1, x2, x3, x4} and Y = {y1, y2, y3} be bases for Z × Z × Z × Z and Z × Z × Z, respectively, with respect to which φ has the given matrix. This means that φ(x1) = n1y1, φ(x2) = n2y2, φ(x3) = n3y3 and φ(x4) = 0. Since the basis X = {x1, x2, x3, x4} generates the domain of φ, we see that {φ(x1), φ(x2), φ(x3)} generates its image H = Im φ. That is, H = h{n1y1, n2y2, n3y3}i. Since Z × Z × Z is abelian, H is a normal subgroup of Z × Z × Z. Consider the homomorphism f : Z × Z × Z → Z × Z × Z/H given by f(α1, α2, α3) = α1y1 + α2y2 + α3y3 + H. Then f is onto, because Y = {y1, y2, y3} is a basis for Z × Z × Z. We claim that the kernel of f is generated by the elements (n1, 0, 0), (0, n2, 0), (0, 0, n3), which will finish the proof by the First Isomorphism Theorem, because then ∼ ∼ Zn1 × Zn2 × Zn3 = Z × Z × Z/ h{(n1, 0, 0), (0, n2, 0), (0, 0, n3)}i = Z × Z × Z/H. Observe that each of these elements is in the kernel of f. For example, f(n1, 0, 0) = n1y1 + H = 0 + H, since n1y1 ∈ H. Hence every combination of the elements (n1, 0, 0), (0, n2, 0), (0, 0, n3) is in the kernel of f. Conversely, if f(α1, α2, α3) = 0 + H, then α1y1 + α2y2 + α3y3 + H = 0 + H. So, α1y1 + α2y2 + α3y3 ∈ H, which implies that α1y1 + α2y2 + α3y3 = γ1n1y1 + γ2n2y2 + γ3n3y3 for some γi ∈ Z. However, since Y = {y1, y2, y3} is a basis, this implies that αi = γini for i = 1, 2, 3. Hence (α1, α2, α3) = γ1(n1, 0, 0) + γ2(0, n2, 0) + γ3(0, 0, n3). Remark. Since 1 divides every integer and every integer divides 0, it might happen that some of the initial ni’s are 1 and that some of the ni’s at the end are 0. Note ∼ that Z1 = {0}, whereas Z0 = Z. Remark. Other scenarios are handled similarly. Suppose, the matrix for a homo- morphism φ : Z × Z × Z → Z × Z × Z × Z reduces to n1 0 0 0 n2 0 , 0 0 n3 0 0 0 ∼ with ni > 0 (and n1|n2|n3). Then Z×Z×Z×Z/Im φ = Zn1 ×Zn2 ×Zn3 ×Z, because X = {x1, x2, x3}, Y = {y1, y2, y3, y4}, and Im φ = h{n1y1, n2y2, n3y3, 0y4}i. 4 Example. Classify Z3 × Z × Z6/ h{([3], 4, [2]), ([2], 0, [5])}i in invariant factor form. ∼ Step 1: G = Z × Z × Z/H with H = h{(3, 0, 0), (0, 0, 6), (3, 4, 2), (2, 0, 5)}i. Step 2: Consider the homomorphism φ : Z × Z × Z × Z → Z × Z × Z given by φ(α1, α2, α3, α4) = α1(3, 0, 0) + α2(0, 0, 6) + α3(3, 4, 2) + α4(2, 0, 5).
Load more

Recommended publications
  • Faithful Abelian Groups of Infinite Rank Ulrich Albrecht
    PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 103, Number 1, May 1988 FAITHFUL ABELIAN GROUPS OF INFINITE RANK ULRICH ALBRECHT (Communicated by Bhama Srinivasan) ABSTRACT. Let B be a subgroup of an abelian group G such that G/B is isomorphic to a direct sum of copies of an abelian group A. For B to be a direct summand of G, it is necessary that G be generated by B and all homomorphic images of A in G. However, if the functor Hom(A, —) preserves direct sums of copies of A, then this condition is sufficient too if and only if M ®e(A) A is nonzero for all nonzero right ¿ï(A)-modules M. Several examples and related results are given. 1. Introduction. There are only very few criteria for the splitting of exact sequences of torsion-free abelian groups. The most widely used of these was given by Baer in 1937 [F, Proposition 86.5]: If G is a pure subgroup of a torsion-free abelian group G, such that G/C is homogeneous completely decomposable of type f, and all elements of G\C are of type r, then G is a direct summand of G. Because of its numerous applications, many attempts have been made to extend the last result to situations in which G/C is not completely decomposable. Arnold and Lady succeeded in 1975 in the case that G is torsion-free of finite rank. Before we can state their result, we introduce some additional notation: Suppose that A and G are abelian groups.
    [Show full text]
  • Algebraic Number Theory
    Algebraic Number Theory Sergey Shpectorov January{March, 2010 This course in on algebraic number theory. This means studying problems from number theory with methods from abstract algebra. For a long time the main motivation behind the development of algebraic number theory was the Fermat Last Theorem. Proven in 1995 by Wiles with the help of Taylor, this theorem states that there are no positive integers x, y and z satisfying the equation xn + yn = zn; where n ≥ 3 is an integer. The proof of this statement for the particular case n = 4 goes back to Fibonacci, who lived four hundred years before Fermat. Modulo Fibonacci's result, Fermat Last Theorem needs to be proven only for the cases where n = p is an odd prime. By the end of the course we will hopefully see, as an application of our theory, how to prove the Fermat Last Theorem for the so-called regular primes. The idea of this belongs to Kummer, although we will, of course, use more modern notation and methods. Another accepted definition of algebraic number theory is that it studies the so-called number fields, which are the finite extensions of the field of ra- tional numbers Q. We mention right away, however, that most of this theory applies also in the second important case, known as the case of function fields. For example, finite extensions of the field of complex rational functions C(x) are function fields. We will stress the similarities and differences between the two types of fields, as appropriate. Finite extensions of Q are algebraic, and this ties algebraic number the- ory with Galois theory, which is an important prerequisite for us.
    [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]
  • Strongly Homogeneous Torsion Free Abelian Groups of Finite Rank
    PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 56, April 1976 STRONGLY HOMOGENEOUS TORSION FREE ABELIAN GROUPS OF FINITE RANK Abstract. An abelian group is strongly homogeneous if for any two pure rank 1 subgroups there is an automorphism sending one onto the other. Finite rank torsion free strongly homogeneous groups are characterized as the tensor product of certain subrings of algebraic number fields with finite direct sums of isomorphic subgroups of Q, the additive group of rationals. If G is a finite direct sum of finite rank torsion free strongly homogeneous groups, then any two decompositions of G into a direct sum of indecompos- able subgroups are equivalent. D. K. Harrison, in an unpublished note, defined a p-special group to be a strongly homogeneous group such that G/pG » Z/pZ for some prime p and qG = G for all primes q =£ p and characterized these groups as the additive groups of certain valuation rings in algebraic number fields. Rich- man [7] provided a global version of this result. Call G special if G is strongly homogeneous, G/pG = 0 or Z/pZ for all primes p, and G contains a pure rank 1 subgroup isomorphic to a subring of Q. Special groups are then characterized as additive subgroups of the intersection of certain valuation rings in an algebraic number field (also see Murley [5]). Strongly homoge- neous groups of rank 2 are characterized in [2]. All of the above-mentioned characterizations can be derived from the more general (notation and terminology are as in Fuchs [3]): Theorem 1.
    [Show full text]
  • Algebra I Homework Three
    Algebra I Homework Three Name: Instruction: In the following questions, you should work out the solutions in a clear and concise manner. Three questions will be randomly selected and checked for correctness; they count 50% grades of this homework set. The other questions will be checked for completeness; they count the rest 50% grades of the homework set. Staple this sheet of paper as the cover page of your homework set. 1. (Section 2.1) A subset X of an abelian group F is said to be linearly independent if n1x1 + ··· + nrxr = 0 always implies ni = 0 for all i (where ni 2 Z and x1; ··· ; xk are distinct elements of X). (a) X is linearly independent if and only if every nonzero element of the subgroup hXi may be written uniquely in the form n1x1 + ··· + nkxk (ni 2 Z, ni 6= 0, x1; ··· ; xk distinct elements of X). The set S := fnlx1 + ··· + nkxk j ni 2 Z; x1; ··· ; xk 2 X; k 2 Ng forms a subgroup of F and it contains X. So hXi ≤ S. Conversely, Every element of S is in hXi. Thus hXi = S. If X is linearly independent, suppose on the contrary, an element a 2 hXi can be expressed as two different linear combinations of elements in X: 0 0 0 0 a = nlx1 + ··· + nkxk = nlx1 + ··· + nkxk; x1; ··· ; xk 2 X; ni; ni 2 Z; ni 6= 0 or ni 6= 0 for i = 1; ··· ; k: 0 0 0 0 0 Then (n1 − n1)x1 + ··· + (nk − nk)xk = 0, and at least one of ni − ni is non-zero (since (n1; ··· ; nk) 6= (n1; ··· ; nk)).
    [Show full text]
  • Group Cohomology
    Group Cohomology I will define group cohomology H*(G, N) for any group G and any G-module N, and relate this to Hilbert Theorem 90. ∗ The context is ExtR(M; N), for a ring R and two (left) R-modules M; N. This ∗ is a long theory to do everything, but one computation of ExtR goes as follows: (1) Choose a free R-module resolution complex F∗ ! M ! 0 (resolution means exact) ∗ (2) Form the cocomplex of abelian groups HomR(F∗;N). Then ExtR(M; N) is ∗ the cohomology H (HomR(F∗;N)) 0 (3) It is easy to see from half-exactness property of Hom that H = HomR(M; N). (4) The codifferentials HomR(Fk;N) ! HomR(Fk+1;N) are given by `adjoints' of the differentials in F∗. Namely, the codifferential of an R-hom z : Fk ! N is the composition zd : Fk+1 ! Fk ! N. In particular, z is a k-cocycle exactly when this composition zd = 0. The k-coboundaries are all compositions Fk ! Fk−1 ! N, where first map is d. (5) Take M = Z with trivial G action and take R = Z[G]. Note R-modules and G-modules are the same thing. Then H∗(G; N) is defined to be Ext∗ ( ;N). Z[G] Z (6) I will write down a free Z[G] module resolution F∗ ! Z ! 0: As a free abelian group, Fk has Z-basis fg0 < g1; ··· gk >g with the gi 2 G.A plain bracket < a; b; ··· ; x > is interpreted to have the identity e 2 G in front.
    [Show full text]
  • Ideals and Class Groups of Number Fields
    Ideals and class groups of number fields A thesis submitted To Kent State University in partial Fulfillment of the requirements for the Degree of Master of Science by Minjiao Yang August, 2018 ○C Copyright All rights reserved Except for previously published materials Thesis written by Minjiao Yang B.S., Kent State University, 2015 M.S., Kent State University, 2018 Approved by Gang Yu , Advisor Andrew Tonge , Chair, Department of Mathematics Science James L. Blank , Dean, College of Arts and Science TABLE OF CONTENTS…………………………………………………………...….…...iii ACKNOWLEDGEMENTS…………………………………………………………....…...iv CHAPTER I. Introduction…………………………………………………………………......1 II. Algebraic Numbers and Integers……………………………………………......3 III. Rings of Integers…………………………………………………………….......9 Some basic properties…………………………………………………………...9 Factorization of algebraic integers and the unit group………………….……....13 Quadratic integers……………………………………………………………….17 IV. Ideals……..………………………………………………………………….......21 A review of ideals of commutative……………………………………………...21 Ideal theory of integer ring 풪푘…………………………………………………..22 V. Ideal class group and class number………………………………………….......28 Finiteness of 퐶푙푘………………………………………………………………....29 The Minkowski bound…………………………………………………………...32 Further remarks…………………………………………………………………..34 BIBLIOGRAPHY…………………………………………………………….………….......36 iii ACKNOWLEDGEMENTS I want to thank my advisor Dr. Gang Yu who has been very supportive, patient and encouraging throughout this tremendous and enchanting experience. Also, I want to thank my thesis committee members Dr. Ulrike Vorhauer and Dr. Stephen Gagola who help me correct mistakes I made in my thesis and provided many helpful advices. iv Chapter 1. Introduction Algebraic number theory is a branch of number theory which leads the way in the world of mathematics. It uses the techniques of abstract algebra to study the integers, rational numbers, and their generalizations. Concepts and results in algebraic number theory are very important in learning mathematics.
    [Show full text]
  • Compact Groups and Fixed Point Sets
    TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 349, Number 11, November 1997, Pages 4537{4554 S 0002-9947(97)02059-X COMPACT GROUPS AND FIXED POINT SETS ALEX CHIGOGIDZE, KARL H. HOFMANN, AND JOHN R. MARTIN Abstract. Some structure theorems for compact abelian groups are derived and used to show that every closed subset of an infinite compact metrizable group is the fixed point set of an autohomeomorphism. It is also shown that any metrizable product containing a positive-dimensional compact group as a factor has the property that every closed subset is the fixed point set of an autohomeomorphism. 1. Introduction AspaceXis defined to have the complete invariance property (CIP) if every nonempty closed subset of X is the fixed point set of a (continuous) self-mapping of X [23]. If this condition holds for autohomeomorphisms of X,thenwesaythat X has the complete invariance property with respect to homeomorphisms (CIPH) [13]. A survey of results concerning CIP for metric spaces may be found in [20], and a number of nonmetric results may be found in [16]. Some spaces known to have CIPH are even-dimensional Euclidean balls [18], compact surfaces and positive- dimensional spheres [19], Menger manifolds [12], the Hilbert cube and metrizable product spaces which have the real line or an odd-dimensional sphere as a factor [13]. Metrizable topological groups are known to have CIP if they are locally compact or contain an arc [15]. In [16] it is shown that an uncountable self-product of circles, real lines or two-point spaces has CIP and that connected subgroups of the plane and compact groups need not have CIP.
    [Show full text]
  • An Introduction to the Cohomology of Groups Peter J
    An Introduction to the Cohomology of Groups Peter J. Webb 0. What is group cohomology? For each group G and representation M of G there are abelian groups Hn(G, M) and Hn(G, M) where n = 0, 1, 2, 3,..., called the nth homology and cohomology of G with coefficients in M. To understand this we need to know what a representation of G is. It is the same thing as ZG-module, but for this we need to know what the group ring ZG is, so some preparation is required. The homology and cohomology groups may be defined topologically and also algebraically. We will not do much with the topological definition, but to say something about it consider the following result: THEOREM (Hurewicz 1936). Let X be a path-connected space with πnX = 0 for all n ≥ 2 (such X is called ‘aspherical’). Then X is determined up to homotopy by π1(x). If G = π1(X) for some aspherical space X we call X an Eilenberg-MacLane space K(G, 1), or (if the group is discrete) the classifying space BG. (It classifies principal G-bundles, whatever they are.) If an aspherical space X is locally path connected the universal cover X˜ is contractible n and X = X/G˜ . Also Hn(X) and H (X) depend only on π1(X). If G = π1(X) we may thus define n n Hn(G, Z) = Hn(X) and H (G, Z) = H (X) and because X is determined up to homotopy equivalence the definition does not depend on X.
    [Show full text]
  • Free Actions of Finite Abelian Groups on the 3-Torus
    CORE Metadata, citation and similar papers at core.ac.uk Provided by Elsevier - Publisher Connector Topology and its Applications 53 (1993) 153-175 1.53 North-Holland Free actions of finite Abelian groups on the 3-torus Kyung Bai Lee Department of Mathematics, University of Oklahoma, Norman, OK 73019, USA Joonkook Shin * Department of Mathematics, Chungnam National University, Taejon 305-764, Korea Shoji Yokura ** Department of Mathematics, University of Kagoshima, I-21-40 Korimoto, Kagoshima 890, Japan Received 3 November 1991 Revised 25 June 1992 and 21 October 1992 Abstract Lee, K.B., J. Shin, and S. Yokura, Free actions of finite Abelian groups on the 3-torus, Topology and its Applications 53 (1993) 153-176. We classify free actions of finite Abehan groups on the 3-torus, up to topological conjugacy. By the works of Bieberbach and Waldhausen, this classification problem is reduced to classifying all normal Abehan subgroups of Bieberbach groups of finite index, up to affine conjugacy. All such actions are completely classified, see Theorems 2.1, 3.4, 4.2, 5.1, 6.1 and 7.1. Keywords: Group actions; Bieberbach groups; Affine conjugacy. AMS CMOS) Subj. Class.: Primary 57825; secondary 57M05, 57S17. Introduction The general question of classifying finite group actions on a closed 3-manifold is very hard. For example, it is not known if every finite action of S3 is conjugate to a linear action. However, the actions on a 3-dimensional torus can be understood easily by the works of Bieberbach and Waldhausen. We shall study only free actions of finite Abelian groups G on a 3-dimensional torus T.
    [Show full text]
  • The Purpose of This Paper Is to Establish the Existence of A
    THE LATTICE OF PROFINITE SUBGROUPS OF PROTORI WAYNE LEWIS ABSTRACT. Compact connected abelian groups, or protori, have intrinsic structural characteristics that present for the entire category. In the case of finite-dimensional torus-free protori, The Resolution Theorem for Compact Abelian Groups sets the stage for demonstrating that the profinite subgroups inducing tori quotients comprise an isogeny class of finitely generated modules over the profinite integers, which is a lattice under intersection (meet) and + (join). The structural results enable the formulation of a universal resolution in the category of protori under morphisms of compact abelian groups. A single profinite subgroup from the lattice in the Resolution Theorem is replaced by the direct limit of the lattice of such subgroups and effects a covering morphism in which the discrete torsion-free Pontryagin dual of the protorus organically emerges as the kernel of the quotient map resolving the protorus. Among other advantages, this enables the study of a finite rank torsion-free abelian group as a canonical subgroup of its dual protorus, with the concomitant topological, analytical, and number-theoretic insights availed by the compact abelian setting. The purpose of this paper is to establish the existence of a covering group for a finite- dimensional protorus G defined in terms of three canonical components: the Lie algebra of G, the Pontryagin dual of G, and the sum of the profinite subgroups of G that induce tori quotients. The result essentially depends upon showing that the collection of profinite subgroups of G which induce tori quotients is a lattice under + (join) and (meet).
    [Show full text]
  • REU 2013: the COHOMOLOGY of GROUPS 1. Introduction This Will
    REU 2013: THE COHOMOLOGY OF GROUPS J.P. MAY 1. Introduction This will be an informal set of notes to be used as a crib sheet in my talks. We are taking up where Inna left off. She took as her motivation the classification of extensions ⊂ q 0 /K /E /Q /1: Here K stands for kernel and Q stands for quotient. She assumed that K is abelian (= commutative), so she wrote its product as + with identity element 0. She did not assume that Q is abelian, so she wrote its product as · with identity element 1. That explains the joke notation with 0 at the left and 1 at the right. The sequence is exact, which means that K is a normal subgroup of E with quotient group Q. She chose a set theoretic section t of q and found that g · k = t(g)kt(g)−1 defines a well-defined action of Q on K. It is convenient to require t(1) = 1. That doesn't change anything seriously, but it leads to a slightly more standard and convenient description of the classification, which she described by saying that isomorphism classes of extensions of Q by K are in bijective correspondence to elements of H2(Q; K). She did not tell you about Hi(Q; K) for any other integers i. This looks kind of bizarre. I'm going to tell you about H0 and H1 and then I'll remind you how Inna described H2. Then I'll begin to try to tell you the whole story and use it to desribe how to partially solve a topological problem: What finite groups can act freely on a sphere Sn? We are going to change notation.
    [Show full text]