Contests in Higher Mathematics: Miklós Schweitzer Competitions, 1962-1991 (Problem Books in Mathematics)

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Problem Books in Mathematics

Edited by P.R. Halmos

Springer New York

Berlin Heidelberg Barcelona Budapest Hong Kong London

Milan Paris Santa Clara Singapore

Tokyo

Problem Books in Mathematics Series Editor: P.R. Halmos

Polynomials by Edward J. Barbeau

Problems in Geometry by Marcel Berger, Pierre Pansu, Jean-Pic Berry, and Xavier Saint-Raymond

Problem Book for First Year Calculus by George W. Bluman

Exercises in Probability by T. Cacoullos

An Introduction to Hilbert Space and Quantum Logic by David W. Cohen

Unsolved Problems in Geometry by Hallard T. Croft, Kenneth J. Falconer, and Richard K. Guy Problems in Analysis by Bernard R. Gelbaum

Problems in Real and Complex Analysis by Bernard R. Gelbaum

Theorems and Counterexamples in Mathematics by Bernard R. Gelbaum and John M.H. Olmsted Exercises in Integration by Claude George Algebraic Logic by S.G. Gindikin

Unsolved Problems in Number Theory (2nd ed.) by Richard K. Guy

An Outline of Set Theory by James M. Henle (continued after index)

Gabor J. Szekely Editor

Contests in Higher Mathematics Miklos Schweitzer Competitions 1962-1991

With 39 Illustrations

Springer

Gabor J. Szekely Department of Mathematics Eotvos Lorand and Technical University Mtizeum krt. 6-8 1088 Budapest, Hungary Series Editor: Paul R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA Mathematics Subject Classification (1991): 15-06, 05-06, 26-06, 51-06

Library of Congress Cataloging-in-Publication Data Contests in higher mathematics:Miklos Schweitzer competitions, 1962-1991 / Gabor J. Szekely (ed.). p. cm. - (Problem books in mathematics) Includes bibliographical references and index. ISBN 0-387-94588-1

1. Mathematics - Competitions - Hungary. 2. Mathematics - Problems, I. Schweitzer, Miklbs, 1923-1945. II. Szekely, Gabor J., 1947- . III. Series. QA99.C62 1995 exercises, etc.

510' .76 - dc20

95-25361

Printed on acid-free paper. © 1996 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Production managed by Hal Henglein; manufacturing supervised by Jeffrey Taub. Camera-ready copy prepared from the editor's LaTeX files. Printed and bound by R.R. Donnelley & Sons, Harrisonburg, VA. Printed in the United States of America.

987654321 ISBN 0-387-94588-1 Springer-Verlag New York Berlin Heidelberg

Preface "I had the opportunity to speak with Leo Szilard about the contests of the Mathematical and Physical Society, and about the fact that the winners of these contests turned out later to be almost identical with the set of mathematicians and physicists who became outstanding ... "

(J. Neumann, in a letter to L. Fejer, Berlin, Dec. 7, 1929)

The solutions to deep scientific problems rarely come to us easily. Thus, it is important to motivate students to begin efforts on these kinds of problems. Scientific competition has proved to be an effective stimulant toward intellectual efforts. Successful examples include the "Concours" for admission to the "Grandes Ecoles" in France, and the "Mathematical Tripos" in

Cambridge, England. At the turn of the century, mathematical contests helped Hungary become one of the strongholds of the mathematical world. With the revolution in 1848 and the Compromise in 1867, Hungary broke free from many centuries of rule by the Turks and then the Hapsburgs, and became a nation on equal footing with her neighbor, Austria. By the end

of the 19th century, Hungary entered a period of cultural and economic progress. In 1891, Baron Lorind Eotvos, an outstanding Hungarian physicist, founded the Mathematical and Physical Society. In turn, the Society founded two journals: the Mathematical and Physical Journal in 1892 and the Mathematical Journal for Secondary Schools in 1893. This latter journal offered a rich variety of elementary problems for high school students. One of the first editors of the journal, Lasz16 Ritz, later became the teacher of John Neumann and Eugene Wigner (a Nobel prize winner in physics). In 1894, the Society introduced a mathematical competition for high school students. Among the winners there were Lipot Fejer, Alfred Haar, Todor

Kirmin, Marcel Riesz, Gibor Szego, Tibor Rado, Ede Teller, and many others who became world-famous scientists. The success of high school competitions led the Mathematical Society to found a college-level contest. The first contest of this kind was organized in 1949 and named after Mikl6s Schweitzer, a young mathematician who died

in the Second World War. Schweitzer placed second in the High School Contest in 1941, but the statutes of the fascist regime of that time prevented his admission to college. Schweitzer Contest problems are proposed and selected by the most prominent Hungarian mathematicians. Thus, V

PREFACE

vi

Schweitzer problems reflect the interest of these mathematicians and some aspects of the mainstream of Hungarian mathematics. The universities of Budapest, Debrecen, and Szeged have alternately been designated by the Society Presidium to conduct the Schweitzer Contests. The jury is chosen by the mathematics departments of the universities in question from among the mathematicians working in the host city. The jury sends out requests to leading Hungarian mathematicians to submit problems suitable for the contest. The list of problems selected by the jury is posted on the bulletin boards of mathematics departments and of local branches of the Mathematical Society (copies are available to anyone interested). Students may use any materials available in libraries or in their homes to solve the

contest problems. In ten days the solutions are due, with the student's name, faculty, course, year, and university or high school recorded on the solution set. The Schweitzer competition is one of the most unique in the world. Winners of the contests have gone on to become world-class scientists. Thus, the Schweitzer Contests are of interest to both math historians and mathematicians of all ages. They serve as reflections of Hungarian mathematical trends and as starting points for many interesting research problems in mathematics. The Schweitzer problems between 1949 and 1961 were previously published under the title Contests in Higher Mathematics, 1949-1961 (Akademiai Kiad6, Budapest, 1968; Chapter 4 of this book summarizes the mathematical work of M. Sch veitzer). Our book is a continuation of that volume.

We hope that this collection of Schweitzer problems will serve as a guide

for many young mathematicians and math majors. The large variety of research-level problems may spark the interest of seasoned mathematicians and historians of mathematics. I wish to close by acknowledging the outstanding work of Dr. Marianna Bolla as Managing Editor. InI addition, without the constant assistance of Dr. Dezso Miklds as Technical Editor, we could not have this book.

Bowling Green, OH August 26, 1995

Gdbor J. Szekely

Contents Preface Chapter 1. Problems of the Contests Chapter 2. Results of the Contests Chapter 3. Solutions to the Problems 3.1 3.2 3.3

3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11

Algebra (J6zsef Pelikan) Combinatorics (Ervin Gyori) Theory of Functions (Janos Bognar and Vilmos Totik) Geometry (Balazs Csik6s) Measure Theory (Janos Bognar) Number Theory (Imre Z. Ruzsa) Operators (Janos Bognar) Probability Theory (Gabriella Szep) Sequences and Series (Jeno Torocsik) Topology (Gabor Moussong) Set Theory (Peter Komjath)

Index of Names

v 1

49 55 55 121 150

244 330 362 388 404 465 517 552

565

Vii

1. Problems of the Contests The letter in parentheses after the text of a problem refers to the section

in Chapter 3 containing its solution. The topics include these areas of mathematics: A: Algebra C: Combinatorics F: Theory of Functions G: Geometry M: Measure Theory N: Number Theory

0: Operators P: Probability Theory S: Sequences and Series T: Topology R: Set Theory Thus, for example, P.3 refers to problem in "Probability Theory" section. When available, the names of proposers are in brackets at the very end of each problem.

1

1. PROBLEMS OF THE CONTESTS

2

1962 1. Let f and g be polynomials with rational coefficients, and let F and G

denote the sets of values of f and g at rational numbers. Prove that F = G holds if and only if f (x) = g(ax + b) for some suitable rational numbers a # 0 and b. (N.1) [E. Fried] 2. Determine the roots of unity in the field of p-adic numbers. (A.1) [L. Fuchs]

3. Let A and B be two Abelian groups, and define the sum of two homomorphisms q and x from A to B by a(17 + x) = a77 + ax

for all a E A.

With this addition, the set of homomorphisms from A to B forms an Abelian group H. Suppose now that A is a p-group (p a prime number). Prove that in this case H becomes a topological group under the topology defined by taking the subgroups pkH (k = 1, 2, ...) as a neighborhood base of 0. Prove that H is complete in this topology and that every connected component of H consists of a single element. When is H compact in this topology? (A.2) [L. Fuchs] 4. Show that

(x2 + y2) = (-1)[] 1<xO

h+k>O

h,k-.0

f (x+h) - f (x-k) h+k

h,k>O

Show that V f and D f are Borel-measurable functions. (M.1) [A. Csaszar]

6. Let E be a bounded subset of the real line, and let S2 be a system of (nondegenerate) closed intervals such that for each x E E there exists an I E S2 with left endpoint x. Show that for every e > 0 there exist a finite number of pairwise nonoverlapping intervals belonging to S2 that cover

E with the exception of a subset of outer measure less than e. (M.2) [J. Czipszer]

7. Prove that the function

f0)

VI_(_X2

dx - 1)(1 -192x2)

(where the positive value of the square root is taken) is monotonically decreasing in the interval 0 < 19 < 1. (F.1) [P. Turin]


3

8. Denote by M(r, f) the maximum modulus on the circle I z I = r of the transcendent entire function f (z), and by M,,, (r, f) that of the nth partial sum of the power series of f (z). Prove the existence of an entire function fo(z) and a corresponding sequence of positive numbers rl < r2 < -> +oo such that lim sup M" (r., fo) _ +00. n-co M(rn, fo)

(F.2) [P. Turan] 9. Find the minimum possible sum of lengths of edges of a prism all of whose edges are tangent to a unit sphere. (G.1) [Miiller-Pfeiffer] 10. From a given triangle of unit area, we choose two points independently with uniform distribution. The straight line connecting these points divides the triangle, with probability one, into a triangle and a quadrilateral. Calculate the expected values of the areas of these two regions. (P.1) [A. Renyi] 1963 1. Show that the perimeter of an arbitrary planar section of a tetrahedron is less than the perimeter of one of the faces of the tetrahedron. (G.2) [Gy. Hajos] 2. Show that the center of gravity of a convex region in the plane halves

at least three chords of the region. (G.3) [Gy. Hajos] 3. Let R = Rl ®R2 be the direct sum of the rings Rl and R2, and let N2 be the annihilator ideal of R2 (in R2). Prove that R1 will be an ideal in every ring R containing R as an ideal if and only if the only homomorphism from R1 to N2 is the zero homomorphism. (A.3) [Gy. Pollak] 4. Call a polynomial positive reducible if it can be written as a product of two nonconstant polynomials with positive real coefficients. Let f (x)

be a polynomial with f (0) # 0 such that f (x') is positive reducible for some natural number n. Prove that f (x) itself is positive reducible. (A.4) [L. Redei]

5. Let H be a set of real numbers that does not consist of 0 alone and is closed under addition. Further, let f (x) be a real-valued function defined on H and satisfying the following conditions:

f(x) < f (y) if x < y and f (x + y) = f (x) + f (y)

(x, y E H).

Prove that f (x) = cx on H, where c is a nonnegative number. (F.3) [M. Hosszu, R. Borges]

6. Show that if f (x) is a real-valued, continuous function on the half-line

0<x 1 and limr . f (r) = +oo, then there exists an infinite set of natural numbers that, for all r, contains r-tuples from at most f (r)

classes. Show that if f (r) 74 +oc, then there is a family of partitions such that no such infinite set exists. (C.2) [P. Erd6s, A. Hajnal] 8. Let n and k be given natural numbers, and let A be a set such that IAI < n(n + 1)

k+1 For i = 1, 2, ... , n + 1, let Ai be sets of size n such that

IAinA;I 2 an is divergent. n=1

(S.10) [P. Turan] 11. Let 1, C2, ... be independent random variables such that m>0 and Var(Cn) = Q2 < 00 (n = 1, 2, ... ). Let {an} be a sequence of positive numbers such that an --+ 0 and F-° 1 an = oo. Prove that Jim

P

n- oo

n

>2 ak4 = oo = 1. k=1

(P.9) [P. Revesz] 12. Let 191 i ... , i9n be independent, uniformly distributed, random variables in the unit interval [0, 1]. Define

h(x) = n #{k: 19k < x}. Prove that the probability that there is an xo E (0, 1) such that h(xo) _ x0, is equal to 1 - n. (P.10) [G. Tusnady] 1971 1. Let G be an infinite compact topological group with a Hausdorff topol-

ogy. Prove that G contains an element g # 1 such that the set of all powers of g is either everywhere dense in G or nowhere dense in G. (A.19) [J. Erdos] 2. Prove that there exists an ordered set in which every uncountable subset contains an uncountable, well-ordered subset and that cannot be represented as a union of a countable family of well-ordered subsets. (N.2) [A. Hajnal]


15

3. Let 0 < ak < 1 for k = 1, 2, .... Give a necessary and sufficient condition for the existence, for every 0 < x < 1, of a permutation iry of the positive integers such that 00

ary (k) 2k

k=1

(S.11) [P. Erdos] 4. Suppose that V is a locally compact topological space that admits no countable covering with compact sets. Let C denote the set of all compact subsets of the space V and U the set of open subsets that are not contained in any compact set. Let f be a function from U to C such that f (U) C_ U for all U EU. Prove that either (i) there exists a nonempty compact set C such that f (U) is not a proper subset of C whenever C C U EU, (ii) or for some compact set C, the set

f-1(C) = U{U E U: f(U) C C} is an element of U, that is, f

(C) is not contained in any compact

set.

(T.7) [A. Mate] 5. Let Al < A2 < ... be a positive sequence and let K be a constant such that n-1

k=1

Prove that there exists a constant K' such that n-1

EAk 2 be an integer, let S be a set of n elements, and let Ai, 1 < i < m, be distinct subsets of S of size at least 2 such that

AinA,#O,AinAk

O,A,nAk#0 imply AinA;nAk#0.

Show that m < 2n-1 - 1. (C.5) [P. ErdBs] 8. Show that the edges of a strongly connected bipolar graph can be oriented in such a way that for any edge e there is a simple directed path from pole p to pole q containing e. (A strongly connected bipolar graph is a finite connected graph with two special vertices p and q having the property that there are no points x, y, x # y, such that all paths from x to p as well as all paths from x to q contain y.) (C.6) [A. Adam] 9. Given a positive, monotone function F(x) on (0, oo) such that F(x)/x is monotone nondecreasing and F(x)/xl+d is monotone nonincreasing for some positive d, let An > 0 and an > 0, n > 1. Prove that if n

00

E AnF an n=1

E AAkn

)

< 00,

k=1 An

or 00o

n=1

AF

(Eak) n + 1) is a point of K for which the volume of the convex hull of P1, . . . , Pk is maximal. Denote this volume by Vk. Decide, for different values of n, about the truth of the statement "the sequence Vn+l, Vn+2,... is concave." (G.19) [L. Fejes-Toth, E. Makai] 10. Let T1 and T2 be second-countable topologies on the set E. We would like to find a real function a defined on E x E such that

0< a(x,y) < +oo, a(x, z) < a(x, y) + a(y, z)

a(x,x) = 0, (x, y, z E E),

and, for any p E E, the sets Vi(p,E) _ {x : a(x,p) < E}

(6>0)

form a neighborhood base of p with respect to T1, and the sets V2(p, E) _ {x : a(p, x) < E}

(s>0)

form a neighborhood base of p with respect to T2. Prove that such a function a exists if and only if, for any p E E and Ti-open set G 3 p (i = 1, 2), there exist a Ti-open set G' and a T3_i-closed set F with p E G' C F C G. (T.8) [A. Csaszir] 11. We throw N balls into n urns, one by one, independently and uniformly. Let Xi = Xi(N, n) be the total number of balls in the ith urn. Consider the random variable

y(N, n) = min 1Xi - N I. 1 oo and N/n3 -+ oo, then

P

(y(N,

- ' 1 - e-'

n) < nyVn

21"

for all x > 0.

(b) If n -* oo and N/n3 < K (K constant), then for any e > 0 there is an A > 0 such that

P(y(N, n) < A) > 1 -

(c) If n->ooand N/n3-4Othen P(y(N, n) < 1) -> 1. (P.11) [P. Revesz] 1973

1. We say that the rank of a group G is at most r if every subgroup of G can be generated by at most r elements. Prove that there exists an integer s such that for every finite group G of rank 2 the commutator series of G has length less than s. (A.21) [J. Erdos] 2. Let R be an Artinian ring with unity. Suppose that every idempotent element of R commutes with every element of R whose square is 0. Suppose R is the sum of the ideals A and B. Prove that AB = BA. (A.22) [A. Kertesz]

3. Find a constant c > 1 with the property that, for arbitrary positive integers n and k such that n > ck, the number of distinct prime factors of (k) is at least k. (N.7) [P. Erdos] 4. Let f (n) be the largest integer k such that nk divides n!, and let F(n) _ max2 v(G) for all sets of the form G = U,=1 (-oo, xi) x (-oo, yz). (P.14) [P. Major] 1975 1. Show that there exists a tournament (T, ->) of cardinality R1 containing no transitive subtournament of size R1. (A structure (T, --*) is a tournament if -+ is a binary, irreflexive, asymmetric, and trichotomic relation. The tournament (T, -4) is transitive if -+ is transitive, that is, if it orders T.) (R.5) [A. Hajnal] 2. Let An denote the set of all mappings f : {1, 2, ... , n} -a {1, 2, ... , n}

such that f -1(i) {1,2,...,i}. Prove

{k

:

f (k) = i} 54 0 implies f -1(j) I'An I =

kn E 2k+1 00

k=0

0, j E


22

(C.9) [L. Lovasz] 3. Let S be a semigroup without proper two-sided ideals, and suppose that for every a, b E S at least one of the products ab and ba is equal to one of the elements a, b. Prove that either ab = a for all a, b E S or ab = b for all a, b E S. (A.24) [L. Megyesi] 4. Prove that the set of rational-valued, multiplicative arithmetical functions and the set of complex rational-valued, multiplicative arithmetical functions form isomorphic groups with the convolution operation f o g defined by

.f(d)g(d)

(f og)(n) _ din

(We call a complex number complex rational, if its real and imaginary parts are both rational.) (N.10) [B. Csakany] 5. Let I fn} be a sequence of Lebesgue-integrable functions on [0, 1] such that for any Lebesgue-measurable subset E of [0, 1] the sequence fE fn is convergent. Assume also that limn fn = f exists almost everywhere. Prove that f is integrable and fE f = limn fE fn. Is the assertion also true if E runs only over intervals but we also assume fn > 0? What happens if [0, 11 is replaced by [0, oo) ? (M.12) [J. Sziics]

6. Let f be a differentiable real function, and let M be a positive real number. Prove that if

If(x + t) - 2f(x) + f(x - t)l <Mt2 for all x and t, then

If'(x+t)- f'(x)I <MItI. (F.20) [J. Szabados]

7. Let a < a' < b < b' be real numbers, and let the real function f be continuous on the interval [a, b'] and differentiable in its interior. Prove that there exist c E (a, b), c' E (a', b') such that

f (b) - f (a) = f ' (c) (b - a),

f(b') - f(a') = f'(cl) (b' - a'), and c < c'. (F.21) [B. Sz6kefalvi-Nagy]

8. Prove that if

m Ean

< Nam

(m=1,2.... )

n=1

holds for a sequence {an} of nonnegative real numbers with some positive integer N, then ai+n > pai for i, p = 1, 2, ... , where iN

ai =n=(i-1)N+1 E an (S.16) [L. Leindler]

(i = 1, 2, ... ).


23

9. Let 10, c, a, g be positive constants, and let x(t) be the solution of the differential equation

([lo+ct,]2x')'+g[l0+cta]sinx=0,

t>0, -2 <x< 2

satisfying the initial conditions x(to) = x0i x'(to) = 0. (This is the equation of the mathematical pendulum whose length changes according to the law 1 = to+ct' .) Prove that x(t) is defined on the interval [to, 00); furthermore, if a > 2 then for every x0 # 0 there exists a to such that liminf I x(t)I > 0. t-100

(F.22) [L. Hatvani] 10. Prove that an idempotent linear operator of a Hilbert space is self-adjoint

if and only if it has norm 0 or 1. (0.4) [J. Szucs] 11. Let X1, X2,..., X,,, be (not necessary independent) discrete random vari-

ables. Prove that there exist at least n2/2 pairs (i, j) such that

H(Xi+Xj) >

1 min {H(Xk)}, 3 1 0

xEG

for some continuous (complex) character X of G, then g is continuous. (F.26) [L. Szekelyhidi]

26


8. Let p > 1 be a real number and R+ = (0, oo). For which continuous functions g : R+ ---> R+ are the following functions all convex?

L

1

g(

)X+1

Z 1g()

n=1,2,... (S.18) [L. Losonczi]

9. Suppose that the components of the vector u = (u0,. .. , u,) are real functions defined on the closed interval [a, b] with the property that every nontrivial linear combination of them has at most n zeros in [a, b].

Prove that if or is an increasing function on [a, b] and the rank of the operator b

A(f) = f u(x)f (x)da(x), a

f E C[a, b],

is r < n, then or has exactly r points of increase. (F.27) [E. Gesztelyi]

10. Let the sequence of random variables {Xm, m > 0}, Xo = 0, be an infinite random walk on the set of nonnegative integers with transition probabilities

pi=P(Xm+1=i+1IXm=i)>0, i>0, Qz=P(Xm+1=i-11 Xm.=i)>0, i>0. Prove that for arbitrary k > 0 there is an ak > 1 such that

Pn(k)=P( max 0<j 2 be an integer, and let X be a connected Hausdorff space such that every point of X has a neighborhood homeomorphic to the Euclidean space R. Suppose that any discrete (not necessarily closed) subspace D of X can be covered by a family of pairwise disjoint, open sets of X so that each of these open sets contains precisely one element of D. Prove that X is a union of at most N, compact subspaces. (T.16) [Z. Balogh]

10. Let P be a probability distribution defined on the Borel sets of the real line. Suppose that P is symmetric with respect to the origin, absolutely continous with respect to the Lebesgue measure, and its density function p is zero outside the interval [-1, 1] and inside this interval it is between the positive numbers c and d (c < d). Prove that there is no distribution whose convolution square equals P. (P.20) IT. F. M6ri, G. J. Szekely] 1982

1. A map F : P(X) -* P(X), where P(X) denotes the set of all subsets of X, is called a closure operation on X if for arbitrary A, B C X, the following conditions hold: (i)

A C F(A);

(ii) A C B = F(A) C F(B); (iii) F(F(A)) = F(A). The cardinal number min{ IAA : A C X, F(A) = X} is called the density of F and is denoted by d(F). A set H C X is called discrete with respect

to F if u V F(H - {u}) holds for all u E H. Prove that if the density of the closure operation F is a singular cardinal number, then for any nonnegative integer n, there exists a set of size n that is discrete with respect to F. Show that the statement is not true when the existence of an infinite discrete subset is required, even if F is the closure operation of a topological space satisfying the T1 separation axiom. (T.17) [A. Hajnal]

2. Consider the lattice of all algebraically closed subfields of the complex field C whose transcendency degree (over Q) is finite. Prove that this lattice is not modular. (A.34) [L. Babai] 3. Let G(V, E) be a connected graph, and let dG(x, y) denote the length of the shortest path joining x and y in G. Let rG(x) = max{dG(x, y) :

y E V} for x E V, and let r(G) = min{rG(x) : x E V}. Show that if


33

r(G) _> 2, then G contains a path of length 2r(G) - 2 as an induced subgraph. (C.17) [V. T. Sos] 4. Let

.f (n) _ E p' PI° 3n/(3n - 1), one can find a covering of the unit square with n homothetic triangles with area of the union less than d. (G.33) 9. Suppose that K is a compact Hausdorff space and K = U°_OAn, where An is metrizable and An C An for n < m. Prove that K is metrizable. (T.18) [Z. Balogh] 10. Let p0, p,.... be a probability distribution on the set of nonnegative integers. Select a number according to this distribution and repeat the selection independently until either a zero or an already selected number is obtained. Write the selected numbers in a row in order of selection without the last one. Below this line, write the numbers again in increasing order. Let Ai denote the event that the number i has been selected and that it is in the same place in both lines. Prove that the events Ai (i = 1, 2, ...) are mutually independent, and P(A2) = pi. (P.21) [T. F. Mori]


34

1983 1. Given n points in a line so that any distance occurs at most twice, show

that the number of distances occurring exactly once is at least [n/2]. (C.18) [V. T. Sos, L. Szekely]

2. Let I be an ideal of the ring R and f a nonidentity permutation of the set {1, 2, ... , k} for some k. Suppose that for every 0 a E R, aI # 0 and Ia # 0 hold; furthermore, for any elements x1i x2, ... , xk E I,

x1x2...xk = xlfx2f ...xkf holds. Prove that R is commutative. (A.35) [R. Wiegandt] 3. Let f : R -+ R be a twice differentiable, 27r-periodic even function. Prove

that if 1

fr(X) + f(x)

f (x + 37r/2)

holds for every x, then f is it/2-periodic. (F.36) [Z. Szabo, J. Terjeki] 4. For which cardinalities n do antimetric spaces of cardinality ,c exist? (X, o) is called an antimetric space if X is a nonempty set, o : X2 ->

[0, oo) is a symmetric map, o(x, y) = 0 holds if x = y, and for any three-element subset {al, a2, a3} of X

o(alf,a2f) + o(a2f,a3f) < o(aif,a3f) holds for some permutation f of {1, 2, 3}. (1Z.8) [V. Totik]

5. Let g : R -> R be a continuous function such that x + g(x) is strictly monotone (increasing or decreasing), and let u [0, oo) --> R be a :

bounded and continuous function such that t

u(t)+

j_ g(u(s))ds

is constant on [1, oo). Prove that the limit limt.,,. u(t) exists. (F.37) [T. Krisztin]

6. Let T be a bounded linear operator on a Hilbert space H, and assume that IIT111 < 1 for some natural number n. Prove the existence of an invertible linear operator A on H such that IIATA-111 < 1. (0.5) [E. Druszt]

7. Prove that if the function f : R2 -+ [0, 1] is continuous and its average on every circle of radius 1 equals the function value at the center of the circle, then f is constant. (F.38) [V. Totik] 8. Prove that any identity that holds for every finite n-distributive lattice also holds for the lattice of all convex subsets of the (n -1)-dimensional Euclidean space. (For convex subsets, the lattice operations are the settheoretic intersection and the convex hull of the set-theoretic union. We call a lattice n-distributive if n

n

x A (V yi) = V (x A( V yi)) i=O

j=0

Ouh and

i(x-h,x+h)n(IR\E)l >uh for all sufficiently small positive values of h. (M.15) [K. I. Koljada]

10. Let R be a bounded domain of area t in the plane, and let C be its center of gravity. Denoting by TAB the circle drawn with the diameter AB, let K be a circle that contains each of the circles TAB (A, B E R). Is it true in general that K contains the circle of area 2t centered at C? (G.34) [J. Sziics] 11. Let Mn C Rn+1 be a complete, connected hypersurface embedded into the Euclidean space. Show that Mn as a Riemannian manifold decomposes to a nontrivial global metric direct product if and only if it is a real cylinder, that is, M" can be decomposed to a direct product of the form Mn = Mk x ]fin-k (k < n) as well, where Mk is a hypersurface in some (k + 1)-dimensional subspace Ek+1 C Rn+l, Rn-k is the orthogonal complement of Ek+1. (G.35) [Z. Szabo] 12. Let X 1 ,.. . , Xn be independent, identically distributed, nonnegative random variables with a common continuous distribution function F. Suppose in addition that the inverse of F, the quantile function Q, is also

continuous and Q(O) = 0. Let 0 = Xo:n < Xl:n < . . . < Xn:n be the ordered sample from the above random variables. Prove that if EX1 is finite, then the random variable 1 [ny]+1

O= sup 0 2 be an integer, and let f 2n, denote the semigroup of all mappings g : {0,1}n -> {0,1}n. Consider the mappings f E 52,,, which have the following property: there exist mappings gi {0,1}2 , {o, 1} (i = 12 n) such that for all (a a a) E {0 1}n :

f(a,,a2i...,an) _

(91(al,a,),92(al,a2),...,9fl(afl-1,an))


44

Let An denote the subsemigroup of 1l, generated by these f's. Prove that An contains a subsemigroup Fn such that the complete transformation semigroup of degree n is a homomorphic image of rn. (A.47) [P. Domosi]

3. Let n1 < n2 < ... be an infinite sequence of natural numbers such that /2k n1k tends to infinity monotone increasingly. Prove that E00 k=11/nk is irrational. Show that this statement is best possible in a sense by giving, for every c > 0, an example of a sequence ni < n2 < ... such that nk/2k > c for all k but EO° k=1 1/nk is rational. (N.19) [P. Erdos] 4. Cancelled. 5. Characterize the sets A C R for which

A+B={a+b:aEA, bEB} is nowhere-dense whenever B C R is a nowhere dense set. (T.21) [M. Laczkovich]

6. Find all functions f : ][83 -> R that satisfy the parallelogram rule f (x + y) + f (x - y) = 2f (x) + 2f (y),

x, y

and that are constant on the unit sphere of RI. (F.47) [Gy. Szabo] 7. Let K be a compact subset of the infinite-dimensional, real, normed linear space (X, II

-

II).

Prove that K can be obtained as the set of

all left limit points at 1 of a continuous function g : [0,1 [-+ X, that is, x belongs to K if and only if there exists a sequence to E [0, 1 [ (n = 1, 2, ...) satisfying limn. to = 1 and limn- 0 11g(tn) - xli = 0. (0.7) [B. Garay] 8. For any fixed positive integer n, find all infinitely differentiable functions f : ][8n -> R satisfying the following system of partial differential equations: n

bask f = 0,

k=1,2,....

i=1

(F.48) [L. Szekelyhidi]

9. Suppose that HTM is a direct complement to a vertical bundle VTM over the total space of the tangent bundle TM of the manifold M. Let v and h denote the projections corresponding to the decomposition TTM = VTM ® HTM. Construct a bundle involution P : TTM -TTM such that Poh = voP and prove that, for any pseudo-Riemannian metric given on the bundle VTM, there exists a unique metric connection V such that

VxPY-VyPX =Poh[X,Y] if X and Y are sections of the bundle HTM, and

VxY - VYX = [X,Y]


45

if X and Y are sections of the bundle VTM. (G.44) [J. Szilasi] 10. Let Y(k), k = 1, 2.... be an m-dimensional stationary Gauss-Markov process with zero expectation, that is, suppose that Y(k + 1) = A Y(k) + e(k + 1),

k = 1,2,...

Let Hi denote the hypothesis A = Ai, and let Pi (0) be the a priori probability of Hi, i = 0, 1, 2. The a posteriori probability P1(k) = P(H1IY(1), ... , Y(k)) of hypothesis H1 is calculated using the assumptions P1(0) > 0, P2(0) > 0, P, (0) + P2(0) = 1. Characterize all matrices A0 such that P{limk . P1 (k) = 1} = 1 if Ho holds. (P.29) [I. Fazekas] 1990 1. Let A be a finite set of points in the Euclidean space of dimension d > 2.

For j = 1,2,...,d, let Bj denote the orthogonal projection of A onto the (d - 1)-dimensional subspace given by the equation xj = 0. Prove that d

IB;l >

IAld-1

j=1

(G.45) [I. Z. Ruzsa] 2. Prove that for every positive number K, there are infinitely many positive integers m and N such that there are at least KN/ log N primes among the integers m + 1, m + 4, ... , m + N2. (N.20) [I. Z. Ruzsa] 3. Let n = pk (p a prime number, k > 1), and let G be a transitive subgroup of the symmetric group Sn. Prove that the order of the normalizer of G in Sn is at most (A.48) [L. Pyber] 4. Let P be a polynomial with all real roots that satisfies the condition P(0) > 0. Prove that if m is a positive odd integer, then lGlk+1.

f k=0

(k)

k!

(0) xk > 0

for all real numbers x, where f = P-"`. (F.49) [J. Szabados] 5. We say that the real numbers x and y can be connected by a b-chain of length k (where 6 : R -+ (0, oo) is a given function) if there exist real numbers xo, X1, ... , xk such that x0 = x, xk = y, and (xi_i+xz) lxi - xi-ll < b

,

i = 1,...,k.

Prove that for every function 6 : R -+ (0, oo) there is an interval in which any two elements can be connected by a 6-chain of length 4. Also, prove that we cannot always find an interval in which any two elements could be connected by a 6-chain of length 2. (F.50) [M. Laczkovich]


46

6. Find meromorphic functions 0 and z#i in the unit disc such that, for any

function f regular in the unit disc, at least one of the functions f - 0 and f - 0 has a root. (F.51) [G. Halasz] 7. Denote by B[O,1] and C[O,1] the Banach space of all bounded functions and all continuous functions, respectively, on the interval [0, 1] with the supremum norm. Is there a bounded linear operator

T : B[0,1] -* C[0,1]

such that T f = f for all f E C[0,1]? (0.8) [G. Halasz] 8. Let Al°), ... , A(°) be a sequence of n >_ 3 points in the Euclidean plane Rz. Define the sequence A111, ... , An (i = 1, 2, ...) by induction as follows: let be the midpoint of the segment where A( Z+1) = A('-1). Show that, with the exception of a set of zero Lebesgue 1 measure, for every initial sequence (Al°), ... , A(°)) E (R2)", there exists a natural number N such that the points AlN)7 ... , (N are consecutive vertices of a convex n-gon. (G.46) [B. Csikos] 9. Prove that if all subspaces of a Hausdorff space X are Q-compact, then X is countable. (T.22) [I. Juhasz] 10. Let X and Y be independent identically distributed, real-valued random variables with finite expectation. Prove that

A)

EJX +YJ > EIX -Y1. (P.30) [T. F. Mori] 1991

1. To divide a heritage, n brothers turn to an impartial judge (that is, if not bribed, the judge decides correctly, so each brother receives (1/n)th of the heritage). However, in order to make the decision more favorable for himself, each brother wants to influence the judge by offering an amount of money. The heritage of an individual brother will then be described by a continuous function of n variables strictly monotone in the following sense: it is a monotone increasing function of the amount offered by him and a monotone decreasing function of the amount offered by any of the remaining brothers. Prove that if the eldest brother does not offer the judge too much, then the others can choose their bribes so that the decision will be correct. (F.52) [V. Totik] 2. Suppose that n points are given on the unit circle so that the product of the distances of any point of the circle from these points is not greater

than two. Prove that the points are the vertices of a regular n-gon. (G.47) [L. I. Szabo] 3. Prove that if a finite group G is an extension of an Abelian group of exponent 3 with an Abelian group of exponent 2, then G can be embedded in some finite direct power of the symmetric group S3. (A.49) [G. Czedli, B. Csakany]


47

4. Let n > 2 be an integer, and consider the groupoid G = (Z U {oo}, o), where

xoy= Ix+1 ifx=yEZn, 00

otherwise.

(Z denotes the ring of the integers modulo n.) Prove that G is the only subdirectly irreducible algebra in the variety generated by G. (A.50) [A. Szendrei]

5. Construct an infinite set H C_ C[0,1] such that the linear hull of any infinite subset of H is dense in C[O, 1]. (F.53) [V. Totik]

6. Let a > 0 be irrational. (a) Prove that there exist real numbers a1, a2, a3, a4 such that the function f : R -> ]E87

f (x) = ex [al + a2 sin x + a3 cos x + a4 COs(ax)1

is positive for all sufficiently large x, and lim inf f (x) = 0. X_+00

(b) Is the above statement true if a2 = 0? (F.54) [T. Krisztin] 7. Given an > an+1 > 0 and a natural number p, such that lim sup n

an < µ, aµn

prove that for all e > 0 there exist natural numbers N and no such that, for all n > no the following inequality holds: Nn

rn

1:ak <EEak. k=1

k=1

(S.27) [L. Leindler] S. Prove that if {ak} is a sequence of real numbers such that 00

00

l ak l /k = oo

and E I

2"-1

n=1

k=1

1/2

k(ak - ak+1 )z/

[0, 00) be a measurable, locally integrable function, and write

H(t) :=

J0

t h(s)ds

(t > 0).


48

Prove that if there is a constant B with H(t) < Bt' for all t, then 00

L

e- H(t)

eH(" )dudt = oo. 0t

(F.56) [L. Hatvani, V. Totik] 10. Consider the equation f'(x) = f (x + 1). Prove that (a) each solution f : [0, oc) --* (0, oo) has an exponential order of growth, that is, there exist numbers a > 0, b > 0 satisfying I f (x) I < a ebx,

x>0;

(b) there are solutions f : [0, oo) -p (-oo, oo) of nonexponential order of growth.

(F.57) [T. Krisztin] 11. Does there exist a bounded linear operator T on a Hilbert space H such

that

00

00

n Tn(H) = {0} but

n Tn(H)-

n_1

n-1

{0},

where - denotes closure? (0.9) [L. Kerchy] 12. Let X1, X2, ... be independent, identically distributed random variables such that, for some constant 0 < a < 1, P {X 1 = 2k/« } = 2-k,

k=1,2,...

Determine, by giving their characteristic functions or any other way, a sequence of infinitely divisible, nondegenerate distribution* functions Gn such that sup

P{

-00<S 2 we get Op = 1 + 7p9+1 + bp23+1 = 1 + cpp'+1 where -y is a p-adic integer, whereas cp, as we can easily see, is a p-adic unit.

This guarantees that 311 is different from 1. We have proved that in the field of p-adic numbers every root of unity is necessarily a (p - 1)th root of unity. Now we show that such roots of unity indeed exist. Let go be a primitive root of unity mod p. We are going to prove that it is possible to determine numbers 0 S ai < p in such a way that the p-adic number 6=go+alp+...+a,,,p"+...

55

3. SOLUTIONS TO THE PROBLEMS

56

should be a (p - 1)th primitive root of unity. It is clear that any power of e with exponent less than p - 1 is different from 1. So it is enough to show that for a suitable choice of the numbers aj, the natural numbers gi = go+alp+ +ajpj satisfy gP-1 EE 1 (mod pi+1) This will be proved by induction. For j = 0 the statement is true, as go is a primitive root. Suppose that the statement holds for some j, that is, gP-1 = 1 + cep3+1

Let aj+1 be the solution of the congruence gox - cj (mod p). Then

1P-1 +1 +1 g3+1 = (gi + aj+17jl )P- = 1 + cjp' 1

1 + (ca -

- gj aj+17 +1

(mod pj+2).

Therefore a and its powers are different (p - 1)th roots of unity; their number is p - 1. There can be no other roots of unity since the polynomial xp-1 - 1 can have at most p - 1 roots in a commutative field. Let us now deal with the case p = 2. Similarly to the case of odd prime

numbers, we can prove that a power with an odd exponent of a dyadic number of the form 1 + can be 1 only in case the number itself is 1. So if a dyadic number is a primitive n-th root of unity then n can have no odd prime divisors. There are two second roots of unity, 1 and the number as well. There can be no more second roots -1 = 1 + 2 + 22 + .. 2'' +

of unity, because the polynomial x2 - 1 can have at most two roots in a commutative field.

We show that there are no other roots of unity in the field of dyadic numbers. Suppose there were another root of unity; then this would be necessarily a primitive root of unity belonging to some power of 2, so there would surely exist a primitive further root of unity, that is, a dyadic number g with 772 = -1. Clearly, g must have the form 77 = 1 + 2' + where r > 1. This gives

-1 - (1 + 2r)2 - 1 + 2''+1 +

22r

(mod 2r+2)

Therefore, -2 =- 2' +1 +2 2''

(mod 4),

that is, 2' + 22' -1

(mod 2),

which is clearly impossible. This concludes the proof.

Problem A.2. Let A and B be two Abelian groups, and define the sum of two homomorphisms q and x from A to B by a(r1 + x) = ar7 + ax

for all a E A.

With this addition, the set of homomorphisms from A to B forms an Abelian group H. Suppose now that A is a p-group (p a prime number).

3.1 ALGEBRA

57

Prove that in this case H becomes a topological group under the topology defined by taking the subgroups pkH (k = 1, 2, ...) as a neighborhood base of 0. Prove that H is complete in this topology and that every connected component of H consists of a single element. When is H compact in this topology?

Solution. H is clearly a commutative group whose 0 element is the homomorphism mapping every element of A to 0 E B, and for i E H, -77 is the mapping defined by a(-77) = -a77. To prove that H is a topological group, we have to check the following: (i) The intersection of the neighborhoods of 0 is 0 alone. Suppose namely

that i E pkH(k = 1, 2.... ). As A is a p-group, the order of any a E A is of the form p" for some n depending on a. Because 77 E p'H, we have 77 = p'X for some X E H, and so a77 = a(p"X) = (p"a)X = OX = 0, proving 77 = 0.

(ii) To any neighborhood U of 0 there is a neighborhood V of 0 such that V + (-V) C U, a suitable choice being V = U. (iii) If a is contained in some neighborhood U of 0, then there exists a neighborhood V of 0 such that a + V C U. Again the choice V = U will do.

(iv) The intersection of any two neighborhoods of 0 contains a neighborhood of 0, because for any two neighborhoods, one of them contains the other. We now come to the proof of completeness. We have to prove that if 771, 772, ... , ?In.... is a Cauchy sequence, then it converges to some element of H. By repeating members of the sequence, if necessary, we can assume that 71i+1 - 77i c p'H, that is, 77i+1 = 77i +p119i. Furthermore, define 770 as 0.

Now define mappings Xi (i = 0, 1, 2, ...) in the following way: If the order of a E A is pk then let aXi = a(19i + pt9i+1 + ...+pk-179i+k-1)

It is easy to check that Xi is indeed a homomorphism from A to B and Xi = 19i + pXi+1

We will show that the limit of the sequence 770, 771, ... , 77,,, ... is Xo. For this it is enough to show Xo - 71i E piH, actually Xo - 71i = p'Xi, which we prove by induction. For i = 0, we have Xo - 770 = Xo - 0 = pOXo. Using our previous observations and the induction hypothesis, we have

Xo - 77i+1 = Xo - 77i - pit9i = pZXi - pil9i = pl(Xi - 7i) =

proving the completeness of H.

pi+1Xi+1,


58

We now prove that the connected component of 0 consists of 0 alone; to prove this, it is enough to show that the intersection of all open-closed sets containing 0 is 0. As p k H is an open subgroup, it is closed as well, and as the intersection of all of them (for k = 1, 2, ...) is already 0 alone, we get the desired result. For compactness we are going to prove the following result: H is compact if and only if the index of pkH in H is finite for every k. Necessity of this condition is easy to establish: the cosets X + pkH(X E H) cover H. If H is compact, then already a finite number of them have to cover H, which is precisely what the condition says. Suppose now that all the subgroups pkH have finite index. We are going to show that if a family of closed subsets of H has the property that any finite number of them have a nonempty intersection, then the whole family has a nonempty intersection. First, we show the following: let UA(A E A) be subsets of the set H such that the intersection of any finite number of them is nonempty.

Suppose that H is the disjoint union of two subsets S and T. Then at least one of the families s n UA(A E A) and T n UA(A E A) inherits the same intersection property. (In particular, either S or T meets all the UA.) Suppose that neither family inherited the property. Then we would have values )\1 i .'2, ... , .\,. and )r+1, )tr+2, , \r+9 such that the intersection of the s n Ua; and also of the T n Ua,,+j would be empty (i = 1, 2, ... , r, j = 1, 2, ... , s). In other words, the sets

tnUa,) n S and (uA+) n T are empty, or equivalently

fl Ua; C T and

(UA+.) C S.

Consequently,

/ c T n s= O,

UA: I (rn+ x=1

a contradiction. We then get the same type of statement for any decomposition of H into a finite number of pairwise disjoint subsets. Let us now consider closed subsets U,\ (A E A) of our topological group H

satisfying the finite intersection condition. Repeatedly applying the above procedure, we see that there exists a sequence 711, 772.... , 77k.... of elements of H for which

,q1+pH

772+p2H2...

1)k+pkH

...

(1)

and such that for any k the intersection of the UA(A E A) with 77k + pkH also satisfies the finite intersection condition; in particular, any UA and r!k +pkH have a common element Xk,A

3.1 ALGEBRA

59

In view of (1) and the completeness of H, the sequence il, iJ2, ... , rik, ... has a limit q for which 7ik + pkH = ri + pkH. For any A we have Xk,a E ri +P k H; thus the limit of the sequence Xl,a, X2,a, . , Xk,a, . is r). Since U. is closed and Xk,A E Ua, we get that r) is contained in every Ua, which completes the proof of the compactness of H.

Problem A.3. Let R = Rl ® R2 be the direct sum of the rings R1 and R2, and let N2 be the annihilator ideal of R2 (in R2). Prove that R1 will be an ideal in every ring R containing R as an ideal if and only if the only homomorphism from R1 to N2 is the zero homomorphism.

Solution. First suppose there is a ring R containing R as an ideal and such that Rl is not an ideal of R. Since R is an ideal in R, we have Rlr C R

and rR1 C_ R for every r E R. Suppose for example, that rR1 C Rl for some f E R. (The case Rlr C Rl can be treated similarly.) Using the element r we will define a nonzero homomorphism from R1 to N2. The direct sum property implies that for every rl E R1 the element rr1 E R can be uniquely decomposed in the form rr1 = g(rl) +h(r1), where g(rl) E Rl and h(ri) E R2. h will be the desired homomorphism. It is clearly defined for all elements of R1. Furthermore, (i) h(ri) E N2 for every r1 E R1. Namely, for every r2 E R2 we have h(rl)r2 = g(rl)r2 + h(rl)r2 = (g(r1) + h(ri))r2 = (rrl)r2 = r(rlr2) = 0

and

r2h(rl) = r2(frl) = (r2r)rl E RR, C R1, but on the other hand, r2h(rl) E R2, proving r2h(rl) = 0, which gives (i).

(ii) h(ri +r'1) = h(ri) +h(ri) for every rl,ri E R1. Since r(r1 + ri) = g(r1 + ri) + h(ri + ri) = rr1 + rr'

= g(rl) + h(ri) + g(ri) + h(ri) = g(r1) + g(ri) + h(r1) + h(ri), where g(rl)+g(ri) E R1 and h(r1)+h(ri) E R2, we get the validity of (ii).

(iii) h(rir') = h(rl)h(ri) for every rl,ri c R1. We will show that both sides of the equality are equal to zero. For the right-hand side, this is clear because h(ri) is an annihilator of R2 and h(ri) E R2. For the left-hand side, we have to consider the product r(rlri): r(rlri) = (rrl)ri = (g(rl) + h(r1))ri = g(rl)ri E R1,

so h(rlri) = 0. This establishes the first part of the statement. For the reverse statement, we use the same idea. We assume that there is a nonzero homomorphism h from Rl to N2, and we construct the element r in such a way that it should produce h.


60

We denote by Ro the zero ring over the additive group of the integers.

The element of R0 corresponding to n E Z will be denoted by n. The additive group of the ring R is defined by R+ = Ro ®Ri ® R2 , and for the elements of R

r=n+rl+ r2,

r'=n'+r'1+ r2,

we define the multiplication by rr' = r1r1 + r2r2 + nh(ri) + n'h(ri).

It is routine to check that R is a ring and R is a subring of it. In fact, R is even an ideal of R in view of RR C R. To verify that R1 is not an ideal of R, take an r1 E R1 with h(ri) # 0.

Then 1r1 = h(rl) V R1 as h(ri) E R2, and this shows that R1 is not an ideal of R.

Problem A.4. Call a polynomial positive reducible if it can be written as a product of two nonconstant polynomials with positive real coefficients. Let f (x) be a polynomial with f (0) # 0 such that f (x') is positive reducible for some natural number n. Prove that f (x) itself is positive reducible.

Solution. Consider the product representation of f (xn):

.f (x") = flgj (x),

(1)

j=1

where the gj (x) are polynomials with positive coefficients. Suppose that some of the polynomials gj(x) contain a term cjkxk with nonvanishing cjk such that n f k. Then the product on the right-hand side of (1) contains the term cjk fli#j gi(0)xk. The coefficient of this term does not vanish, since r['=1 gi (0) = f (0) # 0. Since the polynomials gj (x) all have positive coefficients, the right-hand side of (1) will contain the term xk with a nonvanishing coefficient, which is a contradiction in view of n { k. This means that all the polynomials gj (x) have the form I; gj(x) _ >bjlxmm

(bjl >, 0,l = 0,l,...,lj),

i=O

that is, gj (x) is actually a polynomial of xn: gj (x) = gj (x"). So upon substituting x" = y, we get a

f (y) = f9j (y), j=1

proving the statement of the problem.

3.1 ALGEBRA

61

Problem A.5. Prove that the intersection of all maximal left ideals of a ring is a (two-sided) ideal.

Solution. Let us denote by B the intersection of all maximal left ideals of the ring R (if there are no maximal left ideals, the statement is void). B is clearly a left ideal, so we only have to prove that b E B and r E R imply br E B, or equivalently br V B implies b V B. Since an element is not in B if and only if there is a maximal left ideal not containing it, we actually have to prove the following statement: If for some elements b E B, r E R there is a maximal left ideal M with br V M, then there exists a maximal left ideal N with b V N. Let

N= {xeRI xrEM}. We are going to prove that N has the desired properties. 1. N is a left ideal: if s E R and x E N, then xr E M implies sxr E M (M being a left ideal), which in turn implies sx E N. Also x E N, y E N clearly imply x - y E N.

2. bVNasbrVM. 3. N is maximal. To prove this, we have to show that for any element a of R with a N the only left ideal of R containing both a and N is R itself. We have ar V M since a V N. As M was maximal, every element of R is contained in the left ideal generated by ar and M; in particular, for every c E R, cr can be written in the form

cr = yar + nar + in, where y E R, m E M, and n is an integer. This implies

(c-ya-na)r=inEM, so

d = c - ya - na E N, but this shows that c = ya+na+d is contained in the left ideal generated by a and N, which was to be proved.

Remark. If R has an identity element, then B is of course the Jacobson radical.

Problem A.6. Let R be a finite commutative ring. Prove that R has a multiplicative identity element (1) if and only if the annihilator of R is 0 (that is, aR = 0, a E R imply a = 0).

Solution 1. If R has an identity element e, then aR = 0 (a E R) implies 0 = ae = a, so the annihilator of R is indeed 0.


62

Conversely, suppose R is a finite commutative ring whose annihilator is

This implies that to any element a of R different from 0 there is an element b of R such that ab # 0. If R = (0), then R surely has an identity element. So suppose R :A (0) and ao is an arbitrary element of R different from 0. Then the remark made above shows that for any natural number n we can find an element an E R such that the relations 0.

aoai # 0, aoala2 # 0,..., aoai...an# 0,...

hold true. Since R is finite, we have numbers m and n (0 < m < n) such that anal ... an = (aoal ... am)(am+i ... an) . (1)

This means that the set E of elements e # 0 of the ring R for which there exists a 0# d E R such that de = d

(2)

is nonempty. Choose an element e from E so that the number of elements d E R corresponding to e that satisfy (2) should be maximal. (Such 0 an e exists because R is finite.) We will show that e is an identity element of R.

Suppose, on the contrary, that for some a E R we have ae - a # 0. Then in view of (1) there are elements r, s E R such that 0 # (ae - a)r = (ae - a)rs

(3)

holds. (We allow r to be an empty product.) Equation (3) implies

ar(e - es + s) = ar #0

(4)

consequently, 0 # e - es + s E E E. If de = d, then

d(e-es+s)=de-des+ds=d-ds+ds=d;

(5)

are - ar = (ae - a)r # 0.

(6)

furthermore,

Equations (4), (5) and (6) together show that the element e - es + s E E satisfies condition (2) for more d E R than e. This contradiction shows the existence of an identity element in R.

Solution 2. Suppose the commutative ring R with n(,> 2) elements has annihilator 0. Then R cannot be nilpotent. If R were nilpotent, there

would exist an integer k >, 2 such that Rk = 0, Rk-1 # 0 would be satisfied, but this would imply that Rk-1 is some nonzero annihilator of R, a contradiction.

3.1 ALGEBRA

63

Next, we show that there is an element a of R such that no power of a is equal to 0. Suppose that to every element ai (i = 1, 2, ... , n) of R there exists a natural number li with ai' = 0. Let l be the maximum of the li (i = 1, 2, ... , n). If a product in R does not vanish, then every ai can occur in it at most (1 - 1) times. So every product with n(l - 1) + 1 factors Rn(,-1)+1 = 0, contrary to the nonnilpotency vanishes; in other words, of R.

So suppose a E R is such that all the elements aj (j = 1, 2.... ) are nonzero. Since R is finite, these powers cannot be all different, say ak = ak+l (1 > 0). This implies ak = ak+1 = ak+2i = ak+3i = ... , Choosing m big enough to satisfy ml > k, we have (aml)2 = ami . aml = ak+ml . aml-k = ak . aml-k = aml v

so aml is a nonzero idempotent of R. Let us write aml = e for simplicity. We can write R as a direct sum of the ideals A and B: R = A E B, (7) where A is the set of all elements of the form re (r E R) and B is the set

of all elements of the form r - re (r E R). Clearly, A and B are ideals of R. Any element r of R can be written in the form r = re + (r - re), thus R = A + B. Furthermore, ea = a for all a in A, whereas eb = 0 for all b in B. This implies A n B = 0, so (7) holds indeed. If B = 0 then R = A, so e is the desired identity element of R. If B # 0, then in view of 0 0 e E A, R is a direct sum of two rings, each of which contains fewer than n elements. Consider this case and suppose by induction that all commutative rings with fewer than n elements and with annihilator 0 have an identity element. (The one-element ring clearly has an identity element.) If z E A is such that

zA = 0, then in view of AB = 0 we have zR = z(A + B) = zA + zB = 0, implying z = 0. Similarly, from w E B, wB = 0 we get w = 0. This means that the annihilators of both A and B are 0, so by induction A has an identity element e1 and B has an identity element e2. But then e1 + e2 is an identity element of the ring R = A ® B. Remark. The statement of the problem is true for commutative Artinian rings, too (see R. Baer, Inverses and zerodivisors, Bull. Amer. Math. Soc., 48 (1942), 630-638).

Problem A.7. Let I be an ideal of the ring of all polynomials with integer coefficients such that (a) the elements of I do not have a common divisor of degree greater

than 0, and (b) I contains a polynomial with constant term 1. Prove that I contains the polynomial 1 + x + x2 + natural number r.

+ x''-1 for some

Solution. According to assumption (b), for a suitable f E Z[x] we have

1+xf EI.


64 Because of

(1+xf)xr+xr+l(_f) = xr

for any r > 0, we have xr E

(l+xf,xr+1).

Repeatedly applying this observation, we get (1+xf,xr+1) D (1+x f xr)

...

(1+xf,x)

1,

consequently,

l,x,...,xr E (1

+xf,xr+1),

which in turn means that we can find gr, hr E Z[x] for which 1 + x + ... + xr = (1 + xf)gr + xr+lhr

(1)

holds true. We claim that gr and hr can be chosen in such a way that

deg hr < deg f

(2)

holds. To prove this, observe that the polynomial gr figuring in (1) can be written as gr = xr+lq + p

where p, q c Z[x] and deg p < r. Rom (1) we get

1+x+...+xr = (1+xf)p+xr+1[(1+x f)q+hr], and choosing p (resp., (1 + x f )q + hr) as the new gr (resp., hr) we clearly get an hr satisfying (2) because of degp < r. Suppose s > r and

(1+xf)gs+xs+lh9, where deg hs < deg f. Subtracting from (3) xs-r times (1), we get

1 + x + ... + xs-r-1 = (1 + x.f) (g9 - grxs-r) + xs+l (hs - hr). This means that if there are indices s > r such that

hs - hrEI holds, then 1+x+...+x9-r-1

is also true.

EI

(3)

3.1 ALGEBRA

65

We prove that the ideal contains a nonzero constant polynomial. Take a p E I such that p # 0 and degp is minimal. (Such a p exists in view of

1+xf 54 0.) For any gellet

q=pu+v, where u, v E Q [x] and deg v < degp. Multiplying by a suitable nonzero integer a, we get aq = put + vi, where ul, vi E Z[x] and of course deg vi < degp still. This gives

v1 = aq - pul E I,

so by the choice of p and in view of deg vl < degp, we have v1 = 0. This means that to an arbitrary q E I we can find a u E Q[x] for which q = pu. Let p = opl, where pl is a primitive polynomial and cp E Z. Then q = pl(pu), which in view of Gauss's lemma implies Wu E Z[x]. This in turn implies that for any q E I we have pi q. In view of condition (a), this is only possible if pl = 1; consequently cp E I, thus (cp) C I. Finally, we show that there are values s > r such that I

h9 - hr E (p).

Indeed, any coefficient of a polynomial can take on at most cp values mod W; furthermore, deg hr N for which N I r + 1 and E I.

Problem A.8. Prove that in a Euclidean ring R the quotient and remainder are always uniquely determined if and only if R is a polynomial ring over some field and the value of the norm is a strictly monotone function of the degree of the polynomial. (To be precise, there are two more trivial cases: R can also be a field or the null ring.)


66

Solution. We shall work with the following definition of a Euclidean ring: let R be a ring and N the set of nonnegative integers. R is called a Euclidean ring if there exists a map cp : R -+ N having the following properties: i)

ii) For any a, b E R (b 54 0) there exist q, r E R such that

a=bq+r and

W(r) 1, 1/p+ 1/q = 1, and C(p) is a positive constant depending only on p.

Problem A.13. Consider the endomorphism ring of an Abelian torsionfree (resp. torsion) group G. Prove that this ring is Neumann-regular if and only if G is a discrete direct sum of groups isomorphic to the additive group of the rationals (resp., a discrete direct sum of cyclic groups of prime

3.1 ALGEBRA

73

order). (A ring R is called Neumann-regular if for every a E R there exists

a 0 E R such that a/3a = a.)

Solution.

In the following, "group" will always mean "commutative group", where the operation is written as addition. Instead of Neumannregularity, we shall just write regularity. First, we prove the necessity of the condition. If the group G is torsionfree, call it Go, if it is a torsion group, then it can be written as a (discrete)

direct sum: G = G1 ® ® Gi ®... , where in the group Gi all elements have order a power of pi (pi denotes the ith prime number). Let p be a prime number, and let 4tp be the map of G defined by 4tp: a pa (a E G). cp is clearly an endomorphism. Because of the regularity, there is an endomorphism Tp with (DpPAp = 4'p. Then for any a E G, pa =

p2,P p(a).

(1)

It follows from (1) and the fact that Tp is an endomorphism that if pea = 0, then pa = 0. Therefore every element (# 0) of Gi (i > 0) has order pi. In case G = Go, we can cancel (1) by p, which shows that every element a of Go is divisible by every prime, thus Go is a divisible group. Furthermore - as Go is torsion-free - the quotient is uniquely determined. Let po = 0, and denote by KO the field of rational numbers, whereas for i = 1, 2, ... , denote by Ki the field with pi elements. From our previous observations we see that Gi is a vector space over the field Ki (i = 0, 1, 2.... ). Invoking

Zorn's lemma, we see that Gi has a basis that is just equivalent to the direct sum decomposition formulated in the problem. Sufficiency will follow if we prove the following stronger statement: The endomorphism ring of the direct sum G = Go ® G1® ® Gi ®... is regular if Gi is a vector space over Ki (i = 0, 1, 2.... ). Here "®" can mean either discrete or complete direct sum. First, we prove that the endomorphism ring of each Gi is regular. Let a be an endomorphism of Gi. Then a is a linear transformation of

Gi as a vector space. Since in a vector space every subspace is a direct summand, there exist subspaces Ui, V such that

Gi=Keroz ED UU=U®Ima. It is clear that a maps the elements of Ui onto Im a in a 1-to-l way. Let us define 0: Gi -+ Gi by the conditions /3(a) = 0 for a E V and /3(a) = b for a E Im a, where b is the unique element of Ui with a(b) = a. /3 is clearly an endomorphism of Gi with a/3a = a. This proves the regularity of the endomorphism rings of each of the Gi (i = 0, 1, 2.... ). Now let a be an endomorphism of G. Since for every prime p, if a E G

has order p, then a(a) again has order p or it is 0, a maps Gi to itself if i 3 1. Now let a E Go, and suppose a(a) E Gi (i 3 1). We know pia(a) = 0. As Go is a vector space over KO, there is an x E Go with pix = a. So pl a(x) = 0, thus a(x) E Gi, so pia(x) = a(a) = 0. Applying


74

this argument not only to a itself, but to a and subsequent projections to Gi (i > 1), we see that each component of a(a) belonging to some Gi (i >, 1) is zero, so a maps Go to itself.

These observations imply that every a uniquely determines a vector (ao, al, a2, ...) where ai is an endomorphism of Gi, and conversely every such vector uniquely determines an endomorphism a of G from which we get back the original vector. We clearly have (a,3)i = a10i. Now let a be an endomorphism of G, (ao, al, a2i ...) the corresponding vector. We have already proved the existence of endomorphism Ni of Gi such that ai/iai = ai (i = 0, 1, 2.... ) . The endomorphism / i belonging to (,do, 01, /32, . . . ) clearly satisfies afa = a, so the endomorphism ring of G is regular.

Problem A.14. Let 2t = (A; ...) be an arbitrary, countable algebraic structure (that is, 2t can have an arbitrary number of finitary operations and relations). Prove that 2t has as many as continuum automorphisms if and only if for any finite subset A' of A there is an automorphism 7rA' of 2t different from the identity automorphism and such that (x) 7rA' = x

for every x E A'.

Solution. Suppose first that the condition mentioned in the problem is not satisfied. Let A' be a finite subset of A with the property that every automorphism of 2t fixing A' pointwise is necessarily the identity. Then for any two automorphisms ir1 and ire that satisfy (x)7r, = (x)7r2 for all x E A', irlir2 1 is the identity, so 7r1 = ire. The number of automorphisms of 2t is therefore less than or equal to the number of mappings of A' to A, and so the number of automorphisms is countable. Suppose now that the condition mentioned in the problem is satisfied. We can assume that A = {1, 2.... }. We define recursively an increasing sequence of finite subsets of A: A0 C Al C . . . C_ An C ... and a sequence of automorphisms of %: 7r1, ... , irn, .... Let A0 = 0. Now let n>, 0, and suppose we have already defined An, which is finite, and 7ri for all i < n. Let 7rn be an automorphism of 2t that fixes An pointwise and is not the identity, and let

An+1 = {1, ... , n} U U (An)iri1 ... 7rIn U {min{k I

(k)7rn

k}}

.

(E1,....En eiE{0,1}

Then An+1 is again finite, and by this procedure we have defined An and ir,,, for all n. The sequence so defined has the following properties: a) 1rn is the identity on An but is different from the identity on An+1 ,/a) If ei E {0,1} (i = 1, ... , n), then

n 1 ... (An)-7r1

i nn E

C An+1

3.1 ALGEBRA

75

'y) U°°_1An = A and all the An are finite.

Let (e1.... ,en) ...) be an arbitrary infinite sequence with en E {0, 1} (n = 1, 2, ... ). Let us define the product fn° 1 7rn = 7r in the following way. If k E A, let (k)7r = (k)7r" ...7rnkk,

(1)

where nk denotes the smallest number for which we have k E Ank . Because

of y), it is indeed a map of A into itself. For any n > nk, we have E' E (k)7r = (k)7r1 ...7rn" = ((k)7r1

Enk+1

E1

.

.

Enk

E Enk . 7rnk )ink+1 ... 7rnn = (k)7r1 ...7rnk

(2)

because in view of 0) we have (k)7r11 7rnk+11

7r nkk E Ank+l and in view of a)

, 7rn are all equal to the identity map on Ank+1. Taking into

consideration that 2i contains only finitary operations and relations, we see that in view of (2) it preserves operations and relations. For every 1 E A there exists an n with 1 E An+1. Thus, there exists a k with (k)7ri1 ...7r.Enn = 1. Because of a) we have for any m > n (k)7ri1 ... 7r /2 = 1 so in view of (2) we have (k)7r = 1. Thus it is an automorphism. We are going to show that for (E 1' ... , En, ... ) 7r = 11' 7r' = 11(E'1, ... , E'n ... n=17rEn n n=1 7rn is satisfied. Let n be the smallest number for which En e . We may assume En = 0, E'n = 1. Then 7ri1 .. 7rnn i' = 7r11 ... 7r " i' . Let 1 E An+1 be an element with (l)7rn 1. Let k be the number for which (k)7ri' ... 7rnn it = 1. In view of a) and (2) we have (k)7r = (k)7r1' ... 7rTEyn = (k)7r1' ... 7rnn i' = 1,

so (k)7r' = (l)7rn $ (k)7r = 1. This means that the number of different automorphisms of 21 is at least as much as the number of infinite sequences

of 0's and 1's. Taking into consideration the fact that A is a countable set, we see that the cardinality of the set of automorphisms of 21 is indeed continuum.

Remark. The following example shows that the statement of the problem is no longer true if we allow relations with infinitely many variables. Let 2l = (A, R), where A is the set of natural numbers and the relation R(al,... , an, ...) with countably many variables is true if and only if we have an = n with finitely many exceptions. Then for any automorphism it of 21, (n)7r = n holds with finitely many exceptions. Thus 21 has only countably many automorphisms although the condition for finite subsets formulated in the problem is clearly satisfied by this 21.

Problem A.15. Let G be an infinite group generated by nilpotent normal subgroups. Prove that every maximal Abelian normal subgroup of G is infinite. (We call an Abelian normal subgroup maximal if it is not contained in another Abelian normal subgroup.) Solution. Let the group G be generated by nilpotent normal subgroups, and suppose it has a maximal Abelian subgroup A that is finite.


76

The centralizer C of A in G is a normal subgroup of G that has finite index in G. Indeed, if we map every element g of G to the mapping x " g-1xg of A, then this is a homomorphism from G to the automorphism group of A and the kernel of this homomorphism is C. Therefore G/C is finite.

Suppose B is an Abelian normal subgroup in G. Then we have IBI n, where n = IAI IG : Cl. Indeed, we have C n B = An B, because otherwise the Abelian normal subgroup A(C n B) of G would strictly contain A. Therefore I B : A n BI < IG : CI because B/A n B = B/C n B -_ CB/C G/C, which gives Let H be a nilpotent normal subgroup of G. We claim that the nilpotency class of H is at most 2n - 2. Consider namely the lower central series of H:

H=Hl>H2>...>Hk>Hk+1=1.

It is well known that for the members of the lower central series, [Hi, Hj]
k/2 we have [Hr, Hr] < H2r = Hk+1 = 1, that is, the characteristic subgroup Hr of H is Abelian. By what we said previously, this implies J Hr I < n. This clearly implies that there can be

at most n Hr such that r falls in the range k + 1 >, r > k/2, that is, k+1 < 2n-1 holds. The group G is nilpotent. Indeed, for any elements 91, 92.... ,92n-1 E G, [... [[91,921,931,-

is true, because 91, 92,

,

92n-1] = 1

, 92n-1 belong to some subgroup of G whose nilpo-

tency class is at most 2n - 2. It is true that C = A. If this were not true, then the nilpotent group G/A would contain the nonidentity normal subgroup C/A. By a well-known theorem, this implies that the center of G/A has a nonidentity intersection with C/A; in other words, there is an element g E C \ A for which Ag is contained in the center of G/A thus the subgroup (A, g) would be an Abelian normal subgroup properly containing A. All this would imply that G is a finite group. Indeed, the subgroup C being identical with A is finite and we know that it has finite index in G, is finite.

Problem A.16. Let p > 7 be a prime number, ( a primitive pth root of unity, c a rational number. Prove that in the additive group generated by the numbers 1, (, (2, S3 + (-3 there are only finitely many elements whose norm is equal to c. (The norm is in the pth cyclotomic field.)

Solution. Let L(k)(x) = x1 + (kx2 + b2kx3 + (3k + (-3k )X4

(k = 1,...,p - 1).

It is enough to show that the inequality I L(1) (x) ...

L(P-1)(x)

I < ICI

(1)

3.1 ALGEBRA

77

has only a finite number of rational integer solutions x1i x2, x3, x4. For any such solutions 2P-11cl

>

f

P-1 I L(k>(x) I> 2P-1 JJ I Im L(k) (x)

(r

-2k)x3 - (-k )X2 + ((2k - (-2k

P-1

2P-1

k=1

P-1

= [f l

Sk

k=1

k=1

P-1

fl(r k- (-k) = k=1

P-1 fl(x2+((k+(-k)x3) l

k=1

=P IF(x2,x3)I.

But for any k with p f k we know that rk + C-k is an algebraic number Thue's theorem we see that the homogeneous polynomial F(x2i x3) with rational integer coefficients takes on rational integer values only at finitely many places, so there are only a finite number of possible values for x2 and x3. Therefore, it is enough to show that for any fixed x2 and x3, the number of possible values of x1 and x4 is also finite. For x2 = x3 = 0, the proof of this statement is analogous to the previous proof. If one of x2 and x3 is different from 0, then the absolute value of each of the L(k) (x) is

of degree (p - 1)/2 > 3, so by applying

bounded from below in view of IL (k) (x) I > IIm L(k) (x) I by a positive bound

independent of xl and x4i and as the product of them is bounded from above in view of (1), their absolute value is also bounded from above by a bound independent of xl and x4. But then the system of equations x1 + (r3 + (-3)x4 = L(1) (x) - ((x2 + 2x3), x1 + ((6 + C-6)X4 = L(2) (x) - ((2x2 +(4 X3)

guarantees that Ix1I and Ix4I are bounded from above, so the number of possible pairs xl, x4 is indeed finite.

Problem A.17. We have 2n + 1 elements in the commutative ring R:

a,a1i...,an,P1,...7On Let us define the elements n k

Qk = ka +

aiei i=1

Prove that the ideal (ao, or1, ... , Qk, ...) can be finitely generated.

Solution. First, we are going to show that we can assume that R is a ring with identity. There is a standard way to embed an arbitrary ring R as


78

an ideal into a ring with identity R+: as a set R+ consists of the ordered pairs (a, n) with a E R and n an integer, and the operations are defined in the following way:

(a, n) + (b, m) = (a + b, n + m),

(a, n) (b, m) = (ab + ma + nb, nm).

R can be identified with the ideal of R+ formed by the elements of the form (a, 0), and a subset of R generates the same ideal in R and R+. To verify

this statement, take a subset H of R and denote the ideals generated by H in R and R+ by I and J, respectively. Clearly, I C_ R n J c J. On the other hand, I consists of elements of the form (a, 0), and multiplying these elements by elements of the form (b, m), we again get elements of I so I is an ideal of R+, too. This proves J C I. This shows that the ideal of R mentioned in the text of the problem is an ideal of R+, too, and if it is finitely generated as an ideal of R+, then the same elements clearly generate it as an ideal of R, too. Therefore, from now on we can safely assume that R is a ring with identity. We are going to prove the following generalization of the problem: Theorem. Suppose we have polynomials fi (x) (i = 0, ... , n) with coefficients in the ring (with identity) R. Using these polynomials and fixed elements eo, ... , en of R, we form the elements n (k = 0,1,2,...).

ak = >fi(k)e i=o

Then the ideal (co, a1 i ... , ak, ...) is equal to the ideal (uo, ... , a9) where

s = E(deg fi + 1). We get the original problem as the special case where the polynomials

are fo(x) = x, fi (x) = ai (i = 1,... , n), and go = 1. Proof. Let ri be the degree of fi(x). If n

9-1

i=o

j=o

F(x) = fl(x - ei)r:+1 = x9 - E,Qjx and D is the operator of taking the derivative with respect to x and then multiplying by x, then Dm(xkF(x)) vanishes at the place ei if m f ( s + k)0 i=0 n

+k

ri

=

_Y.,, (s + k)mQ?+k i=0 m=0

r; 8-1 E7'm,ioj (j + k)

8-1

n

ri

j:0j E E 1'm,i (j+ k)m [Ji+k

i=0m=0j=0 8-1

n

j=O

i=om=O

9-1

I:OjI:fi(j + k)e'+k = I:OjO'j+k j=0

i=0

j=0

This means that Qs+k is contained in the ideal generated by Qt with smaller indices, and this proves the theorem.

Problem A.18. Let G and H be countable Abelian p-groups (p an arbitrary prime). Suppose that for every positive integer n, pnG 54 pn+1G.

Prove that H is a homomorphic image of G.

Solution. We say that an element g E G (g : 0) is of infinite height if for every natural number n there is an x E G that satisfies pnx = g. The elements of G that are of infinite height together with 0 form a subgroup A of G. Let G* = G/A be the factor group of G with respect to A. Clearly, G* is a finite or countably infinite Abelian p-group. G* contains no elements

of infinite height. To see this, pick a g* E G* such that pnx* = g* has a solution x* E G* for every natural number n. Take an element g E G which is mapped to g* by the natural homomorphism G -> G* and an x E G which is mapped to x* (for some fixed n). We see that pnx - g is mapped to 0, that is, pn - x E A, so pnx - g = pny for some y c G, which gives g = pn(x - y), and this in turn implies g E A, that is, g* = 0. Now a well-known theorem of Priifer gives that G* is a direct sum of cyclic groups (in the case where G* is finite, this follows more easily from the fundamental theorem of finite Abelian groups). Now we show that for every natural number n, pnG* # pn+1G*

We know that pnG # pn+1G, so for some h E G the equation pnh = pn+lx

has no solutions in G. We want to show that pnh* = pn+ly* has no solutions y* E G* (h* denotes the image of h in G* with the natural homomorphism). Otherwise, for a suitable y E G, pnh - pn+ly would be

80


mapped to 0 E G*, so p1h - pl+ly = pn+lz would hold for some z E G, giving the contradiction p'h = pfl+l (y + z). (This, incidentally, shows that G* cannot be finite.) Suppose the direct sum decomposition of G* is 00

G* = ®Ci, i=1

where Ci is a cyclic group of order pki . Then the set {k1, k2, ... } cannot be bounded. This follows from the fact that if, for some n, all Ci were of order at most p', then pEG* = pn+IG* = 0 would hold, contrary to our previous observation. Suppose ci is an element generating Ci. Then every element g* E G* can be written uniquely in the form g* = EO° 1 aici, where ai is an integer taken modulo pki and, with finitely many exceptions, all the ai

are equal to 0. Now take a finite or countably infinite Abelian p-group H, the elements of which are 0, h1, h2, .... Choose a series cil, ci2,... in such a way that the order of ci, is greater than the order of hi. Then we have a homomorphism of Ci, onto the cyclic subgroup generated by h3 . We map all the other Cl to 0. Since G* is a direct sum of these cyclic subgroups, there is a homomorphism of G* to H extending the above mentioned homomorphisms

of the cyclic subgroups. As every hj is the image of some element of Ci,, this extended homomorphism is clearly onto. The composition of the natural homomorphism G -+ G* and our homomorphism is the required homomorphism of G onto H.

Problem A.19. Let G be an infinite compact topological group with a Hausdorff topology. Prove that G contains an element g 1 such that the set of all powers of g is either everywhere dense in G or nowhere dense in G. Solution. Suppose G is an infinite compact group with the property that the closure of every cyclic subgroup (# 1) has an inner point, that is, every

closed subgroup (# 1) of G is open. We are going to prove that G has a dense cyclic subgroup. First, let G be commutative. The closure of a cyclic subgroup A :/- 1 is open so it has a finite index n in G; therefore, the continuous isomorphism

cp : x , xh (x E G) maps G to this subgroup. Gn # 1 is a compact, thus closed, thus open subgroup of G, therefore Gn n A is dense in Gn. So the cyclic subgroup cp-1(Gn n A) is dense in G. Now we show that the index of every open subgroup of G is a power

of the same prime number. If p is a prime divisor of n, then because GP, (i = 0, 1, 2, ...) form a strictly decreasing GP : G the open subgroups chain, so their intersection is a closed subgroup of infinite index, and so this intersection is the identity. The topology of G that we get by choosing GPi as a base for neighborhoods of 1 is coarser than the the subgroups

3.1 ALGEBRA

81

original compact topology of G, so it is equal to it. Consequently, every open subgroup of G has index a power of p, since it contains a subgroup some Gpi .

In the following, we do not suppose that G is commutative. The center Z of G is an open subgroup. Indeed, the closures of the cyclic subgroups (# 1) of G are open subgroups and cover G, so finitely many of them cover G already, the intersection of these is a subset of Z, and therefore Z is an open subgroup. The group G/Z is a p-group. Indeed, if every open subgroup of Z has index a power of p, and H < G is a subgroup that contains Z as a subgroup of index q (q is prime, therefore H is commutative), then all elements (# 1) of the factor group H9/H12 (0 1) formed using open subgroups of Z have order q; thus p = q and consequently the finite group G/Z is a p-group. Finally, we prove that G is actually commutative. Let V be a subgroup

with Z < V < G such that the center of G/Z is VIZ. Then for every g E G, taking the endomorphism of V defined by x -> [x, g] (x E V), the image of V will be finite (as the kernel of the endomorphism contains a subgroup Z that itself has finite index) so this image is the identity, thus V = Z and thus G = Z, because G/Z is a finite p-group. 0 Remarks.

1. Paragraphs 3, 5 and 6 of the elementary solution described above can be omitted if we use a theorem of Baer which states that if the factor group with respect to the center of a (discrete) group G is finite, then the commutator subgroup of G is also finite. 2. Jozsef Pelikan noticed the following. If we use the fact that every infinite compact group has a proper closed subgroup, then the statement of the problem can be reformulated in the following way:

Statement. If G is an infinite, compact, topological group, every closed subgroup (# 1) of which is open, then G is topologically isomorphic to the topological group of the p-adic integers (p prime).

Problem A.20. Let G be a solvable torsion group in which every Abelian subgroup is finitely generated. Prove that G is finite.

Solution. A finitely generated Abelian torsion group is finite. Therefore, the problem is equivalent to the following one: If in a solvable torsion group

every Abelian subgroup is finite, then G itself is finite. We are going to prove the following stronger statement: If in a solvable torsion group every Abelian normal subgroup is finite, then G itself is finite. First we prove a lemma: If a group G has a commutative normal sub-

group with a commutative factor group, and furthermore every Abelian normal subgroup of G is finite, then G itself is finite. Let H be such a normal subgroup. Using Zorn's lemma, we see that the partially ordered set of all Abelian normal subgroups of G containing

82


H has a maximal element F. If an element g is permutable with every element of F, then F and g generate a commutative normal subgroup (it is normal because G/H is commutative), so by the maximality of F we have g E F, that is, F is its own centralizer. Consequently, G/F is isomorphic to a subgroup of Aut (F) that is itself finite in view of the finiteness of F. Thus G is also finite. Now we prove the statement using induction on the length of the derived series of the group G. If this length is 1, there is nothing to prove. If

the length is 2, the statement follows from the previous lemma. In the general case, consider P, the last term of the derived series, different from the identity. P is a commutative normal subgroup (hence finite) and the length of the derived series of G/P is one less than the length of G. We

only have to check that every commutative normal subgroup of G/P is finite. But if N/P is a commutative normal subgroup of G/P, then N satisfies the conditions of the lemma, so N is finite (hence N/P is also finite), completing the proof. Remark. In the problem, the condition on the solvability (or some weakened form of it) cannot be dropped: Novikov and Adian constructed examples of infinite groups where the order of the elements is finite (in fact bounded) but every Abelian subgroup is finite (actually, cyclic).

Problem A.21. We say that the rank of a group G is at most r if every subgroup of G can be generated by at most r elements. Prove that there exists an integer s such that for every finite group G of rank 2 the commutator series of G has length less than s.

Solution. We start by showing that if G is a finite group of rank 2, N a normal subgroup of it, and N* the intersection of the normal subgroups of N that have prime index, then the following statement is true:

Statement. If in every rank-2 automorphism group of N/N* the length of the commutator series is at most n(> 0), then h(G), the length of the commutator series of G, is at most n + 2. Proof. The condition implies that the group obtained by restricting the inner automorphism of G/N* to N/N* has a commutator series whose length is at most n. In other words, taking the member G(n) of the commutator series G >, G' 3 G" > ... and forming the commutator group [N, G(n)], this will be a subgroup of N*. Now in case N < G(n), applying this observation to N' in place of N we see that the group N'/N'* is cyclic, for if a and b generate N, then the commutators [a, b] and [N', N] generate

N', so N'/N" is cyclic. Thus N'/N"* is commutative, because - again by the first observation - N"/N"* is a subgroup of the center of N'/N"*, which gives N"* = N", and thus N"' = N". Therefore, h(G) n + 2. We now show that for every subgroup A of the group L = GL2 (p) of 2 x 2 invertible matrices over the field of p elements (p prime), we have h(A) < 5 (this bound is not sharp).

3.1 ALGEBRA

83

In the group L of order (p2 - 1)p(p - 1), the matrices of determinant 1 form a normal subgroup S = SL2(p) and the factor group L/S is commutative. If c is a generating element of the multiplicative group of the field with p elements, then the diagonal matrix whose diagonal entries are c and c-1 is an element of order p - 1 in S. By multiplying with a generator element of the multiplicative group of the field with p2 elements, we get an automorphism of order p2 - 1 of the same field; this implies that S has an element of order p + 1. As a consequence, we see that every Sylow subgroup of odd order of S is cyclic, since any odd prime power dividing (p - 1)p(p + 1) divides precisely one of the three factors. Therefore, if A is a subgroup of rank 2 of L, then for every normal subgroup N of An S, any automorphism group of N/N* has a commutator series of length at most 2 (because N/N* is a direct product of a cyclic group of odd order and at most two groups of order 2); therefore, by the statement first proved we have h(A fl s) < 4, and consequently h(A) < 5. Now for every normal subgroup N of a finite group G of rank 2, it is true that every Sylow subgroup of N/N* is either of prime order or the direct product of two groups of prime order. Therefore every automorphism group B of rank 2 of N/N* is isomorphic to a subgroup of the direct product of groups A of the abovementioned type, which implies h(B) < 5, and then, once more using the first proved statement, we get h(G) < 7 (which is again not the exact bound). Remarks. 1. Instead of the straightforward proof described above, we could have obtained the statement of the problem by applying the theorem of Black-

burn on finite p-groups of rank 2 and the theorem of Zassenhaus on solvable matrix groups. 2. The statement of the problem does not remain true if we consider finite groups of rank 3 instead of finite groups of rank 2. A counterexample is provided by the p-Sylow subgroups of the automorphism group of

the direct product of two cyclic groups of order p" (p > 2 prime, n = 1, 2,

).

Problem A.22. Let R be an Artinian ring with unity. Suppose that every idempotent element of R commutes with every element of R whose square is 0. Suppose R is the sum of the ideals A and B. Prove that AB = BA.

Solution. We will use two lemmas. Lemma 1. Every idempotent e of R lies in the center of R.

Proof. Let r be an arbitrary element of R. Then (er - ere)' = 0, and consequently er - ere is permutable with e. Therefore, er - ere = e(er - ere) = (er - ere)e = 0,

er = ere.


84

We get re = ere in a similar fashion. Thus er = re, proving the lemma.

Lemma 2. If A is a nonnilpotent right ideal of a (right) Artinian ring

R, then A can be written as a direct sum A = eR ® N, where e is an idempotent element of A and N is a suitable nilpotent right ideal of R. Proof. The proof of this lemma can be found, for instance, in A. Kertesz,

Vorlesungen fiber artinsche Ringe, Akademiai Kiad6, Budapest, 1968; VEB, Deutsche Verlag, Berlin, 1975, Theorem 6.23, p.155. Turning to the proof of the statement, we see that it is enough to prove that ab E BA for any a E A and b E B. Since R is the sum of the ideals A and B, we have elements ao E A and bo E B such that 1 = ao + bo. This gives

ab = 1 ab = (ao + bo)kab = aoab + Ebiai

(bi E B, ai E A).

If A is nilpotent, then for k sufficiently large we have aoab = 0, that is, ab E BA. If A is not nilpotent, then using Lemma 2, we have A = eR ® N; in particular ao = er + n (r E R, n E N). Using Lemma 1, we get ao = exk + nk for suitable elements xk of R.

Since n is nilpotent, we have nk = 0 for sufficiently large k, that is, ao = exk, and consequently

ab = exkab + Ebiai = b'e + Ebiai E BA

(b' E B)

.

Problem A.23. Let R be an infinite ring such that every subring of R different from {0} has a finite index in R. (By the index of a subring, we mean the index of its additive group in the additive group of R.) Prove that the additive group of R is cyclic. Solution. Choose an element a E R such that the order of a in the additive group R+ is either infinite or a prime number p. Depending on these two cases, let M denote the ring of the integers or the finite field of p elements. Then the subring Ra, generated by a consists of the elements of the form f (a), where f is a polynomial with coefficients from M whose constant term is 0. Suppose f (a) 0 for every such f 0. Then the elements of the form f (a2) form a subring that has infinite index in Ra,, so also in R. Hence,

there exists a polynomial f # 0 over M with constant term 0 such that f (a) = 0. Take such an f whose degree is the least possible. We have f (x) = x(c + h(x)),

where h # 0 (because for c E M, c # 0 we know ca 0) and h is also a polynomial over M with zero constant term. Let b = h(a). Since the degree of f was minimal, we have b # 0. Then

b2 = h(a) h(a) = h(a)(c + h(a)) - c h(a) = -c h(a) = -c b.

3.1 ALGEBRA

85

This implies that the subring Rb generated by b consists of the elements of the form c b (c E M) alone; therefore, the additive group R6 is cyclic. By the assumption of the problem R6 has finite index in R+, so R+ is finitely generated and, as R+ is infinite, R5+ is also infinite. So a could not have finite order in R+, and consequently R+ is torsion-free. By the fundamental theorem of finitely generated Abelian groups, R+ can be written as a direct sum of infinite cyclic groups. But R+ contains an infinite cyclic group of finite index, namely R6 , so the rank of R+ is 1. Thus R+ is an infinite cyclic group.

Problem A.24. Let S be a semigroup without proper two-sided ideals, and suppose that for every a, b E S at least one of the products ab and ba is equal to one of the elements a, b. Prove that either ab = a for all a, b E S

orab=b for alla,bES. Solution 1. S is clearly an idempotent semigroup. Applying a theorem of McLean (see A. H. Clifford, and C. B. Preston, The Algebraic Theory of Semigroups, vol. I., AMS, Providence, R.I., 1961, P. 129), we get that there is a congruence relation O on S such that S/O is a semilattice and each class of the congruence relation O is a rectangular band. It can be seen easily that S/O itself has no proper two-sided ideals, so it consists of one element; in other words, S is a rectangular band. Recalling the definition, this means that there are sets X and Y such that S is isomorphic to the semigroup (X x Y; ), where multiplication is defined by the rule (X 1, yi)(x2, y2) = (x1, y2)

(xi E X, yz E Y).

But then the condition formulated in the problem can hold only in the case where either X or Y consists of a single element, and this proves the statement.

Solution 2. From the second and first conditions of the problem, we see in turn that (1) S is idempotent, (2) for any elements a, b E S, there exist elements x, y E S such that b = xay. Let us define relations L, R on S in the following way:

aLb 44 ab = a, aRb

ab = b.

L is clearly transitive, and in view of (1) it is also reflexive. Suppose for some a, b E S we have aLb. Then using (1) and (2), we see a = ab = axay = axay2 = (axay)y = ay, b = xay = x(ay)(ay) = (xay)(ay) = ba,


86

therefore bLa. Thus, we have proved that L (and similarly R) is an equivalence relation. The definition of L and R and the second condition of the problem give in turn that (3) aLb and aRb are both satisfied only in case a = b, (4) for any elements a, b E S, either aLb or aRb is satisfied. But for equivalence relations L, R, (3) and (4) can be true simultaneously only in the case when one of L and R is the full relation S x S.

Problem A.25. Let Z be the ring of rational integers. Construct an integral domain I satisfying the following conditions:

(a) ZCI; (b) no element of I \ Z is algebraic over Z (that is, not a root of a polynomial with coefficients in Z); (c) I only has trivial endomorphisms.

Solution 1. Choose a transcendental real number a, and let A be the set of numbers of the form f (a)/g(a), where f, g E Z[x] and the polynomial g is primitive, that is, the greatest common divisor of its coefficients is 1. A is an integral domain with identity that satisfies a) A \ Z contains no algebraic number, b) to every a E A (a # 0) there is a 3 E A (/3 0) such that a/3 E Z. Let M be the set of all integral domains I with Z < I < R that satisfy

a) and b). M satisfies the conditions of Zorn's lemma, so it contains a maximal element J. We are going to show that

c) for every a e J (a > 0) there exist elements /3k, /32, ... , I3k E J and 'y1,'y2...... yk E J such that k

a

0ti 2 =

1:n ryj

2

.

Indeed, let a c J, a > 0 be arbitrary. If a is an integer or VG- E J, then the statement holds trivially. Now let a E J \ Z, a # /32 (/3 E J), and look at the integral domain Jam; this satisfies b). If /3 + ry f # 0, (/3, ry E J), then there is a 6 E J (6 0) such that 6(/32 - y2a) E Z holds (we can assume /32 - -y2C, # 0) and so (/3 + -y/) (6/3 - 6\) E Z. Since J J[/, the maximality of J implies that J[/] cannot satisfy a). So there are ,3, ry E J with y # 0 such that /3 + ryvra- is algebraic, that

is, there is a polynomial f (x) = EN0 atixz E Z[x] with f (/3 + ry / ) = 0. We have

N

f (0 +

E ai (0 + 7v/-a-)z = C + D,/-a , i=o

where C and D, being polynomials with integer coefficients of /3, ry and a,

belong to J. If D = 0, then C = 0, so f (/3 - ry /) = C - D/ = 0, that is, /3 - 'yi is also algebraic. This implies that -yam, consequently ry2a, is algebraic, so by a) ry2a = n is an integer, and thus the statement of c)

3.1 ALGEBRA

87

holds for a. If, on the other hand, D 54 0, then C + Dvfa- = 0 implies aD2 = C2, so c) is satisfied again. Now we show that J satisfies the requirements of the problem. Let q' be a nontrivial endomorphism of J. Clearly, 0(1) = 1, and so 0(n) = n

for every n c Z. If a E J (a # 0), then b) implies q(a) 0. If a > 0, then it is easy to deduce from c) that 0(a) > 0, which means that 0 is order-preserving. But then 0 can only be the identity that we easily verify if we extend in an operation-preserving way to the quotient field of J and observe that in this way 0 becomes an order-preserving map, fixing all rational numbers.

Solution 2. We prove that the integral domain

J-zIx1 'x'x+3'x+10 1

1

1

L

satisfies the conditions of the problem. The elements of J \ Z are clearly

transcendental, so we only have to show that J has no nontrivial endomorphism. J is evidently the set of rational functions of the form f (x)/(xk(x + 3)"(x + 10)1), where f (x) E Z[x] and k, n, m >, 0. This shows that the invertible elements of J are precisely the elements of the form exk(x+3)"(x+10)1, where e = +1 and k, n, m are arbitrary integers. Let 0 be an endomorphism of J. If qi is not identically zero, then ¢(1) = 1, so the image of an invertible element is invertible. This implies ¢(x) = e1xk1(x + 3)"1 (x + 10)'"1,

(1)

O(x + 3) = e2xk2(x + 3)"2 (x + 10)m2 = O(x) + 3,

O(x + 10) = e3xk3(x + 3)"3(x + 10)"`3 = O(x) + 10,

and

(2) (3)

where Ej = +1 and ki, ni, mi are suitable integers (i = 1, 2, 3). If k1 < 0, then (1) and (2) imply that k2 = k1 (the right-hand side of (2) has a pole of order -k1 in 0, thus the left-hand side too), which gives 3x-k1 . e2xk2 (x +3) -2 (x + 10)m2 = el (x + 3)"1 (x + (4) 10)'"'1 +

Substituting x = 0, we get E23"210"`2 = E13"110'"

,

which gives E1 = E2, n1 = n2, m1 = m2. But this is impossible in view

of (4); therefore k1 > 0. We get similarly ki > 0, ni >, 0, and mi >, 0 (i = 1, 2, 3).

If nl > 0, then substituting in (3) x = -3 we get n3 = 0 and (-3)k3 713 = 0, which is clearly impossible. Therefore, n1 = 0 and similarly m1 = 0. We can't have k1 = 0, because then ¢(x) = E1 would give E3

O(x + 3) = 2 or 4, contradicting (2). Therefore k1 > 0, and substituting x = 0 in (2), we get k2 = O and E2 3"2 . 1012 = 3. This gives E2 = 1, n2 = 1, and m2 = 0; therefore ¢(x + 3) = x + 3 = O(x) + 3, that is, O(x) = x. Therefore, 0(a) = a holds for all a c J, which was to be proved.


88

Problem A.26. Let p > 5 be a prime number. Prove that every algebraic integer of the pth cyclotomic field can be represented as a sum of (finitely many) distinct units of the ring of algebraic integers of the field.

Solution. Put ( = e27z/r, and let us denote the pth cyclomatic field by K r = Q((). It is well known that 1 + C,1 + (2, ... , 1 + (P-1 and ( are units in K. So E1 = (c + C-1)2 and E2 = -((2 + b-2) are also units. Furthermore, E1 + E2 = 2.

(1)

We show that E1 and E2 are independent, that is, there are no rational integers a and b, at least one of them is nonzero, and such that E1E2 = 1.

(2)

Suppose the contrary. With the notation 4a = a', -2b = b', we deduce from (2) (Sr +

(-1)a' = -(r2 + (-2)b'

(3)

Let d denote the smallest positive integer such that 2d - 1 (mod p). Clearly, 3 < d 5 p - 1. It is well known that KP is a normal extension of Q and the automorphisms of K P are determined by ( - * c (i = 1, 2, ... , p-1). Repeatedly applying the automorphism (-4 r2 to (3), we get (2 r-2)a' ((22 + r _22 )b' (r(r22

+rS5-22)a'

+

=

(S23 +

rSS-23)b' (4)

((2 d-1

d-1

d

)ai

+ S-2

= (S 2

+(-2

d )b'

= (S

+

(-1)bi

(3) and (4) impliy

( + -1)a'd-b'd = 1 . If a'd ; b'd, then ( + C-1 is a root of unity. But ( + (-1 is a real number, so the only way it can be a root of unity is for it to be equal to 1 or -1. But this would imply that ( is a sixth or third root of unity, which is not possible. So a'd = b'd and consequently a' = +b' # 0. But then (3) implies [(S +

(-1)(S2 + -211 a' = 1

(5)

rrS+(-11a = 1

(6)

or

Cb2+ 2/

.

But the left-hand side of both (5) and (6) is a power with exponent a' 54 0 of a real number that can be equal to 1 only if the numbers themselves are equal to 1 or -1. But this would imply that c is a root of a polynomial of degree at most 6 from Z[x], which is different from the seventh cyclomatic polynomial. Since this is impossible, e1 and E2 are indeed independent.

3.1 ALGEBRA

89

As is well known, S1 = 1, (2 = (, ... , (P_1 = (p-2 form an integer basis of KK, which means that every algebraic integer of KK can be represented as a linear combination of these units with integer coefficients. Therefore, every algebraic integer of KK can be written as a sum of units of the form ±(O-I2 (1 < k < p - 1; i, j E Z), and it follows from the previous observations that these units are pairwise different.

Now let a be an arbitrary algebraic integer in Kp, and suppose that

a=

m

p-1-1

q

E E !:aijk(ke1E2,

(7)

i=1j=nk=1

where aijk, 1, m, n, q E Z. We can assume that some aijk $ 0 and that (7) is a representation for which the value of

M =laijkl

(8)

i,j,k

is minimal. Using induction on M, we prove that a has a representation of the form r s p-1

a=

ai jk(kElE22

i=1 j=n k=1

where r, s E Z and aijk = -1, 0, or 1. This is trivially true for M = 1. Suppose M > 2 and that the statement is true for all a such that laijkl < M. i,j,k

Let a be an algebraic integer in KK that can be represented in the form (7) with property (8) (if such an element exists at all). In view of (1), clearly 2(kEjE2 = rkeJ+1d + (keje'+1 . (9)

Applying (9) repeatedly to (7), any aijk with laijkl > 2 can finally be reduced to -1, 0, or 1. Furthermore, by the minimality of M, Ei j k laijkl remains unchanged during the repeated applications of (9). So, after a finite number of steps, we arrive at t

a=

r

p-1

r U

L: a,j k(ke E2 +

j=n k=1

L:

V p-1 E E bljk(k 63102 ,

(10)

i=1+1 j=n k=1

where bijk, t, u, v E Z and a=jk = -1, 0, or 1. In addition,

laijkl+Elbijkl=M j,k

i,j,k

and

laijkl j,k

so we can apply the induction hypothesis to the second term of (10), and this concludes the proof.


90

Problem A.27. Suppose that the automorphism group of the finite undirected graph X = (PE) is isomorphic to the quaternion group (of order 8). P r o v e that the adjacency matrix of X has an eigenvalue of multiplicity at least 4. ( P = {1, 2, ... , n} is the set of vertices of the graph X. The set of edges E is a subset of the set of all unordered pairs of elements of P. The group of automorphisms of X consists of those permutations of

P that map edges to edges. The adjacency matrix M = [mid] is the n x n matrix defined by mil = 1 if {i, j} E E and mi.7 = 0 otherwise.)

Solution. Let 7r be a permutation of the set P, and let A be the corresponding permutation matrix, that is, the matrix that has 1 in the ith row and jth column if i7r = j, and 0 otherwise. We see that An is an orthogonal matrix: AT = A- 1. We also see easily that the element in the ith row and jth column of An 1MAIT is just mi,rjr. This means that (1) it E Aut(X) . A,; 1MA7 = M. We are going to use the following well-known (and trivial) fact:

(2) If AB = BA (A and B are n x n matrices), then the subspace belonging to some eigenvalue of B ("eigensubspace") is an invariant subspace of A.

Let the eigensubspaces of M (in ll ) be V1, ... , V3 and dim Vi = ni. The subspaces V are pairwise orthogonal (with respect to the usual scalar multiplication), and their direct sum is IRT. (This follows from the fact that M is a real symmetric matrix.) So, applying (2) we see that the group of the orthogonal matrices that are permutable with M is a subgroup of the group O (V1) x ... x O (V9) (here x denotes direct product, O (V) the group of orthogonal transformations of the Euclidean space V; later 0(k) will denote the group of k x k real orthogonal matrices). In view of (1) Aut(X) is isomorphic to a subgroup of the group 0(ni) x x 0(n3). Since 0(1) < 0(2) < 0(3), it will be enough to prove the following: (3) The quaternion group is not isomorphic to any subgroup of 0(3) x x 0(3). To prove this, suppose that the quaternion group H = {fl, fi, ±j, ±k} has an embedding

f : H ti 0 (3)x...x0(3), and let us denote by p, the rth projection of the group 0(3) x x 0(3) to 0(3). Now f,. = f o p,.: H --* 0(3) is a homomorphism. We claim the following:

(4) The kernel of any homomorphism h : H -* 0(3) contains -1 E H. This implies (3) as it gives -1 E ker f, for all r, thus -1 E ker f, proving that f cannot be an injection. To prove (4), let us recall the full list of finite subgroups of 0(3): (5) A finite subgroup of 0(3) is isomorphic to one of the following: A5 x Z2, S4 x Z2, A4 x Z2, Z,,, x Z2, D. X Z2, A5, S4, A4, D., Z..,

3.1 ALGEBRA

91

where At denotes the alternating group of degree t, St the symmetric group

of degree t, Zt the cyclic group of order t, and Dt the dihedral group of degree t (thus order 2t) (see H. S. M. Coxeter, Introduction to Geometry, Wiley, New York, 1969, Table III).

(6) Cosequence: 0(3) has no subgroup isomorphic to H. So h : H 0(3) cannot be an embedding. Therefore ker It contains an x E H different

from the identity. But then either x or x2 is equal to -1, so necessarily -1 E kerh. This concludes the proof.

Problem A.28. For a distributive lattice L, consider the following two statements: (A) Every ideal of L is the kernel of at least two different homomorphisms. (B) L contains no maximal ideal. Which one of these statements implies the other? (Every homomorphism cp of L induces an equivalence relation on L: a - b if and only if acp = bcp. We do not consider two homomorphisms different if they imply the same equivalence relation.)

Solution. (A) IMPLIES (B). Assume, by contradiction, that L contains a maximal ideal M. Suppose we have a homomorphism cp whose kernel is equal to M. We will show that the image of L \ M by cp is a single point. Since we also know that under the equivalence relation induced by cp all elements of M are equivalent and no element of L \ M is equivalent to an element of M,

we get that there is at most one homomorphism whose kernel is M. Let a and b be two elem ents of L \ M. The following description of the ideal (M, a) generated by M and a is well known:

(M,a)={xELI ImEM : x 4)j, which maps every (where g* E 4D) to g,* E 1,. Thus A is continuous and element gi E 1-to-1 in view of the regularity and compactness of G, so it produces an automorphism of the topological group SO(2). But there are just two such automorphisms: the identity and (identifying SO(2) with the unit circle of C) the conjugation. These map a rotation with angle a to a rotation with angle a, so - using A(Vi) = cps - we get ai = aj, which was to be proved. This lemma immediately implies

Lemma 2. If some cp E G maps a unit vector to a vector orthogonal to it, then c,2 = -id and cp maps any other vector y to a vector orthogonal to it; furthermore, the vectors y, V(y) span a plane that is invariant with respect to V. Proof. Let I[82m = h1® .. G h,,,, be the decomposition whose existence is

assured by Lemma 1. Suppose that x is orthogonal to cp(x) and x = x1 + +x,,,,, cp(x) = cp(xl)+ +V(x,,,,) the Jordan decomposition with respect


94

to 0 and consider the inner product (x, cp(x)) = Ei'j(xi, cp(xi)) = 0. Lemma 1 guarantees that either (xi, cp(xi)) > 0 for all terms or (xi, cp(xi)) for all terms, therefore necessarily (xi, cp(xi)) = 0 for all terms. Thus, Lemma 1 implies that cp rotates with an angle 7r/2 in every subspace hi, and from this the statements of the present lemma follow immediately. Now the question formulated in the problem is answered by the following Proposition. If the group SO(2m) (2m > 4) has a closed, regular sub-

group G, then 2m = 4 (furthermore, it is true that G is the universal covering group of SO(3), that is, Sp(1)). Proof. Let the unit vectors e0i el be arbitrary, but orthogonal, and let

01 E G be the group element with cpl(eo) = el. Let e2 be a unit vector that is orthogonal to the plane spanned by e0, el, and let e3 = cp1(e2). By Lemma 2, the vectors e0i el, e2, e3 are pairwise orthogonal (the subspace orthogonal to an invariant subspace is itself invariant) and denoting by cpi E G the group element with cpi(eo) = ei, then the following formulas are already true: c01 = -id,

cP2 = -id, cP3 = -id, O1W3 = -P301 = -P2, W2co3 = - 302 = 01 -cpjcpi in the last row follow immediately from the relation (cpicpj ) 2 = -id. In the case 2m = 4, the proof is finished. P1 P2 = The formulas cpicpj

It remains to show that the case 2m > 4 cannot occur. Suppose 2m > 4, and let K be the subspace spanned by e0i e1, e2, e3. If we choose a unit vector e4 that is orthogonal to K and define e5 = cpl(e4), e6 = cP2(e4), e7 = W3(e4), then the vectors e4, e5, e6, e7 are pairwise orthogonal and span a subspace K*, orthogonal to K, consequently e0i e1, ... , e7 is an orthonormal system of vectors. Denoting by cpi E G the group element with cpi (eo) = ei (i = 0, 1, ... , 7), the transformations cpi have the following multiplication table: APO

coo

id

cot

cot

W1

co2

W3

W4

W5

W6

co7

cot

co2

co5

cob

co7

co3

W4

co3 -W2 W2 -co3 -id c1 co2 -cot -id (P3

co5

co4 W4 -W5 -cob -°7

-id

co2

co3

W5 coy

-id

co4

co7

cob

cob

co4

co7

co7

cob -W5

co7

co6 -W7 -co4 c°7

co5

c6 -co5 -c4 co'

cot

co3

-cob -c1 -id W3 -co2 co5 -cot -°3 -id cot co4

-co3

co2

-cot

-id

From this multiplication table, we deduce the following identities: (co4co1co4)co6 = (-co4co4co1)co6 = co1co6 = co1(co2co4) = (co1co2)co4 = co3co4 = W7 w4(co5co6) = co4(co1w4co2co4) = W4(-co1co2co4co4)

_ co4(co1w2) _ co4w3 = -co3co4 = -co7

3.1 ALGEBRA

95

This contradicts the associativity of the multiplication, so the case 2m > 4 cannot occur.

Problem A.31. In a lattice, connect the elements a A b and a V b by an edge whenever a and b are incomparable. Prove that in the obtained graph every connected component is a sublattice.

Solution. Let us denote the fact that two elements a and b belong to the same connected component of the graph by a - b. The statement to be proved is that whenever b1, ... , b, is a path in the graph, we have b1 - b1 V b,,,, which in turn reduces to b1 V b,,,_1 - b1 V bn. We will denote the fact that two elements x and y are incomparable by x 11 y. Lemma. Let x 11 y, a = x V y, b = x A y, z arbitrary. Then one of the following two cases holds:

(a)aVz-bVz; (b) one of a, b, z is - a V z and one of a, b, z is - b V z. This will indeed prove the statement of the problem. In the case b,,,_1 =

x V y and bn = x A y, we can apply the lemma with a = bn - 1, b = bn, z = b1. In case (a), the proof is finished immediately, and in case (b) we get that one of b1, bn_1i bn is - b1 V bn, which suffices to conclude the proof. In the case bn_1 = x A y and bn = x V y, we can proceed similarly.

We will prove the lemma in six steps. First, we deal with that part of the statement that refers to a V z. 1. If a < z, then a V z = a, z so the first part of (b) holds. 2. Suppose a z V y. This gives x z V y since x>, z V y would imply x > y, whereas in case x < z V y we would have a = x V y < z V y. Therefore a V z = x V y V z > x A(y V z). In view of x A a= x, we can 11

further deduce x A (y V z) = x A (a A (y V z)) < x V (a A (y V z)) as soon as we

establish x 11 a A(z V y). To prove the latter statement, suppose first x < a A(z V y). Then in view of y < a A(z V y), we have a = x V y < a A(z V y), contrary to the initial assumption. On the other hand, if x > a A(z V y), then y < a A(z V y) implies y < x, which is not the case. But x V (a A(y V z)) = a since x, a A(y V z) < a and a A(z V y) >, y, thus x V (a A (y V z)) > x V y= a. We arrived at the conclusion a- a V z. The case a z V x can be treated similarly. 3. So we can assume a < z V x, a < z V y, and a 11 z. Thus x < z V y, which implies z V a = z V y. Similarly z V a = z V x consequently z V x = z V y = z V a. We can further assume x I I z and y z. (Suppose namely x < z. This gives a V z = y V z - a. The case y < z is similar, and in case x > z or y > z we get a > z.) Suppose now z 11 (z V b) A a. This implies

zVa-zAa=((zVb)Az)Aa=((zVb)Aa)Az b, and consequently z = z V b. 4. Thus, we can assume z = z V b, z V x = z V y = z V a. If we have

z A x= b, we get z V a= z V x> z A x= b, giving z V a- b, and similarly in the case z A y = b. 5. In the remaining case let t = ((x A z) V (ynz)) . Then z V a = z V x >

z A x= x A t since z> t and z A x< t. Certainly, x 11 t since x< t< z would imply x< z and x > t implies x A y> y A z > b, which is impossible.

So xAt < xVt = xV(ynz). We have x 11 ynz since x < ynz < y is impossible and x y A z gives x A y y A z > b, a contradiction. So, finally, x V (ynz) > x A(y A z) = b since b < z. This gives a V z b, proving the part of the lemma that refers to a V z. 6. Now let us deal with the statement concerning b V z. In case b z, we have b V z = b, z. Otherwise, b V z > b A z, and applying the part of the

statement that has been already proved to the dual lattice, we get that

bAz-tooneofa,b,zorbAz-aAz. In case aZ z, we haveaAz=a,z, whereas in case a 11 z using a V z> a A z we get a V z- b V z, and this finally concludes the proof of the lemma.

Problem A.32. Let G be a transitive subgroup of the symmetric group S25 different from S25 and A25. Prove that the order of G is not divisible by 23.

Solution. We assume that 23 JGI, and we try to reach a contradiction from that assumption. Denote the stabilizer of a point x by Stab (x). By the transitivity and our assumption, all the Stab (x) are isomorphic permutation groups containing a 23-cycle. Suppose that Stab (x) is not transitive. Then taking the fixed point y of Stab (x), we see that the fixed point of Stab (y) is x. That would give a pairing of the 25 points, which is impossible. So G is 2-transitive, and as Stab (x, y) contains a 23-cycle G is actually 3-transitive. We have seen that the 23-cycles of G act transitively so we can take G to be the group generated by them. We know then that G < A25 and G is 3-transitive. By a well-known theorem (see H. Wielandt, Finite Permutation Groups, Academic Press, New York, 1964, 13.10), the order of G cannot be di-

visible by 11. We know that Stab (x, y) has prime degree and is transitive. According to Burnside's theorem (Wielandt, 7.3) it is either 2transitive or isomorphic to a subgroup of the group of all affine mappings: z " az + b (a, b E GF(23)). In the first case, G is 4-transitive, so its order is divisible by 25.24.23.22, therefore by 11, which is excluded. So Stab (x, y) is a subgroup of a group of order 2 11 23. Again 11 is excluded. An element of order 2 would

be of the form z H -z + b, which is a product of 11 transpositions and is thus not contained in A25. Therefore, the order of Stab (x, y) is 23, G

3.1 ALGEBRA

97

is sharply 3-transitive, but then by the theorem of Zassenhaus (Wielandt, 20.5), 25 - 1 = 24 has to be a prime power, which is not the case.

Problem A.33. Let G be a finite group and K a conjugacy class of G that generates G. Prove that the following two statements are equivalent: (1) There exists a positive integer m such that every element of G can be written as a product of m (not necessarily distinct) elements of K.

(2) G is equal to its own commutator subgroup.

Solution. Suppose first that (1) is satisfied, and consider the natural homomorphism cp : G -+ GIG'. Since for any h, k E K there exists a g E G

such that k = g-'hg we have cp(h-1k) = cp(h-1g-'hg) = 1. Thus K is contained in the coset hG' = V(h)

(h E K). h' (h1 .. hm, h'1 .. any two elements of G. By our previous remarks, Now

h' EK)be

p(g) = P(hi) ... cp(hm) = cp(h)m = V(hi) ... V (h') = V(9 ) In other words G/G' has only one element, that is, G = G'. Conversely, suppose (2) is satisfied and let us denote IGI by n. Since G = G', every element g of G can be written as a product of commutators: 9 = a1 -lb1 -la1b1 . . . aT-lbT-laTbT = a1"-lb1"-la1b1

. . .

arT

laTbT(1)

Since K generates G and the inverse of an element of K can be replaced

by its (n - 1)th power, it is true that every ai, bi (i = 1,... , r) can be written as a product of elements of K. Substituting these representations in (1), we can deduce the existence of a natural number k9 such that g can be written as a product of k9 n elements of K. But then for any k > k9, we can also write g as a product of k n elements of K by simply adding a suitable number of factors of the type h'" = 1 (h E K). Let l = max{k9 I g E G}. Then every element of G can be written as a product of m = n l elements of K. Remark. In the proof of (2) =:>. (1) the only property of K that we used was the fact that K is a system of generators of G.

Problem A.34. Consider the lattice of all algebraically closed subfields of the complex field C whose transcendency degree (over Q) is finite. Prove

that this lattice is not modular.

Solution. For any set of numbers H C C, let us denote by A(H) the smallest algebraically closed subfield of C containing H. Let us consider a system of numbers a, ,d, y (t; < w1) that are algebraically independent (over Q). Since the algebraic closure of a countable set of numbers is

countable, such a system does exist. We are going to show that for a


98

A( a,(33 D A(

a () A() O ) Figure A.1.

suitable < wl the following diagram is actually the Hasse diagram of a sublattice of the lattice of algebraically closed subfields of C, and this will prove the statement of the problem. The containments indicated by the lines are evident. Furthermore, A (A({a}) U A({-y£, a + 07C})) = A({a, /3, yg})

since the right-hand side clearly contains the left-hand side, while the left-

hand side contains each of the numbers a, /3, ye. We have A({a}) # A({a, /3}) in view of the original assumption, so all that remains to be proved is to show that for some

we have

A({a, )3}) fl A({N, a + /3'yy}) = A(O).

For this, it is enough to prove A({a, l3}) n (A({^%, a + /3'yC}) \ A(0)) = 0. Suppose

(1)

6. Then the transcendency degree of

A (A({7£1, a + )37£1 }) U

a + )37C2 })) = A({7£1, N2, a, Q})

is equal to 4; therefore, the transcendency degree of A({7f1, a + Q7£1 }) n A({7C2, a + )372 })

is equal to 0, which means (A({7£1, a

+)3-y01 }) \ A(0)) n (A({yC2, a + N7{2 D\ A(0))

Now the set A({a, /3}) is countable and the sets A(0) are pairwise disjoint, so there exists a concludes the proof.

= 0.

a + /3y£}) \

for which (1) holds, and this

3.1 ALGEBRA

99

Problem A.35. Let I be an ideal of the ring R and f a nonidentity permutation of the set {1, 2, ... , k} for some k. Suppose that for every

0 # a E R, al' 54 0 and Ia # 0 hold; furthermore, for any elements x1,x2,...,xk E I, x1x2...xk = xlfx2f...xkf

holds. Prove that R is commutative.

Solution. Let m E {1, 2, ... , k} the smallest number that is not fixed by f. that Clearly, m < mf. We then have x1x2.. Xk

0 for any x1,...,xk E I. Fixing

is,

the elements x2i ... , xk we see that x2 xm_1(xm xk - xm f Xkf) annihilates I so has to be equal to 0. Repeating the argument, we get xm ... xk = xm f ... xk f for any xm, ... , xk E I. Let n = m f . Then for every a E R and xm . . . xk E I, we get

axmxm+l ... Xk = axmfx(m+1)f ... Xkf = xmxm+l ... xn-laxnxn+l ... xk. (We get the second equality by using axmf E I.) Rearranging,

(axm ... xi_1 - xm ... xn_la)xn ... xk = 0 Xk E I. Again applying our previous argument,

for any a E R and Xm we get

axm ... xn-1 = xm ... xn-la

EI.

for

Now let a, b E R and Xm

xn_1 E I be arbitrary elements. Then we

have

(ab)xm ... xn-1 = Xm ... xn_l(ab) = (Xm ... xn_la)b = (axm ... xn-1)b = (axm)xm+l ... xn-lb = b(axm)xm+l ... xn-1 = (ba)xm ... xi_1.

Applying once more our previous argument to the resulting identity

(ab - ba)xm .. xn-1 = 0

(a, b E R, xm . . xn-1 E I),

we get ab - ba = 0 or ab = ba, which was to be proved.

Remark. The proof did not use the fact that I was a two-sided ideal of the ring R, only the fact that I was a left ideal in the multiplicative semigroup of the ring R.


100

Problem A.36. Prove that any identity that holds for every finite ndistributive lattice also holds for the lattice of all convex subsets of the (n 1)-dimensional Euclidean space. (For convex subsets, the lattice operations are the set-theoretic intersection and the convex hull of the set-theoretic union. We call a lattice n-distributive if n

n

x A (V yi) = V (x n( V yi)) j=0

i=0

O 1, if B is a subgroupoid of An with Un C B, then B = An. Proof. Suppose for some n there exists a subgroupoid B of An for which Un C B C An holds, and choose the minimal value of n, for which this can happen. Clearly, n >, 2. Now for an arbitrary u c U, B(xl,... , xn_1, u) is a subgroupoid of An-1, containing Un-1. So, by the minimality of n, we have B(xl,... , xn_1, u) = An-1. So U is a subset of the set

S= {a E A I An-lx{a} C B1,

which implies SI 3 3. On the other hand, in view of B C An, we have S C A. Finally, using the surjectivity of o, we can easily check that S is a subgroupoid. The contradiction proves Lemma 1. Lemma 2. A2 has only the following subgroupoids: (a) subgroupoids with one element;

(b) Jul x A, respectively, A x Jul (u E U); (c) automorphisms of A; (d) A2. Proof. Suppose C is a subgroupoid of A2 not in the list above. Clearly,

priC = A (i = 1, 2); furthermore, for every element u E U, C(x, u) (and similarly C(u, x)) is one of the sets A, {v} (v E U). By Lemma 1, U2 S4 C, so by the previous observations only the following two cases are possible:

C=({a}xA)U(Ax{b})UC'for (1)

some C' C (A \ U)2 and a, be U or

C = {(u, uQ) I u E U} U C' for some

C' C (A \ U)2 and some permutation a of U.

(2)

In case (1), let T = {a E A I (a, a) E C}. Clearly, T is a subgroupoid of A. As T fl U = {a, b}, T is a proper subgroupoid, so DTI = 1 and a = b. By our assumptions, there is an automorphism 7r of A for which a 54 a7r. Consider the following set:

0 = {x E A I I y E A such that (x, y), (x7r-1, y7r-1 E C}

.

3.1 ALGEBRA

109

We can easily see that C is a subgroupoid of A and C fl u = {a, a7r}. Thus 2 < Cl S< JAI, which is impossible. In case (2), let

D= {(x, y) E A2 ] z E A such that (x, z), (y, z) E C} , and denote by i the diagonal of A2, that is, A = {(a, a) la E A}. It is easy to verify that D is a subgroupoid of A2 and D fl U2 C A. Since,

pr1C = pr2C = A and C itself is not a permutation, we have A C D. So the transitive closure of D is a nontrivial congruence of A, which is impossible.

Now we turn to the proof of (*). We prove by induction on n that if B is a subgroupoid of An, then B is isomorphic to some direct power of A. The case n = 1 being trivial, suppose n > 2. If B = An, there is nothing to prove, and if for some index i E n we have IpriBJ = 1, then B -and by induction we are ready. So we can suppose B An and priB = A for every i E n. Choose a minimal index set I C_ n such that pr,B # AI, and let k = III, I = {i1 < < ik}, and C = pr,B. Clearly, k > 1 and prk_{3}C = prI-{i3}B = Ak-1 for every j E k.

(3)

Consider the subgroupoid of A2 of the form

C(u1i... , uk-2, x, y) (u1,.. , uk-2 E U) .

.

Because of (3), C(u1,... , uk_21 x, y) has the property that both projections

of it is A, so in view of Lemma 2 - it is either A2 or an automorphism and our assumption on C imply that Uk 54 C, of A. Since Lemma 1 there exists (ui, ... '42) E Uk-2 for which C(u'l, ... , uk_2, x, y) is an automorphism. Suppose that there exists also some (ui, ... , uk_2) E for which C(ui, ... , uk_2, x, y) = A2. Then in the sequence

C(ui, ... , ui', ui+ll

Uk-2

(i = 0, ... , k - 2)

there are two consecutive terms such that the first is an automorphism of A and the second is A2. So we can assume u2 = u2, ... uk_2 = uk_2. Look at the subgroupoid C' = C(x, u2, ... , uk_2, y, ui) of A2. Clearly

C'(u'1, y) = 1, C'(ui, y) = A. But by Lemma 2 we see that this cannot happen. So we know that

C(u1,... , uk_2, x, y) is an automorphism of A for every u1,... , uk_2 E U. Now let

D ={(x, y) E A2 13 z1i ... , zk-1 E A, such that (z1, ... , zk-1, x), (z1, ..., zk-1, y) E C1.

(4)

110


It is easy to check that D is a subgroupoid of A2 and A C_ D (to prove this we use the following consequence of (3): prkC = A). So by Lemma 2 either D = A or D = A2. Suppose that D = A2, and fix arbitrary elements u, v E U, u v. Since (u, v) E D, there exist elements al, ... , ak_1 E A such that (a1, ... , ak-1, u), (a1, ... , ak_1, v) E C .

So for the subgroupoid C* of Ak-1 defined by

C = Q X1, ... , xk-1, U) n (Xi, ... , xk-1, v), we have (a1, ... , ak_1) E C*. On the other hand, because of (4) we have

C* n Uk-1 = 0,

(5)

which implies that not all of a1, . . . , ak_1 are elements of U. For example, let a1 E A \ U. Since a1 is contained in the subgroupoid pr1C* of A, we have pr1C* = A. So for every u1 E U, C* (u1 i xi, ... , xk_2) # 0 holds. Let 1 < k-2 be the greatest index such that there exist elements u1, ... , ul E U with the property that the set

E = C*(u1,...,ul,x1,...,xk_l-1) is nonempty. In the case 1 < k - 2, the maximality of 1, while in the case l = k - 2 (5) shows that (pr1E) n u = 0. On the other hand pr1E # 0 since E # 0. Thus pr1E is a proper subgroupoid of A which is different from the one-element subgroupoids {u} (u E U). This contradiction proves D = A. The equality D = A shows that the projection C -* prk_{k}C (= Ak-1) is injective, so it is an isomorphism. Consequently, the same is true for the projection B -+ prn-{ik}B. Using the induction hypothesis, we immediately get our claim for B. This concludes the proof of (*). Since in the variety generated by a finite algebra each finitely generated free algebra is isomorphic to a subalgebra of some finite direct power of the original algebra, (*) proves the statement of the problem.

Problem A.44. Let the finite projective geometry P (that is, a finite, complemented, modular lattice) be a sublattice of the finite modular lattice L. Prove that P can be embedded in a projective geometry Q, which is a cover-preserving sublattice of L (that is, whenever an element of Q covers in Q another element of Q, then it also covers that element in L).

Solution. Let us denote by 0 the smallest element of P. Let L' = {a E L 0 < a}. Then L' is a (clearly modular) cover-preserving sublattice of L. Evidently, it is enough to solve the problem for L' in place of L. Let Q be the sublattice of L' generated by those atoms of L' which are less than 1p (1p denotes the greatest element of P). Let 6 be the dimension function

3.1 ALGEBRA

111

on Q. Take a q E Q. Then there is a finite set al, . L' such that 1Q = q V al V V.

. . , an E Q of atoms of V an and such that the value of n is minimal.

Then

b(a1V...Van) =n (Q being modular) and 6(1Q) = 6(q) + n. Using the well-known relation b(x V y) + b(x A y) = b(x) + b(y)

(1)

we get b(q A(al V ...Van)) = 0, that is, a1 V V an is a complement of q, and therefore Q is a projective geometry. Now let Q E Q be an atom of Q. Then there exists an atom a E L' of L' with a < ,3(< lp). So a E Q, and since a was an atom, /3 = a, and thus /3 is an atom in L', too. Now, if q1, q2 E Q and q1 Q Q2 then there is an atom a E Q such that q2 = ql V p (using the fact that Q is a projective geometry), but then the modularity of L' implies that qi - e2 > > ek > 0 (k < n,). Let us denote the order of ai by pdi if i > k and by pei+fi, where fi > 0 if i ek. Suppose, on the contrary that We show that e1 > e2 > ei = ei+1i and let us say fi > fi+1 holds. Then, by replacing ai by ai + ai+1, we would get a base giving fewer nonzero terms in representation

(1). Next we show f1 > f2 >

> fk. Otherwise, assuming fi < fi+1,

replacing ai+1 by p ei -e;+l ai + ai+1 we would get a base giving less nonzero

terms in the representation (1). Now we determine the invariants of G/X (that is, the orders of the cyclic groups whose direct sum is isomorphic to G/X). Take i el-e: al + ... + p e,-i-e. ai-1 + ai (2) ai = P

for i , 1 and z E R, we finally get

1 = -p(p2n-1b2v2)

- z2 + 2z E An+1 + R + R = R,

that is, 1 E R.

Problem A.47. Let n > 2 be an integer, and let Stn denote the semigroup of all mappings g : {0,1}n --+ {0,1}n. Consider the mappings f E Stn, which have the following property: there exist mappings gi : {0,1}2 _ {0,1} (i = 1, 2, ... , n) such that f o r all (al, a2i ... , an) E

0,1 n,

f(al,a2,...,an) = (91(an,al),92(al,a2),...,9n(an-1,an))

Let An denote the subsemigroup of Stn generated by these f 's. Prove that

An contains a subsemigroup Fn such that the complete transformation semigroup of degree n is a homomorphic image of Fn-

Solution. We prove a generalization of the problem, namely the one when {0, 1} is replaced by any group (G, +) of order at least two. (The set {0, 1} with addition mod 2 is such a group.) This generalization causes no additional complication. So let Stn be the semigroup of all mappings Gn -- G' with the operation

f,9 E Stn, a1,...,an E G: (f9)(ai,...,an) =9(f(al,...,an))

3.1 ALGEBRA

115

Let H denote the set of those mappings f E Stn for which there exist mappings gl, . . . , gn : G2 , G such that for every al, ... , an E G, f (al, ... , an) = (gl (an, al), 92(al, a2), ... , gn(an-1, an))

Let On denote the subsemigroup of f2n generated by H. Finally, let Tn denote the semigroup of transformations p: {1, ... , n} -> 11.... , n} with the operation P, q E Tn,

1 1), and let G be a transitive subgroup of the symmetric group S. Prove that the order of the normalizer of G in Sn is at most IGIk+1

Solution. First, we show that if T is a minimal transitive subgroup of G, then T can be generated by k elements. Since T is transitive, we have pk I I T I. Let P be a Sylow p-subgroup of T; we are going to show that P

is transitive, thereby proving that T is necessarily a p-group.

Denoting the stabilizer of a point a in the group G by Gq, we have

IG:G,, I=pk,so pkI IG:G,,

nPI.

Since (IG:PI,p)=1, we get pkI IP:GgnP1,that is,pkl P:P" proving that P is indeed transitive. Now let M be a maximal subgroup of T. Then M is not transitive, so MT,, cannot be transitive either. Therefore MT,,, = M, that is, Tq < M. This shows that Ta is contained in the intersection of all maximal subgroups of T, the Frattini subgroup of T. Therefore I T : 4) (T) I < I T : Ta I = pk,


118

so by a well-known theorem of Burnside, T can indeed be generated by k elements: T = ( 9 1 ,---,9 0We next observe that the elements of N(G), the normalizer of G in S", act by way of conjugation on the elements of G, permuting the elements of G among themselves. The number of possible images of the k-tuple (gl, ... , gk) is at most IGIk. Conjugation by the elements n, m E N(G) has the same effect on (91, ... , gk) if and only if n-1m E C(T), the centralizer of Tin S. Thus IN(G)I IGIk IC(T)l. But the group C(T) is easily shown to be semiregular. Suppose that s e C(T) fixes the point a and for any other

point /3 choose a t E T with at = /3. Then /3 = at = (a)sts-1 = (/3)s-1, that is, /3s = /3, so s is the identity, proving semiregularity. This shows IC(T))I < pk, consequently

IN(G)l < IGIk pk , IGlk+l

Problem A.49. Prove that if a finite group G is an extension of an Abelian group of exponent 3 with an Abelian group of exponent 2, then G can be embedded in some finite direct power of the symmetric group S3. Solution. Applying the Schur-Zassenhaus theorem, we see that the extension is actually a split extension, that is, a semidirect product. Using the fundamental theorem of finite Abelian groups, we get that G actually has C

the form G = Z3 Z2, where is a homomorphism : Z2 -> Aut(Z3) _ GL,n(3). (GL, (3) is the group of automorphisms of Z3 , considered as a vector space over the field of three elements.) Applying Maschke's theorem

to the representation , we see that there is a basis b1, b2, ... , b,,, of the vector space Z3 , the elements of which are eigenvectors of all elements of (Z2). Leta1i a2, ... , an, be a basis in Z2, then bj is an eigenvector of (ai) with eigenvalue ci.7 E {1, 2}, say. Denote the elements (1, 2, 3) and (1, 2) of S3, by a and T, respectively. For an arbitrary b = EE", Aibi E Z3 , let b = (cya1, ... , aam,1, ... 1) E S3 +n

,

and let a. = (,rci1-1 .rCi2-1

m}i

.rcim-1 1

T

1

p 1) E S3 +n

Then B = {b I b E Z3 } is a normal subgroup of S3 isomorphic to Z3 . ,g'3 +,, The mapping ai r--* ai can be extended to a homomorphism Z2

and then A = {a

I

a E Z2} is a subgroup of S3 +n, isomorphic to Z.

Let us consider the homomorphism g : A ---> Aut(B) mapping each x E A

to the conjugation by x. Then for the subgroup BA of S3 +" we have BA ^ B >4 A. On the other hand, for any a E Z2 and b E Z3 we have V(a) = b7(a)

3.1 ALGEBRA

119

(it is enough to check this for basis elements bi and aj). This shows that indeed G ^ BA. Remark. Instead of referring to the Schur-Zassenhaus theorem, we can simply consider a Sylow 2-subgroup.

Problem A.50. Let n > 2 be an integer, and consider the groupoid G = (Zn, U {oo}, o), where

xoy=

(x+1 ifx=yEZn, 1.00

otherwise.

(Z,, denotes the ring of the integers modulo n.) Prove that G is the only subdirectly irreducible algebra in the variety generated by G. Solution. (9 will denote the groupoid G considered as a member of the variety of the problem.) The unary operation taking the constant value oo is a polynomial expression in g, for example, 00 = (x o x) o X.

(0)

(This unary operation will also be denoted by oo.) Furthermore, it is easy to check that for the polynomial expressions

f(x)=xox

and

xVy=f"-1(xoy)

(1)

of G the following hold:

V is a semilattice operation and o0 is the greatest element with respect to it,

f

is an automorphism with respect to

(2)

V,

(3)

and the following identities hold: fn(X) = x,

fk(x) V

fl (X)

=oo if k54l (0 0 k=1

holds in any probability space and for any events Al, ... , An provided that it holds in the trivial probability space. Thus, it is sufficient to verify that the inequality n+1

p(AI U ... U An+l)

(k + 1)2

2

Ep(Ai) - (k + 1)2

E p(Ai n Aj) 1 0,

(i = 1, ... , n+ 1) be the sets given in the problem, and consider the probability space such that S2 = A = Ui 11Ai which holds as well. Now, let Ai

3.2 COMBINATORICS

125

and the probability of each point is 1/JAJ. So, even in this space, 2k + 1 n+1

p(A) = 1 > - (k + 1)2

(k + 1)2 1 3, there are some finite disjoint sets Ek and Fk such that either JAi n EkI = 1 or IAi n FkI = 1,

(1)

if i > k then JAi n EkJ 2). Let xy and uv be an edge of F and F' leaving T, respectively, such that x, u E T. If xy = uv then T' = T+xy is a common subtree and there is a desired sequence of spanning trees by the inductional hypothesis. If y # v, then let F" be a spanning tree containing the tree T+xy+uv as a subgraph. Then T+xy C Ff1F", and so by the inductional hypothesis, there is a sequence F = F0, F1, . . . , F9' = F" of spanning trees such that F i and Fi+1 are neighboring (i = 0, 1, ... , rn-1) and T C T+xy g Fi. Similarly, T+uv C_ F'f1F" and so by the inductional hypothesis, there is a sequence F" = F,,,.+1, ... , Fk = F' of spanning trees such that F i and Fi+1 are neighboring (i = m, m + 1, ... , k - 1) and T C T + uv C Fi, and F = FO, Fl, ... , Fk = F' is the desired sequence of spanning trees. Finally, suppose that xy # uv but y = v. Since G - y is connected, it

has a vertex z ¢ T joined to a vertex w E T. Now, let F" and F"' each be a spanning tree containing T + xy + wz and T + uv + wz, respectively. Then, as above, there is a desired sequence of spanning trees from F to F", from F" to F"', and from F"' to F', and the union of these sequences is a desired sequence from F to F'.

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131

Since G is nonbipartite, it contains a cycle C of odd length. Let x be a vertex of C, and let y and z be its neighbors in C. Let To be a spanning tree of G - x containing C - x. Let F = To + xy and F' = To + xz. Color F so that x is red.

According to the lemma, there is a sequence F = FO, F1, ... , Fk = F' of spanning trees such that Fi and Ft+i are neighboring at some vertex xi # x. Color Fi with colors red and blue so that the colorings of Fi and Fi+1 differ from each other at most in the color of xi. It defines some colorings of the spanning trees F1, . . . , Fk = F' starting with the coloring of FO = F such that x is red in every coloring. Clearly, every vertex got different colors in the colorings of F and F', except x. Let ai denote the number of red vertices in the coloring of Fi. Then lai+1 - ail < 1, ao + ak = 2n + 1.

Thus,

either

ao < n < ak

ak < n < ao,

or

so there is an index 0 < i < k such that ai = n, and this is what we wanted to prove.

Problem C.9. Let An denote the set of all mappings f : {1, 2, ... , n} -> ... , n} such that f -1(i) := {k : f (k) = i} # 0 implies f -1(j) # 0, j E

11, 2,

{1, 2,...,i}. Prove kn

0o

lAnl=Elk+1 k=0

Solution 1. Let an denote the cardinality of the set An. Then the number of mappings f E An such that f (i) = 1 holds for exactly l integers i is equal to (1) an_i. Since 1 is positive for any function f, we have n-1

n

an = t=1

al (7) an-1 = E t=0

forn>1. Using the notation bn = an/n!, we have bo = 1 and n-1

bn = 1=o

1

(n

-l)! bl.

Clearly, bn < 2n, so if Izl < 1/2, then the series E°° 0 bnzn is absolutely


132

convergent. For the sum f (z) of this series, we have 0o n-1

00

1

f(z)=Ebnzn=l+> n=0

(n - l)!

n=1 1=0

00

blzn

00

+E

1

(n - l)!

n=1+1

1=0

zn

b,

00

= 1 + 1:(ez - 1)blzt = 1 + (ez - 1V (z). 1=0

It implies that 00

A z) = 12 1 - 1ez/2 00

k=O

2k+1

00

k=0 n=0

1 kn n z n! 2k+1

0o

kn

1

n=0

00

ekz

(;n;-.l t

k=0

zn

2k+1

Thus, 1

bn

= n!

kn

00

E

k=0

2k+1

which we wanted to prove.

Solution 2. Let sn,i denote the number of surjective mappings of the set {1, 2, ... , n} onto the set {1, 2, ... , i}. Then the cardinality of the set An is

Counting the mappings of the set {1, 2, ... , n} onto the set {1, 2, ... , k}, we get the equality k

kn=

(k)

Sn,i.

Now, we have to prove that

M k

n

i=1

/k\ 2

Sn,i -

2k+1 Sn,i, k=1 i=1

that is, using Sn,n+l = Sn,n+2 = . . . = 0 and supposing that the rearrangement does not change the sum, we have to prove that n i=1

S n,i

-=

k) 2k+1

n i=1

k=i

S n,i.

3.2 COMBINATORICS

133

In the parentheses, we have the sum of the terms of a negative binomial distribution ((k) /2k+1 is the probability of the event that flipping a coin, we get the (i + 1)st head in the (k + 1)st flip), and so the coefficient of every term s,,,,i on the right is 1. Since the series in the parentheses are absolutely convergent, the rearrangement did not change the sum, and the proof is complete.

Problem C.10. Let G be an infinite graph such that for any countably infinite vertex set A there is a vertex p joined to infinitely many elements of A. Show that G has a countably infinite vertex set A such that G contains uncountably infinitely many vertices p joined to infinitely many elements of A.

Solution. We prove the statement by contradiction. Assume that G is a graph with vertex set V such that (1) for any countably infinite vertex set A C V, the set HA of vertices p E V - A joined to infinitely many elements of A is a nonempty, countable set. The set HA is obviously infinite since if not, then for the set B = AUHA,

there is no vertex p c V - B joined to infinitely many elements of B. It implies, that deleting finitely many vertices from G, the resulting graph has property (1). Notice that the vertex set V is clearly uncountable since (1) holds for A=V as well. Now, we show that we can delete finitely many vertices from G so that

the resulting graph G1 has no finite vertex set X covering V with the exception of a countable set. (A vertex set X covers the union of the sets of neighbors of the elements of X.) Suppose it is not the case. Then there is a finite set X0 covering all elements of V except a countable set. Deleting X0, the resulting graph still must have a finite set X1 with the same property, and so on. So if we do not get a desired graph after finitely many steps then we can define the pairwise disjoint finite sets Xn covering V with the exception of some countable sets. Let A = U1 0Xn. Then every element of V (except the elements of a countable set) is joined to infinitely many elements of A, that is, V - HA is countable. Since HA is countable by (1), it implies that V is countable, a contradiction. As we have seen, C1 has property (1) as well, so we may assume that G1 = G. Let A C V be an arbitrary countably infinite set. We show that there is a countably infinite set F(A) C V - A such that HA C F(A) and for any finite subset {a1, . . . , an} of A, the set F(A) has infinitely many elements

not joined to any vertex ai (i = 1, ... , n). Let A = jai, a2.... } and for each n, put infinitely many elements of V - A not joined to any element of Jai,-, an} into F(A). It can be done, as we have seen. Then take the union of HA and the set of the elements obtained.

134


Now, let A0 C V be an arbitrary countably infinite set, and let

Ai = F(A0 u Al u ... u Ai-1)

(i = 1, 2, ... ),

00

A,,, = F U Ai

.

For any element p E & and for any index i, p is joined to finitely many

elements of A. Really, p E A, C V - U°=°0Ai C V - Ai+1, and so p F(Ao uAl u... UAi_1) D HA,. Let u°_oA1 = {al, a2.... }, A,,, = {bl, b2, ... }. Let us define an infinite set C = {c1, c2, ... } such that if C C U°_0Ai and n < m, then cn,, is joined to neither an nor bn. It yields the desired contradiction since p V U°oAi

for p E HC (if p = an, then it is joined to at most n elements c,n), and so, p E HC C HuOOOAi C F (U°OoAi) = A. However, it is a contradiction again, since if p = bn, then it is joined to at most the elements c1i c2, ... , cn of C. Thus, let,cl be an arbitrary element of A0, and suppose that C1, c2, ... , cn are given so that {al, a2i ... , an, C1, C2, ... , Cn} C Ao U Al U U AN for some N. Then, AN+1 has an infinite subset T such that the elements of T are not joined to any of the elements al, a2,. .., an. As we have seen above, each of the elements bl, b2i ... , bn is joined to finitely many elements of T, so by choosing cn+1 out of the rest of T, we meet the requirements.

Remark. Assuming the continuum hypothesis, it is possible to construct a graph G satisfying the conditions of the problem in which every countably infinite set A has a countably infinite subset B such that HB is countable.

Problem C.11. Let f be a family of finite subsets of an infinite set X such that every finite subset of X can be represented as the union of two disjoint sets from R. Prove that for every positive integer k there is a subset of X that can be represented in at least k different ways as the union of two disjoint sets from f. Solution. The solution is based on the following theorem of Ramsey.

Theorem. Let Y be an arbitrary infinite set and let n be an arbitrary natural number. Subdivide the family of all subsets of n elements of Y into two classes. Then, there exists an infinite set Y' C Y such that every subset of n elements of the set Y' belongs to the very same class. For a set A, let [A] n denote the set of subsets of n elements of the set A. For a given natural number k, we show that there is a natural number

m > k and an infinite set Z C Y such that [Z]' C X It implies the statement of the problem, since then any subset of 2m elements of the set Z can be obtained as the union of two disjoint members of f in at least 2n'') > k different ways. We find a desired set Z as follows. Subdivide the set [X] 2 into two classes depending on whether or not the elements are in H. By Ramsey's

3.2 COMBINATORICS

135

theorem above, there exists an infinite set X2 C X such that either

[X2]2 c H

or

[X2]2 n h = 0.

Then classify the subsets of three elements of X2 based on whether or not they belong to fi. Again by Ramsey's theorem above, there exists an infinite set X3 C X2 such that either

[X3]3 C f

or

[X3]3 of = 0.

Proceeding further, we obtain a sequence X D X2 subsets of X with the following property:

either

[X,,,.]' c H

or

X2k of infinite

[X,,,,]' n R = 0

for 2 < m < 2k. For such an m, we naturally have either

[X2k]' C fl

or

[X2k], n x = 0

as well. We are done if we can prove that the second possibility cannot be the case. Actually, choose a subset A of 2k elements of the set X2k. By the conditions of the problem (which are used only at this point), A = B U C for some sets B, C E One of these sets, say B, has at least k elements. Then for m = JBI, we have B E [X2k]m n f, that is, [X2k]m o f # 0, and so [X2k]m C which we wanted to prove. Remarks.

1. We did not use the fact that the finite subsets of X can be obtained as the union of two disjoint sets in fi; this part of the condition can be omitted. Furthermore, it is sufficient to assume that for a fixed integer

c, the finite subsets of X can be obtained as the union of at most c members of fi. 2. Considering natural numbers instead of sets and sum instead of union, we get the following analogous problem: let H be an arbitrary set of natural numbers. Suppose that every natural number is the sum of two elements of H. Does it imply that for any natural number k, there is a number that can be obtained as the sum of two elements of H in at least k different ways? This problem of S. Sidon is open. However, the multiplicative version of the problem (where we say "product" instead of "sum" in all cases) has been answered affirmatively. It follows from the statement of the original problem, as follows: let X be the set of all prime numbers, and assign the set of prime divisors to any natural number in H. Thus, we obtain a family 71 meeting the conditions of the original problem. Thus, there exists a set A C X that can be obtained as the union of two disjoint members of 'H in at least k different ways. Then, the product of the prime divisors in A can be obtained as the product of two elements of H in at least k different ways.


136

Problem C.12. Let the operation f of k variables defined on the set ... , n} be called friendly toward the binary relation p defined on the

{1, 2,

same set if

f(a1,a2,...,ak) p f (bi, b2, . . ., bk) implies ai p bi for at least one i, 1 < i < k. Show that if the operation f is friendly toward the relations "equal to" and "less than," then it is friendly toward all binary relations. Solution. Since f is friendly with the relations "equal to" and "less than," we have

f(1,...,1) < f(2,...,2) 2, and assume that I fl J = 0. We distinguish two cases: Case 1. a b. Let ci = b for i E I and ci = a otherwise. Furthermore, let di = a if i E J and di = b otherwise. Then f (cl, ... , Ck) #

f (al, ... , ak) = a (since ci = b # a = ai if i E I and ci = a # ai

if

i V I), and similarly f (d1i . . . , dk) # f (b1, . . . , bk) = b. Property (*) implies that f (C1, ... , ck) and f (d1, . . . , dk) could only be a or b and thus f (c1 .... ck) = b and f (d1 .... dk) = a. Without loss of generality, we may assume that a < b, that is, f (d1, ... , dk) < f (c1, ... , ck). Since f is friendly with the relation "less than" we must have an i such that di < ci.

On the other hand, however, if i E I or i E J then ci = di; otherwise, ci = a < b = di, a contradiction.

Case 2. a = b. Let ci = c(# a) if i E I and ci = a otherwise; similarly di = c if i E J and di = a otherwise; and finally ei = a if i E J and ei = c otherwise. Similarly to the previous case, we have f(c1,...,ck) f(a1,...,ak) = a, f(d1,...,dk) f(b1,...,bk) = b = a and f(e1,...,ek) # f(d1,...,dk). However, all the values f(c1,...,ck), f (d1i ... , dk) and f (e1, ... , ek) must be either a or c, and thus A C1, ... , ck) = f (d1, ... , dk) = c, and so f (e1, ... , ek) = a. Again, it is enough to see the case when a < b, that is, f (el i ... , ek) < f (cl,... , Ck), which implies the existence of an index i such that ei < ci. However, we

haveci=ei=cifiEI,Ci=ei=aifieJ,andc2=a 1, then pick the edge having the smallest j and among them that having the biggest i. Let g j be the subset of Ga in which the graphs have aiaj as the red edge picked, and let G E 9,;j. The way we chose j ensures that the edges aiaj, a3 a.j_1, ... , a2a1 are edges in G

and furthermore - because of the way we constructed 7r - we can reach ai_1 from the vertex al = 1 through blue edges only. The graph, having the same edge set as G with the exception of the edge alai-1, is strongly connected if and only if G is strongly connected. Thus, we can pair the graphs of g4j so that the pairs differ only in the edge aiai_1. Such a pair contributes 0 to the sum; thus, µ(9',j) = 0 and Eµ(G J) = 0. i,,j

There is exactly one set of red edges not containing any red edge to pick up, namely {aiai-1 1 i = 2, ... , n}, and so µ(Ga) =

(-1)IaI(-1)n-1= (-1)IaI-(n-1).

It is easy to see that any set of edges satisfying property (*) and having all edges going along the permutation can be chosen as the set a of blue edges. Thus, we have (_1)Ia1-(n-1)

r

c. Sgn P2

P3

12=113=1

...

L (P,) PP

1n=1

7

Pi (-1)l'-1

= fT

9=213=1

(-1)13-1

3)

7

=

3.2 COMBINATORICS

139

where Pk denotes the maximal in-degree of vertex k. Since we have n,

E(-1t) t j=1

1

(2) = (-1)

'

(-1)t tj =0

1j)

C-

('))

( -1)(0 - 1)

1,

we get p(G,) = 1, and topthe proof is complete.

Problem C.14. Let T be a triangulation of an n-dimensional sphere, and to each vertex of T let us assign a nonzero vector of a linear space V. Show that if T has an n-dimensional simplex such that the vectors assigned to the vertices of this simplex are linearly independent, then another such simplex must also exist.

Solution 1. Let A = (ao, . . . , an) be a simplex in T such that the vectors . . , v(an) assigned to the vertices of A are linearly independent. We will call an (n - 1)-dimensional simplex in T red if for any index 0 < i < n - 1, only one vector in the set

v (ao), .

Xi = (v(ao),...,v(ai)) - (v(ao),...,v(ai-i))

is assigned to the vertices of it. (Here, (Vi, ... ) vk) denotes the subspace generated by the vectors Vi, ... , vk.) Obviously, (ao, ... , an) is red. Lemma. If the vectors assigned to the vertices of an n-dimensional simplex in T are not linearly independent, then the number of its red faces is 0 or 2.

Proof. If the set v(S) of vectors assigned to the vertices of S does not contain any vector in one of the sets Xo, . . . , Xn_1i then obviously S has no red face. If v(S) contains an element of each of the sets X0, . . . , Xn_1 and U Xn_1 then the vectors assigned to the vertices of a vector not in Xo U S are linearly independent, a contradiction. So, we may assume that v(S) contains an element of each set Xj (j = 0, 1, . . . , n - 1) and that it contains exactly two elements of, say, Xi and exactly one element of the other XD's.

Then S has two red faces by which can be obtained from S, deleting one of the vertices with vectors in Xi assigned to it. The proof of the lemma is complete.

Now, consider the graph G whose vertices are the n-dimensional simplices of T and where two vertices are joined by an edge if they share a common red face. In this graph, the degree of the simplex A is 1. Thus, G contains one more vertex of odd degree. By the above lemma, the vectors assigned to the vertices of the corresponding simplex are linearly independent which we wanted to prove.

Solution 2. We prove the following, slightly more general, statement: If K is an arbitrary n-dimensional complex with boundary 0 (that is, K is a cycle mod 2), and if we assign nonzero vectors of the linear space V to

140


the 0-dimensional simplices (that is, the vertices) of K, then the number of n-dimensional simplices with linearly independent assigned vectors cannot be 1. We prove the statement by induction on n. It obviously holds for n = 1.

Suppose that n > 2. Suppose that dimv(A) = n + 1 for A E K. We have to show that K contains one more such simplex. Let B be an (n - 1)dimensional face of A. If C A is a simplex such that B is a face of it, then v(C) V= v(B) implies that dim v(C) > dim v(B) = n, that is, C has the desired properties. So, we may assume that if C A and B is a face of A, then v(C) C v(B). Let Ko = IS E KIv(S) C_ v(B)} and let K1 = 6(Ko), where 6 denotes the boundary. Obviously, 6(K1) = 6(6(Ko)) = 0, and since A 0 Ko and 6(K) = 0, B is thus the face of an odd number of simplices in K0, that is, B E K1. By the induction hypothesis, there exists a simplex B1 B in K1 such that dimv(B1) = dimv(B), and so v(B1) = v(B). Since B1 E 6(Ko)

but B1 ¢ 6(K), there is a thus simplex Al in K such that B1 is a face of Al but v(Al) V. v(B), and so dimv(Al) > dimv(BI) = n. So, the vectors assigned to the vertices of Al are linearly independent. To make the proof complete, notice that it is impossible that A = Al because if it is the case then A has two faces B and B1 such that v(B) = v(Bi) and so dimv(A) = dimv(B) = n, a contradiction.

Problem C.15. F o r a real number x, let IIxII denote the distance between x and the closest integer. Let 0 < xn < 1 (n = 1, 2, ... ), and let 6 > 0. Show that there exist infinitely many pairs (n, m) of indices such that n 54 m and 1

Ilxn -

xmll <minIE,2In-ml

Solution. Let us fix E > 0. We show that if n is large enough, 0 < xi < 1 (i = 1, ... , n), then there exists a pair (i, j) of indices such that IIxi xjII < min(E,1/over2l i -ii). It suffices since it implies that by subdividing the infinite sequence x1i X2.... into blocks of n terms, we can find a desired

pair of indices in each block and the difference Ii - j I does not change if the indices in the block are replaced by the indices in the whole sequence. Let f : {1, . . . , n} -+ {1, . . . , n} be a permutation such that

0 <xf(1) E. Since the

3.2 COMBINATORICS

141

cyclically taken intervals [xf(1), X f(2)), [x f(2), X f(3)), ... , [xf(n), xf(1)) cover the cyclic interval [0, 1)

1

EE+ - i=1 llxf(i) - xf (i+1) ll ? iEA iA

21f(i) 1- f (i + 1)1'

and JAI < 11E. That is,

1-IAIE>_ 2y If(z)-f(2+1)I Applying the inequality between the arithmetic and harmonic means, we get the inequality _

EIf(i)-f(z+1)I > iV A

2(1

2

-JAII )

In the sum I:n 1 I f (i) - f (i + 1)J, every integer j E {1, ... , n} appears exactly twice with positive or negative signs, so that the total sum of the signs is 0. Obviously, the sum is maximum when the large numbers have coefficient +2 and the small numbers have coefficient -2. Thus, n

If(i)-f(i+1)I < i=1

2

2

when n is even, and

i=1

If(i)-f(i+1)I (1 - IAI/n)2, which is a contradiction if JAI > 1 and n is sufficienly large. If A = 0, then we get 1 = 1, which means that n is even and the numbers If (i) - f (i + 1) I are equal (mean inequality). However, if f (i) > n/2, then f (i) appears twice with positive sign, that is, f (i) > f (i - 1), f (i) > f (i + 1), implying that f (i - 1) = f (i + 1), which is possible only if n < 2. But it is easy to see that equality cannot hold in this case either, and we got the desired contradiction in all cases.

Remark. The existence of a pair (i, j) of indices such that Ilxi - xj II < min(e,1/cli - jI) can be proved if c < f (the proof is much more complicated). The statement is false if c > f since it would imply that for an irrational number a, there are infinitely many rational numbers p/q such that J a - p/qI < 1/cq2.


142

Problem C.16. Consider the lattice L of the contractions of a simple graph G (as sets of vertex pairs) with respect to inclusion. Let n > 1 be an arbitrary integer. Show that the identity n

xn V yi

n

=V

i-0

xA V

yi

0 always holds in equality (1), so what we have to prove is that the inequality < holds if and only if G has no cycle of at least n + 2 vertices. Suppose that v0, ... , vk is a cycle with k > n + 1. If x contracts (v0, vk), yi contracts (vi, vi+1) (0 < i < n), and y,i contracts the vertices vn, vn+1, .... vk, then the left-hand side of (1) is x and the right-hand side of it is not since (vo, vk) is not contracted in the right-hand side.

Conversely, suppose that G has no cycle of at least n+2 vertices, and let (a, b) E x AVa oyi. We show that (a, b) is an element of the right-hand side of (1) as well. The assumption (a, b) E xAV oyi implies that there is a path a = v0, v1, ... , vk = b in G such that (vt, vt+1) E x and (vt, vt+l) E A oyi for 0 < t < k. Let us fix t. There is a path vt = uo, ul, ... , ul = vt+1 such that for every 0 < s < 1, we have (u87 ue+1) E y n for some 0 -

n2/4 if n is even, (n2 - 1)/4 if n is odd.

Subtracting the equality from the double of the inequality, we obtain the desired inequality, d1 > [n/2].

Problem C.19. Let rc be an arbitrary cardinality. Show that there exists a tournament T,, = E,.) such that for any coloring f : E, -+ n of the edge set E,,, there are three different vertices x0, x1, x2 E [/k, such that x0x1, x1x2, x2xo E Ek and

I{f (xoxl), f (xlx2), f (x2xo)}I < 2.

3.2 COMBINATORICS

145

(A tournament is a directed graph such that for any vertices x, y E V,,, x y exactly one of the relations xy E E,,, yx E E,, holds.)

Solution. By a famous theorem of Erdos and Rado, there exists a trianglefree graph G = (V, F) with chromatic number 2". Order the vertex set V of this graph G in an arbitrary way, and let < denote this ordering. Now we define a tournament on V as follows. For x, y c V, x < y, let xy E E if

xy E F and let yxEEifxy¢F. We show that the resulting tournament (V, E) meets the requirements of the problem. Let f : E -> n be an arbitrary coloring of E. For every x E V,

consider the set Ax = If (yx) : y < x, yx E E} c n. Let us fix a vertex x E V. The cardinality of the set {y E V : y < x, yx E F or x < y, xy c F} is greater than 2", so there is a vertex x' such that Ax = A,,, and x'x E E if x' < x, xx' E E if x < x'. Let x1 and x2 denote the smaller and the bigger element from among x and x', respectively. Then f (xlx2) E Ay1 = A,, that is, there exists a vertex xo E V such that xo < xi,

xoxl E E,

and

f(xoxl) = f(xlx2)

However, by the definition of E, xox1, xlx2 E F,

and so x2xo V F

and

x2xo E E,

that is, the vertices xo, x1i x2 have the desired properties. Remark. It is easy to see that we cannot replace 2 by 1 in the statement.

For any tournament T = (V, E) and for any ordering < of V, define a coloring f of E as follows: if x < y, then let f (xy) = 0 if xy E E and let f (yx) = 1 if yx E E. Obviously, there are no three vertices x, y, z E V such that xy, yz, zx E E and f is constant on these three edges.

Problem C.20. Some proper partitions P1,..

.

, P,,

of a finite set S

(that is, partitions containing at least two parts) are called independent if no matter how we choose one class from each partition, the intersection of the chosen classes is nonempty. Show that if the inequality

A

< IP11 ...IPn1

(*)

holds f o r some independent partitions, then P 1 ,. .. , P,, is maximal in the sense that there i s no partition P such that P , P 1 ,.. . , PP are independent.


146

On the other hand, show that inequality (*) is not necessary for this maximality.

Solution. We prove the statement by contradiction. Suppose that (1) holds and that there is a proper partition P0 such that P0, P1, . . . , P, are independent. Let us choose one of the classes of each partition and take the intersection of them. For the sake of brevity, let us call these intersections class intersections. Obviously, the total number of class intersections is IP0 P1 ... IPnI and any two class intersections are disjoint since among the classes constituting any two class intersections there are two classes belonging to the same partition, and so they are disjoint. On the other hand, all the class intersections are nonempty since Po, Pl, . . . , P,' are independent. Thus, we have IPOIIPII ... IPT < ISI.

(2)

Combining (1) and (2) and using the fact that ISI # 0, we obtain that IPnI < 2, which is a contradiction to the assumption that P0 is a proper partition. So, we proved that (1) implies the maximality. On the other hand, the next example shows that (1) is not necessary for the maximality. Let S = {al, . . . , a8}. Let P1={S11,S12},

P2={S21,S22}

be the partitions where Sll = {al, a2, a3, a4}, S12 = {a5, a6, a7, as},

S21 = {al,a5}, S22 = {a2, a3, a4, a6, a7, a8}.

Then Sll n S21 = {al},

(3)

and the other class intersections are not empty either, that is, P1, P2 are independent. Equation (1) does not hold since IP1IIP2I = 4 = ISI/2, and we show that {P1, P2} is maximal. Suppose that P0, P1, P2 are independent

for some proper partition P0 = {S01, S02, ... }. Then the intersection of Sol with the set (3) is nonempty and so al E S01. We similarly obtain al E S02. But then Sol fl S02 0, a contradiction to the assumption that P0 is a partition.

Problem C.21. Show that if k < n/2 and Y is a family of k x k submatrices of an n x n matrix such that any two intersect then

3.2 COMBINATORICS

147

Solution. For any k x k submatrix M E F, let RM and CM denote the k-tuple of its rows and columns, respectively. Obviously, RM and CM determine M in a unique way. The condition of the problem says that RMl n RM2 # 0

and

CM, n CM2 , 0

for any two matrices Ml, M2 E F. Let us take the families

R={RM:ME.F}

and

C={CM:ME.F}.

Then R and C are families of subsets of k elements of a set of n elements such that any two members of R and C have nonempty intersection, respectively. Thus, IR1, 10

(k - 1)

by the famous Erdos-Ko-Rado theorem, and so

(I)2. (n_1)2

Obviously, the bound is sharp since this is exactly the number of k x k submatrices containing a given element.

Problem C.22. Let us color the integers 1, 2, ... , N with three colors so that each color is given to more than N/4 integers. Show that the equation x = y + z has a solution in which x, y, z are of distinct colors.

Solution. We prove the statement by contradiction. Suppose that we can color the integers 1, 2, ... , N with red, green, and blue so that each color is

given to more than N/4 integers and there are no x, y, z of distinct colors, where x = y + z. We may assume that 1 is red. Then by our hypothesis, there are no green and blue integers such that their difference is 1. We will call a nonempty set S C {1, . . . , N} an interval if it consists of consecutive

integers, that is, if there are some integers 1 < a < b < N such that S = Is : a < s < b}. If we delete the red integers, then the remaining integers can be partitioned into intervals. Furthermore, each interval is monochromatic by the observation above, since if an interval contains some integers x < y of distinct colors, then there are two consecutive integers of distinct colors among x, x + 1.... l y. Suppose that there exist intervals of at least two integers in both colors

green and blue. Let A and B denote the longest green and blue interval, respectively.

If a E A and b E B, then ja - bl(= a - b orb - a) is not

red. Hence, the set C = {ja - bI : a C A, b E B} does not contain any red integer. If A = [al, a2], B = [b1, b2], then

r,=

( [b1 - a2, b2 - al] 1. [al - b2, a2 - b1]

if a2 < b1, if b2 < al,

148


so it is an interval of JAI + 1131 - 1 > JAI, JBI integers. The set C does not contain any red integers, so it is monochromatic, as we have seen above. But it is a contradiction to the maximal choice of the intervals A and B. On the other hand, at least one of the colors green and blue contains two consecutive integers, since otherwise the number of red integers would be greater than or equal to the number of green or blue integers. In that case, the number of blue or green integers would be at most N/2, a contradiction to the assumption that each color is given to more than N/4 integers. Suppose now that there are no two consecutive green integers but there

are two consecutive blue integers, that is, JBI > 2 for the longest blue

interval B. If 1 < s < N is green, then s > 2 and s - 1 ands + 1 are red. Suppose that the distance between any two green integers is at least 3. Then the intervals [s - 1, s + 1] are pairwise disjoint for the green integers

s and each of them contains two red integers. If N is not green, then it implies that the number of red integers is at least double the number of the green integers, which is impossible if each color is given to more than N/4 integers. If N is green, then it implies that nr > 2ny - 1, where n, and n9 are the number of red and green integers, respectively. If 2 is

not green, then 1 is not counted in the intervals above, so n,. > 2ny, a contradiction again. If 2 is green, then take a blue interval B = [b1, b2] such that IBI > 2, b2 < N. Now, b2 - 1 is blue, b2 + 1 is red, 2 is green, and they constitute a solution to the equation x = y + z such that x, y, z are of distinct colors, a contradiction. Thus, we may assume that there is a green integer s such that s + 2 is green as well. Either bl > s + 2 or b2 < s, since IBS > 2 and Bf1{s, s + 2} _ 0. We may assume that bl > s + 2. Consider the set

C={b-s:bEB}U{b-s-2:bEB}. Since bl < b2 so C is an interval such that ICI = JBI + 2. The interval C does not contain any red integers, so it is monochromatic. But C is not green because ICI > 2 and not blue because of the maximal choice of B, a contradiction.

Problem C.23. Suppose that a graph G is the union of three trees. Is it true that G can be covered by two planar graphs?

Solution. The answer is no. We construct a graph that is the union of three acyclic graphs but that cannot be covered by two planar graphs. Any acyclic graph can be extended to a tree, so it implies the statement.

Let A and B be a set of n and (3) elements, respectively, where the value of n will be determined later. Let A U B denote the vertex set of the graph. Let us choose three elements of A in all possible ways and join each of these triples to an element of B so that distinct triples are joined to distinct vertices in B. Let G denote the resulting bipartite graph. It can be obtained as the union of three acyclic graphs in the following way: for any vertex b E B, the three edges incident to b are put into three distinct

3.2 COMBINATORICS

149

subgraphs. Any vertex b E B is of degree one in each of the three resulting subgraphs, so these subgraphs do not contain any cycle. Suppose that G is the union of two planar graphs Gl and G2. We may assume that Gl and G2 have no common edge. For every vertex b E B, let us proceed as follows. Delete one of the edges of G incident to b so that the remaining two edges belong to the same graph Gi. (It can be done since two of the edges incident to B always belong to the very same subgraph.) Now, replace the vertex b and the remaining two edges incident to b by one edge joining the end vertices of these two edges in A. If Gl and G2 are planar, then the resulting graphs are as well. The resulting graphs are defined on A. The number of replacements is (3), and an edge is obtained in at most n - 2 different ways since a couple can be extended to a triple in at most this many ways. So, the resulting graph H has at least / l3))

n(n - 1)

n-2

6

edges. It is known that a planar graph of n vertices has at most 3n-6 edges

(if n > 3). Thus, H has at most 6n - 12 vertices. From these estimates, we obtain

n(n-1) 35.


150

3.3 THEORY OF FUNCTIONS Problem F.1. Prove that the function f (0)

dx

T

-

(x2 - 1)(1 - 192x2)

1

(where the positive value of the square root is taken) is monotonically decreasing in the interval 0 < 19 < 1.

Solution 1. Substitute

x2-1

+-

1-192x2

), the value t increases in (0, +oo).

While x increases in the interval 11, 1 19

Differentiating the relation 192t2x2

we obtain

dx dt

- t2 + x2 - 1 = 0, _ t(1 -192x2) x(1 + 02t2)

and, consequently, $

dx

f ('d)

(x2-1)(11)2x2) J1

dx t(1 - 192x2)

J0

dt

J0

dt

1+00

=

+°° dx +00

x(1 + 192t2) =

dt

t(1 - 02x2) dt (1 + t2)(1 + 192t2)

JO

Now, for increasing 19 the integrand decreases. Since the limits of integration are independent of 19, the integral is also monotonically decreasing.

Solution 2. Map the first quadrant of the z-plane (excluding the points z = 1 and z = 1/19 by semicircles open from below) to the w-plane with the help of the function w

do

=

Jo

(1 - (2)(1 - 1922)

(taking the value of the square root that is positive on the positive halfaxis). If, starting from 0, z = x + iy runs through the segment 0 < z < 1, then starting from 0, w = u + iv obviously runs through the segment

0<w< 0

i

dx (1 - x2)(1 - 192x2)

A.

3.3 THEORY OF FUNCTIONS

151

If 1 < z < -, then it is clear that w runs through the segment 79

dx

W

u=A, O k E max I Snt (z) I l=0

jzj=k

We additionally assume that Ck is real, further and

Ck > Ck-1

ck > 1 - 2k.

Finally, we define nk in the following way. Since sn(z) tends to f (z) in the disc IzI < 1, for sufficiently large nk we have max Isnk(z)I < 1.

lzl=Ck

On the other hand, the absolute convergence of the power series of f (z) implies the convergence of E°° o Ianlck; thus, for sufficiently large nk,

E Ianlck < n>nk

In addition to these two inequalities, we require that nk satisfy nk > MkNow consider the power series n2

nl

zln

anzn+n=ni+1 > an \2/ n=1 ((

00

= 1:

nk

1:

k=1n=nk_1+1

+... 00

an

k/n -

L 18n,. k=1

\k/ _ Snk-1 ( Z ) } .

We show that the function fo(z) defined by this power series has the desired property.

Prescribing an arbitrarily large positive R, for k > 2R and IzI < R we have Ian (z/k)ni < Ianl(1/2n). Since E 0 Ianl(1/2n) is convergent, the


153

power series of fo(z) is absolutely convergent in the disc IzI < R. Thus fo(z) is a transcendental entire function. We give an upper estimate for fo(z) on the circle IzI = kck. In view of our previous remarks, k-1

Ifo(z)I
_ lim sup Mmk (rmk , A) >_ lim sup k--+-k-+.

k-2

M(rmk, A)

4

= +oo .

Problem F.3. Let H be a set of real numbers that does not consist of 0 alone and is closed under addition. Further, let f (x) be a real-valued function defined on H and satisfying the following conditions:

f(x)f(y) if x < y

and f (x + y) = f (x) + f (y) (x, y E H).

Prove that f (x) = ex on H, where c is a nonnegative number.

Solution. Let xo be an element of H other than 0, and let c = f (xo)/xo. It can be seen by induction that f (nx) = n f (x) (n = 1, 2, ...) for every x in H. Therefore f (nxo) = cnxo.

(1)

Let y be an arbitrary element of H. Then there is a positive integer no such that (y+noxo)xo > 0. If n is a sufficiently large positive integer, then there exist two positive integers Mn and An such that

mnxo < n(y +noxo) < µnxo and

1 mn - µn1 = 1.

(2)

In view of the monotonicity of the function f (x), we have f (mnxo) < f (n(y + noxo)) < f (µnxo),

so by (1), crnnxo < n f (y + noxo)

It follows that

c1Lnx0.

Mn

c n xo < f(y + noxo) < c

n

xo.

(3)

According to (2),

Mn

nxo - y + noxo,

nxo --> y + noxo.

Thus, from (3) we obtain that f (y + noxo) = c(y + noxo). Consequently, f (y) = f (y+noxo)-f (noxo) = c(y+noxo) -cnoxo = cy. The monotonicity of the function f (x) yields c > 0.


155

Problem F.4. Show that if f (x) is a real-valued, continuous function on the half-line 0 < x < oc, and 00

f2(x)dx < oo,

fo

then the function g(x) = f (x)

-

2e-x

J 0 et f (t) dt 0

satisfies

00

"0 f2 (x)dx.

g2(x)dx =

J0

Solution. We assume that the function f (x) is square integrable in the Lebesgue sense on the half-line (0, oo). The relation

f (x) - g(x) =

2e-'

J

x et f

(t) dt

(1)

0

implies that

(f (x) - g(x))' = f (x) + g(x) almost everywhere.

(2)

Using the Schwarz inequality, we obtain w

e-"

etf(t)dt <e-"

r

w/2

etf(t)dt +

(LW/ezt dt

f w fz (t) dt j/2

(L12 <e

/2

El et f (t) dt

f z (t) dt

J

e2t dt

+e

e-w

OF f(x)dx+ 100/2 f2(t)dt, o

whence it follows that lim

W---00

e

r etf(t)dt=0. Jo w

(3)

Relation (1) assures that f (x) -g(x), and therefore (1/2) (f (x) - g(x))2, is absolutely continuous on each bounded subinterval of (0, oo). By (1) and (2),

(fz(x)

- g2(x)) dx =

"

((f(x)_9(x))2\' J

2

(f (x) -2g(x))21J= 0

Making use of (3), the statement follows.

dx

2e - z"

et f (t) dt (Jow

2


156

Problem F.5. Prove that for every convex function f (x) defined on the interval -1 < x < 1 and having absolute value at most 1, there is a linear function h(x) such that

if(x)-h(x)ldx f (l). Since f (x) is continuous and convex on [-1, 1], there is a largest interval [cl, c21 (-1 < c1 < c2 < 1) on which f (x) is minimal. Introduce the notation min f (x) = p, 1]

x

max11 f (x) = q.

Let 01(y) be the inverse of the restriction to [-1, cl] of the function f (x), and let

f-1(y) if p 0 is arbitrary, it follows that the continuous function f has arbitrarily small periods. Hence f is constant, contrary to the assumption.

161


Problem F.9. Let k be a positive integer, z a complex number, and e < 1/2 a positive number. Prove that the following inequality holds for infinitely many positive integers n:

(n-Wl l e J

zQ

> (2 - e)n .

o<e
2k+1, then Iwk+1zl < 1 < 11 - wI on the circle IwI = 2, so by Rouche's theorem the polynomial considered and the polynomial 1 - w have the same number of zeros inside this circle.

In the case z = -1/4, k = 1, using relation (1), it is easy to see that

an(-1/4) = (n + 1)/2n. This shows that the assertion concerning the lim sup cannot be improved. Remarks.

1. One participant proved the following, stronger, statement: lim sup Ian11/n > k/(k + 1), and this is sharp for every k; (We limsup Ian(z)I1/n assumes its minimum for z = (-kk/(k + omit the proof, which is rather long.) 2. Several participants proved that the an satisfy the following recursive 1)k+1).

definition: ao =

= ak = 1,

an+1 = an + an-kz

if n > k.


162

Problem F.10. Let f (x) be a real function such that lim

f(x = 1

x-+00 ex

and I f"(x)I < cl f'(x) I for all sufficiently large x. Prove that lim f 1(x) = 1. x-.+oo ex

Solution 1. Suppose that I f"(x)I < cl f'(x)I for x > xo. We first show that x > xo and t < 1/c imply

If'(x+t)
I f' (x) 1, since otherwise (1) is trivially fulfilled. Put

to = min{t': t' > 0, If(x+t')I = If (x + t)IJ. Since the function I f'(l;) I is continuous and does not intersect the horizontal line of ordinate I f' (x + t) I for l; E [x, x + to) while it remains under this line at the point x, therefore I f'(6)I < I f'(x+t)I on the whole interval [x, x+to]. Prom the Lagrange mean value theorem,

If'(x+t)I - If'(x)I < If'(x+to)I - If'(x)I =tolf"(6I < tocl f'(6I < toclf'(x + t)I, which gives (1).

From (1) we see that if f'(x) = 0 for some x > xo, then I f'(x') I = 0 for all x' > x; but this is impossible since then e_ ' f (x) would tend to 0. Consequently, f'(x) is either positive for all x > xo or negative for all x > xo. However, because e-x f (x) -> 1, the function f (x) cannot be monotone decreasing. Thus f'(x) > 0

for all x

(2)

x0.

From (1) and (2), we obtain that in case x > xo + 1/c and t < 1/c,

(1-ct)f'(x+u) < f'(x)
(1 - ct) et - 1, t

and therefore

limnff (y)>tlimo1(1-ct)ett 1I=1. From the right-hand side of (3), it similarly follows that lim sup f f (x) ex

X-00

1 - e-t 1 = 11 t-.+0 L 1 - ct t J

< lim

1

1

which proves the statement.

Solution 2. We shall make use of the following theorem, which is well known and easy to prove.

Theorem. Let g be a two times differentiable function such that lima.+C,o g(x) exists and is finite, whereas Ig"(x)I < C if x > xo for suitable real numbers C and x0. Then limx_+0o g'(x) = 0.

Now let g(x) = (f (x)/ex). Then

f(x) 9 (x) =

f'(x)ex

= g"(x)

f"(x) - 2e (x) + f (x)

By assumption, lima-+co g(x) = 1. From the other assumption, it follows that for sufficiently large x the derivative f'(x) is of constant sign. Since limx..+c,. f (x) = +oo, it must be positive. Thus, if xl is sufficiently large and x > x1, then x

If'(x) - f'(xi)I 5 fxl f"(t)l dt < cJxlx f'(t) dt = f(x) - f(xi). Therefore I f"(x) I < c' f (x) and I f'(x) < c' f (x) with a suitable constant c'

and, consequently, Ig"(x)l < (3c' + 1)g(x) if x > x1. It follows that, for some C and x0, Ig"(x) I < C if x > x0.

164


By the theorem stated at the beginning,

lim g'(x) = x-++oo lim fi (x) ex- f

(x)

x-+oo

= 0,

whence limx.+,,o(f'(x)ex) = 1. Remark. In some sense, it is necessary to assume that I f"/ f I and Ig"I are bounded. To show this, let g(x) = 1

Then

g (x) =

+

sin x2

- sin x2 x2

x

+ 2 cos x 2

and

liminf g'(x) = -2 < 2 = limsupg'(x). x-+oo

x-+oo

Problem F.11. Find all continuous real functions f, g and h defined on the set of positive real numbers and satisfying the relation f(x + y) + g(xy) = h(x) + h(y)

for allx>0andy>0. Solution. Choose y = 1 in the equation, and then g(x) = h(x) - f(x + 1) + h(1)

(1)

for x > 0. Substituting (1) into the original equation, we obtain

h(x) + h(y) - h(xy) = f(x + y) - f(xy + 1) + h(1).

(2)

Put H(x, y) = h(x) + h(y) - h(xy). Then H(xy, Z) + H(x, Y) = H(x, yz) + H(y, z)

3)

for any triple of positive numbers x, y, z. By relation (2),

H(x, y) = f (x + y) - f (xy + 1) + h(1); putting this into (3), we find that

f(xy+z)- f(xy+1)+f(yz+1)= f(x+yz)+f(y+z)- f(x+y) (x, y, z > 0).

(4)


165

Since f is continuous on the set of positive numbers, passing to the limit z -+ 0 (z > 0), from (4) it follows that f(xy) - f(xy + 1) + f(1) = f(x) + f(y) - f(x + y)

(5)

Introduce the notations f * (t) = f (t) - f (t + 1) + f (1)

(t > 0)

an d F(x, y) = f(x) + f(y) - f(x + y).

T h en

F(x + y, z) + F(x, y) = F(x, y +- z) + F(y, z)

(x, y, z > 0)

(6)

and, by (5),

F(x,y) = f*(xy)

(7)

Putting (7) into (6), we obtain f*(xz + yz) + f*(xy) = f*(xy +xz) + f*(yz)

(x,y,z > 0).

(8)

Hence, choosing z = 1/y and writing x

u = -, y

v = xy,

(9)

it follows that f*(u + 1) + f*(v) = f*(u + v) + f*(1)

(10)

for all positive values of u and v, since for positive u and v the system of equations (9) can be satisfied by suitable positive values x and y. From (10), interchanging u and v,

f*(u+1)+f*(v)= f*(v+1)+-f*(u), whence, taking v = 1, f*(u + 1) = f*(u) + f*(2) - f*(1).

Substituting this into (10), f*(u+v) + f*(1) = f*(u) + f*(v) + f*(2) - f*(1),

whence, by the continuity of f *,

f*(t)=at+/j,


166

where a and /3 are constants. Then in view of (5),

axy+/3= f(x)+f(y)-f(x+y), and therefore, with the notation j (x) = f (x) = (a/2)x2 - /3, we find that f (x + y) = f (x) + f (y). Thus f (x) = 'yx and

f(x) _ -2x2+yx+/3.

(11)

Putting (11) into (2), we see that the function

h(x) = h(x) + 2 x2 - yx - b (6= 2

- ry - 0 + h(1))

satisfies the equation

(x,y > 0),

h(x) + h(y) = h(xy) which yields h(x) = k In x. Consequently,

h(x) _ -2x2+yx+klnx-6.

(12)

Finally, from (1), using (11) and (12), it follows that

g(x)=kIn x+ax -2b-/3.

(13)

We have shown that the solutions of the equation can only be the functions of the forms (11), (12), and (13). On the other hand, it is easy to see that the functions (11), (12), and (13) are solutions of the equation for any choice of constants a, 0, y, b, and K.

Remark. Most participants first show that f, g, and h are two times continuously differentiable functions and then reduce the problem to a dif-

ferential equation. Several of them note that when using this method it is sufficient to assume the integrability of f, g, and h on every bounded, closed subinterval of the set of positive numbers.

Problem F.12. Let xo be a fixed real number, and let f be a regular complex function in the half-plane Re z > x0 for which there exists a nonnegative function F E Ll (-oo, oo) satisfying If (a + i/3) < F(/3) whenever

a > X0, -00 < /3 < +oo. Prove that

I

a-ioo

f(z)dz = 0.

Solution. Let x0 < al < a2. Let {i3,,} and {y,,} be sequences of real numbers tending to +oo such that F(O,,) -> 0 and F(-y,,) -> 0 . By the Cauchy integral theorem, al+i/jn

az+i(jn

f (z) dz+J i'Yn

ai+io3

-i'Yn

f (z) dz+

Ei)3n

dz+a

i-i'Yn

f (z)

f (z) dz = 0 la2 -iY n


167

(the path of integration is always the connecting segment). Since

faz+ip

Jp01z

f (z) dz a1+iRn

f (a + ion) da 1). Then

B

= pa+boo f(z) dz = i f°° f(a + ij3) d

J

z

Ja-ioo

a > max{1, xo}

a'

a+

°°

is independent of a. Let a -* oo. Since f (a + i13)

a+i/3

< F(3)

(a > 1),

integration and transition to the limit can be interchanged by the theorem of Lebesgue, and we obtain B = 0. Then

A = A - aB

=

j

a

a+ioo

-zoo

f

(z) (1 - z) dz = -

foo

J

00

f (a + i,3) a

13

i'3

d,3.

Here, again, F(Q) is a common majorant of the integrands, and for each fixed value of /3 the integrand tends to 0 as a -a oc. Consequently, by the theorem of Lebesgue, the integral tends to 0, whence A = 0. Remark. All participants first show that the integral under consideration

is independent of a. Then some of them choose the function f (z)/z and the way described above, while others work with the function e-tz f (z), where t > 0 is a real number. They establish similarly (using the fact that this function also satisfies the conditions of the problem and tends to 0 as Re z -+ oo) that

a+i°o

e-"f (z) dz = 0;

a-ioo hence, letting t -+ 0, the desired equation follows.


168

Problem F.13. Let 7rn(x) be a polynomial of degree not exceeding n with real coefficients such that

Iirn(x)I
u > 0).

(3)

170


If y(x) is any continuous function on the interval (-oo, 0], then it is easy to see that y(x) can be uniquely extended to the whole real line so that (3) is valid. In fact, suppose that x0 is the supremum of those values up to which unique extension is possible. Then in the neighborhood of x0 of some radius 6, the function r(t) is greater than some positive E. Denote by y the smaller of the numbers a and 6. Then, by (3), the values of y(x) taken for x < xo -q/2 uniquely determine the values of y(x) in the interval (xo - 77/2, xo + 7)/2). This contradicts the choice of xo.

A similar reasoning shows that if y(x) is positive on (-oc, 0] then it is positive on the whole real line. In this case, with the help of (2) we obtain that y(x) is monotonically increasing on [0, oo). Put

z(x) = y(x)e- fo a(t) dt

(x > 0).

(4)

Differentiating and using (2) we obtain z'(x) = a(x)e- fo a(t) dt [y(x - r(x)) - y(x)]

(5)

To prove the statement of the problem, first suppose that y(x) is positive on (-oo, 0]. Then, as we have seen, y(x) is positive everywhere and increasing for x > 0. Thus, z(x) is positive for all x > 0; further, z(x) is monotone decreasing for sufficiently large x. Equation (1) implies that x - r(x) > 0 if x is large; so from (5) by the monotonicity of y(x) it follows that z'(x) is negative. Since z(x) is positive and decreasing, z(x) exists.

To prove the general case, represent the function y(x) on (-oo, 0] in the form y(x) = y1(x) - y2(x),

(6)

where yl(x) and Y2 (x) are positive, continuous functions. As we have seen, yl(x) and y2(x) can be extended to the whole real line so that they satisfy the differential equation (2) for x > 0. By the uniqueness of the solution of (2) mentioned above, it is also clear that (6) remains valid on the whole real line. Defining the functions zi(x) and z2(x) in analogy with (4), it follows as above that the limits lima-,,,, zl(x) and lim.,. z2(x) exist. This implies the existence of the limit

linnz(x) _ X-co lira (zl(x) - z2(x)) The solution of the problem is complete.

Problem F.15. Let Ai (i = 1, 2, ...) be a sequence of distinct positive numbers tending to infinity. Consider the set of all numbers representable in the form 00

µ= i-1

ni)i,


171

where ni > 0 are integers and all but finitely many ni are 0. Let

L(x) _

M(x) _

and

1

a; <x

1. A <X

(In the latter sum, each p occurs as many times as its number of representations in the above form.) Prove that if lim

then

L(x + 1) L(x)

M(x + 1) - 1

lim Xoo M(x)

Solution. (When in the solution we speak of a p, we mean not only its value but also its representation by a fixed sequence {ni }, keeping in mind that a number can possibly be represented in several ways.)

If pi = >,

Ai and µ2 = E

then let µl Iµ2 mean that

0) (i = 1, 2, ... ). Then

E

µ=

Ai

n>1 integer a; na; Iµ

and

µ<x

Eµ=L: E Ai. µ<x n>1,a, nAiIµ

In the inner sum, p' = p - n)i is also a p-number; let us sum with respect to this. Then

Eµ'_µ'<x E

µ<x

n>1,a; nAi <x-µ'

Ai.

With the notation n> 1,aj nai 0 is a constant depending only on M. Solution 1. It is sufficient to examine the multiplicity of number 1. In fact, if we prove something for 1 then we may apply the result to the polynomial

p(z) = P(az) with Ial < 1, and in this way we obtain the same estimate for all roots lying in the unit disc. The idea of the solution is the following. We consider the integral

F(P) =

log

10

IP(eio)I do 0

and show that it exists and is nonnegative. Then we estimate it from above, once in the neighborhood of 1 with the aid of the multiplicity of 1 and the degree of P, and once at other points using the condition IP(z)I < M. It is sufficient to prove the existence of the integral for polynomials of

the form z - z0. If

n

P(z) = cJJ(z - zi), i=1 then

n

logIP(z)I = loglcl +loglz - zil. i=1

The existence of

I logleio - zoldo 0

1. Next let I zo I = 1. Without loss of generality, we may assume that z0 = 1 (a substitution 0 = 77 + 00 takes them into each other). Then log Ie`0 - 11 = 2 sin 2

is evident if I zo

and

j27r log 12 sin

\

I do

2/

really exists and is equal to 0. Next, compute the integral j27r

f (a) =

log

do

(a # 0)


175

for a # 1. Obviously, its value depends on the absolute value of a only (again, a substitution as above may be applied), so it is the same for the numbers ael, aE2, ... aEn, where the Ej are the nth unit roots. Therefore, n

27r

77f(a) _ E f(aej) = f .9=1

n

log

i-1

0

= f27r

11-aEj eio

do

\\\

eino log 1-

do.

an

0

Now, if lal > 1, then for n - oo the integral on the right-hand side tends

to zero since 1 - eino/an - 1 uniformly; thus f (a) = 0. On the other hand, if Ial < 1 then 27r

n(f(a)+27rloglal)= flog a-end since 1 - IaIn < lan - einol < 1 + Ialn; so in this case f (a) = -27rlog I a I is only possible. In each of the three cases, we have f (a) _> 0. In our case, the relation P(0) = 1 implies

P(z)=f 1- -ziz n

,

j-1

whence

n

F(P) = E f (zj) < 0.

(1)

j_1

Now let P(z) = (z - 1)kQ(z) = ao + a1z + + anzn, where Q(1) # 0. We estimate F(P) with the help of k, n, and M. Let

f = f +f 27r

F(P) =

e

o

Then

F2
aj (z + 1)i = E E aj j=0

j=0 m=0

n

m=k

n

j=m

aj

('m

I zm

(i),

since for m < k the coefficient of zm is 0 by our assumption. Thus

R(z) _

n-k

n

bm = E aj (m j+ k)'

bmzm,

(5)

j =m+k

M=0

Further, by the Cauchy inequalities, jajj = jP(j)(0)j/j! < M. Putting this into (5), we find

n+1 Ibml!5 M - (m+k) _ M(m+k+1) j =m+k

(6)

If jzj = 6 and 8(n - k)/(k + 2) < 1, then in view of (6),

amn+1 (m+k+1 )

IR(z)I
2, then using the relation t! > tte-t we obtain n + 1

k+1)

(n + 1)n(n - 1)...(n - k + 1)

(k+1)! _ (n2 - 1)n(n - 2) ... (n - k + 1) nk+1 < (k+1)! < (k+1)!

en

(k+1)

k+1


177

Thus

log lQ(ei0)I - 0. (ir+e)logM+elog k+1 +eklogk+1

(7)

Now if n = k2/2c, let e = c/k (this fulfills the condition e < ((k + 2)/2(n k))). Then (7) becomes

+ k) log M +

c

2

log c(k

1)

+ clog 2(k

k 1)

> 0.

We only make things worse if we also replace k + 1 by k. Moreover, c/k =

k/2n < 1/2 gives -7r + c/k < 4, c/k log(ek/c) < (c/k) (ek/c) = e < 3, so finally

4log M + 3 > c log 2,

c log 1 = 0.

l

27r

On the other hand, g(z9) < M2 for all 79, and for 1791 < (t7r/2n)

< M2 (c ng)2e

11)22

1W2n(exi0)1 < M2

9(i9) =

(c n l z -

so in this interval log g(V) < 2log M + 2f log

cn19.

We use this relation only for 1191 < (f/ncl) (< (t/n) (7r/2)). At the remaining points, we apply log g(V) < 2log M to obtain

ft

27r

0
8E we have f (x) < (1/E) f (xo). In the problem e < (1 - x0)2, hence Ixoi < 1 - e1/2 and IxI < 1 - 3E1/4, so we can choose a = 3E1/4 and /3 = E1/2, because then a2/3 = 9E > 8E. Then from the preceding estimate

f(x)
0,

r'(x+1) >logx. r(x + 1)

Solution. It is known (see, for example, 1.7. (3) in Erdelyi et al., Bateman Manuscript Project, McGraw-Hill, New York, 1953) that if x > 0, then

I'(x)>0and r, (X) F(X)

1

1

°O

x

(1 v

v_1

x+v

where C is Euler's constant. From this, we get r' (x) C

°°

1

V)

r(x) / r'(x)

F(X)

is strictly increasing. On the other hand,

f

rI (t) dt = log r(x + 1) - log r(x) = log x,

where we used r(x + 1) = xr(x) for all x. Hence, by the mean value theorem, Jlogx=Jx+l

1-

forsomex r()

0

=logx.

Problem F.19. If f is a nonnegative, continuous, concave function on the closed interval [0, 1] such that f (0) = 1, then

r

J

1

Contests in Higher Mathematics: Miklós Schweitzer Competitions, 1962-1991 (Problem Books in Mathematics)

Contests in Higher Mathematics: Miklós Schweitzer Competitions, 1962-1991 (Problem Books in Mathematics)

Contests in Higher Mathematics - Miklós Schweitzer competitions 1962-1991

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Problem-Solving Strategies (Problem Books in Mathematics)

A Problem Seminar (Problem Books in Mathematics)

Exercises in Probability (Problem Books in Mathematics)

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Exercises in Integration (Problem Books in Mathematics)

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Theorems and Counterexamples in Mathematics (Problem Books in Mathematics)

Berkeley Problems in Mathematics (Problem Books in Mathematics)

A Problem Book in Real Analysis (Problem Books in Mathematics)

Problem Book for First Year Calculus (Problem Books in Mathematics)

Exercises in Modules and Rings (Problem Books in Mathematics)

Iranian mathematics competitions, 1973-2007

A primer for mathematics competitions

A primer for mathematics competitions

A Primer for Mathematics Competitions

A Primer for Mathematics Competitions

A Primer for Mathematics Competitions

The contest problem book VII: American mathematics competitions, 1995-2000

A Sequence of Problems on Semigroups (Problem Books in Mathematics)

Higher engineering mathematics

Mathematics as problem solving

Theory of Stochastic Processes: With Applications to Financial Mathematics and Risk Theory (Problem Books in Mathematics)

Mathematics as problem solving

Chinese mathematics competitions and olympiads: 1981-1993

Counterexamples in Analysis (Dover Books on Mathematics)

Mathematics in Western Culture (Galaxy Books)

Counterexamples in Analysis (Dover Books on Mathematics)

Contests in Higher Mathematics: Miklós Schweitzer Competitions, 1962-1991 (Problem Books in Mathematics)