EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS

1 EXERCISES AND SOLUTIONSIN GROUPS RINGS AND FIELDSM ahmut Kuzucuo gluMiddle East Technical TURKEYA pril 18, 2012iiiiiTABLE OF CONTENTSCHAPTERS0. PREFACE .. v1. SETS, INTEGERS, FUNCTIONS .. 12. GROUPS .. 43. RINGS .. 554. FIELDS .. 775. INDEX .. 100ivvPrefaceThese notes are prepared in 1991 when we gave the abstract al-gebra course. Our intention was to help the students by giving themsome EXERCISES and get them familiar with some SOLUTIONS . Some of thesolutions here are very short and in the form of a hint. I would liketo thank B ulent B uy ukbozk rl for his help during the preparation ofthese notes. I would like to thank also Prof.

2 Ismail S . G ulo glu forchecking some of the SOLUTIONS . Of course the remaining errors belongsto me. If you find any errors, I should be grateful to hear from I would like to thank Aynur Bora and G uldane G um u s for theirtyping the manuscript in Kuzucuo gluI would like to thank our graduate students Tu gba Aslan, B u sraC nar, Fuat Erdem and Irfan Kad k oyl u for reading the old versionand pointing out some misprints. With their encouragement I havemade the changes in the shape, namely I put the answers right afterthe , December 2011viM. Kuzucuo glu1. SETS, INTEGERS, a finite set havingnelements, prove thatAhas exactly2ndistinct :Apply Induction |A|= 1, thenAhas exactly two subsets namely theclaim is true forn= hypothesis.

3 For any set having exactlyn 1 elements, thenumber of subsets is 2n nowA={a1,a2, ,an}be a set with|A|= subsetXofAis either contained inB={a1, ,an 1}oran induction hypothesis, there are exactly 2n 1subsets ofAcontained other subsetXofAwhich is not contained inBis of the {an} YwhereYis a subset numberis therefore equal to the number of subsets ofB, 2n thenumber of all subsets ofAis 2n 1+ 2n 1= the given set and relations below, determine which defineequivalence relations.(a)Sis the set of all people in the world today,a bifaandbhave an ancestor in common.(b)Sis the set of all people in the world today,a bifaandbhave the same KUZUCUO GLU(c)Sis the set of real numbersa bifa= b.

4 (d)Sis the set of all straight lines in the plane,a bif a is : b,candd are equivalence relations, but a is two integers. Ifa|bandb|a,then show thata= :Ifa|b,thenb=kafor some |a, thena=`bfor some `b,then we obtainb k`b= impliesb(1 k`) = 0 so eitherb= 0 ork`= 0,thena= 0 and hencea= band we are `= 1,then eitherk= 1 and`= 1 ork= 1 and`= first caseb= a,in the second caseb= ,p2, ,pnbe distinct positive primes. Show that(p1p2 pn) + 1is divisible by none of these :Assume that there exists a prime saypiwherei nsuch thatpidividesp1p2 pn+ 1. Then clearlypi|p1p2 pnandpi|p1p2, pn+ 1 implies thatpi|1 = (p1 pn+ 1) (p1 pn).

5 Which is impossible aspi 2. Hence none of thepi s dividesp1 pn+ that there are infinitely many primes.(Hint: Use the previous exercise .)Solution:Assume that there exists only finitely many primessay the list of all primes{p1,p2,..pn}. Then consider the integerp1p2 pn+ by previous Question none of the primespii= 1,..,ndividesp1p2 pn+ eitherp1p2 pn+ 1 is a prime which isnot in our list or when we writep1p2 pn+ 1 as a product of primeswe get a new primeqwhich does not appear in{p1,p2, pn}.Hencein both ways we obtain a new prime which is not in our list. Hence weobtain a contradiction with the assumption that the number of primesisn.

6 This implies that the number of primes is AND SOLUTIONS IN GROUPS RINGS AND there are integersa,b,s,andtsuch that, the sumat+bs= 1,show thatgcd(a,b) = :We haveat+bs= 1 Assume that gcd(a,b) = by definitionn|aandn|band ifthere existsm|aandm|b, thenm| |awe haven|atandn| |at+ impliesn| that ifaandbare positive integers, thenab=lcm(a,b) gcd(a,b).Solution:Let gcd(a,b) =kandlcm(a,b) =l. Thena=ka1andb=kb1where gcd(a1,b1) = 1 andab= definitiona|landb|l, moreover if there exists an integerssuchthata|sandb|s,thenl| Indeed we havea|ka1b1andb| that there exists an integertsuch thata|tandb|t. Thent=ak1andt=bk2. We havet=ak1=bk2=ka1k1= follows thata1k1=b1k2.

7 Sincea1andb1are relatively prime wehavea1|k2andb1|k1. Thenk2= Then we havea1k1=a1b1u=b1a1cit follows thatu=candt=ak1=ka1b1chencel=ka1b1| KUZUCUO GLU2. any set. Prove that the law of multiplication definedbyab = ais :Letx,y,z want to show thatx(yz) = (xy) (yz) =xy=xby the law of multiplication (xy)z=xz=x,by the same law sox(yz) =x= (xy) that the equationxyz= 1holds in a follow thatyzx= 1?Thatyxz= 1?Justify your :xyz= 1 implies thatx(yz) = Thenwe havexa= 1 and soax= 1 since a is invertible anda 1=x.(Seesolution 6) It follows that (yz)x= the other hand, ifxyz= 1,it is not always true thatyxz= this, letGbe the group of 2 2 real matrices and letx=(1 20 2)y=(0 12 1)andz=( 1/2 3/41 1).

8 Thenxyz=(1 00 1)= (2 25 9/2)6= a nonempty set closed under an associative product,which in addition satisfies:(a) There exists ane Gsuch thatae=afor alla G.(b) Givena G,there exists an elementy(a) Gsuch thatay(a) = thatGmust be a group under this :Givena right inverse exists, there existsy(a) Gsuch thatay(a) = ,y(a) =y(a)e=y(a)(ay(a)) =(y(a)a)y(a).Also, there existst Gsuch thaty(a)t= impliesEXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS5that (y(a)a)y(a)t=ethen (y(a)a)e=eHencey(a)a= everyright inverse is also a left for anya Gwe haveea= (ay(a))a=a(y(a)a) =ae=aaseis a right identity. Henceeis a left a group of even order, prove that it has an elementa6=esatisfyinga2= :Define a relation onGbyg hif and only ifg=horg=h 1for allg,h is easy to see that this is an equivalence relation.

9 The equivalenceclass containinggis{g,g 1}and contains exactly 2 elements if andonly ifg26= ,C2, ,Ckbe the equivalence classes ofGwithrespect to .Then|G|=|C1|+|C2|+ +|Ck|Since each|Ci| {1,2}and|G|is even the number of equivalenceclassesCi,with|Ci|= 1 is even. Since the equivalence class containing{e}has just one element, there must exist another equivalence classwith exactly one element say{a}.Thene6=aanda 1= a finite group, show that there exists a positive integerm such thatam=efor alla :LetGbe finite group and 16=a the seta,a2,a3, ,ak It is clear thatai6=ai+1for some integers from the beginning .SinceGis a finite group there existsiandjsuch thatai=ajimpliesai j= every element has finite order.

10 That is the smallestpositive integerksatisfyingak= 1 (One may assume without loss ofgenerality thati > j). One can do this for eacha leastcommon multiplemof the order of all elements ofGsatisfiesam= 1for alla a group in which(ab)i=aibifor three consecutiveintegersifor alla,b G,show thatGis KUZUCUO GLUS olution:Observe that if there exist two consecutive integersn,n+1 such that(ab)n=anbnand (ab)n+1=an+1bn+1for alla,b G,thenan+1bn+1=(ab)n+1= (ab)nab= we obtainan+1bn+1= multiplying this equation from left byanand from right byb 1weobtainabn= our case takingn=iandn=i+ 1,we haveabi=biaand bytakingn=i+ 1 andi+ 2 we haveabi+1=bi+ shows thatabi+1=bi+1a=bbia=babiand now multiplyingfrom right bybiwe obtainab= a group such that(ab)2=a2b2for alla,b G,thenshow thatGmust be :abab=a2b2applya 1from left andb 1from right.