Chapter 11 These problems implicitly make use of the following lemma (stated casually in the book) Lemma If akß æb is a group and L K is closed under æ then æ is associative in L Proof We know æ is associative in K by definition of a group Now let +ßß- L Since L is closed + æ a æ-bß a+ æ b æ - L However since +ßß- L implies +ßß- K we also know + æa æ-b a+ æ b æ - Hence æ is associative in L Problem 111 Determine which of the following binary operations are associative: (a) the operation æ on defined + æ + (b) the operation æ on defined by + æ ++ (c) the operation æ on defind by + æ + (d) the operation æ on defined by a+ß bæa-ß b a+ -ß b + (e) the operation æ on e! f defined by a æ b Solution (a) Not associative For example a æ bæ! Á æa æ b (b) Associative because a+ æ bæ- a++ bæ- a++ b- a++ b- & 1 + a- -b+ a- -b + a æ-b+ a æ-b + æa æ-b The intermediate steps follow because usual addition and multiplication is associative and commutative in (c) Not associative For example a! æ! bæ& & Á! æa! æ& b (d) Associative because aa+ß bæa-ß bbæa/ß0b a+ -ß bæa/ß0b aa+ -b0 a b/ß a b0b a+ a0b a-0 / bß a0bb a+ß bæa-0 /ß0b a+ß bæaa-ß bæa/ß0bb + - +- Notice we could not say a ß æb is isomorphic to aß b even though intuitively because this would exclude or equal to! (which is encompassed by the former) (e) Not associative For example a æ bæ Á æa æ b Problem 112 Decide which of the binary operations in the preceding exercise are commutative Solution (a) Not commutative For example 1 æ! Á! æ (b) Commutative because + æ ++ ++ æ + due to addition and multiplication being commutative in (c) Commutative because + & + & + æ æ +
due to addition being commutative in (d) Commutative because a+ß bæa-ß b a+ -ß b a-+ß b a-ß bæ a+ß b due to addition and multiplicaiton being commutative in (e) Not commutative For example æ Á æ Problem 113 Prove that the addition of residue classes in defined) Î is associative (you may assume it is well Proof Let +ßß- Î Then + ˆ - +- (by definition--see page 9 in the book) which equals + a-b +- (again by definition) However +- a+ b- +- ˆ + - Problem 114 Prove that the multiplication of residue classes in well defined) Î is associative (you may assume it is Proof Let +ßß- Î Then + ˆ - + - (by definition--see page 9 in the book) which equals + a-b +- (again by definition) However +- a+ b- + - ˆ + - Problem 115 Prove for all that Î is not a group under multiplication of residue classes Proof In the book we've seen a Î b is a group Hence! must be the guilty element of breaking this structure Indeed! has no inverse since! +! +! +! +! for any + Î and we know is the identity since + + + + + for any + Î Since there is no element + such that! +! has no inverse and by definition Î is not a group under multiplication of residue classes Problem 116 Determine which of the following sets are groups under addition: (a) the set of rational numbers (including (b) the set of rational numbers (inculding (c) the set of rational numbers of absolute value (d) the set of rational numbers of absolute value (e) the set of rational numbers with denominators equal to!!î) in lowest terms whose denominators are odd!!î) in lowest terms whose denominators are even together with! or (f) the set of rational numbers with denominators equal to or $ Solution For each respective problem call the group K + - (a) This is a group First if ß K with l Î and l Î (ie both are odd) and a+ß b a-ß b then + - +- K 2
since l Î ; hence we have closure We know it is associative since is on addition and K The! + + +! + identity is!î since for all K Finally each element has an inverse since + +! + for any K (b) This is not a group because it does not have closure For example K but  K since is in lowest terms and the denominator is odd (not even) (c) This is not a group because it does not have closure For example since K Ÿ but  K since k k Î $ $ $ (d) This is not a group since it fails closure For example ß K since k k but a b  K since Î and Á! (e) Assume each rational number is in lowest form This is a group First take ß K with the greatest + - +common divisor of + and and - and equal to 1 Consider Then K Otherwise + - +- K since or since gcd aß+ -b or (if and + - is even the gcd becomes ) Hence K is closed We know it is associative since is on addition and K The! + + +! + identity is!î since for all K Finally each element has an inverse since + +! + for any K By definition K is a group (f) This is not a group because it is not closed For example ß K but  K + - $ $ ' Problem 117 Let K eb l! Ÿ B f and for BßC K let BæC be the fractional part 90 BC ( ie B æc BCcBCd where c+ d is the greatest integer less than or equal to + ) Prove that æ is a well-defined binary operation on K and that K is an abelian group under æ ( called the real numbers mod ) Proof To show the operation is well-defined notice either! Ÿ BC 1 or Ÿ BC In the former case cbcd! so that BæC BCcBCd BC Otherwise cbcd so that BæC BC Hence æ is a well-defined binary operation on K To show closure again consider the two cases mentioned earlier In the former BæC BC and since! Ÿ BC BæC 1 by assumption BæC K In the latter case BæC BC and since Ÿ BC we have! Ÿ BC BæC so again BæC K Therefore K is closed To show it is associative notice for BßCßD K ac Db æ a Cb D abccbcdb D abccbcdbd cbccbcddd Bæ æ æ BCD ccddcbcd ccddd BaC DbcBaC Dbd B æ æ æ The middle equality holds because cbcdcbccbcddd ccddcbcd ccddd which needs to be explicitly justified case-by-case Assume! Ÿ BC Ÿ and! Ÿ CD Ÿ or Ÿ BC and Ÿ CD Then cbcd ccdd so the equation holds Otherwise assume without loss of generality! Ÿ BC Ÿ and Ÿ CD Then cbcd! and ccdd so that cbcdcbccbcddd cbcdd cbcd d ccddcbcdccd dd Hence the operation is associative Furthermore! is the identity since!b c!bd B!cB! d BcBd for any B K Finally each element has an inverse since BaBbcBaBbd abbbcabb Bd! c! d! for each B K Therefore K is a group Finally K is abelian since for any BßC K we have that BCcBCd CBcCBd since addition is commutative in Hence K is an abelian group 3
Problem 11 Let K ed ld for some f (a) Prove that K is a group under multiplication ( called the group of roots of unity of ) (b) Prove that K is not a group under addition Proof (a) To prove closure let AßD K Then bß7 such that A D Then aadb 7 aa b 7 7 ad b 7 and since 7 (the positive integers are closed under multiplication) by definition AD K Hence K is closed Associativity is guaranteed since Ï e! f is a group under multiplication and K Ï e! f (notice!  K since there is no such that! ) The identity is since for we have so that K and furthermore for all D K D D D Finally each element has an inverse since for each D K there is an such that D so that D D D D D Therefore akß b is a group 7 (b) Since K it can not be a group under multiplication since  K as there is no such that (and hence K is not closed) Problem 119 Let K š + È l+ß (a) Prove that K is a group under addition (b) Prove that the nonzero elements of K are a group under multiplication Proof (a) Let + Èß- È K Then Š + È Š - È a+-b a bè is in the group because +-ß (since aß b is a group) Associativity is guaranteed since K and a ß b is a group The identity is!! È since Š!! È Š + È Š + È Š!! È + È for any + È K Finally each element has an inverse since for + È K Š + È Š +a b È Š +a b È Š + È!! È Therefore K is a group under addition (b) Let + Èß- È K Then Š + È Š- È a+- b a+ - bè is in the group because +- ß+ - (since a ß b is a group) Associativity is guaranteed since K and a ß b is a group The identity is! È since Š! È Š + È Š + È Š! È + È for any + È K Finally each element has an inverse since for + È K + È + È + Š Š Š È Š + È! È (this + + + + + È was obtained by solving for - and in +- ß+ -!) We know + + K since + + ß + K (notice the denominator can never be! or that would contradict +ß and neither can both the terms be! since!  KÏ e! f) Therefore KÏ e! f is a group under multiplication Problem 1110 Prove that a finite group is abelian if and only if its group table is a symmetric matrix Proof Assume K e1 ßÞÞÞß1f is a finite abelian group Then the 3 4 entry in its group table is the group element 1314 1413 The 4ß3 entry in its group table is the group element 1413 1314 Hence by definition the group table is a symmetric matrix Now assume K e1 ßÞÞÞß1f is a finite group with a symmetric matrix Then the 3 4 entry is the same as the 4ß3 entry that is 1 1 1 1 However this holds for 3 4 4 3 4
any two elements 13ß14 K so that 1314 1413 for all elements of K This is precisely the definition of an abelian group Problem 1111 Find the orders of each element of the additive group Î Solution The group is Î!ßßßÞÞÞß Then the orders respectively are ' % $ $ß%ß 'ß and Notice these are kkkîgcd abßkkkb Indeed this will be proven later Problem 1112 Find the orders of the following elements of the multiplicative group a Î b : ß & ( ( $ Solution The identity is so the order for B is the smallest e_ f such that B (with _ B ) Respectively these are ßß ß ß % and (since $ ) Problem 1113 Find the orders of the following elements of the additive group Î$' : ß ' *!ß ß ß!ß ) Solution The identity is! so the order for B is the smallest e_ f such that B! (with _B!) Respectively these are $'ß)ß'ß%ß)ß $ß $'ß )ß Problem 1114 Find the orders of the following elements of the multiplicative group a Î$' b : ß ß &ß $ß $ß ( Solution The identity is so the order for B is the smallest e_ f such that B (with _ B ) Respectively these are ß ß 'ß$ß'ß Problem 1115 Prove that a+ + ÞÞÞ+ b + + ÞÞÞ+ Proof Assume a+ + ÞÞÞ+ bb so that B a+ + ÞÞÞ+ b Then + a+ + ÞÞÞ+ bb + so that a+ + ba+ + $ ÞÞÞ+ bb a+ + $ ÞÞÞ+ bb + Similarly a+ $ + % ÞÞÞ+ bb + + Applying this times results in B + + ÞÞÞ+ as desired Problem 1116 Let B be an element of K Prove that B if and only if kb k is either or Proof Assume B If B then k B k Otherwise k B k Á (only the identity has order ) so that kbk by definition since would be the smallest power B need be raised to in order to obtain the identity On the other hand assume kb k is either or If kbk then B as only the identity has order Otherwise kbk so by definition of order B Problem 1117 Let B be an element of K Prove that if kbk for some positive integer then B B Proof Let kbk By definition B Hence B B B B This is precisely the definition of B B Problem 111 Let B and C be elements of K Prove that BC CB if and only if C BC B if and only if B C BC Proof Assume BßC K and BC CB Multiplying by C on the left C BC C CB B Now assume C BC B Multiplying by B on the left B C BC B B Finally assume B C BC 5
Multipling by CB on the left acbbb C BC CaB B bc BC [generalized associativity] ac C bbc BC acbb CB Problem 1119 Let B K and let +ß (a) Prove that B + B + B and ab + b B + + + (b) Prove that ab b B (c) Establish part (a) for arbitrary integers + and ( positive negative or zero) Proof Notice it is obvious B + B + B for all + This is because we can recursively define B + + + + If +! then B Otherwise B B B + + Similarly B B B (a) We will induct on + and using strong induction First notice B B B B [definition] B B 7 7 Now assume B B B for all 7 Ÿ and Ÿ 5 for some 5 Then inductively we show a5 b7 5 7 5 5 5 5 5 5 B B B for all 7 Ÿ 5 First B B B so that B B ˆ B B B B abbb B B 5 $ [definition] B The last step follows because if 5 B B BBB B B Otherwise we use our inductive assumption Since B 5 a B 5 b a we have shown B 5 b B 5 B Now assume a B 5 b ; B 5 B ; a b for some ; Ÿ 5 Then B 5 B ; B 5 B ; B B 5 ; B [inductive a assumption] B 5 b ; a b a b B B 5 ; 7 + + 7 Similarly we can show ab b B First notice ab b B Now assume ab b B for all 7 Ÿ and Ÿ 5 for some 5 Then inductively we show B 7 a b7 a b B for all 7 Ÿ First ab b a B b B 5 a B b 5 5 Ÿ B 5 B 5 Now assume a b for some Then a b a b ab b [part (a)] B a b 5B a b B a b 5 a b B a ba 5 [part (a)] b (b) As in part (a) we can show this inductively First ab b Assume ˆ 5 B B 5 B Then a5 b a5 b 5 5 5 B B B B B ˆ B B ˆ B B [Proposition 5 111(4)] ˆ B + + Hence ab b B in general (c) Let + be any integer Then B +! B!+ B + B! B + B + B! and B! + B +! + ab! b! + ab b! Hence part (a) is valid when + or is zero Otherwise consider when + and are negative Then we a+ b + + + + know B B B B B by part (a) Then ˆ a b B ˆ B B and using part (b) and Proposition 111(4) this yields B a a + bb B + B B a + bb a b a b ˆ so that B + B + B Now without loss of generality assume + is positive and is negative Consider k+ k k k Then + a+ b + a+ b + + with both parts positive Hence B B B B B B B B Now assume k+ k k k Then + a+ b with + negative and k k k+ k so by what we have just proved + + + + + B B B Therefore B B B B B B Problem 1120 For B an element in K show that B and B have the same order Proof Assume kbk By part (b) of the previous exercise ab b B ab b All that remains to be shown is that this is the least Assume there is a 7 such that 7 7 7 and a b Then a b so that ˆ 7 B B ab b 7 B However this would contradict the assumption kbk 7 Hence kb k Now assume kbk _ Suppose B has finite order Then + + + + This is justified because B + B Œ B B B B 3 3 6
a b a b ˆ B B ab b so again B a b However this would mean B has finite order a contradiction Therefore kb k _ Problem 1121 Let K be a finite group and let B be an element of order Prove that if is odd then B ab b 5 for some 5 Proof If B is the identity so the problem is trivial Otherwise let 7 be such that 7 7 7 7 7 Then by the previous exercise ab b so that ab b ab b ab b B 7 7 7 7 Multiplying by B on the right hand side ab b B B ab b ab b B Then Š ab b 7 7 7 7 B B But by the previous exercise B ab b Hence B ab b so that ab b B ˆ 7 a b 7 5 B ab b ab b for 5 7 If we wish to have a positive 5 let < be the least positive 5 5 < 5 < 5< < integer such that < 7 Then B ab b ab b ab b ab b ab b ab b Problem 1122 If B and 1 are elements of the group K prove that kbk k1 B1 k Deduce that k+ k k+ k for all +ß K Proof First we will show a1 B1b 1 B 1 Inductively when we have a1 B1b 1 B 1 1 5 B1 1 B 5 1 1 5 B1 1 5 Otherwise assume a b Then a b a B1b a1 B1b ˆ 1 B 5 1 5 5 5 a1 B1b 1 B a11 bb1 1 B B1 1 B 1 Hence a1 B1b 1 B 1 Then if kbk we see a1 B1b 1 B 1 1 1 Now assume there is a 7 such that 7 and a1 7 B1b 7 7 Then 1 B 1 so left and right multiplication by 1 and 1 respectively yields 11 B 11 11 so that B 7 However this contradicts the assumption kbk Hence k1 B1k kbk Let B + and 1 + Then k+ k ka+ ba+ ba + bk k+ k Problem 1123 Suppose B K and kbk _ If => for some positive integers = and > prove that kb k > = > Proof First it is clear ab = b B => B Assume there is a 7 such that 7 > and = 7 =7 ab b Then B but =7 => so this contradicts the fact kbk Problem 1124 If + and are commuting elements of K prove that a+ b + for all 5 5 5 5 5 5 5 Proof Inductively a+ b + Assume a+ b + Then a+ b a+ b a+ b + + ˆ 5 5 5 5 + + + Notice the penultimate step is justified by commutativity of + and Hence a+ b!!! + for all When! a+ b + Finally when we know a+ b + by what we have just shown but + and commute so a+ b a+ b + Then aa+ b b a+ b a + b a+ b a b + Problem 1125 Prove that if B for all B K then K is abelian Proof Let BßC K Then abcb since BC K Hence BCBC so BC C B However notice that acbb since CB K so CBCB and consequently CB B C Then abcbacbb C B B C This implies BC B C CB Hence BC CB for all BßC K and by definition K is abelian 7
Problem 1126 Assume L is a nonempty subset of akß æb which is closed under the binary operation on K and is closed under inverses ie for all 2 and 5 L hk and 2 L Prove that L is a group under the operation æ restricted to L (such a subset is called a subgroup of K) Proof Closure is given Associativity follows from the lemma at the beginning of this section's solutions For each 2 L K æ 2 2 æ K 2 Hence L K Finally for each 2 L there is an 2 L (by our assumption of closure of inverses) such that 22 2 2 K L Hence L is a group under the operation Problem 1127 Prove that if B is an element of the group K then eb l f is a subgroup of K ( called the cyclic subgroup of K generated by B) Proof Call the set L eb l f Let 2ß5 L Then there are :ß; such that 2 B and : ; : ;: 5 B Then 25 B B B (exercise 19) Since ; : 25 L by definition Therefore L is ; ; ; ; ; closed under the operation of KÞ Finally if 2 B then 22 B B and 2 2 B B Since ; 2 L by definition so L is closed under inverses By the previous exercise L is a subgroup of K Problem 112 Let aeß æb and afß ˆb be groups and let E F be their direct product Verify all the group axioms for E F: (a) prove that the associative law holds: for all a+ 3ß 3b E F 3 ßß$ a+ ß bca+ ß ba+ ß bd ca+ ß ba+ ß bda+ ß b $ $ $ $ (b) prove that aß b is the identity of E F and (c) prove that the inverse of a+ß b is a+ ß b Proof (a) Let a+ ß b E F with 3 ßß$ Then 3 3 a+ ß bca+ ß ba+ ß bd a+ ß ba+ æ+ ß ˆ b a+ æa+ æ+ bß ˆ a ˆ bb $ $ $ $ $ $ aa+ æ+ bæ+ ß a ˆ bˆ b a+ æ+ ß ˆ ba+ ß b $ $ $ $ ca+ ß ba+ ß bda+ ß b $ $ The intermediate step follows because + æa+ æ+ $ b a+ æ+ bæ+ $ and ˆ a ˆ $ b a ˆ bˆ $ by the fact associativity holds for these elements because E and F is a group (b) Let a+ß b E F Then aß ba+ß b a +߈ b a+ß b a+ ߈ b a+ß baß b æ æ (c) Let a+ß b E F Then a+ß ba+ ß b a+ æ+ ߈ b aß b a+ æ+ß ˆ b a+ ß ba+ß b Problem 1129 Prove that E F is an abelian group if and only if both E and F are abelian Proof Assume E F is abelian Then for all + ß+ E and ß F it is true a+ ß ba+ ß b a+ ß ba+ ß b However a+ ß ba+ ß b a+ + ß b and a+ ß ba+ ß b a+ + ß b But then a+ + ß b a+ + ß b and the components must be equal by definition so that + + + + and Hence E and F are abelian Now assume this Let a+ ß bß a+ ß b E F with + ß+ E and ß F Then a+ ß ba+ ß b a+ + ß b a+ + ß b a+ ß ba+ ß b The intermediate step is justified since we assumed E and F are abelian By definition we now know E F is abelian
Problem 1130 Prove that the elements a+ß b and aß b of E F commute and deduce the order of a+ß b is the least common multiple of k+ k and k k Proof We see a+ß baß b a+ ß b a +ß b aß ba+ß b The intermediate step is justified because + E F and E and F are groups so multiplying + or by the identity is commutative Let k+ k and k k 7 Then a+ß b a+ ß b (by a trivial inductive argument) and so a+ß b aß b If there were a 5 5 with 5 such that a+ß b aß b then ˆ + 5 ß 5 ˆ + 5 ß aß b so that + 5 contradicting the fact k+ k Hence ka+ß bk k+ k Similarly kaß bk 7 k k Now let lcm a7ßb with! 7 and for! ß Then a b a b ˆ! 7 7! +ß + ß + ß Š a+ b ß a b ˆ! ß aß b Now assume there is a $ such $ that $ and a+ß b Then ˆ $ $ + ß $ $ so that + Assume l Î$ so that $ :< for some :ß< $!! < + + :< + : + < + : e f with Then a b + < : + < + < However this again contradicts the fact k+ k since < so that l$ Similarly 7l$ But then by definition $ lcm a7ßb a contradiction Hence k+ß k lcm a7ßb Problem 1131 Prove that any finite group K of even order contains an element of order Proof Let kkk If there was an element of order say B we would have B so that B B Let L e1 Kl1 Á 1 f Clearly  L Furthermore consider the following procedure Let L! L and define L3 by removing some pair e13ß13 f from L3 with 13 L3; that is L3 L3Ïe1ß1 f In each iteration we know 1 Á 1 so that two elements are removed Since L is finite (as K is finite) eventually L5 g for some 5 But then kl5 k kl5k kl5 k kl5 k % etc so that kl! k klk 7 for some 7 Therefore L has an even number of elements Since  L klk kkk It is impossible that klk kkk since kkk 1 is odd Hence klk kkk In other words KÏaL e f b Á g But this means there is some B K such that B Á with B B ( B  L by definition) That is kbk Problem 1132 If B is an element of finite order in K prove that the elements ßBßB ßÞÞÞßB are all distinct Deduce that kbk Ÿ kkk Proof Assume B 3 B 4 for some 3 Á 4 (! Ÿ 3ß4 ) Without loss of generality let 3 4 Then B 34! but 34 so this contradicts the fact kbk Hence ßBßB ßÞÞÞßB are all distinct There are of these elements so kk k kbk Problem 1133 Let B be an element of finite order in K 3 3 (a) Prove that if is odd then B Á B for all 3 ßßÞÞÞßÞ 3 3 (b) Prove that if 5 and Ÿ 3 Ÿ then B B if and only if 3 5 3 3 Proof (a) If there is nothing to prove so assume Assume B B for some 3 3 3 Ÿ 3 Ÿ Then B B so that B Clearly 3 Á since 3 is even and is odd If 3 Ÿ there would thus be a contradiction (since kbk 3) Since 3 we then know 3 so that 3 a3b 3 3! 3 However B B B B B In other words we found a positive integer less than such that B to the power of that integer is But this contradicts the fact kbk Hence 3 3 B Á B for all Ÿ 3 Ÿ (b) Let B 3 B 3 for Ÿ 3 Ÿ and assume 3 Á 5 Then B 3 B 3 so that B 3 By assumption 3 Á If 3 Ÿ there would thus be a contradiction (since kbk 3) Since 3 we then know 3 so 9
3 a3b 3 3 that! 3 However B B B B B In other words we found a positive integer less than such that B to the power of that integer is But this contradicts the fact kbk Hence 5 5 5 5 5 3 3 3 5 Now assume 3 5 Since B B B B we have B B or B B Problem 1134 If B is an element of infinite order in K prove that the elements B are all distinct Proof Assume B 3 B 4 for some 3ß4 with 3 Á 4 Then B 34 If 34! this contradicts the assumption B has infinite order If 34! ab 34 b so that B a 34 b B 43 Then 3 4 43! but again this contradicts the assumption B has infinite order Hence B Á B for all 3ß4 with 3 Á 4; that is the elements B are all distinct Problem 1135 If B is an element of finite order in K use the Division Algorithm to show that any integral power of B equals one of the elements in the set eßbßb ßÞÞÞßB f ( so these are all the distinct elements of the cyclic subgroup of K generated by B) Proof Let 5 Then 5 ;< for some ; < with! Ÿ < by the Division Algorithm Hence with! Ÿ < as required 5 ;< ; < ; < ; < < B B B B ab b B B B Problem 1136 Assume K eß+ßß- f is a group of order % with identity Assume also that K has no elements of order % Use the cancellation laws to show that there is a unique group table for K Deduce that K is abelian Proof Assume there are two group tables Q and Q with the same rows and colums both representing ß+ßß and - in that order (so that 1 + 1$ and 1% -) Now assume there is some 3ß4 such that Q ab b Á Q ab b where Q ab b is the 34th entry of group table (matrix) Q 34 34 5 34 5 10