true/false questions 1. the order of the identity element ...may 03, 2012 · true/false questions...

True/False Questions

1. The order of the identity element in any group is 1.

True. n = 1 is the least positive integer such that en = e .

2. Every cyclic group is abelian.

True. Let G be a cyclic group. All elements of G are of the form an , where n∈ . Let x, y∈G : x = ap , y = aq . Then,

xy = apaq = ap+q = aq+p = aqap = yx .

Thus, xy = yx for all x, y∈G . ( p + q = q + p because is an abelian additive group.)

3. Every abelian group is abelian.

False. Let us take U12 = 1[ ], 5[ ], 7[ ], 11[ ]{ }⊂ 12. U12 is abelian group but this group is not

cyclic since the order of elements of U12 can not be 4. In fact, o 1[ ]( ) = 1,o 5[ ]( ) = 2,o 7[ ]( ) = 2, and o 11[ ]( ) = 2

4. If a subgroup H of a group G İS cyclic, then G must be cyclic.

False. G = ,+( ) is a group and H = ,+( ) is a subgroup of G = ,+( ) . H = = 1 is cyclic whereas is not cyclic.

5. Whether a group G is cyclic or not, each element a of G generates a cyclic subgroup.

True. Look at Remark.

6. Every subgroups of a cyclic group is cyclic.

True. Look at Theorem 3.20.

7. If there exists an m∈+ such that am = e , where a is an element of a group G , then o a( ) = m .

False. m must be the least positive integer such that am = e .

Exercises 3.4" Cyclic Groups

1

8. Any group of order 3 must be cyclic

True. Suppose that G is a group of order 3. One of the elements is e . So, G = e,a,b{ } . We

want to show that a2 = b and a3 = e . But these are not that obvius to prove. The thing that we can prove is that

ab = e .Proof: It can not be anything else. If ab = a then b = e , a contadiction. If ab = b then a = e , a contradiction. Next,

a2 = bbecause it can not be anything else. If a2 = e then a2 = ab would give a = b , a contradiction. If a2 = a then a = e , a contradiction. Finally,

a3 = aa2 = ab = eand we are done.

9. Any group of order 4 must be cyclic.

False. Let us take U12 = 1[ ], 5[ ], 7[ ], 11[ ]{ }⊂ 12. Then, the order of this group is 4. But this

group is not cyclic since the order of elements of U12 can not be 4. In fact, o 1[ ]( ) = 1,o 5[ ]( ) = 2,o 7[ ]( ) = 2, and o 11[ ]( ) = 2

10. Let a be an element of a group G. Then a = a−1 .

True. We will show that a ⊆ a−1 and a−1 ⊆ a . For this, let x ∈ a . So x = ak for some

integer k. Moreover, x = ak = a−1( )−k implies that x ∈ a−1 . The first part is complete. Now,

we will show that the second part is true. For this, let x ∈ a−1 . So x = a−1( )k for some

integer k .

Exercises

456 Answers to True/False and Selected Computational Exercises

2. a. subgroup

c. The set {i, 2i} is not a subgroup of G, since it is not closed. We have i ? i 5 21

{i, 2i}.

3. 5 {[0], [2], [4], [6], [8], [10], [12], [14]},

5. a. {[1], [3], [4], [9], [10], [12]},

6. a. ,

c. ,

7. a. ,

9. The set of all real numbers that are greater than 1 is closed under multiplication but isnot a subgroup of G, since it does not contain inverses. (If x . 1, then x21 , 1.)

17. b. 18. a. {1, 21} c. {I3}

27. a.

c.

30. The subgroup d is the set of all multiples of the least common multiple of mand n.

33. Let H 5 {e, s} and K 5 {e, g}.

35. Let H 5 {e, s} and K 5 {e, g}.

Exercises 3.4 Pages 170–174

True or False

1. true 2. true 3. false 4. false 5. true

6. true 7. false 8. true 9. false 10. true

Exercises

1. 8e 95 {e}, 8r 9 5 {e, r, r2}, 8s 9 5 {e, s}, 8g 95 {e, g}, 8d 95 {e, d}

3. The element e has order 1. Each of the elements s, g, and d has order 2. Each of theelements r and r2 has order 3.

5. o(I3) 5 1, o(P1) 5 o(P2) 5 o(P4) 5 2, o(P3) 5 o(P5) 5 3

6. a. o(A) 5 2

8n98m9 CP45 5I3, P46CI3

5 G, CP15 5I3, P16, CP2

5 5I3, P26, CP35 CP5

5 5I3, P3, P56,Ck 5 C2k 5 51, k, 21, 2k6C1 5 C21 5 G, Ci 5 C2i 5 51, i, 21, 2i6, Cj 5 C2j 5 51, j, 21, 2j6,3x4 5 5x6o1 8A92 5 5

8A9 5 b B 324 304304 334R , B 344 304304 314R , B 314 304304 344R , B 334 304304 324R , B 304 304304 304R ro1 8A92 5 38A9 5 b B0 21

1 21R , B21 1

21 0R , B1 0

0 1R r o1 8A92 5 48A9 5 b B0 21

1 0R , B21 0

0 21R , B 0 1

21 0R , B1 0

0 1R ro1 8 344 92 5 6

o1 8 364 92 5 88 364 9o

2.1 = 1{ }, −1 = 1,−1{ }, i = 1,i,−1,−i{ }, −i = 1,−i,−1,i{ }j = 1, j,−1,− j{ }, − j = 1,− j,−1, j{ }, k = 1,k,−1,−k{ }, −k = 1,−k,−1,k{ }.


2. a. subgroup


{i, 2i}.

3. 5 {[0], [2], [4], [6], [8], [10], [12], [14]},

5. a. {[1], [3], [4], [9], [10], [12]},

6. a. ,

c. ,

7. a. ,


17. b. 18. a. {1, 21} c. {I3}

27. a.

c.


33. Let H 5 {e, s} and K 5 {e, g}.

35. Let H 5 {e, s} and K 5 {e, g}.


True or False



Exercises



5. o(I3) 5 1, o(P1) 5 o(P2) 5 o(P4) 5 2, o(P3) 5 o(P5) 5 3

6. a. o(A) 5 2

8n98m9 CP45 5I3, P46CI3

5 G, CP15 5I3, P16, CP2

5 5I3, P26, CP35 CP5



1 21R , B21 1

21 0R , B1 0

0 1R r o1 8A92 5 48A9 5 b B0 21

1 0R , B21 0

0 21R , B 0 1

21 0R , B1 0

0 1R ro1 8 344 92 5 6

o1 8 364 92 5 88 364 9o

4. o 1( ) = 1, o −1( ) = 2, o i( ) = o −i( ) = o j( ) = o − j( ) = o k( ) = o −k( ) = 4.


2. a. subgroup


{i, 2i}.

3. 5 {[0], [2], [4], [6], [8], [10], [12], [14]},

5. a. {[1], [3], [4], [9], [10], [12]},

6. a. ,

c. ,

7. a. ,


17. b. 18. a. {1, 21} c. {I3}

27. a.

c.


33. Let H 5 {e, s} and K 5 {e, g}.

35. Let H 5 {e, s} and K 5 {e, g}.


True or False



Exercises



5. o(I3) 5 1, o(P1) 5 o(P2) 5 o(P4) 5 2, o(P3) 5 o(P5) 5 3

6. a. o(A) 5 2

8n98m9 CP45 5I3, P46CI3

5 G, CP15 5I3, P16, CP2

5 5I3, P26, CP35 CP5



1 21R , B21 1

21 0R , B1 0

0 1R r o1 8A92 5 48A9 5 b B0 21

1 0R , B21 0

0 21R , B 0 1

21 0R , B1 0

0 1R ro1 8 344 92 5 6

o1 8 364 92 5 88 364 9o


2

6. a) n = 2 is the least positive integer such that A2 = I . Therefore, o A( ) = 2. b) n = 4 is the least positive integer such that A4 = I . Therefore, o A( ) = 4.

7. Since o G( ) = 8 , a8 = e .

a) x = 4 is the least positive integer such that a2( )x = e . Therefore, o a2( ) = 4.b) x = 8 is the least positive integer such that a3( )x = e . Therefore, o a3( ) = 8.c) x = 2 is the least positive integer such that a4( )x = e . Therefore, o a4( ) = 2.d) x = 8 is the least positive integer such that a5( )x = e . Therefore, o a5( ) = 8.e) x = 4 is the least positive integer such that a6( )x = e . Therefore, o a6( ) = 4.f) x = 8 is the least positive integer such that a7( )x = e . Therefore, o a7( ) = 8.g) x = 1 is the least positive integer such that a8( )x = e . Therefore, o a8( ) = 1.

8. Since o G( ) = 9 , a9 = e .

a) x = 9 is the least positive integer such that a2( )x = e . Therefore, o a2( ) = 9.b) x = 6 is the least positive integer such that a3( )x = e . Therefore, o a3( ) = 6.c) x = 9 is the least positive integer such that a4( )x = e . Therefore, o a4( ) = 9.d) x = 9 is the least positive integer such that a5( )x = e . Therefore, o a5( ) = 9.e) x = 3 is the least positive integer such that a6( )x = e . Therefore, o a6( ) = 3.f) x = 9 is the least positive integer such that a7( )x = e . Therefore, o a7( ) = 9.g) x = 9 is the least positive integer such that a8( )x = e . Therefore, o a8( ) = 9.h) x = 1 is the least positive integer such that a9( )x = e . Therefore, o a9( ) = 1.

9. a) 1[ ], 3[ ], 5[ ], 7[ ] b) 1[ ], 5[ ], 7[ ], 11[ ] c) 1[ ], 3[ ], 7[ ], 9[ ] d) 1[ ], 2[ ], 4[ ], 7[ ], 8[ ], 11[ ], 13[ ], 14[ ] e) 1[ ], 3[ ], 5[ ], 7[ ], 9[ ], 11[ ], 13[ ], 15[ ] f) 1[ ], 5[ ], 7[ ], 11[ ], 13[ ], 17[ ]

10. a) The divisors of 12 are d = 1,2,3,4,6,12 , so the distinct subgroups of 12 are those subgroups d a[ ] where a[ ] is a generator of 12 . Since 12 = 1[ ] , the generator a[ ] can

be taken 1[ ] . Then,

1. 1[ ] = 12 ; o 12( ) = 12 ,

2. 1[ ] = 2[ ] = 0[ ], 2[ ], 4[ ], 6[ ], 8[ ], 10[ ]{ } ; o 2[ ]( ) = 6,


3

3. 1[ ] = 3[ ] = 0[ ], 3[ ], 6[ ], 9[ ]{ } ; o 3[ ]( ) = 4,4. 1[ ] = 4[ ] = 0[ ], 4[ ], 8[ ]{ } ; o 4[ ]( ) = 3,6. 1[ ] = 6[ ] = 0[ ], 6[ ]{ } ; o 6[ ]( ) = 2,12. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

b) The divisors of 8 are d = 1,2,4,8 , so the distinct subgroups of 8 are those subgroups d a[ ] where a[ ] is a generator of 8 . Since 8 = 1[ ] , the generator a[ ] can be taken

1[ ] . Then,

1. 1[ ] = 8 ; o 8( ) = 8 ,

2. 1[ ] = 2[ ] = 0[ ], 2[ ], 4[ ], 6[ ]{ } ; o 2[ ]( ) = 4,4. 1[ ] = 4[ ] = 0[ ], 4[ ]{ } ; o 4[ ]( ) = 2,8. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

c) The divisors of 10 are d = 1,2,5,10 , so the distinct subgroups of 10 are those subgroups d a[ ] where a[ ] is a generator of 10 . Since 10 = 1[ ] , the generator a[ ] can be taken 1[ ] . Then,

1. 1[ ] = 10 ; o 10( ) = 10 ,

2. 1[ ] = 2[ ] = 0[ ], 2[ ], 4[ ], 6[ ], 8[ ]{ } ; o 2[ ]( ) = 5,5. 1[ ] = 5[ ] = 0[ ], 5[ ]{ } ; o 5[ ]( ) = 2,10. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

d) The divisors of 15 are d = 1,3,5,15 , so the distinct subgroups of 15 are those subgroups d a[ ] where a[ ] is a generator of 15 . Since 15 = 1[ ] , the generator a[ ] can


1. 1[ ] = 15 ; o 15( ) = 15 ,

3. 1[ ] = 3[ ] = 0[ ], 3[ ], 6[ ], 9[ ], 12[ ]{ } ; o 3[ ]( ) = 5,5. 1[ ] = 5[ ] = 0[ ], 5[ ], 10[ ]{ } ; o 5[ ]( ) = 3,15. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

e) The divisors of 16 are d = 1,2,4,8,16 , so the distinct subgroups of 16 are those subgroups d a[ ] where a[ ] is a generator of 16 . Since 16 = 1[ ] , the generator a[ ] can be taken 1[ ] . Then,


4

1. 1[ ] = 16 ; o 16( ) = 16 ,

2. 1[ ] = 2[ ] = 0[ ], 2[ ], 4[ ], 6[ ], 8[ ], 10[ ], 12[ ], 14[ ]{ } ; o 2[ ]( ) = 8,4. 1[ ] = 4[ ] = 0[ ], 4[ ], 8[ ], 12[ ]{ } ; o 4[ ]( ) = 4,8. 1[ ] = 8[ ] = 0[ ], 8[ ]{ } ; o 8[ ]( ) = 2,16. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

f) The divisors of 18 are d = 1,2,3,6,9,18 , so the distinct subgroups of 18 are those subgroups d a[ ] where a[ ] is a generator of 18 . Since 18 = 1[ ] , the generator a[ ] can


1. 1[ ] = 18 ; o 18( ) = 18 ,

2. 1[ ] = 2[ ] = 0[ ], 2[ ], 4[ ], 6[ ], 8[ ], 10[ ], 12[ ], 14[ ], 16[ ]{ } ; o 2[ ]( ) = 9,3. 1[ ] = 3[ ] = 0[ ], 3[ ], 6[ ], 9[ ], 12[ ], 15[ ]{ } ; o 3[ ]( ) = 6,6. 1[ ] = 6[ ] = 0[ ], 6[ ], 12[ ]{ } ; o 6[ ]( ) = 3,9. 1[ ] = 9[ ] = 0[ ], 9[ ]{ } ; o 9[ ]( ) = 2,18. 1[ ] = 0[ ] = 0[ ]{ } ; o 0[ ]( ) = 1.

11-12 a) Since 7 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ], 5[ ], 6[ ]{ } , the order of this group is 6. By

Theorem 3.24, the generators of this group is in the form am iff m,6( ) = 1 . So the values of

m are 1 and 5. Therefore, 7 − 0[ ]{ } = a1 = a5 . Let us find the element a of this group.

Take the element 3[ ] . All of the powers of 3[ ] are

3[ ]1 = 3[ ],3[ ]2 = 2[ ],3[ ]3 = 6[ ],3[ ]4 = 4[ ],3[ ]5 = 5[ ],3[ ]6 = 1[ ].

Therefore, 7 − 0[ ]{ } = 3[ ]1 = 3[ ]5 = 5[ ] .

b) Since 5 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ]{ } , the order of this group is 4. By Theorem 3.24, the

generators of this group is in the form am iff m, 4( ) = 1 . So the values of m are 1 and 3.


5

Therefore, 5 − 0[ ]{ } = a1 = a3 . Let us find the element a of this group. Take the

element 2[ ] . All of the powers of 2[ ] are

2[ ]1 = 2[ ],2[ ]2 = 4[ ],2[ ]3 = 3[ ],2[ ]4 = 1[ ].

Therefore, 5 − 0[ ]{ } = 2[ ]1 = 2[ ]3 = 3[ ] .

c) Since 11 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ], 5[ ], 6[ ], 7[ ], 8[ ], 9[ ], 10[ ]{ } , the order of this group is 10.

By Theorem 3.24, the generators of this group is in the form am iff m,10( ) = 1 . So the

values of m are 1, 3, 7 and 9. Therefore, 11 − 0[ ]{ } = a1 = a3 = a7 = a9 . Let us find the element a of this group. Take the element 2[ ] . All of the powers of 2[ ] are

2[ ]1 = 2[ ],2[ ]2 = 4[ ],2[ ]3 = 8[ ],2[ ]4 = 5[ ],2[ ]5 = 10[ ],2[ ]6 = 9[ ],2[ ]7 = 7[ ],2[ ]8 = 3[ ],2[ ]9 = 6[ ],2[ ]10 = 1[ ].

Therefore, 11 − 0[ ]{ } = 2[ ]1 = 2[ ]3 = 8[ ] = 2[ ]7 = 7[ ] = 2[ ]9 = 6[ ] .

d) Since 13 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ], 5[ ], 6[ ], 7[ ], 8[ ], 9[ ], 10[ ], 11[ ], 12[ ]{ } , the order of this

group is 12. By Theorem 3.24, the generators of this group is in the form am iff m,12( ) = 1 .

So the values of m are 1, 5, 7 and 11. Therefore, 13 − 0[ ]{ } = a1 = a5 = a7 = a11 . Let

us find the element a of this group. Take the element 7[ ] . All of the powers of 7[ ] are


6

7[ ]1 = 7[ ],7[ ]2 = 10[ ],7[ ]3 = 5[ ],7[ ]4 = 9[ ],7[ ]5 = 11[ ],7[ ]6 = 12[ ],7[ ]7 = 6[ ],7[ ]8 = 3[ ],7[ ]9 = 8[ ],7[ ]10 = 4[ ],7[ ]11 = 2[ ],7[ ]12 = 1[ ].

Therefore, 13 − 0[ ]{ } = 7[ ]1 = 7[ ]5 = 11[ ] = 7[ ]7 = 6[ ] = 7[ ]11 = 2[ ] .

e) Since 17 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ], 5[ ], 6[ ], 7[ ], 8[ ], 9[ ], 10[ ], 11[ ], 12[ ], 13[ ], 14[ ], 15[ ], 16[ ]{ } , the

order of this group is 16. By Theorem 3.24, the generators of this group is in the form am iff m,16( ) = 1 . So the values of m are 1, 3, 5, 7, 9, 11, 13, and 15. Therefore,

17 − 0[ ]{ } = a1 = a3 = a5 = a7 = a9 = a11 = a13 = a15 . Let us find the element a of

this group. Take the element 3[ ] . All of the powers of 3[ ] are


7

3[ ]1 = 3[ ],3[ ]2 = 9[ ],3[ ]3 = 10[ ],3[ ]4 = 13[ ],3[ ]5 = 5[ ],3[ ]6 = 15[ ],3[ ]7 = 11[ ],3[ ]8 = 16[ ],3[ ]9 = 14[ ],3[ ]10 = 8[ ],3[ ]11 = 7[ ],3[ ]12 = 4[ ],3[ ]13 = 12[ ],3[ ]14 = 2[ ],3[ ]15 = 6[ ],3[ ]16 = 1[ ].

Therefore,

17 − 0[ ]{ } = 3[ ]1 = 3[ ]3 = 10[ ] = 3[ ]5 = 5[ ] = 3[ ]7 = 11[ ] = 3[ ]9 = 14[ ] = 3[ ]11 = 7[ ] = 3[ ]13 = 12[ ] = 3[ ]15 = 6[ ]

f) Since

19 − 0[ ]{ } = 1[ ], 2[ ], 3[ ], 4[ ], 5[ ], 6[ ], 7[ ], 8[ ], 9[ ], 10[ ], 11[ ], 12[ ], 13[ ], 14[ ], 15[ ], 16[ ], 17[ ], 18[ ]{ } , the order of this group is 18. By Theorem 3.24, the generators of this group is in the form am iff m,18( ) = 1 . So the values of m are 1, 5, 7, 11, 13, and 17. Therefore,

19 − 0[ ]{ } = a1 = a5 = a7 = a11 = a13 = a17 . Let us find the element a of this group. Take the element 3[ ] . All of the powers of 3[ ] are


8

2[ ]1 = 2[ ],2[ ]2 = 4[ ],2[ ]3 = 8[ ],2[ ]4 = 16[ ],2[ ]5 = 13[ ],2[ ]6 = 7[ ],2[ ]7 = 14[ ],2[ ]8 = 9[ ],2[ ]9 = 18[ ],2[ ]10 = 17[ ],2[ ]11 = 15[ ],2[ ]12 = 11[ ],2[ ]13 = 3[ ],2[ ]14 = 6[ ],2[ ]15 = 12[ ],2[ ]16 = 5[ ],2[ ]17 = 10[ ],2[ ]18 = 1[ ].

Therefore,

19 − 0[ ]{ } = 2[ ]1 = 2[ ]5 = 13[ ] = 2[ ]7 = 14[ ] = 2[ ]11 = 15[ ] = 2[ ]13 = 3[ ] = 2[ ]17 = 10[ ] .

13. a) By Corollary 3.23 and Exercise 11, 7 − 0[ ]{ } is a finite cyclic group of order 6 with

3[ ]∈ 7 − 0[ ]{ } as a generator. The distinct subgroups of 7 − 0[ ]{ } are those subgroups

3[ ]d where d is a positive divisor of 6 . Therefore, the distinct subgroups are as follows:

3[ ]1 = 7 − 0[ ]{ },3[ ]2 = 2[ ], 4[ ], 1[ ]{ },3[ ]3 = 6[ ], 1[ ]{ },3[ ]6 = 1[ ]{ }.


9

b) By Corollary 3.23 and Exercise 11, 5 − 0[ ]{ } is a finite cyclic group of order 4 with

2[ ]∈5 − 0[ ]{ } as a generator. The distinct subgroups of 5 − 0[ ]{ } are those subgroups

2[ ]d where d is a positive divisor of 4. Therefore, the distinct subgroups are as follows:

2[ ]1 = 5 − 0[ ]{ },2[ ]2 = 4[ ], 1[ ]{ },2[ ]4 = 1[ ]{ }.

c) By Corollary 3.23 and Exercise 11, 11 − 0[ ]{ } is a finite cyclic group of order 10 with



2[ ]1 = 11 − 0[ ]{ },2[ ]2 = 4[ ], 5[ ], 9[ ], 3[ ], 1[ ]{ },2[ ]5 = 10[ ], 1[ ]{ },2[ ]10 = 1[ ]{ }.

d) By Corollary 3.23 and Exercise 11, 13 − 0[ ]{ } is a finite cyclic group of order 12 with



7[ ]1 = 13 − 0[ ]{ },7[ ]2 = 10[ ], 9[ ], 12[ ], 3[ ], 4[ ], 1[ ]{ },7[ ]3 = 5[ ], 12[ ], 8[ ], 1[ ]{ },7[ ]4 = 9[ ], 3[ ], 1[ ]{ },7[ ]6 = 12[ ], 1[ ]{ },7[ ]12 = 1[ ]{ }.


10

e) By Corollary 3.23 and Exercise 11, 17 − 0[ ]{ } is a finite cyclic group of order 16 with



3[ ]1 = 17 − 0[ ]{ },3[ ]2 = 9[ ], 13[ ], 15[ ], 16[ ], 8[ ], 4[ ], 2[ ], 1[ ]{ },3[ ]4 = 13[ ], 16[ ], 4[ ], 1[ ]{ },3[ ]8 = 16[ ], 1[ ]{ },3[ ]16 = 1[ ]{ }.

f) By Corollary 3.23 and Exercise 11, 19 − 0[ ]{ } is a finite cyclic group of order 18 with



2[ ]1 = 19 − 0[ ]{ },2[ ]2 = 4[ ], 16[ ], 7[ ], 9[ ], 17[ ], 11[ ], 6[ ], 5[ ], 1[ ]{ },2[ ]3 = 8[ ], 7[ ], 18[ ], 11[ ], 12[ ], 1[ ]{ },2[ ]6 = 7[ ], 11[ ], 1[ ]{ },2[ ]9 = 18[ ], 1[ ]{ },2[ ]18 = 1[ ]{ }.

14.

Since Z!7 = ![3]", each subgroup will be generated by [3]m where m is a divisor of 6. Thus

m = 1, 2, 3, 6. Since [3]1 = [3], [3]2 = [2], [3]3 = [6] and [3]6 = [1]. Therefore the subgroupsare as follows:

![1]" = {[1]}![2]" = {[1], [2], [4]}![3]" = Z!

7

![6]" = {[1], [6]}

14. Prove that the set H =!"

1 n0 1

#$$$$ n # Z%

is a cyclic subgroup of GL2(R).

Proof. First let’s prove that H is a subgroup of the invertible 2 by 2 real matrices.

Claim:"

1 10 1

#n

="

1 n0 1

#for all positive integers n.

Clearly the statement holds when n = 1. Assume"

1 10 1

#k

="

1 k0 1

#. Then we have the

following."

1 10 1

#k+1

="

1 10 1

#k "1 10 1

#

="

1 k0 1

#"1 10 1

#

="

1 k + 10 1

#

Thus by induction"

1 10 1

#n

="

1 n0 1


By definition"

1 10 1

#0

="

1 00 1

#.

Suppose n > 0, then we have the following."

1 10 1

#"n

=&"

1 10 1

#n'"1

="

1 n0 1

#"1

="

1 $n0 1

#

Thus for all integers n,"

1 10 1

#n

="

1 n0 1

#.

Now we will prove H is cyclic. Let"

1 10 1

#k

#("

1 10 1

#), then

"1 10 1

#k

="

1 k0 1

## H.

Thus("

1 10 1

#)% H .

Now we will show H %("

1 10 1

#). Let A # H . Then A =

"1 n0 1

#for some n # Z. By

above claim,"

1 n0 1

#=

"1 10 1

#n

#("

1 10 1

#).


11

Since Z!7 = ![3]", each subgroup will be generated by [3]m where m is a divisor of 6. Thus

m = 1, 2, 3, 6. Since [3]1 = [3], [3]2 = [2], [3]3 = [6] and [3]6 = [1]. Therefore the subgroupsare as follows:

![1]" = {[1]}![2]" = {[1], [2], [4]}![3]" = Z!

7

![6]" = {[1], [6]}

14. Prove that the set H =!"

1 n0 1

#$$$$ n # Z%

is a cyclic subgroup of GL2(R).

Proof. First let’s prove that H is a subgroup of the invertible 2 by 2 real matrices.

Claim:"

1 10 1

#n

="

1 n0 1


Clearly the statement holds when n = 1. Assume"

1 10 1

#k

="

1 k0 1

#. Then we have the

following."

1 10 1

#k+1

="

1 10 1

#k "1 10 1

#

="

1 k0 1

#"1 10 1

#

="

1 k + 10 1

#

Thus by induction"

1 10 1

#n

="

1 n0 1


By definition"

1 10 1

#0

="

1 00 1

#.

Suppose n > 0, then we have the following."

1 10 1

#"n

=&"

1 10 1

#n'"1

="

1 n0 1

#"1

="

1 $n0 1

#

Thus for all integers n,"

1 10 1

#n

="

1 n0 1

#.

Now we will prove H is cyclic. Let"

1 10 1

#k

#("

1 10 1

#), then

"1 10 1

#k

="

1 k0 1

## H.

Thus("

1 10 1

#)% H .

Now we will show H %("

1 10 1

#). Let A # H . Then A =

"1 n0 1

#for some n # Z. By

above claim,"

1 n0 1

#=

"1 10 1

#n

#("

1 10 1

#).

Thus H =!"

1 10 1

#$.

28. Let a and b be elments of a finite group G.

(b) Prove that a and bab!1 have the same order.

Proof. Let a, b ! G. To prove a and bab!1 have the same order, we need the followingfact.

Claim: Let n be a natural numbers, then (bab!1)n = banb!1.We will prove this by induction on n. Clearly (bab!1)1 = ba1b!1.Assume (bab!1)k = bakb!1. Then we have the following.

(bab!1)k+1 = (bab!1)k(bab!1) = bakb!1bab!1 = bakab!1 = bak+1b!1

Thus by induction, (bab!1)n = banb!1.Assume |a| = r and |bab!1| = s. So ar = e and (bab!1)s = e. Thus we have thefollowing.

ar = e

barb!1 = beb!1

barb!1 = e

(bab!1)r = e

So since |bab!1| = s the above implies s|r. Moreover we have the following.

(bab!1)s = e

basb!1 = e

b!1basb!1b = b!1eb

as = e

So since |a| = r the above implies r|s.Therefore we have that r = s. In other words, |a| = |bab!1|.

(c) Prove that ab and ba have the same order.

Proof. Assume |ab| = m and |ba| = n, then (ab)m = e and (ba)n = e. Therefore we havethe following.

(ab)m = e

a(ba)m!1b = e

a!1a(ba)m!1b = a!1e

(ba)m!1b = a!1

(ba)m!1ba = a!1a

(ba)m = e

Since |ba| = n, the above gives us that n|m. Similarly to above, you can show that(ba)n = e implies (ab)n = e, and since |a| = m, we can conclude m|n. Therefore n = m.In other words, |ab| = |ba|.

15. a) i) Let us prove that

cosnθ −sinnθsinnθ cosnθ

⎡

⎣⎢

⎤

⎦⎥ =

cosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥

n

.

Clearly the statement holds when n = 1 . Assume

coskθ −sin kθsin kθ coskθ

⎡

⎣⎢

⎤

⎦⎥ =


⎡

⎣⎢

⎤

⎦⎥

k

.

Then we have the following.


12


⎡

⎣⎢

⎤

⎦⎥

k+1

= cosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥

kcosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥

= coskθ −sin kθsin kθ coskθ

⎡

⎣⎢

⎤

⎦⎥


⎡

⎣⎢

⎤

⎦⎥

= coskθ cosθ − sin kθ sinθ −coskθ sinθ − sin kθ cosθsin kθ cosθ + coskθ sinθ −sin kθ sinθ + coskθ cosθ

⎡

⎣⎢

⎤

⎦⎥

=cos k +1( )θ −sin k +1( )θsin k +1( )θ cos k +1( )θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥.

Thus by induction cosnθ −sinnθsinnθ cosnθ

⎡

⎣⎢

⎤

⎦⎥ =


⎡

⎣⎢

⎤

⎦⎥

n

for all integers n .

ii)First let us prove that H is a subgroup of the invertible 2 by 2 real matrices. H is nonempty since

cos0θ −sin0θsin0θ cos0θ

⎡

⎣⎢

⎤

⎦⎥ =

1 00 1

⎡

⎣⎢

⎤

⎦⎥ ∈H for n = 0 .

Take two arbitrary element of H such that

A =cosn1θ −sinn1θsinn1θ cosn1θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥,B =

cosn2θ −sinn2θsinn2θ cosn2θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥∈H for all integers n1 and n2 .

We will show that AB∈H . For this,

AB =cosn1θ −sinn1θsinn1θ cosn1θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

cosn2θ −sinn2θsinn2θ cosn2θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥= cosθ −sinθ

sinθ cosθ⎡

⎣⎢

⎤

⎦⎥

n1cosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥

n2

= cosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥

n1+n2

=cos n1 + n2( )θ −sin n1 + n2( )θsin n1 + n2( )θ cos n1 + n2( )θ

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥∈H .

So H is closed under the matrix multiplication. Now we will show that A−1 ∈H . For this,

A =cosn1θ −sinn1θsinn1θ cosn1θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥⇒ A−1 = 1

detAcosn1θ −sinθsinθ cosn1θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

cosn1θ −sinθsinθ cosn1θ

⎡

⎣⎢⎢

⎤

⎦⎥⎥∈H .

Therefore H is a subgroup of the set of all invertible 2 by 2 real matrices. Now we will prove H is cyclic. By using the trigonometric identity


13

cosnθ −sinnθsinnθ cosnθ

⎡

⎣⎢

⎤

⎦⎥ =


⎡

⎣⎢

⎤

⎦⎥

n

we have H = cosθ −sinθsinθ cosθ

⎡

⎣⎢

⎤

⎦⎥ , as required.

b) H = cos90n −sin90n

sin90n cos90n⎡

⎣⎢

⎤

⎦⎥ :n∈

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 1 0

0 1⎡

⎣⎢

⎤

⎦⎥,

0 −11 0

⎡

⎣⎢

⎤

⎦⎥,

−1 00 −1

⎡

⎣⎢

⎤

⎦⎥,

0 1−1 0

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪.

c)

H = cos120n −sin120nsin120n cos120n

⎡

⎣⎢

⎤

⎦⎥ :n∈

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 1 0

0 1⎡

⎣⎢

⎤

⎦⎥,

− 12

− 32

32

− 12

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

,− 12

32

0 − 12

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

⎧

⎨

⎪⎪

⎩

⎪⎪

⎫

⎬

⎪⎪

⎭

⎪⎪

.

16. To show that multiplication modn is a binary operation on Un , I must show that the product of units is a unit. ! Suppose a,b∈Un . Then a has a multiplicative inverse a−1 and b has a multiplicative inverse b−1 . Now

b−1a−1( ) ab( ) = b−1 a−1a( )b = b−1 e( )b = b−1b = e,

ab( ) b−1a−1( ) = a bb−1( )a−1 = a e( )a−1 = aa−1e.

Hence, b−1a−1 is the multiplicative inverse of ab , and ab is a unit. Therefore, multiplication modn is a binary operation on Un .

I will take it for granted that multiplication mod n is associative.The identity element for multiplication modn is 1, and 1 is a unit in n .Finally, every element of Un has a multiplicative inverse, by definition.Therefore, Un is a group under multiplication modn .

17. ⇐: Let a[ ]∈n . Assume that a and n are reletively prime. Then a,n( ) = 1 . By Theorem 2.12, 1= ax + ny for some integer x, y.Using this fact, we have the following.

1= ax + ny → 1≡ ax modn( )→ 1[ ] = ax[ ] in n → 1[ ] = a[ ] x[ ] in n .

In the last equality, Theorem 2.25 guarantees the existence of a solution s to the congruence as ≡ 1 modn( ) . This shows that a[ ] in n has the multiplicative inverses. By the definition of Un , a[ ]∈Un .


14

:⇒ Let a[ ]∈Un . Suppose first that a[ ] has a multiplicative inverse b[ ] in n . Thena[ ] b[ ] = 1[ ] .

This means that

ab[ ] = 1[ ] and ab ≡ 1 modn( ) .

Therefore,

ab −1= nq

for some integer q , and

a b( ) + n −q( ) = 1.

Answers to True/False and Selected Computational Exercises 457

7. a. 4 c. 2 e. 4 g. 1

8. a. 9 c. 9 e. 3 g. 9

9. a. [1], [3], [5], [7]

c. [1], [3], [7], [9]

e. [1], [3], [5], [7], [9], [11], [13], [15]

10. a. {[0]}, 1; {[0], [6]}, 2; {[0], [4], [8]}, 3; {[0], [3], [6], [9]}, 4; {[0], [2], [4], [6], [8], [10]}, 6; Z12, 12

c. {[0]}, 1; {[0], [5]}, 2; {[0], [2], [4], [6], [8]}, 5; Z10, 10

e. {[0]}, 1; {[0], [8]}, 2; {[0], [4], [8], [12]}, 4;

{[0], [2], [4], [6], [8], [10], [12], [14]}, 8; Z16, 16

11. a. G 5 8[3]9 5 8[5]9c. G 5 8[2]9 5 8[6]9 5 8[7]9 5 8[8]9e. G 5 8[3]9 5 8[5] 9 5 8[6]9 5 8[7]9 5 8[10] 9 5 8[11] 9 5 8[12] 9 5 8[14]9

12. a. [3], [5]

c. [2], [6], [7], [8]

e. [3], [5], [6], [7], [10], [11], [12], [14]

13. a. {[1]}, 1; {[1], [6]}, 2; {[1], [2], [4]}, 3; G, 6

c. {[1]}, 1; {[1], [10]}, 2; {[1], [3], [4], [5], [9]}, 5; G, 10

e. {[1]}, 1; {[1], [16]}, 2; {[1], [4], [13], [16]}, 4;

{[1], [2], [4], [8], [9], [13], [15], [16]}, 8; G, 16

15. c.

18. a. U20 5 {[1], [3], [7], [9], [11], [13], [17], [19]}

C 212

!32

2!32 21

2

SsH 5 cB1 0

0 1R , C21

2 2!32

!32 21

2

S ,

? [1] [3] [7] [9] [11] [13] [17] [19]

[1] [1] [3] [7] [9] [11] [13] [17] [19]

[3] [3] [9] [1] [7] [13] [19] [11] [17]

[7] [7] [1] [9] [3] [17] [11] [19] [13]

[9] [9] [7] [3] [1] [19] [17] [13] [11]

[11] [11] [13] [17] [19] [1] [3] [7] [9]

[13] [13] [19] [11] [17] [3] [9] [1] [7]

[17] [17] [11] [19] [13] [7] [1] [9] [3]

[19] [19] [17] [13] [11] [9] [7] [3] [1]

b. U8 = a[ ]∈n : a[ ], n[ ]( ) = 1{ } = 1[ ], 3[ ], 5[ ], 7[ ]{ }. The Cayley table for U8 is as follows:

• 1[ ] 3[ ] 5[ ] 7[ ]

1[ ] 1[ ] 3[ ] 5[ ] 7[ ]

3[ ] 3[ ] 1[ ] 7[ ] 5[ ]

5[ ] 5[ ] 7[ ] 1[ ] 3[ ]

7[ ] 7[ ] 5[ ] 3[ ] 1[ ]

c. U24 = a[ ]∈24 : a[ ], n[ ]( ) = 1{ } = 1[ ], 5[ ], 7[ ], 11[ ], 13[ ], 17[ ], 19[ ], 23[ ]{ }.


15


c. U24 5 {[1], [5], [7], [11], [13], [17], [19], [23]}

? [1] [5] [7] [11] [13] [17] [19] [23]

[1] [1] [5] [7] [11] [13] [17] [19] [23]

[5] [5] [1] [11] [7] [17] [13] [23] [19]

[7] [7] [11] [1] [5] [19] [23] [13] [17]

[11] [11] [7] [5] [1] [23] [19] [17] [13]

[13] [13] [17] [19] [23] [1] [5] [7] [11]

[17] [17] [13] [23] [19] [5] [1] [11] [7]

[19] [19] [23] [13] [17] [7] [11] [1] [5]

[23] [23] [19] [17] [13] [11] [7] [5] [1]

19. a. not cyclic

c. not cyclic

21. a. f(8) 5 4; a, a3, a5, a7

c. f(18) 5 6; a, a5, a7, a11, a13, a17

e. f(7) 5 6; a, a2, a3, a4, a5, a6

22. a.

c.

e.

23. a. a12

c. a6, a18

24. a. none

c. a5, a10, a15, a20, a25, a30

25. All subgroups of Z are of the form 8n 9 , n a fixed integer.

35. p 2 1

8a9 5 G, 8a79 5 8e9 5 5e68a189 5 8e9 5 5e68a99 5 5a9, a18 5 e68a69 5 8a129 5 5a6, a12, a18 5 e68a39 5 8a159 5 5a3, a6, a9, a12, a15, a18 5 e68a295 8a495 8a895 8a1095 8a1495 8a1695 5a2, a4, a6, a8, a10, a12, a14, a16, a18 5 e68a9 5 G

8a89 5 8e9 5 5e68a49 5 5a4, a8 5 e68a29 5 8a69 5 5a2, a4, a6, a8 5 e68a9 5 G

d. U30 = a[ ]∈ 30 : a[ ], n[ ]( ) = 1{ } = 1[ ], 7[ ], 11[ ], 13[ ], 17[ ], 19[ ], 23[ ], 29[ ]{ }.

• 1[ ] 7[ ] 11[ ] 13[ ] 17[ ] 19[ ] 23[ ] 29[ ]

1[ ] 1[ ] 7[ ] 11[ ] 13[ ] 17[ ] 19[ ] 23[ ] 29[ ]

7[ ] 7[ ] 19[ ] 17[ ] 1[ ] 29[ ] 13[ ] 11[ ] 23[ ]

11[ ] 11[ ] 17[ ] 1[ ] 23[ ] 7[ ] 29[ ] 13[ ] 19[ ]

13[ ] 13[ ] 1[ ] 23[ ] 19[ ] 11[ ] 7[ ] 29[ ] 17[ ]

17[ ] 17[ ] 29[ ] 7[ ] 11[ ] 19[ ] 23[ ] 1[ ] 13[ ]

19[ ] 19[ ] 13[ ] 29[ ] 7[ ] 23[ ] 1[ ] 17[ ] 11[ ]

23[ ] 23[ ] 11[ ] 13[ ] 29[ ] 1[ ] 17[ ] 19[ ] 7[ ]

29[ ] 29[ ] 23[ ] 19[ ] 17[ ] 13[ ] 11[ ] 7[ ] 1[ ]

19. a) If U20 is cyclic, then there exists at least one element a[ ]∈U20 whose order must be 8. But

1[ ]1 = 1[ ]; 3[ ]4 = 1[ ]; 7[ ]4 = 1[ ]; 9[ ]2 = 1[ ]; 11[ ]2 = 1[ ]; 13[ ]4 = 1[ ]; 17[ ]4 = 1[ ]; 19[ ]2 = 1[ ]. b) If U8 is cyclic, then there exists at least one element a[ ]∈U8 whose order must be 4. But

1[ ]1 = 1[ ]; 3[ ]2 = 1[ ]; 5[ ]2 = 1[ ]; 7[ ]2 = 1[ ].


16

c) If U24 is cyclic, then there exists at least one element a[ ]∈U20 whose order must be 8. But

1[ ]1 = 1[ ]; 5[ ]2 = 1[ ]; 7[ ]2 = 1[ ]; 11[ ]2 = 1[ ]; 13[ ]2 = 1[ ]; 17[ ]2 = 1[ ]; 19[ ]2 = 1[ ]; 23[ ]2 = 1[ ].

d) If U30 is cyclic, then there exists at least one element a[ ]∈U20 whose order must be 8. But

1[ ]1 = 1[ ]; 7[ ]4 = 1[ ]; 11[ ]2 = 1[ ]; 13[ ]4 = 1[ ]; 17[ ]4 = 1[ ]; 19[ ]2 = 1[ ]; 23[ ]4 = 1[ ]; 29[ ]2 = 1[ ].

20. U9 = 5[ ]0 = 1[ ], 5[ ]1 = 5[ ], 5[ ]2 = 7[ ], 5[ ]3 = 8[ ], 5[ ]4 = 4[ ], 5[ ]5 = 2[ ], 5[ ]6 = 1[ ]{ } = 5[ ] is a

cyclic group of order 6. The divisors of 6 are 1, 2, 3, and 6, so the subgroups of U9 are

5[ ]1 =U9,

5[ ]2 = 7[ ], 4[ ], 1[ ]{ },5[ ]3 = 8[ ], 1[ ]{ },5[ ]6 = 1[ ]{ },5[ ]4 = 5[ ]2 = 7[ ], 4[ ], 1[ ]{ },5[ ]5 = 5[ ]1 =U9.

21. a) φ 8( ) = 4;a,a3,a5,a7

b) φ 14( ) = φ 2( )φ 7( ) = 1.6 = 6; a,a3,a5,a9,a11,a13

c) φ 18( ) = φ 2( )φ 32( ) = 1.6 = 6; a,a5,a7,a11,a13,a17d) φ 24( ) = φ 3( )φ 8( ) = 2.4 = 8; a,a5,a7,a11,a13,a17,a19,a23

e) φ 7( ) = 6;a,a2,a3,a4 ,a5,a6

f) φ 13( ) = 12; a,a2,a3,a4 ,a5,a6,a7,a8,a9,a10,a11,a12 .

22. a) Let G = a be a cyclic group of order 8. The divisors of 8 are 1, 2, 4, and 8, so the distinct subgroups of G are

a = G,

a2 = a2,a4 ,a6,a8 = e{ },a4 = a4 ,a8 = e{ },a8 = e = e{ }.


17

b) Let G = a be a cyclic group of order 14. The divisors of 14 are 1, 2, 7, and 14, so the distinct subgroups of G are

a = G,

a2 = a2,a4 ,a6,a8,a10,a12,a14 = e{ },a7 = a7,a14 = e{ },a14 = e = e{ }.

c) Let G = a be a cyclic group of order 18. The divisors of 18 are 1, 2, 3, 6, 9, and 18, so the distinct subgroups of G are

a = G,

a2 = a2,a4 ,a6,a8,a10,a12,a14 ,a16,a18 = e{ },a3 = a3,a6,a9,a12,a15,a18 = e{ },a6 = a6,a12,a18 = e{ },a9 = a9,a18 = e{ },a18 = e = e{ }.

d) Let G = a be a cyclic group of order 24. The divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, so the distinct subgroups of G are

a = G,

a2 = a2,a4 ,a6,a8,a10,a12,a14 ,a16,a18;a20,a22,a24 = e{ },a3 = a3,a6,a9,a12,a15,a18,a21,a24 = e{ },a4 = a4 ,a8,a12,a16,a20,a24 = e{ }a6 = a6,a12,a18,a24 = e{ },a8 = a8,a16,a24 = e{ }a12 = a12,a24 = e{ },a24 = e = e{ }.

e) Let G = a be a cyclic group of order 7. The divisors of 7 are 1 and 7 so the distinct subgroups of G are

a = G,

a7 = a7 = e{ }.


18

f) Let G = a be a cyclic group of order 13. The divisors of 13 are 1 and 13 so the distinct subgroups of G are

a = G,

a13 = a13 = e{ }.

23. Since o a( ) = 24 , a24 = e .

a) 2 is the least positive integer such that a12( )2 = a24 = e , so o a12( ) = 2.b) a8,a16

c) a6,a18

d) none

24. Since o a( ) = 35 , a35 = e . a) noneb) a7,a14 ,a21,a28

c) a5,a10,a15,a20,a25,a30

d) none

25. If H ⊂ is a subgroup, then either H = 0{ } is trivial or there exists a unique integer n ≥1 such that H = n = nk : k ∈{ } .

26. All generators of an infinite cyclic group are a and a−1 , so G = a = a−1 . For example,

the additive group is cyclic and = 1 = −1 .

27. First note that a = bn for some n∈ since a∈ b . Now let x ∈ a . This means that

x = am for some m∈ . Thus x = am = bn( )m = bnm ∈ b . Since this is true for all x ∈ a we

have a ⊆ b .

28. a)

ak( )−1 = a−k = a−1( )kak = e

⇒ a−1( )k = e−1 = e⇒ o a−1( ) ≤ o a( ) (1)

Similarly,


19

a−1( )k = e⇒ a−1( )−1( )k = e−1⇒ ak = e−1 = e

⇒ o a( ) ≤ o a−1( ) (2)

By (1) and (2), o a( ) = o a−1( ).

b)

Section 3.4 Solutions 2

3.4.28 Let a and b be elements of a finite group G.b. Prove that a and bab

�1 have the same order.

c. Prove that ab and ba have the same order.

Proof of b. Let e be the identity element of G. First observe that for any x,y 2 G and for any integer m with m �1, (xyx

�1)m = xy

m

x

�1. To prove this proceed by induction on n. For n = 1 we clearly have (xyx

�1)1 = xy

1x

�1.So assume that (xyx

�1)k = xy

k

x

�1 for some integer k � 1. Then(xyx

�1)k+1 = (xyx

�1)k(xyx

�1)= (xy

k

x

�1)(xyx

�1) by the inductive hypothesis= xy

k

eyx

�1

= xy

k

yx

�1

= xy

k+1x

�1. This completes the induction.

Now let n = o(a). This means that a

n = e and n is the least such positive integer. Then we have (bab

�1)n =ba

n

b

�1 = beb

�1 = bb

�1 = e. Thus o(bab

�1) n.Next assume that (bab

�1)m = e for some integer m with 0 < m < n. Then e = (bab

�1)m = ba

m

b

�1. Nowba

m

b

�1 = e

=) a

m

b

�1 = b

�1 multiplying by b

�1 on the left=) a

m = e multiplying by b on the right

However, this contradicts that n is the least integer such that a

n = e. Therefore o(bab

�1) = n = o(a).

Proof of c. First observe that for any x,y 2 G, (xy)m = x(yx)m�1y for all integers m � 1. To prove this proceed

by induction on n. For m = 1 we have (xy)1 = xy = xey = x(yx)0y. So assume that (xy)k = x(yx)k�1

y for someinteger k � 1. Then

(xy)k+1 = (xy)k(xy)= x(yx)k�1

y(xy) by the inductive hypothesis= x(yx)k�1(yx)y by associativity= x(yx)k

y This completes the induction.

Now let n = o(ab). This means that (ab)n = e and n is the least such positive integer. So(ab)n = e

=) a(ba)n�1b = e by the above observation

=) (ba)n�1b = a

�1 multiplying by a

�1 on the left=) (ba)n�1

ba = e muliplying by a on the right=) (ba)n = e.

Thus o(ba) n.Next assume that (ba)m = e for some integer m with 0 < m < n. Then

(ba)m = e

=) b(ab)m�1a = e by the above observation

=) (ab)m�1a = b


�1 on the left=) (ab)m�1

ab = e multiplying by b on the right=) (ab)m = e.

However, this contradicts that n is the least integer such that (ab)n = e. Therefore o(ba) = n = o(ab).

c)

Section 3.4 Solutions 2

3.4.28 Let a and b be elements of a finite group G.b. Prove that a and bab

�1 have the same order.

c. Prove that ab and ba have the same order.

Proof of b. Let e be the identity element of G. First observe that for any x,y 2 G and for any integer m with m �1, (xyx

�1)m = xy

m

x

�1. To prove this proceed by induction on n. For n = 1 we clearly have (xyx

�1)1 = xy

1x

�1.So assume that (xyx

�1)k = xy

k

x

�1 for some integer k � 1. Then(xyx

�1)k+1 = (xyx

�1)k(xyx

�1)= (xy

k

x

�1)(xyx

�1) by the inductive hypothesis= xy

k

eyx

�1

= xy

k

yx

�1

= xy

k+1x

�1. This completes the induction.

Now let n = o(a). This means that a

n = e and n is the least such positive integer. Then we have (bab

�1)n =ba

n

b

�1 = beb

�1 = bb

�1 = e. Thus o(bab

�1) n.Next assume that (bab

�1)m = e for some integer m with 0 < m < n. Then e = (bab

�1)m = ba

m

b

�1. Nowba

m

b

�1 = e

=) a

m

b

�1 = b


�1 on the left=) a

m = e multiplying by b on the right

However, this contradicts that n is the least integer such that a

n = e. Therefore o(bab

�1) = n = o(a).

Proof of c. First observe that for any x,y 2 G, (xy)m = x(yx)m�1y for all integers m � 1. To prove this proceed

by induction on n. For m = 1 we have (xy)1 = xy = xey = x(yx)0y. So assume that (xy)k = x(yx)k�1

y for someinteger k � 1. Then

(xy)k+1 = (xy)k(xy)= x(yx)k�1

y(xy) by the inductive hypothesis= x(yx)k�1(yx)y by associativity= x(yx)k

y This completes the induction.

Now let n = o(ab). This means that (ab)n = e and n is the least such positive integer. So(ab)n = e

=) a(ba)n�1b = e by the above observation

=) (ba)n�1b = a

�1 multiplying by a

�1 on the left=) (ba)n�1

ba = e muliplying by a on the right=) (ba)n = e.

Thus o(ba) n.Next assume that (ba)m = e for some integer m with 0 < m < n. Then

(ba)m = e

=) b(ab)m�1a = e by the above observation

=) (ab)m�1a = b


�1 on the left=) (ab)m�1

ab = e multiplying by b on the right=) (ab)m = e.

However, this contradicts that n is the least integer such that (ab)n = e. Therefore o(ba) = n = o(ab).


20


21

true/false questions 1. the order of the identity element ...may 03, 2012 · true/false questions...

Documents