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 e n = e . 2. Every cyclic group is abelian. True. Let G be a cyclic group. All elements of G are of the form a n , where n ∈. Let x, y ∈G : x = a p , y = a q . Then, xy = a p a q = a p+q = a q+ p = a q a p = 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 U 12 = 1 [] ,5 [] ,7 [] , 11 [ ] { } ⊂ 12 . U 12 is abelian group but this group is not cyclic since the order of elements of U 12 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 a m = e , where a is an element of a group G , then oa () = m . False. m must be the least positive integer such that a m = e . Exercises 3.4 Cyclic Groups 1
21
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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.
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.
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
4. o 1( ) = 1, o −1( ) = 2, o i( ) = o −i( ) = o j( ) = o − j( ) = o k( ) = o −k( ) = 4.
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.
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
Exercises 3.4" Cyclic Groups
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.
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
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
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,
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
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,
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
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
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
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
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
#$$$$ n # Z%
is a cyclic subgroup of GL2(R).
Proof. First let’s prove that H is a subgroup of the invertible 2 by 2 real matrices.
Claim:"
1 10 1
#n
="
1 n0 1
#for all positive integers n.
Clearly the statement holds when n = 1. Assume"
1 10 1
#k
="
1 k0 1
#. Then we have the
following."
1 10 1
#k+1
="
1 10 1
#k "1 10 1
#
="
1 k0 1
#"1 10 1
#
="
1 k + 10 1
#
Thus by induction"
1 10 1
#n
="
1 n0 1
#for all positive integers n.
By definition"
1 10 1
#0
="
1 00 1
#.
Suppose n > 0, then we have the following."
1 10 1
#"n
=&"
1 10 1
#n'"1
="
1 n0 1
#"1
="
1 $n0 1
#
Thus for all integers n,"
1 10 1
#n
="
1 n0 1
#.
Now we will prove H is cyclic. Let"
1 10 1
#k
#("
1 10 1
#), then
"1 10 1
#k
="
1 k0 1
## H.
Thus("
1 10 1
#)% H .
Now we will show H %("
1 10 1
#). Let A # H . Then A =
"1 n0 1
#for some n # Z. By
above claim,"
1 n0 1
#=
"1 10 1
#n
#("
1 10 1
#).
Exercises 3.4" Cyclic Groups
11
Since Z!7 = ![3]", each subgroup will be generated by [3]m where m is a divisor of 6. Thus
m = 1, 2, 3, 6. Since [3]1 = [3], [3]2 = [2], [3]3 = [6] and [3]6 = [1]. Therefore the subgroupsare as follows:
![1]" = {[1]}![2]" = {[1], [2], [4]}![3]" = Z!
7
![6]" = {[1], [6]}
14. Prove that the set H =!"
1 n0 1
#$$$$ 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
#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
#for all positive integers n.
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.
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|.
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
Exercises 3.4" Cyclic Groups
13
cosnθ −sinnθsinnθ cosnθ
⎡
⎣⎢
⎤
⎦⎥ =
cosθ −sinθsinθ cosθ
⎡
⎣⎢
⎤
⎦⎥
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
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 .
Exercises 3.4" Cyclic Groups
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
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