Example 8 N H I K 5 B 5 Y [ G Y 5 B for Some N [ Z

Cyclic Groups - I ! Week 5

Deﬁnition 3.13 Cyclic Subgroup

Let G be a group. For any a ∈G , the subgroup H = x ∈G : x = an for somen ∈ { } is the subgroup generated by a and is denoted by a . This subgroup is called a cyclic subgroup. In particular, G is a cyclic group if there is an element a ∈G such that G = a . 3.3 Subgroups 159

Example* Prove that H in Def. 3.13 is a subgroup of the group G . is the subgroup generated by a and is denoted by a . A given subgroup K of G is a cyclic subgroup if there exists an element b in G such that H I 3.3 Subgroups 159 n K 5 b 5 y [ G y 5 b for some n [ Z . is the subgroup generated by a and is denoted by a . A given subgroup K of G is a cyclic In particular, G is a cyclic group8 9 5if there0 is an element a [ G such6 that G 5 a . subgroup if there exists an element b in G such that 3.3 Subgroups 159 H I Example 8 n H I K 5 b 5 y [ G y 5 b for some n [ Z . Exampleis the subgroup 8 generated by a and is denoted by a . A given subgroup K of G is a cyclic a. The set Z of integers is a cyclic group under addition. We have Z 5 1 and Z 5 21 . Insubgroup particular,if thereG is aexists cyclic an group8 element9 5if thereb in 0G issuch an element that a [ G such6 that G 5 a . b. The subgroup E 8 Z of all even integers is a cyclicH I subgroup of the additive group Z, n H I H I H I generated by 2. HenceK 5 Eb552 y. [ G y 5 b for some n [ Z . Examplec. In Example 8 6, we saw that Ina. particular,The set Z ofG integersis a cyclic is a group 8cyclic9 H 5groupIif there under0 is an addition. element We a [ haveG such Z6 5 that1 and G 5Z 5a .21 .

b. The subgroup E 8 Z of all evenH 5integers 2 , 4is ,a 6cyclic, 8 subgroup8 Z10 of the additive group Z, generated by 2. Hence E 5 2 . H I H IH I Examplec. Inis Examplean abelian 8 6, group we saw with that respect to5 3multiplication.4 3 4 3 4 3 4 6 Since a. The set Z of integers is a cyclicH I group under addition. We have Z 5 1 and Z 5 21 . 2 3 4 b. The subgroup E 8 Z of all2 evenH55 4integers ,2 , 24 , is56 a, cyclic88, 8 2subgroupZ105 6 , of the additive group Z, generated by 2. Hence E 5 2 . H I H I c. isthenIn anExample abelian 6, group we saw with that respect3 4 3to54 3multiplication.4 33 44 3 4 33 44 6 Since3 4 3 4 H I 2 3 4 2 5H 54 , 22H, 455, 682,, 8. 2 8 5Z106 ,

thenis an abelian group with3 respect42 3 4to5 3multiplication.34 43 4 8333 44493 436 Since4 3 4 d. The group S(A) 5 {e, r, r , s, g, d} of Example 3 in Section 3.1 is not a cyclic group. This can be verified by considering a for all possible choices of a in (A). I 2 2 5 4 , H253 52 8., 2 4 5 6 , S

2 H I 83 49 d. Thethen group S(A) 5 {e, r, r3 ,4s, g,3 d4} of3 Example4 3 4 3 in3 Section4 3 4 3.1 is not a cyclic group. ExercisesThis can be verified3.3 by considering a for all possible choices of a in S(A). I H 5 2 . True or False H I 2 83 49 LabelExercisesd. The each group of the S3.3 (followingA) 5 {e, r statements, r , s, g, d as} ofeither Example true or3 infalse, Section where 3.1 His isnot a asubgroup cyclic group. of G. 1. EveryThis can group be verifiedG contains by considering at least two a subgroups.for all possible choices of a in S(A). I True or False d.2. TheThe group identity S( Aelement) = {e,σ }in in a Sectionsubgroup 3.1 H isof a acyclic group group. G must be the same as the identity Label each of the following statements asH eitherI true or false, where H is a subgroup of G. element in G. Exercises1. Every group G3.3contains at least two subgroups. 3. An element x in H has an inverse xϪ1 in H that may be different than its inverse in G. True2. The or identityFalse element in a subgroup H of a group G must be the same as the identity 4. The generator of a cyclic group is unique. Labelelement each ofin theG. following statements as either true or false, where H is a subgroup of G. 5. Any subgroup of an abelian groupϪ is1 abelian. 3.1. AnEvery element group x Gin containsH has an at inverse least two x subgroups.in H that may be different than its inverse in G. 6. If a subgroup H of a group G is abelian, then G must be abelian. 4.2. TheThe generator identity elementof a cyclic in groupa subgroup is unique. H of a group G must be the same as the identity 7.5. AnyTheelement relationsubgroup in G. R ofon an the abelian set of groupall groups is abelian. defined by HRK if and only if H is a subgroup of K is an equivalence relation. 6.3. IfAn a subgroupelement x HinofH ahas group an inverseG is abelian, xϪ1 in thenH that G must may bebe abelian.different than its inverse in G. 8. The empty set [ is a subgroup of any group G. 7.4. TheThe relation generator R onof athe cyclic set of group all groups is unique. defined by HRK if and only if H is a subgroup 9. Any group of order 3 has no nontrivial subgroups. 5. ofAnyK is subgroup an equivalence of an abelian relation. group is abelian. 10. Z under addition modulo 5 is a subgroup of the group Z under addition. 8.6. TheIf5 a emptysubgroup set [H of is aa subgroupgroup G is of abelian, any group then G G. must be abelian. 9.7. AnyThe grouprelation of Rorderon the 3 has set no of nontrivialall groups subgroups. defined by HRK if and only if H is a subgroup 10. Zof5 underK is an addition equivalence modulo relation. 5 is a subgroup of the group Z under addition. 8. The empty set [ is a subgroup of any group G. 9. Any group of order 3 has no nontrivial subgroups.

10. Z5 under addition modulo 5 is a subgroup of the group Z under addition. Cyclic Groups - I ! Week 5 Deﬁnition 3.14 Generator

164 Chapter 3Any Groups element of a of the group G such that G = a is a generator of G .

Remark. If a is a generator of G,thena21 is also, since any element x [ G can be written as x 5 an 5 (a21)2n for some integer n.

Example 1 The additive group

Zn 5 0 , 1 , c, n 2 1 is a cyclic group with generator [1], since5 3 4 3any4 k in3 Zn can4 6be written as k 5 k3 41 where k 1 indicates a multiple of [1] as 3described4 3 4 in Section 3.3. Elements other than [1] may also be generators. To illustrate this, consider the particular case 3 4

Z6 5 0 , 1 , 2 , 3 , 4 , 5 .

The element [5] is also a generator of5 3Z46 since3 4 3 [5]4 3 is4 the3 4 additive3 4 6 inverse of [1]. The following list shows how Z6 is generated by [5]—that is, how Z6 consists of multiples of [5]. 1 5 5 5 2 5 5 5 1 5 5 4 3 354 5 354 1 5 1 5 5 3 4 354 5 324 3 4 3 4 5 354 5 314 3 4 3 4 3 4 6 354 5 304 3 4 3 4 The cyclic subgroups generated by3 4 the3 other4 elements of Z6 under addition are 0 5 0 2 5 2 , 4 , 0 833495 53346, 0 834495 5344, 324, 3046 5 2 . 83 49 53 4 3 46 Thus [1] and [5] are the only elements83 49 that53 4 are3 4 generators3 46 83 49of the entire group. I

Example 2 We saw in Example 8 of Section 3.3 that

H 5 2 , 4 , 6 , 8 8 Z10 forms a cyclic group with respect to multiplication53 4 3 4 3 4 3 and46 that 2 is a generator of H. The element 8 5 2 21 is also a generator of H, as the following computations confirm: 3 4 2 3 4 3 4 3 4 8 5 4 , 8 5 2 , 8 5 6 . I 3 4 3 4 3 4 3 4 3 4 3 4 Example 3 In the quaternion group G 5 {61, 6i, 6j, 6k}, described in Exercise 28 of Section 3.1, we have i2 521 i3 5 i2 ? i 52i i4 5 i3 ? i 52i2 5 1. 164 Chapter 3 Groups

If a is a generator of G,thena21 is also, since any element x [ G can be written as x 5 an 5 (a21)2n for some integer n.

Example 1 The additive group

Z6 5 0 , 1 , 2 , 3 , 4 , 5 .

The element [5] is also a generator of5 3Z46 since3 4 3 [5]4 3 is4 the3 4 additive3 4 6 inverse of [1]. The following list shows how Z6 is generated by [5]—that is, how Z6 consists of multiples of [5]. 1 5 5 5 2 5 5 5 1 5 5 4 3 354 5 354 1 5 1 5 5 3 4 354 5 324 3 4 3 4 5 354 5 314 3 4 3 4 3 4 6 354 5 304 3 4 3 4 The cyclic subgroups generated by3 4 the3 other4 elements of Z6 under addition are 0 5 0 2 5 2 , 4 , 0 833495 53346, 0 834495 5344, 324, 3046 5 2 . 83 49 53 4 3 46 Thus [1] and [5] are the only elements that are generators of the entire group. I Cyclic Groups - I 83 49 53 4 3 !4 3 46 83 49 Week 5

Example 2 We saw in Example 8 of Section 3.3 that

H 5 2 , 4 , 6 , 8 8 Z10 forms a cyclic group with respect to multiplication53 4 3 4 3 4 3 and46 that 2 is a generator of H. The element 8 5 2 21 is also a generator of H, as the following computations confirm: 3 4 2 3 4 3 4 3 4 8 5 4 , 8 5 2 , 8 5 6 . I 3 4 3 4 3 4 3 4 3 4 3 3.44 Cyclic Groups 165 Example 3 In the quaternion group G 5 {61, 6i, 6j, 6k}, described in Exercise 28 of Section 3.1, we have Thus i generates the cyclic subgroup of order 4 given by i2 521 3i 5 2 i, 21, 2i, 1 , i 5 i ? i 52i 3.4 Cyclic Groups 165 although the group G itself is not cyclic.i84 95 i35 ? i 52i2 56 1. I

ThusWhetheri generates a group the cyclicG is cyclicsubgroup or not, of order each element4 given abyof G generates the cyclic subgroup a ,and i 5 i, 21, 2i, 1 , n a 5 x [ G x 5 a for n [ Z . HalthoughI the group G itself is not cyclic.8 9 5 6 I n We shall see that the structure8 9 of 5a depends0 entirely on whether6 or not a 5 e for some Whether a group G is cyclic or not, each element a of G generates the cyclic subgroup Remark.positive integer n. The next two theorems state the possibilities for the structure of a . a ,and H I

n H I H I a 5 x [ G x 5 a for n [ Z . Strategy I The method of proof of the next theorem is by contradiction. A statement p ⇒ q may be n provedWe shall by see assuming that the thatstructurep is8 9 true of 5a anddependsq is0 false entirely and then on whether proving6 or that not this a assumption5 e for some leadspositive to ainteger situation n. The where next some two statementtheorems isstate both the true possibilities and false—a for contradiction.the structure of a . H I H I TheoremStrategy 3.15I TheoremITheInfinite method 3.15of Cyclic proofInﬁnite ofGroup Cyclic the next Group theorem is by contradiction. A statement p ⇒ q may be

proved by assuming that p is true and q nis false and then proving that this assumptionp q Letleadsa tobe a an situation element where in the some group statement G. If a is2 bothe for true every and positive false—a integer contradiction. n, then a 2 a whenever p 2 q in Z, and a is an infinite cyclic group.

TheoremContradiction 3.15 ProofI InfiniteAssume Cyclic that Groupa is H anI element of the group G such that an 2 e for every posi- (p ,q) tive integer n. Having made this assumption, suppose now that Let a be an element in the group G. If an 2 e for every positive integer n, then ap 2 aq ⇒ ,p p q ¿ whenever p 2 q in Z, and a is an infinitea cyclic5 a group. where p 2 q in Z. We may assume that p . q. Then Contradiction Proof Assume that a is H anI element of the group G such that an 2 e for every posi- p q p 2q q 2q (p ,q) tive integer n. Having made thisa 5assumption,a ⇒ a ? supposea 5 anow? a that p2q ⇒ ,p ⇒ ap 5q e. ¿ a 5 a Since p 2 q is a positive integer, this result contradicts an 2 e for every positive integer n. where p 2 q in Z. We may assume that p . q. Then Therefore, it must be that a p 2 a q whenever p 2 q. Thus all powers of a are distinct, and therefore a is an infinite cyclicap 5group.aq ⇒ ap ? a2q 5 aq ? a2q ap2q 5 e. H I ⇒ n Corollary 3.16 ISince p 2 q is a positive integer, this result contradicts a 2 e for every positive integer n. Therefore, it must be that a p 2 a q whenever p 2 q. Thus all powers of a are distinct, and IfthereforeG is a finitea is group an infinite and a cyclic[ G, group. then an 5 e for some positive integer n.

p ⇒ q Proof SupposeH I G is a finite group and a [ G. Since the cyclic subgroup Corollary 3.16 I m a 5 x [ G x 5 a for m [ Z

n p q isIf aG subsetis a finite of G ,groupa must and also a [8 be9G finite., then5 Ita must50 e thereforefor some happen positive that6 integera 5 an. for some integers p and q with p 2 q.ItfollowsfromTheorem3.15thatan 5 e for some positive integer n. p ⇒ q Proof SupposeH IG is a finite group and a [ G. Since the cyclic subgroup

m a 5 x [ G x 5 a for m [ Z p q is a subset of G, a must also8 be9 finite.5 It must0 therefore happen that6 a 5 a for some integers p and q with p 2 q.ItfollowsfromTheorem3.15thatan 5 e for some positive integer n. H I 3.4 Cyclic Groups 165

Thus i generates the cyclic subgroup of order 4 given by i 5 i, 21, 2i, 1 ,

although the group G itself is not cyclic.8 9 5 6 I

Whether a group G is cyclic or not, each element a of G generates the cyclic subgroup a ,and

n H I a 5 x [ G x 5 a for n [ Z . n We shall see that the structure8 9 of 5a depends0 entirely on whether6 or not a 5 e for some positive integer n. The next two theorems state the possibilities for the structure of a . H I H I Strategy I The method of proof of the next theorem is by contradiction. A statement p ⇒ q may be proved by assuming that p is true and q is false and then proving that this assumption leads to a situation where some statement is both true and false—a contradiction.

Theorem 3.15 I Infinite Cyclic Group

Let a be an element in the group G. If an 2 e for every positive integer n, then ap 2 aq whenever p 2 q in Z, and a is an infinite cyclic group.

Contradiction Proof Assume that a is H anI element of the group G such that an 2 e for every posi- (p ,q) tive integer n. Having made this assumption, suppose now that

⇒ ,p p q ¿ a 5 a where p 2 q in Z. We may assume that p . q. Then ap 5 aq ⇒ ap ? a2q 5 aq ? a2q ⇒ ap2q 5 e. Since p 2 q is a positive integer, this result contradicts an 2 e for every positive integer n. Therefore, it must be that a p 2 a q whenever p 2 q. Thus all powers of a are distinct, and therefore a is an infinite cyclic group.

Cyclic GroupsH I - I ! Week 5 Corollary 3.16 CorollaryI 3.16

If G is a finite group and a [ G, then an 5 e for some positive integer n.

p ⇒ q Proof Suppose G is a finite group and a [ G. Since the cyclic subgroup

m a 5 x [ G x 5 a for m [ Z p q 166 Chapter 3is a Groups subset of G, a must also8 be9 finite.5 It must0 therefore happen that6 a 5 a for some integers p and q with p 2 q.ItfollowsfromTheorem3.15thatan 5 e for some positive integer n. H I If it happens that an 2 e for every positive integer n, then Theorem 3.15 states that all the powers of a are distinct and that a is an infinite group. Of course, it may happen that an 5 e for some positive integers n. In this case, Theorem 3.17 describes a completely. H I Theorem 3.17 I Finite Cyclic Group H I Theorem 3.17 Finite Cyclic Group Let a be an element in a group G, and suppose an 5 e for some positive integer n. If m is the least positive integer such that am 5 e, then 0 m 1 2 m21 a. a has order m, and a 5 {a 5 e 5 a , a , a , c, a } b. as 5 at if and only if s ; t (mod m). H I H I p ⇒ q Proof Assume that m is the least positive integer such that am 5 e. We first show that the elements

0 2 m21 a 5 e, a, a , c, a are all distinct. Suppose ai 5 a j where 0 # i , m and 0 # j , m. There is no loss of generality in assuming i $ j. Then ai 5 aj implies ai2j 5 ai ? a2j 5 e where 0 # i 2 j , m. Since m is the least positive integer such that am 5 e,and since i 2 j , m,it must be true that i 2 j 5 0, and therefore i 5 j. Thus a contains the m distinct elements a0 5 e, a, 2 m21 a , c, a . The proof of part a will be complete if we can show that any power of a is equal to one of these elements. Consider anH arbitraryI ak. By the Division Algorithm, there exist integers q and r such that k 5 mq 1 r, with 0 # r , m. Thus ak 5 amq1r 5 amq ? ar by part b of Theorem 3.12 m q r 5 (a ) ? a by part c of Theorem 3.12 5 eq ? ar 5 ar where r is in the set 0, 1, 2, . . . , m 2 1 . It follows that 2 m21 5a 5 e, a, a , c, 6a , and a has order m. p ⇒ (q ⇔ r) To obtain part b, we first observe that if k 5 mq 1 r, with 0 # r , m, then a k 5 ar, 8 9 5 6 8 9 k where r is in the set {0, 1, 2, c, m 2 1}. In particular, a 5 e if and only if r 5 0—that is, if and only if k ; 0 (mod m). Thus as 5 at ⇔ as2t 5 e ⇔ s 2 t ; 0 (mod m) ⇔ s ; t (mod m), and the proof is complete. 3.4 Cyclic Groups 167

We have defined the order o(G) of a group G to be the number of elements in the group.

Definition 3.18 I Order of an Element

The order o(a) of an element a of the group G is the order of the subgroup generated by a. That is, o(a) 5 o a . 3.4 Cyclic Groups 167

AH IB WePart havea of definedTheorem the 3.17 order immediately o(G) of a group translates G to beinto the the number following of elements corollary. in the group. 3.4 Cyclic Groups 167

DefinitionCorollary 3.19 3.18 I OrderFinite Orderof an Elementof an Element We have defined the order o(G) of a group G to be the number of elements in the group. 3.4 Cyclic Groups TheIf o(ordera) is finite,o(a) of then an melement5 o(a a) isof the the least group positive G is the integer order suchof the that subgroup am 5 egenerated. by167 a. Definition 3.18 ThatI Orderis, o(a) of5 ano aElement. WeThe havenext definedexample the illustrates order o( Gthe) of results a group of GTheoremto be the 3.17 number and itsof elementscorollary. in the group. Cyclic Groups - I AH IB ! Week 5 The Partordera ofo( aTheorem) of an element 3.17 immediately a of the group translates G is the into order the followingof the subgroup corollary. generated by a. DeﬁnitionIThat is, o(a 3.18) 5 oOrdera . of an Element Definition 3.18 ExampleOrder of4 anIt canElement be shown (see Exercise 16 at the end of this section) that Corollary 3.19 I Finite Order of an Element The order o(a) ofAH anIB element a of the group G is the order of the subgroup generated by a. Part a of TheoremG 3.175 1immediately, 3 , 5 , 7 ,translates9 , 11 , into13 , the15 following8 Z16 corollary. IfThato(a is,) iso finite,(a) 5 o thena m. 5 o(a) is the least positive integer such that am 5 e. is a group with respect to multiplication53 4 3 4 3 4 3in4 Z3164. The3 4 element3 4 3 346 of G generates a cyclic sub- Corollary 3.19 IgroupFinite of order Order 4 sinceAH ofIB an3 4 Element5 1 , and 4 is the least positive integer m such that 3 m 5 1 . CorollaryThePart nexta 3.19of Theoremexample Finite Order illustrates3.17 immediately of anthe Element results translates of Theorem into 3.17the 3following and4 its corollary. corollary. Thus m If o(a) is finite, then m3 54 o(a3) is4 the least positive integer such that a 5 e. 3 4 3 4 Corollary 3.19 I Finite Order of an Element3 5 3 0 5 1 , 3 , 9 , 11 , Example 4 It can be shown (see Exercise 16 at the end of this section) that The next example illustrates the results of Theorem 3.17 and its corollary. Ifando( thea) is order finite, of thenthe elementm 5 o( a833) 49is is the 4.53 Also,least4 positive powers3 4 3 4 largerinteger3 4 3 than such46 4 thatof 3 am are5 easilye. computed G 5 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 8 Z16 using part b of Therom 3.17. For example, is a group with respectIt can tobe multiplication shown3 4 (see Exercise in Z . The 16 atelement the end 3 of of this G3generates section)4 thata cyclic sub- ExampleThe next 4 example illustrates53 4 3 the4 33 results41913 54 3 16of34 3Theorem3 54 113 4 3.173 46 and its corollary. group of order 4 since 3 4 5 1 , and 4 is the least positive integer m such that 3 m 5 1 . G 5 1 , 3 , 5 , 7 , 9 , 11 , 13 , 153 4 8 Z16 Thussince 191 ; 3 (mod 4). 3 4 3 4 3 4 I It can be3 4 shown3 4 (see Exercise 16 at the end of this section) that3 4 3 4 Exampleis a group with 4 respect to multiplication0 in Z . The element 3 of G generates a cyclic sub- The multiplicative group53 G34 5354 13 3,4 33 5,4 53116,4,733, ,94 9,3 ,11114 ,3 13, 46, 15 # Z16 in Example 4 group of order 4 since 3 4 5 1 , and 4 is the least positive integer m such that 3 m 5 1 . consists of all [a] in ZG165that 1have, 3 multiplicative, 5 , 7 , 9 , inverses.11 , 13 ,This15 group8 Z16 is called the group of and the order of the element 3 is 534. Also,4 3 4 powers3 4 3 4 larger3 4 3 than4 3 3 444 of3 346 are easily computed Thusunits in Z16 and is designated83 by49 the symbol53 4 U3 164. 3 4 3 4 3 46 is a group with respect to3 multiplication4 53 43 34 4 3 4 3in4 Z3164. The3 4 element3 4 3 346 of G generates a cyclic3 4 sub-3 4 usingAs part might b of beTherom expected, 3.17. every For example, subgroup0 of a cyclic group is also a cyclic group. It is group of order 4 since 3 4 53 314 , and5 43 is 5the 1least, 3 positive, 9 , 11 integer, 3 m4 such that 3 m 5 1 . even possible to predict a generator 3of191 the5 subgroup,3 3 5 11 as stated in Theorem 3.20. Thus 3 4 and the order of the element 833 49 is 4.53 Also,4 powers3 4 3 4larger3 4 3 than46 4 of 3 are easily computed since 191 3 (mod 4).3 4 3 4 3 4 3I4 using part; b of Therom 3.17. For3 5 example,3 4 3 0 53 14 , 3 3, 94 , 11 , Strategy I The conclusion of the next theorem3 4 has the form “either a or b.” To prove3 4 this statement, The multiplicative group G 5 1 , 1913 , 5 , 73 , 9 , 11 , 13 , 15 # Z16 in Example 4 andwe canthe assumeorder of that the aelementis false 83 and3 49 is prove 4.53 3Also, 4that5 powers b3 3must4 35 4larger then3114 3be than 46true. 4 of 3 are easily computed consists of all [a] in Z16 that have multiplicative inverses. This group is called the group of usingsince 191part ;b of3 (modTherom 4). 3.17. For example,53 4 3 4 3 4 3 4 3 4 3 4 3 4 3 46 I units in Z16 and is designated by3 4 the symbol3 4 U316.4 3 4 3 4 As might be expected, every subgroup3 191 5 of3 3a5 cyclic11 group is also a cyclic group. It is The multiplicative group G 5 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 # Z16 in Example 4 Theorem 3.20 TheoremevenI Subgrouppossible 3.20 to Subgroup predictof a Cyclic a generator of aGroup Cyclic of the Group subgroup, as stated in Theorem 3.20. sinceconsists191 of; all3 [(moda] in 4).Z16 that have multiplicative inverses. This group is called the groupI of 533 44 3 4 33 44 3 4 33 44 3 4 3 4 3 46 units in Z16 and is designated by the symbol U16. Let GThebe multiplicativea cyclic group group with a G[5G as1 a, generator,3 , 5 , 7 and, 9 ,let11 H,be13 a subgroup, 15 # ofZ G in. Then Example either 4 Strategy I The conclusionAs might be of expected,the next theorem every subgrouphas the form of a“either cyclic a groupor b.” Tois also prove a thiscyclic16 statement, group. It is consistsevena. H possible5 of{ alle} 5[toa] predict ein, Z or16 thata generator have multiplicative of the subgroup, inverses. as statedThis group in Theorem is called 3.20. the group of weunits canin assumeZ and that is designated a is false byand the prove53 symbol4 3that4 3 Ub4 must.3 4 3then4 3 be4 true.3 4 3 46 b. if H 216 {e}, then H 5 a k where k is the16 least positive integer such that a k [ H. As might be Hexpected,I every subgroup of a cyclic group is also a cyclic group. It is Strategy I evenThe conclusion possible to ofpredict the next a generator HtheoremI ofhas the the subgroup, form “either as stated a or bin.” Theorem To prove 3.20.this statement, Theorem 3.20 Iwe canSubgroup assume that of a a Cyclicis false Groupand prove that b must then be true.

Strategy I LetTheG conclusionbe a cyclic of group the next with theorem a [ G as has a generator, the form “either and let aHorbeb a.” subgroup To prove of this G .statement, Then either we can assume that a is false and prove that b must then be true. Theorem 3.20 Ia. HSubgroup5 {e} 5 ofe , a or Cyclic Group b. if H 2 {e}, then H 5 a k where k is the least positive integer such that a k [ H. Let G be a cyclic HgroupI with a [ G as a generator, and let H be a subgroup of G. Then either Theorem 3.20 I Subgroup of a CyclicH IGroup a. H 5 {e} 5 e , or Letb. GifbeH a2 cyclic{e}, group then H with5 a k[whereG as a kgenerator,is the least and positive let H be integer a subgroup such ofthat G .a Thenk [ H either. H I a. H 5 {e} 5 e , or H I b. if H 2 {e}, then H 5 a k where k is the least positive integer such that a k [ H. H I H I Cyclic Groups - I ! Week 5 Corollary 3.21

Any subgroup of a cyclic group is cyclic.