(Cyclic Group). a Group G Is Called Cyclic If a G 9 2 �3� N G = a = a N Z

CHAPTER 4

Cyclic Groups

Properties of Cyclic Groups Definition (Cyclic Group). A group G is called cyclic if a G 9 2 3 n G = a = a n Z . h i { | 2 } We say a is a generator of G. (A cyclic group may have many generators.) 2 1 0 1 2 Although the list . . . , a , a , a , a , a , . . . has inﬁnitely many entries, the set an n Z may have only ﬁnitely many elements. Also, since { | 2 } aiaj = ai+j = aj+i = ajai, every cyclic group is Abelian.

Example. Z under addition is an inﬁnite cyclic group. Z = 1 = 1 (1 and 1 are the only generators). h i h i 0 = 0 1 = 0 ( 1) (interpretation of 10 and ( 1)0). · · For n > 0, n = 1 + 1 + + 1 (interpretation of 1n). n terms· · · For n < 0, n = ( 1) + ( 1) + + ( 1) (interpretation of ( 1)n). | {z· · · } n terms | | Example. Zn =| 0, 1, 2, . . .{z, n 1 with} addition modulo n is a ﬁnite cyclic group. { }

Zn = 1 = n 1 (Note n 1 = 1 mod n). h i h i Other generators are possible depending on n:

Z10 = 1 = 9 = 3 = 7 . h i h i h i h i

46 4. CYCLIC GROUPS 47 Example. For U(12) = 1, 5, 7, 11 , { } 1 = 1 , 5 = 1, 5 , 7 = 1, 7 , 11 = 1, 11 . h i { } h i { } h i { } h i { } Thus U(12) is not cyclic since none of its elements generate the group. Theorem (4.1 — Criterion for ai = aj). Let G be a group and a G. If a = , then all distinct powers of a are distinct group elements (2ai = j | | 1 2 n 1 a i = j). If a < , say a = n, a = e, a, a , . . . , a and ai =(a)j n i j.| | 1 | | h i { } Proof.() | (1) If a = , n Z+ an = e. Then | | 1 6 9 2 3 i j i j a = a a = e i j = 0 i = j. () () () 2 n 1 (2) Assume a < . [To show a = e, a, a , . . . , a .] | | 1 h i { } 2 n 1 i j Clearly, e, a, a , . . . , a are distinct since a = n. For if a = a with i j | | 0 j < i n 1, then a = e, contradicting that   n is the least positive integer such that an = e. Now suppose ak a . Then 2 h i q, r Z k = qn + r, 0 r < n. Then 9 2 3  ak = aqn+r = (an)qar = eqar = ar, k 2 n 1 2 n 1 so a e, a, a , . . . , a = a = e, a, a , . . . , a . 2 { } ) h i { } i j i j (3) (= ) Suppose a = a . [To show n i j.] Then a = e. Now ) | q, r Z i j = qn + r, 0 r < n, and so 9 2 3  i j qn+r n q r q r r e = a = a = (a ) a = e a = a Since n is the least positive integer such that an = e, r = 0, so n i j. | ( =) If n i j, then q Z i j = nq = ( | 9 2 3 ) i j nq n q q i j a = a = (a ) = e = e = a = a . ) ⇤ 48 4. CYCLIC GROUPS Corollary (1 — a = a ). | | |h i| For any group element a, a = a . | | |h i| Corollary (2 — ak = e = a k). ) | || Let G be a group and let a be an element of order n in G. If ak = e, then n divides k. Proof. k 0 Since a = e = a , by Theorem 4.1, n k 0 = n k. ⇤ Note. | ) | Referring to Figure 4.1 on page 80 (shown below),

Theorem 4.1 says that multiplication in a is essentially done by addition modulo n, i.e., if (i + j) mod n = k, then ahiaij = ak. This is for any G and any a. Multiplication in a works like addition in Zn. h i Similarly, if a = , multiplication in a works like addition in Z since aiaj = ai+j. | | 1 h i

Thus Zn and Z are the prototypical cyclic groups.

How can one compute ak from a only? Also, how can one tell when ai = aj ? | | | | h i h i 4. CYCLIC GROUPS 49 Theorem (4.2 — ak = agcd(n,k) and ak = n/ gcd(n, k)). h i h i | | Let G be a group, a G, a = n, and k N. Then 2 | | 2 n ak = agcd(n,k) and ak = . h i h i | | gcd(n, k) Proof. (1) To simplify notation, let d = gcd(n, k) and k = dr. Since ak = (ad)r, ak ad . h i ✓ h i Now s, t Z d = ns + kt. Then 9 2 3 ad = ans+kt = ansakt = (an)s(ak)t = e(ak)t = (ak)t ak . 2 h i Thus ad ak = ad = ak by mutual set inclusion. h i ✓ h i ) h i h i d n n (2) Since d n, (ad)n/d = an = e, so a . But if i N with i < , then | | |  nd 2 d (ad)i = e by the definition of a , so ad = .Then 6 | | | | d n n ak = ak = ad = ad = = . | | |h i| |h i| | | d gcd(n, k) ⇤ Corollary (1 — Orders of Elements in Finite Cyclic Groups). In a finite cyclic group, the order of an element divides the order of the group. 50 4. CYCLIC GROUPS Corollary (2 — Criterion for ai = aj and ai = aj .). h i h i | | | | Let a = n. Then ai = aj gcd(n, i) = gcd(n, j), and ai = aj | | gcd(n, i) =hgcdi(n, jh). i () | | | | () Proof. (1) By Theorem 4.2, ai = agcd(n,i) and aj = agcd(n,j) . h i h i h i h i Then ai = aj agcd(n,i) = agcd(n,j) . h i h i () h i h i ( =) gcd(n, i) = gcd(n, j) = agcd(n,i) = agcd(n,j) = ai = aj . ( ) h i h i ) h i h i (= ) ai = aj = agcd(n,i) = agcd(n,j) = agcd(n,i) = agcd(n,j) = )n h i h ni ) h i h i ) | | | | ) = = gcd(n, i) = gcd(n, j). gcd(n, i) gcd(n, j) ) (2) Follows from the first part and Corollary 1 of Theorem 4.1. ⇤ Corollary (3 — Generators of Finite Cyclic Groups.). Let a = n. Then a = aj gcd(n, j) = 1, and a = aj gcd(|n,|j) = 1. h i h i () | | |h i| () Note. Now, once one generator of a cyclic group is found, all generators can then be easily determined. Example. U(25) = 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24 , { } so U(25) = 20. | | Now U(25) = 2 . By Corollary 3, the following are generators: h i 21 mod 25 = 2 213 mod 25 = 17 23 mod 25 = 8 217 mod 25 = 22 27 mod 25 = 3 219 mod 25 = 13 29 mod 25 = 12 221 mod 25 = 2 211 mod 25 = 23 223 mod 25 = 8 Thus U(25) = 2 = 3 = 8 = 12 = 13 = 17 = 22 = 23 . h i h i h i h i h i h i h i h i 4. CYCLIC GROUPS 51 Corollary (4 — Generators of Zn). An integer k Zn is a generator of Zn gcd(n, k) = 1. 2 () Problem (Page 87 # 10). (1) In Z24, list all generators for the subgroup of order 8. (2) Let G = a and let a = 24. List all generators of the subgroup of order 8. h i | | Solution. (1) 3 = 0, 3, 6, 9, 12, 15, 18, 21 has order 8. h i { } From Corollary 3, the generators are 1 3, 3 3, 5 3, 7 3 or 3, 9, 15, 21. · · · · (2) a3 = e, a3, a6, a9, a12, a15, a18, a21 has order 8. h i { } From Corollary 3, the generators are (a3)1, (a3)3, (a3)5, (a3)7 or a3, a9, a15, a21. ⇤

Classiﬁcation of Subgroups of Cyclic Groups Theorem (4.3 — Fundamental Theorem of Cyclic Groups). Every subgroup of a cyclic group is cyclic. Moreover, if a = n, then the order of any subgroup of a is a divisor of n; and, for|h eiach| positive divisor k of n, the group a hash iexactly one subgroup of order k—namely an/k . h i h i Example. Suppose G = a and G = 42. By the Theorem 4.3, if H G, H = a42/k where kh 42.i Also,| |G has one subgroup of the orders 1, 2, 3, 6, 7, 14, 21h, 42, andi no others.| The proof of the theorem shows how to ﬁnd these subgroups. 52 4. CYCLIC GROUPS Proof. (1) Let G = a and suppose H G. [To show H is cyclic.] If H = e , H is cyclic. Assumeh i H = e . SinceG = a , every element of H has the{ }form t t 6 { } h i t a . If t < 0, a H and t is positive. Thus H contains an element a with t > 0. Let m be2the leastpositive integer such that am H (guaranteed by the Well-Ordering Principle). By closure, am H. 2 h i ✓ [To show am = H.] Suppose b H. Then b G = k G b = ak. h i 2 2 ) 9 2 3 Then q, r Z k = mq + r, 0 r < m. 9 2 3  k mq+r mq r r k mq [To show r = 0.] Then a = a = a a = a = a a . Since k mq m q r ) a = b H and a = (a ) H, a H. But m is the least positive integer 2 am H, and 0 r < m2, so r = 20. Thus 3 2  b = ak = amq = (am)q am , 2 h i so H am . By mutual inclusion, am = H, and so H is cyclic.  h i h i (2) Suppose a = n and H G. Then, from (1), H = am for some |h i|  h i m Z. Since (am)n = (an)m = em = e, from Corollary 2 of Theorem 4.1, am2n. Thus H n. | | | | (3) Let k N k n. [To show an/k is the unique subgroup of a of order k.] By Theorem2 34.2,| h i h i n/k n n a = n = n = k. |h i| gcd(n, k ) k Now let H a with H = k. From (1) and (2), H = am and m n. Then m = gcd(n,mh) andi | | h i | n n k = am = agcd(n,m) = = . | | | | gcd(n, m) m n Thus m = and H = an/k . k h i ⇤ 4. CYCLIC GROUPS 53 n Corollary (Subgroups of Zn). For each k N k n, the set 2 3 | k is the unique subgroup of Zn of order k; moreover, these are the DonlyE subgroups of Zn. Proof.

Apply Theorem 4.3 with G = Zn and a = 1. ⇤ Example.

a = 42 Z42 order |h i| a = e, a, a2, . . . , a41 1 = 0, 1, 2, . . . , 41 42 ha2i ={ e, a2, a4, . . . , a}40 h2i = {0, 2, 4, . . . , 40} 21 ha3i = {e, a3, a6, . . . , a39} h3i = {0, 3, 6, . . . , 39} 14 ha6i = {e, a6, a12, . . . , a36} h6i = {0, 6, 12, . . . , 36} 7 ha7i = {e, a7, a14, . . . , a35} h7i = {0, 7, 14, . . . , 35} 6 ha14i ={ e, a14, a28 } h14i ={ 0, 14, 28 } 3 ha21i = {e, a21 } h21i = {0, 21 } 2 ha42i = {e } h42i = {0 } 2 h i { } h i { } This is an example of isomorphic groups, groups with exactly the same struc- ture.

Definition. The Euler Phi function is deﬁned by (1) = 1 and, for n > 1, (1) is the number of positive integers less than n and relatively prime to n.

Corollary. n > 1, U(n) = (n). 8 | | Note. Some values of (n) are found in Table 4.1 on page 84 of the text. 54 4. CYCLIC GROUPS Theorem (4.4 — Number of Elements of Each Order of a Cyclic Group). If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is (d). Proof. By Theorem 4.3 exactly one subgroup of order d — say a . Then every element of order d9generates a . By Corollary 3 of Theorem 5.2,h ian element ak generates a gcd(k, dh) =i 1. The number of such elements is (d). h i () ⇤ Note. For a ﬁnite cyclic group of order n, this means the number of elements of order d where d n depends only on d. |

Example. Z7, Z490, and Z7000 each has (7) = 6 elements of order 7. Corollary (Number of Elements of Order d in a Finite Group). In a ﬁnite group (not necessarily cyclic), the number of elements of order d is divisible by (d). Proof. Let G be a ﬁnite group. If G has no elements of order d, (d) 0. So suppose a G with a = d. By Theorem 4.4, a had (d) elements of| order d. If all elemen2 ts of order| | d are in a , we are done.h i h i So suppose b G, b = d, b a . Then b also has (d) elements of order d. Thus, if a2and |b| have no62elemenh i ts of horderi d in common, we have found 2(d) elemenh tsi of orderh i d. If c = d and c a b , a = c = b , a contradiction. | | 2 h i h i h i h i h i T Continuing, the number of elements of order d is a multiple of (d). ⇤ 4. CYCLIC GROUPS 55 Subgroup Lattice for Z42.

Problem (Page 89 # 35 (Adjusted)). Determine the subgroup lattices for n Zpn and a with a = p , where p is a prime and n is some positive integer. h i | | 56 4. CYCLIC GROUPS

Subgroup Lattice for Zpq2, where p and q are distinct primes.

Corresponding Subgroup Lattice for a , where a = pq2 and p and q are distinct primes. h i | | 4. CYCLIC GROUPS 57 Example. How many elements of order 4 does D12 have? How many elements of order 4 does D4n have?

Solution. Consider D4n. Since all reﬂections have order 2, elements of order 4 must be rotations and elements of R360/4n where R360/4n = 4n. Since 4 4n, by Theorem 4.4, the number of elemenh ts iof order 4|his (4) =i| 2. For | D12, just let n = 3. ⇤

Note. Cyclic groups will later be shown to be building blocks for all Abelian groups in much the same way as primes are building blocks for the integers. Maple. See cyclic.mw or cyclic.pdf.