Cyclic Groups Part 2

Review Definition: G is cyclic if G = <a> for some a in G. Thm 4.1 If |a| = ∞, ai=aj iff i =j If |a| = n, ai=aj iff n| i – j <a> = {a, a2, … an-1,e} Cor 1: |a| = |<a>| Cor 2: ak = e implies |a| | k

Review (con't) Thm 4.2 If |a| = n, then Proved first part last time. <ak> = <agcd(n,k)> |ak| = n/gcd(n,k) Proved first part last time.

Proof of 4.2 To prove the |ak| = n/gcd(n,k) , we begin with a little lemma. Prove: If d | n = |a|, then |ad| = n/d. Proof: Let n = dq. Then e = an = (ad)q. So |ad| ≤ q. If 0< i < q, then 0 < di < dq = n = |a| so (ad)i ≠ e Hence, |ad| = q which is n/d as required.

Proof that |ak| = n/gcd(n,k) Now let d = gcd(n,k). We have |ak| = |<ak>| by 4.1 cor 1 = |<ad>| by part 1 = |ad| by 4.1 cor 1 = n/d by our lemma. This concludes the proof of 4.2.

Example Suppose G = <a> with |a| = 30. Find |a21| and <a21>. By Thm 4.2, |a21| = 30/gcd(30,21) = 10 Also <a21> = <a3> = {a3, a6, a9, a12,a15, a18, a21, a24, a27, e}

Corollaries to Theorem 4.2 In a finite cyclic group, the order of an element divides the order of the group. Let |a| = n in any group. Then <ai> = <aj> iff gcd(n,i) = gcd(n,j) |ai| = |aj| iff gcd(n,i) = gcd(n,j)

More corollaries to 4.2 Let |a| = n. Then <ai> = aj iff gcd(n,i) = gcd(n,j) 4. An integer k in Zn is a generator of Zn iff gcd(n,j) = 1

Example Find all the generators of U(50) = <3>. |U(50)| = 20 The numbers relatively prime to 20 are 1, 3, 7, 9, 11, 13, 17, 19 The generators of U(50) are therefore 31, 33, 37, 39, 311, 313, 317, 319 i.e. 3, 27, 37, 33, 47, 23, 13, 17

Example In D8, List all generators of <R45º> |R45| = 8 The numbers relatively prime to 8 are 1, 3, 5, 7 The generators are R45, R453, R455, R457 i.e. R45, R135, R225, R315

Fundamental Theorem of Cyclic Groups Every subgroup of a cyclic group is cyclic. If |a| = n, then the order of any subgroup of <a> is a divisor of n For each positive divisor k of n, the group <a> has exactly one subgroup of order k, namely <an/k>

(a) Subgroups are cyclic Proof: Let G = <a> and suppose H ≤ G. If H is trivial, then H is cyclic. Suppose H is not trivial. Let m be the smallest positive integer with am in H. (Does m exist?) ________ By closure, <am> is contained in H. We claim that H = <am>. To see this, choose any b = ak in H. There exist integers q,r with 0 ≤ r < m such that ak = aqm + r (Why?) ___________

Since b = ak = aqma r, we have ar = (am)-q b Since b and am are in H, so is ar. But r < m (the smallest power of a in H) so r = 0. Hence b = (am)q and b is in H. It follows that H = <am> as required.

(b) |H| is a divisor of |a| Proof: Given |<a>| = n and H ≤ <a>. We showed H = <am> where m is the smallest positive integer with am in H. Now e = an is in H, so as we just showed, n = mq for some q. Now |am| = q is a divisor of n as required.

(c) Exactly one subgroup for each divisor k of n (Existence) Given |<a>| = n. Let k | n. Say n = kq. Note that gcd(n,q) = q So |aq| = n/gcd(n,q) = n/q = k. Hence there exists a subgroup of order k, namely <an/q>

(c) Con't. (Uniqueness) Let H be any subgroup of <a> with order k. We claim H = <an/k> From (a), H = <am> for some m. From (b), m | n so gcd(n,m) = m. So k = |am| = n/gcd(n,m) by 4.2 = n/m Hence m = n/k So H = <an/k> as required.

Subgroups of Zn For each positive divisor k of n, the set <n/k> is the unique subgroup of Zn of order k. Moreover, these are the only subgroups of Zn.

Euler Phi Function (n) = the number of relatively prime positive numbers < n |U(n)| = (n) (n) = n*(1-1/p1)(1-1/p2)… where p1, p2 … are prime divisors of n

Theorem 4.4 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 the FTCG, there is a unique subgroup H of order d. Clearly, |a| = d iff a generates H. Choose any generator b. By cor 3 of 4.2, bk generates H iff gcd(k,d) = 1. Hence the number of generators of H is (d).

Example How many elements of order 8 in Z16? (8) = 8(1-1/2) = 4 Find them: In Z16, |2| = 16/2 = 8. Generators of <2> are 2{1,3,5,7} = {2,6,10,14} These are all elements of order 8 in Z16

Another Example How many elements of order 8 in Z800? (8) = 8(1-1/2) = 4 Find them: In Z800 800/8 = 100 has order 8 The generators of <100> are 100{1,3,5,7} = {100, 300, 500, 700} These are all elements of order 8 in Z800

What can we say about all finite groups? Theorem 4.5 In a finite group, the number of elements of order d is a multiple of (d). Proof: Let G be a finite group with n elements of order d. Let b be the number of cyclic subgroups G with order d. Each element of order d belongs to exactly one cyclic subgroup of order d. Thus n = b•(d).

Final Example In U(20) Find the number of elements of order 4 U(20) = {1, 3, 7, 9, 11, 13, 17, 19} <3> = {3, 9, 7, 1} and <13> = {13, 9, 7, 1} U(20) is not cyclic. Elements of order 4 are 3, 7, 13, 9. n = 2•(4) = 4