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.If a is an element of a group G, we define the orderof a to be the smallest positive integer n such that an = e, and we write|a| = n.If there is no such integer n, we say that the order of a is infiniteand write |a| = ∞ to denote the order of a.Example 3.Notice that a cyclic group can have more than a single gen-erator.Both 1 and 5 generate Z6; hence, Z6 is a cyclic group.Not everyelement in a cyclic group is necessarily a generator of the group.The orderof 2 ∈ Z6 is 3.The cyclic subgroup generated by 2 is h2i = {0, 2, 4}.The groups Z and Zn are cyclic groups.The elements 1 and −1 aregenerators for Z.We can certainly generate Zn with 1 although there maybe other generators of Zn, as in the case of Z6.Example 4.The group of units, U (9), in Z9 is a cyclic group.As a set,U (9) is {1, 2, 4, 5, 7, 8}.The element 2 is a generator for U (9) since21=222=423=824=725=526=1.Example 5.Not every group is a cyclic group.Consider the symmetrygroup of an equilateral triangle S3.The multiplication table for this groupis Table 2.2.The subgroups of S3 are shown in Figure 3.1.Notice that everysubgroup is cyclic; however, no single element generates the entire group.Theorem 3.2 Every cyclic group is abelian.58CHAPTER 3CYCLIC GROUPSS3!!aa!Sa!a!a!Sa!a!Sa{id, ρ1, ρ2}{id, µ1}{id, µ2}{id, µ3}a!a!aS!a!aS!a!aaS!!{id}Figure 3.1.Subgroups of S3Proof.Let G be a cyclic group and a ∈ G be a generator for G.If g andh are in G, then they can be written as powers of a, say g = ar and h = as.Sincegh = aras = ar+s = as+r = asar = hg,G is abelian.Subgroups of Cyclic GroupsWe can ask some interesting questions about cyclic subgroups of a groupand subgroups of a cyclic group.If G is a group, which subgroups of G arecyclic? If G is a cyclic group, what type of subgroups does G possess?Theorem 3.3 Every subgroup of a cyclic group is cyclic.Proof.The main tools used in this proof are the division algorithm andthe Principle of Well-Ordering.Let G be a cyclic group generated by a andsuppose that H is a subgroup of G.If H = {e}, then trivially H is cyclic.Suppose that H contains some other element g distinct from the identity.Then g can be written as an for some integer n.We can assume that n > 0.Let m be the smallest natural number such that am ∈ H.Such an m existsby the Principle of Well-Ordering.We claim that h = am is a generator for H.We must show that everyh0 ∈ H can be written as a power of h.Since h0 ∈ H and H is a subgroupof G, h0 = ak for some positive integer k.Using the division algorithm, wecan find numbers q and r such that k = mq + r where 0 ≤ r < m; hence,ak = amq+r = (am)qar = hqar.3.1CYCLIC SUBGROUPS59So ar = akh−q.Since ak and h−q are in H, ar must also be in H.However,m was the smallest positive number such that am was in H; consequently,r = 0 and so k = mq.Therefore,h0 = ak = amq = hqand H is generated by h.Corollary 3.4 The subgroups of Z are exactly nZ for n = 0, 1, 2,.Proposition 3.5 Let G be a cyclic group of order n and suppose that a isa generator for G.Then ak = e if and only if n divides k.Proof.First suppose that ak = e.By the division algorithm, k = nq + rwhere 0 ≤ r < n; hence,e = ak = anq+r = anqar = ear = ar.Since the smallest positive integer m such that am = e is n, r = 0.Conversely, if n divides k, then k = ns for some integer s.Consequently,ak = ans = (an)s = es = e.Theorem 3.6 Let G be a cyclic group of order n and suppose that a ∈ Gis a generator of the group.If b = ak, then the order of b is n/d, whered = gcd(k, n).Proof.We wish to find the smallest integer m such that e = bm = akm.By Proposition 3.5, this is the smallest integer m such that n divides km or,equivalently, n/d divides m(k/d).Since d is the greatest common divisor ofn and k, n/d and k/d are relatively prime.Hence, for n/d to divide m(k/d)it must divide m.The smallest such m is n/d.Corollary 3.7 The generators of Zn are the integers r such that 1 ≤ r < nand gcd(r, n) = 1.Example 6.Let us examine the group Z16.The numbers 1, 3, 5, 7, 9, 11,13, and 15 are the elements of Z16 that are relatively prime to 16.Each ofthese elements generates Z16.For example,1 · 9=92 · 9=23 · 9=114 · 9=45 · 9=136 · 9=67 · 9=158 · 9=89 · 9=110 · 9=1011 · 9=312 · 9=1213 · 9=514 · 9=1415 · 9=7.60CHAPTER 3CYCLIC GROUPS3.2The Group∗CThe complex numbers are defined asC = {a + bi : a, b ∈ R},where i2 = −1.If z = a + bi, then a is the real part of z and b is theimaginary part of z.To add two complex numbers z = a + bi and w = c + di, we just add thecorresponding real and imaginary parts:z + w = (a + bi) + (c + di) = (a + c) + (b + d)i.Remembering that i2 = −1, we multiply complex numbers just like polyno-mials.The product of z and w is(a + bi)(c + di) = ac + bdi2 + adi + bci = (ac − bd) + (ad + bc)i.Every nonzero complex number z = a + bi has a multiplicative inverse;that is, there exists a z−1 ∈∗C such that zz−1 = z−1z = 1.If z = a + bi,thena − biz−1 =.a2 + b2The complex conjugate of a complex number z = a + bi is defined to be√z = a − bi.The absolute value or modulus of z = a + bi is |z| =a2 + b2.Example 7.Let z = 2 + 3i and w = 1 − 2i.Thenz + w = (2 + 3i) + (1 − 2i) = 3 + iandzw = (2 + 3i)(1 − 2i) = 8 − i.Also,23z−1=−i1313√|z| =13z=2 − 3i.There are several ways of graphically representing complex numbers
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