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Abstract Algebra Theory and Applications - Computer Science ...

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3.1 CYCLIC SUBGROUPS 57Proof. The identity is in 〈a〉 since a 0 = e. If g <strong>and</strong> h are any two elementsin 〈a〉, then by the definition of 〈a〉 we can write g = a m <strong>and</strong> h = a n for someintegers m <strong>and</strong> n. So gh = a m a n = a m+n is again in 〈a〉. Finally, if g = a nin 〈a〉, then the inverse g −1 = a −n is also in 〈a〉. Clearly, any subgroup Hof G containing a must contain all the powers of a by closure; hence, Hcontains 〈a〉. Therefore, 〈a〉 is the smallest subgroup of G containing a. □Remark. If we are using the “+” notation, as in the case of the integersunder addition, we write 〈a〉 = {na : n ∈ Z}.For a ∈ G, we call 〈a〉 the cyclic subgroup generated by a. If G containssome element a such that G = 〈a〉, then G is a cyclic group. In this case ais a generator of G. If a is an element of a group G, we define the orderof a to be the smallest positive integer n such that a n = e, <strong>and</strong> we write|a| = n. If there is no such integer n, we say that the order of a is infinite<strong>and</strong> write |a| = ∞ to denote the order of a.Example 3. Notice that a cyclic group can have more than a single generator.Both 1 <strong>and</strong> 5 generate Z 6 ; hence, Z 6 is a cyclic group. Not everyelement in a cyclic group is necessarily a generator of the group. The orderof 2 ∈ Z 6 is 3. The cyclic subgroup generated by 2 is 〈2〉 = {0, 2, 4}. The groups Z <strong>and</strong> Z n are cyclic groups. The elements 1 <strong>and</strong> −1 aregenerators for Z. We can certainly generate Z n with 1 although there maybe other generators of Z n , as in the case of Z 6 .Example 4. The group of units, U(9), in Z 9 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) since2 1 = 2 2 2 = 42 3 = 8 2 4 = 72 5 = 5 2 6 = 1.Example 5. Not every group is a cyclic group. Consider the symmetrygroup of an equilateral triangle S 3 . The multiplication table for this groupis Table 2.2. The subgroups of S 3 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.

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