# More generally for a 1 a k g we define h a 1 a k i z

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More generally, for a 1 , . . . , a k G , we define h a 1 , . . . , a k i := { z 1 a 1 + · · · + z k a k : z 1 , . . . , z k Z } . One also verifies that h a 1 , . . . , a k i is a subgroup of G , and that any subgroup H of G that contains a 1 , . . . , a k must contain h a 1 , . . . , a k i . The subgroup h a 1 , . . . , a k i is called the subgroup generated by a 1 , . . . , a k . An abelian group G is called a cyclic group if G = h a i for some a G , in which case, a is called a generator for G . Multiplicative notation: if G is written multiplicatively, then h a i := { a z : z Z } , and h a 1 , . . . , a k i := { a z 1 1 · · · a z k k : z 1 , . . . , z k Z } . Example 4.34 Z is a cyclic group generated by 1. The only other generator is - 1. 2 Example 4.35 Z n is a cyclic group generated by [1 mod n ]. More generally, h [ m mod n ] i = m Z n , and so is cyclic of order n/d , where d = gcd( m, n ). 2 We can very quickly characterize all cyclic groups, up to isomorphism. Suppose that G is a cyclic group with generator a . Consider the map f : Z G that sends z Z to za G . This map is clearly a surjective homomorphism. Now, ker( f ) is a subgroup of Z , and by Theorem 4.7, it must be of the form n Z for some non-negative integer n . Also, by Theorem 4.19, we have Z /n Z = G . Case 1: n = 0. In this case, Z /n Z = Z , and so we see G = Z . Moreover, by Theorem 4.17, the only integer z such that za = 0 G is the integer 0, and more generally, z 1 a = z 2 a if and only if z 1 = z 2 . Case 2: n > 0. In this case, Z /n Z = Z n , and so we see that G = Z n . Moreover, by Theorem 4.17, za = 0 G if and only if n | z , and more generally, z 1 a = z 2 a if and only if z 1 z 2 (mod n ). The order of G is evidently n , and G consists of the distinct elements 0 · a, 1 · a, . . . , ( n - 1) · a. From this characterization, we immediately have: Theorem 4.22 Let G be an abelian group and let a G . If there exists a positive integer m such that ma = 0 G , then the least such integer is the order of a . Moreover, if G of finite order n , then ord( a ) | n , and in particular na = 0 G . 28

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Proof. The first statement follows from the above characterization. For the second statement, since h a i is a subgroup of G , by Theorem 4.13, its order must divide that of G . Of course, if ma = 0 G , then for any multiple m 0 of m (in particular, m 0 = n ), we also have m 0 a = 0 G . 2 Based on the this theorem, we can trivially derive a classical result: Theorem 4.23 (Fermat’s Little Theorem) For any prime p , and any integer x 6≡ 0 (mod p ) , we have x p - 1 1 (mod p ) . Moreover, for any integer x , we have x p x (mod p ) . Proof. The first statement follows from Theorem 4.22, and the fact that Since Z * p is an abelian group of order p - 1. The second statement is clearly true if x 0 (mod p ), and if x 6≡ 0 (mod p ), we simply multiply both sides of the congruence x p - 1 1 (mod p ) by x . 2 It also follows from the above characterization of cyclic groups that that any subgroup of a cyclic group is cyclic — indeed, we have already characterized the subgroups of Z and Z n in Theorems 4.7 and 4.8, and it is clear that these subgroups are cyclic. Indeed, it is worth stating the following: Theorem 4.24 Let G be a cyclic group of finite order n . Then the subgroups of G are in one-
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