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Case 2 n> 0 in this case z/n z = z n and so we see

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Unformatted text preview: 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 · 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 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 of m (in particular, m = n ), we also have m 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- to-one correspondence with the positive divisors of n , where each such divisor d corresponds to a cyclic subgroup of G d of order d . Moreover: • G d is the image of the ( n/d )-multiplication map (or ( n/d )-power map). • G d contains precisely those elements in G whose order divides d ; i.e., G d is the kernel of the d-multiplication map (or d-power map, for multiplicative groups). • G d ⊃ G d if and only if d | d . Proof. Since G ∼ = Z n , this follows immediately from Theorem 4.8, and the discussion in Exam- ple 4.17. We leave the details to the reader. 2 Example 4.36 Since m Z n is cyclic of order n/d , where d = gcd( m,n ), we have m Z n ∼ = Z n/d . 2 Example 4.37 Consider the group Z n 1 × Z n 2 . For m ∈ Z , then the element m ([1 mod n 1 ] , [1 mod n 2 ]) = ([0 mod n 1 ] , [0 mod n 2 ]) if and only if n 1 | m and n 2 | m . This implies that ([1 mod n 1 ] , [1 mod n 2 ]) has order lcm( n 1 ,n 2 ). In particular, if gcd( n 1 ,n 2 ) = 1, then Z n 1 × Z n 2 is cyclic of order n 1 n 2 , and so Z n 1 × Z n 2 ∼ = Z n 1 n 2 . Moreover, if gcd( n 1 ,n 2 ) = d > 1, then all elements of Z n 1 × Z n 2 have order dividing n 1 n 2 /d , and so Z n 1 × Z n 2 cannot be cyclic.cannot be cyclic....
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Case 2 n> 0 In this case Z/n Z = Z n and so we see that G...

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