# To one correspondence with the positive divisors of n

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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 0 if and only if d | d 0 . 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. 2 Example 4.38 As we saw in Example 4.20, all elements of Z * 15 have order dividing 4, and since Z * 15 has order 8, we conclude that Z * 15 is not cyclic. 2 Example 4.39 The group Z * 5 is cyclic, with [2] being a generator: [2] 2 = [4] = [ - 1] , [2] 3 = [ - 2] , [2] 4 = [1] . 2 Theorem 4.25 If G is a cyclic group, and f : G G 0 is a homomorphism from G to the abelian group G 0 , then f ( G ) is cyclic. 29

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Proof. Exercise. 2 Theorem 4.26 If G is a finite abelian group of order n , and m is an integer relatively prime to n , then mG = G . Proof. Consider the m -multiplication map on G . We claim that the kernel of this map is { 0 G } . Indeed, ma = 0 G , implies ord( a ) divides m , and since ord( a ) also divides n and gcd( m, n ) = 1, we must have ord( a ) = 1, i.e., a = 0 G . That proves the claim. Thus, the m -multiplication map is injective, and because G is finite, it must be surjective as well. 2 Theorem 4.27 If G is an abelian group of prime order, then G is cyclic. Proof. Let | G | = p . Let a G with a 6 = 0 G . Since ord( a ) | p , we have ord( a ) = 1 or ord( a ) = p . Since a 6 = 0 G , we must have ord( a ) 6 = 1, and so ord( a ) = p , which implies a generates G . 2 Theorem 4.28 Suppose that a is an element of an abelian group, and for some prime p and e 1 , we have p e a = 0 G and p e - 1 a 6 = 0 G . Then a has order p e . Proof. If m is the order of a , then since p e a = 0 G , we have m | p e . So m = p f for some 0 f e . If f < e , then p e - 1 a = 0 G , contradicting the assumption that p e - 1 a 6 = 0 G . 2 Theorem 4.29 Suppose G is an abelian group with a 1 , a 2 G such that the a 1 is of finite order n 1 and a 2 is of finite order n 2 , and d = gcd( n 1 , n 2 ) . Then ord( a 1 + a 2 ) | n 1 n 2 /d . Moreover, ord( a 1 + a 2 ) = n 1 n 2 if and only if d = 1 . Proof. Since ( n 1 n 2 /d )( a 1 + a 2 ) = ( n 2 /d )( n 1 a 1 ) + ( n 1 /d )( n 2 a 1 ) = 0 G + 0 G = 0 G , the order of a 1 + a 2 must divide n 1 n 2 /d . On the one hand, if d > 1, then clearly a 1 + a 2 cannot have order n 1 n 2 . On the other hand, if d = 1, then m ( a 1 + a 2 ) = 0 G implies ma 1 = - ma 2 ; since - ma 2 has order dividing n 2 , so does ma 1 ; also, ma 1 has order dividing n 1 , and so we conclude that the order of ma 1
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