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# Is 1 since n 1 and n 2 are relatively prime that is

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is 1, since n 1 and n 2 are relatively prime. That is, ma 1 = 0 G , from which it follows that n 1 | m . By a symmetric argument, one finds n 2 | m , and again, as n 1 and n 2 are relatively prime, this implies that n 1 n 2 | m . That proves that ord( a 1 + a 2 ) = n 1 n 2 . 2 For an abelian group G , the exponent of G is defined to be the least positive integer m such that mG = { 0 G } if such an integer exists, and is defined to be 0 otherwise. We first state some basic properties. Theorem 4.30 Let G be an abelian group of exponent m . 1. For any integer m 0 such that m 0 G = { 0 G } , we have m | m 0 . 2. If G has finite order, then m divides | G | . 3. If m 6 = 0 , for any a G , the order of a is finite, and ord( a ) | m . Proof. Exercise. 2 Theorem 4.31 For finite abelian groups G 1 , G 2 whose exponents are m 1 and m 2 , the exponent of G 1 × G 2 is lcm( m 1 , m 2 ) . 30

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Proof. Exercise. 2 Theorem 4.32 If a finite abelian group G has exponent m , then G contains an element of order m . In particular, a finite abelian group is cyclic if and only if its order equals its exponent. Proof. The second statement follows immeditely from the first. For the first statement, assume that m > 1, and let m = Q r i =1 p e i i be the prime factorization of m . First, we claim that for each 1 i r , there exists a i G such that ( m/p i ) a i 6 = 0 G . Suppose the claim were false: then for some i , ( m/p i ) a = 0 G for all a G ; however, this contradicts the minimality property in the definition of the exponent m . That proves the claim. Let a 1 , . . . , a r be as in the above claim. Then by Theorem 4.28, ( m/p e i i ) a i has order p e i i for each 1 i r . Finally, by Theorem 4.29, the group element ( m/p e 1 1 ) a 1 + · · · + ( m/p e r r ) a r has order m . 2 Theorem 4.33 If G is a finite abelian group of order n , and p is a prime dividing n , then G contains an element of order p . Proof. First, note that if G contains an element whose order is divisible by p , then it contains an element of order p ; indeed, if a has order mp , then ma has order p . Let a 1 , . . . , a n be an enumeration of all the elements of G , and consider the tower of subgroups H 0 := { 0 G } , H i := h a 1 , . . . , a i i ( i = 1 , . . . , n ) . We have n = | H n | / | H 0 | = n Y i =1 | H i | / | H i - 1 | = n Y i =1 | H i /H i - 1 | , and therefore, for some 1 i n , p | | H i /H i - 1 | . Let k = | H i /H i - 1 | . Now, the quotient group H i /H i - 1 is clearly cyclic and is generated by the coset a i + H i - 1 . Let k 0 = ord( a i ). Then k 0 ( a i + H i - 1 ) = k 0 a i + H i - 1 = 0 G + H i - 1 . Therefore, k | k 0 . That proves that p | ord( a i ), so we are done. 2 With this last theorem, we can prove the converse of Theorem 4.26. Theorem 4.34 If G is a finite abelian group of order n , and mG = G , then m is relatively prime to n . Proof. To the contrary, suppose that p is a prime dividing m and n . Then G contains an element of order p by Theorem 4.33, and this element is in the kernel of the m -multiplication map. Therefore, this map is not injective, and hence not surjective since G is finite. Thus, mG 6 = G , a contradiction.
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• Spring '13
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