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Unformatted text preview: We first state some basic properties. Theorem 4.30 Let G be an abelian group of exponent m . 1. For any integer m such that m G = { G } , we have m  m . 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 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 := { G } , H i := h a 1 ,...,a i i ( i = 1 ,...,n ) . We have n =  H n  /  H  = 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 = ord( a i ). Then k ( a i + H i 1 ) = k a i + H i 1 = 0 G + H i 1 . Therefore, k  k . 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 mmultiplication map. Therefore, this map is not injective, and hence not surjective since G is finite. Thus, mG 6 = G , a contradiction. 2 31 4.6 The Structure of Finite Abelian Groups We next state a theorem that characterizes all finite abelian groups up to isomorphism....
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 Spring '13
 MRR
 Math, Algebra, Number Theory

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