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Unformatted text preview: Theorem 4.35 (Fundamental Theorem of Finite Abelian Groups) A finite abelian group (with more than one element) is isomorphic to a direct product of cyclic groups Z p e 1 1 × ··· × Z p er r , where the p i are primes (not necessarily distinct) and the e i are positive integers. This direct product of cyclic groups is unique up to the order of the factors. An alternative characterization of this theorem is the following: Theorem 4.36 A finite abelian group (with more than one element) is isomorphic to a direct product of cyclic groups Z m 1 × ··· × Z m t , where all m i > 1 and m 1  m 2  ···  m t . Moreover, the integers m 1 ,...,m t are unique, and m t is the exponent of the group. Proof. This follows easily from Theorem 4.35 and Example 4.37. Details are left to the reader. 2 Note that Theorems 4.32 and 4.33 follow as easy corollaries of Theorem 4.35; however, the direct proofs of those theorems are much simpler than the proof of Theorem 4.35. The proof of Theorem 4.35 is a bit tedious, and we break it into three lemmas. Lemma 4.37 Let G be a finite abelian group. Then G is isomorphic to a direct product of abelian groups, each of whose exponent is a prime power. Proof. Let m be the exponent of G . If m is a prime power, we are done. Otherwise, write m = m 1 m 2 , where gcd( m 1 ,m 2 ) = 1, and 1 < m 1 ,m 2 < m . Consider the subgroups m 1 G and m 2 G . Clearly, m 1 G has exponent m 2 and m 2 G has exponent m 1 . Since gcd( m 1 ,m 2 ) = 1, m 1 G ∩ m 2 G = { G } . By Theorem 4.21, m 1 G 1 × m 2 G 2 ∼ = m 1 G + m 2 G . Again, since gcd( m 1 ,m 2 ) = 1, there exist integers x 1 ,x 2 such that m 1 x 1 + m 2 x 2 = 1. For any a ∈ G , a = ( m 1 x 1 + m 2 x 2 ) a = m 1 ( x 1 a ) + m 2 ( x 2 a ) ∈ m 1 G 1 × m 2 G, and hence G = m 1 G 1 × m 2 G 2 . Thus, we have an isomorphism of G with m 1 G 1 × m 2 G 2 . The lemma follows by induction on m . 2 Lemma 4.38 Let G be a finite abelian group of exponent p e for prime p and positive integer e . Then there exist positive integers e 1 ,...,e k such that G ∼ = Z p e 1 × ··· × Z p e k . Proof. The proof is a bit long. Consider a sequence of group elements ( a 1 ,...,a k ), with k ≥ 0, along with a corresponding “tower” of subgroups H := { G } , H i := h a 1 ,...,a i i ( i = 1 ,...,k ) . 32 Let us call the sequence of ( a 1 ,...,a k ) “good” if for 1 ≤ i ≤ k , there exists a positive integer e i such that p e i a i = 0 G , p e i 1 a i / ∈ H i 1 , and p e i a ∈ H i 1 for all a ∈ G. (4.1) Let us study some of the properties of a good sequence ( a 1 ,...,a k ). To this end, for 0 ≤ i ≤ k , and for a ∈ G , let us define ord i ( a ) to be the least positive integer m such that ma ∈ H i . Clearly, ord ( a ) = ord( a ), and ord i ( a ) is the order of the coset a + H i in the quotient group G/H i . Since ord( a ) a = 0 G ∈ H i , it follows that ord i ( a )  ord( a )  p e ....
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 Spring '13
 MRR
 Math, Algebra, Number Theory

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