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Unformatted text preview: Q (Exercise 3.5.4 ). Consequently, a linearly independent subset of Z n has at most n elements, and if n m , then the abelian groups Z n and Z m are nonisomorphic (Exercise 3.5.5 ). Lemma 3.5.3. A basis of a free abelian group is a minimal generating set. Proof. Suppose B is a basis of a free abelian group G and B is a proper subset. Let b 2 B n B . If b were contained in the subgroup generated by B , then b could be expressed as a Z linear combination of elements of B , contradicting the linear independence of B . Therefore b 62 Z B , and B does not generate G . n Lemma 3.5.4. Any basis of a nitely generated free abelian group is nite. Proof. Suppose that G is a free abelian group with a (possibly innite) basis B and a nite generating set S . Each element of S is a Z linear...
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 Fall '08
 EVERAGE
 Linear Algebra, Algebra

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