College Algebra Exam Review 177

College Algebra Exam Review 177 - Q (Exercise 3.5.4 )....

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3.5. LINEAR ALGEBRA OVER Z 187 map is injective, and (c) states that the kernel of the map is trivial. So all three assertions are equivalent. n Proposition 3.5.2. Let G be an abelian group and let x 1 ;:::;x n be dis- tinct nonzero elements of G . The following conditions are equivalent: (a) The set B D f x 1 ;:::;x n g is a basis of G . (b) The map .r 1 ;:::;r n / 7! r 1 x 1 C r 2 x 2 C ±±± C r n x n is a group isomorphism from Z n to G . (c) For each i , the map r 7! rx i is injective, and G D Z x 1 ² Z x 2 ² ±±± ² Z x n : Proof. It is easy to see that the map in (b) is a group homomorphism. The set B is linearly independent if, and only if, the map is injective, and B generates G if, and only if the map is surjective. This shows the equiva- lence of (a) and (b). We leave it as an exercise to show that (a) and (c) are equivalent. n Let S be a subset of Z n . Since Z n is a subset of the Q –vector space Q n , it makes sense to consider linear independence of S over Z or over Q . It is easy to check that a subset of Z n is linearly independent over Z if, and only if, it is linearly independent over
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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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