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College Algebra Exam Review 187

# College Algebra Exam Review 187 - 3.6 FINITELY GENERATED...

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3.6. FINITELY GENERATED ABELIAN GROUPS 197 A D G tor . Consequently, B Š G=G tor , so the rank of B is determined. This proves part (b). For part (c), note that any free module is torsion free. On the other hand, if G is torsion free, then by the decomposition of part (b), G is free. n Lemma 3.6.4. Let x be a torsion element in an abelian group, with order a and let p be a prime number. (a) If p divides a , then Z x=p Z x Š Z p . (b) If p does not divide a , then p Z x D Z x . Proof. Consider the group homomorphism of Z onto Z x , r 7! r x , which has kernel a Z . If p divides a , then pZ aZ , and the image of p Z in Z x is p Z x . Hence by Proposition 2.7.13 , Z =.p/ Š Z x=p Z x . If p does not divide a , then p and a are relatively prime. Hence there exist integers s; t such that sp C ta D 1 . Therefore, for all integers r , r x D 1rx D psrx C tarx D psrx (since ax D 0 ). It follows that Z x D p Z x . n Lemma 3.6.5. Suppose G is an abelian group, p is a prime number. and pG D f 0 g . Then G is a vector space over Z p . Moreover, if ' W G ! G is a surjective group homomorphism, then G is an Z p –vector space as well, and
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