College Algebra Exam Review 185

# College Algebra Exam Review 185 - 195 3.6. FINITELY...

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Unformatted text preview: 195 3.6. FINITELY GENERATED ABELIAN GROUPS P i ri xi . Let N denote the kernel of ' . According to Theorem 3.5.13, N is free of rank s Ä n, and there exists a basis fv1 ; : : : ; vn g of Zn and positive integers d1 ; : : : ; ds such that fd1 v1 ; : : : ; ds vs g is a basis of N and di divides dj for i Ä j . Therefore G Š Zn =N D .Zv1 ˚ ˚ Zvn /=.Zd1 v1 ˚ ˚ Z ds vs / The following lemma applies to this situation: Lemma 3.6.1. Let A1 ; : : : ; An be abelian groups and Bi Â Ai subgroups. Then .A1 Bn / Š A1 =B1 An /=.B1 An =Bn : Proof. Consider the homomorphism of A1 An onto A1 =B1 An =Bn deﬁned by .a1 ; : : : ; an / 7! .a1 C B1 ; ; an C Bn /. The kernel of this map is B1 Bn Â A1 An , so by the isomorphism theorem for modules, .A1 Bn / Š A1 =B1 An /=.B1 An =Bn : I Observe that Zvi =Zdi vi Š Z=di Z D Zdi , since r 7! rvi C Zdi vi is a surjective Z–module homomorphism with kernel di Z. Applying Lemma 3.6.1 and this observation to the situation described above gives G Š .Zv1 ˚ ˚ Zvn /=.Zd1 v1 ˚ ˚ Zds vs / Š .Zv1 =Zd1 v1 / Š Z=di Z D Zdi .Zvs =Zds vs / Z=ds Z Zds Z ns Z Zvs C1 Z vn ns : If some di were equal to 1, then Z=di Z would be the trivial group, so could be dropped from the direct product. But this would display G as generated by fewer than n elements, contradicting the minimality of n. We have proved the existence part of the following fundamental theorem: Theorem 3.6.2. (Fundamental Theorem of Finitely Generated Abelian Groups: Invariant Factor Form) Let G be a ﬁnitely generated abelian group. ...
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## This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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