College Algebra Exam Review 189

# College Algebra Exam Review 189 - Similarly Z 24 Š Z 3 ²...

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3.6. FINITELY GENERATED ABELIAN GROUPS 199 product decomposition, we obtain that the number of b i divisible by p is also equal to dim Z p .G=pG/ . If p is an irreducible dividing a 1 , then p divides all of the a i , and hence exactly s of the b i . Therefore, s ± t . Reversing the role of the two decompositions, we get t ± s . Thus the number of factors in the two decompositions is the same. Fix an irreducible p dividing a 1 . Then p divides a j and b j for 1 ± j ± s . Let k 0 be the last index such that a k 0 D p . Then pA j is cyclic of period a j =p for j > k 0 , while pA j D f 0 g for j ± k 0 , and pG D pA k 0 C 1 ² ³³³ ² pA s : Likewise, let k 00 be the last index such that b k 00 D p . Then pB j is cyclic of period b j =p for j > k 00 , while pB j D f 0 g for j ± k 00 , and pG D pB k 00 C 1 ² ³³³ ² pB s : Applying the induction hypothesis to pG (which has period a=p ) gives k 0 D k 00 and a i =p D b i =p for all i > k 0 . Hence, a i D b i for all i > k 0 . But for i ± k 0 , we have a i D b i D p . n The direct product decomposition of Theorem 3.6.2 is called the in- variant factor decomposition . The numbers a 1 ;a 2 ;:::;a s in the theorem are called the invariant factors of G . Example 3.6.6. As in Example 3.1.12 , Z 30 Š Z 5 ² Z 3 ² Z 2
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Unformatted text preview: . Similarly, Z 24 Š Z 3 ² Z 8 . Therefore Z 30 ² Z 24 Š Z 5 ² Z 3 ² Z 2 ² Z 3 ² Z 8 . Regroup these factors as follows: Z 30 ² Z 24 Š . Z 5 ² Z 3 ² Z 8 / ² . Z 3 ² Z 2 / Š Z 120 ² Z 6 . This is the invariant factor decomposition of Z 30 ² Z 24 The invariant factors of Z 30 ² Z 24 are 120;6 . Corollary 3.6.7. (a) Let G be an abelian group of order p n , where p is a prime. Then G is a direct product of cyclic groups, G Š Z p n 1 ² ³³³ ² Z p n k ; where n 1 ± n 2 ± ³³³ ± n k , and P i n i D n , (b) The sequence of exponents in part (a) is unique. That is, if m 1 ± m 2 ± ³³³ ± m ` , P j m j D n , and G Š Z p m 1 ² ³³³ ² Z p m ` ; then k D ` and n i D m i for all i . Proof. This is just the special case of the theorem for a group whose order is a power of a prime. n...
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