College Algebra Exam Review 197

College Algebra Exam Review 197 - p ² 1 Proof Let K ±...

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3.6. FINITELY GENERATED ABELIAN GROUPS 207 and forming the direct product of the chosen groups. There are 6 possibil- ities for the elementary divisor decompositions: Z 8 Z 3 Z 25 Z 7 ; Z 8 Z 3 Z 5 Z 5 Z 7 ; Z 4 Z 2 Z 3 Z 25 Z 7 ; Z 4 Z 2 Z 3 Z 5 Z 5 Z 7 ; Z 2 Z 2 Z 2 Z 3 Z 25 Z 7 ; Z 2 Z 2 Z 2 Z 3 Z 5 Z 5 Z 7 : The corresponding invariant factor decompositions are: Z 4200 ; Z 840 Z 5 ; Z 2100 Z 2 ; Z 420 Z 10 ; Z 1050 Z 2 Z 2 ; Z 210 Z 10 Z 2 : The group of units in Z N The rest of this section is devoted to working out the structure of the group ˚.N / of units in Z N . It would be safe to skip this material on first reading and come back to it when it is needed. Recall that ˚.N / has order '.N / , where ' is the Euler ' function. The following theorem states that for a prime p , ˚.p/ is cyclic of order p 1 . Theorem 3.6.26. Let K be a finite field of order n . Then the multiplicative group of units of K is cyclic of order n 1 . In particular, for p a prime number, the multiplicative group ˚.p/ of units of
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Unformatted text preview: p ² 1 . Proof. Let K ± denote the multiplicative group of nonzero elements of K . Then K ± is abelian of order n ² 1 . Let m denote the period of K ± . On the one hand, m ³ n ² 1 D j K ± j . On the other hand, x m D 1 for all elements of K ± , so the polynomial equation x m ² 1 D has n ² 1 distinct solutions in the field K . But the number of distinct roots of a polynomial in a field is never more than the degree of the polynomial (Corollary 1.8.24 ), so n ² 1 ³ m . Thus the period of K ± equals the order n ² 1 of K ± . But the period and order of a finite abelian group are equal if, and only if, the group is cyclic. This follows from the fundamental theorem of finite abelian groups, Theorem 3.6.2 . n...
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