College Algebra Exam Review 194

# College Algebra Exam Review 194 - C i whose order is a...

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204 3. PRODUCTS OF GROUPS .2/20 ± 1 . mod 3/ ; 12 is congruent to zero mod 3 and 4, and invertible mod 5; .3/12 ± 1 . mod 5/ . Put y 1 D 45 , y 2 D 40 , and y 3 D 36 . For any a;b;c , x D 45a C 40b C 36c is a solution to the simultaneous congruence problem. The solution is unique only up to congruence mod 60 , so we can reduce mod 60 to ﬁnd a unique solution x with 0 ² x ² 59 . For example, let us ﬁnd x congruent to 0 mod 4, congruent to 2 mod 3, and congruent to 4 mod 5. We can take x D .0/45 C .2/40 C 4.36/ D 224 ± 44 . mod 60/ . The Elementary Divisor Decomposition Lemma 3.6.19. Any ﬁnite abelian group is a direct product of cyclic groups, each of which has order a power of a prime. Proof. A ﬁnite abelian group G is a direct product of its subgroups GŒpŁ . Each GŒpŁ is in turn a direct product of cyclic groups of order a power of p , by Corollary 3.6.7 . n Lemma 3.6.20. Suppose a ﬁnite abelian group G is an internal direct product of a collection f C i g of cyclic subgroups each of order a power of a prime. Then for each prime p , the sum of those
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Unformatted text preview: C i whose order is a power of p is equal to GŒpŁ . Proof. Denote by AŒpŁ the sum of those C i whose order is a power of p . Then AŒpŁ ³ GŒpŁ and G is the internal direct product of the subgroups AŒpŁ . Since G is also the internal direct product of the subgroups GŒpŁ , it follows that AŒpŁ D GŒpŁ for all p . n Example 3.6.21. Consider G D Z 30 ´ Z 50 ´ Z 28 . Then G Š . Z 3 ´ Z 2 ´ Z 5 / ´ . Z 25 ´ Z 2 / ´ . Z 4 ´ Z 7 / Š . Z 4 ´ Z 2 ´ Z 2 / ´ Z 3 ´ . Z 25 ´ Z 5 / ´ Z 7 : Thus GŒ2Ł Š Z 4 ´ Z 2 ´ Z 2 , GŒ3Ł Š Z 3 , GŒ5Ł Š Z 25 ´ Z 5 , and GŒ7Ł Š Z 7 . GŒpŁ D for all other primes p . Theorem 3.6.22. (Fundamental Theorem of Finitely Generated Abelian Groups: Elementary Divisor Form). Every ﬁnite abelian group is isomor-phic to a direct product of cyclic groups of prime power order. The number...
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