MT1sol - The Chinese Remainder Theorem (Section 2.8.3)...

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Math 113, Solutions to Midterm Exam # 1 (1) This is part (ii) of Problem 30 on page 45. By the Binomial Theorem, ( a + b ) p = p X i =0 ± p i ² · a i · b p - i , ( * ) where ± p i ² = p · ( p - 1) · ( p - 2) ··· ( p - i + 1) i · ( i - 1) · ( i - 2) ··· 1 . All factors in the denominator are relatively prime from p , and hence p divides ( p i ) provided 0 < i < p . Hence the expression (*) is congruent to a p + b p modulo p . (2) The requirement x 3 (mod 10) implies x 3 (mod 5), and the requirement x 7 (mod 15) implies x 7 2 (mod 5). These two statements are inconsistent. Hence these congruences have no solutions at all: the set of solutions is the empty set. (3) Let G be an abelian group and H any subgroup. For g G and h H we have g - 1 h = hg - 1 and hence g - 1 hg = ( g - 1 h ) g = ( hg - 1 ) g = h ( g - 1 g ) = h H. This shows that H is a normal subgroup of G . (4) (a) The residues 3, 4 and 5 are pairwise relatively prime, and their product is 60.
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Unformatted text preview: The Chinese Remainder Theorem (Section 2.8.3) implies that G is isomorphic to the group Z / 60 Z . (b) By Proposition 2.7.4 (iii), the number of generators is (60) = (3) (4) (5) = 2 2 4 = 16. (b) By Proposition 2.7.4 (iii), the number of elements of order 10 is (10) = (2) (5) = 1 4 = 4. (5) (a) We write the two cycles as permutations: = 1 2 3 4 2 3 1 4 and = 1 2 3 4 2 4 3 1 . Therefore = 1 2 3 4 3 4 1 2 and = 1 2 3 4 4 3 2 1 . (b) In cycle notation, we have = (13)(24) and = (14)(23) ....
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This note was uploaded on 02/03/2011 for the course MATH 113 taught by Professor Ogus during the Spring '08 term at University of California, Berkeley.

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