SolutionsChap1-5 - Solutions Chapters 15 Section 1.1 1....

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Solutions Chapters 1–5 Section 1.1 1. Under multiplication, the positive integers form a monoid but not a group, and the positive even integers form a semigroup but not a monoid. 2. With | a | denoting the order of a , we have | 0 | =1 , | 1 | =6 , | 2 | =3 , | 3 | =2 , | 4 | , and | 5 | =6. 3. There is a subgroup of order 6 /d for each divisor d of 6. We have Z 6 itself ( d = 1), { 0 } ( d =6) , { 0 , 2 , 4 } ( d = 2), and { 0 , 3 } ( d = 3). 4. S forms a group under addition. The inverse operation is subtraction, and the zero matrix is the additive identity. 5. S does not form a group under multiplication, since a nonzero matrix whose deter- minant is 0 does not have a multiplicative inverse. 6. If d is the smallest positive integer in H , then H consists of all multiples of d . For if x H we have x = qd + r where 0 r<d . But then r = x H ,so r must be 0. 7. Consider the rationals with addition mod 1, in other words identify rational numbers that di±er by an integer. Thus, for example, 1 / 3=4 / 3=7 / 3, etc. The group is in²nite, but every element generates a ²nite subgroup. For example, the subgroup generated by 1 / 3is { 1 / 3 , 2 / 3 , 0 } . 8. ( ab ) mn =( a m ) n ( b n ) m = 1, so the order of ab divides mn .Thu s | ab | = m 1 n 1 where m 1 divides m and n 1 divides n . Consequently, a m 1 n 1 b m 1 n 1 = 1 (1) If m = m 1 m 2 , raise both sides of (1) to the power m 2 to get b mn 1 = 1. The order of b , namely n , must divide mn 1 , and since m and n are relatively prime, n must divide n 1 . But n 1 divides n , hence n = n 1 . Similarly, if n = n 1 n 2 we raise both sides of (1) to the power n 2 and conclude as above that m = m 1 . But then | ab | = m 1 n 1 = mn , as asserted. If c belongs to both h a i and h b i then since c is a power of a and also a power of b ,we have c m = c n = 1. But then the order of c divides both m and n , and since m and n are relatively prime, c has order 1, i.e., c =1. 1
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2 9. Let | a | = m , | b | = n .I f[ m,n ] is the least common multiple, and ( ) the greatest common divisor, of m and n , then [ ]= mn/ ( ). Examine the prime factoriza- tions of m and n : m =( p t 1 1 ··· p t i i )( p t i +1 i +1 p t j j )= rr 0 n p u 1 1 p u i i )( p u i +1 i +1 p u j j s 0 s where t k u k for 1 k i , and t k u k for i +1 k j . Now a r has order m/r and b s has order n/s , with m/r (= r 0 ) and n/s (= s 0 ) relatively prime. By Problem 8, a r b s has order mn/rs = mn/ ( )=[ ]. Thus given elements of orders m and n , we can construct another element whose order is the least common multiple of m and n . Since the least common multiple of m , n and q is [[ ] ,q ], we can inductively Fnd an element whose order is the least common multiple of the orders of all elements of G . 10. Choose an element a that belongs to H but not K , and an element b that belongs to K but not H , where H and K are subgroups whose union is G . Then ab must belong to either H or K ,say ab = h H . But then b = a 1 h H , a contradiction. If ab = k K , then a = kb 1 K , again a contradiction. To prove the last statement, note that if H K is a subgroup, the Frst result with G replaced by H K implies that H = H K or K = H K , in other words, K H or H K .
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.

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SolutionsChap1-5 - Solutions Chapters 15 Section 1.1 1....

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