4023f11ps4a

# 4023f11ps4a - Homework #4 Solutions Due: September 21, 2011...

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Homework #4 Solutions Due: September 21, 2011 1. Two results concerning groups are (1): If H and K are subgroups of G , then H \ K is a subgroup (Theorem 0.4.4), and (2): Every subgroup of a cyclic group is cyclic (Theorem 0.4.5). For each of the following subgroups H and K of a cyclic group G = [ a ], identify H \ K by giving a generator. That is, write H \ K = [ b ] for some b 2 G . (a) G = Z , H = 6 Z , K = 45 Z . I Solution. H \ K = f m 2 Z : 6 j m and 45 j m g . Thus H \ K is the set of all common multiples of 6 and 45. Since H \ K is a subgroup of Z it follows, from the Theorem proven in class, (or see the proof of Theorem 0.4.5, Page 25) that H \ K = b Z where b is the smallest positive integer in H \ K . That is, b is the least common multiple of 6 and 45, which is 90. Thus, H \ K = 90 Z . J (b) G = [ a ] with o ( a ) = 20, H = [ a 14 ], K = [ a 15 ]. I Solution. Since o ( a 14 ) = 20 = gcd(20 ; 14) = 10 (Theorem 0.3.4), it follows that the order of the cyclic group H is 10. According to Theorem 0.4.6, there is a unique subgroup of G of order 10, namely [ a 20 = 10 ] = [ a 2 ]. Thus, H = [ a 2 ]. In other words, a 2 is also a generator of H along with a 14 . (In fact, ( a 14 ) 3 = a 2 .) Similarly, o ( a 15 ) = 20 = gcd(15 ; 20) = 20 = 5 = 4. Thus,

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4023f11ps4a - Homework #4 Solutions Due: September 21, 2011...

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