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10Ma5aHw4Sol

# 10Ma5aHw4Sol - M a 5a HOMEWORK ASSIGNMENT 4 SOLUTION FALL...

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Ma 5 a HOMEWORK ASSIGNMENT 4 SOLUTION FALL 2010 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummitt and Foote. Page 65, 16 . (a)Suppose H itself is not maximal, then there is a sub- group H 1 of G such that H ( H 1 . Similarly, if H 1 is not maximal, there is H 2 in G such that H ( H 1 ( H 2 . We get a sequence { H 1 , H 2 , ... } , and for each i , there is some g i H i such that g i / H i - 1 . Note that g i is distinct. Since G is finite, { g i } must also be finite. Therefore, there is a maximal subgroup H m in the sequence, and H ( H m . (b)Let A = { 1 , r, r 2 , r 3 } , then | A | = 4. If there is a subgroup H of D 8 containing A , then | H | | | D 4 | = 8, and 4 = | A | | | H | . But then | H | = 4 or 8, and H = D 8 or H = A . Hence A must be maximal. (c)Suppose H = < x p > . Since p divides n , | H | = n p < n = | G | . Therefore H is a proper subgroup of G . Suppose there is a subgroup H 0 = < x q > G such that H ( H 0 . If p divides q , then q = pk for some k , and x q = ( x p ) k H . Therefore H 0 H . Contradiction. For q not divisible by p , since p is prime, ( p, q ) = 1. Now there exist a , b such that pa + qb = 1. Now ( x

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