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Unformatted text preview: Homework #9 Solutions p 149, #18 Let n > 1. Then n 1 U ( n ) and ( n 1) 2 = n 2 2 n +1 so that ( n 1) 2 mod n = 1. Since n 1 6 = 1, this means that  n 1  = 2 in U ( n ). As the order of any element in a group must divide the order of that group, it follows that 2 must divide the order of U ( n ), i.e. the order of U ( n ) is even. p 149, #22 Let a G , a 6 = e . Then h a i is a nontrivial subgroup of G . Since G has no proper nontrivial subgroup, it must be that G = h a i . That is, G is cyclic. If G is infinite then G = Z , which we know has infinitely many subgroups, and this is a contradiction. Therefore, G must be a finite cyclic group. By the fundamental theorem of cyclic groups, the subgroups of G correspond to the divisors of its order. Since G has no subgroups other than { e } and itself, it must be that  G  is divisible only by 1 and itself, i.e.  G  is prime. p 149, #26 Theorem 1. Let G be a finite group with even order. Then G has an element of order 2. Proof. Since any element and its inverse have the same order, we can pair each element of G with order larger than two with its (distinct) inverse, and hence there must be an even...
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This note was uploaded on 02/29/2012 for the course MATH 3362 taught by Professor Ryandaileda during the Fall '10 term at Trinity University.
 Fall '10
 RyanDaileda

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