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Unformatted text preview: Homework #11 Solutions p 166, #18 We start by counting the elements in D m and D n , respectively, of order 2. If x ∈ D m and  x  = 2 then either x is a flip or x is a rotation of order 2. The subgroup of rotations in D m is cyclic of order m , and since m is even there is exactly φ (2) = 1 rotation of order 2. Therefore, D m contains exactly m + 1 elements of order 2. On the other hand, y ∈ D n can have order 2 only if y is a flip, since the rotations in D n have order dividing n , which is odd. Therefore there are exactly n elements in D n of order 2. We are now in a position to count the elements of order 2 in D m ⊕ D n . Suppose ( x, y ) ∈ D m ⊕ D n and  ( x, y )  = 2. Since  ( x, y )  = lcm(  x  ,  y  ), it must be that either  x  = 2 and  y  = 1 , 2 or  x  = 1 and  y  = 2. In the first case, the preceding paragraph shows that there are m + 1 choices for x and n + 1 choices for y , giving a total of ( m + 1)( n + 1) pairs. In the second case x = e and there are n choices for y , yielding another n pairs. Thus, the total number of pairs with order 2 is ( m + 1)( n + 1) + n. p 166, #28 It is not hard to show that Z 12 ⊕ Z 4 ⊕ Z 15 has no element of order 9, so we won’t be able to find a cyclic subgroup of order 9. We therefore look for the next easiest type of subgroup, namely one of the form H ⊕ K ⊕ J where H ≤ Z 12 , K ≤ Z 4 and J ≤ Z 15 . The order of such a subgroup is  H  ·  K  ·  J  . If this is to equal 9, Lagrange’s theorem tells us that we need  H  = 3,  K  = 1 and  J  = 3. Since Z 12 , Z 4 and Z 15 are all cyclic, they have unique subgroups of these orders. That is, we must take H = h 4 i , K = { } and J = h 5 i , so our subgroup is h 4 i ⊕ { } ⊕ h 5 i p 167, #40 According to Corollary 1 of Theorem 8.2 we have Z 10 ⊕ Z 12 ⊕ Z 6 ∼ = Z 2 ⊕ Z 5 ⊕ Z 12 ⊕ Z 6 ∼ = Z 2 ⊕ Z 60 ⊕ Z 6 ∼ = Z 60 ⊕ Z 6 ⊕ Z 2 . p 167, #44 We have 1 = 8 · 2 mod 15 = 8 · φ (2 , 3) = φ (8 · 2 mod 3 , 8 · 3 mod 5) = φ (1 , 4) which shows that (1 , 4) maps to 1. p 167, #50 Since 165 = 3 · 5 · 11, the Corollary to Theorem 8.3 gives U (165) ∼ = U (3) ⊕ U (5) ⊕ U (11) ∼ = Z 2 ⊕ Z 4 ⊕ Z 10 . p 167, #52 We begin by observing that Aut( Z 20 ) ∼ = U (20) ∼ = U (4) ⊕ U (5) ∼ = Z 2 ⊕ Z 4 . If ( x, y ) ∈ Z 2 ⊕ Z 4 has order 4 then, since  ( x, y )  = lcm(  x  ,  y  ), x is free and y must have order 4. Since Z 4 has φ (4) = 2 elements of order 4, it follows that Z 2 ⊕ Z 4 , and hence Aut( Z 20 ), has 4 elements of order 4. On the other hand, since 4 · ( x, y...
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 Fall '10
 RyanDaileda
 Counting, Abelian group, Subgroup, Cyclic group, Quotient group, G/H

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