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Unformatted text preview: MATH 321 Final Examination Answer Key (1) Give an example of a fact that you have known long before taking MATH 321 but which you recognized in MATH 321 to be a piece of information about groups. There are lots of possible answers here. For example, you may have learned that the wellknown identities cos( a + b ) = cos a cos b sin a sin b sin( a + b ) = sin a cos b + cos a sin b express that the set of rotations is a subgroup of Isom( E ). (2) Denote the residue class of a ∈ Z modulo 18 by ¯ a , modulo 24 by ˜ a . Is the “mapping” ϕ : Z 18→ Z 24 , ¯ a 7→ ˜ a a group homomorphism? We have 18 = 0 but 18 ϕ = f 18 6 = e 0 = 0 ϕ , so ϕ is not well defined and cannot be a group homomorphism. (3) Are Z × 9 and Z × 18 isomorphic groups? In Z × 9 , we have 2 1 = 2 6 = 1, 2 2 = 4 6 = 1, 2 3 = 8 6 = 1, 2 4 = 7 6 = 1, 2 5 = 5 6 = 1, 2 6 = 1, so h 2 i = o (2) = 6 = ϕ (9) =  Z × 9  and Z × 9 = h 2 i is a cyclic group of order 6. In Z × 18 , we have 5 1 = 5 6 = 1, 5 2 = 7 6 = 1, 5 3 = 1 6 = 1, 5 4 = 5 6 = 1, 5 5 = 7 6 = 1, 5 6 = 1, so h 5 i = o (5) = 6 = ϕ (18) =  Z × 18  and Z × 18 = h 5 i is a cyclic group of order 6. Hence Z × 9 ∼ = C 6 ∼ = Z × 18 : the groups Z × 9 and Z × 18 are indeed isomorphic. (4) Can there exist a group G and an element a of G such that o ( a 3 ) = 20 and o ( a 5 ) = 9 ? Either give an example of such a pair G , a , or prove that this is impossible. Suppose that G , a is such a pair. Then a 60 = ( a 3 ) 20 = 1 and o ( a ) is finite, and if fact a divisor of 60. As ( o ( a ) , 3) = 1 or ( o ( a ) , 3) = 3, from 20 = o ( a 3 ) = o ( a ) ( o ( a ) , 3) , we get o ( a ) = 20 or o ( a ) = 60, this yields 9 = o ( a 5 ) = o ( a ) ( o ( a ) , 5) = 20 (20 , 5) = 5 or 9 = o ( a 5 ) = o ( a ) ( o ( a ) , 5) = 60 (60 , 5) = 12, which is impossible. Thus such a pair cannot exist. (5) Let F = h f i be a cyclic group of order 4 and T = h t i a cyclic group of order 10 . Find all group homomorphisms from F into T ....
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 Spring '11
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 Math

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