This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 well-known 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 ....
View Full Document

{[ snackBarMessage ]}