MAT-3110 Midterm: MATH 3110 App State Spring2010 Test2 answer key
Document Summary
Be sure to show your work: (20 points) random group stu fill out the following table: No, 4 does not divide 7 2 = 14. , x6y} where x7 = 1, y2 = 1, and xyxy = 1. Its operation is multiplication (mod n) so the (multiplicative) identity is 1. Z8 | (k, 8) = 1} = {1, 3, 5, 7}. Notice that 32 = 52 = 72 = 1 mod 8, so u (8) has no elements of order 4 and thus is not cyclic. Dn is the dihedral group of order 2n (n 3). The identity of dn is the rotation of zero degrees. Since dn is not abelian, it cannot be cyclic (recall cyclic imples abelian). x4 is a rotation and y is a re ection, so x4y is a re ection (a rotation times a re ection is a re ection). Thus |x4y| = 2 (all re ections have order 2).
