Math 103A 1 a False f False Final Exam Solutions b True g True c False h True d False i False 12 5 05 e True 2 a If a b Z G and x G then xab axb xab Also ax xa implies a 1 xa 1 a 1 xa 1 and so xa 1 a 1 x b Suppose x G Then xZ G xg g Z G gx g Z G Z G x 3 When multiplying permutations odd times odd is even odd times even is odd and even times even is even Hence the product of the permutations 1 k will be even if and only if the number of 1 k which are odd is even Since a cycle is odd if and only it has even length this tells us that a product of cycles is even if and only if the number of even length cycles in the product is even 4 The simplest group is D3 Let a and b be two different flips All flips have order 2 Then ab 6 e and ab is a rotation Hence it has order 3 Another example is Sn with n 3 Then a 12 b 13 and ab 132 In a sense this is the same example since we can view S3 as a subgroup of Sn and S3 D3 5 Since G is abelian gh k g k hk and so gh gh k g k hk g h Challenge One way is to use the structure theorem for finite abelian groups Then h1 hn hk1 hkn This will be an isomorphism if and only if hi 7 hki is a bijection for each i By our study of cyclic groups hk is a generator of hhi if and only if gcd k h 1 In other words hi 7 hki is a bijection if and only if gcd k hi 1 This holds for all i if and only if G has no factors in common with k that is gcd k G 1 6 The possible orders are 1 2 3 4 5 6 7 10 12 The identity has order 1 For k 2 3 4 5 6 7 a k cycle has order k The product of a disjoint 2 cycle and 5 cycle has order 10 The product of a disjoint 3 cycle and 4 cycle has order 12 7 a b c d e f g h i j Z16 Z25 Z400 Z16 Z5 Z5 Z80 Z5 Z8 Z2 Z25 Z200 Z2 Z8 Z50 Z8 Z2 Z5 Z5 Z4 Z4 Z25 Z4 Z4 Z5 Z5 Z4 Z2 Z2 Z25 Z4 Z2 Z2 Z5 Z5 Z2 Z2 Z2 Z2 Z25 Z2 Z2 Z2 Z2 Z5 Z5 8 Let H be the set of matrices in G of determinant 1 If A B H then AB 1 H since det AB 1 det A det B 1 Hence H is a subgroup of G Suppose C G Since det CAC 1 det C det A det C det A 1 CAC 1 H Thus H is normal 1 Math 103A Final Exam Solutions 12 5 05 9 This was discussed in class but even if you were not in class you should have enough knowledge to do it a Since R is contained in C and is closed under multiplication and taking inverses it is a subgroup Every subgroup of an abelian group is normal b aR consists of all points on the half line starting at the origin but not including the origin and passing through a In other words it is all points in C having the same arguments as a c If a point in C has polar coordinates r then r R and R 2 Z Furthermore this correspondence between C and pairs of elements r is a bijection We need to show that the group operations behave correctly The product two complex numbers with polar coordinates r1 1 and r2 2 has polar coordinates r1 r2 1 2 where 1 2 is taken modulo 2 Since we combined r1 and r2 resp 1 and 2 using the operation of R resp R 2 Z we are done 2