Solutions to 100A Final, Dec 6, 2011 (1) A_4 consists of the 12 even permutations (12)(34), (13)(24), (14)(23), (123), (321), (124), (421), (134), (431), (234), (432), and the Identity. D_4 consists of the 4 rotations R, R^2, R^3, R^4 and the 4 flips FR, FR^2, FR^3, FR^4. S_3 consists of the 6 permutations (12), (13), (23), (123), (321), and the Identity. Z(D_4) consists of R^2 and the Identity R^4. Z(S_3) consists only of the identity. *********************** (2) True. Let g be a generator of C_200. That means g has order 200. All generators have the form g^k, where k is relatively prime to 200. It remains to show that phi(200)=80. This is true because phi(200)=phi(25)*phi(8)=20*4=80. ************************ (3) Consider the 3 disjoint cosets H, 3H, and 9H. The first coset consists of all elements of the form 3^k where k = 0 mod 3; the second consists of all elements of the form 3^k where k = 1 mod 3; and the third consists of all elements of the form 3^k where k = 2 mod 3. This exhausts all possible integers k, and so G is the union of these three cosets. Since by definition the index |G:H| equals the number of cosets, this index is 3. ************************* (4) There are (6*5)/2 = 15 transpositions. There are (6*5*4*3)/(2*2*2!)=45 products of two transpositions. There are (6*5*4*3*2*1)/(2*2*2*3!)=15 products of three transpositions. The total is 15 + 45 + 15 = 75. ************************* (5A) For brevity let g' denote the inverse of g. Note that (cd)' = d'c' = c'cd'c' = c'(cd'c') is in CD because (cd'c') is in D by normality of D. (5B) Suppose c1*d1 = c*d. We must show that f(c1*d1) = f(c*d), otherwise f would not be well-defined. By definition of f, we must show that cE = (c1)E, i.e., we must show x:=c'*c1 is in E. From the first sentence, we know that x = d*(d1)', hence x is in both C and D. This shows that x is in E, by definition of E. (5C) True. The kernel of f is the set of elements cd for which f(cd) = the identity coset E. Thus the kernel consists of all elements cd for which cE=E, i.e., all elements cd for which c is in E. The set of elements cd with c in E is precisely the set D, so the kernel of f is D. ************************* (6) The other eight noncyclic abelian groups of order 144 are C3 x C3 x C2 x C2 x C2 x C2 C3 x C3 x C2 x C2 x C4 C3 x C3 x C2 x C8 C3 x C3 x C4 x C4 C9 x C2 x C2 x C2 x C2 C9 x C2 x C2 x C4 C9 x C2 x C8 C9 x C4 x C4 Note that C9 x C16 is not included here, because that group is cyclic. ************************ (7) See pages 335 and 338 of the text. ************************ (8) Let A be the Sylow 3-subgroup and let B be the Sylow 11-subgroup. Note A has 9 elements and B has 11 elements. The number of conjugates k(A) of A is a divisor of 99 which is congruent to 1 mod 3, so k(A)=1. The number of conjugates k(B) of B is a divisor of 99 which is congruent to 1 mod 11, so k(B)=1. Thus both A and B are normal in G (since having only one conjugate is equivalent to normality). Since A and B have relatively prime orders, their intersection is {1}. Hence AB is isomorphic to the direct product A x B, which has 99 elements. Thus G=AB so G is isomorphic to A x B. All groups of order p or p^2 are abelian, so both A and B are abelian. Thus A x B is abelian, so G is abelian. *******************