MATH 103A Midterm Exam II Solutions. November 20, 2013. 1.) Let a = (12345)(
13). Write a2013 as a product of disjoint cycles. Solution: First, we write a itself ...
MATH 103A Midterm Exam II Solutions November 20, 2013
1.) Let a = (12345)(13). Write a2013 as a product of disjoint cycles. Solution: First, we write a itself as a product of disjoint cycles, and we see that a = (145)(23). Thus |a| = lcm(3, 2) = 6. We can compute that the largest multiple of 6 which is ≤ 2013 is 2010, that is, 2013 = 3 mod 6. This implies a2013 = a2010 a3 = a3 = (145)3 (23)3 = (23). 2.) How many elements of order 4 are there in A6 ? Justify your answer. Solution: The answer is 90. We know from homework 5 that any element of order 4 in S6 is of the form σ, where σ is a 4-cycle, or στ , where σ is a 4-cycle and τ is a transposition. Since we are only interested in elements of A6 , that is, even permutations, we exclude the 4 cycles since a cycle of even length is an odd permutation. Thus any element of order 4 in A6 must be of the form στ , where σ is a 4-cycle and τ is a transposition, and we have shown in homework 5 that there are 90 of them. 3.) Let G = R∗ be the group of nonzero real numbers under multiplication. Let H = R+ be the subgroup of G which consists of positive real numbers. What is the index |G : H|? List the cosets. Solution: The answer is |G : H| = 2, but we cannot use the formula |G : H| = |G|/|H| since both G and H are infinite sets. To compute the number of cosets, we see that H itself is a coset. Another coset is (−1)H = {(−1)a|a ∈ H}, which is the set of all negative real numbers. Since H ∪(−1)H = R∗ = G, we conclude that G has exactly two cosets: H and (−1)H. 4.) Find all of the cosets of the subgroup {1, 11} in U (20). 1
Solution: Note that U (20) = {1, 3, 7, 9, 11, 13, 7, 19} under multiplication modulo 20. If H = {1, 11}, then we know that the number of cosets is |U (20) : H| = |U (20)|/|H| = 8/2 = 4. We compute H = {1, 11}, 3H = {3, 13}, 7H = {7, 17} and 9H = {9, 19}. (Note that since U (20) is abelian, by “coset” we mean a left coset aH which is the same as the right coset Ha–however, there are still only 4 cosets total) 5.) Are the groups U (7) and U (9) isomorphic? Justify your answer. Solution: First recall that U (7) = {1, 2, 3, 4, 5, 6} under multiplication modulo 7. We compute the cyclic subgroup generated by 3: < 3 >= {31 = 3, 32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1} = U (7). Thus U (7) is cyclic of order 6. Next we recall that U (9) = {1, 2, 4, 5, 7, 8} under multiplication modulo 9. We compute the cyclic subgroup generated by 2: < 2 >= {21 = 2, 22 = 4, 23 = 8, 24 = 7, 25 = 5, 26 = 1}. Thus U (9) is also cyclic of order 6, and we conclude that the groups U (7) and U (9) are isomorphic. Note: if we want to give an explicit isomorphism f : U (7) → U (9), we should map a generator of U (7) to a generator of U (9). Let us map 3 to 2, so f (3) = 2. Then in order for f to be a homomorphism, we must have that f (3n ) = f (3)n , so we know where the other elements must be mapped as well: f (2) = 4, f (6) = 8, f (4) = 7, f (5) = 5, and f (1) = 1.
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