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MATH1022 ANSWERS TO TUTORIAL EXERCISES III ac a c × = and if a, b, c, and d b d bd 1 are all odd, so are ac and bd. The identity 1 is of the correct form since 1 = , 1 a b a and the inverse of is . Hence G is a group. It is not cyclic, since if is a b a b proposed generator, and p is an odd prime which doesn’t divide either a or b, a then p ∈ G, but p cannot be expressed as a power of . b In {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} under × mod 11 we find that 22 = 4, 23 = 8, 24 = 5, 25 = 10, 26 = 9, 27 = 7, 28 = 3, 29 = 6, 210 = 1. Since every element is equal to a power of 2, this group is cyclic. 1. G is closed under multiplication, since 2. From Exercises II Qus 4, 5 we know that T ET and HEX are groups of order 12. They are non-abelian since in T ET , for example, the products of the rotation through 2π/3 fixing A, and which takes B to C to D and back to B, and that through 2π/3 fixing B, and which takes A to C to D and back to A, are different half-turns depending on which order they are performed, and in HEX, the product of two distinct rotations that interchange top and bottom are different rotations about the vertical axis of symmetry again depending on the order. The two groups are non-isomorphic since HEX has an element of order 6 but T ET does not. 3. {2n : n ∈ Z} is a cyclic group, generated by 2 (or by 1/2), since, by definition, any element of the group is of the form 2n for some integer n. 4. Z2 × Z7 is cyclic, since the element (1,1) has order 14 (we can check (1, 1), (1, 1) + (1, 1) = (0, 2), (1, 1) + (1, 1) + (1, 1) = (1, 3), . . . and find that all group elements arise. The reason is that 2 and 7 are coprime). Since Z14 is also cyclic of order 14, and any two cyclic groups of the same order are isomorphic, Z2 × Z7 ∼ = Z14 . Z24 is not isomorphic to Z4 × Z6 since the latter has no element of order 24 (all its elements have order a factor of 12). 1 0 0 1 2 ω 0 ω 0 = R, = S, 0 ω 2 0 ω ω = B. The group table for the 0 Let us write = I, 0 1 0 ω2 0 = A, = C, 1 0 ω 0 ω2 group in (ii) is then as follows: I R S A B C I I R S A B C R R S I B C A S S I R C A B A A C B I S R B B A C R I S C C B A S R I We check by inspection that this matches up perfectly with the group table given for D3 when we replace I by I, R by R, S by S, A by A, B by B, and C by C throughout. (Note that we had to label B and C in the reverse order to that in which they were given in the question, to achieve the isomorphism.) Hence the groups in (i) and (ii) are isomorphic. Neither is however isomorphic to the group in (iii), since that is abelian (cyclic actually), and they are not. 5. 6. In Z8 the possible generators are 1, 3, 5, and 7, making 4 in all, and in Z9 the possible generators are 1, 2, 4, 5, 7 and 8, making 6. (In general, one just has to count how many numbers between 1 and n − 1 are coprime with n.) 7. Let G be this set of matrices. We first check group. It is that itis a a 0 b 0 ab 0 closed under matrix multiplication, since = , 0 a 0 b 0 ab a 0 the identity matrix lies in G (put a = 1), and the inverse of is 0 a 1/a 0 . An isomorphism from the group of non-zero reals to G is 0 1/a a 0 given by θ where θ(a) = . 0 a 8. We map R to (0, ∞) by θ, where θ(x) = ex . This is 1–1 and onto, and since θ(x + y) = ex+y = ex ey = θ(x)θ(x), θ is an isomorphism. 2