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Algebra 411 Homework 5: hints and solutions Stephen Coughlan As usual, these are only hints and not intended to be a substitute for full solutions. There is often more than one valid method. Also, there may be mistakes! 6.1 We had a theorem in class which stated that the order of an element (g1 , . . . , gn ) in a direct product G1 × · · · × Gn is given by the formula o((g1 , . . . , gn )) = lcm(o(g1 ), . . . , o(gn )). Using this formula and our knowledge of cyclic groups, we get (a) o(4) = 18 hcf(4,18) = 9 and o(9) = 2, so o((4, 9)) = lcm(9, 2) = 18. (b) o(7) = 12 hcf(12,7) = 12 and o(5) = 8 hcf(5,8) = 8 so o(7, 5) = lcm(12, 8) = 24. (c) o(8) = 9, o(6) = 3, o(4) = 2 so o((8, 6, 4)) = lcm(9, 3, 2) = 18. (d) o(8) = 9, o(6) = 17, o(4) = 5 so o((8, 6, 4)) = lcm(9, 17, 5) = 5·9·17 = 765. 6.2 The group Zm × Zn is cyclic if m and n are coprime, i.e. hcf(m, n) = 1. (What is the condition if we have the product of three or more cyclic groups? There are several ways to answer this.) (a) hcf(12, 9) = 3 so not cyclic. (b) hcf(10, 85) = 5 so not cyclic. (c) hcf(4, 6) = 2 so not cyclic. (d) hcf(22, 21) = hcf(22, 65) = hcf(21, 65) = 1 so is cyclic. 6.6 We show that G × H is abelian iff G and H are abelian. The product of more than two groups is the same. We also discussed this a little bit in class. Suppose G and H are abelian, and take elements (g1 , h1 ), (g2 , h2 ) of G × H. We check (g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ) = (g2 g1 , h2 h1 ) = (g2 , h2 )(g1 , h1 ). The middle equality is because G and H are abelian groups. Thus G × H is abelian. To prove G × H is abelian implies that G and H are abelian, just write the same argument in the reverse order! 6.10 Write G = Z2 × Z4 . There are several cyclic subgroups of G, each given by choosing a single element g in G, and taking the cyclic group hgi generated by that element. We list these: h(0, 0)i ∼ = {0}, h(1, 0)i ∼ = Z2 , h(0, 1)i = h(0, 3)i ∼ = Z4 , 1 h(0, 2)i ∼ = Z2 , h(1, 1)i = h(1, 3)i ∼ = Z4 , h(1, 2)i ∼ = Z2 This is eight elements in G, but two pairs of them give the same cyclic group, so we only get six different subgroups. I have also stated above what each subgroup is isomorphic to, although this was not part of the question. There is also the group G itself, which is a subgroup. Finally, we must consider groups generated by two or more elements. The only possible new group that we get here is h(1, 0), (0, 2)i = {(0, 0), (1, 0), (0, 2), (1, 2)} ∼ = Z2 × Z2 . Any other choice of two elements would give one of the groups already listed. (Why?) There is no way to get a new group by choosing more than two generators. (You can check this yourself!) In total, there are 6 + 1 + 1 = 8 subgroups of G. Extra 1 One such isomorphism f : Z6 → Z2 × Z3 is given by the map sending 0 7→ (0, 0), 1 7→ (1, 1), 2 7→ (0, 2), 3 7→ (1, 0), 4 7→ (0, 1), 5 7→ (1, 2). Extra 2 This problem is actually solved in chapter 12 of the textbook. We have to check that f −1 is a bijection, and that f −1 (xy) = f −1 (x)f −1 (y). We check only the second of these conditions. The first is easier, and is left to you. Since f is a bijection, every element of H can be written in the form f (a) for some a in G. In other words, given x, y in H, we can think of x = f (a) and y = f (b) when convenient. Now, we want to check that f −1 (xy) = f −1 (x)f −1 (y), but this can be written in terms of a and b: f −1 (f (a)f (b)) = f −1 (f (a))f −1 (f (b)) = ab, where the last equality is because f −1 is the inverse of f . Then on the left hand side, we note that f (a)f (b) = f (ab) because f is an isomorphism. Hence we need to check that f −1 (f (ab)) = ab. Why is this true? Extra 3 Here are the only two isomorphisms from Z4 to Z4 : 0 7→ 0, 1 7→ 1, 2 7→ 2, 3 7→ 3, and 0 7→ 0, 1 7→ 3, 2 7→ 2, 3 7→ 1. The first of these is also called the identity map. The second of these is actually multiplication by 3 modulo 4! 2 There are no other isomorphisms from Z4 to Z4 . In fact, because f (x + y) = f (x) + f (y), any isomorphism is completely determined by where we send 1 to. In other words, if we choose f (1) = 1, then since f (1 + 1) = f (1) + f (1), it follows that f (2) = 2, and likewise f (3) = 3, f (0) = 0. Similarly if we had started from f (1) = 3, we would get the second isomorphism. What goes wrong if we try f (1) = 2? By the same arguments, isomorphisms from Z8 to Z8 are determined by the value of f (1). Here are the possibilities: 0 7→ 0, 1 7→ 1, 2 7→ 2, 3 7→ 3, 4 7→ 4, 5 7→ 5, 6 7→ 6, 7 7→ 7, 0 7→ 0, 1 7→ 3, 2 7→ 6, 3 7→ 1, 4 7→ 4, 5 7→ 7, 6 7→ 2, 7 7→ 5, 0 7→ 0, 1 7→ 5, 2 7→ 2, 3 7→ 7, 4 7→ 4, 5 7→ 1, 6 7→ 6, 7 7→ 3, 0 7→ 0, 1 7→ 7, 2 7→ 6, 3 7→ 5, 4 7→ 4, 5 7→ 3, 6 7→ 2, 7 7→ 1. Can you see any patterns? Extra 4 (a) The map f (x) = x−1 is one-to-one because every element of G has a unique inverse, i.e. two different elements of G can not have the same inverse. The map is onto because any element x in G is the inverse of x−1 . (b) Be careful with brackets here: f (xy) = (xy)−1 = y −1 x−1 , f (x) = x−1 , f (y) = y −1 , f (x)f (y) = x−1 y −1 . (c) We want to check that f (xy) = f (x)f (y). By part (b), this is equivalent to checking y −1 x−1 = x−1 y −1 . This is only true G is abelian. It may not be true for a general group! 3