Math 122 Pset 2 1 Subgroups of the integers, and their cosets (a) We need to show that nZ contains the identity 0 and is closed under inverses and multiplication: 0 = 0n ∈ nZ shows that nZ contains the identity for any n. Let x, y ∈ nZ. We will next show that x−1 and x + y are in nZ. If x, y ∈ nZ, there exist a, b ∈ Z such that x = an, y = bn. Then x−1 = (−a)n ∈ nZ shows that nZ is closed under inverses, and x + y = an + bn = (a + b)n ∈ nZ shows that nZ is closed under the multiplication operation. (b) Let H be a subgroup of Z. First, note that if H is the zero subgroup, it is already of the form nZ, namely for n = 0. Otherwise, assume that H is not the zero subgroup. First, we argue that H must have at least one positive number. Suppose otherwise. Since H is not the zero subgroup, it must have at least one negative, non-zero number; call this number −a. The inverse of −a must be in H, so (−a)−1 = a ∈ H. But then a > 0, so H must have at least one positive number. Consider the smallest positive number in H; call this number n. We will show that H = nZ. We will quickly show that, since H is closed under multiplication, an and (−a)n must be in H for any positive number a. First, we have the base case, a = 1; then an = n ∈ H, and (−a)n = −n ∈ H, since H must be closed under inverses. For the inductive case, assume that a = N for some fixed N , and that an ∈ H. We wish to show that (a + 1)n ∈ H. But (a + 1)n = an + n; we already know that an, n ∈ H, and H must be closed under multiplication, so we know that (a + 1)n ∈ H. But then −(a + 1)n = (−a − 1)n ∈ H as well, since H must be closed under inverses. Thus, we know that an ∈ H for any a ∈ Z, so nZ ⊂ H. It remains to show that H ⊂ nZ. Suppose otherwise. Then there exists a number x ∈ H such that x = an + b, where a, b ∈ Z, 0 < b < n. Since H is closed under multiplication, −an + (an + b) ∈ H, so (−an + an) + b ∈ H, so b ∈ H. But if b ∈ H, then n is not the smallest positive number in H, and we have a contradiction. Thus, H ⊂ nZ, so we have H = nZ. (c) First, let us deal with the case where n = 0. Then, for any g ∈ Z, gH = {g}, so Z/H = {{0, }, {−1}, {+1}, {−2}, {+2}, · · · }. Next, assume that n 6= 1. Then, for any given g ∈ Z, gH = {g, g ± n, g ± 2n, · · · }. In particular, if |g| ≥ n, there exists an element g 0 ∈ gH such that 0 ≤ g 0 < n (found by adding n or −n to g until the sum is within the correct range). Thus, gH = g 0 H, so the orbits are uniquely defined by elements in the range [0, n). So the elements of Z/H are therefore {{0, ±n, ±2n, · · · }, {1, 1 ± n, 1 ± 2n, · · · }, · · · , {(n − 1), (n − 1) ± n, (n − 1) ± 2n, · · · }}. This set is isomorphic to Z/nZ; the function f : Z/nZ → Z/H, x 7→ {x, x ± n, x ± 2n, · · · } is a bijection.