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Solutions for Assignment 1 Part 1 2. (a) G = Z, a · b = a − b. Not a group. The operation isn’t associative: (1 · 1) · 1 = 0 · 1 = −1, but 1 · (1 · 1) = 1 · 0 = 1. (b) G = {x ∈ Z | x > 0}, with multiplication as operation. Not a group. 2 doesn’t have an inverse. (c) Group. This is the same as the cyclic group of order 7, with ai corresponding to i. (d) G = { pq | p ∈ Z, q is an odd integer}, addition as operation. If p, p0 ∈ Z and q, q 0 ∈ Z are odd integers, then pq 0 + qp0 p p0 + 0 = . q q qq 0 Since q and q 0 are odd, qq 0 is odd, so G is closed under addition. Addition is associative on the real numbers, so it’s associative on this set of rational numbers. For any a ∈ G, a + 0 = 0 + a = a, so 0 is the identity element. For any p ∈ Z and any odd integer q, −p ∈ G and pq + −p = −p + pq = 0, so G has q q q inverses. Since it satisfies all four properties, G is a group. 3. (a) Suppose that f (x) = 2x + 1 and g(x) = −3x + 2. Compute f ∗ g and g ∗ f . f ∗ g is the function (f ∗ g)(x) = f (g(x)) = f (−3x + 2) = −6x + 5. g ∗ f is the function (g ∗ f )(x) = g(f (x)) = g(2x + 1) = −6x − 1. (b) Closure: If f (x) = mx+b and g(x) = m0 x+b0 , then (f ∗g)(x) = m(m0 x+b0 )+b = mm0 x + (mb0 + b). This is an affine function, and if m, m0 6= 0, then mm0 6= 0, and it is an element of the group. Associativity: ((f ∗ g) ∗ h)(x) = (f ∗ g)(h(x)) = f (g(h(x))) (f ∗ (g ∗ h))(x) = f ((g ∗ h)(x)) = f (g(h(x))), so the operation is associative. Identity: Let e(x) = x. Then (f ∗ e)(x) = f (e(x)) = f (x) and (e ∗ f )(x) = e(f (x)) = f (x), so e is the identity. Inverses: If f (x) = mx + b and m 6= 0, then f is an invertible function and f −1 (x) = m−1 (x−b) = m−1 x−m−1 b. By the usual properties of inverse functions, (f −1 ∗ f )(x) = (f ∗ f −1 )(x) = x, so f −1 ∗ f = f ∗ f −1 = e. The identity element is e, the identity function. The group isn’t abelian because f ∗ g 6= g ∗ f when f and g are the two elements in part (a). 1 Part 2 p. 35, #2 First, we consider the case that n ≥ 0. We proceed by induction. Base case: When n = 0, we have (ab)0 = e = a0 b0 . Inductive hypothesis: Suppose that n ≥ 0 and that (ab)n = an bn . Then (ab)n+1 = (ab)n ab = an bn ab (using the inductive hypothesis). Since G is abelian, bn a = abn , so an bn ab = an abn b = an+1 bn+1 , as desired. We conclude that (ab)n = an bn for all a, b ∈ G and all n ≥ 0. Now suppose that n > 0 and consider (ab)−n . By definition, n (ab)−n = (ab)−1 = (b−1 a−1 )n . Since n > 0, the argument above shows that (b−1 a−1 )n = (b−1 )n (a−1 )n = b−n a−n . Since G is abelian, b−n a−n = a−n b−n as desired. p. 35, #3 Suppose that (ab)2 = a2 b2 for all a, b ∈ G. Writing out the exponents, we get abab = aabb. By the cancellation lemma, bab = abb, and ba = ab, as desired. #3 b a V c d H There are four symmetries: the identity (I), the horizontal flip (H), the vertical flip (V ), and the 180◦ rotation through the center (R). Any symmetry f has to send a to one of the vertices. Because the rectangle isn’t a square, once you choose a vertex to be f (a), the images of the other vertices are determined—f (b) is the vertex adjacent to f (a) along a long edge, f (d) is adjacent to f (a) along a short edge, and f (c) is opposite to f (a). So there are four symmetries, one for each vertex. The multiplication table is ◦ I first H V R I I H V R 2 then H H I R V V V R I H R R V H I Since the set consists of all the symmetries of a rectangle, it is closed under composition and inverses. Functions are associative under composition, so the operation is associative. Finally, the identity function is the identity element of the group, so this is a group. #4 Let’s work by the number of elements in a subset. First, subgroups have to be nonempty, so there are no zero-element subgroups. Second, every subgroup contains the identity, so there is one one-element subgroup, {I}. A two-element subgroup contains the identity and one other element. There are three possibilities: {I, H}, {I, V }, and {I, R}. By the table above, H 2 = V 2 = R2 = I, so these are all closed under the operation, so they are subgroups. A three-element subgroup must contain the identity and two other elements. There are three possibilities: {I, H, V }, {I, V, R}, and {I, H, R}. None of these are subgroups, because HV = R, V R = H, and HR = V – none of them are closed. There is only one four-element subgroup, namely the whole group. This is a trivial subgroup. #5 1. For any a ∈ G, we have ae = ea = a by the definition of the identity element. So Z(e) = G. 2. We need to show that Z(a) is closed under the group operation and under taking inverses. Suppose that b, c ∈ Z(a). Then (bc)a = b(ca) = b(ac) = (ba)c = (ab)c = a(bc), so bc ∈ Z(a). Furthermore, we can multiply on the left and right by b−1 to get: ba = ab b · ba = b−1 b−1 ba · b−1 = b−1 ab · b−1 ab−1 = b−1 a. −1 So b−1 ∈ Z(a). By the lemma we proved in class, this implies that Z(a) is a subgroup of G. 3